Advanced Methods of Structural Analysis [2nd ed. 2021] 3030443930, 9783030443931
This revised and significantly expanded edition contains a rigorous examination of key concepts, new chapters and discus
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English Pages 795 [824] Year 2021
Preface to the Second Edition
Preface to the First Edition
Contents
Introduction
The Subject and Purposes of the Structural Mechanics
Modeling of Engineering Structures and Principal Assumptions
Design Diagram
Idealization of Material Properties
Load Idealization
The Location of Points at Which Forces Are Applied to the Body
The Nature of Change of Forces in the Process of Their Application
The Nature of the Dependence of Forces on the Deformation of the Structure
Duration of the Influence of Loads on Structures
Idealization of the Structure as a Whole
Types of Structures
Modeling of Structural Members
Features of Some Loads and Their Load Path
Simplest Industrial Building
Specific Loads
Load Combination
Types of Analysis and Particular Assumptions
Fundamental Approaches of Linear Static Analysis
Part I: Statically Determinate Structures
Chapter 1: Kinematical Analysis of Structures
1.1 Classification of Structures by Kinematical Viewpoint
1.2 Generation of Geometrically Unchangeable Structures
1.2.1 Required and Redundant Constraints
1.2.2 Constraint Replacing
1.3 Analytical Criteria of the Instantaneously Changeable Structures
1.4 Degrees of Freedom
1.5 Null Load Method
Problems
Chapter 2: The Simplest Beams: Theory of Influence Lines
2.1 General
2.2 Analytical Method for Construction of Influence Lines
2.2.1 Influence Lines for Reactions
2.2.1.1 Simply Supported Beam
Influence Line for RA
Influence Line for RB
2.2.1.2 Simply Supported Beam with Overhang
2.2.1.3 Cantilevered Beam
The Vertical Reaction RA
The Moment M0 at Support A
2.2.2 Influence Lines for Internal Forces
2.2.2.1 Bending Moment Mk
Load P = 1 is Located to the Left of Section k
Load P = 1 is Located to the Right of Section k
2.2.2.2 Influence Line Qk
2.2.2.3 Discussion
Bending Moment M1 and Shear Force Q1
Shear Force Q2 and Bending Moment M2 (Fig. 2.10)
2.3 Application of Influence Lines for Fixed and Moving Loads
2.3.1 Fixed Loads
2.3.1.1 Concentrated Loads
2.3.1.2 Uniformly Distributed Load
2.3.1.3 Couple
2.3.1.4 Nonuniform Distributed Load
Summary
2.3.2 Moving Loads
2.3.2.1 Influence Line Forms a Triangle
2.3.2.2 Influence Line Forms a Polygon
2.3.3 Envelope Diagrams of Internal Forces
2.3.3.1 The Beam is Loaded by Single Moving Load P (Fig. 2.16a)
2.3.3.2 The Beam is Loaded by Set of Connected Moving Loads P
2.3.4 Absolute Maximum of Bending Moment
2.4 Indirect Load Application
2.5 Kinematical Method for Construction of Influence Lines
2.5.1 Influence Line for Bending Moment at Section k, IL(Mk)
2.5.2 Influence Line for Shear Force at Section k, IL(Qk)
2.5.3 Conclusion
2.6 Combining of Fixed and Moving Load Approaches
2.7 Properties of Influence Lines
Problems
Chapter 3: Multispan Hinged Beams and Frames
3.1 Generation of Multispan Hinged Beams
3.2 Interaction Diagrams and Load Path
3.3 Fixed Load Method for Beams
3.4 Influence Lines for Reactions and Internal Forces
3.4.1 Static Approach
3.4.1.1 Influence Line for RD
3.4.1.2 Influence Lines for Shear Qk, and Bending Moment Mk at Section k
3.4.1.3 Influence Lines for Shear Qn, and Bending Moment Mn at Section n
3.4.1.4 Influence Lines for Shear Qs, and Bending Moment Ms at Section s
3.4.1.5 Summary
3.4.2 Kinematical Approach
3.4.2.1 Influence Line for RA
3.4.2.2 Influence Line for RB
3.4.2.3 Influence Line for Mi
3.4.3 Indirect Load Application
3.4.3.1 Influence Line for RA (Fig. 3.11c)
3.4.3.2 Influence Line for Mk (Fig. 3.11d)
3.4.3.3 Discussion
3.5 Fixed Load Method for Frames
Summary
Problems
Chapter 4: Plane Trusses
4.1 General
4.1.1 Classification of the Plane Trusses
4.1.2 Assumptions and Design Diagram
Discussion
4.2 The Generation of Statically Determinate Trusses
4.2.1 Simple Trusses
4.2.2 Compound Trusses
4.2.3 Complex Trusses
4.3 Simple Trusses: Fixed Loads
4.3.1 Analytical Methods of Analysis
4.3.1.1 Method of Joint Isolation
4.3.1.2 Method of Through Sections (Ritter´s Method)
4.3.1.3 Special Cases
4.3.2 Maxwell-Cremona Diagram
4.4 Simple Trusses: Influence Line Method-Static Approach
4.5 Trusses with Subdivided Panels
4.5.1 Main and Auxiliary Trusses and Load Path
4.5.1.1 Kinematical Analysis
4.5.2 Single-Tiered Auxiliary Trusses
4.5.3 Two-Tiered Auxiliary Trusses
Discussion
4.6 Special Types of Trusses
4.6.1 Three-Hinged Trusses
4.6.2 Trusses with a Hinged Chain
4.6.2.1 Reaction of Supports and Internal Forces
4.6.2.2 Discussion
4.7 Kinematical Method for Construction of Influence Lines
4.8 Complex Trusses
4.8.1 Substitution Bar Method
Discussion
4.8.2 Closed Section Method
4.8.3 Summary
Problems
Chapter 5: Space Frameworks
5.1 General Assumptions
5.2 Classification of the Space Frameworks
5.2.1 Method of Formation
5.2.2 Three-Dimensional Engineering Rod Structures
5.3 Space Framework Supports
5.4 Kinematical Analysis of Space Frameworks
5.4.1 Attached and Released Frameworks
5.4.1.1 Spatial Trusses Attached to Supports
5.4.1.2 Spatial Trusses Separated from Supports
5.4.2 Improper Connections of 3-D Structures
5.4.3 Meshwork Structures
5.4.3.1 Cauchy Theorem
5.5 Static Analysis of 3-D Structures
5.5.1 General
5.5.1.1 Zero Rod Cases
5.5.1.2 Method of Sections
5.5.2 Meshwork Structures
5.5.3 Compound Space Frameworks
5.5.4 Complex Space Frameworks: The Rod Replacement Method
Discussion
5.5.5 Schwedler Dome
5.5.5.1 Design Diagram of Dome
5.5.5.2 Kinematical Analysis
5.5.5.3 Features of Schwedler Structure
5.5.5.4 Decomposing of Space Framework to Plane Trusses
Problems
Chapter 6: Three-Hinged Arches
6.1 Preliminary Remarks
6.1.1 Design Diagram of Three-Hinged Arch
6.1.2 Peculiarities of the Arches
6.1.3 Geometric Parameters of Circular and Parabolic Arches
6.1.3.1 Circular Arch
6.1.3.2 Parabolic Arch
6.2 Internal Forces
6.2.1 Concept of Substitute Beam
6.2.1.1 Analysis of Formulae (6.8) and (6.10)
6.2.2 Numerical Example
6.2.3 Maximum Economy Arches
6.3 Influence Lines for Reactions and Internal Forces
6.3.1 Influence Lines for Reactions
6.3.2 Influence Lines for Internal Forces
6.3.2.1 Bending Moment
6.3.2.2 Shear Force
6.3.2.3 Axial Force
6.3.2.4 Features of the Influence Lines for Internal Forces
6.3.3 Application of Influence Lines
6.3.3.1 Reactions of Supports
6.3.3.2 Internal Forces in Section k
6.4 Nil Point Method for Construction of Influence Lines
6.4.1 Bending Moment
6.4.2 Shear Force
6.4.3 Axial Force
6.5 Special Types of Arches
6.5.1 Askew Arch
6.5.1.1 Reactions and Bending Moment at Section k
6.5.1.2 Influence Lines for Thrust and Bending Moment Mk
6.5.2 Parabolic Arch with Complex Tie
6.5.2.1 Reactions and Bending Moment at Section k
6.5.2.2 Influence Lines for Thrust and Bending Moment at Section k
Problems
Chapter 7: Cables
7.1 Preliminary Remarks
7.1.1 Direct and Inverse Problems
7.1.2 Fundamental Relationships
7.2 Cable with Neglected Self-Weight
7.2.1 Cables Subjected to Concentrated Load
7.2.1.1 Thrust-Shape Problem
7.2.1.2 Length-Thrust Problem
7.2.2 Cable Subjected to Uniformly Distributed Load
7.2.2.1 Thrust-Shape Problem
7.2.2.2 Length-Thrust Problem
Approximate Solution of Length Determination
Exact Solution of the Length Determination
7.3 Effect of Arbitrary Load on the Thrust and Sag
7.4 Cable with Self-Weight
7.4.1 Fundamental Relationships
7.4.2 Cable with Supports Located at the Same Level
7.4.3 Cable with Supports Located on Different Elevations
7.4.3.1 Saddle Point Within the Span
7.4.3.2 Saddle Point Outside of the Span
7.5 Comparison of Parabolic and Catenary Cables
7.6 Effect of Axial Stiffness
7.6.1 Elastic Cable with Concentrated Load
7.6.2 Elastic Cable with Uniformly Distributed Load
Problems
Chapter 8: Deflections of Elastic Structures
8.1 Introduction
8.2 Double Integration Method
Conclusion
8.3 Initial Parameter Method
8.3.1 Universal Equation of Elastic Curve of a Beam
8.3.2 Beam Subjected to Settlement of Supports
8.3.3 Beams with Intermediate Hinge
8.4 Conjugate Beam Method
Summary
Short History (Bernshtein 1957)
8.5 Strain Energy Methods
8.5.1 Fundamental Concepts
8.5.1.1 The Work Performed by External Forces in Terms of External Forces and Displacements
8.5.1.2 The Work Performed by External Forces in Terms of Internal Forces
8.5.2 Work-Energy Principle
Summary
8.5.3 Castigliano´s Theorem
Short Historical Remarks
8.5.4 Principle of Least Work (Menabrea Principle)
Strain Energy Methods: Summary
Short Historical Remarks
8.6 Maxwell-Mohr Method (Dummy Load Method)
8.6.1 Deflections Due to Fixed Loads
8.6.2 Deflections Due to Change of Temperature
Summary
8.7 Displacement Due to Settlement of Supports and Errors of Fabrication
Discussion
8.8 Graph Multiplication Method
Discussion
8.9 Elastic Load Method
Discussion
Summary
8.10 Reciprocal Theorems
8.10.1 Reciprocal Work Theorem (Betti´s Theorem)
8.10.2 Reciprocal Unit Displacement Theorem (Maxwell´s Theorem)
8.10.3 Reciprocal Unit Reaction Theorem (Rayleigh´s First Theorem)
8.10.4 Reciprocal Unit Displacement and Reaction Theorem (Rayleigh´s Second Theorem)
8.11 Summary
Problems
Part II: Statically Indeterminate Structures
Chapter 9: The Force Method
9.1 Fundamental Idea of the Force Method
9.1.1 Degree of Redundancy, Primary Unknowns, and Primary System
9.1.2 Compatibility Equation in Simplest Case
9.2 Canonical Equations of Force Method
9.2.1 The Concept of Unit Displacements
9.2.2 Coefficients and Free Terms of Canonical Equations
9.3 Analysis of Redundant Beams
9.3.1 Canonical Equation of the Force Method
Summary
9.3.2 The Three-Moment Equation (Clapeyron Theorem)
9.3.2.1 Special Cases
9.3.2.2 Discussion
9.3.2.3 Summary
9.3.2.4 Short Historical Remarks
9.3.3 Focal Ratios Method
Summary
9.3.4 Redundant Beam with Intermediate Hinge
9.4 Redundant Plane Frames
9.4.1 Frames of the First Degree of Redundancy
9.4.2 Frames of the Second and More Degree of Redundancy
Discussion
9.4.3 Frame with Closed Contour. Elastic Center
Discussion
9.4.4 Frame with Elastically Compliant Supports and Joints
9.5 Redundant Trusses
9.5.1 Externally Redundant Truss
Discussion
9.5.2 Internally Redundant Truss
9.5.3 Some Properties of Redundant Trusses
9.6 Redundant Arches
9.6.1 Parabolic Two-Hinged Arch
9.6.2 Circular Arch with Clamped Supports
9.6.3 Analysis of Parabolic Arch on the Basis of Modified Design Diagram
Discussion
9.7 Combined Redundant Structures
9.8 Deflections of Statically Indeterminate Structures
9.9 Settlements of Supports
9.9.1 Internal Forces due to the Settlements of Supports
Discussion
9.9.2 Displacements Due to the Settlements of Supports. Modified Approach
9.10 Temperature Changes
9.10.1 General
9.10.2 Redundant Beams
Discussion
9.10.3 Redundant Trusses
Discussion
9.10.4 Redundant Frames
9.11 Some Features of Redundant Structures
9.12 Comparison of the Redundant and Statically Determinate Structures
Problems
Additional Problems
Chapter 10: The Displacement Method
10.1 Fundamental Idea of the Displacement Method
10.1.1 Kinematical Indeterminacy
10.1.2 Primary System and Primary Unknowns
10.1.3 Compatibility Equation: Concept of Unit Reaction
10.2 Canonical Equations of Displacement Method
10.2.1 Compatibility Equations in General Case
10.2.2 Calculation of Unit Reactions
10.2.3 Properties of Unit Reactions
10.2.4 Procedure for Analysis
10.2.4.1 Continuous Beams
10.2.4.2 Beam with Intermediate Hinge
10.2.4.3 Discussion
10.2.4.4 Summary
10.3 Comparison of the Force and Displacement Methods
10.3.1 Properties of Canonical Equations
10.3.2 Variations of Design Diagrams and Choice of Methods of Analysis
10.4 Sidesway Frames with Absolutely Rigid Crossbars
10.5 Special Types of Exposures
10.5.1 Settlements of Supports
Discussion
10.5.2 Errors of Fabrication
Discussion
10.6 Analysis of Symmetrical Structures: Combined Method
10.6.1 Symmetrical and Antisymmetrical Loading
10.6.2 Concept of Half-Structure
Problems
Chapter 11: Mixed Method
11.1 Fundamental Idea of the Mixed Method
11.1.1 General
11.1.2 Mixed Indeterminacy, Primary Unknowns, and Primary System
11.2 Canonical Equations of the Mixed Method
11.2.1 The Matter of Unit Coefficients and Canonical Equations
11.2.2 Calculation of Coefficients and Free Terms
11.2.3 Computation of Internal Forces
Problems
Chapter 12: Influence Lines Method
12.1 Construction of Influence Lines by the Force Method
12.1.1 General
12.1.2 Continuous Beams
12.1.2.1 Primary System
12.1.2.2 Influence Line for Primary Unknown X1
12.1.2.3 Influence Line for Bending Moment Mk
12.1.2.4 Influence Line for Shear Force Qk
12.1.2.5 Statically Indeterminate Primary System
12.1.3 Hingeless Nonuniform Arches
12.1.3.1 Unit Coefficients
12.1.3.2 Reactions of Support A
12.1.3.3 Bending Moment at Crown C
12.1.3.4 Discussion
12.1.4 Statically Indeterminate Trusses
12.1.4.1 Construction of Function δP1
12.1.4.2 Influence Line for Primary Unknown X1
12.1.4.3 Summary
12.2 Construction of Influence Lines by the Displacement Method
12.2.1 General
12.2.2 Continuous Beams
12.2.2.1 Influence Line for Primary Unknown Z1
12.2.2.2 Influence Line for Bending Moment Mk
12.2.2.3 Influence Line for Shear Force Qk
12.2.2.4 Discussion
12.2.3 Redundant Frames
Discussion
12.3 Comparison of the Force and Displacement Methods
Notes
12.4 Kinematical Method (Müller-Breslau Principle)
12.4.1 General
12.4.2 Continuous Beams: Analytical Solution
12.4.3 Continuous Beams: Models of Influence lines
Summary
Problems
Chapter 13: Matrix Stiffness Method
13.1 Basic Idea and Concepts
13.1.1 Finite Elements
13.1.2 Global and Local Coordinate Systems
13.1.3 Displacements of Joints and Degrees of Freedom
13.2 Auxiliary Diagrams
13.2.1 Joint-Load (J-L) Diagram
13.2.2 Displacement-Load (Z-P) Diagram
13.2.3 Internal Forces-Deformation (S-e) Diagram
Summary
13.3 Initial Matrices
13.3.1 Vector of External Joint Loads
13.3.2 Vector of Internal Unknown Forces
Summary
13.4 Resolving Equations
13.4.1 Static Equations and Static Matrix
13.4.2 Geometrical Equations and Deformation Matrix
13.4.3 Physical Equations and Stiffness Matrix in Local Coordinates
13.5 Set of Formulas and Procedure for Analysis
13.5.1 Stiffness Matrix in Global Coordinates
13.5.2 Unknown Displacements and Internal Forces
13.5.3 Matrix Procedures
13.6 Analysis of Continuous Beams
13.6.1 Fixed Loads
13.6.2 Settlements of Supports
13.6.3 Moving Load (Construction of Influence Lines)
13.6.3.1 Load P = 1 in the First Span
13.6.3.2 Load P = 1 in the Second Span
13.6.3.3 Load P = 1 in the Third Span
13.7 Analysis of Redundant Frame
13.8 Analysis of Redundant Trusses
13.9 Stiffness Matrices: Expanded Forms
13.9.1 Truss Element (Pinned-Pinned Element)
13.9.2 Beam Elements
13.9.3 More General Case of a Finite Element
13.10 Summary
Remarks Concerning Analysis of the Bending Structures by MSM
Problems
Part III: Special Topics
Chapter 14: Plastic Behavior of Structures
14.1 Idealized Stress-Strain Diagrams
14.2 Direct Method of Plastic Analysis
14.2.1 Elastic Analysis
14.2.2 Plastic Analysis
14.3 Fundamental Methods of Plastic Analysis
14.3.1 Kinematical Method
14.3.2 Static Method
14.3.3 Summary
14.4 Plastic Analysis of Continuous Beams
14.4.1 Static Method
14.4.2 Kinematical Method
Discussion
14.5 Plastic Analysis of Frames
14.5.1 Beam Failure
14.5.2 Sidesway Failure
14.5.3 Combined Failure
14.5.4 Limit Combination Diagram
Problems
Chapter 15: Stability of Elastic Systems
15.1 Fundamental Concepts
15.2 Stability of Structures with Finite Number of Degrees of Freedom
15.2.1 Structures with One Degree of Freedom
Discussion
15.2.2 Structures with Two or More Degrees of Freedom
15.3 Stability of Columns with Rigid and Elastic Supports
15.3.1 The Double Integration Method
15.3.1.1 Uniform Clamped-Free Column
15.3.1.2 Uniform Columns with Elastic Supports
15.3.2 Initial Parameters Method
15.3.2.1 Limiting Cases
15.3.2.2 Discussion
15.3.3 Euler´s Solution and Paradox of Critical Load
15.4 Stability of Continuous Beams and Frames
15.4.1 Unit Reactions of the Beam-Columns
Some Features of the Solution
15.4.2 Displacement Method
Special cases
Special Cases
15.4.3 Modified Approach of the Displacement Method
15.5 Stability of Arches
15.5.1 Introduction
15.5.1.1 Forms of the Loss of Stability of the Arches
15.5.2 Circular Arches under Hydrostatic Load
15.5.3 Complex Arched Structure: Arch with Elastic Supports
15.5.4 Parabolic Arch under Gravity Load
Arches with Variable Cross Sections
Problems
Chapter 16: Dynamics of Elastic Systems: Free Vibration
16.1 Fundamental Concepts
16.1.1 Kinematics of Vibrating Processes
16.1.2 Forces Which Arise at Vibrations
16.1.3 Degrees of Freedom
16.1.4 Purpose of Structural Dynamics
16.2 Systems with Finite Number of Degrees of Freedom: Force Method
16.2.1 Differential Equations of Free Vibration in Displacements
16.2.2 Frequency Equation
16.2.3 Mode Shapes of Vibration and Modal Matrix
16.3 Systems with Finite Number of Degrees of Freedom: Displacement Method
16.3.1 Differential Equations of Free Vibration in Reactions
16.3.2 Frequency Equation
16.3.3 Mode Shape of Vibrations and Modal Matrix
16.3.4 Comparison of the Force and Displacement Methods
16.4 Structures with Infinite Number of Degrees of Freedom
16.4.1 Differential Equation of Transversal Vibration of the Beam
16.4.1.1 Boundary Conditions
16.4.1.2 Initial Conditions
16.4.2 Fourier Method
16.4.3 Krylov-Duncan Method
16.4.3.1 Properties of Krylov-Duncan Functions (16.24)
16.4.4 Initial Parameters Method
16.4.5 Transfer Matrices Method
16.4.5.1 Transfer Matrix of Uniform Beam with Distributed Mass
16.4.5.2 Transfer Matrix of Uniform Massless Beam
16.4.5.3 Mathematical Model of a Complex Structure
16.4.6 Displacement Method
16.4.7 Missed (Unaccounted) Frequencies
Problems
Chapter 17: Dynamics of Elastic Systems: Forced Vibration
17.1 Structures with One Degree of Freedom
17.1.1 Differential Equations: Two Classical Approaches
17.1.1.1 Force Method
17.1.1.2 Displacement Method
17.1.2 Types of Excitations
17.1.3 Duhamel Integral and Some Special Types of Excitation
17.1.4 Harmonic Excitation: Equivalent Design Diagram
17.1.4.1 Equivalent Design Diagram
17.1.5 Kinematical Excitation
17.2 Structures with Finite Number of Degrees of Freedom: The Force Method
17.2.1 Resolving Equation of the Force Method
17.2.2 Harmonic Excitation: Reciprocal Theorems
17.2.3 Impulsive Excitation
17.2.4 General Case of Excitation
17.3 Structures with Finite Number of Degrees of Freedom: Initial Parameters Method
17.3.1 Resolving Equations
17.3.2 Steady-State Vibrations
17.4 Structures with Finite Number of Degrees of Freedom: Displacement Method
17.4.1 The Steady-State Vibration
17.4.2 Group Unknowns Method
17.5 Structures with Distributed Parameters
17.5.1 Initial Parameter Method
17.5.2 Displacement Method
17.5.3 Bolotin Approximate Method
Problems
Structures with Finite Number of Degrees of Freedom
Kinematical Excitation
Forced Excitations
Structures with Infinite Numbers Degrees of Freedom
Chapter 18: Special Topics of Structural Dynamics
18.1 Timoshenko-Ehrenfest Beam Theory
18.2 Flexural Vibration of Compressed Beam
18.2.1 Fundamental Equations
18.2.2 Galef´s Formula
18.3 Traveling Load
18.3.1 General
18.3.2 Quasi-Static Loading
18.3.3 Critical Velocity of Moving Load
18.4 Parametric Vibration
18.4.1 Dynamic Stability of Simply Supported Column
18.4.2 Ince-Strutt Diagram
18.5 Vibration Protection
18.5.1 Vibration Protection Methods
18.5.2 Dynamic Absorber
18.5.3 Lumped Vibration Absorber of the Beams
Chapter 19: Nonlinear Structural Analysis
19.1 Introduction and Types of Nonlinearities
19.1.1 Types of Nonlinearities
19.2 Compressed Rods with Lateral Loading
19.2.1 Double Integration Method
19.2.2 Initial Parameters Method
19.2.2.1 Free-Clamped Beam: Precise Solution
19.2.3 P-Delta Analysis
19.3 Static Nonlinearity
19.3.1 Features of the Problem
19.3.2 Transversal Vibration of a Rod
19.4 Physical Nonlinearity
19.4.1 Features of the Problem
19.4.2 Transversal Free Vibration of Uniform Beam
19.5 Geometrical Nonlinearity
19.5.1 General
19.5.2 Stability of a Flexible Rod
Problems
Chapter 20: Conclusion: Once More About Modeling of Structures
20.1 Some Problems of Structural Modeling
20.2 Common Modeling Problems
20.3 Differences Between Structural Analysis and Direct Design
Chapter 21: Fundamental Developments in the History of Structural Mechanics
Appendix
Tabulated Data for Standard Uniform Beams
Bibliography
General Textbooks and Manuals
Handbooks
Related List of Literature
History
Index
Index
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Igor A. Karnovsky Olga Lebed
Advanced Methods of Structural Analysis Second Edition
Advanced Methods of Structural Analysis
Igor A. Karnovsky • Olga Lebed
Advanced Methods of Structural Analysis Strength, Stability, Vibration
Second Edition
Igor A. Karnovsky 811 Northview Pl. Coquitlam, BC, Canada
Olga Lebed 811 Northview Pl. Coquitlam, BC, Canada
ISBN 978-3-030-44393-1 ISBN 978-3-030-44394-8 https://doi.org/10.1007/978-3-030-44394-8
(eBook)
© Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
Ch20. Once More about Modeling of Structures
Ch7. Cables
Ch6. Three-Hinged Arches
Ch5. Space Trusses
Ch4. Plane Trusses
Ch3. Multispan Hinged Beams and Frames
Analysis of different types of statically determinate structures
Ch1. Kinematical Analysis
Book Road Map
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Reciprocal theorems
More detail Section 8.11
Elastic load method
Vereshchagin rule
Mohr integral
Initial Parameters Method
Ch8. Deflection of Elastic Structures
Ch2. General Theory of Influence Lines
Design Diagram
Principal Assumptions
Introduction
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Ch13. Matrix Stiffness Method
Ch12. Influence Line Method
Ch11. Mixed Method
Ch10. Displacement Method
Ch9. Force Method
Different methods of analysis of redundant structures
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Ch19. Nonlinear Structural Analysis
Ch14. Plastic Analysis of Structures
Non-classical Topics
Ch18. Special Topics of Structural Dynamics
Ch17. Forced Vibration of Elastic Systems
Ch16. Free Vibration of Elastic Systems
Ch15. Stability of Elastic Systems
Stability and Vibration
Preface to the Second Edition
The first edition of the book entitled “Advanced Methods of Structural Analysis” (593 pages) was published by Springer Publishing House in 2010 (Karnovsky and Lebed 2010). Authors present the second revised and expanded edition of the textbook under the extended name “Advanced Methods of Structural Analysis: Strength, Stability, Vibration.” The authors retained the general structure of the first edition of the book Advanced Methods of Structural Analysis, but in the second edition of the book key concepts have been added, new chapters have been introduced, and existing chapters have been significantly expanded. These additions relate to the theory, analysis procedures, detailed illustrative examples, and discussion of the results and comments. All additions pursued the following purposes: • Draw reader’s attention to the conceptual provisions that go through the whole course of the theory of structural mechanics. Among them are design diagram of a structure, equilibrium conditions for the whole system and any part of it, correspondence of the stress-deformable state of the system to the superimposed constraints, and invariance of the analysis results from the accepted method of analysis of the structure. • Acquaint the reader with some questions of the theory of structures that are insufficiently covered in the educational literature or even completely absent in them. • Expand the scope of the classical methods used for analyzing the mechanics of rod systems, and show typical examples of the advantages and disadvantages of analysis methods. • Prepare the reader for the idea of presentation of the structure by different design diagrams and justification of its final version that meets various but often contradictory requirements. The following six new chapters have been introduced: Chapter 5: Space Frameworks. The reader is provided with detailed information about the methods of formation of spatial rod structures, the conditions of attachment to the base, and the method of connecting two 3-D structures. Attached and released framework concepts are introduced. Various types of spatial structures (meshwork structures, compound, complex space frameworks) and procedure for their analysis are discussed. Detailed analysis of the Schwedler’s dome is presented. Chapter 17: Dynamics of Elastic Systems. Forced Vibration. This chapter is devoted to the dynamic analysis of structures with finite and infinite number of degrees of freedom subjected to force and kinematical excitations. The steady-state and transitional vibrations are considered. Dynamical analysis is carried out in terms inherent to the deformable structures adopted in the classical course of structural mechanics. The continuity of classic analysis methods (such as force and displacement methods, initial parameter method) to the problems of dynamic behavior of deformable systems is presented. Material covered in new Chapters 5 and 17 is considered within the framework of classical assumptions. Chapter 18: Special Topics of Structural Dynamics*. The peculiarity of this chapter is that each paragraph of this chapter contains an additional assumption that does not contradict the assumptions adopted in the classical structural analysis. This allows for constructing a refined mathematical model of the object, discovering new qualitative aspects of system behavior, and explaining the paradoxes arising under the classical assumptions. Chapter 19: Nonlinear Structural Analysis*. A feature of this chapter is that each paragraph of this chapter contains a rejection of at least one of the conventional assumptions of the classical structural analysis. The need to consider such problems is that in an arbitrary deformable system inevitably there are various kinds of nonlinearity. Taking them into account allows discovering new effects inherent to a real structure. Analytical solutions of a number of nonlinear problems of strength, stability, and vibrations of beams are presented; these solutions can be considered as test result for an alternate method. Chapter 20: Once More About Modeling of Structures. This chapter summarizes modeling deformable structures. The chapter contains requirements and factors influencing the choice of design diagram as well as concepts for transition from a constructive scheme of a structure to its design diagram. It also notes an inevitable trade-off between a detailed design vii
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diagram, for which precise results can be obtained, and a simplified diagram, for which less accurate results can be obtained. This chapter also discusses the difference between analysis of a structure and direct design of a structure. Chapter 21: Fundamental Developments in the History of Structural Mechanics. This chapter covers only two stages— the period of the formation of the science of strength (1638–1826) and the creation of the classical theory of structures itself (from 1826). The following set of issues are discussed: the objective reasons behind deadlock state of the limit state method, the essence of the Claude-Louis Navier reform (1826) and the need to switch from the limit state method to the working state method, the difference between these methods, and the disadvantages and merits of structural analysis utilizing the working state method. The following additions have been made: Introduction is substantially revised and supplemented. The subject of detailed discussion is the concept of design diagram, which is the cornerstone of structural mechanics course, and determines the behavior of the system and the method of its analysis. Chapter 8: A number of classical methods for determining displacements are added (a fictitious beam method, energy methods, the principle of the smallest work, etc.). Chapter 9: Classical methods of beam analysis are added (three moment equations and focal ratio method). Chapter 16: A new material related to the application of the method of initial parameters (16.4.4–16.4.5) and the method of displacements (16.4.6) in problems of dynamics is added. Chapters 2–4: Construction of influence lines by kinematical method for beams and plane trusses is added. Also, examples are added in separate chapters; reference material in the Appendix and the list of bibliography are expanded. Much attention is paid to verifying the results obtained by applying different methods of analysis to the same design diagram. Solving problems of a certain class using various methods allows the reader to compare their effectiveness and evaluate their strengths and weaknesses. This approach is of great methodological significance, since it allows formulating recommendations with an indication of the limits of the effective application of methods. Authors will appreciate comments and suggestions to improve the current edition. All constructive criticism will be accepted with gratitude. Vancouver, Canada
Igor A. Karnovsky Olga I. Lebed
Preface to the First Edition
Theory of the engineering structures is a fundamental science. Statements and methods of this science are widely used in different fields of engineering. Among them are the civil engineering, shipbuilding, aircraft, robotics, space structures, as well as numerous structures of special types and purposes—bridges, towers, etc. In recent years, even micromechanical devices have become objects of structural analysis. Theory of the engineering structures is alive and is a very vigorous science. This theory offers an engineer-designer a vast collection of classical methods of analysis of various types of structures. These methods contain in-depth fundamental ideas and, at the present time, they are developed with sufficient completeness and commonness, aligned in a well-composed system of conceptions, procedures, and algorithms; use modern mathematical techniques; and are brought to elegant simplicity and perfection. We now live in a computerized world. A role and influence of modern engineering software for the analysis of structures cannot be overestimated. The modern computer programs allow providing different types of analysis for any sophisticated structure. As this takes place, what is the role of classical theory of structures with its in-depth ideas, prominent conceptions, methods, theorems, and principles? Knowing classical methods of structural analysis is necessary for any practical engineer. An engineer cannot rely only on the results provided by a computer. Computer is a great help in modeling different situations and speeding up the process of calculations, but it is the sole responsibility of an engineer to check the results obtained by a computer. If users of computer engineering software do not have sufficient knowledge of fundamentals of structural analysis and understanding of physical theories and principal properties of structures, then he/she cannot check obtained numerical results and their correspondence to an adopted design diagram, as well as explain results obtained by a computer. Computer programs “. . . can make a good engineer better, but it can make a poor engineer more dangerous” [Cook R.D., Malkus D.S., Plesha M. E (1989) Concepts and applications of finite element analysis, 3rd edn. Wiley, New York]. Only the knowledge of fundamental theory of structures allows to estimate and analyze numerical data obtained from a computer; predict the behavior of a structure as a result of changing a design diagram and parameters; design a structure which satisfies certain requirements; perform serious scientific analysis; and make valid theoretical generalizations. No matter how sophisticated the structural model is, no matter how effective the numerical algorithms are, no matter how powerful the computers are that implement these algorithms, it is the engineer who analyzes the end result produced from these algorithms. Only an individual who has a deep knowledge and understanding of the structural model and analysis techniques can produce a qualitative analysis. In 1970s, one of the authors of this book was a professor at a structural engineering university in Ukraine. At that time, computers were started to be implemented in all fields of science, structural analysis being one of them. We, the professors and instructors, were facing a serious methodical dilemma: Given the new technologies, how to properly teach the students? Would we first give students a strong basis in classical structural analysis and then introduce them to the related software, or would we directly dive into the software after giving the students a relatively insignificant introduction to classical analysis? We did not know an optimal way to solve this problem. On this subject we have conducted seminars and discussions on a regular basis. We have used these two main teaching models, and many different variations of them. The result was somewhat surprising. The students that were first given a strong foundation in structural analysis quickly learned how to use the computer software, and were able to give a good qualitative analysis of the results. The students that were given a brief introduction to structural analysis and a strong emphasis on the computer software at the end were not able to provide qualitative results of the analysis. The interesting thing is that the students themselves were criticizing the latter teaching strategy. Therefore, our vision of teaching of structural analysis is as follows: in the first step, it is necessary to learn analytical methods, perform detailed analysis of different structures by hand in order to feel the behavior of structures, and correlate their behavior with obtained results; the second step is a computer application of engineering software. Authors wrote the book on the basis of their many years of experience of teaching the structural analysis at the universities for graduate and postgraduate students as well as on the basis of their experience in consulting companies. ix
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This book is written for students of universities and colleges pursuing civil or structural engineering programs, instructors of structural analysis, and engineers and designers of different structures of modern engineering. The objective of the book is to help a reader to develop an understanding of the ideas and methods of structural analysis and to teach a reader to estimate and explain numerical results obtained by hand; this is a fundamental stone for preparation of readers for numerical analysis of structures and for use of engineering software with full understanding. The textbook offers the reader the fundamental theoretical concepts of structural analysis, classical analytical methods, algorithms of their application, comparison of different methods, and a vast collection of distinctive problems with their detailed solution, explanation, analysis, and discussion of results; many of the problems have a complex character. Considered examples demonstrate features of structures, their behavior, and peculiarities of applied methods. Solution of all the problems is brought to a final formula or number. Analyses of the following structures are considered: statically determinate and indeterminate multi-span beams, arches, trusses, and frames. These structures are subjected to fixed and moving loads, changes of temperature, settlement of supports, and errors of fabrication. Also, the cables are considered in detail. In many cases, same structure under different external actions is analyzed. It allows the reader to be concentrated on one design diagram and perform complex analysis of behavior of a structure. In many cases, same structure is analyzed by different methods or by one method in different forms (for example, displacement method in canonical and matrix forms). It allows to perform comparison analysis of applied methods and see advantages and disadvantages of different methods.
Contents
Preface to the Second Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Preface to the First Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xix Part I
Statically Determinate Structures
1
Kinematical Analysis of Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Classification of Structures by Kinematical Viewpoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Generation of Geometrically Unchangeable Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Required and Redundant Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Constraint Replacing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Analytical Criteria of the Instantaneously Changeable Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Null Load Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. 3 . 4 . 4 . 5 . 6 . 6 . 9 . 10 . 12
2
The Simplest Beams: Theory of Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Analytical Method for Construction of Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Influence Lines for Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Influence Lines for Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Application of Influence Lines for Fixed and Moving Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Fixed Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Moving Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Envelope Diagrams of Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.4 Absolute Maximum of Bending Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Indirect Load Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Kinematical Method for Construction of Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Influence Line for Bending Moment at Section k, IL(Mk) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Influence Line for Shear Force at Section k, IL(Qk) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Combining of Fixed and Moving Load Approaches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Properties of Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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15 15 16 16 19 25 25 28 31 32 34 35 37 38 39 40 40 41
3
Multispan Hinged Beams and Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Generation of Multispan Hinged Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Interaction Diagrams and Load Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Fixed Load Method for Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Influence Lines for Reactions and Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Static Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Kinematical Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Indirect Load Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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45 45 46 48 51 51 54 55 xi
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3.5 Fixed Load Method for Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 4
Plane Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Classification of the Plane Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Assumptions and Design Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Generation of Statically Determinate Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Simple Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Compound Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 Complex Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Simple Trusses: Fixed Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Analytical Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Maxwell–Cremona Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Simple Trusses: Influence Line Method—Static Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Trusses with Subdivided Panels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Main and Auxiliary Trusses and Load Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2 Single-Tiered Auxiliary Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.3 Two-Tiered Auxiliary Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Special Types of Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.1 Three-Hinged Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.2 Trusses with a Hinged Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Kinematical Method for Construction of Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Complex Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.1 Substitution Bar Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.2 Closed Section Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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65 65 65 67 68 69 69 70 71 71 76 78 81 82 83 87 90 90 92 95 99 99 102 103 103
5
Space Frameworks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 General Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Classification of the Space Frameworks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Method of Formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Three-Dimensional Engineering Rod Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Space Framework Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Kinematical Analysis of Space Frameworks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Attached and Released Frameworks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Improper Connections of 3-D Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.3 Meshwork Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Static Analysis of 3-D Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Meshwork Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.3 Compound Space Frameworks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.4 Complex Space Frameworks: The Rod Replacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.5 Schwedler Dome . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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107 107 107 107 108 110 112 112 115 116 118 118 120 122 123 125 128
6
Three-Hinged Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Preliminary Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Design Diagram of Three-Hinged Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Peculiarities of the Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Geometric Parameters of Circular and Parabolic Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . .
131 131 131 132 133
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6.2
Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Concept of Substitute Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Numerical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.3 Maximum Economy Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Influence Lines for Reactions and Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Influence Lines for Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Influence Lines for Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Application of Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Nil Point Method for Construction of Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Bending Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Shear Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.3 Axial Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Special Types of Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.1 Askew Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.2 Parabolic Arch with Complex Tie . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
133 134 135 138 139 140 141 144 145 145 146 148 148 149 151 154
7
Cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Preliminary Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 Direct and Inverse Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 Fundamental Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Cable with Neglected Self-Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Cables Subjected to Concentrated Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 Cable Subjected to Uniformly Distributed Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Effect of Arbitrary Load on the Thrust and Sag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Cable with Self-Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Fundamental Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 Cable with Supports Located at the Same Level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 Cable with Supports Located on Different Elevations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Comparison of Parabolic and Catenary Cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Effect of Axial Stiffness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Elastic Cable with Concentrated Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.2 Elastic Cable with Uniformly Distributed Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
159 159 160 160 162 163 164 169 172 172 174 175 180 182 182 183 184
8
Deflections of Elastic Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Double Integration Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Initial Parameter Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Universal Equation of Elastic Curve of a Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Beam Subjected to Settlement of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Conjugate Beam Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Strain Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2 Work-Energy Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Castigliano’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.4 Principle of Least Work (Menabrea Principle) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Maxwell–Mohr Method (Dummy Load Method) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6.1 Deflections Due to Fixed Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6.2 Deflections Due to Change of Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Displacement Due to Settlement of Supports and Errors of Fabrication . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
187 187 189 193 193 200 204 211 211 215 216 223 227 227 232 235
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8.8 8.9 8.10
Graph Multiplication Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Elastic Load Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reciprocal Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.10.1 Reciprocal Work Theorem (Betti’s Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.10.2 Reciprocal Unit Displacement Theorem (Maxwell’s Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . 8.10.3 Reciprocal Unit Reaction Theorem (Rayleigh’s First Theorem) . . . . . . . . . . . . . . . . . . . . . . . . 8.10.4 Reciprocal Unit Displacement and Reaction Theorem (Rayleigh’s Second Theorem) . . . . . . . . . 8.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . .
240 246 250 250 251 252 252 253 255
The Force Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Fundamental Idea of the Force Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Degree of Redundancy, Primary Unknowns, and Primary System . . . . . . . . . . . . . . . . . . . . . . . . 9.1.2 Compatibility Equation in Simplest Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Canonical Equations of Force Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 The Concept of Unit Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Coefficients and Free Terms of Canonical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Analysis of Redundant Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Canonical Equation of the Force Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 The Three-Moment Equation (Clapeyron Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 Focal Ratios Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.4 Redundant Beam with Intermediate Hinge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Redundant Plane Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.1 Frames of the First Degree of Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.2 Frames of the Second and More Degree of Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.3 Frame with Closed Contour. Elastic Center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.4 Frame with Elastically Compliant Supports and Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Redundant Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.1 Externally Redundant Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.2 Internally Redundant Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.3 Some Properties of Redundant Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Redundant Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.1 Parabolic Two-Hinged Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.2 Circular Arch with Clamped Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.3 Analysis of Parabolic Arch on the Basis of Modified Design Diagram . . . . . . . . . . . . . . . . . . . . 9.7 Combined Redundant Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Deflections of Statically Indeterminate Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.9 Settlements of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.9.1 Internal Forces due to the Settlements of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.9.2 Displacements Due to the Settlements of Supports. Modified Approach . . . . . . . . . . . . . . . . . . . 9.10 Temperature Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.10.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.10.2 Redundant Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.10.3 Redundant Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.10.4 Redundant Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.11 Some Features of Redundant Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.12 Comparison of the Redundant and Statically Determinate Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.12.1 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
273 273 273 276 278 279 280 282 282 284 291 299 302 302 305 309 315 318 319 323 325 326 327 329 331 336 338 340 341 345 347 347 348 351 353 356 358 358 364
Part II 9
10
Statically Indeterminate Structures
The Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 10.1 Fundamental Idea of the Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 10.1.1 Kinematical Indeterminacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367
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10.1.2 Primary System and Primary Unknowns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.3 Compatibility Equation: Concept of Unit Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Canonical Equations of Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 Compatibility Equations in General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Calculation of Unit Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Properties of Unit Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.4 Procedure for Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Comparison of the Force and Displacement Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.1 Properties of Canonical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.2 Variations of Design Diagrams and Choice of Methods of Analysis . . . . . . . . . . . . . . . . . . . . . 10.4 Sidesway Frames with Absolutely Rigid Crossbars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Special Types of Exposures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.1 Settlements of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.2 Errors of Fabrication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Analysis of Symmetrical Structures: Combined Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.1 Symmetrical and Antisymmetrical Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.2 Concept of Half-Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . .
369 370 371 371 371 373 373 384 386 386 387 388 389 392 394 394 394 398
11
Mixed Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Fundamental Idea of the Mixed Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Mixed Indeterminacy, Primary Unknowns, and Primary System . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Canonical Equations of the Mixed Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 The Matter of Unit Coefficients and Canonical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2 Calculation of Coefficients and Free Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.3 Computation of Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . .
405 405 405 406 406 406 408 409 410
12
Influence Lines Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Construction of Influence Lines by the Force Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.2 Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.3 Hingeless Nonuniform Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.4 Statically Indeterminate Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Construction of Influence Lines by the Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.2 Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.3 Redundant Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Comparison of the Force and Displacement Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 Kinematical Method (Müller-Breslau Principle) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4.2 Continuous Beams: Analytical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4.3 Continuous Beams: Models of Influence lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
413 413 413 415 421 427 431 431 432 439 441 442 442 443 445 447
13
Matrix Stiffness Methòd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Basic Idea and Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.1 Finite Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.2 Global and Local Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.3 Displacements of Joints and Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Auxiliary Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.1 Joint–Load (J-L) Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.2 Displacement–Load (Z-P) Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.3 Internal Forces–Deformation (S-e) Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Initial Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . .
451 451 451 452 453 453 453 456 457 459
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13.3.1 Vector of External Joint Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3.2 Vector of Internal Unknown Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Resolving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4.1 Static Equations and Static Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4.2 Geometrical Equations and Deformation Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4.3 Physical Equations and Stiffness Matrix in Local Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 13.5 Set of Formulas and Procedure for Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5.1 Stiffness Matrix in Global Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5.2 Unknown Displacements and Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5.3 Matrix Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Analysis of Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.1 Fixed Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.2 Settlements of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.3 Moving Load (Construction of Influence Lines) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7 Analysis of Redundant Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.8 Analysis of Redundant Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.9 Stiffness Matrices: Expanded Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.9.1 Truss Element (Pinned-Pinned Element) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.9.2 Beam Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.9.3 More General Case of a Finite Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Part III
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459 460 461 461 464 465 468 468 469 469 470 471 473 476 480 484 486 487 488 489 491 494
Special Topics
14
Plastic Behavior of Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Idealized Stress-Strain Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Direct Method of Plastic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.1 Elastic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.2 Plastic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Fundamental Methods of Plastic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3.1 Kinematical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3.2 Static Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Plastic Analysis of Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4.1 Static Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4.2 Kinematical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Plastic Analysis of Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.1 Beam Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.2 Sidesway Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.3 Combined Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.4 Limit Combination Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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501 501 504 504 505 506 506 508 508 509 509 511 516 516 517 517 518 519 519
15
Stability of Elastic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Stability of Structures with Finite Number of Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2.1 Structures with One Degree of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2.2 Structures with Two or More Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 Stability of Columns with Rigid and Elastic Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.1 The Double Integration Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.2 Initial Parameters Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.3 Euler’s Solution and Paradox of Critical Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Stability of Continuous Beams and Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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523 523 525 525 529 531 531 535 539 540
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15.4.1 Unit Reactions of the Beam-Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4.2 Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4.3 Modified Approach of the Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5 Stability of Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.2 Circular Arches under Hydrostatic Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.3 Complex Arched Structure: Arch with Elastic Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.4 Parabolic Arch under Gravity Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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540 542 549 551 551 552 557 558 561
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Dynamics of Elastic Systems: Free Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1.1 Kinematics of Vibrating Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1.2 Forces Which Arise at Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1.3 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1.4 Purpose of Structural Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Systems with Finite Number of Degrees of Freedom: Force Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.1 Differential Equations of Free Vibration in Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.2 Frequency Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.3 Mode Shapes of Vibration and Modal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Systems with Finite Number of Degrees of Freedom: Displacement Method . . . . . . . . . . . . . . . . . . . . . 16.3.1 Differential Equations of Free Vibration in Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.2 Frequency Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.3 Mode Shape of Vibrations and Modal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.4 Comparison of the Force and Displacement Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Structures with Infinite Number of Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4.1 Differential Equation of Transversal Vibration of the Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4.2 Fourier Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4.3 Krylov-Duncan Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4.4 Initial Parameters Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4.5 Transfer Matrices Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4.6 Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4.7 Missed (Unaccounted) Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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565 565 565 566 567 569 570 570 571 571 579 579 580 581 586 587 587 588 590 594 600 604 608 609
17
Dynamics of Elastic Systems: Forced Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 Structures with One Degree of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1.1 Differential Equations: Two Classical Approaches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1.2 Types of Excitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1.3 Duhamel Integral and Some Special Types of Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1.4 Harmonic Excitation: Equivalent Design Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1.5 Kinematical Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Structures with Finite Number of Degrees of Freedom: The Force Method . . . . . . . . . . . . . . . . . . . . . . . 17.2.1 Resolving Equation of the Force Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2.2 Harmonic Excitation: Reciprocal Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2.3 Impulsive Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2.4 General Case of Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Structures with Finite Number of Degrees of Freedom: Initial Parameters Method . . . . . . . . . . . . . . . . . 17.3.1 Resolving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3.2 Steady-State Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Structures with Finite Number of Degrees of Freedom: Displacement Method . . . . . . . . . . . . . . . . . . . . 17.4.1 The Steady-State Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4.2 Group Unknowns Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.5 Structures with Distributed Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.5.1 Initial Parameter Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
615 615 615 620 620 626 631 636 637 639 645 646 648 649 652 655 656 664 668 668
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Contents
17.5.2 17.5.3 Problems . . . 17.5.1
Displacement Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bolotin Approximate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........................................................................ Structures with Finite Number of Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . .
675 680 685 686
18
Special Topics of Structural Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 Timoshenko-Ehrenfest Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Flexural Vibration of Compressed Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2.1 Fundamental Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2.2 Galef’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Traveling Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.2 Quasi-Static Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.3 Critical Velocity of Moving Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4 Parametric Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4.1 Dynamic Stability of Simply Supported Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4.2 Ince-Strutt Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.5 Vibration Protection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.5.1 Vibration Protection Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.5.2 Dynamic Absorber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.5.3 Lumped Vibration Absorber of the Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
691 691 693 694 696 697 698 698 701 703 703 705 706 706 707 710
19
Nonlinear Structural Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.1 Introduction and Types of Nonlinearities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Compressed Rods with Lateral Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2.1 Double Integration Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2.2 Initial Parameters Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2.3 P-Δ Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.3 Static Nonlinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.3.1 Features of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.3.2 Transversal Vibration of a Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4 Physical Nonlinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4.1 Features of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4.2 Transversal Free Vibration of Uniform Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5 Geometrical Nonlinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5.2 Stability of a Flexible Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
715 715 716 717 719 723 726 726 727 730 730 730 733 733 734 739
20
Conclusion: Once More About Modeling of Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.1 Some Problems of Structural Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Common Modeling Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.3 Differences Between Structural Analysis and Direct Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . .
741 741 742 744
21
Fundamental Developments in the History of Structural Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745
Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 749 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 787 Name Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791
Introduction
The knowledge of certain principles easily supplies the knowledge of certain facts Claude Adrien Helvétius Essays on the Mind. Essay III, Chap.1, page 196; A Speech lately made in the Court of King’s Bench: The Oxford Magazine, July, 1768, page 11. Entities should not be multiplied without necessity William Ockham’s razor
The Introduction contains general information about important concepts in the course of structural mechanics. This includes the subject and purpose of the course, assumptions, classification of possible types of analysis of engineering structures, and fundamental concept “design diagram of a structure,” which penetrates absolutely in all sections of the course. Also, loadings and modeling of some structural members are discussed.
The Subject and Purposes of the Structural Mechanics The subject and purposes of the theory of structures in the broad sense is the branch of applied engineering that deals with the methods of analysis of structures of different types and purposes subjected to arbitrary types of external exposures. Analysis of a structure implies its investigation from the viewpoint of its static behavior (strength and stiffness), stability, and vibration. The purpose of analysis of a structure from a viewpoint of its strength is determining internal forces, which arise in all members of a structure subjected to the time-independent external exposures, and displacements of specified points of a structure as a result of external exposures. The internal forces produce stresses; the strength of each member of a structure will be provided if their stresses are less than or equal to permission ones. The stiffness of a structure will be provided if its displacements are less than or equal to permission ones. The purpose of stability analysis is to determine the compressive loads on the structure elements, which leads to the appearance of new forms of equilibrium. The transition to a new form usually occurs suddenly and leads to collapse of a structure; the corresponding loads are referred to as critical ones. The stability of a structure will be provided if acting loads are less than the critical ones. The purpose of the free vibration analysis of the structure is to determine the frequencies and corresponding shapes of the vibration. These data are necessary for the analysis of the forced vibration. The purpose of the forced vibration analysis is to determine the response of the structure subjected to arbitrary timedepending excitation, to estimate the effect of such disturbance and to avoid dangerous phenomenon (a resonance), at which a sharp increase in the amplitudes of vibration occurs. Both sections—free and forced vibration analysis—are united by the common title dynamic (or vibration) analysis. These sections—strength, stability, and vibration—constitute the structural mechanics. Other identical names of this science are structural analysis, theory of engineering structures, classic theory of structures, etc. The term “classic” means each section of the course of structural mechanics contains rigorous analytical methods. These methods are based on a number of assumptions and deep ideas that provide analytical solutions. It is important that these solutions can be tested by various methods and therefore can be considered as classical solutions. In the classic course of structural mechanics, the object of research is the rod structures. This choice is easy to justify. First of all, rod systems are widely used in real engineering structures. Then, the rod systems are a simple elastically deformable xix
xx
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object, where new concepts and methods can be easily tested and mastered. Finally, structural mechanics immediately follows the first engineering discipline “Mechanics of Materials,” which considers a single rod as the main object of research. It is in this course that the reader receives the first information about analytical methods for analyzing the stress-strain state of elastic elements subjected to various kind of influences. With the development of the science of strength there are many associated names of eminent scientists. Origins of mechanics of materials and structural mechanics as the strength sciences are linked to names of Leonardo da Vinci (1452– 1519) and Galileo Galilei (1564–1642). Robert Hooke (1635–1703) and Edme Mariotte (1620–1684) independently suggested ratios that are fundamental to the modern classical theory of structures. The further development of these sciences is associated with the names of Jacob Bernoulli (1654–1705), L. Euler (1707–1783), J.L. Lagrange (1736–1813), C.L. Navier (1785–1836), J.C. Maxwell (1831–1879), О. Моhr (1835–1918), G. Lamè (1795–1870), B.P.E. Clapeyron (1799–1864), C.A. Castigliano (1847–1884), Barre de Saint-Venant (1797–1886), J.A.C. Bresse (1822–1883), Lord Rayleigh (1842–1919), and many others. They enriched science with new problem statements, introduced rich fundamental ideas, and developed new theorems and methods for solving problems. Thanks to their works, the modern classical structural mechanics is distinguished by elegant simplicity and perfection. Since the emergence of new concepts is not very frequent, the variability of primary concepts and methods of the classical structural analysis (mechanics) is low. Inquiries of practice lead to the emergence of new materials, structures, new operating conditions, and increased requirements for the reliability of engineering structures. Since the beginning of the twentieth century, a rapid development of various sections of the mechanics of a solid deformable body has occurred. Among them are new sections of science (plasticity, creep, rheology); new materials (nonlinear, composite, anisotropic); new thin-walled structural elements (plates and shells); new directions in science (nonlinear analysis, dynamic stability, theory of reliability, theory of optimal structures, finite element method), etc. A huge team of scientists in different countries is effectively working in these and related areas. In any engineering product, in one form or another, the achievements and results of the science of strength are always present. Characteristic of modern mechanics of deformable body is the use of a serious mathematical apparatus and the extensive use of powerful computers. A number of current trends in the structural mechanics should be noted: • Design and analysis of special-purpose structures (ships, planes, ocean platforms, space technology) • Design and analysis of units with intensive dynamic loads and heavy operating modes (rotary excavators, pipe rolling units) • Analysis of structures under special effects (blast waves, nuclear explosion), taking into account the physics-chemical processes occurring in them (corrosion), under special operating conditions (weightlessness, radiation, and thermal effects), etc. • Protection of structures from harmful effects (radiations, dynamics) • Optimal design of structures • Analysis of structures during their operation under conditions of loading uncertainty This series of actively developed areas is far from complete, but for the reader who begins to explore classical structural mechanics, this short list will be useful.
Modeling of Engineering Structures and Principal Assumptions All types of structural analysis of an arbitrary engineering structure are performed on the basis of replacing the actual structure with its idealized scheme. Idealization of the system is carried out in three main directions. They are (1) idealization of geometric shapes, connections, and supports; the result of these idealizations is the design diagram of structure; (2) idealization of material properties; and (3) idealization of the load. This whole set of issues is called the modeling of engineering structures.
Design Diagram Real structures (residential buildings, bridges and towers, diversified industrial structures, ships and aircrafts, etc.) are characterized, as a rule, by the complexity and variety of constructive forms. It is impossible to carry out an analysis of the structure, taking into account all its design features, and there is no need for this, since many features have an insignificant
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effect on the operation of the structure. Therefore, when analyzing a real structure, its constructive scheme (type of structure, its elements, connections, supports, etc.) is replaced by its design diagram, which is an idealized and simplified model of a structure. This diagram reflects only the basic properties of a real structure. The design diagram of a structure is one of the most important concepts of the theory of structures. The choice of the design diagram and its justification is the first and exclusively responsible stage of the structural analysis. This choice depends on many factors. Among them are the purpose of a structure and the conditions of its operation, the type and objective of the analysis, the required accuracy of computations and available software, etc. A small refinement of the design diagram can lead to significant complication of the analysis, while a simplified design diagram can lead to an unacceptable error. One of the principal requirements for the choice of the design diagram is taking into account the constructive features of a structure that significantly affect its behavior. Among the most important features of any structure are the following: – Types of members of the structure – Types of supports – Type of connections Among characteristics of the secondary importance are shapes of cross sections of members, existence of local reinforcements or holes, size of supports and joints, types of member connections (welding, riveting, bolts), etc. Few general rules of representing a real structure by its design diagram are the following: – – – –
A structure itself is presented as a set of simple structural members. Real supports are replaced by their idealized supports. Any connection between elements of a structure is replaced by idealized joints. Cross section of any member is characterized by it area and/or moment of inertia.
Let us consider each of these elements of the design diagram. 1. Types of members: In many engineering structures, it is possible to discern one-dimensional members, for which one dimension (length) substantially exceeds two other sizes (width and thickness). Such an element is called a rod. An example of rod structures is a bridge truss, a power line pylon, etc. The axis of the rod can be straight as well as curvilinear. In the last case, the curvature of the rod is assumed to be natural, but not the result of deformation of the rod. If the rod experiences bending deformations, then it is called a beam. One-dimensional flexible elements that perceive only axial tensile forces are called cables. They are used as the main element in suspension bridges. It is possible to use two-dimensional elements for which two dimensions (length and width) are significantly larger than the third dimension (thickness). If the surface of a two-dimensional element is flat, then such an element is called a plate, otherwise—a shell. There are shells of various types. Among them are cylindrical shell, spherical shell, cylindrical panel, and others. The rocket body is a cylindrical shell. The hull of the fuselage, wing of the aircraft, and the hull of the ship can be represented as a set of plates and shells with reinforcing one-dimensional elements. The analysis of the stress-strain state of such systems is carried out using the theory of plates and shells. Three-dimensional elements are possible, for which all three dimensions (length, width, and thickness) are comparable. Such elements are called solid (massive). Analysis of the stress-strain state of such bodies is performed by the methods of the theory of elasticity. Structures formed from the above members are called the bar structures, plate (shell) structures, and solid structures (massive), respectively. These structures are presented in Fig. 1. In this book only one-dimensional members (bar structures and cabled structures) are considered. Bar (rod) structures may be planar and space structures. Planar structures contain all members and loads in one plane. Pure planar structures seldom occur in engineering. However, the regular space structures often may be presented as a set of planar structures with loads lying in its plane. A simplest example is a bridge, which contains two braced trusses with a load between them. Analysis of this spatial structure may be performed with sufficient accuracy on a base of one planar truss.
a
b
c
Fig. 1 Structures formed from different members: (a) Bar structure; (b) plate structure; (c) solid structure
xxii
Introduction
On the design diagram, the elements of structure are depicted by axial lines. The design diagram also contains idealized supports and idealized connections of elements (idealized joints), dimensions of the structure, elastic characteristics of the material, and geometric characteristics of the cross section of each element. It is convenient to use relative units: if the bending stiffness of two elements differs twice, then it is convenient to treat the smallest stiffness as 1ЕI, and then the stiffness of the second element will be equal to 2ЕI. This allows us to consider this design diagram for similar structures with different stiffness EI. 2. Types of supports: The supports of any structure may be rigid or contain elastic members. Rigid supports and their kinematical and static characteristics are presented in Table 1. This table also contains the elastic curves near each support; these curves are shown by dotted line. Kinematical characteristics of the support show its ability for displacement; static characteristics of the support show its ability to perceive the load. Table 1 Types of the supports and their characteristics
Hinged movable support (roller) Presentation of supports, displacements and reactions
Kinematical characteristics Static characteristics
Δhor
Hinged immovable support (pinned) H
Clamped support (fixed, built-in)
H
H
θ
θ
R
R
Δvert = 0
Δvert = 0
Sliding support
M R
M
Δvert
Δvert = 0
Δ vert ≠ 0
Δhor ≠ 0
Δhor = 0
Δhor = 0
Δ hor = 0
θ =0
θ =0
θ =0
R≠0 H≠0
R=0 H≠0 M≠0
θ =0 H=0 M≠0
R≠0
• Roller support (hinged movable support): Schematically, this support is represented by one support bar with two ideal hinged ends. This support prevents displacement along the support bar; thus, reaction R of this support passes along support’s element. This support allows two displacements at the end of the member’s structure, such as angular displacement θ and linear displacement Δhor in direction, which is perpendicular to the support bar. • Pinned support (hinged immovable support): Schematically, this support is represented by two support bars with ideal hinges at their ends. This support prevents displacement in two mutually perpendicular displacements and allows only angular displacement θ. Reaction of this support passes through the top hinge, but direction of this reaction remains unknown. This reaction can be resolved into two components of any direction. • Clamped support (fixed support): This support completely excludes linear and angular displacements at the support point. Three unknown reactions (two forces and a moment) arise in this support. • Sliding support: This support allows only vertical displacement Δvert of the end of the member while the horizontal displacement and the angle of rotation are equal to zero. One unknown force and moment arise at this support. The static and kinematic characteristics are interrelated as follows: if displacement (linear, angular) is possible, then there is no corresponding reaction (force, moment) and vice versa, if reaction is possible, then there is no corresponding displacement. Note that the type of reaction (force, moment) depends only on the type of support, but not on the loads, while the value of reaction depends on type, value, and direction of external loads. All constraints, presented in Table 1, are two-side supports. It means that if constraint prevents displacement in any direction, then it prevents displacement in the opposite direction also. The one-side constraint means that the support prevents displacement in one direction and allows displacement in the opposite direction. In classical structural mechanics, only two-side supports are considered. In addition to the marked supports, the elastic supports should be noted; their properties will be considered in the special sections. 3. Types of connection of elements: The members of a structure may be connected together in various ways. In real systems, the connection of two and more rods is elastic compliant; two limiting cases lead to the rigid and hinge joint. In general, three classical types of connections of one-dimensional elements are possible (Fig. 2). 1. Simple hinge (Fig. 2a): One hinge connects two elements in the joint. Note that, in the second sketch, shown in Fig. 2a, the hinge does not divide a vertical element (in the third sketch the hinge does not divide a horizontal element). It means that the hinge does not divide the element (vertical and horizontal) into two separate elements. Therefore, all three sketches in
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Fig. 2a present two-rod joint (connection). Hinged joint can transmit axial and shear forces from one part of the structure to the other; the bending moment at the hinge itself is equal to zero. 2. Multiple hinge (Fig. 2b): The hinge is put into the vertical element, that is, divided into two separate elements. Equivalent presentation of multiple hinge is shown in the second sketch in Fig. 2b. Thus, multiple hinge connects three (or more) elements in the joint. The multiple hinge is equivalent to n1 simple hinges, where n is a number of members connected in the joint. 3. Rigid joint which connects two elements, 1 and 2, is shown in Fig. 2c. The internal forces (axial N, shear Q, and moment M ) in sections infinitely close to the geometric center of the joint are shown in Fig. 2c. From a static point of view, a rigid connection of two elements means the transfer of a bending moment from the end section of one element to the end section of another element without change. From the kinematic point of view, the rigid connection of two elements means that the joint can displace and rotate as a whole; however, mutual angular displacement of two sections located infinitely close to the geometric center of the joint is zero. The properties of the elastic joints will be considered in the special sections of the textbook.
a
b
c
M2 2
N2
1 Q1
Q2 M1
N1
Fig. 2 Types of joints of a structure: (a) simple hinge; (b) multiple hinge; (c) rigid joint and internal forces in the vicinity of joint
The type of connection of the elements, type of supports, and loads determine the nature of the deformation of the entire structure, distribution of internal forces, method, and procedure for its analysis. Figure 3a shows the structure with rigid joints. Elastic line of a structure is shown by dotted line. In the fixed support A, according to Table 1, the angle of rotation (slope) is equal to zero; in the hinge support B the slope φ is not equal to zero. If in the rigid joint C the angle between the elements before load application is α, and then after the deformation it remains the same.
a
b
F
C
F
1
α α φ A
c
F
1
2
R1
R2
F
M1
F
M2
2 R1
R2
B
Fig. 3 Portal frames with different types of supports and joints. (a) Deformation shape; (b, c) forces acting on the cross bar
Let the structures in Fig. 3b, c be subjected to arbitrary vertical force F. In the case of design diagram with hinged joints (Fig. 3b), the state of cross bar does not depend on the presence of columns, and is determined only by a given load. In the case of rigid joints (Fig. 3c), the cross bar is subjected to external load F and moments M1 and M2, which occur at the connection of the cross bar with column. The magnitude of these moments should be determined from the analysis of the entire system as a whole.
Idealization of Material Properties This idealization is associated with the concepts of continuity, homogeneity, ideal elasticity, and isotropy. The concept of continuity means that within geometric boundaries of the body, the whole medium is filled with material completely and continuously. This concept is so universal that it should be attributed rather to the most important principle of
xxiv
Introduction
the mechanics of a deformable body as a whole, but not to the narrower problem associated with the design diagram of a structure. The concept of continuity allows using the methods of analysis of infinitely small. This concept contradicts to the theory of the molecular structure of matter. Since we are exploring objects with sizes significantly exceeding the interatomic distances then the use of this concept is acceptable. The concept of homogeneity of the material means the invariability of the properties of the medium within the geometric boundaries of the body. The continuum medium is endowed with properties that correspond to the properties of a real material. The properties of the medium can be schematized in different ways, depending on the properties of the actual material and the objectives of the study. In the classic course “structural mechanics,” an ideal elastic material model is adopted. This does not exclude other models, for example, elastic-plastic and absolutely rigid. When analyzing a structure with different material models, we do not consider changes in the crystal lattice, but only establish the external manifestations of the processes occurring in it. Reflection of these changes is the nature of the stress-strain diagram and the numerical values of the mechanical characteristics of the material obtained experimentally. Such an approach to the description of material properties is called phenomenological, i.e., “describing the phenomenon.” The concept of ideal elasticity means the ability of the body to fully restore its original shape and size after eliminating the causes which led to deformation of the body. Behavior of a perfectly elastic material obeys Hooke’s law: The deformation of material at each point is directly proportional to the stresses at that point. Hooke’s law is valid for many materials but at stresses not exceeding the proportionality limit. At stresses exceeding the elastic limit, residual (plastic) deformations occur in the material, which remain in the body after the load is removed. The concept of isotropy of a material means that the mechanical properties of the material are the same in all directions. In the classic course of structural mechanics an isotropic material model is adopted. Steel is one of the isotropic materials; among anisotropic materials are wood, reinforced concrete, laminated materials, etc. Constructive anisotropy is possible, which is related to all construction in whole. Such a structure represents a set of similar structural blocks. Constructive anisotropy can be easily implemented in the space framework type of rod slab.
Load Idealization The loads which act on the structure are a very important element of the design diagram. External forces can be classified by several signs described below.
The Location of Points at Which Forces Are Applied to the Body This is one of the most significant signs of force. On this sign, all forces are divided into superficial and volumetric. Volumetric forces include the forces of weight, inertia, magnetic ones, etc. These forces are the result of the interaction of bodies that are not necessarily in contact with each other; they are distributed over the volume of the body, and their dimension is [force/length3]. Superficial forces are the result of the interaction of the given body with other bodies, the flow of liquid, etc. The contact of the bodies always occurs at a certain pad; therefore, all interaction forces are distributed over the surface. It means that there is no existence of the concentrated forces in nature. The unit of superficial forces is [force/length2]. If a contact area is small compared to the size of the body, then the idealized concentrated force means only the resultant of the entire load that acts within the contact area. Idealization in the form of concentrated force is possible only in those cases, when the stress-strain state of the system is studied “as a whole.” This idealization is governed by Saint-Venant principle (1855), according to which in sections sufficiently distant from the places of application of the load, the deformation of the body does not depend on the particular method of loading and is determined only by the static equivalent of the load. This principle is obvious and does not require proofs, but it is illustrated by a large number of examples. The concept of “concentrated force” refers only to the design diagram. Analysis of stresses and strains within a small area is the subject of study of the theory of contact problems. These problems go beyond the classical theory of structural mechanics and are studied in the course of the theory of elasticity. Idealization of a real superficial load allows representing the load distributed over the length; the dimension of such a load is [force/length]. The following idealized loads are concentrated moment and moment distributed along the length. Their dimensions are [force length] and [moment/length], respectively.
Introduction
xxv
The Nature of Change of Forces in the Process of Their Application By this sign, the forces are divided into static and dynamic. Static force means that it is applied so slowly that the arising accelerations of a structure, and consequently the inertia forces, are small and can be neglected. In other words, if the load is such that the kinetic energy of structural members is small compared with the potential energy of elastic deformation, then it can be assumed that at any time the structure is in equilibrium; this assumption leads to the concept of “static load.” Otherwise, the load is considered as dynamic. Dynamic forces can be impact and variable. Impact force means that the change in external force and the state of the body due to this force occurs in a very short period of time. In general, the law of change of force as a function of time can be arbitrary. Of particular importance is the case of harmonic force. Dynamic loads cause vibration of deformable structures.
The Nature of the Dependence of Forces on the Deformation of the Structure There are possible types of loads that not only cause strain but also depend on the strain. In problems of stability of deformable structures, it is necessary to not only specify the magnitude and direction of the force (as is sufficient in the static problems of structural analysis), but also indicate the nature of behavior of the force in the process of deviation of the system from the original equilibrium position. The nature of the load behavior depends on the type of transmitting device between the force and the rod. In the subsequent discussion we assume that the load remains in its original direction and does not depend on the deformations, unless otherwise specified. For more details, see Sect. 15.1. If we talk about the load as a source of stress and strain in the structural elements, then we should mention other factors that cause its stress-strain state. Among them are the displacement of the supports, inaccuracy of manufacturing and installation, effects associated with changes in material properties (e.g., concrete shrinkage), thermal effects, etc.
Duration of the Influence of Loads on Structures By this sign, the forces, depending on the nature of their occurrence, are divided into two large groups. They are the dead and live loads. The dead loads are always static loads, while live loads may be static and dynamic ones. Dead loads have a constant magnitude and fixed position and are permanently attached to a structure. Among them are 1. The load caused by self-weight of all structural elements (columns, beams, overhead slabs, walls, etc.); the special parts of an industrial building (crane girder, skylight, braces, etc.). 2. The load caused by the self-weight of fixed service equipment (ventilating and air-conditioning equipment, plumbing fixtures, electrical cables, etc.). 3. Internal dead loads which arise in the elements of the structure in the absence of external loading: The element subjected to this type of dead load is called a prestressed element. Internal dead loads occur under creating of the specific conditions. Live loads present the extensive class of loads, the duration of which is not permanent. These loads, depending on their nature, may be divided into the following classes: 1. Environmental loads (snow, wind, earthquake, hydrostatic and soil pressure, etc.): Additional environmental loads should be considered, such as the stream current, ice, ocean waves, etc. These types of loads act on the piers of a bridge and on other marine constructions. It is obvious that the same structure in different geographic regions is subjected to different environmental loads. 2. Loads caused by the building being occupied, used, and maintained: Among them are loads in the libraries, loads due to accumulation of people in concert halls, industrial service loads due to the stationary equipment (machine tools, engines, conveyers, etc.), movable incorporated equipment in factories, and moving loads (highway and railroad bridge loads). 3. Special types of loads (loads from supersonic aircraft and blast waves, factory crane loads, to name but a few). The live loads may be static and dynamic (which in turn may be further classified as periodical, impact, etc.). Live loads may have varying directions and points of application. Computation of the dead and live loads and determination of the most unfavorable combinations are governed by special documents specific for region of construction.
xxvi
Introduction
Idealization of the Structure as a Whole We note a number of general assumptions related to the structure as a whole. 1. Displacements of a structure are small. It means that deflections do not change original design diagram of a structure. In this case we say that analysis of the structure is performed on the base of its entire or non-deformable design diagram (Navier assumption, 1826). Analysis of a structure by the deformable design diagram assumes that determination of internal forces of the structure is performed on the basis of corrected design diagram, which takes into account deformations of the structure, caused by applied loads. Such an approach should be applied, in particular, for the analysis of tall and flexible structures. Generally we will consider analysis of structures based on non-deformable scheme. Few examples of analysis of structures based on deformable scheme are considered in special sections. 2. Parameters of a structure are not changed under loading. 3. Superposition principle of the force action is applicable. This principle means that any factor (reaction, internal forces, displacement, etc.), caused by complicated loading which acts simultaneously, can be determined by adding together the required factor due to each load separately. If the structure is loaded by inclined forces then it is convenient to present each of these forces as horizontal and vertical components. In this case the component of required factors is equal to the algebraic sum of this factor due to each load component separately. The superposition principle is based not only on the assumption of a linear relationship between forces and displacements, but also on the assumption of the reversibility of the processes of loading and unloading and the invariability of the design diagram of the structure (Feodos’ev 1975). The superposition principle in the problems of the structural analysis is of fundamental importance. Firstly, this principle allows at various influences to obtain a solution as a result of the addition of particular solutions. Secondly, this principle allows to prove the most important theorems (reciprocity theorems, Mohr integral, Castigliano’s theorem, etc.), on the basis of which fundamental methods for the analysis of structures (force method, displacement method, etc.) were developed. 4. Saint-Venant principle (1855) is valid: If the dimensions of the applied external load area are small (compared to the cross section of the rod), then in sections sufficiently distant from the point of application of the load, the stresses and strains depend little on the way the load is applied. This principle is illustrated in Fig. 4. Cantilevered beam is loaded by the load P (Fig. 4a), but the ways the load is transferred are different. They are using one and two lugs (Fig. 4b, c), respectively, plate connected with beam using bolts or welding (Fig. 4d).
a
P
b
c
d
P
P
P
Fig. 4 (a) Design diagram; (b–d) different ways of load transfer
The Saint-Venant principle significantly expands the scope of application of the basic design formulas, since there is no need to take into account the specific way of application of the load and features of the distribution of forces in the area adjacent to the application of the load. The variety of modern engineering structures, their features, operating conditions, assumptions, types of analysis, and accuracy requirements is so great and varied that it is impossible to unite them within the same discipline. Therefore, a large number of border disciplines emerged that constitute in a broad sense mechanics of deformable solid body. For each of the disciplines included in this cycle of sciences, a certain system of assumptions is characteristic. As an object a separate structural element can be taken, for example, plate and shell. The study of such elements is considered in the theory of plates and theory of shells. Some aspects of the behavior of such elements are considered in the course “stability and vibration of plates and shells.”
Introduction
xxvii
As an object of study the whole structure can be taken (aircraft, ship, bridge, high-rise building, etc.). The study of such structures is considered in the courses “the theory of the aircraft,” “the theory of the ship,” “bridges,” etc. The following disciplines of the general cycle—“the theory of elasticity” and “the theory of plasticity”—solve the problem of analyzing structures and their elements with more precise methods; it is clear that the main assumption in these disciplines is associated with such a property of the material as elasticity and plasticity. The rejection of the isotropic property of a material leads to a new discipline—“the theory of anisotropic plates and shells.” The rejection of the assumptions adopted in the classical course (for example, the material of the structure does not obey Hooke’s law, one or several connections turn out to be one sided, the deformations of the structure are large) leads to an extensive class of nonlinear problems of mechanics of a solid deformable body. For nonlinear systems, the superposition principle in general case is not applicable. The subject of study of such systems is “nonlinear mechanics of deformable bodies.” Structural mechanics as a part of generalized scientific discipline “mechanics of solid deformable body” takes a worthy place. Many of the provisions of structural mechanics find application in each of the scientific disciplines noted above. In any engineering product, in one form or another, the provisions and methods of structural mechanics as a science are invariably present. Even this short list of scientific disciplines related to the analysis of structures shows the importance of classical structural mechanics in creating numerous engineering objects of different types and purposes. It is pertinent to emphasize once again that classical structural mechanics contains a number of fundamental principles, theorems, and in-depth analytical methods for analyzing engineering structures. The basis of these theorems and methods is the set of special assumptions, which lead to linear structural mechanics; a feature of linear mechanics is the validity of the superposition principle. In the future, it is in this sense the term “classical structural mechanics” will be used. It is important that for the adopted design diagram of structure, various methods of classic mechanics lead to identical results. A diverse set of aspects related to the design diagram of structure can be found in an excellent book (Perelmuter and Slivker 2003).
Types of Structures The different combinations of internal forces arise in the cross section of the structural members. They are the bending moments and shear forces; the bending moments, shear, and axial forces; only axial forces (compressive and tensile) or only tensile, etc. Therefore, the behavior of such structures is different. According to the type of internal forces and constructive features of an object, the structures are classified as follows: multispan beams, trusses, frames, arches, cables, and combined structures. Multi-span beams contain intermediate supports (Fig. 5). Therewith beams with intermediate hinges (Fig. 5a) or without them (Fig. 5b) are possible. Multi-span beams are subjected to transverse load and undergo bending deformations. In the cross sections of beams the bending moments and shear forces arise. In the intermediate hinges 1 and 2 (Fig. 5a) the bending moment equals zero. The presence of an intermediate hinge leads to the kinks of the elastic line (EL) of the beam, as shown in Fig. 5a. If there are no intermediate hinges, the elastic line of the beam does not have the kinks (Fig. 5b).
a
2 EL
1
b EL
Fig. 5 Design diagrams of multi-span beams: (a) Statically determinate beam; (b) statically indeterminate beam
The frames are characterized by the presence of at least of one rigid joint; design diagrams of several frames composed of straight members are shown in Fig. 6a–d. In a general case, in the members of a frame arise the bending moments, shear, and axial forces. Axial forces may be tensile and compressive.
xxviii
Introduction
a
b
*
A
*
c
d
C A
*
*
B
e
H A
*
B D
D
H A
* B D
Fig. 6 Plane frames. Different design diagrams and corresponding elastic curves
In any deformable structure, distribution of internal forces and the nature of its deforming depend on the type of supports, the ways of connection of the elements, and the way of loading of a structure. Figure 6a shows a frame with supports which permit a linear (horizontal) displacement of joints (sideway frame). Each rigid joint has an angular displacement; however, the mutual angle of rotation of adjacent elements of each joint is equal to zero. Assume that the frame is subjected to horizontal force. Corresponding deformable shape of the frame is shown by dotted line. The elastic curve within each cross bar has inflection point, which is denoted by the symbol (*). If according to some requirements it is necessary to exclude the bending of the right pillar, then this can be achieved by introducing a movable right support instead of a pinned support (Fig. 6b). New movable support allows the horizontal displacement; therefore, horizontal reaction of this support is absent. As a result, the right vertical element is not bent, and perceives only axial force. Figure 6c represents another design diagram of a frame and corresponding deformable shape; thanks to the clumped support А the joint C of the frame does not have horizontal displacement. For a given load, the rigid joint C of the ACD frame rotates; this leads to the appearance of inflection points within both cross bars. Assume that we need to reduce the influence of one part of the structure, ACD, on the other, CB. This can be achieved by introducing the hinge H, as shown in Fig. 6d. The hinge H is simple because it does not cut in the rigid joint C, but connects only two elements of the frame. They are the Γ-shaped left side ACD of the frame and cross bar HB. It is obvious that the elastic lines in Fig. 6c, d are different. To create greater independence in the work of individual parts of the structure, it is possible to adopt the design diagram of the structure as shown in Fig. 6e. If the hinge H is inserted into the rigid joint, it becomes a multiple, since it now connects the three elements, AH, HB, and HD. The absence of at least one rigid joint turns the frame into a simple connection of three rods, which does not represent the frame. Thus, it becomes obvious that the construction of the design diagram of a structure can be performed both at the stage of analyzing the existing structure and at the stage of designing the structure in order to impart it certain properties. A truss is a structure composed of straight members connected at their ends by hinges (Fig. 7).
a
Fig. 7 Plane trusses. (a) Simple truss; (b) truss with subdivided panel
b
Introduction
xxix
The hinged connection of the truss elements is adopted on the basis of a number of easily realizable assumptions. Their implementation leads to the fact that in the elements of the truss only axial forces arise that are characteristic just for straight elements with the hinges at the ends. On the other hand, these assumptions significantly simplify the procedure of truss analysis. One of the assumptions requires that forces be applied to the truss joints. The corresponding simple truss is shown in Fig. 7a. Truss with subdivided panel (Fig. 7b) contains additional elements. They satisfy the above condition and reduce the length of the panel. The number of different types of trusses is very large. An arch is a curvilinear structure (Fig. 8a, b) in which the vertical load causes not only vertical reactions, but also equal horizontal reactions H, called a thrust. Various types of arches are possible. Among them are three-hinged arch (Fig. 8a), two-hinged arch (Fig. 8b), etc. If the arch is subjected to inclined force (Fig. 8c), then unequal horizontal reactions H1 and H2 occur, which cannot be called a thrust. In the cross section of the arch three types of internal forces occur: bending moment, shear, and axial forces. Axial forces in arches are always compressed.
a
c
b
H
H
H
H
H1
H2
Fig. 8 Plane arches. (a, b) Vertical load: the equal horizontal reactions H are thrust; (c) the force is inclined: the non-equal horizontal reactions H1 and H2 are not the thrust
A cable presents a perfectly flexible element of an engineering structure. Cabled structures contain a cable, which supports a beam (Fig. 9). In the cross sections of cables only axial tensile forces arise.
a
b
Pillar
Mast
Hangers Cable
Guy-rope
Span
Fig. 9 Cabled structures. (a) Suspension bridge; (b) guy-rope of the mast
A combined structure contains the parts subjected to different types of deformations. Two combined structures are presented in Fig. 10. In both of them, element 1 is subjected to bending deformation and elements 2 to axial one.
a
1
b
1 2
2 Fig. 10 Combined structures
xxx
Introduction
Another example of combined structures is presented in Fig. 9a. Indeed, for structure in whole, the different types of internal forces arise in all members (cable, hangers, pillars, and span). By the nature of the reactions, the structures may be classified as thrusted and non-thrusted (thrustless). Non-thrusted structures are such structures, in which the vertical loads lead to vertical reactions only (Fig. 11a, b).
b
a
Fig. 11 Non-thrusted structures
Thrusted structures are such structures, in which the vertical load causes not only the vertical reactions, but also the horizontal reactions as well. The typical thrusted arched structures are presented in Fig. 8. If in the non-thrusted structures in Fig. 11 the movable support is replaced with a pinned or clamped support, then both systems become the thrusted structures (the thrusted frame and arch, respectively). Assume that the structures in Fig. 11a, b are subjected to inclined force. Of course, this will cause a horizontal reaction in the left pillar. However, it cannot be interpreted as a thrust. Structures may also be classified as statically determinate and statically indeterminate. If all reactions of supports and all internal forces can be obtained using equilibrium equations only, then such structures are called statically determinate; simplest statically determined structures are shown in Figs. 5a, 7, 8a, and 11. Otherwise, the structure is called statically indeterminate. If to determine the reactions of a structure the equilibrium equations are not enough, then the system is called externally statically indeterminate. The example of externally statically indeterminate structures is shown in Figs. 5b and 8b, c. If to determine the internal forces in the elements of a structure the equilibrium equations are not enough, then the system is called internally statically indeterminate. The truss shown in Fig. 12 in the central panels contains two diagonal elements which are not connected at the points 1 and 2. This truss is externally statically determinate; however, it is an internally statically indeterminate structure.
1
2
Fig. 12 Externally statically determinate, and internally statically indeterminate truss
Equilibrium equations for a statically determinate structure define unique values of reactions of supports and distribution of internal forces. However, equilibrium equations for statically indeterminate structure do not define unique values of reactions of supports and distribution of internal forces. To determine them, additional conditions should be considered, the so-called conditions of compatibility of deformations. By the arrangement of the rods, the structures are subdivided into flat and spatial. A flat is a structure for which all the elements and the load are located in the same plane. If arch is subject to a horizontal load caused by train braking, then such a system is flat. If arch is subject to a horizontal wind load directed at an angle to the arch plane, then such a system should be considered as a spatial. Behavior of every above structure has its specific features; the methods of their analysis will be considered in corresponding chapters of the book.
Introduction
xxxi
Modeling of Structural Members In some cases, the engineering structure is a system with a high degree of homogeneity. This concept refers to loading, elements of a system, joints, and supports. In this case, the behavior of a separate element of the structure will coincide with the behavior of other adjacent elements, or at least slightly differ from them. The multifaceted concept of “modeling” in the narrow sense of the word means replacing a complex real structure with simple elements, for which analysis turns out to be much simpler than for the system as a whole. A fundamental concept is a tributary area, which is the area that is assumed to load a given member. The modeling of some typical structures is shown below. Modeling of slab. Figure 13 presents a slab supported by two rigid ribs at the opposite sides, free at other opposite sides and loaded by a uniformly distributed load per unit area q [kN/m2]. All strips in the direction of the sides free from supporting are in the same stress-strain state. Therefore, we can model the behavior of the slab by cutting off any strip of unit width along the span a, and then consider this strip as a simply supported beam (or another way depending on the real connection between slab and rigid ribs). This strip is loaded by a uniformly distributed load per unit length q1 [kN/m]. q [kN/m2] 1m
q∙1[kN/m]
l a
a Fig. 13 Modeling of a slab
Modeling of columns. Regular grid of columns is presented in Fig. 14. The columns support a slab, whose weight and live load q [kN/m2] are distributed over the area of the slab. The given structure is a slab on point supports. Suppose the problem is to determine the dimensions of the foundation of the columns. To do this, it is necessary to determine the load on each of the columns. This is a challenging problem of the plate theory. The concept of modeling allows us to simplify this problem. Obviously, presented design diagram of a single column is approximate, since it does not reflect the boundary condition at the top of the column.
A
B a
D
C a
a
qab
1 CL
b
b/2
2
b/2
b
CL
3 a/2 a/2 CL
CL
CL
Fig. 14 Rectangular column grid. Tributary area of different columns and loading of column B2
Each column is subjected to a different load. The tributary area of each column is bounded by the panel centerlines (CL). Tributary area of columns A1, B2, C1, and D2 are shown shaded. For columns B2 and C2 tributary area A ¼ ab; therefore, these column are subjected to load P ¼ qab. Columns B1, C1, B3, C3, A2, and D2 carry the load qab/2. Columns A1, D1, A3, and D3 are subjected to load of qab/4. Supports of the columns are not discussed. Modeling of beams in different structures. Figure 15 presents a rectangular slab of weight q [kN/m2]. The slab is supported by absolutely rigid diaphragms W and two beams B, which are in turn supported by columns C; beams B and diaphragms W are parallel.
xxxii
Introduction
The precise analysis of such a system presents significant difficulties and can be performed using plate theory methods. This 2-D structure can be approximated by two beams oriented in two perpendicular directions. To model this system using 1-D elements, we need to cut out two strips in the directions of x- and y-axes, as shown in Fig. 15a. Such simplification is allowed if the plate is made of a homogeneous material, has no features (local reinforcement, attenuation), and carries the load uniformly distributed over the entire surface of the plate. The tributary area of each strip in direction x (beam B) is bounded by centerlines CL of the local panels; thus, the load on beam B is collected from a strip width a. Each member B is modeled as a simply supported beam loaded by uniformly distributed load intensity of q0a [kN/m] (Fig. 15b). x CL
a
b
CL
qa[kN/m] C
1m
y
x
l
a/2 a/2
q [kN/m2] B
W
C
B
B
W
a
B
B
C
a
y
W
W
1
C
q·1[kN/m]
c
l
a
a
a
a
Fig. 15 Regular beam ceiling. Modeling of regular beam B (b) and slab (c)
In the direction of y-axis, the slab is modeled by a continuous beam of unit width, subjected to uniformly distributed load q1[kN/m]; this beam is supported by two elastic supports B and two rigid supports W (Fig. 15c). If beams B are rigid enough, then supports B should be presented as the rolled supports. It is obvious that all cut-out strips in direction y-axis are in the different stress-strain state. Therefore, the modeling in Fig. 15c well describes the state of the slab only in the case of a significant size l and strip that cut out in the middle zone of the slab, i.e., in the vicinity of l/2. Another beam ceiling system is presented in Fig. 16. The structure Fig. 16a includes four columns, two beams B1, and two beams B2, which are supported by the columns, and a slab with a weight of q0 [kN/m2]; the length of each beam is a, so the total load on the four beams is qa2 [kN]. Bisector rule allows finding a tributary area which for each beam presents a triangle; 1 qa a ¼ qa2 . each beam carries the load as shown in Fig. 16a. The total load on all beams is indeed 4 2 2 a
a
B2 a
B1
a
b
B2 qa/2
B1
a
b
B1
45°
qa/2
B1
b
45° qb/2
qa/2
B2 a
B2 a
Fig. 16 Tributary area of specified beam and corresponding beam design diagrams
If the beams have a different length, then the bisector rule leads to another type of loading beam B2 (Fig. 16b). In any case, the type of supports is not discussed. It is obvious that for each beam the weight of the beam itself should be taken into account.
Introduction
xxxiii
If the slab is supported by the connected beams (B1, B2, B3), which are placed in different directions, then this structure is called the beam cell. Design diagram for each beam depends on how the beams are connected between each other. Below are considered two-leveled and one-leveled connections. Two-leveled connection. The columns support main beams B1. Lateral beams (girders) B2 and B3 are supported by two beams B1 as shown in Fig. 17. The feature of this connection of beams is that the load from the slab is transmitted to the transverse beams B2, B3, and through them to the main beams B1. Tributary area for beam B2 is shaded. Intensity of uniformly distributed load is qa, while for beams B3 the intensity of load is qa/2. Concentrated load on each beam B1 presents the reaction of beam B2 and equals P ¼ qab/2. The total load on all transversal beams B2 and B3 is 3qab.
1 a
a
a
B2,B3
B1
B1
B1 b
B3
B2
B2
2
B3
P/2
1
P
b
B2
2
B1 a/2 a/2
qa
Section 1-1 P/2
P Slab
Section 2-2
B2
B3
a
B3
B2
B1 a
a
Fig. 17 Beam cell, two-level connection. Tributary area of beam B2 and design diagrams of beams B1 and B2
One-leveled connection. The columns support beams B1. Beams B2 and B3 are supported by two beams B1, but unlike the aboe case, all beams are located on one level as shown in Fig. 18. The peculiarity of such a combination of elements is that the weight of the slab is distributed between the main beams B1 and transverse beams (girders) B2 and B3, and then transferred from girders to the main beams in the form of concentrated forces. Thus, the flooring load is transmitted on all beams (main B1 and lateral B2, B3) directly. Tributary areas for all beams are shown below. Slab
a
a
a
B2, B3
B1
2
R
45°
B1
R/2
a/2
a/2
3
1
b
B1
B3
B2
B3
B2
qa
B2
b
4
6
qa/2
B3
b
B1 5
R R
R
R/2
a
B1 a
qa/2
R/2
a
Fig. 18 Beam cell, one-level connection. Tributary areas of beams B1, B2, and B3 and their design diagrams
R/2
xxxiv
Introduction
Intensity of uniformly distributed load for beam B2 is qa, while for beams B3 it is qa/2. 2The tributary area for beam B2 is 1 a a a 1-2-3-4-5-6-1, so the total load on beam B2 is Q ¼ 4 2 2 2 q þ qaðb aÞ ¼ q ab 2 . For beam B1 maximum of 2 distributed load is qa/2. The lumped load R ¼ Q2 ¼ q2 ab a2 presents reaction of beam B2 which acts on the beam B1. It is easy to check that a total load (TL) on the two beams B1 equals 3qab. Indeed, 1 qa q a2 ab ¼ 3qab TL ¼ 2½3ðΔ loadsÞ þ 3ðconc RÞ ¼ 2 3 a þ 3 2 2 2 2 This load equals to the total slab weight. Thus, the design diagram of the beam cell elements essentially depends on the way of their connection with adjacent elements. Modeling of lintels. Let us consider a beam over openings, such as gates or windows. This beam, called a lintel, supports a brick wall (Fig. 19a). We need to create a design diagram of the lintel. The lintel may be presented as a simply supported (or fixed-fixed) beam, depending on the depth of the beam and how the beam is connected to the wall. But what load will act on this beam? If the lintel would be removed, then a part of the brick wall will collapse; according to experiments, it happens as shown by dotted line in Fig. 19b. It means that the remaining part of the wall supports itself, and the lintel is subjected to the collapsed part of the wall only. Thus, the design diagram of the beam is presented in Fig. 19c.
b
a
c ~ 45°
l l Fig. 19 (a, b) Lintel, and collapse of masonry wall; (c) design diagram of lintel
Modeling of rafters. Inclined members (rafters), which support the solid deck, are a typical building covering structure (Fig. 20a). This element is subjected to two types of loads. They are snow load per unit area of horizontal projection, S [kN/m2], and self-weight of the covering per unit area of inclined surface, D [kN/m2]. All inclined rafters are spaced at a [m]. We need to show design diagram of the rafter. The feature of this problem is that both loads, S and D, are gravity loads and therefore they are vertical; however, they are distributed over different planes, horizontal and inclined, respectively. Therefore, the D and S loads cannot simply be added. Snow load S, kN/m2
a Covering load D kN/m2
Rafters, spaced a
b
qhor
θ
Ridge beam
l
l Fig. 20 Rafter under dead and live (snow) loads and its design diagram
Since the length of the inclined beam is l1 ¼ l/ cos θ, a total load (TL) which acts on this beam becomes
Introduction
xxxv
D TL ¼ ðSl þ Dl1 Þa ¼ S þ l a ½kN cos θ The intensity of the load per unit length of the horizontal projection of the beam (Fig. 20b) is TL D ¼ Sþ a ½kN=m qhor ¼ l cos θ Corresponding bending moment in the middle of the beam is M¼
qhor l2 D l2 þ S a ½kN m ¼ cos θ 8 8
ðaÞ
The same result will be obtained if the total vertical load TL is distributed along the sloping length l1. Indeed, qvert inc ¼
TL D a cos θ ¼ ðS cos θ þ DÞa ¼ Sþ l1 cos θ
½kN=m:
This load is directed vertically; its projection on the axis perpendicular to the inclined beam is qinc ¼ qvert inc cos θ ¼ ðS cos θ þ DÞa cos θ
½kN=m
Corresponding bending moment is q l2 ðD þ S cos θÞ a cos θ M ¼ inc inc ¼ 8 8
l cos θ
2
This expression leads to the formula (а). Modeling of arches. The concrete arch bridge is subjected to different types of loads. The principal loads are the vertical loads due to self-weight of the arch and moving load; these loads act in the plane of the arch. Also, the arch is subjected to wind load, which can act in transverse direction (Fig. 21).
a
The fixed loads due by selfweight of the arch in y-direction
b
y
Arch
x
Arch bracing
c
y
x Wind pressure in z-direction
z
x
z Fig. 21 (a) Arch structure; (b) design diagram of structure due to vertical load is arch; (c) design diagram of structure due to transversal load is curvilinear beam
Design diagram of this structure under the vertical fixed and moving loads presents an arch, while design diagram of the same structure in case of transverse wind pressure presents the plane curvilinear beam with load which is perpendicular to the plane x–y.
xxxvi
Introduction
Features of Some Loads and Their Load Path The features of some loads, in particular crane loads, and their transfer to the elements of an industrial structure are discussed. The concept of load combination and design diagrams for some parts of the engineering structure is also considered.
Simplest Industrial Building The constructive scheme of the simplest one-span industrial building and its plan are shown in Fig. 22a, b, respectively. This industrial facility is equipped with overhead traveling cranes (or bridge cranes) for transporting heavy loads. The crane girder is supported by the columns, and the crane bridge is shown by dotted line. The crane rails are placed along the crane girders (under-crane beam) (Fig. 22c) and bridge crane can move along the building, while the trolley with load moves along the bridge, i.e., across the building; directions of movement of bridge and trolley are shown by arrows in Fig. 22a, b. The node, which includes the crane rail, crane beam, and column, is presented in Fig. 22c; the crane beam with a movable crane load and view of structural elements of building on the side are shown in Fig. 22d,e.
a
Truss F1
b
1
2
Bridge
b c
Crane girder
a
a
F2
a
a
A
Trolley motion
Crane wheel
Bridge
Trolley
l
e
Trolley
e
Column
d
B
l A
P1
Crane
c
d
Crane rail
b c
Crane wheel
Crane girder
3 Crane girder
R2
A
P2
2
Column
e
Crane girder
B
R Htr
4
3
a
R3
3 2 7
8 9 Hlon
1- columns, 2 - truss, 3 - slab of covering, 4 - crane girder, 5 - crane, 6 - bottom vertical brace between columns, 7 - upper vertical brace between columns, 8 - vertical brace between trusses, 9 - horizontal brace in bottom panels of the trusses. a,b,c,d are specified sections for analysis of the column
5
4 1
1
6 2
3
4
Fig. 22 (a, b) Elevation and plan of the industrial one-span unit; (c) R and Htr are vertical crane load and transverse braking force of a trolley, respectively; (d) design diagram of the crane girder; (e) elevation of the industrial unit in the longitudinal direction; Tlon longitudinal braking force of a bridge
The principal constructive elements are the covering trusses and stepped columns. The truss may be attached at two points F1 and F2 to side of the upper part of the columns (Fig. 22a). In this case, the vertical distance between the attachment points is significant, and therefore the forces occurring at the attachment points of the truss to the column create a significant couple. This allows us to consider the connection of the truss with the column as rigid.
Introduction
xxxvii
With a soft connection of the truss with a column, the joint should be considered as hinged; such connection occurs, for example, when the truss is supported on the column above. The ways of joining columns and truss of covering depend on many factors and are studied in the special engineering disciplines. Typical design diagram of this structure may be presented as П-shaped frame with rigid (or hinged) joints, fixed supports, and columns of variable stiffness. The stiffness of horizontal member (truss) is taken constant. For static analysis of a frame, we need to have, first of all, geometrical parameters of the frame, such as span and height of both parts of column, and the relative inertia moments for bottom and top parts of column and truss.
Specific Loads Let us adopt the design diagram as shown in Fig. 23. Assume the hinged connection between cross bar and columns. The truss (shown by dotted line) is presented as uniform absolutely rigid cross bar. The structure is subjected to the following loads: the dead load—the weight of the truss, crane girders, columns, etc., and the live load—wind, snow, crane loads, etc. Psnow
Psnow Ptr
Ptr
W+
WRmax
Htr
Htr
Pcg
q+wind
Rmin Pcg
q-wind
l A
B
Fig. 23 Design diagram and loading of the industrial one-span unit
Consider the features of some of these loads and their transfer on the structural elements of industrial building. The dead loads D are the weight of the covering, the weight of the fixed service equipment, and the weight of the truss itself. The weight of covering and service equipment should be presented as concentrated loads in the joints of a truss. Live distributed snow load S should also be presented as concentrated load in the joints of a truss. The tributary area for weight of covering and snow load is bounded by the panel centerlines and the truss centerlines. Snow load: It is required to consider fully and partially located on the truss in order to determine the greatest effect on each element. The columns are subjected to the dead and live loads. The reactions due to the weight of snow Psnow and weight of truss itself Ptr are applied at the section “a” of the column. Also, each column is subjected to dead load Pcg which is reaction of the weight of the crane girder itself. Wind loads W should be considered separately for the height of the truss and the height of the column. Each of them takes into account the positive and negative wind pressure; they act on the windward side and leeward side, respectively. The tributary area is bounded by the column centerlines. The distributed wind load at the height of the truss is reduced to positive and negative concentrated forces W+ and W, which are applied to the columns at the level of the lower chord of the truss. The positive and negative wind loads qþ wind and qwind are distributed within the height of the column to the level of the lower truss belt. For tall structures, it is needed to take into account the change in load intensity over the height. Two directions of the wind action should be considered. The crane loads present a set of concentrated wheel loads; their values and spacing are dependent on the types and rated load of crane used. A peculiarity of this type of live load is that during the service the horizontal braking inertial forces of the bridge crane itself and trolley arise. The following types of loads occur during the operation of bridge cranes:
xxxviii
Introduction
1. Vertical loads caused by the weight of the bridge crane, trolley, and traveling load. Design diagram of the simply supported crane girder between columns 2 and 3 is shown in Fig. 22d. Vertical crane loads P1 and P2 lie in vertical plane of symmetry of the girder, and transmitted on crane girder through the crane wheels. It is obvious that reactions R2 and R3 and internal forces, which arise in the crane girder, depend on the position of the crane within the girder. Moreover, the maximum bending moment in the crane girder and maximum reaction, which is transmitted on the column, correspond to different positions of the crane within the girder. The reaction of the crane girder is transmitted on the columns as vertical load. 2. Horizontal loads caused by braking of the trolley. In case of sharp stopping of the trolley, the transverse braking inertial forces arise. They are directed along transverse axis 1, 2,.., act at the head rail (Fig. 22c), and through special braking beam (it is not shown) are transferred on the adjacent columns. The magnitude of the braking force depends on the lifting capacity of the crane, but the force Htr that is transferred on each column depends on the location of the crane bridge on the crane girder. The greatest force transmitted to the column occurs when the trolley is near one of the columns. 3. Horizontal loads caused by braking of the bridge crane. In case of sharp stopping of the bridge, the horizontal longitudinal braking inertial force Hlon arises; this force acts normally to the plane of the frame, shown in Fig. 22a and perceived by a vertical brace between columns along the axes A and B. Vertical and brake lateral and longitudinal crane loads in the balance of all loads occupy a noticeable and often prevailing place. Some features of the crane and wind loads. The weight Pcg of the crane girder and crane loads R and Htr act on both columns simultaneously. If a trolley is located at the most left end of the crane bridge, and the bridge itself is near the column of row A, then the vertical load Rmax acts on the column of row A and load Rmin—on the column of row B. The total load Rmax and Rmin is equal to the weight of the whole crane and the moving load. Horizontal crane load Htr acts on column A or B in + both directions. Also, the positive qþ wind , W and negative qwind , W wind loads act on the frame simultaneously. The design diagram of a certain part of a structure depends, first of all, on its location in the structure and the way of connecting this part with the adjacent elements. Assume that we need to provide the analysis of the truss only (Fig. 22a). In the case of a hinged connection of the truss with columns, all loads acting on the truss are known; this is dead load (stationary equipment, covering, and truss) and live load (snow, wind). Therefore, the analysis of the truss can be performed independently, that is, without analyzing the whole frame. In the case of a rigid connection of the truss with columns, also unknown moments that occur in a rigid joint should be taken into account in addition to the known loads. In this case, analysis of the truss cannot be performed independently from the frame. Therefore, to perform analysis of the truss, it is necessary to first analyze the entire frame with rigid joints and determine the bending moments at joints. Thus, the role of the design diagram in the computational procedure, its influence on the complexity analysis, and time consuming becomes obvious. Internal forces (normal force N, shear Q, and bending moment M) in the special sections a–d of columns should be determined for each load separately. Internal forces in section d are transmitted to the foundation. To determine the forces in the various points of the frame caused by different loads, the principle of superposition should be applied. This book deals with plane and spatial rod structures. In all cases, we will note the assumptions underlying the analysis and accept for analysis of finished design diagram of the structure. General considerations that should be taken into account when the constructive scheme is replaced by its design diagram are also discussed in Chap. 20. This problem represents a significant part of the analysis of special engineering structures. Special type of loads. These loads mostly include dynamical loads of different nature. They are short-term loads caused by start/stop of equipment, emergency loads caused by blast or failure (for example, breakage of electric power transmission line, failure of guy-wire of a mast), and much more. We note a special type of load that should be considered for arched structure of considerable size—the snow avalanche. This load was taken into account for designing of a New Safe Confinement structure over the old sarcophagus covering the damaged fourth reactor at the Chernobyl nuclear power plant (CNPP), Ukraine. This is a unique multifunctional complex designed to localize radiation. It consists of a number of the most complex engineering substructures. The most important of them is the spatial arched structure with a special double covering. We give some technical characteristics of this unique structure, known to the whole world under the name “Arch.” The span of the arch and its height are 257 and 110 m, respectively. The length of the entire structure is 165 m. The main elements of the structure, the arch belt, are pipes with a diameter of over 800 mm. Coverage area is 86,000 m2. The Statue of Liberty (93 m) and the Colosseum (48 m, 2 ha) freely fit under this arched covering. The total weight of the metalwork is 36,000 tons. This grand structure is mounted on a special
Introduction
xxxix
platform away from the destroyed CNPP power unit, and then along the guides it was moved to 327 m, and finally, sarcophagus was safely isolated. The speed of the snow avalanche at its descent from the arch surface is 40 m/s. Therefore, the sudden descent of snow (avalanche) is in fact a sudden removal of the load with all the ensuing dynamic consequences for the arched structure. Consider some types of loads which occur on high-rise TV tower. Design diagram of a tower may be presented as cantilevered rod of variable cross section (CN Tower, Toronto, Canada; Menara Kuala Lumpur, Malaysia; Ostankino Tower, Russia, etc.). The analysis of the adopted tower structure is performed by taking into account loads in the stage of construction and operation. External dead loads include weight of the tower, weight of the radio and TV equipment, elevators, etc.; internal dead loads are caused by pretensioned reinforcement and ropes. The live loads include wind, seismic, and one-sided temperature heating of the surface of the tower body by solar radiation. Also, it is necessary to take into account a negative effect on the concrete, such as repeated cycles of alternate freezing and thawing of concrete during the year. The intensity of these cycles depends on the lower boundary of the clouds and the air humidity, which are characteristics of the construction site. A possible roll of the tower body, caused by uneven settlement of a foundation, leads to the need to analyze the structure on the deformed scheme. Antenna-mast structures consist of a vertical rod (trunk) and the inclined threads supporting it (guys). One of the variants of the design diagram of such a structure is a deformable rod on intermediate elastic supports. The main loads include the wind, ice on the guys, and sudden discharge of ice.
Load Combination Obviously, all loads, such as dead load, snow and wind, and seismic and crane loads, cannot act simultaneously. Therefore, the concept of “load combinations” is introduced. It allows to take into account the probability of simultaneous occurrence of various loads. Consider this concept as applied to industrial structures in Fig. 22a. Here we note three different parts of the structure—covering truss, crane girder, and column. Unfavorable load combinations for different parts will be different. For truss, the following loads should be included: the weight of the truss itself, truss covering, stationary equipment, wind, and snow. It is necessary to consider separately the snow on the left side of the truss, on the right, and on the entire surface; positive and negative wind pressure; and change in wind direction. Forces in each element of the truss should be determined right right total left from each load separately—longitudinal force due to dead load ND , snow N left S , N S , N S , and wind N W , N W . Under-crane girder perceives its own weight and vertical crane loads. It is obvious that the greatest shear force and bending moment in the girder occur at different positions of the bridge crane on the beam. For column of industrial building in Fig. 22, the following loads should be included: dead loads (truss, covering, equipment, crane girder) and live loads (wind, snow, crane loads). Dangerous cross sections of the columns of stepped variable stiffness are denoted as “a–d” as shown in Fig. 22a, c. The rule signs for forces should be accepted, and for each section a–b for columns determine the bending moment and the longitudinal force MD and ND, MS and NS, etc. caused by each load separately. Each combination of loads leads to a certain force in the specified point. For example, the partial load combination 1.2D + 1.6S + 0.5 W means computation of bending moment, shear, and axial force caused by dead, snow, and wind loads by formulas M ¼ 1:2M D þ 1:6M S þ 0:5M W Q ¼ 1:2QD þ 1:6QS þ 0:5QW N ¼ 1:2N D þ 1:6N S þ 0:5N W In constituting each of the permissible load combinations, the following should be taken into account (Perelmuter and Slivker 2003): 1. Load caused by the own weight necessarily present. 2. Wind, snow, and crane loads act independently of each other. 3. Wind in one direction excludes wind in another direction, one crane braking load excludes all others, and one crane vertical load excludes the other crane load. 4. When considering crane braking loads, crane vertical load should be considered as well. Comparison of various combinations of loads allows to reveal the most unfavorable of them.
xl
Introduction
This course of structural mechanics provides methods for determining of the internal forces in the elements of the rod systems caused by various influences. These forces represent the initial information for the formation of different load combinations. The most unfavorable combination allows to determine the maximum possible internal forces in the elements of structure, on the basis of which the selection of the section of elements is made. Various combinations taking into account different effects are discussed in special regulated documents. There are basic load combinations, load combination including flood load, load combination including atmospheric ice loads, load combination for extraordinary (i.e., low-probability) events (fires, explosions, and vehicular impact), etc. The more types of influences, the more it is necessary to consider the possible combinations. For the case of the abovementioned New Safe Confinement, the number of considered loading options was 572. Emergency situations which occur by various causes, such as the natural disasters and man-made disasters, should also be taken into account. Detailed classification of catastrophes and their analysis are given in the book (Endzhievsky and Tereshkova 2013).
Types of Analysis and Particular Assumptions In the basis of structural analysis as a science are the system of assumptions and deep fundamental ideas. The harmonious theory created on their basis has important practical significance in various fields of technology. This theory provides a design engineer a set of algorithms, procedures, and methods for analyzing structures. Assumptions reflect the purpose of analysis, features of the structure, properties of materials, type of loads and operating conditions, etc. Each type of analysis—statics, stability, and vibration—is based on the general assumptions outlined in previous sections related to material properties and structure in general. In addition to these assumptions, there are particular assumptions related to each type of analysis. And therefore, in problems of statics, stability, and dynamics, assumptions, even for the same object, can be different. Below some particular assumptions relating to different types of analysis are presented. Static linear analysis presumes that the loads act without any dynamical effects. Moving loads imply that only the position of the load is variable. The purpose of this analysis is to determine the internal forces and displacements of a structure due to various time-independent external exposures (loads, displacements of supports, errors of fabrication, change of temperature). Knowing the distribution of internal forces in each structural element of the structure, we can proceed to engineering design. Such design consists of the selection of material and determination of the cross-section dimensions of each individual element to exclude its failure and ensure normal functioning under the prescribed load. The engineering design problem is the essence of the subject of the mechanics of materials. Structural analysis is based on the following principal assumptions: general assumptions, which allow to apply superposition principle (small displacements, elastic properties of material, two-side constraints), and assumption related to the structure in whole (Saint-Venant’s principle). Besides that, for analysis of some special types of structures (trusses, arches, frames, etc.), some partial (local) assumptions should be taken into account. One of the sections of static analysis—plastic behavior of structures—instead of elastic properties of material takes into account the plastic properties of material. Such an assumption allows to reveal additional reserves of bearing capacity of the statically indeterminate structures. Nonlinear static analysis takes into account the different types of nonlinearities. They are physical nonlinearity (material of a structure does not obey Hook’s law), geometrical nonlinearity (displacements of a structure cannot be considered as small ones), structural nonlinearities (structure with gap or constraints are one sided), etc. For such systems, the superposition principle is not applicable. Stability analysis deals with structures which elements are subjected to compressed time-independent forces. Stability analysis includes buckling analysis and P-delta analysis. Buckling analysis. Its purpose is to determine the critical load (or critical load factor) and corresponding buckling mode shapes. Particular assumptions are reduced to a specific behavior of the compressive force in the process of growing deformations. Of particular importance is the case when the direction of the compressive force remains unchanged. Rejection of this assumption leads to a new type of stability problem. P-delta analysis. Particular assumptions are introduced for analysis of tall and flexible structures. For such structures, the transversal displacements may become significant, so assumption of small deformations is not applicable. Therefore, a structural analysis should be performed on the base of the deformed design diagram. It means that we need to take into account the additional bending moments caused by axial compressed loads P on the displacements induced by the lateral loads. Dynamical analysis is required for the structures subjected to time-dependent excitation of forced or kinematical types. The law of change of excitation as a function of time can be arbitrary. Among them are shock, harmonic excitation, as well as
Introduction
xli
moving loads taking into account the dynamical effects. The main parts of classical dynamic analysis are free-vibration and forced vibration (time-history) analysis. Free-vibration analysis. Particular assumptions allow taking into account the forces of inertia of concentrated and distributed masses. The purpose of this analysis is to determine the natural frequencies (eigenvalues) and corresponding mode shapes (eigenfunctions) of vibration. This information is necessary for dynamical analysis of any structure subjected to arbitrary dynamic excitation. In case of stressed free-vibration analysis, the particular assumptions permit to take into account additional axial time-independent forces. Time-history analysis. Particular assumptions relate to the character of the excitations. This can be an instantaneous load application, a load action of finite duration, periodic excitation, specific, harmonic, etc. The purpose of this analysis is to determine the response of a structure, which is subjected to arbitrarily time-varying loads. The rejection of at least one of the assumptions adopted in the classical theory of structures or the introduction of additional assumptions can impart the system new properties and radically change its behavior; naturally, this leads to the need to bring new methods of analysis. For example, the analysis of a beam under the action of a constant transversal force (which is perpendicular to the axis of a beam) is considered in the statics section. If the constant force is directed along the axis of the beam, then the problem is carried to the stability section. If the transversal force changes in time, then the problem goes to the dynamics section. If the duration of this load is short (the term “short” requires additional explanation), then the problem should be considered within the framework of the shock theory. If the force is directed along the axis of the beam, and at the same time varies according to the harmonic law, the problem goes into the section of dynamical stability. Therefore, each type of analysis—statics, stability, and dynamics—is based on certain particular assumptions. Partial assumptions relating to specific types of structures are set out in the relevant sections. The interconnection of the various chapters, paragraphs, and issues is given on the Book Road Map. If the characteristics of the load are clearly defined, then such load is called deterministic; otherwise we have random load. Analysis of structures subjected to the random excitation is performed at the junction of the theory of structures and mathematical theory of probability. Let us briefly summarize the basic concepts adopted in the classic course of structural analysis. The construction is presented in the form of a design diagram. Only rod structures with two-way constraints and ideal connections of elements are considered. Load is conservative (direction unchanged) and determinate, and does not depend on deformations. The systems are linear, the construction material obeys Hooke’s law, the superposition principle is valid, and the analysis is performed according to the non-deformable design diagram. Possible deviations from these assumptions are discussed specifically. Note: In the classic course structural analysis, the student is offered a ready-made design diagram without discussing the procedure for its selection. Only after studying this course the student will be ready for self-made conscious choice of the design diagram and its justification. These issues are studied in special engineering design courses.
Fundamental Approaches of Linear Static Analysis There are two fundamental approaches to the static analysis of any structure. The first approach is related to the analysis of a structure subjected to given fixed loads and is called the fixed load approach. The results of this analysis are diagrams, which show a distribution of internal forces and deflection for the entire structure due to the given fixed loads. These diagrams indicate the most unfavorable point (or member) of a structure under the given fixed loads. The reader should be familiar with this approach from the course of mechanics of material. The second approach assumes that a structure is subjected to unit concentrated moving load only. This load is not a real one but imaginary. Static analysis presumes that both types of load, fixed and moving loads, act without any dynamical effects. Moving loads imply that only the position of the load is variable. The results of the second approach are graphs called the influence lines. Influence lines are plotted for reactions, internal forces, etc. Internal force diagrams and influence lines have a fundamental difference. Each influence line shows distribution of internal forces in the one specified cross section of a structure depending on the location of imaginary unit moving load only. These influence lines indicate the point of a structure where a load should be placed in order to reach a maximum (or minimum) value of the function under consideration at the specified section. It is very important that the influence lines may also be used for the analysis of structure subjected to any fixed loads. Moreover, in many cases they turn out to be a very effective tool of analysis. Influence line method presents the higher level of analysis of a structure than the fixed load approach. Good knowledge of influence line approach will immeasurably increase the understanding of behavior of structure. Analyst, who combines both approaches for analysis of a structure in engineering practice, is capable to perform a complex analysis of its behavior.
xlii
Introduction
Both approaches do not exclude each other. In contrast, in practical analysis both approaches complement each other. Therefore, learning these approaches to the analysis of a structure will be provided in a parallel way. This textbook presents sufficiently full consideration of influence lines for different types of statically determinate and indeterminate structures, such as beams, arches, frames, and trusses.
Part I
Statically Determinate Structures
Chapter 1
Kinematical Analysis of Structures
Kinematical analysis of a structure is necessary for evaluation of the ability of the structure to resist external load. Kinematical analysis is based on the concept of rigid disk, which is an unchangeable (or rigid) part of a structure. Rigid disks may be separate members of a structure, such as straight members, curvilinear, polygonal (Fig. 1.1), as well as their special combination.
Fig. 1.1 Types of the rigid disks
Any structure consists of separate rigid disks. Two rigid disks may be connected by means of link, hinge, and fixed joint. These types of connections and their static and kinematical characteristics are presented in Table 1.1.
Table 1.1 Types of connections of rigid disks and their characteristics
Type of connection
Link
Presentation and description of connection
D1 D2 Rigid discs D1 and D2 are connected by link (rod with hinges at the ends)
Kinematical characteristics
Static characteristics
Hinge D1
Fixed joint D2
D1
D2
Rigid discs D1 and D2 are connected by hinge
Rigid discs D1 and D2 are connected by fixed joint
Mutual displacement of both discs along the link is zero
Mutual displacements of both discs in both horizontal and vertical directions are zeros
Connection transmits one force, which prevents mutual displacement along the link
Connection transmits two forces, which prevent mutual displacements in vertical and horizontal directions
All mutual displacements of both discs (in horizontal, vertical and angular directions) are zeros Connection transmits two forces, which prevent mutual displacements in vertical and horizontal directions, and moment, which prevents mutual angular displacement
The members of a structure may be connected together by a hinge in various ways. Types of connection are chosen and justified by an engineer. 1. Simple hinge: One hinge connects two elements in the joint.
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_1
3
4
1 Kinematical Analysis of Structures
2. Multiple hinge: One hinge connects three or more elements in the joint. The multiple hinge is equivalent to n 1 simple hinges, where n is a number of members connected in the joint. Hinged joint can transmit axial and shear forces from one part of the structure to the other; the bending moment at the hinge joint is zero.
1.1
Classification of Structures by Kinematical Viewpoint
All structures may be classified as follows: • Geometrically unchangeable structure: For this type of structures, any distortion of the structure occurs only with deformation of its members. It means that this type of structures with absolutely rigid members cannot change its form. The simplest geometrically unchangeable structure is triangle, which contains the pin-joined members (Fig. 1.2a). • Geometrically changeable structure: For this type of structures, any finite distortion of the structure occurs without deformation of its members. The simplest geometrically changeable system is formed as hinged four-bar linkage (Fig. 1.2b, c). In Fig. 1.2c, the fourth bar is presented as ground. In both cases, even if the system would be made with absolutely rigid members, it still can change its form. It is pertinent to do the following important remark related to terminology. Sometimes terms “stable” and “unstable” are applied for abovementioned types of structures. However, the commonly accepted term “stable/unstable” in classical theory of deformable systems is related to the concept of critical load, while the term “geometrically unchangeable/ changeable” is related to the way of connection of rigid disks. There is the fundamental difference between kinematical analysis of a structure on the one hand, and analysis of stability of a structure subjected to compressed load, on the other hand. Thus, in kinematical analysis of structures we will use the term “unchangeable structure” instead of “stable structure,” and term “changeable structure” instead of “unstable structure.” Stability analysis is considered in Chap. 15. • Instantaneously changeable structure: This system allows infinitesimal relative displacements of its members without their deformation and after that the structure becomes geometrically unchangeable (Fig. 1.2d). • Instantaneously rigid structure: This system allows infinitesimal relative displacements of its members without their deformation and after that the structure becomes the geometrically changeable (Fig. 1.2e). The term “instantaneously” is related to the initial condition of the structure only.
b
a
d
c
e
Fig. 1.2 Types of structure by kinematical viewpoint
In structural engineering only geometrically unchangeable structures may be accepted (Exceptions to this rule are specifically discussed).
1.2
Generation of Geometrically Unchangeable Structures
In order to produce a rigid structure in a whole, the rigid disks should be connected in a specific way. Let us consider general rules for the formation of geometrically unchangeable structures from two and three rigid disks. If a structure is formed from two disks, then their connections may be as follows: 1. Connection by fixed joint (Fig. 1.3a). 2. Connection by hinge C and rod AB, if the axis of AB does not pass through the hinge C (Fig. 1.3b).
1.2 Generation of Geometrically Unchangeable Structures
5
3. Connection by three nonparallel rods: Point of intersection of any two rods presents a fictitious hinge C0 . In this case, the other rod and fictitious hinge correspond to case 2 (Fig. 1.3c).
b
a
c
C
A
C⬘
B
Fig. 1.3 Geometrically unchangeable structures formed from two disks
If a structure is formed from three disks, then their connections may be as follows: 1. Connection in pairs by three hinges A, B, and C, which do not belong to one line (Fig. 1.4a) 2. Connection in pairs by two (or more) concurrent links, if points of their intersections A, B, and C do not belong to one line (Fig. 1.4b) b
a
c
B
C
C
A D
2
D B
A
A
1
Fig. 1.4 Geometrically unchangeable structures formed from three disks
Case 1.4a may be presented as shown in Fig. 1.4c: additional joint A is attached to rigid disk D by two links 1 and 2. This case leads to a rigid triangle (Fig. 1.2a), which is a simplest geometrically unchangeable structure. This is the main principle of formation of simplest trusses.
1.2.1
Required and Redundant Constraints
All constraints of any structure are divided into two groups. Required constraint of a structure is such a constraint, eliminations of which change a kinematical characteristic of the structure. It means that the entire unchangeable structure transforms into changeable or instantaneously changeable one, instantaneously changeable transforms into changeable, and changeable transforms into changeable with mobility more by unity. Note that constraint assumes not only supports, but elements as well. For example, elimination of any member of truss in Fig. 1.5a (while the hinges remain) transforms this structure into changeable one, so for this structure all the elements are required.
a
b 2
4 3
1
B
C
Fig. 1.5 (a) Structure with required constraints; (b) structure with redundant constraints
Redundant constraint of a structure is such a constraint, eliminations of which do not change a kinematical characteristic of the structure. It means that the entire unchangeable structure remains the unchangeable (Fig. 1.5b), changeable structure remains changeable one, and instantaneously changeable remains instantaneously changeable structure. The structure in
6
1 Kinematical Analysis of Structures
Fig. 1.5a has not redundant constraints. For structure in Fig. 1.5b the following constraints may be considered as redundant: 1 or 2, and 3 or 4, and one of supports—B or C, so the total number of redundant constraints is three.
1.2.2
Constraint Replacing
In case of unchangeable structure, the constraints may be replaced. It means that the required constraint may be eliminated and instead of that another required constraint should be introduced, or the redundant constraint is replaced by another redundant constraint. This procedure allows from a given structure to create a lot of other structures.
1.3
Analytical Criteria of the Instantaneously Changeable Structures
Kinematical analysis of a structure can also be done using the static equations. The following criteria for instantaneously changeable systems may be used: (a) Load has finite quantity but the internal forces have infinite values. (b) Load is absent and the internal forces are uncertain (type of 0/0). Let us discuss these criteria for structure shown in Fig. 1.6a. Internal forces in members of the structure are N ¼ P=2 sin φ. If φ ¼ 0 (Fig. 1.6b), then N ¼ 1. Thus, external load P of
a
j
b
j
P P Fig. 1.6 Kinematical analysis using the static equations
finite quantity leads to the internal forces of infinite values. It happens because the system is instantaneously changeable. Indeed, two rigid disks are connected using three hinges located in one line. Figure 1.7 presents the design diagram of the truss. This structure is generated from simplest rigid triangle; each next joint is attached to previous rigid disk by two end-hinged links. The structure contains three support constraints, which are minimum necessary for plane structure. However, location of these supports may be wrong. Thorough kinematical analysis of this structure may be performed by static equations. a
R
A
b P
Fig. 1.7 Kinematical analysis of the truss
Reaction R of support may be calculated using the equilibrium condition X
MA ¼ 0 ! R a P b ¼ 0 ! R ¼
Pb : a
1. If a ¼ 0, then for any external load P the reaction of the left support is infinitely large (R ¼ Pb 0 ).
1.3 Analytical Criteria of the Instantaneously Changeable Structures
7
2. If a ¼ 0, and P ¼ 0, then reaction R is uncertain (R ¼ 00). Thus, if all lines of support constraints are concurrent at one point (a ¼ 0), then this case leads to instantaneously changeable system. Instantaneously changeable systems may occur if two rigid disks of a structure join inappropriately. Two such connections of rigid disks are shown in Fig. 1.8. If a system may be separated into two rigid disks (shown by solid color) by a section cutting three elements, which are parallel (Fig. 1.8a, elements 1, 2, 3), or concurrent in one point (Fig. 1.8b, point A, elements a, b, c), then the system is an instantaneously changeable one.
a
A
b
3
a
2
c
b
1
Fig. 1.8 Instantaneously changeable systems
However in practice, connection of two rigid disks by two (or more) parallel members may be used in special condition of loading. Figure 1.9 presents the rigid beam (disk D1), which is supported by vertical hinged-end rods 1–3 (disk D2 is a support part). The system may be used if the axial forces in members 1–3 are tensile (Fig. 1.9a). However, the system in Fig. 1.9b cannot be used, because the axial forces in members 1–3 are compressive. P
b
a
D2 1
2
3
1
2
3
D1 P Fig. 1.9 Geometrically changeable systems
Evolution of the structure caused by changing the type of supports is shown in Fig. 1.10. Constraint A (Fig. 1.10a) prevents two displacements, in vertical and horizontal directions. If one element of the constraint A, which prevents horizontal displacement, will be removed, then the structure becomes geometrically changeable (Fig. 1.10b), so the removed constraint is the required one. In case of any horizontal displacement of the structure, all support constraints A, B, and C remain parallel to each other.
a
c A
B
C
A
C B
b A
B
C
Fig. 1.10 (a) Geometrically unchangeable structure; (b) geometrically changeable structure; (c) instantaneously changeable system
8
1 Kinematical Analysis of Structures
The next evolution is presented in Fig. 1.10c. If any support, for example supporting element B, will be longer than other supports, then the structure becomes an instantaneously changeable system. Indeed, in case of any horizontal displacement of the structure, the support constraints will not be parallel any more. It is worth to mention one more static criterion for instantaneously changeable and geometrically changeable structures: internal forces in some element obtained by two different ways are different, or in another words, analysis of a structure leads to contradictory results. This is shown in the example below. Design diagram of the truss is presented in Fig. 1.11a: the constraint support B is directed along the element BC. The system has the necessary minimum number of elements and constraints to be a geometrically unchangeable structure. Let us provide more detailed analysis of this structure. Free-body diagrams for joints A, 1, and B are shown in Fig. 1.11b.
Fig. 1.11 Kinematical analysis of instantaneously changeable system
Equilibrium conditions for these joints and corresponding results are presented below:
P
X ¼ 0 ! NA1 ¼ 0 X P Y ¼ 0 : N C1 sin α P ¼ 0 ! N C1 ¼ sin α X Joint 1: X ¼ 0 : N 1B N C1 cos α ¼ 0 ! N 1B ¼ N C1 cos α ¼ P cot α X Y ¼ 0 : N BC sin β þ RB sin β ¼ 0 ! N BC ¼ RB X Joint B: X ¼ 0 : N 1B N BC cos β RB cos β ¼ 0 ! N 1B ¼ 0 Joint A:
Two different results for internal force N1B have been obtained, i.e., P cot α and zero. This indicates that the system is defective. From mathematical point of view, this happens because the set of equilibrium equations for different parts of the structure is incompatible. From physical point of view, this happens because three support constraints are concurrent in one point C for any angle φ. Any variation of this angle φ keeps the system as instantaneously changeable. If constraint at support C will be removed to point A, or angle of inclination of support B will be different, then system becomes a geometrically unchangeable structure. Let us show this criterion for the system presented in Fig. 1.12. Note that hinges D and E are multiple similarly to hinges C and F.
1.4 Degrees of Freedom
9
Fig. 1.12 Kinematical analysis of geometrically changeable system
P
Pb . aþb Equilibrium conditions lead to the following results:
Reaction RA !
Joint A: Joint C: Joint D:
P
Y ¼0:
P Y ¼0: P Y¼0:
M B ¼ 0 ! RA ¼
RA Pb ¼ sin φ ða þ bÞ sin φ Pb N 4 þ N 1 sin φ ¼ 0 ! N 4 ¼ N 1 sin φ ¼ aþb N4 ¼ 0
RA N 1 sin φ ¼ 0 ! N 1 ¼
We received for internal force N4 two contradictory (or inconsistent) results. Why this happens? To answer this question let us consider the very important concept of “degrees of freedom.”
1.4
Degrees of Freedom
A number of independent parameters, which define the configuration of a system without deformation of its members, is called the degree of freedom. The number of degrees of freedom for any structure may be calculated by Chebyshev’s formula W ¼ 3D 2H S0
ð1:1Þ
where D, H, and S0 are the number of rigid disks, simple hinges, and constraints of supports, respectively. For trusses the degrees of freedom may be calculated by the formula W ¼ 2J S S0
ð1:2Þ
where J and S are the number of joints and members of the truss, respectively. This formula is derived from the following propositions: each joint of the truss has two degrees of freedom on the plane, and each rod connecting two joints as well as support rod is equivalent to one constraint, since it imposes the only one requirement—the constancy of the distances between the joints. Special Cases There are three special cases possible. 1. W > 0: The system is geometrically changeable and cannot be used in engineering practice. 2. W ¼ 0: The system has the necessary number of elements and constraints to be a geometrically unchangeable structure. However, the system still can be inappropriate for engineering structure. Therefore, this case requires additional structural analysis to check if the formation of the structure and arrangement of elements and constraints are correct. This must be done according to rules, which are considered above. For example, let us consider systems, which are presented in Fig. 1.13a–c. Two diagonal members in Fig. 1.13b, d at the point of their intersection do not form a rigid joint.
10
1 Kinematical Analysis of Structures
2
a
1
3
6
5
4
c 2
7
10
8
b
9
d
1
2 1
Fig. 1.13 Different ways of truss formation. For cases (a), (b), and (c) degree of freedom is W ¼ 0, and for case (d) W ¼ 1. Only case (a) may be adopted as engineering structure
Degrees of freedom W for these cases according to formula (1.2) equal to zeros. Indeed: Case a: J ¼ 10, S ¼ 17, S0 ¼ 3, so W ¼ 2 10 17 3 ¼ 0. Case b: J ¼ 10, S ¼ 17, S0 ¼ 3, so W ¼ 2 10 17 3 ¼ 0. Case c: J ¼ 10, S ¼ 16, S0 ¼ 4, so W ¼ 2 10 16 4 ¼ 0. The degrees of freedom for trusses may be calculated also by Chebyshev’s formula (1.1). The truss in Fig. 1.13a contains 17 bars, which are considered as the rigid disks, and 3 support constraints. The number of the equivalent simple hinges H can be calculated as follows: only two hinges are simple (joints 1 and 9); all other hinges are multiple. Each multiple hinge at joints 2, 3, 5, 7, and 10 is equivalent to two simple hinges; each multiple hinge at 4, 6, and 8 is equivalent to four simple hinges. Thus, the total number of equivalent simple hinges is H ¼ 2 1 þ 5 2 þ 3 4 ¼ 24. Chebyshev’s formula leads to the following result: W ¼ 3D 2H S0 ¼ 3 17 2 24 3 ¼ 0. The same results may be obtained for cases (b) and (c). Even if W ¼ 0 for systems in Fig. 1.13a–c, only system (a) may be used as an engineering structure. System (b) has a rigid disk (shown as solid) and geometrically changeable right part. System (c) consists of two rigid disks, which are connected by two members 1 and 2 (while for generation of geometrically unchangeable structures two rigid disks must be connected by three nonconcurrent members), and therefore it is a geometrically changeable system. 3. W < 0: The system has redundant constraints. However, existence of redundant constraints still does not mean that the structure can resist load, because the structure can be generated incorrectly. The system in Fig. 1.13d contains one redundant constraint; indeed, the degree of freedom is W ¼ 2 10 17 4 ¼ 1. However, the assembly of the elements is wrong and therefore this system cannot be considered as an engineering structure. Indeed, the left and right rigid disks (shown by solid) are connected by two elements 1 and 2. Therefore, this structure is geometrically changeable. Let us return to Fig. 1.12. The system contains members with hinges at ends. The number of joints is J ¼ 6, the number of members is S ¼ 8, and the number of support constraints is S0 ¼ 3. Degree of freedom if the system is calculated as for truss is W ¼ 2 6 8 3 ¼ 1. (Chebyshev’s formula leads to the same result, i.e., W ¼ 3 8 2 10 3 ¼ 1, where 10 is a total number of simple hinges.) Therefore, the system does not have a required minimum of elements in order to provide its geometrical unchangeability, and cannot be used as an engineering structure. This system is geometrically changeable, because two rigid disks ACD and BEF are connected incorrectly, i.e., by two members 3 and 5. In other words, this system contains the hinged four-bar linkage CDEF. Therefore, if one more rigid element would be introduced correctly, then the system would become geometrically unchangeable. Such an additional element may connect joints C and E or joints D and F.
1.5
Null Load Method
The concept “degrees of freedom” allows expanding the application of static criterion, previously considered in Sect. 1.3, for determining instantaneous changeability of a system.
1.5 Null Load Method
11
Let us consider structure which has the minimum number of elements required to ensure its geometrical unchangeability; this means that the number of degrees of freedom W ¼ 0. Null load method allows to quickly check the correctness of the connections and placement of the elements of a structure from the point of view of its geometrical unchangeability. Indicators of geometrically unchangeable structure for which W ¼ 0 are as follows: (a) In case of absent load (null load)—the reactions in all constraints are zero. (b) In case of any load on the structure—the reactions in each constraint are not zero; this solution is unique. Let us consider rigid disk D and investigate the effect of the location of the support constraints on geometrical unchangeability of the structure. Assume that the disk D is connected to the foundation by means of three rods 1, 2, and 3 as shown in Fig. 1.14a. The axes of the rods intersect at points A12, A13, and A23. The equilibrium equation Pin the form of a sum of moments of all forces around one of the points of intersection of two rods, for example A23, is S1 ! M A23 ¼ 0. If the system is under the action of null load, then S1 r1 ¼ 0. Since r1 6¼ 0, S1 ¼ 0. Similarly, we will determine that S2 ¼ S3 ¼ 0. This means that if a null load acts on a geometrically unchangeable system, then there are no forces in the system. A13
A
r1
a
A23
b
A23
D
D 1
A
2
3
S1
S3
S2
D 3
1
S1
2
S2
S3
A12 Fig. 1.14 Two ways to connect the rigid disk D to the ground using three rods. (a) Rods do not intersect at one point; (b) rods intersect at one point
If geometrically unchangeable structure (Fig. 1.14a) is subject to any nonzero load, then a definite force arises in each element and the distribution of the reactions and internal forces is unique. Indeed, if the system is loaded by any force F (is not shown), then S1 ¼ Fr23/r1, where r23 is the arm of the force F with respect to the point A23. Now let us consider the rigid disk D which is connected to the foundation by means of three rods 1, 2, and 3, the axes of which intersect at one point (Fig. 1.14b). In case of null load the equilibrium equation becomes ∑MA ¼ 0 : S1r1 þ S2r2 þ S3r3 ¼ 0. Since r1 ¼ r2 ¼ r3 ¼ 0, for the reactions Si, i ¼ 1, 2, 3 we get undefined values. This is a static indicator of an instantaneous changeability of a system. Let a system shown in Fig. 1.14b is subjected to any force F (not shown); the arm of the force F with respect to the point A is r. The equilibrium equation is ∑MA ¼ S1r1 þ S2r2 þ S3r3 þ Fr ¼ 0, where ri is an arm of the force Si around point А. Since r1 ¼ r2 ¼ r3 ¼ 0, S1r1 þ S2r2 þ S3r3 ¼ 0. However, Fr ¼ 6 0, which means that a rigid disk is not in equilibrium. This system performs an infinitesimal turn around point A; that is, this system is instantaneously changeable. After rotation of the system to an infinitesimal angle, the axis of the rods will not intersect at one point, and therefore the reactions S can equilibrate the load F. It means that equation ∑MA ¼ S1r1 þ S2r2 þ S3r3 þ Fr ¼ 0 is satisfied. However, since the arms of the reactions are infinitesimal, the forces in instantaneously changeable structures take infinite values. Therefore these types of structures cannot be used in engineering. Static indicators for investigating the instantaneous changeability of a system are called a null load method. We emphasize that this provision is only used to verify the correct arrangement of elements in the structure which contains the minimum number of rods necessary for its geometric unchangeability. For example, in case of hinged four-bar linkage (Fig. 1.15a) in the absence of loads in the joints, the internal forces in all elements are equal to zero (for this it is sufficient to consider the equilibrium of each of the joints). On this basis, the conclusion that the system is unchangeable is erroneous. We cannot apply the null load method to this system, because the number of degrees of freedom is not zero; indeed W ¼ 2J S S0 ¼ 2 4 4 3 ¼ 1. Or in other words, the system does not have the necessary number of elements to ensure its geometric unchangeability.
a
1
b 4
2
1
c 3
4
2
3
B
Fig.1.15 (a) Hinged four-bar linkage, (b, c) hinged quadrangle with additional element 2–4, and with modified support at joint 3
12
1 Kinematical Analysis of Structures
Let us introduce additional member 2–4 in the hinged four-bar linkage (Fig. 1.15b). Since degrees of freedom W ¼ 2J S S0 ¼ 2 4 5 3 ¼ 0, the new system contains the required number of elements to ensure its geometric unchangeability. Equilibrium equations of unloaded joint 1 lead to internal forces S12 ¼ S14 ¼ 0. We discard the rods 1–2 and 1–4 as zero elements. We get triangle 2–3–4, which is the simplest unchangeable structure. Thus, a structure in Fig. 1.16b is unchangeable, and all elements are arranged properly. Now let us change the orientation of the hinged movable support at the joint 3, as shown in Fig. 1.15c. As before, the number of degrees of freedom W ¼ 0. This structure contains the minimum required number of elements, so we can apply the method of null load. The equilibrium conditions of the unloaded joint 1 lead to internal forces S12 ¼ S14 ¼ 0. We reject the rods 1–2 and 1–4 as zero members. Then consider the equilibrium of the unloaded joint 2. As a result, we get S23 ¼ S24 ¼ 0. We reject the rods 2–3 and 2–4 as zero members. We obtain bar 4–3, and the support rod 3–B. Both bars, 4–3 and 3–B, with hinge ends (4, B) are connected by hinge 3, and all three hinges, 4, 3, and B, are located in one line. This is an instantaneously changeable structure, as was shown in Sect. 1.3, Fig 1.6. Thus, to ensure the geometric unchangeability of the system, it is required not only to use a certain number of elements based on the condition W ¼ 0, but also to ensure their correct arrangement. The structure depicted in Fig. 1.16 contains 7 joints, 11 rods, and 3 support constraints. Therefore, the number of degrees of freedom is W ¼ 2J S S0 ¼ 2 7 11 3 ¼ 0, and the structure contains the minimum necessary number of elements to ensure its geometric unchangeability. To verify the correct arrangement of the rods, we need to consider null load, i.e., the structure should be free from the load. We cut out the joint 1, and projection of all the forces on the vertical axis allows us to find internal force S12 ¼ 0, so the rod 1–2 can be discarded. Then we will sequentially examine the joints 2, 3, 4, 5, A, and B; discard the zero rods; and make sure that the internal forces in all elements of the structure are zero. Consequently, the structure is geometrically unchangeable. 5
3
4
2 A
B
1 Fig. 1.16 Design diagram of the structure
Thus to summarize, null load method is used for quick check if the structure is instantaneously changeable or not. To apply it, we must first verify that the system has zero degrees of freedom and then confirm correct arrangement of the elements.
Problems Perform the kinematical analysis of the following design diagrams:
P1.1–P1.5
a
b
c
d
Fig. P1.1
a
A Fig. P1.2
b
B
A
c
B
A
B
A
f
e
d
B
A
B
A
B
Problems
13
a
b
c
d
Fig. P1.3
b
a
c
Fig. P1.4
a
c
b
Fig. P1.5
P1.6–1.8 Perform the kinematical analysis of the following design diagrams from the point of view of the instantaneous changeability. Apply a null load method.
a
2
b
1 Fig. P1.6
a
b
Fig. P1.7
a
b 1
1 Fig. P1.8
Chapter 2
The Simplest Beams: Theory of Influence Lines
This chapter forms the set of concepts which creates a framework for comprehensive analysis of different statically determinate structures. Concepts of influence lines and direct/indirect load application are introduced. All influence lines for simplest structures (one span simply supported and cantilevered beams) are constructed using analytical expressions for required factor. Applications of influence lines for fixed and moving loads are discussed.
2.1
General
Let us remind the general rules for determining the internal forces, which are known to reader from the course Mechanics of Materials. Consider an arbitrary flat rod system. Suppose that the system is subject to the action of an arbitrary load that lies in the plane of the structure itself. Under these assumptions, the following internal forces arise in each element: bending moment M, shear (transverse) force Q, and longitudinal (axial) force N. To determine them, the section method is used, which allows transferring internal forces into the class of external forces and using equilibrium equations to determine them. In the general case of a structure and its loading, the internal forces are variable along the length of the element. To calculate them, we will use the following definitions. Definitions The bending moment M in a given section is equal to the algebraic sum of the moments of all external forces located on one side of the section relative to the center of gravity of the given section. The shear force Q in a given section is equal to the algebraic sum of the projections of all external forces located on one side of the section onto the axis perpendicular to the longitudinal axis of the element. The normal (axial) force N in a given section is equal to the algebraic sum of the projections of all external forces located on one side of the section on the longitudinal axis of the element. Note: The signs of internal forces do not depend on the location/direction of the coordinate axes, but are determined only by the nature of the deformation of the element. The bending moment is considered positive if the concavity is directed upwards; shear force is considered positive if each force rotates the element around its opposite end in a clockwise direction; normal force is considered positive at which elongation of the element occurs. Signs of positive internal forces are shown in Fig. 2.1.
a Left part
M
M N Q
N a
Q
Right part
M
M
M
+ Extended fibers
N
M N
+ Q
Q
+ +
Q M
Fig. 2.1 Positive signs of internal forces. Section a–a cuts off element onto left and right parts; boundary conditions of both parts are not shown. Dotted line means location of the extended fibers (below on the axial line)
In case of fixed load the change in internal forces along the length of the element is depicted in the corresponding diagrams. There are strict analytic relationships between these diagrams. This is the “bending moment-shear force” ratio dM/dx ¼ Q and © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_2
15
16
2 The Simplest Beams: Theory of Influence Lines
the “shear force-external distributed load” ratio dQ/dx ¼ q. These relationships help to verify the constructed diagrams. In detail, the reader got acquainted with these concepts from the course Mechanics of Materials (Craig Jr 2000). The positive values of shear forces are plotted above the longitudinal axis of a beam. The positive values of bending moments are plotted below the longitudinal axis; it means that the bending moment diagrams in all cases will be drawn on the extended fibers. Shading of diagrams should be directed perpendicular to the longitudinal axis of the element. We will strictly adhere to these rules in the future. In the case of moving loads (for constructing of influence lines) the rules for calculating internal forces and their signs remain same.
2.2
Analytical Method for Construction of Influence Lines
The engineering structures are often subjected to moving loads. Typical examples of moving loads on a structure are traveling cars, trains, bridge cranes, etc. In classical structural analysis the term “moving load” requires some additional comments: this concept means that load position on the structure may be arbitrary. Thus, moving load is considered as static load, without any dynamical effects. It is obvious that internal forces and displacements in the any section of a beam depend on the position of a moving load. An important problem in the analysis of structures is the determination of maximum internal forces in a structure caused by a given moving load and the corresponding most unfavorable position of this load. This problem may be solved using influence lines. Influence line is a fundamental and very profitable concept of structural analysis. Their application allows to perform a deep and manifold analysis of different types of structures subjected to any type of fixed and moving loads. Influence line method becomes an especially effective tool analysis if a structure is subjected to different groups of loads. Definition Influence line is a graph, which shows variation of some particular function Z (reaction, shear, bending moment, etc.) in the fixed cross section of a structure in terms of position of unit concentrated dimensionless load P ¼ 1 on the structure. Each ordinate of influence line means the value of the function, for which influence line is constructed, if the unit load is located on the structure above this ordinate. Therefore, the unit load P, which may have different positions on the structure, is called a moving (or traveling) load. The term “moving load” implies only that position of the load is arbitrary; that is, this is a static load, which may have different positions along the beam. The time, velocity of the moving load, and any dynamic effects are not taken into account. Thus, for convenience, from now on, we will use notion of “moving” or “traveling” load for static load, which may have different positions along the beam. Influence line for any function Z at a specified section of a structure may be constructed using its definition as follows: the required function should be calculated for different positions of unit load within the loaded portion. These values are plotted at the points, which correspond to the position of the load. After that all ordinates should be connected. This approach is correct conceptually and many authors show construction of influence lines using this approach. However such procedure is extremely annoying and cumbersome (especially for statically indeterminate structures). Because of repetitive procedure of construction of influence lines, their advantages are questionable. In this book, the construction of influence lines is performed on the bases of a different approach, i.e., deriving an equation of influence line for required function Z; this equation relates values of Z and position x of the unit load P. Thus the required factor Z is presented as analytical function of the position of the load. Such way of construction of influence lines is called static method. Application of this method for construction of influence lines for reactions is presented below.
2.2.1
Influence Lines for Reactions
The following types of beams are considered: simply supported beam, beam with overhang, and cantilever beam.
2.2.1.1
Simply Supported Beam
The beam AB is loaded by moving load P ¼ 1. The moving load is shown by circle and the dotted line indicates the loaded contour for possible positions of the load on the structure. The distance from the left-hand support to the load is x (Fig. 2.2).
2.2 Analytical Method for Construction of Influence Lines
17
Influence Line for RA Equilibrium equation in the form of moments of all external forces about the support B allows to determine the reaction RA in terms of x: RA !
X
MB ¼ 0 :
RA l þ Pðl xÞ ¼ 0 ! RA ¼
Pðl xÞ l
Reaction RA may be calculated for any location x of the load P. Therefore, last equation should be considered as a function RA(x). This function is called the influence line for RA and denoted as IL(RA). Since P ¼ 1, equation of influence line for reaction RA becomes ILðRA Þ ¼
lx l
ð2:1Þ
If x ¼ 0 (at support A), then ordinate of influence line IL(RA) ¼ 1. If x ¼ l (at support B), then ordinate of influence line IL(RA) ¼ 0. These two points are connected by straight line, since function (2.1) is linear. Influence line for RA is presented in Fig. 2.2.
x
P=1 Loaded contour
A
B 0.25l l
RA
RB
0.75
1
+
0.25
Inf. line RA
+
1 Inf. line RB
Fig. 2.2 Simply supported beam. Influence lines for reactions
We can see that units of influence line ordinates for reaction are dimensionless. In general, units of influence line ordinates for any factor Z are defined as quotient of two units, mainly, unit of the factor Z and unit of the load P [kN]. Thus, unit of influence line for reactions and shear are dimensionless because kN/kN; for bending moment: kNm/kN ¼ m; for linear deflection m/kN; and for angular deflection rad/kN. Influence line IL(RA) can be used for analysis of reaction RA only. Positive ordinates mean that reaction of RA is directed upward for any position of the concentrated load. If load P ¼ 1 is located above point A, then reaction of RA is equal to 1; it means that load P completely transmits on the support A. If load P ¼ 1 is located above point B, then reaction of RA is equal to zero. If load P ¼ 1 has, for example, coordinate x ¼ 0.25l, then reaction of RA is equal to 0.75. Analytical presentation of equation of influence line allows to avoid many times repeated computation of function Z for different locations of the force P; this is a huge advantage of analytical approach for construction of influence lines. Note a following fundamental property: influence lines for reactions and internal forces of any statically determined structures are always presented by straight lines.
Influence Line for RB Equilibrium equation in the form of moments of all the external forces about support A leads to the following expression for reaction RB in terms of position x: RB !
X
MA ¼ 0 :
RB l P x ¼ 0 ! RB ¼
The last equation leads to the following equation of influence line:
Px l
18
2 The Simplest Beams: Theory of Influence Lines
ILðRB Þ ¼
x l
ð2:2Þ
If x ¼ 0 (at support A), then ordinate of influence line IL(RB) ¼ 0. If x ¼ l (at support B), then ordinate of influence line IL(RB) ¼ 1. Influence line for RB is presented in Fig. 2.2. This graph can be used for analysis of reaction RB only. If load P ¼ 1 is located above point A, then reaction of RB is equal to zero. It means that load P does not get transmitted on to the support B, when the load P is situated directly over the left-hand support. If load P ¼ 1 is located above point B, then reaction of RB is equal to 1. If load P ¼ 1 has, for example, coordinate x ¼ 0.25l, then reaction of RB is equal to 0.25.
2.2.1.2
Simply Supported Beam with Overhang
The equilibrium equations and corresponding equations for influence lines of reactions are RA ! RB !
X X
xl xl ! ILðRA Þ ¼ RA l Pðx lÞ ¼ 0 ! RA ¼ P l l x x RB l Px ¼ 0 ! RB ¼ P ! ILðRB Þ ¼ l l
MB ¼ 0 : MA ¼ 0 :
ð2:3Þ
Influence lines of reactions are shown in Fig. 2.3. If load P ¼ 1 is situated at point D (x ¼ l + d), then reaction RA ¼ d=l. The negative sign means that the reaction RA is directed downwards. The maximum positive reaction RA occurs if load P ¼ 1 stands at point A, and the maximum negative reaction RA occurs if load P ¼ 1 stands at point D. x
P=1 B
A
D
1
RB
l
RA
d
+ −
Inf. line RA
1 +
Inf. line RB
Fig. 2.3 Simply supported beam with overhang. Influence lines for reactions
If load P ¼ 1 is situated at point D, then RB ¼ (lþd)/l. This means that reaction RB > P ¼ 1 and is directed upward. The maximum positive reaction RB occurs if load P ¼ 1 stands at point D; the negative reaction RB does not arise. Eqs. (2.1)–(2.3) for influence lines of reactions show that overhang does not change the equations of influence lines; therefore an influence line within the overhang is an extension of influence line within the span. This is a common property of influence lines for any function (reaction, bending moment, and shear). Thus, in order to construct the influence lines for reaction of a simply supported beam with overhang, the influence lines for reaction between supports should be extended underneath the overhang.
2.2.1.3
Cantilevered Beam
At the fixed support A the following reactions arise: vertical and horizontal forces RA and HA, and moment M0; for the given design diagram the horizontal reaction HA ¼ 0. Positive reactions RA and M0 are shown in Fig. 2.4.
2.2 Analytical Method for Construction of Influence Lines
19
The Vertical Reaction RA This reaction may be calculated considering the equilibrium equation in the form of the projections of all external forces on the vertical axis: X RA ! Y ¼ 0 : RA P ¼ 0 ! RA ¼ P: Since P ¼ 1, the equation of influence line becomes ILðRA Þ ¼ 1
ð2:4Þ
It means that reaction RA equals to 1 for any position of concentrated load P ¼ 1.
P=1
M0
A
B x
RA 1
l +
Inf. line RA –
l
Inf. line M0
Fig. 2.4 Cantilevered beam. Design diagram and influence lines for reactions
The Moment M0 at Support A This moment may be calculated considering the equilibrium equation in the form of moment of all the external forces with respect to point A: X M0 ! M A ¼ 0 : M 0 Px ¼ 0 ! M 0 ¼ Px Since load P ¼ 1, equation of influence line is ILðM 0 Þ ¼ x
ð2:5Þ
It means that moment varies according to linear law. If the load P ¼ 1 is located at x ¼ 0 (point A), then the moment M0 at the fixed support does not arise. Maximum moment at support A corresponds to position of the load P ¼ 1 at point B; this moment equals to 1l. The units of the ordinates of influence line for M0 are meters. The positive ordinates of any influence lines are plotted above the reference line.
2.2.2
Influence Lines for Internal Forces
Simply supported beam subjected to moving unit load P is presented in Fig. 2.5. Construction of influence lines for bending moment and shear force induced at section k are shown below.
20
2.2.2.1
2 The Simplest Beams: Theory of Influence Lines
Bending Moment Mk
The bending moment in section k is equal to the algebraic sum of moments of all forces, which are located to the left (or right) of section k, about point k. Since the expression for bending moment depends on whether the load P is located to the left or to the right from the section k, then two positions of the load P need to be considered, i.e., to the left and to the right of section k.
Load P ¼ 1 is Located to the Left of Section k In this case, it is convenient to calculate the bending moment Mk using the right forces. The only reaction RB is located to the right of point k, so the bending moment is X right Mk ! M k ¼ 0 : M k ¼ RB b If position of the load P is fixed, then reaction RB is a number and the bending moment is a number as well. However, if load P ¼ 1 changes its position along the left portion of the beam, then reaction RB becomes a function of position of the load P and, thus, the bending moment is a function too. Thus, the expression for bending moment is transformed to the equation of influence line for bending moment: ILðM k Þ ¼ b ILðRB Þ
ð2:6Þ
So for the construction of influence line for bending moment we need to construct the influence line for reaction RB, after that multiply all ordinates by parameter b, and, as the last step, show the operating range of influence line. Since load P is located to the left of section k, the operating range is left-hand portion of influence line; that is, the above equation of influence line is true when the load P is changing its position on the left portion of the beam. Hatching the corresponding part of the influence line reflects this fact. x
P=1 k
A
B a
b
RA
l ab l
RB
b·IL(RB)
1· b
+
Load P=1 left at section k
Operating range a·IL(RA)
1· a
ab l
+ Operating range
ab l
Load P=1 right at section k
b
a +
Inf. line Mk (m)
Fig. 2.5 Simply supported beam. Construction of influence line for bending moment at section k
2.2 Analytical Method for Construction of Influence Lines
21
Load P ¼ 1 is Located to the Right of Section k In this case, it is convenient to calculate the bending moment Mk using the left forces. The only reaction RA is located to the left of point k, so the bending moment is X left Mk ! M k ¼ 0 : M k ¼ RA a As above, the expression for bending moment is transformed to the equation of influence line for bending moment: ILðM k Þ ¼ a ILðRA Þ
ð2:7Þ
For construction of this influence line, we need to construct the influence line for reaction RA, then multiply all ordinates by parameter a, and finally show the operating range of influence line. Since load P is located to the right of section k, we obtain right-hand portion of influence line as the operating range. Corresponding influence line Mk is presented in Fig. 2.4. Since units of ordinates IL(RA) and IL(RB) are dimensionless, then units of ordinates IL(Mk) are units of length (for example, meter). Ordinate of influence line at section k is ab/l. Sign of influence line for bending moment Mk for simply supported beam is positive, which means that extended fibers at section k are located below longitudinal axis for any position of the load P. Henceforward, all mathematical treatment concerning to construction of influence lines will be presented in tabulated form. The abovementioned discussions are presented in the following table. Load P ¼ 1 left at section k X right Mk ¼ 0 Mk ! M k ¼ RB b ! ILðM k Þ ¼ b ILðRB Þ
Load P ¼ 1 right at section k X left Mk ¼ 0 Mk ! M k ¼ RA a ! ILðM k Þ ¼ a ILðRA Þ
To summarize, in order to construct the influence line for bending moment at section k it is necessary to: 1. Plot ordinates a and b on the left and right vertical lines passing through the left-hand and right-hand support, respectively. 2. Join each of these points with base point at the other support; both lines intersect at section k. 3. Show the operating ranges of influence line. The hatching (operating range) corresponds to the position of the load but not to the part of the beam which is used for computation of internal forces.
2.2.2.2
Influence Line Qk
Since the expression for shear depends on whether the load P is located to the left or to the right from the section k, two positions of the load P need to be considered. The procedure of construction of influence line for shear is presented in tabulated form. In the table we show the position of the load (left or right at section k—this is the first line of the table) and part of the beam, the equilibrium of which is considered. Load P ¼ 1 left at section k X Y right ¼ 0 Qk ! Qk ¼ RB ! ILðQk Þ ¼ ILðRB Þ
Load P ¼ 1 right at section k X Y left ¼ 0 Qk ! Qk ¼ RA ! ILðQk Þ ¼ ILðRA Þ
Using these expressions, we can trace the left-hand portion of the influence line for shear as the influence line for reaction RB with negative sign and right-hand portion of the influence line for shear as influence line for reaction RA (Fig. 2.6). Units of ordinate IL(Qk) are dimensionless. In order to construct the influence line for shear at section k the following procedure should be applied: 1. Plot ordinate +1(upwards) and 1(downwards) along the vertical lines passing through the left-hand and right-hand support, respectively. 2. Join each of these points with base point at the other support. 3. Connect both portions at section k. 4. Show the operating range of influence line: operating range of left-hand portion is negative, and operating range of righthand portion is positive. The negative sign of the left-hand portion and jump at section k may be explained as follows: If the load P ¼ 1 is located on the left part of the beam, then shear Qk ¼ RA P < 0. When load P is infinitely close to the section k to the left, then shear Qk ¼ RB. As soon as the load P ¼ 1 moves over section k, then shear Qk ¼ RA.
22
2 The Simplest Beams: Theory of Influence Lines
x
P=1 k
A
B a
b
RA
RB
l
Load P=1 left at section k
– Operating range
1
−IL(RB) IL(RA)
1 + Operating range
Load P=1 right at section k
IL(RA)
1 +
Inf. line Qk
– 1 −IL(RB)
Fig. 2.6 Simply supported beam. Construction of influence line for shear at section k
It is obvious that the influence line for shear for the section which is infinitely close to support A coincides with influence line for reaction RA, i.e., IL(Q) ¼ IL(RA). If section is infinitely close to support B, then the influence line for shear coincides with influence line for reaction RB with negative sign, i.e., IL(Q) ¼ IL(RB). Now let us consider a simply supported beam with overhangs (Fig. 2.7). Assume that section k is located between supports. In order to construct influence line for bending moment and shear at section k, it is necessary to:
P=1 k a
b l ab l
+
Inf. line Mk
1 +
Inf. line Qk
– 1
Fig. 2.7 Beam with overhangs. Construction of influence lines for bending moment and shear at section k
2.2 Analytical Method for Construction of Influence Lines
23
1. Ignore overhang and construct corresponding influence line for simply supported beam 2. Extend the latter until its intersection with the vertical passing through the end of overhang
2.2.2.3
Discussion
At this point, emphasis must once again be placed on differentiating between influence line for bending moment at point k and bending moment diagram due to load P, which is located at point k (Fig. 2.8). This diagram M is constructed on the tensile fibers. P=1-variable position
P(kN)-fixed position
Load contour
k A
Extended fibers
B
B k a
a
b
RA
b
l
RB
l
ab l
b
P
a +
RB
ab l Bending moment diagram M (kNm)
Inf. line Mk (m)
Fig. 2.8 Influence line for bending moment at section k and bending moment diagram due to load P, which is located at point k
Even though both diagrams look similar, and have same sign (ordinates of bending moment diagram are positive) and same ordinate at section k (if P ¼ 1), they have completely different meanings and should not be confused. The difference between them is presented in Table 2.1. Table 2.1 Comparison of influence line for bending moment and bending moment diagram Influence line Mk This graph shows variations of bending moment at section k only; load P ¼ 1 is moving and has different locations along the beam
Bending moment diagram M This graph shows distribution of bending moment in all sections of the beam; load P is fixed and acts at section k only
The following example generalizes all cases considered above for construction of influence lines of internal forces for different sections in case of simply supported beam with overhangs (Fig. 2.9).
Bending Moment M1 and Shear Force Q1 The influence line of bending moment at the section 1 for simply supported beam without overhangs presents a triangle with maximum ordinate ab/l, where l ¼ a + b. The influence line is extended within the overhang. If load P is located within the span, then bending moment in section 1 is positive; that is, the extended fibers are located below the longitudinal axis. If load P is located outside of the span, then bending moment in section 1 is negative. The influence line for shear in section 1 for the simply supported beam should be extended through support points to the end of overhangs.
24
2 The Simplest Beams: Theory of Influence Lines
P=1
P=1 2
3
1
4
2 3' 3"
5
4 4" 5 '
1
d5
c2 c
a
b
1
d
–
+
a
b +
Inf. line M1
–
–
Inf. line Q2
1 –
c2
Inf. line Q1
1
ab l
1
Inf. line M2
–
Inf. line Q3left
1 + c
–
Inf. line M3
–
1 1
Inf. line M4
–
Inf. line Q3right
Inf. line Q4 left
–
1
d – Inf. line M5 d5
1
+ 1
+
Inf. line Q4right Inf. line Q5
Fig. 2.9 Beam with overhangs. Influence lines for bending moments and shear forces for different sections
Shear Force Q2 and Bending Moment M2 (Fig. 2.10) Since expressions for bending moment and shear at section 2 depend on the position of the load P (to the left or to the right of the section 2), for deriving equations of influence lines two positions of the load should be considered.
P=1 2 x c Fig. 2.10 Construction of influence lines for section 2 on the overhang
2.3 Application of Influence Lines for Fixed and Moving Loads
25
P ¼ 1 left at section 2 X Y left ¼ 0 ! Q2 ¼ P Q2 !
P ¼ 1 right at section 2 X Y left ¼ 0 ! Q2 ¼ 0 Q2 !
ILðQ2 Þ ¼ 1
ILðQ2 Þ ¼ 0
X left M 2 ¼ 0 ! M 2 ¼ Px M2 ! ILðM 2 Þ ¼ x At x ¼ 0 : ILðM 2 Þ ¼ 0 At x ¼ c : ILðM 2 Þ ¼ c
X left M2 ¼ 0 ! M2 ¼ 0 M2 ! ILðM 2 Þ ¼ 0
Note that for any position of the load P ¼ 1 (left or right at section 2) we use the equilibrium equations ∑Yleft ¼ 0 and M left 2 ¼ 0, which take into account forces that are located left at the section. Similarly, for section 5 we will use the forces that are located right at the section 5. Construction of influence lines for bending moments and shear at sections 3, 4, and 5 are performed in the same manner. Pay attention that influence lines of shear for sections 3left and 3right, that are infinitesimally close to support point (as well as for sections 4left and 4right), are different. These sections are shown as 30 and 300 , and 40 and 400 . Influence lines for bending moments and shear forces at all pattern sections 1–5 are summarized in Fig. 2.9. These influence lines are good reference source for practical analysis of one-span beams subjected to any type of loads. Moreover, these influence lines will be used for construction of influence lines for multispan hinged statically determinate beams. P
2.3
Application of Influence Lines for Fixed and Moving Loads
Influence lines, which describe variation of any function Z (reaction, bending moment, shear, etc.) in the fixed section due to moving concentrated unit load P ¼ 1, may be effectively used for calculation of this function Z due to arbitrary fixed and moving loads.
2.3.1
Fixed Loads
Three types of fixed loads will be considered: concentrated loads, uniformly distributed loads and couples (Fig. 2.11a), and nonuniform distributed load (Fig. 2.11b). Let us consider the effect of each load separately.
a
P1
P2
y1
q(x)
b
q
M
Inf. line Z
y2
•
R2
R1 yR1
w
• yR2
Inf. line Z
a
Fig. 2.11 Application of influence line for fixed loads. (a) Loading by Pi, M, and q ¼ const; (b) loading by arbitrary distributed load q(x); R1 is resultant of the nonuniform distributed load which is located over rectilinear portion of influence line for Z; this resultant is applied at the center of gravity
26
2.3.1.1
2 The Simplest Beams: Theory of Influence Lines
Concentrated Loads
If a structure is loaded by a force P (P 6¼ 1), then function Z due to this load is Z ¼ Py, with y the ordinate of influence line for function Z at the point where load P is applied. The sign of Z depends on the sign of ordinate y of influence line. If a structure is loaded by several forces Pi, then according to superposition principle (Fig. 2.11a) X Z ¼ Pi yi ð2:8Þ i
In order to compute the value of any function (reaction, internal forces in any section of the beam, frame or any member of the truss, etc.) arising under the action of several concentrated loads Pi, the corresponding influence line must be constructed; each load should be multiplied by the ordinate of the influence line measured at the load point, and the obtained products must be summed up. It is easy to check that units of ordinate of influence lines, which have been discussed early, lead to the required units for function Z.
2.3.1.2
Uniformly Distributed Load
Value of any function Z due to action of uniformly distributed load q is determined by formula Z ¼ qω, with ω the area of influence line graph for function Z within the portion where load q is applied. If the influence line within the load limits has different signs then the areas must be taken with appropriate signs. The sign of the area coincides with sign of ordinates of influence line.
2.3.1.3
Couple
If a structure is loaded by couple M, then function Z due to this moment is Z ¼ M tan α, where α is the angle between the base line and the portion of influence line for function Z within which moment M is applied. If couple tends to rotate influence line towards base line through an angle less than 90 then sign is positive; if angle is greater than 90 then sign is negative.
2.3.1.4
Nonuniform Distributed Load
If nonuniform distributed load is located over straight portion of an influence line, then the effect of such load is equal to Z ¼ R yR where R is the resultant of this load which is applied at the center of gravity of distributed load (i.e., R is the area of nonuniform load diagram), and yR is the ordinate of influence line under the resultant (Fig. 2.11b).
Summary Influence line for any function may be used for calculation of this function due to arbitrary fixed loads. In a general case, any function Z as a result of application of several concentrated loads Pi, uniform load intensity qj, and couples Mk should be calculated as follows: X X X X Z¼ P i yi þ q jω j þ Rn yRn þ M k tan αk , ð2:9Þ where yi is the ordinates of corresponding influence line; these ordinates are measured at all the load points; ωj is the area bounded by corresponding influence line, the x-axis, and vertical lines passing through the load limits; and αk is the angle of inclination of corresponding influence line to the x-axis. The formula (2.9) reflects the superposition principle and may be applied for any type of statically determinate and indeterminate structures. Illustration of this formula is shown below. Figure 2.12a, b presents a design diagram of a simply supported beam and influence line for reaction RA.
2.3 Application of Influence Lines for Fixed and Moving Loads
a
P
27
M
b
q
q R l/2
RA 1
l/2 y
RB
α
+
l/3
RA 1
y
+
Inf. line RA
RB
Inf. line RA
Fig. 2.12 Computation of reaction RA due to different types of fixed load using influence line
Loading a. The value of reaction RA due to given fixed loads P, q, and M equals RA ¼ Py þ qω þ M tan α ¼ P
1 1 1 P ql M þq 1lþM ¼ þ þ 2 2 l l |{z} |fflfflffl{zfflfflffl} |{z} 2 2 ω
y
tan α
Loading b. The value of reaction RA due to given fixed nonuniform load q equals RA ¼ Ry ¼
1 1 ql ql 1 ¼ 2 3 |{z} |{z} 6 y
R
The change of values of loads and/or their position does not change the procedure of calculation of any factor using corresponding influence line. From this example we observe an advantage of influence line: once constructed influence line may be used for calculation of relevant function due to arbitrary loads. Example 2.1 A design diagram of the simply supported beam with overhang is presented in Fig. 2.13. Calculate shear at section n using influence line. The loads are P1 ¼ 12 kN; P2 ¼ 8 kN; q1 ¼ 3 kN/m; and q2 ¼ 2 kN/m.
P1 q1
P2 q2
n n 6m
4m
3m RB
RA
a 0.4 +
1
Inf. line Qn
–
b
1
0.6
14.1
14
+ 3.9
0.3 8
Shear diagram Q (kN)
–
15.9 Fig. 2.13 Design diagram of the beam. (a) Influence line for shear Qn; (b) shear force diagram
28
2 The Simplest Beams: Theory of Influence Lines
Solution First of all, the influence line for required shear at section n should be constructed (Fig. 2.13a). The shear force at section n caused by the fixed loads is Qn ¼ ∑ Piyi + ∑ qiωi. If load P1 is located infinitely close to the left of the section n, then ordinate y1 ¼ 0.6. If load P1 is located infinitely close to the right of section n, then ordinate y1 ¼ 0.4. Ordinate y2 ¼ 0.3. Areas of the influence line within the distribution loads q1 and q2 are 1 ω1 ¼ 6 0:6 ¼ 1:8 m; 2
1 ω2 ¼ 3 0:3 ¼ 0:45m 2
Note that the area of influence line for reaction and shear has units of length, i.e., [m], while for bending moments length is squared, i.e., [m2]. The peculiarity of this problem is that force P1 is located at the section where it is required to find shear. Therefore, we have to consider two cases when this load is located infinitely close from the left and right sides of the section n. If load P1 is located to the left of section n, then Qn ¼ 12 ð0:6Þ þ 8 ð0:3Þ þ 3 ð1:8Þ þ 2 ð0:45Þ ¼ 15:9 kN If load P1 is located to the right at section n, then Qn ¼ 12 0:4 þ 8 ð0:3Þ þ 3 ð1:8Þ þ 2 ð0:45Þ ¼ 3:9kN In order to understand the obtained results, we need to calculate the shear Qn for section n and show shear force diagram. Reaction of the support A is RA ¼ 14.1 kN. If the load P1 is located to the left of section n, then the force P1 must be taken into account: X Qn ! Y left ¼ 0 ! Qn ¼ RA q1 6 P1 ¼ 14:1 3 6 12 ¼ 15:9kN If the load P1 is located to the right of section n, then the force P1 must not be taken into account: Qn ¼ RA q1 6 ¼ 14:1 3 6 ¼ 3:9kN Shear force diagram Q is presented in Fig. 2.13b. This example shows the serious advantage of influence lines: if we change the type of loading, the amount of the load, and its locations on the beam, then a corresponding reaction (or internal forces) may be calculated immediately. Since the amount of the load and its location do not affect the influence lines, influence lines should be considered as fundamental characteristics of any structure.
2.3.2
Moving Loads
Influence line for any function Z allows calculating Z for any position of a moving load, and that is very important, the most unfavorable position of the moving loads and corresponding value of the relevant function. Unfavorable (or dangerous) position of a moving load is such a position, which leads to the maximum (positive or negative) value of the function Z. The following types of moving loads will be considered: one concentrated load, a set of loads, and a distributed load. The set of connected moving loads may be considered as a model of moving truck. The truck loading, the dimensions between axes of the standard truck, and the distribution of its weight on each axle may be found in a special handbook. The moving distributed load may be considered as a model of a set of containers which may be placed along the beam at arbitrary position. The most unfavorable position of a single concentrated load is its position at a section with maximum ordinate of influence line. If influence line has positive and negative signs, then it is necessary to calculate the corresponding maximum of the function Z using the largest positive and negative ordinates of influence lines. In case of a set of concentrated moving loads, we assume that some of the loads may be connected. This case may be applicable for moving cars, bridge cranes, etc. We will consider different forms of influence line.
2.3 Application of Influence Lines for Fixed and Moving Loads
2.3.2.1
29
Influence Line Forms a Triangle
A dangerous position occurs when one of the loads is located over the vertex of an influence line; this load is called a critical load. (The term “critical load” for problems of elastic stability, Chap. 15, has a different meaning.) The problem is to determine which load among the group of moving loads is critical. After a critical load is known, all other loads are located according to the given distances between them. The critical load may be easily defined by a graphical approach (Fig. 2.14a). Let the moving load be a model of two cars, with loads Pi on each axle. All distances between forces are given. P1
a
P2
P3
P4
A P1
B y1
y2
D
y3
P2
l
b A
q
B
y4 a
b
P3 P4 C Fig. 2.14 Graphical definition of the unfavorable position of load for triangular influence line for function Z. (a) Set of concentrated load; (b) uniformly distributed load of fixed length l
Step 1. Trace the influence line for function Z. Plot all forces P1, P2, P3, and P4 in order using arbitrary scale from the leftmost point A of influence line; the last point is denoted as C. Step 2. Connect the rightmost point B with point C. Step 3. On the base line show the point D, which corresponds to the vertex of influence line and from this point draw a line, which is parallel to the line CB until its intersection with the vertical line AC. Step 4. The intersected force (in our case P2) presents a critical load; unfavorable location of moving cars is presented in Fig. 2.14a. Step 5. Maximum (or minimum) value of relative function is Z ¼ ∑ Pi yi. Referring to the Introduction, Fig. 22, let us suppose that the industrial building is equipped with two bridge cranes. To determine the greatest force that is transferred to the column from two cranes, it is necessary to construct an influence line for the reaction of support, considering the crane girder as a simply supported beam. Then it is necessary to consider the next crane girder and to draw the influence line for the reaction of the same support. For the obtained triangular influence line, the procedure shown in Fig. 2.14 should be applied.
2.3.2.2
Influence Line Forms a Polygon
A dangerous position of the set of moving concentrated loads occurs when one or more loads stand over the vertex of the influence line. Both the load and the apex of the influence line over which this load must stand to induce a maximum (or minimum) of the function under consideration are called critical. The critical apex of the influence line must be convex. In case of uniformly distributed moving load, the maximum value of the function Z corresponds to the location of a distributed load q, which covers maximum one-sign area of influence line. The negative and positive portions of influence line must be considered in order to obtain minimum and maximum of function Z. The special case of uniformly distributed moving load happens, if load is distributed within the fixed length l. In case of triangular influence line the most unfavorable location of such load occurs when the portion ab ¼ l and base AB will be parallel (Fig. 2.14b). Example 2.2 Simply supported beam with two overhangs is presented in Fig. 2.15. Determine the most unfavorable position of load, which leads to maximum (positive and negative) values of the bending moment and shear at section k. Calculate corresponding values of these functions. Consider the following loads: uniformly distributed load q and two connected loads P1 and P2 (a twin-axle cart with different wheel loads).
30
2 The Simplest Beams: Theory of Influence Lines
B
A
C
q
D
k 3m
6m
4m P2
P1 0.3
0.4
1
2m
0.2
Inf. line Qk
– 0.4
0.6
P1
P2 2.4
1.6
1
0.3
P2
P1
+
Inf. line Mk (m)
– 1.2
P2
P1
3m
P1
0.6
1.8
P2 Fig. 2.15 Design diagram of the beam, influence lines, and most unfavorable positions of two connected loads
Solution Influence lines for required functions Z are presented in Fig. 2.15. Action of a uniformly distributed load q ¼ 1.6 kN/m. Distributed load leads to maximum value of the function if the area of influence lines within the distributed load is maximum. For example, the positive shear at the section k is peaked if load q covers all portions of influence line with positive ordinates (portions CA and kB); for minimum shear in the same section the load q must be applied within portion with negative ordinates (portions Ak and BD): 1 1 Qkð max þÞ ¼ 1:6 ð0:3 3 þ 0:4 4Þ ¼ 2 kN; Qkð max Þ ¼ 1:6 ð0:6 6 þ 0:3 3Þ ¼ 3:6kN 2 2 1 1 M kð max þÞ ¼ 1:6 10 2:4 ¼ 19:2 kNm; M kð max Þ ¼ 1:6 ð1:2 3 þ 1:8 3Þ ¼ 7:2kNm 2 2 Positive value of Mk max means that if load is located between AB, the tensile fibers of the beam at section k are located below longitudinal axis of the beam. If load is located within the overhangs, then bending moment at section k is negative; that is, the tensile fibers at section k are located above the longitudinal axis of the beam. Action of the set of loads P1 ¼ 5 kN and P2 ¼ 8 kN. Unfavorable locations of two connected loads are shown in Fig. 2.15. Critical load for bending moment at section k (triangular influence line) is defined by graphical method; the load P2 is a critical one and it should be placed over the vertex of influence line: Qkð max þÞ ¼ 5 0:4 þ 8 0:2 ¼ 3:6 kN; M kð max þÞ ¼ 5 1:6 þ 8 2:4 ¼ 27:2 kNm;
Qkð max Þ ¼ ð5 0:4 þ 8 0:6Þ ¼ 6:8kN M kð max Þ ¼ ð5 0:6 þ 8 1:8Þ ¼ 17:4 kNm
In Fig. 2.15 location of loads P1 and P2 for computation of Qk(max) is not shown. If a set of loads P1 and P2 are modeling a crane bridge then the order of loads is fixed and cannot be changed. If a set of loads P1 and P2 is a model of a moving car then we need to consider the case when a car moves in opposite direction. In this case the order of forces from left to right becomes P2 and P1.
2.3 Application of Influence Lines for Fixed and Moving Loads
2.3.3
31
Envelope Diagrams of Internal Forces
Envelope diagram presents a graph, each ordinate of which is the largest value of the internal force in the corresponding section of the beam, caused by the given set of moving loads. This concept may be applied for any type of structure (beams, arches, etc.), any boundary conditions, internal forces (bending moment, shear force), and moving loads (lumped and distributed). For construction of envelope of internal forces we will use the concept of influence lines. The advantage of the influence line approach is that it allows finding the most unfavorable location of load for any section and a value of corresponding force. The idea of constructing the bending moment diagram envelope (BMDE) is discussed for the simply supported beam of length l. The span of the beam is divided into a number of portions; accuracy of construction of the envelope diagrams depends on the degree of discretization of the beam.
2.3.3.1
The Beam is Loaded by Single Moving Load P (Fig. 2.16a)
The span of the beam is divided into n equal portions. The distances from the section k to the left and right supports are ak and bk; in Fig. 2.16a,b index k for a and b is omitted. For each section 1, 2, . . . , n we will construct the influence line for bending moment. The greatest bending moment in the section k caused by the single moving load P occurs when the load is located over the vertex of the influence line Mk; this moment is Mk ¼ Pakbk/l, k ¼ 1, . . ., n. In each section of the beam 1, 2, . . . , n, we plot the corresponding values of bending moment M1 ¼ Pa1b1/l, M2 ¼ Pa2b2/l, . . . . Mn ¼ Panbn/l, and connect the ends of these ordinates by the curve. This curve is a square parabola; indeed, if the load P is at a distance x from the l bveft support, then the reaction of the left support is equal to RA ¼ (Px) /l and the bending moment in the section x becomes M(x) ¼ RA x. Each ordinate of this graph represents the largest moment that occurs in the marked sections caused by one moving force P. The maximum possible moment in an arbitrary section k of the beam is Mk ¼ Pakbk/l. The maximum moment in the beam MaxM ¼ Pl/4 occurs when x ¼ l/2. The maximum moment at x2 ¼ 0.25l is MaxM2 ¼ 0.25 0.75Pl ¼ 0.1875Pl.
a
P1
x
P=1 k
b k
A
B 1
P2
A 1
n
2
a
RB
l
2
a
b
RA
B n
b
RA P1
+
RB
l
Inf. line Mk (m)
y1
+
l/2
P2 y2 Inf. line Mk (m)
BMDE (kNm)
+ MaxM2
BMDE (kNm)
+
Pl/4
MaxMk =Pab/l
Max Mn MaxM1 MaxMk
Fig. 2.16 Design diagram of simply supported beam and corresponding bending moment diagram envelope (BMDE): (a) case of single lumped load; (b) case of two connected loads P1 and P2
32
2 The Simplest Beams: Theory of Influence Lines
The Beam is Loaded by Set of Connected Moving Loads P
2.3.3.2
Suppose that the beam is loaded by two connected moving loads P1 and P2. As in case 1 we will use influence lines for bending moments constructed for section 1, . . . , k, . . . , n. One influence line, namely for the bending moment at the section k, is shown in Fig. 2.16b. For each influence line we define the critical load (see Example 2.2), and establish the unfavorable positions P of the moving loads. Corresponding maximum of the bending moment is calculated according to general formula M k ¼ Pi yki, where k is a section number, while i is the load number. Thus, in the case of the set of two loads, P1 and P2, the i¼1
maximum moments at points 1, k, and n become M 1 ¼ P1 y11 þ P2 y12 ,
M k ¼ P1 yk1 þ P2 yk2 ,
M n ¼ P1 yn1 þ P2 yn2
Here yk1, yk2 means the ordinates of influence line of the bending moment for section k; these ordinates are taken underneath the loads P1 and P2, respectively. The ends of the ordinates of the maximum bending moments are connected by a smooth curve. Each ordinate of the bending moment diagram envelope indicates the maximum possible value of M in the corresponding point of a beam caused by the given system of moving loads. Let the span l ¼ 10 m of the beam be divided into ten portions, beam be loaded by two forces P1 ¼ 20 kN and P2 ¼ 10 kN, and distance between them be 1 m; then the ordinates of the envelope bending moment diagram are M 1 ¼ 20
19 0:9 8 þ 10 ¼ 26 kNm, 10 9
M 2 ¼ 46kNm,
M 3 ¼ 60 kNm,
The benefits of using influence lines are that this procedure can be used to construct the envelope diagram of any factor (internal forces, displacements) in the case of any structure with arbitrary supports; in other words, we have a general procedure without limitations. Figure 2.15 represents a beam loaded by a system of two connected moving loads P1 and P2. Corresponding calculation in Example 2.2 represents the ordinates of the envelope diagrams for the shear force (positive and negative) and for the bending moment. This calculation is presented for one section only, i.e., for section k. Similarly, the ordinates of envelope diagrams may be calculated for the remaining sections of the beam. The disadvantage of the envelope diagram is as follows. Dividing the beam span into an arbitrary number of sections does not guarantee that an absolute maximum will arise in one of the marked sections.
2.3.4
Absolute Maximum of Bending Moment
Let us consider simply supported beam which is subjected to a set of connected moving loads P1, . . . , Pn (Fig. 2.17). The problem is to determine the specific location of connected loads which produces the maximum possible bending moment in the beam. Winkler’s rule allows arranging this set of loads on the beam in the most disadvantageous way and determine the dangerous cross section in which the maximum bending moment arises. P1
P2
R
x VA
c l
Pn
l-c-x VB
Fig. 2.17 Set of wheel loads P1, . . . , Pn with a resultant R
In accordance with Winkler’s rule, the following two propositions can be formulated. 1. The greatest bending moment occurs only underneath some particular load. The Winkler’s rule does not specify underneath which load precisely the maximum bending moment arises.
2.3 Application of Influence Lines for Fixed and Moving Loads
33
2. The load, underneath which the greatest bending moment occurs, and the resultant of all forces acting on the beam must be located on opposite sides of the middle of the beam at an equal distance. The first proposition is obvious, since in the case of concentrated forces the bending moment diagram has a polygonal outline. The second provision needs proof. Suppose that the maximum bending moment arises underneath specified load 2, which is at a distance x from the left support A. If the resultant of all the forces is R, then the reaction V A ¼ Rðl c xÞ=l. The bending moment underneath load P2 is Mx ¼ VAx Mleft. Here Mleft is the bending moment in the section x caused by forces located to the left of this specified load P2. Since all loads are connected, for any position of the moving loads on the beam the value Mleft remains constant. The maximum bending moment is found from the condition d d Rðlx cx x2 Þ M left ¼ 0 M ¼ l dx x dx R l c ðl c 2xÞ ¼ 0, or finally we get x ¼ : l 2 2 The required bending moment is equal to
This implies
Rðl c xÞ x M left max M ¼ l
ð2:10Þ
x¼ðlcÞ=2
Example 2.3 Simply supported beam with a span of 12 m is subjected to a set of connected moving loads, P1 ¼ 10 kN, P2 ¼ 20 kN, P3 ¼ 25 kN, as shown in Fig. 2.18a. Determine the absolute maximum bending moment which occurs in the beam. CL
a P1
P2
2m
R P3 xc =2m
b
P1
P2
0.5m
3m VA
R xc=2m
P3
0.5m l=12m
VB
Fig. 2.18 (a) Set of wheel loads P1, . . . , Pn with a resultant R; (b) position of set of loads to check maximum bending moment under load P2
Solution The resultant of the loads is R ¼ 55 kN. Position of the resultant can be calculated from the equation R xc ¼ P1 5 þ P2 3, so xc ¼
10 5 þ 20 3 ¼ 2m 55
Assume that the maximum bending moment occurs under the load P2. Corresponding location of the set of loads on the beam is shown in Fig. 2.18b. The centerline CL splits distance between P2 and R in two equal distances of 0.5 m. For this location of the loads, the reaction VA and bending moment at the point where P2 is applied are VA ¼
R 5:5 55 ð5:5Þ ¼ 25:2kN, ¼ 12 l
M 2 ¼ V A 5:5 P1 2 ¼ 25:2 ð5:5Þ 10 2 ¼ 118:6 kNm
A similar procedure must be performed for the load P3: put set of loads in such a way that the centerline CL of the beam divides the distance between R and P3 in half and calculate the bending moment under the load P3. The same procedure should be repeated for load P1 and then select the most unfavorable set location. It should be noted that Winkler’s method is applied for a computation of maximum bending moment only and it is based on the following assumptions: (a) beam is simply supported and (b) all loads remain within the beam. If one or more wheels go beyond the beam, then they must be excluded from consideration, find a new value of the resultant, and reapply the Winkler’s rule.
34
2 The Simplest Beams: Theory of Influence Lines
Note that the absolute maximum found by the Winkler’s rule and using the influence lines may lead to different results (although both methods are exact). Indeed, in the case of bending moment diagram envelope, we arbitrarily nominate a set of sections within the span of a beam for which the maximum bending moments are to be calculated. In the case of absolute maximum by the Winkler’s rule, the dangerous section of a beam is precisely determined (by the values of the loads and the distances between them).
2.4
Indirect Load Application
So far, we have been considering cases when external loads were applied directly to the beams. In practice however, loads are often applied to secondary beams (or stringers) and then are transmitted through them to the main beam (or girder) as shown in Fig. 2.19. Stringers are simply supported beams. Each stringer’s span is called a panel d and each point where the stringer transmits its load onto the main beam is called a panel point or a joint. The load is transmitted from the secondary beams on to the main beam only at panel points. P=1
Rm
Rn
x
Stringer
Floor beam
A
m Main beam
B
n d
ym
yn Inf. line Z
Fig. 2.19 Indirect load application
Fragment of influence line for any function Z in case of direct load application is presented in Fig. 2.19 by dotted line. Ordinates of influence line for parameter Z at the panel points m and n are ym and yn, respectively. If load P ¼ 1 is located at point m or n in case of indirect load application, then parameter Z equals ym or yn. If load P ¼ 1 is located at any distance x between points m and n, then reactions of stringers Rm and Rn are transmitted on the main beam at points m and n. In this case function Z may be calculated as Z ¼ Rm ym þ Rn yn Since Rm ¼
Pðd xÞ d x ¼ , d d
Rn ¼
Px x ¼ , d d
the required parameter Z becomes Z¼
dx x 1 y þ y ¼ ym þ ðyn ym Þx d m d n d
ð2:11Þ
Thus, the influence line for any function Z between two closest panel points m and n is presented by a straight line. This is the fundamental property of influence lines in case of indirect load application. Influence lines for any function Z should be constructed in the following sequence: 1. Construct the influence line for a given function Z as if the moving load would be applied directly to the main beam. 2. Transfer the panel points on the influence line and obtained nearest points connected by straight line, which is called as the connecting line.
2.5 Kinematical Method for Construction of Influence Lines
35
This procedure will be widely used for the construction of influence lines for arches and trusses. Procedure for the construction of influence lines of bending moment and shear at section k for simply supported beam in case of indirect load application is shown in Fig. 2.20. First of all we need to construct the influence line for these functions if load P would be applied directly to the main beam AB. Influence line for bending moment presents triangle with vertex ab/l at the section k; influence line for shear is bounded by two parallel lines with the jump at section k. Then we need to indicate the panel points m and n, which are nearest to the given section k, and draw vertical lines passing through these points. These lines intersect the influence lines at the points m0 and n0 . At last, these points should be connected by a straight line. Stringer Floor beam
P=1
A
m
k
B
n
Main beam
a
b l
RA
RB
Connecting line
b a
m⬘
n⬘
Inf. line Mk
+ n⬘
1
+ m⬘
Connecting line
Inf. line Qk 1
Fig. 2.20 Influence lines for Mk and Qk in case of indirect load application
Pay attention that if a floor beam m will be removed, then the influence line for Qk becomes the positive one-sign function instead of a two-sign function, as presented in Fig. 2.20. If the floor beam n and all following ones (except floor beam at the support B) will be removed, then the influence line for Qk becomes the one-sign function too, but a negative one.
2.5
Kinematical Method for Construction of Influence Lines
Kinematical method allows easily constructing the model of influence line, to transform this model into the real influence line, and determine its specific ordinates. The method is based on the virtual displacement principle. The fundamental concepts of this principle are the virtual displacements, ideal constraints, and virtual work. It is assumed that the reader is familiar with the notion of virtual displacement from earlier courses, such as theoretical mechanics. Virtual displacements of the particles of a system are imaginary infinitesimal displacements which are consistent with the constraints of the system. The small, but real displacement, should not be confused with virtual one. The virtual displacements do not depend on the loading. Virtual displacement at any point of mechanism cannot be defined numerically, because this displacement is mental or imaginary one. For virtual displacement time is not required for its accomplishment. Figure 2.21a shows a slider crank mechanism for a particular position that is fixed by an angle φ. The slider B moves in horizontal direction on a fixed guide. The crank OA rotates around point O. Therefore, virtual displacement δB of the slider B is directed horizontally, while the virtual displacement δA of the point A is directed perpendicular to the crank OA. The virtual displacements δA и δB cannot be calculated separately; we can determine only their ratio. For the rod AB the directions of virtual displacements of points A and B are known (point A is common for both rods, OA and AB). The relation between the virtual displacements of two points belonging to one element obeys the law of distribution velocity for a plane motion of a body: virtual displacements of two points belonging to the same body are related as their instantaneous radii. In
36
2 The Simplest Beams: Theory of Influence Lines
Fig. 2.21 the instantaneous rotation center P of rod AB is the intersection point of the instantaneous radii of AP and BP (each of them forms a right angle with the vector of virtual displacement). Thus δA/δB ¼ KP/BP. This procedure allows determining the direction of a virtual displacement of an arbitrary point K and the relation δK/δB ¼ KP/BP.
P
δA
a
b
B
δB
δK A 0
φ
K ôB
B
RB
Fig. 2.21 (a) Slider crank mechanism and virtual displacements of specified points A, B, and K; (b) concept of ideal constraint
Ideal constraint means that the sum of the work done by all the reactions in virtual displacements is zero. An example of an ideal constraint is the hinged movable support (Fig. 2.21b). The reaction RB of this constraint is directed along the support rod, and the virtual displacement δB is directed perpendicularly to the support rod (just in this case the displacement may occur without disruption of constraint). Therefore the virtual work done by the reaction in the virtual displacement δA ¼ RBδB cos (π/2) ¼ 0. Other examples of ideal constraints are absolutely smooth surface, flexible inextensible constraints. Virtual displacement principle may be stated as follows: For equilibrium of a system of material points with ideal constraints it is necessary and sufficient that the total virtual work done by all the active forces in any virtual displacements is equal to zero. This principle was formulated very similarly but without proof by John Bernoulli (1667–1748). In general form the principle was proved and enunciated by Lagrange in 1788. The principle was generalized for the case of one-side constraints by M.V. Ostrogradsky in 1838–1842. Let us apply this principle to constructing the influence lines of any function in case of simply supported beam with overhang (Fig. 2.22). To construct the influence line of the reaction RA, we need to remove the constraint A, in which the required force arises, and replace the action of eliminated constraint by force X. As the result, the reactive force becomes an active force. Assume that X is directed in positive direction (Fig. 2.22a). Elimination of constraint led to the fact that the nature of the structure has changed—geometrically unchangeable and statically determined structure is transformed into mechanism with one degree of freedom. Just these two factors—mechanism and active forces—are the prerequisite for application of the virtual displacement principle. Next, we should impart a virtual displacement to the system. The virtual displacement is only determined by the type of mechanism and the character of constraints. The bar ABD is fixed at point B; therefore the virtual displacement of the bar may be shown by two ways: rotating around the fixed point B clockwise and counterclockwise. We will show the virtual displacement in such a way that unknown positive force Х in the virtual displacement δX would perform the positive work; advantages of such direction for δX are discussed below. Corresponding virtual displacement diagram of the bar is shown by dotted line. According to virtual displacement principle X δX P δP ¼ 0 δ X¼P P δX
ð2:11Þ
Here δX is the virtual displacement of the point of application of the reaction X in the direction of the discarded constraint, while δP is the virtual displacement of the point of the loading contour where moving load P ¼ 1 is applied. The feature of expression (2.11) is that when constructing the influence line, the term δP is variable, which depends on the position of the load P on the beam, while δX is a constant, since this virtual displacement is determined at the fixed point, where reaction X is applied. Since moving load is unit one, expression for required influence line X in its most general form becomes ILðX Þ ¼
1 ILðδP Þ δX
The expression (2.12) has a deep fundamental character. It shows that
ð2:12Þ
2.5 Kinematical Method for Construction of Influence Lines
37
A shape of influence line of the reaction X of any constraint and the virtual displacement diagram of loading contour δP of system with removed constraint, in which the reaction X arises, are similar.
P=1
x
k
A
B
D
VD diagram of the loading contour
a A
δX
B
δP C
D
X A′
b
δφ1
N
a
K
1 δφ1
δφ2
H
A δP
1
b B
Mk δφ2
a
D
b Q
c 1
K δ2
A δP δ1
B D
k
VD diagram
Q
1
K Fig. 2.22 Simply supported beam. Kinematical construction of influence lines for reaction (a), bending moment Mk (b), and shear force Qk (c) (VD—virtual displacements)
This famous principle was established by Heinrich Müller-Breslau (1851–1925) in 1886. The principle establishes the analogy between the strange, at first glance, two concepts. They are the law of change for force in a certain section of a geometrically unchangeable structure when a unit concentrated force moves and the law of distribution of displacement in the mechanism that is obtained by removing the constraint in which the required force arises. Thus, in each particular case, without making any calculations, it is easy to show the shape of the required influence line, which, in many cases, can be represented without difficulties. In order to transform the shape of the influence line to the influence line itself, let us assume that the scale coefficient δX ¼ 1. Indeed ILðX Þ ¼ ILðRA Þ ¼
ILðδP Þ ðl xÞ l x ¼ , ¼ δX l1 l
which coincides with (2.1). Corresponding influence line is shown in Fig. 2.3. Now let us consider the application of kinematical method for construction of influence lines of internal forces, which arises in any section k of a beam.
2.5.1
Influence Line for Bending Moment at Section k, IL(Mk)
To construct the IL(Mk), we eliminate the constraint that prevents mutual rotation of two cross sections located to the left and to the right from the point k. In other words, at point k we need to create connection that allows a mutual rotation of sections,
38
2 The Simplest Beams: Theory of Influence Lines
and therefore in it a bending moment Mk cannot arise. For this purpose, in the section k of a beam the hinge is inserted, and the action of the eliminated constraint is replaced by two unknown positive moments Mk (Fig. 2.22b). Next, we need to impart a virtual displacement to the system: each of the bars, AK and KBD, connected by hinge K rotates around fixed points А and B, respectively; the angles of rotations of each part of the beam are dφ1 and dφ2. The change of slope of both parts of virtual displacement diagram occurs at the hinge. We plot the virtual displacement diagram above the axis of a beam. In this case the left moment Мk on the virtual displacement δφ1 and the right moment on the virtual displacement δφ2 perform a positive work. The figure AKBD presents a model of influence line for bending moment Mk (with accuracy to sign, since the model which is located above the basic line means the positive ordinates of influence line). Similarly (2.12), the equation of influence line for Mk is ILðX Þ ¼ ILðM k Þ ¼
1 ILðδP Þ, δX
where δP, as before, is a virtual displacement of points of a loaded contour where moving unit load is applied, while δX (which in fact is virtual displacement in direction of M and therefore δX ¼ δM) is a mutual angle of rotation of two bars, AK and KB; this angle ∠A0KA ¼ δM ¼ (δφ1 + δφ2). The scale of influence line is δM ¼ (δφ1 + δφ2) ¼ 1. Just in this case the virtual displacement diagram δP of the load contour becomes the real influence line. Now let us implement condition δX ¼ δM ¼ 1. Considering the small angles, we get tanδφ1 + tan δφ2 ¼ 1. This relationship has a clear geometrical interpretation and allows us to easily construct a real influence line, or in other words indicate its ordinates. To do this, in the model of influence line from the point k we need to plot the horizontal segment equal to unity (in any direction), and then from the point H plot the vertical segment HN ¼ 1. It is easy to verify that the line KN determines the segment AA0 ¼ a on the vertical line passing over the support A. This geometrical construction transforms the model of influence line Mk into the real influence line. Considering triangle AA0 B we can verify that HN ¼ 1 tan δφ1 + 1 tan δφ2 ¼ 1; therefore we get that ordinate of the influence line Мk in the section k indeed equals a(l a)/l. Corresponding IL(Mk) is shown in Fig. 2.5. Condition δM ¼ (δφ1 + δφ2) ¼ 1 allows analytically, without special constructions, to easily determine the ordinate of influence line at section k. Since δφ1 tan δφ1 ¼ δφ1 þ δφ2 ¼
2.5.2
yk y and δφ2 tan δφ2 ¼ k , a b
yk yk ab þ ¼ 1, which immediately leads to the well-known result yk ¼ . l a b
Influence Line for Shear Force at Section k, IL(Qk)
To construct the IL(Qk), we need to eliminate the constraint (this constraint is not shown) that prevents mutual linear displacement of two sections located to the left and to the right from the point k. In Fig. 2.22c two bars, AK and KBD, at section k are connected using two parallel bars with hinges at the ends; such connection of the rods is not capable of perceiving the shear force. The action of the eliminated constraint is replaced by unknown positive shear forces Qk. Next, we need to impart a virtual displacement to the system: two bars, AK and KBD, rotate around points A and B, and virtual displacements of point k in the direction of the introduced shear forces Q are equal to δ1 and δ2. Assume that the rotation of the rods around the centers of rotation A and B occurs in the direction in which each positive force (Qleft directed downward and Qright directed upward), in the corresponding virtual displacement, δ1 and δ2, does the positive work, i.e., Qleftδ1 0, Qrightδ2 0. The virtual displacement diagram presents the model of influence line of shear force Qk. Equation of influence line for Qk is ILðQk Þ ¼ δ1Q ILðδP Þ, where δQ is mutual linear vertical displacement of the bars Ak and kBD at the point k, so δQ ¼ (δ1 + δ2). The scale of influence line equals δQ ¼ (δ1 + δ2) ¼ 1. It means that the left and right portions of the virtual displacement diagram cut off the unit segments on the vertical lines passing through the left and right supports. In other words, the lines AK and KBD are parallel. Corresponding IL(Qk) is shown in Figs. 2.6 and 2.9. Now we will consider a cantilever beam of length l (Fig. 2.23). Briefly discuss the main steps for the construction of influence lines of the reactions and internal forces by kinematical method.
2.5 Kinematical Method for Construction of Influence Lines
39
P=1 MA
A
c
B
Mk
A
k
Shape of Inf. line Mk
δφ
x
VD diagram
RA
l Q
a δx
Virtual displacement diagram
d
VD diagram
δ2
Inf. line RA
RA
A
b MA
δφ
Q
Shape of Inf. line Qk
k
Shape of Inf. line MA
VD diagram
Fig. 2.23 Cantilevered beam. Kinematical construction of influence lines for reactions (a, b) and internal forces (c, d) at section k
To construct the influence line of the vertical reaction X ¼ RA, we need to eliminate constraint in which the required reaction arises, and by a dotted line we show the diagram of virtual displacements δx of the loaded contour (Fig. 2.23a). If the scale factor (the ordinate of the virtual displacement diagram where the reaction RA is applied) is taken equal to unity, then this graph represents the influence line RA. For construction of influence line of the bending moment MA which arises at support A we need to eliminate constraint in which the required bending moment arises. For this purpose, in the section A of a beam the hinge is inserted, and the action of the eliminated constraint is replaced by unknown positive moment MA. Corresponding mechanism and virtual displacement diagram is shown in Fig. 2.23b. If the scale factor (the slope of the VD diagram) is δφ ¼ 1, then ordinate of IL(MA) at point B is equal to l. On the adopted direction of a virtual displacement the required moment MA performs a positive work. Therefore the model of the influence line and real influence line completely coincides. It means that the ordinates of influence line MA are negative, as shown in Fig 2.4. Figure 2.23c, d presents the transformation of the initial cantilever beam into a mechanism for construction of influence line for bending moment (c) and shear (d) at section k, respectively. The peculiarity of the obtained systems (c, d) is that the left part of a beam, Ak, is geometrically unchangeable; that is, this part cannot be considered as a mechanism, and therefore the virtual displacements of this part of a beam are absent. Diagrams of virtual displacements are shown by dotted lines. If a scaled coefficient δφ ¼ 1 (Fig. 2.23c), then ordinate of IL(Mk) at point B is equal to linear dimension kB. If the scaled coefficient δ2 ¼ 1 (Fig. 2.23d), then ordinates of IL(Qk) are constant and equal to unity.
2.5.3
Conclusion
As noted earlier, δX (see Eq. 2.12) is the ordinate of the diagram of the virtual displacements at the point where the required force X is applied. If we put δX ¼ 1, then the model of the influence line presents the real influence line. Thus, the influence line of any function may be constructed as a displacement diagram of a loaded contour caused by unit displacement; character of this displacement must corresponds to the required function. In this formulation we combine two stages, namely the construction of a diagram of virtual displacements as a model of the influence line and a transition from the influence line model to the influence line itself. To construct the influence line of the reaction of support RA, it is necessary to remove the constraint in which this reaction RA occurs, show the positive reaction RA, and impart to system the unit linear displacement in the direction RA (Figs. 2.22a and 2.23a). To construct the influence line of the moment at support MA, it is necessary to remove the constraint in which this moment MA occurs, show the positive moment MA, and impart to system the unit angular displacement in the direction MA (Fig. 2.23b).
40
2 The Simplest Beams: Theory of Influence Lines
To construct the influence line of the bending moment at section k, Mk, it is necessary to remove the constraint in which this moment Mk occurs, show the positive pair of bending moments Mk, and impart to system the unit mutual angular displacement in the direction Mk (Figs. 2.22b and 2.23c). To construct the influence line of the shear force at section k, Qk, it is necessary to remove the constraint in which this shear Qk occurs, show the positive pair of shear forces Qk, and impart to system the unit mutual linear displacement in the direction Qk (Figs. 2.22c and 2.23d). The positive direction of required function Х allows adopting corresponding direction for virtual displacement δX. This procedure automatically leads to the sign of influence line ordinates: if a model of influence line occurs above a basic line, then a real influence line occurs above as well; that is, the ordinates of influence line will be positive. Later we will show the application of the deep idea of Müller-Breslau to the analysis of a redundant structures.
2.6
Combining of Fixed and Moving Load Approaches
So far we showed an application of influence lines Z for analysis of this particular function Z. However, in structural analysis, the application of influence lines is of great utility and we can use influence line for Z1 for calculation of another function Z2. For example, design diagram of the beam and influence line for reaction RA are shown in Fig. 2.24. How can we calculate the bending moment at section k? P 1
B
A
1
k RA 1
a
2
C
RB
+
Inf. line RA yC
Fig. 2.24 Calculation of bending moment Mk using influence line for RA
Using fixed load approach we need to calculate the reaction R1, which arises in beams 1–2; transmit this reaction to joint 1 onto beam ABC; determine reactions RA and RB and, after that, calculate Mk by definition using RA (or RB). Using moving load approach, we need to construct the influence line for bending moment at section k. However, we can combine both approaches. For this show the influence line for reaction RA. According to this influence line, the reaction RA ¼ PyC. Immediately, the bending moment at section k by definition equals Mk ¼ RAa ¼ PyC a. This idea will be used later for analysis of structures. Such combination of two approaches is a very effective way for the analysis of complex statically determinate structures and especially of statically indeterminate structures. Finally, it should be remembered that, first of all, we need the influence lines as a tool for determining reactions and internal forces. Therefore, the construction of the influence line by setting the unit force in different positions and determining the corresponding factor kills the fundamental purpose of influence lines and such an approach should not be considered.
2.7
Properties of Influence Lines
Influence line is a graph inherent to the structure itself, reflects its properties and features, and does not depend on how the structure is loaded. Therefore, they should be considered as a fundamental and reference data for the structure. Influence lines allow finding internal forces and reactions in case of fixed and moving loads as well as the most unfavorable location of the load. The influence lines for reactions and internal forces of statically determinate structures present a
Problems
41
combination of straight segments. In the case of an indirect load application, the influence line between two closest panel points is a straight line. This is true for any factor (reaction, internal forces, displacement), and not just for a statically determinate structure. If we differentiate the equation of the influence line along the abscissa x, then we obtain the equation of the influence line of d the same function caused by a unit moving couple, i.e., ILP¼1 ¼ ILM¼1 . This influence line may be obtained from the dx influence line caused by the moving unit force ILP¼1, if to show a graph of the slopes of its corresponding portions. Unit of ordinate of any influence line of factor Z due by couple M is unit of Z divided by unit of couple M. The procedure for construction of influence lines is developed with sufficient completeness, and consists of several simple well-algorithmic steps. These steps eliminate the need to perform multiple calculations of the desired function (reaction, bending moment, etc.) for different positions of a unit load.
Problems 2.1. The simply supported beam AB is subjected to the fixed loads as shown in Fig. P2.1. Analyze the structure using the fixed and moving load approach. Fixed Load Approach Determine the reaction at the supports and construct the internal force diagrams, caused by given loads. Moving Load Approach (a) Construct the influence lines for reaction at the supports A and B. (b) Construct the influence lines for bending moment and shear force for section k. (c) Construct the influence lines for bending moment and shear force for sections infinitely close to the left and right support B. (d) Calculate the reaction of the supports, bending moment, and shear force for all abovementioned sections using corresponding influence lines. (e) Compare the results obtained by both approaches. P=12 kN
q= 2kN/m B
A
R A = 3.9kN , RB = 14.1kN ,
M k = 23.4kNm, Qkleft = 3.9kN ,
k a=6m
Ans.
b=4m
3m
Qklright = −8.1kN
.
Fig. P2.1
2.2. The simply supported beam is subjected to linearly varying load shown in Fig. P2.2. Determine the reactions of supports and construct the bending moment and shear diagrams. Compute the reaction of the left support using influence line.
q A
B l
Fig. P2.2
Ans. RA =
ql ql 3 2 ; RB = ; M max = ql 6 3 27
42
2 The Simplest Beams: Theory of Influence Lines
2.3. The cantilever beam AB is subjected to the fixed loads as shown in Fig. P2.3. Analyze the structure using the fixed and moving load approach. Fixed load approach Determine the reaction of support B. Construct the internal force diagrams. Moving load approach (a) Construct the influence lines for vertical reaction RB and moment MB at the clamped support. (b) Construct the influence lines for bending moment and shear for section k. (c) Calculate reactions at the support B, bending moment, and shear for section k using corresponding influence lines. (d) Compare the results obtained by both approaches.
P=8kN
q=2kN/m
A
Ans. RB = 14.0kN
B 3m
k
(↑);
M B = 57kNm
3m
Fig. P2.3
2.4.
The beam AB is subjected to the fixed uniformly distributed loads as shown in Fig. P2.4. All dimensions are in meters.
(a) Calculate the bending moment and shear at section k caused by the given fixed load using influence lines for MK and QK, respectively. (b) Calculate the bending moment and shear force at section k caused by the given fixed load, using influence lines for reaction RA. (c) Compare the results. q=2kN/m
A 3
B
k 3
2
Ans. Mk=9 kNm; Qk= 1.8 kN;
2
3
3
Fig. P2.4
2.5. For simply supported beam construct the influence line for reaction of supports, bending moment, and shear at any section caused by moving unit couple M ¼ 1. Explain the units of corresponding ordinates. 2.6. The simply supported beam is subjected to moving connecting loads P1 ¼ 12 kN, P2 ¼ 20 kN, and P3 ¼ 15 kN. Calculate the maximum bending moment at point k (Fig. P2.6). P1
A
k a=5m
Fig. P2.6
P2 P 3 1m 3m
l=12m
Problems
43
2.7. The simply supported beam is subjected to moving connecting loads P1 ¼ 12 kN and P2 ¼ 20 kN. Determine absolute maximum of bending moment and location of corresponding section. P1
P2 2m
A l=10m Fig. P2.7
2.8. For simply supported beam in Fig. P2.7 construct the bending moment diagram envelope (divide beam into five equal portions). Explain the meaning of this diagram. Compare the results with problem 2.7 and explain the difference. 2.9. For cantilevered beam construct the following influence lines using kinematical method: (a, b) vertical reaction and bending moment at support; (c, d) shear force and bending moment at arbitrary section.
Chapter 3
Multispan Hinged Beams and Frames
This chapter is devoted to the analysis of statically determinate multispan beams subjected to fixed and moving loads. Methods for generation of beams and load path, construction of internal force diagrams, construction of influence lines and theirs application are discussed. Also analysis of plane frame subjected to fixed load is presented.
3.1
Generation of Multispan Hinged Beams
Multispan hinged beams are geometrically unchangeable and statically determinate structures consisting of a series of one-span beams with or without overhangs connected together by means of hinges. This type of structures was proposed by W. Gerber (1867) and the theory has been developed by G. Semikolenov (1872) (Bernshtein 1957; Darkov 1989). The simplest Gerber-Semikolenov beams are presented in Fig. 3.1.
b
a
c
Fig. 3.1 Simplest Gerber-Semikolenov beams: (a) Beam with rolled and pinned supports; (b) beam with one clamped support; (c) beam with clamped and slide supports
The following rules of distribution of hinges in beams, which have no clamped ends, provide their unchangeableness and statical determinacy: 1. 2. 3. 4.
Each span may contain no more than two hinges. Spans with two hinges must alternate with spans without hinges. Spans with one hinge may follow each other, providing that first (or last) span has no hinges. One of the supports has to prevent displacement of a whole structure in the horizontal direction. Figure 3.2 shows some examples of the hinge distribution in beams without fixed ends.
a
b
c
d
Fig. 3.2 Distribution of hinges in beams with simply supported ends
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_3
45
46
3 Multispan Hinged Beams and Frames
The distinctive properties of multispan statically determinate beams: 1. Structure with intermediate hinges has less stiffness than structure without intermediate hinges. This leads to substantial reduction of bending moments as compared with continuous beams spanning the same opening. 2. Possibility of control stresses by variation of hinge locations. Advantages of using multispan hinged beams: 1. 2. 3. 4.
Distribution of internal forces does not depend on the stiffness of each part of a beam. Change of temperature and settlements of supports, and imperfect assembly, does not create stresses. Failure of one of the supports may not destroy the entire system. Relatively short members of multispan beams are well suited for prefabrication, transportation, and installation using standard equipment. 5. Multispan hinged beams are usually more economical than a series of disconnected simply supported beams spanning the same opening.
3.2
Interaction Diagrams and Load Path
Gerber-Semikolenov beams may be schematically presented in the form, which shows the interaction of separate parts and transmission of forces from one part of the beam to another. Gerber-Semikolenov beams consist of two types of beams, namely a main (or primary) and suspended (or secondary) beam. A main beam is designed to carry a load, which is applied to this beam as well as to maintain a suspended beam. Therefore, the main beam carries a load, which is applied to this beam, as well as a load, which is transmitted on the main beam as a reaction of the suspended beam. Interaction schemes for simplest Gerber-Semikolenov beams are presented in Fig. 3.3. a A
1
H
2
b A
1
2 1
H
2
B
1
2 1
1
Fig. 3.3 Interaction schemes for simplest Gerber-Semikolenov beams: (a) beam with one primary beam; (b) beam with two primary beams
In these cases beams 1 are primary and beams 2 are secondary ones. Obviously, that failure of the beam 1 causes the collapse of the entire structure while failure of the beam 2 does not cause failure of the beam 1. There is only one way of presenting multispan beams using an interaction diagram. In both cases left support A prevents horizontal displacement; therefore on the interaction scheme the supports at hinge H and support B (Fig. 3.3b) by convention are shown as pinned supports. Interaction schemes allow to clearly indicate the load pass from one part of a structure to another. Also they are helpful for construction of internal force diagrams as well as for more advanced analysis. Five-span beam and its interaction diagram are presented in Fig. 3.4. Since the structure in a whole is restricted against the horizontal displacement owing to support A, then each part of the structure has no displacement in horizontal direction as well. That is why the rolled supports C and E on the interaction diagram are replaced by pinned supports. Restrictions for suspended beams also prevent horizontal displacements. This replacement is conventional. Similar reasoning is related also to structure shown in Fig. 3.3b.
3.2 Interaction Diagrams and Load Path
47
H2
H1 A
B
H3 C
H4
D
E
F
Fig. 3.4 Interaction diagram of Gerber-Semikolenov beam with hinged ends
Beams ABH1 and H1H2 present the main and suspended beams, respectively. It means that collapse of the main beam ABH1 leads to the collapse of suspended beam, while collapse of suspended beam does not affect the main beam. Collapse of the beam H4EF leads to the collapse of the beam H3H4 only, and does not affect other parts of the structure. The entire structure contains five disks, four simple hinges, and seven constraints of supports. Degree of freedom of the structure is W ¼ 3D 2H C0 ¼ 3 5 2 4 7 ¼ 0, so the entire system is a statically determinate and geometrically unchangeable structure. On the other hand, in general case of loads seven reactions of supports arise in the structure. For their calculation we can use three equilibrium equations and four additional conditions; that is, bending moment at each hinge is equal to zero. Therefore, the structure is a statically determinate one. Another four-span beam and its interaction diagram are presented in Fig. 3.5.
H1
H2
P
H3 E
A B
C
P
D
Fig. 3.5 Interaction diagram of Gerber-Semikolenov beam
The entire structure contains four disks, three simple hinges, and six constraints of supports. Degree of freedom of the structure is W ¼ 3D 2H C0 ¼ 3 4 2 3 6 ¼ 0, so the entire system is a statically determinate and geometrically unchangeable structure. On the other hand, in general case of loads six reactions of supports arise in the structure; they are five vertical reactions at supports and one horizontal reaction at support A. For their calculation we can use three equilibrium equations and three additional conditions; that is, the bending moment at each hinge is equal to zero. Therefore, the structure is a statically determinate one. The beam H2H3 is both main for H3E beam and suspended for H1CH2 one. Therefore, the load P, which is applied to the beam H2H3, is transmitted only to beams located below and is not transmitted to the beams located above this one. Some examples of the hinge distribution in multispan statically determinate and geometrically unchangeable beams with one or two clamped supports and corresponding interaction diagrams are shown in Fig. 3.6. On each design diagram, the number of disks D, intermediate hinges H, and number of support links C0 are marked. Each example also contains the interaction diagram.
48
3 Multispan Hinged Beams and Frames
a
b
D=3, H=2, C0=5
D=4, H=3, C0=6
B
A
Fig. 3.6 Distribution of hinges in two-span beams with one (a) and two clamped ends (b)
Design diagram in Fig. 3.6b contains clamped support B which allows horizontal displacement. In this support two reactions arise; this support retains one degree of freedom because it allows horizontal mobility. The constraint, in which the horizontal reaction in the support B arises, is absent, since such a reaction is perceived by the constraint of support A; a similar situation is observed in the simply supported beam. However the interaction diagram contains the conventional clamped support B. Kinematical analysis: The number of degrees of freedom for design diagram in Figs. 3.6a and 3.6b is
Wa = 3D − 2H − C0 = 3 × 3 − 2 × 2 − 5 = 0 Wb = 3D − 2H − C0 = 3 × 4 − 2 × 3 − 6 = 0 Thus, both design diagrams in Fig. 3.6 are statically determinate and geometrically unchangeable. Example of unsatisfactory distribution of hinges is shown in Fig. 3.7.
D=3, H=2, C0=5 A
D H1
B
H2 C
Fig. 3.7 Example of unsatisfactory distribution of hinges H1 and H2
Although the number of degrees of freedom turns out to be zero, one of the parts of the beam, ABH1, turned out to be statically indeterminate (this does not raise objections from the engineering point of view); however the other part, H1H2CD, turned out to be geometrically changeable, which is inadmissible in engineering practice.
3.3
Fixed Load Method for Beams
Static analysis of multispan beams subjected to any fixed load requires definition of reactions of supports and construction of bending moments and shear force diagrams. Such an analysis should be performed in the following order: 1. Perform the kinematic analysis and present the design diagram of multispan beam in the form of an interaction diagram of separate members. 2. Provide analysis of the suspension beam first. Reactions of this beam are transmitted on the underlying (principal) beam. The direction of this force is opposite to the reaction of the suspended beam. 3. Provide analysis of the principal beam subjected to the load applied to the beam itself and to the load that is transferred from the suspended beam. Example 3.1. Design diagram of the hinged beam is shown in Fig. 3.8a. Determine the reactions of supports and construct the shear and bending moment diagrams. Kinematical analysis: Degrees of freedom W = 3D − 2H − C0 = 3 × 2 − 2 × 1 − 4 = 0. The structure is statically determinate and geometrically unchangeable. Structure consists of two beams; they are suspended beam CB and primary beam AC. Interaction diagram is shown in Fig. 3.8b. Suspended beam CB is loaded by uniformly distributed load q and force P1 (Fig. 3.8c). The reactions of supports will be reckoned positive when directed upwards:
3.3 Fixed Load Method for Beams
49
X
10 ¼ 0 ! RB ¼ 11:2 kN 2 X 10 RC ! M B ¼ RC 10 P1 2 þ q 10 ¼ 0 ! RC ¼ 9:8 kN 2
RB !
M C ¼ RB 10 P1 12 q 10
Principal beam AC is loaded by force P2; also it is necessary to take into account the reaction of suspended beam CB which is transferred on the beam AC with opposite direction, i.e., R0C ¼ RC ¼ 9:8 kN. Reactions of the principal clamped-free beam AC are RA ¼ P2 þ R0C ¼ 3:2 þ 9:8 ¼ 13 kN M A ¼ P2 2 þ R0C 4 ¼ 3:2 2 þ 9:8 4 ¼ 45:6 kNm P2=3.2kN
P1=1kN q=2kN/m
a A
D
2m
10m
2m
x
0
b
K
B
C
2m
B
C A
x1
c
P1=1kN
q=2kN/m
C
B 10m
RC=9.8kN
2m RB=11.2kN
P2=3.2kN R′C= RC=9.8kN
d A
C
MA=45.6kNm RA=13kN
2m
2m x0=4.9m
13
e
+
9.8
1.0 Q (kN)
− 45.6
10.2
f -
19.6
2.0
M (kNm)
+ Mmax=24.0 Fig. 3.8 Gerber-Semikolenov beam. (a) Design diagram; (b) interaction diagram; (c, d) suspended and principal beam; (e, f) shear force and bending moment diagrams
After determining the reactions of supports of the principal beam, one should proceed to determining the shear forces and bending moments for the structure. Computations of Q and M can be performed by two different approaches, namely, when the object under consideration is the beam is taken as a whole and when a set of individual beams is adopted. Consider these two approaches.
50
3 Multispan Hinged Beams and Frames
1. The object under consideration is the beam AB in a whole. This beam consists of four portions. The origin will be placed in the left support. Then for each portion the coordinate x varies within the following boundaries: portion АD: 0 ≤ x1 ≤ 2; portion DC: 2 ≤ x2 ≤ 4; portion CB: 4 ≤ x3 ≤ 14; and portion BK: 14 x4 16 (Fig. 3.8a). Shear forces at each portion First and second portions: QAD ¼ RA ¼ 13 kN,
QDC ¼ RA P3 ¼ 13 3:2 ¼ 9:8 kN,
Within these portions a shear is constant. Third portion: QCB = RA − P3 − q(x − 4). Within this portion shear changes by linear law. At the boundary points QC ¼ RA P3 qð4 4Þ ¼ 9:8 kN Qleft B ¼ RA P3 qð14 4Þ ¼ 10:2 kN
Shear at boundary points C and B of the third portion has different signs. Change of sign occurs when QCB ¼ RA P3 q(x 4) ¼ 0. The root of this equation is x ¼ 8.9 m. In this section, the bending moment takes the maximum value. Fourth portion: QBK ¼ RA P3 q(14 4) + RB ¼ 13 3.2 q 10 + 11.2 ¼ 1.0 kN
Bending moments at specified points A, D, and C: M D ¼ M ð2Þ ¼ RA 2 M A ¼ 13 2 45:6 ¼ 19:6 kNm M C ¼ M ð4Þ ¼ RA 4 M A P2 2 ¼ 13 4 45:6 3:2 2 ¼ 0 Third portion: MCB ¼ RAx P3(x 2) MA q(x 4)2/2. Within this portion a shear changes by parabola. At the boundary points ð 4 4Þ 2 ¼0 2 ð14 4Þ2 M B ¼ M ð14Þ ¼ RA 14 3:2ð14 2Þ M A q ¼ 2:0 kNm 2
M C ¼ M ð4Þ ¼ RA 4 3:2ð4 2Þ M A q
Maximum bending moment occurs at section, where QCB ¼ 0, i.e., at x ¼ 8.9 m: M max ¼ M ð8:9Þ ¼ RA 8:9 P3 ð8:9 2Þ M A q
ð8:9 4Þ2 ¼ 24 kNm 2
2. The object under consideration is the separate beams. It is convenient to construct diagrams of internal forces for each beam separately. Suspended beam CBK: The origin of coordinates is placed in the extreme left point of this element, i.e., at the point C (Fig. 3.8c). Shear force at specified sections: Qx1 ¼ RC qx1 ¼ 9:8 2x1 ! QC ¼ 9:8 kN,
QBleft ¼ 9:8 2 10 ¼ 10:2 kN,
Condition Qx1 ¼ RC qx0 ¼ 9.8 2x0 ¼ 0 leads to x0 ¼ 4.9 m (Fig. 3.8e). Bending moments:
QBright ¼ P1 ¼ 1:0 kN
3.4 Influence Lines for Reactions and Internal Forces
M x1 ¼ RC x1 q
51
x21 x2 ¼ 9:8x1 2 1 2 2
M C ¼ M ð0Þ ¼ 0,
M B ¼ M ð10Þ ¼ 9:8 10 2
M max ¼ M ð4:9Þ ¼ 9:8 4:9 2
102 ¼ 2 kNm 2
4:92 ¼ 24:0 kNm 2
Similarly we can consider beam AC subjected to MA ¼ 45.6 kNm (counterclockwise), RA ¼ 13 kN("), P2 ¼ 3.2 kN(#), and R0C ¼ 9:8 kNð#Þ. The final diagrams of shear and bending moments are shown in Fig. 3.8(e, f). Positive shear is shown above the reference line. Positive bending moments are shown below the reference line: this means that bending moment diagram is constructed on the tensile fibers. Note that each line of hatching characterizes corresponding factor (bending moment, shear, influence line, etc.) for each section of a structure; that is, its length represents corresponding value of this factor; therefore, hatching must be perpendicular to the base line of a structure. In some cases, hatching is not shown in some examples in this book so as not to obscure the other features explained. Control: 1. Taking into account the relations between shear, bending moment, and distributed load intensity we have dM 45:6 19:6 ¼ ¼ 13 kN dx 2 dM 19:6 0 QDC ¼ ¼ ¼ 9:8 kN dx 2 9:8 ð10, 2Þ dQ kN ¼ 2 qCB ¼ ¼ 10 dx m QAD ¼
2. The sum of projection of all forces into vertical axis:
∑ Fy = RA − P2 − q × 10 + RB − P1 = 13 − 3.2 − 2 × 10 + 11.2 − 1 = 24.2 − 24.2 = 0 In the case of a large number of elements making up the Gerber-Semikolenov beam, it is more convenient to apply the second approach.
3.4
Influence Lines for Reactions and Internal Forces
To construct influence lines, two methods are used—static and kinematic.
3.4.1
Static Approach
For construction of influence lines of reactions and internal forces for statically determinate multispan beams the following steps are recommended: Step 1: The entire multispan hinged beam should be presented in the interaction diagram form. This helps to classify each element of the structure as primary or secondary beams, and visualize the load path from the secondary beam on the primary one. Step 2: Consider structural element of a multispan beam, which contains a support or section for which the required influence line should be constructed. This element is the primary or secondary simply supported beam with/without overhangs or cantilevered beam. Then we need to construct an influence line for the required function for this separate beam only. Step 3: Take into account the influence of moving load which is located on the adjacent suspended beam and distribute the influence line which is constructed in step 2 along this secondary beam. For this it is necessary to connect an ordinate of influence line at the end point (hinge) with a zero ordinate at the second support of the suspended beam. This procedure called as distribution of influence lines within the suspended beam.
52
3 Multispan Hinged Beams and Frames
Step 4: The beam distribution procedure should be applied for the following suspended beam. To illustrate this procedure for statically determinate multispan beams we will consider a structure in Fig 3.9. It is necessary to construct the influence lines for reactions, bending moment, and shear for indicated sections. We are starting from kinematical analysis of a structure: this beam is a statically determinate and geometrically unchangeable structure.
P=1 A
B H2
H1
s
5m
2m
A
C 2m
B
H1
H3
k 3m
D
3m
n 2m
H2
3m
3m
H3
C
D
1 +
Inf. line RD
0.333
1.5 +
Inf. line Mn (m)
1 0.333
0.5
1
+
Inf. line Qn
– 0.5
1 +
1 Inf. line Qs
0.4 0.8
–
Inf. line Ms (m)
2 Fig. 3.9 Multispan statically determinate beam. Influence lines for reactions, bending moment, and shears at some sections
3.4.1.1
Influence Line for RD
The structural element is beam H3CD. Influence line for reaction RD within the span CD is a straight line with ordinate 1 at support D, zero at support C, and extended within left-hand and right-hand overhangs. If load P ¼ 1 is located on the suspended beam H2H3, then fraction of this force is transmitted to the primary beam H3CD. If load P ¼ 1 is located at point H3 of suspended beam H2H3, then this force is completely transmitted to the primary beam H3CD. Thus, influence line has no discontinuity at point H3. If load P ¼ 1 is located at point H2 on the suspended beam H2H3, then this force is completely transmitted to the primary beam H1BH2 and thus has no influence on the reaction RD. Therefore, influence line of RD has zero ordinate at point H2. If unit load is located within H2H3, then the pressure transmitted from the secondary beam H2H3 to the primary beam H3CD at point H3 varies proportionally to the distance of the unit load from point H2. Therefore, ordinates of influence lines at point H3 should be connected with zero ordinates at point H2. If load P ¼ 1 is located on the beam H1BH2, then force is not transmitted to the suspended beam H2H3; the load does not influence reaction RD. Therefore, ordinates of influence line along AH1BH2 are zeros.
3.4 Influence Lines for Reactions and Internal Forces
53
If load P ¼ 1 (or any load) is located on the beam H2C, then reaction RD is directed downward. Maximum positive reaction RD appears if load P ¼ 1 will be located at the extremely right point of the beam; maximum negative reaction RD appears if load P ¼ 1 will be located at the point H3.
3.4.1.2
Influence Lines for Shear Qk, and Bending Moment Mk at Section k
Suspended beam H2H3 is subjected to loads, which act on this beam only, while a load from other parts of the beam (AH2 and H3D) cannot be transmitted on the beam H2H3. Therefore influence lines for internal forces at section k are same as for simply supported beam without overhangs.
3.4.1.3
Influence Lines for Shear Qn, and Bending Moment Mn at Section n
Influence lines for section n should be constructed as for simply supported beam with overhang H3C. At point H3 this beam supports the beam H2H3, which is a suspended one. Therefore, ordinates of influence lines at point H3 should be connected with zero ordinates at point H2.
3.4.1.4
Influence Lines for Shear Qs, and Bending Moment Ms at Section s
When the load travels along portion AH1, then construction of the influence lines for section s is exactly the same as for a cantilevered beam. If load P ¼ 1 is located on the suspended beam H1H2, then fraction of this force is transmitted to the primary beam AH1. If load P ¼ 1 is located at point H1 of suspended beam H1BH2, then this force is completely transmitted to the primary beam AH1. Therefore, influence line has no discontinuity at point H1. If load P ¼ 1 is located at point B on the suspended beam H1BH2, then this force is completely transmitted to the support B and no part of this force is taken by the support H1 and thus no force is transmitted to the main beam AH1. Therefore, influence lines for shear and moment at section s have zero ordinates at point B. If unit load is located within the H1BH2, then the pressure from the secondary beam on the primary beam at point H1 varies proportionally to the distance of the unit load from point B. Similar discussions should be used, when unit load travels along the suspended beam H2H3. If load travels along the beam H3CD, then no part of this load is transmitted to the beam AH1. Therefore, ordinates on influence lines for Qs and Ms along the part H3CD are zeros. If load is located within the part s-B, then bending moment at section s is negative. It means that the extended fibers at section s are located above the neutral line. If any load will be distributed within BH3, then extended fibers at section s will be located below the neutral line. The presented influence lines make some useful conclusions: 1. If the uniformly distributed load q is located within the H2C portion, then reaction RD is equal to RD ¼ q 8 0.333/2 ¼ 1.33q(kN); the negative sign indicates that the reaction RD is directed downward. 2. If an arbitrary load is applied within portion AH2, then the bending moment in the section n does not arise. 3. If the force P (kN) is located at the point H2, then the bending moment in the section s is equal to Ms ¼ 0.8P(kNm); a positive sign means that the extended fibers in section s are located below the neutral line. 3.4.1.5
Summary
For construction of influence lines for multispan statically determinate hinged beam it is necessary to show interaction diagram, and to show the part of the entire beam, which contains the support or section under consideration; this part is clamped-free or simply supported beam with or without overhangs. Next we need to construct the required influence lines considering the pointed portion of a beam and then distribute influence line along all beams which are suspended with respect to pointed one. Thus construction of influence lines for multispan hinged beams is based on the influence lines for three types of simple beams and does not require any analytical procedures. Influence lines of reactions and internal forces for Gerber-Semikolenov beams, as for any statically determinate structure, are linear.
54
3.4.2
3 Multispan Hinged Beams and Frames
Kinematical Approach
Kinematical method of construction of influence lines, which is based on Müller-Breslau principle, for one span beam had been considered in detail in Section 2.5. The general procedure for construction of influence line of any factor X (reaction, bending moment, shear force) necessitates to perform the following steps: 1. Eliminate the constraint in which the required factor arises and replace it by the corresponding positive force. 2. Impart to system the unit displacement, the character of which corresponds to the required factor, in the direction of this factor, and then construct the displacement diagram of a loaded contour. This diagram presents the required influence line for factor X. Removing one constraint of a statically determinate structure leads to the fact that an unchangeable system turns into a mechanism with one degree of freedom. Therefore, the displacement diagram of the loaded contour of a multispan hinge beam when one constraint is removed and a unit displacement is applied will represent the set of displacements of the non-deformable elements of separate parts of the original structure connected by hinges. This means that the displacement diagram, i.e., influence line, will consist of individual straight line segments. To construct a diagram of the displacement of the loaded contour it is convenient to use the interaction diagram of multispan beams. Application of the kinematic method of constructing the influence lines for Gerber-Semikolenov beam is shown below. Design diagram of multispan hinged beam in shown in Fig. 3.10a. This structure is statically determinate and geometrically unchangeable. Corresponding interaction diagram is presented in Fig 3.10b.
3.4.2.1
Influence Line for RA
In the clamped support A, we remove the vertical rod in which the vertical reaction RA occurs. As a result, we obtain a sliding support. The action of the removed constraint is replaced by the RA reaction (Fig. 3.10c). The sliding support allows vertical displacement of point A, at which the link AH1 performs translational displacement, the link H1BH2 rotates around fixed point B, and link H2C rotates around fixed point C. The corresponding displacement diagram is shown by a dotted line. We accept the displacement in the direction of the eliminated constraint that is equal to unity; therefore this displacement diagram represents the required influence line RA. The accepted positive direction of the reaction leads to the fact that the displacement diagram coincides with the influence line with an accuracy to the sign. Thus, if the load is in the BH2C zone, then the reaction of support RA is directed downwards.
3.4.2.2
Influence Line for RB
It is necessary to eliminate roller constraint B, in which the required vertical reaction arises. Then we need the action of eliminated constraint to replace by a positive reaction RB (Fig. 3.10d) and impart to mechanism the unit linear displacement in the direction RB. Therewith, the link AH1 is fixed, the link H1BH2 rotates around the fixed point H1, and the link H2C rotates around the fixed point C. Since the displacement in the direction of the desired reaction RB is taken equal to unity, the corresponding displacement diagram represents the influence line RB (shown by dotted lines).
3.4.2.3
Influence Line for Mi
It is necessary to eliminate constraint, in which the bending moment Mi occurs. For this, at point i we need to introduce a hinge, replace the action of the deleted constraint by two positive bending moments Mi, and impart to mechanism the mutual unit angular displacement φ ¼ 1 in the direction of positive Mi, as shown in Fig. 3.10e. Obviously, part Ai, as a part of mechanism, does not have a displacement because of the clamped support A. The part iH1 rotates around point i, the part H1BH2 rotates around support B, and part H2C rotates around support C. Since the displacement in the direction of the required function Mi is taken equal to unity, the corresponding displacement diagram represents the influence line for bending moment Mi (shown by dotted lines).
3.4 Influence Lines for Reactions and Internal Forces
a
55
P=1 A
B i
b
H2
H1 B
H1
A
C
C
H2
c 1 B
H2
C
H1
Inf.line RA
RA 1
d
A Inf.line RB
H2
H1 RB Mi
e A
H1 B
i
H2
Inf.line Mi C
φ=1 Fig. 3.10 Multispan hinged beam. (a) Design diagram; (b) interaction diagram; (c–e) kinematical construction of influence lines for RA, RB, and Mi
3.4.3
Indirect Load Application
The construction of the influence lines for stresses in the multispan hinged beams is considered. We assume that the load is transferred to the beam through several nodal points. In this case the construction of influence is recommended to perform in the following order: 1. Consider only the girder itself (without taking into account the floor beams and stringers), for which the interaction diagram of its individual elements should be constructed. 2. Construct the influence line of the required factor assuming that the load moves along the girder. 3. Draw vertical lines passing over all floor beams to the intersection with the influence line. 4. Draw vertical lines from extreme left and right points of the loaded contour to the intersection with the base line. 5. Connect the nearest intersection points. For a clear visualization of the transfer of load from the suspension beams to the main beam, it is recommended to construct an interaction diagram of the multispan hinged beam. Example 3.2 Load is applied to stringers and transmitted to the Gerber-Semikolenov beam ABHCD by floor beams at points m, n, s, and t (Fig. 3.11a). Construct the influence lines for reaction at A and for bending moment at section k.
56
3 Multispan Hinged Beams and Frames
P=1
a
Stringer
Floor beam
Girder
E m k A
n
t
s
H
b0 B
a0
F D
C
b a
c
m
1 e
d
t
+
f
Inf. line RA
f
Inf. line Mk
– n
d
m
k
s
h
a0b0 /l
d t +
e a
– s
n h
Fig. 3.11 Gerber-Semikolenov beam with indirect load application. (a) Design diagram; (b) interaction scheme; (c, d) influence lines for reaction RA and bending moment Mk
Solution First of all, show the interaction scheme for the entire multispan hinged beam (Fig. 3.11b). 3.4.3.1
Influence Line for RA (Fig. 3.11c)
1. Influence line for RA without floor beams and stringers is presented by polygon ahd by dotted line. 2. Draw vertical lines from floor beams (points m, n, s, and t) to the intersection with the influence line. 3. Draw vertical lines from extreme left and right points E and F of the stringers to the intersection with the base line. Corresponding points of intersections have the notations e and f. 4. Connect nearest intersection points by connecting lines. 5. Influence line as a result of indirect load application is bounded by broken line emnstf. 3.4.3.2
Influence Line for Mk (Fig. 3.11d)
If the load is applied to the beam AHD directly, then the influence line for Mk is bounded by akhd points. In case of indirect load application the influence line is bounded by emnstf points.
3.4.3.3
Discussion
1. Because of the indirect load application, the reactions and internal stresses arise in the beam AD even if the load is applied outside the beam AD (close to support E or F). 2. Influence lines are convenient to use for analysis of structures, with modified design diagram. For example, for the given structure the bending moment at section k is positive, if any load is located within the stringer beam Em. It means that the extended fibers in the section k are located under the neutral axis of the beam. However, if a floor beam m will be removed
3.5 Fixed Load Method for Frames
57
and two stringers Em and mn will be replaced by one stringer En, then the points e and n on influence line for Mk should be connected. In this case bending moment at section k becomes negative; it means, the extended fibers in the section k will be located above the neutral line.
3.5
Fixed Load Method for Frames
This paragraph contains analysis of statically determinate plane frame subjected to fixed load. We restrict ourselves to the case when frames contain only rectilinear elements. It is possible to connect the individual parts of the frame using hinges, both simple and multiple. It is assumed that the load is applied in the plane of the frame. A feature of such systems is that three types of internal forces arise in the frame elements: bending moments M, shear forces Q, and normal forces N. The simplest analysis means determining the reactions and internal forces (M, Q, N ) in all elements of frame. For their determination we will use the general rules for computation of internal forces, which are presented in Sect. 2.1, as well as useful relationships between internal forces. Detailed analysis of complex two-span statically determinate frame is presented below. Example 3.5 Find the distribution of internal forces in the two-span frame shown in Fig. 3.12a, if P1 ¼ 12 kN, P2 ¼ 16 kN, q ¼ 3 kN/m, M ¼ 14 kNm, h1 ¼ 8 m and h2 ¼ 6 m.
a
5m
b
P2
P1
16kN
12kN
2
q
a
3
6 5
M
h1
b 4
A
B
c
b
a 3kN/m
A
C
l1=10m
14kNm
1
h2
5m
HA=1.5kN
l2=8m
B
RA=7.8kN
2
3
2
4
2 7.5 1
27
41
6.5
54
5
2
RC=6.75kN
M2-3
+ – M2-1
M (kNm) A
RB=1.45kN
Sign rule
6
C
HB=4.5kN
B
+
C
M2-1= M2-3=2kNm
–
+7.8
d
4
2
3 -8.2 -1.5
A
e
+13.5 6
5
–
-4.5
-6.75
Q (kN)
B
+18
-13.5 4
2
+
-7.8
A
6
-18
N (kN)
C
-8.2
5
B
-1.45
C
-6.75
Fig. 3.12 (a, b) Design diagram of the frame and reactions of supports; (c) bending moment diagram, sign rule, and equilibrium of joint 2; (d–e) shear and normal force diagrams
58
3 Multispan Hinged Beams and Frames
Solution This frame is not the simplest one for which statical determinacy and geometrical unchangeability are obvious. Therefore, the kinematical analysis must be performed first. Kinematical analysis: The structure contains three rigid disks (A-2-4, 4-B, and 5-6-C), two simple hinges (4 and 5), and five constraints of supports. Degree of freedom of the system W ¼ 3D 2H S0 ¼ 3 3 2 2 5 ¼ 0,
ðaÞ
so the structure has minimum number of constraints to be geometrically unchangeable. Now correctness of their arrangement should be checked. Two disks A-2-4 and 4-B are connected accordingly Fig. 1.4c (two unmovable points A and B may be considered as rigid disk A-B). Thus, part A-2-4-B of a whole structure presents a rigid disk. This disk and the next disk 5-6-C are connected by hinge 5 and constraint (support C), the reaction of which does not pass through hinge 5. According to Fig. 1.3b, the system in a whole is rigid disk. So, the entire system is a statically determinate and geometrically unchangeable structure. From another (statical) point of view, five reactions of supports arise in the structure. For their calculation three equilibrium equations and two additional conditions (zero bending moment at the hinges 4 and 5) can be used. Therefore, the structure is statically determinate. Determination of Reactions. 1. Reaction RC (section a-a): A bending moment at the hinge 5 is equal to zero; therefore it is convenient to find RC from the following equation: RC !
X
M right ¼ 0 ! RC l2 q h2 5
h2 ¼ 0 ! RC 8 3 6 3 ¼ 0 ! RC ¼ 6:75 kN, 2
ðbÞ
P where M right is a sum of moments of all forces, which are located to the right of the section passing through hinge 5, about 5 point 5 (these are forces RC and q). Reaction HB (section b-b): The bending moment at the hinge 4 is equal to zero; therefore it is convenient to find HB from the following equation: X
h2 þ2 2 H B 8 þ 6:75 8 3 6 ð3 þ 2Þ ¼ 0 ! H B ¼ 4:5 kN, HB !
M right ¼ 0 ! H B h1 þ RC l2 qh2 4
¼0
ðcÞ
P where M right is a sum of moments of all forces, which are located to the right of the section passing through hinge 4, about 4 point 4 (these are forces HB, RB, RC, and q). Reaction HA: Equilibrium equation for all structure as a whole: X X ¼ 0 ! P1 q h2 þ H B þ H A ¼ 0 ! 12 3 6 þ 4:5 þ H A ¼ 0 ! H A ¼ 1:5 kN ðdÞ HA ! Reaction RA (section b-b): The bending moment at the hinge 4 is equal to zero; therefore RA !
X
M left 4 ¼ 0 ! H A h1 R A l 1 M þ P 2
1:5 8 RA 10 14 þ 16 5 ¼ 0 ! RA ¼ 7:8 kN,
l1 ¼0 2
P left where M 4 is a sum of moments of all forces, acting on the left part of the frame (RA, HA, M, P1, and P2). Reaction RB: Equilibrium equation for all structure as a whole: X Y ¼ 0 ! RA þ RB þ RC P2 ¼ 0 ! 7:8 þ RB þ 6:75 16 ¼ 0 ! RB ¼ 1:45 kN RB !
ðeÞ
ðfÞ
Determination of Bending Moments By definition, bending moments are calculated using one-side forces. Bending moments at hinges are zero: MA ¼ MB ¼ MC ¼ M4 ¼ M5 ¼ 0. The sign rule for bending moments at points of a vertical bar is chosen arbitrarily. Bending moments at specific points using the left forces are
3.5 Fixed Load Method for Frames
59
M left 1A ðsection below point 1Þ : M left 12 ðsection above point 1Þ :
M left 1A ¼ H A 5 ¼ 1:5 5 ¼ 7:5 kNm; M left 12 ¼ H A 5 M ¼ 1:5 5 14 ¼ 6:5 kNm:
Here, M1–A and M1–2 mean the bending moment at point 1 belonging to part 1–A and at point 1 belonging to part 1–2, in other words infinitely close to point 1 from below and above. Thus, both subscripts denote the element, and the first subscript denotes the point of this element where the moment is calculated: M left 21 ðsection below point 2Þ : M left 23 ðsection
right point 2Þ :
M left 21 ¼ H A h1 M ¼ 1:5 8 14 ¼ 2 kNm M left 23 ¼ H A h1 þ M ¼ 1:5 8 þ 14 ¼ 2 kNm:
When calculating the bending moment M left 23 at point 2 belonging to element 2–3, we take into account the left forces and the sign rule related to the horizontal element: M left 32 ðsection left point 3Þ : M right 34 ðsection right point 3Þ :
l1 H A h1 þ M ¼ 7:8 5 1:5 8 þ 14 ¼ 41 kNm 2 M right 34 ¼ RB ð0:5l1 Þ þ H B h1 þ RC ð0:5l1 þ l2 Þ qh2 ðh1 0:5h2 Þ ¼ 1:45 5 þ 4:5 8 þ 6:75 ð5 þ 8Þ 3 6 ð8 3Þ ¼ 41 kNm
M left 32 ¼ RA
M 5B ðsection below point 5 is not shown, forces H B , RB Þ :
M 5B ¼ H B h2 þ RB 0 ¼ 4:5 6 ¼ 27 kNm
or h2 2 ¼ 4:5 6 þ 6:75 8 3 6 3 ¼ 27 kNm
M 54 ðsection above point 5, part B56CÞ :
M 54 ¼ H B h2 þ RC l2 q h2
M 65 ¼ M 6C ðsection left point 6 and below point 6, respectively, Þ :
M 6 ¼ q h2
h2 ¼ 3 6 3 ¼ 54 kNm 2
Bending moment diagram and equilibrium of joint 2 are presented in Fig. 3.12c. All ordinates are plotted on the side of the extended fibers. Let us find out how the bending moment diagram changes if the sign rule for the vertical member is changed (that is, to consider moments to the left of the axis of the strut as negative) and at the same time keep the general rule according to which the diagram of moments is depicted on the extended fibers. With this approach, M left 1A ¼ H A 5 ¼ 1:5 5 ¼ 7:5 kNm, but the extended fibers of the strut caused by the reaction HA still remain on the left, and therefore it should be shown to the left of the axis. The independence of the bending moment diagram from the rule of signs is explained by the fact that the position of the extended fibers is an objective feature of any deformable system, while the sign rule is introduced only for ordering the calculations. Determination of Shear Forces Obviously, the shear forces in the elements of the frame can easily be calculated by definition, i.e., using one-side forces. For example, QA1 ¼ Q12 ¼ HA ¼ 1.5 kN, Q23 ¼ Q34 ¼ RA ¼ 7.8 kN, etc. (Note that here subscripts A–1 and 1–2 means portions of the vertical member A–2.) However, we will take a different approach. For this the free body diagrams for each member of the frame should be considered. The reason of this approach is explained in the summary of this paragraph. Step 1: Let us cut out the portions with constant law of change of bending moments. The cut sections are applied infinitesimally to specified points, such as 1, 2, and 3. Two types of portions are obtained. They are portions with constant slope of bending moment diagram (parts A-1, 1-2, etc.), and portion with curvilinear bending moment diagram (part C-6). Step 2: At the cut sections apply bending moments, which were shown in bending moment diagram, and two unknown shears, which act as a couple. Direction of the bending moments is shown with accordance of position of extended fibers. Direction of unknown shears must be such that a couple, which represents these shears and bending moments, would hold the segment in equilibrium. The free body diagrams and computation of shear forces for each portion of the frame are presented below. A sign of shear for each portion is determined according to general sign rules (Fig. 2.1).
60
3 Multispan Hinged Beams and Frames
For example, for portion A-1 extended fibers are located left of the element; therefore the moment is directed clockwise. This moment is equilibrated by two shear forces; each of them rotates element around opposite end in counterclockwise direction, so shear within this portion is negative. Detailed calculation of shear for some portions of the frame is shown in Fig. 3.13.
Portion A-1
Portion 6-C
Portion 1-2
Q1-A
Q2-1
7.5
Q6-C 54
2.0 5
6
3 6.5
QA-1
q
Q1-2
Q A−1 = Q1− A = −
7.5 = −1.5kN 5
Q1− 2 = −
QC-6 6.5−2 = −1.5kN 3
Q6− C =
54 + 3 × 6 × 3 =18kN 6
Fig. 3.13 Calculation of shear forces
Final shear force diagram is presented in Fig. 3.12d. The reader is invited to determine the shear forces in each element by definition. Determination of Normal Forces Obviously, the normal forces in the elements of the frame can easily be calculated by definition. For example, NA1 ¼ N12 ¼ RA ¼ 7.8 kN, N23 ¼ N34 ¼ P1 HA ¼ 13.5 kN, etc. (Note that the negative sign means that portions A–1 and 1–2 are compressed.) However, as in case of shear forces, we will take a different approach. For calculation of normal forces, which exert in all members of the frame, we have to show free body diagram for all joints and consider equilibrium of each joint. Free body diagrams for some joints are presented in Fig. 3.14. Note that bending moments are not included into the equilibrium equations. Joint 2
Joint 4
N2-3
2
8.2
8.2
Q2-3= 7.8 12
Joint 5 13.5 4
N5-6
N3-4
5 13.5
Q2-A=1.5
N4-5
N2-A SX=0®N2-3= -13.5 SY=0® N2-A= -7.8
N3-4= -13.5 N4-5= -8.2
6.75
4.5 N5-B N5-6= -18 N5-B= -1.45
Fig. 3.14 Calculation of axial forces
Normal force diagram is presented in Fig. 3.12e. The reader is invited to determine the axial forces in each element by definition. Note: Knowing the distribution of internal forces we can verify reactions of supports. For example, at support A we have Q ¼ 1.5 kN, N ¼ 7.8 kN, so reactions are HA ¼ 1.5 kN, RA ¼ 7.8 kN and they are directed as shown in Fig. 3.12b.
Problems
61
Summary 1. Analysis of statically determinate frame subjected to any fixed loads includes the following stags: kinematical analysis, calculation of reactions of supports, construction of internal force diagrams, and their verifications. 2. Internal forces (M, Q, and N ) are calculated by definition (i.e., taking into account the one-sided forces) and in arbitrary order. 3. Correspondence of bending moment and shear force may be checked using the Schwedler theorem Q ¼ dM/dx; that is, at any section of element the slope between the diagram M and axis of the element is equal to shear force at the same section. If to align tangent line of bending moment diagram with the axis of the element requires counterclockwise rotation then the shear force is positive. Correspondence of shear force and intensity of external distributed load q may be checked using the following relationship: q ¼ dQ/dx. In the above example 3.5, the differential relation had been used for determining the shear forces on the base of bending moment diagram. Computation of axial forces had been performed on the base of shear force diagram considering equilibrium of joints. This approach was made to prepare reader for the analysis of statically indeterminate frames. In case of statically indeterminate frames the application of special methods leads, first of all, to determining of bending moments at specified sections, while the reactions of supports are unknown yet. So in this case the calculation of shear and axial forces by definition is not applicable. Therefore, reaction may be calculated only after knowing distribution of all internal forces. This procedure is typical for analysis of redundant frames (see Chaps. 9, 10, 11, 13). Knowing distribution of internal forces, the reactions of the supports may be verified (for statically determinate frames). This idea will be used for the analysis of indeterminate frames, where the order of determining internal forces and reactions is as follows: Bending moment ! Shear force ! Axial force ! Reactions ! Verification
Problems 3.1. Provide kinematical analysis for beams in Fig. P3.1. Construct interaction scheme, point the main and suspended beams, and explain a load path.
Fig. P3.1
a
c
b
d
e
f
g
h
62
3 Multispan Hinged Beams and Frames
3.2 Show that the systems in Fig. P3.2(a, b) have a sufficient number of constraints to ensure its geometric unchangeability; however these systems cannot be used in design practice. Explain why.
a
b
Fig. P3.2
3.3. Vertical settlement of support A of the beam in Fig. P3.3 (a, b) causes appearance of internal forces in the portion. 1. ABC; 2. CDE; 3. Nowhere; 4. Everywhere.
a
B
A
D C
b
F
B
A
E
D
E
C
Fig. P3.3
3.4. The beam in Fig. P3.4(a, b) is subjected to change of temperature. As a result, internal forces appear in the portion. 1. ABC; 2. CDE; 3. Nowhere; 4. Everywhere.
a
B
A
D C
b
F
B
A
E
D
E
C
Fig. P3.4
3.5. Find direction of all reactions for beam in Fig. P3.5, if uniformly distributed load is located within portion CD.
A
B
D
Ans. RA < 0( ); RC >0(); RE = 0.
C
E
Fig. P3.5
3.6. Where concentrated load P should be located (Fig. P3.6) (a) For maximum positive (negative) reaction of support A; (b) In order to reaction RD ¼ 0. B
A
D C
Fig. P3.6
3.7. Analyze the design diagram of two-span Gerber beam is shown in Fig. P3.7. E
C A
B
D
Fig. P3.7
(a) Find the location of distributed load in order to check if reaction at support C will be equal to zero. (b) Find the portion of the beam which is not deformable if concentrated load P is located at point B. (c) Where are located extended fibers within the portion AB if concentrated load P is located at point D.
Problems
63
3.8. Design diagram of the multispan hinged beam is shown in Fig. P3.8. Provide complex analysis of this structure.
P=8kN
q1=2kN/m
q2=3kN/m
B
A H1
K3 1m
H2 5m
2m
M=3kNm D
C
3m
2m
H3
K2 3m
K1 3m
2m
3m
1m
Fig. P3.8
1. 2. 3. 4. 5. 6. 7. 8. 9.
Perform a kinematical analysis of the beam and show interaction scheme of its separate parts. Determine reactions of supports. Construct bending moment diagram. Construct shear force diagram. Construct influence lines for reactions of supports. Construct influence lines for bending moment and shear force for indicated sections. Determine reactions of supports, bending moment, and shear force for indicated sections using influence lines. Compare results obtained by influence line method and fixed load method. Calculate maximum bending moment and shear forces at section K1 caused by two connected forces P1 ¼ 20 kN and P2 ¼ 10 kN apart 1 m. Ans: RA ¼ 3:4 kNð"Þ, M A ¼ 10:2 kNmð\Þ,
RB ¼ 10:6 kNð"Þ,
RC ¼ 13:833 kNð"Þ,
M B ¼ M C ¼ 8:0 kNmð\Þ,
RD ¼ 8:167 kNð"Þ,
M K1 ¼ 6:8 kNmð\Þ,
M K2 ¼ 12:0 kNmð[Þ,
3.9. Design diagrams of frames, loaded by fixed load q ¼ 3 kN/m for case (a) and concentrated load P ¼ 100 kN for case (b), are presented in Fig. 3.9 (a, b), respectively. Draw the internal force diagrams (bending moment, shear force, and normal force), if l ¼ 4 m and h ¼ 5 m. q
a
b
B
k1
B
P C
C x1
h
h
k2 x2
A
A l l
Fig. P3.9
3.10. The redundant three-span beam is loaded by force P at point K. Each span of the beam equals l, and distance BK is l/3. Bending moment diagram is shown in Fig. P3.10. Calculate shear forces applying differential relationship M–Q, and after that determine the reactions of supports. Provide statical verification.
0.0789Pl
0.0543Pl
C
B l
0.1516Pl
Ans. RA = 0.0789P (¯), RB = 0.770P (), D
K
A
Fig. P3.10
P
RC = 0.363P (), RD = 0.0543P (¯)
Chapter 4
Plane Trusses
This chapter is devoted to the analysis of statically determinate plane trusses subjected to fixed and moving loads. Methods for the generation of trusses, static analysis, and construction of influence lines are discussed. The different types of trusses are considered. Among them are simple trusses and trusses with subdivided panels (compound trusses, special types of trusses such as three-hinged trusses, trusses with a hinged chain, and complex trusses).
4.1
General
Definition Truss is a rigid structure composed of straight members, which are connected at their ends by hinges. Points of intersection of members of a truss are called joints. The span of a truss is the distance between axes of its supports. The lower and upper straight members form the lower and upper chords of the truss. Both chords are connected by the web members. The web members may be subdivided into vertical and diagonal ones. The distance between two adjacent joints measured along the chord of a truss is called a panel. The constructive scheme of the simplest truss and the terminology referring to it are presented in Fig. 4.1.
Joint
Panel
Upper chord
Vertical (strut)
Diagonal (tie) Bottom chord Span
Fig. 4.1 Truss terminology
Trusses have the same function as beams; however they are used for covering much larger spans. This is possible because the material of the elements of a truss is used more effectively than the material of a solid beam.
4.1.1
Classification of the Plane Trusses
There are a huge number of types of trusses. According to the way of formation, the trusses may be classified as simple, compound, and complex ones. The simple trusses may be classified, in particular, according to the following criterions: (1) Shape of the upper and lower chords, (2) type of the web, (3) location of supports, (4) level of application of fixed and/or moving load, and (5) purpose of the structure The first criterion permits the subdivision of the trusses into structures with parallel chords (Fig. 4.2d–n), triangular (a) and polygonal (b). Trusses can have curvilinear chord, which means that joints of a chord are located on the curve, but all joints are connected by straight members (Fig. 4.2c). In accordance with the second criterion, trusses may be subdivided in the following way: trusses with triangular pattern of the web (Warren truss), when descending and ascending diagonals alternate (Fig. 4.2d); Pratt and Howe truss, when vertical © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_4
65
66
4
Plane Trusses
and diagonal elements alternate (Fig. 4.2e); trusses with K-pattern of the web, when within one panel, instead of one diagonal, two diagonals of a shorter length are placed so the web members, verticals and diagonals, form a letter K (Fig. 4.2f); and trusses with diamond-shape web (double Warren truss, Fig. 4.2g) and double Whipple truss (Fig. 4.2h).
a
b
c
d
e
g 6
3 5
i
h
1
4
f
2
j
m
k
n
Fig. 4.2 Classification of the plane trusses according to the following criteria: (a–c) shape of the chords; (d–h) type of the web; (i–k) location of supports; (m–n) level of the traveling load
The third criterion is related to trusses which are supported at the both ends (truss of girder type) (Fig. 4.2i), the trusses fixed at one end (cantilevered truss) (Fig.4.2j), and trusses with overhang (cantilever) at one or both supports (Fig. 4.2k). The fourth criterion is related to the bridge trusses. There are possible cases when the railway is located at the bottom level of the bridge (Fig. 4.2n), and the upper level is intended for the motor road (Fig. 4.2m), or both levels. With regard to their purpose, the trusses may be subdivided into bridge trusses; covering trusses of the public and industrial buildings; and trusses used in miscellaneous constructions such as tower cranes, masts, as well as support of the high-voltage power transmission line. The most important classification characteristic is related to the method of formation of trusses. According to this characteristic, the trusses are subdivided into simple trusses (they had been classified above); trusses with subdivided panels; and compound, complex, as well as special types of trusses (three-hinged truss, truss with a hinged chain, etc.). This important characteristic, the way of formation of trusses, may lead to additional type of reactions (thrust), which will be discussed in the relevant paragraphs of this chapter.
4.1 General
4.1.2
67
Assumptions and Design Diagram
A typical constructive scheme of a truss, a method of transferring the load, connections of elements and their centering, as well as supports are shown in Fig. 4.3a,b. Each member of such truss may be made up of two angle cross sections placed back to back and connected (by welding or bolts) to the gusset plates (Fig. 4.3b). The analysis of such a structure taking into account its constructive features is very difficult. In order to adopt an idealized model of the structure shown in Fig. 4.3a, b, i.e., its design diagram, the set of requirements should be formulated. We put forward the following requirements for the truss design diagram: it should accurately reflect the work of a real structure and allow easy obtaining of analytical expressions for internal forces in the elements of the truss, and therewith, the distribution of forces in the elements of the truss should ensure the most effective use of the material.
Purlin
a
3
1
2
b
Joint 2
Joint 1
Joint 3 1-1 Gusset plate
Gusset plate
1 1 Immovable support Abutment
Lower chord
Movable support
2 angles back to back
c
Fig. 4.3 (a) The constructive scheme of a truss with joint loading; (b) typical joints; (c) design diagram of the truss
In order to satisfy the stated requirements, it is necessary, in addition to general fundamental assumptions of the theory of structures, to introduce assumptions inherent just to the theory of plane trusses. They are as follows: 1. All elements of a truss are straight, long, and thin rods. It means the length of the element is much longer than dimensions of its cross section. The longitudinal axes of the rods converging at the joint intersect at its center (Fig. 4.3b). 2. The truss contains ideal supports, such as the rolled and pinned ones. 3. All external loads are applied at joints only and act in the plane of the truss. At the same time, the external load is much larger than the truss’ own weight. 4. The rigidity of the chords and the elements of a web are of the same order. 5. All elements of a truss are connected at their ends by means of ideal hinges (Fig. 4.3c). Assumptions 1–4 are easily implemented in practice, and the set of all assumptions 1–5 leads to the fact that only axial forces (tensile or compressive) arise in the truss elements. Now we need to show that the hinged design diagram in Fig. 4.3c satisfies the stated requirements.
68
4
Plane Trusses
It is known that in the cross section of the bar subjected to an arbitrary load, the bending moments, shear, and axial forces arise. Let us consider the conditions under which only axial (longitudinal) forces arise in the cross sections of each element of the truss. In other words, we are interested in the simplest distribution of internal forces. Just in this case the efficiency of using the material is fully realized. For this, it is obvious that the rod must be rectilinear, its own weight can be neglected, and the load must be applied at the joints. The idealization of rectilinear elements of a truss does not require justification. Real truss support devices (Fig. 4.3b) easily allow their idealization in the form of hinged movable and hinged immovable supports. Load on the joints is easily realized in practice using purlins installed in the joints of the truss (Fig. 4.3a). It is known that the equilibrium of the rod with hinges at the ends with the forces applied at these hinges is possible only if these forces are directed along the bar; this means the appearance in the bar of only axial forces. Thus, in order to ensure the appearance of only longitudinal forces in the elements of the truss, it is necessary to accept the hinged connection of the elements of a truss and take into account only the joint loads. In the case of a significant own weight of the truss, the weight of each rod should be transformed to the joint loads. Now we call attention to the real heavy metal truss of the bridge type. Of course, it is difficult to interpret joints in Fig. 4.3b as hinged. Therefore, the idealized presentation of the joints by hinged connections of elements requires a serious justification. Under adopted assumptions 1–4 the following facts have been discovered experimentally: 1. In a zone remote from joints, the stresses over the cross section are distributed uniformly and along the each member remain practically unchanged. 2. Near the joints in the farthest fibers, the stresses have practically equal values and the same sign. These results allow us to make an important conclusion: the bending moments at the joints are small and exhibited as a minor boundary effect, so the bending moments may be ignored and only axial forces should be taken into account. Thus, the design diagram of a truss is adopted as the structure with straight elements which are hinge connected at the joints as shown in Fig. 4.3c. Such design diagram in case of loading at the joints and assumptions mentioned above describes the behavior of a truss with sufficient accuracy and considerably simplifies its analysis. D.I. Jourawski was, apparently, the first who had adopted the design diagram of a truss with hinged joints, and for its analysis the method of cutting joints was applied (1850, 1856–1857) (Bernshtein 1957).
Discussion 1. This example deals with limited but versatile set of questions that must be taken into account choosing a design diagram of a truss. Since the constructive requirements 1–4 do not have a strict quantitative evaluation, the hinged model of the truss can be considered as a convenient idealized representation of the real truss, and the results of the analysis only as a first approximation. The question of the adequacy of the model and the real object before the performance of the experiment is left open. 2. Rejection of any of the above assumptions leads to the need for a transition from the hinged design diagram of the truss to the new model. For example, assume that the upper chord of the truss has much more rigidity per unit length than the web elements. In this case, it is necessary to adopt a design diagram in which the hinges do not divide the upper chord on the separate elements. This means that the upper chord of the truss works like a continuous beam on elastic supports (Müller-Breslau model, 1908). Such a model leads to an extremely complicated analysis procedure. Therefore, the advantages of the hinged design diagram of the truss, and the simplicity of its analytical analysis, allow considering such model as a classical one. Strictly speaking, the Müller-Breslau model is more a model of the frame, but not a truss, because it has rigid joints related to the upper chord at the points where the web elements adjoin it. 3. The presence of only longitudinal forces in the truss elements is the simplest distribution of forces and leads to the effective use of the material of the rods, in contrast to rods subjected to bending. In the hinged scheme, the analysis procedure, as will be shown below, includes simple algebraic operations. 4. In the following chapters, the choice and justification of the design diagram will not be discussed in detail, but the reader should pay increased attention to all details of the design diagram proposed to the analysis.
4.2
The Generation of Statically Determinate Trusses
The trusses according to their generation are subdivided into simple, compound, and complex ones.
4.2 The Generation of Statically Determinate Trusses
4.2.1
69
Simple Trusses
The simplest geometrically unchangeable system consisting of individual rods is a hinged triangle. Such a structure can change its shape only when its elements are deformed; that is, a rod-jointed triangle is a simplest rigid disk. At their core, simple trusses contain the basic unchangeable hinged triangle. Each additional joint is attached by means of two additional rods, which do not lie on one straight line. It is easy to verify that the formation of each truss, shown in Fig. 4.2, corresponds to the truss of simple type. Truss in Fig. 4.2g includes an original triangle 123. Then joints 4 and 5 are attached to it. Thus, a rigid disk 352143 is obtained, to which the joint 6 is then attached and so on. The relationship between the number of bars (S) and joints (J ) for simple statically determined and geometrically unchangeable trusses is expressed as S ¼ 2J 3:
ð4:1Þ
This formula can be derived using two principally different approaches. 1. The unchangeability condition of a truss. A simple truss has an initial triangle disk and each new joint is attached to the previous rigid disk using two bars (Fig. 1.4c). The total number of bars is S ¼ 3 þ2(J 3). In this formula, the first number represents the three bars used for the initial rigid disk. The 3 within the brackets arises from the three joints of the initial triangle, so J 3 represents the number of joints attached to the initial triangle. The coefficient 2 means the number of bars associated with each additional joint. Therefore, the total number of bars is given by (4.1). P P 2. The statical determinacy condition of a truss. For each joint, two equilibrium equations must be satisfied: X ¼ 0, Y¼ 0. The number of unknown internal forces is equal to the total number of members S. The total number of unknowns, including the reactions of the supports, is S þ 3. The number of equilibrium equations equals 2J. Therefore, a truss is statically determinate if S ¼ 2J 3. Thus, the two different approaches both lead to the same formula (4.1) for the kinematical analysis of a simple truss. Therefore, any simple truss, that is, a truss formed from a rod-hinged triangle by successive attaching of joints, each with two rods not lying on one straight line, is a system that is geometrically unchangeable and statically determinate. If the number of bars S > 2J 3, then the system is statically indeterminate; this case will be considered later in Part 2. If the number of bars S < 2J 3, then the system is geometrically changeable and cannot be used as an engineering structure.
4.2.2
Compound Trusses
A compound truss consists of simple trusses connected to one another. There are different ways to connect simple trusses in order to construct a compound truss. First approach uses three hinges, not arranged in line. Such an interconnection can be considered as a simple triangle on the basis of its three rigid disks. In the examples below, two of the trusses are represented by disks D1 and D2 (Fig. 4.4). Two pinned supports may be treated as rigid disk D3 (the earth). •
D1
C
C D2
•
D1
D2
•
D3
D3 is earth
Fig. 4.4 The generation of a rigid disk by means of three hinges
Another approach uses hinge C and a rod R, which does not pass through the hinge (Fig. 4.5).
70
4
•
C
C
D1
D1
D2
•
D2
•
R
Plane Trusses
R
Fig. 4.5 The generation of a rigid disk by means of a hinge and a rod
A third method uses three rods, which are not parallel and do not all intersect at the same point (Fig. 4.6).
R1
•
D1
R2 •
•
• •
R3
R1
D2 •
D1
R2
D2
R3
Fig. 4.6 The generation of a rigid disk by means of three rods
In addition, compound trusses can be composed of members that are themselves also trusses (Fig. 4.7). Such compound trusses suit for structures carrying considerable loads. F1
F2
F2
F1 Fig. 4.7 A truss with compound members
The analysis of this truss should be performed as if all compounded members are solid rods. The obtained internal forces should be applied to the compound member and then that member considered as a truss. Since each compound member is subjected to two balanced axial forces, the reactions of the supports are each zero. Therefore, the supports of each compound member can be disregarded.
4.2.3
Complex Trusses
Trusses that are complex use other ways to connect two (or three) rigid disks. Figure 4.8a presents a connection of two rigid disks using hinge H, two bars 1 and 2, and an additional support C. This connection arrangement produces a Wichert truss. Figure 4.8b presents a connection of three rigid disks using hinge C and three hinged end bars 1, 2, and 3.
4.3 Simple Trusses: Fixed Loads
71
a
H D1
D2 B 1
b
1
2 C1
2 3
C Fig. 4.8 Complex trusses
4.3
Simple Trusses: Fixed Loads
This section is devoted to the analysis of simple plane trusses subjected to fixed load. The purpose of the analysis is to determine an internal force in the elements of the truss under the action of a fixed load. Analytical and graphical methods are considered.
4.3.1
Analytical Methods of Analysis
General idea of analytical methods is as follows: a required unknown internal force at any member of a truss should be transferred into a class of the external forces. For this, the section method is applied, one of the parts is isolated, and the action of the dissected rods is replaced by their unknown internal forces. Assume that elements in the cut section are in tension, so unknown forces are directed away from the joint along the element. There exist two different types of plane force systems: concurrent and arbitrarily located on a plane. Equilibrium equations are compiled for such force systems. Accordingly, two analytical methods are possible: method of joint isolation (joints method) and method of through sections (cuts or Ritter’s method).
4.3.1.1
Method of Joint Isolation
In this method, equilibrium of each joint is considered separately. In the case of simple truss, the method of joints permits the successive determination of internal forces acting in all the members of a truss. Procedure starts with a joint formed by two bars only (two-member joint). For a system of the plane concurrent forces two equilibrium equations can be constructed. Therefore, the joints should be isolated in such order that there are no more than two unknown forces. This method was introduced by D.I. Jourawski (1850), J.W. Schwedler (1851), and K. Culmann (1851) (Bernshtein 1957).
4.3.1.2
Method of Through Sections (Ritter’s Method)
In a simplest case a truss is divided into two parts by a single cut through three nonconcurrent members. For a system of forces arbitrarily located on the plane, three equilibrium equations can be constructed. Therefore, a section should be passed through the truss in such a way as to cut three nonconcurrent members with unknown internal forces. Equation of equilibrium should be written in such a way so only one unknown would be in the equation. For this, sum of moments about specific point (Ritter’s point) or projection of forces on one of the axes should be considered. Ritter’s point for a given bar is the point where other dissected elements are intersected. Equation of projection should be applied when two of the three sectioned members are parallel to one another and it is required to find internal force in the third member. Method of cuts also can be used in cases when a section cuts more than three nonconcurrent bars (for example, the trusses with complex web), provided that internal forces in all bars except for three are already known. This method was introduced by J.W. Schwedler (1851) and A. Ritter (1862) (Timoshenko 1953).
72
4
4.3.1.3
Plane Trusses
Special Cases
The zero-force members of a truss can generally be determined by inspection of each joint without calculations. 1. If two-member joint has no external force, which is applied at this joint, then internal forces in both members forming the joint are zero. 2. If two members of a three-member joint are collinear and there is no external force, which is applied at this joint, then the internal force in the third member is zero. The forces in these two collinear members are equal and both in tension or in compression. Example 4.1 Design diagram of the simple truss (Howe truss) is presented in Fig. 4.9. Truss is loaded by two concentrated forces F1 ¼ 10 kN and F2 ¼ 20 kN applied at joints 3 and 5, and uniformly distributed load of intensity q ¼ 4 kN/m between panels 7–9 and 9–11. All panels are equal. Geometrical parameters of a truss are d ¼ 3 m and h ¼ 4 m. Compute internal forces of the truss due to indicated fixed load. Apply analytical methods. 6
a
4
8
2
HA
A
γ
j
1
3
F1
5
B 7
F2
d
RA
9
RB
F2
F3
F5
F4
c
d
c
b
1
RA
a
U3-5
1
F1
b F1
e
4
RA
j
r D4-7
j
3
RB
O4-6
2
U3-1 U1-3
r1
V3-2
O1-2
j
12
11
q
F1
RA
h
10
3
5
F2
6
d O6-8
O4-6 •
U5-7 7 c
V6-7
Fig. 4.9 (a) Design diagram of the truss. (b–d) Special part of the truss under consideration. Loads: F1 ¼ 10 kN, F2 ¼ 20 kN, q ¼ 4 kN/m; parameters of truss are d ¼ 3 m, h ¼ 4 m, tanφ ¼ 4/9, and tanγ ¼ 8/9
Forces in the elements of the upper and bottom chords are denoted by the letters O and U, respectively, the vertical V and the diagonal D. Solution Kinematical analysis of a structure. The structure consists of one basic hinge-connected triangle 1–2–3; each additional hinged joint is attached to the previous rigid part by two separate bars not in alignment. The numbers of bars and joints are S ¼ 21 and J ¼ 12. The number of bars in terms of number of joints equals S ¼ 2J 3 ¼ 21. Thus, this truss is a statically determinate and geometrically unchangeable structure. 1. Distributed load should be presented as concentrated forces at joints 7, 9, and 11: d 3 F 3 ¼ F 5 ¼ q ¼ 4 ¼ 6 kN, 2 2 2. Reactions of supports:
F 4 ¼ qd ¼ 4 3 ¼ 12 kN
4.3 Simple Trusses: Fixed Loads
HA ! RA ! RB !
X X
X
73
X ¼ 0 ! HA ¼ 0
M B ¼ 0 ! RA 6d þ F 1 5d þ F 2 4d þ F 3 3d þ F 4 2d þ F 5 d ¼ 0 ! RA ¼ 29:667 kN M A ¼ 0 ! RB 6d F 1 d F 2 2d F 3 3d F 4 4d F 5 5d ¼ 0 ! RB ¼ 24:333 kN
Control: X
Y ¼ RA F 1 F 2 F 3 F 4 F 5 F 6 þ RB ¼ 29:667 10 20 6 12 6 þ 24:333 ¼ 54 54 ¼ 0
3. Computation of internal forces using method of joint isolation and through cuts Joint Isolation Method. By this method, by successively cutting out the joints and considering their equilibrium, it is possible to determine the forces in all elements of the truss. The order of cutting joints: the first joint to be cut and each subsequent one should contain no more than two unknown forces. Force U1–3: To determine the force U1–3 we need to isolate joint 1 (closed section a is shown by dotted line), and consider its equilibrium. Thus, the object under study is joint 1, which is subjected to concurrent forces (Fig. 4.9b). They are known reactions RA ¼ 29.667 kN and HA ¼ 0, and required internal forces U1–3 and O1–2. Equilibrium equations in the form of the sum of projections of all forces on the coordinate axes: X X ¼ 0 ! O12 cos φ þ U 13 ¼ 0 X Y ¼ 0 ! O12 sin φ þ RA ¼ 0 The solution of these equations leads to the following values of the forces in the rods of the truss: RA 29:667 ¼ ¼ 73:053 kN sin φ 0:4061 R 29:667 9 ¼ þ66:75 kN ¼þ A ¼þ 4 tan φ
O12 ¼ U 13
The negative and positive sign means the compressed and tensile internal force, respectively. Next we can isolate the joint 3 (Fig. 4.9c, closed section b). This joint is subjected to concurrent forces. They are known positive force U31 ¼ U13 ¼ þ66.75 kN(directed away from joint 3), active force F1 ¼ 10 kN, and required internal forces U35 and V32. Two forces U13 and U35 are located on one line. Equilibrium equations ∑X ¼ 0 and ∑Y ¼ 0 lead to the following results: U35 ¼ U31 ¼ þ66.75 kN, V32 ¼ þ10 kN. In a similar way, we can sequentially isolate the joints 2, 5, 4, . . .; consider their equilibrium; and determine the forces in the corresponding elements of the truss. The disadvantages of this method are obvious: 1. To determine the forces in all elements of the truss, it is necessary to consider the equilibrium of all joints of the truss. 2. Often, to determine the force in any element, it is required to consider the equilibrium of a certain number of joints that are not directly connected with the rod with the desired force. For example, if it is required to determine the force О6–8, then it is required to consider the equilibrium of other joints (12, 11, 10, 9, 8) first. Through sections method. This method, in the simplest cases, allows to determine the force in a specified element of the truss without preliminary calculations, as in the method of cutting out the joints. Force U5–7: To determine the force U5–7, we need to make a through section c–c, discard one of the parts of the truss, and consider the equilibrium of the remaining part. Object under study is the left part of the section (Fig. 4.9d). This part of truss is subjected to arbitrarily located plane (coplanar) forces. They are known reactions RA and HA, active forces F1 and F2, and unknown forces U5–7, O4–6, and D4–7. For such forces, we can set three equilibrium equations. The solution of this system of equations makes it possible to determine the forces in all three rods. Suppose, it is required to determine the force in one specific rod. In this case it is convenient to compose one equilibrium equation in the form of a sum of moments with respect to the moment point (the Ritter’s point). This point is defined as the point of intersection of the two remaining rods that have been cut by the section c–c. Ritter’s point for bars 5–7 is point 4; therefore U 57 ! U 57 ¼
X
M left 4 ¼ 0 ! F 1 d RA 2d þ U 57 h45 ¼ 0,
2 h45 ¼ h ¼ 2:667 m 3
d 3 ð R 2 F 1 1Þ ¼ ð29:667 2 10 1Þ ¼ þ55:5 kN h45 A 2:667
74
4
Plane Trusses
Force O4–6 (section c–c): Object under study is the left-hand part of the section (Fig. 4.9d). Ritter’s point for bars 4–6 is point 7; therefore X left O46 ! M 7 ¼ 0 ! RA 3d þ F 1 2d þ F 2 d O46 r ¼ 0 tan φ 4 1 36 The arm r ¼ 3d sin φ ¼ 3d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 3 3 qffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffi ¼ 3:655 m 2 9 97 1 þ tan φ 1 þ 16 81 Required internal force: d 3 ¼ 40:22 kN O46 ¼ ðF 1 2 þ F 2 1 RA 3Þ ¼ ð10 2 þ 20 1 29:667 3Þ r 3:655 Internal force is compressed. Force D4–7 (section c–c): Object under study is the left-hand part of the truss (Fig. 4.9d). Ritter’s point for bars 4–7 is point 1; therefore X left D47 ! M 1 ¼ 0 ! F 1 d F 2 2d D47 r 1 ¼ 0 tan γ 8 72 r 1 ¼ 3d sin γ ¼ 3d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 3 3 qffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffi ¼ 5:972 m 2 145 1 þ tan γ 9 1 þ 64 81
D47
1 3 ð10 1 þ 20 2Þ ¼ 25:117 kN ¼ ðF 1 d þ F 2 2d Þ ¼ r1 5:972
Forces O6–8 and V6–7: To determine these forces, you can use a combination of the method of through sections and method of joints. Since force O4–6 has been calculated earlier, now we can isolate joint 6. Object under study is joint 6 (Fig. 4.9e, closed section d ). The equilibrium equations of all forces in the projections onto the coordinate axes are X O68 ! X ¼ 0 ! O46 cos φ þ O68 cos φ ¼ 0 ! V 67
O ¼ O46 ¼ 40:22 kN X68 ! Y ¼ 0 ! V 67 2 O68 sin φ ¼ 0 ! V 67 ¼ 2 O68 sin φ ¼ 2ð40:22Þ 0:4061 ¼ 32:66 kN
Let us consider some particular cases of applying the method of joint isolation and the method of through sections. A truss with parallel chords, with descending and ascending diagonals, is shown in Fig. 4.10. The panel and height of the truss are d and h, respectively.
1 1
b
a
2
6 d
5
e
9
10
12
j
A 3
4 b
c
a
8
7 P1
e
B
11
P2
RA
RB
Fig. 4.10 Design diagram of a truss with parallel chords. Determining of internal forces in some elements
Force O25 (section a–a): Object under study is the left part of a truss. Ritter’s point is point 4: X left O25 ! M 4 ¼ 0 ! RA 2d O25 h ¼ 0 If object under study is the right part of a truss, then O25 !
P
M right ¼ 0 ! RB 4d þ O25 h ¼ 0. 4
4.3 Simple Trusses: Fixed Loads
75
Force D24 (section a–a): Object under study is the left part of a truss. Ritter’s point is in infinity. In this case we need to use another equilibrium equation: X D24 ! Y left ¼ 0 ! RA D24 cos φ ¼ 0 If object under study is the right part of a truss, then D24 ! ∑ Yright ¼ 0 ! RB P1 P2 þ D24 cos φ ¼ 0. Force V23 (section b–b): Object under study is the left part of the truss: X V 23 ! Y left ¼ 0 ! RA þ V 23 ¼ 0 ! V 23 ¼ RA If object under study is the right part of the truss, thenV23 ! ∑ Yright ¼ 0 ! RB P1 P2 V23 ¼ 0. It is obvious that consideration of two parts of the truss, left and right, should have the same result. Force VA-1 (section c–c): Object under study is the joint A: X V A1 ! Y ¼ 0 ! RA þ V A1 ¼ 0 ! V A1 ¼ RA It is obvious that VA1 ¼ V23 ¼ V45 ¼ RA. Force V6–7 (section d–d ): Object under study is the joint 6: X V 67 ! Y ¼ 0 ! V 67 ¼ 0 Let us now consider K-truss with parallel chords (Fig. 4.11a). Suppose we need to determine the force in the central pillar 1–2. This structure belongs to the class of simple trusses. Indeed, we have the basic rigid triangle 1–2–3 and each next joint is attached to the rigid disk using two rods in the following order: 4, 5, 6, A, 7, etc. The relation (4.1) S ¼ 2J 3 is valid, since S ¼ 25, J ¼ 14. Therefore, the system is statically determinate and geometrically unchangeable. However, separate application of the method of joints and the method of through sections cannot be applied. Here their combination is necessary. First, we calculate the support reactions RA and RB. Then we isolate the joint 3 using the closed section (Fig. 4.11a, b). For this joint the equilibrium equation of all forces in the projections on the x-axis X X ¼ D31 cos α þ D32 cos α ¼ 0 leads to the relation D31 ¼ D32.
P1
b
a 4
a
b
8
1
D31
V34 6
3
10
7
A
B 5
2 a
α α
x D32
V35
9
b
P2
RA
RB
c
d
4
5
P1
D31
3
6
V35
A
e
O41
O41 V43
6
RA
3
O14
D32 U52
A
O18
1
h′ 5
U52
D13 2
V12
D17
RA
Fig. 4.11 K-truss. (а) Design diagram of the truss, (b–e) design diagram of special parts of a truss: (b) joint 3; (c, d) sections a–a and b–b; (e) joint 1
76
4
Plane Trusses
Then show a through section a–a (Fig. 4.11a,c). Although this section dissects 4 bars, it is possible to determine the force in the element of the upper (or lower) chord, since three of the intersected bars intersect at one point. For example, forces V43, V35, and U52 intersect at point 5. Therefore O41 !
X
M left 5 ¼ 0 ! RA d þ O41 h ¼ 0 ! O41 ¼ RA
d h
After that we draw a through section b–b (Fig. 4.11a,d). Although this section dissects 4 bars, however, the force in one of them, namely O41, is already known. Therefore X left D31 ! M 2 ¼ 0 ! RA 2d þ O41 h þ D31 h0 ¼ 0, where d, h, and h0 are the panel length, height, and arm of the force D31 with respect to point 2. This equation allows us to find the numerical value of D31 and hence D32. Then we need to perform similar procedures and to determine the force of D17. Finally, we can consider equilibrium of the joint 1 (Fig. 4.11a,e). The equilibrium equation in the projections on the y-axis allows us to find the required force in the vertical member 1–2: X V 12 ! Y ¼ D13 sin α D17 sin α P1 V 12 ¼ 0
4.3.2
Maxwell–Cremona Diagram
Maxwell–Cremona diagram presents the force polygons for all joints of a truss, which are constructed in a single diagram of forces. This simple, elegant, and effective method allows easy determination of the internal forces in all members of a simple truss subjected to any fixed load. This method was introduced by J.C. Maxwell (1864), W.P. Taylor (1869), and L. Cremona (1872). To illustrate this method, we take a truss shown in Fig. 4.12a and previously studied in detail in Example 4.1. For this truss F1 ¼ 10 kN, F2 ¼ 20 kN, F3 ¼ 6 kN, F4 ¼ 12 kN, F5 ¼ 6 kN. To construct Maxwell–Cremona diagram a strict order of approach is very important. The steps are as follows: 1. Determine reactions of the truss RA ! ∑ M12 ¼ 0 ! RA ¼ 29.67 kN, RB ! ∑ M1 ¼ 0 ! RB ¼ 24.33 kN. 2. Choose the scale for the truss dimensions (for example, 1m 1cm) and draw the truss according to scale. 3. Draw given forces, which act on the truss, and reactions, placing all the vectors outside the periphery of the truss, all forces we show in their actual direction (Fig. 4.12a). 4. Denote every region outside the truss between the forces and reactions by capital letters A, B, C, D, . . . and every region inside the truss between the members by small letters a, b, c, d, etc. as shown in Fig. 4.12a. Such labeling goes in the clockwise direction. Start anywhere. 5. Labeling of forces: External forces are named by the letters of two adjacent areas to the member. Their names are read in the clockwise order. Here the force F1 is denoted by FG (not GF), the force F2 is denoted by EF, the force RA is denoted by GA, and RB is denoted by AB. Internal forces are also named by the letters of the two adjacent areas of the member. Their names are also read in the clockwise order. To read the internal force we move clockwise around the joint. Thus, the internal force O4–6 is denoted by Ae if we consider the joint 4 and eA if we consider the joint 6. The arrows are omitted.
4.3 Simple Trusses: Fixed Loads
77
a
A
6
4
8
e 2 1
a
c
f
d
b
10
h
g
k
12
n
3
b
F
F1
G RA= 29.67
F2
E
F3 D
C
F4
F5 B RB= 24.33
n
B h C
k f g
F4 D E A
d
F5 RB=24.33
F3 F2 RA=29.67
e
b
F F1
c a
G
Fig. 4.12 Design diagram of the truss, notation of joints and areas, and corresponding Maxwell–Cremona diagram
6. Construction of the diagram: Choose the scale for the forces (for example, 1cm-10kN). Construct the polygon of external forces (i.e., the active loads F1, F2, . . ., F5 and reactions) to scale laying them off in the order in which they occur clockwise around the periphery of the truss (Fig. 4.12b). (You can start from any force, for example, RA, which has label GA. Next force is reaction RB, which has label AB. Next force is F5 labeled BC, and so on. Thus, we obtained the polygon ABCDEFGA of the external forces acting on the truss (by convention, arrowheads are not shown in the diagram).) Note that in this particular case when all external forces are vertical, this polygon is represented by vertical line. In any case of loading, this polygon must be closed, which means that a system is in equilibrium. Now successively add closed force polygon for each joint, starting from a two-member joint 1 or 12. Joint 1 is in equilibrium under reaction RA (GA) and two unknown internal forces in the elements 1–2 and 1–3; they are Aa and aG. The close force polygon for this joint is constructed by the following procedure: from point A draw line parallel to member A-a and from point G draw a line parallel to member G–a. The intersection of these lines defines the vertex a of the force polygon A–a– G for joint 1. Next we consider joint 3 with two unknown forces in the rods a–b and b–F: from point a draw a line parallel to member a–b and from point F draw a line parallel to member F–b. The intersection of these lines defines the vertex b. Similar construction should be performed for all other joints. Since the errors are accumulated during construction, once we reach the middle of the truss, start from the other end 12. The control of construction is that line e–f must be vertical. The final Maxwell–Cremona diagram for the whole truss is shown in Fig. 4.12b. The magnitude of internal force which arises at any rod presents the corresponding portion of the diagram in the adopted scale for forces. In order to determine the sign of force, say, in vertical member 2–3, we need to consider joint 3; passing clockwise joint 3, the name for required force V23 becomes ab (if we would consider joint 2, then the name of the force would be ba). In the diagram we find that vector ab is directed upwards. Applying this vector to joint 3, we can see that this vector is directed away from joint 3, so force V23 is positive. Advantages of the Maxwell–Cremona diagram: 1. The diagram contains information of magnitude of internal forces and their signs for all members of the truss. 2. This diagram is compact; its construction is very simple and demands small time consumptions.
78
4
Plane Trusses
Now we can create a complete table of forces in all elements of the truss and compare them with the results obtained by the analytical method (see Sect. 4.3.1).
4.4
Simple Trusses: Influence Line Method—Static Approach
Construction of influence line for internal forces in trusses is based on analytical methods used for computing internal forces induced by fixed loads. However, construction of influence line for internal forces in a truss has specific features, which will be considered while using this method. We direct the reader’s attention to the next important and fundamental point: as in the case of beams, the construction of influence lines for trusses will be based not on the repeated calculation of a required factor (reactions, internal forces) for successive position of a unit load as it moves across the span, but on the deriving of the function for the required factor. For construction of influence lines of internal forces we will use the method of sections and method of isolation of joints. The following are some fundamental features of the joints and cuts methods. Using the joint isolation method, three types of position of a unit moving load on a load chord should be considered: 1. A moving load at the considered joint 2. A moving load anywhere in the joint except the considered one 3. A moving load within the dissected panels of a load chord Using the cuts method, three types of position of a unit moving load on load chord should be considered: 1. A moving load on a left-hand part of dissected panel of load chord 2. A moving load on a right-hand part of dissected panel of load chord 3. A moving load within the dissected panel of load chord Design diagram of the simple truss is shown in Fig. 4.13. Influence lines for reactions RA and RB for truss are constructed in the same manner as for reactions for one-span simply supported beam. 6 4 2
h
10
j
1
h = 4m, d = 3m sinj =0.4061
8
P=1
12 3
7
5
d
9
11
l=6d
RA
RB
Fig. 4.13 Design diagram of the triangle truss
The following notation for internal forces will be used: U for bottom chord; O for top chord; V for vertical elements; D for diagonal elements. Influence line for force O4–6 (section 1–1, Ritter’s point 7, Fig. 4.14a). The sectioned panel of the loaded chord (SPLC) is panel 5–7. It is necessary to investigate three positions of unit load on the loaded chord: outside the panel 5–7 (when load P is located to the right of the joint 7 and to the left of joint 5) and within the panel 5–7. Load-bearing contour (or loaded chord) is denoted by dotted line.
1
a
1 O4-6
4
r1
r1
1 5
1 RA
4 r
U5-7
r1=5.972m r=3.655m 10
D4-7
D4-7 3
8
r
2
6
O4-6
12
1 7
U5-7
1
7
9
11
RB
Fig. 4.14 (a) Free-body diagram for left and right parts of the truss; (b) free-body diagram for specified parts of the truss; (c) influence lines for truss with lower loaded chord
b
3
φ
2
6
2
1
U1-3
O6-8
O4-6 V6-7
RA 6
U3-1
3
4
r1
c
V2-3
O1-2
φ
4
8
r
P=1
2
h
10
φ
1
12 3
7
5
d
9
11
RA +
Inf. line O4-6
– Left-hand portion
Right-hand portion
Connecting line
1.231
6d = 3.014 r1
1.5
Connecting line
+
Right-hand portion
1
3d sin φ r
Inf. line D4-7
Connecting line
1.004
Left-hand portion
3d = 2.462 r
Right-hand portion
Left-hand portion
2
h=4m, d=3m r1 =5.972m r =3.655m sin φ =0.4061
Inf. line RA
–
3d h
4
RB
1
3d r
U3-5 3
6d = 45 . h Inf. line U5-7
2
+
3d sin φ = 2 r
Inf. line V6-7 3d = 2.25 h
1.875 Right-hand portion
+
Connecting line
Inf. line U1-3 1 + Inf. line V2-3
Fig. 4.14 (continued)
80
4
Plane Trusses
1. When unit load P is to the right of the sectioned panel, it is more convenient to consider the equilibrium of the left-hand part of the truss: O46 !
X
M left 7 ¼ 0 ! O46 r þ RA 3d ¼ 0 ! O46 ¼
3d R r A
The last equation for force O46 should be transformed into the equation of influence line for O46. When load P has a fixed position, then reaction RA is a specified number and O46 is a number as well. However, if load P is changing its position, then reaction RA is changing its value as well, so it becomes a function (influence line) and therefore internal force O46 becomes a function as well. Thus, equation of influence line for O46 may be expressed in terms of influence line of reaction RA: ILðO46 Þ ¼
3d ILðRA Þ: r
Thus, to construct influence line for O4–6, the following steps are necessary: (a) Show the influence line for RA. (b) Multiply all ordinates by factor 3d/r. (c) Draw a shaded region 7–12, which corresponds to location of the load P (to the right of the sectioned panel), but not part of the truss, which is considered as a free body for equilibrium (!). 2. Similar discussions are applied for the construction of influence line when the unit load P is to the left of joint 5. In this case, it is more convenient to consider the equilibrium of the right-hand part of the truss: O46 !
X
M right ¼ 0 ! O46 r þ RB 3d ¼ 0 ! O46 ¼ 7
3d R r B
The last equation should be transformed in the equation of influence line. When load P has a fixed position, then reaction RB is a number and O46 is a number as well. However, if load P has various positions, then reaction RB becomes a function (influence line) and thus internal force O46 becomes a function as well.Therefore, ILðO46 Þ ¼
3d ILðRB Þ: r
Thus, influence line for O46 is obtained from influence line for RB by multiplying its ordinates by corresponding value (3d/r). Hatching shows regions 1–5, which correspond to the position of the load. Note that the point of intersection of the left-hand portion and right-hand one is always located under the Ritter’s point. If the Ritter’s point is in the infinity, then the left-hand and right-hand portions are parallel. 3. Load P ¼ 1 is located within the sectioned panel 5–7; since external load should be applied at joints only, this case corresponds to indirect load application. Therefore, the influence line connects ordinates of influence lines at points 5 and 7; connecting line is always a straight line. The completed influence line is presented in Fig. 4.14c. As can be seen, the element O4–6 is always compressed for any location of the moving load. Maximum internal force O4–6 occurs when concentrated load is located at point 7. It is very convenient to represent all information related to construction of influence lines in tabulated form as follows: Force O4–6 (section 1–1, SPLC is panel 5–7, Ritter’s point 7): P ¼ 1 left at SPLC X right O46 ! M 7 ¼ 0 ! O46 r þ RB 3d ¼ 0 O46 ¼
3d 3d R ! ILðO46 Þ ¼ ILðRB Þ r B r
P ¼ 1 right at SPLC X left M 7 ¼ 0 ! O46 r þ RA 3d ¼ 0 O46 ! O46 ¼
3d 3d R ! ILðO46 Þ ¼ ILðRA Þ r A r
Analytical expressions for influence lines of internal forces for the remaining elements of the truss are obtained in the same manner as above. From now on we will be presenting analytical expressions for construction of influence lines in tabulated form. The tabulated form contains all necessary information for construction of influence line.
4.5 Trusses with Subdivided Panels
81
Force D4–7 (section 1–1, SPLC is panel 5–7, Ritter’s point 1): P ¼ 1 left at SPLC X right M 1 ¼ 0 ! D47 r 1 þ RB 6d ¼ 0 D47 ! D47 ¼
6d R ! r1 B
ILðD47 Þ ¼
6d ILðRB Þ r1
P ¼ 1 right at SPLC X left D47 ! M 1 ¼ 0 ! D47 r 1 ¼ 0 D47 ¼ 0 ! ILðD47 Þ ¼ 0
The left-hand portion of influence line for D4–7 is obtained by multiplying all ordinates of influence line for RB by factor (6d/r1). Ordinates of the right-hand portion of influence line for D4–7 are zeros. Connecting line connects the joints 5 and 7. Thus, if any load is located within portion 7–12, then internal force in member D4–7 does not arise. Maximum force D4–7 occurs, if load P is placed at joint 5. Force U5–7 (section 1–1, SPLC is panel 5–7, Ritter’s point 4): P ¼ 1 left at SPLC X right 2 U 57 ! M 4 ¼ 0 ! U 57 h RB 4d ¼ 0 3 6d 6d U 57 ¼ RB ! ILðU 57 Þ ¼ ILðRB Þ h h
P ¼ 1 right at SPLC X left 2 M 4 ¼ 0 ! U 57 h RA 2d ¼ 0 U 57 ! 3 3d 3d U 57 ¼ RA ! ILðU 57 Þ ¼ ILðRA Þ h h
The left-hand portion of influence line for U5–7 is obtained by multiplying all ordinates of influence line for RB by factor 6d/h. The right-hand portion of influence line for U5–7 is obtained by multiplying all ordinates of influence line for RA by factor 3d/h. Connecting line runs between points 5 and 7. The force U5–7 is tensile for any location of load on the truss. Force V6–7 (section 2–2, equilibrium of joint 6, Fig. 4.14b): X X ¼ 0 ! O46 cos φ þ O68 cos φ ¼ 0 ! O46 ¼ O68 X Y ¼ 0 ! V 67 2O46 sin φ ¼ 0 ! V 67 ¼ 2O46 sin φ ! ILðV 67 Þ ¼ 2 sin φ ILðO46 Þ The influence line for V6–7 is obtained by multiplying all ordinates of influence line for O4–6 by factor (2 sin φ). The maximum ordinate of this influence line equals to (3d/r) sin φ ¼ 1. This result may be obtained using the other approach: if load P ¼ 1 is located at point 7, then internal forces in all vertical members (except V6–7) and diagonal members are zero; equilibrium equation of joint 7 leads to V6–7 ¼ 1. Force U1–3 (section 3–3, equilibrium of joint 1, Fig. 4.14b): P ¼ 1 applied at joint 1 X left h U 13 ! M 2 ¼ 0 ! U 13 þ 1 d RA d ¼ 0 3 RA ¼ 1 ! U 13 ¼ 0 ! ILðU 13 Þ ¼ 0
P ¼ 1 applied at joint 3 or to the right of joint 3 X left h M 2 ¼ 0 ! U 13 RA d ¼ 0 U 13 ! 3 3d 3d U 13 ¼ RA ! ILðU 13 Þ ¼ ILðRA Þ h h
The right-hand portion of influence line for U1-3 is obtained by multiplying all ordinates of influence line for RA by factor (3d/h). Connecting line connects the joints 1 and 3. All the abovementioned influence lines are presented in Fig. 4.14c.
4.5
Trusses with Subdivided Panels
Trusses with subdivided panels present are simple trusses with additional members. Figure 4.15a shows the Pratt truss with two additional members within each panel; for first panel they are 3–k and k–n. Figure 4.15b shows the Parker truss with three additional members within each panel; they are a–b, a–c, and b–c. In both examples additional elements divide a panel (vertical member k–n in the first case and inclined members a–c and b–c in the second case). In the analysis of such trusses an immediate application of the joints and through section methods might lead to some difficulties.
82
4
4.5.1
Plane Trusses
Main and Auxiliary Trusses and Load Path
Analysis of trusses with subdivided panels can be performed in an orderly fashion using the concepts of the main and secondary (auxiliary) trusses. The main truss is a simple truss, which is enhanced by additional members. The secondary truss is a conventional truss, which contains additional real members (3–k and k–n for truss in Fig. 4.15a) and some imaginary (or concealed) members. These invisible members coincide with members of the main truss. The secondary trusses are shown by dotted lines. The concepts of the main and secondary trusses allow to trace the load pass and to consider trusses with subdivided panels as compound trusses. Combinations of two trusses—a main and sets of secondary trusses—are shown in Fig. 4.15. In all cases the secondary trusses are supported by the joints of the main truss.
b
a 1
2
5
k
2
1
6
a
n
3
7
n
4
7
3 4
Incorrect!
d
c 2
2
1
1 1
6
b
c
Correct
5
1
3
8 5 3
6
4
1 8
7 2
3 7
5 6
4
Fig. 4.15 Trusses with subdivided panels with their main and secondary trusses. (a) Presentation of main and secondary trusses (shown by dotted line); shaded zone is discussed in the kinematical analysis; (a, b) single-tiered secondary trusses; (c) two-tiered trusses; (d) combination of singletiered (joints 5–7–8–6) and two-tiered (joints 1–2–3–4) secondary trusses
How can we distinguish between the main and the secondary trusses? The rule is that elements of the secondary truss are connected with the main truss in such a way that a vertical load, which acts on the joint of the secondary truss, is transmitted only as a vertical load to the joints of the main truss. The secondary trusses in Fig. 4.15 are shown by dotted lines. Figure 4.15a also shows an incorrect presentation of a secondary truss, since in this case the vertical as well as the horizontal reactions are transmitted to the joints of the main truss. The elements of trusses with subdivided panels in case of single-tiered auxiliary trusses can be arranged into three groups as follows: 1. The elements belonging to the main truss only. The internal forces in these elements are not influenced by the presence of secondary trusses. Such elements are labeled #1 in Fig. 4.15c.
4.5 Trusses with Subdivided Panels
83
2. The elements belonging to the secondary system only. The internal forces in these elements arise if a load is applied to the joints of the secondary system. Such elements are labeled #2. 3. The elements belonging simultaneously to both the main and the secondary trusses. The internal forces in such members are obtained by summing the internal forces, which arise in the given element of the main and also of the secondary trusses, calculated separately. Such elements are labeled #3. The elements of trusses with subdivided panels in case of two-tiered auxiliary trusses can be arranged into four groups. In addition to groups mentioned above, the fourth group contains elements belonging to the main truss only for which the influence lines are different depending on the location of moving load on the top- or bottom-loaded chord; this case will be discussed later. The purposes of a secondary truss are to transmit a load, which is applied between the joints of a main truss to the joints of a main truss: 1. Secondary elements (i.e., the elements of the auxiliary truss) transmit loads applied to the lower (or upper) chord to the joints of the same chord of the main truss (single-tiered auxiliary trusses, Fig. 4.15a,b, and truss 5–7–8 in Fig. 4.15d). 2. Secondary elements transmit loads applied to the upper (or lower) chord to the joints of the other chord of the main truss (two-tiered secondary trusses, Fig. 4.15c, and truss 1–2–3–4 in Fig. 4.15d). 3. Vertical elements of an auxiliary truss can divide a compressed member of the upper chord, leading to increased loadcarrying capacity of this element (Fig. 4.15c). Thus, it is possible to have a combination of two types of auxiliary trusses so that a load is transmitted to both chords of the main truss as shown in Fig. 4.15d. In this case hinge 6 belongs to bottom auxiliary truss 5–8–7–6, while hinge 4 belongs to top auxiliary truss 1–2–3–4. Members 3–4 and 7–8 have no common points. If load P is located at point 4, then the load is transmitted to the upper chord (to joints 1 and 2) of the main truss. If load P is located at point 6, then the load is transmitted to the lower chord (to joints 5 and 7) of the main truss. If load P is located between points 4 and 6, then part of the load is transmitted to joint 4 and part to joint 6, and from there it is transmitted from joint 4 to an upper chord and from joint 6 to lower chord.
4.5.1.1
Kinematical Analysis
For kinematical analysis of the truss in Fig. 4.15a, we show the initial rigid disk 1–3–k. Each subsequent hinge is connected using two bars with hinges at the ends; thus part 1–2–4–3 of the truss presents a rigid disk. The next joint cannot be obtained in this way, so instead consider the truss from the right where the base rigid triangle is 6–7–n. Similarly, each subsequent hinge is connected using two bars with hinges at the ends, so part 5–6–7–4 presents a second rigid disk. Both disks are connected by hinge 4 and member 2–5 according to Fig. 4.5; this compound truss is geometrically unchangeable. The trusses in Fig. 4.15b,c can be analyzed in a similar way. For all the design diagrams presented in Fig. 4.15, the relationship S ¼ 2J S0 is satisfied. For example, for structures (a, c) S ¼ 33 and J ¼ 18; for structures (b, d) S ¼ 25 and J ¼ 14. All these structures are statically determinate and geometrically unchangeable compound trusses.
4.5.2
Single-Tiered Auxiliary Trusses
General approach for computation of internal forces in the elements of the truss with subdivided panels is discussed. Detailed construction of influence lines for internal forces by analytical method is presented. Let us consider trusses with subdivided panels (Fig. 4.16). These trusses are strengthened by two additional members in each panel (members 2–4 and 3–4) in Fig. 4.16a, and by three additional members in each panel (2–4, 3–4, and 6–4) in Fig. 4.16b. Removing these rods in all panels leads to a main truss of a simple type. In both cases truss 1–2–3–4 should be treated as a secondary truss. This triangular truss is supported by joints 1 and 3 of the main truss. Analysis of the load path for these structures shows that auxiliary trusses in Fig. 4.16a are single-tiered trusses, while in Fig. 4.16b the two-tiered trusses; indeed, if load P is on joint 6 of the upper chord, then it is transmitted to the joints 1 and 3 of the bottom chord.
84
4
a
1 4
3
b
5
Plane Trusses
6
2 4 4 5
3
2
1
Fig. 4.16 Trusses with subdivided panels. (a) Single-tiered and (b) two-tiered auxiliary trusses
Fundamental Features of the Auxiliary Trusses 1. The elements of axillary truss perceive only local load. 2. If load is applied to the joints of the main truss, the internal forces at elements of auxiliary truss do not arise. Analysis of trusses shown in Fig. 4.16 can be performed by considering the entire structure as a combination of the main truss and a set of additional triangular trusses. There are three groups of elements: the ones that belong to the main truss only, the ones that belong to the secondary truss only, and others that belong to both the main and the secondary trusses. Internal forces in the members of the first group should be calculated considering only the main truss. In this case the load that acts at joint 2 should be resolved into two forces applied at joints 1 and 3 of the main truss and then V 35 ¼ D45 ¼ Dmain V main 35 ; 45 . The internal forces in the elements of the secondary truss should be calculated considering only load, which is applied at the joints of the secondary truss. Thus, in order to find the internal forces in elements 2–4 and 3–4 (due to the load applied at sec joint 2), we need to consider the secondary truss, so the following notation is used: V 24 ¼ V sec 24 and D34 ¼ D34 . Since joints 1 and 3 represent the supports of the secondary truss, if the load is applied at these joints it has no effect on the elements of the secondary truss. Therefore, this load impacts the elements belonging only to the main truss. Internal forces in the members of the third group should be calculated by considering both the main and the secondary trusses and by summing the corresponding internal forces: sec main sec O12 ¼ Omain 13 þ O12 (for truss in Fig.4.16a), U 12 ¼ U 13 þ U 12 (for truss in Fig. 4.16b) main sec D14 ¼ D15 þ D14 (for both truss in Fig. 4.16).
The same principle will be used for the construction of influence lines. Let us consider a detailed analysis of single-tiered subdivided truss shown in Fig. 4.17a. The moving load is applied to the lower chord. As usual, we start from a kinematical analysis. For this we show the initial rigid disk 00 –2–3. Each subsequent hinge is connected using two bars with hinges at the ends; thus part 1–0–4–5 of the truss presents a rigid disk. Similarly we can separate next two rigid disks 5–6–8–9 and 9–10–12–13. The first and second rigid disks are connected by means of hinge 5 and bar 4–6. Similarly connected disks are 2 and 3, i.e., by means of hinge 9 and bar 8–10. Therefore, this truss is geometrically unchangeable. In order for the structure to be statically determinate the number of bars S and joints J must satisfy the formula S ¼ 2J 3. In our case S ¼ 49 and J ¼ 26, so S ¼ 2J 3 ¼ 52 3 ¼ 49. Therefore, this structure is geometrically unchangeable and statically determinate. Once again the necessity of kinematical analysis must be emphasized, especially for trusses that are not simple and contain many elements. For example, what happens if the entire structure in Fig. 4.17a is modified by removing the following members: (a) 40 –50 or (b) 40 –7? Detailed kinematical analysis, as carried out above, shows that this structure remains geometrically unchangeable for case (a). However, for case (b) this structure becomes geometrically changeable and cannot be used as an engineered structure. Therefore, in order to avoid structural collapse an engineer must complete a kinematical analysis as the very first step, and then carefully analyze the results of numerical analysis.
4.5 Trusses with Subdivided Panels 0
a
85 2
4
0⬘ 1
2⬘
6
8
4⬘
10
6⬘
12
8⬘
Entire structure
10⬘ h
a
13 3
1⬘
5
3⬘
5⬘
7
7⬘
9
11
9⬘
11⬘
Combination of main and secondary trusses
b
Secondary truss
1⬘
4⬘
4⬘
Main truss
α
5
3
5
5⬘
7
5⬘
7
d
1
c
+
Inf. line V2-3
d 1 2 sin α
e
Inf. line D4⬘–7 , Dsec 4⬘–5
–
1 2 tan α
+
sec , U sec Inf. line U 5⬘–7 5⬘–5
Left-hand portion
f
+ 1 sin α
1 2 sin α
Inf. line D4ⴕ-5 – Right-hand portion Connecting line for
g
Connecting line for
3d h
1 2 tan α
main D5–6
main U 5–7
Right-hand portion
+
3d h Inf. line U5-5ⴕ
Fig. 4.17 Single-tiered subdivided truss. (a) Entire structure, (b) combination of main and secondary trusses; (c–g) influence lines for internal forces at some members
The entire structure in Fig. 4.17a can be represented as a combination of main and secondary trusses. A feature of this subdivided truss is the following: if the load is located on joints 10 , 30 , 50 , etc. of the secondary trusses on the lower chord, then the reactions of the secondary trusses are transmitted as active loads to the joints of the same chord of the main truss. Influence lines for reactions are constructed in the same manner as in the case of a simple truss. These influence lines are not shown. Now we will consider the construction of influence lines for internal forces of elements belonging to three different groups. These elements are indicated in Fig. 4.17a. The construction of influence lines is presented in tabulated form below. Force V2–3: Vertical member 2–3 belongs to the main truss only; therefore, its influence line is obtained by considering the main truss only. Consider the equilibrium of joint 3 of the main truss:
86
4 P ¼ 1 is applied at joint 3 X Y ¼ 0 ! V 23 P ¼ 0 V 23 !
P ¼ 1 is applied at any joint except 3 X Y ¼ 0 ! V 23 ¼ 0 V 23 !
V 23 ¼ P !
ILðV 23 Þ ¼ 0
ILðV 23 Þ ¼ 1
Plane Trusses
This influence line is shown in Fig. 4.17c. Force D40 –7: Diagonal member 40 –7 belongs to the secondary truss 5–40 –7–5 only (Fig. 4.17). Assume that load P ¼ 1 is located at point 50 . Then the reactions at points 5 and 7 will equal 0.5. 4⬘
α
5
D4′-7 7
5⬘ P=1
U 5sec ⬘– 7
7
0.5
Fig. 4.18 Secondary truss loading
The equilibrium condition of joint 7 (Fig. 4.18) leads to the following results: D40 7 ¼
1 , 2 sin α
U sec 50 7 ¼
1 2 tan α
These influence lines are shown in Fig.4.17d, e. sec sec It is obvious that ILðD40 7 Þ ¼ IL Dsec 40 5 and IL U 50 7 ¼ IL U 50 5 . We can see that if the entire truss is subjected to any load within portions 1–5 and/or 7–13, the internal forces U50 –7 and U50 –5 in the members 50 –7 and 50 –5 of secondary trusses do not arise. Note that if load is located at joint 50 , then reactions arise at points 5 and 7 and internal forces arise at members of the truss 5–40 –7–50 only. However, if the load P is applied at joint 7 then the members of truss 5–40 –7–50 do not work. Force D5–6, main truss: The SPLC is panel 5–7; Ritter’s point is at infinity: P ¼ 1 left at SPLC X Y right ¼ 0 ! Dmain Dmain 56 ! 56 sin α þ RB ¼ 0 1 1 IL Dmain Dmain 56 ¼ sin α RB ! 56 ¼ sin α ILðRB Þ
P ¼ 1 right at SPLC X Dmain Y left ¼ 0 ! Dmain 56 ! 56 sin α þ RA ¼ 0 1 1 main Dmain 56 ¼ sin α RA ! IL D56 ¼ sin α ILðRA Þ
This influence line presents two parallel lines with ordinates 1/ sin α on the supports and a connecting line within panel 5–7 (Fig. 4.17f). Ordinate 1/ sin α on the right support is not shown. Force D40 –5: Diagonal member 40 –5 belongs simultaneously to the main and to a secondary truss. Therefore, an influence line should be constructed considering the main truss and secondary truss 5–40 –7–50 together: main sec sec D40 5 ¼ Dmain 56 þ D40 5 ! ILðD40 5 Þ ¼ IL D56 þ IL D40 5 sec The influence lines for both internal forces Dmain 56 and D540 are already known. The required influence line for D40 5 is 0 presented in Fig. 4.17f. If load P is located at joint 5 or within the portion 50 –7, then the secondary truss within panel 5–7 leads to increasing of compressed force D40 5 .
Force U50 –5: Member 50 –5 belongs simultaneously to the main and to the secondary truss. Therefore, an influence line should be constructed considering the main truss and secondary truss 5–40 –7–50 : main sec sec U 50 5 ¼ U main 57 þ U 50 5 ! ILðU 50 5 Þ ¼ IL U 57 þ IL U 50 5 where IL U sec ¼ IL U sec has already been considered above. The influence line for U 50 5 is presented in Fig. 4.17g. If 50 5 50 7 this truss is subjected to any load within portions 1–5 and/or 7–13, the secondary trusses do not affect internal force in the secondary member U 50 5 .
4.5 Trusses with Subdivided Panels
4.5.3
87
Two-Tiered Auxiliary Trusses
Design diagram of subdivided truss with upper and lower loaded chords is shown in Fig. 4.19a. The structure of the single auxiliary truss 2–4–20 –30 is shown by dashed lines. The feature of truss is as follows: elements of the auxiliary trusses 0–2–00 –10 , 2–4–20 –30 , etc. are connected with joints of the top- and bottom-loaded chords. This means that the load applied to the joints 10 , 30 , 50 , 0 of the lower chord is transmitted to the joints of the upper chord of the main truss. We will consider the construction of influence lines for internal forces of some members. Force D20 –4: Diagonal element 20 –4 belongs to the secondary truss 2–4–20 –30 only. Therefore this element perceives only the local load. If load P ¼ 1 is applied to joint 30 of the lower chord, then reaction which arises at joint 4 of the upper chord equals 1/2. Therefore D20 4 ¼ 1=2 cos β. Corresponding influence line for force D20 –4 is presented in Fig. 4.19b. If load is applied to joints of the upper chord, then D20 4 ¼ 0. Force V4–5: For construction of influence line of internal force in this member two approaches are considered. Approach 1 Entire truss is considered as compound truss, which is a combination of the main and auxiliary trusses. Main truss, section a–a: The SPLC is panel 2–4 (if loaded contour is upper chord) and 5–7 (if loaded contour is lower chord); Ritter’s point is at infinity. For construction of the influence line for internal forces in this element we need to eliminate all auxiliary trusses: P ¼ 1 left at SPLC X Y right ¼ 0 ! V 45 þ RB ¼ 0 V 45 ! V 45 ¼ RB ! ILðV 45 Þ ¼ ILðRB Þ
P ¼ 1 right at SPLC X Y left ¼ 0 ! V 45 þ RA ¼ 0 V 45 ! V 45 ¼ RA ! ILðV 45 Þ ¼ ILðRA Þ
The corresponding influence line contains two parallel lines, IL (RB) and IL(RA), and two connecting lines m–n between joints 2 and 4, and k–s between joints 5 and 7. Parallel straight lines are a consequence of the fact that the truss chords are parallel and Ritter’s point is at infinity. Connecting line m–n corresponds to the upper chord loaded (UCL) (Fig. 4.19c), while connecting line k–s corresponds to the lower chord loaded (LCL) (Fig. 4.19d). Thus, the influence line of internal force in the element 4–5 of the main truss depends on which truss chord (upper or lower) is loaded; this means that the element 4–5 of the main truss belongs to the fourth category (see Sect. 4.5.1). Now we need to take into account load path from two-tiered auxiliary trusses 2–4–20 –30 and 4–6–40 –50 on the rod 4–5. If joints 30 (or 50 ) of the lower chord are loaded by force P, then this force is transmitted to the joints 2 and 4 (or 4 and 6), respectively, of the upper chord (through the elements of auxiliary trusses). In particular, in joint 4, the vertical reaction caused by force P ¼ 1 is equal to 0.5 (if joints 30 and 50 are located in the middle of the corresponding panel). This reaction is directed upwards; therefore a compressive force acts on the rod 4–5. To construct the influence line for required force V4–5, we use the influence lines for this element without considering secondary trusses; these influence lines reflect the different positions of the load contour and are shown in Fig. 4.19c, d. These two influence lines are shown in the combined graph (Fig. 4.19e). Connecting lines m–n and k–s for trusses with upper (UCL) and lower (LCL) load chords are shown by dotted line. Through joint 30 of auxiliary truss we draw a vertical line till crossing with the connecting line m–n; this point is 30 . Through joint 50 of auxiliary truss we draw a vertical line till crossing with the right-hand portion; this point is 50 . Then we connect the nearest joints, 30 , 5, and 50 , by straight lines.
88
4
a
a
b 4
2
0 0′
8
10
9
11
12
6′
2′
1 3′
3
1′
6 4′
β
Plane Trusses
5
5′
7′
7
13
a RB
RA
1/2cosβ
b
Load P=1 moves on the lower chord
+ -
1
IL (RB) Left-hand portion
c
m
Upper chord loaded (UCL)
Connecting line (CL2-4)
1
-IL (R3A)
d
(Main truss)
n k
1/4
1
Right-hand portion
IL (RB)
m(2,3)
k(5)
1
Lower chord loaded
A CL5-7
3′
1 -IL (RA)
CL2-4
5′
n(4)
(8,9)
s(6,7)
Inf. Line V4-5 B
RHP
Upper chord loaded
1/2cosβ
1/2cosβ Load P=1 moves on the lower chord
f g
Inf. Line V4-5 (Main truss)
CL5-7
s
LHP
1
Lower chord loaded (LCL)
-IL (RA)
e
Inf. Line V4-5
Right-hand portion
IL (RB)
Left-hand portion
Inf. Line D2′-4
+ -
Inf. Line D4′-6, and D6′-6 Inf. Line V6-7
LCL
1/2
1/2 UCL
1
Fig. 4.19 Two-tiered subdivided truss. (a) Design diagram, (b–g) influence lines for internal forces (UCL upper chord loaded, LCL lower chord loaded); (c–e) construction of influence line V4–5 (CL2-4 connecting line between joints 2 and 4, LHP left-hand portion, RHP right-hand portion; blue (red) lines—upper (lower) chord loaded; number in brackets means position (joint) of moving load). Ancillary lines are shown by dotted lines
Influence line for the entire truss (taking into account secondary trusses within panels 2–4) when upper chord is loaded by P ¼ 1 is A–m–n–50 –s–B (Fig. 4.19e). Influence line for the entire truss (taking into account secondary trusses within panels 3–5 and 5–7) when lower chord is loaded by P ¼ 1 is A–m–30 –k–50 –s–B. If the load is between joints 0–2 and 1–3, then we need to use the portion A–m of influence line. If the load is between joints 6–12 and 7–13, then portion s–B should be considered. If the load is in joint 30 (50 ) of the truss, then ordinate at point 30 (50 ) of influence line must be considered. If the load is at joint 4 (5) of the truss, then ordinate at point n (k) of influence line must be used.
4.5 Trusses with Subdivided Panels
89
The vertical hatching of the influence lines is conventionally not shown in order not to overload the drawings. Summarize the construction of the influence line for member V4–5. The rod V4–5 has the following features: (i) This rod is a part of the main truss only; (ii) the section a–a through this rod and two other rods (2–4 and 5–7) of the main truss intersects the element 20 –4 of the auxiliary truss also; (iii) the influence lines of the force V4–5 in the main truss are different for upper and lower loaded chord. The procedure for constructing of influence line for internal force for such a rod contains the following steps: 1. Construct the influence for cases of the upper and lower loaded contours; they are presented in Fig. 4.19c and Fig. 4.19d, respectively. 2. On the influence line for case of upper loaded chord (blue line АmnB, Fig. 4.19c,e) we need to show all joints of both loaded contours, the load in which pressure is transferred to the main joint 4 of the upper contour—they are joints 30 and 50 . Then connect the neighboring points (30 , 5, and 50 ) by straight lines. Similarly, on the influence line for case of lower loaded chord (red line АksB, Fig. 4.19d,e) we need to show all joints of the loaded contour, the load in which pressure is transferred to the main joints of the lower contour, and then the neighboring points are connected by straight lines. In our case this step is not performed, because for the given truss there are no auxiliary trusses that transfer the load from the upper contour to the main joints of the lower contour. Approach 2 Entire truss is considered as a whole without its presentation as a combination of the main and auxiliary trusses. Let the load P be located at joint 30 . To determine V4–5, draw a section a–a, and consider the equilibrium of the right part: X V 45 ! Y right ¼ 0 ! V 45 þ RB D20 4 cos β ¼ 0 V 45 ¼ RB D20 4 cos β !
ILðV 45 Þ ¼ ILðRB Þ ILðD20 4 Þ cos β
Thus, to construct the influence line V4–5, one should construct the influence line for element D20 –4 of auxiliary truss and for reaction RB. They are shown in Fig. 4.19b and in Fig. 4.19c (by dotted line), respectively. The operation of subtracting of two influence lines should be performed at three points, 3, 30 , and 5. Since the ordinates of IL (D20 –4) at points 3 and 5 are equal to zero, the ordinates of the required influence line coincide with ordinates IL(RB) at points 3 and 5. 1 1 1 1 Ordinate of IL(D20 –4) at point 30 is ILðV 45 Þ30 ¼ ILðRB Þ30 cos β ¼ ¼ . 2 cos β 4 2 4 Let the load P be located at joint 50 . Influence line for V4–5 within the panel 5–50 –7 may be constructed in a similar way: X V 45 ! Y left ¼ 0 ! V 45 þ RA þ D20 4 cos β ¼ 0 V 45 ¼ RA D20 4 cos β ! ILðV 45 Þ ¼ ILðRA Þ ILðD20 4 Þ cos β Final result of adding of two influence lines, RA (at points 5, 50 , 7) and D20 -4, is reflected by triangle k–s–50 . It can be seen that the first approach for construction of influence line for internal force in element V4-5 allows to easily trace load path from one chord of the truss to another. The second approach suggests a simpler procedure for construction of IL (V45), but at the same time the effect of the load path is presented less clearly.
Discussion 1. Both influence lines (Fig. 4.19d and Fig. 4.19e) show that if the load is to the left of point 2 (3) and to the right of point 6 (7), the amount of force in element 4–5 does not depend on which joints (upper or lower) the main truss is loaded. 2. If load moves along the upper chord, the auxiliary trusses do not have an effect on the force in the element 4–5. 3. If the load moves along the lower chord, then auxiliary trusses in the second and third panels of the main truss will influence the force in the element 4–5. This influence is due to the fact that, if joints 30 and 50 of auxiliary trusses are loaded, then the load is transmitted to the upper joints of the main truss and, as a result, the load is redistributed to the elements of the main truss. Force D40 –6, D60 –6, secondary truss, closed section b. If load P ¼ 1 is located at joint 50 , then reaction which arises at joint 6 of the upper chord equals 1/2. Therefore 0 D4 6 ¼ 1=2 cos β. Corresponding influence line is shown in Fig. 4.19f (left graph). Similarly, if load P ¼ 1 is located at joint 70 , then D660 ¼ 1=2 cos β. Both influence lines are shown on the one basic line (Fig. 4.19f). Force V6–7, main truss, closed section b.
90
4
Plane Trusses
If load P ¼ 1 is located at joint 50 , then element 6–7 is subjected to compressive force 0.5. Indeed, this force presents the reaction at joint 6 which is directed upward and acts on the secondary truss 4–60 –4, so on the member 6–7 this force acts in opposite direction. Thus, if load P ¼ 1 is at joint 50 then V67 ¼ 0.5 . If load P ¼ 1 is located at joint 70 we will get the same result. Final influence line is shown in Fig. 4.19g. If load P ¼ 1 is located at joint 6 of the main truss, then auxiliary trusses do not work and force V6–7 ¼ 1. If load P ¼ 1 is located at joint 7 then V6–7 ¼ 0. If load P ¼ 1 is located at joint 50 (70 ) then V6–7 ¼ 0.5.
4.6
Special Types of Trusses
The following trusses are considered in this section: three-hinged trusses, and trusses with a hinged chain. Each of the abovementioned trusses has some peculiarities.
4.6.1
Three-Hinged Trusses
A three-hinged truss is actually two trusses connected by the hinge C as shown in Fig. 4.20a. Both supports are pinned. The fundamental feature of this structure is that horizontal reactions appear even if the structure is subjected to a vertical load only. These horizontal reactions HA ¼ HB ¼ H are called thrust. Often such types of structures are called thrusted structures. Some examples of three-hinged trusses are presented in Fig. 4.20. Supports may be located at the same or different levels. Truss in Fig. 4.20a contains two pinned supports and the thrust is taken by these supports. A modification of a three-hinged truss is presented in Fig. 4.20b. This structure contains pinned and rolled supports. Instead of one “lost” constraint at support B, we have introduced an additional member that connects both trusses. This member is called a tie; the thrust is taken by the tie.
a
b
C h
h
C
d f
A
d
H
B RB
RA
H
Tie
HB
HA
f
A
B
HA RA
RB
Fig. 4.20 Three-hinged trusses
The structure (a) is geometrically unchangeable. Indeed, two rigid disks, AC and BC, are connected to the ground by two hinges, A and B, and line AB does not pass through the intermediate hinge C. Similarly, the kinematical analysis can be carried out for truss (b). All three-hinged trusses shown in Fig. 4.20 are statically determinate structures. Indeed, the structures in Fig. 4.20a have four unknown reactions, i.e., two vertical reactions, RA and RB, and two horizontal reactions, HA and HB. For scheme in Fig. 4.20b we have three unknown reactions (RA, RB, and horizontal reaction HA) as well as internal force H (thrust) in the tie. For their determination, three equilibrium equations can be formulated considering the whole system. Since the bending moment at hinge C is zero, this provides an additional equation of equilibrium. It means that the sum of the moments of all external forces located on the right or on the left part of the structure with respect to hinge C is zero, i.e., X X M C ¼ 0 or M C ¼ 0: left
right
These four equations of equilibrium determine all four unknowns.
4.6 Special Types of Trusses
91
Three-hinged symmetrical truss is shown in Fig. 4.21a; the height and span of the structure are f and l ¼ 6d, respectively. Assume that the right half of the truss is loaded by fixed uniformly distributed load q. In this case the concentrated forces at joints C-11 are PC ¼ P11 ¼ qd/2, P9 ¼ P10 ¼ qd (these forces are not shown). Reactions: RA !
X
MB ¼ 0 :
HA ¼ H !
X
RA 6d þ PC 3d þ P9 2d þ P10 d ¼ 0 ! RA ¼
M left C ¼0:
RA 3d þ H f ¼ 0 ! H ¼
6qd ql ¼ 8 8
RA 3d ql2 ¼ f 16
Now we will consider construction of influence lines for the reactions and for the internal force in member U3–5. It is obvious that the influence lines for vertical reactions RA and RB are the same as for a simply supported beam. The construction of the influence line for thrust H is presented here in tabulated form. It is evident that HA ¼ HB ¼ H. P ¼ 1 left at joint C X right l H! M C ¼ 0 ! Hf þ RB ¼ 0 2 R l l H ¼ B ! ILðH Þ ¼ ILðRB Þ 2f 2f
P ¼ 1 right at joint C X left l H! M C ¼ 0 ! Hf RA ¼ 0 2 R l l H ¼ A ! ILðH Þ ¼ ILðRA Þ 2f 2f
The left portion of influence line H can be obtained from the influence line for RB by multiplying all ordinates by factor l/2f (Fig. 4.21b). The maximum thrust caused by concentrated load P is Pl/4f. This happens if the force is located at hinge C. P=1
1
a
4
2
d
C
6
9
r
8
7
f
5 H 1
α
11
10
1
3
H B
A RB
RA
b
l 4f
Left-hand portion l IL(RB ) 2f
Right-hand portion
l 2f
+
Inf. line H
5d IL(RB ) r d IL(R A ) r
c d r
d
m
a c b a
– n g
– m
Fig. 4.21 The influence lines of the symmetrical three-hinged truss
5d r Construction of Inf. line U3-5 Inf. line U3-5
n
c
f IL(H ) r
92
4
Plane Trusses
Having the influence line for the thrust allows us to determine the internal force in any member due to arbitrary load. Assume for example that the right half of the truss is loaded by uniformly distributed load q. In this case H ¼ qω ¼
1 l l ql2 11 l ql q¼ and RA ¼ q¼ : 2 4f 2 16f 222 8 |fflffl{zfflffl} ω
After that, finding the internal force in any member is a matter of elementary calculation. So the influence line for thrust should be regarded as a key influence line. Force U3–5 (section 1–1, SPLC is panel 4–6; Ritter’s point is 4): P ¼ 1 left at SPLC X right M4 ¼ 0 U 35 !
P ¼ 1 right at SPLC X left M4 ¼ 0 U 35 !
1 U 35 r þ RB 5d Hf ¼ 0 ! U 35 ¼ ðRB 5d Hf Þ r 5d f ILðU 35 Þ ¼ ILðRB Þ ILðH Þ ðiÞ r r
1 U 35 r RA d þ Hf ¼ 0 ! U 35 ¼ ðRA d Hf Þ r d f ILðU 35 Þ ¼ ILðRA Þ ILðH Þ ðiiÞ r r
The first term in formulas (i) and (ii) corresponds to a common non-thrusted truss (such as a pinned-rolled truss with additional member 7–8). The second term shows that the thrust decreases the internal force in member 3–5; such an influence of the thrust on the distribution of the internal forces is typical for thrusted trusses. For the construction of influence line U3–5 in the case of P ¼ 1 left at SPLC, we need to find the difference 5d f ILðRB Þ ILðH Þ. This step is carried out graphically after both terms in this equation have been plotted (Fig. 4.21c). r r Ordinate ab corresponds to the first term in formula (i), while ordinate bc corresponds to the second term in formula (i). Since 5d f ab > bc, then the difference between the two functions ILðRB Þ and ILðH Þ leads to the positive ordinates for the final r r IL(U3–5) within this part of the truss. d f Similarly, we can find the difference ILðRB Þ ILðH Þ for the case of P ¼ 1 right at SPLC. The specified ordinate nm r r shown in the final influence line is nm ¼ ng mg ¼
d 1 f l d 3d d ¼ ¼ : r 2 r 4f 2r 2r r
Note that the ordinates of the influence line within the right portion are negative. The final influence line for U3–5 is shown in Fig. 4.21d. The influence line for U3–5 shows that this element of the bottom chord in this three-hinged truss may be tensile or compressed, depending on the location of the load (unlike a non-thrust truss of the beam type). This happens because in the structure the thrust H arises, which leads to the negative internal forces in the element 3–5, while external force P leads to the positive internal force in this element.
4.6.2
Trusses with a Hinged Chain
Some examples of trusses with a hinged chain are presented in Fig. 4.22. In all cases these systems consist of two trusses, AC and CB, connected by hinge C and stiffened by additional structures called a hinged chain. The hinged chain may be located above or below the trusses. Vertical members (hangers or suspensions) connect the hinged chain with the trusses. The connections between the members of the chain and the hangers are hinged. In case (c), all the hinges of the hinged chain are located in one line. In cases (a) and (b), a load is applied to the truss directly, while in case (c), the load is applied to the joint of the hinged chain and then transmitted to the truss.
4.6 Special Types of Trusses
a
93
Hanger
Hinged chain
c B
A C
C
b A
C
B
A
B Hinged chain
Hinged chain Fig. 4.22 Trusses with hinged chain
The typical truss with a hinged chain located above the truss is shown in Fig. 4.23. Assume that the parameters of the structure are as follows: d ¼ 3 m, h ¼ 2 m, f ¼ 7 m, and L ¼ 24 m. We need to construct the influence lines for the reactions and the internal forces in hanger Vn. As usual we start with the kinematical analysis of the structure: it shows that the structure is geometrically unchangeable and statically determinate.
4.6.2.1
Reaction of Supports and Internal Forces
Reactions RA and RB for any load can be calculated using the following equilibrium conditions: X X RA ! M B ¼ 0; RB ! M A ¼ 0: For calculation of the internal forces that arise in the members of the hinged chain, we need to show the free-body diagram for any joint n (Fig. 4.23). The equilibrium equation ∑X ¼ 0 leads to the relationship Sn cos α ¼ Sn1 cos γ ¼ H
ð4:2Þ
Thus, for any vertical load acting on the given truss, the horizontal component of the forces, which arise in all members of the hinged chain, is equal. The horizontal component H of the forces Sn and Sn1 is called a thrust. Now we will provide an analysis for the case of a moving load. The influence lines for reactions RA and RB are the same as for a simply supported beam. However, the construction of an influence line for thrust H has some special features. Let us consider them. Thrust H (section 1–1, SPLC is panel 7–C; Ritter’s point is C) Internal force S, which arises in the element m–k of the hinged chain, is denoted as Sleft and Sright. The meaning of the subscript notation is clear from Fig. 4.23. If load P ¼ 1 is located to the left of joint 7, then thrust H can be calculated by considering the right part of the structure. The active forces are reaction RB and internal forces S7-C, S8-C, and Sright. The last force Sright can be resolved into two components: a horizontal component, which is the required thrust H, and a vertical component, which acts along the vertical line C-k. Now we form the sum of the moment of all forces acting on the right part of the truss around point C, i.e., H ! P right M C ¼ 0. In this case the vertical component of force Sright produces no moment, while the thrust produces moment Hf.
94
4
Joint n
Sn
y
α
x x
γ Sn-1
1 Sright k S left m
9⬘
α
n
Vn
f
γ
N
P=1
Vn A
3
5
C
7
2 d
T
α1
9
β RA
4
8
6
B h
D
1 2
Plane Trusses
2
RB
L=24m
4d f
Inf. line H
–
Left portion Connecting line
2d = 0.857 f
(tanγ – tanα )2d
4d IL(RA ) – f (Right portion)
f
+
Inf. line Vn
Fig. 4.23 Truss with a hinged chain: design diagram and influence lines
If load P ¼ 1 is located right at joint C, then thrust H can be calculated by considering the left part of the structure. The active forces are reaction RA and internal forces S7-C, S8-C, and Sleft,. The force Sleft, which is applied at joint m, can be resolved into a horizontal component H and a vertical component. The latter component acts along vertical line m-7. Now we find the sum of the moment of all the forces, which act on the left part of the truss, around point C. In this case, the vertical component of force Sleft produces the nonzero moment around joint C and thrust H has a new arm (m-7) around the center of moments C. In order to avoid these difficulties we translate the force Sleft along the line of its action from joint m into joint k. After that we resolve this force into its vertical and horizontal components. This procedure allows us to eliminate the moment due to the vertical component of S, while the moment due to the horizontal component of S is easily calculated as Hf. Construction of the influence line for H is presented below: P ¼ 1 left at SPLC X right H! M C ¼ 0 ! RB 4d þ Hf ¼ 0 H¼
4d 4d R ! ILðH Þ ¼ ILðRB Þ f B f
P ¼ 1 right at SPLC X left H! M C ¼ 0 ! RA 4d þ Hf ¼ 0 H¼
4d 4d R ! ILðH Þ ¼ ILðRA Þ f A f
The left portion of the influence line for H (portion A-7) presents the influence line for RB multiplied by coefficient 4d=f and the right-hand portion (portion C-B) presents the influence line for RA multiplied by the same coefficient. The connecting line is between points 7 and C (Fig. 4.23). The negative sign for thrust indicates that all members of the hinged chain are in compression. Force Vn: Equilibrium condition for joint n leads to the following result: X Y ¼ 0 : V n þ Sn sin α Sn1 sin γ ¼ 0 ! V n ¼ H ð tan α tan γ Þ Therefore, ILðV n Þ ¼ ð tan α tan γ Þ ILðH Þ: Since α < γ and H is negative, all hangers are in tension. The corresponding influence line is shown in Fig. 4.23.
4.7 Kinematical Method for Construction of Influence Lines
95
The influence line for thrust H can be considered as the key influence line, since thrust H always appears in any cut section for the entire structure. This influence line allows us to calculate thrust for an arbitrary load. After that, the internal force in any member can be calculated simply by considering all the external loads, and the reactions and the thrust as an additional external forces. 4.6.2.2
Discussion
For any location of a load the hangers are in tension and all members of the hinged chain are compressed. The maximum internal force at any hanger occurs if load P is placed at joint C. To calculate the internal forces in different members caused by an arbitrary fixed load, the following procedure is recommended: 1. Construct the influence line for the thrust. 2. Calculate the thrust caused by a fixed load. 3. Calculate the required internal force considering thrust as an additional external force. This algorithm combines both approaches, the methods of fixed and of moving loads, and so provides a very powerful tool for the analysis of complex structures. Example 4.2 The structure in Fig. 4.23 is subjected to a uniformly distributed load q within the entire span L. Calculate the internal forces T and D in the indicated elements. Solution The thrust of the hinged chain equals 1 2d qLd H ¼ qΩH ¼ q L ¼ , 2 f f where ΩH is the area of the influence line for H under the load q. After that, the required force T according to (4.2) is T¼
H qLd ¼ : cos α1 f cos α1
To calculate force D we can use section 2–2 and consider the equilibrium of the right part of the structure: X 1 qL qLd D! Y ¼ 0 : D sin β þ RB þ T sin α1 ¼ 0 ! D ¼ tan α1 sin β 2 f Thus, this problem is solved using the fixed and moving load approaches: thrust H is determined using corresponding influence lines, while internal forces D and T are computed using H and the classical method of through sections.
4.7
Kinematical Method for Construction of Influence Lines
The idea of kinematical method of constructing influence lines is presented in Sect. 2.6 for beams. Let us apply this idea to the construction of the influence lines of internal forces in some elements of the simple statically determinate truss. The following procedure is recommended. 1. The element of a truss, in which the required force arises, should be removed. As a result, the initial statically determined and geometrically unchangeable structure turns into a mechanism with one degree of freedom. The action of the eliminated member is replaced by the required force. It means that the internal force is carried to the class of active forces. Just these two factors—mechanism and active forces—are the prerequisite for application of the virtual displacement principle. 2. Impart the virtual displacement to the mechanism. In this case, each disk of the mechanism will rotate around its instantaneous center of rotation (a point of zero velocity). The virtual displacement (VD) diagram of a loaded contour of a truss presents the model of influence line of the required internal force. 3. To transform the model of influence line into real influence line, the scaled coefficient should be set equal to unity. For this purpose, to the two joints that connected the removed element of a truss, impart the unit mutual displacement in the positive direction of the required internal force. Corresponding displacement diagram represents the influence line of the internal force for removed element. Some peculiarities of the concept of “influence line model”: The virtual displacement diagram of the loaded contour presents model of influence line. For this model ordinates are imaginary, and their magnitude at an arbitrary point of the contour cannot be indicated; we will call it Model 1. Model 2 of influence line presents the displacement diagram of the loaded contour caused by the mutual displacement of joints in the direction of the required force. Both models are similar; however
96
4
Plane Trusses
for model 2 each ordinate of diagram is real, and depends on the magnitude of the mutual displacement, and its magnitude at an arbitrary point of the contour can be indicated. With the change in a value of the mutual displacement of joints, the coordinates of model 2 change as well, but the form of the model is preserved. Only with a unit mutual displacement of joints a model 2 turns into a real influence line. An important step of the analysis is the determination of centers of instantaneous rotation for each element of mechanism. Two special cases are shown in Fig. 4.24. We assume that there are two rigid disks connected by hinge C, while each disk 1 and 2 is connected to the ground by means of pinned and roller supports A and B (Fig. 4.24.). Suppose that three hinges are on the same straight line (Fig. 4.24a). The support A of the disk 1 allows it to rotate around point A, and the right support B allows a horizontal displacement of the point B; the corresponding velocities of points C and B are υC and υB. The center of the velocities of disk 1 is point A; the mutual instantaneous center of velocities of both disks is point C. The velocity υC belongs to both disks. Disk 2 performs a plane motion; therefore for this disk the instantaneous velocity center is defined as the intersection point of the perpendiculars to the velocities of the two points belonging to the disk 2. Thus, the instantaneous center of velocities of disk 2 coincides with point B. υC
b
a υC 1
A
C
2 υB
P C
B
A
υB
B
Fig. 4.24 Connections of two disks and the location of the instantaneous centers of rotation
If three hinges do not lie on the same line, then the determination of the instantaneous velocity center P for disk 2 is shown in Fig. 4.24b. It can be seen that three centers of rotation of two disks (two centers for disks itself and one mutual center of rotation) lie on one straight line (the theorem of three centers of rotation). Application of kinematical method for construction of influence line of internal forces in the simplest truss, shown in Fig. 4.25a, is presented below. Force O4–6: For construction of this influence line, the element 4–6 is removed and its action is replaced by two forces O4–6; assume that these forces are positive; that is, the forces are directed from joints 4 and 6 (Fig. 4.25b). As a result of eliminating of one necessary bar, the original geometrically unchangeable and statically determined truss is transformed into instantaneous changeable system. This system may be represented as two absolutely rigid disks 1 and 2, which are connected by hinge 5, as shown in Fig. 4.24a. Three hinges, 1, 5, and 13, lie on one straight line. Impart a virtual displacement to the system. Disk 1 rotates around fixed point 1, while disk 2 rotates around point 13. The hinge 5 is a mutual center of rotation of disks 1 and 2. Corresponding virtual displacement (VD) diagram of a loaded contour is shown by dotted line; this graph presents a model 1 of influence line for force O4–6 (Fig. 4.25b). In order to construct the model 2 we need to create the mutual displacement δ1 þ δ2 of joints 4 and 6 in the direction of required force O4–6 (Fig. 4.25b). In this case, each of the disks will rotate around their centers of rotation 1 and 13 by an angle φ1and φ2, respectively. We get a graph that coincides in form with model 1. However this is graph of real vertical displacements of the loaded contour. Since a model is located below the reference line, then ordinate of influence line is negative. In order to transform model 2 into real influence line we need to assume that a scaled coefficient (mutual displacement) is equal to unity, i.e., δ1 þ δ2 ¼ 1. Since the relations between components of mutual displacement and the angles of rotation of the disks are δ1 ¼ φ1h and δ2 ¼ φ2h, then the scaled coefficient becomes φ1h þ φ2h ¼ 1. Ordinate of influence line at joint 5 is y5 ¼ φ1 2d ¼ φ2 4d; therefore φ1 ¼ 2φ2, so the scaled coefficient becomes 2φ2h þ φ2h ¼ 1 or φ2 ¼ 1/3h. Finally, for ordinate of influence line at joint 5 we gety5 ¼ φ2 4d ¼ 4d/3h which means that if load P ¼ 1 is located at joint 5, then the value of force O46 ¼ 4d/3h. Verification: If P ¼ 1 is placed at point 5, then reaction of the left support R1 ¼ 2/3. The required force O46 !
X
M left 5 ¼ 0 ! R1 2d þ O46 h ¼ 0 ! O46 ¼
2 2d 4d ¼ 3 h 3h
Combined approach: It can be seen that the kinematical method of constructing an influence line consists of two principal parts: (1) construction of the influence line model as the virtual displacement diagram and (2) its transforming into a real influence line. The first part for simple trusses is not a problem. However, the second step can be cumbersome. Therefore, more effective is a combined approach where in part (2) instead of transforming of influence line model, we compute a specific ordinate of influence line by static approach. For this we need to place the load P ¼ 1 at the point of a loaded contour with specific ordinate of the influence line model and calculate the corresponding internal force.
4.7 Kinematical Method for Construction of Influence Lines
97
Force D5–6: The element 5–6 is removed and its action is replaced by two positive forces D5–6 (Fig. 4.25c). We obtain a system, which consists of two rigid disks 1 and 2, connected by two parallel bars 4–6 and 5–7 with hinges at the ends. This structure is geometrically changeable, as discussed in Sect. 1.4, Fig. 1.13c.
a
0
1
2
4
6
8
10
12
h
α 3
5
b
9
7
δ2
δ1
4
11
d
6
O4-6
1
13
2
1
5
φ1
13 VD diagram (Model 1)
φ2 y5=4d/3h
Inf. Line O4-6
-
Right-hand portion
Left-hand portion
N
c
4
δ2
6
D5-6
1
2
1
VD diagram
δ1
2dsinα
φ2 13
7
5
φ1
4dsinα
4d
M Left-hand portion
y5=1/3sinα
+
Connecting line
Inf. line D5-6
− Right-hand portion
K
y7=1/2sinα φ2
d
6
2
1 VD diagram U5-7
1
φ1
5
δ1
δ2
13
7 y7=3d/2h
Left-hand portion
+ y5
Right-hand portion
Inf. line U5-7
Fig. 4.25 Kinematical method for construction of influence lines for internal forces. (a) Design diagram of the truss; (b–d) structure with removed constraint for required force. 1 and 2 are rigid disks; the virtual displacement (VD) diagram of the loading contour (model 1 of influence line) is shown by dotted line
98
4
Plane Trusses
Impart a virtual displacement to the system. Disk 1 rotates around fixed point 1, while a disk 2 rotates around point 13. Corresponding VD diagram (model 1) is shown by dotted line. The left-hand portion of this diagram is located above the reference line, while the right-hand portion is located below the reference line. The left-hand and right-hand portions are parallel (since the Ritter’s point is in the infinity). Since the character of the influence line is already defined, we can calculate the specific ordinates of influence using the static approach. Assume that load P ¼ 1 is placed at joint 5. Reaction at the left support equals R1 ¼ 1 (4/6) ¼ 2/3. Required forceD56 ! ∑ Yleft ¼ 0, or R1 1.0 þ D56 sin α ¼ 0. Thus, the ordinate of IL(D5-6) at point 5 is y5 ¼ D56 ¼ 1/3 sin α. If P ¼ 1 is placed at joint 7, then R1 ¼ 1/2 and D56 ! ∑ Yleft ¼ 0 ! R1 þ D56 sin α ¼ 0, so y7 ¼ D56 ¼ 1/2 sin α. It means that right-hand portion of influence line cutoff on the vertical line passing over left supports a portion1/ sin α and the left-hand portion cutoff on the vertical line passing over right supports the same portion. Force U5–7: The element 5–7 is removed and its action is replaced by two positive forces U5–7 (Fig. 4.25d). Impart a virtual displacement to the disks 1 and 2. The center of rotation of the disk 1 coincides with the fixed support 1; the center of rotation K of the disk 2 is on the vertical line passing over the movable support 13. According to the theorem about three centers of rotation, it belongs to straight line 1-6-K. The disk 2 rotates around the point K, while the joint 6 is a center of the mutual rotation of the disks 1 and 2. The VD diagram is shown in Fig. 4.25d by dotted line. So model 1 of the influence line U5–7 represents a triangle with a vertex under the Ritter’s point 6 and is located above the reference line; this means a positive sign for ordinates of the influence line. To switch from a model 1 of influence line to a real influence line, we place the load P ¼ 1 at the specific point of influence line (joint 7). In this case the reaction of the left support equals 0.5. The equation of equilibrium is U 57 !
X
M left 6 ¼ 0 ! 0:5 3d U 57 h ¼ 0 ! y7 ¼ U 57 ¼
3d 2h
It means the right-hand portion of influence line cutoff on the vertical line passing over left supports a portion3d/h. From the point of view of kinematics of mechanism, the positive sign means that our assumptions about the direction of rotation of both disks had been done correctly. Force V2–3: The element 2–3 is removed and its action is replaced by two positive forces V2–3 (construction of this influence line is not shown). Therewith we leave hinge 3 because it belongs to the loaded chord. We got two disks 0–2–1–0 and 2–12–13–5. They are connected by rods 1–3 and 3–5, and also by hinge 2. We will create a vertical unit displacement of the joint 3 which belongs to the loaded contour. The hinge chain 1–3–5 is an instantaneously changeable system. The diagram of vertical displacements of the points within the panel 1–5 has the form of a triangle with zero ordinates at joints 1 and 5, and with an ordinate equal to unity at joint 3. The ordinates of the influence line are positive. Assume that load is applied to the upper chord. For construction of influence line for force V2–3 it is necessary to remove member 2–3. Also we remove hinge 3 because it does not belong to the loaded chord. We get rigid disks 0–1–2 and 2–12– 13–5. They are connected by hinge 2 and element 1–5. So these two disks form unchangeable structure, but not mechanism. Therefore, this structure does not have virtual displacements. It means that if load is moved over upper chord, then internal forces in member 2–3 are equal to zero for any load within joints 0–12. So all ordinates of influence line for V2–3 are equal to zero. Features of influence lines for internal forces in the members of simple trusses: These features may be formulated in terms of kinematical method. This allows constructing (or checking) configuration of influence line as the vertical displacement diagram of the loaded contour due to mutual displacement in the direction of removed element. 1. Each disk has corresponding straight line on the displacement diagram of a loaded contour. 2. Each portion of influence line intersects the reference line under the instantaneous center of rotation of the corresponding disk. 3. The left-hand and right-hand portions of the influence line intersect under the mutual center of rotation of both disks (Ritter’s point). 4. If influence line is being constructed for diagonal element, then within the panel of this element, there is a line connecting left and right portions of influence line. Positions 1–4 are shown in Fig. 4.25b–d. Comparison of two methods of constructing the influence lines The influence lines for internal forces in the elements of a truss may be constructed using the static and kinematical approaches. The static method is based on the simple idea—equilibrium equation for determining the required force is applied and then the expression for unknown force is converted into an equation of the influence line. The static method is elegant, simple, and well algorithmized.
4.8 Complex Trusses
99
The kinematical method is based on the deep idea: the shape of the influence line for force in specified element coincides with a virtual displacement diagram of the loaded contour of the system with eliminated element of a truss. This method is based on the virtual displacement principle. The difficulty of the kinematical method is in determining the scale factor, which makes it possible to transform the model of the influence line into the real influence line. To get around the difficulties, the combined approach may be recommended. The kinematical method can be effectively applied to determine the configuration of the influence line and the location of its specific (maximum) ordinates, as well as for the rapid verification of the configuration of the influence line which had been constructed by the static method. Kinematical method for construction of influence lines is discussed in detail by S. Parvanova (2011).
4.8
Complex Trusses
Complex trusses are generated using special methods to connect rigid disks. These methods are different from those used to create the simple trusses, three-hinged trusses, etc., analyzed in the previous sections of this chapter. An example of a complex truss is a Wichert truss.
4.8.1
Substitution Bar Method
Figure 4.26 presents a design diagram of a typical Wichert truss. As before, statical determinacy of the structure can be verified by the formula W ¼ 2J S S0, where J, S, and S0 are the number of hinged joints, members, and constraints of supports, respectively. For this truss we have W ¼ 2 14 24 4 ¼ 0. Thus, the truss in Fig. 4.26 is statically determinate. K A
B 1
2 C
Fig. 4.26 Wichert truss
In this truss two rigid disks are connected using hinge K, members 1 and 2, and rolled support C. If the constraint C was absent, then this structure would be geometrically changeable. However, the connections of both disks to the ground using constraint C lead to a geometrically unchangeable structure. A peculiarity of this multispan truss is that even though this structure is statically determinate its reactions cannot be determined using the three equilibrium equations for the truss as a whole. Therefore, the Wichert truss requires a new approach for its analysis: the replacement bar method, also called the L. Henneberg (1886) method. The main idea behind this method is that the entire structure is transformed into a new structure. For this, the intermediate support is eliminated and a new element is introduced in such a way that the new structure becomes geometrically unchangeable and can be easily analyzed. The equivalent condition of both systems, new and original, allows us to determine the unknown reaction in the eliminated constraint. As an example, let us consider the symmetrical Wichert truss supported at points A, B, and C and carrying a load P, as shown in Fig. 4.27a. Assume that angle α ¼ 60 . P
a
d
A
P
b
K
B
1
A
K
B
45°
Substituted bar L
R
C
L
C N1
α
1 XC Fig. 4.27 (a) Design diagram of a Wichert truss; (b) substituted system
XC
R
100
4
Plane Trusses
To analyze this system, let us replace support C by an additional vertical member and apply external force XC , which is equal to the unknown reaction of support C (Fig. 4.27b). The additional element CK is called a substituted bar. According to the superposition principle, the internal force in the substituted bar CK is FC þ FP; that is, it is the sum of the internal force due to unknown reaction XC of support C and given external load P. Both systems (a) and (b) are equivalent if the internal force in the substituted bar is zero, i.e., FC þ FP ¼ 0. In expanded form, this equation may be written as FX C þ F P ¼ 0,
ð4:3Þ
where FP is the internal force in the substituted bar due to external force P; F is the internal force in the substituted bar due to unit force XC ¼ 1; and FX C represents the internal force in the substituted bar due to unknown reaction XC. Two conditions should be considered. They are P loading and XC loading. Design diagram for both conditions presents a truss simply supported at points A and B. For each condition, the force in the substituted bar can be found in two steps: first, compute the force N1 in the member LC (using section 1–1) and then find required force in substitute bar, considering joint C. This procedure is presented in Table 4.1.
Table 4.1 Computation of internal force in substitute bar
State
Design diagram
Free-body diagram
Substituted system in Fig. 4.27b subjected to given load P(↓); in this case RA=0.5P(↑)
P- loading
K
− R A 3d + N1P cos 30° 2d + N1P sin 30° d = 0
(
)
N1P 2 cos 30° + sin 30° = 3R A RA=0.5P
L
30°
N1P = N1P
3P 2 2 3 2 +1 2
= 0.672 P.
1 FP
N2P
N1P 30 °
∑ X = 0 → N1P = N 2 P = N ∑ Y = 0 → FP + 2 N sin 30° = 0 FP = −2 N sin 30 ° = − 0.672 P
C
N1 X → ∑ M Kleft = 0
1
A Substituted system in Fig.4.27b subjected to unit force Xc=1(↑); in this case RA=0.5(↓)
N1P → ∑ M Kleft = 0
1
A
XC - loading
Equilibrium equations
K
R A 3d + N1 X cos 30° 2d + N1 X sin 30° d = 0
(
RA=0.5
L
30°
N1X
)
N1 X 2 cos 30° + sin 30° = −3R A , 3 1 2 N1 X = − = −0.672 2 3 2 +1 2
1 F
N1X
N2X 30°
C XC=1
∑ X = 0 → N1 X = N 2 X = N X ∑ Y = 0 → F + 2 N X sin 30° + 1 = 0 F = −2 N X sin 30° − 1 = − 0.328
According to (4.3), the unknown reaction of support C equals XC ¼
FP 0:672P ¼ ¼ 2:048P 0:328 F
ð4:4Þ
After this, the calculation of reactions RA and RB presents no further difficulty. As soon as all the reactions are known, the calculation of the internal forces in all the members can be carried out as usual.
4.8 Complex Trusses
101
Discussion 1. For the given Wichert truss, loaded by vertical force P directed downward, the vertical reaction of the intermediate support is also directed downward. This surprising result can be easily explained. Imagine that support C is removed. Obviously, point K would be displaced downward, rigid disk AKL would rotate clockwise around support A, and rigid disk KBR would rotate counterclockwise around support B. Therefore, support C would be displaced upward, so the reaction of support C is directed downward. 2. If the substituted member is a horizontal bar (between joints L and R), then 3 F P ¼ P, 4
F¼
pffiffiffi 3 1 ¼ 0:366 2 2
and for reaction XC we obtain the same result. 3. A Wichert truss is a very sensitive structure with respect to angle α. Indeed, if angle α ¼ 45 , then F P ¼ 0:707P and F ¼ 0, so X C ¼
0:707P 0
This means that the system is simultaneously changeable. If angle α ¼ 30 , then reaction XC is positive and equals 3.55P. Design diagram of another complex plane truss is shown in Fig. 4.28. Even if this structure is statically determinate (W ¼ 2J S S0 ¼ 2 6 9 3 ¼ 0), this truss cannot be analyzed by methods used for analysis of simple trusses. The reactions of truss in Fig. 4.28, in contrast to the Wichert truss (Fig. 4.27), can be determined, but the structure of the truss does not allow determining the internal forces by cutting out joints and through sections. The fact is that unlike simple trusses, the formation of given truss does not involve the original triangle and the subsequent attachment of each next joint by means of two rods is absent. In the truss there is not a single two-rod joint, so the successive isolation of the joints is not possible. Of course, another way to analyze this truss is to compose equilibrium equations for all joints and solve the system of 12 linear algebraic equations with respect to 9 unknown internal forces and 3 reactions of support, but this way is too cumbersome. Any cross section passing through the rod with the required force crosses 4 or even 5 bars and does not lead to such an arbitrary system of forces that would allow establishing the Ritter’s point and composing the corresponding equilibrium equation. In this case, the application of the Henneberg method turns out to be the most effective. In accordance with this method, the original system should be transformed into a statically determinate and geometrically unchangeable structure of a simple type. To do this, we need to remove some rods and replace them by other, but differently located. It means that the eliminated rod may not only support constraint as in Fig. 4.27 for Wichert truss, but also be a lattice element. This procedure can be performed in various ways. Let us determine the internal force which arises in the rod BC. We discard this rod, replace its action with the unknown force X, and introduce a substituted rod AB (Fig. 4.28b). P
a
P
b
D
C
C X
A
B
B
A E
Fig. 4.28 (a) Design diagram of complex truss; (b) substituted system
According to superposition principle, the force SAB in the substituted rod is SAB ¼ SPAB þ SXAB
ð4:5Þ
To determine SPAB it is necessary to take into account the action of the force P only, and consider consecutively the equilibrium of the joints C, D, E, and B. To determine SXAB it is necessary to take into account the action of the force X ¼ 1 only, and consider successively the equilibrium of the same joints C, D, E, and B. Now expression (4.5) can be written in the form
102
4
Plane Trusses
SAB ¼ SPAB þ X SX¼1 AB Since the substituted rod is absent in the initial structure, then equivalency condition of initial and substituted structures becomes SAB ¼ SPAB þ X SX¼1 AB ¼ 0
ð4:6Þ
The required internal force in rod BC equals X¼
SPAB SX¼1 AB
ð4:7Þ
If it turns out that SX¼1 AB ¼ 0 this means that the system is instantaneously changeable (under the condition of its static determinacy). This means that the Henneberg method simultaneously solves another problem, namely, it allows performing a kinematic analysis of a structure by analytical way and decide if the complex statically determinate structure is instantaneously changeable or not.
4.8.2
Closed Section Method
The next example of the complex truss is shown in Fig. 4.29. The reactions of support, as in the case of truss in Fig. 4.28, can easily be determined. However, a way the truss is formed allows avoiding the use of the substitution of bars method and applying a new approach when instead of a through section a closed section is applied (F. Yasinsky approach). The peculiarity of such a section is that some bars are intersected twice (1-B, 2-B) while other rods (A-1, 2-3, and 4-B) are intersected one time. P1 m
2
3
n h2
h1 HA
4
1 A
VA
P2
B l
VB
Fig. 4.29 Design diagram of a complex truss
Equilibrium condition for truss in whole allows to write equilibrium equations for determining the reactions of supports: X X X VA ! M B ¼ 0, V B ! M A ¼ 0, H A ! X¼0
Internal forces in the bars, which are intersected twice, do not exist in the equilibrium equations. The equilibrium conditions of the part of the truss, which is located in the interior of the closed section, allow writing the following equations: X N 1A ! M n ¼ V A l þ H A h1 N 1A l P2 h2 ¼ 0 X N 4B ! M m ¼ H A h1 N 4B l P2 h2 ¼ 0 Solution of this equation is internal forces N1A and N4B. After that we can determine internal forces in other members by the isolation joint method. This example is important in that it allows generalizing the concept of a “section,” namely, the section must completely separate some part of the truss from the rest part.
Problems
4.8.3
103
Summary
Analysis of any truss of the simple and special types can be performed using two different analytical methods: the fixed and moving load approaches. The fixed load approach (methods of joints and through sections) allows calculating the internal force in any selected member of a truss. If it is required to determine the internal forces in all elements of the truss, it is recommended to construct Cremona–Maxwell diagram. The moving load approach requires the construction of influence lines. The following approach for the construction of influence lines may be used: the moving load, P ¼ 1, is placed at each successive joint of the loading chord and for each such loading the required internal force is calculated by the method of joints or method of cuts. This procedure leads to a correct picture of the influence line, but this procedure is repetitive and bothersome; mainly, it is essentially pointless. This harmful inefficient approach is based on the calculation of the required force for the different locations of the load in order to show the influence lines. In this case the whole necessity of the influence line is lost, since by constructing influence lines this way we already determined the internal forces for all positions of the load P ¼ 1. On the contrary, we need the influence line in order to find the value of the required internal force for any location of the load. By no means should this graph be plotted using repeated computation. In principle, such a flawed approach defeats the purpose of the influence line as a powerful analytical tool. This book contains an approach to the construction of influence lines, based on accepted methods of truss analysis: using the cutting out of joints and through section methods we obtain an expression for the required force and after that we transform this expression into an equation for the influence line. The algorithm described above allows the influence line to be presented as a function (in contrast to a set of numerical ordinates for the influence line). Stated as functions, influence lines can realize their full potential as extremely versatile tools for providing different types of analysis. A very important aspect of this approach is that it allows us to find specific ordinates of the influence lines in terms of the parameters of the structure, such as heights, panel dimensions, and angle of inclination of diagonals. This allows us to determine the influence of each parameter on a required force and so provides optimization analysis. Both fundamental approaches, fixed and moving loads, complement each other. This combination is very effective for truss analysis with some peculiarities (especially for special statically determinate trusses). Also, the combination of both approaches is an effective way to analyze the same truss in different load cases (snow, dead, live, etc.). For example, for a thrusted truss with supports on different levels, we can find the thrust by considering the system of equilibrium equations. Then, knowing the thrust, we can find all the required internal forces. This is not difficult, but if we need to do it many times for each loading, then it would be much wiser to construct the influence line for the thrust only once. Then we could find its value due to each loading. It is obvious that in this case the influence line for the thrust should be treated as the key influence line. In the case of a truss with a hinged chain, the internal force in the member of the chain under analysis (or better, the thrust as the horizontal component of such an internal force) should be considered to be the key influence line. For analysis of complex truss the Henneberg method may be recommended.
Problems 4.1a, b.
Provide kinematical analysis for trusses in Fig. P4.1.
a
b
Fig. P4.1
4.2. The simple truss is loaded by two fixed forces 100 and 200 kN as shown in Fig. P4.2; d ¼ h ¼ 4 m. Provide complex analysis of this structure:
104
4
1
2 α
A E
C
3
K
5
6
Plane Trusses
h
4
B
d
100kN 200kN Fig. P4.2
1. Perform kinematical analysis of the truss. 2. Determine internal forces at the members 1, . . ., 6, due to fixed two loads at joints C and K. Apply the most suitable analytical methods. 3. Construct Cremona–Maxwell diagram to determine internal forces in all members of the truss. 4. Construct the influence lines of the internal forces at the indicated members 1–6; use analytical approach. 5. Calculate internal forces at the members 1–6 using influence lines. 6. Compare results obtained by three different approaches. 7. Calculate maximum internal forces at the indicated members 1, . . ., 6, caused by two moving connected forces P1 ¼ 80 kN and P2 ¼ 160 kN apart 2 m. 8. Based on the constructed influence lines check the correct answer: (a) (b) (c) (d)
Force V1 is the maximum tension, if concentrated load P is located at joint E, C, or K. Force V2 is always positive, negative, or zero. Force U3 is maximum tensile, if distributed load q is located within the portion AC, KB, or AB. If load P is located at joint K then the force in members 5 and 6 is D5 > D6, D5 < D6, or D5 ¼ D6.
4.3. The single-span K-truss is shown in Fig. P4.3. Construct the influence lines for the internal forces at the indicated members. Using the constructed influence lines, calculate these forces if uniformly distributed load q is distributed within the three panels on the right and concentrated force P is applied at the middle of the span. 3 qd 2 d +P , 2 h h 3 P V = − qd − , 8 4 3 qd P + D1 = D2 = 8 sinα 4sinα
Ans. U = D1
h
α α
D2
V
U
d
Fig. P4.3
4.4a, b. Two trusses with subdivided panels are presented in Fig. P4.4a, b. For case (a) the diagonal members of the main truss are angled downward in the left part of the truss and angled upward in the right part. In case (b) all the diagonal members of the main truss are angled downward. Construct the influence lines for the indicated members and compare the influence lines for the two cases. Using the constructed influence lines, calculate these forces if load P is placed at joint n and uniformly qd P distributed load q is distributed within the panel A–n. Ans. (b) V ¼ þ 8 4
a
b V
V A
A n
n d
d
Fig. P4.4
4.5.
Design diagram of a truss is presented in Fig. P4.5. Moving loads are applied to the lower chord.
Problems
105
(a) Perform the kinematical analysis; (b) construct the influence lines for the internal forces in the indicated members; (c) using the constructed influence lines, compute the indicated internal forces if external forces P1 and P2 are placed at joints 7 and 9, respectively. 2 D2-5
h
a
1 3
5
7
U7-9
9
d
Fig. P4.5
Ans:U 79 ¼
9d d P þ P ; 4h 1 h 2
D25 ¼
3 1 P þ P 8 cos α 1 2 cos α 2
4.6. Design diagram of a truss with over-truss construction is presented in Fig. P4.6. Pinned supports A and B are located at different elevations. Panel block 1–2–3–4 has no diagonal member. The vertical members (links) are used only to transmit loads directly to the upper chord of the truss. Provide kinematical analysis. Construct the influence line for thrust H. Using the constructed influence line, compute H if forces P1 and P2 are placed at joints 10 and 20 , respectively. What happens to the structure if supports A and B are located at one level? x2
x1 1⬘
2⬘
1
3
l 3l P1 − P2 2h 8h
Ans. H =
2
4 B h
A l Fig. P4.6
4.7. Construct influence line for internal force at members 2–3, 5–6, 5–7, and 9–11. Use kinematical method (Fig. P4.7). Ordinates of influence lines at specified points should be checked by static method. 0
1
2
4
6
8
10
12
h
a
13 3
5
9
7
11
d
Fig. P4.7
4.8.
Construct influence lines for internal force in the indicated members (Fig. P4.8). The loaded chords are shown by dotted lines. U2 V1
Fig. P4.8
V2
V3
V4
106
4
Plane Trusses
4.9. Perform analysis of the complex truss (Fig. P4.9) by Henneberg method. Hint: (1) Delete bar 4–3 (with unknown internal force X1) and substitute it by bar 4–2 (S4–2); bar 1–6 (X2) substitute by bar 2–6 (S2-6); (2) set of algebraic equations for determining X1 and X2:
S4 − 2 = 0 → S4P− 2 + X1S4X−12=1 + X 2 S4X−22=1 = 0 S2 − 6 = 0 → S2P− 6 + X1S2X−16=1 + X 2 S2X−26=1 = 0 P 2
1
A
4
3
B
6
5
Fig. P4.9
4.10. Verify the geometric unchangeability of the truss shown in Fig. P4.10. Apply Henneberg method. Hint: Eliminate the member Be and replace it by a rod eg. Discuss another version of the construction of a replacement system. (Delete the rod ed and introduce the support constraint in the joint g to substitute the deleted rod.) P e
d
c
f
B
A
g
Fig. P4.10
4.11. Discuss analysis of the complex truss (Fig. 4.11) by Henneberg method. Compare solution by the closed section method (Yasinsky approach, see Sect. 4.8.2). P1 2
3 h2
h1
4
1 A
B l
Fig. P4.11
P2
Chapter 5
Space Frameworks
This chapter is devoted to the analysis of statically determinate space trusses subjected to fixed loads. Methods for the generation of trusses and kinematic and static analysis are discussed. The different types of trusses are considered. Among them are meshwork trusses, compound and complex trusses, and Schwedler dome.
5.1
General Assumptions
The term “space framework” indicates three-dimensional through structures, which contain bar elements and loads located in different planes. As in the case of plane trusses, we introduce assumptions about loading, method of connecting elements, and types of supports. 1. All elements of the space framework are straight. 2. For connection of the rods the spherical hinges are used. Each hinge has three degrees of freedom; this means that the elements of a structure can rotate around three mutually perpendicular axes passing through the hinge center. 3. Load is applied only in joints and has an arbitrary direction. 4. The connection of spatial trusses with each other, as well as the attachment of trusses to the base, must ensure their geometrical unchangeability. The analysis of such structures consists of checking its geometrical unchangeability, instantaneous changeability, and determination of reactions and internal forces in all elements of the structure. We shall restrict our consideration to the statically determinate trusses.
5.2
Classification of the Space Frameworks
Three-dimensional bar systems can be classified by the way they are formed and according to their purpose.
5.2.1
Method of Formation
By the method of formation, the space trusses are divided into the following types: simple space trusses, meshwork space trusses, compound space trusses, and complex space trusses. Simple Space Frameworks: The formation of this type of trusses is based on a tetrahedronal nucleus 1–2–3–4 (Fig. 5.1a); the next hinged joint A is connected by three rods that are not in the same plane (they are shown by dashed lines). Meshwork space truss consists of rods that are located only in the planes of the outer faces of a closed polyhedron, forming in each face a triangular unchangeable lattice (Fig. 5.1b). Inside the closed polyhedron of the meshwork space truss the rods are absent. Compound space framework consists of two simple space frameworks which are connected by six bars. The axes of connecting bars do not intersect a single straight line. Complex space framework is a bar structure which cannot be classified as either simple, or meshwork, or compound space framework. © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_5
107
108
5 Space Frameworks
3
a
2
P
b
7 4
P 2
4
8
6 A 1
1
3
5 Fig. 5.1 Formation methods of space frameworks. (а) Simple space framework; (b) meshwork space truss
5.2.2
Three-Dimensional Engineering Rod Structures
Three-dimensional rod systems can be classified according to their purpose. Bridges: The simplest bridge truss is shown in Fig. 5.1b; it consists of two vertical trusses 1234 and 5678. Both trusses are connected along the upper and lower chords by horizontal trusses 2376 and 1485. In addition, there are two diagonal elements 2–5 and 3–8 in the end sections of the truss. There are no diagonal elements inside a closed polyhedron. According to the method of formation, this is a mesh-type truss. Towers are the through-going structures formed from plane trusses and connected with the base at several points. The triangular and tetrahedral towers with a diagonal element are shown in Fig. 5.2a, b.
a
8
7
Side view
5,5′
9
5
4
1
6
6 4
3
2,2′
1′
2′ 3′
4′
5′
6′ 7
8
5
5 2
4,4′
1,1′
9
7
6,6′
3,3′
3
2
1
Top view
7
b
3 1
6 4 2
Fig. 5.2 Types of towers with straight legs, horizontal rod diaphragms, and diagonal elements in side planes. (a) Triangular tower; (b) tetrahedral tower
5.2 Classification of the Space Frameworks
109
Domes: Some types of domes are shown in Fig. 5.3 (front and top views). Schwedler dome contains diagonal elements 2–3, 4–5, etc. in each facet of each tier (Fig. 5.3a). The joints 1, 3, and 5 do not necessary lie on a straight line. Föppl dome contains two elements in each facet (2–5 and 2–6 in facet 1–3–5–6) instead of one diagonal member (Fig. 5.3b).
a
5
3
1
1
6
b
5 6
4
2
3
4
2
3
5
1
1
6 6
5 2
4 2
4
3
Fig. 5.3 Different types of domes: (a) Schwedler dome; (b) Föppl dome
A two-layer geodesic dome represents a bar system whose joints are on the surface of two concentric spheres. The distance between the spheres determines the constructive thickness of the dome. The surface of the sphere is divided into regular spherical triangles. The vertices of a triangle are connected by the shortest distances—geodesic lines that are on arcs of a large circle. All the curvilinear sides of the triangles are replaced by straight lines. This leads to the fact that the dome consists of flat facets. Connection of rods is by a hinged way. There are structures which, despite their spatial bar character, are usually investigated by methods that are inherent to another class of deformable systems. This is possible in the case when the system in whole has incommensurable dimensions. Among them we note masts and rod slabs. Masts present the tall through spatial structures formed from plane trusses with transverse diaphragms. The mast is supported at one point and is equipped by a multitier (in general case) guy-rope system. The transverse dimensions of a mast are substantially less than the height. Therefore the spatial rod nature of the structure is weakened, while a significant linear dimension (mast height) strengthens the properties of one-dimensional beam systems. Generally, design diagram of the mast is a multi-span rod on the lower fixed support and at the points of attachment of the guy-ropes to the mast body on elastically compliant supports. Since the guy-ropes work only on extension on the windward side, these connections are one sided, which inevitably leads to a nonlinear analysis problem. Rod slabs: This term is used for a spatial system that consists of parallel rod grids. These grids are connected by lattice braces, which lie in planes orthogonal or inclined to the planes of grids. For such structures, the homogeneity of the joints and the presence of a repeating structural element, called a crystal or nucleus, are typical. The simplest rod slab with meshes from square cells and with a sparse lower mesh, as well as its crystals, is shown in Fig. 5.4a. Fragments of some structures are presented in Fig. 5.4b, c. In these drawings the thick black line indicates the elements of the upper mesh, the blue dashed lines indicate the elements of the bottom mesh, and the black dashed lines means connecting elements. Many modifications of these systems are possible (Umansky 1973). For example the system in Fig. 5.4b may be complemented by diagonal elements in the bottom mesh, or in the upper mesh, or both at the same time.
110
5 Space Frameworks
a Top view Side view 6
1
7 10 1
4
5
8
Diagonal
2 5
9
9
11
2
8 Crystals
3 1
10
2
11
6 4
3 1 5
Upper mesh
9
7 2 8
Bottom mesh
b
c
Fig. 5.4 Types of rod slabs: (a) system with meshes of square cells with a sparse lower mesh and its crystals. (b) Orthogonal system of square cells without diagonals in the upper and lower meshes. (c) Rod slab structure with equilateral triangular cells
Rod slabs are repeatedly statically indeterminate structures. Distance between grids, i.e., the thickness of a structure, is substantially less than its length and width. Therefore the spatial rod nature of the structure is weakened, while a relatively small thickness strengthens the properties of two-dimensional plate systems. An approximate analysis of such a system reduces to replacing the spatial system by an equivalent homogeneous plate. The parameters of the equivalent plate can be found in the reference literature (Umansky 1973). Below we will consider only statically determinate spatial rod systems.
5.3
Space Framework Supports
Space frameworks are connected to their foundation or any other unyielding structure using the following types of supports: (a) the spherical movable support; (b) the spherical support on cylindrical rollers; and (c) the spherical fixed support. Assume that each support is two sided; that is, if it prevents displacement of a body in some direction then it prevents displacement on the opposite direction as well. (a) The spherical movable support consists of two parallel slabs 1 and ball 2 between them (Fig. 5.5). This support allows body rotation about the x-, y-, and z-axes passing through the center of the ball and any displacement in the xy plane. Thus, this support imposes one constraint on the body displacement. The corresponding reaction Rz of constraint is directed
5.3 Space Framework Supports
111
perpendicular to the rolling plane of the ball. Schematically, such a support is presented in the form of a single rod with hinges at the ends.
B
z
B
1
x
2
1
y
Rz Fig. 5.5 The spherical movable support
(b) The spherical support on cylindrical rollers. This type of support consists of two balancers 1 and a ball 2 located between them in the special recesses (Fig. 5.6a). In the lower balancer 1 and in the plate 3 there are slots in which the cylindrical rollers 4 are inserted. The axes of the rollers 4 are parallel to the x-axis. These rollers allow the displacement in the direction perpendicular to the longitudinal axis of the rollers, while the flanges in the elements 1 and 3 prevent displacements of the rollers along their axes. The body B fixed by such support can rotate about the x-, y-, and z-axes passing through the center of the ball and move translationally parallel to the rolling plane of the rollers 4 along the normal to their axes. The support prevents two displacements: along the z-axis, in a direction perpendicular to the rolling plane and in the direction of the axes of the rollers. Case with rollers parallel to the y-axis is shown in Fig. 5.6b. The flanges in the elements 1 and 3 prevent the rollers 4 from displacement in the y-axis direction. Thus, this support imposes two constraints on the body. Corresponding reactions of constraints are Rx and Rz (Fig. 5.6a) or Ry and Rz (Fig. 5.6b), depending on the position of the rollers. Schematically this support is represented using two bars with hinges at the ends.
a
b
B
z
B
1
Rx
2
2
x
1
1
y
4
3
Rz
4
Ry
3
Rz
Fig. 5.6 The cylindrical movable support: (a) axes of rollers 4 are parallel to the x-axis. (b) Axes of rollers 4 are parallel to the y-axis
(c) The spherical fixed support (or simple spherical support) consists of a ball which is placed between two balancers in the special recess (Fig. 5.7). The lower balancer is fixed on the foundation. The body fixed by such a support can only rotate about the x-, y-, and z-axes passing through the center of the ball. This support imposes on the body three constraints. In such a support arises a reaction with the components Rx, Ry, and Rz which passes through the center of the ball. Schematically this support is represented using three bars with hinges at the ends.
z
B
x y
1 2
B
Rx
1
Ry Fig. 5.7 The spherical fixed support
Rz
112
5.4
5 Space Frameworks
Kinematical Analysis of Space Frameworks
The methods for the formation of statically determinate and geometrically unchangeable rod space structures are considered. Defective attachments of spatial structures to a rigid base as well as defective formation of composite trusses are discussed.
5.4.1
Attached and Released Frameworks
With the kinematic analysis of spatial trusses, it is necessary to distinguish two concepts: attached and released framework. A truss that becomes changeable after its separation from the supports is called attached. The truss which remains unchangeable after its separation from the supports is called released. Consider geometrically unchangeable structure attached to foundation in such a way so that the whole system also would be geometrically unchangeable. The smallest number of constraints necessary for such fastening is determined by the number of possible equilibrium equations for the case of arbitrarily located forces in space. It means that the minimum number of constraints is six. Simplest combination of support constraints, which ensures the invariability of the position of the system relative to the base, is shown in Fig. 5.8a. The following constraints are imposed on the structure: the spherical movable support A, the spherical support on cylindrical rollers B, and the spherical fixed support C.
b
a C B A
Fig. 5.8 Simplest geometrically unchangeable connection of rigid space framework with foundation
The presence of supports B and C only leads to the fact that the axis BC is stationary. However, the presence of only these supports will not prevent rotation of the body around axis BC. If support A will be installed in such a way that the reaction of this support does not intersect the axis BC, then the structure becomes immovable. It means that in the absence of support A, the space structure will have one degree of freedom (rotation around the BC axis). All reactions can be determined from three equations of the projections of all forces on the x-, y-, and z-axes and three equations of moments of all forces around these axes. Another example of the correct attachment of a geometrically unchangeable structure to the ground by means of six rods is given in Fig. 5.8b. It is possible to specify a simple arrangement of six support rods, which always guarantees a fixed attachment of the structure to the base. For this it is sufficient to arrange the support points at the vertices of an arbitrary triangle (Fig. 5.9). In this case, the support rods should be arranged as follows: (1) At point A arrange three rods not belonging to the same plane; (2) through point B show two rods so that their plane does not pass through vertex A; (3) through point C draw the sixth rod so that it does not intersect the straight line AB (Rabinovich 1960). Violation of any of these conditions (for example, at point A arrange three rods belonging to the same plane) does not ensure a fixed attachment of the structure to the foundation. B A C Fig. 5.9 Simple arrangement of six support rods for geometrically unchangeable connection of rigid space framework with foundation
5.4 Kinematical Analysis of Space Frameworks
113
If six support rods are set correctly, then the system is externally statically determinate. If the separated spatial structure is changeable, then in order to impart the entire system geometrically unchangeable, the number of support constraints should be increased. Suppose the hinged rectangle ABCD is attached to the base by means of three two-rod supports (the spherical support on cylindrical rollers) at joints A, B, and C (Fig. 5.10a). Such connection of the vertices of the rectangle and superimposed constraints allows the displacement of the point D in the vertical direction (DN); this leads to fold around of the diagonal AC. Besides this, change of the shape of rectangle in its plane is possible; this is due to the fact that point A has the ability to move in the direction perpendicular to the plane of the two rods that make up the support A. Thus, the straight line AD occupies the new position A0 D0 . Therefore, this system has two degrees of freedom. To eliminate two permissible degrees of freedom, we add two constraints at joint D as shown in Fig. 5.10b. Immovability of the joint A is provided by two rods of support A and a horizontal constraint of support D. Fastening of the joint B is provided by two constraints of the support B and a horizontal constraint of the support A. Immovability of joints C and D may be checked by a similar way.
a
N
c
b
D′ C
D
A
D
C
D
C
A′ B
A
B
A
B
Fig. 5.10 Hinged rectangle ABCD and its different connections with foundation. (a) Although connection contains six support constraints, hinged set ABCD has two degrees of freedom. (b) Geometrically unchangeable structure with two additional constraints at point D. (c) System with one degree of freedom contains one vertical constraint at point D and one horizontal constraint at point C—this connection allows the horizontal displacement along line AD
Let us return to Fig. 5.10a. Now two additional constraints will be introduced in another way: at joint D introduce one vertical constraint, and horizontal constraint at joint C along the line BC (Fig. 5.10c). In this case, one degree of freedom (rotation around the diagonal) is eliminated, but the second (the horizontal displacement of the joints A and D along the line AD) remains. On the other hand, since a horizontal constraint at the support B already exists, then a horizontal rod at the joint C is not necessary. Thus, special attention should be paid to the arrangement of additional constraints. Let us compose the analytical relations that determine the most important characteristics of the system, such as its static determinacy (indeterminacy) and geometrical unchangeability (changeability). First consider the case of trusses connected with supports and trusses separated from supports.
5.4.1.1
Spatial Trusses Attached to Supports
Suppose that the truss contains J joints, S0 supporting constraints, and S elements of truss itself. Each joint is considered as a point which, in the absence of any connections, has three degrees of freedom. If two points are connected by a rod of length d, then the total number of degrees of freedom of two points is decreased by 1, since the six coordinates of two points are related by (x1 x2)2 + (y1 y2)2 + (z1 z2)2 ¼ d2. This means that each rod is equivalent to one constraint. A total number of constraints is equal to S + S0. The number of degrees of freedom of a system is the difference between the number of degrees of freedom of joints and the total number of constraints: W ¼ 3J ðS þ S0 Þ
ð5:1Þ
The difference between the total number of constraints and the number of degrees of freedom of joints is a number of redundant constraints n, or degree of redundancy: n ¼ W ¼ ðS þ S0 Þ 3J If W > 0 (n < 0) then the system is changeable, since it does not have the necessary number of rods.
ð5:2Þ
114
5 Space Frameworks
If W ¼ 0 (n ¼ 0) then the system does not have redundant rods. The system is statically determinate, but it can be geometrically unchangeable or changeable. If W < 0 (n > 0) then the system has redundant rods. The system is statically indeterminate, but it can be geometrically unchangeable or changeable. In the last two cases it is necessary not only to count the number of rods but also to analyze their arrangement. If n ¼ 0 then a necessary but not sufficient indication of statical determinacy is the following condition: 3J ¼ S þ S0
ð5:3Þ
Spatial truss attached to the ground by means of 7 support constraints is shown in Fig. 5.11a.
a
b C
D
A
B
Fig. 5.11 Design diagram of a spatial structure. (a) Truss connected with foundation; degrees of freedom of this structure W ¼ 0. (b) Structure separated from support; degree of freedom of this structure is W ¼ 1 (n ¼ 1)
This structure contains 8 joints, 17 bars, and 7 support constraints, so the number of degrees of freedom is W ¼ 3J ðS þ S0 Þ ¼ 3 8 ð17 þ 7Þ ¼ 0 Thus, this structure contains the number of rods and support constraints necessary to ensure static determinacy and geometrical unchangeability. The presence of two supports A and B and only one vertical constraint at support C allows horizontal displacement of element CD. However, the introducing of the horizontal constraint at the support C directed along the element CD excludes such displacement. Thus, this spatial framework is statically determinate and geometrically unchangeable. Below we will consider only statically determinate spatial rod structures.
5.4.1.2
Spatial Trusses Separated from Supports
Suppose that the rod system is separated from the supports and at the same time it remains geometrically unchangeable; it means that we are dealing with released structure. To ensure the geometric unchangeability of the structure in whole (i.e., for the structure itself + supports), six support constraints are required. Therefore, a number of redundant bars of a released truss according to (5.2) is expressed by the formula n ¼ S þ 6 3J
ð5:4Þ
Since for statically determinate truss n ¼ 0, the necessary indication of geometrically unchangeable and statically determinate released truss becomes 3J ¼ S þ 6
ð5:5Þ
Figure 5.11b presents a space framework separated from its supports. This structure contains 8 joints and 17 elements, so the number of redundant bars is n ¼ S + 6 3J ¼ 17 + 6 3 8 ¼ 1. In other words, this structure has one degree of freedom. Therefore, the structure in Fig. 5.11b is attached, but cannot be released. It is for this reason we compensated one missing bar of the truss itself by setting the horizontal rod at the support C as shown in Fig. 5.11a.
5.4 Kinematical Analysis of Space Frameworks
115
If it is required to ensure the geometrical unchangeability of the structure separated from the supports, then one additional rod, for example, as the diagonal of the lower base, should be inserted into the system in Fig. 5.11b; just in this case the truss becomes a released one.
5.4.2
Improper Connections of 3-D Structures
Let us consider some cases of improper arrangement of support constraints and improper formation of compound space framework (connection of two trusses) which leads to structural changeability. The unconstrained rigid body in space has six degrees of freedom. Each superimposed constraint eliminates one degree of freedom. Therefore, the minimum number of constraints that is required to ensure the immobility of the body is six. Connection of two bodies with the help of six rods has to ensure the geometrical and instantaneous unchangeability of the compound structure as a whole. It is assumed that one of the bodies is immovably attached to the foundation (to the ground). Therefore, the problem of formation of geometrically unchangeable connection of two bodies and the problem of immobile attachment of the body to the earth are equivalent. Some examples of the arrangement of six support constraints that lead to a changeable or instantaneously changeable system are shown in Fig. 5.12. It is assumed that the separated structure is geometrically unchangeable (released), and we are talking only about its connection to the base with the help of six constraints (Kiselev 1960). Six rods cross one straight line. This straight line will be the axis of an infinitesimal or finite rotation of one body relative to the other (earth) (Fig. 5.12a). Here four cases are possible. 1. If there are two points A and B, in each of which three support rods intersect, then the line AB is the axis of rotation of the body, which is intersected by all the rods (Fig. 5.12b). 2. The rods are located in two planes. The intersection line n–n represents the axis of rotation; this axis is crossed by all the rods (Fig. 5.12c). 3. Constraints are located in parallel planes; constraints 1–2, 3–4, and 5–6 satisfy this condition (Fig. 5.12d). Parallel planes intersect at infinity along a straight line which is intersected by all the constraints. In other words, the parallel planes allow displacements of all points which belong to these planes. 4. In one plane there are four constraints (Fig. 5.12e). Constraints 1–4 lie in the plane of П. The remaining two constraints 5 and 6 intersect the plane П at points A and B. Constraints 1–4 cross the line AB at points 10 –40 . Thus, line AB is intersected by all the constraints. This line represents the axis of rotation. Three support rods intersect at one point and lie in one arbitrarily oriented plane Π. Three rods of support A (Fig. 5.12f) satisfy these two conditions. These rods exclude two translational movements in the plane of the rods П, i.e., three rods of support A eliminate two degrees of freedom. To eliminate the four degrees of freedom, the remaining three rods of supports B and C are insufficient. More than three rods intersect at one point. Here two cases are possible: 1. The rods are not parallel (Fig. 5.12g). Such rods eliminate three degrees of freedom. The remaining rods (less than three) are not enough to eliminate three degrees of freedom. 2. The rods are parallel. Four or more parallel rods eliminate three degrees of freedom: translational movement in their direction and two rotations around axes which belong to the plane perpendicular to constraints (Fig. 5.12h, dotted arrows). The other three degrees of freedom (translational movement in any direction in a plane perpendicular to the rods and rotation around an axis parallel to it) cannot be eliminated by two or less rods (Fig. 5.12h, solid arrows).
116
5 Space Frameworks
Axis of rotation
•• • •
a
• •
c
b
n B
••
n A
••
••
Axis of rotation
6
Rigid foundation
e Displacement line (DL)
d
4
5
3
B
DL
DL
2 1
6 4
2 1
•
5
•
4′
П
A
3 Axis of rotation
f
• 2′
• 1′
• 3′
•
g A
C
B
П
h Dotted arrows – movement is not possible Solid arrows - movement is possible
Fig. 5.12 Some cases of geometrically changeable connection of unyielding space framework with foundation. (a–e) Six rods cross one straight line. (f) Three support rods intersect at one point and lie in one arbitrarily oriented plane. (g, h) More than three rods intersect at one point
5.4.3
Meshwork Structures
The simplest example of mesh structure is shown in Fig. 5.13. The bridge truss of the mesh type is shown in Fig. 5.1b. In these trusses, all the rods are located only in the planes of the outer faces of a closed polyhedron, forming in each face a triangular unchangeable lattice.
5.4 Kinematical Analysis of Space Frameworks
117
Fig. 5.13 Meshwork released structure
If into the structure shown in Fig. 5.11b introduce diagonal element connecting the joints of the upper and lower faces, then the structure turns out to be geometrically unchangeable, but will not satisfy the formation condition of meshwork. If the additional element is positioned as a diagonal of the lower face, then the structure becomes a meshwork type. For meshwork it is easy to check whether the system is statically determinate and geometrically unchangeable. Since each rod is the edge of two adjacent faces, the relation between the number of faces F of a system and the number of rods S is S ¼ 3F/2. On the other hand, according to Euler theorem, the relationship between the number of rods S, joints J, and faces F is S ¼ J + F 2. Eliminating the number of faces F from these relations, we obtain S ¼ 3J 6
ð5:6Þ
This formula establishes a relationship between the number of rods and the number of joints for statically determinate and geometrically unchangeable released spatial truss of the meshwork type. The above relations are easy to verify; for the structure shown in Fig. 5.13 we have S ¼ 12, J ¼ 6, F ¼ 8. It should be noted that the bottom base is divided by two bars and therefore consists of three faces.
5.4.3.1
Cauchy Theorem
Now we formulate the Cauchy theorem (1813) useful for structural analysis of space frameworks of a special type. The bar system in the form of a polyhedron with triangular faces without internal rods (meshwork) satisfies the conditions of static determinacy and geometrically unchangeability (Rabinovich 1960)
Let us analyze the space framework shown in Fig. 5.14a. This structure, although it does not have internal elements, nevertheless does not belong to a meshwork-type structure. Indeed, the structure does not represent a closed polyhedron with triangular faces, the structure does not have diagonal elements in the low 1–2–3–4 and upper 5–6–7–8 bases, and four rods between spherical fixed supports A, B, C, and D are absent. Despite this, the original structure has the necessary number of constraints to ensure its geometric unchangeability. Indeed, the total number of elements is S ¼ 2 4 + 8 + 8 ¼ 24, total number of joints (including A and C) and support constraints are J ¼ 12, and S0 ¼ 4 3 ¼ 12. For the original structure in Fig. 5.14a, which is not separated from support constraints, in accordance with the formula (5.2), the number of degrees of freedom is W ¼ 3J (S + S0) ¼ 3 12 (24 + 12) ¼ 0. This means that the structure has the required number of rods in order to be geometrically unchangeable; we note, however, that the question of the arrangement of rods is not considered. What would happen if this structure is required to be transported from manufacturer to construction site? How to ensure geometrical unchangeability of the structure during transportation? In this case structure without supports (or separated structure) should be considered. The number of degrees of freedom of the structure separated from supports equals W ¼ 3J S ¼ 3 10 24 ¼ 6; that is, this structure is changeable (W ¼ 6). To ensure its unchangeability, six temporary rods must be added (shown by bold dotted line in Fig. 5.14b). After attachment of the structure to supports A, B, C, and D, all these temporary rods may be removed. As a result, the attached structure is unchangeable. This unchangeability is achieved by using six necessary constraints for rigid body (as shown in Fig. 5.9) and six additional supports (which replaced temporary rods from separated structure).
118
5 Space Frameworks
8
a
7
4 5
4
3
5
6 C
D 1
A
2
B
7
8
b
3 6
D 1
A
C
2
B
Fig. 5.14 (a) Attached 3-D structure. (b) Introduced elements (they are shown by thick dotted lines) transform the set of members separated from supports into released meshwork structure
It is easy to notice that separated structure with six introduced rods represents the released meshwork structure. Indeed, we obtained a closed polyhedron with triangular faces and without internal rods. For this structure we have 12 joints and total number of elements is S ¼ 5 4 + 8 + 2 ¼ 30. According to Cauchy formula S ¼ 3J 6 ¼ 3 12 6 ¼ 30. Therefore this structure is statically determinate and geometrically unchangeable. Same can be confirmed by condition S ¼ 3F/2 (for this mesh structure S ¼ 30, F ¼ 20). In some cases the Cauchy theorem allows to perform a quick kinematical analysis of the spatial trusses. For example, the released rod structure in Fig. 5.15 is a meshwork structure and therefore is geometrically unchanged. For this structure, the number of faces is F ¼ 10, the number of joints is J ¼ 7, and the number of rods is S ¼ 15. It is easy to check that the structure is geometrically unchangeable and does not have redundant rods. Indeed, for a given structure the relations S ¼ 3J 6 and 2S ¼ 3F are satisfied.
Fig. 5.15 Released meshwork structure
5.5
Static Analysis of 3-D Structures
There are various ways to determine the internal forces in elements of space frameworks. We consider the method of sections (the method of joints, the method of through section), the method of decomposing of a space truss into the flat systems, and the method of rod replacement.
5.5.1
General
Important for the theory of the space trusses, and in particular for determining the forces in the elements of the truss, is the problem of the resolving force into three directions, if these directions and given force intersect at one point. Concurrent space forces satisfy the equilibrium equations ∑X ¼ 0, ∑ Y ¼ 0, ∑ Z ¼ 0.
5.5 Static Analysis of 3-D Structures
119
If the three concurrent directions lie in one plane, and the force P does not belong to this plane, the force resolving problem is insoluble. This case has a clear physical interpretation: if the three concurrent rods lie in the same plane, then the joint is unable to perceive a force that is outside this plane. If three directions lie in one plane and the force P belongs to this plane, then the resolving problem of the force is uncertain. The problem turns out to be uncertain in cases when the force is resolved into four or more directions. The problem becomes certain in the sole case, when the three directions do not lie in the same plane. Solvability of the problem follows from the condition of geometrical unchangeability and immobility of the system consisting of three rods. The terms “certain–uncertain” are related only to the concept of the possibility of obtaining a unique solution mathematically. Without stopping on the procedure for the force resolving, we note important special cases of a balanced system of forces that do not lie in one plane and intersect at one point with a given force. Definitions A zero-force rod is the element with internal force that equals to zero. A single rod of a joint is a rod that does not belong to the plane in which all the other rods of this joint are located. 5.5.1.1
Zero Rod Cases
1. If a joint with no load contains a single rod then it is a zero rod. 2. If a joint contains a number of rods lying in one plane, a load in the same plane (or is absent), and a single rod, then it is a zero rod. 3. If a joint with no load contains three rods that do not lie in one plane, then all these rods are zero rods. 4. If the joint is in equilibrium and all concurrent forces which act on the joint are located in two planes, then the resulting forces of each of the planes lie on the line of intersection of the planes, equal and opposite in direction. 5.5.1.2
Method of Sections
The essence of this method is as follows: carry out a section through a certain number of rods in which the internal forces should be determined. By this, internal forces are transferred into the class of external forces. Then one of the parts of the truss is chosen, the acting longitudinal forces are applied to the intersected bars, and after that the equilibrium equations are formed for the forces applied to the cutoff part of the truss. The following versions of the method of sections are considered: joint isolation method, the moments method, and the rod replacement method. 1. Joint isolation method. This method means that the section separates a joint, for which the forces in the rods form a concurrent spatial force system. For such system of forces, it is possible to write three equilibrium equations for projections of forces on coordinate axes. Therefore, the determination of forces in the elements of the truss begins with a three-rod joint, whose rods do not lie in the same plane. If there is no external load in this joint, then the forces in all the rods are zero. Then we need to look for a new joint containing no more than three rods with unknown forces, which do not lie in the same plane: X X X F x ¼ 0, F y ¼ 0, Fz ¼ 0 ð5:7Þ 2. Through section method. This method means that the section separates one part of the truss from the other in such a way that the unknown forces in the rods form an arbitrary spatial force system. For such a system of forces, it is possible to make up only six equations of equilibrium (three equations of the projections of forces on the coordinate axes and three equations in the form of moments of forces with respect to the coordinate axes): X X X F y ¼ 0, Fz ¼ 0 F x ¼ 0, X X X ð5:8Þ M x ¼ 0, M y ¼ 0, Mz ¼ 0 3. The rod replacement method (Henneberg method). This method is applicable to the analysis of complex trusses for which it is impossible to isolate a joint with three unknown forces or pass a section cutting only six rods with unknown forces. The essence of Henneberg method is that a complex truss is transformed into a simple truss. For this purpose, one rod is removed, replacing it with an unknown force, and an additional rod is inserted in such a way that determining the forces in
120
5 Space Frameworks
it due to a given load and unknown force would not cause difficulties. An unknown force in the eliminated rod is determined from the following condition: the force in the inserted rod caused by a given load and unknown force is zero.
5.5.2
Meshwork Structures
Let us consider simplest space truss of the pyramidal form. The base of the truss is a square with side a, the height of the truss is h, and the side faces are the identical triangles. The bottom face contains the diagonal element BD. The supports of the truss are spherical fixed support A, spherical support on cylindrical rollers B, and spherical movable support C. Constraint 1 is directed along the side AB, constraints 2 and 4 are directed along the sides AD and BC, respectively, and constraints 3, 5, and 6 are vertical. The truss is subjected to a force P, which is parallel to x-axis, as shown in Fig. 5.16. E
P
h
C
D
6 z R1
1
2
a
A
δ
3
a
4
B 5
x
R4
y R2
R6
φ
R3
R5
Fig. 5.16 Design diagram of space meshwork framework
The conditions for the formation of this truss (internal rods are absent, all the joints are located on the outer surfaces of the polyhedron, the faces are rigid triangles) meet the requirements of the meshwork trusses. The attached truss contains 5 joints, 9 rods, and 6 support rods. Number of degrees of freedom is W ¼ 3J (S + S0) ¼ 3 5 (9 + 6) ¼ 0. Thus, the system has sufficient number of rods to ensure its geometric unchangeability. For a meshwork truss separated from the foundation J ¼ 5, S ¼ 9, W ¼ 3J S 6 ¼ 3 5 9 6 ¼ 0. This means that the structure freed from the foundation is statically determinate and geometrically unchangeable, i.e., released truss. The truss contains two three-rod joints, A and C, the rods of which do not lie in the same plane. However, these joints are loaded with unknown reactions R1, R2, R3, and R6. Therefore, the problem of determining the unknown internal forces is preceded by the problem of determining the reactions; note that this is not a general rule. We assume that the reactions are positive if they stretch the support rod. The unknown reactions of supports are determined from the following equilibrium equations: X R1 ! F x ¼ 0 : R1 þ P ¼ 0 ! R1 ¼ P; X R6 ! M x ¼ 0 : R6 a ¼ 0 ! R6 ¼ 0; X R5 ! M y ¼ 0 : R5 a þ Ph ¼ 0 ! R5 ¼ Ph=a; X R3 ! F z ¼ 0 : R3 R5 ¼ 0 ! R3 ¼ R5 ¼ Ph=a; X R4 ! M z ¼ 0 : R4 a þ Pa=2 ¼ 0 ! R4 ¼ P=2; X R2 ! F y ¼ 0 : R2 þ R4 ¼ 0 ! R2 ¼ R4 ¼ P=2: Determination of internal forces in the elements of the truss starts from joint C. Since R6¼0, stresses SEC ¼ SBC ¼ SDC ¼ 0. Analysis of the three-rod non-loaded joint D allows us to conclude that SAD ¼ SBD ¼ SDE ¼ 0. Then we consider the equilibrium of the joints E, B, and A.
5.5 Static Analysis of 3-D Structures
121
X
Joint E :
X
F z ¼ SAE sin φ SBE sin φ ¼ 0 ! SAE ¼ SBE ; Fsx ¼ SAE cos φ cos δ þ SBE cos φ cos δ þ P ¼ 0
Take into account δ ¼ 45 . Solution of these equations leads to the following results: pffiffiffi pffiffiffi P 2 P 2 ðcompressed Þ, SAE ¼ ðtensileÞ: SBE ¼ 2 cos φ 2 cos φ
Joint B: Equation ∑Fx ¼ SBE cos φ cos 45 SAB ¼ 0 leads to SAB ¼ P/2. Equation ∑Fz ¼ SBE sin φ R5 ¼ 0 serves to verify the forceSBE. Joint A: Equations X F z ¼ R3 SAE sin φ ¼ 0 X F y ¼ R2 SAE cos φ sin 45 ¼ 0 X F x ¼ R1 SAE cos φ cos 45 þ SAB ¼ 0 serve to verify previously determined internal forces and reactions. There are possible situations when the joint isolation method should be combined with the method of through cuts. As an example, consider the meshwork truss shown in Fig. 5.17. The numbers of joints and rods of a released truss are J ¼ 8 and S ¼ 18 , respectively. The number of degrees of freedom of separated truss is W ¼ 3J S 6 ¼ 3 8 18 6 ¼ 0. Since released truss is connected by foundation by six support rods, the structure in Fig. 5.17 is a statically determinate and geometrically unchangeable one.
3
4
P
z h 2
1
D
C
П SA1
δ
b
SAD
φ
A
Ax
SAB
Ay y
Az
a By
B
Cz
x
Bz
Fig. 5.17 Design diagram of space framework. The equilibrium objects are the joints 2 and A, and the upper cutoff part of the truss
The analysis of the truss begins with the definition of support reactions. Positive reactions Ax, Ay, Az, By, Bz, and Cz are considered to be directed from the support joint. To determine them, we formulate the following equilibrium equations: X X M x ¼ C z b ¼ 0 ! Cz ¼ 0; Ax ! F x ¼ Ax þ P ¼ 0 ! Ax ¼ P; Cz ! X X Bz ! M y ¼ Bz a C z a Ph ¼ 0 ! Bz ¼ Ph=a; Ay ! F y ¼ Ay þ By ¼ 0 ! Ay ¼ Pb=a; X X By ! M z ¼ By a Pb ¼ 0 ! By ¼ Pb=a; Az ! F z ¼ Az Bz C z ¼ 0 ! Az ¼ Bz ¼ Ph=a:
122
5 Space Frameworks
In the truss there is a joint 2 in which three rods converge. This joint is not loaded; therefore the internal forces S21 ¼ S2B ¼ S23 ¼ 0. Equilibrium equation for the next joint A allows finding the forces SAB, SAD, and SA1. As before, the positive internal forces in the elements of the truss are directed from the joints: X F x ¼ Ax þ SAB ¼ 0 ! SAB ¼ P; X F y ¼ Ay SAD ¼ 0 ! SAD ¼ Pb=a; X F z ¼ Az þ SA1 ¼ 0 ! SA1 ¼ Ph=a: It is easy to see that each next cut-out joint contains more than three unknown forces, and as a result, the joints method cannot be applied. Therefore we will apply the through section method. The secant plane П passes across all the vertical rods A1, B2, C3, D4 and the diagonals of the side faces 1B, B3, C4, D1. Solid and dotted arrows denote forces in visible and invisible crossed rods, respectively. There are eight rods in the through section, but in two of them, A1 and 2B, the internal forces have already been found. Then, we discard one part of the truss, for example, the lower one, and compose six equilibrium equations for the upper part of the truss with respect to the forces in the dissected rods of the truss. These equations are presented in Table 5.1. The column for S2B ¼ 0 is omitted. Table 5.1 Equilibrium equations for space truss in Fig. 5.17 ∑Fx¼ ∑Fy¼ ∑Fz¼ ∑Mx¼ ∑My¼ ∑Mz¼
SA1 0 0 SA1 0 0 0
SC3 0 0 SC3 SC3 b SC3 a 0
SD4 0 0 SD4 SD4 b 0 0
S1B S1B cosφ 0 S1B sinφ 0 S1B cosφ h 0
SB3 0 SB3 cosδ SB3 sinδ SB3 cosδ h SB3 sinδ a SB3 cosδ a
SC4 SC4 cosφ 0 SCE sinφ SC4 sinφ b SC4 cosφ h SC4 cosφ b
SD1 0 SD1cosδ SD1 sinδ SD1cosδ h 0 0
P P 0 0 0 Ph 0
¼0 ¼0 ¼0 ¼0 ¼0 ¼0
The force SA1 (in table it is shown by bold letter) is already known. A system of six linear algebraic equations with respect to six unknown forces in the dissected rods is obtained. Its solution presents no difficulty using computer. A sequential examination of the joints, B, D, 1, etc., makes it possible to determine the forces in the elements of the structure belonging in the lower and upper planes. Considering joint C, we have one unknown force in the element DC. This internal force can be found from the condition ∑Fx ¼ 0. To verify the calculations, it is also possible to determine force SDC by considering the equilibrium of the joint D.
5.5.3
Compound Space Frameworks
Let us briefly consider the formation and procedure of analysis of compound structure consisting of two meshwork space trusses A and B, which are connected by means of six rods (Fig. 5.18). The freed structure B presents the released structure (geometrically unchangeable itself), and has six support constraints necessary for geometric unchangeability of the structure B in whole. In means that truss B presents the rigid basis for other structure A that should be attached to B. Both simple 3-D frameworks are connected by using six bars 1–6, the axes of which do not intersect a single straight line. Compound framework in whole contains 14 joints, 12 + 18 + 6 elements, and 6 support constraints, so the degree of freedom of the compound truss is W ¼ 3J (S + S0) ¼ 3 14 (36 + 6) ¼ 0; we use this formula instead of W ¼ 3J S 6 because compound structure in whole does not belong to the meshwork truss. Thus, the structure under consideration is statically determinate and geometrically unchangeable. Note that if structure A would not be a meshwork truss, then the number of connecting rods must be increased.
5.5 Static Analysis of 3-D Structures
123
f P
A
e
a
Truss A. J = 6, S = 12 W = 3J – S – 6 = 3×6 – 12 – 6 = 0
c d
3 1
6
b 4
5
Truss B. J = 8, S = 18 W = 3J – S – 6= 3×8 – 18 – 6 = 0
2
B z x y
Fig. 5.18 Design diagram of the space compound framework
For the analysis of compound spatial trusses, the combination of the method of through sections with the joint isolation method is effective. It is obvious that we can easily determine all reactions of support. To determine the forces in elements 1–6, the internal forces S1–S6 should be transferred in the class of external forces. For this purpose we draw the section through the elements 1–6; assume that the unknown forces are positive, i.e., they are directed from the joint; and then consider the equilibrium of one of the parts of the truss. Since all the reactions of supports are applied to the truss B, it is more convenient to consider the upper part of the structure. In this case, the reactions of supports do not enter into the equilibrium equations. Equations of equilibrium have the form (5.8); it is recommended to represent the equilibrium equations in tabular form. After determining the internal forces S1–S6, it is possible to provide the analysis of the trusses A and B separately. For the truss A, the sequence of cutting out the joints is as follows: a, c, b, etc.
5.5.4
Complex Space Frameworks: The Rod Replacement Method
Methods for the formation of complex and simple spatial trusses are different. Therefore, the methods of analysis of simple trusses in application to complex trusses are difficult or cannot be applied at all. The rod replacing method allows replacing the original complex truss by a simple truss and from the condition of their equivalence determines the force in one or several rods of a complex truss. Further analysis of the truss is carried out by known methods that are used for the analysis of simple truss. Let us consider the spatial complex truss shown in Fig. 5.19a. The rods of this truss form an octahedron; that is, the truss contains eight faces. In the visibility zone are the faces ACB, AC3, A31, and A1B. In the zone of invisibility there are 2CB, 2C3, 231, and 2B1. Assume that each of the face presents the isosceles triangle with bases CB, B1, 13, and 3C. The structure has geometric symmetry with respect to vertical plane which passes through the BC and the horizontal rod 1–3. Therefore, the joints 2 and A are located on the horizontal line. The support constraints are imposed at the vertices of the triangle ABC, the plane of which takes an oblique position. Thus, the support A and the supports C and B are located at a different level. The truss does not contain internal elements, and each face represents a triangle; that is, the truss is a mesh type.
124
5 Space Frameworks
1
3
a
2
A
1 (3)
P
b 2
A
X
3
A
1
P
2
C
C B
B
B (C)
Fig. 5.19 Complex space meshwork truss. (a) Design diagram and side view. (b) Substituted structure: bars 1–3 and A–2 are replaced and substituted bars, respectively
Kinematical analysis. For the structure as a whole, the number of joints is J ¼ 6, the number of rods is S ¼ 12, and the number of support constraints is S0 ¼ 6. Number of degrees of freedom is W ¼ 3J (S + S0) ¼ 3 6 12 6 ¼ 0. If the concept of released mesh truss is used, then a number of redundant bars of a released truss according to (5.4) is n ¼ S + 6 3J ¼ 12 + 6 3 6 ¼ 0. So this structure is statically determinate and geometrically unchangeable. Static features of a structure. This truss has no joint in which the three elements meet. Therefore, it is impossible to apply the method of consecutive isolation of the joints. Formally, it is possible to apply the isolation joint method, and compose three equilibrium equations for each joint. The set of equations will include 12 unknown forces in the elements of the truss and 6 unknown support reactions, totaling 18 unknowns. The total number of equations is 3J ¼ 3 6 ¼ 18. Thus, the problem is fundamentally solvable. Also, it is possible to apply the method of through sections. However, for only one such section we obtain a set of six connected algebraic equations. To determine the remaining unknown forces we need to apply the method of joint isolation. Both methods of analysis are possible in principle, but are extremely cumbersome. The rod replacement method. This method allows to avoid the difficulty of analyzing the complex structure caused by the special way of the rod connection. This method is based on the following principle: Any complex statically determinate structure can be converted into a simple one by replacing one or more elements (Henneberg method, 1886) (Darkov 1989) According to this principle, the following analysis procedure can be recommended: 1. The original structure is replaced by the substituted structure (Fig. 5.19b). For this, it is necessary to choose one of the rods, for example 1–3: eliminate it, and replace it with a replacement rod, for example A–2; these rods are called replaced and substituted rod, respectively. Internal force in the replaced rod is denoted as X. As a result of this procedure, the original system is replaced by the truss in which there are three-rod joints (3 and 1); analysis of such a truss does not cause difficulties. In this simplest example, it is sufficient to remove only one rod and, accordingly, to enter one substituted rod. 2. Form the expression for the force N in the substituted rod. According to the superposition principle N ¼ N P þ N X ¼ N P þ XN X ,
ð5:9Þ
where NP and NX are the forces in the substituted rod A-2 caused by the external load P and the unknown force X; N is the force in the substituted rod caused by the force X ¼ 1. Calculation NP (Fig. 5.19b): Isolate joint 1 and determine the forces in elements 1–A, 1–2, and 1–B. Then we cut out joint 3. Since there is no external load in joint 3, the internal forces at elements 3–2, 3–A, and 3–C are zero. Finally, we isolate joint 2. There are five rods in this joint, but in two rods, 3–2 and 1–2, the forces are already defined. Equations of equilibrium for concurrent forces in space allow to determine NP. Calculation N X : This force can be represented in the form ð 1Þ
ð3Þ
NX ¼ Nx þ Nx , ð1Þ
ð3Þ
ð5:10Þ
where N x and N x are the forces in the rod A-2 caused by force X ¼ 1 applied at the joint 1 and at the joint ð1Þ ð3Þ ð1Þ 3, respectively. Since plane of symmetry coincides with vertical contour 1BC3, N x ¼ N x ; therefore N X ¼ 2N x . In ð1Þ ð3Þ general, the definition of the forces N x and N x does not cause difficulties.
5.5 Static Analysis of 3-D Structures
125
3. Form the equivalence conditions for original and substituted structures. Since in the initial structure the rod A-2 is absent, the force in the substituted element should be equal to zero. Thus, the equivalence condition takes the form N ¼ N P þ XN X ¼ 0
ð5:11Þ
4. Unknown force in the substitute element A-2: X¼
NP NX
ð5:12Þ
5. The force in an arbitrary k-th rod is determined by the formula N k ¼ N kP þ XN kX ,
ð5:13Þ
where NkP and N kX are forces in rod k caused by given load P and force X ¼ 1, respectively.
Discussion Henneberg method makes it easy to conduct a kinematic analysis of the system. Since while performing kinematic analysis we do not take into account the external loads, NP ¼ 0. Now the formula (5.12) becomes X ¼ 0=N X; this expression allows us to draw two important conclusions: 1. If N X ¼ 0, then the force in the substituted rod takes an undefined value X ¼ 0/0. This means that the system is instantly changeable. 2. If N X 6¼ 0, then the force in the substituted rod takes the value X ¼ 0. Therefore, expression (5.13) becomes zero, since both terms are equal to zero. This means that the system is geometrically unchangeable. These arguments can be considered as a proof of the null-load method (see Chap. 1).
5.5.5
Schwedler Dome
Schwedler cupola presents a spatial rod structure in the form of a regular truncated pyramid. Below are considered a way for the formation of a structure, the conditions for its supporting, and methods for its analysis.
5.5.5.1
Design Diagram of Dome
Spatial structure in Fig. 5.20a presents a truncated two-tier pyramid. In the base this structure has a regular hexagon. The side faces of pyramid consist of flat trapezoids with one diagonal. The visible side faces of lower tier are A–4–3–B, B–3–e–C, . . ., etc. The invisible side faces of lower tier are A–4–5–F, F–5–6–E, . . ., etc. The side faces of the upper tier are 4–1–d–3, 3–d–c– e, . . . etc. The edges A–4 and 4–1, forming a vertical plane, have a kink in the joint 4. The visible elements are shown by solid lines, and visible joints are denoted by bold letters. The structure is subjected to load P at joint 1. This force has arbitrary orientation.
126
5 Space Frameworks
P
a
b
a,c
2,d
1
5-a
6,e
3,5
4
f E-f
A-5 F-6
A
B,F
b
C,E
F
E 6
5
Supporting contour
a
2 P A
d
3 B
f
b
1
4
D
D
c
e C
Fig. 5.20 Design diagram of Schwedler dome. (a) Front view, visible joints are denoted by bold letters, invisible members are shown by dashed line; (b) top view
If the entire set of rods are separated from the supports, we get a geometrically changeable system, i.e., the set of the rods presents the attached structure. If this structure would have six elements of connected points A, B, . . ., F of the base contour and three diagonal elements in each top and bottom base, then the system would be bounded by triangular faces. In other words, we would have a released mesh structure. The absence of 6 elements on the contour level and 6 diagonal elements on the top and bottom bases is compensated by the 12 support constraints. Therefore six inclined legs at points A, B, . . ., F are fixed using spherical fixed supports; each support in Fig. 5.20a has three constraints. So total support constraints is 18, where 6 of them are required minimum for geometrically unchangeable connection of rigid space framework with foundation (Fig. 5.9).
5.5.5.2
Kinematical Analysis
The entire structure contains J ¼ 18 joints, S ¼ 36 rods, and S0 ¼ 6 3 ¼ 18 support constraints. Number of degrees of freedom is W ¼ 3J (S + S0) ¼ 3 18 36 18 ¼ 0. It means that the structure has the required number of constraints to ensure geometrical unchangeability. Computation of Internal Forces Two different methods are considered: joints method and decomposing of space framework to plane trusses. Joints Method: Let us consider joint 2 which contains four rods; the member 2-a is a single rod (see definition in Sect. 5.5.1), since it lies outside the plane of the trapezium 1–2–5–4. Since joint 2 has no external force rod 2-a is a zero-force element; such elements are denoted by thin solid line. Next consider joint a, which is not loaded. In it rod 2-a is a zero element so it can be virtually removed; also three bars converge in this joint, a-5, a-6, and a-b; so each of them is a zero-force element.
5.5 Static Analysis of 3-D Structures
127
Now consider no-loaded joint b. If we virtually remove rod a-b (as zero-force element), then three rods, b–6, b–f, and b–c, which do not lie in one plane, are zero-force elements. Then the joints c and d are successively considered similarly. After virtually removing zero elements a–6 and b–6, in joint 6 we have four converge bars, 6–5, 6–F, 6–E, and 6–f. The last rod is single; therefore rod 6–f is a zero-force one. After discarding it, we get in the no-loaded joint 6 three rods, which lie in the same plane. Therefore, joint 6 should be considered after determining the force in one of the rods 6–5, 6–F, or 6–E. Let us move to the no-loaded joint f. Discarding the rod 6–f, we obtain three rods, f–E, f–D, and f–e, not lying in the same plane; therefore they are zero-force members. Next, three rods e–D, e–C, and e–3 converge at joint e. Since joint e is no loaded, the internal forces in these rods are zero. Thus, we obtain all zero-force elements. To determine the forces in the remaining rods, we should consider the equilibrium of each joint in the order of 1, 2, . . ., 6. In joints 1, 3, 4, and 5 we have three unknown forces that do not lie in the same plane, while in joints 2 and 6 we have two unknown forces. For joints with three elements it is necessary to compose three equations of equilibrium, and for joints 2 and 6, two equations in the form of a sum of the projections of forces on the coordinate axes. After determining the internal forces in all elements of the structure, it is easy to determine the reactions of support. In each support joint A, B, . . ., F arises three components of reaction.
5.5.5.3
Features of Schwedler Structure
1. If the force P is moved to joint 2, and its inclination to the rods in the new joint remains the same as when loading in joint 1, then the forces in all the rods of the truss are obtained by cyclic permutation. 2. A particularly simple solution is obtained if all the joints of each contour are loaded with the same vertical forces. In this case, all the rods of the same contour experience equal forces. Equal forces arise in all meridians, while the forces in the diagonal elements are zero. This allows restricting ourselves to cutting out joints of only one single meridian, for example, joints 1 and 4.
5.5.5.4
Decomposing of Space Framework to Plane Trusses
It is possible to decompose the space framework into separate plane trusses, if they satisfy the following conditions: 1. The truss is geometrically unchangeable and fixed in its plane. 2. A load is applied to the joints of this truss and lying in the plane of the truss. In this case the load is perceived only by the rods of the plane truss, and the internal forces in all other rods of the space framework are zero. This theorem makes it easy to find internal forces in some trusses, in particular those having the form of a prism or a truncated pyramid. Schwedler dome can be considered as this type of structures, if the lateral legs are rectilinear, i.e., do not have kinks at the points 4, 3, e, etc. (Fig. 5.21). In this case we can isolate and consider plane trusses fixed in their planes separately (Fig. 5.21b).
a
F
2
A
B
D
U
V
c
T 3
3
e B
B
3
f
b d V
U
4
a
1
4
d
1
6
5
A
b
E
d
c
e C
C
Fig. 5.21 Space framework with rectilinear meridian legs (A–4–1, B–3–d, etc.). (a) Design diagram (top view). (b) Decomposition of space framework into plane trusses
128
5 Space Frameworks
Assume that the force P is applied to the joint d. This force is resolved into the components T, U, and V directed along the three bars of this joint (Fig. 5.21a). The force T causes compression of the rod B-d and will be balanced by the same reaction applied at the support B. All other rods of the space framework will not exert any stresses due to the T force. The force U is perceived by the plane truss A–1–d–B, and the force V by the plane truss B–d–c–C (Fig. 5.21b). The elements B-3 and 3-d are involved into two plane trusses. They are A-1-d-B and B-d-c-C trusses. Final internal forces in the B-3 and 3-d rods are equal to the algebraic sum of forces caused by T-force along d-3 rod and the forces in the B-3 and 3-d rods of two plane trusses caused by the U and V forces separately: V S3d ¼ ST3d þ SU 3d þ S3d
Here internal force ST3d ¼ T is caused by force T, while the remaining terms are caused by forces U and V, and they should be determined considering the equilibrium of two plane trusses; their design diagrams are presented in Fig. 5.21b.
Problems 5.1. Describe the types of supports of the space structures. 5.2. Explain the concepts of released and attached structures. 5.3. Describe the general case of geometrically changeable connection of unyielding space framework with foundation and special cases. 5.4. Describe the simple arrangement of six support rods for geometrically unchangeable connection of rigid space framework with foundation. 5.5. Describe the mesh structure and its features. Explain the essence of Cauchy theorem for the spaced frameworks of the meshwork types. 5.6. Provide the kinematical analysis of the space structures shown in Fig. P5.6. Apply nil load method. 3
a
b 1
4
2
6
5 9
7 8 Fig. P5.6
5.7. Explain why the structure in Fig. P5.7 cannot be classified as released three-dimensional meshwork truss. Transform it into the released space truss of meshwork type.
Fig. P5.7
Problems
129
5.8. The joints A, B, and C of the space structure are rigidly fixed by means of six support constraints to the plane yOz (types of supports are not shown); OC ¼ OB ¼ a (Fig. P5.8). Classify this structure. Provide kinematical analysis. Determine internal forces in members 1, 2, and 3.
z
2
Ans. N1 = N 2 =
A B
3
α O a
P P ; N3 = 2 sin α sin β
a
β
x
α
P
C
1
y
Fig. P5.8
5.9. The joints of the spatial rod structure are located at the vertices of the prism with sides a, b, and h (Fig. P5.9). Provide kinematical analysis. Determine the internal forces in the vertical members 1–4 and diagonals 5–8 of side surfaces. Hint: Combine the joint isolation method with method of through cuts.
7
P
3
4
6
8
5
1
h
2 b
a Fig. P5.9
5.10. The cantilevered space framework of meshwork type has three vertical planes 147, 258, and 369; last face is rigidly fixed by means of six support constraints to the vertical plane A at joints 3, 6, and 9 (Fig. P5.10a). The structure is loaded by two horizontal forces P, which are parallel to the rods 47 and 58. Provide the analysis of the structure using two approaches: (a) method of sections and (b) reduction to plane trusses (Fig. P5.10b, c). Prove that members 4–7, 5–8, 4–8, and 5–9 do not work. Compare results using two approaches. Is it possible to apply approach (b) for determining internal force in 4–7 and 4–8? Explain your conclusion.
130
5 Space Frameworks 3
A
a P
6 P
1
7
P1
b
5 8
4
1
9
7
P
P2 α
Fig. P5.10
Ans: S23 ¼ 0, S35 ¼ S38 , S56 ¼ S89 :
P1
P1
c
2
3
2
8
9
Chapter 6
Three-Hinged Arches
The arches are widely used in modern engineering. Arches permit to cover a larger span. The greater is the span than an arch becomes more economical than a truss. From esthetic point of view the arches are more attractive than trusses. Materials of the modern arches are concrete, steel, and wood. The body of the arch may be solid or consist of separate members. The arches are classified as three-hinged, two-hinged, and arch with fixed supports. A three-hinged arch is a geometrically unchangeable statically determinate structure which consists of two curvilinear members, connected together by means of a hinge, with two-hinged supports resting on the abutment. This chapter is devoted to analysis of only three-hinged arches. Two-hinged arch and arch with fixed supports will be considered in Part 2.
6.1 6.1.1
Preliminary Remarks Design Diagram of Three-Hinged Arch
The arch with overarched structure is shown in Fig. 6.1a. The arch contains three hinges. Two of them are located at the supports and third is placed at the crown. These hinges are distinguishing features of the three-hinged arch. Design diagram also contains information about the shape of the neutral line of the arch. Usually this shape is given by the expression y ¼ f(x). Each post which connects the beams of overarched structure with arch itself has the hinges at the ends. It means that in the poles only axial force arises. Idealized design diagram of the three-hinged arch without overarched members is shown in Fig. 6.1b.
C
y
a
f
y=f(x)
j
b
f
A
B
x HB
HA l
l RA
RB
Fig. 6.1 (a) Design diagrams for deck-arch bridge. (b) Design diagram of three-hinged arch
Degree of freedom of three-hinged arch according to Chebyshev formula (1.1) is W ¼ 3D 2H S0 ¼ 3 2 2 1 4 ¼ 0 so this structure is geometrically unchangeable. Indeed, two rigid disks AC and BC are connected with the ground by two hinges A and B and line AB does not pass through the intermediate hinge C. This structure has four unknown reactions, i.e., two vertical reactions RA and RB and two horizontal reactions HA and HB. For their determination, three equilibrium equations can be formulated considering the structure in whole. Since bending moment at the hinge C is zero, this provides additional equation for equilibrium of the part of the system. It means that the sum of the moments of all external forces, which are located on the right (or on the left) part of the structure with respect to hinge C, is zero:
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_6
131
132
6 Three-Hinged Arches
X
MC ¼ 0
or
left
X
MC ¼ 0
ð6:1Þ
right
These four equations of equilibrium determine all four reactions at the supports. Therefore, three-hinged arch is a geometrically unchangeable and statically determinate structure.
6.1.2
Peculiarities of the Arches
The fundamental feature of arched structure is that horizontal reactions appear even if the structure is subjected to vertical load only. These horizontal reactions HA ¼ HB ¼ H are called a thrust; such types of structures are often called as thrusted structures. One type of thrusted structures (thrusted frame) was considered in Sect. 4.6.2. It will be shown later that at any cross section of the arch the bending moments, shear, and axial forces arise. However, the bending moments and shear forces are considerably smaller than corresponding internal forces in a simply supported beam covering the same span and subjected to the same load. This is the fundamental property of the arch thanks to thrust. Thrusts in both supports are oriented towards each other and reduce the bending moments that would arise in beams of the same span and load. Therefore, the height of the cross section of the arch can be much less than the height of a beam to resist the same loading. So the three-hinged arch is more economical than simply supported beam, especially for large-span structures. Both parts of the arch may be connected by a tie. In this case in order for the structure to remain statically determinate, one of the supports of the arch is being rolled (Fig. 6.2a, b).
a
y
A
b
C
B
Tie
RA
RB
x
y
C Elevated tie
A
RA
B
x
RB
Fig. 6.2 (a, b) Design diagram of three-hinged arch with tie on the level supports and elevated tie
Prestressed tie allows controlling the internal forces in arch itself. Tie is an element connected by its ends to the arch by mean of hinges; therefore the tie is subjected only to an axial internal force. So even if horizontal reactions of support equal zero, an extended force (thrust) arises in the tie. Thus, the arch is characterized by two fundamental markers such as a curvilinear axis and appearance of the thrust. The structure in Fig. 6.3 presents the curvilinear trustless simply supported element; that is, this is just an element with curvilinear axis, but it is not arch. It is obvious that the distribution of bending moments for this structure and for beam of the same span and load will not be different. Thus, the fundamental feature of the arch (decreasing of the bending moments due to the appearance of the thrust) for this structure is not observed.
Р A RA Fig. 6.3 Simply supported thrustless curvilinear member
B
RB
6.2 Internal Forces
6.1.3
133
Geometric Parameters of Circular and Parabolic Arches
Distribution of internal forces in arches depends on a shape of the central line (or axis) of an arch. The concept “axis of the arch” was introduced by L. Salimbeni (1787) and is a major step in the construction of the design diagram of the arch. Equation of the central line and some necessary formulae for circular and parabolic arches are presented below. For both cases, origin of coordinate axis is located at point A (Figs. 6.1b and 6.2).
6.1.3.1
Circular Arch
Ordinate y of any point of the central line of the arch is defined by the formula sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 l y ¼ R2 x R þ f; 2
R¼
f l2 þ 2 8f
ð6:2Þ
where x is the abscissa of the same point of the central line of the arch; R the radius of the arch; f and l the rise and span of the arch. The angle φ between the tangent to the center line of the arch at point (x,y) and horizontal axis is shown in Fig. 6.1b. Trigonometric functions for this angle are as follows: sin φ ¼ ðl 2xÞ
6.1.3.2
1 ; 2R
cos φ ¼ ðy þ R f Þ
1 R
ð6:3Þ
Parabolic Arch
Ordinate y of any point of the central line of the arch: y ¼ 4fxðl xÞ
1 l2
ð6:4Þ
Trigonometric functions of the angle between the tangent to the center line of the arch at point (x, y) and a horizontal axis are as follows: dy 4f ¼ ðl 2xÞ; dx l2 1 cos φ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 þ tan 2 φ tan φ ¼
sin φ ¼ cos φ tan φ
ð6:5Þ
For the left-hand half-arch the functions are sin φ > 0, cos φ > 0, and for the right-hand half-arch the functions are sinφ < 0 and cosφ > 0. l2 Radius of curvature of the arch at the crown is R0 ¼ . 8f
6.2
Internal Forces
Design diagram of a three-hinged symmetrical arch with intermediate hinge C at the highest point of the arch and with supports A and B on one elevation is presented in Fig. 6.4. The span and rise of the arch are labeled as l and f, respectively; equation of central line of the arch is y ¼ y(x).
134
6.2.1
6 Three-Hinged Arches
Concept of Substitute Beam
Determination of internal forces, and especially construction of influence lines for internal forces of the three-hinged arch, may be easily and attractively performed using the conception of the “reference (or substitute) beam.” The reference beam is a simply supported beam of the same span as the given arch and subjected to the same loads, which act on the arch (Fig. 6.4a).
a P1
y
P2
b
Tangent C k jk f
yk
Mk
Pn
P1
P2
k
Nk jk
yk
y(x)
Qk
B
A HA
Tangent
H
x
HB xk l
RA P1
RB
P2
Pn
RA
x1 x2 xk
C Reference beam
R A0 A
x1 x2
R B0 l
A
Fig. 6.4 Three-hinged arch. (a) Design diagram and reference beam. (b) Positive internal forces at any section k
The following reactions arise in the arch: RA, RB, HA, HB. The vertical reactions of three-hinged arches carrying the vertical loads have same values as the reactions of the reference beam: RA ¼ R0A ;
RB ¼ R0B
ð6:6Þ
The horizontal reactions (thrust) at both supports of three-hinged arches subjected to the vertical loads are equal in magnitude and opposite in direction: HA ¼ HB ¼ H
ð6:7Þ
Bending moment at the hinge C of the arch is zero. Therefore, by definition of the bending moment l l l M C ¼ RA P1 x1 P 2 x2 H A f ¼ 0 2 2 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} M 0C
Underlined set of terms is the bending moment acting over section C of the reference beam (this section is located under the hinge of the arch). Therefore last equation may be rewritten in the form M 0C H A f ¼ 0 which allows immediately to calculate the thrust:
6.2 Internal Forces
135
H¼
M 0C f
ð6:8Þ
Thus, the thrust of the arch equals to the bending moment at section C of the reference beam divided by the rise of the arch. In any section k of the arch, the following internal forces arise: the bending moment Mk, shear Qk, and axial force Nk. The positive directions of internal forces are shown in Fig. 6.4b. Internal forces acting over a cross section k may be obtained considering the equilibrium of free-body diagram of the left or right part of the arch. It is convenient to use the left part of the arch. By definition M k ¼ RA xk
X
Pi ðxk xi Þ Hyk ;
left
Qk ¼
RA
X
! P cos φk H sin φk ;
left
ð6:9Þ
! X N k ¼ RA P sin φk H cos φk left
where Pi are forces which are located at the left side of the section k; xi are corresponding abscises of the points of application; xk and yk are coordinates of the point k; and φk is the angle between the tangent to the center line of the arch at point k and a horizontal. These equations may be represented in the following convenient form: M k ¼ M 0k Hyk Qk ¼ Q0k cos φk H sin φk N k ¼ Q0k sin φk H cos φk
ð6:10Þ
Here expressions M 0k and Q0k represent the bending moment and shear force at the section k for the reference beam (beam’s bending moment and beam’s shear).
6.2.1.1
Analysis of Formulae (6.8) and (6.10)
1. Thrust of the arch is inversely proportional to the rise of the arch. 2. In order to calculate the bending moment in any cross section of the three-hinged arch, the bending moment at the same section of the reference beam should be decreased by value Hyk. Therefore, the bending moment in the arch is less than in the reference beam. This is the reason why the three-hinged arch is more economical than simply supported beam, especially for large-span structures. 3. In order to calculate shear force in any cross section of the three-hinged arch, the shear force at the same section of the reference beam should be multiplied by cosφk and this value should be decreased by Hsinφk. 4. Unlike beams loaded by vertical loads, there are axial forces, which arise in arches loaded by vertical loads only. These axial forces are always compressed. Analysis of three-hinged arch subjected to fixed loads is presented below. This analysis implies determination of reactions of supports and construction of internal force diagrams.
6.2.2
Numerical Example
Design diagram of the three-hinged circular arch subjected to fixed loads is presented in Fig. 6.5. The forces are P1 ¼ 10 kN, P2 ¼ 8 kN, and q ¼ 2 kN/m. It is necessary to construct the internal force diagrams M, Q, and N.
136
6 Three-Hinged Arches
q=2kN/m
P1=10kN Design diagram
k 3
y
C 4
n
5 f=8m
1
Circle
6
2
A
P2=8kN
7
B
x
H
H 4m
4m
l=32m
RA P1
RB P2
q
Reference A beam
B C
R A0 14.5
R B0 4.5
+
Q0 (kN) – 19.5
11.5
M0 (kNm) 58
78 116
125
124 152
154
Fig. 6.5 Three-hinged circular arch. Design diagram, reference beam, and corresponding internal force diagrams
Solution Reference beam: The reactions are determined from the equilibrium equations of all the external forces acting on the reference beam: X R0A ! M B ¼ 0 : R0A 32 þ P1 24 þ q 8 12 þ P2 4 ¼ 0 ! R0A ¼ 14:5 kN, X R0B ! M A ¼ 0 : R0B 32 P1 8 q 8 20 P2 28 ¼ 0 ! R0B ¼ 19:5 kN: The values of the two reactions just found should be checked using the equilibrium equation X
Y ¼ R0A þ R0B P1 q 8 P2 ¼ 14:5 þ 19:5 10 2 8 8 ¼ 34 34 ¼ 0
The bending moment M0 and shear Q0 diagrams for reference beam are presented in Fig. 6.5. At point C (x ¼ 16 m) the bending moment is M 0C ¼ 152 kNm. Three-hinged arch: The vertical reactions and thrust of the arch are RA ¼ R0A ¼ 14:5 kN,
RB ¼ R0B ¼ 19:5 kN,
H¼
M 0C 152 ¼ 19 kN ¼ 8 f
For construction of internal force diagrams of the arch, a set of sections has to be considered and for each section the internal forces should be calculated. All computations concerning geometrical parameters and internal forces of the arch are presented in Table 6.1.
6.2 Internal Forces
137
Table 6.1 Internal forces in three-hinged circular arch (Fig. 6.5); (RA ¼ 14.5 kN; Section 0 A 1 2
x (m) 1 0.0 4 8
y (m) 2 0.0 4 6.330
sinφ 3 0.8 0.6 0.4
cosφ 4 0.6 0.8 0.9165
k 3 4 (C) 5 6 n 7
10 12 16 20 24 26 28
7.0788 7.596 8.0 7.596 6.330 5.3205 4
0.3 0.2 0.0 0.20 0.40 0.50 0.6
0.9539 0.9798 1.0 0.9798 0.9165 0.8660 0.8
B
32
0.0
0.8
0.6
M 0x
(kNm) 5 0 58 116
RB ¼ 19.5 kN;
H ¼ 19 kN)
Hy (kNm) 50 0.0 76 120.27
Mx (kNm) 6 0 18 4.27
125 134 152 154 124 101 78
134.497 144.324 152 144.324 120.27 101.089 76
29.497 10.324 0.0 9.676 3.73 0.089 2
0
0.0
0
Q0x (kN) 7 14.5 14.5 14.5 4.5 4.5 4.5 4.5 3.5 11.5 11.5 11.5 19.5 19.5
Qx (kN) 8 6.5 0.2 5.6892 3.4757 21.4074 0.6091 4.5 0.3707 2.9397 0.459 2.2 4.2 3.5
Nx (kN) 9 23 23.9 23.213 19.213 219.474 19.516 19.00 19.316 22.013 22.204 22.1 26.9 27
Notes: Values in nominator and denominator (columns 8 and 9) mean value of the force to the left and to the right of corresponding section. Values of discontinuity due to concentrated load equal Pcosφ and Psinφ in shear and normal force diagrams, respectively.
The column 0 contains the numbers of sections. For specified sections A, 1–7, and B the abscissa x and corresponding ordinate y (in meters) are presented in columns 1 and 2, respectively. Radius of curvature of the arch is R¼
f l2 8 322 þ ¼ þ ¼ 20 m 2 8f 2 8 8
Coordinates y are calculated using the following expression: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l y ¼ yð xÞ ¼ R 2 x R þ f ¼ 400 ð16 xÞ2 12 2 Columns 3 and 4 contain values of sinφ and cosφ, which are calculated by formulae sin φ ¼
l 2x 32 2x ¼ , 2R 40
cos φ ¼
y þ R f y þ 12 ¼ R 20
Values of bending moment and shear for reference beam, which are presented in columns 5 and 7, are taken directly from corresponding diagrams in Fig. 6.5. Values for Hy are contained in column 50 . Columns containing separate terms for Q0cos φ, Q0 sin φ, H cos φ, H sin φ are not presented. Values of bending moment, shear, and normal forces for threehinged arch are tabulated in columns 6, 8, and 9. They have been computed using Eq. 4.10. For example, for section A we have QA ¼ Q0A cos φA H sin φA ¼ 14:5 0:6 19 0:8 ¼ 6:5kN, N A ¼ Q0A sin φA H cos φA ¼ 14:5 0:8 19 0:6 ¼ 23 kN The final internal force diagrams for arch are presented in Fig. 6.6. Bending moment diagram is shown on the side of the extended fibers; thus the signs of bending moments are omitted. As for beam, the bending moment and shear diagrams satisfy Schwedler’s differential relationships. In particular, if at any point a shear changes sign, then a slope of the bending moment diagram equals zero; that is, at this point the bending moment has local extreme (for example, points 2, 7, etc.).
138
6 Three-Hinged Arches
q=2kN/m
P1=10kN k 3
y
C 4
n
5
6
2 f=8m
1
A
P2=8kN
7
B
x
H
H 4m
4m
l=32m
RA
RB
18 4.27 9.49 10.32 0.089 M (kNm) 3.73
2.0
9.676 P1cos φ 2
5.69
4.5
0.2
2.2 +
3.5 Q (kN)
– 6.5
1.4 2.94
3.47
4.2 N (kN)
–
23 23.21 19.21
19.47 P1sinj 2
22.1 26.9
27
Fig. 6.6 Design diagram of three-hinged circular arch. Internal force diagrams
It can be seen that the bending moments which arise in cross sections of the arch are much less than in a reference beam.
6.2.3
Maximum Economy Arches
Assume that a three-hinged arch with known dimensions l and f is subjected to known load. The following problem may be formulated: Is it possible to make an arch so that bending moments do not arise in the cross sections? The analytic condition for the absence of bending moments in a cross section of the arch has the form M k ¼ M 0k Hyk ¼ 0 The bending moment in the reference beam M 0k and the thrust H are completely determined by the general dimensions l and f of the arch and the load. They do not depend on the character of axial line of the arch, while the parameter yk is determined by the equation of its central line. Thus, the condition for the absence of bending moments in the arch becomes yk ¼ Corresponding axis of the arch is called as rational axis.
M 0k H
6.3 Influence Lines for Reactions and Internal Forces
139
The thrust H ¼ M 0C =f is a constant value for a given bending moment M 0C and rise f. Thus, at any section of a three-hinged arch the bending moments do not arise, if the ordinates yk of axis of the arch are proportional to the ordinates of the bending moments in the reference beam. Assume that three-hinged arch is subjected to uniformly distributed load q. Bending moment at any section k in the reference beam is M 0k ¼
ql qx2 q x ¼ xðl xÞ 2 2 2
Equation of axial line of parabolic arch is yk ¼
4f xð l xÞ l2
It can be seen that both functions M 0k and yk are proportional, because they contain the same function x(l x), while the proportional coefficient M 0k =yk ¼ ql2 =8f does not depend on abscissa x. Therefore at three-hinged parabolic arch under uniformly distributed load of any intensity q the bending moments do not arise. Thus, design of cross sections of such economy arch is performed without consideration of bending moments.
6.3
Influence Lines for Reactions and Internal Forces
Equations 6.6, 6.8, and 6.10 can be used for deriving of equations for influence lines. Vertical reactions: The equations for influence lines for vertical reactions of the arch are derived from Eq. (6.6). Therefore the equations for influence lines become ILðRA Þ ¼ IL R0A ; ILðRB Þ ¼ IL R0B ð6:11Þ Thus, influence lines for vertical reactions of the arch do not differ from influence lines for reactions of the reference simply supported beam. Thrust: The equation of influence line for thrust is derived from Eq. (6.8). Since for given arch a rise f is a fixed number, the equation for influence line becomes ILðH Þ ¼
1 IL M 0C f
ð6:12Þ
Thus, influence line for thrust H may be obtained from the influence line for bending moment at section C of the reference beam, if all ordinates of the latter will be divided by parameter f. Internal forces: The equations for influence lines for internal forces at any section k may be derived from Eq. (6.10). Since for given section k the parameters yk, sinφk, and cosφk are fixed numbers, the equations for influence lines become ILðM k Þ ¼ IL M 0k yk ILðH Þ; ILðQk Þ ¼ cos φk IL Q0k sin φk ILðH Þ; ILðN k Þ ¼ sin φk IL Q0k cos φk ILðH Þ
ð6:13Þ
In order to construct the influence line for bending moment at section k, it is necessary to sum two graphs: one of them is influence line for bending moment at section k for reference beam and second is influence line for thrust H with all ordinates of which have been multiplied by a constant factor (yk). Equation of influence line for shear also has two terms. The first term presents influence line for shear at section k in the reference beam, all the ordinates of which have been multiplied by a constant factor cos φk. The second term presents the influence line of the thrust of the arch, all ordinates of which have been multiplied by a constant factor ( sin φk). Summation of these two graphs leads to the required influence line for shear force at section k. Similar procedure should be applied for construction of influence line for axial force. Note that both terms for axial force are negative. Figure 6.7 presents the arched structure of the arch itself and overarched construction, which includes the set of simply supported beams and vertical posts with hinged ends. Unit load, which moves along the horizontal beams, is transmitted over the posts on the arch at discrete points. Thus, this design diagram corresponds to indirect load application. Parameters of the arch are same as in Fig. 6.5.
140
6 Three-Hinged Arches
P=1
k 2 A
C
3
5
6
1
7
B
H
H ak= 10m RA
bk= 22m
aC= 16m
1
bC= 16m 0.75
0.5
+
RB
0.25
0.125
Inf. line RA
acbc = 8m l + Inf. line MC0 (m)
0.25
acbc =1 lf
0.5
0.5 0.25
+ Inf. line H
1.0 +
Inf. line Qk0 1.0
ak bk = 6.875 l
5.0 + Inf. line Mk0 (m)
Fig. 6.7 Three-hinged arch. Design diagram and influence lines for reactions of supports and internal forces at section k for substitute beam
6.3.1
Influence Lines for Reactions
According to (6.11), influence lines for vertical reactions RA and RB of the arch do not differ from influence lines for reaction of supports of a simply supported beam. Influence line for thrust may be constructed according to Eq. (6.12); the maximum ordinate of influence line for bending moment at section C of the reference beam is equal to ac bc =l ¼ l=4 ¼ 8 (m). Therefore the maximum ordinate of influence line for thrust H of the arch becomes 1 ac bc l 32 ¼1 ¼ ¼ f 4f 4 8 l
6.3 Influence Lines for Reactions and Internal Forces
141
Influence lines for reactions of supports of the arch (RA, H ) and internal forces for reference beam M 0C , Q0k , . . . M 0k are shown in Fig. 6.7.
6.3.2
Influence Lines for Internal Forces
The section k is characterized by the following parameters: ak ¼ 10 m,
6.3.2.1
bk ¼ 22 m,
yk ¼ 7:0788 m,
sin φk ¼ 0:30,
cos φk ¼ 0:9539
Bending Moment
Influence line for M at section k may be constructed according to Eq. (6.13): ILðM k Þ ¼ IL M 0k yk ILðH Þ
ð6:13aÞ
Step 1. Influence line for bending moment at section k of reference beam M 0k presents the triangle with maximum ordinate ak bk =l ¼ 10 22=32 ¼ 6:875 (m) at section k and 5.0 m at section C (Fig. 6.7). Step 2. Influence line for thrust H presents triangle with maximum ordinate l=4f ¼ 1 at section C. Term yk IL(H ) presents the similar graph; the maximum ordinate is yk 1 ¼ 7.0788 m. So the specified ordinates of graph yk IL(H ) at section k and C are 4.42425 and 7.0788 m, respectively. Step 3. Procedure(6.13a) is presented in Fig. 6.8, Construction Inf. Line Mk. Since both terms in (6.13a) have different signs, both graphs IL M 0k and yk IL(H ) should be plotted on the one side on the basic line. The ordinates of required IL(Mk) will be located between these both graphs. Specified ordinates of final graph (6.13a) at section k and C are 6:875 4:42425 ¼ 2:45075 m and
5:0 7:0788 ¼ 2:0788 m
P=1 k C
3 2
yk
1
B
A H
H ak= 10m
bk= 22m RB
RA 7.0788
6.875
yK .IL(H)
–
+
5.0
IL(MkO)
Construction Inf. line Mk
4.42425 1.96
+
2.4507
Connecting line
*
Inf. line Mk (m)
– 2.0788
1.0394
0.5194
Fig. 6.8 Three-hinged arch. Design diagram and construction of influence line for bending moment at section k of the arch
142
6 Three-Hinged Arches
Step 4. Influence line between joints 2 and 3 presents a straight line because of indirect load application. Final influence line IL (Mk) is presented in Fig. 6.8; the connected line between joints 2 and 3 is shown by solid line.
6.3.2.2
Shear Force
This influence line may be constructed according to equation ILðQk Þ ¼ cos φk IL Q0k sin φk ILðH Þ
ð6:13bÞ
Step 1. Influence line for shear at section k for reference beam is shown in Fig. 6.7; the specified ordinates at supports A and B equal 1.0. The first term cos φk IL Q0k of Eq. (6.13b) presents a similar graph with specified ordinates cos φk ¼ 0.954 at supports A and B, so ordinates at the left and right of section k are (0.298) and 0.656, while at crown C it is 0.477. Step 2. Influence line for thrust is shown in Fig. 6.7; the specified ordinate at crown C equals 1.0. The second term sin φk IL(H ) of Eq. (6.13b) presents a similar graph with specified ordinates 0.3 1.0 ¼ 0.3 at crown C. Specified ordinate at section k is 0.1875. Step 3. Procedure (6.13b) is presented inFig. 6.9. As in case for bending moment, both terms in (6.13b) have different signs; therefore both graphs cos φk IL Q0k and sin φk IL(H ) should be plotted on the one side on the basic line. Ordinates between both graphs present the required ordinates for influence line for shear. Specified ordinates of final graph (6.13b) left and right at section k are 0:298 þ 0:1875 ¼ 0:4855
and
0:656 0:1875 ¼ 0:4685
Fig. 6.9 Three-hinged arch. Design diagram and construction of influence line for shear at section k of the arch
6.3 Influence Lines for Reactions and Internal Forces
143
At crown C ordinate of influence line Qk is 0.477 0.3 ¼ 0.177. Step 4. Influence line between joints 2 and 3 presents a straight line; this connected line is shown by solid line. Final influence line IL(Qk) is presented in Fig. 6.9.
6.3.2.3
Axial Force
This influence line may be constructed according to equation ILðN k Þ ¼ sin φk IL Q0k cos φk ILðH Þ
ð6:13cÞ
Step 1. Influence line for shear at section k for reference beam is shown in Fig. 6.7. The first term sin φk IL Q0k of Eq. (6.13c) presents a similar graph with specified ordinates sinφk ¼ 0.30 at supports A and B, so at the left and right of section k ordinates are 0.09375 and 0.20625, while at crown C there is 0.15.
P=1
2
k φk 3
C
1
B
A 0.9539
0.59618
cosj k. IL(H)
0.09375 –
sinj k=0.3 0.20625
*
Construction
0.3 Inf. line Nk 0.15
sinj k. IL(QkO) Inf. line Nk
sinj k
– 0.40194
0.2759 0.5519
0.50243 1.1039 0.80243 Connecting line
Fig. 6.10 Three-hinged arch. Design diagram and construction of influence line for axial force at section k of the arch
Step 2. Influence line for thrust is shown in Fig. 6.7; the specified ordinate at crown C equals 1.0. The second term cosφk IL(H ) of (6.13c) presents a similar graph with specified ordinates 0.9539 1.0 ¼ 0.9539 at crown C. Specified ordinate at section k is 0.59618. Step 3. Procedure 0 (6.13c) is presented in Fig. 6.10. Both terms in (6.13c) have same signs; therefore both graphs, sin φk IL Qk and cosφk IL(H ), should be plotted on the different sides on the basic line. Ordinates for required IL(Nk) are located between these both graphs. Specified ordinates of final graph (6.13c) left and right at section k are ð0:59618 0:09375Þ ¼ 0:50243 and
ð0:59618 þ 0:20625Þ ¼ 0:80243
144
6 Three-Hinged Arches
At crown C ordinate of influence line Nk is (0.9539 + 0.15) ¼ 1.1039. Step 4. Influence line between joints 2 and 3 presents a straight line; this connected line is shown by solid line. Final influence line IL(Nk) is shown in Fig. 6.10.
6.3.2.4
Features of the Influence Lines for Internal Forces
1. Influence line for bending moment has significantly less ordinates than for reference beam. This influence line contains the positive and negative ordinates. It means that at section k extended fibers can be located below or above the neutral line depending on where the load is placed. 2. Influence line for shear, as in case of reference beam, has two portions with positive and negative ordinates; all ordinates are significantly less than in the reference beam. 3. Influence line for axial force has only negative ordinates. So in case of arbitrary load the axial forces in arch are always compressed.
6.3.3
Application of Influence Lines
Assume that arch is subjected to fixed loads as shown in Fig. 6.5. The reactions of supports and internal forces caused by fixed load may be calculated by the formula Z¼
X
Pi yi þ
X
q jΩ j,
where Z is any factor, for which the influence line is constructed; y is the ordinate of influence line under the concentrated load P; and Ω is the area of influence line graph under the distributed load q.
6.3.3.1
Reactions of Supports
Ordinates of influence line for RA at the points of application of the loads P1 and P2 are 0.75 and 0.125, respectively. The area of the influence line under the uniformly distributed load is Ω¼
0:5 þ 0:25 8 ¼ 3:0 2
Therefore, the reaction RA ¼ P1 0.75 + q 3 + P2 0.125 ¼ 14.5 kN. The thrust H of the arch using influence line equals H ¼ P1 0:5 þ q
1 þ 0:5 8 þ P2 0:25 ¼ 19 kN 2
These values of reactions coincide with those computed previously in Sect. 6.2.2.
6.3.3.2
Internal Forces in Section k
The internal forces can be found in a similar way, using the relevant influence lines. They are the following:
6.4 Nil Point Method for Construction of Influence Lines
M k ¼ P1 1:96 q
145
2:0788 þ 1:0394 8 P2 0:5194 ¼ 9:500 kNm 2
Qk ¼ P1 0:3883 þ q
0:177 þ 0:0885 8 þ P2 0:04425 ¼ 1:405 kN 2
N k ¼ P1 0:40194 q
1:1039 þ 0:5519 8 P2 0:2759 ¼ 19:473 kN 2
The magnitudes of just found internal forces Mk, Qk, and Nk coincide with those computed in Sect. 6.2 and presented in Table 6.1. Example 6.1 Let us consider design diagram of the arch in Fig. 6.7. It is necessary to find bending moment in section 3 due to the force P ¼ 10 kN, applied at point 7. Solution The feature of this problem is as follows: we will compute the bending moment at the section 3 without influence line for M3 but using influence line for reaction. As the first step, we obtain the vertical reaction and thrust, which are necessary for the calculation of internal forces. Step 1. Find H and RA from previously constructed influence lines, presented in Fig. 6.7: RA ¼ P yR ¼ 10 0:125 ¼ 1:25 kN H ¼ P yH ¼ 10 0:25 ¼ 2:5 kN where yR and yH are ordinates of influence line for RA and H, respectively, under concentrated force P. Step 2. The bending moment in section 3, considering left forces, becomes M 3 ¼ H y3 þ RA x3 ¼ 2:5 7:596 þ 1:25 12 ¼ 3:99 kNm where x3 and y3 for section 3 are presented in Table 6.1. This example shows that one of the advantages of influence line is that the influence lines for reactions and thrust constructed once may be used for their computation for different cases of arbitrary loads. Then, knowing reactions and thrust, the internal forces at any point of the arch may be calculated by definition without using influence line for that particular internal force. This idea is the basis of complex usage of influence lines together with fixed load approach, which will be effectively applied for tedious analysis of complicated structures, in particularly for statically indeterminate ones.
6.4
Nil Point Method for Construction of Influence Lines
Each influence line shown in Figs. 6.8, 6.9, and 6.10 has the specified point labeled as (). These points are called nil (or neutral) point of corresponding influence line. Such points of influence lines indicate a position of the concentrated load on the arch, so internal forces M, Q, and N in the given section k would be zero. Nil points may be used for simple procedure for construction of influence lines for internal forces and checking the influence lines which were constructed by analytical approach. This procedure for symmetrical three-hinged arch of span l is discussed below.
6.4.1
Bending Moment
Step 1. Find nil point (NP) of influence line Mk. If load P is located on the left half of the arch, then total reaction of the support B passes through crown C. Bending moment at section k equals zero, if total reaction of support A passes through the point k. Therefore, NP(Mk) is point N of intersection of lines BC and Ak (theorem about three concurrent forces). The nil point () is always located between the crown C and section k. Step 2. Lay off along the vertical passing through the support A, the abscissa of section k, i.e., xk.
146
6 Three-Hinged Arches
Step 3. Connect this ordinate with nil point and continue this line till a vertical passing through crown C and then connect this point with support B. Step 4. Take into account indirect load application; connecting line between joints 2 and 3 is not shown (Fig. 6.11). P=1 NP(Mk)
k
C
3
2
6 f
yk
1
5
RA A
RB
α
7 β
B
xk uM
l - uM l1
l2 l
xk
+
NM
*
uM
–
Inf. line Mk
Fig. 6.11 Construction of influence line Mk using nil point method
Location of NP(Mk) may be computed by the formula uM ¼
l f xk yk l 2 þ xk f
Indeed, from triangles ANN1 and BNN1 we have NN 1 ¼ uM tan α ¼ uM
yk , xk
and NN 1 ¼ ðl uM Þ tan β ¼ ðl uM Þ
f l2
Solution of this equation with respect to uM immediately leads to the above formula.
6.4.2
Shear Force
Step 1. Find nil point (NP) of influence line Qk. If load P is located on the left half of the arch, then reaction of the support B passes through crown C. Shear force at section k equals zero, if reaction of support A will be parallel to tangent at point k. Therefore, NP (Qk) is the point of intersection of line BC and line which is parallel to tangent at point k. For given design diagram and specified section k the nil point () is a fictitious one (Fig. 6.12a).
6.4 Nil Point Method for Construction of Influence Lines
a
147
P=1 Tangent at k k C
3
5
6
2
NP(Qk)
1
RA φk
A
RB
7
b
B
Parallel to tangent at k
cosφ k
RHP-1 RHP-2
+
*NQ
–
Inf. line Qk
LHP
uQ
b
P=1
Tangent at k
NP(Qk)
k
φk
A
C Parallel to tangent at k
cosφ k
b
B
RHP-1
+ LHP
Inf. line Qk
NQ
*
–
RHP-2
uQ Fig. 6.12 Construction of influence line Qk for three-hinged arch using nil point method. (a) Symmetrical arch. The nil point NQ is fictitious; (b) nonsymmetrical arch. The nil point NQ is real
Step 2. Lay off along the vertical passing through the support A, the value cos φk. Step 3. Connect this ordinate with nil point. A working zone of influence line is the portion between section k and vertical passing through crown C—right-hand portion 1 (RHP-1). Then connect the point under crown C with support B—righthand portion 2 (RHP-2). Step 4. Left-hand portion (LHP) is parallel to right-hand portion 1 and connects two points: zero ordinate at support A and point under section k. Step 5. Take into account indirect load application; connecting line between joints 2 and 3 is not shown.
148
6 Three-Hinged Arches
Figure 6.12b presents a nonsymmetrical three-hinged arch with real nil point for influence line Qk; this point is located within the span of the arch. Therefore we have one portion with positive shear and two portions with negative shear. l tan β Location of NP(Qk) (Fig. 6.12a,b) may be computed by the formula uQ ¼ . tan β þ tan φk
6.4.3
Axial Force
Step 1. Find nil point (NP) of influence line Nk. If load P is located on the left half of the arch, then reaction of the support B passes through crown C. Axial force at section k equals zero, if reaction of support A will be perpendicular to tangent at point k. The nil point () is located beyond the arch span (Fig. 6.13). Step 2. Lay off along the vertical passing through the support A, the value sin φk. Step 3. Connect this ordinate with nil point and continue this line till vertical passes through crown C. A working zone is a portion between section k and vertical passing through crown C (right-hand portion 1—RHP-1). Then connect the point under crown C with support B (right-hand portion 2—RHP-2). Step 4. Left-hand portion (LHP) is parallel to right-hand portion 1 and connects two points: zero ordinate at support A and point under section k. Step 5. Take into account indirect load application; connecting line between joints 2 and 3 is not shown. Location of NP(Nk) may be computed by the formula uN ¼
NP(Nk)
l tan β . tan β cot φk
Perpendicular to tangent at k
P=1 Tangent at k k 3
jk
RA
C
5
2
6
1
RB
A NP(Nk)
*
uN
7
b
Inf. line Nk
sin jk
– LHP
RHP-2 RHP-1
Fig. 6.13 Construction of influence line Nk using nil point method
6.5
Special Types of Arches
This section is devoted to the analysis of special types of arches. Among them are arch with support points located on the different levels and parabolic three-hinged arch with complex tie.
6.5.1
Askew Arch
The arch with support points located on the different levels is called askew (or rising) arch. Three-hinged askew arch is a geometrically unchangeable and statically determinate structure. Analysis of askew arch subjected to the fixed and moving loads has some features.
6.5 Special Types of Arches
149
Design diagram of three-hinged askew arch is presented in Fig. 6.14. Let the shape of the arch be parabola, span of the arch l ¼ 42 m, and support B at Δ ¼ 3.5 m higher than support A. The total height of the arch at hinge C is f + f0 ¼ 8 m. The arch is loaded by force P ¼ 10 kN. It is necessary to calculate the reactions and bending moment at section k, construct the influence lines for thrust and bending moment Mk, and apply influence lines for calculation of bending moment and reactions due to fixed load. Equation of the axis of parabolic arch: y ¼ 4ð f þ f 0 Þ ð L x Þ
x L2
ð6:14Þ
where the span for arch A–C–B0 with support points on the same level is L ¼ l + l0 ¼ 48 m. For x ¼ 42 m (support B) the 42 Δ 3:5 ¼ 0:0833, cosα ¼ 0.9965, sinα ¼ 0.08304. ordinate yð42Þ ¼ Δ ¼ 4 8 ð48 42Þ 2 ¼ 3:5 m, so tan α ¼ ¼ l 42 48 Other geometrical parameters are f0 ¼ aC tan α ¼ 24 tan α ¼ 2:0 m ! f ¼ 8 2 ¼ 6 m ! h ¼ f cos α ¼ 6 0:9965 ¼ 5:979 m:
ðaÞ
For x ¼ 6 m (section k) the ordinate yk ¼ 3.5 m.
6.5.1.1
Reactions and Bending Moment at Section k
Reactions of supports: It is convenient to resolve total reaction at point A into two components. One of them, R0A, has vertical direction and the other, ZA, is directed along line AB. Similarly resolve the reaction at support B. These components are R0B and
Fig. 6.14 Design diagram of an askew three-hinged arch
ZB. The vertical forces R0A and R0B present a part of the total vertical reactions. These vertical forces may be computed as for reference beam: R0A ! R0B !
X X
MB ¼ 0 :
R0A 42 þ P 12 ¼ 0 ! R0A ¼ 2:857 kN
MA ¼ 0 :
R0B 42 P 30 ¼ 0 ! R0B ¼ 7:143 kN
ðbÞ
Since a bending moment at crown C is zero, ZA !
X
M left C ¼ 0:
ZA ¼ ZB ¼ Z
Z A h M 0C ¼ 0 ! Z A ¼
M 0C 2:857 24 ¼ 11:468 kN ¼ 5:979 h
where M 0C ¼ R0A l=2 is a bending moment at section C for reference beam.
ðcÞ
150
6 Three-Hinged Arches
Thrust H presents the horizontal component of the Z, i.e., H ¼ Z cos α ¼ 11:468 0:9965 ¼ 11:428 kN
ð6:15Þ
The total vertical reactions may be defined as follows: RA ¼ R0A þ Z sin α ¼ 2:857 þ 11:468 0:08304 ¼ 3:809 kN RB ¼ R0B Z sin α ¼ 7:143 11:468 0:08304 ¼ 6:191 kN
ð6:16Þ
Bending moment at section k: M k ¼ M 0k Hy ¼ 3:809 6 11:428 3:5 ¼ 17:144 kN
6.5.1.2
ð6:17Þ
Influence Lines for Thrust and Bending Moment Mk
Thrust: Since H ¼
M 0C cos α, the equation of influence line for thrust becomes h ILðH Þ ¼
cos α IL M 0C h
ð6:18Þ
The maximum ordinate of influence line occurs at crown C and equals cos α aC bC 0:9965 24 18 ¼ 1:71428 ¼ h 5:979 24 þ 18 l Bending moment: Since M k ¼ M 0k Hyk , the equation of influence line for bending moment at section k becomes ILðM k Þ ¼ IL M 0k yk ILðH Þ
ð6:19Þ
Influence line may be easily constructed using the nil point method. Equation of the line Ak is (Fig. 6.15) y¼
3:5 x ¼ 0:5833x 6
ðdÞ
Equation of the line BC is y yC ¼ m ð x xC Þ ! y 8 ¼
4:5 ðx 24Þ ! y ¼ 14 0:25x 18
ðeÞ
where m is a slope of the line BC, and x and y are coordinates of any point of the line BC. The nil point NP(Mk) of influence line for Mk is the point of intersection of lines Ak and BC. Solution of Eqs. (d) and (e) is x0 ¼ 16.8 m. Influence lines for H and Mk are presented in Fig. 6.15. Maximum positive and negative bending moment at section k occurs if load P is located at section k and hinge C, respectively. If load P is located within portion x0, then extended fibers at the section k are located below the neutral line of the arch. The thrust and bending moment at the section k may be calculated using the relevant influence lines H ¼ Py ¼ 10 1:1428 ¼ 11:428 kN M k ¼ Py ¼ 10 ð1:7143Þ ¼ 17:143 kNm These values coincide exactly with those calculated by formulas (6.15) and (6.17).
6.5 Special Types of Arches
151
As before, the influence lines for thrust constructed once may be used for its computation for different cases of arbitrary loads. Then, knowing the vertical reactions and thrust, the internal forces at any point of the arch may be calculated by definition without using influence line for that particular internal force. P=10kN
y
NP(Mk) xk= 6m
12m
C
k
B
h yk= 3.5m
A
3.5m • B⬘
a
H
H
ac=24m
8m x
bc=18m 1.71428 1.1428 +
xk=6m
Inf. line H
2.5714
+
1.7143
Inf. line Mk (m)
–
x0= 16.8m
Fig. 6.15 Three-hinged askew arch. Design diagram and influence lines
6.5.2
Parabolic Arch with Complex Tie
Analysis of such a structure subjected to fixed and moving load has some features. Design diagram of the symmetrical parabolic arch with complex tie is presented in Fig. 6.16. The arch is loaded by vertical uniformly distributed load q ¼ 2 kN/m. We need to determine the reactions of the supports, thrust, and bending moment at section k (ak ¼ 18 m, yk ¼ 11.25 m, cosφk ¼ 0.970, sinφk ¼ 0.2425) as well as to construct the influence lines for abovementioned factors.
q=2kN/m
1 k C
E
D
Parabolic arch
y
f=12m F
A⬘ A
H
H
B⬘ f0=2m
B
1
HA ak=18m RA Fig. 6.16 Design diagram of the arch with complex tie
12m l=48m
RB
x
152
6.5.2.1
6 Three-Hinged Arches
Reactions and Bending Moment at Section k
The vertical reactions are determined from the equilibrium equations of all the external forces acting on the arch: RA ! RB !
X X
MB ¼ 0 :
RA 48 þ q 12 6 ¼ 0 ! RA ¼ 3 kN
MA ¼ 0 :
RB 48 q 12 42 ¼ 0 ! RB ¼ 21 kN
ðaÞ
Horizontal reaction at support A is HA ¼ 0. The thrust H in the tie (section 1-1) is determined from the following equation: H!
X
M left C ¼0:
M 0C l RA þ H ð f f0 Þ ¼ 0 ! H ¼ ¼ 7:2 kN 2 f f0
ð6:20Þ
Equilibrium equations of joint F lead to the axial forces at the members of AF and EF of the tie. Internal forces at section k for a reference simply supported beam: M 0k ¼ RA xk ¼ 3 18 ¼ 54 kNm, Q0K ¼ RA ¼ 3 kN
ðbÞ
Internal forces at the point k for three-hinged arch are determined as follows: M k ¼ M 0k H ðyk f0 Þ ¼ 54 7:2 ð11:25 2Þ ¼ 12:6 kNm Qk ¼ Q0k cos φk H sin φk ¼ 3 0:970 7:2 0:2425 ¼ 1:164 kN N k ¼ Q0k sin φk þ H cos φk ¼ ð3 0:2425 þ 7:2 0:970Þ ¼ 7:711 kN
ð6:21Þ
Note that the discontinuity of the shear and normal forces at section E left and right at the vertical member EF are NEF cos φ and NEF sin φ, respectively.
6.5.2.2
Influence Lines for Thrust and Bending Moment at Section k
Vertical reactions: Influence lines for vertical reactions RA and RB for arch and for reference simply supported beam coincide, i.e., ILðRA Þ ¼ IL R0A ;
ILðRB Þ ¼ IL R0B :
Thrust: According to expression (6.20), the equation of influence line for thrust becomes ILðH Þ ¼
1 IL M 0C f f0
ð6:22Þ
The maximum ordinate of influence line for H at crown C: 1 l 48 ¼ 1:2 ¼ 4 4 ð12 2Þ ð f f0 Þ Influence line for thrust may be considered as key influence line.
ðcÞ
Special Types of Arches
153
1 k y
C
yk=11.25m
f=12m F
A⬘ A
H
B⬘
H f0=2m
HA
x
B
1 12m
ak=18m
l=48m
RA
RB
1.2 0.6
+
1125 .
Inf. line H
111 . (yK-f0) . IL (H)
+ 8.325
9.0
Construction Inf. line MK
IL (MK0)
(m)
11.25-8.325=2.925
+ – NP(MK)
1.05
Inf. line MK (m)
11.1-9.0=2.1
Fig. 6.17 Three-hinged arch with complex tie. Influence lines for H and Mk
Bending moment: According to expression (6.21) for bending moment at any section, the equation of influence line for bending moment at section k becomes ILðM k Þ ¼ IL M 0k ðyk f0 Þ ILðH Þ ¼ IL M 0k 9:25 ILðH Þ
ð6:23Þ
ak bk 18 30 ¼ ¼ 11:25 m at section k, so the ordinate at 48 l crown C equals 9 m. Influence line for thrust H presents the triangle with maximum ordinate 1.2 at crown C. Ordinate of the graph (yk f0) IL(H ) at crown C equals (11.25 2) 1.2 ¼ 11.1 m, so ordinate at section k equals 8.325 m. Detailed construction of influence line Mk is shown in Fig. 6.17. Since both terms in (6.23) have different signs,they should be plotted on the one side on the basic line; the final ordinates of influence line are located between two graphs IL M 0k and 9.25 IL(H ). Maximum bending moment at section k occurs if load P is located above section k and crown C. Bending moment at section k may be positive, negative, and zero. If load P is located within the portion A-NP(Mk), then extended fibers at section k are located below neutral line of the arch. Figure 6.17 also presents the construction of influence lines for bending moment using nil points; pay attention that construction of this point must be done on the basis of conventional supports A0 and B0 . History of the arched structures may be found in Timoshenko (1953), Todhunter and Pearson (1960), Benvenuto (1991), Rabinovich (1960), Bernshtein (1957). More detailed analysis of arches is discussed in Karnovsky (2012). Influence line M 0K presents a triangle with maximum ordinate
154
6 Three-Hinged Arches
Problems 6.1. Design diagram of symmetrical three-hinged parabolic arch ACB with overarched members is presented in Fig. P6.1. This structure is loaded by two fixed forces which are applied at the points 20 and 50 and transformed to the arch itself at joints 2 and 5. Provide complex analysis of arch, using the steps below. The intermediate and final results for some steps are convenient to present in tabulated form. 1. Perform kinematical analysis of the arched structure. 2. Construct the reference beam, determine the reactions of supports, and construct the bending moment and shear force diagrams. 3. Determine the reactions of supports of the arch. 4. Determine internal forces (bending moments, shear, and axial forces) at the sections A, 1, . . .7, B due to fixed loads at joints 2 and 5. Apply the analytical formulas. 5. Construct the influence lines of the internal forces at sections 2 and 6; use analytical approach. 6. Calculate internal forces at sections 2 and 6 using influence lines. 7. Compare results obtained by two different approaches (fixed and moving loads). 8. Calculate maximum internal forces at sections 2 and 6, caused by two moving connected forces P1 ¼ 80 kN and P2 ¼ 160 kN apart 2 m. 100kN
200kN
2′
5′
2 1
C 5
3
6 7
f=12m
A
B l=48m
Fig. P6.1
6.2. Design diagram of symmetrical three-hinged arch is presented in Fig. P6.2. Construct the influence lines for internal forces at section n using the nil point method.
P=1 C n
Fig. P6.2
Problems
155
6.3. Design diagram of nonsymmetrical three-hinged arch is presented in Fig. P6.3. Construct the following influence lines: (a) for vertical reactions at arch supports A and B; (b) for thrust; (c) for internal forces at sections k and n. Use the nil point method. Take into account indirect load application. P=1
C k 1
A
2
n
fc
3
ac
B
bc
Fig. P6.3
6.4.
Design diagram of three-hinged arch is presented in Fig. P6.4.
(a) Find the location of uniformly distributed load so tensile fibers at section k would be located below the neutral line and as this takes place, the bending moment would be maximum. (b) Find the location of concentrated force P so shear at section k would be positive and maximum value. (c) Find a portion of the arch where concentrated clockwise moment M should be placed so shear at section k would be positive and maximum value. P=1
C
2 n
k
3
1 A
B
Fig. P6.4
6.5.
Design diagram of three-hinged arch is presented in Fig. P6.5.
1. For the axial force at section n to be maximum, the uniformly distributed load should be located (a) left at the section n; (b) right at the section n; (c) within all span AB. 2. For the axial force at section n to be maximum a concentrated force P should be located (a) left at the section n; (b) right at the section n; (c) at the hinge C; (d) left at hinge C; (e) right at hinge C. 3. The vertical settlement of support A leads to appearance of internal forces at the following part of the arch: (a) portion AC; (b) portion CB; (c) all arch AB; (d) nowhere. 4. Decreasing of external temperature within the left half-arch AC leads to appearance of internal forces at the following part of the arch: (a) portion AC; (b) portion CB; (c) all arch AB; (d) nowhere.
C n A Fig. P6.5
B
156
6 Three-Hinged Arches
6.6. The three-hinged parabolic nonsymmetrical arch span l is subjected to concentrated force Р at the hinge C (Fig. P6.6). Determine the effect of the location of the hinge C on the value of the thrust H. (Hint: y ¼ 4f0 xðl xÞ1=l2 , H ¼ M 0C =fC .)
P C
y
Ans. H =
f0
fC
x
B
A
Pl 4 f0
l/2 a
b
Fig. P6.6
6.7. The three-hinged arch with tie is subjected to concentrated force Р at the hinge С (Fig. P6.7). Determine the effect of the location f0 of the tie on the axial force H in the tie.
P Ans. H =
C
Pl 4( f – f 0 )
f B
A
f0
l/2 Fig. P6.7
6.8. The three-hinged askew arch span l is subjected to concentrated force Р at the hinge С (Fig. P6.8). Determine the effect of the parameter f0 on the thrust of the arch. (Hint: H ¼ Z cos α.) P Pl Ans. H = 2(2 f – f 0 ) Z
C
A
a
Z R⬘A Fig. P6.8
l/2
a f h
B f0 R⬘B
Problems
157
6.9. The three-hinged symmetrical arch with span l and rise f has supports A and B on one level. Derive the equation of the rational axis of the arch for following cases of loading: (a) uniformly distributed load through the entire span (Fig. P6.9a); (b) distributed load according to triangle law (Fig. P6.9b); (c) distributed load forming two symmetrical triangles (Fig. P6.9c). Ans: ðaÞ y ¼ 4f
x x 8 x x2 x 4 x2 1 ; ð bÞ y ¼ f 1 2 ; ðcÞ y ¼ 3f 1 2 l l 3 l l 3 l l
a
b
q
q
C
y
C
y
f x
B
A
f B
A
x
l/2
l/2 q
c C y f B
A
x
l/2 Fig. P6.9
6.10. Three-hinged symmetrical arch is loaded by the radial distributed load q. Show that the rational axis of the arch presents the circle. 6.11. Prove that for three-hinged parabolic arch the total area of the influence line for bending moment at arbitrary section k is equal to zero. 6.12. Is it possible to apply the nil point method for construction of influence lines of internal forces in the following cases: (a) nonsymmetrical three-hinged arch; (b) arch with complex tie; and (c) arbitrary equation of the central line of the arch? 6.13. Is it possible to arrange vertical elements of overarched structure so that influence line of bending moment at some section would be one signed only?
Chapter 7
Cables
This chapter is devoted to the analysis of cables under fixed loads of different types. Among them are concentrated loads, and uniformly distributed load along the horizontal line and along the cable itself. Important formulas for analysis of the cable subjected to arbitrary loads are derived; they allow determining the changing of reactions, internal forces, and shape due to any additional live loads. Relationships between the thrust, internal forces, and total length of a cable are established. The influence of elastic properties of a cable is discussed.
7.1
Preliminary Remarks
The cables as the permanent members of the load-bearing structures are used extensively in modern engineering. Some examples of cabled structures are suspension bridges (Fig.7.1), anchoring systems of different objects such as guy-rope of the masts, sea drilling platforms, stadium covering, cableways, floating breakwaters, etc. Cables are also used as the temporary guys during erection of the structures. Cables are made from high-strength steel wires twisted together, and present a flexible system, which can resist only axial tension. The cables allow covering very large spans; modern suspension bridges permit coverage of spans hundreds of meters in length. This may be explained by two reasons. (1) In axial tension, the stresses are distributed uniformly within all areas of cross section, so the material of a cable is utilized in full measure; (2) cables are made from steel wires with very high ultimate tensile strength (σ υ ffi 1860 MPa, while for structural steel ASTM-A36 σ υ ffi 400 MPa). Therefore the own weight of a load-bearing structure becomes relatively small and the effectiveness of application of cables increases with the increasing of the spans. A cable as a load-bearing structure has several features. One of them is the vertical loads that cause horizontal reaction, which, as in case of an arch, is called a thrust. However, the behavior of the cable and the arch is fundamentally different. Bending moments, shear, and axial compressive forces arise in the arch, and only tensile axial forces in the cable.
Pillar Hangers
Sag
a)
Anchorage Anchor span
Main span
Fig. 7.1 Suspension bridge
Another feature of cables is the high sensitivity of their shape, depending on the total length of the cable, type of the load, as well as a load location along the span. In other words, a cable is a flexible system. The fundamental feature of a cable is its unknown shape in advance. Defining the shape of a cable is one of the important problems. The following types of cables are considered: © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_7
159
160
7 Cables
1. Cable with self-weight ignored (a cable subjected to concentrated loads, to uniformly distributed load, as well as to arbitrary dead load and after that additional any live load) 2. Cable with self-weight (with or without external load) Under these loads the cable shape takes different forms. Shape of a cable in case of concentrated loads only is defined by slopes of straight portions of a cable. Shapes of a cable in case of distributed load along horizontal and along cable itself are parabola and catenary, respectively. Note that the intensity of the load due to weight of cable itself along its axis is constant, while the intensity of the same weight along horizontal projection of the cable is variable. Assumptions: 1. The cable is an inextensible one (elastic properties of the cable will be considered in Sect. 7.6). 2. The cable is perfectly flexible; that is, the cable does not resist to shear and bending. In this case the internal force in any point of a cable is tensile, which is directed along the tangent at this point of a cable. Horizontal component of a tension force is called a thrust. It will be shown later that thrust remains constant along the length of the cable.
7.1.1
Direct and Inverse Problems
The simplest problem is finding thrust, internal forces, and total length of a cable if the shape of a cable is given. Solution of this problem is well known. However, for practice this type of problem does not have a reasonable sense because shape of the cable cannot be known in advance. Instead, an engineer knows allowable horizontal force, which a tower can resist at the top. Based on this, we will consider the following two fundamental practical problems for cables under fixed loads: 1. Find the shape of a cable and internal forces, if a thrust is given. This type of problem is called the thrust-shape problem. 2. Compute thrust and internal forces, if a total length of a cable is given. This type of problem is called the length-thrust problem.
7.1.2
Fundamental Relationships
Design diagram of a cable with supports without their mutual displacements is shown in Fig. 7.2a. The cable is loaded by some vertical concentrated loads Pi (the distance from left support is xi) and arbitrary load q distributed along horizontal axis. Since the cable is perfectly flexible the axial force Nx at any section is directed along tangent at this section. Axial forces NA and NB at the supports A and B are resolved into two directions. They are the vertical direction and direction A-B. These forces are denoted as rA and hA for support A and rB and hB for support B. Summing of all forces on the horizontal axis leads to the relationship hA ¼ hB ¼ h. The vertical component of the reaction at B: ð X X rB ! M A ¼ 0 : rB l Pi xi qðxÞxdx ¼ 0 2 3 ð 1 6X 7 Pi xi þ qðxÞxdx5 rB ¼ 4 l
l
ð7:1Þ
l
This expression coincides with general formula for reaction of simply supported beam. Therefore if axial forces at supports are resolved as shown in Fig. 7.2a, then a vertical component of axial force at support equals to corresponding reaction of the reference beam.
7.1 Preliminary Remarks
161
a
b
c
Fig. 7.2 (a) Cable subjected to arbitrary load; (b) free-body diagram of the part of the cable; (c) free-body diagram of the elementary portion of the cable
Now we can calculate ordinate of the cable at section C with abscissa c. Since the bending moment at any section C is zero, M C ¼ r A c hA f1C 2 1 f1C ¼ 4r A c h
X
ðc Pi ðc xi Þ qðxÞðc xÞdx ¼ 0 0
X
ðc
3
ð7:2Þ
Pi ðc xi Þ qðxÞðc xÞdx5 0
where f1C is perpendicular to the inclined chord AB. It is obvious that the expression in bracket (7.2) presents the bending moment at section C of the reference beam. Therefore f1C ¼
M 0C h
ð7:3Þ
To determine the vertical coordinate of the point C we need to modify this formula. For this we need to introduce a concept of a thrust H, which is a horizontal component of the axial force at any section. From Fig. 7.2b we can see that equilibrium equation ∑X ¼ 0 leads to the formula H ¼ const for any point of the cable subjected to arbitrary loads. At any support h ¼ H / cos φ. Since f1C ¼ fC cos φ, the expression (7.3) can be rewritten as fC cos φ ¼ M 0C =ðH= cos φÞ, which leads to the fundamental formula fC ¼
M 0C H
ð7:4Þ
where fC is measured vertically from the inclined chord AB. This parameter is called the sag of the cable. Formula (7.4) defines the shape of the cable subjected to any vertical load. The sag is proportional to the bending moment of the “reference beam”; proportional coefficient is 1/H. The sag and y-coordinate are equal if supports are located on the same elevation. Now the total vertical reactions of supports may be determined in terms of thrust as follows:
162
7 Cables
RA ¼ r A þ h sin φ ¼ r A þ H tan φ RB ¼ r B h sin φ ¼ r B H tan φ
ð7:5Þ
The concept of “shear” at any section of the reference beam will be helpful. The portion length of x is loaded by uniformly distributed load q, concentrated force P, reaction RA, thrust H, and axial force N(x) as shown in Fig. 7.2b. The beam shear is Q(x) ¼ RA P qx. At section x (point K ) three concurrent forces act. They are Q(x), N(x), and H. It is obvious that for any section N ð xÞ ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H 2 þ Q 2 ð xÞ
ð7:6Þ
Now let us derive the differential relationships between load q and ordinate of the cable y. The angle θ between the axial line of a cable and x-axis (Fig. 7.2b) obeys the equation tan θ ¼
dy Q ¼ dx H
ð7:7Þ
To derive a relationship between Q and intensity of the load q, consider a free-body diagram of the elementary portion of the cable (Fig. 7.2c). The axial forces at the left and right ends of a portion are N and N þ dN, respectively. Their components are H and Q at the left end, while at the right end H and Q þ dQ. Equilibrium equation ∑Y ¼ 0 leads to the following relationship: dQ/dx ¼ q. Taking into account (7.7), the required relationship between y and q becomes d 2 y qð x Þ ¼ H dx2
ð7:8Þ
The first and second integrating of Eq. (7.8) leads to the expressions for slope and y-ordinate, respectively. The constant of integration should be determined from the boundary conditions. The determination of stresses in cables in case of their various loadings may be facilitated using the influence line method. We return to relation (7.4). Suppose that the supports are at the same level. It is obvious that in the case of loading by concentrated force in the middle of the span, the greatest deflection fC is at point C under the force. We can transform the IL M 0C . expression for the thrust H to the equation of influence line for the thrust: ILðH Þ ¼ fC Influence line of the bending moment at the middle point C of the substitute beam has the form of a triangle with a vertex at the point C; the ordinate of the vertex is l/4. Therefore, the influence line of the thrust has the same form with the ordinate l/4f, where f is the cable sag in the middle. Let the cable be loaded with a uniformly distributed load q along the entire span. In this case, the thrust H becomes H l ¼ qωl ¼ q2
1 l l ql2 ¼ 2 2 4f 8f
If the load occupies the left (right) half of the span, then the thrust is H l=2 ¼ qωl=2 ¼ q
7.2
1 l l ql2 ¼ 2 2 4f 16f
Cable with Neglected Self-Weight
This section contains analysis of the perfectly flexible inextensible cables subjected to concentrated loads, as well as distributed loads along horizontal axis through the entire span. For both loading (concentrated and distributed) of a cable the thrust-shape and length-thrust problems are considered. Often, these problems are referred as direct and inverse problems.
7.2 Cable with Neglected Self-Weight
7.2.1
163
Cables Subjected to Concentrated Load
In this case of the loading each portion of the cable between two adjacent forces presents the straight segment. The simplest design diagram of a cable, subjected to one concentrated load, is presented in Fig. 7.3a. Parameters of the system are a ¼ 10 m, l ¼ 25 m, and P ¼ 20 kN. y
a
k A
a0
H RA
x
B f
NA-1 RA H
P=20kN A
k A
RB
P=20kN b=15m l=25m
a=10m
c
b H
a1
1
k
H
B
C
k
a0
a0
RA
NA-1
120kNm Fig. 7.3 (a) Design diagram of the cable; (b) free-body diagram; (c) reference beam and corresponding bending moment diagram
7.2.1.1
Thrust-Shape Problem
Determine a shape of the cable, if the thrust of the system is given as H ¼ 24 kN. Vertical reactions of the cable: RA ! RB !
X X
MB ¼ 0 : MA ¼ 0 :
Pðl aÞ 20ð25 10Þ ¼ ¼ 12 kN, l 25 Pa 20 10 ¼ ¼ 8 kN RB ¼ l 25 RA ¼
We can see that the vertical reactions do not depend on the value of the thrust H. It happens because supports A and B are located on the same elevation. The forces acting on the segment at support A and corresponding force polygon are shown in Fig. 7.3b. For assumed x-y-coordinate system the angles α0 and α1 belong to the fourth and first quadrant, respectively, so a shape of the cable is defined as follows: RA 12 1 2 ¼ ¼ ! cos α0 ¼ pffiffiffi , H 24 2 5 RB 8 1 3 ¼ ! cos α1 ¼ pffiffiffiffiffi tan α1 ¼ ¼ 24 3 H 10
tan α0 ¼
The y-ordinate of the cable at the location of the load P is y ¼ a tan α0 ¼ 10 0:5 ¼ 5 m: The negative sign corresponds to the adopted x-y-coordinate system. The sag at the point C is f ¼ 5 m. Tensions in the left and right portions of the cable may be presented in terms of thrust as follows: pffiffiffi pffiffiffiffiffi H 24 5 H 24 10 ¼ ¼ N A1 ¼ ¼ 26:83 kN, N B1 ¼ ¼ 25:30 kN cos α0 2 cos α1 3 Increasing of the thrust H leads to decreasing of the angle α0 and α1; as a result, the sag of the cable is decreasing and tension in both parts of the cable is increasing.
164
7 Cables
The tensions in the left and right portions of the cable can be determined using expression (7.6). Reference beam shear forces are QA1 ¼ RA ¼ 12 kN, Q1B ¼ RA P ¼ RB ¼ 8 kN. Therefore qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Q2A1 þ H 2 ¼ 122 þ 242 ¼ 26:83 kN; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Q21B þ H 2 ¼ ð8Þ2 þ 242 ¼ 25:30 kN
N A1 ¼ N 1B
Now let us find sag using the concept of the reference beam (Fig. 7.3c). The bending moment at point C equals M0 Pab ð20 10 15Þ 120 ¼ 120 kNm. According to (7.4) the sag f at point C is fC ¼ C ¼ MC ¼ ¼ ¼ 5 m. Obtained result 25 l 24 H presents the distance between chord AB and cable and it is not related to adopted x-y-coordinate system.
7.2.1.2
Length-Thrust Problem
Determine a thrust of the cable, if total length L of the cable is given. Length of the cable shown in Fig. 7.3a equals L¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a la þ ¼ a 1 þ tan 2 α0 þ ðl aÞ 1 þ tan 2 α1 cos α0 cos α1
According to force triangle (Fig. 7.3b), tanα0 ¼ RA/H ¼ (P/H ) (l a)/l ; therefore length of the cable in terms of active force P and thrust H may be presented as follows: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ffi P2 l a P2 a2 þ ð l aÞ 1 þ 2 2 L¼a 1þ 2 l H H l Solving this equation with respect to H leads to the following expression for thrust in terms of P, L, l, and a: 2l0 δð1 δÞ ffi, H ¼ P qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4 2 l0 2l0 2δ 2δ þ 1 þ 4δ2 4δ þ 1
L l0 ¼ , l
δ¼
a l
Let δ ¼ 0.4 and L ¼ 1.2l ¼ 30 m, so l0 ¼ 1.2. In this case the thrust equals 2 1:2 0:4ð1 0:4Þ H ¼ P qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:7339P 4 2 1:2 2 1:2 2 0:42 2 0:4 þ 1 þ 4 0:42 4 0:4 þ 1 After that, the shape of the cable and internal forces in the cable may be defined easily. The special case a ¼ 0.5l leads to the following results for thrust and sag: P H ¼ qffiffiffiffiffiffiffiffiffiffiffiffi , 2 l20 1 qffiffiffiffiffiffiffiffiffiffiffiffi Pl l ¼ l2 1 f ¼ 4H 2 0 L There are some interesting numerical results. Assume that l0 ¼ ¼ 1.01. In this case f ¼ 0.071 l, so if the total length of the l cable L exceeds the span l only on 1%, then sag of the cable comprises 7% of the span.
7.2.2
Cable Subjected to Uniformly Distributed Load
Distributed load within the horizontal projection of the cable may be considered for case when the hangers (Fig. 7.1) are located very often. Design diagram of flexible cable under uniformly distributed load is presented in Fig. 7.4.
7.2 Cable with Neglected Self-Weight
165
y A
x
B y(x)
H
H q
RA
RB
x l Fig. 7.4 A cable under uniformly distributed load
7.2.2.1
Thrust-Shape Problem
Thrust H of the cable is given. Determine the shape of the cable and calculate the distribution of internal forces. For solving of this problem we will use two approaches: 1. Integration of fundamental equation of a flexible cable (7.8) leads to the following expressions for slope and shape of the cable: dy q ¼ x þ C1 dx H q x2 y¼ þ C1 x þ C2 H 2 Constant of integration is obtained from boundary conditions: at x ¼ 0 (support A) y ¼ 0 and at x ¼ l (support B) y ¼ 0. These conditions applied to equation y(x) lead to the following constants of integration: C2 ¼ 0 and C1 ¼ ql/2H. Now the shape of a cable and slope in terms of load q and thrust H are described by equations ql2 x x2 yð xÞ ¼ ð7:9aÞ 2H l l2 dy ql x 2 1 ð7:9bÞ tan θ ¼ ¼ dx 2H l Equation y(x) presents symmetrical parabola. At x ¼ l/2, maximum y-coordinate equals to ymax ¼
ql2 8H
ð7:10Þ
The tension N at any section x in terms of thrust H is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H ¼ H 1 þ tan 2 θ ¼ H N ð xÞ ¼ cos θ
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 q 2 l2 x 1þ 2 2 1 l 4H
This equation may be obtained from (7.6), where shear Q ¼ RA qx. At the lowest point (x ¼ l/2) a tension N ¼ H. The maximum tension occurs at supports rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q2 l 2 N max ¼ H 1 þ 4H 2
ð7:11Þ
ð7:12Þ
2. The concept of the reference beam leads to the following procedure. The bending moment for reference beam at any section x is M 0x ¼ ðql=2Þx ðqx2 =2Þ. According to (7.4) and taking into account the direction of the y-axis (Fig. 7.4) we immediately get the expression (7.9a) for y(x), and as result, the formulas (7.9b, 7.10, 7.11, 7.12) for slope, ymax, tension N(x), and maximum tension Nmax.
166
7.2.2.2
7 Cables
Length-Thrust Problem
Expression for total length L of the cable is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 dy L¼ 1þ dx dx ðl
ð7:13Þ
0
Since the sag of the cable is f ¼ ql /8H, expression (7.9b) for slope at any x in terms of sag f may be presented as 2
dy ql 2x 4f 2x ¼ 1 ¼ 1 dx 2H l l l Therefore, the total length L of the cable in terms of sag according to (7.13) becomes sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðl 2 4f 2x 1þ L¼ 1 dx l l
ð7:14Þ
ð7:13aÞ
0
Approximate Solution of Length Determination It is known that the radical can be represented in a series as pffiffiffiffiffiffiffiffiffiffiffi 1 1 1 þ ε ffi 1 þ ε ε2 þ . . . 2 8
ðε 0). 2. Secondary displacements δik, i 6¼ k may be positive, negative, or zero. 3. Secondary displacements satisfy to the reciprocal displacement theorem δik ¼ δki
ð9:5aÞ
It means that unit displacements symmetrically placed with respect to principal diagonal of canonical equations are equal. 4. The determinant composed of the unit displacements of the canonical equations is strictly positive: D > 0. In particular, for a structure with two redundant constraints D ¼ δ11δ22 δ12δ21 > 0. The unit of displacements δik presents the ratio of unit for displacement according to index i and units for force according to index k. Construction of internal force diagrams. Solution of (9.4) is the primary unknowns Xi, i ¼ 1, . . ., n. After that the primary system may be loaded by determined primary unknowns and given load. Internal forces may be computed as for usual statically determinate structure. However, the following way allows once again an effective use of the bending moment diagrams in primary system. The final bending moment diagram MP may be constructed by formula M P ¼ M 1 X 1 þ M 2 X 2 þ . . . þ M n X n þ M 0P
ð9:6Þ
Thus in order to compute the ordinates of the resulting bending moment diagram, it is necessary to multiply each unit bending moment diagrams M k by corresponding primary unknown Xk and summing up with bending moment diagram due to applied load in the primary system M 0P. This formula expresses the superposition principle. Advantage of formula (9.6) is that it may be effectively presented in tabulated form.
9.2 Canonical Equations of Force Method
281
Shear forces may be calculated on the basis of bending moment diagram using relationship dM/dx ¼ Q and axial forces on the basis of shear force diagram consideration of equilibrium of joints of the structure. Finally, having internal force diagrams, all reactions are easy to determine. Procedure for analysis. The following procedure provides analysis of statically indeterminate beams and frames using the canonical equations of the force method: 1. Provide the kinematical analysis and define the degree of statically indeterminacy n of a structure. 2. Choose the primary system and replace the eliminated redundant constraints by corresponding primary unknowns Xi, i ¼ 1, . . ., n. 3. Formulate the canonical equations of the force method. 4. Apply the successive unit forces X1 ¼ 1, X2 ¼ 1, . . ., Xn ¼ 1 to primary system and for each unit primary unknown construct corresponding bending moment diagram M 1 , M 2 , . . . , M n . 5. Calculate the unit coefficients δik. 6. Construct the bending moment diagram M 0P due to applied load in primary system and calculate the load terms ΔiP of Eqs. (9.4). 7. Solve the system of equations with respect to primary unknowns X1, X2, . . ., Xn . 8. Construct the bending moment diagrams by formula (9.6), next compute the shear and construct corresponding shear force diagram, and lastly compute axial forces and construct the corresponding axial force diagram. 9. Having internal force diagrams, calculate the reactions of supports. Other way is consider primary system subjected to determined primary unknowns and given load and provide computation of all internal forces by definition; this way is less effective. 10. Provide the static control for all structure (or any of its part). 11. Provide the kinematical control (9.2) for displacements of an entire structure in direction of primary unknowns. Intermediate checking of computation These verifications are recommended to be performed before solving canonical equations for determining primary unknowns Xi, i.e., after steps 5 and 6 of the algorithm above. For control of unit displacements and free terms it is necessary to construct summary unit bending moment diagram M Σ ¼ M 1 þ M 2 þ þ M n . The following types of controls are suggested as follows: (a) Row verification of unit displacements: Multiply summary unit bending moment diagram M Σ on a primary bending moment diagram M i : ð MΣ Mi X ds M 1 þ M 2 þ þ M n M i ¼ δi1 þ δi2 þ . . . þ δin ¼ EI EI The result of this multiplication equals to the sum of unit displacements of the i-th equation. (b) Total verification of unit displacements: Multiply the summary unit bending moment diagram M Σ on itself. n X MΣ MΣ ¼ δ11 þ þ δ1n þ δ21 þ þ δ2n þ þ δn1 þ þ δnn ¼ δik |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} EI i, k¼1 1th equation
2nd equation
nth equation
The result of this multiplication equals to the sum of all unit displacements of canonical equations. It is quite sufficient to perform only total control; however, if errors occur, then the row control should be performed for tracking the wrong coefficient. (c) Verification of the load displacements: Multiply summary unit bending moment diagram M Σ on a bending moment diagram M 0P . ð M Σ M 0P X ds M 1 þ M 2 þ þ M n M 0P ¼ Δ1P þ Δ2P þ . . . þ ΔnP ¼ EI EI The result of this multiplication equals to the sum of all free terms of canonical equations.
282
9.3
9 The Force Method
Analysis of Redundant Beams
Analysis of the redundant beams may be performed by different methods. Among them are the force method in canonical form, three-moment equations, and focal ratios method. Below all of these methods are considered.
9.3.1
Canonical Equation of the Force Method
Applying this method to the analysis of continuous beams is based on the canonical equations of the force method in the form (9.4). Let us consider two-span continuous beam on rigid supports (Fig. 9.8a). Figure 9.8b shows one version of the primary system, which presents the set of statically determinate beams. The primary unknown X1 is bending moment at intermediate support. P
a
0
1
EI=constant
2 n a=0.4l
b=0.6l
l
b
l P
X1
0
2
X1=1
c
M1 1
P
d
M P0 Pab = 0.24 Pl l P
e
M1
2 n 0.4l
l
l
R2=0.304P
0.096Pl
f
MP 0.1824Pl
g
0.696P 0
1
0.096P
2
Q kN
n
R0
P
+
R1
0.304P
R2
Fig. 9.8 (a, b) Design diagram of a beam and primary system; (c, d) bending moment diagrams in the primary system due to unit primary unknown (M 1 ) and given load (M 0P ); (e) primary system loaded by primary unknown and given load; (f) final bending moment diagram; (g) shear force diagram and reactions of supports
9.3 Analysis of Redundant Beams
283
Canonical equation of the force method is δ11X1 þ Δ1P ¼ 0, where δ11 is displacement in the direction of the first primary unknown due to unit primary unknown X1 ¼ 1; Δ1P is displacement in the same direction due to applied load. This canonical equation shows that for the adopted primary system the mutual angle of rotation at support 1 caused by primary unknown X1 and the given load P is zero. For calculation of displacements δ11 and Δ1P it is necessary to construct the bending moment diagrams in the primary system caused by unit primary unknown X1 ¼ 1 and acting loads; they are shown in Figs. 9.8c and 9.8d, respectively. For calculation of unit displacement δ11 we need to multiply the bending moment diagram M 1 by itself δ11 ¼
M1 M1 1 1 2 2l rad 1l 1¼ ¼2 EI 2 3 3EI kNm EI
For calculation of free term we need to multiply the bending moment diagram M 0P by M 1 Δ1P ¼
M 1 M 0P l Pl2 ½1 ð2 0:4Þ þ 0 ð1 þ 0:4Þ 0:24Pl ¼ 0:064 ¼ ðradÞ 6EI EI EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Table A2, line 5
Primary unknown (bending moment at support 1) is X1 ¼
Δ1P ¼ 0:096PlðkNmÞ: δ11
Construction of bending moment diagram. The bending moment diagram can be constructed considering each simply supported beam separately under action of applied load and obtained bending moment at support, as shown in Fig. 9.8e; constraints at support 1 are shown apart for convenient. Final bending moment diagram is shown in Fig. 9.8f. Computation of shear. Having the bending moment diagram we can calculate shear forces using differential relationships dM Q¼ . In our case we get dx 0:096Pl ¼ 0:096P l 0:1824Pl ð0:096PlÞ ¼ 0:696P ¼ 0:4l 0:1824Pl ¼ 0:304P ¼ 0:6l
Q01 ¼ Q1n Qn2
Final shear force diagram is shown in Fig. 9.8g. Reactions of supports. Having shear force diagram we can calculate the reaction of all supports. They are the following: R0 ¼ 0:096P,
R1 ¼ Qright Qleft 1 1 ¼ 0:696P ð0:096PÞ ¼ 0:792P,
R2 ¼ 0:304P
Static verification. Equilibrium condition for all structure in whole is X Y ¼ 0:096P þ 0:792P þ 0:304P P ¼ 1:096P þ 1:096P ¼ 0
Summary 1. Adopted primary system as a set of simply supported beams leads to the simple (triangular) shape of bending moment diagrams in each unit state. Therewith the bending moment diagram for primary unknown Mk is distributed only within two adjacent spans, left and right of support k. The bending moment diagram due to given load is located only within each loaded span. Therefore, computation of coefficients and free terms of canonical equations is elementary procedure. Moreover, the set of canonical Eq. (9.4) occurs incomplete. In more detail this problem is discussed in Sect. 9.3.2. 2. Canonical equations of the force method allow easy to take into account the different bending stiffness for each span.
284
9.3.2
9 The Force Method
The Three-Moment Equation (Clapeyron Theorem)
These equations realize the idea of the force method in the application to a continuous beam. The method is especially effective in the case of a beam with a large number of spans. Three-moment equations establish relationships between the bending moments on three successive supports. Let us consider continuous beam with hinged supports at the ends (Fig. 9.9a). The number of spans, their lengths, and the way of loading the beam are not specified. Numbering of supports is from left to right. The index of span coincides with the index of the right support.
a
P
0
1
4
3
2 EC
l1
b
l2 M1
0 c
P
M2
d
EC
θnright θnleft n
l4
l3
Mn-1 n-1
M3
4
q
Mn
ln
θnleft
n
Fig. 9.9 (a) Design diagram of continuous beam; (b) primary system; (c) elastic curve at the n-th support; (d) fragment of the primary system: the factors which affect on the slope at the n-th support are Mn1, Mn and arbitrary load q
The degree of redundancy of this structure is equal to the number of intermediate supports. As the primary unknowns, we will adopt the bending moments at supports. The primary system is a set of single-span statically determinate beams loaded with support moments and a load at the span (Fig. 9.9b). The problem is to determine the moments at supports. The arbitrary load leads to the deformation of a beam; the fragment of an elastic curve (EC) is shown by dotted line. Since the elastic curve is smooth and without kink, then slope infinitesimally close left and right to the any support n are equal (Fig. 9.9c) right θleft n ¼ θn
ð9:7Þ
Now we will consider the simply supported n-th span, which is subjected to given load q and unknown moments Mn1 and Mn at supports n 1 and n, respectively. Elastic curve within span ln and slope at the left of support n, θleft n , are shown in Fig. 9.9d. According to superposition principle M n1 q n θleft þ θM n þ θn , n ¼ θn
ð9:8Þ
M n1 ln M n M l , θn ¼ n n are slopes at support n caused by moments Mn1 and Mn. Negative sign means that 6EI n 3EI n section at n-th support (this section belongs to the n-th span) rotates counterclockwise. For computation of slope θqn caused by given load q we will apply the conjugate beam method. First of all we need to construct the bending moment diagram M due to the given load q only. Then we need to construct the fictitious (conjugate) beam (for given simply supported beam the fictitious beam is also simply supported), which should be loaded by the bending moment diagram of the given beam (Fig. 9.10). The resultant load ωn (which is numerically equal to total area of the bending moment diagram) passes through the center gravity of the diagram M. Fictitious reaction on the n-th support, fictitious shear force, and actual slope at support n are n1 where θM ¼ n
9.3 Analysis of Redundant Beams
285
q
a n-1
n
ln
b
M
c ωn Af
an
bn
Bf
Fig. 9.10 (a, b) Simply supported n-th span subjected to given load q, and corresponding bending moment diagram; (c) fictitious beam
Bf ¼
Qf ωn a n ω a ω a , Q f ¼ B f ¼ n n , θn ¼ ¼ n n ln ln EI ln EI n
Thus we get a final expression for slope left of support n, θleft n , considering the n-th span is θleft n ¼
Bfict M n1 ln M n ln ωn an M l M l ¼ n1 n n n n 6EI n 3EI n ln EI n 6EI n 3EI n EI
ð9:9Þ
The last term of (9.9) presents the slope at the right support of span n caused by load within this span. , considering the (n þ 1) span: Similarly, the final expression for slope right of support n, i.e., θright n fict
θright ¼ n
M n lnþ1 M nþ1 lnþ1 ωnþ1 bnþ1 M n lnþ1 M nþ1 lnþ1 Anþ1 þ þ ¼ þ þ 3EI nþ1 6EI nþ1 lnþ1 EI nþ1 3EI nþ1 6EI nþ1 EI
ð9:9aÞ
The last term of (9.9a) presents the slope at the left support of the span n þ 1 caused by load within this span. The positive sign means that section at n-th support (this section belongs to the (n þ 1) span) rotates clockwise. For uniform continuous beam (In ¼ Inþ1 ¼ const), which is made from one material (E ¼ const), the condition (9.7) leads to the following expression M n1 ln þ 2M n ðln þ lnþ1 Þ þ M nþ1 lnþ1
ωn an ωnþ1 bnþ1 ¼ 6 þ þ Afict ¼ 6 Bfict n nþ1 ln lnþ1
ð9:10Þ
f means the left reaction of fictitious beam for span n þ 1, while Bnf means the right fictitious reaction for span n. Here Anþ1 So both of these reactions arise at support n. Equation (9.10) establishes a relationships between bending moments on three successive supports, n 1, n, and n þ 1. For a beam with hinged supports at the ends, the number of equations of type (9.10) is equal to the number of intermediate supports. The author of this remarkable equation is outstanding scientist B.P.E. Clapeyron (1857).
286
9 The Force Method
The complex 6Afict and 6Bfict for different loading are presented in Table 9.3. Table 9.3 Components of three-moment equation for specified loading; A fict and B fict are the left and right reactions of fictitious beam (Rabinovich 1960, part 1)
Continuous beam is widely used in modern engineering. Also, continuous beam is best for testing each new method of analysis or modification of the existing methods. The reason for such a preference of the continuous beam over the other structures is that of all the static indeterminate systems, the continuous beam has the greatest simplicity and regularity of formation for any number of redundant unknowns.
9.3.2.1
Special Cases
1. Assume the left end of the beam is clamped (Fig. 9.11a). In this case we need to introduce additional non-loaded simply supported span A-0 of length ladd ¼ l0 (Fig. 9.11b). Three-moment equation for supports A–0–1 is ω0 a 0 ω1 b 1 þ M A ladd þ 2M 0 ðladd þ l1 Þ þ M 1 l1 ¼ 6 l0 l1
9.3 Analysis of Redundant Beams
287
a
q
M 1
0 l1
b A
q 0 ladd=l0
l1
M
3 l3
l2
1
F
P 2
a
P 3 M 3
2
l2
l3
Fig. 9.11 Continuous beam with clamped support and loaded cantilever
Since ladd ¼ 0 and ω0 ¼ 0, then this equation becomes 2M 0 l1 þ M 1 l1 ¼ 6
ω 1 b1 ¼ 6Afict 1 l1
2. Let beam have a loaded cantilever (Fig. 9.11a). In this case the moments at last support 3 is known. Equation for bending moments at supports 1–2–3 becomes ω2 a 2 ω3 b 3 þ M 1 l2 þ 2M 2 ðl2 þ l3 Þ þ M 3 l3 ¼ 6 , where M 3 ¼ Fa l2 l3
9.3.2.2
Discussion
Let us discuss the case when for primary unknowns of continuous beam are taken the reactions of supports. In this case, the primary system is a simply supported beam with cantilever. Following the canonical equations of the force method, in order to determine the unit and loaded terms of equations, it is necessary to plot the bending moment diagrams in the primary system caused by the unit primary unknowns and the given load. The peculiarity of these diagrams is that they are distributed throughout the primary system. Therefore, all the coefficients and the free terms turn out to be nonzero. In other words, the system of canonical equations of the method of forces turns out to be complete, i.e., each equation includes all unknowns. This is a serious drawback in comparison with three-moment equations, since Clapeyron equations turn out to be three-term. If we take the support moments as the primary unknowns and apply the procedure of the method of forces in the canonical form, then, as in the Clapeyron method, we obtain a system of three-term equations. This is explained by the fact that each individual unit bending moment diagram is distributed along two spans only. Thus, the choice of the primary system significantly affects the laboriousness of the calculations and, as a result, the accuracy of the calculations. If for continuous beam the primary system is adopted as a set of simply supported beams, then canonical equation of the force method and Clapeyron equation express the same idea: mutual angle on the intermediate support is equal to zero. As result, each equation which are composed by these remarkable methods are occurs three-terms ones. Example 9.2 Design diagram of uniform three-span continuous beam is shown in Fig. 9.12a. Calculate the bending moment at the supports. Solution The structure is statically indeterminate to the second degree. The primary unknowns are the bending moments at supports 1 and 2, M1 and M2 (they are not shown). The primary system is a set of three simply supported beams. The bending moment diagrams in the primary system due by given loads is shown in Fig. 9.12b.
288
9 The Force Method
P
l2/2
0
a
q
2
1 l1
l2
l3 P
0
B1fict A2fict
A1fict
c M1=5.8kNm
1
0
R0
l=8m
R1left
q
M1
3
2
1
b
3
ω2
B2fict A3fict
ω3
B3fict
P=15kN
M2=21.8kNm 2
1
R1right
l/2=4m
R2left
M2
2 R2right
q=3kN/m
3 l=8m
R3
Fig. 9.12 (a) Design diagram. (b) Primary system—three simply supported beams: bending moment diagrams and computation of right parts of three moment of equations; (c) reaction of each separate span and total reactions of support
Three-moment equations for supports 1 and 2 are
ω1 a 1 ω 2 b 2 þ þ Afict M 0 l1 þ 2M 1 ðl1 þ l2 Þ þ M 2 l2 ¼ 6 ¼ 6 Bfict 1 2 l1 l2 ωa ω b M 1 l2 þ 2M 2 ðl2 þ l3 Þ þ M 3 l3 ¼ 6 2 2 þ 3 3 ¼ 6 Bfict þ Afict 2 3 l2 l3 In our case we have M0 ¼ M3 ¼ 0. fict Span 1. Since load is absent, then ω1 ¼ 0, Afict 1 ¼ B1 ¼ 0 Span 2. The span is loaded by force P: Pl2 Pl2 1 Pl ω2 ¼ l2 2 ¼ 2 ! Afict ¼ Bfict ¼ 2: 2 2 2 4 8 16 The right side of the first Eq. (a) Pl22 3 fict fict 6 B1 þ A2 ¼ 6 0 þ ¼ Pl22 8 16 ql2 2 ql2 ql2 Span 3: The span is loaded by uniformly distributed load q: ω3 ¼ l3 3 ¼ 3 ! Afict ¼ Bfict ¼ 3 3 3 3 8 12 24 The right side of the second Eq. (a) 2 ql2 Pl2 ql23 3 fict fict þ ¼ Pl22 þ 3 6 B2 þ A3 ¼ 6 8 16 24 4
ðaÞ
9.3 Analysis of Redundant Beams
289
Three-moment equations are 3 2M 1 ðl1 þ l2 Þ þ M 2 l2 ¼ Pl22 8 ql2 3 M 1 l2 þ 2M 2 ðl2 þ l3 Þ ¼ Pl22 3 8 4
ðbÞ
If l1 ¼ l2 ¼ l, then solution of this system becomes 3 ql2 Pl þ 40 60 3 ql2 M 2 ¼ Pl 40 15
M1 ¼
ðcÞ
The signs show us that stretched fibers at support 2 are located above axial line of a beam, while stretched fibers at support 1 caused by load q are located below. Reactions of supports are conveniently determined numerically. Assume l ¼ 8 m, P ¼ 15 kN, q ¼ 3 kN/m. In this case the moments at supports are M1 ¼ 5.8 kNm and M2 ¼ 21.8 kNm. Now we need to consider each span separately subjected to the moments at supports and the load in span (Fig. 9.12c). X R0 ! M 1 ¼ R0 l M 1 ¼ 0 ! R0 ¼ 0:725kN ð#Þ X Span 01 left M 0 ¼ Rleft Rleft 1 ! 1 l M 1 ¼ 0 ! R1 ¼ 0:725kN ð"Þ
Span 12
Span 23
X Rright ! M 2 ¼ 0 ! Rright ¼ 5:5kN ð"Þ 1 1 X M 1 ¼ 0 ! Rleft Rleft 2 ! 2 ¼ 9:5kN ð"Þ X Rright ! M 3 ¼ 0 ! Rright ¼ 14:725kN ð"Þ 2 2 X M 2 ¼ 0 ! R3 ¼ 9:275kN ð"Þ R3 !
The total reaction of supports R0 ¼ 0:725kN ð#Þ;
right R1 ¼ Rleft ¼ 0:725 þ 5:5 ¼ 6:225kN ð"Þ 1 þ R1
right ¼ 9:5 þ 14:725 ¼ 24:225kN ð"Þ, R2 ¼ Rleft 2 þ R2
R3 ¼ 9:275kN ð"Þ
ðdÞ
Equilibrium condition for entire structure in whole X Y ¼ R0 þ R1 þ R2 þ R3 P ql ¼ 0:725 þ 6:225 þ 24:225 þ 9:275 15 3 8 ¼ ¼ 39:725 39:725 ¼ 0
9.3.2.3
Summary
1. The advantage of the Clapeyron equations (three-moment equations) is that they turn out to be three-term while the method of forces in canonical form in general case leads to a complete system of equations, where each equation contains all the unknowns. 2. If continuous beam is performed from one material, then the moments at supports, internal forces, and reactions do not depend on the modulus of elasticity. It means, two beams which are made from different materials and subjected to same loads experience the same forces
290
9 The Force Method
3. Three-moment equations allow considering beams with elements of variable bending rigidity EI. In this case, the rigidity of one span should be taken as the basic EI0, and the rigidity of the other spans should be expressed in terms of basic rigidity. Example 9.3 Design diagram of uniform four-span continuous beam with equal spans is shown in Fig. 9.13. The beam is loaded by force P in the first span. Calculate the bending moment at the supports. P 0
1 l
l/2
4
3
2 l
l
fict B1fict A2
Fig. 9.13 Continuous beam with pinned supports
Solution The structure is statically indeterminate to the third degree. Since bending moments M0 ¼ M4 ¼ 0, all spans are equal and only first span is loaded, then three-moment equations become 3 fict fict M 0 l þ 2M 1 ðl þ lÞ þ M 2 l ¼ 6 Bfict þ A þ 0 ¼ Pl ¼ 6 B 1 2 1 8 fict fict M 1 l þ 4M 2 l þ M 3 l ¼ 6 B2 þ A3 ¼ 0 M 2 l þ 4M 3 l þ M 4 l ¼ 0 fict Here Bfict 1 means the right fictitious reaction of span 1, while A2 means the left fictitious support of the span 2. Thus, the fict fict value B1 þ A2 means the total fictitious reactions of support 1. After simplification we get
3 4M 1 þ M 2 ¼ Pl 8 M 1 þ 4M 2 þ M 3 ¼ 0 M 2 þ 4M 3 ¼ 0 Solution of this set of equations is 45 Pl ¼ 0:1004Pl 32 14 3 M2 ¼ Pl ¼ 0:0268Pl 8 14 2 3 M3 ¼ Pl ¼ 0:00670Pl 4 8 14 M1 ¼
These results show the features of the distribution of the support moments of a continuous beam when only one span is loaded. 1. The signs of the bending moments at support alternate. 2. With the distance from the loaded span, the bending moments at supports quickly decrease. In our case M 1 M 2 M 3 ¼ 0:1004 0:0268 0:00670 ¼ 1 0:267 0:0667 9.3.2.4
Short Historical Remarks
The continuous beam as a research object attracted the attention of many eminent scientists. First of all, we note the contribution of Euler, who considered an absolutely rigid two-span beam on flexible supports. The paradoxical result (the greatest pressure is transferred to one of the extreme supports at any load) led him to the conclusion that it is necessary to take into account the beam deformation.
9.3 Analysis of Redundant Beams
291
Eytelwein (1808) and later Navier (1826) considered a continuous two-span elastic beam on rigid supports. They took the support reactions as the primary unknowns. Navier for each of the spans compiled the equations of the elastic line and determined the constants of integration from the four boundary conditions. The solutions and formulas obtained by these authors were very cumbersome. At this time, the practice was little interested in solving this problem. The next stage in the history of continuous beam refers to 1848. During the battles of the French people with government forces, a five-span continuous bridge over the Seine at Asnières was seriously damaged. The prominent French scientist B.P. Clapeyron (1799–1864) was involved in the restoration of the bridge. He was not familiar with the problem of continuous beam analysis. Application of the Navier method led to enormous difficulties and Clapeyron proposed to adopt a set of one-span beams as a primary system, and the bending moments at the points over supports as primary unknowns. Unfortunately, on this progressive new path he did not perform simple intermediate procedure (excluding the angle of rotation on the support); as a result for analysis of the n-span beam, 2n three-term equations should be compiled. However, this result in comparison with the Navier method represented a significant step in the problem of the analysis of a continuous beam. And only nine years later he saw the possibility of further simplification of the equations, which led to the famous equations of three moments. Clapeyron easily shared his ideas with his colleagues, and was in no hurry to publish new results, although his friend Gabriel Lamè strongly encouraged him to do so and promoted the scientific results of his long-time colleague. Therefore, the ideas and scientific achievements of Clapeyron in his surroundings were known long before they appeared in print. The author of the first publication devoted to the continuous beam problem was engineer Henri Bertot (1855). This material was published in the minutes of the meetings of the Society of Civil Engineers of Paris (15.06 and 22.07, 1855) and contained the verbal formulation of the equation of three moments for a beam of constant section under uniformly distributed load without any justification (Bernshtein 1957, page 208). Timoshenko in his History has noted the following: “In this work, Bertot refers to Clapeyron’s idea but he does not derive the theory, giving only the method of solving a system of these equations” (Timoshenko 1953, page 145). For the first time, the reader becomes acquainted with Clapeyron’s ideas and his initial results through publications of L. Molinos and C. Pronnier (1857, Paris) as well as F. Laissle and A. Schübler (1857, Stuttgart). After their publications in the same year on the pages of the most authoritative French magazine “Comptes Rendus de l’Acad. Des Sciences de Paris” (pages 1076–1080) Clapeyron’s article appears where the final version of three-moment equation had been presented. In 1858, Heppel, a railroad engineer in India, independently derived an equation of three moments without having at his disposal literature and published his work (Proc. Of the Royal Society, 1870, vol XIX, 123, June 1870). Truly, if the idea is claimed, then a way to implement it can be found in different parts of the globe. Thus, in the period of 1855–1857, the three-moment equation became well known, immediately entered into engineering practice, and the name “Clapeyron’s equation” was assigned to this equation. In the period 1860–1880, the three-moment equation was generalized to various cases of great practical importance. Among them, beams of variable stiffness, with different spans, are under the action of arbitrary loads, temperature effects, settlement of supports, etc. For more detailed information, see Rabinovich 1960 part 2; Bernshtein 1957; Timoshenko 1953.
9.3.3
Focal Ratios Method
This method allows effective performing analysis of continuous beams with a large number of spans, provided that a beam contains only one loaded span. Let us consider design diagram shown in Fig. 9.14. Continuous five-span beam is loaded in one span, for example, in the first span. The expected bending moment diagram in this span is shown by dotted line. A bending moment M1 arises at support 1. It is obvious that the sole source of stress-deformable state of non-loaded spans is M1; expected bending moment diagram within non-loaded spans is shown by thin line. M1
P 0
F2 N
2
F3
l1 Fig. 9.14 Continuous beam having one loaded span
F4
4
5
3
1 l2
l3
l4
l5
292
9 The Force Method
Assume within the span 0–1 the type of the loading has changed, also the location of the load, and its value. How these changes will effect on the bending moment diagram within the non-loaded spans? Obviously, in the span 0–1 the character of the distribution of the bending moments and, of course, the magnitude of the support moment M1 will change. It can be said that, in comparison with the initial loading, the support moment M1 will be the k times larger (smaller), and maybe changes the sign. This means that all ordinates of the bending moment diagram at the sections outside the loaded span, i.e., in the non-loaded spans, will be changed by the same factor. Consequently, at those points where the ordinates were equal to zero before changing the load, remain equal to zero after changing the load. Points of non-loaded spans in which the bending moments are zero are called the foci. At these points, the elastic line of the beam undergoes an inflection. The point F5 is a special focal point for support 5; the point N cannot be called as focus since it belongs to the loaded span. Each span of a beam contains the left and right focal points (or simply, foci). If all unloaded spans are located to the left of the loaded span, then the bending moment diagram in all left spans is passing through left focal points only. These focal points are indicated as F Ln , F Ln1 (Fig. 9.15a). If non-loaded spans are located to the right of the loaded span, then the bending moment diagram in all right spans is passing through right focal points only, which is indicated as F Rn , F Rnþ1 (Fig. 9.15b). The left and right focal ratios relate consecutive support moments as follows: M n1 , M n2 M ¼ n , M nþ1
kLn1 ¼ kRnþ1
Mn , M n1 M kRn ¼ n1 , Mn kLn ¼
M nþ1 Mn M n2 ¼ M n1
k Lnþ1 ¼ kRn1
a
Mn Mn-2 FnL−1
*
FnR−1
Loaded span
n-1
*
*
FnL
n-2
Mn-1
ln-1
b
ð9:11Þ
*
FnR
n
Elastic curve
ln
Mn-1 Loaded span
FnR
*
*
FnL n-1
n
FnL+1
*
FnR+1
*
n+1
Mn ln
Mn+1
ln+1
Fig. 9.15 Explanation of the foci points: (a) left foci points; (b) right foci points
The concepts of “foci” and “focal ratios” had been introduced by famous French scientist J.A.C. Bresse (1859). Now we establish relationship between the focal ratios of two consecutive spans. Assume we have uniform beam which is made from one material, so bending stiffness EI for beam is constant. In this case the three-moment equation for two spans, n and n þ 1, is M n1 ln þ 2M n ðln þ lnþ1 Þ þ M nþ1 lnþ1 ¼ 0 We divide both sides of the equation into Mn M n1 M l þ 2ðln þ lnþ1 Þ þ nþ1 lnþ1 ¼ 0 Mn n Mn Now, taking into account formulas (9.11), we can present this equation in terms of focal ratio
9.3 Analysis of Redundant Beams
293
1 ln þ 2ðln þ lnþ1 Þ kLnþ1 lnþ1 ¼ 0 k Ln
This equation allows presenting the left recursive focal ratio kLnþ1 for (n þ 1) span in terms of the previous left focal ratio kLn as follows kLnþ1 ¼ 2 þ
ln
2
lnþ1
1 kLn
ð9:12Þ
The right recursive relationship for previous (n 1)-th span in terms of the next right focal ratio kRn is kRn1 ¼ 2 þ
ln ln1
2
1 k Rn
ð9:12aÞ
To be able to apply these relationships, we need to determine the focal ratios for the first span. For this, we will consider two particular cases of boundary conditions of a beam. Case a. If the very left support 0 is pinned, then left focus for the first span coincides with support 0 and the left focal ratio M becomes kL1 ¼ 1 ¼ 1 (Fig. 9.16a). M0
a
b
M1 F1L 0
A
*
1
l1
M0
ladd=l0
M1
F1L
0
*
1 l1
Fig. 9.16 Focal ratios for pinned and fixed supports
Similarly, if the very right support is pinned, then for last span the right focal ratio is kRlast ¼ 1. Case b. Let the very left support 0 be clamped (Fig. 9.16b). In this case the three-moment equation for additional span A-0 of length ladd ¼ l0 ! 0 and for span 0–1 of length l1 becomes: 2M0l1 þ M1l1 ¼ 0. This equation immediately leads to the left M focal ratio k L1 ¼ 1 ¼ 2. It means that if a very left support is clamped and first span is non-loaded, then a moment at the M0 clamped support is twice less than moment at the next pinned support. Similarly, if a last right support is clamped and last span is non-loaded, then a moment at the clamped support is twice less than in previous pinned support. Thus, focal ratios are positive numbers in the range of two to infinity, 2 k 1. If load is applied only within one span ln, then distribution of bending moments is shown in Fig. 9.17. This figure also contains the left foci points on left non-loaded spans and right foci points on right non-loaded spans. The nil points c1 and c2, which are located within a loaded span, are not foci. Mn
Mn-1 FnL−2
*
ln-2
L n-2 Fn−1
*
n-1 ln-1
*c1
*c2 ln
FnR+1
n
n+1
FnR+2 n+2
*
* ln+1
ln+2
Fig. 9.17 Continuous beam. Notation, bending moment diagram in the given structure due to applied load, left and right foci points
Now let us determine the moments Mn1 and Mn which arise at supports n 1 and n of the loaded span ln; in fact these moments are transferred to the non-loaded left and right parts of the beam, respectively. For this purpose, we write the following three-moment equations (Fig. 9.17).
294
9 The Force Method
fict M n2 ln1 þ 2M n1 ðln1 þ ln Þ þ M n ln ¼ 6 Bfict þ A , n n1 fict M n1 ln þ 2M n ðln þ lnþ1 Þ þ M nþ1 lnþ1 ¼ 6 Bfict n þ Anþ1 ,
Bfict n1 ¼ 0; Afict nþ1 ¼ 0
In order to determine bending moments Mn1, Mn, which arise at supports of the loaded n-th span, the ratios M M M n2 ¼ Ln1 and M nþ1 ¼ R n should be substituted into equations above; solution of these equations are as follows kn1 k nþ1 A f kR B f M n1 ¼ 6 n L n R n ln k n k n 1
ð9:13Þ
B f kL A f M n ¼ 6 n L n R n ln k n k n 1
Here kLn , kRn are left and right focal ratios for n-th span; Anf , Bnf are fictitious left and right reactions for n-th loaded simply supported beam. Expressions of (9.13) are called the Bresse formulas. These formulas are not interconnected and does not depend on the total number of spans. Knowing the bending moments of a loaded span the bending moments at other supports are easily determined as follows: M n1 M ; M n3 ¼ Ln2 ; . . . k Ln1 k n2 Mn M nþ1 ¼ R ; M nþ2 ¼ R ; . . . knþ1 knþ2
M n2 ¼ M nþ1
Note, that after all focus ratios are calculated, before proceeding to formulas (9.13), they should be verificated as follows. In the case of a continuous beam with a constant cross section, different spans, and with the hinge ends, the relationship between the left and right focal ratios has the form (Rabinovich 1960, part 2). L L L k 2 k3 k4 . . . kLm lm ¼ kR1 kR2 k R3 . . . kRm1 l1 ð9:14Þ For continuous beam with left hinged end and with right clamped end we have the following relationship L L
k2 k3 . . . k Lm1 2kLm 1 lm ¼ 2kR1 kR2 . . . k Rm1 l1
ð9:14aÞ
Summary The focal ratios method is based on the idea of the force method, which is realized in the form of the Clapeyron equations, and is supplemented by the new concepts. They are the foci and focal relations. These concepts characterize the continuous beam itself, and do not depend on the nature of the load. Example 9.4 The uniform beam with equal spans (li ¼ l ¼ const) is loaded by concentrated force at the first span (Fig. 9.18). Determine the bending moments at supports. Apply the focal ratios method.
P F2R 2
0 1 l1/2 Fig. 9.18 Continuous beam with pinned end support
*
* l2
F3R
*
* l3
3
F4R *4
* l4
9.3 Analysis of Redundant Beams
295
Solution 1. Computation of the left and right focal ratios. General formula for left focal ratios is kLnþ1
¼2þ
ln lnþ1
1 2 L kn
For computation of left focal ratios we start from left span of the beam and perform them in the following order kL1 ! kL2 ! kL3 ! kL4 The left focal ratios for all spans are l 1 1 ¼ 4, kL2 ¼ 2 þ 1 2 L ¼ 2 þ 1 2 1 l2 k1 l 1 1 k L3 ¼ 2 þ 2 2 L ¼ 2 þ 1 2 ¼ 3:75, 4 l3 k2 l 1 1 k L4 ¼ 2 þ 3 2 L ¼ 2 þ 1 2 ¼ 3:733 3:75 l4 k3
k L1 ¼ 1,
General formula for right focal ratios is kRn1
1 ¼2þ 2 R ; ln1 kn ln
For computation of right focal ratios we start from right span of the beam and perform them in the following order k R4 ! kR3 ! kR2 ! kR1 l4 1 1 R R ¼ 4, k4 ¼ 1, k3 ¼ 2 þ 2 R ¼2þ1 2 1 l3 k4 l 1 1 kR2 ¼ 2 þ 3 2 R ¼ 2 þ 1 2 ¼ 3:75, 4 l2 k3 l 1 1 kR1 ¼ 2 þ 2 2 R ¼ 2 þ 1 2 ¼ 3:733 3:75 l1 k2 Control kL2 kL3 k L4 . . . kLm lm ¼ 4 3:75 3:733l4 ¼ 55:995l4 , kR1 kR2 k R3 . . . kRm1 l1 ¼ 3:733 3:75 4l1 ¼ 55:995l1 , ðl1 ¼ l4 Þ 2. Determination of the support moments of loaded span. The area of the bending moment diagram in the first span 1 Pl Pl2 ω1 ¼ l ¼ 2 4 8 The reactions of conjugate beam of the loaded span 1 Pl2 fict Afict 1 ¼ B1 ¼ ω1 2 ¼ 16 The support moments of the loaded span l1 are:
296
9 The Force Method
A f kR B f A f kR B f M n1 ¼ 6 n L n R n ! M 0 ¼ 6 1 L 1 R 1 ¼ 0, ln k n k n 1 l1 k 1 k 1 1
kL1 ¼ 1
B f kL A f B f kL A f M n ¼ 6 n L n R n ! M 1 ¼ 6 1 L 1 R 1 ln k n k n 1 l1 k 1 k 1 1 The peculiarity of loading of the first span with hinged support is as follows: the expression for M0 contains a factor kL1 ¼ 1 in denominator only (therefore M0 ¼ 0), while for M1 contains the same factor in the numerator and denominator. To disclose the uncertainty, we divide the numerator and denominator by k L1 ¼ 1. As a result, we get M 1 ¼ 6
B1f Pl2 1 ¼ 0:1004Pl ¼ 6 R 16 3:733l l1 k1
3. Determination of moments on subsequent supports Mn M 0:1004Pl ! M 2 ¼ R1 ¼ ¼ 0:0268Pl; R 3:75 k nþ1 k2 M M 0:0268Pl ¼ Rnþ1 ! M 3 ¼ R2 ¼ ¼ 0:00669Pl 4 knþ2 k3
M nþ1 ¼ M nþ2
The same results were obtained earlier using three-moment equations (Example 9.3). If the first span of the beam is loaded, then a bending moment diagram in the non-loaded spans passes through the right focal points, as shown in Fig. 9.18. Example 9.5 Uniform four-span continuous beam with different span lengths is loaded by concentrated force P in the second span (Fig. 9.19). Provide static analysis of the beam, if length l2 ¼ 10 m, u ¼ 0.4 (υ ¼ 0.6). P
a 0
1 l1=6m
1.0918
b
υl2=6m
ul2=4m
F1L
P
0.545
0.6766 F3R
0.6415
F4R
*
* P
4 0.2727
R0
2
Q(kN) Factor P
3 0.03354
0.3586 R1
Factor P
0.2930
0 1
M(kNm)
0.1677
1.475
c
l4=5m
l3=8m
k
*
4
3
2
R2
R3
R4
Fig. 9.19 Analysis of continuous beam by focal ratio method. (a) Design diagram of the beam; (b) bending moment diagram and corresponding elastic curve; (c) shear force diagram and reactions of supports
Solution Numeration of the supports and spans is shown in Fig. 9.19a. 1. Calculating the left and right focal ratios Left focal ratios. General formula for left focal ratios and these ratios for spans l1 and l2 are
9.3 Analysis of Redundant Beams
297
1 ¼2þ 2 L ; lnþ1 kn l 1 6 1 1 L L 2 ¼ 2:90 2 L ¼2þ k1 ¼ 2, k2 ¼ 2 þ 10 2 l2 k1
k Lnþ1
ln
Right focal ratios. General formula for right focal ratios and these ratios for spans l2, l3, and l4 are l 1 k Rn1 ¼ 2 þ n 2 R ln1 k n l 1 5 1 kR4 ¼ 1, kR3 ¼ 2 þ 4 2 R ¼ 2 þ 2 ¼ 3:25, 8 1 l3 k4 l 1 8 1 kR2 ¼ 2 þ 3 2 R ¼ 2 þ 2 ¼ 3:3538 10 3:25 l2 k3 Reactions at conjugate beam of the loaded span, according to Table 9.3, for u ¼ 0.4, υ ¼ 0.6 are 1 2 1 2 2 Afict 2 ¼ 6 Pl2 uυð1 þ υÞ ¼ 6 Pl2 0:4 0:6ð1 þ 0:6Þ ¼ 0:064Pl2 1 2 1 2 2 Bfict 2 ¼ 6 Pl2 uυð1 þ uÞ ¼ 6 Pl2 0:4 0:6ð1 þ 0:4Þ ¼ 0:056Pl2 General formulas for bending moment Mn1 and Mn at the left and right supports of the loaded span ln: A f kR B f B f kL A f M n1 ¼ 6 n L n R n , M n ¼ 6 n L n R n , ln kn k n 1 ln kn kn 1 Thus, the bending moments M1 and M2 at the left and right supports 1 and 2 of loaded span l2 become: A f kR B f 0:064 3:3538 0:056 2 M 1 ¼ 6 2 L 2 R 2 ¼ 6 Pl ¼ 1:0918P kNm l2 ð2:90 3:3538 1Þ 2 l2 k 2 k 2 1 B f kL A f 0:056 2:9 0:064 2 M 2 ¼ 6 2 L 2 R 2 ¼ 6 Pl ¼ 0:6766P kNm l2 ð2:9 3:3538 1Þ 2 l2 k2 k2 1 The moment at supports of the non-loaded spans are M0 ¼
M 1 1:091P ¼ 0:545P kNm, ¼ 2 k L1
M3 ¼
M 2 0:545P ¼ 0:1677P kNm, ¼ 3:25 kR3
M4 ¼ 0
Bending moment diagram is shown in Fig. 9.19b; the positive ordinates are plotted below. Corresponding elastic curve is shown by dotted line. Inflection points coincide with foci points. Location of the extended fibers is shown by dotted line in Fig. 9.19a. For computation of bending moment under the force we need to consider simply supported beam subjected to force P and moments M1 ¼ 1.0918P kNm and M2 ¼ 0.6766P kNm. As a result we get Mk ¼ 1.475P kNm (Fig. 9.19b). dM Shear forces on the different portions are determined by differential relationship Q ¼ dx ð0:545 þ 1:091ÞP ¼ 0:2727P, 6 ð1:091 þ 1:475ÞP ð1:475 þ 0:6766ÞP ¼ 0:6415P, Qk2 ¼ ¼ 0:3586P, ¼ 4 6 ð0:6766 þ 0:1677ÞP 0:1677P ¼ 0:2930P, Q34 ¼ ¼ 0:03354P ¼ 8 5
Q01 ¼ Q1k Q23
The shear force diagram is shown in Fig. 9.19c; the positive ordinates are plotted above. Reaction of supports are
298
9 The Force Method
R0 ¼ 0:2727Pð#Þ, R1 ¼ ð0:2727 þ 0:6415ÞP ¼ 0:9142Pð"Þ, R2 ¼ ð0:3586 þ 0:2930ÞP ¼ 0:6516Pð"Þ, R3 ¼ ð0:2930 þ 0:03354ÞP ¼ 0:3265Pð#Þ,
R4 ¼ 0:03354Pð"Þ
All reactions are shown in Fig. 9.19c. Verification of results X Y ¼ R0 þ R1 þ R2 þ R3 þ R4 P ¼ ð0:2727 þ 0:9142 þ 0:6516 0:3265 þ 0:03354 1ÞP ¼ ð1:5992 þ 1:5993ÞP 0 Let the force P occupy a new position, with new parameters u and υ. In this case the focal ratios does not change (these parameters inherent to the structure itself), but the support moments of the loaded span will change. The further procedure of analysis remains same. Locations of zero points of the bending moment diagram for non-loaded spans remain unchanged. It can be seen, the focal ratio method allows easy construction of the influence line of the moments at supports for continuous beam. It is obvious that in case of loading of the several spans, the superposition principle may be applied. One of the advantages of the focal ratios method is that with an arbitrary loading of one of the spans of a continuous beam, the zone of the stretched and compressed fibers of the beam becomes immediately known (Fig. 9.19a, b). Comments 1. The focal ratios method does not use in the explicit form the concepts inherent to the force method. Among them are the degree of static indeterminacy, primary system and primary unknowns, and compatibility equations. However, the fundamental formulas of the focal ratios method (9.12) and (9.13) are derived on the basis of three-moment equations. 2. Method of the focal ratios does not require forming and solution of algebraic equations. Instead, it is necessary to compute the moments at the ends of the loaded span (for this two independent algebraic expressions should be used), and then, applying the focal ratios (9.11) to distribute these moments along non-loaded spans. The increase in the number of intermediate supports practically does not affect the laboriousness of the analysis. The need to develop a new method is largely related to calculations, namely, the desire to reduce cumbersome calculations, and to provide the necessary accuracy of results. Therewith a fundamental question inevitably arises: with what accuracy should calculations be done and how much can you trust the results obtained? These questions are closely related to the concept of the design diagram of a structure and adopted primary system (primary unknowns) (Perelmuter and Slivker 2003). If for a continuous beam the support reactions are chosen as the primary unknowns, then we get a complete system of algebraic equations, i.e., each of the equations contains all the unknowns. If for a continuous beam the moment at supports are chosen as the primary unknowns, then we obtain a system of algebraic equations, each of which contains the moments on three successive supports. It is clear that for each primary system there is a corresponding certain system of equations and accuracy of calculations. At the same time, the predictable question arises: whether or not the adopted primary system ensures the stability of solution? In other words, would the small changes in the coefficients and/or free terms lead to small changes in the solutions as well? As an example, let us consider a system of linear algebraic equations, which corresponds to some primary system 800X 1 þ 399X 2 ¼ 200 399X 1 þ 200X 2 ¼ 100 The exact solution of these equations is X1 ¼ 100, X2 ¼ 200. Suppose that the refined system of equations is 800X 1 þ 399X 2 ¼ 200 399X 1 þ 201X 2 ¼ 100 With a smallest change in one of the coefficients (200 ! 201) one would expect only a small change in the solution. However, solution of these equations is X1 ¼ 50.1(100), X2 ¼ 99.9(200); in brackets are shown solution for entire set of equation. Small changes in unit displacement led to cardinal changes in X1 and X2. This is the result of ill-conditioned matrix of the system, when determinant of a system presents a difference of close values. In our case, the determinant of the entire primary system turned out to be equal 800 200 3992 ¼ 160000 159201. In this case the solution is very sensitive to the change coefficient of algebraic equations.
9.3 Analysis of Redundant Beams
299
Different primary systems in force method inevitably lead to different values of the basic unknowns. The force method requires justified choice of the primary system. Unsuccessful primary system can lead to a poorly conditioned matrix of a system of equations. The method of focal ratios is not connected with the choice of the primary system. Therefore, in this method there is no choice element, the relations (9.13) are inherent only to the structure itself and these formulas are independent. As a result, the focal ratios method does not require a solution of algebraic equations and the question related to the degree of conditionality of the matrix does not arise. The founder of the method of focal ratios is famous French scientist Jacques A.C. Bresse (1859). Later this method was developed in the works E. Winkler (1862), К. Culmann (1866, sec. edn 1875), М. Levy (1886), B.N. Zhemochkin (1929), Ya. M. Rippenbejn (1933), I.M. Rabinovich (1936) (Bernshtein 1957).
9.3.4
Redundant Beam with Intermediate Hinge
In engineering practice there are statically indeterminate beams with peculiarities. Among them is redundant beam with the intermediate hinges. Figure 9.20 presents two design diagrams of the corresponding structures.
a
b
X1
X1
X2
Fig. 9.20 (a, b) Redundant beams with intermediate hinges and simplest version of the primary system
A feature of this class of beams: the presence of a hinge leads to a discontinuity in slope. In this case, the recommended method of analysis is the forces method in canonical form. The primary system is selected as usual, by removing redundant constraint and replacing them with primary unknowns (Fig. 9.20a, b). The primary system is the Gerber-Semikolenov beam. It may be presented in the form of an interaction diagram, which allows visualizing the scheme of transmitting the reactions of a suspended beam to the main beam. This fact is important in determining the coefficients and the free terms of the canonical equations. Example 9.6 The uniform one-span fixed-fixed beam АB with intermediate hinge C is subjected to uniformly distributed load q (Fig. 9.21a). Provide analysis of this structure and determine the reaction of supports. Solution The beam under consideration is redundant to the second degrees. Let the primary unknowns be the moment X1 and the horizontal reaction X2 at support B. Corresponding primary system and interaction diagram are shown in Fig. 9.21b. This diagram contains the main beam AC and secondary beam CB. Canonical equations of the force method are δ11 X 1 þ δ12 X 2 þ Δ1P ¼ 0, δ21 X 1 þ δ22 X 2 þ Δ2P ¼ 0, where δ11 and δ12 are displacements in the direction of first primary unknown due to unit primary unknowns X1 ¼ 1 and X2 ¼ 1, respectively; Δ1P is displacement in the same direction due to applied load; δ21, δ22 and Δ2P are the displacements in direction of second primary unknown due to unit primary unknowns X1 ¼ 1, X2 ¼ 1, and to applied load, respectively. 1 4 ¼ arises, which is transferred to the beam AC. In the unit state (X1 ¼ 1) at support C the reaction RC ðX 1 Þ ¼ 3l=4 3l Bending moment diagrams in the first unit condition is shown in Fig. 9.21c. Bending moment diagram is located within two beams, AC and CB. Therefore the unit displacement δ11 ¼ δ11(AC) þ δ11(CB).
300
9 The Force Method
1 1 þ M CB M CB ¼ EI EI 1 1 l 2 1 1 1 3l 2 1 7l þ 1 1 ¼ 2 3 4 3 3 EI 2 4 3 EI 27EI |fflfflfflffl{zfflfflfflffl} |{z} |fflfflffl{zfflfflffl} |{z}
δ11 ¼ M AC M AC
ΩAC
ΩCB
yAC
yCB
q
a A
B C
l/4
l X1
b
A
X2 C B
C
A l/4
3l/4 RC(X1)=4/3l C
c
X1=1
B
MCB
1/2
Unit State
1/3
yc2
RC(X)=4/3l
yc1
1 y c1 = 2 . 1 , y c2 = 3 . 1 3 3 4 3
MAC
A
C l/4
3l/4 RC(q)=3ql/8
d C
*
Loaded State
MA(RC)=3ql2/32 A xc1
RC(q)=3ql/8
* C
xc2
*
MCBmax(q)=9ql2/128
MAC(RC)
MA(q)=ql2/32 A
B MCB(q)
MAC(q)
x c1 = 1 . l , x c2 = 1 . l 3 4 4 4
C
Fig. 9.21 One-span redundant beam with intermediate hinge. (a) Design diagram; (b) primary system and its interaction diagram; (c, d) unit and loaded states and corresponding bending moment diagrams
The second primary unknown Х2 is axial force. If we take into account only flexural deformations, then we obtain δ12 ¼ δ21 ¼ δ22 ¼ 0; therefore X2 ¼ 0. This conclusion can be done from another viewpoint. Since support A prevents longitudinal displacement, it is not necessary to install constraint at point B, which also prevents axial displacement. Therefore, the clamped support B can be replaced by the sliding support, which prevents the vertical and angular displacement of support B, while allows its longitudinal displacement.
9.3 Analysis of Redundant Beams
301
3l is loaded by the uniformly distributed load q. Reaction 4 2 1 3l 3ql q 3l 9 2 . The bending moment in the middle of the beam CB is M CB max ¼ ql . ¼ of support RC ðqÞ ¼ q ¼ 2 4 8 8 4 128 3ql which is transmitted from the beam CB, and uniformly The main beam AC is loaded by the reaction RC ðqÞ ¼ 8 distributed load q, which acts on beam AC itself. The bending moment diagrams caused by each of these loads are denoted by MAC(RC) and MAC(q) (Fig. 9.21d). The asterisk * means the position of the center of gravity of the bending moment diagram in the loaded state, xC1, xC2; the corresponding ordinates of the bending moment diagrams in the unit state are yC1, yC2.
Free term of canonical equation Δ1P. Suspended beam CB of length
The free term of canonical equations caused by the bending of two beams CB and AC AC Δ1P ¼ ΔCB 1P þ Δ1P , M CB ðqÞ M CB 2 9ql2 3 1 1 9 ql3 l 1 ¼ ¼ ¼ ΔCB 1P 3|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} 2 EI 16 32 EI EI 128 4 |{z} Ω
ΔAC 1P
¼
ΔAC 1P ðRC Þ
þ
yc
ΔAC 1P ðqÞ:
Displacement ΔAC 1P caused by reaction RC and distributed load q are, respectively, ΔAC 1P ðRC Þ ¼
M AC ðRC Þ M AC 1 3 2 l 2 1 1 ql3 ¼ ql ¼ 2|fflfflfflfflfflfflffl 32ffl{zfflfflfflfflfflfflfflffl4} |ffl{zffl} 3 3 EI 32 12EI EI Ω
ΔAC 1P ðqÞ ¼
yc1
M AC ðqÞ M AC 1 1 2 l 3 1 1 ql3 ¼ ql ¼ EI 3|fflfflfflfflfflfflffl 32ffl{zfflfflfflfflfflfflfflffl4} |ffl{zffl} 4 3 EI 32 48EI Ω
yc2
So, the free term which is caused by the bending of the beam AC becomes AC AC ΔAC 1P ¼ Δ1P ðRC Þ þ Δ1P ðqÞ ¼
5 ql3 32 48 EI
The total free term of canonical equation
Δ1P ¼
ΔCB 1P
þ
ΔAC 1P
3 9 5 ql 1 ql3 ql3 þ ð27 þ 5Þ ¼ ¼ ¼ 32 16 32 48 EI 32 48 EI 48EI
Primary unknown 3 27ql2 9 2 Δ1P ql 7l X1 ¼ ql ¼ ¼ ¼ 112 7 48 δ11 48EI 27EI
ðclockwiseÞ
Further analysis of the structure of the difficulties does not cause. X
2
9 2 qð3l=4Þ 3 27 ql þ RB l ¼ 0 ! RB ¼ qlð"Þ; 112 4 56 2 X 27 29 RA ! Y ¼ ql þ ql þ RA ¼ 0 ! RA ¼ qlð"Þ, 56 56 2 X left 29 l qðl=4Þ 11 2 ql MA ! M C ¼ ql þ MA ¼ 0 ! MA ¼ 56 4 112 2
RB !
M right ¼ C
This result was previously obtained by the method of initial parameters (Chap. 8, Example 8.9). The reader is invited to compare both methods in terms of their effectiveness, explain the signs in the expressions for the reactions, and show a sketch of the elastic line of the beam.
302
9.4
9 The Force Method
Redundant Plane Frames
In this section are shown the detailed procedures of the force method in canonical form applied to the frames subjected to fixed loads. The analysis includes the construction of diagrams of internal forces, determination of displacements, and description of the nature of the deformation, as well as the corresponding verifications. The distinctive feature of the framed structures is presence, at least, of one rigid joint. Such joint, when the frame is deformed, can move in an arbitrary direction and rotate. However, the initial angle between the rods at the junction point remains unchanged. In other words, the mutual rotation angle of the rods forming the rigid joint is zero. Section 9.4.4 deals with the analysis of frameworks with elastically flexible joints and supports. In frames of various types the predominant type of deformation is bending. Therefore, in calculation of coefficients and free terms of the canonical equations in the Mohr–Maxwell integral (8.46), the terms that take into account the axial and shear forces are omitted. For computation of coefficients and free terms of canonical equations is recommended to apply procedure of multiplication of bending moment diagrams. If the frame element has variable stiffness along its length, it is recommended to divide it into a series of portions with constant stiffness within each part and apply the Vereshchagin rule. If the frame elements have different flexural rigidity, then it is convenient to take the rigidity of one element as the base one EI0, and express the rigidity of the remaining elements through the base one, EIi ¼ kiEI0. After determining the basic unknowns, the calculation of internal forces and reactions are performed in the following order: Bending moments ! Shear forces ! Axial forces ! Reactions. In the case of a frame with closed contour, it is necessary to distinguish external and internal statical indeterminacy. If three support constraints are imposed on such structure, it turns out externally statically determinate and internally statically indeterminate. For frames with closed contour the notion of “primary system” is extended. It is shown that three unknown forces, which are statically equivalent to the primary unknowns, can be located at an arbitrary point in the plane; this allows reducing a number of secondary unit displacements to zero.
9.4.1
Frames of the First Degree of Redundancy
Design diagram of the simplest frame is presented in Fig. 9.22a. The flexural stiffness for all members is EI. It is necessary to construct the bending moment diagram, to show sketch of the elastic curve of the frame, and calculate a horizontal displacement of the crossbar. Primary system and primary unknown. The structure has four unknown reactions, so the degree of redundancy is n ¼ 4 3 ¼ 1. Let us choose the primary unknown X1 as a vertical reaction at point 1. The primary system is obtained by eliminating of support 1 and replacing it by X1 (Fig. 9.22b). Canonical equation of the force method is δ11X1 þ Δ1P ¼ 0. This equation shows that for the adopted primary system the vertical displacements of the left rolled support caused by primary unknown X and the given load q is zero. Bending moment diagrams in the primary system caused by unit primary unknown X1 ¼ 1 and given load is shown in Fig. 9.22c, d. These graphs also show the displacements along eliminated constraint. The unit displacement δ11 is obtained by “multiplying” the M 1 graph by itself, i.e., δ11 ¼
M1 M1 1 1 2 1 4 a3 m aa aþ aaa¼ ¼ EI 2 3 EI 3 EI kN EI
Displacement in the primary system due to applied load is Δ1P ¼
M 1 M 0P 1 1 qa2 3 1 qa2 5 qa4 ¼ a a aa¼ ðm Þ EI 3 2 4 EI 2 8 EI EI
The negative sign at each term means that ordinates of the two bending moment diagram M 1 and M 0P are located on different sides of the neutral line of the corresponding members of the frame.
9.4 Redundant Plane Frames
303
a
b
q
q 1
Design diagram
X1
a
3
2
EI
3¢
Primary system
EI
4 a
c
d
a
d 11
qa 2 8
q
a
D1P
X1=1 Unit
e
15 2 qa 32
a/2
X1
Loaded state
state
M P0
M1
f
qa 2 2
M 1 . X1
qa 2 32
q
g P=1
7qa 64
2
qa2/32 M
MP= M 1 . X 1 + M P0
h
1·a q
X1
X1=1 1 qa2/8 1 M1
M P0
Fig. 9.22 (a, b) Design diagram of a frame and primary system, (c, d) unit and loaded states and corresponding bending moment diagrams, (e, f) construction of final bending moment diagram using the superposition principle, (g) unit state for calculation of horizontal displacement, (h) new version of primary system and corresponding bending moment diagrams
The primary unknown is X1 ¼
Δ1P 15 ¼ qa ðkNÞ 32 δ11
The positive sign shows that the chosen direction for the primary unknown coincides with its actual direction. Construction of bending moment diagram. The final bending moment diagram will be constructed using the superposition principle M ¼ M 1 X 1 þ M 0P . The first term presents the bending moment diagram due to actual primary unknown X1 ¼ 15qa/32 (Fig. 9.22e), and the second term presents the bending moment diagram due to given load in primary system as presented in Fig. 9.22d. The ordered calculation of bending moments at specified points of the frame is presented in Table 9.4. Signs of bending moments are chosen arbitrarily and used only for convenience of calculations; these rules do not influence the final bending moment diagram. In our case, signs are accepted as shown below.
304
9 The Force Method Table 9.4 Calculation of final bending moments
Points 1 2 3 3⬘ 4 Factor
M P0 0 +1/8 +1/2 -1/2 -1/2 qa2
M 1 ∙ X1 0 -15 /64 -15 /32 +15 /32 +15 /32 qa2
M1 0 -a/2 -a +a +a
M 1 ∙ X 1 + M P0 0 -7 /64 + 1/32 - 1/32 - 1/32 qa2
Signs of bending moments + −
+
−
The final bending moment diagram is presented in Fig. 9.22f. As earlier, the bending moment diagram is shown on the extended fibers. This diagram allows tracing a corresponding elastic curve of the frame; this curve is shown by dotted line. Rolled support at point 1 does not prevent horizontal displacement of the crossbar, i.e., this structure presents the frame with sidesway. All ordinates of bending moment diagram for vertical member 3–4 are located on one side (the extended fibers for this member are located to the right from the axial line). Therefore, elastic curve for this member has no point of inflection, the joint is shifted to the left from the initial position and rotated counterclockwise. For crossbar in the zone of negative bending moments, the extended fibers are located below axial line, while in a small zone of positive bending moments (in the vicinity of the rigid joint), they are located above the axial line. Elastic curve for crossbar has point of inflection; this point coincides with section where M ¼ 0. Static verification of bending moment diagram. Rigid joint of the frame has to be in equilibrium. In our case (Fig. 9.22f) M 3 M 30 ¼ ðqa2 =32Þ ðqa2 =32Þ ¼ 0. Direction of the bending moment at point 3 and 30 is shown according to the location of the extended fibers. Kinematical verification. Displacement in the direction of constraint 1 in the original system has to be zero. This displacement may be calculated by multiplication of two bending moment diagrams, i.e., one is final bending moment diagram for given structure MP (Fig. 9.22f), and the second is bending moment diagram M 1 in unit state (Fig. 9.22c). Δ1 ¼
MP M1 a 1 7 1 1 1 qa4 qa4 0 0 þ 4 a qa2 a qa2 a a qa2 ¼ ¼ ¼0 6EI 2 ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl 64 32 32 EI 32EI 32EI |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl} EI vertical element
horizontal portion, Simpson rule
Horizontal displacement of the crossbar. Unit state is shown in Fig. 9.22g. Required displacement is Δhor ¼
MP M 1 1 qa2 qa4 ¼ a1a ¼ EI EI 2 32 64EI
The positive result means that crossbar shifted from right to left. Another version of the primary system and corresponding bending moment diagrams M 1 and M 0P are shown in Fig. 9.22h. In this 4 a qa3 qa2 , Δ1P ¼ . A primary unknown X 1 ¼ ; this is bending moment at the rigid joint, as shown in Fig. 9.22f. case δ11 ¼ 3 EI 24EI 32 Property of Statically Indeterminate Frames of the First Degree of Redundancy Let us consider important property of any statically indeterminate frames of the first degree of redundancy. Design diagram of the frame is presented in Fig. 9.23. Two primary systems and corresponding bending moment diagrams for the unit conditions are presented in Fig. 9.23 and denoted as versions 1 and 2. For version 1 the primary system is obtained by eliminating the support constraint, and for version 2 by introducing a hinge at joint C. Therefore, the primary unknown in version 1 is reactions of support; primary unknown in the version 2 is bending moment at joint C, so the dimensions of δ11 is m/kN for version 1 and rad/kNm for version 2.
B
C
X1=1
h
X1=1
1/ h
1
1 h
q
h/l Version 2
Version 1
A
1
1/ h
l
h/l Fig. 9.23 Property of a primary system for structure of the first degree of redundancy
1/ l
1/ l
9.4 Redundant Plane Frames
305
The unit bending moment diagrams for any primary system are similar. This is a general property for a statically indeterminate structure with the first degree of redundancy. After bending moment diagram is constructed, the kinematical verification must be performed. This procedure involves multiplication of the final and unit bending moment diagram in any primary system. It is obvious that multiplication of the final and unit bending moment diagram for each version of primary system will be equal to zero. However, meaning of these multiplications will be different. For example, M M vers1 ¼ 0 means that horizontal displacement at point B is zero; M M vers2 ¼ 0 means that mutual angle of rotation of two sections at joint C is zero.
9.4.2
Frames of the Second and More Degree of Redundancy
Detailed analysis of statically indeterminate frame using canonical equations of the force method is presented in Example 9.7; this design diagram will be analyzed in more detail at a later time (settlements of supports, change of temperature, other methods of analysis). Example 9.7 A frame is clamped at point A and rolled at points B and C as presented in Fig. 9.24a. The frame is loaded by force P ¼ 8 kN and uniformly distributed load q ¼ 2 kN/m. The relative flexural stiffness of each element is shown in a circle, i.e.,EIAD ¼ EIDC ¼ 1EI, while EIDB ¼ 2EI. Construct the bending moment, shear, and normal force diagrams. Determine the reactions of supports. Solution 1. Primary system and primary unknowns. The structure has five unknown reactions. The degree of indeterminacy is n ¼ 5 3 ¼ 2. One version of the primary system with primary unknowns X1 and X2 (vertical and horizontal reactions at points B and C) is presented in Fig. 9.24b. Canonical equations of the force method are δ11 X 1 þ δ12 X 2 þ Δ1P ¼ 0 δ21 X 1 þ δ22 X 2 þ Δ2P ¼ 0
ðaÞ
These equations show that for the adopted primary system the vertical displacement at support B and horizontal displacement at support C caused by both primary unknowns and the given loads are zero. 2. The unit displacements and free terms of canonical equations. Figure 9.24c–e presents the bending moment diagrams M 1, M 2 in the unit states and diagram M 0P caused by applied load in the primary system; also these diagrams show the unit and loaded displacements δik and ΔiP. The graph multiplication method for determination of unit displacements and free terms leads to the following results: M1 M1 1 1 1 2 666:67 m 10 5 10 þ 10 10 10 ¼ ¼ 1EI 2EI 2 3 EI kN EI M2 M2 1 1 2 170:67 m δ22 ¼ 88 8¼ ¼ 1EI 2 3 EI kN EI M M2 1 3þ8 275 m δ12 ¼ δ21 ¼ 1 5 10 ¼ ¼ 1EI 2 EI kN EI M 1 M 0P 1 1 1 Δ1P ¼ 32 5 10 25 5 10 ¼ 1EI 1EI 3 EI 1 4 2294 ð2 10 þ 6Þ 32 ¼ ðmÞ 2EI 6 EI 0 M2 MP 1 3þ8 Δ2P ¼ 5 32 ¼ 1EI 2 EI 1 5 1161:25 ð8 25 þ 4 5:5 6:25 þ 3 0Þ ¼ ðmÞ 1EI 6 EI δ11 ¼
ðbÞ
ðcÞ
306
9 The Force Method
a
C
P=8kN B
A
3
X1
Primary system
6m
δ22
d
X1=1
M P0
M2
8
q=2
25 32
X2=1
f
3
3 13
D
D
6
10
Δ1P
6.25
5.5
M1
P=8
32
δ12
3
10
Δ2P
e
X2=1
δ11
6
10
5m
7
2
6
1
4m
δ21
.
q
4
8
6 3
Design diagram
1
P
4
2
q=2kN/m
X2
5
3m
1
D
c
b
ΣMD = 0
10
X1=1
MΣ
13
18
g
C
13.085 1.817
11.349
13.085
D
q EC
1.817
ΣMD = 0
A
8.036
i
1.817 3
3.756
6.108
H3
3.756 A
HA
k
6.108
− A
6
3.756
N6-7
3
N3-A
j C D
D
1.892 Q kN
+ 6.244
6.108
4
3.756
+
q
8.036
B
D
11.268 1.324 M kNm
h
11.268
P
P=8kN
3.756
B 7
q=2kN/m N kN
1.892 6.244
8.036 6.108
Fig. 9.24 (a, b) Design diagram of a frame and primary system, (c–e) bending moment diagrams in primary system caused by unit primary unknowns and given load, (f) summary unit bending moment diagram, (g) final bending moment diagram, elastic curve (EC) and static control for joint D, (h) separated element A-3 for calculation of shear, (i) shear force and free-body diagram for joint D, (j) axial force diagram; (k) reactions of supports
9.4 Redundant Plane Frames
307
3. Verification of coefficients and free terms of canonical equations. The unit and loaded displacement should be checked before solving canonical equations (a). For this purpose we need to construct the summary unit bending moment diagram M Σ ¼ M 1 þ M 2 (Fig. 9.24f). For joint D we have ∑MD ¼ 3.0 þ 10 13 ¼ 0. (3a) First row control. The sum of coefficients in the first canonical equation must be equal to the result of multiplication of the summary unit bending moment diagram M Σ by a primary bending moment diagram M 1 . Indeed, 666:67 275 941:67 þ ¼ , EI EI EI M M1 1 13 þ 18 1 1 2 941:67 while Σ ¼ 5 10 þ 10 10 10 ¼ EI 1EI 2 2EI 2 3 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} δ11 þ δ12 ¼
Portion AD
Portion DB
Therefore, the first row control is satisfactory. The second row control may be performed similarly, i.e., δ21 þ δ22 ¼
MΣ M2 EI
(3b) Simultaneous control. The total sum of all coefficients of all canonical equations must be equal to the result of multiplication of summary unit bending moment diagram by “itself” 1 1387:34 ð666:67 þ 275 þ 275 þ 170:67Þ ¼ , EI EI MΣ MΣ 1 1 2 1 5 33 3þ ð2 13 13 þ 2 18 18 þ 13 18 þ 18 13Þþ ¼ 1EI 2 3 1EI 6 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} δ11 þ δ12 þ δ21 þ δ22 ¼
Portion DB
Portion CD
1 1 2 1387:34 þ 10 10 10 ¼ 2EI 2 3 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Portion AD
(3c) Free terms control. The sum of the loaded displacements is Δ1P þ Δ2P ¼
1 3, 455:25 ð2, 294 þ 1, 161:25Þ ¼ EI EI
Multiplication of the summary unit bending moment diagram M Σ by a bending moment diagram M 0P due to applied load in the primary structure M Σ M 0P 1 5 13 þ 18 18 57 þ 13 32 þ 4 38:25 ¼ 1EI 6 2 EI 1 4 3, 455:25 ð2 10 þ 6Þ 32 ¼ 2EI 6 EI Therefore, the coefficients and free terms of Eq. (a) are computed correctly. 4. Primary unknowns. Canonical equations for primary unknowns X1 and X2 become 666:67X 1 þ 275X 2 2, 294 ¼ 0 275X 1 þ 170:67X 2 1, 161:25 ¼ 0
ðdÞ
All coefficients and free terms contain factor 1/EI, which can be canceled. It means that primary unknowns of the force method depend only on relative stiffnesses of the elements. The solution of these two equations leads to X 1 ¼ 1:8915kN,
X 2 ¼ 3:7562kN
5. Internal forces. The bending moment diagram of the structure will be readily obtained using the expression M ¼ M 1 X 1 þ M 2 X 2 þ M 0P
ðeÞ
308
9 The Force Method
Location of the specified points 1–8 is shown in Fig. 9.24b. Corresponding calculation is presented in Table 9.5. The sign of bending moments is chosen arbitrarily for summation purposes only. Table 9.5 Calculation of bending moments at the specified points
Points 1(A) 2 3 4 5 6 7 8
M1 -10 -10 -10 0.0 0.0 -10 -6.0 0.0
M 1 ∙ X1
-18.915 -18.915 -18.915 0.0 0.0 -18.915 -11.349 0.0
M2 -8.0 -5.5 -3.0 -3.0 0.0 0.0 0.0 0.0
M 2 ∙ X2
M P0
M
-30.049 -20.659 -11.268 -11.268 0.0 0.0 0.0 0.0
+57.0 +38.25 +32.0 0.0 0.0 +32.0 0.0 0.0
+8.036 -1.324 +1.817 -11.268 0.0 +13.085 -11.349 0.0
Signs of bending moments + −
+
−
The resulting bending moment diagram and corresponding elastic curve are presented in Fig. 9.24g. The position of the extended fibers in the vicinity of joint D is shown by dotted lines. Statical verification of bending moment diagram. The free-body diagram of joint D is presented in Fig. 9.24g. Equilibrium condition of this joint is X M ¼ 11:268 þ 1:817 13:085 ¼ 0 Kinematical verification of bending moment diagram. Displacement in the direction of any primary unknown in the given system must be equal to zero. This condition is verified by multiplying bending moment diagram MPin the actual state by bending moment diagram M i in any primary system. Displacement in the direction of the first primary unknown is Δ1 ¼
M1 MP 1 5 ð8:036 10 1:817 10 þ 4 1:324 10Þþ ¼ 1EI 6 EI 1 4 þ ð2 13:085 10 þ 2 11:349 6 13:085 6 þ 10 11:349Þ 2EI 6 1 1 2 þ 11:349 6 6 ¼ 2EI 2 3 1 ð195:453 195:511Þ 0 EI
The relative error is 0.029%. Similarly, it is easy to check that displacement in the direction of the second primary M MP unknown is zero, i.e., Δ2 ¼ 2 ¼ 0. EI dM Shear forces may be calculated using differential relationships Q ¼ . This formula leads to the following results: dx 13:085 ð11:349Þ ¼ 6:1085 kN 4 11:349 11:268 ¼ 1:8915 kN; Q45 ¼ ¼ 3:756 kN ¼ 6 3
Q67 ¼ Q78
The portion A-3 subjected to load q and couples at points A and 3 is shown in Fig. 9.24h. Shear forces are calculated from the following equations X HA ! M 3 ¼ 0 : H A ¼ 6:2438 kN ! QA ¼ þ6:2438 kN X H3 ! M A ¼ 0 : H 3 ¼ 3:756 kN ! Q3 ¼ 3:756kN The final shear force diagram is shown in Fig. 9.24i.
9.4 Redundant Plane Frames
309
Axial forces may be derived from the equilibrium of rigid joint D; a corresponding free-body diagram is presented in Fig. 9.24i. The shear at point 3 is negative, so this force, according to the sign law, should rotate the body counterclockwise. It is assumed that unknown forces N6–7 and N3–4 are tensile. X X ¼ 0 : N 67 ¼ 0 N 67 ! X N 3A ! Y ¼ 0 : N 3A ¼ 6:108 kN Final diagram for axial force N is presented in Fig. 9.24j. Since the member DB has no constraints for its horizontal displacement, a normal force in this member is zero. Since support C does not prevent vertical displacement, the axial force NDC ¼ 0. 6. Reactions of supports. Internal force diagrams M, Q, and N allow us to show all reactions of the supports. The negative shear 3.756 means that horizontal reaction at support C equals 3.756 kN and directed to the left. The positive shear 6.244 means that horizontal reaction RA at clamped support A equals 6.244 kN and this reaction produces the positive shear. All reactions of supports are shown in Fig. 9.24k. Now we can perform verification of the obtained reactions of supports. Equilibrium equations for the frame in whole are X X ¼ q 5 6:244 3:756 ¼ 10 10 ¼ 0 X Y ¼ P þ 6:108 þ 1:892 ¼ 8 þ 8 ¼ 0 X M 7 ¼ 2 5 2:5 þ 8:036 6:108 4 6:244 5 þ 3:756 3 þ 1:892 6 ¼ 55:656 55:652 ffi 0 Therefore, equilibrium of the structure in a whole is held. It is left as an exercise for the reader to check the equilibrium of some parts of the system (for example, use a cut section through left/or right at point 7 and consider the equilibrium of the either part of the structure). Discussion 1. For any statically indeterminate structure subjected to action of arbitrary external load, the distribution of internal forces (bending moment, shear and normal forces), as well as reactions of supports, depends only on relative stiffnesses of the elements, and does not depend on their absolute value of flexural stiffness EI. 2. If we accept the rule of signs for bending moments different from those given in Table 9.5, then the signs of the resulting bending moments will change; however the bending moment diagram remains unchanged, because ordinates are plotted on the extended fibers.
9.4.3
Frame with Closed Contour. Elastic Center
Among the structures, whose degree of static indeterminacy is equal to three, it is especially worth noting structures with the closed contour. Such structures find wide application in modern engineering. Some of these systems are shown in Fig. 9.25. The structures with closed contour have specific feature, which allows expanding the concepts of “primary system and primary unknowns,” and avoid the accumulation of computational errors while solving the canonical equation.
Fig. 9.25 Types of structures with close contour
310
9 The Force Method
Let us consider a flat closed frame of a rectangular contour as shown in Fig. 9.26a. Geometry of a frame and stiffness of all members are symmetrical with respect to vertical axis. All joints of the frame are absolutely rigid; the supports are pinned and rolled ones.
a
EI
c
b
EI1
AS
EI
X2
X1 h
MX1=1
MX2=1
d
X3 MX3=1
EI2 A
B l
Fig. 9.26 (a) Design diagram of frame with close contour; (b–d) bending moments diagrams Mi in the primary system caused by primary unknowns Xi; AS—axis of symmetry
This structure is externally statically determinate and internally statically indeterminate (third degree). In other words, the support reactions can be determined using only the static equilibrium equations, while to determine internal forces we also need to take into account the compatibility conditions for deformations. The primary system is obtained by cutting the frame along the axis of symmetry (AS). In this case, the basic unknowns are the internal forces at sections at AS; they are the bending moment X1, and axial and shear forces, X2 and X3, respectively. Bending moment diagrams caused by unit primary unknowns are shown by thin lines. It can be seen the bending moment diagrams M1 and M2 are symmetrical while the bending moment diagram M3 is antisymmetrical. Unit coefficients and free terms are Xð M M X ð Mi M0 i k P δik ¼ ds, ΔiP ¼ ds EI EI It is obvious that multiplication of symmetrical and antisymmetrical diagrams equals zero; therefore δ13 ¼ δ31 ¼ M 1 M 3 ¼ 0, δ23 ¼ δ32 ¼ M 2 M 3 ¼ 0 Canonical equation of the force method becomes
δ11 X 1 þ δ12 X 2 þ Δ1P ¼ 0 δ21 X 1 þ δ22 X 2 þ Δ2P ¼ 0
and
δ33 X 3 þ Δ3P ¼ 0
Thus the system of canonical Eq. (9.4) with three unknowns has split into two separate systems: two simultaneous equations with unknowns X1, X2 and one equation with unknown X3. Let us assume that loading of the frame is symmetrical. In this case, the bending moment diagram is symmetrical; therefore Δ3P ¼ M 0P M 3 ¼ 0, and, as a result, the antisymmetrical unknown X3 ¼ 0. Assume that loading of the frame is antisymmetric. In this case the bending moment diagram is antisymmetrical; therefore Δ1P ¼ M 0P M 1 ¼ 0, Δ2P ¼ M 0P M 2 ¼ 0, and, as a result, the symmetrical unknowns X1 ¼ X2 ¼ 0. Rigid cantilevered concept. Simplification of the system of equations is associated with the features of the choice of the primary unknowns. We show that this choice can be widely varied. The concept of rigid cantilevers serves this purpose. Let us consider redundant frame shown in Fig. 9.27a. This frame using new concept may be transformed into new redundant structure which is statically equivalent to the given structure. For this we need to perform the following steps: at any point C draw the through section, separate one part of the frame from another, attach two absolutely rigid cantilever, and connect them by means of three rods which do not intersect in one point. In case of such connection the mutual displacements (angular and linear displacements in the vertical and horizontal directions) of two sections located infinitely close to the left and right of the section C would be equal to zero. It means that both structures, original and new, are statically equivalent. The through section a–a across three introduced constraints transfers internal forces in these constraints into class of external forces. Just three introduced rods may be treated as redundant constraints, and internal forces in these constraints will
9.4 Redundant Plane Frames
311
be considered as the primary unknowns. For structure in Fig. 9.27c the primary unknowns are Xi, i ¼ 1, 2, 3. Corresponding primary system is shown in Fig. 9.27d, and bending moment diagrams shown in Fig. 9.27e–g. The fundamental difference between structures in Figs. 9.26 and 9.27 is as follows: the primary unknowns for original structure are internal forces M, N, and Q in the section C, while in modified structure the primary unknowns are independent forces Xi, which are applied at different points. However, the set of internal forces at section C and set of three forces Xi in modified structure are statically equivalent. Thus, the introduced absolutely rigid together connected cantilevers illustrate a profound idea: As the primary unknowns it is possible to adopt three forces in inserted rods, and these rods can be placed on the plane arbitrary, provided they do not intersect at one point (or not parallel). This conclusion significantly expands the concept of primary system and primary unknowns. Let us discuss how to use this new concept for structural analysis. Assume the connection of the rigid cantilevers is performed using two horizontal rods 1, 2, and one vertical rod 3 (Fig. 9.27c). The vertical axis of symmetry passes through point C. The primary unknowns are force X1 acting at the end of cantilever (its length is still not defined), the force X2 which acts along crossbar, and shear force X3 at same section C (Fig. 9.27d). It is obvious the rods 2 and 3 prevent mutual displacements of both parts of the frame in horizontal and vertical directions, while both rods 1 and 2 prevents mutual angular displacement of both half-frames. Arbitrary length of the cantilever gives us opportunity to change the bending moment diagram caused by force X1. It should be remembered that this force is not the internal force at section C, but introduced only for ensuring the equivalency of both systems, primary and original ones.
a
b
q D
c
C
EI1
1
EI
h
A
d
a
C
K1
X1
X1
K2
X3
X3
B
l
X2
2
K1 3 a
K2
X2
AS
e
f
g
X2
2h/3
l/2
y0 X1
X1
● X3
MX1=1
h/3
h
h
q
ql2/8
MX2=1
MX3=1
i
q
MD
MC M0P
H
A
MP MA
B
H
Fig. 9.27 (a) Design diagram of the portal frame; (b, c) concept of the absolutely rigid cantilevers and different ways of the rigid connection of two parts; (d) primary system and primary unknowns Xi, i ¼ 1, 2, 3; (e–g) bending moment diagrams caused by unit primary unknowns; (h) bending moment diagram in the primary system caused by the given load; (i) final bending moment diagram
312
9 The Force Method
Unit bending moment diagram is shown in Fig. 9.27e–g. Since the introduced cantilevers are absolutely rigid (EI ¼ 1), then bending moment diagram within cantilevers, caused by unit forces, is not shown. The bending moment diagrams М1 and M2 are symmetrical, while М3 is antisymmetrical; therefore δ13 ¼ δ31 ¼ δ23 ¼ δ32 ¼ 0. Let us choose the length of the cantilevers y0 ¼ 2h/3. In this case opposite to the center of gravity of diagram M2 will be M M2 zero ordinate from diagram M1 as shown in Fig. 9.27e, f. It means that δ12 ¼ δ21 ¼ 1 ¼ 0. Thus, all secondary unit EI 1 displacements are equal to zero. Therefore the system of canonical Eq. (9.4) split into three independent equations each with one primary unknown. δ11 X 1 þ Δ1P ¼ 0 δ22 X 2 þ Δ2P ¼ 0 δ33 X 3 þ Δ3P ¼ 0 Transformation of complete canonical equations into the set of unconnected equations with one unknown in each allows eliminating the accumulation of errors that inevitably arise when solving joint equations. Principal unit displacements are M1 M1 h 4h2 1h2 h 2h 1 2h l 2h δ11 ¼ ¼2 þ2 2 2 þ 2 ¼ 6EI 1 3 3 EI 2 3 2 3 EI 9 9 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Crossbar, Vereshchagin rule
Vertical membr, trapezoid rule
¼
3
2
2
2 h 4h l 2h l þ ¼ ð2 þ k Þ, 9 EI 1 9 EI 2 9 EI 2
k¼
EI 2 h EI 1 l
1 2 1 2 h3 δ22 ¼ 2 h h h ¼ 2 3 EI 1 3 EI 1
Coefficient 2 in both expressions means two symmetrical parts of the primary system. Free terms are 1 2h 2h h h ql2 1 ql2 l 2h 1 þ Δ1P ¼ M 1 M 0P ¼ 2 ¼ 2 3 3 3 3 8EI 1 3 8 2 3 EI 2 ql2 h2 ql3 h ql3 h 3k þ 2 þ ¼ 24EI 1 36EI 2 12EI 2 6 1 ql2 1 ql2 h2 ¼ M 2 M 0P ¼ 2 h h ¼ 2 8 EI 1 8EI 1 ¼
Δ2P Primary unknowns are
Δ1P ql2 3k þ 2 ¼ δ11 16h 2 þ k Δ2P 3ql2 ¼ X2 ¼ δ22 16h
X1 ¼
Since load is symmetrical, then Δ3P ¼ M 3 M 0P ¼ 0, and antisymmetrical unknown X3 ¼ 0. Bending moments at arbitrary point j of the given structure are determined by formula j
j
M j ¼ X 1 M 1 þ X 2 M 2 þ M 0,j P where M 0,j P is bending moment at j-th point caused by external load in the primary system.
9.4 Redundant Plane Frames
313
ql2 3k þ 2 2 ql2 3k þ 2 h ¼ 3 16h 2 þ k 24 2 þ k ql2 3k þ 2 2 ql2 ql2 1 D 0,D h þ ¼ MD ¼ X1M1 þ MP ¼ 3 16h 2 þ k 8 6 2þk 2 ql 3k þ 2 h 3ql2 ql2 A A M A ¼ X 1 M 1 þ X 2 M 2 þ M 0,A P ¼ 16h 2 þ k 3 16h ðhÞ þ 8 C
MC ¼ X1M1 ¼
¼
ql2 1 12 2 þ k
Horizontal reaction X
X¼0:
H X1 þ X2 ¼ 0 ! H ¼
ql2 1 4h 2 þ k
Elastic Centre. This concept is related to arbitrary closed structure with third degrees of redundancy. This concept means the following: There is a point exist on the plane of structure, where all primary unknowns which arise at specified section of a structure should be placed in order to all secondary unit displacements of canonical equation of the force method would be zero, i.e., δik ¼ 0, i 6¼ k. This point is called the elastic center of structure. Coordinate of elastic center may be calculated by formulas (Klein 1980) P l i xi EI x0 ¼ P i , li EI i
P li yi EI y0 ¼ P i li EI i
ð9:15Þ
In these formulas, the numerators represent the static moment of magnitudes li/EIi with respect to the y and x axes, and the denominator is the sum of these magnitudes. If the frame has at least one axis of symmetry, one unknown force in the elastic center should be directed along this axis, and the other force should be perpendicular to it. Position of the elastic center depends on the adopted primary unknowns. Idea of elastic center for a closed frame with polygonal contour is shown in Fig. 9.28a.
y
a X3
X1 X3
X1 A
X2
b
X2
A
A
x
y
C 1.5EI
3m X2
AS
1.0 EI
X3
X3
2.0EI C’
B
* X2
7m y0
A
X1
X1
D
*
8.5m 3.5m x
0
8m Fig. 9.28 (a) Concept of elastic center; (b) design diagram of symmetrical structure with closed contour and determining of elastic center location
314
9 The Force Method
Let us show the application of formulas (9.15) to the symmetrical frame with rigid joints shown in Fig. 9.28b. Assume the primary unknowns will be internal forces (bending moment X1, and forces X2, X3) at point C on the axis of symmetry. These internal forces may be transferred to elastic center, which is located at the ends of absolutely rigid cantilevers (EI ¼ 1). Let y-axis coincide with axis of symmetry, x-axis coincide with rod AB. The x-coordinate of elastic center is zero. The ycoordinate of elastic center becomes 2 3 X X 7 2 4 li li 2 64 7 5 7 5 7 y0 ¼ yi þ þ 0 3:5 8:5 þ þ ¼ 6 5 EI 2:0 1:0 1:5 ¼ 4:2839 ðmÞ EI 42:0 1:5 EI i EI i |fflfflffl{zfflfflffl} 1:0 |fflfflfflfflffl{zfflfflfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} AC 0
DC
AD
Factor 2 in nominator and denominator takes into account two symmetrical parts of the frame. If internal primary forces will be placed at the elastic center, then complete set of canonical equations are split on three independent equations each with only one unknown force. In case of curvilinear contour the coordinates of the elastic center can be determined by formulas (Darkov 1989) ðl
ðl
0
0
ds x0 ¼ x Ix
ds , Ix
ðl
ðl
0
0
ds y0 ¼ y Ix
ds , Ix
ð9:15aÞ
where s is curvilinear coordinate of the axis of the arch. Assume we need to find the position of the elastic center for symmetrical nonuniform parabolic arch with fixed supports (Fig. 9.29). The span and height of the arch are l and f, respectively. The neutral line of the arch is determined by equation 4f y ¼ 2 ðl xÞx; the origin is located with the left fixed support. Let the primary unknown be internal forces at crown C. l Assume that the cross-sectional moments of inertia vary inversely to cosφx by expression Ix ¼ IC / cos φx, where Ix is a moment of inertia of an arbitrary cross section with coordinates x and y, IC is the moment of inertia of the cross section at the crown of the arch, φx is the angle formed by the tangent line of the arch at point x with the horizontal line. y
φx
C IC
Ix
y(x)
f
* M(x0,y0)
y
x C′
x
l/2
l/2
Fig. 9.29 Redundant arch. Computation of position of the elastic center M(x0,y0)
Now let us express the curvilinear coordinate s by the rectilinear coordinate x by formula ds ¼ dx/cos φx. Taking into account this expression and relationship Ix ¼ IC / cos φx, the formulas (9.15a) become ðl
ðl
0
0
dx x0 ¼ x IC
dx , IC
ðl
ðl
0
0
dx y0 ¼ y IC
dx IC
In case of symmetrical arch the elastic center M is located on the axis of symmetry x0 ¼ l/2, while for y0 coordinate we have Ðl 4f y0 ¼
0
2
l
ðl xÞx Ðl dx 0 IC
dx IC
2 ¼ f 3
9.4 Redundant Plane Frames
315
Thus, if the primary unknowns (bending moment, axial and shear forces) are placed at elastic center (MC0 ¼ (2/3)f ), then all secondary unit displacements are equal to zero.
Discussion 1. In this section two ways of simplifying the system of canonical equations are considered; both approaches can be applied only to structures with a closed contour—one-span frames of arbitrary configuration with clamped supports, arches with fixed supports, rings, boxes, and similar structures. 2. The first approach consists in replacing the unknown internal forces (bending moment and two forces) at chosen section with a statically equivalent system of three unknown forces. This set of the forces may be applied at any points of the plane. This approach allows extending the concept of a primary system: the primary system is a statically determinate system obtained from a given one by transferring internal forces (M, Q, N) into a class of external forces and replacing them with a statically equivalent system of three forces (Fi, i ¼ 1, 2, 3) applied at an arbitrary points of the plane. The second approach consists in transferring unknown internal forces from chosen section to a certain point on the plane, i.e., to the elastic center of the system. Both approaches use the concept of absolutely rigid cantilevers which provide compatibility conditions of deformations. 3. The location of the elastic center is determined by parameters of the closed structure (geometry, and moment of inertia of each member) only and does not depend on loading. The elastic center is a concept related to a section where primary unknowns where initially chosen (and then transferred to elastic center). 4. The problem of simplifying the system of equations is considered under the assumption that only the bending moments are taken into account in the Mohr–Maxwell integral, while the axial and shear forces are neglected. 5. Determination of the position of the elastic center does not cause difficulties (although in the case of a curvilinear contour, an integration operation should be used). It is especially effective to use the elastic center concept for analysis of symmetric closed structures. In the asymmetrical systems additional difficulties arise: it is required to determine not only the position of the elastic center but also the directions along which the primary unknown forces must be directed. 6. Representing canonical equations in the form of three independent equations (using elastic center), each with one unknown, simplifies the computational procedure and allows avoiding the accumulation of errors in the numerical solution of equations. Most importantly, this approach does not consider the primary system from the point of view of ill-conditioned matrix of canonical equations. In addition, such a representation will be used for analysis of the frame with elastically compliant joints and supports (Sect. 9.4.4).
9.4.4
Frame with Elastically Compliant Supports and Joints
This section is devoted to calculation of coefficients of canonical equations of the force method in case of elastically compliant supports and joints. In the rod structures, various types of elastic constraints are possible. Their main types are given in Fig. 9.30.
a
c
b 1 r
r
d
1 r
r
1
R Fig. 9.30 Types of the elastic constraints: (a, c) elastically compliant supports; (b, d) elastically compliant connection of two members
In cases (a, b), the parameter r (kN/m) means the rigidity of an elastically compliant constraint in the linear mutual displacement of its ends. In other words, stiffness r is a force which should be applied to elastic member in order to mutual
316
9 The Force Method
linear displacements of its ends would be equal to unity. When the rod is elongated (shortened) by y(m), the reactive force, which arises in the elastic constraint, is R ¼ ry (kN). In the cases (c, d), the parameter r (kNm/rad) means the rigidity of an elastically compliant constraint in the angular mutual displacement. In other words, stiffness r is a couple which should be applied to the elastic member in order to angular displacements of support would be equal to unity. When the rods are rotated by θ (rad), the reactive couple, which arise in the elastic constraint is M ¼ rθ (kNm). If the rod system is deformed, the reactions of the elastically compliant constraints perform work that should be considered as the work of internal forces. Therefore, the expression for displacement along any i-th direction takes the form (Rabinovich 1960, part2) X ð Mi Mp X Ri Rp Δip ¼ ds þ ð9:16Þ EI r This expression takes into account the flexural deformations of the rods (first term), and reactions in elastically deformed constraints only, namely, Rp and Ri are reactions in the loaded and unit states, respectively. Let us show the application of this formula to the simplest problems. The cantilever beam of the length l and flexural rigidity EI is loaded with the force P. The elastically compliant support of the beam has a rigidity r (Fig. 9.31a). It is required to determine the angle of rotation at the right free end A.
a
Pl
P
r
A
B
b
MP
l
1 M1 X1=1
Fig. 9.31 (a) Design diagram of a cantilever beam with elastically clamped support and bending moment diagram MP; (b) unit state and bending moment diagram M1
The bending moment diagram in the loaded state, MP, is shown in Fig. 9.31а. Reaction Rp in elastically compiled support is the bending moment at support B in the loaded state, i.e., Rp ¼ Pl. Therefore the angle of rotation of the elastically clamped support is Δp ¼ φB ¼ Pl/r. The unit state and corresponding bending moment diagram Mi ¼ M1 are shown in Fig. 9.31b. Reaction R1 in elastically compiled support is the bending moment at support B in the unit state, i.e., R1 ¼ 1. In both states the reactive moment at elastically clamped support is directed counterclockwise. The slope of the free end А of the beam becomes Δ1P ¼ φA ¼
X ð M1 MP EI
ds þ
R1 Pl2 1 Pl2 Pl Pl2 2EI Rp ¼ þ Pl ¼ þ ¼ 1þ r rl r 2EI r 2EI 2EI
Thus, the slope at the section A is determined as a sum of angle of rotation caused by deformation of the beam itself and the angle of rotation of a beam as absolutely rigid body caused by rotation of support through angle φB ¼ Pl/r. Figure 9.32a presents the uniform one-span beam of length l. The right elastic support has stiffness r. The beam is loaded by uniformly distributed load q. Determine vertical displacement of the beam at the middle of a span. In the loaded state in the elastic support arise the reactive force Rp ¼ RB ¼ ql/2. Corresponding displacement of point B is Δp ¼ ΔB ¼ ql/2r. In the unit state the beam is loaded by unit force X ¼ 1 at point C (Fig. 9.32b), corresponding reaction of elastic support is R1 ¼ 1/2. In the loaded and unit states the bending moment diagrams Mp and M1 are presented. They serve for computation of displacement at point C in case of absolutely rigid supports.
9.4 Redundant Plane Frames
317
a
q
B r
A
Rp=ql/2
X=1 b
MP
C
M1 l/2
l/2
R1=1/2 Fig. 9.32 (a) Design diagram of a beam with elastic support and bending moment diagram MP; (b) unit state and bending moment diagram M1
The required displacement becomes Δ1P ¼ yl=2 ¼
XðM M R 5ql4 1 ql 5ql4 ql 1 P ¼ ds þ 1 Rp ¼ þ þ EI r 384EI 2r 2 384EI 4r
Portal frame with compliant supports and joints. Let us show the application of this procedure for analysis of portal symmetrical frame in Fig. 9.33a. Assume that the compliance of each support is characterized by coefficients α1, α2, α3. Here α1 is a horizontal displacements of support caused by a horizontal unit force; α2 is the angle of rotation of support caused by the unit couple, and α3 is the vertical displacement of supports caused by the unit vertical force applied to support. Parameter α4 means the mutual angled deformation of the joint at the upper end of the column caused by the unit couple, which is applied to this joint (Fig. 9.33b); note that compliance αi and stiffness ki are related as follows: αi ¼ 1/ki. The primary system is shown in Fig. 9.33c (see Sect. 9.4.3, Fig. 9.27d–g, y0 ¼ 2h/3). Unit states and corresponding bending moment diagrams are shown in Fig. 9.33d–f. These diagrams also indicate reactions corresponding to all compliances for each unit state.
a D
c
b
q EI2
X2
X2
X1
X1
α4
EI1
h
AS α2
α1
B
A
X3
X3 AS
l α3
d
f
e X2=1
2h/3 R42
R41
M1 h/3
R21
R43
X1
X1=1 R11
l/2
R31
R11=1, R21=h/3, R31=0, R41=2h/3.
M2
R12
R13
X3=1 M3
h R22
R23 R32
R12=1, R22=h, R32=0, R42=0.
R33
R13=0, R23=l/2, R33=1, R43=l/2.
Fig. 9.33 Portal frame with elastically compliant supports and joints. (a, b) Design diagram and numbering compliances for support and joint; (c) primary system; (d–f) bending moment diagrams caused by unit primary unknown and notation of the reactions at the compliant supports and joints
318
9 The Force Method
Reactive forces with first index 1 and 3 (R11, R31, R12, R32, R13, R33) correspond to the linear compliances α1, α3, while a reactive moments with first index 2 and 4 (R21, R41, R22, R42, R23, R43 correspond to the angular compliance α2, α4. The second index 1, 2, and 3 means the number of the primary unknown forces Х1,2,3. The problem is to determine the additional terms to the principal unit displacements δ11, δ22, δ33. The required additional terms are calculated by formula δ0ii ¼ 2
4 X Rji Rji , rj j¼1
i ¼ 1, 2, 3
ð9:17Þ
Here coefficient 2 takes into account the symmetry of the frame. Index i means the state, while index j ( j ¼ 1, . . ., 4) means the reaction in compliant connection. Since we calculate only primary unit displacement due to primary unknown, then reaction in compliant connections (Rji) appears twice in nominator in expression (9.17). 0 δ011 ¼ 2
4 X
R j1 R rj j¼1
j1
1
B h h 2h 2h C C¼ ¼ 2B 1 α1 |{z} 1 þ α2 þ α4 @|{z} |{z} |{z} |{z} |{z} |{z} 3 3 3 3 A |{z} |{z} R11
"
R11
1=r1
R21
2 2 # h 2h ¼ 2 α1 þ α2 þ α4 3 3 δ022 ¼ 2
4 4 X X R j2 R2 ¼2 R j2 α j R rj j¼1 i¼1
δ033
j2
1=r2
R21
R41
1=r4
R41
¼ 2ðR12 α1 R12 þ R22 α2 R22 þ R32 α3 R32 þ R42 α4 R42 Þ ¼
¼ 2½1 α1 1 þ hα2 h þ 0 α3 þ 0 α4 ¼ 2 α1 þ h2 α2 4 4 X X R j3 R j3 l l l l ¼2 ¼2 R j3 α j R j3 ¼ 2 0 þ α2 þ 1 α3 1 þ α4 ¼ 2 2 2 2 rj j¼1 i¼1 " 2 # 2 l l α2 þ α3 þ α4 ¼2 2 2
Finally, the unit coefficients on the main diagonal of canonical equation becomes δ11 ¼ δ011 þ δ011 ,
δ22 ¼ δ022 þ δ022 ,
δ33 ¼ δ033 þ δ033 ,
where δ011 , δ022 , δ033 means the unit coefficients in case of absolutely rigid supports and joints. Note, these primary unit displacements had been calculated in Sect. 9.4.3. Procedure for computation of free terms remains same.
9.5
Redundant Trusses
Statically indeterminate trusses are geometrically unchangeable structures for which all the reactions and all internal forces cannot be determined using only the equilibrium of equations. Figure 9.34 presents three types of statically indeterminate trusses - externally, internally, and mixed. a
b
Fig. 9.34 Statically indeterminate trusses and types of their redundancy
c
9.5 Redundant Trusses
319
The truss in Fig. 9.34a contains one redundant support member and does not contain the redundant members in the web. This structure presents the first degree of externally statically indeterminate truss. In cases (b) and (c) the diagonal members do not have a point of intersection, i.e., these members are not connected (nor hinged, nor fixed) with each other. Case (9.34b) presents the internally statically indeterminate truss to six degrees of redundancy. Case (9.34c) presents the internally statically indeterminate truss to six degrees of redundancy and externally to the first degree. Assume, the intersected diagonals do not form a common rigid joint. The analysis of statically indeterminate trusses may be effectively performed by the force method in canonical form. The primary system is obtained by elimination of redundant constraints; to replace the eliminated constraints, their reactions, i.e., primary unknowns, must be introduced into the structure. As in the general case (9.4) of canonical equations, Xn are primary unknowns; δik are unit displacements of the primary system in the direction of i-th primary unknown due to unit primary unknown Xk ¼ 1; the loaded terms ΔiP are displacements of the primary system in the direction of i-th primary unknown due to acting load. For computation of coefficient and free terms of canonical equations we will use the second term of the Maxwell–Mohr integral (8.46), so δik ¼
X Ni Nk l n
ΔiP ¼
EA
X Ni N0 l P EA n
ð9:18Þ
where l is length of n-th member of the truss; N i , N k are axial forces in n-th member due to unit primary unknowns Xi ¼ 1, Xk ¼ 1; and N 0P is axial force in n-th member of the primary system due to acting load. Summation procedure in (9.18) should be performed on all members of the truss (subscript n is omitted). Solution of (9.4) is the primary unknowns Xi (i ¼ 1, . . ., n). Internal forces in the members of the truss may be constructed by the formula N ¼ N 1 X 1 þ N 2 X 2 þ . . . þ N 0P
ð9:19Þ
Kinematical verification of computed internal forces may be done using the following formula X
NN
l ¼0 EA
ð9:20Þ
If a primary unknown is the reaction of support, then this equation means that displacement in the direction of the primary unknown in the original structure is zero. If the primary unknown is the internal force in any redundant member of a web, then (9.20) means that a mutual displacement of points of application of primary unknown (in the direction of this unknown) in the given structure is zero.
9.5.1
Externally Redundant Truss
The following example presents detailed analysis of the externally statically indeterminate truss. Two versions of the primary system and the result of increasing the rigidity of the conditionally necessary element are discussed. Example 9.8 Symmetrical statically indeterminate truss that carries two equal forces P ¼ 120 kN is shown in Fig. 9.35a. The axial stiffness for all members is EA. This structure is first degree of external statical indeterminacy. The reaction of the intermediate support is being as the primary unknown. The primary system is shown in Fig. 9.35b.
320
9 The Force Method
1
2
Primary system
3
b
a 3m 7
0
5
4
6 P 4m
c
4m
4m
0.833 0.0
4m
63.71 120
120
H0
g
1
0
α
α
U0-7=63.71 U6-7=63.71 6
U6-5=63.71
R6
R0
120
2
X1
3
Second version of primary system
6
0 P=120
D6-3=120.25
D0-1=79.75 D6-1=120.25
63.71
P=120
120
V6-2=0.0
f
0.0
P=120
P=120
120
Final axial forces
120
0.0
P=120
0.5
-120.25
0.0
160
160
32.43
Loaded state
-160
-200
-0.667
e -79.75
d
-0.833 0.0
X1=1
0.5
P X1
Unit state
1.333
-0.667
P
P
4 P=120
Fig. 9.35 (a, b) Externally redundant truss and primary system, (c, d) unit and loaded states, (e) final axial forces diagram, (f) free-body diagrams of joints 0 and 6, (g) second version of primary system
Solution Canonical equation of the force method and primary unknown are δ11 X 1 þ Δ1P ¼ 0 ! X 1 ¼
Δ1P δ11
For calculation of δ11, Δ1P it is necessary to show the unit and loading states (Fig. 9.35c, d) The analysis of this statically indeterminate truss is presented in the tabulated form (Table 9.6). Column 1 contains the flexibility for each member; the factor 1/EA is omitted. Internal forces for all members in unit and loaded states are presented in columns 2 and 3, respectively.
9.5 Redundant Trusses
321
Table 9.6 Calculation of internal forces in the members of an externally statically indeterminate symmetric truss, (Fig. 9.35), EA ¼ const, X1 ¼ 144.36 kN (X1 is determined after completing columns 4 and 5 and calculating δ11 and Δ1P)
Members
l EA
N1
N P0
N1 . N P0
l EA
2 l N1 . EA
N1 . X1
N= N P0
+ N1 X 1 7
N1 . N
l EA
1
2
3
4
5
6
1-7 2-6 3-5 0-7 7-6 6-5 5-4 1-2 2-3
3 3 3 4 4 4 4 4 4
0 0 0 -0.667 -0.667 -0.667 -0.667 1.333 1.333
120 0 120 160 160 160 160 -160 -160
0 0 0 -426.88 -426.88 -426.88 -426.88 -853.12 -853.12
0 0 0 1.774 1.774 1.774 1.774 7.107 7.107
0 0 0 -96.288 -96.288 -96.288 -96.288 192.43 192.43
120 0 120 63.71 63.71 63.71 63.71 32.43 32.43
0 0 0 -169.98 -169.98 -169.98 -169.98 172.92 172.92
0-1 1-6 6-3 3-4 Factor
5 5 5 5
0.833 -0.833 -0.833 0.833
-200 0 0 -200
-833 0 0 -833
3.469 3.469 3.469 3.469
120.25 -120.25 -120.25 120.25
-79.75 -120.25 -120.25 -79.75
-332.16 500.84 500.84 -332.14
1 EA
1 EA
Diagonals
Upper chord
Lower chord
Vertical members
0
1 EA
8
1 EA
For computation of loaded displacement Δ1P the entries of the column 4 should be summated. Similar procedure should be performed for computation of unit displacement δ11 (column 5). X l 5, 079:76 ¼ ½426:88 4 853:12 2 833 2 ¼ N 1 N 0P Δ1P ¼ EA EA X 2 l 35:186 δ11 ¼ ¼ ½1:774 4 þ 7:107 2 þ 3:469 4 ¼ N1 EA EA Δ1P ¼ 144:36 kN . Column 6 contains the computation of the first term of (9.19). δ11 Computation of final internal force in each element of the truss is provided according to formula (9.19) and shown in column 7. Final axial forces diagram is shown in Fig. 9.35e. Since structure is symmetrical, Fig. 9.35c–e presents internal forces within the left part of the truss only. The primary unknown is X 1 ¼
Reaction of supports. Knowing the internal forces we can calculate the reaction R of any support. Free-body diagrams of joints 0 and 6 are presented in Fig. 9.35f. Equilibrium equations for joint 0 is X Y ¼ 0 : R0 79:75 sin α ¼ 0 ! R0 ¼ 47:85 kN It is obvious that R4 ¼ R0 ¼ 47.85 kN. Equilibrium equation for joint 6 X Y ¼ 2 120:25 sin α þ R6 ¼ 0 leads to the following reaction at the intermediate support of the truss: R6 ¼ 2 120:25 0:6 ¼ 144:3kN Pay attention, this result has been obtained early as the primary unknown X1, so it can be used for control. Static verification. For truss in whole the equilibrium equations X Y ¼ 2 47:85 þ 144:3 2 120 ¼ 240 240 ¼ 0, X X ¼ 0 : H0 ¼ 0 The last result may be checked. For joint 0 the equilibrium equation ∑X ¼ 0 (Fig. 9.35f) leads to the following result: H0 þ 63.71 79.75 0.8 ¼ 0 ! H0 ¼ 63.8 63.71. The relative error is 0.14%. Kinematical verification. Displacement in the direction of the primary unknown due to primary unknown and given loads is zero. According to (9.20) for computation of this displacement we need to multiply the entries of column 7, 2, and 1; this result is presented in column 8. The sum of entries of this column equals
322
9 The Force Method
X
N1 N
l ¼ 4 169:98 þ 2 172:92 2 332:16 þ 2 500:84 ¼ 1, 344:24 þ 1, 347:52 ¼ 3:28: EI
The relative error is
3:28 100% ¼ 0:24%. 1, 344
Lukin’s Theorem Lukin's Theorem (1929) states: In any statically determinate and redundant truss of a beam type subjected to vertical forces which are applied at the line connecting support points, the internal forces satisfy to condition (Rabinovich 1960) n X
N i li ¼ 0
ð9:21Þ
i¼1
In other words, the sum of product of internal force and corresponding length of the rod is equal to zero. Here Ni and li are internal force at the i-th element of a truss and the length of corresponding member. Summation should be performed over all elements of a truss. In case of truss in Fig. 9.35a (Table 9.6) we have X N l ¼ 2 120 3 þ 4 63:71 4 þ 2 32:43 4 2 79:75 5 2 120:25 5 ¼ 1, 998:8 2, 000:0 ¼ 1:2 1:2 100% ¼ 0:06% 2, 000 If load is applied below (above) of the line connecting support points, then formula (9.21) becomes The relative error is
n X i¼1
N i li > 0,
n X
N i li < 0, respectively
i¼1
Discussion 1. This truss is externally statically indeterminate. As the primary unknown, the reaction R6 of the truss in the intermediate support is adopted. Corresponding primary system is one span truss 0–1–3–4–0. Support 6 presents the conditionally necessary external constraint. The canonical equation means the vertical displacement of joint 6 in primary system caused by primary unknown X1 ¼ R6 and given load is zero. However as the primary unknown we can take the internal force, for example, at member 1–2. Removal of this rod leads the original system to two separate unchangeable statically determinate trusses 0–1–6–0 and 6–2–3–4–6 (Fig. 9.35g). In other words, rod 1–2 (as well as rod 2–3) is conditionally necessary. Corresponding canonical equation means the horizontal mutual displacement of joints 1 and 2 in primary system caused by primary unknown X1 ¼ N1–2 and given load is zero. 2. Assume the elements of a statically indeterminate truss have different axial rigidities EAi. In this case, it is expedient to select the rigidity of one of the elements, EA, as the basic one, and express the stiffness of the remaining elements through the basic rigidity, i.e., EAi ¼ kiEA, where ki is some positive number. After that, each term of canonical equation of the force method contains factor 1/EA which can be canceled. Therefore if a statically indeterminate truss is subjected to any loads, then distribution of internal forces depends on relative axial stiffness EA of the members and does not depend on their absolute value EA. 3. Suppose that the rigidity of conditionally necessary elements of the upper chord of a truss in Fig. 9.35a has changed and becomes 2EA, while the rigidity of the other elements has remained unchanged, EA. In this case, the line for member 1–2 in Table 9.6 should be recalculated. Finally, the loaded term, unit displacement, and primary unknown X1 ¼ R6 become EAΔ1P ¼ 4, 226:64, EAδ11 ¼ 28:096, X 1 ¼ Δ1P =δ11 ¼ 150:44 ð144:36Þ kN In brackets here and below are original values from Table 9.6 for truss with equal rigidity for elements. Internal forces in vertical members V1–7 and V3–5 of the truss remained unchanged since N 1 for these members is zero, term N 1 X 1 turns to zero, and only the term N 0P remains. The forces in the elements of the upper chord of the truss are O12 ¼ O23 ¼ 40.66 kN (while in original truss with equal EA for all elements, it is 32.43 kN). Internal forces at the other members of the truss are
9.5 Redundant Trusses
323
U 07 ¼ U 76 ¼ U 65 ¼ U 54 ¼ 59:66 ð63:71Þ kN, D01 ¼ D34 ¼ 74:68ð79:75Þ, D16 ¼ D36 ¼ 125:32ð120:25Þ An increase in the rigidity of the conditionally necessary element leads to an increase in the primary unknown and a redistribution of forces in the remaining elements of the truss. In this case, a redistribution of reactions occurred: R0 ¼ D01 sin α ¼ 74:68 0:6 ¼ 44:81ð47:85Þ kN R6 ¼ ðD61 þ D65 Þ sin α ¼ 2 125:32 0:6 ¼ 150:38ð144:3Þ kN The reaction R0 decreased, while the reaction R6 as the primary unknown increased. The equation of equilibrium for the whole truss is satisfied. X Y ¼ 2R0 þ R6 2P ¼ 2 44:81 þ 150:38 2 120 ¼ 240 240 ¼ 0
9.5.2
Internally Redundant Truss
Design diagram of externally statically determinate and internally redundant truss is presented in Fig. 9.36a. The truss is loaded by force P ¼ 12 kN at joint 1; axial stiffness EA is constant for all members. We need to compute the axial forces in all members of the truss. The reactions of supports may be determined using the equilibrium equations only. They are R0 ¼ 8 kN, and R3 ¼ 4 kN. Let the primary unknown of the force method be the internal force in member 1–4. Primary system is shown in Fig. 9.36b. P=12kN
a 1
y
P
b 1
2
2 X1
3m 3
0
4m
x
4m
c
Loaded state -10.667 2
1
2 -13.333
N1
-0.6
-0.6 1.0
3
0 5
P=12
d
Unit state
X1=1
-0.8
4
5
4m
-0.8
1
3
0
4
5
Primary system
-6.667
6.667
-4.0
0.0
3
0 5.333
-10.667 5
4
4
5.333
R0=8 P=12
e 1 -13.333
Final axial forces, (kN)
-8.00
2
-3.333 -2.00
2.00
10.667
R0=8
5
8.00
R3=4 V1-5 = 2.00
f
D5-2=3.334
NP
-6.667
α
3
3.334
0
4
0
NP
5.333
U0-5=10.667
5
U5-4= 8.00
R3=4
Fig. 9.36 (a, b) Internally redundant truss and primary system, (c, d) axial forces in the unit and loaded states, (e) final axial force diagram, (f) freebody diagram for joint 5
324
9 The Force Method
Canonical equation of the force method is δ11X1 þ Δ1P ¼ 0. Computation of δ11, Δ1P, and the internal forces are presented in Table 9.7, columns 1–7; the column 8 serves for kinematical verification of internal forces. For calculation of δ11 and Δ1P it is necessary to show the unit and loading states and corresponding internal forces (Fig. 9.36c, d), Table 9.7, columns 2 and 3. In the unit state, we have a self-balanced system of two forces X1, so the truss reactions are equal to zero, and internal forces occur only in the rods of the loaded panel 1–2–4–5–1. In the remaining elements, the forces in the unit state are zero. Table 9.7 Calculation of internal forces in the members of an internally redundant truss; EA ¼ constant, X1 ¼ 3.334 kN
Members 0
Other members
Rods of loaded contour 1-2-4-5
V1-5 V2-4 O1-2 U5-4 D2-5 D1-4 U0-5 U4-3 D0-1 D2-3 Factor
l EA
N P0
N1
N1 . N P0
l EA
2 l N1 . EA
N1 . X1
N= N P0
+ N1 X 1
N1 . N
l EA
1
2
3
4
5
6
7
8
3 3 4 4 5 5 4 4 5 5 1 EA
-0.6 -0.6 -0.8 -0.8 1.0 1.0 0.0 0.0 0.0 0.0
-4.0 0.0 -10.667 5.333 6.667 0.0 10.667 5.333 -13.333 -6.667
7.20 0.0 34.134 -17.065 33.335 0.0 0.0 0.0 0.0 0.0 1 EA
1.08 1.08 2.56 2.56 5.0 5.0 0.0 0.0 0.0 0.0 1 EA
2.000 2.000 2.667 2.667 -3.334 -3.334 0.0 0.0 0.0 0.0
-2.000 2.000 -8.000 8.000 3.333 -3.334 10.667 5.333 -13.333 -6.667
3.60 -3.60 25.60 -25.60 16.665 -16.665 0.0 0.0 0.0 0.0 1 EA
The loaded and unit displacements are
δ11
X
l 1 57:604 ¼ ð7:20 þ 34:134 17:065 þ 33:335Þ ¼ ðmÞ, EA EA EA X 2 l 17:28 m ¼¼ ð1:08 þ 2:56 þ 5:0Þ 2 ¼ ¼ N1 EA EA kN
Δ1P ¼
N 1 N 0P
Δ1P 57:604 ¼ 3:334 kN. ¼ 17:28 δ11 Internal force in any bar may be calculated by general formula
The primary unknown becomes X 1 ¼
N ¼ N 1 X 1 þ N 0P Corresponding computation is show in Table 9.7, column 7. The final axial force diagram is shown in Fig. 9.36e. Verification of results. 1. Free-body diagram of joint 5 is shown in Fig. 9.36f. Equilibrium equations for this joint are X X ¼ 10:667 þ 8:00 þ 3:334 cos α ¼ 10:667 þ 10:667 ¼ 0 X Y ¼ 2 þ 3:334 sin α ¼ 2 þ 2 ¼ 0 2. Kinematical verification. It is easy to check that mutual displacements of joints 1 and 4 are zero; indeed from column 8 we have
P
N1 N
l ¼ 0: EA
3. Umansky formula This relationship generalizes the Lukin’s formula (9.20) if the location of external forces, including reactions, are arbitrary (Klein 1980, p. 231, 233)
9.5 Redundant Trusses
325 n X
N i li ¼
X
Px x þ Py y ,
ð9:21aÞ
i¼1
where n is the number of elements of a truss, x and y are coordinates of the external applied forces, Px and Py are projection of external forces onto axis x and y. In our case the left part of (9.21) n X i¼1
N i li ¼ ð13:333 3:333 þ 3:334 6:676Þ 5 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Diagonal members
þð10:667 þ 8:00 8:00 þ 5:333Þ 4 þ ð2:00 þ 2:00Þ 3 ¼ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Horizontal members
Vertical members
¼ 100:00 þ 64 ¼ 36 The negative sign corresponds to the external load which applied at joint located above the line connecting support points. The right part of (9.21a) is ∑(Pxx þ Pyy) ¼ R0 0 þ R3 0 P 3 ¼ 36 If all forces are vertical (Px ¼ 0) and all coordinates y ¼ 0, then (9.21a) presents the Lukin’s formula (9.21).
9.5.3
Some Properties of Redundant Trusses
We note a number of fundamental properties of statically indeterminate trusses (Rabinovich 1960, Bernshtein 1957). 1. For statically determinate truss it is always possible to find for every element of a truss such cross section where maximum allowable stresses arise. In other words, the material of cross sections of all the rods would be fully utilized. In the general case, such problem for statically indeterminate systems is unsolvable, i.e., in a redundant truss it is impossible to establish such cross sections so that the allowable stresses would occur in all truss elements. It means that material in cross sections in all elements of a redundant truss could not be fully utilized. The best solution of this problem is as follows: it is possible to ensure that the maximum allowable stresses occur in all the rods of the selected primary system and to reconcile with the fact that the stresses in the redundant elements of the truss are lower than the allowable ones. The essence of this affirmation is Maurice Lévy (1873) and Victor Kirpichev (1902) theorem. 2. Assume the structure has one internal redundant constraint of length l1 and axial stiffness ЕА1, where A1 is a cross-sectional area. Let the primary unknown X1 be the force in this constraint. For its determining the canonical equation of the force method δ11X1 þ Δ1P ¼ 0 serves, so X1 ¼ Δ1P/δ11. The free term Δ1P is independent of cross-sectional area of the redundant member 1. A different situation for unit displacement δ11. It may be presented in the form δ11 ¼ C þ ðl1 =EA1 Þ: Here the first term C means the displacement δ11 caused by the compliances of all elements of a truss, except redundant member 1, while the second term takes into account compliance of redundant rod (in Table 9.7 term C presents the sum of all entries of the column 5 except for rod D1–4). It is obvious, C is strictly positive, C > 0. The force in the redundant member equals X1 ¼
Δ1P C þ l1 =EA1
ðaÞ
If the cross-sectional area of redundant rod is A1 ¼ 0, then primary unknown becomes X1 ¼ 0. If the area A1 is insignificant, then with a gradual increase of A1 it is possible to neglect C, and the force in the redundant rod 1 increases almost in proportion to the A1. With further increase of A1, a nonlinear effect appears (Fig. 9.37). If A1 ! 1, then X1 ! Δ1P/C. This limit is equal to the force that would have arisen in the redundant rod under the condition of its absolute rigidity. This means that increasing of the cross-sectional area of the redundant element has its own objective boundaries. In other words, beyond these limits, further consumption of material in order to reduce the stress of the overloaded rod is useless. This feature is especially pronounced when analyzing stresses in the redundant rod
326
9 The Force Method
σ1 ¼
X1 Δ1P ¼ A1 CA1 þ l1 =E
ðbÞ
According to (b) in the case of a large cross-sectional area A1 of a redundant rod, the stress σ 1 varies almost inversely with the value A1, while in case of small values A1, the stress is almost independent of the area A1. If A1 ¼ 0, then σ 1 ¼ EΔ1P/l1 6¼ 0. In other words, an infinitely thin rod, although it will take an infinitesimal small force, will, however, experience a finite stress. This result can be explained from a geometric point of view: a change in the distance between the end joints of the redundant rod 1 is finite. These fundamental properties of the redundant trusses are shown in Fig. 9.37.
X1 A1 σ1 A1 Fig. 9.37 Force X1 and stress σ 1 in the redundant element of the truss vs. its cross-sectional area A1
3. From formula (a) for X1 the next conclusion can be made: any change of rigidity EA1 of redundant element of the truss cannot change the sign of internal force in this element. 4. The change in the cross-sectional area of the redundant rod results in a change in the force in this rod. For a truss as a whole, changing the force in one redundant rod is equivalent to adding two equal and oppositely directed forces ΔX applied to the ends of this redundant element. It is obvious that in some rods of the truss this additional load causes additional tensile forces, while in others elements - the compressive forces. Therefore, when the cross-sectional area of the redundant rod changes, a redistribution of forces in the remaining rods of the truss takes place, therewith a change in the sign of the force is possible in individual rods (not redundant) of the truss. 5. Upon differentiating the expression for the primary unknown X over the cross-sectional area A1 of the redundant rod, and taking into account (a), we obtain ∂X 1 Δ1P l1 l1 ¼ ¼ X1 2 2 ð CEA ∂A1 1 þ l1 ÞA1 ðC þ l1 =EA1 Þ EA1 This means that the increment of the force X1 due to a given change of the cross-sectional area A1 depends not only on the value of A1 but also on the magnitude of the force X1. In other words, a change of the cross-sectional area of heavily loaded rod causes a significant change of the force in the rod itself, and hence in all other rods. Conversely, a change of the cross section of a weakly loaded rod insignificantly changes the stress state of the elements of the truss. 6. It is important to point the element of a truss which will turn out to be the most sensitive to the change in the cross-sectional area of the redundant element. The following assertion is true: of all the rods of the truss the most sensitive to changing the cross-sectional area in the ith redundant rod is an element, in which the greatest force caused by the force Xi ¼ 1 arises. In particular, for the truss in Fig. 9.36, the rod 2–5 is the most sensitive to the change in the cross-sectional area in the redundant rod 1–4, the least sensitive are the vertical members 1–5 and 2–4. This statement helps to choose a strategy for finding rods in which the cross-sectional area should be changed to obtain the desired effect.
9.6
Redundant Arches
Different types of statically indeterminate arches are presented in Fig. 9.38. They are two-hinged arches with or without tie (Fig. 9.38a, b), one-hinged arch (Fig. 9.38c), and arch with fixed ends (Fig. 9.38d).
9.6 Redundant Arches
327
a
d
c
b n=1
n=1
n=2
n=3
Fig. 9.38 Types of the redundant arches
The most effective method for analytical analysis of statically indeterminate arches is the force method in canonical form (9.4): for two-hinged arches (a, b) we have one primary unknown, two unknowns for one-hinged arch (c), and three unknowns for hingeless arch (d). The degree of redundancy n is shown for each design diagram. As for three-hinged arch (Chap. 6), the distribution of internal forces for redundant arches depends on a shape of the neutral line (parabola, circular, etc.). This should be taken into account when calculating unit coefficients and free terms of canonical equations. In general case, these coefficients and free terms depend on bending moments, axial and shear forces ð ð ð Qi Qk ds M i M k ds N i N k ds δik ¼ þ þμ EI EA GA ðsÞ
ð
ΔiP ¼ ðsÞ
M i M 0P ds þ EI
ð
ðsÞ
N i N 0P ds þμ EA
ðsÞ
ð
ðsÞ
Qi N 0P ds GA
ðsÞ
where M i , Qi , N i and M k , Qk , N k are bending moment, shear and axial forces in the i-th and k-th unit states; and M 0P , Q0P , N 0P are bending moment, shear and axial force caused by given load in primary system. This paragraph presents an analysis of a two-hinged and hingeless arches subjected to fixed load. Also, the detailed analysis is presented for case of modified design diagram, which allows applying the Vereshchagin rule for computation of unit and loaded displacements. This approach has important methodological significance because it can be applied to analysis of the arches with features (arbitrary change in flexural rigidity, the presence of supports at various levels, etc.).
9.6.1
Parabolic Two-Hinged Arch
Two-hinged uniform symmetrical arch is shown in Fig. 9.39a. The flexural stiffness of the cross section of the arch is EI. The 4f equation of the neutral line of the arch is y ¼ 2 ðl xÞx. The arch is subjected to uniformly distributed load q within all the l span. It is necessary to find the distribution of internal forces. Let us regard the horizontal component of the reaction, thrust, as redundant unknown (Fig. 9.39b). Canonical equation is δ11 X 1 þ Δ1P ¼ 0 Complete expressions for unit and loaded displacements take into account bending moments, shear and axial forces. Assumptions 1. When calculating displacements, the influence of normal and shear forces in the arch are neglected. Therefore, the expressions for δik and ΔiP become truncated. 2. Infinitesimal element of the arch axis ds can be taken equal to its projection on the horizontal x axis, i.e., ds dx. It means the integration along the curvilinear axis of the arch is replaced by the integration along the horizontal projection of the arch axis within 0-l.
328
9 The Force Method
a
b
q=2kN/m
φ
φ
f=6m
EI=const y(x)
y X1
x l=24m
Primary system
x
d
c
q=2kN/m
N1
M1
3
2
4 A
1
φ
Q1
y
0 A
H
φ
5
6
f=6m
y
q=2kN/m
7
l/8 l/8
B
8
H
H=1 l=24m
RB
30.00
26.64
24.73
24.00
24.73
26.64
33.93
RA
N8
33.93
N0
30.00
x
N (kN)
−
Fig. 9.39 (a, b) Two-hinged parabolic arch. Design diagram and primary system, (c) positive directions of internal forces, (d) design diagram with notation of the specified points 0–8; the final axial force diagram and reaction of supports
With these assumptions we have ð δ11 ¼
2
M 1 ds EI
ðsÞ
ð Δ1P ¼
M 1 M 0P ds EI
ðsÞ
ðl
2
M 1 dx EI
ðaÞ
M 1 M 0P dx EI
ðbÞ
0
ðl 0
For design diagram of parabolic arch in Fig. 9.39a, b we have M 1 ¼ 1 y ¼
4f ql x2 qx ðl xÞx, M 0P ¼ x q ¼ ðl xÞ, 2 2 2 2 l
where M 1 and M 0P are bending moments at section x in the primary system due to unit primary unknown (X1 ¼ 1) and given load, respectively. Thus, unit and loaded displacements should be calculated according to (a) and (b) as follows: δ11
16 f 2 ¼ 4 l EI
ðl
16 f 2 ðl xÞ x dx ¼ 4 l EI
0
Δ1P
ðl
2 2
1 ¼ EI
2 2 16 f 2 l5 8 f 2l , ¼ l x 2lx3 þ x4 dx ¼ 4 l EI 30 15 EI
0
ðl 0
4f qx 2q f 2 ðl xÞx ðl xÞdx ¼ 2 2 l l EI
ðl x2 ðl xÞ2 dx ¼ 0
ql3 f : 15EI
9.6 Redundant Arches
The primary unknown X 1 ¼
329
Δ1P ql2 ¼ . For given parameters q, δ11 8f X1 ¼
l, f we have
ql2 2 242 ¼ ¼ 24kN 8f 86
After determining the primary unknown X, the calculation of all internal forces does not cause difficulties. It is obvious, the horizontal reaction at left support is H ¼ X1 (Fig. 9.39d). The bending moment at any section x is qx2 ql ql2 4f qx2 ql lx x2 x2 x þ 2 ðl xÞx ¼ x ¼
0 M x ¼ RA x X 1 y 2 2 l 2 8f 2 l l l |{z} |fflfflfflfflfflffl{zfflfflfflfflfflffl}
ðcÞ
y
X1
Thus, if the parabolic two-hinged uniform arch is loaded by uniformly distributed load along entire span, then in all sections of the arch the bending moments are equal to zero. Consequently, the shear forces are also equal to zero. Thus, the feature of this design diagram is that only axial forces arise at the sections of the arch. Notes: 1. Assume the arch is loaded only within left (or right) half of the span. Since the arch is symmetrical, the loaded displacement ql3 f ql2 will be equal to Δ01P ¼ , and final expression for thrust becomes H ¼ . 30EI 16f 2. Let one of the arch supports be movable, and both half-arches are connected by the tie at the level of the supports as shown in Fig. 9.38a; arch is loaded by uniformly distributed load along entire span. The cross-sectional area of the tie is Аt. Material of the arch and tie is same. Suppose that the primary unknown X1 is a force N1 in the tie. In this case the expression of unit displacement should be written as follows ðl δ11 ¼
2
M 1 dx þ EI
0
ðl
2
N 1 dx , EAt
N1 ¼ 1
ðdÞ
0
Since N 1 ¼ 1, then formula (d) leads to the following result δ11
1 ¼ δ11 ðM Þ þ EAt
ðl dx ¼
8 f 2l l þ , 15 EI EAt
0
while loaded term Δ01P remains unchanged. The required primary unknown is X1 ¼
ql2 : 15 I 8f 1 þ 8 f 2 At
Note, this formula includes the inertial moment I of the cross section of the arch itself. Limiting cases: 1. If At ! 0 , then X1 ! 0 ; ql2 2. If At ! 1, then X 1 ! . 8f
9.6.2
Circular Arch with Clamped Supports
This paragraph provides an analysis of a circular hingeless uniform arch subjected to uniformly distributed load. The concept of elastic center is applied. The analytical solution is presented. Influence of the axial and shear forces on the displacements of the arch is not taken into account.
330
9 The Force Method
Design diagram of uniform circular symmetrical arch of radius r and central angle π is shown in Fig. 9.40a. The arch is subjected to uniformly distributed load q within the entire span. We need to calculate the reactions of supports and internal forces at crown C, located on the axis of symmetry (AS). q
a
y
b C
EI=const X1
r AS A
X2
α
A
B
X2
r
2r
c
x
X1 0 y0
B r
0.046
0.05
MP
factor qr2
α0
A
B
0.108 Fig. 9.40 Arch with fixed ends. (a, b) design diagram and primary system; (c) bending moment diagram
This arch is the third degree of redundancy. Let the primary unknown be internal forces at the crown C. They are axial force X1, bending moment X2, and shear force X3. Since actual load is symmetrical, then antisymmetrical unknown X3 is equal to zero; therefore this unknown is not shown. Thus, we have two coupled algebraic equations with respect to unknowns X1 and X2. Primary unknowns X1 and X2 are transferred to the elastic center of the arch (point 0); in this case the secondary unit displacements δ12 ¼ δ21 ¼ 0 and each of the two equations will contain only one unknown X1 and X2 . The coordinates of the elastic center of a half-circular uniform arch coincide with the center of gravity of semicircle x0 ¼ 0;
y0 ¼ 2r=π ¼ 0:637r:
Coordinates of an arbitrary section of the arch and the differential of the curvilinear coordinate ds x ¼ r cos α;
y ¼ r sin α y0 ¼ r sin α
2r ; π
ds ¼ rdα
Bending moments in the loaded state in the primary system M 0P ¼ qx2 =2 ¼ qr 2 =2 cos 2 α Bending moments in the unit states in the primary system M 1 ¼ 1 y ¼ ðr sin α y0 Þ ¼ r sin α þ 2r=π M 2 ¼ 1 Unit displacements caused by X1 ¼ 1and X2 ¼ 1 are
9.6 Redundant Arches
331
ð δ11 ¼
2
M1 2 ds ¼ EI EI
π=2 ð
ðsÞ
0
π=2 ð 2r 2 2r 3 4 4 r sin α þ |{z} rdα ¼ sin 2 α sin α þ 2 dα π ffl} π EI π |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl
ds
2
0
M1
2r 3 α sin 2α 4 cos α 4απ=2 π 4 r3 þ þ 2 ¼ 4 π 2 π EI EI 2 π 0 π=2 ð 2 ð M2 2 πr ¼ ð1Þ2 rdα ¼ ds ¼ EI EI EI ¼
δ22
ðsÞ
0
Loaded displacements in primary system ð Δ1P ¼
M 1 M 0P 2 ds ¼ EI EI
ðsÞ
π=2 ð
0
2r qr 2 2 r sin α þ rdα cos α |{z} π ffl} 2 |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} ds M1
qr 4 ¼ EI
π=2 ð
cos 2 α sin α
M 0P
2 cos 2 α dα π
0
¼
qr 4 EI
π=2 ð
2 2 qr 4 ; sin α sin 3 α þ sin 2 α dα ¼ π π 6EI
0
ð Δ2P ¼
M 2 M 0P 2 ds ¼ EI EI
ðsÞ
π=2 ð
qr 2 cos 2 α πqr 3 rdα ¼ ð1Þ 2 4EI
0
Canonical equations of the force method δ11X1 þ Δ1P ¼ 0 and δ22X2 þ Δ2P ¼ 0. Primary unknowns are Δ1P qr 4 1 qr ¼ ¼ ¼ 0:560qr δ11 6 ðπ=2 4=π Þr 3 6ðπ=2 4=π Þ Δ πqr 3 1 qr 2 ¼ ¼ 0:250qr 2 X 2 ¼ 2P ¼ δ22 4 πr 4
X1 ¼
Bending moments at supports and crown C qr 2 þ X 1 y0 X 2 ¼ 0:108qr 2 2 M C ¼ X 1 ðr y0 Þ X 2 ¼ 0:046qr 2
MA ¼ MB ¼
Bending moment diagram is plotted on the tensile fibers and shown in the Fig. 9.40. Maximum negative bending moment equals Mmin ¼ 0.05qr2 and occurs at α0 ¼ 34 100. The vertical reactions and thrust are V A ¼ V B ¼ qr,
9.6.3
H A ¼ H B ¼ X 1 ¼ 0:560qr
Analysis of Parabolic Arch on the Basis of Modified Design Diagram
In some cases, computation of coefficients and free terms of canonical equations of the force method by integration leads to some difficulties. To avoid them an approximate procedure is recommended when modified design diagram should be used: curvilinear axis of the arch is approximated by set of rectilinear segments.
332
9 The Force Method
Procedure for analysis of statically indeterminate arches is as follows: 1. Accept the modified model of the arch. Calculate the geometrical parameters of the arch at specified points. 2. Choose the primary system of the force method. 3. Calculate the unit and loaded displacements, using truncated formulas (Sect. 9.6). Computation of these displacements may be performed using the graph multiplication method. 4. Find the primary unknown using canonical Eq. (9.4) of the force method. 5. Construct the internal force diagrams; for structure with first degree of redundancy the following formulas may be applied: M ¼ M 1 X 1 þ M 0P ðeÞ
Q ¼ Q1 X 1 þ Q0P N ¼ N1X1 þ
N 0P
where X1 is primary unknown; M 1 , Q1 , N 1 are bending moment, shear and axial force caused by primary unknown X1 ¼ 1; and M 0P , Q0P , N 0P are bending moment, shear and axial force caused by given load in primary system. 6. Calculate the reactions of supports and provide their verifications. In order to compare the approximate results with known solution, we consider the arch in Fig. 9.39a from Sect. 9.6.1. As the primary unknown X1 we will accept the same force, i.e., the thrust of the arch. Modified design diagram means the curvilinear axis of the arch is replaced by a group of straight line segments. In this case for computation of unit and loaded displacements the Vereshchagin’s rule may be applied. Specified points of the arch. The span of the arch is divided into eight parts with equal horizontal projection; the specified points are labeled 0–8 (Fig. 9.39d). Parameters of the arch for these sections are presented in Table 9.8; the following formulas for calculation of trigonometric functions of the angle φ between the tangent to the arch and x-axis have been used: tan φ ¼ y0 ¼
4f ðl 2xÞ , l2
1 cos φ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , 1 þ tan 2 φ
sin φ ¼ cos φ tan φ
Table 9.8 Geometrical parameters of parabolic arch
Points 0 1 2 3 4 5 6 7 8
Coordinates, m x y 0.0 0 2.625 3 4.500 6 5.625 9 6.000 12 5.625 15 4.500 18 2.625 21 0.0 24
tanj 1.00 0.75 0.50 0.25 0.0 -0.25 -0.5 -0.75 -1.00
cosj 0.7070 0.800 0.8944 0.9701 1.0 0.9701 0.8944 0.800 0.7070
sinj 0.7070 0.6000 0.4472 0.2425 0.0 -0.2425 -0.4472 -0.6000 -0.7070
The length of the chord between points n and n 1 equals qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ln,n1 ¼ ðxn xn1 Þ2 þ ðyn yn1 Þ2
ð9:22Þ
The chord lengths of each portion of the arch are presented in Table 9.9. Table 9.9 The chord length of each portion of the arch
Portion Length (m)
0-1 3.9863
1-2 3.5377
2-3 3.2040
3-4 3.0233
4-5 3.0233
5-6 3.2040
6-7 3.5377
Internal forces in the unit state. The arch is subjected to unit primary unknown X1 ¼ 1 (Fig. 9.39b).
7-8 3.9863
9.6 Redundant Arches
333
Horizontal reaction H ¼ 1 and the positive directions of internal forces are shown in Fig. 9.39c. M 1 ¼ 1 y ð9:23Þ
Q1 ¼ 1 sin φ N 1 ¼ 1 cos φ Internal forces at specified points in the unit state according to (9.23) are presented in Table 9.10.
Table 9.10 Internal forces of the arch in the unit state
Points
M1
Q1
N1
0 1 2 3 4 5 6 7 8
0.0 -2.625 -4.50 -5.625 -6.00 -5.625 -4.50 -2.625 0.0
-0.7070 -0.6000 -0.4472 -0.2425 0.0 0.2425 0.4472 0.6000 0.7070
-0.7070 -0.8000 -0.8944 -0.9701 -1.0000 -0.9701 -0.8944 -0.8000 -0.7070
The unit displacement caused by primary unknown X ¼ 1 equals X ð M1 M1 δ11 ¼ ds EI
ðfÞ
ðsÞ
Thus, the only column M 1 (Table 9.10) will be used for calculation of unit displacement; the columns Q1 and N 1 will be used for computation of final shear and axial forces as indicated at step 5 of procedure. Internal forces in the loaded state. Displacement in the primary system caused by applied load equals X ð M1 M0 P Δ1P ¼ ds, ðgÞ EI ðsÞ
where M 0P is bending moments in the arch in the primary system due to given load q. In these formulas, the integral is calculated for each straight line segment, summation is performed over all segments. The reactions of supports of the primary system in the loaded state are R0A ¼ R0B ¼ 24kN ; this state is not shown. Expressions for internal forces are as follows (0 x 24) x2 M 0P ¼ R0A x q ¼ 24x x2 0 2 0 QP ¼ RA qx cos φ ¼ ð24 2xÞ cos φ N 0P ¼ R0A qx sin φ ¼ ð24 2xÞ sin φ Internal forces at specified points in the loaded state of the primary system are presented in Table 9.11.
334
9 The Force Method Table 9.11 Internal forces of the arch in the loaded state
Points 0 1 2 3 4 5 6 7 8
M P0
QP0
N P0
kNm 0.0 63 108 135 144 135 108 63 0.0
kN 16.968 14.400 10.7328 5.8206 0.0 -5.8206 -10.7328 -14.400 -16.968
kN -16.968 -10.800 -5.178 -1.455 0.0 -1.455 -5.178 -10.800 -16.968
Computation of unit and loaded displacements. For calculation of displacements, the Simpson’s formula is applied. Unit and loaded displacements are M 1 M 1 X li 2 ¼ a1 þ 4c21 þ b21 EI 6EI 1 n
δ11 ¼
Δ1P ¼
ðhÞ
n M 1 M 0P X li ða1 aP þ 4c1 cP þ b1 bP Þ ¼ EI 6EI 1
where li is the length of the i-th straight portion of the arch (Table 9.9); n is the number of the straight portions of the arch (in our case n ¼ 8); a1, aP are the ordinates of the bending moment diagrams M 1 and M 0P at the extreme left end of the portion; b1, bP are the ordinates of the same bending moment diagrams at the extreme right end of the portion; and c1, cP are the ordinates of the same bending moment diagrams at the middle point of the portion. Thus, procedure of integration (t) and (g) is changed by procedure of summation (h). Calculation of the unit and loaded displacements is presented in Table 9.12. Section “Unit state,” columns a1, c1, and b1 contain data from column M 1 , Table 9.10. Section “Loaded state,” columns aP, cP, and bP contain data from column M 0P , Table 9.11. As an example, for portion 12 (l12/6 ¼ 3.5377/6 ¼ 0.5896), the entries of columns 6 and 10 are obtained by the following way h i 0:5896 ð12Þ ð2:625Þ2 þ 4ð3:5625Þ2 þ ð4:50Þ2 ¼ 45:9335=EI δ11 ¼ EI 0:5896 ð12Þ Δ1P ¼ ½ð2:625Þ 63 þ 4ð3:5625Þ 85:5 þ ð4:50Þ 108 EI ¼ 1, 102:40=EI Table 9.12 Calculation of coefficient and free term of canonical equation
Portion 1 0-1 1-2 2-3 3-4 4-5 5-6 6-7 7-8 Factor
li 6 EI 2 0.6644 0.5896 0.5340 0.5039 0.5039 .0.5340 0.5896 0.6644 1 EI
a1 3 0.0 -2.625 -4.500 -5.625 -6.000 -5.625 -4.500 -2.625
Unit state c1 b1 4 -1.3125 -3.5625 -5.0625 -5.8125 -5.8125 -5.0625 -3.5625 -1.3125
5 -2.625 -4.500 -5.625 -6.000 -5.625 -4.500 -2.625 0.0
M1 ´M1 EI
6 9.1563 45.9335 82.4529 102.1815 102.1815 82.4529 45.9335 9.1563 1 EI
aP 7 0.0 63 108 135 144 135 108 63
Loaded state cP bP 8 31.5 85.5 120.5 139.5 139.5 121.5 85.5 31.5
9 63 108 135 144 135 108 63 0.0
M 1 ´ M P0 EI
10 -219.75 -1102.40 -1978.87 -2452.35 -2452.35 -1978.87 -1102.40 -219.75 1 EI
9.6 Redundant Arches
335
Unit displacement and loaded term are h i XM M 1 479:4484 m 1 1 δ11 ¼ ¼ ð9:1563 þ 45:9335 þ þ 9:1563Þ ¼ EI EI EI kN X M1 M0 1 11, 506:74 P Δ1P ¼ ½m ¼ ð219:75 þ 1102:40 þ þ 219:75Þ ¼ EI EI EI Canonical equation and primary unknown (thrust) are 479:4484 11506:74 X1 ¼0 EI EI
X 1 ¼ 24:00ðkNÞ
Construction of internal force diagrams. Internal forces, which arise in the entire structure, may be calculated by formulas (e). Calculation of internal forces in the arch due to given fixed load is presented in Table 9.13; internal forces M 1, Q1, and N 1 due to unit primary unknown X1 ¼ 1 are presented earlier in Table 9.10. Corresponding axial force diagram is presented in Fig. 9.39d.
Table 9.13 Calculation of internal forces at specified points of the arch
Points 0 1 2 3 4 5 6 7 8
M
Q
1
2
3
4
5
6
1+4
2+5
3+6
0.0 -63 -108 -135 -144 -135 -108 -63 0.0
-16.968 -14.400 -10.733 -5.820 0.0 5.820 10.733 14.40 16.968
-16.968 -19.2 -21.466 -23.282 -24.00 -23.282 -21.466 -19.20 -16.968
0.0 63 108 135 144 135 108 63 0.0
16.968 14.400 10.733 5.820 0.0 -5.820 -10.733 -14.40 -16.968
-16.968 -10.80 -5.178 -1.455 0.0 -1.455 -5.178 -10.800 -16.968
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
-33.936 -30.0 -26.644 -24.737 -24.0 -24.737 -26.644 -30.0 -33.936
M 1 X1
Q1 X1
N 1 X1
M P0
Q P0
N P0
N
Knowing the internal forces at points 0 and 8 we can calculate the reactions at supports A and B. Axial force N0 ¼ 33.936 kN is shown at support A (Fig. 9.39d). Reactions of this support are RA ¼ N 0 sin φ0 ¼ 33:936 0:707 ¼ 24kN, H ¼ N 0 cos φ0 ¼ 24 kN For primary unknown X1 we have obtained the same result. Verification of results. (a) For arch in whole ∑Y ¼ RA þ RB q 24 ¼ 0 (b) The bending moment at any point of the arch M ðxÞ ¼ RA x Hy
qx2 4f qx2 ¼ 24x 24 2 xðl xÞ 2 2 l
for given parameters f ¼ 6m, l ¼ 24m, and q ¼ 2kN/m indeed equals to zero for any x. Also, Q ¼ dM/dx ¼ 0.
Discussion 1. Comparison of Table 9.13 and results for arch considered in Sect. 9.6.1 (Fig. 9.39) shows that approximation of curvilinear axis of the arch by the set of rectilinear segments leads to very good results. It is important that the analysis based on the modified design diagram retains the main features of the structure (absence of bending moments and shear in all sections of the arch, the magnitude of the axial force in the crown, and the increase in axial force from the crown to the support).
336
9 The Force Method
At the same time, high accuracy of axial force calculation was achieved. Indeed, since Qx ¼ 0, then equilibrium equation ∑X ¼ 0 for arbitrary portion of the arch of length x leads to the following expression for axial force N(φ) ¼ H/cos φ. The values of axial forces, approximate and precise (shown in brackets) for some specified sections, are presented below. N 0 ¼ 24=0:707 ¼ 33:946ð33:936Þ, N 2 ¼ 26:834ð26:644Þ, N 3 ¼ 24:740ð24:737Þ 2. Procedure for the analysis of nonuniform arch remains the same. However, in this case Table 9.8 must contain additional column with parameter EI for each point 0–8, Table 9.9 must contain parameter EIi for middle point of each portion, and column 2 of Table 9.12 should be replaced by column li /6EIi.
9.7
Combined Redundant Structures
One of the case of combined structure is when it contains elements of two types: elements subjected to bending, and elements which perceive longitudinal forces only, for example, the arch with tie. The feature of the analysis of such structure is that the coefficients and free terms of canonical equations should be calculated taking into account those type of forces that occurs in specific structural elements. Design diagram of combined redundant symmetric structure that carries one lumped force P is shown in Fig. 9.41a. The structure consists of beam AB of length l and the elements of strengthening A-1, 1–2, 1–4, 3–4, and 4-B. The beam AB is subjected to bending deformations, while the strengthening elements are subjected to axial deformations. The bending and axial stiffnesses of the beam AB are EI and EA ¼ 1, respectively. The axial stiffness of strengthening elements is shown in the design diagram. We need to calculate the internal force in the tie 1–4. The structure under investigation is statically indeterminate of the first degree. The primary system is shown in Fig. 9.41b; the primary unknown X1 is the force in the tie 1–4. Canonical equation of the force method is δ11X1 þ Δ1P ¼ 0. The primary unknown X1 ¼ Δ1P/δ11. Internal forces in the unit state (Fig. 9.41c, d). The structure is subjected to unit unknown X1 ¼ 1. pffiffiprimary ffi Equilibrium conditions of joint 1 lead to the following axial forces N A1 ¼ 2 and N 12 ¼ 1. Corresponding axial forces are shown in Fig. 9.41c. The forces N 12 ¼ N 34 ¼ 1 are transferred on the beam АB at points 2 and 3, respectively. Corresponding bending moment diagram for beam AB is shown in Fig. 9.41d. The unit displacement presents the mutual displacement of points 1 and 4; for calculation of this displacement we take into account the bending moments at beam AB and axial forces in all the parts of a structure, i.e., δ11 ¼ δ11 ðM Þ þ δ11 ðN Þ
ð9:24Þ
The unit displacement produced by bending moments in the beam 2 δ11 ðM Þ ¼
X M1 M1 EI
¼
3
3 7 1 6 62 1 l 1 l 2 1 l þ l 1 l 1 l 7 ¼ l 4 5 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 24 4 3 4 2|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl 4 24EI ffl4} Portions A2, 3B
Portion 23
Since axial stiffness of the beam EAAB ¼ 1, then unit displacement δ11(NAB) caused by axial forces in the beam is equal to zero. The unit displacement, which is produced by axial forces in the strengthening elements, is 2 δ11 ðN Þ ¼
3
XN N 6 1 1 l l 1 pffiffiffi l pffiffiffi pffiffiffi7 1 1 2 1 þ 26 ð1Þ ð1Þ þ 2 27 ¼ 4 5¼ E 1 A0 2 E 1 A1 4 E 1 A2 4 EA |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflffl ffl{zfflfflfflfflfflfflffl} Portion 14
Portion 12
pffiffiffi l 1 1 2 2 ¼ þ þ 2 E1 A0 E 1 A1 E 1 A2
Portion
A1
9.7 Combined Redundant Structures
a
337
P
l/2 l/4 45° E 1A 2
B
EA=∞
EI 2 E 1A 1
A
3 E1A0
1
b
4
P
B
A
2
3 X1 4
1
c
Unit state X1=1
A
Joint 1
B
2
N1
3
N 1-2 = 1
-1
√2
X1=1
NA-1=√2
4
1
X1=1 1
1×l/4
d
B
A 1
2
3
N1-2=1
1
Loaded state
e A
1
P C
2
M1
3
B
0
MP
Pl/8 Pl/4 4
1
Fig. 9.41 Combined structure. (a, b) Design diagram and primary system; (c) unit state, axial forces in the strengthening elements; N1–2 is compressed; (d) unit state, and bending moment diagram for beam AB; (e) loaded state, and bending moment diagram for beam AB; (c–e) diagrams are related to the primary system
The loaded displacement 2 Δ1P ¼
X M1 M0
P
EI
¼
3
6 7 1 61 l l 2 Pl 1 Pl Pl l l7 11 Pl3 26 þ þ 1 7¼ EI 42|fflfflfflfflfflfflfflfflfflfflfflfflfflffl 4 ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl 4 3 8ffl} 2 8 4 4 45 384 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Portion A2
Portion 2C
The total unit displacement δ11 ðM Þ þ δ11 ðN Þ ¼ Primary unknown
pffiffiffi l3 l 1 1 2 2 þ þ þ 24EI 2 E1 A0 E 1 A1 E1 A2
338
9 The Force Method
X1 ¼
Δ1P 11 ¼ δ11 384
P 1 24
þ
1 2 k0
þ
1 2 k1
11 P pffiffiffi ¼ pffiffiffi 16 þ 2 k2 1 þ 12k 0 þ 12k1 þ 24 2 k2
Here dimensionless parameters are k0 ¼
EI , l E1 A0 2
k1 ¼
EI , l E1 A1 2
k2 ¼
EI l E 1 A2 2
Axial force X1 in the tie depends on its axial stiffness. With increasing EA0, the dimensionless parameter k0 ¼ (EI)/l2EA0 decreases; it leads to increasing of force X1 in the tie. If stiffness EA0 ¼ 0, then force X1 in the tie is zero. Indeed, such stiffness means that the strengthening elements does not work, i.e., the original statically indeterminate structure becomes statically determinate. After determining the primary unknown X1 the internal forces in the structure should be calculated by formulas M ¼ M 1 X 1 þ M 0P N ¼ N 1 X 1 þ N 0P Bending moment diagram M 1 is caused by force X1 ¼ 1 (which is represented by two compressed forces N12 ¼ N34 ¼ 1 applied to the beam AB) as shown in Fig. 9.41d. Since extended fibers of the beam caused by these forces and P are located above and below of the reference line, respectively, then bending moment at the point where force P is applied l Pl M ¼ X1 þ 4 4 The force in member N12 ¼ 1 X1 þ 0 ¼ X1(compressed) The peculiarity of the behavior of the combined system is that the reinforcement elements have an unloading effect on the beam AB. This is indicated by the negative sign of the first term. Note: Analysis of two hinged arch with tie as combined structure briefly considered in Sect. 9.6.1, Note 2.
9.8
Deflections of Statically Indeterminate Structures
As it is known, for calculation of deflection of any bending structure it is necessary to construct the bending moment diagram in the actual state, then construct unit state and corresponding bending moment diagram, and finally both diagrams must be multiplied ð MPM M M Δk ¼ ds ¼ P ð9:25Þ EI EI s
Here MP is a bending moment diagram of a given statically indeterminate structure due to applied load, while a bending moment diagram M is pertaining to unit state. The construction of bending moment diagram MP for continuous beams, frames, arches have been discussed in the previous sections. Now the following principal question arises: how to construct the unit state for determining M ? It is obvious that unit load must correspond to the required deflection. But which structure (original or somehow modified) must carry this unit load? If unit load is applied to the given statically indeterminate structure, then for construction of bending moment diagram in the unit state the additional analysis of redundant structure is required. Therefore computation of deflections for redundant structure becomes cumbersome. However, solution of this problem can be significantly simplified, taking into account the following fundamental concept. Bending moment diagram MP of any statically indeterminate structure can be considered as a result of application of two types of loads to a statically determinate structure. They are the given external loads and primary unknowns. It means that a given redundant structure may be replaced by any statically determinate structure subjected to a given load and primary unknowns, which are now treated as external forces. It does not matter which primary system has been used for construction of final bending moment diagram, since on the basis of any primary system the final bending moment diagram will be the same. Therefore, the unit load (force, moment, etc.), which corresponds to required displacement (linear, angular, etc.), should be applied in any statically determinate (!) structure, obtained from a given structure by elimination of any redundant constraints. This fundamental idea is applicable for arbitrary statically indeterminate structures.
9.8 Deflections of Statically Indeterminate Structures
339
Calculation of deflection. Simplest redundant beam is shown in Fig. 9.42. It is necessary to calculate the slope θA at support A. For this we need to show bending moment diagram MP in entire structure and apply unit moment at support A in any statically determinate structure. Two versions of unit states are shown in Fig. 9.42a: version 1 is the unit state for free-clamped beam, and version 2 is the unit state for simply supported beam.
a
b
q B
A RA =
q
ql 8
3 ql 8
MP
RA =
ql 2 8
A
ql 2 8
MP
M
A
1 M=1
2
ql 2 16
Version 1
A
ql 8
3 ql 8
ql 2 16
M=1
B
A
2
P=1
M
l/2
l
l
Version 2
M
1/2 1 Fig. 9.42 (a) Computation of slope θA for two versions of the primary systems; (b) kinematical verification (Computation of vertical displacement at support A)
Computation of the slope leads to the following results: MP M l ql2 ql2 ql3 Version 1 θA ¼ ¼ 01þ4 1 1 ¼ EI 6EI 16 8 48EI 2 2 M M l ql 1 ql ql3 Version 2 θA ¼ P ¼ 01þ4 0 ¼ EI 6EI 16 2 8 48EI It easy to check that superposition principle leads to the same result. Indeed, in case of simply supported beam subjected to uniformly distributed load q and support moment MB ¼ ql2/8, the slope at support A equals B θA ¼ θqA þ θM A ¼
ql3 M l ql3 ql2 l ql3 ¼ B ¼ 24EI 6EI 24EI 8 6EI 48EI
Also this concept may be effectively applied for verification of the resulting bending moment diagram. Since displacement in the direction of the primary unknown is zero, then ð MPM M M Δk ¼ ds ¼ P ¼ 0, ð9:26Þ EI EI s
where M is bending moment diagram due to unit primary unknown. This is called a kinematical control of the resulting bending moment diagram. Equations (9.25) and (9.26) are applicable for determination of deflections and kinematical verification for any flexural system. Kinematical verification for structure in Fig. 9.42 is shown below. For given structure the vertical displacement of support A is zero. We can check this fact using above theory. Unit state is constructed as follows: support A is eliminated and unit load P ¼ 1 is applied at point A. Two bending moment diagrams, MP and M, are shown in Fig. 9.42b. Their multiplication leads to the following results: M MP l ql2 l ql2 ver 00þ4 ¼ l ¼0 ΔA ¼ 6EI EI 16 2 8 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Simpson rule
340
9 The Force Method
Indeed, the vertical displacement of support A is zero. The reader is invited to check that slope at support B is zero. For multiplication of both bending moment diagrams, it is recommended to apply the Simpson’s rule. Modified computation of deflection. There is another option for the representation of the bending moment diagrams in unit and loaded states. Bending moment diagram caused by given load may be shown in the any primary statically determinate system (this diagram is denoted M sds P ); in this case the bending moment diagram caused by unit force which corresponds to the required red displacement (unit state) should be shown in the original redundant structure (this diagram is denoted M ) (Kiselev 1960). Δk ¼
M sds P M EI
red
ð9:27Þ
This modified procedure is shown in Fig. 9.43. For redundant beam AB (Fig. 9.43a) we need to calculate the slope at the rolled support A. Let the primary structure be obtained from the original structure by removing support A; thus, we obtained the free-clamped beam. This beam should be loaded by the given uniformly distributed load q along the span l. Corresponding bending moment diagram is denoted by M sds P (Fig. 9.43b). q a
B
A q
b
ql 2 2
ql
2
8
M Psds
l/2 c
1/2
A
M
red
1/4
M=1
Fig. 9.43 Modified computation of displacement. (a) Design diagram of redundant structure; (b) given load in the primary state—statically determinate system (SDS) and corresponding bending moment diagram M sds P , (c) unit state—original redundant structure and corresponding bending moment diagram M
red
The unit state presents the original redundant rolled-clamped beam loaded by unit couple at support A. Corresponding red bending moment diagram, M , is shown in Fig. 9.43c. This diagram may be easily constructed taking into account the concept of the focal ratios; thus, the specified ordinates at points A and B are 1 and 1/2. Multiplication of both bending moment diagrams, using Simpson’s rule, leads to the following result θA ¼
M sds P M EI
red
¼
l ql2 1 ql2 1 ql3 014 ¼ þ 6EI 8 4 2 2 48EI
Verification of this result may be easily done using initial parameter method. Modified approach may be recommended if an user has a handbook with tabulated data which can be used for construction of bending moment diagram for entire redundant structure subjected to unit generalized force (unit state), or easily construct the bending moment diagram by the other appropriate way.
9.9
Settlements of Supports
If any statically indeterminate structure is subjected to settlement of supports, then internal forces arise in the members of the structure. Analysis of such structures may be effectively performed by the force method in canonical form. The primary system and primary unknowns should be adopted as in case of the fixed loads.
9.9 Settlements of Supports
9.9.1
341
Internal Forces due to the Settlements of Supports
Let us consider any statically indeterminate structure with n redundant constraints. Some of supports have linear and/or angular displacements di. The problem is to determine inertial forces which arise in the elements of the structure caused by known settlements of supports. Canonical equations are δ11 X 1 þ δ12 X 2 þ þ δ1n X n þ Δ1s ¼ 0 δ21 X 1 þ δ22 X 2 þ þ δ2n X n þ Δ2s ¼ 0 δn1 X 1 þ δn2 X 2 þ þ δnn X n þ Δns ¼ 0
ð9:28Þ
where free terms Δks (k ¼ 1, 2, . . ., n) represent displacements of the primary system in the direction of primary unknowns Xk due to settlements of the supports. For calculation of these terms we need to use the theorem of reciprocal unit displacements and reactions (Rayleigh second theorem). Let support i have the unit displacement δi. The displacement at any point k may be calculated using the abovementioned theorem, i.e., δki ¼ rik. So, the displacement in direction k due to unit displacement in direction i may be calculated as reaction at support i caused by unit load in direction k. If support i has nonunity displacement di, then the displacement Δk at any point k may be calculated using formula Δki ¼ Rik di , where Rik presents the reaction at support i due to unit load at direction k. In fact, it means, that both parts of formula δki ¼ rik are multiplied by di. In case of several displacements di of the supports, the free terms of canonical equation of the force method are calculated using the following expressions (index s for “settlements” at Δks is omitted): X Δk ¼ Rik di ð9:29Þ i
In this formula Rik is reaction of the shifted constraint in the direction of a given displacement di ; this reaction is caused by unit primary unknown Xk ¼ 1. In other words, Ri1 and Ri2 are reactions in the primary system due to primary unknowns X1 ¼ 1 and X2 ¼ 1, respectively; these reactions are determined in supports, which are subjected to displacement. After calculation of the primary unknowns Xi construction of internal forces is performed as usual. The final bending moment diagram can be constructed by formula Ms ¼ M1 X1 þ M2 X2 þ . . . þ Mn Xn Note that unlike the analysis of structures subjected to loads, the term M 0s in (9.24) is absent. Kinematical verification of the final bending moment diagram may be performed by the following expression X XðM M Σ ds þ Δis ¼ 0 EI
ð9:30Þ
ð9:31Þ
where M Σ is a summary unit bending moment diagram. Procedure for analysis of redundant structures subjected to the settlement of supports is as follows: 1. Provide the kinematical analysis, determine the degree of redundancy, choose the primary system of the force method, and formulate the canonical Eq. (9.28). 2. Construct the unit bending moment diagrams and calculate the unit displacements. 3. Calculate the free terms of canonical equations. 4. Solve the canonical equation with respect to primary unknowns Xi. 5. Construct the internal force diagrams. 6. Calculate the reactions of supports and provide their verifications. Let us consider a two-span uniform beam with equal spans. The middle support 1 is shifted by Δ as shown in Fig. 9.44a. It is necessary to calculate the bending moment at support 1. The degree of static indeterminacy of the structure equals one. The primary system is the set of two simply supported beams (Fig. 9.44b). Bending moment diagram due to unit primary unknown is shown in Fig. 9.44c.
342
9 The Force Method
X1=1 0
EI
a
2
1
b Δ
l
c
l
M1 1
d
0
Δ l
Х1=1
X1=1
Δ l
2
e Δ
Δ Δ1s = −2
Δ l 1/l
1/l
Fig. 9.44 Two-span beam subjected to settlement of support. (a–c) Design diagram, primary system, and bending moment diagram due to X1 ¼ 1; (d, e) two approaches for computation of free term of canonical Eq. (9.28). (d) Mutual displacement in direction X1 due by settlements of support; (e) reaction in shifted support caused by primary unknown X1
The canonical equation and primary unknown are δ11X1 þ Δ1s ¼ 0, X1 ¼ Δ1s/δ11. The unit displacement δ11 ¼
M1 M1 1 2 1 2l ¼2 l1 ¼ 2 3 EI 3EI EI
The free term of canonical equation Δ1s is a displacement in direction of the primary unknown X1 (i.e., the mutual angle of rotation at support 1 caused by the given settlements of support). In this simple case, the Δ1s can be calculated by procedure shown in Fig. 9.44d. If support 1 is shifted, then slope at supports 0 and 2 are Δ/l; therefore the mutual angle of rotation Δ Δ Δ Δ1s ¼ þ ¼ 2 l l l Negative sign means that each moment X produces the negative work on the angular displacement Δ/l due to the settlement of support. The free term can be calculated using a general algorithm (9.29) as follows: let us apply X1 ¼ 1 and compute the reaction in direction of given displacement. As a result, two reactions each 1/l arises at support 1, as shown in Fig. 9.44e. Expression 1 (9.29) leads to the same result. Indeed, Δ1s ¼ 2 Δ. l 2l Δ X 2 ¼ 0. The primary unknown, i.e., the bending moment at support 1 equals Canonical equation becomes 3EI 1 l X1 ¼ M1 ¼ 3
EI Δ l2
A positive sign shows that extended fibers are located below the longitudinal axis of the beam. Note that bending moments at supports of uniform continuous beams with equal spans, caused by vertical displacements of one of its supports, are presented in Table A.14. It can be seen that in case of settlements supports, the distribution of internal forces depends not only on relative stiffness but also on absolute stiffness EI as well. Example 9.9 The design diagram of the redundant frame is the same as in Example 9.7. No external load is applied to the frame, but the frame is subjected to settlement of fixed support A as presented in Fig. 9.45a. The relative flexural stiffness of each element is shown in a circle. Construct the internal force diagrams and calculate reactions of supports. Assume that the vertical, horizontal, and angular settlements are a ¼ 2cm, b ¼ 1cm, and φ ¼ 0.01 rad ¼ 340 3000 , respectively. Solution If the primary system is chosen as in Example 9.7, then the primary unknowns are reactions X1 and X2 (Fig. 9.45b).
9.9 Settlements of Supports
a
343
Design diagram
C
b
X2
5
Primary system
Δ 2s
3m
1
B
4 3
2
1
8
6
Δ1s X1
5m
A a
1
a
φ
b
δ 22
δ21
φ
b X2=1
δ11
10
a
6
3
b
10m
c
4 φ
δ 12
3
10 X1=1
M2
M1 R φ1 = 10
Rb1 = 0
R a2 = 0
Ra1 = 1
d 1.1190
R φ2 = 8
8
Rb2 = 1
e
2.1228
X2=1 3 13
2.1228 1.1190
1.0038 M kNm (factor 10-3EI)
10
X1=1
M∑
1.0038 (factor 10-3EI)
4.5418
6
18
Fig. 9.45 Redundant frame subjected to settlement of fixed support A. (a) Design diagram; (b) primary system and pictorial presentation of free terms of canonical equations; (c) bending moment diagrams in unit states (M 1 and M 2 ) and reactions at the shifted support A; (d) resulting bending moment diagram; (e) summary unit bending mom
Canonical equations of the force method are δ11 X 1 þ δ12 X 2 þ Δ1s ¼ 0
ðaÞ
δ21 X 1 þ δ22 X 2 þ Δ2s ¼ 0 where δik are unit displacements, which have been obtained in the Example 9.7. They are equal to δ11 ¼
666:67 , EI
δ12 ¼ δ21 ¼
275 , EI
δ22 ¼
170:67 EI
ðbÞ
The free terms Δ1s and Δ2s present the displacements in the direction of primary unknowns X1 and X2 in a primary system due to settlement of support (Fig. 9.45b). A feature of the system is that the points of application of the primary unknowns X1 and X2 and the given displacements of support A do not coincide. According to (9.29) we have: X Δ1s ¼ Ri1 di ¼ Ra1 a þ Rb1 b þ Rφ1 φ X ðcÞ Δ2s ¼ Ri2 di ¼ Ra2 a þ Rb2 b þ Rφ2 φ where a, b, and φ are the vertical, horizontal, and angular displacements of support A. The unit reactions R related to support, which is subjected to the given settlement. Each reaction has two indexes. The first index (a, b, or φ) corresponds
344
9 The Force Method
to given direction of the settlement of the support. The second index (1 and 2) means the primary unknown (X1 or X2) which leads to appearance of unit reactions. Thus, Ra1, Ra2 are reactions in direction of a-displacement due to primary unknowns X1 ¼ 1 and X2 ¼ 1; Rb1, Rb2 are reactions in direction of b-displacement due to primary unknowns X1 ¼ 1 and X2 ¼ 1; Rφ1, Rφ2 are reactions in direction of φ-displacement (i.e., the moments at the shifted support A) due to the same primary unknowns. The bending moment diagrams, unit displacement δik, and reactions Ri1 Ri2 in the primary system caused by the unit primary unknowns X1 and X2 are shown in Fig. 9.45c. The expressions (c) in an expanded form are Δ1s ¼ Ra1 a þ Rb1 b þ Rφ1 φ ¼ ð1 a þ 0 b þ 10 φÞ ¼ ð1 0:02 þ 0 0:01 þ 10 0:01Þ ¼ 0:12 m, Δ2s ¼ Ra2 a þ Rb2 b þ Rφ2 φ ¼ ð0 a þ 1 b þ 8 φÞ ¼
ðdÞ
ð0 0:02 þ 1 0:01 þ 8 0:01Þ ¼ 0:09 m In the expression (d) displacements a and b must have the same units (meters) as the dimensions of the structure. Canonical equations become 666:67X 1 þ 275X 2 0:12EI ¼ 0, 275X 1 þ 170:67X 2 0:09EI ¼ 0
ðeÞ
The roots of these equations are X 1 ¼ 1:119 104 EI,
ðfÞ
X 2 ¼ 7:076 104 EI The bending moment diagram is constructed using the superposition principle according to (9.30) M ¼ M1 X1 þ M2 X2
ðgÞ
The corresponding calculations are presented in Table 9.14. Table 9.14 Calculation of bending moments
Points 1 3 4 5 6 8 Factor
M1 -10 -10 0.0 0.0 -10 0.0
M 1 ∙ X1
1.119 1.119 0.0 0.0 1.119 0.0 10-3EI
M2 -8.0 -3.0 -3.0 0.0 0.0 0.0
M 2 ∙ X2
M
-5.6608 -2.1228 -2.1228 0.0 0.0 0.0 10-3EI
-4.5418 -1.0038 -2.1228 0.0 1.1190 0.0 10-3EI
Signs of bending moments + −
+
−
The resulting bending moment diagram is presented in Fig. 9.45d. Static verification. A free-body diagram for rigid joint of the frame is shown in Fig. 9.45d. The extended fibers are shown by dotted lines. Equilibrium of the rigid joint X M ¼ ð2:1228 1:0038 1:1190Þ 103 EI ¼ ð2:1228 2:1228Þ 103 EI ¼ 0
9.9 Settlements of Supports
345
Kinematical verification. The summary unit bending moment diagram M Σ ¼ M 1 þ M 2 is shown in Fig. 9.45e. The formula (9.31) leads to the following result 2
3
6 1 7 M MΣ X 1 2 1 1 2 3 3 3 2:1228 10 10 1:1197 þ Δis ¼ 6 4 5 10 EIþ 1EI 2 3 2EI 2 3 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} portion 68
portion 45
1 5 ð2 13 1:0038 þ 2 18 4:5418 þ 13 4:5418 þ 18 1:0038Þ 103 EI 1EI 6 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} portion 13
þð0:12Þ þ ð0:09Þ ¼ 0:2286 0:2286 ¼ 0 |fflfflfflffl{zfflfflfflffl} |fflfflfflffl{zfflfflfflffl} Δ1s
Δ2s
After verification for bending moment diagram, we can construct the shear and axial force diagrams, find reactions of supports. These procedures have been described in Example 9.7.
Discussion In the case of settlements of supports, the primary unknowns as well as the reactions and internal forces (bending moment, shear, and normal force) depend on both the relative and absolute stiffnesses EI. This is the common property of any statically indeterminate structure subjected to settlement of supports.
9.9.2
Displacements Due to the Settlements of Supports. Modified Approach
The procedure of determining displacement of any redundant structure caused by settlements of supports as discussed in Sect. 9.8 can be significantly simplified. For this purpose the unit generalized force, which corresponds to the required displacement, should be applied to the entire redundant structure; then reaction of the displaced supports should be determined. The general formula for required displacement may be presented as follows X Δir ¼ R0i Δ ð9:32Þ The left part of (9.32) is required displacement in the i-th direction of the given entire statically indeterminate structure, R0i are reactions of the original redundant structure caused by unit load which corresponds to the required displacement Δir; the given settlement of support is denoted as Δ. This idea is illustrated by the following example. The uniform beam of length l with two clamped support is subjected to the angular displacement Δ of the left support. We need to compute the vertical displacement Δir at the middle of the beam (Fig. 9.46). Δ
Δir=?
A
B l/2
l/2
P=1
M′A
R′A
l/2
M′B
R′B
Fig. 9.46 Calculation of displacement in the redundant structure caused by the angular displacement of the left support
346
9 The Force Method
The force P ¼ 1 which corresponds to the required displacement Δir should be applied to the entire redundant structure. The reactive forces R0A ¼ R0B ¼ 1=2 and the reactive couples M 0A ¼ M 0B ¼ l=8 are shown accordingly in Table А.4, row 3. Only reactive moment M 0A performs work on displacement Δ; this work is negative. The required vertical displacement becomes X l lΔ 0 Ri Δ ¼ Δ ¼ Δir ¼ 8 8 Example 9.10 Design diagram of two-span beam ABC with equal spans of l is shown in Fig. 9.47a. Calculate the vertical displacement at point K, if each support is shifted down vertically by ΔA, ΔB, and ΔC. Solution For computation of displacement ΔK we need, first of all, to apply force P ¼ 1 at section K where displacement should be determined, and calculate all reactions which produce a work within the given displacements ΔA, ΔB, and ΔC; these reactions are RA, RB, and RC. This can be done by different methods, such as by three-moment equation, focal ratio method, canonical equation of the force method, or influence line method, which is presented in Table A.16a. If beam is loaded by force P at point K, then according to this table the bending moment at support A is 0.0926Pl and the final bending moment diagram and reaction of support are presented in Fig. 9.47b.
a
K
B
A
ΔA
ΔC
ΔK=?
ΔB l
l/3
2l/3 P
0.0926Pl
b
C
M B
RA=0.0926P
0.1605Pl
RB=0.8519P
RC=0.2407P
Fig. 9.47 Redundant beam subjected to settlement of support B. (a) Design diagram of the redundant beam; (b) reactions and bending moment diagram due by force P at section K
If P ¼ 1, then reactions of each shifted support become RA0 ¼ 0.0926, According to (9.29), the required displacement of point K becomes ΔK ¼
3 X
RB0 ¼ 0.8519,
RC0 ¼ 0.2407
Ri 0 Δi ¼ ½0:0926ΔA þ ð0:8519ÞΔB þ ð0:2407ÞΔC
i¼1
Term (0.8519)ΔB means that the reaction RB directed upward performs negative work on the displacement ΔB. Term [ (0.8519)ΔB ] ¼ 0.8519ΔB means that displacement of point K caused by settlements of support B occurs at the same direction as ΔB. If all settlements of supports are equal, ΔA ¼ ΔB ¼ ΔC ¼ Δ, then ΔK ¼ ð0:0926 1:0926ÞΔ ¼ 1:0Δ It means that beam is shifted as absolutely rigid body. Notes. The advantage of formula (9.29) is the simplicity of its application. In addition, there are extensive tabulated data for reactions of redundant structures subjected to different loads. Problems of this type (settlement of supports) are encountered in the analysis of forced vibrations of the redundant structures subjected to kinematical excitation (see Chap. 17).
9.10
9.10
Temperature Changes
347
Temperature Changes
Generally, if an arbitrary statically indeterminate structure is subjected to change of temperature, then the internal forces arise in the members of the structure. Analysis of such structure may be effectively performed on the basis of a canonical equation of the force method. A primary system and the primary unknowns of the force method are chosen as usual. Below is presented analysis of some redundant structures (beam, frame, and truss) subjected to the temperature changes.
9.10.1 General The canonical equations of force method for structure with n redundant constraints are δ11 X 1 þ δ12 X 2 þ þ δ1n X n þ Δ1t ¼ 0 δ21 X 1 þ δ22 X 2 þ þ δ2n X n þ Δ2t ¼ 0 δn1 X 1 þ δn2 X 2 þ þ δnn X n þ Δnt ¼ 0
ð9:33Þ
where Xn are primary unknowns; the δik and Δit are displacements of the primary system in the direction of i-th primary unknown caused by the unit primary unknown Xk ¼ 1 and by the change of temperature, respectively. The unit displacements are calculated as usual. The free terms Δit (i ¼ 1, 2, . . ., n) should be calculated using the following expression Xð X ð Δt Δit ¼ αt av N i ds þ α M i ds ð9:34Þ h where α is the coefficient of thermal expansion; h is a height of a cross section of a member; N i , M i are normal force and bending moment in a primary system due to action of unit primary unknown Xi. The average temperature (temperature at axial line) and temperature gradient are t av ¼
t1 þ t2 , 2
Δt ¼ jt 1 t 2 j
where t1 and t2 are changes of temperature on the top and bottom fibers of the member; the average temperature tav and temperature gradient Δt are related to uniform and nonuniform temperature changes, respectively. In case of constant α, tav, Δt, and h within each member ð X X Δt ð αt av N i ds þ α M i ds ð9:35Þ Δit ¼ h Ð Ð The integrals N i ds, M i ds present area of corresponding diagram in the primary system. Thus, for computation of Δ1t, the unit primary unknown X1 ¼ 1 should be applied to the primary system and then procedures (9.34) or (9.35) should be performed; the procedure of summation is related to all members. It can be seen that the free term of canonical equation is calculated taking into account the bending moment and axial forces. Solution of (9.33) is the primary unknowns Xi. Bending moment diagram is constructed using the formula Mt ¼ M1 X1 þ M2 X2 þ . . . þ Mn Xn Kinematical control of the final bending moment diagram are performed using the following expression X ð Mt MΣ X ds þ Δit ¼ 0 EI
ð9:36Þ
ð9:37Þ
where M Σ is the summary unit bending moment diagram and Mt is the resultant bending moment diagram caused by change of temperature. To summarize, the procedure for analysis of redundant structures subjected to the change of temperature is as follows:
348
9 The Force Method
1. Provide the kinematical analysis, determine the degree of redundancy, choose the primary system of the force method, and formulate the canonical Eq. (9.33). 2. Construct the unit bending moment and axial force diagrams. 3. Calculate the unit displacements. In case of bending structures take into account only bending moments. 4. Calculate the free terms of canonical equations. For this: (a) Calculate the average temperature and temperature gradient for each member of a structure. (b) Apply formula (9.35). 5. Solve the canonical Eq. (9.33) with respect to primary unknowns Xi. 6. Construct the internal force diagrams according to (9.36). 7. Calculate the reactions of supports and provide their verifications (9.37).
9.10.2 Redundant Beams One-span redundant beam. Let us consider fixed-rolled uniform beam (the height of the cross section of the beam is h), subjected to the following change of temperature: the temperature of the above beam is increased by t1 , while below of the beam increased by the t2 , t1 > t2 (Fig. 9.48).
a
h
A
B
EI, l
b
t1 t2
X1
c d
M1 1 Mt M1
Fig. 9.48 Change of temperature. (a, b) Design diagram and primary system; (c) unit bending moment diagram; (d) final bending moment diagram
The canonical equation is δ11X1 þ Δ1t ¼ 0. The unit displacement is δ11 ¼
M1 M1 1 2 1 l ¼ l1 ¼ 2 3 EI 3EI EI
The axial force N 1 in primary system due to the primary unknown is zero, so the first term of (9.34) is zero. Ð The second term of (9.34) contains the expression M 1 ds, which is the area of the bending moment diagram in the unit state. In our case the free term may be presented as ð Δt t t2 1 al Δ1t ¼ α M 1 ds ¼ a 1 1 l ¼ ðt 1 t 2 Þ h 2 2h h |fflfflfflfflffl{zfflfflfflfflffl} area M 1
The negative sign means that bending moment diagram is plotted on the more cold fibers. Canonical equation and required bending moment at the clamped support are l al X ðt t 2 Þ ¼ 0 3EI 1 2h 1 3aEI ðt t 2 Þ X1 ¼ M1 ¼ 2h 1 Final bending moment diagram Mt is constructed on the basis of (9.36) and shown in Fig. 9.48d. Kinematical control leads to the following result
9.10
Temperature Changes
349
ð
Mt M1 1 1 2 3aEI al ðt 1 t 2 Þ ðt 1 t 2 Þ ¼ 0 ds þ Δ1t ¼ 1 l EI |fflfflfflfflffl{zfflfflfflfflffl} 2 3 2h 2h EI |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Ω
yC
Discussion Let the height of rectangular cross section of the beam be increased by n-times. In this case the bending moment at clamped support increases by n2. Indeed M1 ¼
3 3aE bðnhÞ 3aEI Δt n2 Δt ¼ 2h 2ðnhÞ 12
It can be seen that change of length of the beam does not influence on the final bending moment diagram. Continuous beam. Let us consider a continuous beam, which allows deformation in the axial direction. This occurs if all movable supports have parallel constraints. In this case the uniform change of temperature leads to axial change of the length of the beam since the movable supports do not prevent axial displacement of the beam and, as result, the internal thermal forces do not arise. However, if a beam is subjected to a temperature gradient across its depth (nonuniform change of temperature), then thermal internal forces arise. In this section we assume that temperature along the longitudinal axis of a member is constant and temperature across the depth of the element changes linearly. Analysis of continuous beams may be effectively performed using three-moment equations. The primary system presents a set of simply supported beams and primary unknowns are the bending moments at the supports. In case of temperature changes, the standard three-moment equation for n-th support is written as follows M n1 l0n þ 2M n l0n þ l0nþ1 þ M nþ1 l0nþ1 ¼ 6EI 0 Δtemp n
ð9:38Þ
Here Mn1, Mn, and Mn+1 are bending moments at three consecutive supports; l0n is reduced length of the n-th span l0n ¼ ln EI 0 =EI n, EI0 is a flexural stiffness of any specified span, which is assumed as basic one. The mutual angle of rotation in the primary system at support n caused by change of temperature equals ¼ αt ðt 1 t 2 Þ Δtemp n
ln l þ nþ1 2dn 2d nþ1
ð9:39Þ
where α is coefficient of thermal expansion; t1 and t2 are change of temperature under and above the beam, respectively; dn and dn+1 are heights of cross section of a beam for n-th and n+1 spans. Sign (+) corresponds to case, when t1 > t2. Example 9.11 Design diagram of a continuous nonuniform beam is presented in Fig. 9.49a. Construct the internal force diagrams caused by a nonuniform change of temperature. The height of a cross section of the beam dn ¼ 0.05ln ; the temperature gradients are t1 ¼ +20 and t2 ¼ 20 for lower and upper fibers of the beam, respectively. Solution Let the stiffness of the first span be the basic one, i.e., EI1 ¼ 1 EI0. Reduced lengths are l01 ¼ l1
EI 0 ¼ 6 m, EI 1
l02 ¼ l2
EI 0 1 ¼ 4 ¼ 2 m, 2 EI 2
l03 ¼ l3
EI 0 1 ¼ 4 ¼ 2m 2 EI 3
Three-moment Eq. (9.38), starting from additional pinned support (with index 1) (Fig. 9.49a), become 2M 0 l0 þ l01 þ M 1 l01 ¼ 6EI 0 Δ0t M 0 l01 þ 2M 1 l01 þ l02 þ M 2 l02 ¼ 6EI 0 Δ1t M 1 l02 þ 2M 2 l02 þ l03 þ M 3 l03 ¼ 6EI 0 Δ2t Since additional span l0 ¼ 0, then the three-moment equations become 2M 0 6 þ M 1 6 ¼ 6EI 0 Δ0t M 0 6 þ 2M 1 ð6 þ 2Þ þ M 2 2 ¼ 6EI 0 Δ1t M 1 2 þ 2M 2 ð2 þ 2Þ þ M 3 2 ¼ 6EI 0 Δ2t
ðaÞ
350
9 The Force Method
a
0
2EI0
l1=6m l1⬘=6m
l2=4m
l3=4m
l2⬘=2m
l3⬘=2m
t2= - 20° t1=+20°
2EI0
2
1
d
3
EI1=1∙ EI0
0
-1
2
1
2m
dn=0.05ln
3
l0
b
552 192
104
Mt (factor EI0α)
c
+
138 Qt (factor EI0α)
−
14.67
90
d
M1=1 M1
1 Fig. 9.49 (a) Design diagram of the beam and scheme with additional span l0; (b, c) bending moment and shear force diagram due to temperature change; (d) bending moment diagram in primary system caused by M1 ¼ 1
Temperature displacements according to (9.39) ¼ αt ðt 1 t 2 Þ Δtemp n
ln l þ nþ1 2dn 2d nþ1
Let for each span dn ¼ 0.05ln. In this case the free terms become 6 ¼ 400α Δ0t ¼ α ½20 ð20Þ 0 þ 2 0:05 6 6 4 þ ¼ 800α Δ1t ¼ α ½20 ð20Þ 2 0:05 6 2 0:05 4 4 4 þ ¼ 800α Δ2t ¼ α ½20 ð20Þ 2 0:05 4 2 0:05 4
ðbÞ
Since M3 ¼ 0, then canonical equations become 12M 0 þ 6M 1 ¼ 6EI 0 400α 6M 0 þ 16M 1 þ 2M 2 ¼ 6EI 0 800α
ðcÞ
2M 1 þ 8M 2 ¼ 6EI 0 800α Roots of these equations are M0 ¼ 104EI0α, M1 ¼ 192EI0α, M2 ¼ 552EI0α. Corresponding bending moment diagram is presented in Fig. 9.49b. All ordinates must be multiplied by factor EI0α. In case of change of temperature, the bending moment diagram does not reflect the elastic curve and therefore does not provide a picture of where tensile fibers are located.
9.10
Temperature Changes
351
Shear forces are obtained using formula Q ¼
dM : dx
192 104 EI 0 α ¼ 14:67EI 0 α 6 552 192 EI 0 α ¼ 90EI 0 α ¼ 4 552 ¼ EI 0 α ¼ 138EI 0 α 4
Q01 ¼ Q12 Q23
ðdÞ
Corresponding shear force diagram is presented in Fig. 9.49c. Reactions of supports are R0 ¼ Qright ¼ 14:67EI 0 αð#Þ 0 Qleft R1 ¼ Qright 1 1 ¼ 90 ð14:67Þ ¼ 75:34EI 0 α
ðeÞ
Qleft R2 ¼ Qright 2 2 ¼ 138 ð90Þ ¼ 228EI 0 αð"Þ R3 ¼ Qleft 3 ¼ 138EI 0 α Reactions and distribution of internal forces depend not only on relative stiffness but also on absolute stiffness EI0. Control: 1. Static test. Equilibrium condition for reactions is X Y ¼ ð14:67 75:34 þ 228 138ÞEI 0 α ¼ ð228:01 þ 228ÞEI 0 α 0 2. Kinematical test. Mutual angle of rotation at support 1 is Mt M1 6 4 þ Δ1t ¼ ð2 1 192 1 104ÞEI 0 α þ ð2 1 192 1 552ÞEI 0 α 6EI 0 6 2EI 0 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} span 01
span 12
þ 800α ¼ 800α þ 800α ¼ 0 where M 1 is the bending moment diagram in unit state, shown in Fig. 9.49d.
Discussion 1. A continuous beam can be considered as one-span beam for which the external load is only the reactions in redundant supports. Therefore the bending moment diagram caused by temperature changes is a polygon with vertices above the supports. 2. At temperature exposure the bending moments, shear forces and reactions of supports are proportional to the flexural rigidity EI. Therefore, a change in the modulus of elasticity of a beam material (or cross-sectional inertia moment of a beam) leads to a proportional change in internal forces and reactions. 3. The proportional change in the length of all spans of the beam does not affect the bending moments, but it does affect the shear forces and reactions. The increase in spans n times leads to a decrease in shear forces by n times. 4. Resultant of all reactions is equal to zero.
9.10.3 Redundant Trusses If any members of a truss are subjected to change of temperature, then the unit displacements and free terms of canonical equations of the force method (9.33) should be calculated by the formulas δik ¼
X Ni Nk l EA
j
,
X Δit ¼ α N i Δt l j
ð9:40Þ
352
9 The Force Method
where l is a length of j-th member of the truss; N i , N k are internal forces in j-th member due to unit primary unknowns Xi ¼ 1, Xk ¼ 1; Δt is a thermal gradient; and α is a coefficient of thermal expansion. In expressions (9.40) summation is done on all members of the truss (subscripts j are not shown). Axial force in members of the truss using the superposition principle is determined by the formula N ¼ N1 X1 þ N2 X2 þ . . . þ Nn Xn
ð9:41Þ
where n is degree of statical indeterminacy. Example 9.12 Design diagram of the truss is presented in Fig. 9.50. Axial stiffness for diagonal and vertical members equals EA, for upper and lower chords equal 2EA. Determine the reaction of the middle support and internal forces in all members of the truss caused by temperature changes on Δt degrees. Consider two cases: the temperature gradient is applied to (a) all members of the truss and (b) the lower chord only. Design diagram
2⬘
2EA
d=4m
h=3m
EA
Primary system 4⬘ 3⬘
EA 1
2EA
5 2
3
4
X1
Fig. 9.50 Externally statically indeterminate truss. Design diagram and primary system
Solution The primary system is obtained by eliminating the middle support. The primary unknown X1 is the reaction of the middle support (Fig. 9.50). Canonical equation of the force method is δ11X1 þ Δ1t ¼ 0. The primary unknown is X 1 ¼
Δ1t . Final internal forces δ11
may be calculated by formula N t ¼ N 1 X 1 Results of the analysis are presented in Table 9.15. The column 1 contains lengths of all members of the truss. Internal forces in the primary system caused by unit primary unknown X1 ¼ 1 are presented in column 2. The column 3 serves for calculation of the free term of the canonical equation for the case when all members of the truss are subjected to temperature gradient Δt. In this case, the sum of all entries is: Δ1t ¼ 18:994 19:002 ðrelative error equals 0:04%Þ, and for primary unknown we get X1 ¼ 0. It means that if the change of temperature is related to all members of the truss, then internal forces in the members of the truss do not arise. The column 4 serves for calculation of free term of the canonical equation if only members of the lower chord of the truss are subjected to temperature change Δt. In this case, Δ1t ¼ 10:672α Δt Column 5 is used for calculation of unit displacement of the canonical equation. In our case, δ11 ¼
24:548 . EA
These results yield the primary unknown X1 ¼
Δ1t 10:672α Δt ¼ ¼ 0:435αEA Δt δ11 24:548=EA
Column 6 contains internal forces in all members of the truss caused by temperature changes of the lower chord only. These forces, according to (9.41), are equal to N ¼ N 1 X 1 . Column 7 serves for control of analysis: the sum of all terms of this column equals to Δ1t with the opposite sign.
9.10
Temperature Changes
353
Table 9.15 Statically indeterminate truss subjected to temperature change – calculation of internal forces in all members of the truss
Discussion 1. The structure under consideration has one absolutely necessary constraint, the reaction of which may be determined from the equilibrium equation. This constraint is the support bar at joint 1, which prevents horizontal displacement. If the truss is externally statically indeterminate, then the temperature gradient, which is related to all members of the truss, induces a displacement in the direction 1–5 of the absolutely necessary constraint and internal forces in all members of the truss induced by temperature gradient are equal to zero. 2. If any member of the truss has been made by Δ units longer than required, then this error of fabrication may be treated as a thermal expansion, i.e., Δ ¼ α l Δt , where l is the length of a member; for all other members α l Δt ¼ 0. Canonical equation becomes δ11X1 þ Δ1t ¼ 0. Free term according to second formula (9.40) becomes Δ1t ¼ αlt Δt N 1, where N 1 is the force induced in the same member by a force X1 ¼ 1.
9.10.4 Redundant Frames The framed structures are analyzed on the basis of the canonical equations of the force method (9.33). It is assumed that the deformation of all frame elements occurs in the plane of a frame. The unit displacements are calculated as usual by the formulas (9.5), (9.5a), while the free terms—by formula (9.35) Δt ¼ Δt ðN Þ þ Δt ðM Þ, ð ð Δt Δt ðN Þ ¼ αt av N i ds, Δt ðM Þ ¼ α M i ds h Here first and second terms take into account the axial forces and bending moments, respectively. Example 9.13 Design diagram of the redundant frame is same as in Example 9.7. Stiffnesses for vertical and horizontal members are 1EI and 2EI, respectively. Heights of the cross section h (cross section size in the plane of bending) for vertical and horizontal members are 0.60m. The frame is subjected to temperature changes as presented in Fig. 9.51a. Construct the internal force diagrams and calculate reactions of supports.
354
9 The Force Method
Design diagram
C
t2=0
3m
B
h
t1= −10°C 1 h
2
A
4
8
6
tav=+5° Δt=30°
tav=+10° Δt=+20°
1
b = 0.4m
10m
b = 0.2m
Primary system
3
5m
t3=+20°C
X2
5
tav= -5° Δt=10°
h = 0.6m
a
b
X1
tin= +20°
Δ t=30° tout= - 10°
h = 0.6m
tav=+5° h
c
δ22
δ21 δ11
10
X2=1
δ12
3
10 X1=1
1.0
X1=1
+
N1
M1
M2 8
d
e
14.628
82.99
X2=1 3 13
82.99
68.37
10
14.628 Mt kNm
6
68.37
X1=1
MΣ
(factor αEI)
18
43.99
Fig. 9.51 (a) Design diagram of redundant frame and temperature distribution; (b) primary system; (c) bending moment and axial force diagrams in unit states, N 2 ¼ 0 ; (d) final bending moment diagram and static verification; (e) summary unit bending moment diagram
Solution Let us accept a primary system as shown in Fig. 9.51b. Canonical equations of the force method are δ11 X 1 þ δ12 X 2 þ Δ1t ¼ 0 δ21 X 1 þ δ22 X 2 þ Δ2t ¼ 0
ðaÞ
where δik are unit displacements and Δ1t and Δ2t are displacements in a primary system in the direction of primary unknowns X1 and X2, respectively, due to change of temperature. The unit states and corresponding bending moment and axial force diagrams caused by unit primary unknowns are presented in Fig. 9.51c. All normal forces due to X 2 ¼ 1 are zeros, i.e., N 2 ¼ 0. The unit displacements δik for this primary system have been obtained in Example 9.7; they are δ11 ¼
666:67 , EI
δ12 ¼ δ21 ¼
275 , EI
δ22 ¼
170:67 EI
ðbÞ
For calculation of free terms we will use expressions (9.35). These expressions contain the average temperature tav and temperature gradient Δt for each member; they are
9.10
Temperature Changes
355
20 þ ð10Þ ¼ 5∘ ; Δt ¼ j10 ðþ20Þj ¼ 30∘ , 2 0 þ ð10Þ ¼ 5∘ ; ¼ Δt ¼ j10 0j ¼ 10∘ , 2 0 þ 20 ¼ 10∘ ; Δt ¼ j0 ðþ20Þj ¼ þ20∘ ¼ 2
For portion 1‐3
t av ¼
for portion 4‐5
t av
for portion 6‐8
t av
ðcÞ
These parameters for each element of the frame are shown on the primary system in Fig. 9.51b. Detailed calculation of free terms is presented in Table 9.16. Table 9.16 Calculation of free terms Δit of canonical equations according to formula (9.35)
∫
α tav N i ds for portion
α
1-3
4-5
6-8
Δ1t
α × 5 ×1× 5 = 25α
0
0
α
Δ2t
0
0
0
30 3 + 8 × ×5 = 0.6 2 1375α
Δt M i ds for portion h
1-3
4-5
30 ×10 × 5 = 0.6 2500α
α×
∑
∫
0
α×
10 1 × ×3× 3 0.6 2
6-8
20 1 × ×10 ×10 0.6 2 = 1666.7α
α×
0
4191.7 α
1450 α
75 α
The first term in (9.35) is positive if normal force in the unit state and temperature of the neutral fiber tav has the same sign (for example, for element 1–3, normal force in the first unit state equals +1 and average temperature equals +5). The second term in (9.35) is positive if the bending moment diagram in the unit state is located on the side of more heated fibers. For member 1–3, the first component is Δ1t ¼ α 5 1 5, where 1 5 presents the area of the axial force diagram N 1 along 30 this member, while for this member the second component is α 10 5, where 10 5 presents the area of the bending 0:6 moment diagram M 1 . Canonical equations become 666:67X 1 þ 275X 2 þ 4191:7α EI ¼ 0 275X 1 þ 170:67X 2 þ 1450α EI ¼ 0
ðdÞ
Roots of these equations are X1 ¼ 8.2988α EI, X2 ¼ 4.8759α EI. Bending moment diagram is constructed using formula (9.36). The corresponding calculation is presented in Table 9.17. Table 9.17 Calculation of bending moments
Points 1 3 4 5 6 8 Factor
M1 -10 -10 0.0 0.0 -10 0.0 -
M 1 ∙ X1 +82.998 +82.998 0.0 0.0 +82.998 0.0 α EI
M2 -8.0 -3.0 -3.0 0.0 0.0 0.0 -
M 2 VX 2 -39.008 -14.628 -14.628 0.0 0.0 0.0 α EI
Mt +43.99 +68.37 -14.628 0.0 +82.99 0.0 α EI
Signs of bending moments + −
+
−
356
9 The Force Method
The resulting bending moment diagram is presented in Fig. 9.51d. Static verification (equilibrium of rigid joint). X M ¼ ð68:37 þ 14:628 82:99Þ αEI ¼ ð82:998 82:99Þ αEI 0 Kinematical verification (total displacement in direction of the primary unknowns X1 and X2). The summary unit bending moment diagram M Σ is shown in Fig. 9.51e. The expression (9.34) leads to the following result: Xð M M X t Σ ds þ Δit EI 5 13 þ 18 43:99 þ 68:37 ¼ 18 43:99 þ 4 þ 13 68:37 α EI 1EI 6 2 2 1 1 2 1 1 2 3 3 ð14:628Þα EI 10 10 82:99α EI 1EI 2 3 2EI 2 3 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl} portion 68
portion 45
þ4191:7α |fflffl{zfflffl} ¼ |fflfflfflffl{zfflfflfflffl} þ 1450α Δ2t
Δ1t
¼ 5686:32α þ 5685:58α ffi 0 Summary Distribution of internal forces (bending moment, shear force and normal force) as well as reactions depends on both relative and absolute stiffnesses, as well as on coefficient of thermal expansion. This is the property of any statically indeterminate structure subjected to change of temperature.
9.11
Some Features of Redundant Structures
For any statically indeterminate structures some fundamental properties may be discovered considering the primary unit displacements δ11, δ22, , δnn (Rabinovich 1960, part 2). 1. Increasing stiffness of i-th redundant constraint in which the primary unknown Xi arises leads to the decreasing of corresponding unit displacement δii and increasing of the primary unknown Xi. This idea is illustrated below for simple redundant combined structure shown in Fig. 9.52a. Let the primary unknown be axial force X1 which arises in the crossbar. The corresponding unit displacements δ11 is determined by the flexural stiffness EI of the vertical member and axial stiffness EA of a horizontal rod, i.e., δ11 ¼ h3/3EI þ l/EA. Here I, A are the area moment of inertia of the column and cross-sectional area of the horizontal element, respectively. It can be seen, that increasing the cross-sectional area A leads to decreasing in unit displacement δ11. Obviously, the change in area A does not affect the loaded term Δ1P, and therefore an increase in the rigidity of the redundant constraint leads to an increase in the primary unknown X1 ¼ Δ1P/δ11.
a E, A
X1
b
1
3
5
P l h E, I
2
4
6
1
3
5
X1
c
q
2
X2 4
6
k l Fig. 9.52 Design diagrams and primary system for different types of structures. (a) Combined structure, (b) internally statically indeterminate truss, (c) redundant beam with elastic support
9.11
Some Features of Redundant Structures
357
Figure 9.52b presents the internally redundant truss. Let the primary unknowns be the internal forces in the conventionally necessary elements 1–4 and 4–5. Each of these primary unknowns presents the self-equilibrated system of the forces. It means that with the action X1 ¼ 1 the reactions of supports are equal to zero, the reactive forces arise only in the contour elements 1–3, 3–4, 4–2, 2–1, and in the diagonal rod 2–3. Therefore δ11 consists of six terms, of which four terms relate to the elements of contour, one term to the diagonal element 2-3, and the last term to the element 1–4. Thus, unit displacement δ11 may be presented as follows: l14
δ11 ¼ ðl13 =EA13 þ l34 =EA34 þ l42 =EA42 þ l21 =EA21 Þ þ l23 =EA23 þ δ14 11 ¼ δ11 þ EA |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} 14: contour elements
ðaÞ
member 23
Similarly, the unit displacement δ22 may be presented as follows l45
δ22 ¼ ðl35 =EA35 þ l56 =EA56 þ l64 =EA64 þ l43 =EA43 Þ þ l35 =EA35 þ δ45 22 ¼ δ22 þ EA |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 45
ðbÞ
δ 22
As may be seen from the above formulas, with any change in the stiffness of the redundant elements 1–4 and 4–5 values δ 11 and δ 22 remain unchanged, while change in the stiffness of element 1–4 will only affect the value δ11, and a change in the stiffness of element 4–5 will only affect the value δ22. The loaded term is defined in the primary system. Therefore the change in the stiffness of those elements in which the primary unknowns arise does not affect the magnitudes Δ1P and Δ2P. Thus, detailed analysis of the canonical equations of the force method shows the following: 1. For any redundant structure with arbitrary number degrees of redundancy any change in the stiffness of one redundant element i leads to change of the unit displacement δii only; with this, all other coefficients and free terms of the canonical equations remain unchanged. 2. From expressions above it can be seen also that no change in stiffness of the redundant constraint can lead to a change in the sign of the force in this element. 3. As the stiffness of redundant element increases, the primary unknown X increases in absolute magnitude and retains its sign. This force reaches a maximum with the greatest rigidity of this constraint and vanishes with a rigidity equal to zero. With an infinite increase in the rigidity of the constraint, the primary unknown tends to the limiting value. Therefore, an attempt to significantly increase the magnitude of the redundant force X by excessively increasing the rigidity of the conditionally necessary constraint may not be very effective since there exists the objective limiting values for X. For example, for beam in Fig. 9.52c the reaction in elastic support (i.e., the primary unknown X1) equals (Table A.6, case 5) 3 1 X 1 ¼ ql 8 1 þ 3EI=kl3 Minimum of primary unknown X1 ¼ 0 becomes at k ¼ 0, i.e., if elastic support is absent. Maximum of primary unknown becomes 3ql/8 at k ¼ 1). However, it should be borne in mind that there is a principal possibility of changing of the primary unknown within certain limits by changing the stiffness of the corresponding redundant constraint. 4. In the case of an analysis of a structure that is subjected to external load, all coefficients and free terms of the canonical equations contain the moduli of elasticity E and G. If we replace both moduli with new values of kE and kG, where k is an arbitrary number, then the values of the primary unknowns remain unchanged. This means that replacing the material of a structure by another with proportional elastic moduli does not cause a change in the magnitudes of the primary unknowns. If the shear phenomenon is not taken into account, then each term of canonical equations includes only the Young’s modulus E. Therefore, replacing one homogeneous elastic material with another homogeneous and elastic material does not change the magnitudes of the reactions and internal forces which arise in any structure. Of course, we are talking about structures that are subjected to any type of mechanical loading. Such a conclusion in the case of change of temperature, settlements of supports, errors of fabrication, and shrinkage of the material does not hold. 5. There is a class of transient systems that formally have redundant constraints, but their rigidity is substantially less than the rigidity of absolutely necessary constraints. In these cases, the forces in the redundant constraints are negligible and the system actually works as statically determinate one.
358
9.12
9 The Force Method
Comparison of the Redundant and Statically Determinate Structures
1. If the statically determinate structure (DS) is subjected to any loading, then there is one possible distribution of internal forces that satisfies the equilibrium conditions of the original structure. Statically indeterminate (redundant) structure (RS) allows an infinite number of options for the distribution of internal forces that satisfy the equilibrium conditions, but only one of them will satisfy the condition of compatibility of deformations. It is this option that represents the true distribution of reactions and internal forces. The RS compared with the DS is more rigid because it contains the redundant constraints. The loss of one or even all redundant constraints leads to a new geometrically unchangeable system, i.e., does not lead to the destruction of the entire structure. The loss of even one constraint in DS leads to a geometrically changeable system, which is inadmissible for the structure. 2. In the case of DS, the settlements of supports, change of temperature, and errors of fabrication (inaccuracy of the assembly) of a structure lead to some displacements. However, all displacements are developed as free or non-constrained ones. Therefore, these effects do not cause the appearance of forces in the elements of a structure. Internal forces in DS arise if and only if the structure is subjected to the external loads; distribution of internal forces does not depend on the stiffness of the individual elements and the elastic properties of the material. However, in the case of RS, the settlement of supports, change of temperature, manufacturing errors, and shrinkage of the structure material lead to the appearance of non-free, limited displacements and, as a result, nonzero internal forces. Their distribution depends on the distribution of the stiffness of individual elements, i.e., on the shape and size of cross sections of the individual elements and the physical properties of the materials. If RS is made of one material, then the distribution of internal forces depends on the cross sizes of individual elements and does not depend on physical properties of the material. 3. The forces that occur in the system in the absence of an external load are called initial. Initial forced can occur only in the statically indeterminate structures. The reason for their occurrence is the settlement of supports, errors of fabrication, assembly and installation of parts of structure, change of temperature, as well as shrinkage of concrete. The property of statically indeterminate structures to respond to these actions is important for engineering. If the actions that lead to the initial forces are of natural or man-made origin, then they are not subject to an engineer during the operation of the structure. As a rule, they act as adverse factors “taking away” from structures the significant part of their bearing capacity. However, this property of the redundant structures can be used for the project needs. The forces that arise in such systems caused by the given external load can be adjusted, changed, and even eliminated in the individual elements by the artificial way. For this purpose, it is possible to apply the settlement of supports, create design inaccuracies in the manufacture of the structure, create initial temperature fields in the process of assembling and installing structures, etc., and, as a result, obtain desired distribution of internal forces. 4. In statically determinate structures, overstress of at least one element is unallowable. This conclusion also applies to absolutely necessary elements of RS, while overstress of conditionally necessary elements of the structure is permissible. Therefore, estimates of the vitality and durability of statically determinate and redundant structures are different. For the two classes of structures, DS and RS, the safety factors of the structures in whole are also different.
Problems Problems 9.1 through 9.6 are to be solved by superposition principle. The flexural rigidity EI is constant for each beam. 9.1. The continuous two-span beam supports the uniformly distributed load q (Fig. P9.1). Find the reaction of supports and construct the bending moment diagram.
q A
C
B l
Fig. P9.1
Ans. RB =
l
5 ql; 4
M B = - 0.125ql 2 ;
M max (0.375l ) = 0.0703ql 2 .
Problems
359
9.2. The clamped-pinned beam of length l is subjected to couple M0 at the roller support B (Fig. P9.2). Find the reaction of supports and construct the bending moment diagram. M0
Ans. M A = − 0.5 M 0
B
A
RA = −RB =
MA
3 M0 2 l
Fig. P9.2
9.3. The fixed-pinned beam of length l is subjected to concentrated force P at the midspan (Fig. P9.3). Find the reaction of supports and construct the bending moment diagram. P A
Ans. M A = −
0.5l
MA
MC =
B
C
3 Pl 16
.
5 5 Pl , RB = P 32 16
Fig. P9.3
9.4. The beam AB with clamped support A and elastic support B is subjected to uniformly distributed load q; the stiffness coefficient of elastic support is k (Fig. P9.4). The deflection λ of support B, reaction RB of this support, and stiffness coefficient R k are related as λ ¼ B . Determine the reaction of supports. Consider two special cases: k ¼ /, k ¼ 0. k q B
A
3 8
Ans. RB = ql ∙
λ
1 3EI , α= 3 1+ α kl
k l Fig. P9.4
9.5. Calculate the reaction and bending moment at support B (Fig. P9.5), if the vertical settlement of this support is Δ.
A
Ans. RB =
C
B l
6EI l3
Δ.
M B = RAl =
l
3EI l2
Δ
Fig. P9.5
9.6*. Pinned-pinned-pinned uniform beam is subjected to distributed load q (Fig. P9.6). Construct the bending moment diagram. Consider two versions of primary system. Primary unknown is: (1) Reaction R1; (2) Bending moment at support 1. Calculate the deflection at the middle of the first span. Hint: Version 1. Compatibility equation y1(q) þ y1(R1) ¼ 0. q 1
0
2
k l
l
Ans. M1 = −
yk =
ql 2 , 16
7 ql 4 768 EI
Fig. P9.6 right left left right Version 2. Compatibility equation θleft , θleft ¼ θright ðM 1 Þ 1 ¼ θ1 1 ¼ θ 1 ðqÞ þ θ 1 ðM 1 Þ, θ 1 1
360
9 The Force Method
Problems 9.7 through 9.8 are to be solved using the force method in canonical form. 9.7a, b. Design diagram of the uniform clamped-pinned beam of length l is shown in Figs. P9.7a, b. Calculate the reaction of supports and construct the internal force diagrams. Show the elastic curve of the beam. For scheme P9.7b construct the influence lines for RA, MA, and MC.
a
b
q
P
A
C
A
B
(
)
2 Pl MA = υ 1− υ 2 , 2
(
υl
ul
l
B
5 ql 2 Ans. (a) R A = ql; M A = ; 8 8 Pυ 3 − υ2 , (b) RA =
MC =
)
Plu 2υ (3 − u ) 2
Fig. P9.7
9.8a, b. Design diagrams of the uniform beam with elastic support B are shown in Fig. P9.8a, b. For elastic support the ratio between reaction R of support, deflection λ of elastic support, and a stiffness coefficient k of elastic support is R ¼ λk. Calculate the reaction of supports and construct the internal force diagrams. Show the elastic curve of the beam.
q
a
B k
A l
3 1 Ans. (a) RB = ql ; 8 1 + 3EI kl3
P
b
C
A
B k
υl
ul
(b) RB =
Pu 2 1 (3 − u ) 2 1 + 3EI kl 3
Fig. P9.8
Problems 9.9 through 9.12 are to be solved by three-moment equations. The flexural rigidity, EI, is constant for each beam. 9.9. The two-span uniform continuous beam with hinged supports is subjected to concentrated force P (Fig. P9.9). Provide total analysis of structure (primary system and primary unknown, bending moment and shear force diagrams, reactions of support, static and kinematic verification). Assume l1 ¼ l2 ¼ l P 0
EI=const
1
Ans. M1 = − 0.096 Pl
2 a=0.4l
b=0.6l
l1
l2
Fig. P9.9
9.10. The continuous beam with clamped left support and cantilever at the right is presented in Fig. P9.10. Compute the bending moments at supports 1 and 2.
q=2kN/m 2
1
F=1kN
P=12kN EI=const
3 n
l1=8m
a=ul=6m l2=10m
Fig. P9.10
b=υl=4m c=2m
Ans. M1 = −8.013kNm M 2 = −15.975kNm
Problems
361
Problems 9.11–9.13 are to be solved by foci method. The flexural rigidity, EI, is constant for each beam. 9.11. The uniform beam is loaded arbitrary in the first span (Fig. P9.11). Calculate the ratio M1/M3. Assume l4/l3 ¼ 0.8, l3/l2 ¼ 1.2. Show the elastic curve and location of the tensile fibers for all spans. q F2R
0
*
1 l1
*
l2
Ans. M1 M 3 = 14.64
F4R
F3R
2
*4
3
l3
l4
Fig. P9.11
9.12. The three-span beam with overhang is loaded by the concentrated force P at the free end (Fig. P9.12). Calculate the bending moments at the supports. Show the elastic curve, and the inflection points. F=10kN 0
1
EI=const
2
3
Ans. M 2 = − (−20 / 3.76) = 5.32 kNm M 1 = −1.25 kNm
M 0 = 0.62 kNm
l2=8m
l1=6m
2m
l3=8m
Fig. P9.12
9.13. The three-span beam with overhang is loaded by the uniformly distributed load q in the first span (Fig. P9.13). Calculate the bending moments at the supports. Show the elastic curve. Hint: For computation of M0 and M1 use Bresse formula. Ans. M 0 = −10.98 kNm
q=3kN/m 0
1
EI=const
2
M1 = −5.06 kNm,
3
M 2 = 1.26 kNm
l2=8m
l1=6m
l3=8m
2m
Fig. P9.13
For next problems the most appropriate method should be used 9.14. The frame is subjected to uniformly distributed load q, as shown in Fig. P9.14. Construct the internal force diagrams. Calculate the reactions of supports. Show the elastic curve of the frame. Determine the horizontal displacement of joint C. q A
C
EI a
EI B a
Fig. P9.14
Ans. M B =
15 qa 2 qa 4 , RA = qa , ΔChor = 32 64 EI 32
( )
362
9 The Force Method
9.15. The frame is subjected to uniformly distributed load q, as shown in Fig. P9.15; parameter n is any positive number. Construct the bending moment diagram. Explain relationship between moments at extreme left and right points of the crossbar. Show the elastic curve of the frame. Explain the influence of parameter n.
nEI
q
B
Ans. R A = h
EI A
3 qh2 . 4 l
l
Fig. P9.15
9.16. The portal frame is loaded by uniformly distributed load q, as shown in Fig. P9.16. Calculate the axial force X1 in the member AB. Construct the bending moment diagram. Determine the horizontal displacement of the crossbar. A
B
Ans. X1 = −
q EI
EI
C
D
MD =
h
3 5 2 qh, M C = qh , 16 16
3 2 qh , 16
Δ=
qh4 16 EI
Fig. P9.16
9.17. The portal frame is subjected to horizontal force P. The stiffness of vertical members is EI, and horizontal member is nEI, where n is any positive number (Fig. P9.17). Calculate horizontal reaction HB at support B. Construct the internal force diagrams. Analyze the influence parameter n on distribution of internal forces. Ans. H B = − P
nEI δ11 =
EI
h
A
δ11
,
2h 3 3 a 1+ , 3EI 2n h
δ 1P = −
B
Δ1P
Ph3 3 a 1+ 3EI 2n h
a Fig. P9.17
9.18a, b. Construct the bending moment diagram for uniform beam with clamped ends, loaded as shown in Figs. P9.18a, b. For both design diagrams calculate maximum deflection. For case (b) construct influence lines for reactions at supports A and B and for displacement at point C. Ans. (a) M A = -
a
b
q A
B l
P C
A a=ul
b= υl l
Fig. P9.18
B
(b) RA =
ql 2 , 12
Pb 2 l3
y
l ql 4 ; = 2 384EI
( 3a + b ) ,
M A = Plu υ2 , M C = 2 Plu 2υ 2 ,
fC = P
a3b3 3EI l 3
.
Problems
363
9.19. The frame in Fig. P9.19 is subjected to concentrated force P and uniformly distributed load q within the vertical element. The relative flexural stiffnesses are shown in circle; h ¼ 5 m, l ¼ 10 m, a ¼ 4 m, b ¼ 6 m, P ¼ 8 kN, q ¼ 4 KN/m. Construct the bending moment diagram using different primary systems. Provide the kinematical verification. Calculate shear and axial forces and determine the reactions of supports. Trace the elastic curve and show inflection points. P B
2 1
M 2 = − 13.93kNm
2
3
q
Ans. M1 = 10.842kNm ( ∪) h
1 A
b
a
Fig. P9.19
9.20. Design diagram of the frame is shown in Fig. P9.20. The flexural stiffness EI for all members are equals. Construct the internal force diagrams and trace the elastic curve. Provide the statical and kinematical verification. q=10kN/m D
A
B
Ans. M DA = 56.1 kNm , M DB = 24.2 kNm , M DC = 31.9 kNm
6m C 8m
8m
Fig. P9.20
9.21. Symmetrical rod structure is subjected to load P, as shown in Fig. P9.21. The cross-sectional area of the vertical member is A, and for inclined members are kA, where k is any positive number. The length of inclined members is l. Calculate the internal force in the vertical member.
kA
A α α
P Fig. P9.21
Ans. Nvert = kA, l
P 1 + 2k cos3 α
.
364
9 The Force Method
9.22. Design diagram of a two-hinged parabolic arch with the tie is presented in Fig. P9.22. The equation of the neutral line of the arch is y ¼ (4f/l2)x(l x). The cross-sectional moments of inertia vary by law Ix ¼ ICcosφx [m4]. The flexural stiffness in the highest section of the arch is EIC. The cross-sectional area of the tie is At [m2]. Modulus of elasticity of the material of the arch and the tie are E [kN/m2} and Et, respectively. Calculate internal axial force X1 in the tie. Ans.
q=2kN/m
P=8kN
y
X1 =
xP=6m f=6m
EIC
8558.67 517.082 + 24 EIC Et At
xq=12 E tA t
x l=24m
Fig. P9.22
9.23. Design diagram of a rigid horizontal weightless bar is shown in Figs. P9.23a, b. The stiffness EA is constant for all vertical rods. Determine the axial forces in members 1–3.
a
Ans. (a) N1 = 0.833P ,
b d 1
d 2
d
l 3
1
d 2
d 3
l 4
N 2 = 0.333P, N 3 = − 0.167 P;
(b) N1 = 0.4 P, N 2 = 0.3P , N 3 = 0.2 P, N 4 = 0.1P .
P
P Fig. P9.23
Additional Problems 9.24. The tie ab of a truss is subjected to thermal gradient Δt, degrees (Fig. P9.24). Axial stiffness of the upper and lower chords of the truss is 2EA; for all other members of the structure axial stiffness is EA. Calculate the axial force X1 at the member ab. 2EA 2m h=3m
EA 2EA EA d=4m Fig. P9.24
EA a d
Δt
b
Ans. X 1 = - 0.152 α ∙ Δt ∙ EA
Problems
365
9.25. The beam with clamped support A and elastic support B is subjected to unity angular displacement of clamped support (Fig. P9.25). The flexural stiffness EI is constant; stiffness of elastic support is k (RB ¼ kΔB) and the length of the beam is l. Construct the bending moment diagram. Consider limiting cases (k ¼ 0, and k ¼ /). Apply the force method in canonical form.
φ =1 EI, l
A
Ans. M A =
B
3EI 1 3EI , α= 3 . ∙ l 1+ α kl
k Fig. P9.25
9.26. The portal frame is subjected to settlements a, b, and φ of support B (Fig. P9.26). Calculate the free terms of canonical equation of the force method for different versions of a primary system.
Design diagram
h
X3 Version 2
Version 1
A
X1
B
X3
a l
X1
X1
X2
X3
φ
b
Version 3
X2
X2
Ans:
Version of primary system 1 2 −
3
Δ 1d
Δ 2d
Δ 3d
-b +b
+a -a+lφ
-φ +φ
a b − l 2h
b h
a b − l 2h
Fig. P9.26
9.27. Uniform clamped-pinned beam of length l is subjected to the angular displacement Δ of the clamped support. Calculate the vertical displacement Δir of section K, located as shown in Fig. P9.27. Hint: for calculation of M 0A use Table A.3, row 3. Δ A ul Fig. P9.27
M′A
Δir
P=1
B
K υl
l Ans. Δ ir = υ 1 − υ 2 Δ 2
(
R′A
υl
R′B
)
366
9 The Force Method
9.28. Uniform pinned-clamped beam of length l is subjected to the vertical displacement Δ of the pinned support (Fig. P9.28). Calculate the vertical displacement Δir at the middle of the beam and the angular displacement φir at the left support of the beam. Hint: Use Table A.3, row 3 (for computation of Δir) and line 4 for computation of φir Δir Ans. Δ ir =
φir Δ
5Δ 3Δ , φ ir = 16 2l
l/2
Fig. P9.28
9.29. Uniform three-span beam ABCD is subjected to vertical displacement Δof support D (Fig. P9.29). Each span of the beam is l. Bending stiffness is EI. Determine the displacement at point 1. Hint: Reaction at support D caused by force P ¼ 1 applied at point 1—see Table A.17, line 10; Reaction at support D caused by force P ¼ 1 applied at point 2—see Table A.11, row 5. 2
C
B
A
1
l/2
l/3
D
Ans. Δ1Δ = − 0.0543 , Δ
Δ 2Δ = 0.25Δ .
Fig. P9.29
9.30. Determine a total area of influence line for bending moment at arbitrary section of two-hinged uniform symmetrical parabolic arch. Hint: Use result of Sect. 9.6.1.
Chapter 10
The Displacement Method
The displacement method is a powerful method for analysis of statically indeterminate frames subjected to different external exposures. The displacement method is effective especially for analyzing sophisticated structures with a large number of redundant constraints. Two forms of the displacement method are applied in structural engineering. One uses the expanded form and the other the canonical form equations. In this book, only canonical form is considered. The displacement method was introduced in 1826 by C. L. Navier (1785–1836). A. Clebsch (1833–1872) presented the displacement method for trusses in expanded form (1862). This method was further developed by H. Manderla (1880), E. Winkler (1892) and O. Mohr (1892). The application of the displacement method to frames was presented by A. Bendixen (1914) and was called the slope-deflection method. Later this method was developed and presented in canonical form by A. Ostenfeld (1926) (Rabinovich 1954). A significant contribution in the development of the canonical form was added by S. Leytes (1936) (Umansky 1973). The displacement method in canonical form contains a deep fundamental idea and is brought to elegant simplicity. This form offers a unified and rigorous convenient algorithm for analysis of different statically indeterminate frames. Among them are the single- or multi-span as well as multileveled frames with deformable or infinitely rigid crossbars, etc. Canonical form can be effectively applied for analyses of structures subjected to different loads, temperature changes, settlements of supports, and errors of fabrication. Application of the displacement method in canonical form for the construction of influence lines is very effective and leads to a simple and clear procedure. Moreover, the displacement method in canonical form is a very effective tool for the special parts of structural analysis, such as stability and structural dynamics.
10.1
Fundamental Idea of the Displacement Method
This section contains the fundamental concepts of the displacement method. Among them are degree of the kinematical indeterminacy, primary system, primary unknowns, and idea of the displacement method.
10.1.1 Kinematical Indeterminacy In the case of the force method, the unknowns are forces at the redundant constraints. Knowing these forces we can find the distribution of internal forces and after that, displacements at any point of a structure. The fundamental approach in the displacement method is the opposite: initially we calculate the displacements at the ends of the members and then the internal forces in the members. Thus, the primary unknowns in the displacement method are the displacements of the joints. Analysis of a structure by the displacement method is based on the following partial assumptions: 1. The deformations of the members caused by axial and shear forces can be neglected. 2. The difference between the length of the deformable element and its initial length can be neglected. Analysis of any statically indeterminate structure by the displacement method begins with determining the degree of kinematical indeterminacy. This fundamental characteristic of the displacement method for any structure is determined by the formula © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_10
367
368
10
The Displacement Method
n ¼ nr þ nd
ð10:1Þ
where nr is the number of unknown angles of rotation of the rigid joints of a structure and nd is the number of independent linear displacements of the joints. In general, the degrees of kinematical and statical indeterminacy are not equal. To calculate the number of linear displacements, nd, we need to introduce the concept of the hinged system (or scheme). A hinged system is obtained from the original structure by introducing hinges at all rigid joints and supports while considering all members of the hinged scheme to be absolutely rigid. The degree of mobility of a hinged system is determined by the number of additional members, which would transform a hinged system into a geometrically unchangeable structure. The degree of mobility in turn determines the number of independent linear joint displacements, nd. Let us consider the structure shown in Fig. 10.1a; the elastic curve is shown by a dotted line. Since axial deformations are neglected and support 1 is unmovable, the joints of the entire structure have no horizontal displacements. Since the number of rigid joints equals two, then nr ¼ 2. To show the hinged scheme we introduce hinges at all rigid joints; this scheme represents a geometrically unchangeable structure. Indeed, the structure 1–2–3 presents a rigid disc and joint 4 is connected with this rigid disc and with the ground at support 5. So nd ¼ 0 and the total degree of kinematical indeterminacy equals two.
b
a 1
2
4
D4
D2
D1 2
4
j2 j 4
j2
3 1
Hinged scheme
1
5 2
3
4
3 1
5
Hinged scheme
5 2
3
4
5
Fig. 10.1 (a, b) Design diagrams of frames, deflected shapes and hinged schemes
Due to the action of the external load, rigid joints 2 and 4 rotate by angles φ2 and φ4, respectively. These angular displacements, φ2 and φ4, determine completely the deformable shape of the structure and represent the unknowns of the displacement method. Figure 10.1b presents the same frame, but with a movable support 1. The number of unknown angles of rotation of rigid joints is, as before, nr ¼ 2; angular displacements φ2 and φ4 are not shown. In the case of rolled support 1 the structure has the linear displacements Δ of joints 1, 2, and 4 as well. According to conditions 1 and 2 of the assumptions, these linear displacements Δ are the same for all joints. So the number of independent linear displacements is nd ¼ 1. Indeed, the hinged structure presents a geometrically changeable system because support 1 is movable. To obtain a geometrically unchangeable structure, only one additional member needs to be introduced into this system (as shown by line 3–4), thus nd ¼ 1. Therefore, the total degree of kinematical indeterminacy equals three. The unknowns of the displacement method are angular displacements φ2 and φ4 and linear displacement Δ. Some statically indeterminate frames and their hinged schemes are presented in Fig. 10.2. In case (a), only one additional member (shown by the dashed line) transforms the hinged system into a geometrically unchangeable structure, so nd ¼ 1; the number of rigid joints is 2. Therefore, the degree of kinematical indeterminacy of the entire structure equals 3. The unknowns of the displacement method are the angular displacements of rigid joints and the horizontal displacement of the crossbar. In case (b), the number of rigid joints equals 3 and the number of independent linear displacements nd ¼ 2, so the degree of kinematical indeterminacy equals 5. Case (c) presents a multistorey frame. Joint C shows that member CD is connected with vertical member AE by a hinge, while the member ACE does not have a hinge at point C. To construct the hinged scheme it is necessary to introduce hinges at
10.1
Fundamental Idea of the Displacement Method
369
joints B, D, and C for member ACE (note that hinge C includes the original hinge at joint C and the introduced hinge), i.e., the hinge C is multiple one. For the given structure, the number of rigid joints is 3 (joints B, D, and C for member ACE) and the number of independent linear joint displacements is 2, so the total number of unknowns by the displacement method is 5. Case (d) also presents a multistory frame, in which the crossbars are absolutely rigid. Since the joints cannot rotate, the number of unknown angles of rotation is zero and the number of independent linear joint displacements is 2.
Design diagram
a
c
Hinged scheme
b
A
B C
D
d C*
E Fig. 10.2 Calculation of independent linear joints displacements
10.1.2 Primary System and Primary Unknowns The primary system of the displacement method is obtained from the given one by introducing additional constraints to prevent rotation of all rigid joints and all independent displacements of various joints. These introduced constraints are shown by the shaded squares and the double lines, respectively. Primary systems for some structures are shown in Fig. 10.3. 1
a
2
b
2
1
3
5 4
3 6 1
c 2
4
d
1
2
3 5
Fig. 10.3 Design diagrams and primary systems of the displacement method
Primary unknowns Zi (i ¼ 1, 2, . . ., n) represent displacements of introduced constraints (angles of rotations and/or linear displacements of various joints of a frame). The number of primary unknowns, n, for each structure equals to the degrees of its kinematical indeterminacy. In case 10.3a, the primary unknowns Z1 and Z2 are angular displacements of introduced rigid joints 1 and 2; the primary unknown, Z3, is the linear displacement of introduced constraint 3. In case 10.3b, the primary unknowns are angular
370
10
The Displacement Method
displacements Z1, . . ., Z4 and linear displacements Z5, Z6. In case 10.3c, the primary unknowns are angular displacements Z1, Z2, Z3 and linear displacements Z4 and Z5. In case 10.3d, the primary unknowns are linear displacements Z1 and Z2. It can be seen that a primary system of the displacement method consists of a number of single-span redundant beams. In most cases they are fixed-fixed and fixed-pinned uniform beams; fixed-guided beams and beams with elastic support as well as nonuniform beams are also possible. The next sections show how to use the primary system and tabulated data (Tables A.3– A.7) for the analysis of statically indeterminate structures by displacement method in canonical form.
10.1.3 Compatibility Equation: Concept of Unit Reaction The fundamental idea of the displacement method presentation in canonical form is explained below by considering the simplest frame in Fig. 10.4a. 1. The elastic curve of the frame caused by given load P is shown by the dashed line; rigid joint 1 rotates in a clockwise direction by some angle Z1. 2. The primary system is obtained by introducing additional constraint 1 (Fig. 10.4b). This additional constraint prevents angular displacement. Therefore, the reactive moment due to external load P arises in the introduced constraint and deformation occurs in the horizontal element only. The elastic curve and reactive moment R1P at constraint 1 in the primary system are shown in Fig. 10.4c. The difference between the given structure and the primary one is obvious: in the primary system the reactive moment arises at the additional constraint while in entire structure in Fig. 10.4a such constraint in joint 1 is absent. 3. In order to eliminate the difference between the given and primary structures, constraint 1 must be rotated by some angle Z1. In this case, the reactive moment arises in constraint 1. If the angle of rotation is a unit angle (Z1 ¼ 1), then the reactive moment is called a unit reaction and denoted as r11. The term “unit” means that this moment is caused by a unit displacement of the introduced constraint. The first numeral in r11 is the label of the constraint where the reactive moment arises and the second numeral is the label of the constraint, which is rotated (in this example, they are the same). Z1
a
P
P
b
c
P
Entire structure
d
1
1
1
Primary system
R1P
Z1=1 1
r11
Fig. 10.4 (a, b) Design diagram and primary system. (c, d) The primary system is subjected to given load P and unit rotation of introduced constraint 1
Sign rule. If constraint 1 is rotated clockwise and after that it is released, then this constraint tends to rotate back counterclockwise. Therefore, the reactive moment will act in an opposite direction, i.e., clockwise. Thus, the positive reaction coincides with the positive displacement of the introduced constraint. Figure 10.4d shows the positive directions for displacement Z1 and unit reaction r11. 4. The total reaction caused by rotation of the introduced constraint and the given load is r11Z1 þ R1P, where the first term r11Z1 is the reactive moment in constraint 1 due to the real angle of rotation Z1. The second term represents the reactive moment in constraint 1 due to the actual load. Lowercase letter r means that this reaction is caused by the unit displacement (rotation) of the introduced constraint, while capital letter R means that this reaction is caused by the real external load. If this total reactive moment (due to both the given load and the angular displacement of the introduced constraint) is equal to zero, then the behavior of the given and primary structures is identical. This statement may be written in the following form: r 11 Z 1 þ R1P ¼ 0
ð10:2Þ
Equation (10.2) presents the displacement method in canonical form for a structure of the first degree of kinematical indeterminacy.
10.2
Canonical Equations of Displacement Method
371
The primary unknown is obtained as Z1 ¼ R1P/r11. Knowing the primary unknown Z1 we can consider each element of the frame to be a standard beam due to the action of found angle Z1 and the external load applied to this particular element. Moments of supports for these standard beams are found in Tables A.3–A.4. Using the principle of superposition, the final bending moment diagram for the entire frame is constructed by the formula M P ¼ M 1 Z 1 þ M 0P
ð10:3Þ
where M 1 is the bending moment diagram caused by the unit primary unknown; term M 1 Z 1 is the bending moment diagram caused by the actual primary unknown Z1, and the bending moment diagram in the primary system caused by the given load is denoted as M 0P .
10.2
Canonical Equations of Displacement Method
Canonical equations of displacement method offer the unified and effective procedure for analysis of continuous beams and statically indeterminate frames of different types. The calculation of the coefficients of canonical equations is discussed. Detailed analysis of the structures subjected to fixed loads are considered.
10.2.1 Compatibility Equations in General Case Now we consider an arbitrary n-times kinematically indeterminate structure. The primary unknowns Zi (i ¼ 1, . . ., n) are displacements (linear and/or angular) of introduced constraints. The canonical equations of the displacement method are written as follows: r 11 Z 1 þ r 12 Z 2 þ . . . þ r 1n Z n þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ . . . þ r 2n Z n þ R2P ¼ 0 : :
: : :
: :
: :
: : :
:
ð10:4Þ
r n1 Z 1 þ r n2 Z 2 þ . . . þ r nn Z n þ RnP ¼ 0 The number of canonical equations is equal to number of primary unknowns of the displacement method. Interpretation of the canonical equations: Coefficient rik represents the unit reaction, i.e., the reaction (force or moment), which arises in the i-th introduced constraint (first letter in subscript) caused by unit displacement Zk ¼ 1 of k-th introduced constraint (second letter in subscript). The term rikZk represents the reaction, which arises in the i-th introduced constraint due to the action of real unknown displacement Zk. Free term RiP is the reaction in the i-th introduced constraint due to the action of the applied loads. Thus, the left part of the i-th equation represents a total reaction, which arises in the i-th introduced constraint due to the actions of all real unknowns Z as well as the applied load. The total reaction in each introduced constraint in the primary system caused by all primary unknowns (the linear and angular displacements of the introduced constraints) and the applied loads is equal to zero. In this case, the difference between the given structure and the primary system vanishes, or, in other words, the behavior of the given and the primary systems is the same.
10.2.2 Calculation of Unit Reactions The frame presented in Fig. 10.5a allows angular displacement of the rigid joint and horizontal displacement of the crossbar. So the structure is twice kinematically indeterminate. The primary system of the displacement method is presented in Fig. 10.5b. Constraints 1 and 2 are additional introduced constraints that prevent angular and linear displacements. In constraint 1, which prevents angular displacement, only the reactive moment arises; in constraint 2, which prevents only linear displacement, only the reactive force arises. The corresponding canonical equations are
372
10
The Displacement Method
r 11 Z 1 þ r 12 Z 2 þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ R2P ¼ 0 To determine coefficients rik of these equations we consider two states. State 1 presents the primary system subjected to unit angular displacement Z1 ¼ 1. State 2 presents the primary system subjected to unit horizontal displacement Z2 ¼ 1. For both states we will show the bending moment diagrams. These diagrams caused by the unit displacements of the introduced constraints Z1 ¼ 1 and Z2 ¼ 1 are shown in Fig. 10.5c, d, respectively. The elastic curves are shown by dashed lines; the asterisk () denotes the inflection points of the elastic curves. For member 1–2 the extended fibers are located below the neutral line (Fig. 10.5c). Bending moment diagrams are plotted on the extended fibers.
a
b
EI2
1
2
h
EI1
l
c
Z1 =1
Extended fibers
r11 1
4EI1 /h
State 2
State 1 2
d
1 *
Elastic curve
g
r11
6EI1 /h2
6EI1 /h2
2EI1 /h
e
r22
2
3EI2 /l *
Z2 =1
Z2 =1
r12
r21
r12
3EI2/l
1
1
6EI1 /h2
4EI1 /h
h
f 6EI1 /h
2
6EI1 /h2 2EI1 /h
2
1
1
r21 12EI1 /h
4EI1 /h
6EI1 /h2
6EI1 /h2
12EI1 /h3
2
r22
3
12EI1 /h3 6EI1 /h2
Fig.10.5 (a, b) Design diagram and the primary system of the displacement method; (c, d) unit states, corresponding bending moment diagrams and calculation of the unit reactions; (e, f) free-body diagrams for joint 1 and crossbar, state 1; (g, h) free-body diagrams for joint 1 and crossbar, state 2
First unit state (Z1 ¼ 1): Rigid joint 1 rotates clockwise (positive direction) through angle Z1 ¼ 1. The reactions for both members (fixed-pinned and fixed-fixed) in the case of angular displacement of the fixed supports are presented in Tables A.3 and A.4. In this case, the bending moment at joint 1 of the horizontal member equals 3EI2/l (Table A.3, row 1); for the vertical member the specified ordinates are 4EI1/h for joint 1 and 2EI1/h at the bottom clamped support (Table A.4, row 1). As a result
10.2
Canonical Equations of Displacement Method
373
of the angular displacement, the unit reactive moment r11 arises in constraint 1 and reactive force r21 arises in constraint 2; all unit reactions are shown in the positive direction. Second unit state (Z2 ¼ 1): If constraint 2 has horizontal displacement Z2 ¼ 1 from left to right (positive direction), then introduced joint 1 has the same displacement and, as a result, the vertical member is subjected to bending. Unit reactive moment r12 and unit reactive force r22 arise in constraints 1 and 2, respectively. The specified ordinates for the vertical member at the bottom and at point 1 are equal to 6EI1/h2 (Table A.4, row 2). Unit reactions r11, r12 represent the reactive moments in constraint 1 for both states and unit reactions r21, r22 represent the reactive forces in constraint 2 for both states. Unit reactions rii, located on the main diagonal of the canonical equations, are called the main reactions (r11, r22); other unit reactions are called secondary ones (r21, r12). To calculate all the unit reactions, we need to consider the free-body diagrams for joint 1 and for crossbar 1–2 for each state. The free-body diagram for joint 1 in state 1 is shown in Fig. 10.5e. The direction of moments 4EI1/h and 3EI2/l corresponds to the location of the extended fibers in the vicinity of joint 1 (Fig. 10.5e); the extended fibers are shown by dashed lines. Positive unit reactive moment r11 is shown by the direction of unit displacement Z1. The equilibrium condition for joint 1 is ΣM ¼ 0; therefore we get r 11 ¼
4EI 1 3EI 2 þ h l
To calculate unit reaction r21 we need to consider the free-body diagram for the crossbar. The shear force infinitely close to joint 1 is found by considering the equilibrium of the vertical element through the following steps. Moments at the end points of the vertical element shown in Fig. 10.5f are taken from the bending moment diagram in Fig. 10.5c. The moments 4EI1/h at the top and 2EI1/h at the support are equilibrated by two equal forces, 6EI1/h2. The upper force is transmitted to the crossbar with opposite sign. After that, unit reaction r21 is found by considering the equilibrium equation of the crossbar, (ΣFx ¼ 0), so 1 4EI 1 2EI 1 6EI r 21 ¼ þ ¼ 21 h h h h It is obvious, this reaction can also be taken directly from Table A.4, row 1. Similarly, equilibrium equations ΣM ¼ 0 for the free-body diagrams for joint 1 in state 2 (Fig. 10.5g) and equilibrium ΣFx ¼ 0 for the crossbar in state 2 (Fig.10.5h) lead to the following unit reactions: r 12 ¼
6EI 1 , h2
r 22 ¼
12EI 1 h3
10.2.3 Properties of Unit Reactions 1. Main reactions are strictly positive (rii > 0). Secondary reactions rik may be positive, negative, or zero. 2. According to the reciprocal reactions theorem, the secondary reactions satisfy the symmetry condition rik ¼ rki, i.e., r12 ¼ r21. 3. The dimensions of unit reaction rik are determined by the following rule: the dimension of reaction (force or moment) at index i is divided by the dimension of the displacement (linear or angular) at index k. In our case [r11] ¼ kNm/rad ¼ kNm, [r12] ¼ kNm/m ¼ kN, [r21] ¼ kN/rad ¼ kN, [r22] ¼ kN/m.
10.2.4 Procedure for Analysis For analysis of statically indeterminate continuous beams and frames by the displacement method in canonical form, the following procedure is applied: 1. Define the degree of kinematical indeterminacy and construct the primary system of the displacement method. 2. Formulate the canonical equations of the displacement method (10.4). 3. Apply successively unit displacements Z1 ¼ 1, Z2 ¼ 1, . . ., Zn ¼ 1 to the primary structure. Construct the corresponding 1, M 2, . . . , M n using Tables A.3–A.6. bending moment diagrams M 4. Calculate the main and secondary unit reactions rik.
374
10
The Displacement Method
5. Construct the bending moment diagram M 0P due to the applied load in the primary system and calculate the free terms RiP of the canonical equations. 6. Solve the system of equations with respect to unknown displacements Z1, Z2, . . ., Zn. 7. Construct the final bending moment diagrams by formula 1 Z 1 þ M2 Z 2 þ . . . þ Mn Z n þ M 0P MP ¼ M
ð10:5Þ
n Z n represents a bending moment diagram due to actual displacement Zn. The term M 0P represents a The term M bending moment diagram in the primary system due to actual load. 8. Compute the shear forces using the relationship Q ¼ dM/dx considering each member due to the given loads and the end bending moments and construct the corresponding shear diagram. 9. Compute the axial forces considering the equilibrium of joints of the frame and construct the corresponding axial force diagram. 10. Calculate reactions of supports and check them using the equilibrium conditions for an entire structure as a whole or for any separated part.
10.2.4.1
Continuous Beams
Let us show the application of this algorithm to the analysis of a uniform continuous two-span beam A-1-B (Fig. 10.6a). According to (10.1), this continuous beam is kinematically indeterminate to the first degree. Indeed, support A prevents linear displacement of joints A, 1, and B, and there is one rigid joint at support 1; therefore there is only angular displacement at this support. Thus, the primary unknown is the angular displacement Z1 at support 1. The primary system is obtained from a given structure by introducing constraint 1 at middle support 1 (Fig. 10.6a); this constraint prevents angular displacement at support 1. The canonical equation of the displacement method is r 11 Z 1 þ R1P ¼ 0
ðaÞ
To calculate unit reaction r11 we need to rotate the introduced constraint clockwise by angle Z1 ¼ 1. The corresponding elastic curve, the location of the extended fibers, and the bending moment diagram caused by the unit primary unknown are 3EI shown in Fig. 10.6b. The bending moment at the fixed support for a fixed-pinned beam is (Table A.3 row 1). The freel body diagram of joint 1 from diagram M 1 is shown in Fig. 10.6c. According to the elastic curve, the extended fibers in the vicinity of joint 1 are located above the neutral line to the left of point 1 and below the neutral line to the right of point 1. The moments 0.375EI and 0.3EI are shown according to the location of the extended fibers. Unit reactive moment r11 is shown assuming its positive direction (clockwise). Equilibrium condition ∑M ¼ 0 leads to unit reaction r11 ¼ 0.675EI (kNm/rad). To calculate free term R1P of the canonical equation we need to construct the bending moment diagram in the primary system caused by the given load. This diagram is shown in Fig. 10.6d; each element is considered as a separate beam; the location of the extended fibers is shown by the dashed line. The extended fibers in the vicinity of joint 1 are located above the neutral line to the left and right of joint 1. The bending moment at the fixed support for the left span subjected to uniformly distributed load q, according to Table A.3 row 5, equals M 01A ¼
ql21 q 82 ¼ ¼ 16 ðkNmÞ 8 8
The bending moment at the specified points for the right span subjected to concentrated force P, according to Table A.3, row 3, equals 12 10 Pl2 0:4 1 0:42 ¼ 20:16 ðkNmÞ υ 1 υ2 ¼ 2 2 Pl2 2 12 10 0 Mk ¼ u υ ð 3 uÞ ¼ 0:62 0:4ð3 0:4Þ ¼ 20:736 ðkNmÞ 2 2
M 01B ¼
The free-body diagram of joint 1 from diagram M 0P is shown in Fig. 10.6e. According to the location of the extended fibers, the moment of 16 kNm is shown to be counterclockwise and moment of 20.16 kNm is clockwise. Reactive moment R1P is assumed to have a positive direction, i.e., clockwise. Equilibrium condition ∑M ¼ 0 leads to R1P ¼ 4.16 kNm.
10.2
Canonical Equations of Displacement Method
375
Canonical Eq. (a) becomes 0.675EI Z1 4.16 ¼ 0. The root of this equation, i.e., the primary unknown is Z1 ¼
a
6:163 ðradÞ EI
ðbÞ
P=12 kN
q=2kN/m 1
A
k
B
EI
ul2=4 m
ul2=6 m l1=8 m
l2=10 m
Z1
q
P
1
b
3EI = 0.375 EI l1
Z1=1 1
c
ul2=4 m
r11
0.12EI
1
M1 0.3EI
0.375EI Extended fibers
d
0 M 1A
Elastic curve
3EI = 0.3EI l2 0 M 1B
r11=0.675EI
k
R1P
e
P
q
1 M P0
16
R1P= – 4.16 (kNm)
M k0
f
M1
q
P k
Extended fibers
20.16
MP
Mk
Fig. 10.6 (a) Design diagram and primary system. (b, c) Bending moment diagram caused by unit angular displacement and calculation of r11; (d, e) bending moment diagram of a primary system caused by a given load and calculation of free term R1P; (f) final bending moment diagram and location of extended fibers
The bending moments at the specified points can be calculated by the following formula M P ¼ M 1 Z 1 þ M 0P In our case we have 6:163 16 ¼ 18:31 ðkNmÞ, EI 6:163 20:16 ¼ 18:31 ðkNmÞ ¼ 0:3EI EI
M 1A ¼ 0:375EI M 1B
ðcÞ
376
10
M k ¼ 0:12EI
The Displacement Method
6:163 þ 20:736 ¼ 21:475 ðkNmÞ EI
Of course, M1A and M1B are equal. The negative sign indicates that the extended fibers at support 1 are located above the neutral line. The final bending moment diagram MP is presented in Fig. 10.6f; the location of the extended fibers is shown by the dotted line.
10.2.4.2
Beam with Intermediate Hinge
The next example presents an application of the displacement method in canonical form for detailed analysis of a redundant beam with intermediate hinge. This structure was analyzed earlier by the initial parameter method and by the force method in canonical form. Example 10.1 Design diagram of the one-span redundant beam of length l is shown in Fig. 10.7a. Both parts of the beam, AC and CB, are connected by hinge C. The bending stiffnesses of the left and right parts of the beam are EI and kEI, where k is any positive number. The beam is subjected to uniformly distributed load q. Determine the bending moment at the supports. q
a
A
EI
B
kEI
C
l2=3l/4
l1=l/4 l q
b
A
i1 1
B
i2
Z1
M2=3i2/l2
M1=3i1/l1
c
Extended fibers
Z1=1 RA=3i1/l12
RC1
RC1
d
M1
Elastic curve
RC1=RA
RC2
r11 MB(q)=ql22 /8
RC1(q) RC2(q)
f
q
e
RC2=RB
RB=3i2/l22
r11 MA(q)=ql12 /8
RC2
RC1(q) RC2(q)
MP0 RA(q)=5ql1/8
R1P
RB(q)= 5ql2/8
RC2(q)
RC1(q) R1P
Fig. 10.7 (a, b) Design diagram of the redundant beam and primary system of the displacement method; (c, d) bending moment diagram caused by unit linear vertical displacement and calculation of r11; (e, f) bending moment diagram caused by a given load in primary system and calculation of free term R1P
Solution The system has one unknown of the displacement method; it is the vertical displacement Z1 of hinge C. The introduced constraint 1 prevents the vertical displacement of joint C (Fig. 10.7b). The bending stiffnesses per unit length for left and right parts of structure are
10.2
Canonical Equations of Displacement Method
i1 ¼
377
1EI EI 4EI ¼ ¼ , l1 l l=4
i2 ¼
kEI 4kEI ¼ l2 3l
Canonical equation of the displacement method is r 11 Z 1 þ R1P ¼ 0, where the primary unknown Z1 is the vertical displacement of joint C, r11 and R1P are reactions of the introduced constraint 1 caused by the unit displacement of this constraint and given load q, respectively. Calculation r11: The bending moment diagram M 1 caused by the unit displacement Z1 ¼ 1 is shown in Fig. 10.7c; elastic curve and corresponding extended fibers are shown by the dashed lines. The stretched fibers are located above of the axial line of the beam. Therefore, the bending moment in support A is directed counterclockwise, while in support B clockwise. To equilibrate the moment M1, the shear force RC1 which is located at section infinitely close to support C of part 1 (portion AC) should be directed downwards. Thus, the shear forces RA and RC1 are directed upwards and downwards, respectively, what is more RA ¼ RC1 ¼ 3i1 =l21 . Direction of the reactions and their magnitudes for fixed-hinged beam are presented in Table A.3. Similarly, the magnitude and direction of the shear forces of the portion 2 are determined: RB ¼ RC2 ¼ 3i2 =l22 . Now we need to select the infinitely small element of the beam in the vicinity of the introduced support C and transfer the found shear forces RC1 and RC2 to this element, as shown in Fig. 10.7d. The equilibrium equation of the selected part and the corresponding unit reaction are X Y ¼ 0 ! r 11 RC1 RC2 ¼ 0 3i1 3i2 4EI 1 4kEI 1 þ 2 ¼3 þ3 ¼ 2 2 l 3l ð3l=4Þ2 l1 l2 ðl=4Þ 3 4 16EI k 192EI k ¼ 1þ 1þ ¼ 3 27 27 l3 l
r 11 ¼ RC1 þ RC2 ¼
Calculation R1P: The bending moment diagram in the primary system caused by the given load q, MP0, is shown in Fig. 10.7e. For clamped-pinned beam AC the bending moment at support A is equal to M A ðqÞ ¼ ql21 =8 (Table A.3, line5). This moment is directed counterclockwise; the extended fibers near clamped support are located above of the longitudinal axes. Shear force at section which is located infinitely close left to C for portion 1 is directed upwards; this shear force is RC1(q) ¼ 3ql1/8. Now we need to transfer this shear force to the selected item, as shown in Fig. 10.7f, and to perform the same operation for the right part 2. Thus, two forces RC1(q) ¼ 3ql1/8 and RC2(q) ¼ 3ql2/8 appear on the selected element; both of them are directed downwards (Fig. 10.7f). Equilibrium equation for the introduced isolated support C and corresponding loading term are X Y ¼ 0 ! R1P þ RC1 ðqÞ þ RC2 ðqÞ ¼ 0 h i 3 3 3 R1P ¼ ½RC1 ðqÞ þ RC2 ðqÞ ¼ ql1 þ ql2 ¼ ql 8 8 8 The primary unknown Z1 ¼
R1P 3 ¼ ql 8 r 11
l3 ql4 ¼ k k 192EI 1 þ 512EI 1 þ 27 27
To compute the bending moments in the original system, the formula (10.5) should be applied. The bending moments at support А and B are ql2 3i1 ql4 1¼ l1 512EI ð1 þ k=27Þ 8 2 4 4EI 1 ql q l ql2 324 ¼ 3 ¼ 1þ l l=4 512EI ð1 þ k=27Þ 8 4 27 þ k 128
M A ¼ M 1 Z 1 þ M 0P ¼
ql2 3i2 ql4 2¼ l2 512EI ð1 þ k=27Þ 8 2 4 4kEI 1 ql q 3l ql2 4k 27 ¼ 3 þ9 ¼ 3l 3l=4 512EI ð1 þ k=27Þ 8 4 128 3 27 þ k
M B ¼ M 1 Z 1 þ M 0P ¼
378
10
The Displacement Method
If assume the beam is uniform (k ¼ 1), then expressions for vertical displacement of hinge C (i.e., Z1) and the moments MA and MB at the clamped supports are presented as follows: ql4 ql4 ¼ 1:2656 672EI k 512EI 1 þ 27 11 2 9 2 ql , M B ¼ ql MA ¼ 112 112 Z1 ¼
The negative sign means the extended fibers in vicinity of the clamped support are located above the longitudinal axis of the beam. These expression had been derived early by the initial parameter method (Chap. 8, Example 8.10, Table 8.4) and the force method in canonical form (Chap. 9, Example 9.6). The reader is invited to compare these methods for the given design diagram. Now we will consider the analysis of some frames by the displacement method in canonical form. Example 10.2 The crossbar of the frame in Fig. 10.8a is connected with vertical members by means of hinges. The bending stiffness is EI for all vertical members and 2EI for the crossbar; their relative stiffnesses, 1 and 2, are shown in the circles. Concentrated force P acts horizontally at the level of the crossbar. Construct the internal force diagrams. Solution The system has two unknowns of the displacement method: the angular displacement of joint 1 and the linear displacement of the crossbar. The primary system is shown in Fig. 10.8b. The introduced constraint 1 is related only to the horizontal member, but not the vertical one. The primary unknowns of the displacement method are angular displacement Z1 of constraint 1 and linear displacement Z2 of constraint 2. The bending 1EI 2EI stiffnesses per unit length for the vertical and horizontal members are ivert ¼ ¼ 0:2EI, ihor ¼ ¼ 0:333EI . The 5 6 canonical equations of the displacement method are r 11 Z 1 þ r 12 Z 2 þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ R2P ¼ 0
ðaÞ
Calculation of unit reactions: To calculate coefficients r11 and r21 we need to construct the bending moment diagram M 1 in the primary system due to the unit rotation Z1 ¼ 1 of introduced constraint 1 and to determine the reactions which arise in the introduced constraints 1 and 2. Elastic curve due to rotation of Z1 is shown by dashed line. It can be seen that the vertical elements are not deformed because of the hinged connections of the vertical elements with the horizontal bar (Fig. 10.8c). The crossbar in the primary system presents two pinned-clamped beams. The bending moment at the introduced clamped support for both portions of the crossbar are 3i ¼ 3 (2EI/6) ¼ 1 EI (Table A.3, row 1). Bending moments in the vertical elements do not arise. The positive unknown unit reactions are shown by the dotted arrow. The equilibrium of joint 1 from bending moment diagram M 1 leads to r 11 ¼ 3i þ 3i ¼ 6i ¼ 2EI ðkNm=radÞ: The equilibrium of the crossbar, considering the bending moment diagram M 1 , leads to r21 ¼ 0. Indeed, since within the vertical members the bending moments and shear forces do not arise, then equation ∑X ¼ 0 for the crossbar leads to the above result. To calculate coefficients of r12 and r22 it is necessary to construct the bending moment diagram M 2 in the primary system due to the unit linear displacement Z2 ¼ 1 of the introduced constraint 2 and to determine the reaction r12 and r22, which arise in the introduced constraints 1 and 2 (Fig. 10.8d). Elastic curve due to linear displacement of Z2 is shown by dashed line. It can be seen that the crossbar is not deformed because of the hinged connections of the crossbar and vertical elements. The bending moment at the bottom of the vertical members is 3i/h ¼ 3 0.2EI/5 ¼ 0.12EI (Table A.3, row 2). The equilibrium equation of joint 1 leads to r12 ¼ 0. The free-body diagram for the crossbar is presented in Fig. 10.8e, where shear in each vertical member due to unit displacement Z2 ¼ 1 is Q¼
0:12EI 0:12EI ¼ ¼ 0:024EI h 5
Therefore, the equilibrium equation ∑X ¼ 0 for the crossbar leads to the following result: r 22 ¼ 3 0:024EI ¼ 0:072EI
ðkN=mÞ
10.2
Canonical Equations of Displacement Method
a
379
1
P
1
P
c
h=5m
Design diagram
l=6m
2
i=0.333EI
2 1
Z2
Z1
b
Primary system
i=0.2EI
6m r11
Z1=1
3i=1EI
2
1
r21
=
3i
1 3i
r11 3i=1EI
1
2
M1
Z2=1
r12
d
r21
1
r22
2
M2 3i . EI = 012 h
0.12EI
0.12EI
e
r12 0.024EI
0.024EI
0.024EI
Z2=1 1
R1P P
r22 Q=0.024EI
M=0.12EI
M
M
f
2
0.024EI 0.024EI
1
2
R2P
SX = 0:
R2 P = – P
g P MP
1.667P
P 3
1.667P
P 3
1.667P
P 3
Fig.10.8 (a, b) Design diagram of the frame and primary system. (c) Unit bending moment diagram due to Z1 ¼ 1 and calculation of r11 and r21. (d, e) Unit bending moment diagram due to Z2 ¼ 1 and calculation of r12 and r22. (f) Free-body diagram of the crossbar in the loaded condition. (g) Final bending moment diagram and reactions of supports
380
10
The Displacement Method
Calculation of free terms: The force P is applied at the level of the crossbar; therefore there is no bending of the horizontal elements in the primary system. Since constraint 2 prevents displacement in a horizontal direction, then bending of the vertical members of the frame does not occur either. Thus, in the primary system there are no elements subject to bending. But this does not imply that all reactions of the introduced constraints 1 and 2 are zero. Indeed, R1P ¼ 0, but R2P ¼ P. The last expression is obtained from the equilibrium of the crossbar subjected to the given load P (Fig. 10.8f). The canonical equations become 2EI Z 1 þ 0 Z 2 ¼ 0 0 Z 1 þ 0:072EI Z 2 P ¼ 0
ðbÞ
P ðmÞ. The result Z1 ¼ 0 means that the crossbar is not deformed, 0:072EI but displaced in the horizontal direction as an absolutely rigid element. This is because the crossbar is connected to the vertical elements by means of hinges. The bending moment diagram can be constructed using the principle of superposition: The roots of these equations are Z 1 ¼ 0,
Z2 ¼
M P ¼ M 1 Z 1 þ M 2 Z 2 þ M 0P
ðcÞ
Since Z1 ¼ 0 and acting load P does not cause bending of the members in the primary system, then formula (c) becomes M P ¼ M 2 Z 2 . The resulting bending moment diagram is presented in Fig. 10.8g. The bending moments at the clamped supports are 0:12EI
P ¼ 1:667P 0:072EI
ðkNmÞ
The shear force at vertical members Q¼
10.2.4.3
1:667P 1:667P P ¼ ¼ h 5 3
ðkNÞ
Discussion
1. Even if the primary system has a nonzero bending moment diagram within the crossbar (diagram M 1 ), the resulting bending moment diagram along the crossbar is zero. This happens because Z1 ¼ 0, M 2 ¼ 0, and M 0P ¼ 0. 2. The active load acts at the level of the crossbar and the bending moments in the primary system due to this load do not arise. However, this is not to say that all free terms of canonical equations are zero. Introduced constraints prevent angular and linear displacements of the frame. So free terms present the reactive moment and force in the introduced constraints. The reactive moment R1P ¼ 0, while the reactive force of introduced constraint R2P 6¼ 0. 3. If the flexural stiffness of the crossbar is increased by n times, then unit reaction r11 becomes r11 ¼ 2EI n, while all other coefficients and free terms remain the same. So Eq. (b) lead to the same primary unknowns and the resulting bending moment diagram remains same. This happens because for given design diagram the bending deformations for crossbar are absent. 4. The length l of the span has no effect on the bending moment diagram. 5. If the frame, shown in Fig. 10.8a, is modified (the number of the vertical members being k), then the reaction of each support is P/k. The next example presents a detailed analysis of a frame by the displacement method in canonical form. This frame was analyzed earlier by the force method. Therefore, the reader can compare the different analytical approaches to the same structure and see their advantages and disadvantages. Example 10.3 A design diagram of the frame is presented in Fig. 10.9a; the specified section of the frame is shown in Fig. 10.9b. The flexural stiffnesses for the vertical member and crossbar are EI and 2EI, respectively; their relative flexural stiffnesses are shown in circles. The frame is loaded by force P¼8kN and uniformly distributed load q ¼ 2 kN/m. Construct the bending moment diagram.
Canonical Equations of Displacement Method
a
381
b
P
3m
10.2
1
P
2
1
7 5m
1 4m
i4-5
8
2
q
c
P
5
q
i6-8
2
4
1
3
q
6
i1-3
6m
State 1: The unit angular displacement Z1=1
d
Z1=1 3i4-5=1.0EI
r11
3i4-5 0.333EI
r21 4i1-3=0.8EI
3i4 – 5 = 0.333EI l4 – 5 0.24EI
0.24EI
r21
4i1-3
3i6-8=0.6EI *
M1
2i1-3=0.4EI
6i1– 3 = 0.24 EI l1– 3
2i1-3
e State 2: The unit linear displacement Z2=1 3i4 – 5 = 0.333EI l4 – 5
3i4 – 5 l4 – 5
Z2=1
r12
3i4 – 5 l42– 5
= 0.111EI
0.111EI
r22 *
0.096EI
6i1– 3 = 0.24 EI l1– 3
0.096EI 6i1– 3 l1– 3
6i1– 3 l1– 3
M2
f
12i1– 3 l12– 3
r22
= 0.096 EI
Loaded state
M6=15.36
ul
ul
R1P M3=4.1667
*
*
5.0 5q/2=5
R2P
M7=9.984 M2=2.0833
q
M P0
*
g
R2P
5.0
13.084
P=8kN
1.814
11.349
11.27 1.324
q=2kN/m
M kNm
8.037 Fig. 10.9 (a–c) Design diagram, the specified sections and primary system. (d) State 1: bending moment diagram due to unit angular displacement Z1 ¼ 1 and free-body diagram for the calculation of r21. (e) State 2: bending moment diagram due to unit linear displacement Z2 ¼ 1 and free-body diagram for the calculation of r22. (f) Bending moment diagram in the loaded state and free-body diagram for the calculation of free term R2P. (g) Final bending moment diagram M (kNm)
382
10
The Displacement Method
Solution It is easy to check the number of independent linear displacements nd ¼ 1 (the hinged scheme is not shown) and the total number of unknowns of the displacement method is 2. The primary system is obtained by introducing two additional constraints, labeled 1 and 2 (Fig. 10.9c). Constraint 1 prevents only angular displacement of the rigid joint and constraint 2 prevents only linear displacement of the crossbar. The flexural stiffness per unit length for each element of the structure is as follows: i13 ¼
1EI ¼ 0:2EI; 5
i68 ¼
2EI ¼ 0:2EI; 10
i45 ¼
1EI ¼ 0:333EI 3
where the subscript of each parameter i indicates an element of the frame (Fig. 10.9b). The canonical equations of the displacement method are r 11 Z 1 þ r 12 Z 2 þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ R2P ¼ 0 To calculate unit reactions rik, it is necessary to construct the unit bending moment diagrams M 1 , M 2 in the primary system. To construct diagram M 1 , it is necessary to rotate introduced constraint 1 by angle Z1¼1 clockwise. To construct diagram M 2, it is necessary to shift introduced constraint 2 to the right by distance Z2 ¼ 1. The bending moment diagrams for states 1 and 2 as well as the free-body diagrams for calculation of unit reactions rik are shown in Fig. 10.9d, e and Tables 10.1 and 10.2. The positive reactions are shown by dashed arrows. Table 10.1 Calculation of unit and loaded reactions at introduced constraint 1
Diagram
Free-body diagram of joint 1
Equilibrium equation
Reaction
1.0EI M1
0.6EI
r11
∑ M =0
kNm rad
– r11 + (10 . + 0.6 + 0.8) EI = 0
r11 = 2.4 EI
∑
r12 = 0.093EI
0.8EI 0.333EI M2
r12
– r12 – 0.24 EI + 0.333EI = 0
0.24EI 15.36 M P0
M =0
R1P
∑
M =0
– R1P – 1536 . + 4.1667 = 0
kNm m
R1 P = – 111933 . kNm
4.1667
To calculate free terms RiP, it is necessary to construct the diagram caused by the applied load in the primary system (loaded state). This state and corresponding bending moment diagram are shown in Fig. 10.9f, as well as in Tables 10.1 and 10.2. The ordinates of the bending moment diagrams for standard uniform elements due to different loads are presented in Tables A.3–A.7. The ordinates of the M 0P diagram at the specified sections according to Table A.4 (for element 1–3) and Table A.3 (for element 6–8) are M1 ¼ M3 ¼ M2 ¼
ql213 ¼ 4:1667 kNm; 12
ql213 ¼ 2:0833 kNm; 24
10.2
Canonical Equations of Displacement Method
383
8 10 Pl υ 1 υ2 ¼ 0:6 1 0:62 ¼ 15:36 kNm; 2 2 Pl 8 10 0:42 0:6ð3 0:4Þ ¼ 9:984 kNm M 7 ¼ u2 υð3 uÞ ¼ 2 2 M6 ¼
All the bending moment diagrams are plotted on the extended fibers of the frame (Fig. 10.9d–f). Elastic curves are shown by dashed lines. The asterisks () on the elastic curves show the points of inflection. To calculate reactive moments r11, r12, and R1P, it is necessary to consider the free-body diagrams of joint 1 using M 1 , M 2 , and M 0P diagrams. The ordinates of the bending moments infinitely close to the joint are taken from the corresponding diagrams. The direction of each of these moments must correspond to the location of the extended fibers in each diagram. The calculations of unit reactions r11, r12, and R1P are presented in Table 10.1. Positive reactive moments are directed clockwise; locations of extended fibers are shown by dashed lines. To calculate reactive forces r21, r22, and R2P, it is necessary to consider the equilibrium of horizontal member 6–8. For this we need to cut off element 6–8 from diagrams M 1 , M 2, and M 0P by sections infinitely close to joint 1 from above and below. These diagrams present the free-body diagram for crossbar using closed contours. The calculations of unit reactions r21, r22 and loaded reaction R2P are presented in Table 10.2. Table 10.2 Calculation of unit and loaded reactions at introduced constraint 2
Diagram
Portion (crossbar) of the structure 0.333EI
M1
r21
0.24EI
Equilibrium equation
S X =0
r21 = 0.093EI
kN rad
. EI – 0.096 EI = 0 r22 – 0111
r22 = 0.207 EI
kN m
S X =0
R2 P =–5kN
r21 + 0.24 EI – 0.333EI = 0
S X =0
0.111EI
M2
r22
0.096EI M P0
R2P
5
Result
R2 P +5 = 0
Let us consider in detail the procedure for the calculation of reactive forces r21 (Fig. 10.9d). The crossbar should be separated from the vertical members 4–5 and 1–3. When a section is passed across member 4–5 infinitely close to joint 1 from above, the moment 3i4–5 is applied. Since the extended fibers are located to the right of member 4–5, then moment 3i4–5 is directed clockwise. Moment 3i4–5 is equilibrated by two forces, 3i4–5/l4–5 ¼ 0.333EI; (the force at support 5 is not shown). Force 0.333EI is transmitted to the horizontal member and has an opposite direction. Member 1–3 should be considered in a similar way. It is necessary to draw section infinitely close to joint 1 from below, then apply two bending moments 4i1–3 at the top of the member and 2i1–3 at the bottom of the member (the direction of the moments should correspond to the location of the extended fibers) and then equilibrate them by two forces, 6i1–3/l1–3 ¼ 0.24EI. As the last step, this force is transmitted to the horizontal member in the opposite direction. The canonical equations of the displacement method become: 11:1933 ¼0 EI 5 0:093Z 1 þ 0:207Z 2 ¼ 0 EI 2:4Z 1 þ 0:093Z 2
The roots of these equations present the primary unknowns and they are equal to Z1 ¼
3:794 ðradÞ, EI
Z2 ¼
22:450 ðm Þ EI
The final bending moment diagram is constructed using the principle of superposition:
ðaÞ
384
10
The Displacement Method
M P ¼ M 1 Z 1 þ M 2 Z 2 þ M 0P The corresponding calculations for the specified points (Fig. 10.9b) are presented in Table 10.3. The signs of the bending moments in the unit conditions are conventional.
Table 10.3 Calculation of bending moments
Points 1 2 3 4 5 6 7 8 factor
M1 -0.4 +0.2 +0.8 -1.0 0.0 -0.6 -0.36 0.0 EI
M 1 · Z1
-1.5176 +0.7588 +3.0352 -3.794 0.0 -2.2764 -1.3658 0.0
M2 +0.24 0.0 -0.24 -0.333 0.0 0.0 0.0 0.0 EI
M 2 · Z2
+5.388 0.0 -5.388 -7.476 0.0 0.0 0.0 0.0
M P0
M kNm
4
+4.1667 -2.0833 +4.1667 0.0 0.0 +15.36 -9.984 0.0
+8.037 -1.324 +1.814 -11.27 0.0 +13.084 -11.349 0.0
3
6
Signs of bending moments + + – –
The final bending moment diagram M is presented in Fig. 10.9g. The same diagram was obtained by the force method (Example 9.7). The reader is invited to compare these methods for the given design diagram. Construction of shear and axial force diagrams, as well as computation of the reactions, is described in detail in Example 9.7.
10.2.4.4
Summary
The presentation of the equation of the displacement method in canonical form is conveniently organized and also prescribes a well-defined algorithm for the analysis of complex structures. The special parts of structural analysis can be carried out more easily with the canonical form of the displacement method. These include the construction of influence lines (Chap. 12), Stability (Chap. 15) and Vibration (Chaps. 16 and 17).
10.3
Comparison of the Force and Displacement Methods
The force and displacement methods are the principal analytical methods in structural analysis. Both of these methods are widely used, not only for static analysis but also for stability and dynamical analysis. Below we will provide a comparison pertinent to the two methods’ presentation in their canonical forms. Both methods use the concept of the primary system. Both methods require construction of bending moment diagrams for unit exposures (forces or displacements). In both methods, a difference between the primary system and the original one is eliminated using the set of canonical equations. Fundamental differences between these methods are presented in Table 10.4. It can be easily seen that these methods are dual, i.e., one column of the table can be obtained from the other by linguistic restatement.
10.3
Comparison of the Force and Displacement Methods
385
Table 10.4 Fundamental differences between the force and displacement methods Comparison criteria Primary system (PS) Primary unknowns (PU) Number of PU Number of PS and way of obtaining PS
Canonical equations
Meaning of equations Character of canonical equations Matrix of coefficients of canonical equations
Meaning of unit coefficients Meaning of free terms Dimensions of unit coefficients
Force method Obtained by eliminating redundant constraints from a structure Uses forces (forces and moments), which simulate the actions of eliminated constraints. Reactions of eliminated constrains are PU. Equals to the degree of statical indeterminacy Nonunique. PS can be chosen so that all redundant constraints must be eliminated and replaced by corresponding reactions (forces and/or moments). Which redundant constraints should be eliminated is a matter of choice, but the obtained PS should be statically determinate.a δ11 X 1 þ δ12 X 2 þ . . . þ Δ1P ¼ 0 δ21 X 1 þ δ22 X 2 þ . . . þ Δ2P ¼ 0 : : : : : : : : : : Number of canonical equations equals to the number of PU Total displacement in the direction of eliminated constraints caused by the action of all primary unknowns (forces and/or moments) and applied forces is zero Kinematical: the left part of canonical equations represents displacements.
Displacement method Obtained by introducing additional constraints to a structure Uses displacements (linear and angular), which neutralize the actions of introduced constraints. Displacements of introduced constraints are PU. Equals to the degree of kinematical indeterminacy Unique. PS must be constructed so that in every rigid joint an additional constraint is introduced to prevent angular rotation; and for every independent linear displacement an additional constraint is introduced to prevent linear displacement. PS presents a set of standard statically-indeterminate beams. r 11 Z 1 þ r 12 Z 2 þ . . . þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ . . . þ R2P ¼ 0 : : : : : : : : : : Number of canonical equations equals to the number of PU Total reaction in the introduced constraints caused by the action of all primary unknowns (linear and/or angular displacements) and applied forces is zero Statical: the left part of canonical equations represents reactions.
3 δ11 δ12 δ1n 7 6δ 6 21 δ22 δ2n 7 A¼6 7, detA > 0 4 5 δn1 δn2 δnn where A is the flexibility matrix Unit displacement δik presents displacement in the direction of i-th eliminated constraints due to primary unknown (force) Xk ¼ 1 Displacement ΔiP presents displacement in the direction of i-th eliminated constraint due to applied forces δik—Dimension of displacement at i (linear or angular) is divided by dimension of action (force or moment) at k
3 r 11 r 12 r 1n 7 6r 6 21 r 22 r 2n 7 R¼6 7, detR > 0 4 5 r n1 r n2 r nn where R is the stiffness matrix Unit reaction rik presents reaction in the i-th introduced constraints due to primary unknown (displacement) Zk ¼ 1
2
2
Reaction RiP presents reaction in the i-th introduced constraint due to applied forces rik—Dimension of force at i (force or moment) is divided by dimension of action (linear or angular displacement) at k
a
We are providing only the classical approach; however, an experienced reader may choose to designate the primary system of the force method as statically indeterminate, providing that he/she has all the necessary formulas for calculating the accepted statically indeterminate primary system
Both methods are based on deep physical ideas. Despite the fundamental differences, the methods are based on the same assumptions, such as Hooke’s law, linear differential equation of the elastic line of the beam, small deformations, and superposition principle. Therefore, the application of these methods to the analysis of engineering structures leads to the same results. Obviously, the sphere of effective application of these methods is different and depends on the type of structure and its design diagram. The most important advantages of the fundamental methods: 1. The force and displacement methods allow performing an analysis of the structure on the basis of the single computational procedure, which excludes any iterative or repeated clarifying calculations. 2. Both the methods may be applied to the problems of analysis of deformable structures under the various influences, such as the loads, change of temperature, settlements of supports, and errors of fabrication. At the same time, the idea of canonical equations, the properties of the coefficients of the equations, and the procedure of their calculation remain unchanged. Minor modifications of the methods reflect the nature of the action and relate only to the calculation of the free terms of the canonical equations. 3. Both the methods may be applied for construction of the influence lines (Chap. 12). Both the methods have extensive and effective application to the problems of stability (Chap. 15), and to free and forced vibration (Chaps. 16 and 17) of the deformable structures.
386
10
The Displacement Method
10.3.1 Properties of Canonical Equations 1. The main coefficients of the canonical equations of the force and displacement methods are strictly positive: δii > 0; rii > 0. 2. The matrix of coefficients of the canonical equations is symmetrical with respect to the main diagonal: δik ¼ δki; rik ¼ rki. These coefficients may be positive, negative, or zero. 3. The coefficients of the canonical equations depend only on the type of structure, but do not depend on external loads, settlements of supports, temperature changes, or errors of fabrication. 4. The determinant of the matrix of coefficients of the canonical equations is strictly positive. This condition describes an internal property of structures based on a fundamental law of elastic systems: the potential energy of a structure subjected to any load is positive. Since D 6¼ 0, the solution to the canonical equations of any redundant structure, subjected to any load, change of temperature, or settlements of supports, is unique.
10.3.2 Variations of Design Diagrams and Choice of Methods of Analysis It is time to ask a question: when is it more convenient to apply the force method and when the displacement method? The general answer is the following: the more rigid the system due to given constraints, then the more efficient the displacement method will be. This can be illustrated by considering different structures. Figure 10.10a presents a frame with fixed supports. The frame, according the force method, has nine unknowns (FM-9), while by the displacement method only one unknown (DM-1): the angle of rotation of the only rigid joint. Analysis of this structure by the displacement method is a very simple problem. Should this structure be modified by adding more elements connected at the rigid joint, nevertheless, the number of unknowns by the displacement method is still the same, while the number of unknowns by the force method is increased with the addition of each new element. Another frame is shown in Fig. 10.10b. The number of unknowns by the force method is one, while by the displacement method it is six (four rigid joints and two linear independent displacements). Therefore the force method is more preferable.
a
c
b
e
f FM-4 DM-1
g
EI= ∞
P
DM-1
1 FM-3 DM-3
FM-6 DM-1
FM-3 DM-6
FM-1 DM-6
FM-9 DM-1
d
2 3
Fig. 10.10 Different types of structures that can be analyzed by either the force method (FM) or the displacement method (DM)
A statically indeterminate arch with fixed supports has three unknowns by the force method (Fig. 10.10c). In order to analyze this structure by the displacement method its curvilinear axis should be replaced by a set of straight members. One version of such segmentation of the arch is shown by dashed lines; in this case the number of unknowns by the displacement method is six.
10.4
Sidesway Frames with Absolutely Rigid Crossbars
387
Obviously, the analysis of the modified design diagram cannot be considered as precise one. However, if the arch is the part of any complex structure, as shown in Fig. 10.10d, and curvilinear part is not loaded, then such a structure can be easily analyzed by the displacement method in canonical form. Indeed, the number of unknowns by the force method is six, while by the displacement method it is just one (the angular displacement of the rigid joint). Detailed tables for parabolic uniform and nonuniform arches are presented in Tables A.17 and A.18. Figure 10.10e shows statically indeterminate beam. The number of unknowns by the force and displacement methods equals are four and one, respectively. It is obvious that the displacement method is more preferable in this case. Figure 10.10f shows statically indeterminate frame with an absolutely rigid crossbar. By the displacement method this structure has only one unknown, no matter how many vertical elements it has; analysis of frames of this type is considered in Sect. 10.4. Figure 10.10g presents a statically indeterminate frame and primary system of the displacement method. For this frame the number of unknown by the force and displacement method are same and equals three. This frame has the horizontal displacement and an inclined element. It is the combination of these two features that brings additional difficulties to the analysis of the frame. These difficulties are connected with the following fact: the unit horizontal displacement of the introduced constraint 3 leads to the same displacement of joint 1 in the horizontal direction. However, this displacement for an inclined bar is not tabulated, since it is not directed perpendicular to the axis of the inclined bar. This feature of the frame leads to additional cumbersome computations of the reactions in the introduced constraints. Therefore, for structures with similar features the force method is more effective than the displacement method.
10.4
Sidesway Frames with Absolutely Rigid Crossbars
So far it has been assumed that the each rigid joint of the frame can rotate. Such joint corresponds to one unknown of the displacement method. This section describes analysis of a special type of frame: sidesway frames with absolutely rigid crossbars (flexural stiffness EI ¼ 1). Let us consider the frame shown in Fig. 10.11a. The connections of the crossbar and the vertical members are rigid. A feature of this structure is that the crossbar is an absolutely rigid body; therefore, even if the joints are rigid, there are no angles of rotation of the rigid joints. Thus, the frame has only one primary unknown, i.e., the linear displacement Z1 of the crossbar. The primary system is shown in Fig. 10.11b. Introduced constraint 1 prevents horizontal displacement of the crossbar. Flexural stiffness per unit length is i ¼ EI/h. The canonical equation of the displacement method is r11Z1 þ R1P ¼ 0. The bending moment diagram caused by unit linear displacement Z1 is shown in Fig. 10.11c. Unit reaction r11 is calculated using the equilibrium equation for the crossbar. The bending moment for the clampedclamped strut due to lateral unit displacement is M ¼ 6i/h; therefore the shear force for vertical elements is Q ¼ 2M/h ¼ 12i/h2. This force is transmitted to the crossbar (Fig. 10.11d) and after that the unit reaction can be calculated as follows: r 11 !
X
X ¼ 3
12i 36i þ r 11 ¼ 0 ! r 11 ¼ 2 h2 h
Applied load P does not produce bending moments in the primary system. Nevertheless, the free term is not zero and should be calculated using the free-body diagram for the crossbar (Fig. 10.11e). Equilibrium condition ∑X ¼ 0 leads to the following result: R1P ¼ P. The primary unknown becomes Z1 ¼
R1P Ph3 ¼ r 11 36EI
The bending moment at each of the specified sections of the frame is calculated by the formula M P ¼ M 1 Z 1 þ M 0P Since applied load P does not produce bending moments in the primary system, then M 0P ¼ 0 and the resulting bending moments due to applied load P are determined as M P ¼ M 1 Z 1. The bending moment at the top and bottom of each vertical member is M0 ¼
6i Ph3 Ph ¼ h 36EI 6
388
10
The Displacement Method
b
a P
1
P
R1P
EI=∞ EI
c
EI
h
EI
6i h
i
Z1=1
d
Z1=1 1
1
r11 Q
Q M 6i h
M1
6i h
6i h
e
Q
Q
EI=∞
f
M0
P
M0
Ph 6
R0
M=6i/h
M
Q
Q=12i/h2
M=6i/h
R1P
M0 Deflection curve Inflection point
MP M0 =
Q
Q
1
P
Q
M
M
r11
M0
R0
M0
R0=P/3
Fig. 10.11 Frame with infinitely rigid crossbar: (a, b ) Design diagram and primary system. (c, d) Bending moment diagram in unit state and the free-body diagram of the crossbar. (e) Free-body diagram of the crossbar in loaded state. (f) Final bending moment diagram
The corresponding bending moment diagram and the reactions of supports are presented in Fig. 10.11f. Shear within all vertical members is Q¼
M0 þ M0 P ¼ 3 h
The reaction of all supports R0 ¼ Q ¼ P=3. Figure 10.11f shows the final deflection curve of the frame and inflection point of all vertical members. Note that a frame with an absolutely rigid crossbar has only one unknown of the displacement method for any number of vertical members.
10.5
Special Types of Exposures
The displacement method in canonical form can be effectively applied for analysis of statically indeterminate frames subjected to special types of exposures such as settlements of supports and errors of fabrication. For both types of problems a primary system of the displacement method should be constructed as usual.
10.5
Special Types of Exposures
389
10.5.1 Settlements of Supports For a structure with n degrees of kinematical indeterminacy, the canonical equations are r 11 Z 1 þ r 12 Z 2 þ þ r 1n Z n þ R1s ¼ 0 r 21 Z 1 þ r 22 Z 2 þ þ r 2n Z n þ R2s ¼ 0
ð10:6Þ
r n1 Z 1 þ r n2 Z 2 þ þ r 2n Z n þ Rns ¼ 0 where the free terms Rjs ( j ¼ 1,2, . . ., n) represent the reaction in the j-th introduced constraint in the primary system due to the settlements of supports. These terms are calculated using Tables A.3–A.7, taking into account the actual displacements of a support. Unit reactions rik should be calculated as before. The final bending moment diagram is constructed by the formula M s ¼ M 1 Z 1 þ M 2 Z 2 þ þ M n Z n þ M 0s
ð10:7Þ
where M 0s is the bending moment diagram in the primary system caused by the given settlements of supports. Example 10.4 The redundant frame in Fig. 10.12a is subjected to the following settlement of fixed support A: a ¼ 2 cm, b ¼ 1 cm, and φ ¼ 0.01 rad ¼ 340 3000 . This figure contains the numeration of the specified points 1–8 of the frame. Construct the bending moment diagram. (Note: this problem was considered early by the force method (Example 9.10). Solution The primary system of the displacement method is presented in Fig. 10.12b. The primary unknowns are the angle of rotation Z1 and linear displacement Z2. The canonical equations of the displacement method are r 11 Z 1 þ r 12 Z 2 þ R1s ¼ 0 r 21 Z 1 þ r 22 Z 2 þ R2s ¼ 0 The unit reactions rik were obtained early in Example 10.3; they are r 11 ¼ 2:4EI r 21 ¼ 0:093EI
ðkNm=radÞ, ðkN=radÞ,
r 12 ¼ 0:093EI
ðkNm=mÞ,
r 22 ¼ 0:207 ðkN=mÞ
Free terms R1s and R2s represent reactions in both introduced constraints 1 and 2 in the primary system due to the settlements of support A. They are calculated using the bending moment diagrams caused by the displacements a, b, and φ of support A. Diagrams Ma, Mb, and Mφ (Fig. 10.12c–e) present the bending moments in the primary system due to separate displacements of support A based on given values a ¼ 2 cm, b ¼ 1 cm, and φ ¼ 0.01 rad ¼ 340 3000 , respectively. In the case of the vertical a-displacement of support A the elastic curve is shown by a dashed line (Fig. 10.12c). It is obvious the vertical members 1–3 and 4–5 are not subjected to the bending deformation and the bending moments arise only within member 6–8. To construct the bending moment diagram Ma, we use Table A.3, row 2. The bending moment at point 6 equals 3i68 3 0:2EI 0:02 ¼ 0:0012EI. a¼ 10 l68 In the case of the horizontal b-displacement, only member 1–3 deforms (Fig. 10.12d); in order to construct bending moment diagram Mb, we use Table A.4, line 2. In the case of the angular, φ—displacement of support A, only member 1–3 of the frame in the primary system deforms (Fig. 10.12e). This happens because introduced constraint 1 prevents distribution of the deformation. Bending moment diagram Mφ is shown according to Table A.4, row 1. The values presented in Tables A.3 and A.4 correspond to unit displacements; therefore, in order to obtain the reactions according to the given displacements it is necessary to multiply the tabulated values by corresponding values a, b, and φ. The total bending moment diagrams Ma, Mb, and Mφ is presented in Fig. 10.12f. To calculate the free terms of canonical equations it is necessary to consider the free-body diagrams for the introduced constraint 1 (rigid joint) and for the horizontal element (Fig. 10.12g, h). The equilibrium condition of the rigid joint 1 yields R1s ¼ 0:0064EI þ 0:0012EI ¼ 0:0076EI
390
10
a
b
5 3m
i4-5= 0.333EI
8
4
A
a
j
b
2
i1-3= 0.2EI
5m
1
i6-8= 0.2EI
1
2 1
The Displacement Method
6
A a
3
j
b
10m
d
c
6i1-3 6 . 0.2 EI . b= 0.01 = 0.0024 EI l1-3 5
R1a
a
R2a
´
Ma
A
R2b
R1b
3i6-8 3. 0.2 EI . a= 0.02 = 0.0012 EI l6-8 10
A
Mb
0.0024EI b
e
f 2i1-3. j =2. 0.2EI. 0.01=0.004EI
(0.0024+0.004)EI=0.0064EI
R1j
R2 j
´
R2s
R1s 0.0012EI
Mj
j
(0.0024+0.008)EI=0.0104EI
A A
Ma +Mb +Mj
a
4i1-3. j =0.008EI
j
b
g
h 0.0012EI R1s
0.00336EI
1
M3-1=0.0064EI
6
8
R2s 0.00336EI
0.0064EI 0.00336EI
i
M1-3=0.0104EI
0.112
0.212 0.100 M kNm (factor 10-2EI)
0.454 Fig. 10.12 Settlements of support. (a, b) Design diagram and primary system. (c–e) Bending moment diagram Ma, Mb, and Mφ, in primary system caused by settlements a, b, φ of support A. (f) Total bending moment diagram Ma + Mb + Mφ. (g, h) Free-body diagrams for rigid joint 1 and crossbar. (i) Final bending moment diagram M
10.5
Special Types of Exposures
391
The shear force within portion 1–3 is equal to M 31 þ M 13 0:0064 þ 0:0104 EI ¼ 0:00336EI: ¼ 5 l13 These forces rotate the member counterclockwise. Force 0.00336EI is transmitted to horizontal member 6–8 in the opposite direction, i.e., from left to right. The equilibrium condition of the horizontal element is used to calculate R2s X X ¼ 0 ! R2s ¼ 0:00336EI, The canonical equations become 2:4Z 1 þ 0:093Z 2 þ 0:0076 ¼ 0 0:093Z 1 þ 0:207Z 2 0:00336 ¼ 0 The roots of these equations are Z 1 ¼ 0:38629 102 ðradÞ Z 2 ¼ 1:79674 102 ðmÞ The final bending moment diagram (Fig. 10.12i) is constructed by the following expression using the principle of superposition M s ¼ M 1 Z 1 þ M 2 Z 2 þ M 0s Here M 0s ¼ M a þ M b þ M φ is the bending moment in the primary system caused by the settlements of the support; this diagram is shown in Fig. 10.12f. The corresponding computation is presented in Table 10.5. Table 10.5 Calculation of bending moments Ms Z1 ¼ 0.38629 102 (rad), Z2 ¼ 1.79674 102 (m)
Points 1 3 4 5 6 8 Factor
M1 -0.4 +0.8 -1.0 0.0 -0.6 0.0 EI
M 1 ∙ Z1
0.15452 -0.30903 0.38629 0.0 0.23177 0.0 10-2 EI
M2 +0.24 -0.24 -0.3333 0.0 0.0 0.0 EI
M 2 ∙ Z2
0.43122 -0.43122 -0.59885 0.0 0.0 0.0 10-2 EI
Ma+MB+M φ
Ms
-1.040 +0.640 0.0 0.0 -0.120 0.0 10-2 EI
-0.45426 -0.10025 -0.21256 0.0 0.11177 0.0 10-2 EI
Signs of bending moments + + − −
The final bending moment diagram due to the settlements of the support is presented in Fig. 10.12i. This same diagram was obtained previously by the force method (Example 9.9); the discrepancy according to the force and displacement method for point 1 is approximately 0.05%.
Discussion 1. In case of the structure subjected to external loading, the internal force distribution depends only on the relative stiffness of the elements, while in case of the settlements of support, the distribution of internal forces depends on both the relative and the absolute stiffness of the elements. 2. In the case of settlements of a support, the calculation of free terms is an elementary procedure using the displacement method, while by the force method, free terms are more difficult to calculate. Therefore, in cases of settlement of supports, the displacement method is more preferable than the force method.
392
10
The Displacement Method
3. The structure of expressions for unit displacements δik and unit reactions rik is different: indeed, with increasing flexural rigidity EI, unit displacements decrease, and unit reactions increase. 4. The reader is invited to trace the units of the intermediate and final results of the analysis of the structure using the force and displacement methods. Consider separately the cases of a frame subjected to any external load and settlements of the supports.
10.5.2 Errors of Fabrication In cases of fabrication errors, analysis of the frame can be effectively performed by the displacement method in canonical form. Construction of the primary system and calculation of unit reactions should be performed as usual. Canonical equations should be presented in form (10.4), but the free terms RiP should be replaced by Rie. These free terms present the reactions of the introduced constraints caused by the errors of fabrication. Example 10.5 A design diagram of the frame is presented in Fig. 10.13a. Member AB has been fabricated Δ ¼ 0.8 cm too long. The moment of inertia of all cross sections is I ¼ 9.9895105 m4, and the modulus of elasticity is E ¼ 2 108 kN/m2 (EI ¼ 19979 kNm2). Calculate the angle of rotation of joint B and construct the bending moment diagram.
a
D A
b
1
B
EI
D
EI
h = 2.8m
C l = 5m
4 EI l
c
Z1 =1
d
D A
B
r11
2 EI l
3EI
3EI h
h2 M eo
M1
R1e D
C
26.1
e
* 13.05 M (kNm)
Fig. 10.13 Errors of fabrication. (a, b) Design diagram and primary system. (c, d) Bending moment diagrams for unit and loading states. (e) Resulting bending moment diagram
Solution The primary system of the displacement method is shown in Fig. 10.13b. The primary unknown is the angle of rotation of included constraint 1. The canonical equation of the displacement method is r11Z1 þ R1e ¼ 0. The primary unknown becomes Z 1 ¼ R1e =r 11 . The bending moment diagram in unit state is presented in Fig. 10.13c. The loaded state (Fig. 10.13d) presents Δ-displacement of joint B. The member AB is not subjected to bending; corresponding elastic curve is shown by a dashed
10.5
Special Types of Exposures
393
line. The moment R1e is a reactive moment at the introduced constraint caused by displacement Δ. Table A.3 presents the reactions caused by the unit displacement; therefore the specified ordinate of the bending moment diagram is (3EI/h2)Δ. It is obvious that r 11 ¼
4EI 3EI þ , l h
R1e ¼
3EI Δ h2
The negative sign means that adopted direction of R1e (clockwise) is opposite to the real direction, which can be determined by location of the extended fibers (they are shown by dashed line). The required angle of rotation of joint B is Z1 ¼
3 0:008ðmÞ 3Δ ¼ 0:001635ðradÞ ¼ 4 3 2:82 h2 þ 4 ð m Þ þ 3 2:8 ð m Þ l h 5
The resulting bending moments at the specified points of crossbar are 4EI 4EI Z ¼ 0:001635ðradÞ ¼ 0:001308EI ðkNmÞ; l 1 5:0ðmÞ ¼ 0:5M BA
M BA ¼ M AB
Coefficient 0.5 reflects the position of the focal point () on the member AB. The resulting bending moments at the specified point B of vertical member BC is 3EI 3EI Z 2 Δ h 1 h 3EI 3EI 0:001635ðradÞ ¼ 0:008ðmÞ ¼ 0:001309EI ðkNmÞ: 2:8ðmÞ 2:82 ðm2 Þ
M BC ¼
Here the first term is shown in the diagram (c) on the left side of the element and by convention is taken with positive sign. Therefore the negative sign of result means that this ordinate should be shown on the right side of the element as shown in diagram (e). For given stiffness EI ¼ 19979 kNm2 bending moment at specified points are M BA ¼ 0:001309 19979 ¼ 26:1ðkNmÞ; M AB ¼ 0:5M BA ¼ 13:05 ðkNmÞ The corresponding bending moment diagram is presented in Fig. 10.13e. It can be seen that joint B is in the equilibrium. Construction of shear and axial forces and computation of reactions of supports should be performed as usual.
Discussion We can see that an insignificant error of fabrication leads to significant internal forces in the structure. This fact lays the basis of the inverse problem, which can be formulated as follows: find the initial displacement of specified points of a structure for obtaining the required distribution of internal forces. For example, let us consider the two-span continuous beam subjected to a uniformly distributed load shown in Fig. 10.14a. q
a l
l
b D Fig. 10.14 Controlling of the stresses in the beam by initial displacement of middle support. (a) Design diagram of a beam subjected to load q. (b) Kinematical loading of the same beam for reducing of the bending moment at the intermediate support
394
10
The Displacement Method
In this case, the extended fibers at the intermediate support are located above the neutral line and the corresponding moment at the support is ql2/8 (Table A.10, row 1). If the external load is absent and the intermediate support is placed below the beam level on Δ, then the extended fibers at the support will be located below the neutral line and the corresponding moment at the support will be ð3EI=l2 ÞΔ (Table A.14, row 1). If beam is subjected to load q and displacement Δ simultaneously, then we can determine parameter Δ for the required distribution of internal forces. For example, what Δ should be in order that the bending moment at the intermediate support be zero, etc.
10.6
Analysis of Symmetrical Structures: Combined Method
Symmetrical structures are used very often in structural engineering. Symmetrical structures mean their geometrical symmetry, symmetry of supports, and stiffness symmetry of the members. For the analysis of symmetrical statically indeterminate frames subjected to any loads, the combined method (the combination of the force and displacement methods) can be effectively applied.
10.6.1 Symmetrical and Antisymmetrical Loading Combined method is based on the concept of resolving a total load on symmetrical and antisymmetrical components. Symmetrical frame subjected to horizontal load P is shown in Fig. 10.15a. This load may be presented as a sum of symmetrical and antisymmetrical components (Fig. 10.15b,c). In general, any load may be presented as the sum of symmetrical and antisymmetrical components.
a
b
P
P/2
Symmetrical load
P/2
Antisymmetrical load P/2
+
= Axis of symmetry (AS)
c
P/2
AS
AS
Fig. 10.15 Presentation of the load P as a sum of the symmetrical and antisymmetrical components
Symmetrical frame subjected to symmetrical and antisymmetrical components allows simplifying the entire design diagram. This simplification is based on the change of the entire design diagram by its equivalent half-frame (the physical basis of such changing will be discussed below). This procedure leads to the following fact: the degree of statical (or kinematical) indeterminacy for a half-frame is less than for an entire frame. Equivalent half-frames should be analyzed by classical methods. It will be shown below that for analysis of the half-frame, different methods (Force and Displacement methods) should be applied for symmetrical and antisymmetrical components. Therefore, the method under consideration is called the combined method.
10.6.2 Concept of Half-Structure In case of symmetrical structure subjected to symmetrical and antisymmetrical loads, the elastic curve at the point on the axis of symmetry (AS) has specific properties. These properties allow replacing an entire frame by two equivalent half-frames
10.6
Analysis of Symmetrical Structures: Combined Method
395
separately for symmetrical and antisymmetrical loading. In case of nonsymmetrical change of temperature or settlements of supports, these exposures could also be replaced by symmetrical and antisymmetrical components. At any section of a member the following internal forces arise: symmetrical unknowns, such as bending moment M and axial force N, and antisymmetrical unknown shear force Q (Fig. 10.16). Q M
M N
N
Q Fig. 10.16 Symmetrical and antisymmetrical internal forces
At the point of the axis of symmetry the following displacements arise: the vertical Δv, horizontal Δh, and angular φ. Considering these displacements at the AS (point A) allows us to construct an equivalent half-frame. Depending on loading (symmetrical or antisymmetrical) the different displacements will be at the point A and, as result, the different support conditions for equivalent half-frame. Let us consider the arbitrary symmetrical frames (Table 10.6). Elastic curves in the neighborhood of the point A on the axis of symmetry are shown by dotted line. Assume that number of spans is odd. In the case of symmetrical load, the horizontal and angular displacements at the point A are zeros, while the vertical displacement occurs. Therefore, an equivalent half-frame must contain a support at point A, which would model corresponding displacements, i.e., which would allow the vertical displacement and does not allow the horizontal and angular displacements. Only the slide support corresponds to these types of displacements. In the case of antisymmetrical load, the horizontal and angular displacements at the point A exist, while a vertical displacement is zero. It means that an equivalent half-frame at point A must contain a support, which would allow such displacements, i.e., the horizontal and angular displacements and does not allow a vertical one. Only roller support corresponds to these types of displacements. Table 10.6 also contains the mathematical conditions for replacing the given frame by its equivalent half-frame. In both cases, the support at the point A for equivalent half-frame simulates the displacement at the point A for entire frame. Now assume that the number of spans is even. It means that additional vertical member is placed on the axis of symmetry. In the case of symmetrical load, the horizontal and angular displacements at the point A, as in case of odd spans, are zero, while the vertical displacement of A is zero because of the vertical member at axis of symmetry. Therefore, an equivalent halfframe must contain a support at point A, which would model corresponding displacements. Only clamped support is related with displacements described above. In case of anti-symmetrical loading, the half-frame contains the member at the axis of symmetry with bending stiffness 0.5EI. Fundamental properties of internal force diagrams for symmetrical structures: 1. In case of symmetrical loading, the internal force diagrams for symmetrical internal forces (M, N ) are symmetrical, while for antisymmetrical unknown (Q) is antisymmetrical. 2. In case of antisymmetrical loading, the internal force diagrams for symmetrical unknowns (M, N) are antisymmetrical, and for antisymmetrical unknown (Q) is symmetrical. Table 10.6 also contains the number of unknowns for entire frame and for half-frame by force (FM) and displacement (DM) methods. The rational method is shown by bold. We can see that in case of symmetrical loading, the more effective is displacement method, while in the case of antisymmetrical loading, the more effective is the force method. From this table, we can see advantages of the combined method. Bending moment diagram is constructed for each case and then summated (on the basis of superposition principle) to obtain final bending moment diagram for original frame. The following procedure may be recommended for analysis of symmetrical structures: 1. 2. 3. 4.
Resolve the entire load into symmetrical and antisymmetrical components. Construct the equivalent half-frame for both types of loading. Provide the analysis of each half-frame using the most appropriate method. Find the final distribution of internal forces using superposition principle.
396
10
The Displacement Method
Table 10.6 Symmetric frame and corresponding half-frame for symmetric and antisymmetric loading
Number of spans
Symmetrical loading
Antisymmetrical loading
A
A
1,3,5,.. A Half-frame
Dv ¹ 0 Dh = 0 j =0
Number of unknowns Entire frame Half-frame FM: 9 FM: 5 DM: 9 DM: 4
Half-frame
A
Dv = 0 Dh ¹ 0 j ¹0
Number of unknowns Entire frame Half-frame FM: 9 FM: 4 DM: 9 DM: 5
A
A EI
2,4,6,… Half-frame
Internal force diagrams
A
Dv = 0 Dh = 0 j =0
Number of unknowns Entire frame Half-frame FM: 12 FM: 6 DM: 10 DM: 4 Bending moment diagram – symmetrical. Normal force diagram – symmetrical. Shear force diagram – antisymmetrical.
Half-frame
A 0.5EI
Dv = 0 Dh ¹ 0 j ¹0
Number of unknowns Entire frame Half-frame FM: 12 FM: 6 DM: 10 DM: 6 Bending moment diagram– antisymmetrical. Normal force diagram – antisymmetrical. Shear force diagram – symmetrical.
The principal stages of the combined method for analysis of the redundant symmetrical frame are illustrated below. The portal frame subjected to horizontal force P is shown in Fig. 10.17a. This structure has three unknowns of the force method, as well as the same number of unknowns of the displacement method. Entire load P is resolved into symmetrical and antisymmetrical components (P ¼ P/2). Corresponding design diagrams are shown in Figs. 10.17b and 10.17c, respectively. The corresponding deformation schemes are symmetrical and antisymmetrical. Each of these design diagrams are presented by the equivalent half-frame. For case of symmetrical load the half-frame is shown in Fig. 10.17d. This structure has one unknown of the displacement method. Corresponding primary system is shown in Fig. 10.17f. For case of antisymmetrical load the half-frame is shown in Fig. 10.17e. This structure has one unknown of the force method. Corresponding primary system is shown in Fig. 10.17g. Analysis of each half-frame in Figs. 10.17f, g may be performed by elementary way. Bending moment diagram for halfframe in case of symmetrical loading (displacement method, DM) should be constructed by formula M DM ¼ M Z 1 ¼1 Z 1 þ M 0PðDM Þ
Analysis of Symmetrical Structures: Combined Method
397
According to #1 of fundamental properties, the bending moment diagram caused by symmetrical loads, for entire frame (Fig. 10.17b), is symmetrical one. Bending moment diagram for half-frame in case of antisymmetrical loading (force method, FM) should be constructed by formula M FM ¼ M X 1 ¼1 X 1 þ M 0PðFM Þ According to #2 of fundamental properties, the bending moment diagram caused by antisymmetrical loads (Fig. 10.17c), for entire frame, is antisymmetrical one.
Symmetric loading
a
Antisymmetric loading
c
b
A
P
P
A
*
P*
P
A P*
*
Axis of symmetry
A
d Half-frame
P
Dv ¹ 0 Dh = 0 j =0
*
f Primary system
e
P
Z1
A Dv = 0 Dh ¹ 0 j ¹0
P*
g
A
*
A
P*
Canonical equation
r11Z1 + R1P = 0
Primary unknown
Z1 = - R1P r11
X1
d 11 X 1 + D1P = 0
X1 = - D1P d 11
h
MB
MD
a
P
h H
Fig. 10.17 Combined method: (a–c) Design diagram of symmetrical frame and resolving of entire load on symmetrical and antisymmetrical components. (d, f) Half-frame for symmetrical loading and primary system of displacement method. (e, g) Half-frame for antisymmetrical loading and primary system of force method. (h) Final bending moment diagram
The final bending moment diagram (Fig. 10.17h) for given load P according to the superposition principle presents the sum of two bending moment diagrams shown in Fig. 10.17b, c. Having the bending moment diagram for entire structure, it is easy to construct the shear and axial force diagrams for entire structure and after that compute the reactions of support. Note, the final internal force diagrams neither symmetrical nor antisymmetrical ones. Final formulas for this problem may be found in Problem 10.21.
398
10
The Displacement Method
Problems Problems 10.1 through 10.6 provide complete analysis by the displacement method, including the following: 1. 2. 3. 4.
Determine the bending moment at the support and construct the internal force diagrams. Calculate the reactions of the supports and provide static control. Provide kinematical control (check the slope and vertical displacement at the intermediate support). Compute the vertical displacement at the specific points and show the elastic curve.
10.1.
The two-span uniform beam with two fixed ends is subjected to fixed load (Fig. P10.1). Provide complete analysis. P
Ans. M 1 = -
1 l
Pl . 16
0.5l 0.5l
Fig. P10.1
10.2. The two-span uniform beams with pinned and rolled supports are subjected to uniformly distributed load in the left span (Fig. P10.2a), and lumped force P applied in the right span of the beam (Fig. P10.2b). Provide complete analysis for each design diagram. For case (b) construct the influence line for M1. q
a
P
b
1
1
Ans. (a) M1 = -
C l
l
l
ul
ul
(b) M 1 = -
(
ql 2 16
;
)
Pl u 1 -u 2 . 4
Fig. P10.2
10.3. The two-span uniform beams with fixed left and rolled right support are subjected to uniformly distributed load q (Fig. P10.3a), and lumped force P (Fig. P10.3b). Provide complete analysis for each design diagram.
a
Ans.
q 1
0 l
P
b l
ql 2 ql 2 , M1 = ; 28 14 2 (b) M 1 = - Plu 1 - u2 7
(a) M 0 =
1 l
ul
(
ul
)
Fig. P10.3
10.4. The two-span nonuniform beam with fixed ends is loaded as shown in Fig. P10.4. Parameter n is any positive number. Construct the bending moment and shear force diagrams. Show the elastic curve.
0
1 EI l
Fig. P10.4
2 nEI l
Ans. M1 =
ql 2 , M 0 = - 0.5M1 , 12 (1+ n )
M2 = -
ql 2 . 2 + 3n 24 1 + n
Problems
399
10.5. Design diagram of a two-span uniform beam is shown in Fig. P10.5. Calculate the slope at support 1 and compute the bending moments at the supports. F=1kN
P=12kN
q=2kN/m 1
0
10.616 rad EI M 0 = 8.013kNm
2
Ans. Z1 =
EI=const a=6m l1=8m
b=4m
c=2m
M1 = 15.975kNm
l2=10m
Fig. P10.5
10.6. The two-span uniform beam with fixed and pinned end supports is subjected to vertical settlements of support, as shown in Fig. P10.6. Provide complete analysis Ans. M1 = -
1
12 EI D. 7 l2
D l
l Fig. P10.6
10.7. The two-span uniform beam with fixed ends is subjected to uniformly distributed load q and vertical settlements of the intermediate support, as shown in Fig. P10.7. Define the value of the settlement in order that: (a) the bending moment at the intermediate support would be equal to zero; (b) the bending moment at the fixed supports would be equal to zero; (c) the bending moment at the midpoint of the span would be equal to zero. q 1
D
Ans. (a) D =
l
l
ql 4 ¯ 72EI
()
Fig. P10.7
10.8. Design diagram of a frame is presented in Fig. P10.8. Supports 2, 3, and 4 are fixed, sliding, and rolled, respectively. Bending stiffnesses are EI and 2EI for vertical and horizontal members, respectively; their relative stiffnesses are shown in the circles. Compute the angle of rotation of joint 1. Construct the internal force diagrams; q ¼ 3kN/m
q=3kN/m 3
1
2
2
Fig. P10.8
8m
24 (rad ), M13 = 58 ( kNm ) EI
4
Z1 = -
6m
M 31 = - 38 ( kNm) , M14 = 42 ( kNm)
2
1 8m
Ans.
400
10
The Displacement Method
10.9. The frame is subjected to uniformly distributed load q, as shown in Fig. P10.9. The stiffness of the right elastic support is k (kN/m). The relative stiffnesses for all members are shown in the circles. Calculate the angle of rotation of joint 1. Construct the bending moment diagram. Check the limiting cases k¼1 and k¼0. Trace the elastic curves for both limiting cases. Is it possible to find a stiffness of elastic support such that rigid joint 1 would not rotate under the given load?
q 1
2
Ans. Z1 = 2
k
a= 6m
1 8m
12q . 24a - 32 3 EI 11 + 9a
1 1 + 6 EI 512k
,
.
8m
Fig. P10.9
10.10. Design diagrams for four Г-frames with different boundary conditions at support 3 are shown in Figs. P10.10a–d. All frames are subjected to uniformly distributed load q. For all frames the span is l and height h ¼ l. The flexural rigidity is EI for all members. Determine the primary unknowns of the displacement method for each case (note that for case (d) structure has two unknowns of the displacement method). Construct the internal force diagrams. Trace the elastic curves and show the inflection points. Explain the differences in structural behavior caused by the different types of supports 3.
a
q 1
3
c
q
b
1
1 3
3
d
q
Ans. (a) Z1 = -
q 1
3
b) Z1 = h
l
2
2
2 l
l
2
(c)
ql 3 96EI
ql 3 15EI
Z1 = -
l
;
;
ql 3 56EI
Fig. P10.10
10.11. The sidesway frame is subjected to horizontal load P along the crossbar (Fig. P10.11). The connections of the crossbar with the left and right vertical members are hinged while the connection with the intermediate vertical member is rigid. The bending stiffness of all vertical members is EI and for the crossbar it is 2EI. Will the axial forces in the crossbar be equal to zero or not? Verify your answer by constructing the internal force diagrams. Trace the elastic curve of the frame. Calculate the reactions of supports. Provide static and kinematical verification. 5 2 1 2
1 l=6m Fig. P10.11
Ans. Q1 = Q3 = + 0.1945 P ,
6
3 6m
h=5m
4
P
Q2 = + 0.611P
Problems
401
10.12. The structure shown in Fig. P10.12 contains a nonuniform parabolic arch. The moment of inertia at the arbitrary section of the arch is Ix ¼ IC / cos α, where IC is the moment of inertia of the cross section at the crown C, and α is the angle between the horizontal line and the tangent at any section of the arch. Uniformly distributed load q ¼ 4kN/m is placed within the left half-span of the arch. Construct the bending moment diagram. Solve this problem in two versions: (a) The axial forces in the cross section of the arch are ignored and (b) the axial forces in the cross section of the arch are taken into account. The moment of inertia and the area of the cross section at the crown C are IC ¼ 894 106 mm4, AC ¼ 33,400 mm2. Hint: The reader can find all the required data for parabolic arches as standard members in Appendix, Table A17 (uniform arches) and Table A18 (nonuniform arches). Compare the results obtained by each version. q
IC=2EI
a
C
1
A
D
8.1 (rad ) , M1A = 8.1kNm , EI M1B = 4.05kNm, M 1C = 12.15kNm
Ans. (a). Z1 =
4m
Ix 3EI
(b). Z1 =
8m
EI
8.4996 (rad ), EI
B
9m
18m
Fig. P10.12
10.13. The frame with an infinitely rigid crossbar is subjected to horizontal load P. The connections between the vertical members and the crossbar are hinged (Fig. P10.13). Calculate the support moments and the reactions. Construct the internal force diagram. Show the elastic curve.
EI=¥
P
Ans. M 0 = Ph 3 , R0 = P 3 . EI
EI
EI
h
Fig. P10.13
10.14. Design diagram of a frame with columns having a step-variable cross section is presented in Fig. P10.14. The frame is subjected to uniformly distributed load q ¼ 5 kN/m within the first span. Construct the bending moment diagram using the displacement method
C
7m
H=10m
3m
B
EI
4EI
2EI
Ans. M A = 12.41kNm, M B = 21.02 . M D = 8.83, M CB = 47.24 ,
2EI
M CD = 14.95, M CE = 32.29
D
A l=12m Fig. P10.14
EI
E 4EI
l=12m
.
402
10
The Displacement Method
10.15. Design diagram of a two-story frame is presented in Fig. P10.15. The crossbars are absolutely rigid. Connections between the vertical members and the crossbars are rigid. Construct the internal force diagrams. Determine the reactions of support. EI=¥
P
EI
EI EI= ¥
2P
Ans. QA = QB = 1.5P h
2.5Ph l
N A = - NB = M A = 0.75 Ph
2EI
2EI
h
B
A l Fig. P10.15
10.16a–d. Resolve the given load into symmetrical and antisymmetrical components. Axis of symmetry (AS) is shown by dotted line (Fig. P10.16)
a
d
c
b
q
q2
q1
q
q
Fig. P10.16
10.17. The symmetrical portal frame is loaded by the force P, as shown in Fig. P10.17. Resolve the given load into symmetrical and antisymmetrical components and calculate the bending moments at specified points for corresponding loading. Calculate the horizontal reaction H. Determine the horizontal displacement of the crossbar and show the elastic curve.
EI2
B
Ans. H =
D
P EI1 A
h
AS C
H l
Fig. P10.17
H
P , 2
MB =
3k Ph 2 (1 + 6k )
MA =
1 + 3k I h Ph; k = 2 . . 2 (1+ 6k ) I1 l
Problems
403
10.18. The portal frame in Fig. P10.18 is subjected to horizontal uniformly distributed load q. Construct the bending moment diagram by the combined method. Determine the horizontal displacement of the crossbar and show the elastic curve. EI ¼ constant for all members; l ¼ 4 m, h ¼ 8 m. D
B q
Ans. M Asym = 3.335q; M Bsym = 1.335q h
M Ctotal = 7.746 q; M Dtotal = 6.257 q
C
A l Fig. P10.18
10.19. Design diagram of the uniform symmetrical portal frame and a sketch of the bending moment diagram are shown in Fig. P10.19. Compute the bending moments at joints B and D, the bending moment and horizontal reaction at support C. Apply the combined method. Check whether the sketch of the bending moment diagram is correct. Show the elastic curve and points of inflections. Specify the direction of the joints rotation.
B
MD
I2
a
P
MB
D
I1
h
I1
P
l C
Fig. P10.19
±
+
+
HC
Chapter 11
Mixed Method
This chapter is devoted to analysis of special types of structures. They are the frames with different flexibility ranges of their separate parts. These structures have specific properties which allow simplify their analysis.
11.1
Fundamental Idea of the Mixed Method
11.1.1 General In engineering practice the statically indeterminate frames with specified features may be found: one part of a structure contains a small number of reactions and a large number of the rigid joints, while another part contains a large number of reactions and a small number of the rigid joints. Two examples of this type of structures are presented in Fig. 11.1a, b. Part 1 of a structure (a) contains only one support constraint (one vertical reaction at point A) and three rigid joints, while part 2 contains nine unknown reactions and two rigid joints. We can say that parts 1 and 2 are “soft” and “rigid,” respectively. The frame (b) may be considered also as a structure with different ranges of flexibility of their separate parts 1 and 2. Part 1 is soft while part 2 is rigid one.
b
a
1
2
1
2
Z2
c
Z3
A
X1
X1
d Z3
X2
Z4
X1
Fig. 11.1 (a, b) Design diagrams of frames with different ranges of flexibility of their separate parts; (c, d) Primary systems of the mixed method
For analysis of these types of structures the mixed method should be applied. In this method, some unknowns represent unknowns of the force method and some represent unknowns of the displacement method. The mixed method was introduced and developed by prof. A. Gvozdev in 1927 (Rabinovich 1954). © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_11
405
406
11 Mixed Method
11.1.2 Mixed Indeterminacy, Primary Unknowns, and Primary System It is convenient to apply the force method to the “soft” part of the structure and the displacement method to the “rigid” part of the structure. The primary system of the mixed method is obtained from a given structure by eliminating the redundant constraints in the “soft” part of a structure and introducing additional constraints at rigid joints in the “rigid” part of a structure (Fig. 11.1c, d). Thus, the primary unknowns of the mixed method are forces and displacements simultaneously. For scheme (a) the primary unknowns are the force X1 and displacements Z2 and Z3. For scheme (b) the primary unknowns of the force method are labeled as X1, X2, while the primary unknowns of the displacement method are Z3, Z4. A numbering of the unknowns is through sequence. For frames with different ranges of flexibility of their separate parts the number of unknowns of the mixed method is less than a number of unknowns of the force or displacement methods. This is an advantage of the mixed method. The mixed method may also be presented in canonical form.
11.2
Canonical Equations of the Mixed Method
Let us consider a frame shown in Fig. 11.2a; the flexural rigidity of vertical and horizontal elements are EI and 2EI, respectively. For given structure the number of unknowns by the force method equals 4, and the number of unknowns by the displacement method also equals 4 (angles of rotation of rigid joints 2, 4, 5 and linear displacement of the crossbar 4-5). Both parts of the frame have different ranges of flexibility: the right part is “rigid” and left part is “soft.” Indeed, the right part 1-2-3 has six constraints of supports at the points 1 and 3, while the left part 2-4-5-A of the frame has only one constraint support A. Therefore, for the left part of the frame it is convenient to apply the force method, while for right part the displacement method. The primary system of the mixed method is obtained from the given structure by eliminating the left vertical constraint A and introducing the additional constraint at rigid joint 2, simultaneously (Fig. 11.2b). Primary unknowns are X1 and Z2, where X1 is unknown of the force method and Z2 is unknown of the displacement method. Compatibility conditions for structure in Fig. 11.2 may be presented in canonical form δ11 X 1 þ δ012 Z 2 þ Δ1P ¼ 0 r 021 X 1 þ r 22 Z 2 þ R2P ¼ 0
ð11:1Þ
Thus, the mixed method for this particular design diagram reduces the number of unknowns to two. The first equation means that displacement in the direction of the eliminated constraint 1 must be zero. This displacement is caused by reaction of this constraint (primary unknown X1 of the force method), rotation of the introduced constraint 2 (primary unknown Z2 of the displacement method), and applied load. The second equation means that reaction in the introduced constraint 2 must be zero. This reaction is caused by reaction X1 of eliminated constraint 1, rotation Z2 of introduced joint 2, and applied load.
11.2.1 The Matter of Unit Coefficients and Canonical Equations Equations (11.1) contain the unit coefficients, which belong to four different groups, namely: δ11 represents a displacement due to the unit force δ012 represents a displacement due to the unit displacement r 021 represents a reaction due to the unit force r22 represents a reaction due to the unit displacement So, the coefficients δ11 and r22 are unit coefficients of the classical force and displacement method, respectively. The essence of the unit coefficients δ012 and r 021 is different from the coefficients δ11 and r22: coefficient δ012 is unit displacement caused by the unknown of the displacement method, and coefficient r 021 is unit reaction caused by the unknown of the force method. As before, the term “unit” means that the cause (force, displacement) is equal to unity.
11.2
Canonical Equations of the Mixed Method
407
a
b 5
4
2EI
q=6kN/m
2
4
EI
q 3
Primary system
1
1
A
12 m
Z2
2
3
2EI
8m
EI
b
4m
a)
8m
X1
c
d 12
Z2=1
r⬘21
12
M21= 0.5EI ⬘ d12
M1
d 11
M23= 1EI
M2
0.25EI
X1=1
q=6kN/m
f
q
e
3.2
M2-3=32
41.6
12.8 9.6
C R2P MC=16 Δ1P
0.5EI
r22
M P0
MC=20.8
kNm
MP kNm
A
4.8 3.2
g 12
12.8
12 16
20
9.6
M1 X1=1 Fig. 11.2 (a, b) Design diagram and primary system of the mixed method; (c, d) Bending moment diagrams caused by force X1 ¼ 1 and displacement Z2 ¼ 1; (e) Bending moment diagram in the primary system due to given load; (f) Resulting bending moment diagram and free body diagram for joint 2; (g) Unit bending moment diagram in primary system of the force method.
Each term of equations (Eq. 11.1) has the following meaning: First equation: δ11X1 ¼ displacement in the direction of the eliminated constraint (point A, vertical direction) caused by the unknown force X1 δ012 Z 2 ¼ displacement of point A in the same direction caused by the unknown angle of rotation Z2 Δ1P ¼ displacement of point A in the same direction caused by the applied load.
408
11 Mixed Method
Second equation: r 021 X 1 ¼ reaction of the introduced constraint 2 due to the unknown force X1 r22Z2 ¼ reaction of the same constraint 2 due to the unknown angle of rotation Z2 R2P ¼ reaction of the same constraint 2 due to the applied loads. Thus, the first equation (11.1) has the kinematical character because each term is related to displacement, while the second equation (11.1) has static character since each term is related to reaction.
11.2.2 Calculation of Coefficients and Free Terms The general properties of elastic structures and reciprocal theorems provide the following properties and relationships between coefficients of canonical equation of the mixed method δnn > 0,
r kk > 0,
δnm ¼ δmn ,
r nm ¼ r mn ,
r nm ¼ δmn
ð11:2Þ
Bending moment diagrams due to the primary unknowns X1 ¼ 1 and Z2 ¼ 1 are presented in Fig. 11.2c, d. Construction of bending moment diagram caused by primary unknown X1 ¼ 1 is obvious. For construction of bending moment diagram caused by primary unknown Z2 ¼ 1, we need to rotate introduced constraint 2 through angle Z2 ¼ 1. In this case, members of the right part only will sustain deformation, while left part will move as absolutely rigid body. Corresponding deformable position is shown by dashed line. Unit displacements (δ11, δ012 ) and unit reactions (r22, r 021 ) are shown in the corresponding diagrams. Ordinates of bending moment diagram for clamped-clamped beam due to the unit rotation of joint 2 are M 23 ¼
4EI 23 2EI ¼ 1:0EI, ¼4 8 l23
M 21 ¼
4EI 21 1EI ¼ 0:5EI: ¼4 8 l21
Having the unit bending moment diagrams due to X1 ¼ 1 and Z2 ¼ 1 we can calculate the coefficients and free terms of canonical equations of the mixed method. Multiplication of bending moment diagram M 1 by itself leads to coefficient of the force method Xð M M 1 12 12 2 1 864 1 1 δ11 ¼ 12 þ 12 4 12 ¼ ðm=kNÞ ds ¼ EI 2EI 2 3 1EI EI Equilibrium condition of introduced constraint 2 from bending moment diagram M 1 leads to r 021 ¼ 12 ðmÞ The units of coefficients of canonical equations of the mixed method can be defined as for the force and displacement methods. In general, for r 0ik , the unit of reaction which corresponds to the index i, should be divided by unit of factor, which 0 kNm corresponds to the index k. In our case we get r 21 ¼ ¼ ðmÞ. kN Equilibrium condition of introduced constraint 2 from bending moment diagram M 2 leads to coefficient of the displacement method r 22 ¼ 1EI þ 0:5EI ¼ 1:5EI ðkNm=radÞ Theorem of reciprocal displacements and reactions leads to δ012 ¼ r 021 ¼ 12 (m/rad). to the index i, should be divided by The unit of δ0ik is defined by following rule: the unit of displacement, whichcorresponds 0 m unit of factor, which corresponds to the index k. In our case we get δ21 ¼ . rad For calculation of free terms of canonical equations we need to construct the bending moment diagram in the primary system due to distributed load q; this diagram is presented in Fig. 11.2e. Specific ordinates of M 0P diagram are as follows: M 23 ¼ M 32 ¼
ql2 6 82 ¼ ¼ 32 ðkNmÞ, 12 12
From this diagram we get R2P ¼ 32 (kNm) and Δ1P ¼ 0. Canonical equations of the mixed method becomes:
MC ¼
ql2 ¼ 16 ðkNmÞ 24
11.2
Canonical Equations of the Mixed Method
409
864 X þ 12Z 2 ¼ 0 EI 1 12X 1 þ 1:5EIZ 2 32 ¼ 0 Roots of these equations are X 1 ¼ 0:266 ðkNÞ,
Z2 ¼
19:2 ðradÞ EI
11.2.3 Computation of Internal Forces Resulting bending moments acting at different cross sections of the frame are calculated by formula M P ¼ M 1 X 1 þ M 2 Z 2 þ M 0P
ð11:3Þ
Corresponding calculation for specified points is presented in Table 11.1. Resulting bending moment diagram is presented in Fig. 11.2f. Table 11.1 Calculation of bending moments; X 1 ¼ 0:266 ðkNÞ,
Z2 ¼
19:2 ðradÞ EI
MP
Points
M1
M 1 · Z1
1 2-1 2-3 3-2 2-4 4-2 4-5 5 C Factor
0.0 0.0 0.0 0.0 +12.0 +12.0 -12.0 0.0 0.0
0.0 0.0 0.0 0.0 -3.2 -3.2 +3.2 0.0 0.0
M2 -0.25 +0.5 -1.0 +0.5 0.0 0.0 0.0 0.0 -0.25
M 2 · Z2
-4.8 +9.6 -19.2 +9.6 0.0 0.0 0.0 0.0 -4.8
M P0
0.0 0.0 +32 +32 0.0 0.0 0.0 0.0 -16
kNm -4.8 +9.6 +12.8 +41.6 -3.2 -3.2 +3.2 0.0 -20.8
Signs of bending moments + –
+
–
EI
Verification. The static and kinematical verifications may be considered. They are: 1. Free-body diagram for joint 2 is presented in Fig. 11.2f. Direction of the moments corresponds to location of the extended fibers, which are shown by dotted lines. Equilibrium equation ∑M2 ¼ 3.2 + 9.6 12.8 ¼ 0 is satisfied. 2. Displacement in the direction of eliminated constraint A in the original system must be zero. This displacement may be computed by multiplication of two bending moment diagrams: one of them is the resultant bending moment diagram MP for the entire given structure and second is bending moment diagram caused by the X1 ¼ 1 in any primary system of the force method. One version of the primary system of the force method and corresponding unit bending moment diagram M 1 are presented in Fig. 11.2g. The vertical displacement of point A for entire structure equals X ð MP M1 1 1 2 1 ΔA ¼ 12 12 3:2 3:2 4 12þ ds ¼ 2EI 2 3 1EI EI 8 ð12:8 12 þ 4 20:8 16 41:6 20Þ ¼ 887:46 þ 887:47 ffi 0 6 2EI Construction of shear and axial force diagrams, computation of all reactions, and their verifications should be performed as usual.
410
11 Mixed Method
Problems 11.1. Describe the difference between the combined and mixed methods. Describe peculiarities of the mixed method: area of application (types of structures), primary unknowns and primary systems, canonical equations and their physical meaning, types of unit coefficients and their units. 11.2. Design diagram of the frame is shown in Fig. P11.2. 1. 2. 3. 4. 5. 6.
Determine the number of unknowns by the force method (FM), displacement (DM), and mixed (MM) methods. Show the primary system of the mixed method and setup of corresponding canonical equations. Explain the meaning of primary unknowns and canonical equations of the mixed method. Define the unit of coefficients and free term of canonical equations. Describe the way of calculation of all coefficients of canonical equations and free terms. Explain the advantage of the mixed method with comparison of the force and displacement methods.
Ans. FM-11 unknowns; DM-9; MM-4.
P
Fig. P11.2
11.3. Design diagrams of the nonsymmetrical frames are presented in Fig. P11.3a–e. Choose the best method (force, displacement or mixed) of analysis for each frame. Explain your decision. Choose primary unknowns, show primary system, write the canonical equations, and trace way for calculation of coefficients and free terms.
a
d
Fig. P11.3
c
b
e
Ans. (a) FM-1; (b) DM-1; (c) FM-2; (d) MM-2; (e) MM-6; DM-6.
Problems
411
11.4. The frame is shown in Fig. P11.4. The bending stiffness for each member is EI . Construct the bending moment diagram and perform its static and kinematical verifications. Hint: primary unknowns are the reaction X1 of support 1 and the angle of rotation Z2 of joint 2. Fig. 11.4
P 6m 3m
1
2 5m 4m
4m
11.5. Construct the bending moment diagram for structure in Fig. P11.5. Perform the static and kinematical control of resulting bending moment diagram. Hint: primary unknowns are the reaction X1 of support A and the angle of rotation Z2 of joint D.
R
6m
1EI q=6 kN/m
Ans. MB=4.84 kNm,
D 1EI
1EI
A
MC = 41.75 kNm. C
8m
2EI
B 2R=12m
8m
Fig. 11.5
11.6.
Describe the difference between combined method (Sect. 10.6.1) and mixed one.
Chapter 12
Influence Lines Method
This chapter is devoted to construction of influence lines for different statically indeterminate structures. Among them are continuous beams, frames, nonuniform arches, and trusses. Analytical methods based on the force and the displacement methods are applied. Also, kinematical method of construction of influence lines is discussed. This method allows tracing the models of influence lines.
12.1
Construction of Influence Lines by the Force Method
This paragraph contains the fundamental concepts for construction of influence lines of the statically indeterminate structures by the force method. The procedure for their const//ppdys-fs02/Springer_I/WOMAT/Production/ArtFinal/0000000053/ 188856_2_En/978-3-030-44393-1/0004792791/188856_2_En_12_Fig23_Print.epsruction is illustrated in detail for a continuous beam, parabolic nonuniform hingeless arch, and externally redundant truss. Also, combined approach (using moving and fixed load) is shown.
12.1.1 General Let us consider a first-degree statically indeterminate structure. In case of a fixed load, the canonical equation of the force method is δ11 X 1 þ Δ1P ¼ 0
!
X1 ¼
Δ1P δ11
ð12:1Þ
In this equation, the free term Δ1P represents displacement caused by given fixed load. Since moving load is unit one, then let us replace free term Δ1P by the unit free term δ1P; this free term presents a displacement in primary system in the direction of X1 caused by load P ¼ 1. Then the primary unknown X1 ¼
δ1P δ11
Unit displacement δ11 presents displacement caused by primary unknown X1 ¼ 1. Therefore, δ11 is some number, which depends on the type of a structure, its parameters, and chosen primary system; it does not depend on the load. However, δ1P depends on unit load location. Since load P ¼ 1 is traveling, then δ1P becomes a function of position of this load, and as a result, the primary unknown becomes a function as well: ILðX 1 Þ ¼
1 ILðδ1P Þ δ11
ð12:2Þ
Influence line for bending moment. In case of fixed load, a bending moment at any section k equals M k ¼ M k X 1 þ M 0k
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_12
ð12:3Þ
413
414
12
Influence Lines Method
where X1 is the primary unknown of the force method; M k is the bending moment at section k in a primary system due to unit primary unknown X ¼ 1, and M 0k is the bending moment at section k in a primary system due to given load. Now we need to transform (12.3) for the case of moving load. The bending moment M k at any section k presents the number, because this moment is caused by the unit primary unknown X1; the second multiplier X1 according to (12.2) presents a function. The last component, the bending moment M 0k , is caused by the given load, which is considered as moving load now; therefore, the bending moment M 0k also becomes a function of the position of this load. As a result, the bending moment at any section k becomes a function: ILðM k Þ ¼ M k ILðX 1 Þ þ IL M 0k
ð12:4Þ
Influence line for shear force. In case of fixed load, the shear at any section k is a number Qk ¼ Qk X 1 þ Q0k
ð12:5Þ
Similarly in case of traveling load, the shear at any section k becomes a function; therefore ILðQk Þ ¼ Qk ILðX 1 Þ þ IL Q0k
ð12:6Þ
For the n-times statically indeterminate structure the canonical equations of the force method in case of a fixed load P ¼ 1 are δ11 X 1 þ δ12 X 2 þ þ δ1n X n þ δ1P ¼ 0
ð12:7Þ
δn1 X 1 þ δn2 X 2 þ þ δnn X n þ δnP ¼ 0 Unit displacements δik which are caused by unit primary unknowns present numbers. Displacements δiP are caused by unit moving load. Fundamental feature of system (12.7) is that the free terms δiP are some functions of position of unit load P ¼ 1. Therefore, a solution of the system (12.7) leads to the primary unknowns Xi as the functions of the load position, in fact, to IL(Xi), i ¼ 1,. . ., n. In this case the bending moment at the any specified section k becomes function, so influence line should be constructed by formula ILðM k Þ ¼ M k1 ILðX 1 Þ þ M k2 ILðX 2 Þ þ þ IL M 0k
ð12:8Þ
Influence lines expressions for shear and axial force at any section may be constructed similarly. In case of statically indeterminate truss of the first degree of redundancy, the expression for influence line of internal force, which is induced at any member k of the truss, may be constructed by formula ILðSk Þ ¼ Sk ILðX 1 Þ þ IL S0k ð12:9Þ where Sk is the internal force in k-th member of a truss caused by the primary unknown X1 ¼ 1 in the primary system and IL S0k is influence line for internal force in k-th member of a truss in the primary system. Construction of influence lines for internal forces in statically indeterminate structures includes important step-construction of influence lines for primary unknowns IL(X1), IL(X2), . . . . Influence lines for reactions and internal forces may be constructed by the force method using the following procedure: 1. Adopt the primary unknowns and primary system. For continuous beams it is recommended to choose a primary system as a set of simply supported beams. 2. Write the canonical equations of the force method. 3. Compute the coefficients δik of canonical equations. These unit displacements present some numbers. 4. Compute the free terms δiP of canonical equations. These load terms present the displacement in the direction of i-th primary unknown due to the unit moving load and therefore δiP presents some functions. 5. Solve the canonical equations and construct the influence lines for primary unknowns. 6. Construct the influence lines for reactions and internal forces at specified section. Construction of influences line for some redundant structures are presented below. Among them are continuous beam, hingeless nonuniform parabolic arch, and truss.
12.1
Construction of Influence Lines by the Force Method
415
12.1.2 Continuous Beams Let us consider analysis of continuous uniform two-span beam ABC with equal spans l. Analysis includes construction of influence lines for primary unknown, internal forces at specified section k (Fig. 12.1a), and application of these influence lines for analysis of the beam subjected to fixed loads.
Fig. 12.1 (a, b) Design diagram and primary system; (c) Bending moment diagram in primary system due to X1 ¼ 1; (d, e) Location of the load P ¼ 1 in the left and right span, elastic curve, corresponding expression for loaded term δ1P, and influence line for δ1P; (f) Influence line for primary unknown X1
416
12
12.1.2.1
Influence Lines Method
Primary System
The structure is the statically indeterminate of the first degree. The primary system of the force method presents two simply supported beams (Fig. 12.1b). The primary unknown X1 is the bending moment at the middle support.
12.1.2.2
Influence Line for Primary Unknown X1
1 ILðδ1P Þ ,where δ11 is a δ11 mutual angle of rotation in direction of X1 due to primary unknown X1 ¼ 1 and δ1P is a slope at the middle support in primary system due to traveling load P ¼ 1.
Equation of influence line for primary unknown is presented by formula (12.2), ILðX 1 Þ ¼
1. For calculation of unit displacement δ11 we need to construct the bending moment diagram M 1 in primary system due to primary unknown X1 ¼ 1 (Fig. 12.1c). Graph multiplication method leads to the following result: δ11 ¼
M1 M1 1 2 1 2l ¼2 l1 1 ¼ 2 3 EI 3EI EI
2. For calculation of displacement δ1P we need to place the moving load at the left and right spans as shown in Fig. 12.1d. Let the load P ¼ 1 be located within the left span of the primary system. The angle of rotation δ1P at the right support B of the simply supported beam is presented in terms of dimensionless parameter u, which defines the position of the load. If the load P ¼ 1 travels along the right span, then the angle of rotation δ1P at the left support of the simply supported beam (i.e., the same support B) is presented in term of dimensionless parameter υ. Expressions for δ1P in terms of parameters u and υ are taken from Table A.15, pinned-pinned beam. Each span is subdivided into five equal portions and the angle of rotation δ1P is calculated for each point (u ¼ 0.0, 0.2, 0.4, 0.6, 0.8, 1.0) (Table 12.1). Parameter u is reckoned from the left support of each span, parameters u and υ satisfy the following condition: u + υ ¼ 1. Corresponding function of δ1P , i.e., the influence line for δ1P , is shown in Fig. 12.1e. Table 12.1 Calculation of δ1P (factor l2/EI) and primary unknown X1 (factor l )
P=1 in the left span Point
Parameter u
1(A) 2 3(k) 4 5 6(B) Factor
0.0 0.2 0.4 0.6 0.8 1.0
P=1 in the right span u -u3 l 2 6 EI 0.0 0.032 0.056 0.064 0.048 0.00 l 2 EI
d 1P =
IL(X1)
Point
Parameter
0.0 -0.048 -0.084 -0.096 -0.072 0.0 l
6(B) 7 8 9 10 11(C)
1.0 0.8 0.6 0.4 0.2 0.0
d 1P =
- 3 l2 6 EI 0.0 0.048 0.064 0.056 0.032 0.0 2 l EI
IL(X1) 0.0 -0.072 -0.096 -0.084 -0.048 0.0 l
12.1
Construction of Influence Lines by the Force Method
417
3. Influence line of primary unknown X1 is obtained by dividing the ordinates of influence line for δ1P by δ11 ¼ 2l/3EI and result should be taken with negative sign. For example, for point 2 ordinate of influence line X1 equals δ 0:032 l ¼ 0:048l (Table 12.1). Corresponding influence line for X1 is presented in Fig. 12.1f; all X ¼ 1P ¼ δ11 2=3 ordinates must be multiplied by parameter l. Thus, for any location of the load P the bending moment at support B is negative. It means the extended fibers in vicinity of support B are located above the neutral line.
12.1.2.3
Influence Line for Bending Moment Mk
This influence line should be constructed using formula (12.4). Bending moment at section k in the primary system due to primary unknown X1 ¼ 1 is M k ¼ 0:4 (Fig. 12.1c), therefore ð12:10Þ ILðM k Þ ¼ 0:4 ILðX 1 Þ þ IL M 0k Construction of IL(Mk) step by step is presented in Fig. 12.2.
6
7
8
9
10
0.0192
5
0.0384
4
0.0288
3
0.0288
2
0.0384
1
0.0336
P=1
a
11
0.0336
0.0192
k
–
b
–
Load P=1 in the left span
0.4 IL(X1) (factor l)
Load P=1 in the right span
Mk0=0 (see primary system in Fig. 12.1.b)
Inf. line Mk0 (factor l)
0.08
0.16
0.24
0.12
+
( ) ( )
( )
c
0.0512
0.1216
0.2064
0.1008
–
0.0192
0.0336
+
0.0384
0.0288
IL M k = 0.4 IL X 1 + IL M k0
Inf. line Mk (factor l)
Fig. 12.2 Construction of influence line for bending moment at section k. (a, b) The first and second terms of (12.10); (c) Final influence line Mk
The first term 0.4 IL(X1) of (12.10) is presented in Fig. 12.2a. The second term IL M 0k is the influence line of bending moment at section k in the primary system. If load P ¼ 1 is located in the left span, then influence line presents the triangle with maximum ordinate at the section k; this ordinate equals to ab 0:4l 0:6l ¼ ¼ 0:24l: l l
418
12
Influence Lines Method
If load P ¼ 1 is located in the right span, then the bending moment at section k does not arise, and the influence line has zero ordinates. It happens because the primary system two separate beams. Influence line for M 0k is shown in Fig. 12.2b. presents 0 Summation of two graphs 0.4IL(X1) and IL M k leads to the final influence line for bending moment at section k; this influence line is presented in Fig. 12.2c. The maximum bending moment at the section k occurs when the load P is located at the same section. The positive sign means that if a load is located within the left span, then extended fibers at the section k will be located below the neutral line.
12.1.2.4
Influence Line for Shear Force Qk
This influence line should be constructed using formula (12.6). Reaction at support A in primary system caused by primary unknown X1 ¼ 1 equals RA ¼ 1=lð"Þ (Fig. 12.1c), so the shear in primary system at section k due to primary unknown X1 ¼ 1 is Qk ¼ RA ¼ 1=l. Therefore 1 ILðQk Þ ¼ ILðX 1 Þ þ IL Q0k l
ð12:11Þ
Construction of IL(Qk) step by step is presented in Fig. 12.3.
5
6
7
8
9
10
0.048
4
0.084
3
0.096
2
0.072
1
0.072
P=1
0.096
a
11
0.084
0.048
k
1 IL ( X1 ) l
–
0.6
–
0.2
0.4
Inf. line Qk0
–
0.480
0.084
0.072 0.128
0.484
–
0.304
+
0.096
1 IL(Qk ) = IL( X 1 ) + IL(Qk0 ) l
0.516
0.248
c
0.4
+
0.2
b
Inf. line Qk
Fig. 12.3 Construction of influence lines for shear force at section k. (a, b) The first and second terms of (12.11); (c) Final influence line Qk
12.1
Construction of Influence Lines by the Force Method
419
The first term of (12.11) is the influence line for X1 scaled by 1/l, so ordinates of this graph (Fig. 12.3a) are dimensionless. The second term IL Q0k is the influence line of shear force at section k in the primary system. If load P ¼ 1 is located in the left span, then influence line of shear force for simply supported beam is shown in Fig. 12.3b. If load P ¼ 1 is located in the right span, then shear at section k does not arise, and influence line has zero ordinates. 1 Summation of two graphs ILðX 1 Þ and IL Q0k leads to the required influence line for shear force at section k; this influence l line is presented in Fig. 12.3c. It can be seen the following important property of influence line for Qk: the sum of the absolute values of the ordinates at the left and right of section k is equal to unity. Indeed, 0.516 (0.484) ¼ 1. Discussion Obtained influence lines allow calculating bending moment and shear for any section of the beam. Let P ¼ 10 kN be located at section 9. In this case, the moment M k ¼ 0:0336 l 10 ¼ 0:336 l ðkNmÞ It means, that reaction of the left support is RA ¼ M k =ð0:4lÞ ¼ 0:336 l=ð0:4lÞ ¼ 0:84 ðkNÞ. Negative sign shows that reaction is directed downward. Now construction of bending moment and shear diagrams has no problem. For example, the shear is Q1 ¼ . . . ¼ Qleft 6 ¼ 0:84 ðkN Þ. The bending moments at specified points are M 2 ¼ 0:84 0:2l ¼ 0:168 ðkNmÞ, :: . . . , M 5 ¼ 0:84 0:8l ¼ 0:672 ðkNmÞ It is obvious that same results may be obtained using influence line for X1. If load P ¼ 10kN is located at section 9, then moment at support B is X1 ¼ 0.84l. Negative sign indicates that extended fibers in vicinity of support B are located above the neutral line, i.e., this moment acts on support B of the left span clockwise and on the same support B of the right span in counterclockwise direction. In this case the reaction of support A equals RA ¼ X 1 =l ¼ 0:84 ðkNÞ. Influence line for primary unknown X1 should be treated as a fundamental characteristic for given structure. This influence line allows us to construct the bending moment diagram in case of any fixed load. Example 12.1 The uniform continuous beam with two equal spans l ¼ 10m is loaded by two forces P1 ¼ 10 kN and P2 ¼ 20 kN, which act at points 2 and 4 (Fig. 12.4). It is necessary to construct the bending moment diagram. Use the influence line for bending moment X1 at the middle support B (Fig. 12.1f). P1=10kN P2=20kN B
A 2
0.048
0.084
Inf. line X1 (factor l)
–
– P1
2
0.096
10m 0.072
0.096
4m
0.084
0.048
4m
0.072
2m
C
4
P2
M=24kNm
4
RA
24
M kNm 27.2 41.6
Fig. 12.4 Construction of bending moment diagram using influence line for primary unknown X1
420
12
Influence Lines Method
Solution The bending moment at the middle support B using corresponding influence line is X Py ¼ 10 ð10 0:048 20 0:096Þ ¼ 24kNm X1 ¼ MB ¼ l After that, the initial statically indeterminate continuous beam may be considered as a set of two statically determinate simply supported beams subjected to given load and moment at support B. In other words, the statical indeterminacy has been disclosed. Reaction at support A is RA ¼
10 8 þ 20 4 24 ¼ 13:6 kN 10
Bending moment at specified points are M 2 ¼ 13:6 2 ¼ 27:2 kNm M 4 ¼ 13:6 6 10 4 ¼ 41:6 kNm M B ¼ 13:6 10 10 8 20 4 ¼ 24 kNm Discussion: Influence line for X1 should be considered as a reference data for analysis of beams subjected to different sets of fixed loads. If we need to analyze a structure once due to given set of loads, we can use any classical method or use influence line as a referred data. If we need to analyze the same structure many times, and each time the structure is subjected to different set of loads, then it is much more convenient to construct influence line once and then use it as reference data for all other sets of loadings. Thus, combination of two approaches, i.e., moving and fixed load, is extremely effective for analysis of structures.
12.1.2.5
Statically Indeterminate Primary System
Design diagram of two-span continuous beam with left clamped support is presented in Fig. 12.5a. The degree of statical indeterminacy is equal to two. Therefore for construction of influence lines of two primary unknowns X1 and X2 of the force method the equations (12.7) should be applied. The computational procedure can be significantly simplified if to ignore the condition that the primary system should be statically determinate. Assume the only primary unknown X1 being the bending moment at the middle support B. The primary system as before consists of two separate beams, AB and BC, as shown in Fig. 12.5b. However the adopted primary system has the following fundamental peculiarity: one part of the primary system, beam AB, is statically indeterminate. Such an approach is expedient and efficient if at our disposal we have reference data related to the analysis of the statically indeterminate part of the original structure (in particular, the tabulated data for clamped-pinned beam AB are presented in Table A.15). The bending moment diagram M 1 in primary system due to primary unknown X1 ¼ 1 is shown in Fig. 12.5c. Since the left focal ratio for clamped-pinned beam is equal to two, then bending moment at support A equals to 0.5. Unit displacement becomes δ11 ¼
M1 M1 l l 7l ð2 0:5 0:5 þ 2 1 1 0:5 1 0:5 1Þ þ ¼ ¼ 6EI 3EI 12EI EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |{z} Beam AB
Beam BC
The expressions of slope δ1P at support B for different location of moving load P are presented in Fig. 12.5d. These expressions are taken from Table A.15. Corresponding influence line for δ1P is shown in Fig. 12.5e. To obtain the ordinates of influence line for primary unknown the ordinates of influence line for δ1P should be divided by 7l and taken the negative sign. factor δ11 ¼ 12EI
12.1
Construction of Influence Lines by the Force Method
421
P=1
a
EI=constant
B A
C l
l
P=1
X1
b
c
Primary system
X1=1
0.5
M1 1.0
d u 2 (1 – u ) l 2 . d1P = 4 EI
1
Load P=1 in the left span
Load P=1 in the right span
P=1
P=1
2
3
4
6
7
B
ul
ul
8
9
10
+
8
9
10
0.1097
– 0.0823
0.0548
0.0617
0.0411
7
B
– 0.0137
0.032
6
u –u3 . l2 6
EI
Inf. line d 1P (factor l 2/EI )
11 Inf. line X1 (factor l)
0.0548
5
0.056
4
0.0960
0.032
3
0.064
0.036
2
0.048
0.024
+
0.008 1
d 1P =
C
ul
ul
e
f
11
d1P
d1P
*
A
5
Fig. 12.5 (a, b) Design diagram of continuous beam and primary statically indeterminate system (part AB); (c) Bending moment diagram in primary system due to X1 ¼ 1; (d, e) Location of the load P ¼ 1 in the left and right span, elastic curve, corresponding expression for loaded term δ1P, and influence line for δ1P; (f) Influence line for primary unknown X1
If load P ¼ 100 kN is located at point 8, then bending moment at support B equals to X1 ¼ MB ¼ 10.97l (kNm) and at the clamped support MA ¼ 0.5 10.97l ¼ 5.48l (kNm) Note that the influence lines for primary unknowns, internal forces, and reactions for statically indeterminate structures are bounded by the curved lines, unlike statically determinate structures.
12.1.3 Hingeless Nonuniform Arches Let us apply the general procedure for analysis of symmetrical parabolic nonuniform arch with clamped ends shown in Fig. 12.6a. The equation of the neutral line is y¼
4f ð l xÞ x l2
422
12
a
y
b
P=1
a=ul
C
Ix
jx
ds
IC B
x
P=1
Primary system
C 0.1l
MA u=0.1 H
X2
X3
dx = ds cosj
l
ul
C
dx
x
c
X1
X1 P=1 X2
j dy
f
y
A
0.2
A
Influence Lines Method
0.3
0.4
0.5
f
MB 0.6
0.7
l/2
0.8 l/2
0.9 H
B RB
RA
d 0.1320 0.0960 0.0610 0.0305 0.0085
0.2160 0.1940 0.1654
0.1320 0.1654 0.1940 0.2160 0.2295 0.2344 0.2295
0.0085 0.0305 0.0610 0.0960
+
Inf. line H (factor l/f)
e + 1.000 0.993 0.972 0.930 0.896 0.844 0.784 0.718 0.648 0.576 0.500 0.425 0.352 0.282 0.216 0.156 0.104 0.061 0.028 0.007
Inf. line RA
0.4 l 0.0368 0.0184
0.6l
–
0.0215 0.0113 0.0032
+ 0.0174 0.0312 0.0418 0.0480 0.0498 0.0473 0.0410 0.0320
0.0395 0.0607 0.0678 0.0640 0.0528
f
Inf. line MA (factor l)
0.0080 0.0246 0.0468 0.0246 0.0080
0.0016 0.0052 0.0090 0.0120 0.0127 0.0102 0.0034 –
+
0.0034 0.0102 0.0127 0.0120 0.0090 0.0052 0.0016
0.132 l 0.132 l
g
–
Inf. line MC (factor l)
Fig. 12.6 Parabolic non-uniform arch with clamped ends. (a–c) Design diagram, primary system, and general notation; (d–f) Influence line for reactions of the support H, RA, and MA; (g) Influence line for bending moment at the crown MC
12.1
Construction of Influence Lines by the Force Method
423
Assume that the cross-sectional moment of inertia varies by law Ix ¼
Ic cos φx
where IC corresponds to the highest point of the arch (crown C); this function corresponds to increasing the moment of inertia from crown to supports. It is necessary to construct the influence lines for reactions of support A and bending moment at crown C. This arch is statically indeterminate to the third degree. The general notation (geometry, reactions, position of moving load, and abscises u of specified points of the arch) is shown in Fig. 12.6a. Let us accept the primary system presented in Fig. 12.6b, so the primary unknowns are the bending moment X1, the normal force X2, and shear X3 at crown C of the arch. Canonical equations of the force method are δ11 X 1 þ δ12 X 2 þ δ13 X 3 þ Δ1P ¼ 0 δ21 X 1 þ δ22 X 2 þ δ23 X 3 þ Δ2P ¼ 0
ð12:12Þ
δ31 X 1 þ δ32 X 2 þ δ33 X 3 þ Δ3P ¼ 0 The unit coefficients can be calculated by formula ðl δik ¼
Mi Mk ds: EI x
0
Since X1 and X2 are symmetrical unknowns, then the unit bending moment diagrams M 1 and M 2 are symmetrical, while M 3 diagram is antisymmetrical. It is obvious that all displacements computed by multiplying symmetrical diagram by antisymmetrical ones equal to zero. Therefore, δ13 ¼ δ31 ¼ 0, δ23 ¼ δ32 ¼ 0, and the canonical equations (12.12) fall into two independent systems
δ11 X 1 þ δ12 X 2 þ Δ1P ¼ 0 δ21 X 1 þ δ22 X 2 þ Δ2P ¼ 0
δ33 X 3 þ Δ3P ¼ 0
and
ð12:13Þ
It is obvious that for further simplification of the system of equations we can to apply a concept of elastic center. However the detailed analysis below is provided on the basis of a set (12.13). Coefficients and free terms of canonical equations are calculated taking into account only bending moments, which arise in the arch. The expression for bending moments in the left part of the primary system for unit and loaded states (the force P ¼ 1 is located within the left part of the arch) are presented in Table 12.2. Table 12.2 Bending moments due to unit primary unknowns and given unit load
12.1.3.1
Unit Coefficients ðl δ11 ¼ 0
M1 M1 ds ¼ EI x
ðl 0
1 1 cos φx dx ¼ cos φx EI C
ðl 0
dx l ¼ EI C EI C
424
12
ðl δ12 ¼ δ21 ¼
ðl ðl M1 M2 dx 4f dx fl ds ¼ 1 ð f yÞ ¼ f 2 ðl xÞ x ¼ EI C EI 3EI EI x C C l
0
ðl δ22 ¼
Influence Lines Method
0
ðl
M2 M2 dx ds ¼ ð f yÞ2 ¼ EI C EI x
0
0
ðl
0
4f f 2 ðl xÞ x l
2
dx f2l ¼ EI C 5EI C
0
ð l 2
δ33 ¼ 2
ð2 l
M3 M3 ds ¼ 2 EI x
0
l x 2
2
dx l3 ¼ EI C 12EI C
0
Free Terms Since P ¼ 1, then the free terms are denoted through δiP ða δ1P ¼ 0 ða
δ2P
ða M 1 M 0P dx a2 ds ¼ 1 ða xÞ ¼ EI C EI x 2EI C
M2 ¼ EI x
M 0P
0 ða
ða dx 4f dx ds ¼ 1 ð f yÞ 1 ða xÞ ¼ 1 f 2 ð l xÞ x ð a xÞ ¼ EI C EI C l
0
δ3P
0
0
a2 f 1 2 1 ¼ þ u u2 EI C 2 3 3 ða ða
0 3 M3 MP l dx l 2 1 u ¼ ds ¼ ¼ u x ð a xÞ 2 EI C EI C 4 6 EI x 0
0
Canonical equations (12.13) become 8 2 > < lX 1 þ f l X 2 a ¼ 0 3 2
> : f l X þ 1 f 2 lX þ fa2 1 þ 2 u 1 u2 ¼ 0 1 2 3 5 2 3 3
and
l3 1 u ¼0 X 3 þ l3 u 2 4 6 12
The solution of these equations leads to the following expressions for the primary unknowns in terms of dimensionless parameter u ¼ a/l, which defines the location of the unit force P: 8 3 5 5 2 > 2 > X þ u u ¼ u l > 1 > 4 2 4 > > < 15 l ð12:14Þ X 2 ¼ u2 ð1 uÞ2 > 4 f > > >
> > : X 3 ¼ 12u2 1 þ u 4 6 These formulas should be applied for 0 u 0.5. Since X1 and X2 are symmetrical unknowns, then the expressions for these unknowns for the right part of the arch (0.5 u 1) may be obtained from expressions (12.14) by substituting u!1u. Since X3 is antisymmetrical unknown, the sign of expression for X3 should be changed and parameter u should be substituted by 1u. Influence lines for the primary unknowns X1, X2, and X3 are not presented. After computation of the primary unknowns we can calculate the reaction and internal forces at any section of the arch. 12.1.3.2
Reactions of Support A
The following reactions should be calculated: thrust, vertical reaction, and moment. Thrust: H ¼ X2 ¼
15 2 l u ð1 uÞ2 4 f
for
0 u 1:0
This formula presents the thrust of the arch as the function of the dimensionless parameter u, i.e., this expression is the influence line for H (Fig. 12.6d). Maximum thrust is Hmax ¼ 0.2344(Pl / f ) and it occurs when the force P is located at crown C. This formula shows that decreasing of the rise f leads to increasing of the thrust H.
12.1
Construction of Influence Lines by the Force Method
425
Vertical reaction:
1 u RA ¼ X 3 þ 1 ¼ 12u2 þ þ1 4 6
for
0 u 0:5
Since X3 is antisymmetrical unknown, then for the right part of the arch it is necessary to change sign on opposite and make the change u!1u. Therefore, if unit load P is located on the right part of the arch, then reaction RA is
1 1u RA ¼ X 3 ¼ 12ð1 uÞ2 þ 4 6
0:5 u 1:0
for
Corresponding influence line is presented in Fig. 12.6e. Moment at support A: l 9 5 M A ¼ 1 ul þ X 1 þ X 2 f X 3 ¼ u 1 þ u 6u2 þ u3 l for 0 u 0:5 2 2 2 l 5 u2 u l for 0:5 u 1:0 M A ¼ X 1 þ X 2 f X 3 ¼ ð 1 uÞ 2 2 2 Corresponding influence line is presented in Fig. 12.6f.
12.1.3.3
Bending Moment at Crown C 3 5 5 M C ¼ X 1 ¼ u 2 þ u u2 l 4 2 4
for
0 u 0:5
Since X1 is symmetrical unknown, then for the right part of the arch it is necessary to make the change u ! 1 u. Therefore, if unit load P is located on the right part of the arch, then bending moment at crown is 3 5 5 M C ¼ X 1 ¼ ð1 uÞ2 þ ð1 uÞ ð1 uÞ2 l for 4 2 4
0:5 u 1:0
Influence line for MC is presented in Fig. 12.6g.
12.1.3.4
Discussion
1. If load P is placed in the portion of 0.132l in both sides from crown C, then the extended fibers at C are located below the neutral line of the arch. If load P is placed within 0.4l from the left support, then bending moment at A acts clockwise. 2. For given parabolic nonuniform arch we obtained the precise results. It happens because a cross section moment of inertia is changed by law Ix ¼ IC / cos φx. Since dx ¼ ds cos φ, then ds/EIx ¼ dx/EIC, and all integrals are presented in analytical form. 3. The accepted relationship Ix cos φx ¼ IC means increasing of bending stiffness of the arch from crown to supports; this can be achieved by increasing not only the height of the section (as shown in Fig. 12.7a) but also the width.
a
I Ix = C cosj x
IC
x
b
I x = I C cosj x
IC
x
Fig. 12.7 Types of non-uniform arches. (a, b) The bending stiffness of the arch increases (decreases) from crown to supports
4. In arch with pinned supports, the zero bending moments arise at supports. For this case the following law for moment of inertia of cross section may be taken: IC cosφx ¼ Ix. This expression corresponds to decreasing of bending stiffness of the arch from crown to supports. Both types of arches are presented in Fig. 12.7. Thus, it can be observed that shape (Fig. 12.7a) is not wise to use for pinned type of supports, while the shape (Fig. 12.7b) is dangerous to use in case of
426
12
Influence Lines Method
clamped supports. It is obvious that the laws for moment of inertia of cross section in real structures are not limited to two considered cases above. 5. For parabolic hingeless arch the ordinates of influence lines may be computed by formulas presented in Tables A.17 and A.18, line 2. Example 12.2 Design diagram of symmetric nonuniform parabolic arch with clamped ends is presented in Fig. 12.8. The cross-sectional moment of inertia varies by law Ix ¼ IC /cos φ. The arch is subjected to concentrated load P ¼ 30 kN and uniformly distributed load q ¼ 2 kN/m. Calculate the reactions of support A and internal forces (shear and bending moment) at crown C. Use the influence lines obtained above; the following factors should be taken into account: l/f ¼ 4, and l ¼ 24 m. 0.25l=6m
q=2kN/m
P=30kN C 0.1l
MA u=0.1 A
H
0.2
0.3
0.4
0.5
f=6m
MB 0.6
0.7
l/2 =12m
0.8 l/2=12m
0.9 H
B
RA
RB
Fig. 12.8 Design diagram of parabolic non-uniform arch
Solution Influence lines for reactions and bending moment at crown C are shown in Fig. 12.6d–g. Internal forces may be defined using the corresponding influence lines by formula S ¼ P y + q Ω, where y is the ordinate of influence line under concentrated force, Ω is the area of influence line within acting distributed load. The area of curvilinear influence line may be calculated approximately by replacing curvilinear segments between two neighboring ordinates by straight lines (Fig. 12.9).
yn
yn+1 h
yn+2 h
ym-1
……
ym h
Fig. 12.9 Approximation for calculation of area of curvilinear influence line
If a horizontal distance h, which separates these ordinates, remains constant, then the area bounded by two ordinates yn and ym is given by formula
y ym n Ωm ðaÞ n ¼ h 2 þ ynþ1 þ ynþ2 þ . . . þ ym1 þ 2 Ordinates of influence lines in Fig. 12.6d–g are presented over 0.1l ¼ 1.2m. Dividing this portion into two parts to use the formula (a), we can calculate all reactions at support A due to fixed load P and q. Thrust: l 0:2344 0:2160 H ¼ P 0:1320 þ q þ 0:2295 þ 1:2 ¼ 20:20 kN f 2 2 Vertical reaction: 0:5 0:352 þ 0:425 þ 1:2 ¼ 27:36 kN RA ¼ P 0:844 þ q 2 2 Moment at support:
12.1
Construction of Influence Lines by the Force Method
427
h
i 0:0312 0:048 M A ¼ l P 0:0528 þ q þ 0:0418 þ 1:2 ¼ 33:32 kNm 2 2 Obtained values of reactions at support A (as well as the influence lines for primary unknowns Xi) allow calculating all internal forces at any section of the arch. For this it is necessary to eliminate all constraints at the left end of the arch and replace them by the reactive forces just founded, i.e., to consider the given arch as statically determinate curvilinear rod, which is clamped at B only and is subjected to given load and reactions at support A. For example, bending moment at crown C, by definition, equals l l M C ¼ RA H f þ M A P 0:25l ¼ 27:36 12 20:20 6 33:32 30ð12 6Þ ¼ 6:2 kNm 2 2 Here we again use the fixed and moving load approaches in parallel way. The bending moment at crown C using the influence line 0:0468 0:008 M C ¼ l P 0:0127 þ q þ 0:0246 þ 1:2 ¼ 6:15 kNm 2 2 Relative error is 6:2 6:15 100% ¼ 0:8%: 6:2 This error is due to approximate calculation of the area of influence lines. Shear force at crown C is obtained projecting all forces, located to the left of this section, on the vertical: QC ¼ RA P ¼ 27:36 30 ¼ 2:64 kN Discussion Influence lines for reactions of supports have the fundamental meaning, since they allow easy calculations of reactions of statically indeterminate arch, subjected to arbitrary fixed load. After that, calculating internal forces at any section of the arch is performed as for statically determinate structure. Therefore, the influence lines presented in Fig. 12.6 may be treated as the reference data for a nonuniform parabolic arch of arbitrary length l and height f.
12.1.4 Statically Indeterminate Trusses In case of trusses the general procedure for construction of influence lines has the fundamental features. Let us consider the statically indeterminate truss shown in Fig. 12.10a. The axial stiffness of all members equals EA. It is required to construct the influence line for reaction at the middle support. The structure under consideration is externally statically indeterminate truss of the first degree of redundancy. Let the primary unknown X1 be a reaction of the middle support; the primary system is shown in Fig. 12.10b. Force method leads to the following expression for the primary unknown X1 ¼
δ1P , δ11
where δ1P is displacement in direction of 1-th primary unknown due to traveling load P ¼ 1. Principal concept. Expression for X1 may be modified. According to the reciprocal displacements theorem, δ1P ¼ δP1 and so expression for X1 becomes X1 ¼
δ1P δ ¼ P1 δ11 δ11
ð12:15Þ
This equation contains the in-depth fundamental idea: Instead of calculation of displacement δ1P in the direction of primary unknown X1 due to the moving load P = 1, we will calculate displacement δP1 at points of application of load P = 1 due to unit primary unknown X1 = 1 (Fig. 12.10c). Unit state (X1 ¼ 1) and corresponding internal forces in all members of the truss are shown in diagram N (Fig. 12.10d).
428
12
Influence Lines Method
a 3m
P=1
d=4m 1⬘
b
2⬘
3⬘
4⬘
5⬘
4
5
P=1 0
6 2
1
3 X1
P=1
c d 1P = d P1 d P1
d
d1P +1.333
+1.333
+1.333
+0.833
-0.833
+0.833
+1.333 -0.833
-1.0
+0.833
-0.666
X1=1
-1.333
-0.666
-1.333
X1=1
0.5
+0.833
-0.666
N P = N1
-0.666 0.5
e +5/12
+5/12
1 -1/3
-0.5
N1
-1/3
0.25
0.25 0.25
0.25 +1/3
f
+1/3 –5/12
–5/12 +0.25
+0.25
N2
2 0.25
0.25 0.25
0.25
Fig. 12.10 Statically indeterminate truss. Fundamental concept. (a, b) Design diagram and primary system; (c) Reciprocal displacemets theorem δ1P ¼ δP1: displacement δ1P in the direction of the primary unknown X1; displacement δP1 is taken under the travelling load. (d) Unit state: internal forces N caused by unit primary unknown X1 ¼ 1; (e, f) Design diagram for calculation of the first and second elastic loads; (g) Fictitious truss subjected to elastic loads and corresponding bending moment diagram for fictitious beam; (h) Influence line for primary unknown X1; (i) Truss subjected to fixed load P1 ¼ 60 kN and P2 ¼ 20 kN
12.1
Construction of Influence Lines by the Force Method
429
86.7
69.648
g
d11
69.648
38.374
38.374
– A
B 1.776
3.555
R Af
8.525
3.555 1.0
0.803
h
(Factor 1/EA)
1.776
R Bf
0.803
0.443
M f = d P1
0.443
+ Inf. line X1 a 2′
1′
3m
i A
2
1 d=4m
C
4
B
5
a P1=60kN
RA
RC=X1
P2=20kN
Fig. 12.10 (continued)
12.1.4.1
Construction of Function δP1
Since the lower chord of the truss is loaded by traveling load, we have to find the vertical displacements of the joints of the lower chord caused by the primary unknown X1 ¼ 1. Thus, this unit force plays the role of external influence. The most effective procedure for calculation of δP1 is the elastic load method (see Sect. 8.9). Elastic load W1. According to this method we need to apply two unit couples of opposite directions to the members, which are located to the left and to the right from joint 1. Let the left and right couples rotate counterclockwise and clockwise, respectively. Each couple should be presented as two forces at joints 0 and 1 for left couple, 1 and 2 for right couple. These forces have a vertical direction and their values are 1/d ¼ 1/4 ¼ 0.25. The group of four loads at joints 0, 1, 2 is shown in Fig. 12.10e. These four forces present a self-equilibrate set of forces; therefore the reactions of supports are zeros. Corresponding internal forces in each element of the truss in the primary system caused by these four applied forces (0.25 each) are shown on diagram N 1 . The first elastic load is determined by formula W1 ¼
XN N l 1 P EA
ð12:16Þ
where NP ¼ N1 is internal forces in each element of the truss in the primary system caused by given external load, i.e., the unit primary unknown X1 ¼ 1; these forces are shown in Fig. 12.10d. Summation is performed by all elements of the truss. The first elastic load according to Eq. (12.16) becomes
5 1 1:776 1 1 5 1 ¼ þ 0:833 5 þ ð0:833Þ 5 W 1 ¼ 2 ð0:666Þ 4 EA EA 3 EA 12 EA 12 members 01, 12
member 010
member 210
Elastic load W2. The loads 1/d ¼ 1/4 ¼ 0.25 are applied at joints 1 and 2, 2 and 3 as shown in Fig. 12.10f; corresponding internal forces are shown on the diagram N 2 . The second elastic load becomes X N2 NP l 1 1 5 1 5 1 3:555 W2 ¼ ¼ 2 1:333 4 ð0:833Þ 5 0:833 5 þ þ ¼ EA 3 EA 12 EA 12 EA EA members 10 20 , 20 30
members 10 2
members 230
430
12
Influence Lines Method
Elastic load W3. Similarly procedure leads to the following result for third elastic load W3 ¼ 8.525/EA. Since the structure is symmetric, then the elastic loads W4 ¼ W2 and W5 ¼ W1. Fictitious beam presents the simply supported beam, which is loaded by elastic loads (Fig. 12.10g). The elastic load W1 is positive, so this load must be directed as forces at joint 1, i.e., upward (Fig. 12.10e). Similarly, all the elastic loads should be directed upward. Reactions at the left and right supports of fictitious beam are 9.5935/EA. The corresponding bending moment diagram is presented in Fig. 12.10g; factor 1/EA for all ordinates is not shown. The ordinates of this diagram present the displacements δP1 of corresponding joint of the truss due to unit primary unknown X1 ¼ 1. It is obvious that the unit displacement becomes δ11 ¼ 86.7/EA.
12.1.4.2
Influence Line for Primary Unknown X1
This key influence line is obtained by dividing all ordinates of diagram δP1 (Fig. 12.10g) for fictitious beam by (δ11). Corresponding influence line X1 is presented in Fig. 12.10h. In case of arbitrary external load the internal force in any member can be calculated using the influence line for primary unknown X1. For example, the truss is loaded by concentrated force P1 ¼ 60 kN at joint 2 and P2 ¼ 20 kN at joint 5 as shown in Fig. 12.10i. For this loading the primary unknown X1 which is the reaction at support C becomes X Pi yi ¼ P1 0:803 þ P2 0:443 ¼ 57:04 kN RC ¼ X 1 ¼ Now we can calculate the reaction at support A: X RA ! M B ¼ 0 : RA 6d þ P1 4d RC 3d þ P2 d ¼ 0 ! RA ¼ 34:81 kN After that we can calculate any internal force. For example, for force U1-2 (section a-a) we get: U 13 !
12.1.4.3
X
M left 10 ¼ 0 :
RA d þ U 12 h ¼ 0 ! U 12 ¼
34:81 4 ¼ 46:41 kN 3
Summary
1. The reciprocal displacements theorem allows considering δP1 as the model of vertical displacements of joints of the truss (within a constant multiplier which equals 1/δ11). 2. For calculation of δP1 the elastic loads method have been applied. This method is more effective in comparison to MaxwellMohr integral by the following reason. When a set of four loads is applied, then the internal forces arise only in the members, which form the two adjacent panels of the truss. Therefore, procedure of summation is related only to elements, which belongs to these two panels of the truss, while using the Maxwell-Mohr integral, a summation is related to all elements of the truss. 3. The elastic loads method also allows calculating unit displacement δ11. It is obvious that δ11 could be calculated using Maxwell-Mohr formula δ11 ¼
XN N l EA
However, since in the unit state practically in all members of the truss internal forces arise (Fig. 12.10d), then MaxwellMohr procedure becomes cumbersome. The elastic load method is very effective and moreover allows calculating δP1 for all joints of the loaded contour and δ11 in one step, as shown in Fig. 12.10g. It is obvious that when load P is placed at point application X1, then δP1 ¼ δ11.
12.2
Construction of Influence Lines by the Displacement Method
431
4. Influence line of internal force, which is induced at any member k of the truss, may be constructed by formula (12.9). 5. Influence line for primary unknown X1 should be treated as key influence line. Indeed, using this influence line we can calculate X1 in case of arbitrary fixed load. After that, the truss should be considered as statically determinate one, which is subjected to the given loads and the force X1.
12.2
Construction of Influence Lines by the Displacement Method
This paragraph contains the fundamental concepts of construction of influence lines using the displacement method. The procedure of their construction is illustrated in detail by the example of a continuous beam with hinged ends and frames with non-shifted joints. It is noted the significance of influence lines as a powerful tool of the combined approach (moving and fixed load) to the analysis of statically indeterminate structures.
12.2.1 General Let us consider a first-degree kinematically indeterminate structure. In case of a fixed load, the canonical equation of the displacement method is r 11 Z 1 þ R1P ¼ 0 In this equation the free term R1P represents reaction caused by given actual load. Since moving load is unit one, then (for the sake of consistency notations) let us replace a free term R1P by the unit free term r1P; this free term presents a reaction in primary system in the introduced constraint 1 caused by load P ¼ 1. The primary unknown Z1 ¼
r 1P r 11
Now we need to transform this equation for the case of moving load. The unit reaction r11 presents reaction in the introduced constraint caused by unit displacement of this constraint. Therefore, r11 is some specific number, which depends on the type of a structure and its parameters, and the type of displacement (Tables A.3–A.7). However, r1P depends on unit load location. Since load P ¼ 1 is traveling, then r1P becomes a function of position of this load, and as a result, the primary unknown becomes a function as well: ILðZ 1 Þ ¼
1 ILðr 1P Þ r 11
ð12:17Þ
The function IL(r1P) may be constructed using Tables A.3–A.5 (line 3), A.6 (line 4); non-dimensionless parameters u and υ (note that u + υ ¼ 1) denote the position of load P. In case of fixed load, a bending moment at any section k may be calculated by formula M k ¼ M k Z 1 þ M 0k where Z1 is the primary unknown of the displacement method; M k is the bending moment at section k in a primary system due to unit primary unknown Z ¼ 1; and M 0k is bending moment at section k in a primary system due to given load. Now we need to transform this equation for the case of moving load. The bending moment M k at any section k presents the number, because this moment is caused by the unit primary unknown Z1; the second multiplier Z1 presents a function. The last component moment M 0k is caused by the moving load; therefore, the bending moment M 0k also becomes a function of the position of unit load. As a result, the bending moment at any section k becomes a function, which can be presented as follows ILðM k Þ ¼ M k ILðZ 1 Þ þ IL M 0k In case of fixed load, the shear at any section k is a number
ð12:18Þ
432
12
Influence Lines Method
Qk ¼ Qk Z 1 þ Q0k Similarly, in case of traveling load, the shear at any section k becomes a function; therefore ILðQk Þ ¼ Qk ILðZ 1 Þ þ IL Q0k
ð12:19Þ
For the n-times kinematically indeterminate structure, the canonical equations of the displacement method in case of a fixed load P ¼ 1 are r 11 Z 1 þ r 12 Z 2 þ þ r 1n Z n þ r 1P ¼ 0
ð12:20Þ
r n1 Z 1 þ r n2 Z 2 þ þ r nn Z n þ r nP ¼ 0 Unit reactions rik are caused by unit primary unknowns. They can be calculated using the typical procedures of the displacement method; these reactions are presented as the specific numbers. Reaction riP can be calculated using Tables A.3–A.6. Fundamental feature of system (12.20) is that the free terms riP are some functions of position of unit load P ¼ 1. Therefore, a solution of the system (12.20) leads to the primary unknowns Zi as the functions of the load position, in fact, to IL(Zi), i ¼ 1,. . ., n. Bending moment at any section k in case of fixed load is calculated by formula M k ¼ M 1 Z 1 þ M 2 Z 2 þ þ M 0P In case of moving load, the bending moment at the any specified section k becomes function, so influence line should be constructed by formula ILðM k Þ ¼ M k1 ILðZ 1 Þ þ M k2 ILðZ 2 Þ þ þ IL M 0k ð12:21Þ Influence lines for shear and axial force at any section can be constructed similarly. Similarly to the force method, the construction of influence lines for internal forces in statically indeterminate structures by the displacement method starts from construction of influence lines for primary unknowns IL(Z1), IL(Z2), . . ., IL(Zn). The following procedure is recommended for construction of influence lines for statically indeterminate structures by the displacement method: 1. Determine the degree of kinematical indeterminacy, construct the primary system, and formulate the canonical equations of the displacement method. 2. Compute the coefficients of canonical equations (12.20); unit coefficient rik is the reaction at introduced constraint i caused by unit displacement of introduced constraint k. These unit reactions present some specific numbers. 3. Derive the expressions for free terms of canonical equations. Each load term riP presents reaction at introduced constraint i due to load, which depends on location of unit load and therefore presents some functions. 4. Solve the canonical equations; since the free terms are functions, the primary unknowns also present the functions of location of unit load. 5. Construct the influence lines for primary unknowns. This step presents the central and very important part of procedure. 6. Construct the influence lines for reactions and internal forces (bending moment and shear) at specified section. The detailed procedure of construction of influence line for primary unknown and internal forces for continuous beam and frame is illustrated below.
12.2.2 Continuous Beams Let us demonstrate the above procedure for uniform two-span continuous beam ABC with equal spans l. We will construct influence lines for primary unknown as well as for bending moment and shear at a specified section k (Fig. 12.11a). Primary system. The structure is kinematically indeterminate of the first degree. The primary system of the displacement method is shown in Fig. 12.11b. The primary unknown Z1 is the angle of rotation of introduced constraint at the middle support.
12.2
Construction of Influence Lines by the Displacement Method
12.2.2.1
433
Influence Line for Primary Unknown Z1
1. For calculation of r11 it is necessary to plot the bending moment diagram due to primary unknown Z1 ¼ 1 (Fig. 12.11c). It is obvious that r 11 ¼ 6EI=l.
a
P=1
0.4l
EI=constant
B A
C k l
l
b
Z1
P=1
Primary system
c
3EI l
Mk
l
r11
r11 Unit state
M1
3EI l
0.4l
3EI
Z1=1
2
3EI l
3EI l
d Load P=1 in the left span
(
l u 1 - u2 2 r1P
P=1
ul
ul
)
(
r1P =
l u 1 - u2 2
4
5
(
)
l u 1 - u2 2
(
)
10
11
l u 1 - u2 2 r1P
)
r1P = -
Load P=1 in the right span
P=1
ul
ul
P=1
e 1
2
3
6
7
8
9
k
0.016
0.028
0.032
+ 0.024
-
0.024
l 0.032
0.028
0.016
l
Inf. line Z1 (factor l 2 /EI)
Fig. 12.11 Continuous beam. (a, b) Design diagram of the beam and primary system; (c) Bending moment diagram caused by unit primary unknown and calculation of r11;. (d) Calculation of r1P; (e) Specified section of continuous beam and influence line for primary unknown Z1
2. For calculation of r1P it necessary to consider the load P ¼ 1 located within the left and right span separately. Bending moment diagrams due to traveling load P ¼ 1 are shown in Fig. 12.11d.
434
12
Influence Lines Method
Expressions for reactive moment r1P for different beams are presented in Table A.3. Note that length υl is measured from the pinned (roller) support. Ordinates of influence line for r1P in terms of position υ of the load P ¼ 1 in the left and right spans are presented in Table 12.3. Table 12.3 Calculation of r1P and ordinates of influence line for Z1
Moving load P=1 at the left span Inf. Line Z1
Point 1 2 3 4 5 6 Factor
Moving load P=1 at the right span
0 0.2 0.4 0.6 0.8 1.0
0 0.096 0.168 0.192l 0.144l 0 l
0.0 -0.016 -0.028 -0.032 -0.024 0.0 l2/EI
Inf. Line Z1
Point 6 7 8 9 10 11
1.0 0.8 0.6 0.4 0.2 0
0 -0.144 -0.192 -0.168 -0.096 0 l
0.0 0.024 0.032 0.028 0.016 0.0 l2/EI
3. Influence line of primary unknown Z1 is obtained according to formula (12.17): all ordinates of influence line for r1P must be divided by r 11 ¼ 6EI=l. Corresponding influence line for Z1 is presented in Fig. 12.11e; all ordinates must be multiplied by parameter l 2 /EI. If load is located on the right span, then angular displacement at support B is positive, i.e., this displacement occurs clockwise.
12.2.2.2
Influence Line for Bending Moment Mk
This influence line should be constructed by formula (12.18). The bending moment in primary system at section k due to primary unknown Z1 ¼ 1 is shown in Fig. 12.11c and equals to Mk ¼
3EI EI 0:4 ¼ 1:2 ; l l
the negative sign means that extended fibers at section k are located above the neutral line. Therefore ILðM k Þ ¼ 1:2
EI ILðZ 1 Þ þ IL M 0k l
ð12:22Þ
EI ILðZ 1 Þ, is presented in Fig. 12.12a. The second term, IL M 0k , presents the influence line l of bending moment at section k in primary system. For construction of this influence line we need to consider a load P ¼ 1 placed within the left and right spans (Fig. 12.12b). The position of the load P ¼ 1 is determined by dimensions ul and υ l. According to Table A.3 (row 3) they are measured from the clamped and pinned support, respectively. Dimensionless parameters u and υ satisfy condition u + υ ¼ 1 . Load P ¼ 1 in the left span. Reaction of the left support is The first term of (12.22), 1:2
R¼
u2 ð 3 uÞ 2
The bending moment at section k depends on the location of the load P ¼ 1 with respect to section k. Two positions of moving load should be considered. They are load P ¼ 1 is located to the left and to the right of section k. Detailed calculation of bending moment at section k in primary system caused by unit moving load P, which is located at the points 1, 2, 3(k),. . ., 6, is presented in Table 12.4. Load P ¼ 1 in the right span. In this case the bending moment at section k does not arise (because of the introduced constraint), and influence line for M 0k has zero ordinates. Influence line M 0k is shown in Fig. 12.12b.
12.2
Construction of Influence Lines by the Displacement Method
435
Table 12.4 Calculation of IL M 0k . Load P ¼ 1 is located in the left span
The final influence line for bending moment at section k is constructed using expression (12.18) and shown in Fig. 12.12c. The same result had been obtained early by the force method (Fig. 12.2).
4
5
6
7
8
9
10
0.0192
3
0.0336
2
0.0384
1
0.0288
P=1
a
11
k
b Influence line for Mk0 Load P=1 in the left span
P=1 1
2
M k0 depends of position P
3
P=1
P=1 4
5
6
7
8
9
10
υl
ul
Load P=1 in the right span for any position P
0.0512
0.1216
−
0.0192
0.0336
+
0.0384
0.0288
c 0.2064
Inf. line Mk0 (factor l)
0.0224
0.1728
0.0832
+ 0.0816
11
ul
υl
0.1008
EI ∙ IL( Z 1 ) l
M k0 = 0
υl R
− 1.2
(All ordinates must be multiplied by factor l)
k
the left or right at k
R = u2 (3-u) / 2
−
0.0288
0.0384
0.0336
0.0192
+
Final Inf. line Mk (factor l )
Fig. 12.12 Construction of influence line for bending moment Mk. (a) Specified sections of a beam and the first terms of (12.22); (b) The second terms of (12.22); (c) Final influence line Mk
436
12
12.2.2.3
Influence Lines Method
Influence Line for Shear Force Qk
This influence line should be constructed by formula (12.19). According to Fig. 12.11c, the reaction at support 1 due to Z1 ¼ 1 is directed downward, so the shear in primary system at section k due to primary unknown Z1 ¼ 1 is Qk ¼ 3EI=l2. Therefore, (12.19) becomes ILðQk Þ ¼
3EI ILðZ 1 Þ þ IL Q0k 2 l
ð12:23Þ
3EI ILðZ 1 Þof the (12.23) is presented in Fig. 12.13a. The second term presents influence line of shear at l2 section k in the primary system. For calculation of this term we need to consider location of the load P ¼ 1 in both spans separately. Load P ¼ 1 in the left span. Shear force at section k depends on the location of the load P ¼ 1 with respect to section k (to the left or to the right). Calculation of shear at section k is presented in Table 12.5. The first term
Table 12.5 Calculation of IL Q0k . Load P ¼ 1 is located in the left span
Load P = 1 is in the right span. In this case the shear at section k does not arise, and influence line for Q0k has zeros ordinates. The final influence line for shear Qk is constructed using the expression (12.23). This influence line is presented in Fig. 12.13c. The same result had been obtained early by the force method.
5
6
7
8
9
10
0.048
4
0.084
3
0.096
2
0.072
1
0.072
a
0.096
P=1 11
k
0.084
3EI l2
( )
· IL Z1
0.048
0.084
0.096
+
–
Inf. line Qk0
0.072
0.056 0.128
–
0.208
0.248
0.516
c
–
–
+ 0.568
0.296
0.432
b
0.484 0.304
0.048
+
Final Inf. line Qk
Fig. 12.13 Construction of influence line for shear force Qk. (a) Specified sections of a beam and the first terms of (12.23); (b) The second terms of (12.23); (c) Final influence line Qk
12.2
Construction of Influence Lines by the Displacement Method
12.2.2.4
437
Discussion
1. Influence lines for M 0k and Q0k for the force method are bounded by the straight lines, because the primary system is a set of statically determinate simply supported beams. The same influence lines in the displacement method are bounded by the curved lines, because the primary system is a set of statically indeterminate beams. 2. If a load is located in the right span, then M 0k and Q0k for section k in the left span are zero. In the force method it happens because the primary system presents two separate beams; therefore, the load from one beam cannot be transmitted to another. In the displacement method it happens because the introduced support in the primary system does not allow transmitting of internal forces from one span to another. 3. The influence line for shear Qk has the following important property: the sum of the absolute values of the shears at the left and right at section k is equal to unity: 0.484 + 0.516 ¼ 1.0. 4. Having the influence lines for primary unknown of the force (or displacement method), or influence lines of internal forces for any section of a beam, we can easily construct the internal force diagrams for any fixed loads. This is shown on the following example. Example 12.3 The two-span beam with equal spans l is subjected to force P as shown in Fig. 12.14a. The beam is divided into 10 equal portions; the numbering of points is presented in Figs. 12.11e, 12.12a and 12.13a. Find the reaction at the middle support B. Solve this problem by three different ways: (i) use the influence lines for Mk and Qk; (ii) use the influence lines for primary unknown of the displacement method; (iii) use the influence lines for primary unknown of the force method.
P
a
k
A
B
C 8 0.4l
0.4l
RA
0.6l
RB
l P
0.096P 1
b
6
3
11 8
k 0.0384Pl
0.096P
RB Z1
c
RC
P
A
C B
8 ul=0.6l
ul=0.4l
RA
RC
RB P
d 1
2 l/5
RA
B
3
11 8
X1
X1 X1/l
X1/l+0.6P
0.6l RC
Fig. 12.14 (a) Design diagram of the beam. (b–d) Calculation of reaction at support B using the following approaches: (b) the influence line for internal forces at section k; (c) the influence line Z1 of the displacement method, Z1 ¼ 0.032 l2/EI ; (d) the influence line X1 of the force method, X1 ¼ 0.096 Pl
438
12
Influence Lines Method
Solution (i) Since a load P is located at point 8, then internal forces at section k according to Figs. 12.12c and 12.13c are Mk ¼ 0.0384Pl and Qk ¼ 0.096P. Now we can make a cut section of the beam at the point 3(k) and show corresponding internal forces at this section (Fig. 12.14b). Direction of the bending moment and shear are shown according to their signs. The free-body diagram for part 1-3 allows finding the reaction of support A (0.096P downward) and verifying the calculation X M A ¼ 0:096P 0:4l 0:0384Pl ¼ 0 Free-body diagram for part 3-11 allows calculating all reactions and constructing internal force diagrams, which correspond to given location of the force P. For example, X MC ¼ 0 RB ! RB l P 0:6l 0:096P ð0:6l þ lÞ 0:0384Pl ¼ 0 ! RB ¼ 0:792Pð"Þ (ii) If force P is located at point 8, then primary unknown Z1 of the displacement method (Fig. 12.11e) equals Pl2 Z 1 ¼ 0:032 . Now we can consider two beams. The left pinned-clamped beam AB is subjected to angular displacement EI Z1 only and the right clamped-rolled beam BC is subjected to angular displacement Z1 and force P (Fig. 12.14c). The vertical reaction at support A for the left beam (according to Table A.3) equals RA ¼
3EI 3EI Pl2 Z 1 ¼ 2 0:032 ¼ 0:096Pð#Þ 2 EI l l
The vertical reaction at support C for the right beam (according to Table A.3, rows 1 and 3) equals RC ¼
3EI Pu2 3EI Pl2 P 0:42 Z þ 0:032 ð 3 u Þ ¼ þ ð3 0:4Þ ¼ 0:096P þ 0:208P ¼ 0:304Pð"Þ 1 2 EI 2 l2 l2
Required reaction at support B X RB ! Y ¼ 0 : RA þ RB P þ RC ¼ 0 ! 0:096P þ RB P þ 0:304P ¼ 0 ! RB ¼ 0:792Pð"Þ (iii) Reaction of support B using influence line X1 of the force method is calculated by the following way. If force P is located at point 8, then primary unknown X1 (bending moment at support B) equals X1 ¼ MB ¼ P 0.096l (Fig. 12.1f). Now we can consider two simply supported beams (left beam is subjected to moment MB ¼ X1 only and the right beam is subjected to MB ¼ X1 and force P) as shown in Fig. 12.14d. All reaction should be calculated as follows X1 ¼ 0:096Pð#Þ, l X1 1 P 0:6l RB ¼ 2 þ 0:6P ¼ P 0:096l 2 þ ¼ 0:792P, l l l X RC ¼ 1 þ 0:4P ¼ 0:304Pð"Þ l
RA ¼
Using the above reactions, we can calculate the bending moments for all sections of the beam l ¼ 0:0192Pl, 5 2l M 3 ¼ 0:096P ¼ 0:0384Pl, . . . 5
M 2 ¼ 0:096P
Thus, having the influence lines for primary unknown of the force (or displacement method), or influence lines of internal forces for any section of a beam, construction of the internal force diagrams for any fixed loads is done very easy.
12.2
Construction of Influence Lines by the Displacement Method
439
12.2.3 Redundant Frames Construction of influence lines for statically indeterminate frames may be effectively performed using the displacement method. As in case of the fixed load, the displacement method is more effective than the force method for framed structures with high degree of the static indeterminacy. The following example illustrates the construction of influence line for primary unknown of redundant frame. As have been shown early, this influence line should be treated as key influence line. Figure 12.15a presents a design diagram of statically indeterminate frame. Bending stiffness EI is constant for all portions of the frame. We need to construct the influence line for angle of rotation of the rigid joint. The primary system of the displacement method is presented in Fig. 12.15b. The primary unknown Z1 is the angle of rotation of the rigid joint. 1 Equation of influence line for primary unknown Z1 is ILðZ 1 Þ ¼ ILðr 1P Þ. r 11
P=1
a
Z1
b
P=1
1 i
i l
l
l
r11
c
i=EI/l
Z1=1
Mk
r11 2i
3i
4i
3i k
3i 3i
4i M1
d
P in left span
P in right span
r1P
r1P
P
P
ul
ul
0.0192
0.0144
–
5
6
7
8
9
10
Inf. line Z1 Factor l2/EI
+ 0.0032
4
0.0096
3
0.0144
2
0.0128
1
0.0168
0
0.0096
Z1
e
Fig. 12.15 (a, b) Design diagram and primary system; (c) Unit state and corresponding bending moment diagram; (d) Calculation of free term r1P; (e) Influence line for primary unknown Z1
440
12
Influence Lines Method
1. Calculation of unit reaction r11. Bending moment diagram M 1 caused by unit rotation of introduced constraint 1 and freebody diagram of this constraint is shown in Fig. 12.15c. Equilibrium condition of rigid joint leads to r11 ¼ 10i. 2. Calculation of free term r1P. It is necessary to consider two positions of moving load P ¼ 1: the load is traveling along the left and right spans. The position of load P ¼ 1 in each span is indicated by parameters u and υ (u+υ ¼ 1), where parameter u for both spans is reckoned from left support. Bending moment diagrams are shown in Fig. 12.15d. Ordinates of bending moment diagrams for pinned-fixed and fixed-fixed beams are taken from Tables A.3 (row 1) and A.4 (row 3), respectively. Note, in our case the left span of the frame is a mirror of those presented in Table A.3; however a distance from the left pinned support is labeled as ul. This should be taken into account, i.e., in formula for MA (Table A.3) instead of parameter υ we must write the parameter u. Therefore, if load P ¼ 1 is located in the left span, then the free term of canonical equation is ul r 1P ¼ 1 u2 . If load P ¼ 1 is located in the right span, then we can directly apply formula from Table A.3, so 2 r1P ¼ uυ2l ¼ u(1 u)2l. 3. Influence line for primary unknown. Having expressions for r1P in terms of position P and r11 ¼ 10i, the expressions for primary unknown Z1 is presented as follows: P ¼ 1 in the left span ILðZ 1 Þ ¼
l2 1 ul u 1 u2 ¼ 1 u2 10i 2 20 EI
ð12:24Þ
1 u l2 uð1 uÞ2 l ¼ ð1 uÞ2 10i 10 EI
ð12:25Þ
P ¼ 1 in the right span ILðZ 1 Þ ¼
Left and right spans are divided for five equal portions. Calculation of ordinates of influence line of primary unknown at specified points 1-10 is presented in Table 12.6. Table 12.6 Ordinates of influence line of primary unknown of displacement method (factor l2/EI).
The final influence line for primary unknown Z1 is presented in Fig. 12.15e. Discussion 1. Once constructed influence line for primary unknown presents the fundamental data, because it carries comprehensive and important information about structure. This key influence line can be used for analysis of structure subjected to arbitrary load placed along the loaded counter. 2. The use of tabular data involves the use of dimensionless parameters u and υ fixing the position of the moving load. As shown in Figs. 12.12b and 12.15d these parameters are introduced by different ways. This difference is particularly noticeable in Fig. 12.11d. Therefore, the using of tabular formulas demands the special care and, if necessary, they should be modified.
12.3
Comparison of the Force and Displacement Methods
12.3
441
Comparison of the Force and Displacement Methods
For summarizing, let us compare two fundamental analytical methods for construction of influence lines. This comparison is presented in Table 12.7; it develops and complements Table 10.4 for case of moving load. As in the case of the dead loads, construction of influence lines for any statically indeterminate structure starts from determining of number and types of unknown, and presentation of corresponding primary system of the force and displacement methods. Table 12.7 Comparison of the force and displacement methods for construction of influence lines
442
12
Influence Lines Method
Notes Construction of influence lines for primary unknowns and internal forces by the force and displacements methods leads, as would be expected, to the identical results. These influence lines disclose the static (kinematic) indeterminacy of a structure. After this, the determination of the internal forces caused by a fixed load, as well as the computation of the maximum possible forces due to the system of moving loads, is an elementary procedure. Of course, the construction of influence lines for statically indeterminate structures may look as a complex procedure. However, the effectiveness of the influence line method for the analysis of a structure subjected to action of a moving load, as well as the possibility of a combined approach to the analysis of a structure, (moving and fixed approach) allows considering the influence line method as an extremely powerful tool for structural analysis.
12.4
Kinematical Method (Müller-Breslau Principle)
The method allows showing the shape of the influence line, indicate location of its specific ordinates, and compute their values. This paragraph contains the fundamental concepts which are necessary for the construction of influence lines by the kinematical method. The procedure for constructing of influence lines is considered in detail for continuous beam with hinged ends.
12.4.1 General The shape of influence line is often referred as a model of influence line. This model allows finding the most unfavorable position of the load. The model of influence lines may be constructed by kinematical method using Müller-Breslau principle, which is considered below. Let us consider n times statically indeterminate continuous beam. It is required to construct an influence line for any reaction (or internal force) X at any section of a beam. Released system of the force method is obtained by eliminating constraint, which corresponds to the required force X1 and replacing this constraint by force X1. Released system presents (n1) times statically indeterminate structure. Canonical equation for specified unknown X1 in case of unit load P is presented in the form δ11X1 + δ1P ¼ 0. Influence line for primary unknown X1 becomes ILðX 1 Þ ¼
1 ILðδ1P Þ, δ11
where δ11 is displacement in the direction of primary unknown X1 caused by unit primary unknown X1 ¼ 1; δ1P presents displacement in the direction of primary unknown caused by moving unit load P. According to reciprocal displacements theorem, δ1P ¼ δP1, where δP1 presents displacement at the point of application of the moving load P caused by unit primary unknown X1. Therefore, the influence line for primary unknown may be constructed by formula ILðX 1 Þ ¼
1 ILðδP1 Þ δ11
ð12:26Þ
The ordinates of influence line for any function X (reaction, bending moment, etc.) are proportional to ordinates of the elastic curve due to unit force X, which replaces the eliminated constraint where the force X arises (Müller-Breslau principle, 1887). This principle with elastic loads method was effectively applied previously for analytical construction of influence lines for statically indeterminate trusses. Now we illustrate the Müller-Breslau principle for two types of problems. They are analytical computation of ordinates of influence lines and construction of models of influence lines. Both of these problems are referred as kinematical method. Our consideration of this method will be limited only for the continuous beams. In order to construct the model of influence line for a certain factor X (reaction, bending moment, shear force) by a kinematical method, the following steps must be performed: 1. Indicate the constraint or section, in which factor X arises. 2. Show the new system (released system) by eliminating constraint where factor X arises.
12.4
Kinematical Method (Müller-Breslau Principle)
443
3. Apply the X ¼ 1 instead of eliminated constraint. 4. Show the elastic curve (diagram of the vertical displacement) due to X ¼ 1 in new system. This diagram is a model for influence line of factor X. Fig. 12.16 presents elimination of constraint k where factor X arises, and replacing this constraint by corresponding force X.
a
k
c
b
k
k
X=Q
X=M=1
X=Rk=1
k
k
X=Q=1
Fig. 12.16 Elimination of constraint and replacing it by corresponding force X ¼ 1. Case (a) should be used for construction of influence line for reaction at any support of continuous beam; the cases (b, c) used for construction of influence lines for bending moment and shear at any section k, respectively.
In case (a) the rolled support is eliminated and unit reaction is applied. In case (b) we introduce hinge k which allows mutual angular displacement of both parts of a beam at given section k, but relative horizontal and relative vertical displacements are absent. In case (c) we introduce special device (two parallel rods with hinges at the ends) which allows mutual vertical displacement of both parts of a beam at given section k, but the relative horizontal and relative angular displacements are absent.
12.4.2 Continuous Beams: Analytical Solution Figure 12.17a presents two-span uniform continuous beam. It is necessary to construct the influence line for bending moment X1 at middle support. According to Fig. 12.16b we need to eliminate constraint where moment X1 arises, replace eliminated constraint by hinge, and show two couples X1. Thus, the primary system presents two separate simply supported beams as shown in Fig. 12.17b. Also this figure shows the elastic curve due to unit force P ¼ 1. The angle of rotation at middle support caused by P ¼ 1 is denoted as δ1P. Elastic curve of the primary system caused by unit unknown X1 ¼ 1 is shown in Fig. 12.17c. Both elastic curves in Fig. 12.17b, c illustrate the reciprocal displacement theorem. According to this theorem δ1P ¼ δP1. Equation of influence line for X1 is described by (12.26). The feature of this expression is as follows: Instead of calculation of a slope of elastic curve at middle support δ1P for different position of P ¼ 1, we will calculate the vertical displacement δP1, which occurs under the force P caused by unknown X1 ¼ 1, as shown in Fig. 12.17c. Elastic curve δP1 caused by fixed unit couple X1 ¼ 1 may be easily constructed by the initial parameter method. For the left span of a beam (Fig. 12.17d) according to (8.10) we have EIy ¼ EIδP1 ¼ EIy0 þ EIθ0 x
Rðx 0Þ3 , 3!
where R is the reaction of the left support. Since R ¼ 1/l, and vertical displacement at initial point y0 ¼ 0 (point 1), the equation of elastic curve becomes EIy ¼ EIδP1 ¼ EIθ0 x
3 1 ð x 0Þ l 6
Initial parameter θ0 may be calculated using boundary condition at the right support (point 6): EIyðlÞ ¼ EIδP1 ðlÞ ¼ EIθ0 l
1 l3 l ¼ 0 ! θ0 ¼ l 6 6EI
444
12
Influence Lines Method
P=1
a 1
2
3
4
5
6
7
8
9
l
11
l
P=1
b
10
P=1
X1
M=X1
Primary system
d1P
d1P
Elastic curve due to P=1
X1=1
c d P1
d
dP1
Elastic curve due to X1=1
R=1/l
y
X1=1 Left span
x
dP1
X1=1
e
M1 1.0
f
Inf. line X1 (factor l) 0.048
0.084
0.072
0.096
– 0.072
0.096
0.084
0.048
–
Fig. 12.17 Two span continuous beam. (a) Design diagram. (b, c) Primary system and elastic curves due to P ¼ 1 and X1 ¼ 1. (d) Computation of δP1 due to X1 ¼ 1 as a function of position P ¼ 1. (e, f) Unit bending moment diagram and influence line for bending moment M ¼ X1 at middle support
Finally, displacement in the direction P caused by unit primary unknown X1 ¼ 1 may be written as follows l2 x x2 1 2 δP1 ¼ 6EI l l
ð12:27Þ
The position of the moving load P is determined by parameter x. Equation (12.27) presents the model of influence line for M ¼ X1 at middle support. Now we need to compute the unit displacement δ11. Bending moment diagram caused by unit primary unknown X1 is shown in Fig. 12.17e. Using the graph multiplication method and taking into account two identical portions 1-6 and 6-11, we get δ11 ¼
M1 M1 1 2 1 2 l ¼2 l1 1 ¼ 2 3 EI 3 EI EI
Equation (12.26) leads to the following equation for real influence line for bending moment at support 1: 1 l x x2 ILðX 1 Þ ¼ 1 2 ILðδP1 Þ ¼ δ11 4 l l
ð12:28Þ
12.4
Kinematical Method (Müller-Breslau Principle)
445
l For point 2 (x ¼ 0.2l) we have ILðX 1 Þx¼0:2l ¼ 0:2 1 0:22 ¼ 0:048l 4 Influence line for bending moment at middle support (point 6) is presented in Fig. 12.17f. It is obvious that this influence line is symmetrical. Discussion The same influence line had been obtained early in Sect. 12.1.2, Fig. 12.1f. The fundamental difference between both solutions is related to the free term of canonical equation. In Sect. 12.1.2 we calculated a free term δ1P, which is the angle of rotation at support 1, caused by moving unit load. In Sect. 12.4.1, Fig. 12.17f, we calculated a free term δP1, which is the vertical displacement at the point of application of the moving load caused by moment X1 ¼ 1 at the middle support. Now it is evident that for the construction of influence line for truss, as shown in Sect. 12.1.4, we have also used the Müller-Breslau principle (12.15); however for computation of displacement of the joints, which belong to the loaded contour, the elastic load method was applied (Sect. 8.9).
12.4.3 Continuous Beams: Models of Influence lines In Sect. 12.4.2 the precise analytical method has been applied for calculation of δP1. As a result, influence line was presented with computed numerical ordinates at specified points using formula (12.26). Without consideration of constant factor (1/δ11), this formula allows constructing the model of influence line. Thus, the model of influence line is constructed using only function δP1, which presents the vertical displacement diagram due to unit primary unknown which arise in the eliminated constraint. In other words, if at any point A we apply the unit factor X (force, bending moment), then the elastic curve presents the model of the influence line X at point A. If at point A we apply the unit displacement, which corresponds to the required factor X, then the elastic curve presents genuine influence line X at point A. Indeed, in this case the unit displacement δ11 is taken into account automatically. Displacement, which corresponds to the required factor X, means the linear displacement for required reaction, mutual angular displacement for bending moment, and mutual vertical displacement for shear. In more details this question had been discussed for statically determinate structures in Sect. 2.4. Example 12.4 Design diagram for continuous beam is presented in Fig. 12.8. Construct the models of influence lines for reactions and internal forces. Solution Influence line for reaction R1. In this case we need to eliminate a constraint, where the vertical reaction R1 arises and apply the positive force R1. Elastic curve due to unit force R1 in the new system is a model of influence line for R1 (this model is not shown). The specified ordinate at support 1 can be calculated by two ways. If load P ¼ 1 is located above support 1, then reaction R1 ¼ 1; therefore ordinate of influence line at this support equals to 1. Assume that δ11 ¼ 1. In this case, according to (12.26), a model of influence lines transforms into the real influence line. Therefore, for construction of the real influence line the point of application R1 should be displaced by unity in the direction of R1. It can be seen that if load is located on portions 1-2 and 3-4, then ordinates of influence lines for R1 are positive, i.e., the vertical reaction R1 is directed upward. Influence line for bending moment at support M3. In this case we need to introduce the hinge at support 3, to apply two positive unit couples M3 and show the corresponding elastic curve. If a load is located within portions 2-3 and 3-4, then ordinates of influence line are negative, i.e., the extended fibers at support 3 is located above the longitudinal axis of the beam. In order to transform this model into real influence line it is necessary to accept the mutual slope at support 3 equals to unity; this mutual slope is not shown. Influence line for bending moment Mk in the span. In this case we need to introduce the hinge at the section k, to apply two positive unit couples Mk and show the elastic curve. The positive ordinates of influence line show that if load is located over these ordinates, then the extended fibers at the section k are located under neutral axis of the beam. The model of influence line may be transformed into real influence line if a mutual angle of rotation at section k is equal to unity, as shown in Fig. 10.19, Inf. line Mk (see also Sect. 2.4, Fig. 2.22).
446
12
Influence Lines Method
P=1 1
2
FL
3
FR
5
k
n 1
4
+
Inf. line R1
R1 M3 Inf. line M3 –
– 1 + Inf. line Mk Mk + Inf. line Mn –
Mn +
Inf. line MFR
Qk + – Qk
Inf. line Qk 1
1 +
Inf. line Q3R
– +
Inf. line Q3L
–
– 1
Fig. 12.18 Continuous beam. Design diagram and kinematical construction of models of influence lines for reactions and internal forces; FL and FR are left and right focal points
For each span we can determine two specific points, which are called the left and right focal points. They are labeled as FL and FR. Formulas for computation of location FL and FR are presented in Sect. 9.3.3, formulas (9.12, 9.12a). Depending on where section is located (between two focal points, between support and focal point, and exactly at focal point), there may be three different shapes of influence line for bending moment. Figure 12.18 presents these points for span 3-4 only.
Problems
447
If the section k is located between two focal points FL and FR, then ordinates of influence line within corresponding span are positive (influence line for Mk). If a section n is located between left support and focal point FL, then ordinates of influence line within the corresponding span are positive and negative (influence line for Mn). The same conclusion will be done if the section is located between FR and right support. Assume, the section under consideration coincides with FR. In this case, if load P is located in the first and second spans, the bending moment does not arise in the section FR (influence line for MFR). Therefore, for construction of influence line for bending moment at any section within the span it is necessary to find location of focal points for given span and then define which of the above cases occurs. Influence line for shear force Qk. In this case we need to eliminate the constraint, which corresponds to the shear force at the section k and apply two positive shears Qk as shown in Fig. 12.16c. Elastic curve in the new system due to forces Qk ¼ 1 is a model of influence line for Qk. The features of the real influence line are the following: 1. Mutual vertical displacement of the two points at section k which belong to the left and right parts of a beam is equal to unity. 2. Two tangents at the same points are parallel. Shape of influence line for shear forces at the sections, which are infinitely close to the support, may be obtained as limiting cases, when the section is located within the span. It is obvious that the construction of influence lines for statically determinate multispan beams (Sect. 3.4.2, Fig. 3.10) using interaction scheme reflects the Müller-Breslau principle.
Summary The purpose of influence lines and their application for statically determinate structures in case of the fixed and moving load have been discussed in Chaps. 2–4. Of course, this remains also for statically indeterminate structures. However, in case of the statically indeterminate structures, the importance of influence lines and their convenience are sharply increased. Although the construction of influence lines for statically indeterminate structures is not as simple as for statically determinate ones, the consumption of a time is paid off by their advantages. Additional and very important advantage of influence lines for statically indeterminate structure is as follows: influence lines for primary unknown and any factor (reaction, bending moment, the angle of rotation, etc.) allows calculating not only this unknown and corresponding factor but also finding a distribution of internal forces for any types of fixed loads. It may be done combining the fixed and moving load approaches. Ability of an engineer to apply both methods separately and together increases his opportunity of analysis and allows performing in-depth qualitative and quantities investigation of structural behavior.
Problems 12.1. (a) (b) (c) (d) (e)
Continuous beam is shown in Fig. P12.1. Trace the models of the following influence lines:
Reaction of all supports. Bending moments at all supports and at sections n, and k. Bending moment for sections which coincide with the left and right foci points (FL, FR). Shear for sections n and k and for sections 1 and 4. Shear for sections which are placed infinitely close left and right to support 2.
P=1 1
FL
2
n Fig. P12.1
FR k
3
4
448
12.2.
12
Influence Lines Method
Trace the models of the following influence lines:
a) b) c) d)
Reaction of all supports. Bending moments at all supports and at section k. Shear for section k and shear for sections 1 and 4. Shear for sections which are placed infinitely close left and right to support 2. For problems (a)–(d) take into account indirect load application. e) Will be changed the shape of the influence line models in case of nonuniform continuous beams? f) Is it possible to change the sign of influence line as the result of the changing of the stiffness of the structure?
P=1
k
1
2
3
4
Fig. P12.2
12.3.
The uniform clamped-pinned beam is shown in Fig. P12.3.
a) Construct the influence lines for reactions RA and RB, moment at clamped support A, bending moment, and shear at section k (u ¼ 0.4). b) Construct the bending moment diagram if force P ¼ 10 kN is placed at point u ¼ 0.6. Use the above constructed influence lines.
P=1
0.4l
Ans. (b) RB=4.32 kN, MA=1.68l (kNm)
B A
k
MA
ul
ul RA
RB
Fig. P12.3
12.4. Design diagram of a frame is presented in Fig. P12.4. The relative flexural stiffnesses are shown in the circle; a = b = 0.5l. 1. Construct the influence line for horizontal reaction H at support A and bending moment at section k. 2. Construct the bending moment diagram if force P ¼ 100 kN is placed at point x/l ¼ 0.75. Use the influence line for primary unknown. P=1
x
Ans. H = 11.718kN R A = 30.859kN,
B 2
1 A
a
b l=10m
Fig. P12.4
M k = 95.7kNm h=5m
k
;
Problems
449
12.5. Analyze each design diagram and choose the most effective method for analytical construction of influence line for bending moment at section k.
a
d
c k
k
k
b k Fig. P12.5
12.6. Design diagram of a uniform semicircular two-hinged arch subjected to traveling load P ¼ 1 is presented in Fig. P12.6. Flexural stiffness of the arch is EI, axial stiffness of the tie is EtAt. Derive expression for truss H in terms of the angle φP. Take into account displacement of the arch due to bending deformation of the arch itself and axial deformation of the tie. Analyze limiting case (Et At ¼ 0, and Et At ¼ 1). P=1 y EI=const R Tie
A
O
B
Et At
x
l=2R Fig. P12.6
12.7. Design diagrams of uniform two-span beams with equal spans are presented in Fig. P12.7a,b; the flexural stiffness EI is constant. For both diagrams construct the influence line for angle of rotation Z at the middle support. Construct the bending moment diagram, if force P ¼ 100 kN is located at point 8; use the influence line for primary unknown.
a 1 2 3 4 5 6 7 8 9 10 11 l
l
b 2
4 l
Fig. P12.7
6
8
10 l
450
12
Influence Lines Method
12.8. The pinned-pinned-pinned beam is subjected to moving concentrated load P (Fig. P12.8). Construct the influence line for angle of rotation at the middle support and calculate the bending moment at support 1 due to force P in case of ul ¼ 0.4l. Calculate the vertical displacement at section k and show the elastic curve. P 0
1
2
k 0.5l
ul
ul l
l
Fig. P12.8
12.9. The frame is subjected to fixed load P (Fig. P12.9). The bending stiffness for each member is EI ¼ const. Construct the influence line for vertical reaction at point A. Use this influence line for constructing the bending moment diagram. Calculate the horizontal displacement at support A and angle of rotation of rigid joint due to fixed load P. P A k a
0.5a a Fig. P12.9
12.10. Design diagram of continuous three-span beam is shown in Fig. P12.10. Force P is applied at point 8. Construct the bending moment diagram. Use the influence line method (Table A.16b).
0
2
4
l Fig. P12.10
6
8
10
l
12
14
16
l
18
Ans.
Chapter 13
Matrix Stiffness Method
Matrix stiffness method (MSM) is a modern powerful method of analysis of engineering structures. Its effective and widespread application is associated with availability of modern computers and effective computer programs. The MSM allows performing detail analysis of any sophisticated 2D and 3D engineering structure and takes into account different features of a structure and loading. The method demands a set of new concepts. They are finite elements, global and local coordinate systems, possible displacements of the ends, ancillary diagrams, initial matrices, stiffness matrix of separate element and structure in whole, etc. This method uses the idea of the displacement method and contains its further development: arbitrary structure should be presented as a set of finite elements and three aspects of any problem—static, geometrical, and physical—should be presented in matrix form. The MSM does not require constructing bending moment diagram caused by unit primary unknowns in the primary system. Instead it is necessary to prepare few initial matrices according to strong algorithms and perform matrix procedures by computer using the standard programs. This chapter contains the detailed discussion of MSM; all auxiliary diagrams and initial matrices are constructed by hand. Such presentation of material allows reader to see the internal logic and the features of the method, to understand physical meaning of each step and to find the corresponding result in analysis of a structure by displacement method in canonical form. At the present time MSM is developed with great detail. The reader can find out in literature the different presentations of MSM. In our book we will consider this method in simplest form (Anokhin 2000) and apply it for analysis of the planar bar structures only. Among them are the statically indeterminate beams, frames, trusses, and combined structures subjected to different external exposures (loads, settlements of supports, change of temperature, influence lines).
13.1
Basic Idea and Concepts
The fundamental concepts of MSM are the following: the finite elements; the possible displacements of the ends and degree of kinematical indeterminacy (degree of freedom); the global and local coordinate systems.
13.1.1 Finite Elements Each structure may be subdivided into separate elements of simple geometrical configuration called the finite elements. This step has no theoretical justification. The result of the stress-strain analysis for each finite element is presented in existing handbooks. For presentation of the given structure as a set of the finite elements, the features of the structure as well as required accuracy of analysis should be taken into account. Engineering experience is an important factor for choosing the type and number of the finite elements. In case of a truss, the separate members of the truss may be adopted as finite elements. Therefore, the discrete model of the truss in terms of finite elements coincides with design diagram of the truss. In case of the frame with uniform members, separate members of the frame may also be adopted as the finite elements (Fig. 13.1a). If the frame contains the member with variable cross section, then this member may be divided into several portions with constant stiffness along each element (Fig. 13.1b).
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_13
451
452
13
a
P
P
b
1
Matrix Stiffness Method
1
2
2
3
Fig. 13.1 Frames and their presentation by the set of finite elements
The uniform beams in Tables A.3–A.7 may be considered as the simplest finite elements. The finite elements can be one, two, or three dimensional. This chapter deals with planar bar structures only, so the finite elements are straight thin bars with different types of constraints at the ends. They are hinged-hinged (truss member), fixed-pinned, and fixed-fixed (frame member). General idea of MSM. At the end points of each finite element, some displacements and interaction forces arise. For structure in whole these forces are internal, while for each finite element these forces should be considered as the external loads. For all finite elements we can write three groups of equations: (1) equilibrium, (2) physical, and (3) geometrical equations. Equilibrium equations take into account external forces for each finite element. Physical equations relate forces and displacements at end points of each element. Geometrical equations describe continuity conditions between ends of the elements. Solving of these equations allows determining displacements and forces at end points of each element.
13.1.2 Global and Local Coordinate Systems The local coordinate system is referred to specified element, while global system is related to the whole structure. To understand these concepts, let us consider a truss, subjected to force P (Fig. 13.2). x1
x3 a
y3
Y
1
y2
y1 4
X
3 x2
A
2
D
P
Fig. 13.2 Local x-y and global X-Y coordinate systems
The members 1,2,3,4 met at the joint A. Internal forces Ni of the elements 1–4 do not coincide with given external load P. Axial deformation Δi of each element does not coincide with vertical displacement Δ of joint A. Therefore, we need to distinguish between the internal and external forces, deflection of separate members, and displacements of joints of the structure. For this purpose, we introduce two coordinate systems. Each element of the structure, corresponding internal force and deformation may be referred to the local coordinate system x-y. The origin of this system coincides with initial point of this element and one of the axis directed along the element itself as shown for members 1–3; for member 4 the local coordinates are not shown. The structure in whole, the external force, and displacements of the joints are referenced to global coordinates X-Y. The difference between the local displacements Δi of each element and the global vertical displacement Δ of joint A can be seen from the following relationships: x1 ¼ Δ, x2 ¼ 0,
y1 ¼ 0; y1 ¼ 0;
x3 ¼ Δ sin α,
y3 ¼ Δ cos α, . . .
13.2
Auxiliary Diagrams
453
13.1.3 Displacements of Joints and Degrees of Freedom Once a structure is presented as a set of finite elements, we need to identify the possible displacements of the ends of each member. In fact these displacements present unknowns of the displacement method (angular displacements of the rigid joints and their independent linear displacements). These displacements are called the possible angular and linear displacements of the joints. The term possible means that in given structure such displacement is possible, but not necessarily realized. For example, in case of two-span continuous beam a section at the middle support generally rotates; however, if the beam and its loading are symmetrical, then this angle of rotation is zero. In summary, we can see that any possible displacement of the joint is a displacement in global coordinates. Degree of kinematical indeterminacy n of a structure is determined by formula n ¼ nr þ nd, where nr is a number of unknown angles of rotation of the rigid joints of a structure and nd is a number of independent linear displacements of joints. Parameter n defines the degrees of freedom of a structure. The beam in Fig. 13.3a has one unknown of the displacement method, i.e., the degree of kinematical indeterminacy equals one. So the degree of freedom n ¼ 1. The frame in Fig. 13.3b has three unknowns of the displacement method (two angular displacements Z1 and Z2 of the rigid joints and one linear displacement Z3 of the crossbar). So the number of degrees of freedom n ¼ 3.
a
Z1
b
Z1
Z2
Z3
Fig. 13.3 Elastic curves and concept of degrees of freedom
Thus, the term “degrees of freedom” implies possibility of angular and linear displacements of the joints are caused by deformation of the structure. It can be seen that for kinematical analysis of a structure (Chap. 1) the term “degrees of freedom” had other meaning: in this case the concept “degrees of freedom” implies independent displacements of the system which contains the absolutely rigid discs.
13.2
Auxiliary Diagrams
To present a structure in such form which can be accepted by a modern computer, the entire design diagram should be expanded by three auxiliary diagrams. They are joint–load (J-L), displacement–load (Z-P), and internal forces–deformation (S-e) diagrams.
13.2.1 Joint–Load (J-L) Diagram This diagram presents the transformation of the arbitrary load into the equivalent joint loads. For construction of joint–load diagram, the following steps must be completed: 1. Identify the possible angular and linear displacements of the joints. 2. Construct the bending moment diagram M 0P in the primary system of the displacement method ( first state) due to external exposures (loads, settlements of supports, temperature change). 3. Present the moments and forces as the joint load in direction of the possible displacements (second state). Continuous beam in Fig. 13.4 has one unknown angular displacement at support 1. We need to convert the given load to the equivalent moment at the support 1. The first state is the bending moment diagram M 0P in primary system. The fixed-end moment equals to M 1 ¼ ql21 =8 ¼ 16 ðkNmÞ. We can see the joint moment 16 kNm acts on the introduced support 1 counterclockwise. Corresponding extended fibers in vicinity of joint 1 are located above the longitudinal axis; they are shown by dashed line. This moment should be transported on the joint–load diagram counterclockwise; this diagram presents the second state.
454
13
a
c
q=2kN/m
1
1
l1=8m
l2
16kNm
M1 = 16kNm
q
b
1-state
Matrix Stiffness Method
d
M P0
2-state
1
J-L
Mj1= 16kNm Fig. 13.4 (a, b) Design diagram of continuous beam and bending moment diagram in primary system; (c) Transformation of ending moment on the support 1; (d) Equivalent joint load diagram (J-L)
Figure 13.5a presents design diagram of the frame; P1 ¼ 10 kN, P2 ¼ 16 kN, q ¼ 3 kN/m. Degree of kinematical indeterminacy equals 4, where nr ¼ 3 and nd ¼ 1. The primary system is shown in Fig. 13.5b; all introduced constraints are labeled as 1–4 and the specified sections as 5–8. Also this figure contains the bending moment diagram in primary system M 0P (first state). Bending moments which act on joints 1, 2, 3 are shown in Fig. 13.5c. ql215 3 42 ¼ ¼ 6 kNm, 8 8 3 M 72 ¼ P2 l27 ¼ 18 kNm, 16 P l 10 8 M 23 ¼ 1 23 ¼ ¼ 10 kNm, 8 8 M 32 ¼ M 23 ¼ 10 kNm
M 15 ¼
b
a P1
q
4m
4m
c
2 1 P2
3m 5m
P1 M 3-2 4 3
M 2-3
M1-5
3m
P2
5
q
6
M P0
7
8
M7-2
4m
Joint 3
Joint 2
Joint 1
M1-5 1
2
3 M2-3
d
M3-2
e R1-5 1
Cross bar 1-2-3-4 2
Mj2=10kNm
Mj3=10kNm
3 R2-7
Pj4= 12.5kN Mj1=6kNm
Joint-load diagram
Fig. 13.5 (a) Design diagram of the frame; (b) Bending moment diagram in the primary system of the displacement method; (c) Computation of the fixed-end moments; (d) Computation of equivalent joint load Pj4; (e) Joint-load diagram (state 2)
13.2
Auxiliary Diagrams
455
Figure 13.5d shows the horizontal forces which arise in the vertical loaded members 1–5 and 2–7. These forces are equal 5ql15 5 3 4 ¼ 7:5 kN; ¼ 8 8 5P 5 16 ¼ 2¼ ¼ 5 kN 16 16
R15 ¼ R27
Both of these forces are transmitted on the crossbar, so final horizontal joint load equals Pj4 ¼ R15 þ R27 ¼ 7.5 þ 2.5 ¼ 12.5. This force may be applied at any joint of the frame (1, 2 or 3). Thus, the entire load may be presented as equivalent moments Mj1, Mj2, Mj3 at joints 1–3 and force Pj4 in horizontal direction; subscript j means that entire loads are transformed to joint load. The final J-L diagram (second state) is shown in Fig. 13.5e. M j1 ¼ 6 kNm;
M j2 ¼ 10 kNm;
M j3 ¼ 10 kNm;
P
j4
¼ 12:5 kN
Note again that joint load presents the equivalent bending moments and forces, which are merely transported on the joints and on the crossbar on the same direction, but they are not reactions as in displacement method. In case of truss all loads are applied at the joints; therefore the joint–load diagram coincides with entire design diagram. Example 13.1 The continuous beam is subjected to change of temperature as shown in Fig. 13.6a. Construct the joint–load diagram.
a
EI, a
-24° b
a
h=0.4m
c
M1-0
b 0
d
M3-2 M3-4
M1-2
1
2
3
4
c d
Joint 3
Joint 1
Joint-load diagram
M1-2 M3-2
M1-0
M3-4 3
1
30aEI
30aEI
Fig. 13.6 Continuous beam subjected to change of temperature. (a, b) Design diagram of the beam and bending moment diagram in primary system (first state); (c, d) Fixed-end moments and joint-load diagram (state 2)
Solution The primary system of the displacement method and bending moment diagram caused by given temperature exposure is shown in Fig. 13.6b (first state). The bending moments for pinned-fixed beam and fixed-fixed beam are 3EIα Δt 3EIα 24 ¼ ¼ 90αEI, ðTable A:3, row 8Þ; 2h 2 0:4 EIα Δt ¼ ¼ 60αEI, ðTable A:4, row 7Þ h
M 10 ¼ M 34 ¼ M 12 ¼ M 32
Figure 13.6c presents the bending moments which act on the joints 1 and 3. The equivalent (resulting) joint moments equal M j1 ¼ M 10 M 12 ¼ 90αEI 60αEI ¼ 30αEI ðkNmÞ counterclockwise; M j2 ¼ 0, M j3 ¼ M 34 M 32 ¼ 90αEI 60αEI ¼ 30αEI ðkNmÞ clockwise: The final joint–load diagram or second state is shown in Fig. 13.6d. Example 13.2 The uniform continuous beam is subjected to the angular displacement φ of fixed support 0 and vertical displacement Δ of rolled support 2 (Fig. 13.7a). Construct the joint–load diagram.
456
13
Matrix Stiffness Method
Solution The primary system of the displacement method and bending moment diagram caused by given settlements of supports is shown in Fig. 13.7b (first state).
a
j j=0.05rad
D=0.032m 2
1
M1-2
M1-0
b
M P0
D
0
l1=5m
l2=4m
d
Joint 1
c M1-0
Joint-load diagram
1
M1-2 1
Mj1=0.014EI (kNm)
Fig. 13.7 Settlement of supports of redundant beam. (a, b) Design diagram of the beam and bending moment diagram in primary system (first state); (c, d) Fixed-end moments and joint-load diagram (state 2)
According to tabulated data, the bending moments for fixed-fixed and pinned-fixed beams are 2EI 2EI 0:05 ¼ 0:02EI ðkNmÞ, ðTable A:4, row 1Þ, φ¼ l1 5 3EI 3EI ¼ 2 Δ ¼ 2 0:032 ¼ 0:006EI ðkNmÞ, ðTable A:3, row 2Þ 4 l2
M 10 ¼ M 12
Computation of equivalent moment at the joint 1 and corresponding J-L diagram (first state) are shown in Fig. 13.7c–d.
13.2.2 Displacement–Load (Z-P) Diagram This diagram shows a numeration of angular and independent linear displacements Z at the joints and their positive direction. Also this diagram contains type of load, which corresponds to displacements Z, and their positive directions. Sign rule. Assume that the positive angular displacement Z occurs clockwise; the positive horizontal linear displacement occurs from left to right, and vertical linear displacement occurs upward. Direction of the positive load coincides with positive direction of the displacement. Design diagram of a frame and corresponding primary system of the displacement method are shown in Fig. 13.8a, b. Introduced constraints 1–3 prevent angular displacements Z1–Z3; corresponding loads are moments M1–M3. Constraint 4 prevents linear displacement Z4, so corresponding load is P4.; these loads and all displacements are shown in positive directions. The Z-P diagram contains only skeleton of the entire scheme of the structure, the type of end displacements, its positive direction, and corresponding loads (Fig. 13.8c). Geometrical parameters, stiffnesses, type of loading, and any numerical data are not shown on the Z-P diagram. So this diagram reflects only conceptual character of structure.
a
b Z1, M1
c Z2, M2
Z3, M3 Z4, P4
1
2
3
1
2
3 4
4 Z-P diagram
Fig. 13.8 Redundant frame. (a) Design diagram. (b) Primary system, types of unknowns Zi and corresponding type of loads; (c) DisplacementLoad (Z-P) diagram
13.2
Auxiliary Diagrams
457
Construction of Z-P diagram for truss is shown in Fig. 13.9a. Support A prevents two displacements; therefore, this joint does not have possible displacements as well as corresponding joint loads. Joint C prevents horizontal displacement, so possible displacement of this joint and corresponding load is directed vertically (labeled 1). The possible displacement of joint B and corresponding load are directed horizontally (4). All other joints of frame have two possible displacements (vertical and horizontal) and two corresponding joint loads.
a
b
3
1
C
2 B
A
4
A 6
5
Fig. 13.9 Redundant truss. (a, b) Design diagram and corresponding Z-P diagram
It is obvious that Z-P diagram (Fig. 13.9b) shows the positive possible displacements and corresponding external loads in global coordinates. Pay attention that the symbol P is generalized notation for bending moments as well as for forces. Similarly, the symbol Z is generalized notation for angular and linear displacements. Also note that both, the joint–load (J-L) and displacement–load (Z-P) diagrams, use the term “load.” However, to distinguish this term, various symbols are used, such as L and P. The reason for this lies in the various concepts of the term “load.” Joint–load (J-L) diagram deals with numerical values of equivalent joint loads, while displacement–load (Z-P) diagram deals with type of a load only, which corresponds to character of introduced constraints according to displacement method.
13.2.3 Internal Forces–Deformation (S-e) Diagram This diagram shows a numeration and positive directions of all unknown internal forces S, which correspond to deformations e at the end points. Figure 13.10a shows three bar elements from any truss. Since all connections are hinged, then for all members only axial deformations e are possible; corresponding internal forces are axial forces. The concept of S-e diagram for truss is shown in Fig. 13.10b. Each member contains three parts for S-e diagram. They are two parts adjacent to joints and one intermediate part. Positive internal force which arises in the member of a truss is directed away from the joints. The positive internal forces which act on the intermediate part are shown by bold arrows. i-th bar
a
j-th bar
Si
b Sj
Sk
Fig. 13.10 Concept of S-e diagram for truss
Figure 13.11a presents the design diagram of the truss (loading is not shown) and numeration 1-8 of the members. The corresponding S-e diagram is shown in Fig. 13.11b. According to Fig.13.10b, directions of positive internal forces of the intermediate part of i-th bar are shown for each element of a truss. In other words, for each element the arrows are directed towards the joints. Since this arrow is related to intermediate part, then S-e diagram means the positive force Si in i-th member and corresponding positive axial deformation ei.
458
13
a
b
2 1
6
7 5
2
3
8
Matrix Stiffness Method
1
6
7
4
3
8
5
4
Fig. 13.11 Redundant truss. (a, b) Design diagram and corresponding S-e diagram
For beams we will denote the points with unknown bending moments. These points are located infinitely close to the left and right of the supports. Figure 13.12a shows a continuous beam. Each element has three portions; they are intermediate portion and two portions which adjacent to the supports (Fig. 13.12b). The positive moments (they are shown by solid arrows) rotate the intermediate portion of element around opposite end in clockwise direction. These moments are transmitted on the adjacent portion with opposite directions (shown by dotted arrows). Corresponding S-e diagram is presented in Fig. 13.12c. This diagram shows unknown positive internal moments S. These moments are shown by dotted arrows and act at the points located infinitely close to the left and right of the support. The ordinates of S-e diagram are plotted on the extended fibers.
a
b
1
2
3
c
S-e diagram
S2
S1
4
5
S4
S3
S5
Fig. 13.12 Redundant beam. (a) Design diagram; (b) Positive bending moments which act on the intermediate portion (solid arrows) and moments which act at the vicinity of supports (dotted arrows); (c) Concept of S-e diagram; the S-e diagram is plotted on the extended fibers
For frames, as for beams, we will denote the points with unknown internal forces. The required internal forces are bending moments at the ends of each member. Figure 13.13a shows the frame; the joints are labeled as i, j, k, where j and k are related to the end points of the crossbar and strut, respectively.
b
a i
Si
c
Sj
Sj
j k
Sk
Si
Sk
Fig. 13.13 (a) Design diagram of redundant frame; (b, c) Construction and final S-e diagram
Construction of S-e diagram is presented in Fig. 13.13b. This diagram shows the type of internal loads, i.e., the bending moments. The final S-e diagram of positive bending moments is shown in Fig. 13.13c.
Summary The joint–load (J-L), displacement–load (Z-P), and internal forces–deformation (S-e) diagrams deal with similar concepts. For this concept, the J-L and Z-P diagrams uses the term load, while for S-e diagram term forces. For these concepts the different symbols are used; they are L, P, and S. The symbol L is used for presentation of the given loading by equivalent joint loads. The symbol P is used for presentation of the type of the load, which corresponds to possible displacements Z. The symbol S is used for presentation of the type of the unknown internal forces S.
13.3
13.3
Initial Matrices
459
Initial Matrices
The first group of initial matrices describes the loading of structure and unknown internal forces.
13.3.1 Vector of External Joint Loads The arbitrary exposure on a structure (fixed or moving load, settlement of supports, change of temperature) should be presented in mathematics form. Vector P of the external joint loads serves for this purpose. This vector may be constructed on the basis of the joint–load (J-L) and the displacement–load (Z-P) diagrams. Let us consider two cases how to create vector of external joint loads. The truss is loaded as shown in Fig. 13.14a; the Z-P diagram is shown in Fig. 13.14b.
a
P1=10 kN C
b
3
1
2
P2=8 B P4=20
A
4
A
P6=15
6
5
Fig. 13.14 (a, b) Design diagram of the redundant truss and corresponding Z-P diagram
Since the truss has six possible displacements, then vector of external load will contains six entries, i.e., 2 3 P1 ! 6 7 P ¼ 6 7: 6 7 6 P6 7 For convenience, this vector we will present in transposed form as !
P ¼ bP1
P2
P 6 cT ; !
the symbol Т is reserved for the matrix transpose procedure. Considering Fig. 13.14a, b we can compose the vector P of joint loads: P¼ b 10
8 0
20
0 15 cT
The first entry 10 means that in the direction 1 (Fig. 13.14b) there acts the force P1 ¼ 10. The negative sign at the fourth entry indicates that external force P4¼20 is directed opposite the positive fourth possible displacement. Zeros for third and fifth entries show that in directions 3 and 5 of Z-P diagram the external forces are absent. Now let us compose the vector of external forces for frame shown in Fig. 13.15a. For this frame the joint–load and Z-P diagrams are shown in Fig. 13.15b, c. Comparing the J-L diagram with Z-P diagram we can form the following vector of external loads !
P¼ b 6
10
10
12:5 cT
460
13
a
Matrix Stiffness Method
c P1
3kN/m
4m
Mj2=10kNm
3m
P2
Pj4= 12.5kN
3m
Mj1=6kNm
d
5
M 2-3
q
M 3-2
P1
1
2
1
4
2
M1-5
J-L diagram
4m
4m
5m
b
Mj3=10kNm
3 4
3
P2 6
M P0
7
Z-P diagram
8
M7-2 Fig. 13.15 Redundant frame. (a) Design diagram; (b) Primary system and corresponding bending moment diagram; (c, d) Joint-Load (J-L) and ! Displacement-Load (Z-P) diagrams. Both diagrams serves for composing of the external load vector P
13.3.2 Vector of Internal Unknown Forces The required internal forces may be presented in the ordered mathematics form. Matrix-vector of unknown internal forces S serves for this purpose. The entries of this vector are the axial forces for end-hinged members for trusses and the bending moments at the ends of the bending members. This vector can be constructed having the joint–load (J-L) and internal forces– deformation (S-e) diagrams. Let us consider the frame in Fig. 13.15a. Skeleton of this frame (without external load) and numeration of the end sections 1–10 are shown in Fig 13.16a. The positive directions of the end moments for each element of the frame and corresponding S-e diagram are shown in Fig. 13.16b, c. Pay attention that the sections where the moments are zeros have been eliminated from consideration. They are points at the upper rolled support A and at hinge H that belongs to the second column (Fig. 13.16b).
b
a
c
A 3 2 1
4
5
7
H 6
8
7
5
3
8
M3
M2
9
2
10
1
4
M4
9 6
10
M1
M8
M5 M9
M7
M6
M10
Fig. 13.16 Redundant frame. (a) Skeleton of the frame with numeration!of sections; (b, c) Positive unknown internal forces and corresponding S-e diagram; this diagram serves for composing of the internal force vector S for specified sections 1–10
The bending moment diagram in primary system M 0P (Fig. 13.15b, 13.15b) and S-e diagram allows constructing the vector of the moment at the indicated sections 1–10 (Fig. 13.16a, b) in primary system (state 1) due to given load. This vector becomes S1 ¼ b 0 Number of sections, Fig. 13.16a
1
0 2
6 0 3
4
0 5
18 10 6
7
10
0
0 cT
8
9
10
13.4
Resolving Equations
461
The signs of the moments are established on the basis of the M 0P and S-e diagrams. For example, since a moment M15 ¼ 6 kNm on the bending moment diagram M 0P is plotted left (Fig. 13.15b), and the positive moment M3 at the same section on the S-e diagram is plotted right (Fig. 13.16c), then third entry of the S vector has a negative sign. The vector of required internal moments at the specified sections 1, . . ., 10 is !
S ¼ b M1
M2
M9
M 10 cT
This vector will present a result of analysis of the frame by a computer. If that occurs, for example, the first entry (bending moment M1) will be positive, then according to S-e diagram (Fig. 13.16c), the extended fibers at the section 1 (bottom of the left column) will be located at the right. Therefore the internal moment M1 acts on the intermediate part of the strut as shown in Fig. 13.16b, i.e., clockwise.
Summary The following combination of two diagrams leads to the initial vectors: !
1. Joint–load (J-L) diagram þ displacement–load (Z-P) diagram ! Vector of external joint loads P . !
2. M 0P diagram þ (internal force–deformation (S-e) diagram) ! Vector of internal forces S 1 in the first state.
13.4
Resolving Equations
Matrix stiffness method considers three sides of problem: they are static, geometrical, and physical. On their basis the second group of initial matrices may be constructed. They are static matrix, deformation matrix, and stiffness matrix in local coordinates. These matrices describe the different features of structure: configuration of structure, its geometry and supports, stiffness of each element, and the order of their connection.
13.4.1 Static Equations and Static Matrix Assume that structure is m-times kinematically indeterminate. It means that structure has m external joint loads. These loads, ! according to Z-P diagram, may be presented as vector P ðm1Þ ; the number of rows is m. Let the structure have n unknown !
internal forces; these unknowns according to S-e diagram may be presented as vector S ðn1Þ ; the number of rows is n. Both vectors are connected by static matrix A(mn) by formula !
!
P ¼ AS
ð13:1Þ
This relationship is called the static matrix equation. Matrix A expresses the vector of external forces Pi in terms of required internal forces Sj. The number of rows m of the static matrix A equals to the number of the possible displacements; the number of columns n of the matrix A equals to the number of the unknown internal forces. If m > n, then a structure is geometrically changeable; if m ¼ n, then a structure is statically determinate; if m < n, then a structure is statically indeterminate. In fact, the matrix Eq. (13.1) describes the structure, its supports, type of joints, and order of the elements connections. The static matrix A(mn) may be constructed on the basis of a set of equilibrium conditions for specific parts of a structure in ordered form. Equilibrium equations for frames must be constructed for each joint which have the angular displacement and for part of the frame which contains the joints with linear displacements. In case of possible angular and/or linear displacement the equation ∑M ¼ 0 and/or ∑X ¼ 0, respectively, should be used. In case of trusses the equilibrium of each joint in form ∑X ¼ 0 and/or ∑Y ¼ 0 should be considered. Each possible load (each component of the vector Р) must be presented in terms of all unknowns (all components of the ! vector S ). Left part of each equation of equilibrium should contain only a possible load, while the right part—unknown internal forces. The type of possible load corresponds to the type of possible displacement according to diagram Z-P. If a
462
13
Matrix Stiffness Method
possible displacement is the angle of rotation, then corresponding load is a moment. If possible displacement is linear, then a corresponding load is a force. Let us consider few examples for construction of static matrix. Figure 13.17a presents the continuous beam. Unknown angular displacements Z of intermediate supports A and B and corresponding moments P1 and P2 are labeled on the Z-P diagram by 1 and 2, so m ¼ 2 (Fig. 13.17b). Positive unknown internal moments Mi (i ¼ 1–5) at the ends of each portions are labeled on the S-e diagram by 1–5, so n ¼ 5 (Fig. 13.17c). F
a A
c
1
2
3
4
5
B S-e
b 2
1
Z-P M1
M2
M4 M3
d
M5
P2
P1 Joint A
M2
M3
P1 = M 2 + M 3
Joint B M4
M5
P2 = M 4 + M 5
Fig. 13.17 Continuous beam. (a, b) Design diagram and corresponding Z-P diagram. (c) S-e diagram. (d) Construction of static matrix A for beam
Equilibrium equations for intermediate supports A and B lead to the following results (Fig. 13.17d) X M A ¼ 0 : P1 ¼ 0 M 1 þ 1 M 2 þ 1 M 3 þ 0 M 4 þ 0 M 5 X M B ¼ 0 : P2 ¼ 0 M 1 þ 0 M 2 þ 0 M 3 þ 1 M 4 þ 1 M 5 In matrix form (Eq. 13.1) these equations can be rewritten as follows 3 2 M1 7 6 6 M2 7 7 P1 0 0 1 1 0 0 6 7 , Static matrix A M ¼ ¼ 6 37 ð25Þ 6 0 0 0 0 1 1 6 P2 7 6 M4 7 7 6 6 M5 7
1
1
0 0
0
0
1 1
ðaÞ
Each row of matrix A presents the corresponding equilibrium equation; the entries of the matrix are coefficient at Mi. Note again that equilibrium equations are formulated for possible loads P, which corresponds to possible displacements, but not for given load F. The truss in Fig. 13.18a contains 4 possible linear displacements and corresponding external loads P, labeled as 1-4 (m ¼ 4) (Fig. 13.18b). Support A has no linear displacements; therefore, Z-P diagram does not contain the vectors of displacement at support A. The possible displacements 1 and 4 describe the rolled supports B and D as well as their orientation. Numeration of elements is shown in Fig. 13.18c, so n ¼ 5.
13.4
Resolving Equations
a
463
B
C
F1=16kN F2=10kN
b
1
3
c
a 4m
1
d
S-e
P3
P1 S2
Joint B
4
Z-P
3m
S3
S2
S4
P2
Joint C
S1
3
5
4
D
A
3m
2
2
Joint D
P4
S3
S5
Fig. 13.18 Redundant truss. (a–c) Design diagram and corresponding Z-P and S-e diagrams; (d) Construction of static matrix A for truss; Si are internal forces and Pi are possible joint loads which corresponds to the possible displacements
Equilibrium equations should be formulated for each joint with possible load (Fig. 13.18d) X Joint B Y ¼ 0 : P1 ¼ S1 Joint C
Joint D
X X X
X¼0:
P2 ¼ S2 S3 cos α þ S5 cos α
!
Y ¼0:
P3 ¼ S3 sin α þ S5 sin α
P3 ¼ 0:8S3 þ 0:8S5
X¼0:
P4 ¼ S4 þ S3 cos α
!
!
P2 ¼ S2 0:6S3 þ 0:6S5
P4 ¼ 0:6S3 þ S4
These equations in the ordered form can be rewritten as follows P1 ¼ 1 S1 þ 0 S2 þ 0 S3 þ 0 S4 þ 0 S5 P2 ¼ 0 S1 þ 1 S2 0:6 S3 þ 0 S4 þ 0:6 S5 P3 ¼ 0 S1 þ 0 S2 þ 0:8 S3 þ 0 S4 þ 0:8 S5 P4 ¼ 0 S1 þ 0 S2 þ 0:6 S3 þ 1 S4 þ 0 S5 !
These equations allow presenting the vector of possible joint forces P ð41Þ ¼ bP1 P2 P3 unknown internal forces Sð51Þ ¼ bS1 S2 S3 S4 S5 cT using the static matrix A(45) 2 3 S1 2 3 2 3 1 0 0 0 0 P1 6 7 S2 7 6 P 7 6 0 1 0:6 0 0:6 7 6 7 6 27 6 7 6 6 6 7¼6 7 6 S3 7 6 P3 7 4 0 0 0:8 0 0:8 5 6 7 7 6 7 6 S4 7 6P 7 6 7 0 0 0:6 1 0 4 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 6 S5 7
P4 cT in terms of the vector of
Að45Þ
Each row of this matrix presents the corresponding equilibrium equation; the entries of the matrix are coefficient at Si. We can see that static matrix describes the structure. First of all, this structure is statically indeterminate, since the number of rows m ¼ 4 while the number of columns n ¼ 5. Moreover, the matrix A describes the ways of connections of different members. For example, second row shows that members 2, 3, and 5 are connected together, while the last row shows that members 3 and 4 are connected. At last, this matrix also describes the inclination of different members. Therefore, the static matrix A is a very informative matrix. Example 13.3 The frame is shown in Fig. 13.19. Construct the static matrix. Solution This structure has two angular displacements of the rigid joints and one linear displacement. The ancillary Z-P diagram is presented in Fig. 13.19b. The arrows 1 and 2 show possible angular displacements and corresponding possible external loads (moments, kNm); the arrow 3 shows possible linear displacements and corresponding possible external load !
(force, kN). The vector of possible joint loads is P 31 ¼ bP1
P2
P 3 cT .
464
13
c
b
a 3 2
4
1
6
2
3
5
3
4
A
6 B
M4
C
2
h
d
S-e diagram
1
M3
M5
M6
M1
P2
P1 Joint A
M2 M3
5
Z-P diagram
1
Matrix Stiffness Method
P1 = M 2 + M 3
Joint B
M6
M4
P2 = M 4 + M 5 + M 6
M5
M2 P3
Cross bar A-C M1/h+M2/h
SX = 0:
P3 = -
M1 + M 2 M 5 h h
M5/h
Fig. 13.19 Redundant frame. (a) Design diagram and notation of specified sections; (b, c) Ancillary Z-P and S-e diagrams; (d) Equilibrium conditions for rigid joints, crossbar and construction of static matrix A
The ancillary S-e diagram is presented in Fig. 13.19c. The vector of unknown internal moments is
! S ð61Þ
¼ bM 1 M 2 M 6 cT . Since m < n, then frame is three times statically indeterminate since n m ¼ 3. Equilibrium equations for rigid joints A, B and for crossbar ABC are presented in Fig. 13.19d. In matrix form these static relationships can be presented as follows 3 2 M1 7 2 3 2 3 6 6 M2 7 P1 0 1 1 0 0 0 7 6 6 7 6 7 6 M3 7 7 0 0 1 1 15 6 6 P2 7 ¼ 4 0 6M 7 6 7 6 47 6 P3 7 1=h 1=h 0 0 1=h 0 7 6 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 6 M 5 7 7 6 Að36Þ 6M 7 6
Note again, the static matrix defines the structure itself (supports, connections of the members), and does not depend on the type of external exposure.
13.4.2 Geometrical Equations and Deformation Matrix These equations present the relationships between deformations e of the elements and displacements Z of the joints. The required relationships is !
!
e ¼ BZ
ð13:2Þ
It is obvious the concepts “deformation” and “displacement” are related to the end points of the finite element. Vector of deformation is e ¼ b e1 e2 en cT . The entries of this vector are deformations of the elements at the ðn1Þ
sections with unknown internal forces, i.e., at the end points. Therefore, dimension n of this vector equals to the number of !
unknown internal forces S. Vector of joint displacement is Z ¼ b Z 1 ðm1Þ
Z2
Z m cT ; dimension m of this vector equals
to the number of primary unknowns Z of displacement method. Matrix B(nm) presents the matrix of deformation. The entry bij (i-th row and j-th column) means deformation in direction of unknown internal force Si caused by displacement Zj. In fact, the matrix equation (13.2) describes the conditions of the continuity of deformations. Let us show the construction of matrix deformation B for continuous beam (Fig. 13.20a). The primary system, the positive angular displacements Z, and positive direction of the unknown bending moments Mi (i ¼ 1, . . ., 5) at the ends of each member are shown in Fig. 13.20b.
13.4
Resolving Equations
a
465
Z1
F 1
2 3
M2
b
4 5
Z2 M4
1
2
M3
M5
M1
Fig. 13.20 Construction of the deformation matrix. (a) Design diagram of a beam and numeration of specified points; (b) Primary system of displacement method and S-e diagram
If introduced constraint 1 has angular displacement Z1, then deformation (angle of rotation) at the section 1 will be equal, e1 ¼ 0 Z1. If introduced constraint 2 has angular displacement Z2, then deformation at the section 1 will be zero, also. Thus, the deformations in direction Mi are 2 3 2 3 0 0 e1 e1 ¼ 0 Z 1 þ 0 Z 2 6 7 6 7 6 e2 7 6 1 0 7 & ’ e2 ¼ 1 Z 1 þ 0 Z 2 6 7 6 7 Z1 6 7 6 7 7 6 6 e3 ¼ 1 Z 1 þ 0 Z 2 or in matrix form 6 e3 7 ¼ 6 1 0 7 7 Z2 6 7 6 7 e4 ¼ 0 Z 1 þ 1 Z 2 6 e4 7 6 0 1 7 6 7 4 5 6 7 e5 ¼ 0 Z 1 þ 1 Z 2 6 e5 7 0 1 |fflfflfflfflffl{zfflfflfflfflffl} Bð25Þ
The second equation shows that if introduced constraint 1 has angular displacement Z1, then corresponding deformation (angle of rotation) at the section 2 will be same, i.e., e2 ¼ 1 Z1 . If introduced constraint 2 has angular displacement Z2, then corresponding deformation at the section 2 will be zero, because here is placed the introduced constraint 1. Note that deformation matrix B and static matrix A are connected as follows ð13:3Þ
B ¼ AT This is general rule, so for calculation of matrix B we can apply two approaches.
13.4.3 Physical Equations and Stiffness Matrix in Local Coordinates These equations present the relationships between unknown internal forces S and deformations e of the elements. The required relationships is !
!
S¼e ke
ð13:4Þ
!
!
where S ðm1Þ ¼ bS1 S2 Sm cT is a vector of unknown internal forces; e ðm1Þ ¼ be1 e2 em cT is a vector of deformation; e k is a stiffness matrix of the system. In general form e k is diagonal matrix 2 3 k1 0 0 6 0 k 0 7 2 6 7 e k¼6 ð13:5Þ 7 4 5 0
0
km
The diagonal entry ki is a stiffness matrix of i-th finite element of a structure. Each diagonal entry ki is called the internal stiffness matrix or stiffness matrix in local coordinates for specified member i; matrix (Eq. 13.5) in whole is internal stiffness matrix or stiffness matrix in local coordinates for all structure. For any truss element (bar with hinges at the ends), a deformation is
466
13
e¼
Sl , EA
so S ¼
Matrix Stiffness Method
EA e l
Thus, internal stiffness matrix for truss element contains only one entry and presented as EA ½ 1 l
k¼
ð13:6Þ
This expression allows determining the axial force S if axial deformation of element e ¼ 1. The symbol [1] means a matrix with the sole entry equals 1. Below are presented some examples for creation of stiffness matrix. Let us form the stiffness matrix for truss shown in Fig. 13.21; this figure contains the numeration of the members. Assume that for all members EA is constant. Stiffness matrices for each member are shown below. EA EA EA EA ½1; k2 ¼ ½1; k 3 ¼ ½1; ½ 1 ¼ l1 3 4 6 EA EA EA ½1; k6 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½1 ¼ pffiffiffiffiffi ½1 k4 ¼ k5 ¼ 2 2 5 52 6 þ4
k1 ¼
1 2
6
4
5
4m
3
3m
3m
Fig. 13.21 Design diagram of redundant truss and numeration of the members
The internal stiffness matrix of the truss becomes 2 1=3 0 6 0 1=4 6 6 6 0 0 e k ¼ EA6 6 0 0 6 6 4 0 0 0
0 0
0 0
0 0
0 0
1=6
0
0
0
0 0
1=5 0
0 1=5
0
0
0
0 0 pffiffiffiffiffi 1= 52
0
3 7 7 7 7 7 7 7 7 5
For bending elements we can use Table A.3–A.4. If a uniform fixed-pinned beam is subjected to angular displacement e of 3EI the fixed support, then a bending moment at this support equals M ¼ e, so the stiffness matrix of such element in local l coordinates is k
f p
¼
EI ½3 l
ð13:7Þ
If a fixed-fixed uniform beam is subjected to unit angular displacements e1 and e2 of the fixed ends, then the following bending moments arise at the both supports: EI ð4e1 þ 2e2 Þ l EI M 2 ¼ ð2e1 þ 4e2 Þ l M1 ¼
These formulas may be presented in matrix form (Eq. 13.4), i.e.,
13.4
Resolving Equations
467
M1 M2
¼
EI 4 l 2
2 e1 4 e2
ð13:8Þ !
where the left part of Eq. (13.8) bM 1 M 2 cT presents the vector of internal forces, i.e., S ¼ bM 1 M 2 cT , vector of angular ! displacements at the ends of element e ¼ be1 e2 cT and the stiffness matrix in local coordinates for uniform beam with fixed ends becomes EI 4 2 k f f ¼ ð13:9Þ l 2 4 Let us form the stiffness matrix in local coordinates for frame shown in Fig. 13.22. Assume that bending stiffness equals EI for horizontal members 2 and 4, and for vertical members 3EI. According to primary system of the displacement method we have the fixed-fixed members 1 and 2, and fixed-pinned members 3 and 4. Stiffness matrices for each member in local coordinates are " # " # " # 2 1 EI 1 4 2 3EI 4 2 k1 ¼ ¼ ¼ EI 6 2 4 l1 2 4 1 2 " # " # " # 1 0:5 EI 2 4 2 EI 4 2 k2 ¼ ¼ ¼ EI 4 2 4 l2 2 4 ð13:9aÞ 0:5 1 EI 3 3EI ½3 ¼ EI ½1:5 ½ 3 ¼ 6 l3 EI EI k4 ¼ 4 ½ 3 ¼ ½3 ¼ EI ½0:4 7:5 l4 k3 ¼
4
2 1
6m
3
4m
7.5m
Fig. 13.22 Design diagram of the frame and numeration of the members; EI2 ¼ EI4 ¼ EI; EI1 ¼ EI3 ¼ 3EI
Each of these stiffness matrices is presented in form which contains general multiplier EI. The internal stiffness matrix of the frame becomes
2 1 0 ~ k = EI 0 0 0
1
0
0
0
0
2 0
0 1
0 0.5
0 0
0 0
0 0.5 0 0
1 0
0 1.5
0 0
0
0
0
0.4
0
Internal stiffness matrix for combined structure (e.g., the frame with tie) can be constructed similarly.
468
13.5
13
Matrix Stiffness Method
Set of Formulas and Procedure for Analysis
The behavior of any structure can be described by following three groups of equations: !
Equilibrium equations. This matrix equation establishes relationships between external possible joint loads P and unknown ! internal forces S !
!
P ¼ AS
ð13:10Þ
where A is a static matrix. !
Geometrical equations. This matrix equation establishes relationships between deformation of elements e and possible ! displacements Z of the joints !
!
!
e ¼ BZ ¼ AT Z
ð13:11Þ
where B is a matrix of deformation. !
Physical equations. This matrix equation establishes relationships between required internal forces S and deformation of ! elements e !
!
S¼e ke
ð13:12Þ
where e k is a stiffness matrix of a structure in local coordinates (internal stiffness matrix). These three groups of equations completely describe any structure (geometry, distribution of stiffness of separate members, types of connections of the members), types of supports, and external exposure.
13.5.1 Stiffness Matrix in Global Coordinates Rearrangement of Eqs. (13.10)–(13.12) allows obtaining the equation for vector of unknown internal forces S. For this purpose, let us apply the following procedure. The vector e may be eliminated from Eqs. (13.11) and (13.12). For this Eq. (13.11) should be substituted into Eq. (13.12). This procedure allows us to express the vector of unknown internal forces S in terms of vector of unknown displacements Z !
!
!
S¼e ke ¼ e k AT Z
ð13:13Þ
Now vector S can be eliminated from Eqs. (13.10) and (13.13). For this Eq. (13.13) should be substituted into Eq. (13.10); this procedure allows us to express the vector of possible joint external loads P in terms of unknown displacements Z !
!
!
P ¼ AS ¼ A e kAT Z
ð13:14Þ
This equation may be rewritten in form !
!
P ¼ KZ
ð13:15Þ
The matrix K presents the stiffness matrix of a structure in global coordinates (external stiffness matrix) K ¼ Ae kAT
ð13:16Þ
The entries of the stiffness matrix K ¼ Ae kAT present the unit reactions of the displacement method in canonical form. Therefore, this matrix is a symmetrical one and it has the strictly positive entries on the main diagonal. Dimension of this matrix is (mm), where m is a number of unknowns of the displacement method. Thus, for calculation of stiffness matrix K in global coordinates we need to know static matrix A and stiffness matrix e k (Eq. 13.5) of a structure in local coordinates.
13.5
Set of Formulas and Procedure for Analysis
469
13.5.2 Unknown Displacements and Internal Forces Equation (13.15) allows to calculate the vector of ends displacements !
!
Z ¼ K1 P
ð13:17Þ
where K1 is inverse stiffness matrix in global coordinates. Knowing vector Z we can calculate, according to Eq. (13.13), the unknown internal forces of the second state ! S2
!
¼e kAT Z
ð13:18Þ !
!
The formula (13.10) may be used for verification of obtained internal forces, i.e., AS 2 ¼ P . Final internal forces may be calculated by formula ! S fin
!
!
¼ S1 þ S2
ð13:19Þ
!
!
where the vector S 1 presents the internal forces at specified sections in the first state. It is obvious that for trusses S 1 is a zero! vector. For beams and frames the entries of S 1 are bending moments at the end sections. This vector forms on the basis of the M 0P diagram (1 state) and S-e diagram. If final ordinate M at any section is positive, then this ordinate should be plotted at the same side of the member as on the S-e diagram. It is now appropriate to discuss analogies between the method of displacement in canonical form (Chap. 10, formulas 10.4 and 10.5) and the formulas (13.17)–(13.19) of the matrix stiffness method. The entries of matrix K present the unit reactions of the displacement method. The entries of vector P mean the loaded terms (free members) of canonical equations taken with the opposite sign. r 11 Z 1 þ r 12 Z 2 þ . . . þ r 1n Z n þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ . . . þ r 2n Z n þ R2P ¼ 0 : :
: : :
: :
: :
: : :
:
r n1 Z 1 þ r n2 Z 2 þ . . . þ r nn Z n þ RnP ¼ 0 The formula (13.17) means the solution of the system of canonical equations (10.4). In other words, the entries of vector Z mean joints displacements in the global coordinate system. The bending moments at the end points of the elements of the primary system of the displacement method are determined by the formula (10.5) M P ¼ M 1 Z 1 þ M 2 Z 2 þ . . . þ M n Z n þ M 0P !
The expression (13.18) means the procedure M 1 Z 1 þ M 2 Z 2 þ . . . þ M n Z n , i.e., S 2 takes into account the bending ! moments caused by actual displacements Z of the joints, while the expression S 1 means term M 0P, i.e., the bending moments in the primary system caused by external load only.
13.5.3 Matrix Procedures The following procedures for analysis of any structure by matrix stiffness method may be proposed: 1. Define the degree of kinematical indeterminacy and type of displacement for each joint. 2. Calculate the fixed end moments, and construct the J-L diagram; this step for trusses is not applicable. 3. Numerate the possible displacements of the joints, then construct the Z-P diagram, and taking into account J-L diagram !
form the vector P of external joint loads. 4. Numerate the unknowns internal forces S (for truss it is a number of the elements, for beams and frames they are nonzero bending moments M at the ends of each elements), and then construct the S-e diagram. !
!
5. Compose the vector S 1 of internal forces. For this use the first state M 0P , J-L and S-e diagrams; for truss S 1 ¼ 0.
470
13
Matrix Stiffness Method
6. Consider the equilibrium conditions for each possible displacement of the joint and construct the static matrix A; the number of the rows and columns of this matrix equals to degree of kinematical indeterminacy and to the number of the unknown internal forces, respectively. 7. Construct the stiffness matrix for each member ki and for entire structure e k in local coordinates. 8. Perform the following matrix procedures: Compute the intermediate matrix complex e kAT (this complex will be used in the next steps) Compute the stiffness matrix in global coordinates K ¼ Ae kAT and its inverse matrix K1 !
!
Calculate the vector of joint displacements Z ¼ K1 P Calculate the vector of unknown internal forces of the second state S2 ¼ e kAT Z !
!
!
Calculate the vector of final internal forces S fin ¼ S 1 þ S 2 . All matrix procedures (13.16)–(13.19) may be performed by standard programs using computer. For trusses the procedure (13.19) leads to the axial forces at each member. For frames this formula leads to nonzero bending moments at the ends of each finite element. To plot the final bending moment diagram, the signs of obtained final moments should be consistent with S-e diagram. The shear forces can be calculated on the basis of the bending moment diagram considering each finite element subjected to given loads and the end bending moments. The axial forces can be calculated on the basis of the shear diagram considering the equilibrium condition for joints of the frame. Finally, having all internal force diagrams, we can show the reactions of supports and check them using equilibrium conditions for an entire structure as a whole or for any separated part. Notes 1. In different textbooks algorithm above (or modified algorithm) is referred differently. They are the matrix displacement method, finite element method, and stiffness method. All of them realize the one general idea: presentation of the structure as a set of separate elements with necessary requirements for consistency of deformations. 2. Generally speaking, a vector of internal unknown forces S may be constructed by different approaches. In this textbook, the vector S is presented in the simplest form. This vector contains only nonzero bending moments (for beams and frames) at the ends of each separate element. This choice of the vector S leads to the very compact stiffness matrix. Indeed, for fixedpinned beam, this matrix contains only one entry, for fixed-fixed standard member, this matrix has dimension (22). Certainly, we can expand the vector state S including, for example, the shear and axial force. This leads to the stiffness matrix with expanded dimensions. Even if the final result would contain more complete information, observing over all matrices and procedures over them is difficult. Therefore, we limited our consideration of the matrix stiffness method only for simplest presentation of vector S. This leads to the simple and vivid of intermediate and final results, and significant simplification of numerical procedures.
13.6
Analysis of Continuous Beams
This section presents a detailed analysis of statically indeterminate beams subjected to different types of exposures: fixed load, settlements of supports, and moving load. In the simplest version of the matrix stiffness method the vector of unknowns in the beams and framed structures present the nonzero bending moments in the end sections of all the finite elements. Structural analysis begins with the construction of auxiliary J-L, Z-P, and S-e diagrams. The vector of required unknowns is equal to the sum of two vectors; they are the bending moments in two states of the system. The first state presents the primary system of displacement method subjected to given external loads. Ordinates of bending moment diagram M 0P at the ends of finite elements are elements of the required vector of the first state. The second state represents a given system loaded by real rotation and linear displacements of introduced constraints only.
13.6
Analysis of Continuous Beams
471
13.6.1 Fixed Loads Design diagram of the uniform two-span beam subjected to fixed load is shown in Fig. 13.23a. This structure may be presented as a set of two finite elements: A-1 and 1-B. To transmit the given load to the joint load, we need to construct the primary system of the displacement method and calculate the fixed end moments at support 1 (Fig. 13.23b). They are equal to ql21 2 82 ¼ ¼ 16 ðkNmÞ, 8 8 Pl 12 10 0:4 1 0:42 ¼ 20:16 ðkNmÞ ¼ 2 υ 1 υ2 ¼ 2 2
M 01A ¼ M 01B
Thus, the equivalent moment equals 20.1616 ¼ 4.16 (kNm); this couple acts in clockwise direction (Fig. 13.23c) and corresponding joint–load diagram (first state) is shown in Fig. 13.23d.
a
P=12 kN
q=2kN/m
b
1
A
0 M 1A
0 M 1B
P
q
B
k
EI ul2=6 m l1=8 m
c
υl2=4 m M k0
l2=10 m 4.16
d
1
16
M P0
Joint load diagram
20.16
f
e 1
S2
S1
S-e diagram
Z-P diagram
P1
S1 Positive bending moments
S1
S2
g
q
P
18.31
k
S2
MP
Mk Fig. 13.23 Redundant beam under fixed load. (a, b) Design diagram, and bending moment diagram M 0P in primary system; (c, d) Bending moments at support 1 and corresponding equivalent joint-load Z-P diagram; (e) Displacement-load (Z-P) diagram and presentation of joint load P1 in terms of internal forces Si; (f) Force-deformation (S-e) and positive bending moments diagrams; (g) Final bending moment diagram
The beam has one unknown angular displacement at support 1 and corresponding one possible external joint load. The displacement–load (Z-P) diagram is presented in Fig. 13.23e. Having the joint–load and Z-P diagrams we can construct the ! vector of external joint moments; in this simplest case the vector P ¼ ½4:16, so this vector has only one entry. Positive sign means that equivalent moment at joint load and Z-P diagrams act at one direction (Fig. 13.23d, e).
472
13
Matrix Stiffness Method
Unknown internal forces (moments S1 and S2) and their positive directions are shown on S-e diagram (Fig. 13.23f). To construct the vector S of internal forces in the state 1 we need to take into account bending moment diagram M oP (Fig. 13.23b). This vector is ! 16 S1 ¼ 20:16 The signs of the entries correspond to S-e diagram. A static matrix A can be constructed on the basis of the Z-P and S-e diagrams. Figure 13.23e shows free-body diagram for joint 1 subjected to unknown internal “forces” S1 and S2 in vicinity of joint 1 and load P1, which corresponds to possible displacement of the joint 1. Since displacement of joint 1 is angle of rotation, then corresponding loads S1 and S2 are the moments. Equilibrium condition leads to the equation P1 ¼ S1 þ S2. So the static matrix becomes A ¼ b 1 1 c. Stiffness matrices for left and right fixed-hinged spans are 3EI 1 3EI 3EI ½1 ¼ ½ 5 ½1 ¼ 8 40 l1 3EI 2 3EI 3EI ½1 ¼ ½ 4 k2 ¼ ½1 ¼ 10 40 l2
k1 ¼
Internal stiffness matrix for entire beam in local coordinates is k1 0 3EI 5 e k¼ ¼ 40 0 0 k2
0 4
Stiffness matrix for entire structure in global coordinates and its inverse are 3EI 5 0 27 40EI 1 T ~ K ¼ AkA ¼ b 1 1 c ¼ ! K1 ¼ 1 40 0 4 40EI 27 The angular displacement of the joint 1 !
!
Z ¼ K1 P ¼
40EI 6:163 ðradÞ 4:16 ¼ 27 EI
Vector of internal forces of the second state ! S2
!
¼e kAT Z ¼
3EI 5 40 0
0
4
1 1
6:163 ¼ EI
2:3111
ðkNmÞ
1:8489
Final internal forces ! S fin
¼
! S1
! þ S2 ¼
16 20:16
þ
2:3111 1:8489
¼
18:31 18:31
ðkNmÞ
The negative sign means that the final bending moment at point 2 (infinitely close to the right of support 1) according to S-e diagram is directed in opposite direction, so extended fibers right at the joint 1 are located above the neutral line. Final bending moment diagram is shown in Fig. 13.23g. Equilibrium condition ∑M1 ¼ 0 for joint 1 is satisfied. The bending moment at point k (Figs. 13.23b, g) may be calculated considering second span as simply supported beam subjected to force P and moment 18.31 kNm counterclockwise. Note 1. Let us discuss the physical meaning of the most important matrix procedures. !
!
6:163 ðradÞ is the value of the primary unknown of the displacement method; equivalent EI expression in classical displacement method is Z1 ¼ R1P/r11.
a. Procedure Z ¼ K1 P ¼
13.6
Analysis of Continuous Beams
473
2:3111 b. Procedure ðkNmÞ means calculating bending moments to the left and right at support 1; 1:8489 equivalent expression in classical displacement method is M Z 1 ¼ M 1 Z 1 . ! ! 18:31 ! ðkNmÞ means calculating bending moments at end points of two finite c. Procedure S fin ¼ S 1 þ S2 ¼ 18:31 members over support 1; equivalent expression in classical displacement method is M P ¼ M 1 Z 1 þ M 0P . ! S2
!
¼e kAT Z ¼
2. Computation of shear forces may be performed on the basis of the MP diagram. Reactions of supports can be calculated on the basis of the shear diagram. Reaction of intermediate support is R1 ¼ Qright Qleft 1 . 1 3. The same example was considered in Sect. 10.2.4 by classical displacement method. The reader is asked to compare the procedure and the results of the analysis of the classical displacement method and the stiffness matrix method.
13.6.2 Settlements of Supports Example 13.4 Design diagram of a nonuniform continuous beam is presented in Fig. 13.24a. The angle of rotation of the clamped support is φ ¼ 0.01 rad and the vertical displacement of the support 2 is Δ ¼ 0.04 m. Construct the bending moment diagram.
a
φ EI1=1EI0
B
2EI0 D Δ
C
2EI0
A l1=6m
b
MBA
MBC
c
MAB
B
MCB − MCD= 0.015EI0
Joint-load diagram
C
A
l3=4m
l2=4m
M Δ0
MCD
MCB
MBC − MBA= 0.02667EI0
e
d Z-P diagram
S1
2
1
S2
S3
S-e diagram
S1
f
96
M, kNm (factor 10-4EI0)
98 174
S2
S3
g A
B
S5
S4
S4
S5
Reactions, (factor 10-4EI0) D C
MA=98 RA=32.33
RB=99.83
RC=111
RD=43.5
Fig. 13.24 Redundant beam, settlements of supports. (a, b) Design diagram and bending moment diagram in primary system; (c) Converting effect due to settlement of support to the joint-load moments; (d, e) Z-P and S-e diagrams; (f, g) Final bending moment diagram and reactions of supports
474
13
Matrix Stiffness Method
Solution Degree of kinematical indeterminacy equals two. Both joints at supports B and C have angular possible displacements. Now we need to present the effect of the settlements of supports in the form of moments at the joints B and C. The primary system and corresponding bending moment diagram M 0Δ is shown in Fig. 13.24b. The fixed end moments are 4EI 4EI 0 φ¼ 0:01 ¼ 0:00667EI 0 lAB 6 2EI 2EI 0 M BA ¼ φ¼ 0:01 ¼ 0:00333EI 0 lAB 6 6EI 6 2EI 0 M BC ¼ M CB ¼ 2 Δ ¼ 0:04 ¼ 0:03EI 0 42 lBC 3EI 3 2EI 0 M CD ¼ 2 Δ ¼ 0:04 ¼ 0:015EI 0 42 lCD M AB ¼
Here MBA means the bending moment at section infinitely close to the left of section B, i.e., this section belongs to element BA. Corresponding joint–load (J-L) diagram are shown in Fig. 13.24c. The positive possible angular displacements of the joints and corresponding possible external forces are presented on the Z-P diagram (Fig. 13.24d). The S-e diagram and positive bending moments diagram are shown in Fig. 13.24e. The vector of external joint loads, on the juxtapose the J-L and Z-P diagrams, is P¼ EI 0 b 0:02667 0:015 cT : Vector of fixed end moments in the first state, according to the M 0Δ diagram, is 6 7T 6 7 6 7 ! 4 5 S 1 ¼ EI 0 0:00667 |fflfflfflffl{zfflfflfflffl} 0:00333 |fflfflfflffl{zfflfflfflffl} 0:03 |fflffl{zfflffl} 0:03 |fflffl{zfflffl} 0:015 |fflffl{zfflffl} M AB
M BA
M BC
M CB
M CD
Static matrix. Considering the Z-P and S-e diagrams we get X M B ¼ 0 : P1 ¼ S2 þ S3 X M C ¼ 0 : P2 ¼ S4 þ S5 so the static matrix becomes Að25Þ ¼
0
1
1
0 0
0
0
0
1 1
Stiffness matrix. Stiffness matrices for finite elements AB, BC, and CD in local coordinates are " # " # " # 0:667 0:333 EI 1 4 2 EI 0 4 2 k1 ¼ ¼ ¼ EI 0 l1 2 4 6 2 4 0:333 0:667 " # " # " # 2 1 EI 2 4 2 2EI 0 4 2 k2 ¼ ¼ ¼ EI 0 l2 2 4 4 1 2 2 4 k3 ¼
EI 3 2EI 0 ½ 3 ¼ ½3 ¼ EI 0 ½1:5 l3 4
Stiffness matrix for structure in whole in local coordinates is 2 0:667 0:333 6 0:333 0:667 6 6 e 0 kð55Þ ¼ 6 6 0 6 0 4 0 0
0
0
0
0
0
0
2 1
1 2
0
0 1:5
3
7 0 7 7 0 7 7EI 0 7 0 5
13.6
Analysis of Continuous Beams
475
Matrix procedures. Intermediate matrix complex 2 0:667 0:333 6 6 0:333 0:667 6 e T ¼ EI 0 6 0 0 kA 6 6 0 4 0 0
0
0 0
0 0
2 1
1 2
0
0
3 2 0 0 7 6 0 7 61 7 6 6 0 7 7 61 7 6 0 5 40 0
1:5
3 2 0 0:333 7 6 07 6 0:667 7 6 6 2 ¼ EI 07 06 7 7 6 15 4 1 1
3 0 7 0 7 7 1 7 7 7 2 5 1:5
0
Stiffness matrix for structure in whole in global coordinates and inverse stiffness matrix are 3 2 0:333 0 7 6 6 0:667 0 7 7 6 2:667 1 0 1 1 0 0 T e 7 6 K ¼ A kA ¼ 1 7 ¼ EI 0 EI 0 6 2 1 3:5 0 0 0 1 1 7 6 2 5 4 1 0 1:5 Inverse matrix K1 ¼
0:42 1 EI 0 0:12
0:12
0:32
In this simplest case of (22) for matrix K, the following relationship had been used: d c 1 0 a c 1 1 1 , detK ¼ ad bc. It is easy to check K K ¼ If K ¼ , then K ¼ detK b a 0 1 b d If matrix K has dimensions (33) or more, then to obtain the inverse matrix a computer should be used. Vector of joint displacements !
!
Z ¼ K1 P ¼
0:42 1 EI 0 0:12
0:12 0:32
EI 0
0:02667 0:015
¼
0:0094 0:0016
ðradÞ
Vector of unknown bending moments (kNm) of the second state is 3 2 3 2 0:333 0 0:00313 7 6 7 6 6 0:00627 7 6 0:667 0 7 7 6 7 6 ! ! 0:0094 7 6 1 7 kAT Z ¼ EI 0 6 S2 ¼ e 7 0:0016 ¼ EI 0 6 0:0204 7 6 2 7 6 7 6 2 5 6 0:0126 7 4 1 7 6 6 0:0024 7 0 1:5 The final vector of required bending moments (kNm) is 3 3 3 2 2 2 0:00667 0:00313 98 7 7 7 6 6 6 6 0:00333 7 6 0:00627 7 6 96 7 7 7 7 6 6 6 ! ! * 4 7 7 7 6 6 S ¼ S1 þ e kAT Z ¼ EI 0 6 6 0:03 7 þ EI 0 6 0:0204 7 ¼ EI 0 6 96 7 10 7 7 7 6 6 6 6 0:03 7 6 0:0126 7 6 174 7 7 7 7 6 6 6 6 0:015 7 6 0:0024 7 6 174 7 Corresponding final bending moment diagram is presented in Fig. 13.24f. Bending moment diagram allows computing shear force. According to formula Q ¼ dM/dx, within the each span 1-3 the shear forces is determined as slope of corresponding portion of bending moment diagram. Below shear are presented for each portion: (with factor β ¼ 104EI0)
476
13
Matrix Stiffness Method
96 þ 98 β ¼ 32:333β ðkNÞ 6 174 ð96Þ β ¼ 67:5β ðkNÞ Q2 ¼ 4 0 174 Q3 ¼ β ¼ 43:5β ðkNÞ 4 Q1 ¼
Reaction of i-th support may be determined by formula Ri ¼ Qright Qleft i i . RA ¼ Q1 ¼ 32:333β ðkNÞð#Þ RB ¼ Qright Qleft B B ¼ ½67:5 ð32:33Þ β ¼ 99:83β ðkNÞð"Þ Qleft RC ¼ Qright C C ¼ ð43:5 67:5Þβ ¼ 111:0β ðkNÞð#Þ RD ¼ Q3 ¼ 43:5β ðkNÞð"Þ All reactions are shown in Fig. 13.24g Verification. Equilibrium equations for entire beam are X Y ¼ ð32:333 þ 99:83 111:0 þ 43:5Þβ ¼ ð143:33 þ 143:33Þβ ¼ 0 X M D ¼ ð98 32:333 14 þ 99:83 8 111:0 4Þβ ¼ ð896:64 896:66Þβ ffi 0
13.6.3 Moving Load (Construction of Influence Lines) Example 13.5 Design diagram of the uniform three-span continuous beam is presented in Fig. 13.25a. Construct the influence lines for bending moments at the supports B and C. Solution Each span of the beam is divided into equal portions and specified sections are numerated (0–18). Next we need to show the displacement–load (Z-P) and internal forces–deformation (S-e) diagrams. Unknown moments S1, S2 arise at support B and moments S3, S4 at support C (Fig. 13.25a). Static matrix. The Z-P and S-e diagrams allow constructing the following equilibrium equations: X M B ¼ 0 : P1 ¼ S1 þ S2 X M C ¼ 0 : P2 ¼ S3 þ S4 so the static matrix of the structure is
1
1
0
0
0
0
1
1
Stiffness matrix. Stiffness matrices for each finite element are EI EI 4 k1 ¼ ½3, k2 ¼ l l 2
2
Að24Þ ¼
4
k3 ¼
,
EI ½ 3 l
Stiffness matrix e k of entire structure in local coordinates and intermediate complex e kAT are 2
3
6 EI 6 0 e k¼ 6 l 40 0
0
3
0
0
4 2
2 4
07 7 7, 05
0
0
3
2
3
6 EI 6 0 e kAT ¼ 6 l 40 0
0 0
0
3
2
1 0
3
2
3
0
0
3
4 2 2 4
6 7 6 07 7 6 1 0 7 EI 6 4 7¼ 6 76 l 42 05 40 15
0 0
3
0 1
3
27 7 7 45
13.6
Analysis of Continuous Beams
a
477
Points numeration
Sections numeration 0
EI
6 2
10
8
4
B
l
C
1
3
4
l S-e diagram
2
Z-P diagram
P=1
b
Load P=1 in the first span
2
16
14
l
1
18
12
(
l u 1 − u2 2
S1
S3
S2
S4
) M P0
υl
ul
Joint-load (J-L) diagram
(
l u 1 − u2 2
) P=1
u υ 2l
u 2 υl
c
M P0
υl
ul l
Load P=1 in the second span
Joint-load (J-L) diagram
u 2 υl
u υ 2l
2
10
8
4
5.426
C 7.897
7.893
9.872
B Final influence lines
18
12
6
14
16
1.973
0
2.468
d
IL(MB)
7.893
7.897
9.872
factor 0.01l
5.426
2.468
1.973
−
− e
IL(MC) factor 0.01l
Static verification of results
B 0.07893l
C 0.07893l
0.01973l
0.01973l
M 2B = −0.07893l M 2C = 0.01973l
Fig. 13.25 Construction of influence lines. (a) Design diagram of continuous beam, sections and points numeration, Z-P and S-e diagrams; (b) Load P¼1 in the first span and corresponding J-L diagram; (c) Load P ¼ 1 in the second span and corresponding J-L diagram; (d) Influence lines for bending moments at supports B and C; (e) Equilibrium condition for joints B and C, if load P ¼ 1 is placed at section 2. Note: Sections numeration (0–18) means position of load P ¼ 1; points numeration (1–4) means location of the required end bending moments (see S-e diagram)
478
13
Matrix Stiffness Method
The stiffness matrix K of entire structure in global coordinates and inverse stiffness matrix K1 are 2 3 3 0 6 7 1 1 0 0 EI 6 4 2 7 EI 7 2 ¼ K¼Ae kAT ¼ 6 , 7 l 42 45 l 2 7 0 0 1 1 0 3 7 2 l K1 ¼ 45EI 2 7 To construct the matrices P and S1 we need to consider the unit moving load in all spans separately. For each specified location of unit load we need to construct the bending moment diagram in primary system of displacement method and show the corresponding joint–load diagram.
13.6.3.1
Load P = 1 in the First Span
The bending moment diagram in the primary system of displacement method and equivalent joint–load diagram are shown in Fig. 13.25b. On the basis of the joint–load (J-L) and displacement–load (Z-P) diagrams the vector P of the external joint loads for any position u of unit load P in the first span is ! 0:5uð1 u2 Þ P¼l ðaÞ 0 This vector has two entries, because we have two angular displacements 1 and 2 and corresponding two reactive moments at the introduced constraints, as shown on Z-P diagram (Fig. 13.25a). Expression for first entry is taken from Table A.3 (row 3, formula for MA) with change υ ! u. Negative sign means the moment at support B (Fig. 13.25b) is directed opposite to the arrow 1 at B on Z-P diagram (Fig. 13.25a). The second zero entry means that in the primary system the bending moment in introduced constraint at support C does not arise, because of the constraint at support B (Fig. 13.25b). On the basis of the M 0P and S-e diagrams the vector S1 of unknown internal forces S1–S4 in the first state for any position P becomes 3 2 2 3 S1 0:5uð1 u2 Þ 7 6 6S 7 ! 0 7 6 6 27 ðbÞ S 1 ¼ 6 7 ¼ l6 7 7 6 6 S3 7 0 7 6 6 7 7 6 6S 7 0 4
The entries of vector S mean the bending moments at points which are shown on S-e diagram (Fig. 13.25a). First entry is taken from Fig. 13.25b. Positive sign is because this ordinate on the Fig. 13.25b and ordinate S1 on S-e diagram (Fig. 13.25a) are both located on one side of the reference line. The next zero entries mean the bending moments at points 2,3,4 caused by load P¼1 in the first span do not arise (Fig. 13.25b). Matrices (a) and (b) can be composed for any values of u, which determines the position of the moving load and is measured from the left pinned support. The procedure for construction of influence lines is presented in Table 13.1. This table is related only for the case when the movable load P ¼ 1 is in the first span. The number of columns in this table is equal to the number of sections in which the load may be located; we limit ourselves to only two sections, namely 2 and 4. Table 13.1 contains three characteristic parts. The first part shows the vector P of the external joint load and vector S1 in the first state. For their formation we need to use formulas (a) and (b). The entries of vector S1 are bending moments caused by 0 load P ¼ 1 in the points 1, . . ., 4 of the primary system of displacement method, i.e., the ordinates from diagram MP . Numeration of specified sections with required internal forces is presented on the S-е diagram (Fig. 13.25a). The second part of Table 13.1 contains matrix procedures for computation of two vectors. Vector Z presents the values of the angle of rotation of introduced constraint, Z1 and Z2, if load P ¼ 1 is placed at section 2 (left column) and section 4 (right column). Vector S2 presents the bending moments at points 1–4 caused by rotation of introduced constraints 1 and 2. In fact this vector presents the following expression M1 ¼ M1Z1 þ M1Z2 M2 ¼ M2Z1 þ M2Z2
13.6
Analysis of Continuous Beams
479
Table 13.1 Computation of ordinates of influence lines of bending moment at supports B and C. Load P ¼ 1 is located in the first span (sections 2 and 4)
P=1 at the section 2 (u=0.333, υ =0.667)
P=
P1 − 0.1480 =l , P2 0
1. Vector P and S1 0.1480
M1 M2 S1 = M3 M4
P=1 at the section 4 (u=0.667, υ =0.333)
=l 1
M1 M2 S1 = M3 M4
−0.1851 P =l , 0
0 0 0
0.1851 0 , =l 0 0 1
2. Matrix procedures
Z=
Z1 − 0.1480 7 −2 l = K −1 P = ×l = Z2 0 45 EI − 2 7
Z=
Z1 7 −2 − 0.1851 l = K −1 P = ×l = Z2 0 45 EI − 2 7 −7 l2 243 .11EI 2
−7 l2 304 .05 EI 2
M1 M2 S2 = M3 M4
2
3 ~ EI 4 = kA T ∙ Z = l 2 0
0 −7 2 l2 ∙ = 4 304 .05 EI 2 3
M1 M2 S2 = M3 M4
= kAT 4
3 EI 4 Z= l 2 0
0 2 4 3
=7 l2 = 243.11EI 2
−0.08638 −0.09872 l −0.02468
− 0.06907 − 0.07893 l − 0.01973 0.01973
0.02468
3. Bending moment at support B (points 1 and 2) and support C (points 3 and 4) S = S1 + S 2 . Numeration of sections corresponds to S-e diagram
M1 0.1480 −0.06907 0.07893 M2 − 0.07893 0 −0.07893 =l +l =l S= M3 −0.01973 − 0.01973 0 M4 0 0.01973 0.01973 M P0
M 1Z1 + M 2 Z 2
Explanation signs at final vector S (bending moments)
Support B: M left = M1 > 0 M right =M2 0
0.1851 0
S=l
0 0
− 0.08638 − 0.09872 +l =l − 0.02468 0.02468
0.09872 − 0.09872 − 0.02468 0.02468
M final
P=1 0
B S1
C
M
S3
S-e S2
S4
480
13
Matrix Stiffness Method
Third part of Table 13.1 presents computation of final bending moments at supports B and C for two positions of load P ¼ 1 within first span (sections 2 and 4). According to final vector S, if load P¼1 is located within first span, then ordinates of influence lines at supports B and C are shown in Fig. 13.25d. According to Table 13.1 for u ¼ 0.333 (load P at section 2), the last vector S shows that equilibrium condition for supports B and C is hold (Fig. 13.25e); similarly same conclusion for u ¼ 0.667 (load P at section 4). Location of extended fibers in vicinity of support are shown by dotted line. The signs of the bending moments should be treated according to S-e diagram and general rules of bending moments. If the load P ¼ 1 is placed at section 2, then ordinates of influence lines for bending moments at sections 6 (support B) and 12 (support C) are the following: M B2 ¼ 0:07893l; M C2 ¼ 0:01973l (Fig. 13.25e). If the load P ¼ 1 is placed at section 4, then ordinates of influence lines for bending moments at the supports B and are the following: M B4 ¼ 0:09872l; M C4 ¼ 0:02468l. Negative sign means if load is placed at section 2 or 4, the extended fibers in vicinity of support B are located above.
13.6.3.2
Load P = 1 in the Second Span
The bending moment diagram in the primary system of displacement method and equivalent J-L diagram are shown in Fig. 13.25c. Now for any position u of unit load P in the second span, we can compose the vector P of the external joint loads (using the J-L and Z-P diagrams) and vector S1 of unknown bending moments in the first state (using the M 0P and S-e diagrams). For fixed-fixed beam according to Table A.4 , case 3, line 1 we have 3 2 0 6 uυ2 7 ! ! uυ2 7 6 P¼l , S 1 ¼ l6 2 7 2 6 uυ 7 u υ 7 6 6 0 7 After that we will determine the entries of these both matrices when moving load P is placed at the sections 8 and 10 and perform corresponding matrix procedures.
13.6.3.3
Load P = 1 in the Third Span
This case can be considering elementary taking into account the symmetry of the beam and loading in the first span. The final influence lines for moments at the supports B and C are shown in Fig. 13.25d. If load P ¼ 1 is placed at sections with odd numbers 1,3,. . ., then ordinates of influence lines MB and MC may be computed by similar way.
13.7
Analysis of Redundant Frame
This paragraph contains a detailed analysis of a frame structure by matrix stiffness method (MSM). Analysis of this frame has been performed early by the both classical methods (Examples 9.7 and 10.3), so this frame may be treated as the etalon one. The reader can compare the procedures and results of the analysis. Special attention should be paid to the physical meaning of matrix procedures. It is easy to do by considering both methods (the displacement and MSM) in parallel. The simplified version of the MSM method presented here makes it easy to analyze structures with a high degree of kinematic indeterminacy. Example 13.6 Design diagram of a frame is shown in Fig. 13.26a. Provide analysis of this structure by matrix stiffness method. Comparing with displacement method in canonical form will allow us to understand a physical meaning of an each matrix procedure. Solution The frame has two unknowns of the displacement method. They are the angular displacement at joint 1 and linear displacement of crossbar 1-C. The first state (primary system and M 0P diagram) is shown in Fig. 13.26b.
13.7
Analysis of Redundant Frame
481
b
a EI
P=8kN 1
C
2 1
2EI 5m
q=2kN/m
P
15.36 4.1667
3m
B
EI
M P0
q First state
A 6m
4m
c
e
d
Joint-load (J-L) diagram
15.36kNm
11.1933
1
5.0
5.0 5q/2=5
4.1667kNm q
5.0
f
g S-e diagram S4 state
Z-P diagram
1 state
2
S2 S3
S1
i
h
P1 S4
S4
1
S3
S4/h2
P2
(S1+S2)/h1 S2
S2
S1
S1/h1+ S2/h1
(S1+S2)/h1
Fig. 13.26 Etalon frame. (a) Design diagram; (b) First state -Primary system and bending moment diagram; (c, d) Calculation of equivalent joint moment and force; (e–g) J-L, Z-P and S-e diagrams; (h, i) Construction of a static matrix A-presentation of Pi in terms of Sj
The finite elements are A-1, 1-C, and 1-B. The fixed end moments at joint 1 are ql2 M A ¼ M 1A ¼ 1A ¼ 4:1667 kNm; M 1B ¼ 0; 12 8 10 Pl 0:6 1 0:62 ¼ 15:36 kNm M 1C ¼ υ 1 υ2 ¼ 2 2 The equivalent moment at joint 1 is M ¼ 15.36 4.1667 ¼ 11.1933 kNm, clockwise (Fig. 13.26c); the equivalent force for crossbar is P ¼ 5 kN (Fig. 13.26d). In order to construct the J-L diagram we need to transmit the equivalent moment and force on the initial design diagram, as shown in Fig. 13.26e. Auxiliary Z-P diagram (Fig. 13.26f) shows that the structure has one possible angular displacement of joint 1 and corresponding possible external joint moment, as well as one horizontal displacement 2 of crossbar and corresponding force. The joint–load J-L and displacement–load Z-P diagrams allow constructing the vector of external equivalent joint loads
482
13
P¼ b 11:1933
5:0 cT
Matrix Stiffness Method
ðcÞ
The entries of this vector present the free terms, which equal to free terms of canonical equations of the displacement method but taken with the opposite sign (compare with Example 10.3). Unknown internal forces are nonzero bending moments at the end points of each finite element: A, 1-A, 1-C, and 1-B. These moments are denoted S1 and S2 for member 1-A, S3 for member 1-C, and S4 for element 1-B, and their positive directions are shown on S-e diagram (Fig. 13.26g). The vector of fixed-end moments (vector of internal forces of the first state) at sections 1, 2, 3, 4 on the basis of the M 0P and S-e diagrams becomes !
S ¼ b 4:1667 1
4:1667
0 cT :
15:36
Static matrix. This matrix is constructed on the basis of the Z-P and S-e diagrams. Figure 13.26h shows free-body diagram for joint 1 subjected to three unknown internal forces in vicinity of joint 1 (bending moments) and moment P1. Equilibrium condition ∑M1 ¼ 0 for joint 1 is P 1 ¼ S2 þ S3 þ S4 Equilibrium condition ∑X ¼ 0 for crossbar 1-C is P2 ¼
S1 S2 S4 þ h1 h1 h2
Indeed, the positive moments S1 and S2 at the ends of the member A-1 may be equilibrated by two forces S1/h1 + S2/h1 (Fig. 13.26i). Then this force should be transmitted on the crossbar; similar procedure should be done for member 1-B. Both expressions for P1 and P2 may be presented in matrix form 2 3 S1 6 7 ! P1 0 1 1 1 6 S2 7 P¼ ¼ 6 7 7 1=h1 1=h1 0 1=h2 6 P2 6 S3 7 6S 7 4 !
where vector P is a vector of external forces. The number of entries of this vector equals to degree of kinematical ! indeterminacy, in our case number of entries of vector P equals two. !
!
According to equation P ¼ AS (13.10) the static matrix becomes 0 1 1 A¼ 1=h1 1=h1 0
1 1=h2
For given parameters h1 and h2 we get A¼
0 0:2
1
1
0:2 0
1
0:333
Stiffness matrices of the elements in the local coordinates. The stiffness matrix for each member in local coordinates is EI 4 2 EI 4 2 kA1 ¼ k1 ¼ 1 , ¼ 5 2 4 l1 2 4 EI 2EI EI k1C ¼ k2 ¼ 2 ½3 ¼ ½3 ¼ ½3, 10 5 l2 EI 3 EI EI ½3 ¼ ½3 ¼ ½5 k1B ¼ k3 ¼ 3 5 l3 For whole structure the stiffness matrix in local coordinates is
13.7
Analysis of Redundant Frame
483
2
k1 6 e k¼4 0
0 k2
0
0
2 3 4 0 62 7 EI 6 0 5¼ 6 5 40 k3 0
0 3
3 0 07 7 7 05
0 0
5
2 0 4 0
Matrix procedures. For whole structure the stiffness matrix in global coordinates 3 2 3 2 0 0:2 4 2 0 0 7 6 7 6 2:4 0 1 1 1 EI 6 2 4 0 0 7 6 1 0:2 7 K ¼ Ae kAT ¼ 6 7 ¼ EI 76 5 40 0 3 05 41 0 5 0:093 0:2 0:2 0 0:333 1 0:333 0 0 0 5
0:093
0:207
The entries of the matrix K are unit reactions of the displacement method in canonical form (Example 10.3). The determinant of this matrix is det K ¼ 0.48815, so the inverse matrix 1 0:4241 0:1905 K1 ¼ EI 0:1905 4:9165 !
!
The matrix resolving equation KZ ¼ P allows finding the vector of unknown displacements !
Z¼
Z 1 ðradÞ Z 2 ðmÞ
1 0:4241 ¼K P¼ EI 0:1905 1
!
0:1905
4:9165
11:193
5
1 3:7944 ¼ EI 22:452
!
The entries of matrix Z present the angle of rotation of the rigid joint and linear displacement of the crossbar; they have been obtained previously by the displacement method (Example 10.3). ! ! ! Vector of internal unknown bending moments is S fin ¼ S 1 þ S2 , where for second state we have 3 2 3 2 3 2 0 0:2 4 2 0 0 3:8707 6 2:3529 7 7 6 7 6 ! ! EI 6 2 4 0 0 7 6 1 0:2 7 1 3:7944 7 6 S2 ¼ e ¼6 kAT Z ¼ 6 7 7 76 5 40 0 3 05 41 EI 7 6 5 0 22:452 6 2:2766 7 6 1 0:333 0 0 0 5 11:2709 7 In fact this vector reflects the following procedure of the displacement method in canonical form: M i ðZ Þ ¼ M i1 Z 1 þ M i2 Z 2 : !
Control: AS ¼ P . In our case 3 3:8707 6 7 11:1946 1 6 2:3529 7 6 7¼ 6 2:2766 7 4:9979 0:333 7 6 6 11:2709 7 2
0
1
1
0:2
0:2
0
Comparing with the vector (c), we note that the error of calculations is very insignificant. Final bending moments at specified sections 1–4 are 3 2 3 2 3 2 4:1667 3:8707 8:0374 6 4:1667 7 6 2:3529 7 6 1:8138 7 ! ! ! 7 6 7 6 7 6 S fin ¼ S 1 þ S 2 ¼ 6 7þ6 7¼6 7 6 15:36 7 6 2:2766 7 6 13:0834 7 7 6 7 6 7 6 7 6 11:2709 7 6 11:2709 7 6 0 |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflffl {zfflfflfflfflfflfflfflfflfflffl} Pffl{zfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflM 0 MP
MiZi
Note, the bending moment diagram of the second state is not shown separately.
P
484
13
Matrix Stiffness Method
The meaning of each vector is explained according to the displacement method in canonical form. This vector allows constructing the bending moment diagram. All ordinates should be plotted according to S-e diagram (Fig. 13.26g). For example, M1 ¼ 8.037 kNm should be plotted at the support A left at neutral line. Final bending moment diagram is presented in Fig. 10.9g. Note that stiffness matrix method is precise and some disagreement with data obtained previously is a result of the rounding. The reader is advised to compare the procedure and results of the analysis by the classical displacement method and by the stiffness matrix method, and to find correspondences between the main stages of the two methods of analysis.
13.8
Analysis of Redundant Trusses
Below is presented the simplest internally statically indeterminate truss. Let us compute the displacements of all the joints, and calculate the internal forces and reactions. Design diagram of truss is shown in Fig. 13.27a. Assume that the stiffness EA for all members are equal.
a
b
F1=4kN D
Z-P diagram
F2=2kN
B
c
2 3
S-e diagram 1 3 2
d 5 A
d
4
Joint B
Joint A
S2
S4
P1
S3 P3
S1 S4
e
Joint C
P2 S6
6
C 1
d
5 4
Joint D
YD
S5 P5
S6
XD
S1
P4
S5
S2
S3
Fig. 13.27 Redundant truss. (a–c) Design diagrams and corresponding Z-P and S-e diagrams; (d) Free-body diagram for each joint; (e) Computation of reactions at support D
First let us construct the Z-P diagram (Fig. 13.27b). This diagram shows possible joint displacements and corresponding possible loads. After that we can construct the vector of external forces P¼ b 0
4
2
0 0 cT :
This vector has five enters because the given structure allows five possible joint displacements. The first, fourth, and fifth enters of this matrix equal to zero, because in the possible direction 1, 4, and 5 the active forces are absent. The negative sign for second and third entries mean that in the possible directions 2 and 3 the external forces F1 and F2 have direction opposite to positive directions 2 and 3. To construct the static matrix let us show the S-e diagram (Fig. 13.27c). Then we need to consider free-body diagram for joints which have possible displacements (all joints except the pinned support D) and express the possible forces P1-P5 in terms of unknown internal forces S1–S6. This step is presented in Fig. 13.27d.
13.8
Analysis of Redundant Trusses
485
X
Joint A :
X
Joint B :
X X
Joint C :
X
Y ¼ 0 ! P1 ¼ S2 0:707S4 Y ¼ 0 ! P2 ¼ 0:707S4 þ S5 X ¼ 0 ! P3 ¼ S1 þ 0:707S4 Y ¼ 0 ! P4 ¼ 0:707S3 S5 X ¼ 0 ! P5 ¼ 0:707S3 þ S6
Thus, the static matrix becomes 2
Að56Þ
0
6 60 6 ¼6 61 6 40 0
1
0
0:707
0
0 0
0 0
0:707 0:707
1 0
0 0
0:707 0:707
0 0
1 0
0
3
7 07 7 07 7 7 05 1
The stiffnesses of each member in local coordinates are EA EA EA ½1; k2 ¼ k5 ¼ k6 ¼ k1 ¼ ½1; ¼ l1 d d EA EA k3 ¼ k4 ¼ pffiffiffi ½1 ¼ ½0:707 d d 2
k1 ¼
so stiffness matrix of the entire structure in local coordinates is 2 1 0 0 0 60 1 0 0 6 6 6 0 0 0:707 0 EA 6 e k¼ 6 d 60 0 0 0:707 6 40 0 0 0 0 0 0 0
0 0
0 07 7 7 0 07 7 0 07 7 7 1 05 0 1
Stiffness matrix of the entire structure in global coordinates is 2 1:3534 0:3534 0:3534 6 0:3534 1:3534 0:3534 6 EA 6 T e 6 K ¼ AkA ¼ 0:3534 0:3534 1:3534 d 6 6 0 1 0 4 0
0
0
3
0 1 0 1:3534 0:3534
0
7 7 7 7 0 7 7 0:3534 5 1:3534 0
As expected, a symmetric matrix with positive elements on the main diagonal is obtained. Inverse matrix may be calculated by computer using a standard program. This matrix is 3 2 0:8965 0:5 0:1035 0:3965 0:1035 7 6 2:4149 0:5 1:9149 0:5 7 6 0:5 7 6 d 6 K1 ¼ 0:1035 0:5 0:89655 0:3965 0:1035 7 7 6 EA 6 7 2:313 0:6035 5 4 0:3965 1:9149 0:3965 0:1035 0:5 0:1035 0:6035 0:8965 It is easy to verify that KK21 = K21K = E, where E is unit matrix.
3
486
13
Matrix Stiffness Method
Vector displacements 2
Z1
3
2
2:2071
3
7 6 7 6 6 Z2 7 6 8:6594 7 7 7 6 6 ! ! d 6 1 7 7 Z¼6 6 Z 3 7 ¼ K P ¼ EA 6 0:2071 7 7 6 7 6 6 Z4 7 6 6:8665 7 7 6 7 6 6 Z5 7 6 1:7929 7 The negative sign for Z1 indicates that the joint A according to Z-P diagram has a negative displacement, i.e., downwards. Vector of internal forces 3 2 2 3 S1 0:2071 6 2:2071 7 6S 7 7 6 6 27 7 6 6 7 7 6 6 S3 7 ! ! T 7¼e 6 2:5360 7 kA S¼6 Z ¼ 6 3:1217 7 6S 7 7 6 6 47 7 6 6 7 6 1:7929 7 6 S5 7 7 6 6 7 6 1:7929 7 6S 7 6
The negative sign for S4 according to S-e internal forces are kN. 2 0 6 60 6 ! ! Control: AS ¼ P . In our case AS ¼ 6 61 6 40 0
diagram means that the diagonal member 4 is compressed. The units for all 1 0
0 0
0:707 0:707
0 1
0
0
0:707
0
0 0
0:707 0:707
0 0
1 0
3 2 3 2 3 0:2071 0 0 6 2:2071 7 7 7 6 7 6 07 6 7 6 4 7 7 7 6 2:5360 7 6 7 7 6 6 07 7 6 3:1217 7 ¼ 6 2 7 7 7 6 7 6 05 6 7 6 0 7 7 6 1:7929 7 6 7 6 0 7 6 1 6 1:7929 7
The above confirms that equilibrium conditions for all joints are hold. As the last step, for calculation of reaction of supports we need to consider equilibrium conditions for rolled and pinned supports. For example, the free-body diagram for pinned support D and corresponding equilibrium equations are shown in Fig. 13.27e. X X ¼ 0 ! X D ¼ S1 þ 0:707S3 ¼ 0:2071 þ 0:707 2:5360 ¼ 2:0 kN X Y ¼ 0 ! Y D ¼ S2 þ 0:707S3 ¼ 2:2071 þ 0:707 2:5360 ¼ 4:0 kN Note: For this externally statically determinate structure all reactions may be determined considering the structure in whole. However this example shows how to calculate reactions using typical approach for any statically indeterminate structure.
13.9
Stiffness Matrices: Expanded Forms
Behavior (or state) of any finite element is described by set of forces and set of displacement at the end points of an element. A stiffness matrix relates a vector of forces on both ends with vector of displacements on the same ends. Stiffness matrix is defined by the type of the finite element. In previous sections, truncated or simplest stiffness matrices had been used for analysis of different structures (beam, frame, and truss). This made it possible to significantly reduce the volume of processed information and present the essence of computational procedures in a form which has physical meaning inherent to the displacement method. In particular, for truss member and bending fixed-pinned member, we used stiffness matrix, which according to Eq. (13.9a) is a scalar. For fixedfixed beam member a stiffness matrix is (2 2); for this element the relationships between the bending moments at the ends and angular displacements at the ends according to Table A.4 is described by Eq. (13.9). Assume it is required to develop more informative model of finite element. For this we need to take into account other possible forces at the ends of the element and the corresponding displacements at the ends.
13.9
Stiffness Matrices: Expanded Forms
487
Let us consider creation of stiffness matrix for linear finite element for different classical cases. Among them are truss element subjected to axial displacement of the end points, beam element subjected to the linear and angular displacements at the end points, and case which combines two previous cases. Now we consider stiffness matrices in local coordinate system for some partial cases.
13.9.1 Truss Element (Pinned-Pinned Element) The simplest case concerning to uniform pinned-pinned truss element, which is loaded by two axial forces F1 and F2 at points 1 and 2, is shown in Fig. 13.28. Corresponding axial displacements are d1 and d2. It is necessary to establish relationships between axial forces and displacements. For this two states of elements should be considered. x
Forces F2
2
2
State 2
State 1
Displacements
F2= −F1, d2=0
d2
d2
F2
E,A l
d1 1
1
d1
F1
F1
F1= −F2, d1=0
Fig. 13.28 Pinned-pinned truss element
State 1. This state represents the case when element is subjected to active axial force F1 and axial displacement d2 is prevented F l by introduced constraint at point 2. The axial displacement of section 1 according to Hook’s law is d1 ¼ 1 . Now we can EA express the active force F1 and corresponding reaction F2 in terms of d1 F1 ¼
EA d l 1
F 2 ¼ F 1 ¼
EA d l 1
ð13:20Þ
State 2. This state represents the case when element is subjected to active axial force F2 and axial displacement d1 is F l prevented by introduced constraint at point 1. The axial displacement of section 2 is d 2 ¼ 2 . The active force F2 and EA corresponding reaction F1 in terms of d2 are F2 ¼
EA d l 2
F 1 ¼ F 2 ¼
EA d l 2
By superposition, both equations (13.20) and (13.21) may be combined and presented in the following form EA EA d d l 1 l 2 EA EA d þ d F2 ¼ l 1 l 2 F1 ¼
A matrix presentation of these equations is
ð13:21Þ
488
13
!
!
F ¼ kloc d
!
!
F¼
or
F1 F2
¼
Matrix Stiffness Method
d1 EA 1 1 l 1 1 d2
ð13:22Þ
!
where entries of the vector F and d are axial forces and displacements at end points of the rod. The corresponding expanded stiffness matrix in local coordinate system for uniform pin-connected element becomes k11 k12 EA 1 1 kloc ¼ ð13:23Þ ¼ l 1 1 k21 k22
13.9.2 Beam Elements The next type of a finite element is a uniform beam. We will consider the plane bending. It means all the displacements and reactions are lying in the same plane x-y. Figure 13.29 presents two types of standard elements (fixed-fixed and pinned-fixed beams), the notation 1 and 2 of the ends sections, displacements of supports and corresponding reactions. All displacements and reactions are shown in the positive directions. The direction of positive reaction R1 may be explained by the following way: support 1 from a new position tends to transfer into initial position, then a corresponding reaction will be directed downward. Similarly may be explained the positive directions for all reactions. Thus, this signs rule for positive reactions is same for reactions in introduced constraints of the displacement method. 1
2
EI
φ2
φ1
y1
1
x y2
2
EI
y1
M2
M1 R1
x
φ2
y2
M2
R2
R1
l
R2 l
y
y
Fig. 13.29 Positive displacements of supports and corresponding positive reactions for fixed-fixed and pinned-fixed beams
Assume, for fixed-fixed beam the displacements at both supports are known, while reactions at the same supports are unknown. The following sequences for displacements and reactions are adopted: y1, φ1, y2, φ2 and R1, M1, R2, M2 According to Table A.4, cases 1, 2, 3 the force-displacement relationships can be presented in the following form 3 2 2 3 2 3 2 3 y1 y1 R1 6=l2 3=l 6=l2 3=l 6M 7 6φ 7 6 6 3=l 7 φ 2 3=l 1 7 6 17 7 6 17 6 1 7 2EI 6 ð13:24Þ 7 ¼ k f f 6 7 ¼ 6 6 76 7 l 4 6=l2 3=l 6=l2 3=l 5 6 y2 7 6 R2 7 6 y2 7 7 6 6 7 6 7 6M 7 6φ 7 6φ 7 3=l 1 3=l 2 2 2 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} k
f f
Thus, for beam with clamped ends we have the (4 4) extended stiffness matrix. This matrix equation allows presenting any force at the left and/or right end of the fixed-fixed finite element in terms of displacements on the both ends. For example,
13.9
Stiffness Matrices: Expanded Forms
489
R1 ¼
12EI 6EI 12EI 6EI y 1 þ 2 φ1 3 y 2 þ 2 φ2 l3 l l l
This formula describes the superposition principle on the base of cases 1 and 2 from Table A.4. Now a meaning of each 12EI 6EI entry of the stiffness matrix (13.24) becomes evident. For example, the entries k11 ¼ 3 and k 12 ¼ 2 (coefficient 2EI/l is l l taken into account) present reaction R1 caused by unit displacements y1 and φ1, respectively. Thus, each entry for row M1 (Eq. 13.24) means the moment M1, caused by corresponding unit displacement y1, φ1, y2, or φ2. Similarly, for pinned-fixed beam we have 2 3 2 3 2 3 2 3 R1 y1 y1 1=l2 1=l2 1=l 6 7 6 7 3EI 6 7 6 7 2 2 ð13:25Þ 6 R2 7 ¼ kpf 6 y2 7 ¼ 4 1=l 1=l 1=l 5 6 y2 7 l 6 7 6 7 6 7 6 M2 7 6φ 7 6 7 φ2 1=l 1=l 1 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} kpf
This matrix equation allows presenting any force at the left and/or right end of the pinned-fixed finite element in terms of displacements on the both ends. For example, R1 ¼
3EI 3EI 3EI y 1 3 y 2 þ 2 φ2 l3 l l
This formula describes the superposition principle for cases 1 and 2 from Table A.3. The matrices in Eqs. (13.24) and (13.25) are presented in local coordinates. Both matrices kff and kpf are square, symmetrical, and the main entries kii are positive. Tables A.5 and A.6 allow constructing a stiffness matrix for fixed-sliding and clamped beam with elastic support. Stiffness matrices for different finite elements may be found in special handbooks devoted to Finite Element Analysis.
13.9.3 More General Case of a Finite Element The uniform element is subjected to the axial displacements d, vertical and angular displacements y and φ at the end points 1 and 2. As a result, the reactions of supports are F1, M1, R1 at point 1 and F2, M2, R2 at point 2. All the displacements of the end points of the member and reactions belong to x-y plane (Fig. 13.30). Suppose the uniform element has length l, crosssectional moment inertia Iz ¼ I, and cross section area A. The modulus of elasticity of material is E. x
M2 l
F2 2
M1
E,I ,A R2 1
F1 d1 y1 φ1
d2 y2 φ2
z R1 y
Fig. 13.30 Linear finite element: The forces and displacements at end points
The problem is to construct a stiffness matrix for given finite element. For this purpose we need to combine results (13.23) and (13.24), so general relationship between forces and displacements can be written as follows:
490
13
Matrix Stiffness Method
* * S ¼ k D,
ð13:26Þ
* * * where k is stiffness matrix, S and D are vector of forces and D displacements, respectively, at the end points. 3 2 EA EA 0 0 0 0 l 7 6 l 3 6 2 12EI 6EI 12EI 6EI 7 7 2 d1 3 6 F1 0 3 7 6 0 3 2 2 l l l l 7 6 7 6R 7 6 6 y1 7 Initial point 6 17 6 6EI 4EI 6EI 2EI 7 7 6 7 6 0 7 6 0 7 6 2 2 6 M1 7 6 6 φ l l 7 6 17 l l 7¼6 7 6 ð13:27Þ 7 7 6 F 7 6 EA EA 7 6 d 2 7 6 2 7 6 6 0 0 0 0 7 7 7 6 l l 7 6 6 R2 7 6 Final point y2 7 7 6 7 6 6 6 12EI 6EI 12EI 6EI 7 6 7 7 6M 7 6 0 0 7 6 φ 2 2 6 l3 l2 l3 l2 7 5 4 6EI 2EI 6EI 4EI 0 2 0 l l l2 l * Thus, the vector of forces S is composed according to the principle of belonging of forces to points 1 and 2 separately. In other words, first are recorded the forces F, R, M for point 1 and then same forces for point 2. Same principle is related to * vector of displacements D. * * The vectors S and D may be presented in another form, for example, d F 1 F 2 R1 M 1 R2 M 2 et and * d d 1 d2 y1 φ1 y2 φ2 et , respectively. It means the vector S is composed according to the principle of belonging of forces to different types of deformations, i.e., the axial and bending ones. In this case, one part of the force vector S takes into account axial forces of the finite element at points 1 and 2, while the other part of the same vector corresponds to forces which are inherent to the bending deformations, i.e., shear and bending moment of the same points. Same is related to vector of * displacements D. This leads to the following matrix relationships
F1 F2 R1 = M1 R2 M2
EA l EA − l 0 0
EA l EA l
−
0
0
0
0
0
0
0
0
0
12EI
6 EI
3
2
l 6 EI
0
0
0
0
0
2
−
l 12EI l3 6 EI l2
l 4 EI l 6 EI − l2 2 EI l
−
12EI
6 EI
3
2
−
l 6 EI 2
l 12EI
−
l3 6 EI l2
l 2 EI l 6 EI − l2 4 EI l
d1 d2 y ∙ 1 φ1
Axial components (13.28) Bending components
y2 φ2
In this case, the stiffness matrix contains two square stiffness sub-matrix related to axial behavior of the finite element and bending state of the same element. Both matrices (13.27) and (13.28) are symmetrical and the entries on the main diagonal are positive. It is possible to have another order of presentation of entries for matrices of forces and displacements, which leads to new stiffness matrix. It is obvious that we can construct a stiffness matrix for prismatic rod with different additional features. For example, beam subjected to torques, beam on elastic foundation, and their different combinations. The matrix stiffness method may be successfully applied to analysis of frames with inclined members. Since we represent the reactions in term of the ends displacements (13.24) and (13.25), then we use the FEM in the form of displacements. Of course, it is possible to present the FEM in the form of forces; in this case the displacements of the ends should be represented in terms of the forces at the ends. Corresponding matrix is called flexibility matrix. The flexibility matrix may be easily formulated. Above we had considered only FEM in the form of displacements. The finite element method is more general than the displacement method in canonical form, and it does not require construction of the unit bending moment diagrams. However, on the first stage of learning of FEM and for methodical purposes only the FEM in relation with displacement method had been shown.
13.10
13.9 Stiffness Matrices: Expanded Forms
491
Decision of what form of the stiffness matrix to use, i.e., truncated or expanded, should be based on few thoughts. For example, if structure is fairly simple, then it is easy to use truncated matrix for determining bending moments only and then calculate shear forces and reactions by hand. However, if structure is complex and we have access to power computers, then it is worthwhile to construct stiffness matrix in the expanded form, then solution of the problem will lead to all internal forces and reactions without extra calculations by hand.
13.10
Summary
The matrix stiffness method (MSM) is modern effective method for analysis of deformable structures subjected to arbitrary actions. Among them are external loads, change of temperature, and settlements of supports. For trusses, beams, and frames with straight uniform members the MSM leads to the exact results.
Remarks Concerning Analysis of the Bending Structures by MSM 1. As the first step of analysis by MSM we need to construct the primary system of displacement method. Finite elements of entire structure coincide with separate elements of the primary system. The unknown forces, in the simplest version of the MSM, are bending moments at the fixed ends of each finite element. The bending moment diagram in the primary system of the displacement method presents the first state. 2. External exposure should be replaced by the equivalent joint loads (moments and forces) and be presented in the form of the joint–load (J-L) diagram. The equivalent joint loads are taken from the first state diagram and transmitted to the joints of the original design diagram. 3. Displacement–load (Z-P) diagram shows the character of primary unknowns of displacement method (i.e., character of possible displacements of the joints) and corresponding type of forces. This diagram has the conceptual character and does not contain any numerical data. 4. The S-e diagram presents the end sections of each finite element with nonzero bending moments, their numeration and the positive unknown bending moments location at these sections. 5. The static matrix A connects the possible external loads P and required internal forces S. The member aik is coefficient at
!
unknown force Sk in an equation for Pi. The entries aik may be positive, negative, or zero. The dimensions of the vectors P !
6.
7.
8.
9. 10. 11.
and S equal to the degree of kinematical indeterminacy and the number of unknown terminal moments, respectively. Deformation matrix B connects the end deformation of each finite element in the primary system of the displacement method and unit displacement of introduced constraints. The member bik is displacement in direction of unknown force Si due to unit displacement of introduced constraint k. The number of rows and columns of matrix B equals to the number of unknown internal forces S and the number of the introduced constraints of displacement method, respectively. The deformation and static matrices obey to equation B ¼ AT. The stiffness matrix k of finite element in local coordinates connects unknown internal force S and displacement of the end of the element. These matrices may be presented in truncated or in expanded form. In truncated form, the stiffness matrices for fixed-pinned member presents a scalar, while for fixed-fixed member is (2 2) matrix. The stiffness matrix in local coordinates for whole structure e k presents matrix which contains on the principal diagonal the stiffness matrices of separate elements ki. This matrix is square, symmetrical and all entries are positive. Juxtapose Z-P and J-L diagrams lead to the vector of equivalent joint loads P. In fact this vector presents the free terms (with opposite signs) of canonical equation of the displacement method. Thus, the number of entries of this vector equals to the number of primary unknowns of the displacement method. If a structure is subjected to different types of external loads (fixed loads, change of temperature, settlements of supports), then P will present a matrix. In such case, each column represents single type of loading. The stiffness matrix K = Ae kAT of whole structure in global coordinate is symmetrical square (m m) matrix, where m is the number of primary unknowns of the displacement method. The members of this matrix are the unit reactions rik. The vector of joints displacements, i.e., the vector of primary unknowns, is Z ¼ K1P. The final results for required unknown internal forces (bending moments) may be presented by formula S ¼ S1 þ S2. Juxtapose S-e and M 0P diagrams lead to the vector of internal forces S1 of the first state. Thus, vector S1 corresponds to the
492
13
term M 0P in general formula M ¼ M 0P þ
m P
Matrix Stiffness Method
M i Z i, where m is a number of primary unknowns of the displacement method.
i¼1
The vector of unknown internal forces of the second state is S2 ¼ e kAT Z. This vector corresponds to the second term in the formula above. The number of entries of both vector S1 and S2 equals to the number n of required unknown bending moments at the ends of finite elements. Thus, the joint displacements and the distribution of internal forces of any structure are defined only by three matrices. kðnnÞ of a structure in local coordinates, and the vector of They are the static matrix A(mn) of a structure, stiffness matrix e !
equivalent joint loads P ðm1Þ . The deformation matrix B, since B ¼ AT, is reflected in the analytical formulas using the transposed matrix A. 12. In order to plot the final bending moment diagram, the signs of obtained final moments should be juxtaposed with S-e diagram. The shear force can be calculated on the basis of the bending moment diagram; the axial forces can be calculated on the basis of the shear diagram. Reactions of supports can be calculated on the basis of the axial and shear forces and bending moment diagrams. Flowchart of the numerical procedure of the matrix stiffness method and comparison of classical displacement method and matrix stiffness method are presented in Fig.13.31 and Table 13.2 respectively.
a Entire system and finite elements
Primary unknowns of displacement method
Primary system, and bending moment diagram MP0
b Displacementload, Z-P
Joint-load, J-L
Forcesdeformation, S-e
c Possible displacement vector and external
Static matrix A
forces P
Vector of unknown forces of the first state S1
d Stiffness matrix in local coordinates ~ ki, k
Stiffness matrix in global coordinates K = AkA T
Vector of required internal forces S = S1+S2, final diagrams, and reactions of supports
Vector of unknown displacements
Vector of unknown forces of the second ~ state S 2 = kA T ∙ Z
Z = K −1 P
Fig. 13.31 Numerical procedure of matrix stiffness method. (a) Design diagram, primary unknowns of displacement method and bending moment diagram due to given exposure in the primary system; (b) Auxiliary diagrams; (c) Initial matrices; (d) Matrix procedures, and final result of analysis
13.10
13.9 Stiffness Matrices: Expanded Forms Table 13.2 Comparison of procedures of analysis by displacement and matrix stiffness methods (N/A–not applicable)
493
494
13
Matrix Stiffness Method
Analysis of truss by MSM has some features. They are as follows: 1. The concepts of primary system, equivalent joint loads, J-L diagram, and “first state” are not applicable; therefore the vector of internal forces S1 ¼ 0. 2. Finite elements coincide with separate elements of entire structure. The unknown forces are axial forces in the members of a truss. 3. The Z–P diagram shows linear displacements of joints and corresponding forces along these displacements. 4. The S-e diagram contains numeration of axial forces and their positive direction (Figs. 13.10 and 13.11) 5. In truncated form, the stiffness matrix for truss member presents a scalar. 6. The number of entries of vector P equals to the number of the possible displacements of joints of a truss.
Problems 13.1a–c. The uniform two-span beam is subjected to fixed load as shown in Fig. P13.1. The flexural rigidity of the beam is EI. Determine the angle of rotation at support 1 and the bending moments at specified points.
a
b
q EI
0
1
0
1
(a) Z1 = −
ql 3 ql 2 , M1 = − ; 14 56EI
(b) Z1 = −
ql 2 ql 3 , M1 = − ; 28 84EI
2
EI
2
l
Ans.
q
l
l
l
c
P 1
0
(c) Z1 =
3Pl 2 . 112EI
2
EI
D
l
l
Fig. P13.1
13.2. The uniform three-span beam with different spans is subjected to uniformly distributed load q (Fig. P13.2). The flexural rigidity of the beam is EI. Determine the angle of rotation at supports and the bending moments at specified points. Compare the result with data in Appendix, Table A13a. q
Ans. M1 = − 0.074ql12 , M 2 = − 0.023ql12
EI 0
1
2 l2
l1=1.2l2
l3=0.6 l2
Fig. P13.2
13.3. The uniform three-span beam with equal spans l is subjected to the settlement of support 1 (Fig. P13.3). The flexural rigidity of the beam is EI. Determine the angle of rotation at support 1 and the bending moments at specified points. Compare with data at the Appendix, Table A14.
EI 0
Fig. P13.3
1
Δ
Ans. M1 = 3.6
2 l
EI l
2
Δ , M 2 = −2.4
EI l2
Δ
Problems
495
13.4. Design diagram of the uniform two-span continuous beam is presented in Fig. P13.4. Construct the influence line for angle of rotation and bending moment at the support B (section 6). Compare with data at the Appendix, Table A16a.
6 10
8
4
2
B
l
l
Fig. P13.4
13.5a,b. The frames in Fig. P13.5 are subjected to uniformly distributed load q. Construct the bending moment diagrams. Present your answer in terms of EI, l, and q. Compare distribution of bending moment within the columns and explain their difference.
a
b
q b EI
b
2EI
ql 2 ql 2 ; , Mb = 72 36
(b) M a = M b =
ql 2 56
EI
h=l
2EI
Ans. (a) M a = −
q
a
a
l
l Fig. P14.5
13.6a,b. The combined structures are subjected to load as shown in Fig. P13.6a,b. Construct the internal force diagrams. Calculate the reactions of support and provide check of results. For vertical members are used the steel wide-flange shape A W150x22 (I ¼ 12.1106 mm4, for tie rod ∅4.3 cm ( ¼ 125 m2 ). I
b B
C
2EI
EI
EA
h=8m
F=100kN EI
A
q=10kN/m B
2EI
EI
EA
C EI
A
D l=6m
h=8m
a
Ans. (a) N A−C = 165.8 (kN) (b) N A−C = 4.663 (kN)
D l=6m
Fig. P13.6
13.7a,b The portal frames with absolutely rigid crossbar are subjected to load as shown in Fig. P13.7a,b. Calculate the horizontal displacement of the crossbar. Construct the internal force diagrams. Calculate the reactions of support and axial force S in the crossbar. Provide your answer in terms of EI, h, and given load.
a
b
A Fig. P13.7
B
EI D
q
Ans. (a) M A = M D = Fh 2, S = F 2 ;
C
EA= ∞
5 2 3 2 qh , M D = qh , 16 16 3 qh (compr). S= 16
(b) M A = EI A
EI
h
EI
C
EA= ∞
h
B
F
D
496
13
Matrix Stiffness Method
13.8. The combined structure with absolutely rigid crossbar BC is subjected to load F. Both diagonal ties AC and BD are not connected at point K (Fig. P13.8a,b). Calculate the horizontal displacement Z of the crossbar. Construct the internal force diagrams. Calculate the reactions of support and axial force S in the crossbar. Assume A/I ¼ 60 m2. B
F=10kN
C
EA=∞
Ans. Z BC = 2.308 EI , M A = M D = 0.108kNm , EI
h=8m
K EI
EA
A
S AC = − S BD = 8.31kN , S BC = F 2
D l=6m
Fig. P13.8
13.9. The combined structure with absolutely rigid crossbar BC is subjected to load F. Both diagonal ties AC and BD are connected by hinge at point K (Fig. P13.9). Calculate the horizontal displacements of the crossbar and joint K. Construct the internal force diagrams. Calculate the reactions of support and axial force S in the crossbar. Assume A/I ¼ 60 m2. B
F=10kN
C
EA=∞
h=8m
K EI
EI
EA
A
Ans. Z BC = 2.308 EI , Z Khor = 1.154 EI M A = M D = 0.108kNm, S KC = S AK = 8.31kN
D l=6m
Fig. P13.9
13.10. Determine the displacement of the joints for truss shown in Fig. P13.10. Calculate internal forces in all member of the truss and then reactions of supports. Check these reactions using equilibrium conditions for truss in whole; EA ¼ constant for all elements, d ¼ 1 m. F1=4kN 1 3 2
5
4 6 d
Fig. P13.10
F2=2kN
Ans. S1 = 0.6212; S2 = S5 = S6 = −1.3788 , d
S3 = 1.9503; S 4 = − 3.7074
Problems
497
13.11. The externally and internally statically indeterminate truss is subjected to two groups of external loads. The first group is F1 ¼ 12 kN and F2 ¼ 5 kN; the second group is N1 ¼ 16 kN and N2 ¼ 4 kN as shown in Fig. P13.11. For each set of load calculate internal forces in each element. Knowing internal forces compute the reactions of supports. EA ¼ constant for all elements. F1 A
C
4
10
5
1
3
6
L 2
B
F2
E
9
8
α
14
13
11
3m
Ans. Set F: S1 = − 6.2910; S 2 = 10.0328,.... S15 = − 0.4672. RD = 8.57606 kN
G D
7
12
N1
N2
F
4m
15
Fig. P13.11
13.12. Design diagram of frame is shown in Fig. P13.12, l ¼ h, EI is constant. The frame has two unknowns of the displacement method. Determine the vector of angular displacement of rigid joint and linear displacement of crossbar and vector of bending moments at the end points of each finite elements. C
P
h=l
Ans. P = 0 P
T
, A=
0 1 1 , −1 h −1 h 0
A l
4 2 0 ~ k1 0 EI = 2 4 0 , k= 0 k2 h 0 0 3 −5 Z1 ( rad ) Ph Ph3 6 h T −3 , S fin = kA Z = Z= = Z2 ( m) 48EI 7 8 3
Hint:
Finite elements C 2
P
P M=0
1 A Fig. P13.12
J-L diagram
1 Z-P diagram
S-e diagram
2
2 3 1
498
13
Matrix Stiffness Method
13.13. Construct the influence lines for internal force in each member of the internally statically indeterminate truss shown in Fig. P13.13; EA ¼ constant for all elements. The columns 1, 2, and 3 correspond to location P ¼ 1 at joint B, D, and F, respectively.
4
10
5
1
3 2
6
B
Ans. ordinates of influence lines
9 8
α 7
D
14
13
11 12
F
15
4m
Fig. P13.13
3m
S1 − 1.250 − 0.8333 − 0.4167 S 2 1.000 0.6667 0.3333 ... ... ... ... . S= S 7 0.7608 0.9412 0.4941 ... ... ... ... S15 0.3333 0.6667 1.000
Part III
Special Topics
Chapter 14
Plastic Behavior of Structures
This chapter is devoted to the analysis of a structure, taking into account the plastic properties of material. Such analysis allows utilizing the reserves of strength of material, which remains unused considering material of structure as elastic. Therefore plastic analysis allows defining the limit load on the structure and thus design a more economical structure. Fundamental idea of the plastic analysis is discussed using the direct method. Kinematical and statical methods of calculation of the limit loads are considered. Detailed plastic analysis of beams and frames are presented.
14.1
Idealized Stress-Strain Diagrams
In the previous chapters we considered structures taking into account only elastic properties of materials for all members of a structure. Analysis of a structure based on elastic properties of material is called the elastic (or linear) analysis. Elastic analysis does not allow finding out the reserve of strength of the structure beyond its elastic limit. Also this analysis cannot answer the question: what would happen with the structure if the stresses in its members will be larger than the proportional limit? Therefore, a problem concerning to the actual strength of a structure cannot be solved using elastic analysis. The typical stress-strain diagram for the specimen of structural steel is presented in Fig. 14.1a. Elastic analysis corresponds to the initial straight portion of the σ-ε diagram. If a specimen is loaded into the proportional limit (or below) and then released, then material will unload along the loading path back to the origin. So, there are no residual strains. This property of unloaded specimen to return to its original dimensions is called elasticity, and material in this region is called the linearly elastic. Within the elastic region a relationship between stress and strain obey to Hooke’s law σ ¼ Eε. Let the specimen be loaded up to the elastic limit. The stress at this point slightly exceeds the proportional limit. From this point the material unloads along the line which is parallel to straight portion of the diagram and thus, the material has the very small residual strain. Plastic behavior starts slightly beyond the elastic limit. The region CD is referred as the perfect plastic zone. In this region, the specimen continues to elongate without any increase in stress. Above the yield plateau, starting from point D, the behavior of the specimen is described by nonlinear relationships σ-ε. If the specimen will be unloaded at point A (Fig. 14.1b), then unloading line will be parallel to the load straight line, so the specimen returns to its original length only partially. Total strain of the specimen is ON, while the strain MN has been recovered elastically and the strain OM remains as residual one. If the material remains within the elastic region, it can be loaded, unloaded, and loaded again without significantly changing the behavior. However, when the load is reapplied in a plastic region, the internal structure of material is altered, its properties change, and the material obeys to Hook’s law within the straight line MA; it means that the proportional limit of the material has been increased. This process is referred to as the strain-hardening.
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_14
501
502
14
a
True fracture stress
Ultimate stress
σ
Yield plateau C
Plastic Behavior of Structures
D
Fracture stress
Yield stress Elastic limit Proportional limit Unloading line
O Elastic Yielding region Elastic behavior
b
ε
Strain hardening
Necking
Plastic behavior
σ
A
Unloading path
O
M Residual strain
N
ε
Elastic strain
Total strain Fig. 14.1 (a) Typical stress-strain diagram for structural steel. (b) Loading-unloading diagram
For plastic analysis, we change the typical diagram by its idealized diagram. Different idealized diagrams are considered in engineering practice. Some of idealized models are presented in Table 14.1. Table 14.1 Idealized σ-ε diagrams for axially loaded members
Material Elasto-plastic
Elasto-plastic with linear hardening
Rigid-plastic
σ
σ A
α
σy
σ
σ
α0
σy
σy α
α ε
Rigid-plastic with linear hardening
ε
α0 σy
α ε
ε
For further analysis we will consider idealized elasto-plastic material and rigid-plastic material. We start from elasto-plastic material; corresponding diagram is called Prandtl diagram. This diagram has two portions—linear “stress-strain” part and the yield plateau. Elastic properties of material hold up to yield point stress σ y. The yield plateau shows that displacement of material can become indeterminate large under the same stress. Idealized elasto-plastic material does not have effect of hardening. This diagram may be applicable for a structural steel and for reinforced concrete. Structural analysis on the basis of
14.1
Idealized Stress-Strain Diagrams
503
idealized diagram is referred as the plastic analysis. The quantitative results of plastic analysis are much closer to the actual behavior of a structure than the results obtained on the basis of elastic properties of material. In case of statically determinate structure, yielding of any member leads to the failure of the structure as a whole. Other situation occurs in case of statically indeterminate structure. Assume that for all members of the structure, the Prandtl diagram is applicable. In the first stage, when loads are small, behavior of all members follows the first portion of the Prandtl diagram. Proportional increase of all loads leads to the yielding in the most loaded member. It means that the degree of statical indeterminacy is decreased by one. The following proportional increase of all loads leads to the following effect: the internal force in the yielding member remains the same, while the forces in the other members will be increased. This effect will continue until the next member starts to yield. Finally, the structure becomes statically determinate and yielding of any member of this structure immediately leads to the failure of the structure, since it is transformed into a mechanism. In general, if the structure has n redundant constrains, then its failure occurs when the number of yielding member becomes n + 1. Its means that capability of a structure to carry out the increasing load has been exhausted. This condition is called limit equilibrium condition. In this condition, the limit loads and internal forces satisfy to equilibrium condition. The following increase of a load is impossible. In this condition the displacement of the structure becomes undefined. While the linear portion of typical stress-strain diagram leads to linear problems of structural analysis (elastic problems), the Prandtl diagram leads to nonlinear problems of plastic behavior of structures. Indeed, the design diagram of a structure is changed upon different levels of loads. Transition from one design diagram to another happens abruptly. Let us consider plane bending of a beam of a rectangular (b h) cross section. In the elastic region of the stress-strain diagram, the normal stresses are distributed within the height of a cross-section of the beam linearly. The maximum tensile and compressive stresses are located at the extreme fibers of the beam. The stress σ y corresponds to yield plateau (Fig. 14.2a). Increasing of the load leads to appearance and developing of the yield zone and decreasing of the “elastic core” of the section of the beam. Diagrams on Fig. 14.2(b,c) correspond to partially plastic bending of a beam, which means that the middle part of the cross-section is in elastic condition, while the bottom and top parts of the beam are in plastic condition. Further increasing of load leads to complete plastic state (Fig. 14.2d), which corresponds to the limit equilibrium, i.e., we are talking about appearance of so-called plastic hinge (Fig. 12.2d,e). It is obvious, that all other sections of the beam are in the different states. Defining of the location of the plastic hinge is an additional problem of plastic analysis. This problem will be considered below.
a
b
σy
c
σy
σy Plastic zone
h
Elastic core Plastic zone
b
d
e
σy
Plastic hinge
F F
Fig. 14.2 Distribution of normal stresses within the height of a beam
What’s the difference between plastic and ideal hinge? First of all the plastic hinge disappears if the structure is unloaded, so the plastic hinge may be considered as fully recoverable or one-sided hinge. Second, in the ideal hinge, the bending moment equals to zero, while plastic hinge is characterized by appearance of bending moment, which is equal to the limit (or plastic) moment of internal forces F ¼ σ y(bh/2) (Fig. 14.2d). A bearing capability of a structure is characterized by plastic moment h bh2 M p ¼ F ¼ σy 2 4 Plastic analysis involves determination of limit load, which structure can resist before full failure due to yielding of some elements. The limiting load does not depend on settlements of supports, errors of fabrication, prestressed tension, and temperature changes; this is fundamental difference between plastic and elastic analysis. In the following sections we will consider different methods of determining limit loads, after which the load carrying capacity of the structure is exhausted, and the structure becomes a mechanism.
504
14.2
14
Plastic Behavior of Structures
Direct Method of Plastic Analysis
The fundamental concept of plastic analysis of a structure may be clear presented using the direct method. Let us consider the structure, shown in Fig. 14.3a subjected to load P at point K. The horizontal rod is absolutely rigid. All hangers have constant stiffness EA. The plastic analysis must be preceded by elastic analysis.
14.2.1 Elastic Analysis This analysis should be performed on the basis of any appropriate method of analysis of statically indeterminate structures. Let us consider structure shown in Fig. 14.3a (Rzhanicyn 1982). Omitting this analysis, which is familiar for reader and presents no difficulties, the distribution of internal forces in members 1–4 of the structure is as follows (Fig. 14.3b): N 1 ¼ 0:4P;
N 2 ¼ 0:3P;
N 3 ¼ 0:2P;
a
N 4 ¼ 0:1P
ðSee problem 9:23a, bÞ
b d 1
d 2
d 3
0.4P
l
0.3P
0.2P
0.1P
N 3(2)
N 4(2)
4
K P
P
c
d Ny
0.75Ny
0.5Ny
Ny
0.25Ny
N 2(2)
ΔP2=0.3 Ny
P=ΔP1=2.5Ny
e
f Ny
Ny
0.6Ny
Ny
0.20Ny
P2=P1+ ΔP2=2.8Ny
Ny
N 3(3)
N 4(4)
ΔP3=0.2Ny
h P 3.0 2.8 2.5
g Ny
Ny
Ny
P3= P2+ΔP3=3.0Ny
N4=0
0.75 1.0
2.0
ΔK
Fig. 14.3 (a, b) Design diagram and distribution of internal forces according to elastic analysis; (c) Step 1—Plastic state in the member 1 and internal forces in the rest members; (d, e) Step 2—Internal forces in the members 2–4 due to load ΔP2 and plastic state in the members 1 and 2; (f, g) Step 3—Internal forces in the members 3–4 due to load ΔP3 and plastic state in the members 1, 2 and 3; (h) P ΔK diagram in plastic analysis
14.2
Direct Method of Plastic Analysis
505
14.2.2 Plastic Analysis Step 1: At this stage, the system works as completely elastic. Increasing of load leads to appearance of the yield stresses. They are reached in the most highly stressed member. In our case this member is element 1. Let N1 become equal to limit load, i.e. N1 ¼ Ny. Since N1 ¼ 0.4P, then it occurs if external load would be equal to P ¼ ΔP1 ¼ N y =0:4 ¼ 2:5N y Expression P ¼ ΔP1 ¼ 2.5Ny means the full (maximum) load on the structure at the end of the first stage of loading. For this load, the limit tension will be reached in the first hanger. Internal forces in the another members are (Fig. 14.3c) N 2 ¼ 0:3P ¼ 0:3 2:5N y ¼ 0:75N y , N 3 ¼ 0:2P ¼ 0:2 2:5N y ¼ 0:5N y , N 4 ¼ 0:1P ¼ 0:25N y Equilibrium equation in the end of the first stage is (1 + 0.75 + 0.5 + 0.25)Ny 2.5Ny ¼ 0. Step 2: If load P ¼ ΔP1 ¼ 2.5Ny, the initial for the second stage will increase by ΔP2 then N1 ¼ Ny remains without changes. It means that additional load ΔP2 will be distributed between three members 2, 3, and 4, i.e., the design diagram had been changed (Fig. 14.3d). This structure is once statically indeterminate. Elastic analysis of this structure due to load ΔP2 leads to the following internal forces ð2Þ
N 2 ¼ 0:833ΔP2 ;
ð2Þ
N 3 ¼ 0:333ΔP2 ;
ð2Þ
N 4 ¼ 0:167ΔP2
These forces in the elements 2, 3, 4 are caused by the increment of the load ΔP2 in the second stage. Since first hanger is already in yield condition (and cannot resist any additional load), the most highly stressed member due to load ΔP2 is the second hanger. The total load in this element equals N 2 ¼ 0:75N y þ 0:833ΔP2 In this formula the first term corresponds to ultimate load at the end of the first loading stage P ¼ 2.5Ny (Fig. 14.3c), while the second term corresponds additional load ΔP2 at which the second stage of loading is completed. In this case, the maximum force in the second hanger is N2 ¼ Ny. Thus equation N2 ¼ 0.75Ny + 0.833ΔP2 ¼ Ny leads to the following value for increment of load ΔP2 ¼ 0.25Ny/0.833 ¼ 0.3Ny. Thus, the value ΔP2 ¼ 0.3Ny represents additional load, which is required so that the second hanger reaches its yielding state. Therefore, if load on the second stage reach value P2 ¼ 2:5N y þ 0:3N y ¼ 2:8N y , then both members 1 and 2 reach their limit state. As this takes place, the internal forces in hangers 3 and 4 (Fig. 14.3e) are following N 3 ¼ 0:5N y þ 0:333 0:3N y ¼ 0:6N y ;
N 4 ¼ 0:25N y 0:167 0:3N y ¼ 0:20N y
Control (1.0 + 1.0 + 0.6 + 0.20)Ny 2.8Ny ¼ 0. Step 3: Since internal forces in hangers 1 and 2 reached the limit values, then the following increase of the load by value ΔP3 (Fig. 14.3f) affects the members 3 and 4 only. Elastic analysis of this statically determinate structure leads to the following internal forces in members 3 and 4 due to load increment ΔP3: ð3Þ
ð3Þ
N 3 ¼ 2ΔP3 and N 4 ¼ ΔP3 Similarly as above, the limit state for this case occurs if internal force in hanger 3 reaches its limit value N 3 ¼ 0:6N y þ 2ΔP3 ¼ N y Here first term 0.6Ny means force in hanger 3 caused by load P ¼ 2.8Ny. This equation leads to the following value for increment of the load ΔP3 ¼ 0.4Ny/2 ¼ 0.2Ny. The total value of external force (Fig. 14.3g)
506
14
Plastic Behavior of Structures
P3 ¼ 2:5N y þ ΔP2 þ ΔP3 ¼ 2:5N y þ 0:3N y þ 0:2N y ¼ 3:0N y The first term in this formula corresponds to limit load in the first member; increment of the force by 0.3Ny leads to the limit state in the second member. The following increment of the force by 0.2Ny leads to the limit state in the third member. After that the load carrying capacity of the structure is exhausted. From the equilibrium equation for the entire structure, we can see that on this stage N4 ¼ 0 (Fig. 14.3g). All forces satisfy to equilibrium condition. Now let us consider corresponding plastic displacements of structure in Fig. 14.3. If force in the some of the elements reached its limiting value and the load continues to increase, then we cannot determine displacements of the system using only elastic analysis. However, plastic analysis allows calculating displacements of a structure on the each stage of loading. Let us show the graph of displacement of the point application of force P (point K ). On the first stage loading (P ¼ 2.5Ny), the internal force in second element equals 0.75Ny (Fig. 14.3c) and vertical displacement of point K is ð1Þ
ΔK ¼ 0:75
Nyl EA
On the second stage of loading (P ¼ 2.8Ny) the internal force in second element equals Ny (Fig. 14.3e) and vertical displacement of point K is ð2Þ
ΔK ¼
N yl EA
On the third stage of loading (P ¼ 3.0Ny) the internal force in third element equals Ny (Fig. 14.3g) and deflection of this element equals Nyl/EA. Since internal force in fourth element equals zero, its deflection is zero and required displacement at point K is ð3Þ
ΔK ¼ 2
Nyl EA
Corresponding P ΔK diagram is shown in Fig. 14.3h; the factors Nyl/EA and Ny for horizontal and vertical axis, respectively. This diagram shows that plastic analysis is nonlinear analysis. Plastic behavior analysis in this particular case leads to the increment of the limit load by ((3 2.5)/2.5)100 % ¼ 20%.
14.3
Fundamental Methods of Plastic Analysis
Analysis of the plastic behavior of a structure may be performed also by kinematical and static methods. Both these methods are exact and much easier than the direct method. The first of these methods deals with various forms of failure, while the second method deals with various distributions of internal forces satisfying equilibrium conditions. The idea of these methods is explained below for structure which had been considered in previous section: absolutely rigid rod is suspended by four hangers 1–4 as shown in Fig. 14.3a. The axial stiffness of all members EA is constant. The problem is to find the limit load considering both methods.
14.3.1 Kinematical Method This method requires consideration of different forms of failure of a structure. For each form of failure, there is a corresponding well-defined value of the failure load. The actual limit load is a minimum load among all possible.
14.3
Fundamental Methods of Plastic Analysis
507
Let us consider all possible forms of the failure of the structure; they are shown in Fig. 14.4a. Each form 1–4 presents position of the rigid rod for different scheme of failures. Assume that elastic displacements of hangers are much less then plastic ones and may be ignored. Therefore, each failure position is obtained by rotation of the rod around the point of connection of a hanger and the rod. Internal forces, which arise in each hanger, are Ny.
a Kinematical Method Ny
Ny
Ny
Ny
Ny
K
2.
1.
Ny
P=6Ny
P=∞ Ny
Ny
Ny
Ny
3.
Ny
Ny D
4. P=4Ny
P=3Ny
b
Statical Method Ny
Ny
Ny
Ny
Ny
Ny N3= − Ny
N4=0 1.
2. P Ny
P Ny
Ny
3.
Ny
Ny
Ny
N1=3 Ny
N2= 4.
K P
P
Fig. 14.4 (a) Kinematical method. Different forms of failure of the structure; (b) Static method. Different stress conditions of the structure
Now for each scheme of failure, we need to find the load P using equilibrium condition. Scheme 1: P ¼
N y ðd þ 2d þ 3dÞ ¼ 6N y d
Scheme 2: Force P is not present in the moment equation ∑MK ¼ 0, since the moment arm of the force P with respect to point of rotation K is zero. Therefore this equation holds for an arbitrarily large force P; Scheme 3 : P ¼
Scheme 4 : P ¼
N y ð2d þ d þ d Þ ¼ 4N y d
N y ð3d þ 2d þ dÞ ¼ 3N y 2d
Corresponding values of P is shown on the schemes 1–4. The minimum value of P ¼ 3Ny. This case corresponds to rotation of the rod around the point D. This result coincides with result, which was already obtained by the direct method.
508
14
Plastic Behavior of Structures
14.3.2 Static Method According to this method, it is necessary to find all possible distributions of internal forces in statically indeterminate structure; we assume that for each distribution the internal forces do not exceed the limit load. For each distribution of internal forces there is a corresponding well-defined value of the external load P. Actual limit load is a maximum load among all possible limit loads. The various distributions of internal forces in the structure are shown in Fig. 14.4b. For all these schemes, the following two conditions should be met: 1. Since the structure contains four rods, then forces in any three members have to be equal to the limit load Ny, which means that these three members correspond to horizontal part of Prandtl diagram. 2. The force in the remaining forth member must be less than Ny. There are possible the following cases. N1 N1 N1 N2
¼ N2 ¼ N2 ¼ N3 ¼ N3
¼ N3 ¼ N4 ¼ N4 ¼ N4
¼ Ny ¼ Ny ¼ Ny ¼ Ny
The force in the remaining forth member should be defined from equilibrium condition. The sum of the moments with respect to point of application of the force P leads to the following results: X M K ¼ N y d N y d N 4 2d ¼ 0 ! N 4 ¼ 0, X M K ¼ N y d N y 2d N 3 d ¼ 0 ! N 3 ¼ N y , X M K ¼ 0 ! N 2 any value, X M K ¼ N 1 d þ N y d þ N y 2d ¼ 0 ! N 1 ¼ 3N y Unknown force N2 does not appear in the equation for case 3. It means that this force may have any value, in particularly N2 > Ny. Thus cases 3 and 4 should be omitted, because internal forces are greater than Ny. Now we need to consider cases 1 and 2 in more detail. Case 1. Since N4 ¼ 0, then equilibrium equation ∑Y ¼ 0 leads to P ¼ 3Ny. Case 2. Since N3 ¼ Ny, then the same equilibrium equation leads to result: P ¼ 3Ny Ny ¼ 2Ny. The maximum load which the system can resist is 3Ny; thus, the actual limit plastic load corresponds to Case 1. This result have been obtained earlier by direct and kinematical methods.
14.3.3 Summary For any statically indeterminate structure, there are a number of various forms of failure, which are possible. The kinematical theorem states that the true form of collapse is that one that corresponds to the minimum value of the limit load. For any statically indeterminate structure, there are a number of internal force distributions satisfying equilibrium conditions. The static theorem states that the distribution of internal forces, which occurs at the maximum value of the limit load, corresponds to exhausted bearing capacity of a structure. Both the methods express two extreme properties of plastic load for statically indeterminate structure, if the property of material of all members obey to Prandtl diagram.
14.4
Plastic Analysis of Continuous Beams
509
Note, the magnitude of the limit loads does not depend on the presence in the system of initial stresses caused by the displacement of the supports, initial imperfections, and inaccuracy of assembly. This is a fundamental difference from the analysis of the system in the elastic stage.
14.4
Plastic Analysis of Continuous Beams
So far we have discussed plastic analysis of a structure in case of ideal elasto-plastic material. Now we will consider the ideal rigid-plastic material (column 2, Table 14.1). When this stress-strain diagram may be adopted? Since elastic displacements of a structure are significantly less than plastic displacements, then these elastic displacements may be ignored. In this case, the material of the structure is called as idealized rigid-plastic material, and the structure is called as rigid-plastic one. This material is not real, but using this material, the procedure for plastic analysis of elasto-plastic structures may be simplified. This simplification is based on the following fact: if two structures, which are made from elasto-plastic and rigid-plastic materials, have the same limit plastic load, then the limit condition of the elasto-plastic structure asymptotically approaches the limit condition of the rigid-plastic structure. Let a structure is subjected to different loads simultaneously. Assume that each load may increase independently of each other. In this case the limit conditions may be approached under different values of loads. What will be limit load in this case? Concept of “limit load” becomes unclear. Therefore, let us assume that the loading is simple. It means that if the structure is subjected to different loads P1 and P2 ¼ λP1 acting simultaneously, then these loads are increasing together and parameter λ of the load (coefficient between loads) remains constant during entire loading process. A plastic analysis of idealized rigid-plastic structures may be performed using two principal methods, namely static and kinematical methods. Fundamental condition for both the methods is that in the limit plastic state, the bending moments at all plastic hinges equal to yielding moment (or plastic moment) My, which is a characteristic of material and cross-section of the beam. For plastic analysis by both methods it is necessary to show a collapse mechanism first. Let us consider application of static and kinematical methods for plastic analysis of continuous beams.
14.4.1 Static Method Two-span beam of constant cross-section subjected to two equal forces P is shown in Fig. 14.5a. It is required to determine the limit load. First, let us consider the behavior of structure subjected to given load and the failure mechanism. First stage. In the elastic condition the bending moment diagram is presented in Fig. 14.5b. Increasing of the loads leads to the increasing of bending moment ordinates. Since maximum bending moment occurs at the support 1 (M1 ¼ 3Pl/16), then a material of the beam begins to yield at this support. Spreading of the plastic zone at the support 1 during the increasing loading is shown in Fig. 14.2. Thus, the first plastic hinge appears at support 1 (it is shown by solid circle, Fig. 14.5c), and the entire two-span continuous beam is transformed into two simply supported beams (Fig. 14.5d); (in fact, Figs. 14.5c and d are equivalent). The maximum possible bending moment at the support 1 is equal to the plastic bending moment My. Corresponding bending moment diagram is presented in Fig. 14.5e. Second stage. Each of these simply supported beams is subjected to force P and plastic moment My at support 1 (Fig. 14.5d). Bending moment diagrams caused by plastic moment My and force P are shown in Fig. 14.5e,f. Bending moment at the point of application of force equals to l Pl M y M ¼ ð14:1Þ 2 4 2 Again, we will increase the loads P. It is clear that the maximum moment occurs at the point of application of load P. Finally, a new plastic hinge occurs within the span (in this simplest case, this plastic hinge will be located at the point of application of the load). As the result, three hinges will be located on the each span of the beam; they are—hinge at support A, hinge under the point of application of the load P and hinge at support 1. The nature of these hinges is different. Hinge at the point A is ideal one, which represents support, while two other hinges are plastic ones, and they are the result of exhausted bearing capability of the beam. The same situation is with the second beam 1-B. Even though the hinges are of different nature, but since they are located on one line, this leads to the failure of the structure.
510
14
a
P
P l
A l/2
Plastic Behavior of Structures
B
l/2
l/2
l/2
M1
b
P
c
P
M1
B
A
P
d
P
M1
My
e
M1
f
MP
Pl 4
g
My 2
Pl 4 − M y 2
P
h
My
Pl 4
P
My
φ l l/2
My
My l/2
l/2
l l/2
Fig. 14.5 Plastic analysis of redundant beam. (a, b) Design diagram and bending moment diagram in elastic state; (c, d) Plastic hinge at the support 1, and presenting the continuous beam as two separate beams; (e, f) Bending moments diagrams due to the limit plastic moment My at support 1 and given load P; (g) Finding of limit load by graphical method; (h) Calculation of limit load by kinematical method
14.4
Plastic Analysis of Continuous Beams
511
In the limit state, the bending moment at the point of application of force P should be equal to plastic moment, i.e., Pl M y ¼ My 4 2
ð14:2Þ
This equation leads to the limit load Plim ¼
6M y l
ð14:3Þ
Limit load can be also found using graphical procedure. For this we need to use superposition of two bending moment diagrams M1 and MP. These bending moment diagrams are shown in Fig. 14.5e,f, while the computation of the left part of (14.2) in Fig. 14.5g. If to assume that this part of (14.2) is equal My, then we immediately get formula (14.3).
14.4.2 Kinematical Method This method is based on the following idea: in the limit state, the total work done by unknown limit loads and all plastic bending moments My is zero. This method consists of the following steps: 1. Identify the location of the potential plastic hinges. These hinges may be located at supports and at points of concentrated loads. In case of distributed load, plastic hinge may appear at point of zero shear. Also, plastic hinges may occur at the joints of the frame. Thus, we identify possible failure mechanisms. 2. Equilibrium equations should be written for failure mechanism, which is following: plastic hinge first appears at the middle support and thus, turning the original statically indeterminate beam into two statically determinate simply supported beams (Fig. 14.5h). With further increase of loads P, plastic hinges appear at the points of application of these forces. As the result, three hinges are located on one line and thus, the system becomes instantaneously changeable, which corresponds to limit state. Assume that in the limit state the vertical displacement at the each force P equals unity. Then the angle of inclination of the left and right parts of each mechanism is φ¼
1 2 ¼ l l=2
The work done by limit loads equals W(P) ¼ 2Plim 1. The work W1 produced by plastic moments My at the intermediate support equals W1 ¼ 2Myφ. The work W2 produced by plastic moments at the points of load P equals W2 ¼ 4Myφ (these plastic moments are shown according to location of extended fibers in elastic stage). Since W(P) + W1 + W2 ¼ 0, then this condition can be rewritten as follows: 2Plim 1 2M y φ 4M y φ ¼ 0 The limit load becomes Plim ¼
6M y l
As it was expected, results obtained by both static and kinematical methods are identical. Some typical examples of plastic analysis of statically indeterminate beams are shown below. Example 14.1 Design diagram of pinned-clamped beam, which is subjected to concentrated load P is presented in Fig. 14.6a. Calculate the limit load P. Solution From the elastic analysis of the pinned-clamped beam we know that the maximum moment occurs at the support B. Therefore here will be located the first plastic hinge; the corresponding plastic moment is My (Fig. 14.6b).
512
14
Plastic Behavior of Structures
P
a A
B l/2
l/2 P
b A
B P
RA
c
My
A
B
My
My
d Pl 4
My My/2
Fig. 14.6 Plastic analysis and graphical calculation of limit load for pinned-clamped beam
Now we have simply supported beam, i.e., the appearance of the plastic hinge at B does not destroy the beam but led to the changing of design diagram; reaction of such beam is RA ¼ P/2 My/l. For this beam we can increase the load P until the second plastic hinge appears at the point of application of the load. Thus we have three hinges (Fig. 14.6c), which are located on the one line, so this structure becomes instantaneously changeable. Using superposition principle, the moment at the point of application of force is Pl M y M ðPÞ þ M M y ¼ 4 2 In the limit condition, the moment at the point of application of force equals to plastic moment: Plim l M y ¼ My 4 2
ðaÞ
This equation leads to the limit load P Plim ¼
6M y l
The procedure (a) may be presented graphically as shown in Fig. 14.6d. Obviously, in the case of an arbitrary position ξ ¼ ul of the force P on the beam, the limit load is a function of position ξ. Example 14.2 Design diagram of a pinned-clamped beam is presented in Fig. 14.7. Calculate the limit load q and find the location of a plastic hinges. Solution It is obvious that the first plastic hinge appears at the clamped support B; corresponding plastic moment is My. Now we have simply supported beam AB (this design diagram is not shown), so we can increase the load q until the second plastic hinge appears. Location of this plastic hinge will coincide with position of maximum bending moment of simply supported beam subjected to plastic moment at the support B and given load q.
14.4
Plastic Analysis of Continuous Beams
513
q A
B
MB
x l
RA
RB
q B
A
MB=My
x0 l
RA
RB
Fig. 14.7 Plastic analysis of pinned-clamped beam
For this beam, the general expressions for shear and bending moment at any section x are ql M B QðxÞ ¼ RA qx ¼ qx, 2 l qx2 ql M B qx2 ¼ x M ð xÞ ¼ R A x 2 2 l 2 In these expressions, the moment MB at the support B equals to plastic moment My. The maximum moment occurs at the point where Q(x) ¼ 0. This condition leads to x0 ¼ (l/2) (MB/ql). Corresponding bending moment equals 2 ql M B l MB q l MB M max ðx0 Þ ¼ 2 2 2 2 l ql ql The limit condition becomes when this moment and the moment at the support B will be equal to My, i.e., Mmax ¼ My. This condition leads to the following equation M 2y 3M y ql2 þ
q2 l4 ¼0 4
ðaÞ
If we consider this equation as quadratic with respect to My, then solution of this equation is My ¼
pffiffiffi ql2 3 8 2
Minimum root is My ¼
pffiffiffi ql2 3 2 2 ¼ 0:08578ql2 2
The limit load becomes qlim ¼
2M y My pffiffiffi ¼ 11:657 2 l 3 2 2 l2
pffiffiffi The plastic hinge occurs at x0 ¼ l 2 1 ¼ 0:4142l. If we consider equation (a) as quadratic with respect to q, then solution of this equation is pffiffiffi 2M y qmax ¼ 2 3 þ 2 2 l This result coincides with (b).
ðbÞ
514
14
Plastic Behavior of Structures
The following example is related to the problem of determining the limit load in the case of a simple loading of a beam by lumped force P and distributed load q. Simple loading means that in the process of increasing loads the ratio between the loads P and q remains unchanged, P ¼ kq, where k is any fixed parameter. Example 14.3 Two-span beam with overhang is subjected to force P and uniformly distributed load q as shown in Fig. 14.8a. The loading of the beam is simple; assume that relationship between loads is always P ¼ 2ql2. Determine the limit load, if a ¼ 3 m, b ¼ 4 m, l2 ¼ 6 m, l3 ¼ 2 m, and bearing capacity of all cross sections within the beam is My ¼ 60 kNm. P
a
q A
K a=3m
b=4m
P=2ql2 l2=6m
l1=7m
b
P
My
My
q
l3=2m My D
K
c
D
C
B
LPM qliml22 8
Plimab l1
My My
LPM Fig. 14.8 Plastic analysis of continuous beam. (a) Design diagram; (b) Beam with plastic hinges at supports; (c) Bending moment diagram inscribed between two limit plastic moments (LPM) lines
Solution The given structure has two redundant constraints. There is possible the different schemes of structural failure. Let us consider one of them. The progressive increase of the loads leads to the appearance of the plastic hinge at one of the supports, so the structure becomes statically indeterminate of the first degree. Further increase of the loads leads to the appearance of the plastic hinge at another support. Finally, plastic hinge happens at the last support. It means that the entire continuous beam is being transformed into two simply supported beams subjected to given loads and plastic moments My at the supports A, B and C as shown in Fig. 14.8b. Direction of moments My is shown according to location of extended fibers in elastic state. The order of appearance of the plastic hinges on supports depends on the relationships between force P and load q as well as geometrical parameters of the beam. This failure mechanism allows developing the theory of plastic analysis of continuous beams subjected to several loads. However, it does not mean that exactly the above sequence of formation of plastic hinges will be realized. For example, if load q is small, then plastic hinges can appear first of all at the supports A, B and at point K, and after that at support C and in the second span. Real emergence sequence of plastic hinges may be defined only after determination of limit load as shown below. Reaction of support A and bending moment at the point K of the first span caused by force P as well as plastic moments at supports A and B equals
14.4
Plastic Analysis of Continuous Beams
515
A B Pb M y M y Pb RA ¼ RA ðPÞ þ RA M Ay þ RA M By ¼ þ ¼ l1 l1 l1 l1 Pab A M K ¼ M K ðR A Þ M K M y ¼ My l1
ðaÞ
Bending moment at the middle point of the second span caused by distributed load q and plastic moments at supports B and C equals M¼
ql22 My 8
ðbÞ
The following increase of the load leads to the appearance of the plastic moments within the first and second spans. In the limit state these bending moments must be equal to plastic moments; therefore expressions (a) and (b) should be rewritten for both spans as follows: Pab Pab M y ¼ M y or ¼ 2M y l1 l1 2 2 ql2 ql2 For second span M y ¼ M y or ¼ 2M y 8 8 For first span
ðcÞ ðdÞ
Equation (c) and (d) show that in limit state, the limit bending moment caused by external load equals twice plastic moment. This can be presented geometrically as shown in Fig. 14.8c. Limit plastic moments (LPM) are shown by two horizontal dotted lines. These lines show that limit moments for supports and for any section of the beam are equal, may be negative or positive, however, they cannot be more than My. Now we can fit a space between two LPM lines by bending moment diagrams caused by external load for each simply supported beam. This procedure is called equalizing of bending moments and can be effectively applied for any continuous beam. For the first span, (c) allows calculating the limit force P 2M y l1 2 60ðkNmÞ 7ðmÞ Plim ab ¼ 2M y ! Plim ¼ ¼ ¼ 70 kN l1 ab 3ð m Þ 4ð m Þ
ðeÞ
For the second span, (d) allows calculating the limit distributed load q 16M y 16 60ðkNmÞ qlim l22 ¼ 2M y ! qlim ¼ 2 ¼ ¼ 26:67 kN=m 8 36ðm2 Þ l2 Now we need to take into account the condition of simple loading as well as limiting force P ¼ 70 kN and load q ¼ 26.67 kN/m. Knowing the distributed load we can calculate, according condition P ¼ 2ql2, corresponding limit force P Plim ¼ 2:0 26:67ðkN=mÞ 6ðmÞ ¼ 320 kN > 70 kN
ðfÞ
For part CD 2M y 2 60ðkNmÞ qlim l23 ¼ 30 kN=m, ¼ M y ! qlim ¼ 2 ¼ 2 22 ðm2 Þ l3 which leads to Plim ¼ 2:0 30 6 ¼ 360 kN > 70 kN
ðgÞ
Thus, limit distributed load q in both cases leads to the limit force P which can not be accepted. The final limit load is governed by minimum value given by formulae (e), (f) and (g), so the limit load Plim ¼ 70 kN, and corresponding qlim ¼
Plim 70 ¼ 5:83 kN=m ¼ 26 2l2
516
14
Plastic Behavior of Structures
Discussion On the basis of obtained numerical results we can explain the order of appearance of the plastic hinges. The limit load P ¼ 70 kN and corresponding load q are determined from the conditions of appearance of plastic hinges at the supports A, B and at point K of application force P. Thus, the failure of the structure in whole is defined by a failure of the span AB because this simply supported beam is being transformed in mechanism. In this case, with further increase of the load q, the relationship P ¼ 2ql2 is not held anymore, since load P cannot reach the value greater than Plim. It is obvious that span BC still can resist to increased load q, however, the structure in whole is differ from the original one. The problem of determination of limit load for continuous beam with given bearing capacity has unique solution.
14.5
Plastic Analysis of Frames
A frame can be failed by different ways. The different schemes of failure are presented in Fig. 14.9. They are following: beam mechanism of failure (B1, B2, B3), mechanism of sidesway failure (S), joint failure (J ), framed (F) and different combined mechanisms.
Design diagram
B1
B2
B3
S
F
J
B3+S+J
Fig. 14.9 Design diagram of the frame and possible mechanisms of failure
The type of mechanism of failure, which will occur, is not known in advance. This is the principal difficulty for plastic analysis of frames. Therefore, for each mechanism of failure and their different combinations, the equilibrium conditions should be considered and then that mechanism of failure should be adopted, which occurs at the minimum load.
14.5
Plastic Analysis of Frames
517
Now we will consider the determination of limit load for different types of failure.
14.5.1 Beam Failure Design diagram of the portal frame is presented in Fig. 14.10a. The loading of the frame is simple. Assume that Q ¼ 2P and vert the limit moments for vertical and horizontal members satisfy to condition M hor y ¼ 2M y . This scheme is characterized by failure of its horizontal element only; the horizontal displacement of the frame is absent (Fig. 14.10b). In this case the plastic hor hinges appear at the joints C (and D) as well as at the point K. Corresponding plastic moments are denoted as M vert y and M y . Their directions are shown according to the location of extended fibers; location of such fibers are easy to find if bending moment diagram is plotted in elastic state. The inclination of the parts CK and KD are denoted by α. Equation of limit equilibrium for beam scheme of failure can be found using kinematical method (in the limit condition the total work done by unknown limit loads and all plastic bending moments My equals to zero): Qlim
αl vert 2M hor y α 2M y α ¼ 0 2
vert We can see that beam mechanism is realized only by load Q. Since M hor y ¼ 2M y , then solution of this equation leads to the following limit load Qlim ¼ 12M vert y =l. Since Q ¼ 2P, then corresponding limit load P is following:
Plim ¼
6M vert y l
ð14:4Þ
14.5.2 Sidesway Failure This scheme of failure of the structure is characterized by appearance of plastic hinges at supports A and B as well at joints C and D; a failure of horizontal element is absent (Fig. 14.10c). In this case the structure is undergoing the horizontal displacement only. The inclination of the vertical members AC and BD are denoted by β. Equation of limit equilibrium using kinematical method is following: Plim βh 4M vert y β ¼ 0 We can see that sidesway mechanism of the failure is realized only by load P. Since l ¼ 2 h, then solution of this equation Plim ¼
4M vert 8M vert y y ¼ h l
ð14:5Þ
14.5.3 Combined Failure This scheme of failure of a structure is characterized by appearance of plastic hinges at supports A and B, at joint D as well at point K (Fig. 14.10a). In order to the system becomes a mechanism, the total number of plastic hinges must be n + 1, where n is a degree of redundancy. In our case, the system becomes a mechanism when the number of plastic hinges is 4 (Fig. 14.10d). The inclination of the vertical members AC and BD are denoted by θ. Since a joint C remains a rigid ones, then inclination of the horizontal members are also θ.
518
14
b
l/2 P
Qlim
M yvert
αl/2
Q=2P
a
D
C
Plastic Behavior of Structures
M yvert α
h=l/2
K
M yhor
B
A l
d
βh Plim
Plim
M yvert
β
M yvert
M yhor
θ
M yvert
M yvert
e
M yvert
θ
β
M yvert
Qlim
θh θl /2
c
θ
M yvert
k2 16
12
B B+S
10.6 6 8
4
S 0
0
4 5.33
8
12
116 6
k1
Fig. 14.10 Plastic analysis of redundant portal frame. (a) Design diagram of the frame; (b) Beam mechanisms of failure (B); (c) Sidesway k2 ¼ mechanisms of failure (S); (d) Combined mechanisms of failure (B + S); (e) Graph of limit combination of loads, k1 ¼ Plim l=M vert y , Qlim l=M vert y
As before, the equation of limit equilibrium using kinematical method is following: Qlim
θl hor þ Plim θ h 4M ver y θ 2M y θ ¼ 0 2
vert We can see that combined mechanism is realized by both loads Q and P. For M hor y ¼ 2M y and l ¼ 2h, last equation may be rewritten as
Qlim l Plim l þ vert ¼ 16 M vert My y
ð14:6Þ
Problems
519
14.5.4 Limit Combination Diagram The result of the plastic analysis may be presented by diagram shown in Fig. 14.10e. The designations of the axis are k1 ¼ Plim l=M vert and k 2 ¼ Qlim l=M vert y y . For beam failure Qlim ¼ 12M vert y =l. This case is shown on the Fig. 14.10e by solid line B, which is parallel to horizontal axis because the beam mechanism is realized only by load Q. For sidesway failure Plim ¼ 8M vert y =l. This case is shown on the Fig. 14.10e by solid line S, which is parallel to vertical axis because the sidesway mechanism is realized only by load P. For combined failure relationships between limit loads is given formula (14.6). Corresponding line is shown in Fig. 14.10e by solid line B + S. Since Q ¼ 2P, then Plim ¼
16M vert 5:33M vert y y ¼ 3l l
ð14:7Þ
Formulae (14.4), (14.5) and (14.7) present limit load P, which corresponds to different mechanisms of failure. Failure is governed by minimum load Plim ¼
5:33M vert y , l
which corresponds to combined mechanism of failure; therefore, this type of failure will happens. Note that increasing of all geometrical dimensions of the frame (l, h) by n-times leads to decreasing of the limiting loads by n-times.
Problems 14.1. The straight rod with cross sectional area A is located between two rigid supports M and N and subjected to axial load P (Fig. P14.1). The yield stress of material is σ y. Define the limit load P for two design diagrams: (a) structure in Fig. P14.1a does not have a gap and (b) element is placed between supports with gap Δ. Solve these problems by direct and static methods. Explain obtained results.
a
b
P M
N a
M
b
Ans. Plim = 2 σ y A
P N a
b
Δ
Fig. P14.1
14.2. The symmetrical rod structure is subjected to load P as shown in Fig. P14.2. The cross sectional area of the vertical member is A, and for rest members are kA, where k is any number. Yielding stress for all members is σ y. Perform the elastic analysis and calculate Pallow. Calculate the limit load and determine ratio Plim/Pallow. Use the static method.
kA
1
α
A
α
P Fig. P14.2
Ans. Plim = σ y A(1 + 2k cos α ) .
520
14
Plastic Behavior of Structures
14.3. Absolutely rigid rod is suspended by vertical hangers as shown in Fig. P14.3. The axial stiffness of all members EA is constant. The limit internal force for each element is Ny. Consider different mechanisms of failure and determine the limit load P for entire design diagram.
Ans. P = d
d
d
1
d
2
3
N y (2d + d + d ) 3d
= 1.33N y
l 4
P Fig. P14.3
14.4. Concentrated force P acts on the uniform beams as shown in Fig. P14.4. The limit moment is My. Determine the limit load Plim for pinned-clamped beam (case Fig. P14.4a) and for clamped-clamped beam (case Fig. P14.4b). Sketch the graphs Plim ¼ Plim(ξ) and determine the position of the force on the beam and the corresponding value of P at which the value of the limit load is smallest. Ans.
P
a
P
b
(a) Plim =
Plimmin ( 0.41) = 5.82
ξl
ξl
(1 + ξ ) M y , ξ ∙ (1 − ξ ) l My l
l
l
(b) Plim =
My
2
ξ ∙ (1 − ξ ) l
Fig. P14.4
14.5. Determine the limit distributed load q for beam with clamped supports. q Ans. qlim =
16M y l2
l Fig. P14.5
14.6. The two-span uniform beam ABC is subjected to uniformly distributed load as shown in Fig. P14.6. Find a location of the plastic hinge. Calculate the limit load. q B
C
A x0 l Fig. P14.6
l
( ) (6 + 4 2 ) = 11.657 Ml
Ans. x0 = l 2 − 1 ; qlim =
My l
2
y 2
Problems
521
14.7. The two-span uniform beam ABC is subjected to uniformly distributed load and concentrated load P as shown on design diagram. Both loads are satisfy to condition P ¼ kql. Find the parameter k, which leads to plastic conditions at the both spans at once. P
q
Ans. k=0.5147.
B
C
A l
l/2
l/2
Fig. P14.7
14.8. Design diagram of the frame is presented in Fig. P14.8. Analyze the plastic behavior and find the mechanism of failure (MF), if the loading of the frame is simple; assume that Q ¼ 2P and limit moments for vertical and horizontal members satisfy vert to condition M hor y ¼ 0:5M y .
6m P
Q=2P
Ans. Beam MF Plim = D
C
Sidesway MF: Plim =
h=6m
K
B
A
3M vert y l
8M vert y
Combined MF: Plim =
l
;
;
5.33M vert y l
.
l=12m Fig. P14.8
14.9. Some statically indeterminate structure is subjected to load, settlements of supports and change of temperature at the same time. How the settlements of supports and change of temperature influence on the value of the limit load?
Chapter 15
Stability of Elastic Systems
Theory of structural stability is a special branch of structural analysis. This theory explores very important phenomenon that is observed in the behavior of the structures subjected to compressed loads, namely the abrupt change of initial form of equilibrium. Such phenomenon is called loss of stability. As a rule, the loss of stability of a structure leads to its collapse. Engineering practice knows a lot of examples when ignoring this feature of a structure led to its failure. This chapter is an introduction to stability analysis of engineering structures subjected to compressed loads. Among them are structures that contain nondeformable members as well as beams, frames, and arches. Classical methods of analysis will be discussed.
15.1
Fundamental Concepts
We will differentiate two types of structures, mainly, the structures consisting of absolutely rigid bodies connected by elastic constrains and structures consisting of deformable members; it is possible to combine both types of members in one structure. Some examples of these structures under compressed loads are shown in Fig. 15.1. Structures that contain absolutely rigid members (EI ¼ 1) are shown in Fig. 15.1a,b; these design diagrams present structures with elastic joints. Elastic joint means that the angle between two adjacent members is changed upon load application. Figure 15.1c,d presents structures, which contain deformable elements; structure in Fig. 15.1e contains the absolutely rigid part AC and deformable part CB.
a
P EI= ∞
b
c
P
P
P
1
P
EI= ∞
EI
k d P
P
k
k EI= ∞
A
B EI=∞ C
EI
2
e P
EI= ∞
P
P
EI
EI
Fig. 15.1 Types of structures and forms of buckling. (a, b) Structures with absolutely rigid members; (c, d) Structures with deformable members only; (e) Structure with absolutely rigid and deformable members
Structures, which are subjected to compressed loads, may be either in stable or in unstable equilibrium. Stability is a property of a structure to keep its initial position or initial deformable shape. Stable structure will regain to its original state if any disturbed factor, which changed the initial state, is removed. A structure subjected to compressed loads may be disturbed from initial equilibrium state by, for example, a small lateral load. After removing this disturbance, a structure can return to the initial state, or tends to return to the initial state, or remains in the new state, or even tends to switch into new state. Behavior of a structure after removing a disturbance depends on the © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_15
523
524
15
Stability of Elastic Systems
value of compressed loads. For small compressed load, a structure will return to the initial state, i.e., this equilibrium state is a stable one, while for larger compressed load, a structure will not return to the initial state, i.e., this equilibrium state is unstable. However, what does “small and large” load mean? It is obvious that the behavior of a structure after removing a disturbance load depends not only on the value of the compressed load but also on the types of supports, the length of compressed members, and their cross sections. Any compressed load for tall column with small cross section may be treated as a large load, while for short column with large cross section the same load may be treated as a small one. The theory of elastic stability gives the precise quantitative characteristics of compressed loads for different types of structures, which leads to well-defined state of a structure, and allows understanding the influence of parameters of a structure (boundary conditions, cross section, and properties of material) on the value of this load and corresponding behavior of a structure. Let us introduce the following definitions: Stable equilibrium state means that if the structure under compressed load is disturbed from an initial equilibrium state and after all disturbing factors are removed, then the structure returns to the initial equilibrium state. This is concerning to the elastic structures. If a structure consists of plastic or elasto-plastic elements, then a complete returning to the initial state is impossible. However, equilibrium state is assumed to be stable, if a structure even tends to return to the initial equilibrium state. In case of absolutely rigid bodies, we are talking about stable position of a structure, while in case of deformable elements, we are talking about stable equilibrium form of a deformable state. In all these cases, we say that the acting compressed load is less than the critical one. Definition of the critical load will be given later. Unstable equilibrium state means that if a structure under compressed load is disturbed from an initial equilibrium state and after all disturbing factors are removed, then the structure does not return to the initial equilibrium state. In this case, we say that the acting compressed load is larger than the critical one. Change of configuration of a structure under action of compressed load is called a loss of stability of the initial form of equilibrium or a buckling. If the compressed load is a static one, then this case is referred as the static loss of stability. In this chapter, we will consider absolutely rigid and ideal elastic structures subjected to static compressed loads only. If a structure switches to other state (as a result of loss of stability) and remains in this state in equilibrium, then this new equilibrium state is called the adjacent form of equilibrium. The static load may be of two types: conservative and nonconservative. The work done by conservative forces is determined only by the initial and final position of points of application of a force. Example of a conservative force is a force, which keeps its direction (Fig. 15.2a).
a
P
P>Pcr
b
P
P
Fig. 15.2 The column under conservative and tracking force
The work done by nonconservative forces is determined by trajectory between the initial and final position of points of application of the force. Example of nonconservative force is a tracking force whose orientation depends on the slope of the elastic curve at the point of application of the force (Fig. 15.2b). The corresponding systems are called conservative and nonconservative. In this chapter, only conservative systems are considered. The critical force Pcr is the maximum force at which the structure holds its initial equilibrium form (the structure is still stable) or minimum force at which the structure no longer returns to the initial state (the structure is already unstable) if all disturbing factors are removed. The state of a structure that corresponds to critical load is called the critical state. The switching of a structure into new state occurs suddenly and as a rule leads to the collapse of a structure. The theory of static stability of the structures is devoted to methods of calculation of critical loads. Degree of freedom is a fundamental concept of stability analysis. Degrees of freedom present independent parameters, which define the structure’s configuration. The structures, which contain only rigid elements, have the finite number of degrees of freedom. Each deformable element should be considered as a member with distributed parameters, so the structures that contain only deformable elements have infinite number of degrees of freedom. The structures presented in Fig. 15.1a, b have one degree of freedom, while the structures presented in Fig. 15.1c, d have infinite number of degrees of freedom.
15.2
Stability of Structures with Finite Number of Degrees of Freedom
525
Generalized coordinates are independent parameters, which uniquely defines configuration of a system in new arbitrary state. A structure with n degrees of freedom has n generalized coordinates. A structure with n degrees of freedom has n critical loads. Each critical load corresponds to one specified form of equilibrium. For structure with one degree of freedom, there exists only the unique form of the loss of stability (Fig. 15.1a, b) and its corresponding unique critical load. For structure with infinite number of degrees of freedom, there exist infinitely many critical loads and corresponding forms of loss of stability. Figure 15.1d shows only the first buckling form of a frame. In Fig. 15.1c the numbers 1 and 2 indicate the first and second forms of the loss of stability of a beam. It is very important to define the smallest critical load, because it leads to the loss of stability according to first form, i.e., to the failure of the structure. The second and following forms may be realized upon the special additional conditions. There exist precise and approximate methods for calculating of critical loads. Precise methods are static, energy method and dynamical ones. These methods reflect the fact that the concepts “stable or unstable state of equilibrium” may be considered from different points of view. Static method (or equilibrium method) is based on the consideration of equilibrium of a structure in a new configuration. The critical force is such minimum force, which can hold the structure in equilibrium in the adjacent condition, or maximum force, for which initial straight form of equilibrium is yet possible. Energy method requires consideration of total energy of a structure in a new configuration. This energy U equals to stress energy U0 and potential W of external force U ¼ U0 þ W
ð15:1Þ
The potential W of external force equals to work, which is produced by external force on the displacement from final state into initial one. The stable equilibrium of the structure corresponds to a minimum of the total energy. For system with n degrees of freedom, the critical load may be calculated from the set of equations ∂U ¼ 0; ∂q1
∂U ¼ 0; ∂q2
∂U ¼ 0, ∂qn
ð15:2Þ
where qi are the generalized coordinates of the structure. This method is equivalent to the virtual displacement method, according to which the sum of work done by all forces on any virtual displacements is zero. The static and energy methods are considered below.
15.2
Stability of Structures with Finite Number of Degrees of Freedom
This section is devoted to calculation of critical load for structures containing only absolutely rigid bars, with elastic constraints. Three types of elastic constraints will be considered. They are the following: 1. Elastic support, which allows an angular displacement. Rigidity of support is krot. The reactive moment of such support and its angular displacement θ are related as M ¼ krotθ. 2. Elastic support, which allows a linear displacement. Rigidity of support is k. Reaction of support and its linear displacement f are related as R ¼ kf. 3. Elastic connections between absolutely rigid members (elastic hinged joint) of a structure. Rigidity of connection is krot. The moment of this joint and mutual angular displacement θ of two adjacent absolutely rigid bars are related as M ¼ krotθ. From methodical point of view, it is reasonable that all structures with finite number of degrees of freedom are divided into two large groups—those with one degree of freedom and structures with two or more degrees of freedom. Two different methods for structures with finite number of degrees of freedom are presented below. They are the static and energy methods.
15.2.1 Structures with One Degree of Freedom Elastic supports. A simplest structure with one degree of freedom is shown in Fig. 15.3a. Absolutely rigid vertical weightless column is placed on rigid supporting plate. Since a foundation is flexible, the column in whole may rotate around fixed point О. The column is subjected to axial compressed static force P. Assume that an angular rigidity of elastic support is krot; such
526
15
Stability of Elastic Systems
type of elastic support is presented as two springs of equal stiffness. Within their deformation, the forces that arise in both springs create the elastic couple.
a
b
P
l
EI=∞ O
c
P f
P
θ
θ
Δ W = −P∆
O U 0 = k rot
krot
θ2 2
M=krot θ Fig. 15.3 Absolutely rigid vertical cantilever bar on the elastic support. (a) Design diagram; (b) Static method; (c) Energy method
To determine the critical load by static or energy method, first of all, we need accept a generalized coordinate. Let an angular displacement θ of the supporting plate be the generalized coordinate. This parameter describes completely a perturbed configuration of the structure. The structure in the strained state is shown by dotted line. Static method (Fig. 15.3b). The moment that is produced in the support is M ¼ θ krot. The moment due to external load N with respect to support point O is Pf . We assume that angular displacement θ is small, and therefore f ¼ l sin θ lθ. Equilibrium equation ∑M0 ¼ 0 leads to the stability equation Pf θ krot ¼ Plθ θ krot ¼ 0
ð15:3Þ
This equation is obtained on the basis of linearization procedure sinθ θ; therefore the equilibrium equation (15.3) is called linearized stability equation. Equation (15.3) is satisfied at two special cases: 1. The angle θ ¼ 0. It means that initial vertical form of the column remains vertical for any force P. This is a trivial solution, which corresponds to the initial form of equilibrium. 2. The angle θ 6¼ 0. It means that the strained form of equilibrium is possible for load Pcr ¼ krot =l. This load is critical. If the given structure is subjected to load P < Pcr, then initial vertical position of the column is the only equilibrium position; therefore, if the structure would be disturbed, then it returns to its initial position. Pay attention that the value of the angle θ cannot be determined on the basis of linearized stability equation. Energy method (Fig. 15.3c). Since the vertical displacement of the point of application of the force P is Δ ¼ lð1 cos θÞ ffi
lθ2 , 2
then the potential of external load N is W ¼ PΔ ¼ Pl
θ2 2
This expression has negative sign since the work of the force P should be calculated on the displacement Δ from final to initial state. Since the rigidity of support is krot, then the strain energy of elastic support is U 0 ¼ krot The total energy of the system is
θ2 2
15.2
Stability of Structures with Finite Number of Degrees of Freedom
527
U ¼ U 0 þ W ¼ k rot
θ2 θ2 Pl 2 2
Condition (15.2) ∂U ¼ krot θ Plθ ¼ 0 ∂θ leads to the same critical load Pcr ¼ krot =l. Example 15.1 The structure contains two absolutely rigid bars (EI ¼ 1), which are connected by hinge C, and supported by elastic support at this point; the rigidity of elastic support is k. The structure is subjected to axial compressed force P (Fig. 15.4a). Calculate the critical force by the static and energy methods.
a
P
C k
l
b
C
c
a P
P
P
U 0 = kf 2 2
ka/2
l
f P
θ R=ka
P
EI= ∞
ka/2
k
W = −P∆ ΔC
Δ
Fig. 15.4 Absolutely rigid 2-elements structure with elastic support. (a) Design diagram; (b) Static method; (c) Energy method
Solution Static method (Fig. 15.4b). The structure has one degree of freedom. Let a vertical displacement a of the hinge C be a generalized coordinate. The reaction of elastic support is R ¼ ka and reactions of pinned and rolled supports are ka/2. Bending moment at hinge C equals zero; therefore stability equation becomes M left C ¼
ka l Pa ¼ 0 2
This equation immediately leads to the critical force and becomes Pcr ¼ kl=2. Energy method (Fig. 15.4c). Let an angular displacement θ at the support points be a generalized coordinate. The vertical displacement of hinge C equals f ¼ l sin θ ffi lθ, while the horizontal displacement at the same point C is ΔC ¼ l(1 cos θ), so horizontal displacement at the point of application P becomes Δ ¼ 2ΔC ¼ 2lð1 cos θÞ ffi 2l
θ2 2
Strain energy accumulated in elastic support U 0 ¼ Rf =2, where reaction of elastic support is R ¼ kf, so U0 ¼ k
f2 l2 θ 2 ¼k 2 2
The potential of external force is W ¼ PΔ ¼ Plθ2, so the total energy is U ¼ U0 þ W ¼ k
l2 θ 2 Plθ2 2
Derivative of this expression with respect to generalized coordinate equals to zero, i.e.,
528
15
Stability of Elastic Systems
∂U ¼ kl2 θ 2Plθ ¼ 0, ∂θ which leads to above-determined critical load.
Discussion 1. The static method requires calculation of reactions of all supports, while energy method requires calculation of reactions only for elastic supports. 2. Why potential of external force equals W ¼ PΔ, while the energy accumulated in elastic support contains coefficient 0.5? If a structure switches into the new position, a compressed force P remains the same; that is why potential of this force equals W ¼ PΔ. The internal forces in the elastic constrain increase from zero to maximum values; that is why the expression for accumulated energy contains coefficient 0.5. Elastic joint. So far we have considered only two types of joints, mainly hinged and rigid joints. Let us introduce a concept of elastic joint. If a structure is subjected to any loading, then for such joint the initial angle between members forming the joint changes. Each elastic joint is characterized by the rigidity of the elastic hinge krot. In the simplest case, the moment which arise at elastic connection C is M ¼ krotθ, where θ is a mutual angular displacement of the members at the elastic joint. Let us show an application of static and energy methods for determination of critical load for case when two absolutely rigid rods are connected by elastic hinge C and loaded by axial static force P (Fig. 15.5a).
a
c θ=2α
P α
krot
l
M=krot θ
b
C
P
P
U 0 = k rot
P
α
θ2 2
α
f ΔC
P l
θ=2 α α
P Δ
W = −P ∆
Fig. 15.5 Absolutely rigid elements with elastic joint C. (a) Design diagram; (b) Static method; (c) Energy method
The structure has one degree of freedom. Take angle α as a generalized coordinate. A new form of equilibrium is shown by solid line. Static method (Fig. 15.5b). The vertical displacement of the joint C is f ¼ l sin α ffi lα. The mutual angle of rotation of the bars is θ ¼ 2α, so the moment, which arises in elastic connection at point C, equals M ¼ krotθ ¼ 2krotα. The vertical reactions of supports are zero. The bending moment at the hinge C equals zero, so the stability equation becomes X left M C ¼ 0 : Pf krot θ ¼ 0 ! Plα 2krot α ¼ 0 Nontrivial solution of this equation is Pcr ¼
2krot l
Energy method (Fig. 15.5c). Horizontal displacement of the point of application of the force P equals Δ ¼ 2ΔC ¼ 2lð1 cos αÞ 2l
α2 , 2
15.2
Stability of Structures with Finite Number of Degrees of Freedom
529
so potential of the external force is W ¼ PΔ ¼ Plα2. Mutual angle of rotation at joint C is θ ¼ 2α, the strain energy accumulated in elastic connection is krot ð2αÞ2 ¼ 2krot α2 , 2 so the total energy is U ¼ Plα2 + 2krotα2. Condition
∂U ¼ 0 leads to the following stability equation ∂α
2Plα þ 4krot α ¼ 0, and for critical force we get the same expression Pcr ¼
2krot . l
15.2.2 Structures with Two or More Degrees of Freedom For stability analysis of such structures first, it is necessary to define the number of degrees of freedom, sketch the structure in any arbitrary position, and chose the generalized coordinates. As for structures with one degree of freedom, the static method requires consideration of equilibrium conditions. Energy method requires the calculation of potential of external forces and energy accumulated in elastic constraints. The static method requires calculation of reactions of all supports, while energy method requires calculation of reactions only for elastic supports. Static equations (for static method) as well as conditions for energy method (15.2) lead to n algebraic homogeneous equations with respect to unknown generalized coordinates. Trivial solution corresponds to initial (or unperturbed) state of equilibrium. To obtain nontrivial solution, it is necessary to equate the determinant of coefficients before generalized coordinates to zero. This equation serves for calculation of critical loads. Their number equals to the number of degrees of freedom. Each critical load corresponds to specified shape of loss of stability. Example 15.2 The structure contains three absolutely rigid bars (EI ¼ 1), which are connected by ideal hinges C1 and C2, and supported by elastic supports at these points; the rigidity of elastic supports is k. The structure is subjected to axial compressed force N as shown in Fig. 15.6a. Calculate the critical load.
a C2
C1
EI=∞
N
k
b N
a1
C1
C2
c a2
f1 N
N
k l
l
l C1
C2
d
a1=1
R2=ka2
N
B β2
R1=kf1 R1=ka1
f2
A β1
k (2a1 + a2 ) 3
N
R2=kf2
∆
k (a1 + 2a2 ) 3
e
N= N1cr N
N= N2cr a1=1
a2=1 N
a2= −1 Fig. 15.6 Structure with two degrees of freedom. (a) Design diagram; (b) Static method; (c) Energy method; (d, e) First and second form of the loss of stability
530
15
Stability of Elastic Systems
Solution Let us consider this problem by the static and energy methods. Static method. The structure has two degrees of freedom. Displacements of the hinges C1 and C2 are a1 and a2; they are considered as generalized coordinates. The reactions of elastic supports are R1 ¼ ka1 and R2 ¼ ka2; reactions of the left and right supports are shown in Fig. 15.6b Bending moments at hinges C1 and C2 are equal to zero; therefore taking into account the reactions at the left and right supports we get k ð2a1 þ a2 Þ l ¼ 0, 3 kða1 þ 2a2 Þ l¼0 ¼ Na2 3
M left 1 ¼ Na1 M right 2
This system may be rewritten as homogeneous algebraic equations with respect to unknown generalized coordinates a1 and a2 a1 ð3N 2klÞ a2 kl ¼ 0, a1 kl þ a2 ð3N 2klÞ ¼ 0
ðaÞ
The trivial solution a1 ¼ 0, a2 ¼ 0 corresponds to initial nondeformed configuration of the structure. Nontrivial solution of this system occurs if determinant of the system equals zero: 3N 2kl kl ¼ 0 ! ð3N 2klÞ2 ðklÞ2 ¼ 0, D¼ kl 3N 2kl or 3N 2 4N k l þ ðk lÞ2 ¼ 0 The roots of this equation present the critical loads; they are N 1cr ¼
kl , 3
N 2cr ¼ kl
ðbÞ
Energy method. Let the generalized coordinates be angles β1 and β2 (Fig. 15.6c). The angle between portion C1C2 and horizontal line is β1 β2, so horizontal displacement of the point of application force N is Δ ¼ lð1 cos β1 Þ þ lð1 cos β2 Þ þ lð1 cos ðβ1 β2 ÞÞ h i l ffi β21 þ β22 þ ðβ1 β2 Þ2 ¼ l β21 þ β22 β1 β2 2 The potential of the force N is W ¼ NΔ. The vertical displacements of points C1 and C2 equal to f1 ¼ l tan β1 ffi lβ1 and f2 ¼ l tan β2 ffi lβ2, respectively. Therefore, reactions of elastic supports are R1 ¼ kf1 ¼ klβ1 and R2 ¼ kf2 ¼ klβ2. The energy accumulated in elastic supports is X Ri fi 2
k k ¼ ðlβ1 Þ2 þ ðlβ2 Þ2 2 2
The total energy U ¼ NΔ þ
X Ri fi
k k ¼ Nl β21 þ β22 β1 β2 þ l2 β21 þ l2 β22 2 2 2
Derivative of the total energy with respect to generalize coordinate (according to 15.2) leads to the following equations ∂U ¼ 2Nlβ1 þ Nlβ2 þ kl2 β1 ¼ 0 ∂β1
ðk l 2N Þβ1 þ Nβ2 ¼ 0 ðcÞ
or ∂U ¼ 2Nlβ2 þ Nlβ1 þ kl2 β2 ¼ 0 ∂β2 Nontrivial solution of homogeneous system (c) occurs if
Nβ1 þ ðk l 2N Þβ2 ¼ 0
15.3
Stability of Columns with Rigid and Elastic Supports
531
k l 2N D ¼ N
N ¼0 k l 2N
Stability equation becomes (kl 2N )2 N2 ¼ 0. This equation leads to the same critical loads (b). Each critical load corresponds to a specified shape of equilibrium. Both critical loads should be considered. 1. Let N ¼ N 1cr ¼ kl=3. Substituting it in the first equation of system (a), we obtain kl a1 3 2kl a2 kl ¼ 0, 3 and relationship between generalized coordinates is a1 a2 ¼ 1, which determines the first form of a loss of stability; considering the second equation of system (a) we will get the same result. Corresponding equilibrium form is presented in Fig. 15.6d. 2. Let N ¼ N2cr ¼ kl. Substituting it in the first equation of system (a), we obtain a1 ð3 kl 2klÞ a2 kl ¼ 0, and the second form of the loss of stability is defined by relationship between generalized coordinates as a1 a2 ¼ + 1. Corresponding equilibrium form is presented in Fig. 15.6e. Note that for each critical load we cannot define the displacements a1 and a2 separately. However, the shape of the loss of stability is defined by their relationships.
15.3
Stability of Columns with Rigid and Elastic Supports
Elastic bar is element with infinite number of degrees of freedom. Structures which contain such elements are called the structures with distributed parameters. Their stability analysis may be effectively performed on the basis of differential equation of the elastic curve and initial parameter method. Both methods are presented below.
15.3.1 The Double Integration Method Stability analysis of the uniform compressed columns is based on the moment-curvature equation of beam EI
d2 y ¼ M ðxÞ, dx2
ð15:4Þ
where x and y are the coordinate of any point of the beam and corresponding lateral displacement; EI is the flexural rigidity of the beam; M (x) is the bending moment at the section x of the beam caused by given loads. This equation allows finding exact value of the critical load for deformable columns with rigid and/or elastic supports. To apply this equation, we need to show a column in deflected state, then it is necessary to construct the expression for bending moment in terms of lateral displacement y of any point of the column, and to write the differential equation (15.4). As a result we get the ordinary differential equation, which could be homogeneous or nonhomogeneous. Then for each specific problem, we need to integrate this equation and find the constant of integration using the boundary conditions. For typical supports they are the following: Pinned support: y ¼ 0 and y00 ¼ 0 (lateral displacement and bending moment are zero) Clamped support: y ¼ 0 and y0 ¼ 0 (lateral displacement and slope are zero) Sliding support: y0 ¼ 0 and y000 ¼ 0 (slope and shear force are zero) Free end: y00 ¼ 0 and y000 ¼ 0 (bending moment and shear force are zero) A detailed procedure for analyzing the stability of uniform columns with various boundary conditions is discussed below.
532
15.3.1.1
15
Stability of Elastic Systems
Uniform Clamped-Free Column
Let the column be subjected to axial compressed force P (Fig. 15.7). Elastic curve of the column is shown by dotted line. x P
P
f
y x
l
0
y Fig. 15.7 Buckling of clamped-free column
If the lateral displacement of the free end is f, then the bending moment is M(x) ¼ P( f y). Corresponding differential equation of the compressed column becomes EI
d2 y ¼ Pð f yÞ or dx2
EI
d2 y þ Py ¼ Pf dx2
ð15:5Þ
This equation may be transformed to the form d2 y þ n2 y ¼ n2 f , dx2
rffiffiffiffiffi P 1 n¼ EI length
ð15:6Þ
Equation (15.6) is nonhomogeneous linear differential equation of order two in one variable x with constant coefficient n2. Therefore, the solution of this equation should be presented in the form y ¼ A cos nx þ B sin nx þ y where A and B are constants of integration. The partial solution y we will find in the form of the right part of (15.6). Since the right part is constant (does not depend on y), then suppose y ¼ C. Substitution of this expression into (15.6) leads to n2C ¼ n2f ! C ¼ f. Therefore, the general solution of (15.6) and corresponding slope are y ¼ A cos nx þ B sin nx þ f y0 ¼ An sin nx þ Bn cos nx To determine unknown integration parameters A and B, let us consider the following boundary conditions: 1. At x ¼ 0 (fixed end) the slope y0 ¼ 0. Expression for slope leads to the B ¼ 0; 2. At x ¼ 0 the displacement y ¼ 0. Expression for y leads to the A ¼ f. Thus, the displacement of the column becomes y ¼ f ð1 cos nxÞ At x ¼ l (free end) the displacement y ¼ f. Therefore f ¼ f(1 cos nl), which holds if cos nl ¼ 0 This equation is called the stability equation for given column; the smallest root equals n l ¼ π/2. The value n ¼ π/2l is called the critical parameter. Thus, the smallest critical load for uniform clamped-free column becomes Pcr ¼ n2cr EI ¼
15.3.1.2
π 2 EI 4l2
Uniform Columns with Elastic Supports
Now let us consider the compressed column with elastic supports at both the ends (Fig. 15.8a). The flexural stiffness of the column is EI ¼ constant; the stiffness coefficients of elastic supports are k1 [kN/m] and k2 [kNm/rad] for linear and angular displacements, respectively.
15.3
Stability of Columns with Rigid and Elastic Supports
533
The cross section at support A can rotate through angle φ while the support B has a linear displacement f. Thus, reactions of elastic supports are moment at the supports A and force at support B. They are equal MA ¼ k2φ and RB ¼ k1f. Deformable state, elastic curve and all reactions are shown in Fig. 15.8b. The bending moment at any section x equals M ðxÞ ¼ Pð f þ yÞ þ k 1 f ðl xÞ Substituting this expression into (15.4) leads to the buckling differential equation of the column
a
b
x
x
P
P
B
RB=k1f
f k1
l y
y
EI x
A
φ
y
RA
k2 MA=k2 φ
P
Fig. 15.8 (a) Design diagram of a column with elastic supports; (b) Deformable shape of the column and reactions of supports
EI
d2 y þ Py ¼ f ½k 1 ðl xÞ P dx2
ð15:7Þ
This is the second-order nonhomogeneous differential equation. In case of k1 ¼ 1 the transversal displacement f ¼ 0, so we get homogeneous differential equation. The partial solution of Eq. (15.7) is y. By direct substitution, it is easy to verify that the partial solution k y¼ f 1 ðl xÞ 1 satisfies equation (15.7). P The general solution of differential equation (15.7) and corresponding slope are P EI dy f k1 ¼ y0 ¼ C 1 n sin nx þ C 2 n cos nx dx P y ¼ C1 cos nx þ C 2 sin nx þ y ,
n2 ¼
Unknown parameters C1, C2, and f may be determined using the following boundary conditions: 1: yð0Þ ¼ 0;
2: y0 ð0Þ ¼ φ;
3: yðlÞ ¼ f
1. The first boundary condition leads to equation k1 l 1 ¼0 C1 þ f P 2. At point A (x ¼ y ¼ 0) the reactive moment equals MA ¼ RBl Pf ¼ k1fl Pf ¼ f(k1l P); thus the angle of rotation at A is φ¼
MA f ¼ ðk1 l PÞ k2 k2
Therefore the second boundary condition y0(0) ¼ φ leads to the following equation C2 n f
k1 k1 l P þ k2 P
3. The third boundary condition leads to the following equation
¼0
534
15
Stability of Elastic Systems
C1 cos nl þ C 2 sin nl ¼ 0 Conditions 1–3 may be rewritten in the form of the homogeneous algebraic equations with respect to unknowns C1, C2, and f. Equation for critical load is presented as determinant from coefficients at these unknowns, i.e., k1 l 1 0 1 P k k l P ¼ 0 D¼ 0 n 1þ 1 P k2 cos nl sin nl 0 or 1 D ¼ 0 1
0 n tan nl
k1 l 1 P k k l P ¼ 0 1þ 1 k2 P 0
This equation may be rewritten as the transcendental equation with respect to parameter n k1 l 1 2 EI n tan nl ¼ nl ðk l n2 EI Þ l k1 l þ 1 2 k2 n EI For given parameters l, EI, k1, and k2 of the structure, solution of this equation leads to parameter n of critical load. The critical load is Pcr ¼ n2EI. Table 15.1 presents the columns with specified supports and corresponding stability equations. The stability equations for cases 4-6 contain dimensionless parameter (α or β); the roots for these cases can be calculated numerically for specified α (or β). Table 15.1 Limiting cases for columns with elastic supports (the length of column l and flexural stiffness EI)
1
2
k1=0
3
k1=∞
k1=∞
Case k2=∞
k2=0
cos nl = 0
tan nl = nl
sin nl = 0
/ 2
4.493
k2=∞ Stability equation Root (nl)min 4
k1=0
5
6
k1=∞
k1
Case k2
Stability equation
tan nl = EI α= k 2l
1 nlα
k2=∞
k2
tan nl =
nl
n 2l 2 α + 1 EI α= k 2l
,
(
tan nl = nl 1 − n 2 l 2 β EI β= 3 k1l
)
15.3
Stability of Columns with Rigid and Elastic Supports
535
Limiting case. For case 4 the stability equation tan nl ¼
1 EI , α¼ nlα k2 l
may be presented as
tan nl k2 ¼ nl Pl
For absolutely rigid body EI ¼ 1, and n ¼
pffiffiffiffiffiffiffiffiffiffiffi P=EI ! 0. Since lim
nl!o
tan nl ¼ 1, nl
then a critical force becomes Pcr ¼ k2/l. This result has been obtained in Sect. 15.2.1.
15.3.2 Initial Parameters Method This method may be effectively applied for stability analysis of the columns with rigid and elastic supports and columns with step-variable cross section, for beams with overhang, intermediate hinges, sliding, and elastic supports. Moreover, this method allows deriving the useful expressions which will be applied for stability analysis of the frames. Let us consider a beam with constant cross section. The beam is subjected to axial compressed force P. Differential equation of the beam is EI
d2 y þ Py ¼ 0, dx2
where y is a lateral displacement. Twice differentiation of this equation leads to fourth-order differential equation rffiffiffiffiffi 2 d4 y d2 y d4 y P 2d y þn ¼ 0, n ¼ EI 4 þ P 2 ¼ 0, or EI dx dx dx4 dx2
ð15:8Þ
Solution of Eq. (15.8) may be presented as yðxÞ ¼ C1 cos nx þ C 2 sin nx þ C 3 x þ C 4 ,
ð15:8aÞ
where Ci are unknown coefficients. It is easy to check that this solution satisfies (15.8). The expressions for slope, bending moment, and shear are φðxÞ ¼ y0 ðxÞ ¼ C 1 n sin nx þ C 2 n cos nx þ C3 M ðxÞ ¼ EIy00 ðxÞ ¼ EI C 1 n2 cos nx þ C2 n2 sin nx QðxÞ ¼ EIy000 ðxÞ ¼ EI C1 n3 sin nx C 2 n3 cos nx At x ¼ 0 the initial kinematical and static parameters become y ð 0Þ ¼ y 0 ¼ C 1 þ C 4 φð0Þ ¼ φ0 ¼ C2 n þ C3 M ð0Þ ¼ M 0 ¼ C1 n2 EI Qð0Þ ¼ Q0 ¼ C 2 n3 EI where y0, φ0, M0, Q0 are lateral displacement, angle of rotation, bending moment, and shear at the origin (Fig. 15.9). From these relations the constants Ci may be expressed in terms of kinematical initial parameters (displacement y0 and slope 000 dy φ0 ¼ y0 ¼ Þ and static initial parameters (bending moment M 0 ¼ EIy000 and shear Q0 ¼ EIy0 ) as follows dx C1 ¼
M0 M ¼ 0, P n2 EI
C2 ¼
Q0 Q ¼ 0, n3 EI nP
C 3 ¼ φ0
Q0 , P
C 4 ¼ y0
M0 P
Substitution of these constants Ci in 15.8a leads to the following expressions in terms of initial parameters
536
15
1 cos nx nx sin nx Q0 P nP n sin nx 1 cos nx 0 y ð xÞ ¼ φ 0 M 0 Q0 P P sin nx M ðxÞ ¼ M 0 cos nx þ Q0 n QðxÞ ¼ M 0 n sin nx þ Q0 cos nx
Stability of Elastic Systems
yð xÞ ¼ y0 þ φ 0 x M 0
ð15:9Þ
These equations present the first form of the initial parameter method for uniform compressed columns. In these equations Q0 is directed perpendicular to the tangent of an elastic line of the beam at the initial point of the rod. It can be seen that, in spite of the external lateral load being absent, the shear Q(x) is variable along the column. In order to eliminate this it is suitable to replace the real shear force Q0 by the “conventional” shear force Q0 which is directed perpendicular to the initial straight line of the rod (Fig. 15.9). Y x
M0
y0
P
φ0 Q0
Q0
Fig. 15.9 Initial parameters of a beam, Q0 and Q0 are real and “conventional” shear forces
According to equilibrium equation ∑Y ¼ 0 (axis Y is directed along Q0) we have Q0 ¼ Q0 cos φ0 þ P sin φ0 ffi Q0 þ Pφ0 Then, substitution of this expression into the first equation (15.9), and after the first and second differentiation we get the following result nx sin nx 1 cos nx Q0 þ Pφ0 ¼ P nP 1 cos nx nx sin nx nx sin nx ¼ y 0 þ φ0 x M 0 Q0 Pφ0 ¼ P nP nP sin x 1 cos nx nx sin nx ¼ y 0 þ φ0 M0 Q0 , n P nP dy n sin nx 1 cos nx φðxÞ ¼ ¼ φ0 cos nx M 0 Q0 , dx P P d2 y n2 cos nx n sin nx Q0 ¼ M ðxÞ ¼ EI 2 ¼ EI φ0 n sin nx M 0 P P dx sin nx ¼ φ0 nEI sin nx þ M 0 cos nx þ Q0 n
y ð x Þ ¼ y 0 þ φ0 x M 0
ð15:10Þ
In second form (15.10) the shear Q(x) along the beam is constant, since an external lateral load is absent. Indeed, it is easy to check that dM ðxÞ dφðxÞ QðxÞ ¼ P ¼ Q0 : dx dx Both forms (15.9) and (15.10) are equivalent. Let us calculate the critical load for uniform clamped-free column (Fig. 15.7); EI ¼ constant. The origin is placed at the clamped support. The geometrical initial parameters are y0 ¼ 0 and φ0 ¼ 0. The third equation of system (15.9) becomes M ðxÞ ¼ M 0 cos nx þ Q0
sin nx n
Since the bending moment for free end of the column (x ¼ l) is zero, then
15.3
Stability of Columns with Rigid and Elastic Supports
537
M l ¼ M 0 cos nl þ Q0
sin nl ¼0 n
It is obvious that Q0 ¼ 0 and M0 6¼ 0; therefore the stability equation becomes cosnl ¼ 0. This result had been obtained using integration of the differential equation (15.4). Now we illustrate the application of the set (15.10) for stability analysis of the stepped fixed-free column shown in Fig. 15.10; the upper and lower portions of the column be l1 and l2, respectively. The bending stiffness for both portions are EI1 and EI2. The column is loaded by two compressed axial forces P1 ¼ P and P2 ¼ βP, where β is any positive number. It means that growth of all loads up to critical condition of a structure occurs in such way that relationships between all loads remain constant (simple loading). Let l2 ¼ αl (in this case l1 ¼ (1 α)l). Let the origin 01 be located at the free end (Fig. 15.10). Initial parameters (at free end) for upper portion are φ0 ¼ y00 6¼ 0, M 0 ¼ 0; Q0 ¼ 0. The slope and bending moment at the end of the first portion (at the x ¼ l1) according to (15.10) are φ1 ðx ¼ l1 Þ ¼ φ0 cos n1 l1 M 1 ðx ¼ l1 Þ ¼ φ0 n1 EI 1 sin n1 l1 rffiffiffiffiffiffiffi P1 n1 ¼ EI 1
ðaÞ
For the second portion of the column the origin 02 is placed at the point where force P2 is applied. Initial parameters for this portion coincide with corresponding parameters at the end of the first portion (at x ¼ l1); they are φ1 ¼ y01 6¼ 0, M 1 6¼ 0; Q1 ¼ 0. The slope at the end of the second portion (at the x ¼ l2) according to second equation (15.10) can be presented as rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n sin n2 l2 P1 þ P2 ðbÞ , n2 ¼ φ2 ðx ¼ l2 Þ ¼ φ1 cos n2 l2 M 1 2 P1 þ P2 EI 2 In this equations φ1 and M1 are initial parameters for portion 2. Substituting (a) into (b) yields n2 sin n2 l2 φ2 ðx ¼ l2 Þ ¼ φ0 cos n1 l1 cos n2 l2 n1 EI 1 sin n1 l1 P1 þ P2 P1=P
P 01
l1
EI1 P2= βP
l
Initial parameters for first portion
P2 φ0 02
φ0 ≠ 0, M 0 = 0, Q0 = 0
Initial parameters for second portion
φ1 ≠ 0, M 1 ≠ 0, Q = 0 1
l2 EI2 x Fig. 15.10 Stepped clamped-free column, l2 ¼ αl, l1 ¼ (1 α)l
For a clamped support the slope φ2(x ¼ l2) ¼ 0. Since φ0 6¼0, then stability equation becomes cos n1 l1 cos n2 l2 n1 EI 1 sin n1 l1
n2 sin n2 l2 ¼0 P1 þ P2
After a simple rearrangement, this equation may be presented as tan n1 l1 tan n2 l2
n1 ð1 þ β Þ ¼ 0 n2
538
15
Stability of Elastic Systems
This equation may be presented in other form . Let the total length of column l1 + l2 ¼ l and l2 ¼ α l, where α is any positive number. In this case l1 ¼ (1 α) l and stability equations becomes tan ½n1 ð1 αÞ l tan n2 α l
15.3.2.1
n1 ð1 þ β Þ ¼ 0 n2
ðcÞ
Limiting Cases
1. Let α ¼ 0. This case corresponds to uniform column of length l ¼ l1, stiffness EI1 and loaded by one force P1 at free end. In n 1þβ this case the stability equation may be presented in form tan n1 l ¼ 1 ¼ 1, so tann1l ¼ 1, the root of equation is n2 tan n2 αl (n1l)min ¼ π/2 (Table 15.1, case 1), and the required critical load Pcr1 ¼
π 2 EI 1 4l2
2. Let α ¼ 1. This case corresponds to the uniform column of length l ¼ l2, stiffness EI2 and loaded by two forces P1 ¼ P and P2 ¼ βP at the free end. Stability equation becomes tann2l ¼ 1, so the complex of critical load Pcr2 ¼ ðP þ βPÞcr2 ¼
π 2 EI 2 4l2
Transcendental equation (c) contains two unknown parameters of critical load; they are n1 and n2. In general case, this equation should be solved numerically. For this purpose (c) should be presented using one unknown critical parameter λ ¼ n1l. In terms of basic parameters α ¼ l2/l, β ¼ P2/P1, k ¼ EI2/EI1, equation (c) becomes rffiffiffiffiffiffiffiffiffiffiffi ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þβ λ ¼ k ð1 þ β Þ tan ½ð1 αÞλ tan α k pffiffiffiffiffiffiffiffiffiffiffiffiffiffi The smallest root of this transcendental equation is λ ¼ n1 l ¼ l P1 =EI 1 which allows to calculate the corresponding critical force Pcr1 ¼ λ2EI1/l2. Let α ¼ 0.5, β ¼ 3, k ¼ 2. In this case the stability equation becomes tan(0.5λ) tan (0.705λ) ¼ 2.82. Smallest root of this equation and corresponding critical load are λ ¼ 1:69,
P1cr ¼
λ2 EI 1 2:856EI 1 ¼ , P2cr ¼ 3P1cr , l2 l2
Summary Initial parameters method may be effectively applied for stability analysis of the stepped columns subjected to several forces. In this case, the origin should be shifted for each following portion. Initial parameters for each following portion coincide with parameters at the end of the previous portion. Example 15.3 Design diagram of the uniform beam with elastic support at the left end is subjected to compressed force P as shown in Fig. 15.11. The flexural stiffness of the beam is EI. Stiffness coefficient of elastic support is k [kNm/rad]. Derive the stability equation. k P
y(x)
P
φ0 x
M0 R Fig. 15.11 Beam with elastic support
l R
x
15.3
Stability of Columns with Rigid and Elastic Supports
539
Solution Suppose that the section at the left end of the beam rotates clockwise. Corresponding reactions M0 (which arise in elastic support) and R of the beam are shown in Fig. 15.11. The lateral displacement of the beam according to the first equation of the system (15.10) y ð x Þ ¼ φ0
sin nx 1 cos nx nx sin nx M0 Q0 n P nP
ðaÞ
Initial parameters are M0 ¼ kφ0, and shear Q0 ¼ R ¼
kφ0 l
Their signs are accepted according to Fig. 15.9. Thus, (a) may be rewritten as y ð x Þ ¼ φ0
sin nx 1 cos nx kφ0 nx sin nx þ kφ0 n P nP l
ðbÞ
Boundary condition: at x ¼ l y ¼ 0. Therefore sin nl 1 cos nl k nl sin nl y ð l Þ ¼ φ0 þk ¼0 n P l nP Since φ0 6¼ 0, then sin nl 1 cos nl k nl sin nl þk ¼0 n P l nP This expression leads to the following stability equation tan nl ¼ where dimensionless parameter α ¼
15.3.2.2
nl , n2 l 2 α þ 1
ðcÞ
EI . Equation (c) had been derived earlier (Table 15.1, case 5). kl
Discussion
1. It may appear that we used only one boundary condition. However, it is not true. There are two boundary conditions used: the second boundary condition is M(l) ¼ 0 and this condition allows writing expression for R ¼ kφ0 =l. 2. If k ¼ 1 (α ¼ 0), then the stability equation becomes tannl ¼ nl. This case corresponds to clamped-pinned beam (Table 15.1, case 2). If k ¼ 0 (simply-supported beam), then the stability equation becomes tannl ¼ 0, or sinnl ¼ 0 (Table 15.1, case 3).
15.3.3 Euler’s Solution and Paradox of Critical Load Let us consider the stability problem of simply supported uniform column and discuss the Euler’s formula for critical load (Table 15.1, case 3). pffiffiffiffiffiffiffiffiffiffiffi 00 00 The initial equation EIy ¼ Py may be presented in form y + n2y ¼ 0, n ¼ P=EI. Solution of this equation (equation of elastic line) is yðxÞ ¼ C1 cos nx þ C 2 sin nx Boundary conditions are y(0) ¼ y(l) ¼ 0. From the first boundary condition y(0) ¼ 0 we have C1 ¼ 0, so the elastic line of deformable rod becomes yðxÞ ¼ C 2 sin nx From the second boundary condition y(l) ¼ 0 we get yðlÞ ¼ C2 sin nl ¼ 0
540
15
Stability of Elastic Systems
This equation have two possible solutions: C2 ¼ 0 or sinnl ¼ 0. The first solution, C2 ¼ 0, is trivial, because it corresponds to the rectilinear form of rod equilibrium. The second solution, sinnl ¼ 0, presents the stability equation and its solution nl ¼ π leads to the critical force Pcr ¼ π 2EI/l2. This solution contains obvious ambiguities. First of all, the magnitude of the maximum displacement C2 remains unknown and it is unclear how this displacement depends on the force P. Also, it is unclear what happens when the force P occurs slightly greater than the first critical force. The matter is that, the critical force P was found from the condition nl ¼ π, while if nl 6¼ π, then sinnl 6¼ 0. Therefore, from condition y(l) ¼ C2 sin nl ¼ 0 we need to accept C2 ¼ 0. Hence, the general expression y(x) leads to the fact that the rod is straight (!). Thus, we get paradoxical result: if P ¼ Pcr the rod loses stability and assumes a curvilinear shape; however if P > Pcr, the rod again becomes a rectilinear (paradox of critical load) (Feodosiev, 2016, page 416). 00 The reason for this paradox is as follows. The initial differential equation EIy ¼ Py is approximate, which is used in classical linear theory for structures subjected to bending. The precise equation has the form EI
1 y00 ¼ EI h i3=2 ¼ Py ρ 1 þ ð y0 Þ 2
If the force exceeds the critical one, then the displacements grow so rapidly that the value (y0)2 in the denominator cannot be neglected. It is appropriate to quote the apt expression of Academician L.I. Mandelstam “idealization avenges itself” (Panovko & Gubanova, 1973). This example shows the importance of assumptions related to the design diagram. A field of application of the accepted assumptions is always limited and the violation of the boundaries of their application often leads to paradoxical results. It should be remembered that the correct mathematical model will be one that not only describes the qualitative assessment of the phenomenon but also provides the basis for a quantitative assessment of stability (Feodos’ev, 1975, page 118]. Detailed solution of stability problem on the basis of nonlinear equation is presented in Sect. 18.4.
15.4
Stability of Continuous Beams and Frames
This section is devoted to stability analysis of compressed continuous beams and frames. We assume that beams are subjected to axial forces only. Frames are subjected to external compressed loads, which are applied at the joints of a frame. If several different compressed loads Pi act on a structure, we assume that all loads can be presented in terms of one load P. It means that growth of all loads up to critical condition of a structure occurs in such a way that relationships between all loads remain constant, i.e., the loading is simple. Both the classical methods can be applied for stability analysis of continuous beams and frames. However, for such structures, the displacement method often occurs more effective than the force method. The primary system of the displacement method must be constructed as usual, i.e., by introducing additional constraints, which prevent angular and linear displacements of the joints. The primary unknowns are linear and angular displacements of the joints. However, calculation of the unit reactions has specific features.
15.4.1 Unit Reactions of the Beam-Columns A primary system of the displacement method contains the one-span standard (pinned-clamped, clamped-clamped, etc.) members. These members are subjected not only to the settlements (angular or linear) of the introduced constraints but also to the axial compressed load P, i.e., we have case of the compressed-bending rod. For stability analysis this is a fundamental feature. As earlier, we need to have the reactions of standard elements, which are subjected to unit settlements of constraints and axial compressed load. Calculation of reactions for typical member is shown below.
15.4
Stability of Continuous Beams and Frames
541
The pinned-clamped beam is subjected to unit angular displacement of support B and axial compressed force P. The length of the beam is l, EI ¼ const. Figure 15.12 presents the elastic curve (EC) and positive unknown reactions R and M. For their determination we can use the initial parameters method. 1 φ0 EC
A
P
M
B
P
l
Q0 = RA
Fig. 15.12 Reaction of compressed beam subjected to unit angular displacement at support B
The origin is placed at support A. Since y0 ¼ 0 and M0 ¼ 0, then (15.10) becomes sin nx nx sin nx Q0 , n nP 1 cos nx y0 ðxÞ ¼ φ0 cos nx Q0 P
yðxÞ ¼ φ0
For calculation of two unknown initial parameters φ0 y0(l) ¼ 1, therefore
and
Q0 ¼ R we have two boundary conditions: y(l ) ¼ 0 and
sin nl nl sin nl Q0 ¼0 n nP 1 cos nl y0 ðlÞ ¼ φ0 cos nl Q0 ¼1 P
yðlÞ ¼ φ0
Solution of these equations is Q0 ¼
3i υ2 tan υ , l 3ð tan υ υÞ
where i ¼ EI/l; the dimensionless parameter of critical force is
rffiffiffiffiffi P υ ¼ nl ¼ l EI
Since Q0 is negative, then reaction RA is directed downward. Reactive moment at support B is M ¼ RA l ¼ 3i
υ2 tan υ 3ð tan υ υÞ
The negative sign means that the reactive moment acts clockwise; indeed direction of moment coincides with direction of angular displacement of clamped support.
Some Features of the Solution 1. It can be seen the reactions RA and M of the compressed-bending rods depend of parameter υ, which take into account the compressive force. In case of υ ¼ 0 (i. e., P ¼ 0) we obtain the well-known formulas of classical displacement method (see Tables A.22 and A.23). Indeed, it is easy to show that lim
υ!0
υ2 tan υ ¼1 3ð tan υ υÞ
542
15
Stability of Elastic Systems
2. Note the reactions are nonlinearly dependent on the compressive force P. Additional nonlinear problems of this type (compressed rods with lateral load) is considered in Sect. 19.2. 3. Bending moment at any section x is M(x) ¼ RAx Py(x), so the bending moment diagram is curvilinear. The expressions for unit reactions for standard members subjected to compressive force P and special types of displacements can be derived similarly; for typical uniform beams they are presented in Table A.20. In all cases, the length of the beam is l, the flexural stiffness is EI, bending stiffness per unit length is i ¼ EI/l. Corresponding elastic curve is shown by dotted line; the graphs present the real direction of reactions; the bending moment diagrams are plotted on the tensile fibers. Row 3 of Table A.20 presents the bending moment diagrams when a clamped support rotates and switches in transversal direction simultaneously. While the angle of rotation is fixed as Z ¼ 1, the displacement Δ does not require indication of its value. These cases may be used for analysis of compressed beams with elastic supports. Thus, in case of frames with sidesway, the primary system is obtained by introducing constraints, which prevents only angular displacements, and bending moment diagrams should be traced as for member with elastic supports. This approach is discussed in Sect. 15.4.3. It is recommended to show elastic curves and remember that bending moment has one-sign ordinates. Expressions of special functions in two forms are presented in Table A.22; the more preferable is form 2. Also this table contains approximate presentation of these special functions in the form of Maclaurin series. Numerical values of these functions in terms of dimensionless parameter υ are presented in Table A.23.
15.4.2 Displacement Method Canonical equations of the displacement method for stability analysis of structures with n unknowns Zj, ( j ¼ 1,2,. . .,n) are r 11 Z 1 þ r 12 Z 2 þ þ r 1n Z n ¼ 0 r 21 Z 1 þ r 22 Z 2 þ þ r 2n Z n ¼ 0 r n1 Z 1 þ r n2 Z 2 þ þ r nn Z n ¼ 0
ð15:11Þ
Features of the set of equations (15.11): 1. Since the forces Pi are applied only at the joints, then the canonical equations are homogeneous ones. 2. Bending moment diagram caused by unit displacements of introduced constrains within the compressed members are curvilinear. Reactions of constraints depend on axial forces in the members of the frame, i.e., contain parameter υ of critical load. If a frame is subjected to different forces Pi, then critical parameters should be formulated for each compressed member υi2 ¼ Pili2/(EI)i and after that all of these parameters should be expressed in terms of parameter υ for specified basic member. Thus, the unit reactions are functions of parameter υ, i.e., rik(υ); expressions for reactions contain specific functions φ(υ) and η(υ) depending on the boundary condition of the element and the type of displacement (linear, angular). The trivial solution (Zi ¼ 0) of (15.11) corresponds to initial nondeformable design diagram. Nontrivial solution (Zi 6¼ 0) corresponds to the new form of equilibrium. This occurs if the determinant, which is consisting of coefficients of unknowns, equals zero, i.e., r 11 ðυÞ r 12 ðυÞ r 1n ðυÞ r ðυÞ r ðυÞ r ðυÞ 22 2n 21 ð15:12Þ D¼ ¼0 r ðυÞ r ðυÞ r ðυÞ n1
n2
nn
Expression (15.12) is called the stability equation of a structure in the form of displacement method. For practical engineering, it is necessary to calculate the minimum root of the above equation. This root defines the smallest parameter υ of critical force or smallest critical force. It is obvious that condition (15.12) leads to transcendental equation with respect to parameter υ. Since the functions φ(υ) and η(υ) are tabulated (Tables A.22 and A.23), then a solution of stability equation may be obtained by graphical method keeping in mind that determinant is very sensitive with respect to parameter υ. The displacement method is effective for stability analysis of stepped continuous beams on rigid supports with several axial compressed forces along the beam and for frames with/without sidesway.
15.4
Stability of Continuous Beams and Frames
543
Let us derive the stability equation and determine the critical load for frame shown in Fig. 15.13a. This frame has one unknown of the displacement method. The primary unknown is the angle of rotation of rigid joint. Figure 15.13b shows the primary system, elastic curve, and bending moment diagram caused by unit rotation of introduced constraint. The bending moments diagram for compressed vertical member of the frame is curvilinear. The ordinate for this member is taken from Table A.20, row 1.
a
b
P
r11 Z=1
EI2
4i1φ 2 (υ1 )
EI1
l1
4i2
Elastic curve
l2 Fig. 15.13 Stability analysis of redundant frame. (a) Design diagram; (b) Primary system of the displacement method and unit bending moment diagram
The bending moment diagram yields r 11 ¼ 4i1 φ2 ðυ1 Þ þ 4i2 , where parameter of critical load rffiffiffiffiffiffiffi P υ1 ¼ l1 EI 1 Note, that subscript 2 at function φ , i.e., φ2, is related to the clamped-clamped member subjected to angular displacement of the one support (Table A.20, row 1), while the subscript 1 at the parameter υ is related to the compressed-bent member 1. Canonical equation of the displacement method is r 11 ðυ1 ÞZ ¼ 0: Nontrivial solution of this equation leads to equation of stability r11 ¼ 0 or in expanded form r 11 ðυ1 Þ ¼
4EI 1 4EI 2 φ ðυ Þ þ ¼0 l1 2 1 l2
Special cases 1. Assume that l2 ! 0. In this case, the bending stiffness per unit length for crossbar increases, the second term 4EI 2 =l2 ! 1, rigid joint is transformed to clamped support, and the initial frame in whole is transformed into the vertical clampedclamped column. Stability equation becomes φ2(υ1) ¼ 1. Root of this equation (Table A.23) is υ1 ¼ 2π and critical force becomes Pcr ¼
υ1 2 EI 4π 2 EI π 2 EI ¼ 2 ¼ , 2 l1 l1 ð0:5l1 Þ2
where μ ¼ 0.5 is effective-length factor for clamped-clamped column. 2. Assume EI2 ! 0. In this case, the rigid joint is transformed to hinge and the initial frame is transformed into the vertical clamped-pinned column. Stability equation becomes φ2(υ1) ¼ 0. Root of this equation is υ1 ¼ 4.488 (Table A.23) and critical force
544
15
Pcr ¼
υ1 2 EI 4:4882 EI π 2 EI ¼ ¼ , 2 2 l1 l1 ð0:7l1 Þ2
Stability of Elastic Systems
μ ¼ 0:7
3. If l1 ¼ l2, EI1 ¼ EI2, then stability equation becomes φ2(υ1) + 1 ¼ 0. The root of this equation is υ1 ¼ 5.3269 and critical load equals Pcr ¼
υ1 2 EI 28:397EI ¼ l21 l21
It is seen, the softer the structure, the less critical force. Now let us consider a nonuniform two-span continuous beam shown in Fig. 15.14a. We need to derive a stability equation and determine the critical load.
a
b EI1
EI2 l2
l1
3EI 1 φ υ 1( 1 ) l1
P
r11 Z=1 Elastic curve
3EI 2 φ υ 1( 2 ) l2
Fig. 15.14 Continuous compressed beam. (a) Design diagram; (b) Primary system and unit bending moment diagram
The axial compressed forces in both spans are equal, so the dimensionless parameters υ1 and υ2 for both spans are rffiffiffiffiffiffiffi rffiffiffiffiffiffiffi P P ; υ2 ¼ l2 υ1 ¼ l1 EI 1 EI 2 Let the left span be considered as the basic member, so υ1 ¼ υ. In this case, the parameter υ2 can be presented in terms of the basic parameter υ as follows rffiffiffiffiffiffiffi rffiffiffiffiffiffiffi l2 EI 1 l2 EI 1 ¼ υα, α ¼ , υ1 ¼ υ υ2 ¼ υ1 l1 EI 2 l1 EI 2 The primary unknown of the displacement method is the angle of rotation at the intermediate support. The primary system and bending moment diagram due to unit rotation of the introduced constrain are presented in Fig. 15.14b. Since both spans are compressed, then the bending moment diagrams are curvilinear. According to Table A.20, the moment for clampedpinned beam in case of angular displacement has the multiple φ1(υ). Unit reaction, which arises in introduced constraint, is r 11 ðυÞ ¼
3EI 1 3EI 2 3EI 1 3EI 2 φ ðυ Þ þ φ ðυ Þ ¼ φ ðυ Þ þ φ ðαυÞ l1 1 1 l2 1 2 l1 1 l2 1
As before, the subscript 1 at the function φ1(υ) reflects the type of the beam and type of displacement (Table A.20), while subscripts 1 and 2 at the parameter υ denote the number of the element (span). Canonical equation of the displacement method is r11(υ)Z ¼ 0. Nontrivial solution of this equation leads to equation of a critical force r 11 ðυÞ ¼
3EI 1 3EI 2 φ ðυÞ þ φ ðαυÞ ¼ 0 l1 1 l2 1
Special Cases 1. Assume l2 ! 0. In this case, the bending stiffness per unit length for second span increases, intermediate pinned support is transformed into clamped support, and initial beam becomes one-span pinned-clamped beam with length l1. Stability equation becomes φ1(υ1) ¼ 1. Root of this equation is υ1 ¼ 4.488 and critical force
15.4
Stability of Continuous Beams and Frames
545
Pcr ¼
υ1 2 EI 4:4882 EI π 2 EI ¼ ¼ , 2 2 l1 l1 ð0:7l1 Þ2
μ ¼ 0:7
2. Assume l1 ¼ l2 ¼ l and EI1 ¼ EI2 ¼ EI. In this case parameter α ¼ 1, the stability equation becomes φ1(υ) ¼ 0, and parameter of critical load υ ¼ π. So the critical load Pcr ¼ π 2 EI=l2 . This critical load corresponds to column with pinnedrolled supports. rffiffiffiffiffiffiffi l EI 1 3. Assume l2 ¼ 2l1 and EI1 ¼ EI2 ¼ EI. In this case parameter α ¼ 2 ¼ 2 and the stability equation becomes l1 EI 2 3:869EI φ1(υ) + 0.5φ1(2υ) ¼ 0; parameter of critical load υ ¼ 1.967. So the critical load Pcr ¼ . It can be seen that l21 increasing of the one span by two times has profound effect on the critical load (coefficient 3.869 instead of π2 for two-span beam with equal spans). Below the displacement method in canonical form is applied for stability analysis of the multistory frame and frames with sidesway. Example 15.4 The two-story frame in Fig. 15.15a is subjected to compressed axial forces P and αP. Geometrical parameters of the frame are h and l ¼ β h, the bending stiffness of the members are EI for members of the second level and kEI for column of the first level. Derive the stability equation in terms of arbitrary positive numbers α, β, and k.
a
c
b
P EI
i/β
1 h
αP EI EI
i
kEI
2iφ 3 (υ)
Z1=1 r11
4iφ 2 (υ )
4i β
r21
2iβ 3 (υ )
r12
ki A
Z2=1 4iφ 2 (υ )
r22
i/β
2 h
d
4i β
4kiφ2 (υ2 )
M2
M1
l=βh Fig. 15.15 Stability analysis of two-story redundant frame. (a, b) Design diagram and primary system; (c, d) Unit bending moment diagrams
Solution The primary system is presented in Fig. 15.15b. Let us assign member 1-2 as basic one; its flexural stiffness per unit length equals i ¼ EI/h. The flexural stiffness per unit length for each member are shown in Fig. 15.15b. Parameters of critical force for member 1-2 and 2-A are determined as follows: rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P P þ αP 1þα υ1 ¼ h ¼ υ, υ2 ¼ h ¼υ EI kEI k Bending moment diagrams caused by Z1 ¼ 1 and Z2 ¼ 1 are shown in Fig. 15.15c,d, respectively. Unit reactions are r 11 ¼ 4iφ2 ðυÞ þ
4i ; β
r 12 ¼ r 21 ¼ 2iφ3 ðυÞ;
r 22 ¼ 4iφ2 ðυÞ þ 4kiφ2 ðυ2 Þ þ
Stability equation in general and expanded forms are r 11 r 12 ¼0 r r 22 21 1 1 4 φ2 ðυÞ þ φ2 ðυÞ þ kφ2 ðυ2 Þ þ φ23 ðυÞ ¼ 0 β β Let α ¼ 3,
β ¼ 1,
k ¼ 4. In this case υ2 ¼ υ and stability equation becomes
4i β
546
15
Stability of Elastic Systems
4½φ2 ðυÞ þ 1 ½5φ2 ðυÞ þ 1 φ23 ðυÞ ¼ 0 The root of this equation is υ ¼ 4.5307. The critical load is Pcr ¼
υ2 EI 4:53072 EI ¼ h2 h2
Example 15.5 The frame in Fig. 15.16a is loaded by two forces at the joints. Derive the stability equation and find the critical load P. P
a
1.4P
b 1 i=0.4EI
2EI i=0.1EI
h=10m
EI
EI
2
υ1-3= υ
υ 2-4 =1.1832 υ 3
4
l=5m
c
Z2=1
Z1=1
r11
r21
R
R
21
R
M13
Z2=1
r12
d
r22
R1 M13
M13
M12
R2
M2
M1 R M31
M31
M31
M42
M 13 = 6i φ4 (υ ) = 0.6 EIφ 4 ( υ)
M13 = 4iφ 2 ( υ ) = 0.4 EI φ 2 (υ )
12i η2 ( υ) = 0.012 EI η2 (υ ) l2 3i R2 = 2 η1 (1.1832 υ ) = 0.003 EI η1 (1.1832υ ) l R1 =
R = 6i φ4 ( υ ) = 0.6 EI φ4 ( υ ) M12 = 3i = 1.2 EI
Fig. 15.16 Stability analysis of portal frame with sidesway. (a, b) Design diagram of the frame and primary system; (c, d) Unit bending moment diagrams
Solution The frame has two unknowns of the displacement method. They are the angle of rotation Z1 of rigid joint and horizontal displacement Z2 of the crossbar. The primary system is presented in Fig. 15.16b. Bending moment diagrams caused by unit displacement of the introduced constraints are presented in Fig. 15.16c,d; elastic curves are shown by dotted line. The diagrams within members 1-3 and 2-4 are curvilinear. Direction of R for M 1 diagram is explained for the left column 1-3 (Fig. 15.16c). Similarly may be explained directions for R1 and R2 (on M 2 diagram). Notation M13 means the bending moment at point 1 which belongs to member 1-3. Parameters of a critical load are υ13
rffiffiffiffiffi P ; ¼υ¼h EI
Canonical equations and unit reactions are
υ24
rffiffiffiffiffiffiffiffiffiffi 1:4P ¼ 1:1832υ ¼h EI
15.4
Stability of Continuous Beams and Frames
547
r 11 Z 1 þ r 12 Z 2 ¼ 0 r 21 Z 1 þ r 22 Z 2 ¼ 0,
ðaÞ
where r 11 ¼ ½0:4φ2 ðυÞ þ 1:2EI; r 21 ¼ r 12 ¼ 0:6EIφ4 ðυÞ; r 22 ¼ ½0:012η2 ðυÞ þ 0:003η1 ð1:1832υÞEI Stability equation becomes: 0:4φ2 ðυÞ þ 1:2 0:6φ4 ðυÞ ¼0 0:6φ ðυÞ 0:012η2 ðυÞ þ 0:003η1 ð1:1832υÞ 4 Solution of this equation leads to parameter of critical load υ ¼ 2.12. The critical load is Pcr ¼
υ2 EI 4:49EI ¼ h2 h2
Thus, the frame becomes unstable if it will be loaded by two forces Pcr and 1.4Pcr simultaneously. Example 15.6 Design diagram of the multispan frame is presented in Fig. 15.17a. Derive the stability equation and calculate the critical force. P
a
h
P
b
Z1=1
EI
3i h
3i 3i φ1 (υ ) φ1 (υ ) h h
3i h
3i h
r11
c Q1
d
r11
P
P
P
P
P
Q2
e
P
Q3
Q4
Q5
P
P
P
P
EI=∞
h
EI
EI
h l
l
Fig. 15.17 Stability analysis of a multispan frame with sidesway. (a–c) Design diagram, primary system and free-body diagram of a cross-bar; (d, e) Multispan regular frames; a number of columns is k
Solution The primary system is shown in Fig. 15.17b. The introduced constraint 1 prevents linear displacement. Canonical equation is r11Z1 ¼ 0, so stability equation is r11 ¼ 0. Bending moment diagram caused by unit displacement of the introduced support is presented in Fig. 15.17b. Within the second and third columns the bending moment diagrams are curvilinear. Free body diagram of a crossbar is shown in Fig. 15.17c. Shear forces for compressed-bent members are presented in rffiffiffiffiffi P Table A.20. Parameter of critical load υ ¼ h . Shear forces for each vertical members are EI Q1 ¼ Q4 ¼ Q5 ¼ Q2 ¼ Q3 ¼
3i ; h2
3i η1 ðυÞ h2
548
15
Stability of Elastic Systems
Unit reaction equals r 11 ¼ 3
3i 3i þ 2 2 η 1 ðυ Þ 2 h h
ðaÞ
The stability equation becomes 3 + 2η1(υ) ¼ 0. The minimum positive root of this equation, i.e., the parameter of lowest critical load, is υ ¼ 2.4521. Critical load Pcr ¼
υ2 EI 6:0128EI ¼ h2 h2
Discussion. Analysis of this frame allows easy considering of two important typical cases. Both cases are related to the regular multispan sidesway frames with k columns loaded by equal forces P at each joint. Figure 15.17d presents regular frame with hinged joints, while the Fig. 15.17e shows the frame with absolutely rigid crossbar and fixed joints. In both cases, the frame has one unknown of the displacement method. Introduced constraint prevents linear displacement; this constraint is not shown. For case of frame with hinged joints (Fig. 15.17d), the horizontal reaction of introduced constraint (Table A.20, pinnedclamped beam, row 2) is r 11 ¼ k
3i η1 ðυÞ h2
Stability equation becomes η1(υ) ¼ 0 and the lowest root is υ ¼ π/2 (Table A.23). The first (smallest) critical load is equal to Pcr ¼
υ2 EI π 2 EI ¼ h2 ð2hÞ2
This case corresponds to single fixed-free column (Table 15.1, case 1); the effective-length factor is 2 (Craig Jr., 2000). The form of loss of stability is shown by dotted line. It means that when a compressed force P increases up to first critical value, loss of stability is characterized by horizontal displacement of the crossbar. For case of frame with absolutely rigid crossbar and fixed joints (Fig. 15.17e), the unit horizontal reaction of introduced constraint (Table A.20, clamped-clamped beam, row 2) is r 11 ¼ k
12i η2 ðυÞ h2
Stability equation becomes η2(υ) ¼ 0 and the lowest root is υ ¼ π (Table A23). The critical load is equal to Pcr ¼
π 2 EI h2
This case corresponds to single column with fixed ends while the one fixed support permit the horizontal displacement. Conclusion. For both cases of regular multispan frames, the loss of stability is characterized by the horizontal displacement of the crossbar, and the critical load P does not depend on the number of columns k and length l of each span. Let us return to either of structures, shown in Fig. 15.17d,e. It is important that in any case the calculated critical force allows determining the effective-length factor. Now let us consider the frame in Fig. 15.17d, but in the absence of one of any force. In this case, the bending moment diagram along the unloaded column caused by unit displacement of introduced constraint 1 turns out to be linear, i.e., different from the other columns, and the unit reactions becomes r 11 ¼ 3
3i 3i η 1 ðυ Þ þ 2 h2 h
Of course, the root of the equation r11(υ) ¼ 0 will differ from υ ¼ π/2 received earlier. Naturally, changing the loading scheme leads to a new critical force, and as a result the effective-length factor changes. Therefore, any change in loading scheme (eliminating or introducing of one or some compressed forces), changing the relationship between the forces, etc., leads to a new equation of stability. Thus, we obtain different values of the critical parameter υ.
15.4
Stability of Continuous Beams and Frames
549
Consequently, the effective-length factor (Craig Jr., 2000) of any rod of a frame depends not only on the geometric dimensions, the ratio of the stiffness of separate members, the ways of connecting the elements and the types of support, but also on the loading scheme of the frame and relationships between separate loads (!) (Smirnov, 1947, p.198).
15.4.3 Modified Approach of the Displacement Method In general case, the displacement method requires introducing constraints, which prevent angular displacement of rigid joints and independent linear displacements of joints. However, in stability problems of a frame with sidesway it is possible some modification of the classical displacement method. Using modified approach, we can introduce a new type of constraint, mainly the constraint, which prevents angular displacement, but simultaneously has a linear displacement. This type of constraint is presented for pinned-clamped and clamped-clamped members in Table A.20, row 3. Example below presents analysis of the frame with sidesway (Fig. 15.18a) using two approaches. The first approach corresponds to classical primary system of the displacement method; the primary system contains two introduced constraints, one of which prevents angular and the other prevents linear displacements (Fig. 15.18c,d). The second approach corresponds to modified primary system of the displacement method; the primary system contains one introduced constraint, which prevents angular displacement and allows linear displacement Δ (Fig. 15.18e–g). We will derive the stability equation using both approaches and calculate the critical load.
a
b
P
Primary system 1
kEI
1
EI
2
h
l
c
d Z1=1
r11
3EI φ1 ( υ ) h
3kEI l
r21
Z2=1 r22
r12
Elastic curve
3EI φ1 ( υ) h2
Elastic curve
M2
M1
r22
r21
3EI φ υ 1( ) h2
e
3EI η υ 1( ) h3
f
Z1=1
Primary system 2
1
Δ
g
r11
1 Elastic curve
3kEI l
EI υ tan υ h
M1 Fig. 15.18 Stability analysis of redundant frame. (a) Design diagram of the frame; (b–d) First approach – classical primary system 1 and corresponding unit bending moment diagrams; (e–g) Second approach—modified primary system 2, elastic curve and corresponding unit bending moment diagram
550
15
Stability of Elastic Systems
First approach. The primary unknowns are angle of rotation of rigid joint and linear displacement of a crossbar. Figure 15.18b–d shows the primary system (b), elastic curve and bending moment diagrams caused by the unit rotation of introduced constraint 1 (c) and unit linear displacement of introduced constraint 2 (d). Bending moment diagrams within compressed vertical member of the frame are curvilinear. The ordinates are found in accordance with Table A.20. The bending moment diagrams yield 3EI 3kEI 3EI φ ðυÞ þ ; r 12 ¼ 2 φ1 ðυÞ h 1 l h 3EI 3EI ¼ 2 φ1 ðυÞ; r 22 ¼ 3 η1 ðυÞ h h
r 11 ¼ r 21
pffiffiffiffiffiffiffiffiffiffiffi where parameter of stability υ ¼ h P=EI . Again, the subscript 1 at functions φ and η is concerning to the pinned-clamped member subjected to angular and linear displacements of the clamped support, respectively. Nontrivial solution of canonical equations of the displacements method leads to the following stability equation r 11 r 12 ¼ 0 or r 11 ðυÞr 22 ðυÞ r 2 ðυÞ ¼ 0 12 r r 21 22 The stability equation in the expanded form 2 3EI 3kEI 3EI 3EI η1 ðυÞ φ1 ð υ Þ ¼ 0 φ ðυÞ þ h 1 l h3 h2 If we assume that l ¼ h and according to Table A.22 we will take into account the expressions for functions φ1(υ) and η1(υ) ¼ φ1(υ) υ2/3, then stability equation becomes
ðφ1 ðυÞ þ kÞ φ1 ðυÞ υ2 =3 φ21 ðυÞ ¼ 0 After some cumbersome rearrangements we get υ3 ð3k υ tan υÞ ¼ 0 The smallest positive root of this equation is υ ¼ 0; this root corresponds to the initial state of the frame and means the trivial solution. The nontrivial solution occurs if 3k υ tan υ ¼ 0 This transcendental stability equation allows calculating the critical parameter υ for any value of k. Some results are presented in Table 15.2. Table 15.2 Critical load in terms of parameter k; Pcr ¼ υ2
Parameter k 1 10
Root of equation 1.193 1.521 1.57
EI h2
Critical load Pcr (factor EI/h2) 1.423 2.313 2.465
Second approach. In this case only constraint 1, which prevents angular displacement, is introduced (Fig. 15.18e). However, this constrain allows the linear displacement Δ. This case is presented in Table A.20, row 3. Elastic curve caused by unit angular displacements (if linear displacement Δ occurs) and corresponding bending moment diagram are shown in Fig.15.18f,g. Unit reaction r 11 ¼ so the stability equation becomes
EI 3kEI υ tan υ þ , h l
15.5
Stability of Arches
551
h υ tan υ þ 3k ¼ 0 l If l ¼ h, then this stability equation is the same as was obtained above. The second approach is significantly more effective than the first one.
15.5
Stability of Arches
This section is devoted to stability analysis of the plane uniform arches with different boundary conditions. Two types of arches are considered. They are circular arch subjected to uniform distributed radial load (hydrostatic load) and parabolic arch subjected to uniformly distributed gravity load. Precise analytical solutions are presented.
15.5.1 Introduction Stability analysis of the different types of arches is based on a solution of a differential equation. Precise analytical solution may be obtained only for specific arches and their loading. Design diagram of any arch include the following data: type of the arch, equation of axial line of arch, and the type of loading. Among the classical types of arches we note the hingeless arch with elastic-fixed supports and its special cases (twohinged and hingeless arches), as well as three-hinged arch. Among the various equations of the centerline of the arch we note circular and parabolic arches. The main types of loads for in-plane stability analysis of the arch are presented in Fig. 15.19. Among them are: (а) Tracking load. For this load, an angle δ between the load and deformable axis of an arch remains constant (Fig. 15.19a); (b) Hydrostatic load, which is directed perpendicular to the deformable axis of an arch. This load is a special case of the tracking load at δ ¼ π/2 (Fig. 15.19b); (c) Polar (radial for circular arch) load directed to a fixed center (Fig. 15.19c); (d) Gravity load. Direction of this load does not depend on the deflections of the arch (Fig. 15.19d).
a
δ
δ
b
Tracking load
c
Tangent π/2
P π/2 P
Hydrostatic load
d Polar load
Gravity load
Fig. 15.19 Types of loads on the arches
15.5.1.1
Forms of the Loss of Stability of the Arches
Stability analysis of the arches includes the deriving of the stability equations, determining of the smallest critical load and corresponding form of the loss of the stability. Loss of stability of typical symmetric arches can occur in a symmetric and antisymmetric forms (Fig. 15.20). As it is shown by analytical analysis and experiments, the smallest critical load for two-hinged and hingeless arches corresponds to antisymmetrical buckling form (Fig. 15.20b), while for three-hinged arches corresponds to symmetrical buckling form (Fig.15.20a).
552
15
Hingeless arch C
Two-hinged arch C
a φA≠0 A
δC≠0
φA=0 A
B
b
Three-hinged arch C
φC=0 δC≠0
φC=0 δC≠0
φA≠0
B
A
B
φC≠0 δC=0
φC≠0 δC=0
φA≠0
Stability of Elastic Systems
φC≠0 δC=0 φA≠0
φA=0
Fig. 15.20 Arches with different boundary conditions. (a, b) Symmetrical and antisymmetrical forms of the loss of stability
Assumptions for arches which will be considered in this textbook 1. 2. 3. 4.
Arches are symmetrical. The arch material is linearly elastic (Hooke’s law applies). The axial deformation is neglected. The critical stresses are less than the yield stress.
Method of analysis. Generally, solution to the issue of stability of arches of various types is based on the integration of differential equations of the arch. Type of equation takes into account different peculiarities of the arch. Among them are constant or variable stiffness, specified equation of the axial line, type of load, etc.
15.5.2 Circular Arches under Hydrostatic Load In this section we will consider a plane uniform symmetrical arch with constant radius R of curvature (circular arch), which is subjected to uniformly distributed pressure normal to the axis of the arch (hydrostatic load). Assume the arch has the elasticfixed supports; their rotational stiffness coefficient is k [kNm/rad]. The central angle of the arch is 2α; the flexural rigidity is EI and the intensity of the radial uniformly distributed load is q (Fig. 15.21a).
a
c
qRdθ
q dθ H 2α
k
R
b φ
θ
k N=q R
V
w(θ) ds
d dθ
N
α φ N
M0=k φ
sin θ M0 sin α
m
AS
w(θ) θ
EC
A
A M0
N=q R M0=k φ
m=
m
φ θ
N=qR
θ w(θ)
N=q R M0=k φ M0
Fig. 15.21 Stability analysis of the circular arch with elastic supports. (a) Design diagram; (b) Antisymmetrical buckling form, reactions and distribution of bending moments caused by two reactive moments M0; (c) Elementary load qds ¼ qRdθ and computation of total reactions H and V; (d) Free-body diagrams for left and right portions of the arch (load q is not shown) serve for computation of bending moment at point A of deformable axis; AS axis of symmetry, EC elastic curve
15.5
Stability of Arches
553
For stability analysis of the arch we will use the following differential equation d2 w w M þ ¼ , EI ds2 R2
ð15:13Þ
where w is a displacement point of the arch in radial direction (Fig. 15.21b), and M is bending moment which is produced in the cross sections of the arch when it losses stability. Let ds present the arc, which corresponds to central angle dθ. Since dw dw dθ 1 dw ¼ ¼ , ds dθ ds R dθ
and
d2 w 1 d2 w ¼ , ds2 R2 dθ2
then equation (15.13) may be presented in polar coordinates in the following form d2 w M þ w ¼ R2 , 2 EI dθ
ð15:13aÞ
which is called the Boussinesq’s equation (1883). Fundamental feature of the adopted design diagram is that the axial compressive force of the arch caused by uniformly distributed hydrostatic load q is constant for any cross section: N ¼ qR. Indeed, the total load within the portion ds (Fig. 15.21b) equals qds ¼ qRdθ, and all load which acts on the left half-arch is perceived by the left support, so the horizontal and vertical components of reaction N (Fig. 15.21c) are ðα
ðα
H ¼ qR sin θ dθ ¼ qR cos α, 0
V ¼ qR cos θ dθ ¼ qR sin α,
so
N¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H 2 þ V 2 ¼ qR
0
Thus, the considered problem, as all previous problems in this chapter, has a general feature—a structure before buckling is subjected to only compressed load. The slope φ at the elastic support and corresponding reactive moment M0 are related as M0 ¼ kφ. Distribution of the bending moments caused by two antisymmetrical angular displacements φ of elastic supports (or reactive moments M0) is presented in Fig. 15.21b. Bending moment at section A with central angle θ caused by only reactive moments M0 equals m (Fig. 15.21b). Let us turn to the calculation of the bending moment M in an arbitrary section A of the arch; the position of this section is determined by the angle θ (Fig. 15.21d). Of course, this moment can be determined by the forces located to the left of section A. However, this will lead to cumbersome transformations. It is evident that this moment depends on two factors, N and m, which act at section A. The first factor takes into account the constant axial force N ¼ qR which produces the moment on the displacement w. The second factor takes into account the moment m, which is the portion of reactive moment M0 that occurs in the elastic support (Fig. 15.21b). Therefore, the total bending moment at any section, whose position is determined by the angle θ, equals M ðθÞ ¼ Nw m ¼ qR w
sin θ kφ sin α
ð15:14Þ
The radial displacement of this section is w (θ) (Fig. 15.21d). For two-hinged arch the second term, which takes into account moment due to elastic supports, should be omitted. Thus, differential equation (15.13a) becomes d2 w qR3 kφ R2 þ 1 þ w ¼ sin θ ð15:15Þ EI sin α EI dθ2 Denote
qR3 EI kφ R2 C¼ EI sin α
n2 ¼ 1 þ
Differential equation (15.15) may be rewritten as follows
ð15:15aÞ ð15:15bÞ
554
15
Stability of Elastic Systems
d2 w þ n2 w ¼ C sin θ dθ2
ð15:16Þ
Pay attention that C is unknown, since the angles of rotation φ of the supports are unknown. Solution of this equation is w ¼ A cos nθ þ B sin nθ þ w
ð15:17Þ
The partial solution w should be presented in the form of the right part of (15.16), mainly w ¼ C0 sin θ, where C0 is a new unknown coefficient. Substituting of this expression into (15.16) leads to equation C 0 sin θ þ n2 C 0 sin θ ¼ C sin θ,
so
C0 ¼
C n2 1
Thus, the solution of equation (15.15) becomes w ¼ A cos nθ þ B sin nθ þ
C sin θ n2 1
ð15:17aÞ
Unknown coefficients A, B, and C may be obtained from the following boundary conditions: 1. For point of the arch on the axis of symmetry (θ ¼ 0), the radial displacement w ¼ 0 (because the antisymmetrical form of the loss of stability); this condition leads to A ¼ 0; 2. For point of the arch at the support (θ ¼ α) the radial displacement is w ¼ 0, so B sin nα þ
C sin α ¼ 0 n2 1
ð15:18Þ
3. Using expression (13.17a), the slope is dw C cos θ ¼ Bn cos nθ þ 2 dθ n 1
ð15:19Þ
dw ¼ φ ds
ð15:19aÞ
The slope at the support is
the negative sign means the reactive moment M0 and angle φ have the opposite directions. On the other hand dw dw ¼ , so ds Rdθ dw ¼ Rφ dθ According to (15.15b) we get φ¼C so substitution into (15.19a) leads to relationships If θ ¼ α, then the expression (15.19) becomes
EI sin α , kR2
dw EI sin α ¼ C . dθ kR
Bn cos nα þ
C cos α EI sin α ¼ C kR n2 1
After rearrangements this expression may be presented in form
15.5
Stability of Arches
555
cos α EI sin α ¼0 Bn cos nα þ C 2 þ kR n 1
ð15:20Þ
Equations (15.18) and (15.20) are homogeneous linear algebraic equations with respect to unknown parameters B and C. The trivial solution B ¼ C ¼ 0 corresponds to state of the arch before the loss of stability. Nontrivial solution occurs if the following determinant is zero: sin α sin nα 2 n 1 ð15:21Þ D¼ ¼0 cos α EI sin α n cos nα; þ kR n2 1 If the entries of the first and second columns are divided by cosnα and sinnα /(n2 1), respectively, then we get tan nα 1 ¼ 0 D¼ EI 2 n n 1 cot α þ kR This relationship immediately leads to the stability equation: tan nα ¼
n cot α þ ðn2 1Þ
ð15:22Þ
EI kR
Here the second term in denominator takes into account flexibility of supports. Solution of transcendental equation (15.22) for given α and dimensionless parameter k0 ¼ kR=EI is the critical parameter n. According to (15.15a) the critical load becomes EI qcr ¼ n2 1 3 R
ð15:23Þ
If the central angle 2α ¼ 60 , then the roots of equation (15.22) for different k0 are presented in Table 15.3. Table 15.3 Critical parameter n for circular arch with elastic clamped supports, 2α ¼ 60
k0 n
0.0 6.0000
1.0 6.2955
10 7.5294
100 8.4628
1000 8.6051
105 8.621
Limiting cases The general stability equation (15.22) allows us to consider some specific arches. 1. Two-hinged arch. In this case the stiffness k ¼ 0 and stability equation (15.22) is tannα ¼ 0. The minimum root of this equation is nα ¼ π, so n ¼ π=α and corresponding critical load equals qcr min
2 π EI ¼ 1 α2 R3
ð15:24Þ
Critical load for α ¼ π=2 (half-circular arch with pinned supports) equals qcr min ¼ 3
EI : R3
2. Arch with fixed supports. In this case the stiffness k ¼ 1 and stability equation (15.22) becomes tan nα ¼ n tan α In case α ¼ π/2 (half-circular arch) this equation can be presented in the form
ð15:25Þ
556
15
cot
nπ ¼ 0, 2
so
nπ π ¼ , 2 2
Stability of Elastic Systems
3π , 2
Solution n ¼ 1 is trivial because this solution, according to expression (15.15a), corresponds to q ¼ 0. Thus, minimum root is n ¼ 3, so for a half-circular arch with clamped supports the critical load equals qcr min ¼ 8
EI R3
Roots n of stability equation (15.25) for different angle α are presented in Table 15.4. Table 15.4 Critical parameter n for circular arch with fixed supports
30° 8.621
n
45° 5.782
60° 4.375
90° 3.000
It is worth to present the critical load for three-hinged uniform symmetrical arch under hydrostatic load. The critical load for antisymmetric buckling form coincides with critical load for two-hinged arch. In case of symmetrical buckling form, the critical load should be calculated by formula 2 4u EI qcr ¼ 1 , ð15:26Þ 2 α R3 where parameter u is a root of transcendental equation ð tan α αÞ tan u u ¼4 u3 α3
ð15:27Þ
Roots of this equation are presented in Table 15.5. Table 15.5 Three-hinged circular uniform arch. Critical parameter u for symmetrical buckling form
u
30° 1.3872
45° 1.4172
60° 1.4584
90° /2=1.5708
For all above cases, the critical load may be calculated by formula qcr ¼ K
EI , R3
where parameter K is presented in Table 15.6.
Table 15.6 Parameter K for critical hydrostatic load of circular arches with different boundary conditions; symmetrical (SF) and antisymmetrical (ASF) buckling forms
Types of arch and buckling form Hingeless, ASF
a =15° 294
30° 73.3
45° 32.4
60° 18.1
75° 11.5
90° 8
Two-hinged, ASF
143
35
15
8
4.76
3
Three-hinged, SF
108
27.6
12
6.75
4.32
3
15.5
Stability of Arches
557
Table 15.6 indicates that for three-hinged arch the lowest critical hydrostatic load corresponds to loss of stability in symmetrical form. For arches with elastic supports, factor К satisfies condition K1 K K2 where K1 and K2 are related to hingeless and two-hinged arches.
15.5.3 Complex Arched Structure: Arch with Elastic Supports In practical engineering, the stiffness coefficient k of the elastic supports is not given; however, in special cases it can be determined from an analysis of adjacent parts of the arch. Let us consider a structure shown in Fig. 15.22a. The central part of the structure presents the circular arch; supports of the arch are rigid joints of the frames. The arch is subjected to uniformly distributed hydrostatic load q. Assume that R ¼ 20 m and the central angle 2α ¼ 60 . The stiffness of all members of the structure is EI. Since the left and right Γ-shaped frames are deformable structures, then supports A and B of the arch should be treated as elastic ones. Therefore the stability equation should be adopted in the form (15.22) tan nα ¼
n cot α þ ðn2 1Þ
ðaÞ
EI kR
To determine the rotational stiffness k of supports A and B we need to consider parts of structure which support arch itself, i.e., both the Γ-shaped frames. Since the joints A and B are rigid, so the angle of rotation for frame and arch are same. Therefore, for calculation of the stiffness k of the arch we have to calculate the couple M, which should be applied at the rigid joint A of the frame in order to rotate this joint by angle φ ¼ 1. In order to solve this problem, we can use the displacement method. The frame subjected to unknown moment M ¼ k is shown in Fig. 15.22b. Primary system of the displacement method is shown in Fig. 15.22c.
a
q B
A EI
8m
R=20m
6m
6m α =30°
b
c
d
M=k=? i1=EI/6
r11
e Z1=1
1
R1P M=k=?
3i1 i2=EI/8
4i2 2i2
Fig. 15.22 Complex arched structure. (a) Design diagram of the structure; (b–e) Calculation of the stiffness k of the elastic supports of the arch
Canonical equation is r 11 Z 1 þ R1P ¼ 0 Displacement Z1 ¼ φ ¼ 1 and corresponding bending moment diagram is shown in Fig. 15.22d. The unit reaction
558
15
Stability of Elastic Systems
r 11 ¼ 3i1 þ 4i2 ¼ 0:5EI þ 0:5EI ¼ 1:0EI ½kNm=rad The primary system subjected to external unknown couple M is presented in Fig. 15.22e, so R1P ¼ M. The canonical equation becomes 1EI Z1 M ¼ 0. If the angle of rotation Z1 ¼ 1 [rad], then M ¼ k ¼ 1.0EI. For given parameters R and α the stability equation of the structure becomes tan nα ¼
Rn R cot α þ n2 1
Numerical solution of this transcendental equation presents the required parameter n of critical load (Table 15.7). Table 15.7 The roots n of the stability equation for different angle α, critical load (n2 – 1) and limiting cases for circular arches: hingeless/two-hinged arches (the central angle of arch is 2α)
Root n of stability equation 7.955 5.4647 4.191 -
Angle α
π/6 π/4 π/3 Factor
Critical load n2 - 1
Limiting cases for circular arches
62.282 28.863 16.564 EI/R3
73.3 - 35 32.4 - 15 18.1 - 8.0 EI/R3
The critical load is EI EI qcr ¼ 7:9552 1 3 ¼ 62:282 3 R R Control. According to Table 15.6 the critical load for arch with fixed supports and for two-hinged arch (the central angle in both cases is 2α ¼ 60∘) are qcr ¼ 73:3EI=R3 and qcr ¼ 35EI=R3, respectively. All above calculated critical loads are located between two limiting cases (Table 15.7, column 4). Note. The special case occurs for half-circular arch (α ¼ 90∘). In this case the total reaction of the arch is directed vertically, and therefore Γ-shaped frames are not deformable. Therefore, term ðn2 1ÞEI=kR from equation (a) should be omitted, and stability equation becomes tannα ¼ n tan α.
15.5.4 Parabolic Arch under Gravity Load Design diagram of parabolic uniform arch is shown in Fig. 15.23. The span and height of arch are l and f, respectively, and the moment of inertia of the cross section is I ¼ const. q 0
x
f R(φ)
y R0
φ l
● ●α ●
Fig. 15.23 Arch geometry and loading. Boundary conditions are not shown
φ
15.5
Stability of Arches
559
It is known that if parabolic arch is subjected to uniformly distributed vertical load, then until the loss of stability in the all sections of the arch, only compressive forces arise. Assumptions: 1. Axial deformation caused by compressive force are neglected. 2. The arch material works in a linearly elastic region. The equilibrium equation for a parabolic arch in a deformed state is 2 d d 3 dM þ1 cos φ þm M sec 4 φ ¼ 0, dφ dφ dφ2
m¼
qR30 EI
ð15:28Þ
where R0 is radius of curvature of parabola at the vertex; the angle φ is measured from the axis of symmetry, and the central angle 2α. This is differential equation of third order with variable coefficients with respect to bending moment M which arises in the arch when the loss of stability occurs. This equation was derived by A.N. Dinnik in 1946. The analytical solution of equation (15.28), apparently, is unknown. However, this equation may be integrated numerically. Thus, stability analysis of parabolic arch is generally essentially more complicated than the corresponding analysis of the circular arches. EI The required intensity of vertical load qcr ¼ m 3 may be presented in terms of the arch bending stiffness EI and R0 2 l l2 geometrical parameters l and f of the arch. Indeed, since equation of parabola x2 ¼ 2R0y, then ¼ 2R0 f ! R0 ¼ 2 8f and expression for critical intensity becomes 3 ð8f Þ3 EI EI 8f EI EI qcr ¼ m 3 ¼ m ¼m ¼K 3 l R0 l3 l l6
ð15:29Þ
Stability problem of parabolic arches under the given assumptions reduces to the determining the smallest positive parameter m, for which a nonzero solution of equation (15.28) for bending moment M may exist and satisfy the boundary conditions for M. Boundary conditions are as follows: for two-hinged arch the bending moment at left and right supports are M(α) ¼ M(α) ¼ 0. For antisymmetrical form of the loss of stability (in this case bending moment M is the odd function) the moment at the axis of symmetry is M(0) ¼ 0; this is the second boundary condition (Fig. 15.23) Numerical procedure. For given ratio f / l ¼ 0.1, 0.2, . . . we adopt any dimensionless parameter m and integrate Dinnik’s equation (15.28). If it turns out that the condition M(0) ¼ 0 on the axis of symmetry is satisfied, the solution procedure is completed. In otherwise case, we should adopt another value m and repeat a procedure. Thus, the numerical solution of the stability problem of parabolic arch is reduced to a complicated problem known in mathematics as the two-point boundary value problem. For hingeless arch we have the same condition with additional boundary conditions, which means that the angle of rotations at left and right supports are zero. Table 15.8 contains parameter K as final result for parabolic uniform symmetrical arches subjected to uniform distributed tracking load q; for conservative load parameter K is shown in parenthesis (Umansky, 1973). Dashes represent unavailable data. The critical load qcr ¼ KEI / l3.
Table 15.8 Parameter K for parabolic uniform arch f l
0.1 0.2 0.3 0.4 0.5 0.6 0.8 1.0
Hingeless arch
Two-hinged arch
Antisymmetric form
Antisymmetric form
60.7 101.0 115.0 111.0 97.4 83.8 59.1 43.7
28.5 (28.8) 45.4 (46.1) 46.5 (48.4) 43.9 (45.0) 38.4 (-) 30.5 (31.7) 20.0 (-) 14.1 (15.4)
One-hinged Arch
33.8 59.0 84.0 96.0 87.0 80.0 63.0 48.0
Three-hinged arch Antisymmetric form
28.5 (28.8) 45.4 (46.1) 46.5 (48.4) 43.9 (45.0) 38.4 (-) 30.5 (31.7) 20.0 (-) 14.1 (15.4)
Symmetrical form
22.5 (22.7) 39.6 (40.2) 47.3 (49.8) 49.2 (54.5) 43 (-) 38.0(-) 28.8(-) 22.1(-)
560
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Stability of Elastic Systems
For two-hinged and three-hinged arches at antisymmetric forms of the loss of stability coefficients K coincides. This fact was discussed earlier. Comparing the tabulated data, we conclude that a smaller value of the parameter K occurs with a tracking load. In other words, if the arch is loaded by the uniformly distributed tracking load q, then such way of loading leads to the smallest value of critical load. Although the question of its implementation of such loading remains questionable, the adoption of this method of applying the load goes to the safety margin of the structure. In case of three-hinged arch the coefficients К are presented for symmetrical (1) and anti-symmetrical (2) form of the loss of stability vs. dimensionless parameter f / l is presented in Fig. 15.24. For very shallow arches ( f / l < 0.3) the symmetrical form of the loss of stability is realized, because of the following inequality KSF < KASF.
K 50 40 2 30
1
20 0.1
0.2
0.3
0.4
f/l
Fig. 15.24 Stability coefficient K for three-hinged arch vs parameter f/l; 1, 2-symmetrical and antisymmetrical forms of loss of the stability
Arches with Variable Cross Sections In case of such arches it is necessary to specify not only how the moment of inertia changes along the axis of the arch but also to introduce an additional assumption about the distribution of material across cross section. Let parabolic arch be loaded by uniformly distributed load through the entire span. This is rational arch, because only axial compressive force arises at any cross section (see Chap. 9, Table 9.12). The cross section of the arch at the highest point is a rectangle with sides b0 (width) and h0 (height). Axial force in the arch varies according to the law N(φ) ¼ N0 / cos φ (Tables 9.7 and 9.12). Therefore the arch becomes of the equal strength, if the cross-sectional area changes according to the law A(φ) ¼ A0 / cos φ. Since the axial force along the arch axis turns out to be variable, then the cross section also has variable dimensions, i.e., b(φ) and h(φ). The set of possible relations between the sides of the cross section, b and h, allows specifying two limiting cases. This is a case of a constant width along the axis of the arch and the case of a constant height of the cross section. Aφ 1 A0 In the first case, bφ ¼ b0 ¼ const, we have hφ ¼ ¼ . The moment of inertia of the cross section if the loss of cos φ b0 b0 stability occurs in the plane of the arch b0 h3φ b0 ¼ Iφ ¼ 12 12
1 A0 cos φ b0
3
b ¼ 0 12
1 b0 h0 cos φ b0
3 ¼
I0 cos 3 φ
Aφ 1 A0 ¼ , and the moment of inertia cos φ h0 h0 bφ h30 h30 h30 1 A0 1 b0 h0 I Iφ ¼ ¼ ¼ ¼ 0 12 12 cos φ h0 12 cos φ h0 cos φ
In the second case, hφ ¼ h0 ¼ const, we have bφ ¼
Stability coefficient K for different types of symmetrical parabolic arches for both partial cases vs. parameter f / l are presented in (Karnovsky 2012). This book also contains some specific topics of the stability problems of the arches. Among them are partial loading, arch with tie, out-of-plane loss of stability, etc. In conclusion we note that a deep consideration of the fundamental concepts of the theory of stability and their analysis can be found in the excellent book (Feodos’ev 1975). Among them, the concept of choosing the design diagram of the structure, the identification of two concepts “loss of stability” and “the existence of arbitrarily close forms of equilibrium,” the relativity of the concepts of stability depending on the magnitude of possible disturbances, as well as numerous interesting and enlightening in-depth examples illustrating the principal features of the problem.
Problems
561
Problems 15.1. The two absolutely rigid bodies (EI ¼ 1) connected by hinge at point C (Fig. P15.1). Stiffness coefficient of each elastic support is k. Derive the stability equation, find critical forces, and show corresponding shapes of loss of stability.
P
C
EI=∞
k
k l
Ans. P1cr = P2cr =
l
3+ 5 kl = 0.3819kl , 2
3+ 5 kl = 2.6180kl 2
Fig. P15.1
15.2. The simply supported beam with elastic support A is subjected to axial compressive force P (Fig. P15.2). The restoring moment in rotational spring is M ¼ kφ0. Derive the stability equation. Apply two approaches: (a) double integration method; (b) initial parameter method. k
φ0
P
Ans. tan λ l =
A
λl 2 2
λ l α +1
, λ=
EI P , α= kl EI
l Fig. P15.2
15.3. Absolutely rigid rod is loaded by two forces P1 ¼ P and P2 ¼ αP as shown in Fig. P15.3; α is any positive number. Determine a critical force.
P2= αP
P1=P k1
EI = ∞
Ans. Pcr =
k2 l1
k1 (l1 + l2 )2 + k 2l22 l1 + l2 (1 + α )
l2
Fig. P15.3
15.4. Design diagram of the column is presented in Fig. P15.4. The total length of the column is l, while the length of the bottom part is αl. The column is subjected to forces N1 and N2. Relationship between these forces is N2 ¼ βN1. Parameters α and β are the given positive numbers. Stiffness rigidity of the portions are EI1 and EI2. Derive the stability equation. Consider two limiting cases: α ¼ 0 (uniform column with stiffness EI1) and α ¼ 1 (uniform column with stiffness EI2). Apply double integration method.
N1=N
EI1 N2= β N l
n1 =
αl
Fig. P15.4
Ans. tan (1 − α ) n1l ∙ tan n2αl −
EI2
N1 , n2 = EI1
(1 + β ) N EI 2
n1 (1 + β ) = 0 , n2
562
15
Stability of Elastic Systems
15.5. Design diagram of two-story frame with elastic constraints between adjacent absolutely rigid members is shown in Fig. P15.5. Calculate the critical load using energy method.
P k1
Ans. Pcr =
h
k2
k1 + k 2 + k3 2h
k3 h l Fig. P15.5
15.6. Derive the stability equation for uniform columns subjected to axial compressive force. Consider the following supports: (a) clamped-free; (b) clamped-pinned; (c) pinned-pinned. Apply the initial parameter method. Compare the results with those presented in Table 15.1, Textbook. 15.7. The clamped beam with elastic support is subjected to axial force N (Fig. P15.7). Derive the stability equation. (Hint: Reaction at the right support is ky1, where y1 is vertical displacement at the right support. The moment at the left support should be calculated taking into account the lateral displacement y1 of axial force N, i.e., M0 ¼ ky1l Ny1). Consider limiting cases (k ¼ 0, k ¼ 1).
N
EI
x
k
M0
Ans. tan nl = nl 1 −
l Q0
n 2l 2 EI kl 3
, n=
N EI
ky1
Fig. P15.7
15.8. The uniform beam with overhang is subjected to axial compressive force P (Fig. P15.8). Derive the stability equation. Use the initial parameter method. Consider limiting case a ¼ 0 P
EI l
a
Ans. nl (tan nl + tan na) = tan nl · tan na , n =
P EI
Fig. P15.8
15.9. The uniform pinned-clamped beam with intermediate hinge C is subjected to axial compressive force P (Fig. P15.9). Derive the stability equation by initial parameter method. Calculate the critical load and trace the shape of the loss of stability. Consider two special cases (a ¼ 0, a ¼ l). C
P
EI
a l Fig. P15.9
Ans. sin na nl cos n ( l − a )− sin n ( l − a ) = 0
Problems
563
15.10. The uniform pinned-pinned beam with elastic support is subjected to axial compressed force P (Fig. P15.10); stiffness coefficient of elastic support is k. The total length of the beam is l ¼ l1 + l2. Derive the stability equation. Use the double integration method. Consider the special cases; (a) k = 0; (b) l1 ¼ l2; (c) k ¼ 1. EI
P k
l1
Ans. sin nl1 sin nl2 = nl sin nl · l2
l1l2 P , n= l2 kl
P EI
Fig. P15.10
For problems 15.11 through 15.15 it is recommended to apply the Displacement method. All stability functions φ1, φ2, . . . are presented in Appendix, Tables A22 and A23. 15.11. Design diagram of the uniform continuous beam is presented in Fig. P15.11. Derive the equation for critical load in terms of parameter α. Consider special case for α ¼ 0.5. EI
P
Ans.
4φ
α
2
(αυ 0 ) +
3 φ 1 (1− α )υ 0 = 0 1−α
(
)
υ1 = l1 P = α L P = αυ 0 , l1= α L
EI
l2=(1- α)L L
EI
υ 2 = (1 − α) L P = υ0 (1 − α) EI
Fig. P15.11
15.12. The two-span beam of spans l1 and l2 ¼ βl1 is subjected to axial forces P and αP (Fig. P15.12). The flexural rigidity for the left and right spans are EI and kEI. Derive the equation for critical load in terms of parameters α, β, and k. Consider the special cases: (a) α ¼ 3, β ¼ 1 and k ¼ 4; (b) α ¼ 0, k ¼ 1; (c) α ¼ 0, k ¼ 1, β ¼ 1. Explain obtained results. k Ans. φ1 ( υ1) + φ1 (υ 2 ) = 0 , β
αP
P
EI
kEI
υ 1 = l1
l2= β l1
l1
P P + αP 1+ α , υ 2 = l2 = υ1 β EI kEI k
Fig. P15.12
15.13. Design diagram of the frames with deformable crossbar are presented in Fig. P15.13. Derive the equation for critical load for boundary conditions shown in Fig. 15.13a and in Fig. 15.13b
a
b
P
P
kEI
Ans. (a) 3k kEI
υ h + = 0, l tan υ P ; EI 2
υ =h
EI
h
EI
h (b) υ tan υ = 3k
l Fig. P15.13
l
h l
564
15
Stability of Elastic Systems
15.14. The frame with absolutely rigid crossbar is presented in Fig. P15.14. Derive the stability equation and calculate the critical load.
P EI=∞ Ans. η1 (υ ) + 2k = 0, υ = h EI
kEI
h kEI
P . EI
2 If k=1, then υ =2.67, Pcr = 7.13EI/ h
l2
l1 Fig. P15.14
15.15. Design diagram of the frame is presented in Fig. P15.15. Relationship between two forces is constant. Derive the stability equation and find critical load.
P 3EI
Ans.
2
EI
EI B
A l=6m Fig. P15.15
3EI h=4m
1
4P
l=6m
φ 2 ( υ) + 2 1.0
υ1A = υ = h υ min=2.097.
1.0 = 0, 0.75φ1 (2υ ) + 3.5
P 4P ; υ2B = h = 2υ , EI EI
Chapter 16
Dynamics of Elastic Systems: Free Vibration
Structural dynamics is a special branch of structural analysis, which studies the behavior of structures subjected to dynamical loads. Such loads develop the dynamical reactions, dynamical internal forces, and dynamical displacements of a structure. They all change with time and maximum values often exceed static ones. Dynamical analysis of a structure is based on the free vibration analysis. This chapter is devoted to linear free vibration analysis of elastic structures with lumped and distributed parameters. The fundamental methods of structural analysis (force and displacement methods) are applied for calculation of frequencies of the free vibration and corresponding mode shape of vibration. They are inherent to the structure itself and called the eigenvalues and eigenfunctions.
16.1
Fundamental Concepts
This paragraph contains brief information necessary for the analysis of dynamic systems. Among them are the kinematics of vibrational processes, the types of forces that are the source of vibration and the classification of forces inherent vibrations, as well as the most important concept of the degrees of freedom of dynamic systems.
16.1.1 Kinematics of Vibrating Processes The simplest periodic motion can be written as yðt Þ ¼ A sin ðω t þ φ0 Þ where A is the amplitude of vibration; φ0 is initial phase of vibration; t is time. This case is presented in Fig. 16.1a. The initial displacement y0 ¼ A sin φ0 is measured from the static equilibrium position. The number of cycles of oscillation during 2π 2π (radians per second or s1), T(second) is seconds is referred to as circular (angular or natural) frequency of vibration ω ¼ T the period of vibration. Figure 16.1b, c presents the damped and increased vibration with constant period.
c
b
a y
y0 A
y
y T
T
T t
T
T t
T t
Fig. 16.1 Types of oscillatory motions. (a) Harmonic vibration with constant amplitude; (b, c) Harmonic vibration with variable amplitudes
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_16
565
566
16
Dynamics of Elastic Systems: Free Vibration
16.1.2 Forces Which Arise at Vibrations During vibration a structure is subjected to different forces. These forces have a different nature and exert a different influence on the vibrating process. All forces may be divided into the following groups: disturbing forces, positional (restoring) forces, resisting forces, and forces of the mixed character. 1. Disturbing forces may be of the following types: (a) Immovable periodical loads produced by stationary units and mechanisms with moving parts. These loads have a periodical, but not necessary a harmonic character and generally do not depend on the elastic properties of the structure. (b) Impact (impulsive) loads are produced by falling weights or collision of bodies. Impulsive loads are characterized by very short duration of their action and depend on the elastic properties of the structure, which is subjected to such loads. (c) Moving loads act on the structures through wheels of a moving train or truck. The availability of the rail joins on the railway bridge or irregularities of the deck on the car bridge lead to appearance of inertial forces. This type of loads should be distinguished from moving one, which has been studied in the sections “Influence lines” because unit moving load P ¼ 1 had been considered without dynamical effects. (d) Seismic loads arise due to earthquakes. The reason of the seismic load on the structure is acceleration of the supports caused by acceleration of the ground. This type of disturbance is called kinematical. The acceleration of supports leads to the acceleration of the separate parts of the structure, and as a result inertial forces act on these parts. Seismic forces, which arise in the members of the structure, depend on the type and the amount of ground acceleration, distribution of the mass within the members of the structure, and their elastic properties. 2. Restoring forces depend on the displacement of the structure, arise due to deviation of system from a static equilibrium position, and tend to return the system to its initial position. Restoring properties of a system are described by its elastic characteristic P ¼ P( y), where P is a static force which is applied to the structure. Characteristic P-y may be linear or nonlinear. Some types of characteristics P-y are presented in Table 16.1; in all cases y is the displacement at the point of P. Table 16.1 Types of elastic members and their statical characteristics
Design diagram P
Characteristic P-y P
P y
y P y0
Design diagram
y0 P
P y0
y
Characteristic P-y P y0
y
P y
3. Resisting forces. The forces of inelastic resistance (friction or damping forces) depend on the velocity υ of motion, R ¼ R(υ), and always act in the opposite direction of velocity. These forces are the result of resisting properties of media and internal friction in the material of a structure and/or in the connections of a system. Different types of forces acting on a structure lead to different types of vibration, namely free and forced vibration. Vibrations of a system in which disturbing forces are absent are called free vibrations. At free vibration the system is subjected to forces inherent to the system itself, i.e., the restoring and resisting forces. To impose free vibrations, nonzero initial conditions should be created, which means some initial displacement and/or initial velocity. Free vibration may be linear or nonlinear depending on the characteristics of restoring and resisting forces. Absence of resisting forces leads to the free undamped vibrations; in this case the system is subjected only to a restoring force. Vibration of a system caused by any disturbing forces is called a forced vibration. Absence of resisting forces leads to the forced undamped vibration. Just as the free vibration, the forced vibration may be linear and nonlinear. Free and forced vibrations are the subject of study of this and the subsequent chapter, respectively.
16.1
Fundamental Concepts
567
16.1.3 Degrees of Freedom The fundamental difference of the concept of “degrees of freedom” in static and structural dynamics in spite of the same definition (a number of independent parameters, which uniquely defines the positions of all points of a structure) is the following. In statics, the number of degrees of freedom is related to a structure consisting of absolutely rigid discs. If the degree of freedom is greater than or equal to one, then the system is geometrically changeable and cannot be assumed as an engineering structure; when the degree of freedom equals to zero it means that a system is geometrically unchangeable and statically determined. In structural dynamics the number of degrees of freedom is determined taking into account the deformation of the members. The number of degrees of freedom of a dynamic system equals to the number of degrees of freedom of various masses located on it. If the degree of freedom equals to zero, then a system presents an absolutely rigid body and all displacements in space are absent. All structures may be divided into two principal classes according to their degree of freedom. They are the structure with concentrated and distributed parameters. Members with concentrated parameters assume that the distributed mass of the member itself may be neglected in comparison with the lumped mass, which is located on the member. The structure with distributed parameters is characterized by uniform or nonuniform distribution of mass within its parts. From mathematical point of view the difference between the two types of systems is the following: the systems of the first class are described by ordinary differential equations, while the systems of the second class are described by partial differential equations. Figure 16.2a, b shows a massless statically determinate and statically indeterminate beam with one lumped mass. These structures have one degree of freedom, since transversal displacement of the lumped mass defines position of all points of the beam. Note that these structures, from point of view of their static analysis, have the number of degrees of freedom W ¼ 0 for statically determinate beam (a) and W ¼ 3D 2H S0 ¼ 3 1 2 0 5 ¼ 2 for statically indeterminate beam (b) with two redundant constraints. It is obvious that a massless beam in Fig. 16.2c has three degrees of freedom. It can be seen that introducing of additional constraints on the structure increase the stiffness of the structure, i.e., increase the degrees of static indeterminacy, while introducing additional masses increase the number of degrees of freedom.
a y1
b
x
e
y1
c d
y1
Pontoon
f C
y3
y2 x
z
y
y Fig. 16.2 (a-f) Design diagrams of structures
Figure 16.2d presents a cantilevered massless beam carrying one lumped mass. However, this case is not a plane bending, but bending combined with torsion, because mass is not applied at the shear center. That is why this structure has two degrees of freedom, such as the vertical displacement and angle of rotation in y-z plane with respect to the x-axis (if horizontal displacement can be neglected). A structure in Fig. 16.2e presents a massless cantilever with an absolutely rigid body at the end. The structure has two degrees of freedom, such as the lateral displacement y of the body and angle of rotation of the body in y-x plane.
568
16
Dynamics of Elastic Systems: Free Vibration
Figure 16.2f presents a bridge, which contains two absolutely rigid bodies. These bodies are supported by a pontoon. Corresponding design diagram shows two absolutely rigid bodies connected by hinge C with elastic support. So this structure has one degree of freedom. The plane and spatial bars structures and plane truss are presented in Fig. 16.3. In all cases we assume that all members of a structure do not have distributed masses. The lumped mass (Fig. 16.3a) can move in vertical and horizontal directions; therefore, this structure has two degrees of freedom. Similarly, the statically indeterminate structure shown in Fig. 16.3b, has two degrees of freedom. However, if we assume that horizontal member is absolutely rigid in axial direction (axial stiffness EA ¼ 1), then the mass can move only in vertical direction and the structure has one degree of freedom. Structure 16.3c has three degrees of freedom; they are displacement of the lumped mass in x-y-z directions. Introducing additional deformable but massless members does not change the number of degrees of freedom.
a
c d
b
B
A
Fig. 16.3 Design diagrams of bar plane structures (a, b, d) and (c) space structure
The plane truss (Fig. 16.3d) contains five concentrated masses. The mass at joint A is fixed, the mass at joint B can move only in horizontal direction, and the other masses can move in horizontal and vertical directions. So, this structure has seven degrees of freedom. If we assume that horizontal displacements of the joints may be negligible in comparison with vertical displacements, then this truss may be considered as a statically determinate structure with three degrees of freedom. If additional members will be introduced in the truss (shown by dotted lines), then this truss should be considered as two times statically indeterminate structure with three degrees of freedom. Thus, the number of degrees of freedom is determined not by the number of concentrated masses but by the number of independent generalized coordinates. Figure 16.4 presents plane frames, column, and arches. In all cases, we assume that all members of a structure do not have distributed masses. The lumped mass M in Fig. 16.4a can move in vertical and horizontal directions, so these structures have two degrees of freedom. If the lumped mass is replaced by a body with a mass M and a moment of inertia J, then the system will have three degrees of freedom. They are the vertical and horizontal displacement, and the angle of rotation of the body (Fig.16.4b). Figure 16.4c shows the two-story frame containing absolutely rigid crossbars (the total mass of each crossbar is M). This frame may be presented as column in Fig. 16.4d.
a
m q2
c
q1
b
q3
M,J
M EI=¥ M EI=¥
q2
d
M
e
M
f
q1
Fig. 16.4 Design diagrams of frames (a-c), column (d), and arches (e-f)
Arches with one and three lumped masses are shown in Fig. 16.4e, f. Taking into account their vertical and horizontal displacements, the number of their degrees of freedom will be two and six, respectively. For gently sloping arches the
16.1
Fundamental Concepts
569
horizontal displacements of the masses may be neglected; in this case, the arches should be considered as structures having one and three degrees of freedom in the vertical direction. All cases shown in Figs. 16.2–16.4 present the design diagrams for systems with lumped (concentrated) parameters. Since masses are concentrated, the configuration of a structure is defined by displacement of each mass as a function of time, i.e., y ¼ y(t), and behavior of such structures is described by ordinary differential equations. It is worth discussing the term concentrated parameters for cases 16.2f (pontoon bridge) and 16.4с (two-story frame). In both cases the mass, in fact, are distributed along the correspondent members. However, the bending stiffness of these members is very large, and therefore position of these members is defined by only one coordinate. For the structure in Fig 16.2f, such coordinate may be the vertical displacement of the pontoon or the angle of inclination of the span structure, and for two-story frame (Fig. 16.4c), it may be the horizontal displacements of the each crossbar. The structures with distributed parameters are more difficult for analysis. The simplest structure is a beam with distributed mass m. In this case a configuration of a system is determined by displacement of each elementary mass as a function of time. However, since masses are distributed, then a displacement of any point is a function of a time t and location x of the point, i.e., y ¼ y(x, t), so behavior of the structures is described by partial differential equations. It is possible a combination of members with concentrated and distributed parameters. Figure 16.5 shows a frame with a massless strut ВF (m ¼ 0), members АВ and ВС with distributed masses m and absolutely rigid member СD (EI ¼ 1). The simplest form of vibration is shown by the dotted line.
A
B
EI, m
C
EI=
D
EI, m EI, m=0 F
Fig. 16.5 Frame with distributed and concentrated parameters
The behavior of this structure is described by set of ordinary and partial differential equations. Thus, the number of degrees of freedom is dependent on the assumptions related to the design diagram of the structure. The number of degrees of freedom is primary characteristic for dynamical analysis of any structure. The reader can familiarize himself with the case of a system with non-integer number of degrees of freedom (in particular, 1.5 degrees of freedom) in a book (Karnovsky and Lebed 2016).
16.1.4 Purpose of Structural Dynamics Fundamental problem of dynamical analysis of the structures is determination of the dynamic forces and displacements of specified points of a structure, which is subjected to different dynamic excitations. The term “dynamic” means that the forces and displacements changes in the time. The solution of this problem is based on the determining of the important characteristics of a structure—its frequencies and shapes of free vibration. Free vibration of a structure occurs with some frequencies. These frequencies depend on only the structure itself and its properties (type of structure, boundary conditions, distribution of masses and stiffnesses within the members, etc.) and does not depend on the reason of vibration. Therefore these frequencies are often called as eigenfrequencies (eigenvalues), because these frequencies are inherent to a given structure. The number of frequency vibrations coincides with the number of degrees of freedom. The structure with distributed parameters has infinite number of degrees of freedom. The set of frequency vibration presents the frequencies spectrum of a structure. Each mode shape of vibration shows the form of elastic curve, which corresponds to specific frequency. This chapter contains only free vibration analysis which is based on the general assumptions of the classical structural analysis and on the specified assumptions related to the structural dynamics. Among them are: 1. Only linear vibrations are considered. 2. Damping effects are ignored. 3. Stiffness and inertial effects of the structure are time independent.
570
16.2
16
Dynamics of Elastic Systems: Free Vibration
Systems with Finite Number of Degrees of Freedom: Force Method
Behavior of such structures may be described by two types of differential equations. They are equations in the form of the force method (which contain displacements) and equations in the form of the displacement method (which contain reactions). In both cases we will consider only the undamped vibration.
16.2.1 Differential Equations of Free Vibration in Displacements The essence of the first approach consists of expressing the forces of inertia as function of unit displacements. For the derivation of differential equations, let us consider a structure with concentrated masses (Fig. 16.6). In the first unit state we show inertial force F in 1 ¼ 1 and corresponding displacements δi1, i ¼ 1, , n; similarly for other states.
F1in =1
F2in m1
y1
Fnin
m2
y2 mn
yn
F2in =1 m1
m2
m1
d 11 d 21
n-th unit state
Second unit state
First unit state
F1in
mn
d12 d22
d n1
m1
m2 mn
d n2
m2
d1n d 2n
Fnin =1 mn
d nn
Fig. 16.6 Design diagram and unit states
In case of free vibration, each mass is subjected to forces of inertia only. Displacement of each mass may be presented as in in y1 ¼ δ11 F in 1 þ δ12 F 2 þ þ δ1n F n in in y2 ¼ δ21 F in 1 þ δ22 F 2 þ þ δ2n F n
:
: :
: : :
: :
: : :
ð16:1Þ
in in yn ¼ δn1 F in 1 þ δn2 F 2 þ þ δnn F n
where δik is displacement in ith direction caused by unit force acting in the kth direction. Each equation of (16.1) means displacement yi (i ¼ 1, . . ., n) in direction of masses m1, m2, . . ., mn caused by inertial forces of all masses; for writing of this system the superposition principle is used. •• Since the force of inertia of mass mi is F in i ¼ mi yi , then the differential equations (16.1) become ••
••
••
δ11 m1 y1 þδ12 m2 y2 þ . . . þ δ1n mn yn þy1 ¼ 0 : : : : : : : : : : : : : •• •• •• δn1 m1 y1 þδn2 m2 y2 þ . . . þ δnn mn yn þyn ¼ 0:
ð16:2Þ
Each equation of (16.2) presents the compatibility condition. The differential equations of motion are coupled dynamically because the second derivatives of all coordinates appears in each equation. •• Unknown inertial forces (F in i ¼ mi yi ) may be considered as primary unknowns of the force method. Therefore, hereafter (16.2) will be called the differential equations of free undamped vibration in displacements or canonical equations in the form of the force method.
16.2
Systems with Finite Number of Degrees of Freedom: Force Method
571
16.2.2 Frequency Equation Solution of system of differential equations (16.2) is y1 ¼ A1 sin ðω t þ φ0 Þ, . . . , yn ¼ An sin ðω t þ φ0 Þ,
ð16:3Þ
where Ai are the amplitudes of the corresponding masses mi and φ0 is the initial phase of vibration. The second derivatives of these displacements over time are ••
y1 ¼ A1 ω2 sin ðω t þ φ0 Þ, . . . ,
••
yn ¼ An ω2 sin ðωt þ φ0 Þ
By substituting (16.3) and (16.3a) into (16.2) and reducing by ω2 sin (ω t + φ0) we get m1 δ11 ω2 1 A1 þ m2 δ12 ω2 A2 þ . . . þ mn δ1n ω2 An ¼ 0 m1 δ21 ω2 A1 þ m2 δ22 ω2 1 A2 þ . . . þ mn δ2n ω2 An ¼ 0 : :
: :
: : :
: :
: : : : m1 δn1 ω2 A1 þ m2 δn2 ω2 A2 þ . . . þ mn δnn ω2 1 An ¼ 0
ð16:3aÞ
ð16:4Þ
The equations (16.4) are homogeneous algebraic equations with respect to unknown amplitudes Ai. Trivial solution Аi ¼ 0 corresponds to the system at rest. Nontrivial solution (nonzero amplitudes Ai) is possible if the determinant of the coefficients of amplitude is zero 2 3 m2 δ12 ω2 ... mn δ1n ω2 m1 δ11 ω2 1 6 m δ ω2 m2 δ22 ω2 1 . . . mn δ2n ω2 7 1 21 6 7 ð16:5Þ D ¼ det6 7¼0 4 5 ... ... ... ... m1 δn1 ω2
m2 δn2 ω2
...
mn δnn ω2 1
This equation is called the frequency equation in terms of displacements. Solution of this equation ω1, ω2, . . ., ωn presents the eigenfrequencies of a structure. The number of the frequencies of free vibration equals to the number of degrees of freedom. The frequency equation (16.5) in the equivalent form is (Kiselev 1980) 2 3 m1 δ11 λ m2 δ12 ... mn δ1n 6 m δ m2 δ22 λ . . . mn δ2n 7 1 21 6 7 D¼6 ð16:5aÞ 7¼0 4 5 ... ... ... ... m1 δn1
m2 δn2
...
mn δnn λ
where term λ ¼ 1/ω2 is called characteristic number or eigenvalue, while equation (16.5a) is called secular equation. This equation was first obtained in astronomy by Leverrier (1846). The characteristic numbers that represent the squares of the periods of motion of the planets are measured by large numbers—by centuries; it is here that the origin of the name of the equation.
16.2.3 Mode Shapes of Vibration and Modal Matrix The set of equations (16.4) are homogeneous algebraic equations with respect to unknown amplitudes A. This system does not allow us to find each of these amplitudes separately. However, we can find the ratios between different amplitudes. If a structure has two degrees of freedom, then the system (16.4) becomes m1 δ11 ω2 1 A1 þ m2 δ12 ω2 A2 ¼ 0 ð16:4aÞ m1 δ21 ω2 A1 þ m2 δ22 ω2 1 A2 ¼ 0: From these equations we can find the following ratios
572
16
Dynamics of Elastic Systems: Free Vibration
A2 m δ ω2 1 A m1 δ21 ω2 ¼ 1 11 or 2 ¼ 2 A1 A m2 δ12 ω m2 δ22 ω2 1 1
ð16:6Þ
If we substitute the first frequency of vibration ω1 into any of the two equations (16.6), then we can find ðA2 =A1 Þω1 . Then we can assume that А1 ¼ 1 and calculate the corresponding А2 (or vice versa). The numbers А1 ¼ 1 and А2 defines the distribution of amplitudes at the first frequency of vibration ω1; such distribution is referred as the first mode shape of vibration. This distribution is presented in the form of vector-column φ1, whose elements are A1 ¼ 1 and the calculated A2; this column vector is called the first eigenvector φ1. Thus, the set of equations (16.4a) for ω1 defines the first eigenvector within an arbitrary constant. Second mode shape of vibration or second eigenvector, which corresponds to the second frequency vibration ω2, can be found in a similar manner. After that we can construct a modal matrix Φ ¼ bφ1 φ2c. For system with two degrees of freedom for verification of numerical results the following relationships may be recommended: 1 1 þ ¼ m1 δ11 þ m2 δ22 ω21 ω22 A2 A2 m ¼ 1 A1 ω1 A1 ω2 m2
ð16:7Þ ð16:8Þ
If a structure has n degrees of freedom, then the modal matrix Φ ¼ bφ1 φn c Example 16.1 Design diagram of the frame is shown in Fig. 16.7a. Find eigenfrequencies and mode shape of vibration.
P1=1 m
2EI
a
P2=1
1 .l q2 q1
h
EI
M2
M1 1 .h l
b
m
m
1.0
1.0 1.1328 w1- first mode
0.8828 w2 - second mode
Fig. 16.7 Free vibration analysis of frame with two degrees of freedom. (a) Design diagram of the frame and unit states; (b) Mode shapes of vibration
Solution The system has two degrees of freedom. Generalized coordinate are q1 and q2. We need to apply unit forces in direction of q1 and q2, and construct the bending moments diagram. Unit displacements are Xð M M M M1 1 1 2 1 l3 l2 h 1 1 1ll 1lþ 1lh1l¼ ; dx ¼ 1 ¼ þ δ11 ¼ 2EI 2 3 EI EI EI 6EI EI M2 M2 1 1 2 h3 ¼ 1hh 1h¼ ; EI 2 3 EI 3EI M M2 1 1 h2 l : ¼ δ21 ¼ 1 ¼ 1hh1l¼ EI 2 2EI EI
δ22 ¼ δ12
16.2
Systems with Finite Number of Degrees of Freedom: Force Method
573
l3 . In this case δ11 ¼ 13δ0; δ22 ¼ 16δ0; δ12 ¼ δ21 ¼ 12δ0. 6EI Equation for calculation of amplitudes (14.4) 13δ0 mω2 1 A1 þ 12δ0 mω2 A2 ¼ 0 12δ0 mω2 A1 þ 16δ0 mω2 1 A2 ¼ 0
Let h ¼ 2l and δ0 ¼
Let λ ¼
ðaÞ
1 6EI ¼ . In this case equation (a) may be rewritten δ0 mω2 mω2 l3 ð13 λÞA1 þ 12A2 ¼ 0 12A1 þ ð16 λÞA2 ¼ 0
ðbÞ
Frequency equation becomes D¼
13 λ
12
12
16 λ
¼ ð13 λÞð16 λÞ 144 ¼ 0
Roots in descending order are λ1 ¼ 26.593; λ2 ¼ 2.4066 Eigenfrequencies in increasing order are rffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi 6EI EI , ¼ 0:4750 ω1 ¼ λ1 ml3 ml3 Control 1: According to (16.7) we have
rffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi 6EI EI ω2 ¼ ¼ 1:5789 λ2 ml3 ml3
1 1 þ 2 ¼ m1 δ11 þ m2 δ22 . In our case 2 ω1 ω2
1 1 ml3 ml3 ml3 ml3 þ 2¼ ¼ 4:8332 þ ¼ ð4:4321 þ 0:4011Þ 2 2 2 EI EI ω1 ω2 0:4750 EI 1:5789 EI l3 ml3 m1 δ11 þ m2 δ22 ¼ m 13δ0 þ m 16δ0 ¼ 29mδ0 ¼ 29m ¼ 4:8333 6EI EI Mode shape of vibration may be determined on the basis of equations (b). For first mode (λ1 ¼ 26.593) ratio of amplitudes are A2 13 λ 13 26:593 ¼ ¼ 1:1328 ¼ 12 12 A1 A2 12 12 ¼ ¼ 1:1328 ¼ 16 λ 16 26:593 A1 Assume that A1 ¼ 1, so the first eigenvector φ becomes φ ¼ b φ11
φ21 cT ¼ b 1
For second mode (λ2 ¼ 2.4066) ratio of amplitudes are A2 13 2:4066 ¼ 0:8828 ¼ 12 A1 A2 12 ¼ ¼ 0:8828 16 2:4066 A1 The modal matrix Φ is then defined as
1 Φ¼ 1:1328
1 0:8828
1:1328 cT
574
16
Dynamics of Elastic Systems: Free Vibration
Corresponding mode shapes of vibration are shown in Fig. 16.7b. It easy to check the scalar multiplication of two vectors φT1 Mφ2 ¼ 0 , where M is diagonal mass matrix (Kiselev 1980). In our case we have m 0 1 ¼ ð1 1:000036Þm ffi 0 b 1 1:1328 c 0 m 0:8828 This result shows that both forms of vibrations are orthogonal. It means that the force which corresponds to the first mode does not produce a work on the displacements of the second mode, and vice versa. Control 2: Relationships (16.8) for our case leads to the following results A2 A2 ¼ 1:1328ð0:8828Þ ¼ 1:0 A1 ω1 A1 ω2
and
m1 ¼ 1 m2
Example 16.2 Design diagram of structure containing two hinged-end is shown in Fig. 16.8a. Modulus of elasticity pffiffimembers ffi E and area of cross section A are constant for both members; l and l 3 are length of the members, α ¼ 60 , β ¼ 30 . Find eigenvalues and modal matrix and present the mode shapes.
a
α
E,A,l
b
β
S1 = 3 2
l 3
m
First state
Second state
S2 = 1 2
c
P2=1
P1=1
y1
Second mode
First mode
1.0 3
S2 = − 3 2
S1 = 1 2
y2
1.0 1
3
Fig. 16.8 Free vibration analysis of rod structure with two degrees of freedom. (a) Design diagram of the structure; (b) Unit states; (c) Mode shapes vibrations
Solution The structure has two degrees of freedom. The first and second unit states and corresponding internal forces S for each member are shown in Fig. 16.8b. Equations (16.4a) for unknown amplitudes mδ11 ω2 1 A1 þ mδ12 ω2 A2 ¼ 0 ðbÞ mδ21 ω2 A1 þ mδ22 ω2 1 A2 ¼ 0 Unit displacements pffiffiffi pffiffiffi XðS S pffiffiffi
3 3 1 1 1 pffiffiffi l 1 1 lþ l 3 ¼ 3þ 3 ds ¼ 2 EA 2 2 2 4EA EA pffiffiffi pffiffiffi XðS S p ffiffi ffi pffiffiffi
3 3 1 1 1 l 2 2 l 3 ¼ lþ 1þ3 3 ¼ ds ¼ 2 2 EA 2 2 4EA EA pffiffiffi pffiffiffi
X ð S1 S2 p ffiffi ffi 3 1 3 1 1 l pffiffiffi l l 3 ¼ 33 ds ¼ ¼ δ21 ¼ EA 2 2 2 2 4EA EA
δ11 ¼ δ22 δ12 Let us denote λ ¼
1 l , then ,δ ¼ mδ0 ω2 0 4EA pffiffiffi
δ11 ¼ δ0 3 þ 3 ,
pffiffiffi
δ22 ¼ δ0 1 þ 3 3 ,
δ12 ¼ δ21 ¼ δ0
pffiffiffi
33
16.2
Systems with Finite Number of Degrees of Freedom: Force Method
and equation (b) becomes
pffiffiffi
pffiffiffi 3 þ 3 λ A1 þ 3 3 A2 ¼ 0
pffiffiffi
pffiffiffi 3 3 A1 þ 1 þ 3 3 λ A2 ¼ 0
575
or
ð4:7320 λÞA1 1:2679A2 ¼ 0 1:2679A1 þ ð6:1961 λÞA2 ¼ 0
Frequency equation D¼
4:7320 λ 1:2679 ¼0 1:2679 6:1961 λ
Roots (in decreasing order) of frequency equation and corresponding (in decreasing order) eigenvalues are 1 4EA EA ¼ 0:5774 ¼ λ1 mδ0 6:9280 ml ml 1 EA EA 2 ffi ¼ 0:9999 λ2 ¼ 4:0002 ! ω2 ¼ λ2 mδ0 ml ml
λ1 ¼ 6:9280 ! ω21 ¼
Mode shape of vibration may be determined on the basis of equation (b). For first mode (λ1 ¼ 6.9280) the ratio of amplitudes is pffiffiffi pffiffiffi A2 3 þ 3 λ1 4:7320 6:9280 ¼ ¼ pffiffiffi ¼ 1:73 ¼ 3 1:2679 A1 33 pffiffiffi pffiffiffi 33 A2 1:2679 pffiffiffi ¼ ¼ ¼ 3 A1 1 þ 3 3 λ1 6:1961 6:9280 pffiffiffi T Assume that A1 ¼ 1, so the first eigenvector φ becomes φ ¼ b φ11 φ21 cT ¼ 1 3 For second mode ( λ2 ¼ 4.0002) the ratio of amplitudes are A2 4:7320 4:0002 1 ¼ ¼ 0:577 ¼ pffiffiffi 1:2679 A1 3 A2 1:2679 1 ¼ 0:577 ¼ pffiffiffi ¼ A1 6:1961 4:0002 3 The modal matrix Φ is then defined as
Φ¼
1 pffiffiffi 3
1 pffiffiffi : 1= 3
Corresponding mode shapes of vibration are shown in Fig. 16.8c. Orthogonality condition is (Kiselev 1980) pffiffiffi m 0 1pffiffiffi ¼0 1 3 0 m 1= 3 Control 1.
1 1 þ ¼ m1 δ11 þ m2 δ22 : ω21 ω22 1 1 ml ml ml þ ¼ 2:7320 þ ¼ EA ω21 ω22 0:5774 EA 0:9999 EA m1 δ11 þ m2 δ22 ¼ m
A2 A2 m Control 2. ¼ 1 A1 ω1 A1 ω2 m2
pffiffiffi
pffiffiffi
l l ml 3þ 3 þm 1 þ 3 3 ¼ 2:7320 4 EA 4 EA EA
ðcÞ
576
16
A2 A1
A2 A1
ω1
ω2
pffiffiffi 1 ¼ 3 pffiffiffi ¼ 1, 3
Dynamics of Elastic Systems: Free Vibration
m1 ¼ 1 m2
Example 16.3 The beam in Fig. 16.9a carries three equal concentrated masses mi. The length of the beam is l ¼ 4a, and flexural stiffness beam EI. The mass of the beam is neglected. It is necessary to find eigenvalues and mode shape of vibrations.
a
EI
m1
m2
a
m3
a
a
a
P1=1
b M1 l 16
2l 16
3l 16
P2=1
M2
l 8
l 8 l 4
P3=1
M3
l 16
2l 16
3l 16
c l1 = 31.5563 w1 = 4.9333
m1
m2
m3
EI ml 3
y11=1.0
y21=1.4142
d
y31=1.0
y32=-1.0
l2 = 2.0
m1
EI w 2 = 19.5959 ml 3
m2
m3
y22=0.0 y12=1.0 y23= -1.4142
e l3 = 0.44365
m1
m2
m3
EI w 3 = 41.6064 ml 3 y13=1.0
y33=1.0
Fig. 16.9 Free vibration analysis of a beam with three degrees of freedom. (a) Design diagram of the beam; (b) Unit bending moment diagrams; (c) Mode shape vibration which corresponds to fundamental (lowest) frequency; (d,e) Second and third modes of vibration
16.2
Systems with Finite Number of Degrees of Freedom: Force Method
577
Solution The beam has three degrees of freedom. The bending moment diagrams caused by unit inertial forces are shown in Fig. 16.9b. Multiplication of corresponding bending moment diagrams leads to the following results for unit displacements ð M1M1 1 1 l 3l 2 3l 1 3l 3l 2 3l 9 l3 þ ¼ δ11 ¼ dx ¼ , EI 2 4 16 3 16 2 4 16 3 16 768 EI EI ð ð M2M2 16 l3 M3M3 9 l3 dx ¼ , δ33 ¼ dx ¼ , δ22 ¼ 768 EI 768 EI EI EI ð ð M1M2 11 l3 M1M3 7 l3 δ12 ¼ δ21 ¼ dx ¼ , δ13 ¼ δ31 ¼ dx ¼ , 768 EI 768 EI EI EI ð M1M3 7 l3 δ13 ¼ δ31 ¼ dx ¼ , 768 EI EI 11 l3 δ23 ¼ δ32 ¼ δ12 ¼ δ21 ¼ 768 EI Let δ0 ¼
l3 . Matrix of unit displacements F (flexibility matrix) is 768EI 2 3 2 3 9 11 7 δ11 δ12 δ13 6 7 6 7 F ¼ ½δik ¼ 4 δ21 δ22 δ23 5 ¼ δ0 4 11 16 11 5 ¼ δ0 F0 δ31
δ32
δ33
7
11
9
Equations (16.4) with unknown amplitudes Ai of mass mi are m1 δ11 ω2 1 A1 þ m2 δ12 ω2 A2 þ m3 δ13 ω2 A3 ¼ 0 m1 δ21 ω2 A1 þ m2 δ22 ω2 1 A2 þ m3 δ23 ω2 A3 ¼ 0 m1 δ31 ω2 A1 þ m2 δ32 ω2 A2 þ m3 δ33 ω2 1 A3 ¼ 0 In our case all masses mi ¼ m. Divide by mδ0ω2 and denote λ ¼
ðaÞ
1 . Equations for amplitudes Ai mδ0 ω2
ð9 λÞA1 þ 11A2 þ 7A3 ¼ 0, 11A1 þ ð16 λÞA2 þ 11A3 ¼ 0,
ðbÞ
7A1 þ 11A2 þ ð9 λÞA3 ¼ 0 Frequency equation becomes 2
9λ 11 6 D ¼ det4 11 16 λ 7
11
3 7 7 11 5 ¼ 0 9λ
Eigenvalues in descending order λ1 ¼ 31:5563, λ2 ¼ 2:0,
ðcÞ
λ3 ¼ 0:44365 Verification: 1. The sum of the eigenvalues is λ1 + λ2 + λ3 ¼ 31.5563 + 2.0 + 0.44365 ¼ 34; on the other hand, the trace of the matrix F0 (i.e., the sum of diagonal members) is Tr(F0) ¼ 9 + 16 + 9 ¼ 34. 2. The multiplication of the eigenvalues is λ1 λ2 λ3 ¼ 31.5563 2.0 0.44365 ¼ 28; on the other hand, detF0 ¼ 28. (Note that determinant of unit displacement matrix is strictly positive, i.e., det F > 0).
578
16
Dynamics of Elastic Systems: Free Vibration
Frequencies of the free vibration in increasing order rffiffiffiffiffiffiffi 1 768 EI EI EI , ¼ ¼ ¼ 24:337 3 ! ω1 ¼ 4:9333 λ1 mδ0 31:5563 ml3 ml ml3 rffiffiffiffiffiffiffi 1 768 EI EI EI ω22 ¼ , ¼ ¼ 384 3 ! ω2 ¼ 19:5959 λ2 mδ0 2:0 ml3 ml ml3 rffiffiffiffiffiffiffi 1 768 EI EI EI 2 ω3 ¼ ¼ ¼ 1731:09 3 ! ω3 ¼ 41:6064 λ3 mδ0 0:44365 ml3 ml ml3
ω21
ðdÞ
For each ith eigenvalue the set of equation (b) for calculation of amplitudes is ð9 λi ÞA1 þ 11A2 þ 7A3 ¼ 0, 11A1 þ ð16 λi ÞA2 þ 11A3 ¼ 0,
ðeÞ
7A1 þ 11A2 þ ð9 λi ÞA3 ¼ 0 Equations (e) are divided by A1. Let ρ2 ¼
A2 , A1
ρ3 ¼
A3 . A1
Equations for modes become ð9 λi Þ þ 11ρ2i þ 7ρ3i ¼ 0, 11 þ ð16 λi Þρ2i þ 11ρ3i ¼ 0, 7 þ 11ρ2i þ ð9 λi Þρ3i ¼ 0
ðfÞ
Assuming A1 ¼ 1 we can calculate ρ2 and ρ3 for each calculated eigenvalue. For their calculation we can consider set of any two equations. 1. Eigenvalue λ1 ¼ 31.5563 ð9 31:5563Þ þ 11ρ2 þ 7ρ3 ¼ 0 11 þ ð16 31:5563Þρ2 þ 11ρ3 ¼ 0 Solution of p these ffiffiffi equations is ρ2 ¼ 1.4142, ρ3 ¼ 1.0. Therefore the first (principal) mode is defined as y11, y21 ¼ 2 y11 , y31 ¼ y11 . If we assume that y11 ¼ 1, then the eigenvector φ1 which corresponds to the frequency T pffiffiffi ω1 is φ1 ¼ 1:0 2 1 . Corresponding mode shape of vibration is shown in Fig. 16.9c. 2. Eigenvalue λ2 ¼ 2.0. In this case ð9 2:0Þ þ 11ρ2 þ 7ρ3 ¼ 0 11 þ ð16 2:0Þρ2 þ 11ρ3 ¼ 0 Solution of these equations is ρ2 ¼ 0.0, ρ3 ¼ 1.0, and second eigenvector becomes φ2 ¼ b 1:0 shape of vibration is shown in Fig. 16.9d.
0:0
1 cT ; this mode
3. Eigenvalue λ3 ¼ 0.44365. In this case ð9 0:44365Þ þ 11ρ2 þ 7ρ3 ¼ 0 11 þ ð16 0:44365Þρ2 þ 11ρ3 ¼ 0 Solution of these equations is ρ2 ¼ 1.4142, vibration is shown in Fig. 16.9e.
ρ3 ¼ 1.0, and eigenvector φ3 ¼ 1:0
T pffiffiffi 2 1 . Third mode shape of
We can see that the number of the nodal points of the mode shape of vibration is one less than the number of the mode. The modal matrix is defined as
16.3
Systems with Finite Number of Degrees of Freedom: Displacement Method
2
Φ ¼ bφ 1
φ2
579
φ11 6 φ3 c ¼ 4 φ21
φ12 φ22
3 2 φ13 1 7 6 pffiffiffi φ23 5 ¼ 4 2
3 1 1 pffiffiffi 7 0:0 2 5,
φ31
φ32
φ33
1
1
ðgÞ
1
where the ith and kth indexes at φik mean the number of mass and number of frequency (or form), respectively. The vectors φ1 and φ2 are orthogonal. Indeed
φ1T φ2 ¼ 1:0
pffiffiffi 2
2 3 pffiffiffi 1 7 1 6 6 0:0 7 ¼ 1 1 þ 2 0:0 þ 1 ð1Þ ¼ 0 6 1 7
Similarly, it is easy to check the orthogonality conditions for other form of vibration: φT1 φ3 ¼ φT2 φ3 ¼ 0. Above procedure may be applied for analysis of free vibration of the arbitrary linear deformable structures. The detailed examples of free vibration of different types of uniform arches (three-hinged, hingeless, circular, parabolic, etc.) are presented in book (Karnovsky 2012).
16.3
Systems with Finite Number of Degrees of Freedom: Displacement Method
Now we will consider a dynamical analysis of the structures with finite number of degrees of freedom using the concept of unit reactions. For some types of structures, the displacement method is more preferable than the force method.
16.3.1 Differential Equations of Free Vibration in Reactions In essence, this method consists in expressing the forces of inertia as a function of unit reactions. According to the displacement method, we need to introduce additional constraints which prevent each displacement of the mass. Thus, the total number of introduced constraints equals to the number of degrees of freedom. Let us consider a structure with concentrated masses mi, i ¼ 1, . . ., n (Fig. 16.10a). This structure has n degrees of freedom. They are lateral displacements of the frame at the points where each mass is located. Primary system of the displacement method is shown in Fig. 16.10b. In the first unit state we show displacement Z1 ¼ 1 and corresponding reactions ri1, similarly for other states (Fig.16.10c). m1
a
m2
m1
b
1
m2 2
n
mn mn
c
m1
n-th unit state
Second unit state
First unit state
m1
m2
m1
m2
m2 Zn=1
mn
Z1=1 r11
r21
rn1
1 r12 r22
rn2 mn
mn r1n
r2n
Fig. 16.10 (a) Design diagram of the frame; (b) Primary system; (c) Unit states. Elastic curves are shown by dotted line
rnn
580
16
Dynamics of Elastic Systems: Free Vibration
Displacement of each lumped mass (or displacement of each introduced constraints) is yi. Assume that the positive displacements and accelerations of mass as well as the reactions of introduced constraints are directed downward (or right in •• case of mass located on vertical element). The inertial forces of each mass, mi yi , may be presented in terms of unit reactions rik as follows ••
m1 y1 ¼ r 11 y1 þ r 12 y2 þ . . . þ r 1n yn : : : : : : : : : : : : •• mn yn ¼ r n1 y1 þ r n2 y2 þ . . . þ r nn yn The condition of the absence of a reaction in kth introduced constraint has the form ••
m1 y1 þr 11 y1 þ r 12 y2 þ . . . þ r 1n yn ¼ 0 : : : : : : : : : : : : •• mn yn þr n1 y1 þ r n2 y2 þ . . . þ r nn yn ¼ 0
ð16:9Þ
The system (16.9) represents the canonical equations for determining displacements of mass m1, . . ., mn ; each equation describes the equilibrium condition for corresponding mass. The coefficient rik presents the reaction in ith introduced constraint caused by unit displacement of kth introduced constraint. The term rikyk means reaction in ith introduced constraint caused by real displacement of kth introduced constraint. This set of differential equations in the form of displacement method describes the undamped free vibration of the multidegree of freedom structure. These equations are coupled statically, because the generalized coordinates appear in each equation.
16.3.2 Frequency Equation Solution of system (16.9) y1 ¼ A1 sin ðω t þ φ0 Þ,
y2 ¼ A2 sin ðω t þ φ0 Þ,
...
y3 ¼ A3 sin ðω t þ φ0 Þ,
ð16:10Þ
where Ai is amplitude of the displacement of mass mi; φ0 is the initial phase of vibration. Substituting (16.10) into (16.9) leads to algebraic homogeneous equations with respect to unknown amplitudes of lumped masses r 11 m1 ω2 A1 þ r 12 A2 þ . . . þ r 1n An ¼ 0 r 21 A1 þ r 22 m2 ω2 A2 þ . . . þ r 2n An ¼ 0 ð16:11Þ : : : : : : : : : : r n1 A1 þ r n2 A2 þ r nn mn ω2 An ¼ 0 Nontrivial solution (nonzero amplitudes Ai) is possible if the determinant of the coefficients to amplitude of (16.11) is zero 2 3 r 12 ... r 1n r 11 m1 ω2 6 7 r 21 r 22 m2 ω2 . . . r 2n 6 7 ð16:12Þ D ¼ det6 7¼0 4 5 ... ... ... ... r n1
r n2
...
r nn mn ω2
This equation is called the frequency equation in the form of the displacement method. Solution of this equation presents the eigenvalues of a structure. The number of the frequencies of free vibration is equal to the number of degrees of freedom.
16.3
Systems with Finite Number of Degrees of Freedom: Displacement Method
581
16.3.3 Mode Shape of Vibrations and Modal Matrix Equations (16.11) are homogeneous algebraic equations with respect to unknown amplitudes Ai of each mass. This system does not allow us to find these amplitudes. However, we can find the ratios between different amplitudes. If a structure has two degrees of freedom, then system (16.11) becomes r 11 m1 ω2 A1 þ r 12 A2 ¼ 0 r 21 A1 þ r 22 m2 ω2 A2 ¼ 0 From these equations we can find the following ratios A2 r m 1 ω2 ¼ 11 or A1 r 12 A2 r 21 ¼ A1 r 22 m2 ω2
ð16:13Þ
If we assume A1 ¼ 1, then entries b 1 A2 cT define for each eigenfrequency the corresponding column φ of the modal matrix Ф. The formulas obtained from displacement and force methods (16.13 and 16.6) lead to the same result. Let us show application of the displacement method for free vibration analysis of a beam with three equal lumped masses (Fig. 16.11a); previously this structure had been analyzed by force method (Example 16.3). It is necessary to find eigenvalues and modal matrix. The introduced constraints 1, 2, 3 which prevent displacement yi are shown in Fig. 16.11a. For calculation of unit reactions, we need to construct bending moment diagram due to unit displacements of each introduced constraint. For first state bending moment diagram M 1 is presented in Fig. 16.11b. It is constructed using Table A.14, line 3 and column B. Elastic curve is shown by dotted line, bending moment diagram is shown on the extended fibers. Now we need to show free-body diagram for each introduced constraint. The sections are passes infinitely close to the left and right of introduced constraints. It means that each span consists of three portions—two portions which adjoin to the left and right supports and intermediate portion. Next we need to show bending moments which act on the intermediate portion. For direction of these moment we need to take into account location of the extended fibers. These bending moments may be equilibrated by two shear forces which act at the end points of the intermediate portion. Finally, these forces should be transmitted on the portion of a beam which adjoin support with opposite sign. According to bending moment diagram M 1 for first span shear force is Q1 ¼ 3:6429
EI 1 EI ¼ 3:6429 3 a2 a a
Considering second span in similar manner, shear force Q2 is found as Q2 ¼ ½3:6429 ð2:5714Þ
EI 1 EI ¼ 6:2143 3 a2 a a
Equilibrium equation for first introduced constraint leads to the expression for unit reaction r11 (Fig.16.11b) r 11 ¼ ð3:6429 þ 6:2143Þ
EI EI ¼ 9:8572 3 a3 a
Considering the second state we need to construct the bending moment diagram M 2 (Fig.16.11d) and on its base calculate r12, r22, r32. Finally, from equilibrium condition for joint 1 (second state, Fig. 16.11d), we get EI EI r 12 ¼ ð2:5714 þ 6:8571Þ 3 ¼ 9:4285 3 . Since a structure is symmetrical, then bending moment diagram caused by a a unit displacement constraint 3 is not shown. All shear and unit reactions have multiplier EI/a3. Omitting the intermediate transformations, the final stiffness matrix the elements of which are expressed through the span l ¼ 4a becomes
582
16
2
9:4285 13:7142
9:8572 64EI 6 S ¼ 3 4 9:4285 l 3:8572
a
9:4285
3 3:8572 7 9:4285 5
EI
m2
y1
a
ðaÞ
9:8572
m1
EI
Dynamics of Elastic Systems: Free Vibration
y2
a
m3 y3
a
m1
m2
m3
1
2
3
a
b c
3.6429 6.2143 M1 M1 M2 1
1 1
2
r11
2
3
r21
r31
2.5714
M1 Q1
Q2
0.6429 3.6429
r11=9.8572 M1 = 3.6429
M2 = 2.5714 2.5714
d 6.8571
1
1
3 2
r12
r32
r22 2.5714
2.5714
1
M2 r12=-9.4285 4.2857
Fig. 16.11 (a) Design diagram of the beam and primary system; (b) First unit state and corresponding bending moment diagram (factor EI/a2); (c) Calculation of r11 (factor EI/a3); (d) Second unit state, corresponding bending moment diagram and calculation of r12
Since all masses are equal, then Eq. (16.12) may be rewritten as 2 3 9:8572 λ 9:4285 3:8572 6 7 4 9:4285 13:7142 λ 9:4285 5 ¼ 0, 3:8572 9:4285 9:8572 λ
ðbÞ
where parameter λ ¼ mω2. Note that expressions for eigenvalue λ for displacement and force methods ðλ ¼ 1=mδ0 ω2 Þ are different. The eigenvalues present the roots of equation (b); in increasing order they are
16.3
Systems with Finite Number of Degrees of Freedom: Displacement Method
λ1 ¼ 0:3804
EI , a3
λ2 ¼ 6:0
EI , a3
583
λ3 ¼ 27:0482
EI a3
ðcÞ
Now we can calculate the frequencies of vibration which correspond to eigenvalues. EI EI EI ¼ 0:3804 43 3 ¼ 24:345 3 , ma3 ml ml EI EI EI EI 2 2 λ2 ¼ 6:0 3 ¼ mω2 ! ω2 ¼ 6:0 3 ¼ 6:0 64 3 ¼ 384 3 , a ma ml ml EI EI EI EI 2 2 λ3 ¼ 27:0482 3 ¼ mω3 ! ω3 ¼ 27:0482 3 ¼ 27:0482 64 3 ¼ 1731:08 3 a ma ml ml λ1 ¼ 0:3804
EI ¼ mω21 ! a3
ω21 ¼ 0:3804
ðdÞ
Same frequencies have been obtained by the force method. For each i-th eigenvalue the set of equation for calculation of amplitudes is ð9:8572 λi ÞA1 9:4285A2 þ 3:8572A3 ¼ 0 9:4285A1 þ ð13:7142 λi ÞA2 9:4285A3 ¼ 0 3:8572A1 9:4285A2 þ ð9:8572 λi ÞA3 ¼ 0 Equations (e) divide by A1. Let ρ2 ¼ A2 =A1 ,
ðeÞ
ρ3 ¼ A3 =A1 . Equations for modes become
ð9:8572 λi Þ 9:4285ρ2i þ 3:8572ρ3i ¼ 0 9:4285 þ ð13:7142 λi Þρ2i 9:4285ρ3i ¼ 0
ðfÞ
3:8572 9:4285ρ2i þ ð9:8572 λi Þρ3i ¼ 0 Eigenvalue λ1 ¼ 0.3804 ð9:8572 0:3804Þ 9:4285ρ2 þ 3:8572ρ3 ¼ 0 9:4285 þ ð13:7142 0:3804Þρ2 9:4285ρ3 ¼ 0 3:8572 9:4285ρ2 þ ð9:8572 0:3804Þρ3 ¼ 0 To determine two coefficients ρ2 and ρ3, which define amplitudes distribution, it is enough to consider any two equations. Solution of the first and second equations: ρ2 ¼ 1.4142, ρ3 ¼ 1.0. Verification. With the found values ρ2 and ρ3, the third equation becomes 3:8572 9:4285 1:4142 þ 9:4768 1:0 ¼ 13:3340 13:3337 ffi 0 The same procedure should be repeated for λ2 ¼ 6.0 and λ3 ¼ 27.0482 The modal matrix Φ is defined as 2 3 1 1 1 6 7 Φ ¼ 4 1:4142 0:0 1:4142 5 1 1 1
ðgÞ
Same mode shape coefficients have been obtained by force method. Example 16.4 Design diagram of multistore frame is presented in Fig. 16.12a. The crossbars are absolutely rigid bodies; their masses are shown in design diagram. Flexural stiffness of the vertical members are EI and masses of the struts are ignored. Calculate the frequencies of vibrations and find the corresponding mode shapes. Solution The primary system is shown in Fig. 16.12b. For computation of unit reactions, we need to construct the bending moment diagrams due to unit displacements of the introduced constraints, and then for calculation of unit reactions consider the equilibrium condition for each crossbar. Bending moment diagram caused by unit displacement of the constraint 1 is shown in Fig. 16.12c. Elastic curve is shown by dotted line. Since crossbars are absolutely rigid members, then joints cannot be rotated and each vertical member should be considered as fixed-fixed member. In this case, specified ordinates are 6i/h (Table A.4, Case 2). Now we need to show free-
584
16
Dynamics of Elastic Systems: Free Vibration
body diagram for each horizontal member using the closed sections (shown by dotted lines in Fig. 16.12b). Bending moments at cut sections are 6i/h. Both moments may be equilibrated by two forces 12i/h2. These forces should be transmitted on both crossbars. Positive unit reactions r11, r21, and r31 are shown by dotted arrows.
a
b
m 1= m
d
1
1.00
EI = h
EI m2= 2m 2
EI = h
0.866
EI m3= 2m
0.5
3
EI = h
EI
Z1=1
c
r11
1
6i h
12i h 2
6i h r21
6i h
r11
1
12i h 2
r21
6i h 2
2 r31
r31
3
3
M1 Fig. 16.12 (a) Design diagram of the frame; (b) Primary system; (c) Bending moment diagram caused by unit displacement of the constraint 1 and calculation of unit reactions r11, r21, and r31; (d) First (principal) mode of vibration
Equilibrium condition for each crossbar leads to the following unit reactions r 11 ¼ 2
12i i ¼ 24 2 , h2 h
r 21 ¼ 24
i , h2
r 31 ¼ 0,
i¼
EI h
Similarly, considering the second and third unit displacements, we get r 12 ¼ 24
i , h2
r 13 ¼ 0,
r 23 ¼ 24
r 22 ¼ 48 i , h2
i , h2
r 32 ¼ 24
r 33 ¼ 48
i , h2
i h2
Let r 0 ¼ 24i=h2 . Equations (16.11) become
r 0 mω2 A1 r 0 A2 þ 0 A3 ¼ 0 r 0 A1 þ 2 r 0 mω2 A2 r 0 A3 ¼ 0 0 A1 r 0 A2 þ 2 r 0 mω2 A3 ¼ 0
The frequency equation is
ðaÞ
16.3
Systems with Finite Number of Degrees of Freedom: Displacement Method
2
r 0 mω2 6 D ¼ det4 r 0
r 0 2r 0 2mω2
0 r 0
r 0
2r 0 2mω2
0 If eigenvalue is denoted as λ ¼
585
3 7 5¼0
mω2 mω2 h2 ¼ , then the system (a) may be rewritten as r0 24i ð1 λÞA1 A2 ¼ 0 A1 þ 2ð1 λÞA2 A3 ¼ 0
ðbÞ
0 A2 þ 2ð1 λÞA3 ¼ 0 Eigenvalue equation is 2
1λ
6 D ¼ det6 4 1 0
3
1
0
2ð 1 λ Þ
1
1
2ð 1 λ Þ
4ð1 λÞ3 3ð1 λÞ ¼ 0
or
7 7¼0 5
h i ð 1 λ Þ 4ð 1 λ Þ 2 1 ¼ 0
The eigenvalues in increasing order are pffiffiffi 3 λ1 ¼ 1 , 2
λ2 ¼ 1,
pffiffiffi 3 λ3 ¼ 1 þ 2
Corresponding frequencies of free vibration (eigenfrequencies) pffiffiffi 3 EI EI ω21 ¼ 24 1 , ω22 ¼ 24 3 , 2 mh3 mh
pffiffiffi 3 EI ω23 ¼ 24 1 þ 2 mh3
Mode shapes of vibration. Now we need to consider the system (b) for each calculated eigenvalue. If we denote ρ2 ¼ A2/A1 and ρ3 ¼ A3/A1, then system (b) may be rewritten as ð 1 λ Þ ρ2 ¼ 0 1 þ 2ð1 λÞρ2 ρ3 ¼ 0 0 ρ2 þ 2ð1 λÞρ3 ¼ 0 This system should be solved with respect to ρ2 and ρ3 for each eigenvalue. pffiffiffi 3 First mode ω1 : λ1 ¼ 1 . Equations (c) become 2 ð 1 λ 1 Þ ρ2 ¼ 0 1 þ 2ð1 λ1 Þρ2 ρ3 ¼ 0 ρ2 þ 2ð1 λ1 Þρ3 ¼ 0 pffiffiffi Solution of the first and second equations are ρ2 ¼ 3=2, ρ3 ¼ 1=2. The last equation is satisfied. Indeed pffiffiffi pffiffiffi 3 3 1 þ2 1 1 ¼0 2 2 2 pffiffiffi 3 The same procedure should be repeated for λ2 ¼ 1.0 and λ3 ¼ 1 þ . 2 The modal matrix Φ is defined as
ðcÞ
586
16
2
1 6 pffiffiffi Φ ¼ 4 3=2
1 0
3 1 pffiffiffi 7 3=2 5
1
1=2
Dynamics of Elastic Systems: Free Vibration
1=2
First (principal) mode of vibration, which corresponds to smallest frequencyω1, is presented in Fig. 16.12d. Verification. Orthogonality conditions of the first and second modes of vibration should be written as follows: φT1 Mφ2 ¼ 0 where φT1 is a transposed first column of the matrix Φ, and M is the mass matrix. In our case we have j
φT1 Mφ2 ¼ 1
2
k
1
pffiffiffi 6 3=2 1=2 m4 0 0
0
0
3 2
1
3
j
7 7 6 05 6 0 7 ¼ 1 6 7 6 1 7 2
2 0
Similarly, it is easy to check that φT1 Mφ3 ¼ 0, and
pffiffiffi 3=2
2
3
pffiffiffi 3 1 6 7 0 2¼0 1=2 m6 0 7 ¼ 1 1 þ 2 2 6 7 6 2 7 k
1
φT2 Mφ3 ¼ 0
16.3.4 Comparison of the Force and Displacement Methods Some fundamental data about application of two fundamental methods for free vibration analysis of the structures with finite number of degrees of freedom is presented in Table 16.2. Table 16.2 Comparison of the force and displacement methods for free vibration analysis Force method (Analysis in terms of displacements) Coupled differential equations Canonical form
••
••
Solution Equations for unknowns amplitudes Ai 1. Canonical form 2. Matrix form
Frequency equation 1. Canonical form 2. Matrix form
••
••
δ11 m1 y1 þδ12 m2 y2 þ . . . þ δ1n mn yn þy1 ¼ 0
m1 y1 þr 11 y1 þ r 12 y2 þ . . . þ r 1n yn ¼ 0
:
:
:
:
:
: :
:
••
Type of coupling Matrices
Displacement method (Analysis in terms of reactions)
:
:
: :
••
:
: ••
: :
:
:
:
:
: :
:
:
:
••
δn1 m1 y1 þδn2 m2 y2 þ . . . þ δnn mn yn þyn ¼ 0 Dynamic Flexibility matrix 3 2 2 3 δ11 δ12 . . . δ1n y1 7 6 6y 7 6 δ21 δ22 . . . δ2n 7 6 27 7 6 F¼6 7, Y ¼ 6 7 6...7 6... ... ... ...7 6 7 5 4 6y 7 n δn1 δn2 . . . δnn
mn yn þr n1 y1 þ r n2 y2 þ . . . þ r nn yn ¼ 0 Static Stiffness matrix 3 2 2 3 r 11 r 12 . . . r 1n y1 7 6 6y 7 6 r 21 r 22 . . . r 2n 7 6 27 7 6 S¼6 7, Y ¼ 6 7 6...7 6... ... ... ...7 6 7 5 4 6y 7 n r n1 r n2 . . . r nn
M is diagonal mass matrix Y ¼ A sin (ωt + ϕ0) 2 m1 δ11 1=ω A1 þ m2 δ12 A2 þ . . . þ mn δ1n An ¼ 0
M is diagonal mass matrix Y ¼ A sin (ωt + ϕ0) 2 r 11 m1 ω A1 þ r 12 A2 þ . . . þ r 1n An ¼ 0 : : : : : : : : : : r n1 A1 þ r n2 A2 þ r nn mn ω2 An ¼ 0
:
: :
:
:
:
:
: : : : : : m1 δn1 A1 þ m2 δn2 A2 þ . . . þ mn δnn 1=ω2 An ¼ 0 h i 1 FM 2 I A ¼ 0, I is unit matrix ω m1 δ11 1=ω2 m2 δ12 ... mn δ1n 2 m δ m δ 1=ω . . . m δ 1 21 2 22 n 2n . . . . . . . . . . . . m1 δn1 m2 δn2 . . . mn δnn 1=ω2 ¼ 0 FM ð1=ω2 ÞI ¼ 0
[S ω2M] A ¼ 0 r 11 m1 ω2 r 21 ... r n1
r 12 r 22 m2 ω2 ... r n2
... r 2n ¼0 ... ... . . . r nn mn ω2 ...
S ω2 M ¼ 0
r 1n
16.4
Structures with Infinite Number of Degrees of Freedom
587
Generally, for nonsymmetrical beams, the force method is more effective than the displacement method. However, for frames especially with absolutely rigid crossbar, the displacement method is beyond the competition.
16.4
Structures with Infinite Number of Degrees of Freedom
The more precise dynamical analysis of engineering structure is based on assumption that a structure has distributed masses. In this case, the structure has infinite number of degrees of freedom and mathematical model presents a partial differential equation. This section considers the simplest mathematical model of a plane vibration of uniform beam taking into account the bending moments only.
16.4.1 Differential Equation of Transversal Vibration of the Beam Bernoulli-Euler theory presents the simplest mathematical model of transversal vibration of the beam (classical theory) based on the following assumptions. 1. Bernoulli’s plane section hypothesis: cross sections, which are plane and perpendicular to the axis of the nondeformed beam, remain plain and perpendicular to the elastic curve of the deformed beam. It means that only axial stresses are considered while shear deformation is not taken into account. 2. The beam is inextensible, displacements and deformations are small. 3. The dimensions of the beam section are small compared with the length of the beam or radius of curvature of the beam axis (a case of the long beams). 4. The behavior of the material is subject to the Hooke’s law. Differential equation of the uniform beam is EI
d4 y ¼ q, dx4
ð16:14Þ
where y is the transverse displacement of a beam, E is the modulus of elasticity, I is the moment of inertia of the cross section about the neutral axis, and q is the transverse load per unit length of the beam. In case of free vibration, the load per unit length is q ¼ ρA
d2 y , dt 2
ð16:15Þ
where ρ is the mass density and A is the cross-sectional area. Equations (16.14) and (16.15) lead to the following differential equation of the transverse vibration of the uniform Bernoulli-Euler beam 4
EI
2
∂ y ∂ y þ ρA 2 ¼ 0 ∂x4 ∂t
ð16:16Þ
Thus, the transverse displacement of a beam depends on the axial coordinate x and time t, i.e., y ¼ y(x, t).
16.4.1.1
Boundary Conditions
The behavior of the system substantially depends on the type of support constraints. Boundary conditions determine the parameters characterizing its state at the boundary of the element (beam, rod). Boundary conditions have geometric, forces or combined character. In the case of plane bending, the boundary conditions of the first group include the transverse displacement of the rod and the angle of inclination of the tangent to the elastic line of the rod (slope), while the bending moment and the shear force describe the boundary conditions of the second group. The classical boundary conditions for the transversal vibration of a beam are presented in Table 16.3.
588
16
Dynamics of Elastic Systems: Free Vibration
Table 16.3 Classical boundary conditions for plane bending: displacements y and θ are transversal deflection and slope; M and Q are bending moment and shear force
Free end (Q=0, M=0)
Clamped end (y=0, θ=0)
x
x y = 0;
θ=
∂y =0 ∂x
Q = EI
Pinned end (y=0, M=0)
∂3 y ∂2 y = 0; M = EI =0 3 ∂x ∂ x2
Sliding end (Q=0, θ =0)
x y = 0; M = EI
x 3
∂2 y =0 ∂ x2
Q = EI
∂ y 0; θ ∂ y 0 = = 3 = ∂x ∂x
It can be seen that for the clamped end the boundary conditions are geometrical type, for the free end are the force type, while for the hinged end and for the sliding support the boundary conditions are mixed. In other cases, the boundary conditions are expressed in a more complicated way. For example, in the case of elastic fixing of the end of the rod, the boundary condition should take into account the nature of the possible displacements at the end of the rod and the resulting elastic restoring forces.
16.4.1.2
Initial Conditions
These conditions present the initial distribution of the displacement and the initial distribution of the velocities of each point of a beam at t ¼ 0 dy • ðx, 0Þ ¼ yðx, 0Þ ¼ υðxÞ dt
yðx, 0Þ ¼ uðxÞ;
16.4.2 Fourier Method A solution of differential equation (16.16) may be presented in the form yðx, t Þ ¼ Y ðxÞT ðt Þ
ð16:17Þ
where Y(x) is the space-dependent function (shape function, mode shape function, eigenfunction);T(t) is the time-dependent function. The shape function Y(x) and time-dependent function T(t) depend on the boundary conditions and initial conditions, respectively. Plugging the form (16.17) into Eq. (16.16), we get ••
EIY IV T þ ¼0 mY T
ð16:18Þ
••
It means that both terms are equal but have opposite signs. Let TT ¼ ω2 , where ω still remains to be undetermined; then for functions T(t) and Y(x) may be written the following differential equations ••
T þω2 T ¼ 0 Y ðxÞ k 4 Y ðxÞ ¼ 0, IV
ð16:19Þ ð16:20Þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where k ¼ 4 mω2 =EI and m ¼ ρA is mass per unit length of the beam, ρ is density of material, A is area of cross section of a beam. Thus, instead of one partial differential equation (16.16) containing two independent parameters (time t and coordinate
16.4
Structures with Infinite Number of Degrees of Freedom
589
х) we obtained two uncoupled ordinary differential equations with respect to unknown functions Y(x) and T(t). This procedure is called the separation of variables method. The solution of Eq. (16.19) is T(t) ¼ A1 sin ω t + B1 cos ω t, where ω is a frequency of vibration. This equation shows that displacements of vibrating beam obey to harmonic law; coefficients A1 and B1 should be determined from initial conditions. The general solution of equation (16.20) is Y ðxÞ ¼ A cosh kx þ B sinh kx þ C cos kx þ D sin kx where A, B, C, and D may be calculated using the boundary conditions. The natural frequency ω of a beam is defined by equation rffiffiffiffiffi rffiffiffiffiffi λ2 EI 2 EI ¼ 2 , where λ ¼ kl ω¼k m m l
ð16:21Þ
ð16:22Þ
To obtain frequency equation using general solution (16.21), the following algorithm is recommended: Step 1. Represent the mode shape in the general form (16.21), which contains four unknown constants. Step 2. Determine constants using the boundary condition at x ¼ 0 and x ¼ l. Thus, the system of four homogeneous algebraic equations is obtained. Step 3. The nontrivial solution of this system represents a frequency equation. Example 16.5 Calculate the frequencies of free vibration and find the corresponding mode shapes for simply supported beam. The beam has length l, mass per unit length m, modulus of elasticity E, and moment of inertia of cross-sectional area I. Solution The shape of vibration may be presented in form (16.21). For pinned-rolled beam displacement and bending moment at x ¼ 0 and at x ¼ l equal zero. Expression for bending moment (accurate to constant EI) is Y 00 ðxÞ ¼ k2 ðA cosh kx þ B sinh kx C cos kx D sin kxÞ Conditions Y(0) ¼ 0 and Y00 (0) ¼ 0 lead to the equations AþC ¼0 AC ¼0 Thus, A ¼ C ¼ 0. Conditions Y(l) ¼ 0 and Y00 (l) ¼ 0 lead to the equations B sinh kl þ D sin kl ¼ 0 B sinh kl D sin kl ¼ 0 Thus, B ¼ 0 and D sin kl ¼ 0. Nontrivial solution occurs, if sinkl ¼ 0. This is the frequency equation. Solution of this equation is kl ¼ π, 2π, . . . Thus, the frequencies of vibration of uniform simply supported beam are rffiffiffiffiffi EI ω¼k , m 2
3:14162 ω1 ¼ l2
rffiffiffiffiffi EI , m
6:28322 ω2 ¼ l2
iπ x, l
i ¼ 1, 2, 3, . . .
rffiffiffiffiffi EI m
The mode shape of vibration is Y i ðxÞ ¼ D sin ki x ¼ D sin
This means that in the case of vibration of a beam with a frequency ω1, the curved axis of the beam represents a sin function with one half-wave, and vibration with a second frequency ω2 corresponds to the form of vibration with two half-waves, and so on. Thus, the transition to higher frequencies leads to a violation of the third assumption, since the beam from the category of long and thin goes into the category of short and thick. Therefore, Bernoulli-Euler theory cannot be applied for determination of higher frequencies. Different theories for beams (Rayleigh, Bresse, Timoshenko, etc.) are discussed in (Karnovsky and Lebed 2001, 2004b).
590
16
Dynamics of Elastic Systems: Free Vibration
16.4.3 Krylov-Duncan Method A general solution of differential equation (16.20) may be presented in the form Y ðxÞ ¼ C 1 SðkxÞ þ C2 T ðkxÞ þ C 3 U ðkxÞ þ C 4 V ðkxÞ,
ð16:23Þ
where Y(x) is the general expression for mode shape; Ci are constant coefficients; S(kx), T(kx), U(kx), V(kx) are the KrylovDuncan functions (Krylov 1936; Duncan 1943). They represent the combination of trigonometric and hyperbolic functions. 1 SðkxÞ ¼ ð cosh kx þ cos kxÞ 2 1 T ðkxÞ ¼ ð sinh kx þ sin kxÞ 2 1 U ðkxÞ ¼ ð cosh kx cos kxÞ 2 1 V ðkxÞ ¼ ð sinh kx sin kxÞ 2
ð16:24Þ
Each combination (16.22) satisfies to equation of the free vibration of uniform Bernoulli-Euler beam.
16.4.3.1
Properties of Krylov-Duncan Functions (16.24)
1. Krylov-Duncan functions and their derivatives result in the unit matrix at x ¼ 0. 000
Sð0Þ ¼ 1 T ð 0Þ ¼ 0
S0 ð 0Þ ¼ 0 T 0 ð 0Þ ¼ 1
S00 ð0Þ ¼ 0 T 00 ð0Þ ¼ 0
U ð 0Þ ¼ 0 V ð 0Þ ¼ 0
U 0 ð 0Þ ¼ 0 V 0 ð 0Þ ¼ 0
U 00 ð0Þ ¼ 1 U ð0Þ ¼ 0 000 V 00 ð0Þ ¼ 0 V ð0Þ ¼ 1
S ð 0Þ ¼ 0 000 T ð 0Þ ¼ 0
ð16:25Þ
000
2. Krylov-Duncan functions and their derivatives satisfy to circular permutations (Fig. 16.13, Table 16.4).
S(x) • •T(x)
V(x)• • U(x) Fig. 16.13 Krylov-Duncan functions circular permutation
3. Some relations of Krylov-Duncan functions Table 16.4 Krylov-Duncan functions and their derivatives
Function
First derivative
Second derivative
Third derivative
Fourth derivative
S(x)
kV(x)
k2 U(x)
k3 T(x)
k4 S(x)
T(x)
kS(x)
k2 V(x)
k3 U(x)
k4 T(x)
U(x)
kT(x)
k2 S(x)
k3 V(x)
k4 U(x)
V(x)
kU(x)
k2 T(x)
k3 S(x)
k4 V(x)
16.4
Structures with Infinite Number of Degrees of Freedom
591
Different combination of Krylov-Duncan functions may be presented in terms on trigonometric and hyperbolic functions (Karnovsky and Lebed 2001, 2004a). 1 ST UV ¼ ð cosh kx sin kx þ sinh kx cos kxÞ 2 1 TU SV ¼ ð cosh kx sin kx sinh kx cos kxÞ 2 S2 U 2 ¼ cosh kx cos kx; T 2 V 2 ¼ 2 SU V 2 ¼ sinh kx sin kx 1 1 U 2 TV ¼ ð1 cosh kx cos kxÞ; S2 TV ¼ ð1 þ cosh kx cos kxÞ 2 2 1 2 2 T SU ¼ SU V ¼ sinh kx sin kx; 2SU ¼ T 2 þ V 2 2
ð16:26Þ
The properties of the Krylov-Duncan functions (16.25) may be effectively used for deriving of the frequency equation and determining of the mode shape of free vibration for uniform beams. The relationships (16.26) will be used for determining the mode vibration and free vibration analysis for more complicated design diagrams. To obtain frequency equation using Krylov-Duncan functions, the following algorithm is recommended: Step 1. Represent the mode shape in the form that satisfies boundary conditions at x ¼ 0. This expression will have only two Krylov-Duncan functions and, respectively, two constants. The decision of what Krylov-Duncan functions to use is based on (16.25) and the boundary condition at x ¼ 0. Step 2. Determine constants using the boundary condition at x ¼ l and Table 16.4. Thus, the system of two homogeneous algebraic equations is obtained. Step 3. The nontrivial solution of this system represents a frequency equation. Example 16.6 The simply supported beam of length l has mass per unit length m, modulus of elasticity E, and moment of inertia of cross-sectional area I (Fig. 16.14). Calculate the frequency of vibration and find the mode shapes of vibration. m, EI
x
l i=1
ω1
i=2
ω2
Fig. 16.14 Design diagram of the beam and first (i ¼ 1) and second (i ¼ 2) mode shapes vibration
Solution At the left end (x ¼ 0) deflection and the bending moment are zero: 1: Y ð0Þ ¼ 0; 2: Y 00 ð0Þ ¼ 0 It means at x ¼ 0 the Krylov-Duncan functions and their second derivatives equal zero. According to properties (16.25), only T (kx) and V (kx) functions satisfy these conditions. So, the expression for the mode shape is Y ðxÞ ¼ C 2 T ðkxÞ þ C 4 V ðkxÞ 00
Constants C2 and C4 are calculated from boundary conditions at x ¼ l: Y(l) ¼ 0; Y (l ) ¼ 0 Y ðlÞ ¼ C 2 T ðklÞ þ C 4 V ðklÞ ¼ 0 Y 00 ðlÞ ¼ k2 ½C 2 V ðklÞ þ C 4 T ðklÞ ¼ 0 The second equation (a) is compiled on the basis of Table 16.4.
ðaÞ
592
16
Dynamics of Elastic Systems: Free Vibration
A nontrivial solution of the above system is the frequency equation T ðklÞ V ðklÞ
V ðklÞ ¼ 0 ! T 2 ðklÞ V 2 ðklÞ ¼ 0 T ðklÞ
According to (16.26), the last formula may be presented as sinhkl sin kl ¼ 0. Since sinhkl 6¼ 0, then the frequency equation is sinkl ¼ 0. The roots of this equation are λ ¼ kl ¼ π, 2π, . . . Parameter rffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffi 4 mω2 4 mω2 mω2 k¼ ! λ ¼ kl ¼ l ! λ4 ¼ l4 , EI EI EI so the frequencies of the free vibration are λ2 ωi ¼ 2i l
rffiffiffiffiffi EI , m
π2 ω1 ¼ 2 l
rffiffiffiffiffi EI , m
4π 2 ω2 ¼ 2 l
rffiffiffiffiffi EI , ... m
The mode shape of vibration is C4 Y ðxÞ ¼ C2 T ðkxÞ þ C 4 V ðkxÞ ¼ C2 T ðki xÞ þ V ðki xÞ C2 Since the ratio C4/C2 from first and second equations (a) are T ðk i l Þ V ðki lÞ C4 ¼ ¼ , C2 V ðk i lÞ T ðk i l Þ then ith mode shape (eigenfunction) corresponding to ith frequency of vibration (eigenvalue) is
T ðk i lÞ V ðk i l Þ Y ðxÞ ¼ C2 T ðkxÞ þ C 4 V ðkxÞ ¼ C2 T ðki xÞ V ð k i xÞ ¼ C 2 T ð k i xÞ V ð k i xÞ V ðk i l Þ T ðk i lÞ
ðbÞ
Since the Krylov-Duncan functions T(π) ¼ V(π), T(2π) ¼ V(2π),... so the mode shapes according to (16.24) are X i ðxÞ ¼ C½T ðk i xÞ V ðk i xÞ ¼ C sin k i x ¼ C sin
iπ x, l
i ¼ 1, 2, . . .
ðcÞ
The first and second modes are shown in Fig. 16.14. Table of the numerical values of the Krylov-Duncan functions S, T, U, V in terms of parameter kx (0 kx 10.0, Δkx ¼ 0.1) may be found in (Karnovsky and Lebed 2001, 2004a]. Example 16.7 Calculate the frequency of vibration and find the mode of vibration for uniform clamped free beam (Fig 16.15). m, EI l
i=2
x
i=1
Fig. 16.15 Design diagram for clamped-free beam and mode shapes vibration for i ¼ 1 and i ¼ 2
16.4
Structures with Infinite Number of Degrees of Freedom
593
Solution At the left end (x ¼ 0) deflection and slope are zero: 1: Y ð0Þ ¼ 0; 2: Y 0 ð0Þ ¼ 0 It means, at x ¼ 0 the Krylov-Duncan functions and their first derivative equal zero. According to properties (16.25), only U (kx) and V (kx) functions satisfy these conditions. So, the expression for the mode shape is Y ðxÞ ¼ C 3 U ðkxÞ þ C 4 V ðkxÞ Constants C3 and C4 are calculated from boundary conditions at x ¼ l: the bending moment and shear force at free end are 00 000 Y (l) ¼ 0 and Y (l) ¼ 0 ; therefore Y 00 ðlÞ ¼ k2 ½C 3 SðklÞ þ C 4 T ðklÞ ¼ 0
ðaÞ
000
Y ðlÞ ¼ k 3 ½C3 V ðklÞ þ C4 SðklÞ ¼ 0 A nontrivial solution of the above system is the frequency equation SðklÞ T ðklÞ 2 V ðklÞ SðklÞ ¼ 0 ! S ðklÞ T ðklÞV ðklÞ ¼ 0 According to (16.26), this leads to equation cosh kl cos kl þ 1 ¼ 0 The roots of the equation are λ ¼ kl ¼ 1:875,
4:6964,
7:855, . . .
rffiffiffiffiffi EI , m
4:6942 ω2 ¼ l2
so the frequencies of the free vibration are λ2 ωi ¼ 2i l
rffiffiffiffiffi EI , m
1:8752 ω1 ¼ l2
rffiffiffiffiffi EI , ... m
The mode shape of vibration is C Y ðxÞ ¼ C 3 U ðkxÞ þ C 4 V ðkxÞ ¼ C3 U ðki xÞ þ 4 V ðki xÞ C3 Since the ratio C4/C3 from first and second equations (a) are Sð k i l Þ V ðk i l Þ C4 ¼ ¼ , C3 T ðk i l Þ Sðk i lÞ then ith mode shape (eigenfunction) corresponding to ith frequency of vibration (eigenvalue) is Sðk i lÞ V ðki lÞ Y ðxÞ ¼ C3 U ðkxÞ þ C4 V ðkxÞ ¼ C 3 U ðk i xÞ V ðki xÞ ¼ C 3 U ðk i xÞ V ð k i xÞ T ðk i lÞ Sð k i l Þ
ðbÞ
For calculated eigenvalues kil ¼ 1.875, 4.6964, . . . the values of the functions S(kil ), T(kil), V(kil) are determined by formulas (16.24). The first and second modes are shown in Fig. 16.15. Fundamental data for one-span uniform beams with classical boundary conditions are presented in Table A.24. This table contains the frequency equation, and the first, second, and third eigenvalues. For the nodal points of the mode shapes of a free vibration the origin is placed on the left end of the beam. Detailed data for eigenfunctions and their derivatives for uniform beam with classical boundary conditions may be found in the following books (Karnovsky and Lebed 2001, 2004b).
594
16
Dynamics of Elastic Systems: Free Vibration
16.4.4 Initial Parameters Method This method allows simplifying solution procedure of the eigenvalues and eigenfunction problems for uniform beams. The method makes it easy to take into account the elastic supports and lumped mass. Let us consider the ordinary differential equation (16.20) and corresponding general expression for displacement (16.23) Y ðxÞ ¼ C1 SðkxÞ þ C 2 T ðkxÞ þ C3 U ðkxÞ þ C4 V ðkxÞ In order to find the constants Ci we need on the basis of this equation to derive formulas for angle of rotation 00 θ(x) ¼ dY/dx ¼ Y0, bending moment M ¼ EIY and shear force Q ¼ EIY000). They are in terms of Krylov-Duncan functions become θðxÞ ¼ C 1 kV ðkxÞ þ C 2 kSðkxÞ þ C 3 kT ðkxÞ þ C 4 kU ðkxÞ, M ðxÞ ¼ EI C1 k2 U ðkxÞ þ C 2 k 2 V ðkxÞ þ C 3 k 2 SðkxÞ þ C4 k2 T ðkxÞ , QðxÞ ¼ EI C 1 k 3 T ðkxÞ þ C 2 k 3 U ðkxÞ þ C 3 k 3 V ðkxÞ þ C 4 k 3 SðkxÞ
ð16:27Þ
Initial parameters are the following: at x ¼ 0 we have Y ¼ Y0, θ ¼ θ0, M ¼ M0, Q ¼ Q0. According to (16.25), only S(0) ¼ 1, while T(0) ¼ U(0) ¼ V(0) ¼ 0. Therefore, substitution of the initial parameters in (16.23) and (16.27) leads immediately to the values of unknown coefficients in terms of initial parameters C 1 ¼ Y ð 0Þ ¼ Y 0 ,
1 θ C 2 ¼ X 0 ð0Þ ¼ 0 , k k
C3 ¼
1 M 0 X 0 ð 0Þ ¼ 2 0 , k 2 EI k EI
C4 ¼
Q 1 00 X 0 ð 0Þ ¼ 3 0 k3 EI k EI
The fundamental solution of Equations (16.23) and (16.27) in terms of Krylov Duncan functions becomes Q 1 M Y ðxÞ ¼ Y 0 SðkxÞ þ θ0 T ðkxÞ þ 2 0 U ðkxÞ þ 3 0 V ðkxÞ; k k EI k EI Q0 M0 T ðkxÞ þ 2 U ðkxÞ; θðxÞ ¼ Y 0 kV ðkxÞ þ θ0 SðkxÞ þ kEI k EI Q 2 M ðxÞ ¼ EIY 0 k U ðkxÞ þ EIθ0 kV ðkxÞ þ M 0 SðkxÞ þ 0 T ðkxÞ; k QðxÞ ¼ EIY 0 k3 T ðkxÞ þ EIθ0 k 2 U ðkxÞ þ M 0 kV ðkxÞ þ Q0 SðkxÞ
ð16:28Þ
The typical procedure for determining the frequencies of free vibration is as follows. From system (16.28) we have to choose two equations that become equal to zero at the right end of the beam. These two equations contain four initial parameters, of which the two initial parameters vanish under the constraint conditions at the left end of the beam. Thus, for right end we obtain a system of two linear algebraic homogeneous equations with two unknown initial parameters. Its nontrivial solution leads to the equation of the frequencies of free vibration. Let us consider some classical problems. Example 16.8 Derive the frequency equation for uniform pinned-clamped beam (Fig. 16.16)
Initial parameters:
EI, l, m
x=0
Y0=0, θ0≠0 M0=0, Q0≠0
x
Parameters at x=l:
Y=0, θ=0 M≠0, Q≠0
l
Q0
ω1 N
ω2
0.44l Fig. 16.16 Design diagram for pinned-clamped beam; the first and second forms of vibration are shown by dotted line
16.4
Structures with Infinite Number of Degrees of Freedom
595
Solution The initial parameters and kinematical conditions are shown in Fig.16.16. At the right support the displacement and slope are zero. Therefore we need to use expressions for displacement and slope at any section x Q 1 M Y ðxÞ ¼ Y 0 SðkxÞ þ θ0 T ðkxÞ þ 2 0 U ðkxÞ þ 3 0 V ðkxÞ; k k EI k EI Q M θðxÞ ¼ Y 0 kV ðkxÞ þ θ0 SðkxÞ þ 0 T ðkxÞ þ 2 0 U ðkxÞ; kEI k EI
ðaÞ
Taking into account initial conditions (Y0 ¼ 0, M0 ¼ 0) expressions for right fixed support are T ðklÞ V ðklÞ þ Q0 3 ¼ 0 k k EI U ðklÞ θðlÞ ¼ θ0 SðklÞ þ Q0 2 ¼0 k EI Y ðl Þ ¼ θ 0
ðbÞ
Thus, the homogeneous linear algebraic equations with respect to θ0 and Q0 are obtained. This system has a nontrivial solution if and only if the following determinant, which represents the frequency equation, is zero 2 3 T ðklÞ V ðklÞ 6 k k3 EI 7 6 7 ¼ 0 ! T ðklÞU ðklÞ SðklÞV ðklÞ ¼ 0 ðcÞ 4 U ðklÞ 5 SðklÞ k2 EI According to Krylov-Duncan relationships (16.26, second row), we can present this equation in the trigonometric form cosh kl sin kl sinh kl cos kl ¼ 0 ! tan kl ¼ tanh kl The roots of this transcendental equation are λ1 ¼ k1l ¼ 3.9266, λ2 ¼ k2l ¼ 7.0686, rffiffiffiffiffi rffiffiffiffiffi EI λ2i EI 2 The frequencies vibration are ωi ¼ ki ¼ 2 , i ¼ 1, 2, :Corresponding first and second modes (eigenfunctions) m m l are shown in Fig. 16.16. The nodal points N of eigenfunctions for the first mode are located at the supports; for second mode— at the supports and additional point N in the span. For third form the nodal points at the span located at x/l ¼ 0.308 and 0.616 (Table A.24). Example 16.9 Design diagram of uniform cantilevered beam with intermediate support is shown in Fig. 16.17. Parameters of beam are the total length l, mass per unit length m, flexural stiffness EI ¼ const. Derive the frequency equation (Babakov 1965). Conditions:
Initial parameters:
Y0=0, θ0=0 M0≠0, Q0≠0
M(l)=0 Q(l)=0
y(l)=0
0 EI, m l/2
M0 Q0
x
l/2 R
Fig. 16.17 Design diagram of uniform cantilevered beam and first mode shape of vibration
Solution The origin of the coordinates is placed on the extreme left support, the x axis is directed to the right. Initial parameters and conditions for intermediate support at x ¼ l/2 and for free end are shown in Fig. 16.17. According to (16.28), the total expression for vertical displacements is
596
16
Dynamics of Elastic Systems: Free Vibration
Q0 M0 R l Y ðxÞ ¼ 2 U ðkxÞ þ 3 V ðkxÞ þ 3 V k x 2 k EI k EI k EI The structure of the terms which take into account the reaction R and shear force Q0 are same, i.e., they contain the same 1 coefficient 3 , and the same function V; the difference is that the function V for the reaction R has a shifted argument k EI x l/2. It means that the last term should be take into account only for x > l/2. At the intermediate support we have Q0 l M0 kl kl Y ¼ 2 U þ 3 V ¼0 ðaÞ 2 2 2 k EI k EI The total expression for bending moments and shear are Q0 R l M ðxÞ ¼ M 0 SðkxÞ þ ; T ðkxÞ þ T k x k k 2 l QðxÞ ¼ M 0 kV ðkxÞ þ Q0 SðkxÞ þ RS k x 2 so at the free end we have Q0 R l ¼ 0; T ðklÞ þ T k k 2 k kl ¼0 QðlÞ ¼ M 0 kV ðklÞ þ Q0 SðklÞ þ RS 2
M ðlÞ ¼ M 0 SðklÞ þ
ðbÞ
The unknowns of equations (a) and (b) are M0, Q0, and R. The frequency equation we get from a condition of the nontrivial solution of homogeneous algebraic equations (a) and (b) is 1 λ 1 λ 0 V 3 k2 EI U 2 2 k EI 1 1 λ D¼ T ðλÞ T SðλÞ ¼ 0, λ ¼ kl k k 2 λ kV ðλÞ Sð λ Þ S 2 The frequency equation in the expanded form is U λ=2 T λ Sλ=2 þ V λ=2 T λ=2 V λ Sλ T λ=2 U λ=2 Sλ=2 Sλ V λ=2 ¼ 0 The frequencies equation in the elementary functions becomes sin λ cosh λ sinh λ cos λ ¼ sinh λ sin λ To determine the interval in which the roots of the equation will be located, the following theorem can be used (Babakov 1965, pages 134, 279). If a linear constraint is imposed on a system with n-degrees of freedom, then the frequencies of the modified system with n 2 1 degrees of freedom are located between the frequencies of an initial system. This theorem may be extended to structures with distributed parameters. For given Example 16.9, the clamped-free beam of length l should be treated as initial structure. For this beam the three eigenvalues are kl ¼ 1.875, 4.694, 7.855, . . . (Table A.24). Therefore, the first and second eigenvalues λ1 and λ2 of cantilevered beams with additional constraint (modified system) satisfied to conditions 1:875 λ1 4:694 λ2 7:855
16.4
Structures with Infinite Number of Degrees of Freedom
597
This wide range should be narrowed down. The definition of the boundaries of the refined range is based on a change in the sign of the rffiffiffiffiffi D. The refined interval for the first eigenvalue is 3.1 < λ1 < 3.2. The first frequency vibration is rffiffiffiffiffideterminant 2 λ EI EI ¼ 21 . Corresponding first mode of vibration is shown in Fig. 16.17. ω1 ¼ k21 m m l Initial parameters method may be successfully applied to beam with elastic support, beam with distributed mass together with lumped mass, etc. Example 16.10 Derive the frequency equation for uniform beam with elastic support (stiffness coefficient c) (Fig. 16.18). Initial parameters:
Y0≠0, θ0≠0 M0=0, Q0=-cY0
Conditions
EI, m
x=0
y(l)=0 θ (l)=0 x
c l Q0 Fig. 16.18 Design diagram of uniform beam with elastic support.
Solution Expressions for displacement and slope at x ¼ l Q θ0 T ðklÞ þ 3 0 V ðklÞ ¼ 0, k k EI Q θðlÞ ¼ Y 0 kV ðklÞ þ θ0 SðklÞ þ 2 0 U ðklÞ ¼ 0 k EI yðlÞ ¼ Y 0 SðklÞ þ
ðaÞ
Initial displacement and shear force are connected as Y0 ¼ Q0/c; the negative sign means the Y0 and Q0 are directed opposite. Therefore formulas (a) may be rewritten as T ðklÞ V ðklÞ SðklÞ θ0 þ 3 Q0 ¼ 0, k c k EI ðbÞ U ðklÞ kV ðklÞ SðklÞθ0 þ 2 Q0 ¼ 0 c k EI We obtained two algebraic homogeneous equations with respect to unknown θ0 and Q0. The frequency equation is D ¼ 0, where D is determinant of the coefficients of unknown initial parameters θ0 and θ0. T ðklÞ V ðklÞ SðklÞ k c k3 EI ¼ 0: D ¼ U ðklÞ kV ðklÞ SðklÞ c k2 EI In the expanded form the frequency equation becomes c S2 TV ¼ 3 TU SV k EI Multiplication of this equation by factor λ3 ¼ k3l3 leads to the frequency equation in terms of Krylov–Dunkan functions λ3
S2 ðλÞ T ðλÞV ðλÞ cl3 ¼ EI T ðλÞU ðλÞ SðλÞV ðλÞ
Frequency equation in the elementary functions is
598
16
λ3
Dynamics of Elastic Systems: Free Vibration
cosh λ cos λ þ 1 cl3 ¼ cosh λ sin λ sinh λ cos λ EI
rffiffiffiffiffi rffiffiffiffiffi EI λ2 EI The root λ of this equation allows calculating the frequency of free vibration ω ¼ k ¼ 2 . m m l 2
Special Cases 1. Let c ¼ 0. In this case we have free-clamped beam. The frequency equation is S2 ðλÞ T ðλÞV ðλÞ ¼ 0 In the elementary function coshλ cos λ + 1 ¼ 0. The roots of this equations are λ1 ¼ 1:8751,
λ2 ¼ 4:6941,
The frequencies of free vibration are ω21
λ2 ¼ 21 l
rffiffiffiffiffi rffiffiffiffiffi EI 1:87512 EI ¼ , m m l2
2. Let c ¼ 1. In this case we have pinned-clamped beam. The frequency equation is T ðλÞU ðλÞ SðλÞV ðλÞ ¼ 0 In the elementary function we have coshλ sin λ sinh λ cos λ ¼ 0 or λ1 ¼ 3.9266, λ2 ¼ 7.0685, . The frequencies of free vibration are ω21
λ2 ¼ 21 l
rffiffiffiffiffi rffiffiffiffiffi EI 3:92662 EI ¼ , m m l2
tan λ ¼ tanh λ. The roots of this equation are
Numerical solution of the frequency equation is presented in Table 16.5. Table 16.5 Frequency parameter λ ¼ kl for different dimensionless coefficient c ¼ cl3/EI
c*
0.0(*)
5.0
10
20
40
60
λ1 λ2
1.875 4.694
2.367 4.743
2.639 4.794
2.968 4.897
3.303 5.103
3.474 5.295
c*
80 3.575 5.466
100 3.541 5.616
200 3.781 6.128
400 3.854 6.566
500 3.869 6.668
∞(*) 3.927 7.068
λ1 λ2 (*)
Case c* = 0 and c* = ∞ corresponds to fixed-free beam and fixed-pinned beam respectively
Example 16.11 Uniform cantilevered beam of length l carries uniformly distributed mass m per unit length and lumped mass M at free end (Fig. 16.19). Derive the equation for frequency of free vibration. The ratio of a lumped mass M and total mass ml is a given number M/ml ¼ ψ. Solution General expression for mode shape of vibration is yðxÞ ¼ Y 0 SðkxÞ þ
Q θ0 M T ðkxÞ þ 2 0 U ðkxÞ þ 3 0 V ðkxÞ k k EI k EI
16.4
Structures with Infinite Number of Degrees of Freedom
599
a
Initial parameters:
Conditions:
Y′′(l)=0, Y′′′(l )=My(l)ω2
Y0=0, θ0=0 M0≠0, Q0≠0
Fin EI, m
x=0
l
x M
Fig. 16.19 Design diagram of cantilevered beam with distributed mass m and lumped mass M; a is acceleration; Fin is inertial force which acts on the mass (this force acts on the beam in the opposite direction)
Since Y0 ¼ 0 and θ0 ¼ 0, then displacement of a beam at free end should be calculated using functions U and V Q M yðlÞ ¼ 2 0 U ðλÞ þ 3 0 V ðλÞ, k EI k EI
λ ¼ kl,
rffiffiffiffiffiffiffiffiffi 4 mω2 , k¼ EI
ðaÞ
The bending moment and shear force at free end are known. They are M ðl Þ ¼ 0
and
QðlÞ ¼ MyðlÞω2 ¼ ψml yðlÞω2
ψ ¼ M=ml
where y(l) is displacement of a beam at point where mass M is attached. The nature of shear force presents the inertial force of lumped mass M. If the acceleration of the mass M is directed in the positive direction (upward), then the inertia force is directed downwards. This means that force acts on the beam in the opposite direction (upward) T ðλÞ Q0 ¼ 0 k QðlÞ ¼ kV ðλÞM 0 þ SðλÞQ0 ¼ ψml yðlÞω2 M ðlÞ ¼ SðλÞM 0 þ
ðbÞ ðcÞ
Substitution of (a) into (c) leads to the following result QðlÞ ¼ kV ðλÞM 0 þ SðλÞQ0 ¼ ψml
Q0 M0 ω2 U ð λ Þ þ V ð λ Þ k 2 EI k3 EI |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
ðdÞ
yðlÞ
Coefficient
mlω2 mω2 l ¼ kl ¼ λ. Relationships (c) after simplification becomes ¼ EI k3 k3 EI ½kV ðλÞ þ kψλ U ðλÞM 0 þ ½SðλÞ þ ψλ V ðλÞQ0 ¼ 0
ðeÞ
Relationships (b) and (e) present a set of two linear homogeneous algebraic equations with unknowns initial bending moment M0 and shear force Q0. Nontrivial solution is possible if determinant of coefficients is zero. kV ðλÞ þ kψλ U ðλÞ SðλÞ þ ψλ V ðλÞ ¼0 T ðλÞ SðλÞ k In terms of Krylov-Duncan function the frequency equation is S2 ðλÞ V ðλÞT ðλÞ ¼ ψλ T ðλÞU ðλÞ SðλÞV ðλÞ
ðfÞ
This equation in elementary functions becomes 1 þ cosh λ cos λ ¼ ψλ cosh λ sin λ sinh λ cos λ
ðgÞ
600
16
Dynamics of Elastic Systems: Free Vibration
Assume the total mass of a beam ml is twice more thanrlumped mass M, i.e., ψ ¼ 0.5. Smallest root of this equation is ffiffiffiffiffi 1:412 EI λ1 ¼ 1.41, so the smallest frequency vibration ω1 ¼ 2 . m l Special Case Let ψ ¼ 0; it means that M ¼ 0 and we have clamped-free beam with uniformly distributed mass m. Corresponding frequency equation becomes coshλ cos λ ¼ 1. Frequencies and shapes of vibrations for different types of beams (one and multispan beams, uniform and nonuniform beams, beams with classical and nonclassical boundary conditions, with lumped and rotational masses, etc.) as well as for frames may be found in handbook (Karnovsky and Lebed 2004a). Also this book contains the general equations of initial parameters method taking into account lumped masses along the beam.
16.4.5 Transfer Matrices Method The method of transfer matrices allows to significantly expand the scope of application of the method of initial parameters. This method is well algorithmic and is very effective for analyzing free and forced vibrations of complex dynamic systems. Real engineering structures have a number of peculiarities. Among them it should be noted the presence of elements with step-variable cross sections, the presence of elastic supports, lumped bodies with mass and moment of inertia, as well as arbitrary combinations of these peculiarities. Fundamental concept of the transfer matrices allows presenting a dynamic model of a complex deformable system as an ordered set of matrices of individual elements (Babakov 1965). It is assumed that the dynamic system consists of elastic elements and is described by linear relationships. The method allows the combination of elements subjected to various types of vibrations—bending, longitudinal, and torsional. Below the idea of the transfer matrices is applied to the free bending vibration only.
16.4.5.1
Transfer Matrix of Uniform Beam with Distributed Mass
The set of linear equations (16.28), which describe the bending vibration of uniform inertial rod with arbitrary boundary condition, may be presented in matrix form YðxÞ ¼ AðxÞY0
ð16:29Þ
Transfer matrix of uniform portion with distributed mass m is 2
1 T ðkxÞ k
6 6 6 6 kV ðkxÞ AðxÞ ¼ 6 6 6 6 EIk 2 U ðkxÞ 4
SðkxÞ
1 U ðkxÞ 2 k EI 1 T ðkxÞ kEI
EIkV ðkxÞ
SðkxÞ
EIk 3 T ðkxÞ
EIk2 U ðkxÞ
kV ðkxÞ
SðkxÞ
3 1 V ð kx Þ 7 k 3 EI 7 7 1 7 U ð kx Þ 2 7 k EI 7 7 1 T ðkxÞ 7 5 k SðkxÞ
ð16:29aÞ
mω2 , m is distributed mass per unit length, EI is a bending stiffness of the rod, ω is frequency vibration. The EI state of a rod in arbitrary section x and in initial section x ¼ 0 are determined by the following vectors where k4 ¼
2
Y ð xÞ
3
6 θ ð xÞ 7 6 7 YðxÞ ¼ 6 7, 4 M ð xÞ 5 Q ð xÞ
2
Y0
3
6θ 7 6 0 7 Y0 ¼ 6 7 4 M0 5
ð16:30Þ
Q0
According to (16.29), matrix A(x) transforms the vector state Y0 in the initial section into the vector state Y(x) in the arbitrary section х. This matrix is called the transfer matrix of uniform portion with distributed mass m.
16.4
Structures with Infinite Number of Degrees of Freedom
601
The matrix relation (16.29) is equivalent to four linear algebraic equations for the variables Y(x), θ(x), M(x), and Q(x), which determine the state of the beam in section x. Thus, with bending vibration, the state of system is determined by four components. They are transversal displacement, slope, bending moment, and shear force (with corresponding coefficients). Equation (16.29) allows writing relationships between the states of a rod in two specific sections, at x ¼ 0 and x ¼ l 2 2
Y
3
6θ7 6 7 6 7 4M 5 Q
1 T k
S
x¼l
6 6 6 6 kV ¼ AY0 ¼ 6 6 6 6 EIk 2 U 4 EIk3 T
1 U k2 EI 1 T k EI
S EIkV
S
EIk 2 U
kV
3 1 V 2 3 k3 EI 7 Y0 7 7 1 7 U7 6 6 θ0 7 6 k 2 EI 7 7 4M 7 05 7 1 T 7 5 Q0 k S
ð16:31Þ
The transfer matrix A for x ¼ l is called the transfer matrix of span. Relationships (16.31) contains all Krylov-Duncan functions S, T, U, V of argument kl.
16.4.5.2
Transfer Matrix of Uniform Massless Beam
From transfer matrix A of uniform rod with distributed mass m we can derive a transfer matrix for massless uniform beam of length l and bending stiffness EI . For this purpose, it is necessary to determine the values of all the elements of the matrix A from (16.31) if m ¼ 0, k ! 0. According to (16.25), on the main diagonal we get S(0 ) ¼ 1. If k ! 0, then all elements below the main diagonal are zero. For elements containing a parameter k in the denominator, the limit should be found at k ! 0. For example, a12 ¼ lim
k!0
1 1 1 0 T ðklÞ ¼ lim ð sinh kl þ sin klÞ ¼ k 2 k!0 k 0
In order to uncover uncertainty, it is needed to apply L’Hôpital’s rule a12 ¼
1 1 1 l cosh kl þ l cos kl lim ð sinh kl þ sin klÞ ¼ lim ¼l 2 k!0 k 2 k!0 1
Similarly, V ðklÞ 1 sinh kl sin kl l3 lim ¼ ¼ 2EI k!0 6EI k!0 k 3 EI k3
a14 ¼ lim
Here L’Hopital’s rule is applied three times. Finally the transfer matrix of the uniform massless element becomes 2
60 6 Am¼0 ¼ 6 40
l
l2 =2EI
1 0
l=EI 1
0
0
0
1
l3 =6EI
3
l2 =2EI 7 7 7 l 5 1
If we assume that l ¼ 0 and lm ! M, then similarly we can derive the matrix of the lumped mass M
ð16:32Þ
602
16
2
1 6 0 6 AM ¼ 6 4 0 Mω2
16.4.5.3
0 1
3 0 07 7 7 05
0 0
1
0 0 1 0
Dynamics of Elastic Systems: Free Vibration
ð16:33Þ
Mathematical Model of a Complex Structure
Figure 16.20 contains design diagram of the beam AB of the total length L. The beam has the following peculiarities: each portion of length li (i ¼ 1, 2, 3) is characterized by the stiffness EIi and distributed mass mi ; the portion l1 is massless (m1 ¼ 0). The boundary points of the portions are denoted as C1 and C2. Lumped mass M is attached to the beam at point C2. The origin is placed at the right end of the beam, axis x is directed left. The numbering of the portions from right to left, i.e., l1, l2, l3. Yx=L EI3 m3
x B
l3
M C2
EI2 m2 l2
C1
Y0 EI1 A m1=0 l1
Fig. 16.20 Design diagram of the beam. Numeration of the adjacent portions
The compilation of a mathematical model begins with the extreme right portion AC1. According to the general relation (16.29), the state vector at the final point C1 of the portion l1 is determined by formulaYC1 ¼ A1 Y0 . Since the portion l1 is massless, then the transfer matrix A1 of the first portion should be written in form (16.32). For portion l2 the vector YC1 describes the state at the initial point; therefore Yright C2 ¼ A2 YC 1 ; the symbol “right” means the point C2 is located right from mass M . Since for portion l2 distributed mass m2 6¼ 0, then transfer matrix A2 should be written in form (16.31). right Accounting for lumped mass M is performed according to the formulaYleft C 2 ¼ AM YC2 ; transfer matrix of mass M, AM, is presented in (16.33). For last portion l3 the fundamental relationships (16.29) is Yx¼L ¼ A3 Yleft C 2 . The transfer matrix A3 should be written in form (16.31). Now we need to connect two state vectors at the initial and final points of the structure Y0, and Yx ¼ L. This procedure is presented below: right Yx¼L ¼ A3 Yleft C 2 ¼ A3 AM YC 2 ¼ A3 AM A2 YC 1 ¼ A3 AM A2 A1 Y0
ð16:34Þ
The set (16.34) Yx ¼ L ¼ A3AMA2A1 Y0 presents the mathematical model of a structure in Fig. 16.20. This model takes into account all peculiarities of a structure and allows expressing the state vector at the extreme left end of a beam in terms of the initial state vector. It is assumed that each element of the system and the entire system as a whole are described by linear relationships. Thus, in general case, mathematical model contains four equations which determine the total state of the structure (displacement, slope, bending moment and shear) at section x ¼ L. The set of equations like 16.34 contains four initial parameters. However, under the conditions of support at the right end, the two initial parameters vanish. To determine the frequencies of free vibration, we need from the system of equations to choose such two equations that equal zero at the left end. This choice of equations leads to a system of two linear homogeneous algebraic equations. The condition for a nontrivial solution of such a system leads to the frequency equation.
16.4
Structures with Infinite Number of Degrees of Freedom
603
Example 16.12 The simply supported uniform massless beam (m ¼ 0) of length L ¼ 2l and flexural stiffness EI ¼ const have one lumped mass M at the middle of the span (Fig. 16.21). Calculate the frequency of free vibration. Yx=2l x
M
EI
B
Y0 A
C l
l=L/2 Fig. 16.21 Design diagram of uniform massless beam with lumped mass M
Solution The origin is placed at the right end of the beam, axis x is directed to the left. The numbering of the portions is from right to left, i.e., AC and CB. Vector of initial parameters (right support) and vector of a state at left support are 2
3 2 3 0 y0 6 φ 7 6φ 7 6 07 6 07 Y0 ¼ 6 7 ¼ 6 7; 4 M0 5 4 0 5
2
Yx¼2l
Q0
Q0
3 2 3 0 yB 6 φ 7 6φ 7 6 B7 6 B7 ¼6 7¼6 7 4 MB 5 4 0 5 QB
QB
Relationships between initial parameters and parameters at the left end (final point) can be constructed as multiplication of corresponding matrices (initial parameters and transfer matrices) considering them successively from right to left. The concept of the transfer matrix is presented by the following relationship between the initial (point A) and final (point B) states Yx¼2l ¼ ACB AM AAC Y0 Here ACB and AAC are matrices of massless portions CB and AC, while AM is a matrix of lumped mass M. In expanded form, this matrix ratio is written as 3 2 0 1 6 φ 7 60 6 B7 6 6 7¼6 4 0 5 40 2
QB |fflfflffl{zfflfflffl}
3 2 3 3 2 1 0 0 0 0 1 l l2 =2EI l3 =6EI 7 6 7 6 6 7 2 2 1 0 0 7 6 0 1 l=EI 1 l=EI l =2EI 7 6 0 l =2EI 7 6 φ0 7 7 76 76 7 76 0 1 05 40 0 0 1 l 5 4 0 1 l 5 4 0 5 Mω2 0 0 1 Q0 0 0 0 1 0 0 0 1 ffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflffl{zfflffl l
YL
l2 =2EI
l3 =6EI
3 2
AM
ACB
Y0
AAC
The result of the intermediate operation of multiplication is 2
0
3
2
6φ 7 6 6 B7 6 6 7¼6 4 0 5 4 QB |fflfflffl{zfflfflffl} YL
1 þ Mω2 l3 =6 EI
l
l2 =2EI
Mω2 l2 =2EI Mω2 l
1 0
l=EI 1
l3 =6EI
3 2
φ0 l þ Q0 l3 =6EI
6 2 l2 =2EI 7 7 6 φ0 þ Q0 l =2EI 76 Q0 l l 5 4
3 7 7 7 5
Mω2 Q0 0 0 1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ACB AM
AAC Y0
Since at the left support B are yB ¼ 0 and MB ¼ 0, then the first and third equations lead to set of two homogeneous equations with respect to unknown initial parameters φ0 and Q0. After isolation of these unknown initial parameters, we obtain the following system of homogeneous equations
604
16
Dynamics of Elastic Systems: Free Vibration
3 Mω2 l4 l Mω2 l3 7 l3 2l þ φ0 þ 1þ þ Q ¼0 6 EI 0 6EI 6EI 6EI Mω2 l4 Mω2 l2 φ0 þ 2l þ Q0 ¼ 0 6EI Nontrivial solution occurs if determinant of coefficients at unknowns is equal to zero, so the frequency equation becomes
Mω2 l4 2l þ 6EI
2
"
3 2 # 8 l3 l 2 Mω l þ Mω ¼ 0: 6 EI 6EI 2 2
This equation may be presented as follows ð2l þ kÞ2 kð8l þ k Þ ¼ 0,
Mω2 l4 ¼k 6EI
rffiffiffiffiffiffiffiffi 6EI Solution of this equation is k ¼ l, so the frequency of free vibration becomes ω ¼ . In terms of the total span L ¼ 2l M l3 rffiffiffiffiffiffiffiffiffiffi 48EI we get well-known result ω ¼ . ML3 Obviously, this result can be obtained in other simpler ways. However, the transfer matrix method is out of competition in the numerical analysis of complex systems. Example of some complex system may be a rotor. Its design diagram is a continuous beam, which contains portions of varying length, different stiffness, lumped and distributed masses, as well as some additional features, for example, dynamical absorber, etc. When analyzing such systems using the transfer matrix method, you should select a specific portion with basic parameters (l, ЕI, m) and similar parameters for the remaining portions should be expressed in terms of basic parameters. All matrix procedures are performed using modern software. Detailed catalog of the transfer matrices is presented in book (Ivovich 1981).
16.4.6 Displacement Method This section is devoted to determining the frequencies of free vibrations of the structures taking into account distributed mass of elements, i.e., we will consider structures with infinite number of degrees of freedom. As in case of system with finite number of degrees of freedom (Sect. 16.3.1), we will use the concept of unit reactions. For bending structures, such as continuous beams and frames the displacement method is more preferable than the force method. According to this method the primary unknown are linear and angular displacement of the joints of the structure. The primary system of this method is obtained from the given one by introducing additional constraints to prevent rotation of the rigid joints and all independent linear displacements of joints. Unknown dynamic reactions of the introduced constraints are Z 1 ðt Þ ¼ Z 1 sin ω t,
Z 2 ðt Þ ¼ Z 2 sin ω t, , Z n ðt Þ ¼ Z n sin ω t
Canonical equation of displacement method will be written as follows: r 11 Z 1 þ r 12 Z 2 þ þ r 1n Z n þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ þ r 2n Z n þ R2P ¼ 0 r n1 Z 1 þ r n2 Z 2 þ þ r nn Z n þ RnP ¼ 0
ð16:35Þ
This system contains the amplitude of vibrational displacements Zi for unknown variables. In case of free vibration we need to assume RiP ¼ 0. The unit reactions rik are determined depending on the type of the boundary condition of a simplest element of the primary system (beam), the type of unit displacement (linear, angular), and the type of reactions (force, couple). The peculiarity of unit reactions is as follows: effect of inertial forces of distributed masses is taken into account by correction
16.4
Structures with Infinite Number of Degrees of Freedom
605
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 functions. Numerical values of these functions depend on the parameter λ ¼ kl ¼ mω2 l4 =EI , where m is mass per unit length, l and EI are the length of rod and its bending stiffness, ω is frequency vibration. Corresponding reactions may be presented in two forms: in terms of Krylov-Duncan functions (Table A.25), and in the form of hyperbola-trigonometric Smirnov’s functions (Table A.28). The numerical values of the functions can be found in the Handbook (Karnovsky and Lebed 2001, 2004a). It is easy to obtain expressions for reactions caused by unit displacements of supports using the method of initial parameters in the form of a system of equations (16.28). In this case, put the geometric initial parameters Y0 ¼ 1 or θ0 ¼ 1. This leads to a system of inhomogeneous algebraic equations for unknown initial force parameters M0 and Q0. In order to derive the frequency equation, we have to compile the determinant, consisting of coefficients with the unknown variables in canonical equations, set it equal to zero, expand it, and find parameter λ ¼ kl. Example 16.13 Determine the frequencies of free vibration for the frame shown in Fig. 16.22a. Each element of the frame has uniformly distributed mass m, the length l, and the bending stiffness EI ¼ const.
a
b B
1
Z1=1 r11
1
m, l, EI
M1B
M1A A
Fig. 16.22 (a) Design diagram; (b) The primary system, unit primary unknown Z1 ¼ 1, and corresponding bending moment diagram
Solution The primary system of the displacement method and the bending moment diagram due to the unit angular displacement Z1 ¼ 1 are given in Fig. 16.22b. Since the distributed inertial forces are taken into account, then the bending moment diagram within each element is curvilinear. Reaction of the introduced constraint 1 is presented in Кrylov-Duncan form. According to Table A.25 line 1 we have M 1A ¼ M 1B ¼
4EI f ðλÞ, l 3
f3 ¼
λ TU SV , 4 U 2 TV
4EI f ðλÞ. Canonical equation is r11Z1 ¼ 0. Since r11 6¼ 0, then nontrivial solution occurs if l 3 r11 ¼ 0. Thus, the frequency equation is r11 ¼ TU SV ¼ 0. In the equivalent form, taking into account (16.26), and after simplest transformation, the frequency equation becomes so the unit reaction is r 11 ¼ 2
tan kl ¼ tanh kl The roots of this equation are λ ¼ kl ¼ 3.926, 3:9262 ω1 ¼ l2
ðaÞ
7.0685, . . . The exact frequencies are rffiffiffiffiffiffi EI , m0
7:06852 ω2 ¼ l2
rffiffiffiffiffiffi EI , ... m0
It is worth mentioning that the Smirnov’s functions are very effective for analyzing the free vibrations of continuous beams and frame systems using the displacement method. These functions refer to uniform single-span beams. Table A.28 contains beam reactions due to force and kinematic harmonic excitation. A feature of the functions is that they are composed taking into account the distributed mass of the beam. These reactions are presented in an analytical form through trigonometric functions, and tabulated.
606
16
Dynamics of Elastic Systems: Free Vibration
Example 16.14 Determine the frequency of free vibration for frame shown in Fig. 16.23a. Mass per unit length for all members is m.
a
b
4EI
EI
EI
l
1
4i 0.707u
i, u
2
i,u
l Z1=1
c
r12 r11
1
2
Z2=1
J12
r21
r22
1
4iψ2(u)
6(i/l)ψ5(u)
12iψ1(0.707u)
2
M2
M1 6(i/l)ψ6(u)
2iψ3(u)
6(i/l)ψ7(u)
Fig. 16.23 (a, b) Design diagram of the frame and primary system; (c) First and second unit states and corresponding bending moment diagrams; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 u ¼ mω2 l4 =EI
rffiffiffiffiffiffiffiffiffiffiffiffi 4 4 mω2 l Solution Parameter of the corrected functions ψ for vertical bar is equal u ¼ , while for crossbar the parameter EI becomes rffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffirffiffiffi 4 4 4 mω2 l 4 mω2 l 4 1 ¼ ¼ 0:707u 4 4EI EI Primary system is shown in Fig. 16.23b. Bending moment diagrams caused by unit primary displacements Z1 ¼ 1 and Z2 ¼ 1 are shown in Fig. 16.23c. For calculation of unit reaction the Smirnov’s functions can be used. Unit reactions according to Table A.28 are r 11 ¼ 4iψ 2 ðuÞ þ 3 4iψ 1 ð0:707uÞ ¼ 4iψ 2 ðuÞ þ 12iψ 1 ð0:707uÞ,
r 12 ¼ r 21 ¼
6i ψ ð uÞ l 5
When calculating the unit reaction r22, one should take into account not only the shear forces in the vertical bar but also the inertial forces of the distributed masses of the crossbar. If introduced constraint 2 executes a harmonic linear displacement with a frequency of free vibration ω, then inertial force of the distributed masses of the cross-rod 1-2 is equal J 12 ¼ mlω2 ¼ ml
u4 EI i ¼ 2u 4 ml l
Therefore, the total reaction r22 is equal to r 22 ¼ The frequency equation
12i 3i i ψ 10 ðuÞ þ 2 ψ 12 ðuÞ 2 u l2 l l
16.4
Structures with Infinite Number of Degrees of Freedom
r 11 D ¼ r 21
607
r 12 ¼ r 11 r 22 r 212 ¼ 0 r 22
In expanded form, the frequency equation becomes 2 12i 3i i 6i D ¼ ½4iψ 2 ðuÞ þ 12iψ 1 ð0:707uÞ 2 ψ 10 ðuÞ þ 2 ψ 12 ðuÞ 2 u ψ 5 ð uÞ ¼ 0 l l l l After division on i2/l2 we get D ¼ ½4ψ 2 ðuÞ þ 12ψ 1 ð0:707uÞ ½12ψ 10 ðuÞ þ 3ψ 12 ðuÞ u ½6ψ 5 ðuÞ2 ¼ 0 For solution of this equation the analytical expressions of Smirnov’s function (Table A.28) may be used, or tables of numerical data of these functions (Karnovsky and Lebed 2004a). The smallest root of this equation is umin ¼ 1.68. Fundamental frequency of free vibration u2 ω¼ 2 l
rffiffiffiffiffi rffiffiffiffiffi EI 2:82 EI ¼ 2 m m l
It is useful to discuss some aspects of the analysis due to the change in the design diagram shown in Fig. 16.23a. (a) Modified design diagram contains additional massless element 1-A of length l1, bending stiffness EI1, and with roller support A (Fig.16.24a). All supports of the structure allow horizontal displacement of the crossbar. Primary system contains two introduced constraints 1 and 2 (Fig. 16.24b). The bending stiffness per unit length of element 1-A is EI1/l1 ¼ i1.
a A
c
b EI1, 1 m=0
4EI, m EI, m
EI, m
A l
1
ki i, u
4i 0.707u i,u
2
Z1=1 3ki 1
A
r11
2
r21
M12
M1B
M1 l1
l
B
MB1
Fig. 16.24 (a, b) Design diagram of the frame and the primary system; (c) Bending moment diagram M 1 , M1B ¼ 4iψ 2(u), qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 M12 ¼ 12iψ 1(0.707u), u ¼ mω2 l4 =EI
MB1 ¼ 2iψ 3(u),
Let the stiffness i1 of additional member 1-A and stiffness i of basic vertical element be related as i1 ¼ ki, where к is some positive number. Since element 1-A is massless, then in case of displacement Z1 ¼ 1 of introduced constraint 1 the bending moment diagram within portion 1-A is linear (Fig. 16.24c). Therefore, the unit reaction r11 takes an additional term 3ki (Table A.3, row 1) and becomes r11 ¼ 4iψ 2(u) + 12iψ 1(0.707u) + 3ki. The remaining coefficients of canonical equations do not change. (b) Let support A of modified structure be adopted as pinned one. In this case the structure does not have a side displacement; therefore the number of unknowns of displacement method is equal to one. The primary system contains only one introduced constraint 1, so the canonical equation of displacement method becomes r11Z1 ¼ 0. The frequency equation is r11 ¼ 0, or r11 ¼ 4ψ 2(u) + 12ψ 1(0.707u) + 3k ¼ 0. In this case the parameters of the right vertical bar (EI, m, l) are not included in the frequency equation.
608
16
Dynamics of Elastic Systems: Free Vibration
16.4.7 Missed (Unaccounted) Frequencies The general property of frequencies of an arbitrary deformable system is as follows: imposition of additional constraints on the system leads to increase of the frequencies of free vibration. In other words, the greater the rigidity of a structure, the higher the frequency of free vibration. However, this obvious conclusion contradicts the data contained in Table 16.6 (Babakov 1965). Table 16.6 Frequencies of free vibration of uniform beams with different boundary condition, ωi ¼ k 2i
pffiffiffiffiffiffiffiffiffiffiffi EI=m
Roots of frequency equations ##
1
2
3
4
5
Design diagram of a beam
Number of constraints
Frequency equation
(kl)1
(kl)2
(kl)3
6
coshkl coskl = 1
4.730
7.853
10.996
tanh kl = tan kl
3.927
7.068
10.210
coshkl coskl = −1
1.875
4.694
7.854
tanh kl = tan kl
3.927
7.068
10.210
coshkl coskl = 1
4.730
7.853
10.996
4
3
2
0
The table contains the design diagrams of one-span homogeneous beams with different boundary conditions. For the first scheme, the number of supported constraints is 6, for Scheme 5 there are no supporting constraints (floating beam). The beams are arranged in decreasing order of a number of constraints (6,4,3,2,0). It can be seen, the table has full symmetry with respect to the third scheme. Symmetry is related to the difference in number of constraints for adjacent schemes, the frequency equations, and the roots of frequency equations. The paradox is that decrease of a number of support constraints (i.e., decrease in the rigidity of a structure) leads to decrease in eigenvalues only for Schemes 1–3, while for Schemes 4–5 there is an increase in eigenvalues (!). This paradox can be explained by the fact that the data in Table 16.6 refers only to bending vibrations. Analysis of the nature of the beam vibration allows explaining this phenomenon (Panovko and Gubanova 1973). Cases 1–3 lead to solutions that determine the shapes of vibration, which are accompanied by only the deformations of a beam. In cases 4 and 5 we have different situation. Here additional solutions appear that determine the shapes of vibration not only as a deformable body but also as an absolutely solid body. In Scheme 4, such a movement corresponds to the angular displacement of a beam with a center of rotation on the left support, while in Scheme 5, two movements of the beam as an absolutely rigid body are possible—the linear displacement and rotation about certain point. Each such movement of the beam occurs with zero frequency of bending vibrations. Таble 16.7 contains the ordered set of frequencies vibration for uniform rod, taking into account its possible movements as an absolutely rigid body.
Problems
609 Table 16.7 Frequencies of free vibration of uniform beams with different boundary condition
Roots of frequency equations #scheme
(kl)1
(kl)2
(kl)3
1 2 3 4 5
4.730 3.927 1.875 0.000 0.000
7.853 7.068 4.694 3.927 0.000
10.996 10.210 7.864 7.068 4.730
For Scheme 4, the first form is the rotational motion, and for Scheme 5, the first and second forms are the translational and rotational motion of the body. For Scheme 4, the first frequency (kl)1 ¼ 0 means the rod vibrates as an absolutely rigid body. For Scheme 5, the rod vibrates as an absolutely solid body in two forms (kl)1 ¼ 0 and (kl)2 ¼ 0. Thus, the contradiction formally disappears, and the table corresponds to the nodes theorem on its own forms (Babakov 1965). According to this theorem, the quantity of nodes N is one less than the number of a harmonic (shape mode) (Table A.24). For example, for Scheme 5, the vibration shapes are shown in Fig. 16.25.
i=1 (N=0) N 2 (N=1) N
3 (N=2) 4 (N=3)
Fig. 16.25 The first four forms of vibrations for a free-floating beam (Scheme 5); i is a number of the shape vibration, N is amount of nodal points
The possibility of the appearance of zero frequencies should always be borne in mind.
Problems 16.1. Statically determinate beam carried one lumped mass. Determine the frequency of free vibration. M
a
EI, l
M
b
Ans. (a) ω = a
b
c
l
a
2EI
EI
l Fig. P16.1
l
M
(c) ω =
3lEI ; a 2b 2 M 2 EI . 3l 3 M
610
16
Dynamics of Elastic Systems: Free Vibration
16.2. Statically indeterminate beam carried one lumped mass (Fig. P16.2). Determine the frequency of free vibration.
a
b
M
EI b
EI
M Ans. (a) ω =
l/2
a
l/2 (b) ω =
c
d
M
EI
EI
b=l-a
a
M
(c) ω =
b=l-a
a
(d) ω =
12 EI a (3b + 4a )M 2
768 EI 7l 3 M
;
;
12l 3EI 2 3
Mb a ( 3a + 4b )
;
3l 3EI Ma3b3
Fig. P16.2
16.3. Symmetrical frame with absolutely rigid crossbar of total mass M is shown in Fig. P16.3. Find the frequency of free horizontal vibration. EI=∞, M
24 EI
Ans. ω = h
h3M
EI
Fig. P16.3
16.4. Symmetrical statically indeterminate frame carried one lumped mass as shown in Fig. P16.4. Determine the frequency of free horizontal and vertical vibrations. M Ans. ω hor =
EI2 h
EI1
12 EI 2
h3M ( 2β + k )
ω vert = 48EI 2 · 12k + 8β , β = I 2 , k = l ; 3 l M
3k + 8β
I1
h
l/2 Fig. P16.4
16.5. Calculate the frequency of free vertical vibration of symmetrical truss presented in Fig. P16.5. Axial rigidity for diagonals and vertical elements EA, for lower and top chords is 2EA. A lumped mass M is placed at joint 6.
1
3
2EA
5
7
9 Ans. ω = h=3
EA 2
4 2EA
M 6
10
8 d=4m
Fig. P16.5
(
)
1 EA = sec −1 . mδ11 24.555M
Problems
611
16.6. A massless beam carried two lumped mass (Fig. P16.6). Calculate the frequency of free vibration, find the modal matrix, and show the shapes of vibrations.
M EI a
a
Ans. ω 1 = 0.8909
2M ω 2 = 3.6887
a
EI 3
a M
EI a3M
,
1 1 1.0483 − 0.4769
,Φ =
Fig. P16.6
16.7. The massless beam carried body of mass M and moment of inertia Jx with respect to x0-axis which pass over centroid of mass M (Fig. P16.7), radius of gyration is ρ ¼ J x0 =M l2 . Calculate the frequencies of the free vibration and find the modal matrix.
z
ρ
EI x
x0
l
M Jx
Ans. ω12 = y λ1, 2 =
3EI λ1Ml 3
, ω 22 =
3EI λ 2 Ml 3
,
1+ 3 ρ ± 1+ 3ρ + 9ρ 2 2
Fig. P16.7
16.8. The massless uniform beam carried two lumped mass as shown in Fig. P16.8. Calculate the frequencies of free vibrations and find the modal matrix. (Hint: For construction of bending moment diagram in unit states use the influence lines).
M2=2M
M1=M y1
EI
y2 l/3
l/3
l
Ans. ω1 = 6.379 ω2 = 11.047
EI 3
EI l 3M
,
, Φ=
l M
1 1 − 1.4142 0.3462
l
Fig. P16.8
16.9. The uniform circular rod is clamped at point B and carrying the lumped mass M at free end A (Fig. P16.9). Calculate the frequencies of free vibrations, find the modal matrix, and show the shapes of vibrations. The first and second generalized coordinates are directed in vertical and horizontal direction, respectively. Ans. ωi2 = EI R R Fig. P16.9
EI
λi · 0.7854MR3 λ1 = 1.4195, λ2 = 0.03395
M
B
Φ =
1 1 0.659 - 1.517
612
16
Dynamics of Elastic Systems: Free Vibration
16.10. The symmetrical frame with absolutely rigid crossbars of total mass M and 2M and massless columns is shown in Fig. P16.10. Calculate the frequencies of free vibrations, find the modal matrix, and show the shapes of vibrations. y1
EI=∞, M EI
h
y2
EI=∞, 2M h
EI EI , ω2 = 6.9282 3 , h M h3 M 1 1 Ф= 0.5 − 1
Ans. ω1 = 3.4641
2EI
Fig. P16.10
16.11. The symmetrical frame with absolutely rigid crossbars of total mass M and 2M and massless columns is shown in Fig. P16.11. Find the frequencies of free vibrations and the modal matrix.
y1
EI=∞, M h EI=∞, 2M h
EI EI , ω22 = 41,7366 3 , 3 h M h M 1 1 A= 0.6765 − 0.7390
Ans. ω12 = 7.7628
EI
EI
y2 EI
Fig. P16.11
16.12. Derive the frequency equation, calculate the frequencies of vibration, and find the mode shape of vibrations for beam (length l, flexural stiffness EI, mass per unit length m) with the following classical boundary conditions: rffiffiffiffiffi (a) Fixed-fixed beam; λ2i EI (b) Clamped-pinned beam; (c) Guided-clamped beam. Apply Krylov-Duncan functions, ωi ¼ 2 . m l rffiffiffiffiffi 4:732 EI 2 Ans: ðaÞ U ðklÞ T ðklÞV ðklÞ ¼ 0 ! cosh kl cos kl ¼ 1 ! λ ¼ kl ¼ 4:7300; . . . ω1 ¼ 2 , ... m l U ðk i lÞ T ðk i lÞ X i ð xÞ ¼ C U ð k i xÞ V ð k i xÞ ¼ C U ð k i xÞ V ð k i xÞ V ðk i l Þ U ðk i l Þ ðbÞ tan kl ¼ tanh kl ! kl ¼ 3:92266; 7:0768; 10:2102, . . . ðcÞ tan kl þ tanh kl ¼ 0 ! kl ¼ 2:3650;
5:4987, ::
16.13. Derive the frequency equation, for beam with fixed left support and elastic translational right support; the length of the 000 beam is l, flexural stiffness EI, mass per unit length m, and stiffness of elastic translational support is c, Q(l) ¼ EIX (l ) ¼ cX(l). Ans: λ3
cosh λ cos λ þ 1 cl3 þ ¼0 cosh λ sin λ sinh λ cos λ EI
16.14. Compare the values of the fundamental frequency of free vibration of uniform massless beam with lumped mass M in the middle of the span for three different types of beam: (1) simply supported beam, (2) clamped-supported beam, and (3) clamped-clamped beam. Hint: Take into account Problems 16.1a and 16.2b,d. Ans: ω1 ω2 ω3 ¼ 1 1:51 2:0
Problems
613
16.15. Design diagram of uniform two span beam with distributed mass is shown in Fig. P16.15. (a) Derive the frequency equation; apply initial parameters method. (b) Determine the interval in which the roots of the equation are located; the theorem about introduced linear constraint can be used (Sect. 16.4.4).
y(l)=0 x=0 EI, m l
M0
x
Ans. (a) U lV2l Tl + Vl 2 S 2l − U lVl T2l − U 2lVl Tl = 0 ; Ul = U ( kl ) , V2l = V ( 2kl ) ,···
l
(b), 3.926 ≤ λ1 ≤ 7.068 ≤ λ 2 ≤ 10.210 .
R
Q0 Fig. P16.15
16.16. Design diagram of uniform two span beam with uniformly distributed mass m is shown in Fig. P16.16. Derive the frequency equation. Apply the initial parameters method. Show the symmetrical and nonsymmetrical mode shape of vibrations. Present the frequency equation in general form and in the elementary functions.
EI, m l
l
1 1 0 EI T (k l) V (k l) k k3 1 1 1 Ans: EI T (2k l) V (2k l) V (k l) = 0 , k k3 k3 1 1 EIk V(2k l) T (2k l) T (k l) k k sin kl (tan kl − tanhkl) = 0
Fig. P16.16
16.17. Derive the frequency equation of uniform cantilever beam of length l; bending stiffness EI ¼ const. Apply the concept of transfer matrix. (Fig. P16.17).
x=0 x=l M(l)=0 Q(l)=0 Fig. P16.17
Ans. 1 + cosh kl cos kl = 0 . kl = 1.875, 4.694, 7.855,···
l, EI Y0=0 θ0=0
ω i = ki2
EI m
614
16
Dynamics of Elastic Systems: Free Vibration
16.18. Design diagram of symmetrical frame is shown in Fig. P16.18. All members of length l and the bending stiffness EI ¼ const have uniformly distributed masses m0. Derive frequency equation, and solve it for l ¼ 6 m. Apply displacement method. Hint: Take into account the symmetry of structure. For analysis of symmetrical vibration apply group of unit angular displacements: both introduced constraints at joints 1 and 2 rotate in the opposite directions. Considering joint 1 don’t forget a term due to rotation of joint 2. 1
l, EI, m0
2 Ans. 3iψ1 ( u ) + 8iψ2 (u ) − 2iψ3 (u ) = 0 ; m0ω 2 l 4 , EI u1 = 3.34, u2 = 4.25, u3 = 4.73,··· u=4
Fig. P16.18
16.19. Determine the frequencies of free vibration for the frame shown in Fig. 16.19. Each element of the frame has uniformly distributed mass m0, the length l, and the bending stiffness EI ¼ const. Apply the displacement method and Smirnov functions. 1
B m, l, EI
Ans. ψ2 (u ) = 0 , u = l
4
mω 2 EI ,
A Fig. P16.19
16.20. Design diagram of frame with absolutely rigid crossbar of total mass M is shown in Fig. 16.20. r Mass ffiffiffiffiffi per unit length for 1:92 EI . vertical members is m. Determine ratio M/ml that provides the frequency of free vibration ω ¼ 2 m l EI=∞, M Ans. 24ψ10 ( u ) = EI, m,l
EI, m,l
M 4 u , u = l 4 mω 2 EI , ml
M = 1.0889 ml
Fig. P16.20
16.21. The simply supported beam of length l has a lumped mass m. How will the frequency of free vibrations change if the lumped mass m is moved from position x ¼ 0.5l to the position x ¼ 0.25l? ω Ans: 0:25l ¼ ω0:5l
rffiffiffiffiffiffiffiffiffi 85:3 ¼ 1:333 48
Chapter 17
Dynamics of Elastic Systems: Forced Vibration
This chapter is devoted to analysis of forced vibration of linear deformable structures. The structures with finite and infinite number of degrees of freedom, subjected to forced and kinematic excitation are considered. Vibration analysis is performed in the terms inherent to the deformable structures adopted in the classical course “Structural analysis.” The continuity of classic analysis methods, such as the force and displacements methods and the initial parameters method, to dynamic problems of deformable systems is presented. If a structure is disturbed by any reason from its rest condition and then this reason is eliminated, then the structure will perform free vibration; this type of vibration had been considered in Chap. 16. If there is an excitation source that continuously feed energy to the structure, then the structure will accomplish forced vibrations. When analyzing free vibration, we classified systems by the type of differential equation of motion (linear, nonlinear) and by the number of degrees of freedom (with one degree of freedom, finite and infinite number of degrees of freedom). In the case of forced vibration, the nature of the excitation should be taken into account in addition. First of all, we will classify disturbance as force and kinematic excitation. The force excitation means the disturbance of a system by loads (lumped and distributed forces and moments), which are applied to the masses of the system and/or to the structure itself. The kinematic excitation means the disturbance of a system by displacements (linear, angular) of the supports of a structure. In both cases, the law of change of excitation should be indicated as a function of time. When a structure is subjected to any type of excitation, its response presents a combination of free and forced vibrations. Vibrations, which are caused by the initial conditions and disturbing excitation of any nature simultaneously, are called the transient vibrations. Due to the inevitable resistances, a free vibration of a system will subside with time. Vibration of structure that take place after the damping of free vibration and are supported only by external periodic excitation are called the steady-state vibrations.
17.1
Structures with One Degree of Freedom
This paragraph is devoted to deriving the differential equation of vibration of an arbitrary linear deformable system with one degree of freedom. Two fundamental approaches are applied—the force and displacement method. Restrictions on the type of system, degree of statical indeterminacy, and the type of supports are not imposed. A general method for solving the differential equation using the Duhamel’s integral is considered and its various applications in the case of typical excitations are shown.
17.1.1 Differential Equations: Two Classical Approaches Let us consider a massless beam with arbitrary boundary conditions. The beam carries one lumped mass m which can move in vertical direction only. So this system has one degree of freedom. The static equilibrium position determines the position of the elastic line of the beam due to the body weight P0 ¼ mg (Fig. 17.1a). A beam is subjected to a force P(t) that acts at an arbitrary point of the beam and causes movement of a mass m. In the process of vibration an elastic line of the beam occupies an intermediate position, and corresponding position of mass m is measured from the static equilibrium position by coordinate y (Fig. 17.1b). Assume that the positive direction for displacement, velocity, and acceleration are directed downwards. In position A, we separate the mass m from the beam and show the forces acting on the beam: they are the force of interaction R between the beam and mass, and disturbing force P(t) (Fig. 17.1c). In the same position A the forces which act on © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_17
615
616
17
Dynamics of Elastic Systems: Forced Vibration
the mass m are interaction force R, resisting force Pdirected opposite of the velocity, and inertial force m€y, which is directed opposite of the acceleration (Fig. 17.1d).
a
a
b
yst m
SEP
a
yst
c
P(t)
SEP
y
P(t)
R IP
IP
A
A
d
mÿ P* m R
Fig. 17.1 The massless beam with one lumped mass m. (a) Design diagram; (b) Intermediate position (IP), SEP—static equilibrium position; (c, d) The forces acting on the beam, and on the mass in the position A
It is obvious that this system has one degree of freedom. However, the problem is to present the equation of motion of mass m in terms that are inherent for an arbitrary deformable system. Interaction force R may be presented in two forms. According to D’Alembert principle R ¼ m€y þ P
ð17:1Þ
On the other hand, displacement of the mass m may be presented in terms of displacement of the beam as follows y ¼ δ11 R þ Δ1P ðt Þ ! R ¼
y Δ1P ðt Þ δ11
ð17:2Þ
Here δ11 is unit displacement of the beam at point where mass m is located caused by force P¼1 which applied at the mass and acts in direction of motion of mass (the negative sign means that interaction force R, which acts on the beam, and displacement of mass m have different direction (Fig 17.1b)); Δ1P is displacement of the same point of the beam caused by the given load P(t). Thus, interaction force R may be presented in terms of acceleration €y of mass m (17.1) or in terms of displacement y of a beam (17.2) at point where m is attached. Two expressions (17.1) and (17.2) for force R can be combined in two ways. This important circumstance leads to the possibility of analyzing forced vibrations by classical forces and displacements methods (Kiselev 1980). 1. We will explore the equilibrium state of mass m by the expression (17.1) for which the force R is determined by the formula (17.2). For this we need R according to (17.2) to insert into (17.1). In this case we have which leads to equation €y þ
y Δ1P ðt Þ ¼ m€y þ P , δ11
P 1 1 1 y¼ Δ ðt Þ. If we denote ¼ ω2 , finally we get þ mδ11 1P mδ11 m mδ11 €y þ
P þ ω2 y ¼ ω2 Δ1P ðt Þ ¼ F ðt Þ, m
Δ1P ðt Þ ¼ δ1P Pðt Þ,
ð17:3Þ
where δ1P is displacement in direction of motion of the mass m caused by the disturbing force P ¼ 1. 2. Now we will explore the displacement of mass m by the expression (17.2) for which the force R is determined by the formula (17.1). For this we need R according to (17.1) to insert into (17.2). In this case we have y ¼ δ11 R þ Δ1P ðt Þ ¼ δ11 ðm€y þ P Þ þ Δ1P ðt Þ: This expression immediately leads to equation (17.3). It turns out there is a fundamental possibility of obtaining equations of motion in two ways. For systems with one degree of freedom two approaches for deriving differential equation of forced vibration leads to the same result. In case of structures with finite number of degrees of freedom the differential equations of vibrations take the different forms. The advantage of equation (17.3) is that it contains parameters δ11 and Δ1P. Therefore, it allows presenting an arbitrary massless deformable system with one degree of freedom in the form of linear oscillator (mass on the spring) and to consider free and forced vibration in terms inherent to the structure itself and take into account its peculiarities. Among them are type of
17.1
Structures with One Degree of Freedom
617
structure (beam, frame, truss, arch, etc.), boundary conditions, geometrical parameters, stiffness, etc. Therewith the location of the mass and the disturbing force P(t) can be arbitrary. Note that, taking into account the properties of the deformable system, the real perturbing force P (t) is replaced by the conventional or equivalent force F(t) ¼ ω2δ1PP(t). The presented derivations of equation (17.3) have two disadvantages. First, there is absent visual link between the dynamic system in Fig. 17.1a (arbitrary deformable structure with one lumped mass) and typical mass-spring model. Secondly, the difficulty of applying these approaches to deriving equations in the case of a system with a number of degrees of freedom more than one becomes obvious. To eliminate these shortcomings we modify the original deformable system. Modification of the original dynamic system. The equation of forced vibrations (17.3) of a mass located on an elastically deformable structure can be obtained by the classical force and displacement methods. To use these methods, we first modify the original dynamic system: at the point where mass m is located, we introduce additional linear elastic constraint k, the reaction of which ky equals the interaction force between the mass and the beam (Fig. 17.2a). R ¼ ky ¼ m€y þ P
ð17:4Þ
The fundamental idea of this modification is as follows. 1. Original dynamical structure may be presented as a simple oscillator. The stiffness of the elastic element is k¼
m€y þ P y
ð17:5Þ
The introduced elastic constraint increases the degree of redundancy by one. 2. Introducing the elastic support reduces the dynamic problem to a static one: it is necessary to determine the reaction of the elastic support, which is equal to the dynamic force acting on the beam at the point where mass m is attached.
a
b a
P(t) m
k
b
b P(t) m a X1
Fig. 17.2 (a) Modified design diagram of a beam with elastic support of stiffness k; (b) Force method: X1 is reaction of elastic support
Now, to derive the differential equation of motion of the original system, two classical methods can be applied to the modified design diagram.
17.1.1.1
Force Method
Reaction of this elastic support X1 can be considered as a primary unknown of the force method. Corresponding primary system is shown in Fig. 17.2b. This unknown is determined from the canonical equation δ11 X 1 þ Δ1P ¼
R k
It is obvious that primary unknown X1 and interaction force R between mass and beam, shown in Fig 17.1c and 17.2b, are related as follows: X ¼ R. Therefore the previous equations may be presented in the form
618
17
δ11 X 1 þ Δ1P ¼
Dynamics of Elastic Systems: Forced Vibration
X1 1 ! δ11 þ X 1 þ Δ1P ¼ 0 k k
ð17:6Þ
This equation has fundamental feature. Since the displacement y(t) of mass m is unknown, then Eq. (17.6) involves unknown spring stiffness k. Moreover, since Eq. (17.5) involves stiffness k in the differential form, then Eq. (17.6) is not algebraic. According to Eq. (17.4) we have X ¼ R ¼ ðm€y þ P Þ
ð17:7Þ
Substitute the expression (17.5) for stiffness of elastic support and expression (17.7) for unknown reaction X into (17.6) δ11 þ
y ðm€y þ P Þ þ Δ1P ¼ 0 m€y þ P
This equation relates the required function y(t), the unit δ11, and loaded Δ1P displacements, and immediately leads to Eq. (17.3).
17.1.1.2
Displacement Method
In order to get the primary system of the displacement method for modified design diagram, we introduce an additional constraint in the direction of mass displacement (Fig. 17.3a). Displacement Z of introduced constraint coincides with required displacement y(t) of mass m, i.e., y ¼ Z1 .
a
P(t) m
b
r11 1
Fig. 17.3 Analysis of modified design diagram of beam. (a) Primary system of displacement method; (b) Unit displacement and reaction in the introduced constraint
Unit displacement Z1 ¼ 1 of introduced constraint (i.e., the mass m) leads to reaction r11 in this constraint. The positive directions of the reaction r11 and displacement Z1 ¼ 1 coincides. Canonical equation of displacement method r 11 Z 1 þ R1P ðt Þ ¼ 0,
ð17:8Þ
where R1P is reaction of introduced constraint caused by disturbing force P(t). Unit reaction r11 consists of two quantities. The first term r 11 is restoring force, which is determined by elastic response of the beam; the reaction r 11 and unit displacement relates as r 11 ¼ 1=δ11. The second term is reaction which is caused by deformation of introduced elastic constraint of stiffness k; thus, unit reaction is r 11 ¼ r 11 þ k. If expression for r11 and expression (17.5) for k are substituted into Eq. (17.8), we get r 11 þ k Z 1 þ R1P ðt Þ ¼ 0 or
r 11 þ
m€y þ P y þ R1P ðt Þ ¼ 0 ! r 11 y þ m€y þ P þ R1P ðt Þ ¼ 0 y
ð17:9Þ
17.1
Structures with One Degree of Freedom
619
This is the required differential equation of the motion of lumped mass m in the form different from Eq. (17.3); it connects the desired function y(t), unit r11, and loaded R1P reactions. Now we need to compare the differential equations which have been obtained by the force and displacement methods. Since r 11 ¼
Δ ðt Þ 1 , and R1P ðt Þ ¼ 1P , δ11 δ11
then Eq. (17.9) becomes Δ ðt Þ 1 y þ m€y þ P 1P ¼ 0, δ11 δ11 which immediately leads to Eq. (17.3). Thus, for systems with one degree of freedom, both approaches and both methods lead to the same differential equation (17.3) of motion. It is shown above that the analysis of linear deformable structures with a finite number of degrees of freedom can be performed by two classical analytical methods—the force and displacement methods, which were previously widely used in problems of strength and stability of the rod structures. Both forms of representation of the dynamics of systems with a finite number of degrees of freedom will be considered in Sect. 17.2. Example 17.1 The uniform clamped-free beam of length l and stiffness EI carries lumped mass m at the free end and subjected to disturbing force P(t) (Fig. 17.4). Compose the equation of forced vibration of mass m; the resisting forces may be ignored. P(t) m
A l/2
1
Fig. 17.4 Design diagram of structure with one degree of freedom
Solution The mass on the beam performs vertical vibration in direction 1. To determine the displacement δ11, it is necessary to construct the bending moment diagram M 1 due to force P ¼ 1 applied to the mass m. To determine the displacement δ1P, we need to construct the bending moment diagram M P due to force P ¼ 1 applied at point A. The required displacements are determined by multiplying the corresponding diagrams as follows δ11 ¼
M1 M1 l3 ¼ , EI 3EI
δ1P ¼
M1 MP 5l3 ¼ EI 48EI
Frequency of free vibration is rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi 1 3EI ω¼ : ¼ mδ11 ml3 According to (17.3), we have €y þ ω2 y ¼ ω2 δ1P Pðt Þ, so the differential equation of mass m in the form of linear oscillator becomes €y þ
3EI 5 y¼ Pðt Þ 16m ml3
Equation €y þ ω2 y ¼ ω2 δ1P Pðt Þ has great generality, since it holds not only for arbitrary deformable rod systems but also for two-dimensional structures such as plates with one lumped mass.
620
17
Dynamics of Elastic Systems: Forced Vibration
17.1.2 Types of Excitations The behavior of a dynamic system essentially depends on the character of excitation (force and kinematical), the type of external excitation (periodic, impulsive, etc.), and the law of change of the excitation as a function of time (harmonic, stepped, etc.). Generally, disturbing forces may be of the following types. 1. Immovable periodical loads produced by stationary units and mechanisms with moving parts. Such loads are located at a fixed point of a structure, and generally do not depend on its elastic properties. The character of the periodic law can be continuous and discontinuous. Of special interest is harmonic load F(t) ¼ F sin θt, which occurs with a constant frequency excitation θ and amplitude F of the disturbing force F(t). Such an excitation leads to steady-state vibration of the structure, which happens with a frequency of excitation. If the frequency of excitation and the frequency of free vibration of a structure are equal (θ ¼ ω), then vibration of the structure occurs with infinitely increasing amplitudes. This is a dangerous phenomenon for engineering structures and is called as resonance. 2. Impulsive or short-duration loads are produced by a blast, falling weights (pile drivers, hammers, etc.), or collision of bodies. Such loads are characterized by rapid development and rapid disappearance, i.e., by very short duration of their action, and depend on the elastic properties of the structure which is subjected to such loads. Impact loading is characterized by a sharp change in the velocity of the body being struck in a short period of time. It often occurs as shock periodic load (Harris and Piersol 2002). 3. Moving loads affect structures through the wheels of a moving train or truck. If we take into account the mass of the moving load, then the design diagram should include inertial forces. This type of moving load should be distinguished from moving load, which was studied in the “Influence Lines” sections, where the term “moving load” implies that only the load position can be arbitrary, i.e., it is a static load, for which inertial effects are not taken into account (Filippov 1970). 4. Seismic loads arise due to earthquakes. The reason for the presence of seismic load on a structure is acceleration of the supports caused by acceleration of the ground. This type of excitation is kinematical. The acceleration of supports leads to the acceleration in the individual parts of the structure, and as a result the inertial forces act on these parts. Seismic forces which arise in the individual elements of the structure are dependent on the type and the amount of ground acceleration, distribution of the mass within the elements of the structure, as well as their elastic properties (Clough and Penzien 1975). 5. Deterministic and random loads. If forces which act on the structure are characterized by the preciseness of their parameters, then the corresponding vibrations are called deterministic vibrations. There exists a series of dynamical loads which are characterized by the lack of preciseness of their parameters. These include loads created by wind, loads which arise on account of irregularities of the deck on car bridges, seismic loads, etc. In all these cases, it is impossible to set factual parameters to these loads. Such loads are called nondeterministic and the corresponding vibrations are random vibrations (Bolotin 1964). By the type of deformation of elastic structures, vibrations can be divided into bending, longitudinal, and torsional ones (Nowacki 1963). Below we limit ourselves to bending vibrations of linear systems with a finite and infinite number of degrees of freedom and, at the same time, we will neglect energy dissipation (dissipative forces), unless otherwise specified.
17.1.3 Duhamel Integral and Some Special Types of Excitation This section is devoted to analysis of any linear deformable structure subjected to some typical excitations: the constant-force excitation, pulse of duration τ, impulse and harmonic excitations. The forced undamped vibration of a linear oscillator subjected to an arbitrary disturbing force F(t) is described by the equation m€y þ ky ¼ F ðt Þ,
ð17:10Þ
where y(t) is the displacement of the mass m measured from its static equilibrium position, k is the stiffness coefficients. The general solution of equation (17.10) is yðt Þ ¼ y0 cos ω t þ
υ0 sin ω t þ y ðt Þ ω
ð17:11Þ
17.1
Structures with One Degree of Freedom
621
The first two terms of expression (17.11) describe free vibration of a mass mffi with its initial displacement y0 and initial pffiffiffiffiffiffiffiffi velocity υ0. These vibrations occur with frequency of free vibration ω ¼ k=m . Last term in (17.11), y(t), describes the forced vibration which depends on the disturbing force F(t) (Newland 1989) 1 y ðt Þ ¼ ωm
ðt F ðτÞ sin ωðt τÞdτ
ð17:12Þ
0
Expression (17.12) is called Duhamel’s integral (1843), or alternatively convolution integral or Green’s integral. This formula is derived on the basis of superposition of the responses of the oscillator (17.10) to a sequence of impulses. The derivation of this formula can be found in numerous textbooks, in particular (Thomson 1981; Timoshenko 1972). In case of nonzero initial conditions y0 and υ0 the expressions (17.11), (17.12) describe the transient vibrations of oscillator (17.10). The equations (17.10, 17.11, and 17.12) do not take into account the resisting forces. It may be explained as follows: because of the inevitable presence of various damping forces in the structure, the free vibration in the (17.11) rapidly disappears and only the purely forces component y(t) remains. To simplify further transformations, we assume that purely forced vibrations y(t) occur without taking into account the resistance forces. Therefore, formula (17.12) allows determining the response of the steady-state vibration of linear system (17.10) with a single degree of freedom in the case of an arbitrary disturbing force F(t) and ignoring the resisting forces. Now we need to present Duhamel integral (17.12) in the form, which can be applicable to an arbitrary deformable structure, which is considered as the elastic structure with one degree of freedom. Assume that at moment t ¼ 0, the plane bar structure is subjected to distributed load (Clough and Penzien 1975) qðx, t Þ ¼ qðxÞ φðt Þ,
ð17:13Þ
where q(x) and φ(t) are arbitrary functions of coordinate x and time t, respectively. Approximate analysis of a bar structure, subjected to arbitrary loads (impact, harmonic excitation, etc.), is based on the following assumptions: 1. Vibration of the bar structure occurs as in a system with a single degree of freedom. 2. The mode of shape of vibration X(x) coincides with elastic line which corresponds to static load q(x). Upon these assumptions, Duhamel’s integral, which is modified for deformable structure, may be presented as follows (Rabinovich 1960), ðt yðx, t Þ ¼ ω X ðxÞ φðτÞ sin ω ðt τÞdτ
ð17:14Þ
0
where X(x) is mode of vibration, ω is the frequency of free vibration, and φ(t) is a function which characterizes the law of change of load in time. The formula (17.14) has a high accuracy for the smallest frequency, which corresponds to the simplest form X(x) of vibration. The importance of the Eq. (17.14) for response y(x, t) is as follows: the features of the deformable structure, such as the type of structure, its supports and degrees of redundancy, the type of load and its position, etc., are taken into account by the form of vibration X, and the structure itself is considered as a system with one degree of freedom. Now we will show an application of formula (17.14) for some standard excitations. Constant-Force Excitation (Harris and Piersol 2002) An arbitrary load is suddenly applied to the deformable structure and then permanently remains on it (Fig. 17.5a). In this case, at t > 0 φ(t) ¼ 1, so expression (17.14) becomes ðt yðx, t Þ ¼ ω X ðxÞ 1 sin ω ðt τÞdτ ¼ X ðxÞð1 cos ω t Þ ¼ ystat ð1 cos ω t Þ 0
υðx, t Þ ¼
dyðx, t Þ ¼ ωX ðxÞ sin ω t dt
ð17:15Þ
622
17
Dynamics of Elastic Systems: Forced Vibration
It means, the structure executes the harmonic vibration around the static elastic curve X(x). Dynamic coefficient y(x, t)/ystat ¼ 1 cos ω t is a function of time, and for any point of the structure the maximum displacement is twice more than the static displacement at the same point, ymax ¼ 2ystat. For linear oscillator (17.10) the graph of vibration is presented in Fig. 17.5b.
b
a F(t)
y(t)
F0
ystat t
t
Fig. 17.5 Constant force excitation and the corresponding response
Solution (17.15) can be applied only in the case when the load being suddenly applied to the structure remains on it for an unlimited time. The frequency of free vibration ω and the form of vibration X(x) is determined by the type of construction, its supports, and the static nature of the load q(x). For example, a simply supported beam of length l is subjected to suddenly applied the uniformly distributed load q along entire span and then this load remains on the beam. The exact expression for elastic curve caused by static load is X ðxÞ ¼ ystat ¼
q 3 l x 2lx3 þ x4 , 24EI
where x is measured fromrthe ffiffiffiffiffi left support. The frequency of a free vibration of a beam with uniformly distributed mass π 2 EI m ¼ q/g equals ω ¼ 2 . Thus, we assume that the mode of vibration X(x) coincides with elastic curve caused by m l uniformly distributed load q. Displacement at the middle of the beam and the slope at support are l 5 ql4 y , t ¼ X ðxÞjx¼l=2 ð1 cos ω t Þ ¼ ð1 cos ω t Þ 2 384 EI dX ðxÞ ql3 ð1 cos ω t Þ ¼ ð1 cos ω t Þ θð0, t Þ ¼ θstat ð1 cos ω t Þ ¼ dx x¼0 24 EI Let us consider arbitrary deformable structure ignoring the mass of its elements. Assume that this structure contains a lumped mass m at point 1. Force F0 is applied at an arbitrary point of the structure and then remains there. Loading of structure by force F0 may be treated as introducing into a system the additional mass М ¼ F0/g. This leads to an increase in the number of degrees of freedom by one. Let mass M be satisfied to condition M m. In this case for determining the response of the structure—only mass m should be taken into account, i.e., consider this structure as a system with one degree of freedom. Thus, displacement of mass m according to (17.15) becomes yðt Þ ¼ ystat ð1 cos ω t Þ ¼ F 0 δ1P ð1 cos ω t Þ,
ð17:16Þ
where δ1P is displacement of point 1 caused by force F0. For example, fixed-free uniform beam of length l and bending stiffness EI carried a lumped mass m at the free end. Force F0 is applied at point at distance a from fixed support, and remains on the beam. In this case the displacement of an arbitrary point х of cantilevered beam becomes F 0 x2 ð3a xÞ ð1 cos ω t Þ for x a 6EI 2 F a yðx, t Þ ¼ 0 ð3x aÞ ð1 cos ω t Þ for x a 6EI
yðx, t Þ ¼
Displacement of a free end is
17.1
Structures with One Degree of Freedom
623
yðl, t Þ ¼
F 0 a2 ð3l aÞ ð1 cos ω t Þ 6EI
Pulse of Duration τ The load q(x) suddenly appeared on the structure at the time t ¼ 0. This load remains constant during the time t ¼ τ and then suddenly disappeared. This type of load is called as the rectangular pulse (Fig. 17.6a). Within the interval t < τ the displacement of structure is described by (17.15). Assume that duration of load actions τ and period of free vibration T satisfy the condition τ < 0.25 T or ωτ π/2. In this case for t < τ we have yðx, t Þ < X ðxÞ,
∂y >0 ∂t
ð17:17Þ
It means, within time interval 0 τ (τ π/2ω), when the load yet is on the structure, the deformation of the structure did not have time to reach the static value X, and the speed is still positive. In other words, a maximum displacement occurs after disappearing of the load (Rabinovich 1960). The rectangular pulse in Fig. 17.6a can be presented as a sum of two infinite in time loads. One of them, a given load q(x), is applied at moment t ¼ 0, аnd second, a compensated load, q(x), is applied at moment t ¼ τ (Fig. 17.6b). Their sum leads to the given stepped load within time interval 0 < t < τ.
a
b
q(t) q(x) 0
q(t) q(x)
τ
t
0
τ
t
Fig. 17.6 (a, b) Stepped excitation of finite duration, and its transformation to the two constant-force excitations
Since each of these loads act non-limiting time, it is possible to apply formula (17.15). yðx, t Þ ¼ X ðxÞ ½1 cos ω ðt 0Þ X ðxÞ ½1 cos ω ðt τÞ Here, each term corresponds to the load that is attached to the structure and remains on it for an unlimited time. The first term reflects the effect of the load q(x) that is applied at the moment t ¼ 0, and the underlined term reflects the effect of the load q(x), which is directed opposite q(x), and is applied at moment t ¼ τ . Thus, we get yðx, t Þ ¼ X ðxÞ½ cos ωðt τÞ cos ωt
ð17:18Þ
The expression (17.18) may be rewritten in the following form yðx, t Þ ¼ 2X ðxÞ sin
τ ωτ sin ω t 2 2
ð17:18aÞ
Thus, the structure executes vibration around unloaded state with frequency of the free vibration ω and amplitude ωτ 2X ðxÞ sin . Therefore, 2 y max , min ¼ 2X ðxÞ sin
ωτ 2
ð17:19Þ
Since the frequency of free vibration is ω ¼ 2π/T, then dynamical coefficient is μdyn ¼
ymax ymax πτ ¼ 2 sin ¼ T ystat X
Thus, the effect of the short-term load depends on the dimensionless parameter τ/T, where T is period of free vibration. Maximum of the dynamical coefficient is equal to 2. For τ < T/6 dynamical coefficient is μdyn < 1 (Table 17.1)
624
17
Dynamics of Elastic Systems: Forced Vibration
Table 17.1 Dynamical coefficient vs. parameter τ/T
t T mdyn
0.01
0.02
0.05
0.10
0.167
0.20
0.30
0.40
0.50
0.0628
0.126
0.313
0.618
1.000
1.175
1.617
1.902
2.000
Impulse Excitation A bar structure is subjected to a distributed load q(x). This load acts within a very small time interval τ. Elementary load and its elementary impulse are q(x)dx and τ q(x)dx. Now assume that the same structure is subjected to distributed impulse s(x), so the elementary impulse becomes s(x)dx. Thus, elementary impulse may be presented in two forms: in terms of distributed load q(x), elementary portion dx and time τ, and from other hand, as distributed impulse s(x)and portion dx. Therefore τqðxÞdx ¼ sðxÞdx, thus qðxÞ ¼ sðxÞ=τ Thus, the action of a distributed impulse s(x) is formally reduced to the action of a distributed load q(x). To determine the maximum displacement, the relation (17.19) can be used. If we multiply and divide its right side by ωτ/2, we get y max , min ¼ X ðxÞ ωτ
sin ðωτ=2Þ ωτ=2
Since the X(x) is an elastic curve, which corresponds to a static load q(x), then X(x) ωτ is the elastic curve due to the static load ω τ q(x) ¼ ω s(x). Thus, with respect to effect, the action of any distributed impulse s(x) is equivalent to a static distributed load qeq(x) with the intensity at any point of a structure being qeq ðxÞ ¼ ω sðxÞ
sin ðωτ=2Þ ωτ=2
ð17:20Þ
The unit of distributed impulse s(x) is [kNsec/m]. If impulse S [kNsec] is applied at a single point, then equivalent static force Feq at the same point is F eq ¼ ω S
sin ðωτ=2Þ ωτ=2
ð17:20aÞ
sin α sin α < 1 and lim ¼ 1, then for two impulses with equal S but different τ, the impulse with smaller τ is more α α!0 α dangerous. Therefore, the instantaneous impulse is most dangerous. Equivalent distributed and concentrated loads for this case are Since
qeq ðxÞ ¼ ω sðxÞ F eq ¼ ω S Two fundamentally different cases should be distinguished. They are the effect of a constant force and the effect of a constant impulse; both force and impulse acts over time τ. With the decrease of time τ, effect of the constant force F trends to zero, but the effect of the constant impulse S trends to a maximum. Indeed, in the case of constant force F, the impulse of a force S ¼ Fτ with decreasing τ trends to zero. In the second case, the impulse S for any τ retains its magnitude: this may happen in the case when with a decrease in time τ the force F increases (Rabinovich 1960; Karnovsky 2012). The peculiarity of the action of short-term and instantaneous impulses (τ ! 0) on the structure is the following. According to (17.20, 17.20a), the equivalent static load depends not only on the values of the impulse but also on the properties of the structure. Indeed, the mechanical properties of the structure will be determined by its boundary conditions, geometry, material, the stiffness of separate elements, and the value of the distributed mass of elements. The greater the stiffness and the lighter the structure, the greater is its frequency ω, and thus the greater is the equivalent load.
17.1
Structures with One Degree of Freedom
625
Example 17.2 Determine the effect of instantaneous impulse s which acts on the uniform simply supported beam. This impulse is uniformly distributed within the entire span l. Parameters of the beam are: the flexural rigidity EI, mass per unit length m rffiffiffiffiffi π 2 EI Solution The frequency of the free vibration is ω ¼ 2 . Equivalent static load on the beam is m l rffiffiffiffiffi π 2 s EI qeq ¼ ω s ¼ 2 , ðaÞ m l so the reactions of the beam are R max , min
qeq l π2s ¼ ωs ¼
¼ 2 2l
rffiffiffiffiffi EI m
ðbÞ
Bending moment and shear force are rffiffiffiffiffi qeq l2 π2s EI M max , min ¼
¼
8 m 8 rffiffiffiffiffi 2 qeq l π s EI Q max , min ¼
¼
2 2l m
ðcÞ
Displacement of the beam at the middle span is ymax ¼
4 5 qeq l 5π 2 sl2 pffiffiffiffiffiffiffiffiffi ¼ 384 EI 384 EI m
ðdÞ
Maximum normal stresses at the extreme fibers of the beam are M M π2 s e σn ¼ e ¼ 2 e ¼
I 8 r Ar
rffiffiffiffiffiffiffi E , Am
ðeÞ
where e is the distance from the neutral line to an extreme fiber of the beam, r is the radius of inertia, and A is the area of the cross section of the beam. Maximum shear stress at the neutral line is QU π2 s U στ ¼
¼
bI 2l br
rffiffiffiffiffiffiffi EI , Am
ðfÞ
where b is the width of the cross section at the neutral line and U is the first moment of the cross-sectional area above (or below) of the neutral axis with respect to the neutral axis. Discussion Comparison of the static loading and impulsive excitation has fundamental differences (Rabinovich 1960; Karnovsky 2012): 1. Even if the entire beam presents a statically determinate structure, in the case of impulsive excitation the reactions and internal forces depend on the flexural rigidity EI of the beam, while in case of static loading these parameters do not depend on the rigidity EI. 2. Increasing of flexural rigidity EI of a beam by n times leads to decreasing of displacement by n times for any static loading pffiffiffi and by n times at the impulsive excitation. 3. In the case of impulsive excitation, the expressions for the reactions, internal forces, stresses, and displacements have a pffiffiffiffi factor 1= m, while in the case of static loading, these expressions have a factor m. Indeed, the uniformly distributed load q may be presented in terms of the mass per unit length as q¼mg, where g is the acceleration due to gravity. 4. The bending moment due to impulsive loading does not depend on the length of the beam, while the shear force is inversely proportional to the length of the beam. In the case of a simply supported beam of length l, subjected to static uniformly distributed load, the bending moment and shear are proportional to l2 and l, respectively.
626
17
Dynamics of Elastic Systems: Forced Vibration
These comparisons show that the usual conceive about the behavior of structures under static loading cannot be formally transferred to the case of impulsive disturbing of engineering structures.
17.1.4 Harmonic Excitation: Equivalent Design Diagram Assume that the structure is subjected to a harmonic distributed load q(x, t) ¼ q(x)cos θ t or a concentrated force F(t) ¼ F0 cos θ t, where θ is a frequency of an external excitation. This structure is considered as having a single degree of freedom. Duhamel’s integral (17.14) in case of undamped deformable system leads to the following result ðt yðx, t Þ ¼ ωX ðxÞ
cos θτ sin ωðt τÞdτ ¼ ð17:21Þ
0
X ðxÞ ω2 ¼ X ð xÞ 2 ð cos θ t cos ω t Þ ¼ ð cos θ t cos ω t Þ, 2 ω θ 1 ðθ=ωÞ2 where X(x) is the elastic curve which is caused by the static action of a given load q(x) or F0. Expression (17.21) presents combination of forced vibration which occurs with frequency θ of disturbing force (the first term in brackets) and free vibration with frequency ω (the second term). This type of vibration is called as free concomitant (or accompanying) vibration (Babakov 1965). Note, if instead of periodic disturbance function cosθ τ to take sinθ τ, then Duhamel’s integral (17.21) should be written as follows: ðt yðx, t Þ ¼ ωX ðxÞ 0
X ð xÞ sin θτ sin ωðt τÞdτ ¼ 1 ðθ=ωÞ2
θ sin θ t cos ω t ω
In case of transient vibration also there is a third type of vibration caused by nonzero initial conditions; they are described by the first and second terms in (17.11). These vibrations, as concomitant vibration, occur with frequency of free vibration. In practice, due to the inevitable presence of various damping forces, the natural vibration with frequency ω rapidly disappears. The steady-state forced vibration of the simplified model, when a concomitant vibration in (17.21) are ignored, becomes yðx, t Þ ¼ A cos θ t ¼
X ð xÞ cos θ t 1 ðθ=ωÞ2
ð17:21aÞ
This expression describes a sustained periodic motion around the state of static equilibrium; this movement occurs with an amplitude A and a frequency θ of the disturbing force. By other words, the disturbing force imposes to the system its own vibration frequency. The factor μdyn ¼
A 1 ¼ X ðxÞ 1 ðθ=ωÞ2
ð17:21bÞ
is called the dynamic coefficient. This coefficient is the ratio of maximum displacement A of any point x of the structure and displacement at x due to the static load q(x). If λ ¼ θ/ω ! 1, the amplitude of forced vibrations sharply increases. If θ/ω ¼ 1, the dynamic coefficient becomes equal μdyn ¼ 1. This phenomenon is called resonance. In a real system, vibrations with infinite amplitude are impossible. Indeed, if θ/ω ! 1, the simplified model (17.21a, b), which does not take into account the dissipation of energy, does not reflect the actual phenomenon. In this case, we cannot neglect the free concomitant vibrations with a frequency ω and the exact solution (17.21) should be considered. X ð xÞ 0 If θ ! ω, then lim ð cos θ t cos ω t Þ ¼ . Evaluating uncertainty 0/0 (using L0 Hôpital rule), we can obtain 0 θ!ω 1 ðθ=ωÞ2 the precise expression for displacement of the structure in the resonance regime yðx, t Þ ¼ X ðxÞ
ωt ωt π sin ω t ¼ X ðxÞ cos ω t 2 2 2
ð17:22Þ
17.1
Structures with One Degree of Freedom
627
The peculiarity of this solution is that it describes nonperiodic motion. In the resonance zone (θ ! ω), vibrations increase unlimited over the time. This is indicated by the coefficient that contains time t as a multiplier. The expression with such coefficient is called secular term. From Eq. (17.22) it can be seen that at resonance the displacement of a structure and the disturbing force F(t) ¼ F0 cos θ t (or q(x, t) ¼ q(x) cos θ t) have a phase shift T/4, where T is a period of vibration. This shift occurs relatively to the disturbing force F(t) ¼ F0 cos ω t (it should be remembered that we consider the case when θ ffi ω). It means that the displacements of a structure become extreme at the moments when disturbing force is zero. Note the discrepancy between the initial assumption (small vibration) and the result (large vibration). This means that the mathematical model of the system in zone of resonance should be refined. However, in any case, the conclusion about the increase in vibration at resonance remains valid. If we assume that the energy dissipates in the system and the resistance force is proportional to the first power of the velocity F dis ¼ βy_ , then the differential equation and corresponding dynamic coefficient become β , m 2n γ¼ ω
€y þ 2n_y þ ω2 y ¼ P0 cos θt 2n ¼ 1 μdyn ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , 2 2 1 λ þ γ 2 λ2
The derivation of this formula can be found in numerous textbooks on the theory of vibration, in particular (Timoshenko et al. 1974). Thus, μdyn is a function of two dimensionless parameters, detuning λ ¼ θ/ω and damping parameter γ ¼ β/mω, where m is a lumped mass of a mechanical model of the deformable structure with one degree of freedom. For this model the frequency of free vibration ω is determined by the type of deformable structure, the boundary conditions, and geometrical and stiffness peculiarities. Graphs of dynamic coefficient for certain values of λ and parameter γ are given in Fig. 17.7; the dotted lines are related to case when a structural damping is neglected.
μdyn 5.0 4.5 γ=0.0 4.0 3.5
0.0
0.2
3.0 0.3 2.5 2.0 1.5 (√2, 1)
0.5 1.0 0.5 0.0
λ 0.25
0.5
0.75
1.0
1.25
1.5
1.75
2.0
Fig. 17.7 Steady state vibration of deformable structure with one degrees of freedom. Dynamic coefficient μdyn vs. detuning λ ¼ θ/ω; μdyn ¼ A/δ0; damping parameter γ ¼ β/mω. Dotted lines are related to undamped vibration
628
17
Dynamics of Elastic Systems: Forced Vibration
1. If λ ¼ θ/ω < 1, then the amplitude of forced vibration is A > δ0 ¼ P0/ω2. However, if the frequency ratio λ is very small (θ ω), then the amplitude of forced vibration is approximately equal to the static displacement of the mass m, i.e., A ffi δ0. Starting approximately from θ/ω ¼ 0.25, the dynamical coefficients rapidly increase. Note that an increase of the damping parameter γ leads to a decrease of dynamic coefficient. 2. At resonance (λ ¼ 1), the amplitude of forced damped vibration is Ares ¼ δ0/γ. The maximum values of dynamic coefficients are shifted left from the vertical line λ ¼ 1. The damping forces significantly decrease the amplitudes at the resonance. 3. If the frequency ratio is very large (θ ω), the dynamic coefficient is μdyn 1, i.e., the amplitude A of forced vibration is very small. For example, if λ ¼ θ/ω ¼ 2.5, and γ ¼ 0, then A ¼ 0.2δ0. This case is of special interest for the problem of the suppression of the forced vibrations in structures (Karnovsky and Lebed 2016) 4. The smaller the coefficient of resistance β (or dimensionless parameter γ ¼ β/mω), the larger the amplitude A of forced vibration.
17.1.4.1
Equivalent Design Diagram
Let an arbitrary deformable structure with one degree of freedom be subjected to harmonic excitation. If dissipation of energy is neglected, then for analysis of steady-state vibration may be effectively applied the following fact: Displacement of the lumped mass m caused by harmonic force F(t) ¼ F0 sin θt, which acts at an arbitrary point A of a massless structure, as well as by the inertial force J(t) ¼ J sin θ t of the lumped mass, is equal to the displacement of the mass caused by its weight G ¼ mg and force F(t) factored by dynamic coefficient μdyn and applied at the same point А (Fig. 17.8) (Kiselev 1980).
This assertion is written in the form δ1F F ðt Þ þ δ11 J ðt Þ ¼ μdyn δ1F F ðt Þ,
ð17:23Þ
where index 1 means the vertical displacement of mass m; dynamic coefficient μdyn is determined by formula (17.21b); δ1F and δ11 are displacement of mass m caused by unit force F which acts at point A and unit force J at point 1, respectively. Equivalent structure is subjected to force μdynF0 sin θt and the weight of the mass G¼mg; in this case the inertial force of the mass is not taken into account. m
a
F0sinθt A
1
J sinθt
a
μdyn F0sinθt
m
b G=mg Fig. 17.8 Transformation of design diagram of structure with one degree of freedom: (а) Initial design diagram, (b) Equivalent design diagram
The advantage of this transformation is that the analysis of the transformed design diagram in Fig. 17.8b is much simpler because it can be done using purely static methods. This conclusion also is valid with force F(t) ¼ F0 cos θt. Assume that a simply supported massless beam AB of length l carries lumped mass m and is subjected to harmonic couple M(t) ¼ M0 sin θt. In this case the relationship (17.23) should be written as follows δ1M M ðt Þ þ δ11 J ðt Þ ¼ μdyn δ1M M ðt Þ
ðaÞ
Let the mass m be located at the middle span, and couple M(t) be applied at the right support B. In this case we can write the following expressions for displacement in the direction 1 caused by unit couple and unit inertial force, as well as the frequency of free vibration
17.1
Structures with One Degree of Freedom
629
δ1M ¼
l2 , 16EI
δ11 ¼
l3 , 48EI
ω¼
rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffi 1 48EI ¼ mδ11 ml3
ðbÞ
Now analysis of the initial dynamic system is reduced to the static analysis of problem: the simply supported beam is loaded in the middle by the force P ¼ mg and couple M(t) ¼ M0 sin θ t/(1 θ2/ω2) applied on support B. The dynamic support reaction RA, bending moment under the mass, and the mass displacement are calculated as follows: P M P M sin θt RA ðt Þ ¼ þ ¼ þ 0 ; 2 l 2 l 1 θ2 =ω2 l l ¼ RA ðt Þ ; M 2 2 l Pl3 M l2 sin θt y ¼ þ 0 2 48EI 16EI 1 θ2 =ω2 Example 17.3 Design diagram of structure with one lumped mass m is shown in Fig. 17.9a. The mass is subjected to harmonic force F(t) ¼ F0 sin θt (Fig. 17.9a). Calculate the reaction RA, bending moment, and displacement at point where mass is attached. Consider the steady-state vibration. F0sinθt
a m
EI
l/2 μF0sinθt
b
m
RA
G=mg
RB
Fig. 17.9 Design diagram of structure and its transformation for static analysis
Solution The frequency of free vibration and dynamic coefficient neglecting dissipation of energy are rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffi 1 48EI ω¼ , ¼ mδ11 ml3
μ¼
1 1 θ2 =ω2
The mass m in the original design diagram (Fig. 17.9a) is subjected to the following forces: the weight G ¼ mg, where g ¼ 981 cm/s2 is the acceleration of gravity, harmonic force F0 sin θt, and inertial force J ¼ mθ2ysin θt applied to the mass m (this force by convention is not shown). Here, y is displacement of mass m. Equivalent “static” design diagram is shown in Fig. 17.9b. The following forces are applied to the mass m: the weight G ¼ mg, and harmonic force μ F0 sin θt. Dynamical reaction of support A and its amplitude value are 1 1 1 Gþ F 0 sin θt , RA ¼ ðG þ μF 0 sin θt Þ ¼ 2 2 1 θ2 =ω2
1 Rampl ¼ ðG þ μF 0 Þ A 2
Dynamical bending moment at the middle of a beam and dynamical displacement of the mass are M
l l l ¼ RA ¼ ðG þ μF 0 sin θt Þ , 2 2 4
630
17
y
Dynamics of Elastic Systems: Forced Vibration
ðG þ μF 0 sin θt Þl3 l ¼ 2 48EI
It can be seen that the reaction RA, bending moment M, and displacement y are a function of time t. The presence of the term μF0 sin θt in brackets in the expressions for RA, M, and y indicates that their amplitude values exceed the static values. Despite the fact that the problem under investigation is statically determinate, the reaction and bending moment depend on the rigidity of the structure. Indeed, with an increase in the rigidity EI of the structure, the frequency of free vibration ω increases, the dynamic coefficient μ decreases. Note: If the deformable structure has one degree of freedom and a harmonic force F0 sin θ t is applied to the mass m, then at steady state of vibration a force of inertia J arises which satisfies the relation (Kiselev 1980) F0 þ J ¼ F0
1 1 θ2 =ω2
Indeed, in case of these assumptions δ1F ¼ δ11 and expression (17.23) leads to the following result δ11 F ðt Þ þ δ11 J ðt Þ ¼ μdyn δ11 F ðt Þ ! F ðt Þ þ J ðt Þ ¼ μdyn F ðt Þ, This allows interpreting the expression F0/(1 θ2/ω2) as “effective static force” and applying the method of influence lines for analyzing the steady vibration of systems with one degree of freedom. Example 17.4 Fixed-fixed beam of length l contains lumped mass m and subjected to harmonic excitation (Fig. 17.10). Dissipation of energy is neglected. Provide total dynamical analysis for case of steady-state vibration. Apply the influence lines concept (Kiselev 1980). Psinθt A
MA
m
B
MB
C
a=ul
b=l-a=υl
RA
RB
J
Fig. 17.10 Design diagram of fixed-fixed beam. Harmonic force is applied to the mass; the mass and force together have arbitrary location
Solution According to Table A.4, row 3, or Table A.15, column 3, we have M A ¼ uυ2 Pl,
M B ¼ u2 υPl,
RA ¼ υ ð1 þ 2uÞP, 2
M C ¼ 2u2 υ2 Pl
RB ¼ u ð1 þ 2υÞP 2
Since a given structure has one degree of freedom, and force is applied to mass, then at point C acts the force P/(1 θ2/ω2) which is treated as a “static” one. Therefore the expressions for influence lines of dynamical reactions should be written as follows P sin θt ab2 P sin θt ¼ , 1 θ2 =ω2 l2 1 θ2 =ω2 P sin θt b2 P sin θt ¼ ð3a þ bÞ , ILðRA Þ ¼ υ2 ð1 þ 2uÞ 2 1 θ =ω2 l3 1 θ2 =ω2 P sin θt a2 b2 P sin θt ¼ 2 M ðaÞ ¼ 2u2 υ2 l 2 1 θ =ω2 l3 1 θ2 =ω2
ILðM A Þ ¼ uυ2 l
The amplitude values of reactions and bending moments are
17.1
Structures with One Degree of Freedom
MA ¼
631
ab2 P , l2 1 θ2 =ω2
Correspondingly, M B ¼
RA ¼
a2 b P , l2 1 θ2 =ω2
b2 P ð3a þ bÞ , 1 θ2 =ω2 l3
RB ¼
M ðaÞ ¼ 2
a2 b2 P l3 1 θ2 =ω2
a2 P ða þ 3bÞ 1 θ2 =ω2 l3
17.1.5 Kinematical Excitation Let us consider a beam subjected to the displacement of support Δ(t). Corresponding dynamic elastic curve of the beam is denoted by ydyn, while displacement of mass m from its initial position is y(t). In case of static displacement of support the inertial force of mass m is absent; corresponding elastic curve is yst.. If static displacement Δ ¼ 1, then corresponding unit displacement of mass m is δmΔ (Fig. 17.11) Static elastic curve yst
m Δ(t)
y(t) Dynamic elastic curve ydyn
1 δmΔ Fig. 17.11 Notation for kinematical excitation
According to superposition principle, the general expression for displacement of the mass m is: yðt Þ ¼ δmΔ Δðt Þ þ δ11 ðm€y β y_ Þ
ð17:24Þ
Here, δ11 means the vertical displacement of mass m caused by the unit statical force applied to the mass. The terms m€y and βy_ are the inertial and resistant forces which are applied to the mass m. The product δ11 m€y means the displacement of mass m caused by the force of inertia applied to the mass. After simplification, the equation (17.24) takes the form €y þ 2h y_ þ ω2 y ¼ ω2 δmΔ Δðt Þ ¼ F 1 ðt Þ β 1 g g 2h ¼ , ω2 ¼ ¼ ¼ m δ11 m δ11 P ystat
ð17:25Þ
Thus, the following assertion is true: The effect of an arbitrary kinematical excitation Δ(t) on the deformable structure with one degree of freedom coincides with the influence of a force ω2δmΔΔ(t) acting on the same system with a fixed support. It means that for analysis of forced vibration of the system with one degree of freedom subjected to kinematic excitation, it is possible to apply all formulas related to vibrations under the force excitation, if appropriate replacement will be done. Further, suppose that there are no resistance forces. The total solution (17.25) under initial conditions y(0) ¼ y0, y_ ð0Þ ¼ υ0 is yðt Þ ¼ y0 cos ωt þ
ðt υ0 sin ωt þ ωδmΔ ΔðτÞ sin ωðt τÞdτ ω
ð17:26Þ
0
Here the last term represents the Duhamel integral for the case of kinematic excitation. In case of harmonic excitation Δ(t) ¼ Δ sin θ t and Δ(t) ¼ Δ cos θ t this integral becomes
632
17
Dynamics of Elastic Systems: Forced Vibration
δmΔ Δ θ if Δðt Þ ¼ Δ sin θ t yð t Þ ¼ sin θ t sin ω t ω 1 ðθ=ωÞ2 δmΔ Δ yð t Þ ¼ ð cos θ t cos ω t Þ if Δðt Þ ¼ cos θ t 1 ðθ=ωÞ2
ð17:27Þ
Due to the inevitable resistances, the free vibration which occurs with the frequency ω will disappear and only the steadystate vibration with the frequency θ will remain. δmΔ Δ sin θ t 1 ðθ=ωÞ2 δ Δ cos θ t yðt Þ ¼ mΔ 1 ðθ=ωÞ2
yð t Þ ¼
if
Δðt Þ ¼ Δ sin θ t
if
Δðt Þ ¼ Δ cos θ t
In case of steady-state vibration, the inertial force equals J ¼ mθ2y. To compute displacement δmΔ of the arbitrary point of deformable structure subjected to the unit linear or angular displacement of support we need to apply algorithm (9.23), which had been considered in Sect. 9.9. After determining the displacement y, the structure is considered taking into account the inertia force. Example 17.5 The uniform fixed-rolled beam with one lumped mass m is subjected to the angular harmonic displacement Δ(t) ¼ φ sin θ t of the fixed support (Fig. 17.12a). The length of a beam l, bending stiffness EI, and dimensionless ratio θ/ω are given. Compute the dynamical bending moments and shear forces at support A and under mass m in case of the steady-state vibration. Dissipative forces are neglected.
a
c
φ(t)=φsinθt A ul
φ=1
MA(φ) =3EI/l
δmφ
m
QA(φ) =3EI/l2
υl
B Mφ=1
A
B
l/2
l/2
d
b P=1
M′A
MA(J) =3J*l/16
J* MJ=1
R′A
ul
υl
R′B
QA(J) =11J*/16
Fig. 17.12 Kinematical excitation, steady state vibration. (a) Design diagram; (b) Unit state for computation of deflection δmφ caused by excitation φ ¼ 1; (c, d) Bending moment diagrams caused by unit rotation of support A and inertial force J
Solution First of all, we need to determine the vertical displacement of mass m caused by the static rotation of the left support through angle φ. Unit state presents the given statically indeterminate structure loaded by force P ¼ 1. This force should be applied at the point with required displacement; corresponding bending moment diagram is shown in Fig. 17.12b. The values of l reactions for unit state P ¼ 1 may be found in Table A.3: M 0A ¼ υ 1 υ2 . According to Sect. 9.9, required displacement is 2 X l l 0 2 Ri φ ¼ υ 1 υ φ ¼ υ 1 υ2 φ Δmφ ¼ 2 2 Here R0i means the reactions of the redundant structure in the unit state. In the subsequent discussion we assume u ¼ υ ¼ 0.5. In this case, the displacement mass m due to static angular displacement φ equals
17.1
Structures with One Degree of Freedom
Δmφ ¼
633
l 1 1 3l 1 φ ¼ φ: 22 4 16
If φ ¼ 1, then δmφ ¼
3l 16
This result may be obtained using Таble A.9, row 5. The equation of motion of the mass m is yðt Þ ¼ δmφ
φð t Þ 3 φ sin θt ¼ l 1 ðθ=ωÞ2 16 1 ðθ=ωÞ2
7 l3 Inertial Force. Unit displacement of the point x ¼ l/2 caused by force P ¼ 1 applied at the same point is δ11 ¼ . The 768 EI 1 768 EI 768 EI ¼ , so a mass in terms of frequency is m ¼ . Inertial force in square of frequency of free vibration ω2 ¼ mδ11 7 ml3 7 ω 2 l3 case of steady-state vibration becomes J ¼ mθ2 yðt Þ ¼
768 EI 3 φ sin θ t 144 EI θ2 φ sin θ t θ2 l ¼ 3 2 2 7 ω l 16 1 ðθ=ωÞ 7 l2 ω2 1 ðθ=ωÞ2
According to superposition principle, the bending moment at support A caused by angular displacement φsinθt and inertial force J (Table A.3, row 3) are M A ¼ M AðφÞ þ M AðJ Þ ¼
3EI 3 φ sin θ t J l l 16
The signs of bending moments at support A are established on the basis of Fig. 17.12c, d. In expanded form, this equality is written as 3EI 3 144 EI θ2 φ sin θ t φ sin θ t l¼ l 16 7 l2 ω2 1 ðθ=ωÞ2
3EI θ2 9 θ2 φ sin θ t 3EI 16 θ2 φ sin θ t 1 2 1 ¼ l 7 ω2 1 ðθ=ωÞ2 l 7 ω2 1 ðθ=ωÞ2 ω
MA ¼
ðaÞ
The shear force and reaction at support A 3EI 11 3EI 11 144 EI θ2 φ sin θ t φ sin θ t þ ¼ φ sin θ t þ ¼ J 16 16 7 l2 ω2 1 ðθ=ωÞ2 l2 l2
3EI θ2 33 θ2 φ sin θ t 3EI 40 θ2 φ sin θ t ¼ 2 1 2 ¼ 2 1 , RA ¼ QA ð#Þ 2 2 2 7 7 ω ω ω l l 1 ðθ=ωÞ 1 ðθ=ωÞ2
QA ¼ QAðφÞ þ QAðJ Þ ¼¼
ðbÞ
Bending moment under the mass m
l 3EI 1 4 θ2 φ sin θ t M ðl=2Þ ¼ M A RA ¼ þ 2 l 2 7 ω2 1 ðθ=ωÞ2
ðcÞ
The shear force at support B QB ¼
M ðl=2Þ 3EI 8 θ2 φ sin θ t , ¼ 2 1þ 7 ω2 1 ðθ=ωÞ2 l=2 l
RB ¼ jQB j ð"Þ
ðdÞ
It can be seen that the found expressions for bending moments, shear forces, and reactions change according to a harmonic law with the frequency of external excitation θ. Coefficients in all expressions include the stiffness of the ffi beam EI, its length l, pffiffiffiffiffiffiffiffiffiffiffiffiffiffi the amplitude value of the rotation angle, and the dimensionless parameter θ/ω, where ω ¼ 1=mδ11 is the frequency of free vibration of a beam. To calculate the amplitude values of the force, we should put sinθ t ¼ 1 and take the specific value of the detuning θ/ω.
634
17
Dynamics of Elastic Systems: Forced Vibration
Static Verification. Equilibrium equation X
3EI 40 θ2 144 EI θ2 3EI 8 θ2 φ sin θ t Y ¼ RA J þ RB ¼ 2 1 þ 2 1þ 2 2 ω2 2 7 7 7 ω ω l l l 1 ðθ=ωÞ2 3EI 40 θ2 48 θ2 8 θ2 φ sin θ t ¼ 2 1 þ 1 ¼0 2 2 7 ω 7 ω 7 ω2 1 ðθ=ωÞ2 l
Note: Expressions for bending moment MA and shear force QA of massless uniform beams AB with one lumped mass, subjected to the different harmonic excitation, are presented in Таble А.27. The results (a) and (b) obtained above are presented in Table A.27, row 5. Example 17.6 Uniform two-span beam ABC with lumped mass m is subjected to harmonical vertical displacement Δ(t) ¼ Δ0 sin θ t of support C (Fig. 17.13a). Each span of the beam is l. Bending stiffness is EI. The dissipative forces are neglected. Determine dynamical reactions of supports and bending moment under the lumped mass. Solution Bending moment diagram and reactions of supports caused by given static displacement Δ of support C are presented in Fig. 17.13b; for this tabulated data had been used (Table A.14, row 1, column C). Next, for this loading we need to determine the vertical displacement ΔmΔ of the mass m. To do this, in the initial redundant (!) structure we apply a vertical unit force P ¼ 1 at the point where it is necessary to determine the displacement (Fig. 17.13c). Then we determine the reaction in the support that is shifted (support C); this reaction according to Table A.10, row 2(!) is equal to R0C ¼ 0:406ð"Þ . Corresponding bending moment diagram is shown in Fig. 17.13c.
a
m
B
A
C
Δ(t)
l
l/2 MB=1.5k1
b
C
A
MΔ
k1=EIΔ/l2,
Δ
k2=EIΔ/l3.
B ΔmΔ RA=1.5k2
RB=3.0k2 RC=1.5k2
c
0.094l
P=1 C
A B 0.094
red M P=1
0.203l
0.688
R′C =0.406
d
P=1
A
C
B
det M P=1
0.25l Fig. 17.13 Kinematical excitation of redundant structure. (a) Design diagram; (b) Vertical displacement of support C, elastic curve and static bending moment diagram and reactions; (c) Unit state in redundant system and corresponding bending moment diagram; (d) Unit state in determinate system and corresponding bending moment diagram
According to (9.29, Sect. 9.9.1), the displacement of the mass m caused by static displacement Δ of support C is ΔmΔ ¼ R0C Δ ¼ ð0:406ΔÞ ¼ 0:406Δ:
If Δ ¼ 1
then δmΔ ¼ 0:406
The minus sign in parentheses means that the force R0C on displacement Δ performs negative work.
17.1
Structures with One Degree of Freedom
635
Thus, the equation of motion of the mass m becomes yðt Þ ¼ δmΔ
Δ ðt Þ Δ sin θ t ¼ 0:406 0 1 ðθ=ωÞ2 1 ðθ=ωÞ2
The next step of analysis consists of computation of inertial force which arises at periodic motion of the lumped mass m. First of all, we need to determine the frequency of the free vibration. The unit displacement in the entire structure M red M det P¼1 det δ11 ¼ P¼1 . Here, M red P¼1 and M P¼1 present the bending moment diagram caused by P ¼ 1 in entire redundant EI structure (Fig. 17.13c) and any statically determinate (!) primary system, which is obtained from given structure. In particular, Fig. 17.13d shows a statically determinate system obtained by removing support A. Calculation of δ11 is performed according to the trapezoid rule. δ11 ¼
l l3 ½2 0:203l 0:25l 0:094l 0:25l þ 2 0:203l 0:25l ¼ 0:01496 2 6EI EI
Here, the first and second terms are related to the left portion of the span BC, and the third member to the right portion of the same span. It is obvious, the computation of the unit displacement by multiplication of diagram using the formula red EIδ11 ¼ M red P¼1 M P¼1 leads to the same result. The frequency of free vibration and mass in terms of frequency are rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 EI 66:845EI !m¼ ¼ ω¼ mδ11 0:01496ml3 ω2 l 3 Inertial force which occurs at the steady vibration is J ¼ mθ2 yðt Þ ¼
2 66:845EI 2 0:406Δ0 sin θ t 27:139EI θ Δ0 θ ¼ sin θ t 2 3 ω ω2 l3 l 1 ðθ=ωÞ 1 ðθ=ωÞ2
If positive direction of displacement and acceleration of mass m is directed as given displacement Δ, then force J acts on the mass upward and force which acts on the beam is directed in opposite direction, i.e., downward. Of course, having value for mass m, it is possible to get the expression for the inertia force J directly by the formula J ¼ mθ2y(t); however, the above procedure allows introducing detuning θ/ω into consideration. Finally we can calculate dynamic reactions and bending moments. According to superposition principle, computation of reaction RA should be performed by formula RA ¼ RΔ¼1 Δðt Þ þ Rred P¼1 J
where RΔ ¼ 1 and Rred P¼1 are reaction RA of the original system caused by the static displacement Δ ¼ 1 and the static load P ¼ 1 applied to the mass, respectively. Therefore, the first and second terms represent the force RA caused by a given displacement Δ(t) and inertial load J, respectively. Amplitude values of dynamic reaction of supports are (factor sinθt is omitted) RA ¼ RA Δðt Þ þ
RAP¼1 J
2 EI 27:139EI θ Δ0 ¼ 1:5 3 Δ0 þ 0:094 ¼ 3 |ffl ffl {zffl ffl } ω l l 1 ðθ=ωÞ2 |fflffl{zfflffl} red from M Δ
from M P¼1
" # 1:7ðθ=ωÞ2 1 þ 0:7ðθ=ωÞ2 EI EI ¼ 1:5 3 Δ0 1 þ Δ ð#Þ; ¼ 1:5 0 l l3 1 ðθ=ωÞ2 1 ðθ=ωÞ2 2 EI 27:139EI θ Δ0 B ¼ RB ¼ RB Δðt Þ þ RP¼1 J ¼ 3:0 3 Δ0 þ 0:688 ω 1 ðθ=ωÞ2 l l3 " # 6:224ðθ=ωÞ2 1 þ 5:224ðθ=ωÞ2 EI EI ¼ 3:0 3 Δ0 1 þ Δ ð"Þ; ¼ 3:0 0 l l3 1 ðθ=ωÞ2 1 ðθ=ωÞ2
636
17
Dynamics of Elastic Systems: Forced Vibration
2 EI 27:139EI θ Δ0 Δ ð " Þ0:406 ¼ 0 ω 1 ðθ=ωÞ2 l3 l3 " # 7:346ðθ=ωÞ2 1 8:346ðθ=ωÞ2 EI EI ¼ 1:5 3 Δ0 1 ð#Þ: ¼ 1:5 3 Δ0 2 l l 1 ðθ=ωÞ 1 ðθ=ωÞ2
RC ¼ RC Δðt Þ þ RCP¼1 J ¼ ð#Þ1:5
Amplitude values of bending moment at support B and at lumped mass m are 2 EI 27:139EI θ Δ0 M B ¼ M B Δ ðt Þ þ ¼ 1:5 2 Δ0 0:094 ¼ 3 ω l l 1 ðθ=ωÞ2 " # 1:7ðθ=ωÞ2 1 þ 0:7ðθ=ωÞ2 EI EI ¼ 1:5 2 Δ0 1 þ Δ ¼ 1:5 0 l l2 1 ðθ=ωÞ2 1 ðθ=ωÞ2 2 l EI l 27:139EI θ Δ0 m M m ¼ RC Δðt Þ þ M P¼1 J ¼ 1:5 3 Δ0 þ 0:203l ¼ 3 2 2 ω l l 1 ðθ=ωÞ2 " # 7:3456ðθ=ωÞ2 1 8:3456ðθ=ωÞ2 EI EI ¼ 0:75 ¼ 1:5 2 Δ0 1 Δ 0 2l l2 1 ðθ=ωÞ2 1 ðθ=ωÞ2 M BP¼1 J
For dynamic reactions we need to introduce factor sinθt. Static Verification l 1. The bending moment at support B calculated by the right forces M B ¼ RC l J leads to the above result. 2 2. Let us introduce the notation α ¼ θ2/ω2. Then we have X
Y ¼ RA þ RB RC J ¼ EIΔðt Þ ½1:5ð1 þ 0:7αÞ þ 3ð1 þ 5:224αÞ 1:5ð1 8:346αÞ 27:139α sin θt ¼ l ð1 αÞ2 EIΔðt Þ ¼ 3 ½ð1:5 þ 3:0 1:5Þ þ αð1:05 þ 15:672 þ 12:519 27:139Þ sin θt ¼ l ð1 αÞ2 EIΔðt Þ ¼ 3 ½0 þ αð28:189 þ 28:191Þ sin θt ffi 0 l ð1 αÞ2 ¼
3
Note. It is possible case when any deformable structure with one degree of freedom is subjected to force and kinematical excitation simultaneously. If at steady-state vibration the resulting force acting on the mass will be equal to zero, then the system will be at rest. This means that the influence of the kinematic action can be neutralized by force action and vice versa (Karnovsky and Lebed 2016).
17.2
Structures with Finite Number of Degrees of Freedom: The Force Method
This paragraph is devoted to the forced vibrations of the elastic structures of different types with finite number of degrees of freedom. The basis of dynamic analysis of deformable systems is the force method. Corresponding resolving equations are presented. These equations are applied for analysis of vibration caused by harmonic excitation (case of steady-state vibration is considered), as well as the impulsive and stepped excitation. In all cases, the structures are considered under the following assumptions: (a) the superposition principle is true, (b) the mass of the structure itself and dissipative forces are neglected. Dynamic analysis involves computation of the amplitudes for inertial forces, reactions, internal forces, displacements, as well as testing the structure for possible resonance.
17.2
Structures with Finite Number of Degrees of Freedom: The Force Method
637
17.2.1 Resolving Equation of the Force Method The arbitrary deformable elastic structure with lumped masses mi (i ¼ 1, . . ., n) is subjected to several harmonic disturbing loads Pk(t) ¼ Pk sin θ t. Assume that all the disturbing forces have the same frequency excitation θ and act in the same phase, while amplitudes Pk of these forces may be different. In this section we will consider steady-state undamped vibration. In this case, the displacements and acceleration, as well as the inertial forces, reactions, and internal forces, occur according to harmonic law yi(t) ¼ ai sin θ t, €yi ðt Þ ¼ ai θ2 sin θ t. It means all of them approach their extreme values simultaneously. The conditional elastic structure (beam, frame, truss, etc.) with lumped masses and their positive displacements yi(t) and accelerations ai ðt Þ ¼ €yi ðt Þ are shown in Fig. 17.14a. Each mass mi is subjected to inertial force X i ðt Þ ¼ mi€yi , (Fig. 17.14b). Dynamic analysis of structure with a finite number of degrees of freedom contains two large parts. The first one contains the determination of the frequencies of free vibration, and the second part contains the determination of the amplitude values of inertial forces and the construction of dynamic diagrams of internal forces.
P(t)
a
mi
m1 y1 a1
yi ai P (t)
b
mi
m1 X1
Xi
Fig. 17.14 (a) Force excitation of a structure with a finite number degrees of freedom; (b) Loading diagram of a structure
On the basis of superposition principle, the displacement of any mass mi at time t is given by the set of formulas y1 ¼ δ11 X 1 þ δ12 X 2 þ þ δ1n X n þ Δ1P ðt Þ y2 ¼ δ21 X 1 þ δ22 X 2 þ þ δ2n X n þ Δ2P ðt Þ yn ¼ δn1 X 1 þ δn2 X 2 þ þ δnn X n þ ΔnP ðt Þ
ð17:28Þ
where Xi are unknown inertial forces of the corresponding masses, δ11, δ12, . . ., δ1n are displacements in the directions of the force X1 caused by unit forces X1, X2, . . ., Xn. We will consider case when free term ΔiP(t) is displacement in the direction of Xi caused by the harmonic load, i.e., ΔiP(t) ¼ ΔiP sin θ t. Since displacements yi, inertial forces yi, and free terms yi change according to the law which is characterized by a factor sinθ t to their amplitude values, then this factor is presented in all members of the system of equations (17.28). After reduction by this factor, all unknown and free terms should be considered as amplitude values of the corresponding variables. The system (17.28) is mixed: each equation contains the displacement y and the inertial forces X. We transform this system into two forms, one of which is written in inertial forces X, and the other in displacements of system y. The Set of Equations (17.28) in Form with Respect to Inertial Forces X Since inertial force Xi of any mass mi is equal to X i ¼ mi€yi ðt Þ ¼ mi θ2 yi, then any equation of the set (17.28) may be rewritten in the form Xi ¼ δi1 X 1 þ δi2 X 2 þ þ δii X i þ þ δin X n þ ΔiP mi θ 2 or in equivalent form 1 δi1 X 1 þ δi2 X 2 þ þ δii X i þ þ δin X n þ ΔiP ¼ 0: mi θ 2
638
17
Dynamics of Elastic Systems: Forced Vibration
Finally, we get a set of equations to calculate the amplitudes of the inertial forces Xi δ11 X 1 þ δ12 X 2 þ þ δ1n X n þ Δ1P ¼ 0, δ21 X 1 þ δ22 X 2 þ þ δ2n X n þ Δ2P ¼ 0, :
: : :
: :
: : :
: :
: : :
ð17:29Þ
δn1 X 1 þ δn2 X 2 þ þ δnn X n þ ΔnP ¼ 0 Here, δik is the displacement along the direction of inertial force Xi due to action of unit unknown Xk which is applied to mass mk; ΔkP is the displacement of mass mk caused by amplitude values of the vibrational forces. The diagonal (principal) elements are calculated by formula δii ¼ δii 1= mi θ2 . Feature of set (17.29). This set of canonical equations (17.29) for dynamic analysis of elastic structures and canonical equation for analysis of redundant structures (Chap. 9) coincide in form. However, the fundamental difference is that here the unknowns X are not the reactions of redundant constraints, but the amplitudes of the inertial forces of the masses that arise when the system vibrates. However, if we assume that the original system is modified by introducing additional elastic constraints that prevents the movement of each mass and the inertia force arising in each introduced constraint is considered as the principal unknown, then the coincidence is not only in form but also in essence. Therewith it should be noted the difference which is related to the structure of the coefficients for unknowns standing on the main diagonal. It is important to note here that the system of equations (17.29) remains unchanged both for statically determinate and indeterminate structures subjected to dynamical excitation. The solution of system (17.29) is Xi ¼ Di/D, where D is determinant composed of coefficients with unknowns δ11 δ 21 D¼ ... δ n1
δ12
...
δ22 ...
... ...
δn2
...
δ1n δ2n . . . δnn
ð17:30Þ
The determinant Di is obtained from D if the ith column is replaced by column of free terms. Once the amplitudes of the inertial forces Xi are calculated from (17.29), dynamic forces S (bending moment, shear, reactions, etc.) at the specified section of a structure according to superposition principle may be calculated by formula S ¼ S1 X 1 þ S2 X 2 þ þ S2 X 2 þ SP where Si is force caused by the Xi ¼ 1; SP is force caused by amplitude value P of the disturbance force. The Set of Equations (17.28) in Form with Respect to Displacement y The expression X k ¼ mk €yk ðt Þ we substitute into (17.28) δ11 m1€y1 δ12 m2€y2 δ1n mn€yn y1 þ Δ1P ¼ 0 δ21 m1€y1 δ22 m2€y2 δ2n mn€yn y2 þ Δ2P ¼ 0
δn1 m1€y1 δn2 m2€y2 δnn mn€yn yn þ ΔnP ¼ 0 Since €yi ðt Þ ¼ θ2 yi then we get
δ11 m1 θ2 1 y1 þ δ12 m2 θ2 y2 þ þ δ1n mn θ2 yn þ Δ1P ¼ 0 δ21 X 1 m1 θ2 y1 þ δ22 m2 θ2 1 y2 þ þδ2n mn θ2 yn þ Δ2P ¼ 0 δn1 m1 θ2 y1 þ δn2 m2 θ2 y2 þ þ δnn mn θ2 1 yn þ ΔnP ¼ 0
ð17:31Þ
17.2
Structures with Finite Number of Degrees of Freedom: The Force Method
639
This set of equations (17.31) allows finding the amplitude of displacement yi of lumped masses as a function of frequency excitation θ. Required displacements yi become, yi ¼ Di/D, where D is determinant composed of coefficients with unknowns δ11 m1 θ2 1 δ12 m2 θ2 δ1n mn θ2 δ m θ2 δ22 m2 θ2 1 δ2n mn θ2 21 1 ð17:31aÞ D¼ δ m θ2 δ m θ2 δ m θ2 1 n1
1
n2
2
nn
n
The determinant Di is obtained from D if the ith column is replaced by column of free terms. If we put all ΔiP ¼ 0 with the replacement θ to ω, we obtain the equation of the frequencies of free vibration which coincides with (16.5) δ11 m1 ω2 1 δ m ω2 21 1 D¼ δ m ω2 n1
1
δ12 m2 ω2 δ22 m2 ω2 1 δn2 m2 ω2
¼0 δnn mn ω2 1
δ12 mn ω2 δ2n mn ω2
ð17:32Þ
The first equivalent form of the frequency equation is obtained if each element of determinant (17.32) is divided into ω2 1 δ11 m1 2 ω D ¼ δ21 m1 δn1 m2
δ12 m2 1 δ22 m2 2 ω δn2 m2
δ2n mn ¼ 0 1 δnn mn 2 ω
δ12 m2
ð17:32aÞ
The second equivalent form of the frequency equation is obtained if the first column of the determinant (17.32) is divided by m1ω2, the second column is divided by m2ω2, etc. δ11 1 2 δ δ 12 12 m ω 1 1 δ δ δ 21 22 2n 2 ð17:32bÞ D¼ ¼0 m ω 2 1 δn1 δn2 δnn m ω2 n
If the frequency of the excitation θ coincides with any of the frequencies of free vibration ω, the determinant of system (17.31a) is equal to zero. This means the realization of the resonance phenomenon, in which the amplitudes of vibration increase unlimitedly. The number of resonant regimes coincides with the number of degrees of freedom.
17.2.2 Harmonic Excitation: Reciprocal Theorems Among the vast class of systems with finite number of degrees of freedom a special place is occupied by dynamic systems with two degrees of freedom. Their particular importance lies in the fact that it is these structures that make it possible to easily trace all the main stages of analysis characteristic of systems with a finite number of degrees of freedom, and to obtain analytical results. Below is a detailed analysis of the steady-state vibration of a deformable system with two degrees of freedom. The force method is applied. For dynamic analysis of steady-state vibration it is necessary to perform the following fundamental steps: 1. Select of generalized coordinates, calculate the unit displacements, and determine the frequencies of free vibration. 2. Compute of the free terms of the canonical equations, form the canonical equation (17.29) of the force method, determine amplitude values of the inertial forces, and construct diagrams of dynamic bending moments, shear forces, etc.
640
17
Dynamics of Elastic Systems: Forced Vibration
It should be remembered that unit displacements that are necessary for determining the frequencies of free vibration and inertial forces which depend on the harmonic disturbing forces are related to the original structure. In case of original redundant structure, the unit displacements are determined as explained in Sect. 9.8. In the case of structures which are subjected to bending deformation, the displacement is determined by a single-term formula that takes into account only the bending moments; the procedure for multiplication of bending moment diagrams may be recommended. Reciprocal Theorems In the case of steady-state vibration of deformable structure subjected to harmonic excitation, the important reciprocity relations remain. It is assumed that the superposition principle for structure is hold (Kiselev 1980). Theorem of Reciprocal Unit Dynamic Displacement Let arbitrary deformable structure in the state 1 be subjected to forced excitation Psinθt, and in the state 2 to forced excitation of the same frequency Ksinθt. Assume that amplitudes of forced excitation are P ¼ K ¼ 1. Denote δ12 as the amplitude of displacement in the direction of the load of the 1st state caused by harmonic load of unit amplitude of the 2nd state. Similarly, δ21 is the amplitude of displacement in the direction of the load of the 2nd state caused by harmonic load of unit amplitude of the 1st state. In this case we can assert that in any elastic system, amplitude of displacement δ12 equals to amplitude of displacement δ21. In general case, δik ¼ δki. Illustration of this theorem is presented in Fig. 17.15a. Obviously, the nature of the loads may be different, for example, P is a lumped force, K is a couple. In this case, the displacement in P direction is measured in units of length, the displacement in the K direction is measured in radians. Theorem of Reciprocal Unit Dynamic Reactions Let arbitrary deformable structure in the state 1 be subjected to harmonic kinematical excitation Δ1 sin θt, and in the state 2 to harmonic kinematical excitation of the same frequency Δ2 sin θt. Similarly as above, assume amplitudes of kinematical excitations are Δ1 ¼ Δ2 ¼ 1. Denote r12 as the reaction that arises in the constraint in the 1st state due to harmonic displacement of unit amplitude of the constraint in the 2nd state. Similarly, r21 is the reaction that arises in the constraint in the 2nd state due to harmonic displacement of unit amplitude of the constraint in the 1st state. In this case we can assert that in any elastic system amplitude of reaction r12 equals amplitude of reaction r21. In general case, rik ¼ rki. Obviously, the nature of the displacements may be different, for example, Δ1 may be a linear displacement, Δ2 may be an angular displacement. In this case, the reaction in Δ1 direction is measured in units of force, the reaction in the Δ2 direction is measured in units of couple. Illustration of this theorem is presented in Fig. 17.15b.
Fig. 17.15 Dynamic reciprocal theorems. (a) Theorem of reciprocal unit displacements; (b) Theorem of reciprocal unit reactions
17.2
Structures with Finite Number of Degrees of Freedom: The Force Method
641
Let arbitrary deformable structure in the first state be subjected to excitation by force P1 sin θt and in the second state be subjected to kinematic excitation Δ2 sin θt of constraint. If the frequency of both excitations is equal, and amplitudes of excitation are equal to unity, i.e., P1 ¼ 1 and Δ2 ¼ 1, then the following relation is hold δ12 ¼ r21. Here, δ12 is the amplitude of displacement in the direction of the force of the first state caused by the amplitude of kinematical harmonic excitation of the second state; r21 is the amplitude of reaction which arises at the constraint of the second state caused by the amplitude of harmonic force excitation of the first state. Example 17.7 Uniform weightless simply supported beam with overhang beam carries two identical lumped mass m, and a harmonic load P(t) ¼ P sin θ t is applied to one of them, as shown in Fig. 17.16. The bending stiffness of the beam is EI. Consider the steady-state vibration and construct the dynamic bending moment diagram.
a
P (t) m l/2
m l/2
l/2
X1=1
b
M1 l/4 l/4
l/2
X2=1 M2
c
Pl/4
Pl/2
P MP
d
0.25
A
0.5448
1
0.9672 0.50
B
2
Mdyn (factor Pl)
X2 =0.9344P
X1 =0.245P RA= 1.0896P
P
RB= 2.7790P
Fig. 17.16 Harmonic excitation of a structure with two degrees of freedom. (a) Design diagram, (b) Unit bending moment diagrams caused by unit inertial forces X1 ¼ 1 and X2 ¼ 1; (c) Bending moment diagram caused by amplitude value P of the disturbance force P(t) ¼ Psinθt; (d) Amplitude values of dynamic bending moment diagram; the dotted line presents the bending moment diagram due the amplitude value P in the static state
Solution This structure is statically determinate and has two degrees of freedom. The generalized coordinates are vertical displacements of each mass m. Step 1. To determine the frequencies of free vibration, we plot the bending moment diagrams M 1 and M 2 caused by unit inertial forces X1 ¼ 1 and X2 ¼ 1, which are applied to the masses (Fig. 17.16b). The unit displacements δik are determined by “multiplication” of corresponding bending moment diagrams. 1 1l l 2l 2 l3 l3 ¼ ¼ 0:0208 ; M1 M1 ¼ EI 2 2 4 3 4 EI 48EI EI 1 l3 l3 ¼ M2 M2 ¼ ¼ 0:125 ; EI 8EI EI 1 l3 l3 ¼ δ21 ¼ M 1 M 2 ¼ ¼ 0:03125 EI 32EI EI
δ11 ¼ δ22 δ12
642
17
Dynamics of Elastic Systems: Forced Vibration
The frequency equation of free vibration in the form (17.32a) 1 δ11 m 2 ω δ21 m
¼0 1 δ22 m 2 ω δ12 m
ðaÞ
Disclosing of the determinant leads to quadratic equation with respect to 1/ω2
1 ω2
2
ðδ11 þ δ22 Þm
1 þ δ11 δ22 δ212 m2 ¼ 0 2 ω
After substitution of unit displacements δ we get
1 ω2
2
7 ml3 1 5 l6 m2 2þ ¼0 48 EI ω 3072 ðEI Þ2
Solution of this equation is 1 7 ml3 ¼
2 96 EI ω
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2 h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii 7 ml3 5 l6 m2 7 ml3 7 ml3 ¼ 1
1 0:30612 ¼ ð1 0:833Þ 2 96 EI 3072 ðEI Þ 96 EI 96 EI
ðbÞ
The roots of the equation and the corresponding frequencies of free vibration are rffiffiffiffiffiffiffi EI ; ml3 rffiffiffiffiffiffiffi 1 ml3 EI ! ω ¼ ¼ 9:0619 2 ω22 82:118EI ml3 1 ml3 ¼ ! ω1 ¼ 2:7353 2 ω1 7:482EI
Verification:
ðcÞ
1 1 þ ¼ m1 δ11 þ m2 δ22 . In our case ω21 ω22 1 1 ml3 ml3 ml3 þ 2¼ þ ¼ 0:1458 2 EI ω1 ω2 7:482EI 82:118EI m1 δ11 þ m2 δ22 ¼
ml3 ml3 ð0:0208 þ 0:125Þ ¼ 0:1458 EI EI
Step 2. For computation of inertial forces X1 and X2 we need to construct the bending moment diagram MP caused by amplitude of the disturbing force P and compute the free terms of the equations (17.31). 1 Pl3 ; M1 MP ¼ EI 32EI 1 Pl3 ¼ M2 MP ¼ ; EI 8EI
Δ1P ¼ Δ2P
ðdÞ
The system of canonical equations (17.29) with respect to amplitude values f the inertial forces X is δ11 X 1 þ δ12 X 2 þ Δ1P ¼ 0,
δ21 X 1 þ δ22 X 2 þ Δ2P ¼ 0 1 1 , δ22 ¼ δ22 2 : mθ2 mθ The term 1=mθ2 may be calculated by different ways as follows.
where δ11 ¼ δ11
1. The simplest method is direct calculation with a known mass m and excitation frequency θ. However, in this case the role of the frequency of free vibration is not obvious.
17.2
Structures with Finite Number of Degrees of Freedom: The Force Method
643
2. The second method allows taking into account the system detuning θ/ω. To clarify this approach, we return to the expression for the frequency of free vibration 1 ml3 7:482EI ¼ !m¼ 2 ω1 7:482EI ω21 l3 Then the additional term and diagonal unit displacements of the set (17.29) become 1 l3 ¼ , mθ2 7:482EI ðθ=ωÞ2 δ11
" # 1 l3 6:4154 1 ¼ δ11 2 ¼ , 48EI mθ ðθ=ωÞ2
δ22
" # 1 l3 1:0692 1 ¼ δ22 2 ¼ 8EI mθ ðθ=ωÞ2
ðeÞ
Further calculations will be carried out for a specific value (θ/ω)2 ¼ 0.5. In this case δ11 ¼ 0:246
l3 , EI
δ22 ¼ 0:142
l3 EI
Canonical equations for determining the inertial forces X after reduction by a factor l3/EI become 0:246X 1 0:0312X 2 ¼ 0:0312P
ðfÞ
0:0312X 1 0:142X 2 ¼ 0:125P
Solution of these equations yields X1 ¼ 0.245P; X2 ¼ 0.9344P. Computation of ordinates of the dynamic bending moment diagram is presented in Table 17.2; points 1 and B are shown in Fig. 17.16d Table 17.2 Computation of final bending moments, (θ/ω)2 ¼ 0.5
Points 1 B
M1 -l/4 0
M 1 X1 0.0612Pl 0
M2 l/4 l/2
M 2X2 0.2336Pl 0.4672Pl
MP
Final Mdyn
Pl/4 Pl/2
0.5448 Pl 0.9672 Pl
Corresponding amplitude values of the inertial forces, bending moment diagram, and reaction are presented in Fig. 17.16d. All of these functions of steady-state vibration change by harmonic law with frequency θ of external excitation. Bending moments caused by the static action of the amplitude value of the disturbing force P on a massless beam is shown by a dotted line (Fig. 17.16d). The dynamic coefficient for points 1 and B are μ1 ¼ 0.5448/0.25 ¼ 2.18 and μB ¼ 0.9672/0.5 ¼ 1.93. Shear forces within each portion are: QA1 ¼ 1.0896P, Q1B ¼ 0.8446P, QB2 ¼ 1.9344P. Corresponding reactions are shown in Fig. 17.16d. Verification of results: X Y ¼ ð1:0896 þ 0:245 þ 2:7790 0:9344 1:0ÞP ¼ ð3:0240 þ 3:0240ÞP ¼ 0 3. Let us discuss another way of calculating the term 1/mθ2 inq expressions ffiffiffiffiffiffiffiffiffiffiffiffiffiffi for δ11 , δ22. Let us assume that the structure of the
expression for frequency θ is the same as for ω, i.e., θ ¼ k arbitrary value for k, we thereby fix the detuning θ/ω.
EI=ml3, where k is a free dimensionless parameter; taking an
Now the term 1/mθ2 and the diagonal coefficients δ11 , δ22 at inertial forces X may be presented as follows: 1 1 l3 ¼ ¼ , mθ2 mk 2 EI k 2 EI 3 ml 1 l3 1 1 , δ11 ¼ δ11 2 ¼ EI 48 k2 mθ
δ22
1 l3 1 1 ¼ δ22 2 ¼ EI 8 k 2 mθ
ðgÞ
644
17
Dynamics of Elastic Systems: Forced Vibration
Canonical equations for determining the inertial forces X after reduction by a factor l3/EI become 1 0:020833 2 X 1 0:03125X 2 ¼ 0:03125P k 1 0:03125X 1 þ 0:125 2 X 2 ¼ 0:125P k
ðhÞ
This system allows splitting the frequency range of the excitation into three sub-bands: before the first resonance, between the first and second resonances, and after the second resonance. The boundary values of the parameter k are defined as follows: rffiffiffiffiffiffiffi EI ω1 ¼ 2:7353 ! k2 ¼ 7:480; ml3
rffiffiffiffiffiffiffi EI ! k2 ¼ 82:120 ω2 ¼ 9:0619 ml3
In general, inertial forces may be determined as described in Sect. 17.2.1 by formula D ¼ Di/D. The values of inertial forces (factor P) for different k are presented in Table 17.3. The value k ¼ 0 (θ ¼ 0) means static load application; as expected, in this case the inertial forces X1 ¼ X2 ¼ 0. When computing inertial forces near resonance, calculations should be carried out with high accuracy, since the matrix of coefficients for unknowns turns out to be ill-conditioned. This leads to very small values of the determinant D, and at resonance we get the case of a degenerate matrix, for which the determinant D ¼ 0. Table 17.3 Inertial forces X1 and X2 as a function of dimensionless parameter k; factor P for X1 and X2 is omitted
Xi
k=0
k=0.5
k=2.5
X1 X2
0.0 0.0
-0.08094
-1.28047
0.032323
4.71287
k=2.735311 1st resonance
∞ ∞
k=3
k=8
1.55612
1.19654
-5.50433
-0.80153
k=9.06187 2nd resonance
∞ ∞
k=10
k=20
k=∞
-1.16176
-0.061499
0.0
-1.40214
-1.03607
-1.0
The graph of changes in the inertial forces X1 and X2 vs. dimensionless parameter k is shown in Fig. 17.17; factor P for X1 and X2 is omitted. In the first subrange 0 k 2.7353 there is a continuous growth of inertial forces from zero to infinity. In this subrange both masses moves in the opposite directions. When passing through the first resonance k ¼ 2.7353 a change in the sign of inertial forces occurs. In the second interval 2.7353 k 9.06187, in the vicinity of the first resonance, a sharp drop in inertia forces is observed, and when approaching the second resonance, a sharp increase in inertial forces occurs, while in the zone remote from both resonances, the inertia forces remain almost constant. At the beginning of the third interval k > 9.06187, the inertial forces change the sign and decrease, and then remain constant. When k > 10, the inertial force X2 is equal to the amplitude of the excitation force P, and when k > 12, the inertial force X1 is equal to zero.
Fig. 17.17 Inertial forces Х1 и Х2 as a function of dimensional parameter к
17.2
Structures with Finite Number of Degrees of Freedom: The Force Method
645
Comments General statements of the current paragraph may be applied to any linear deformable structure with finite number of degrees of freedom for which superposition principle is applicable. In case of trusses, the displacements are determined by the singleterm formula (8.48), which takes into account only the axial forces, in the case of combined systems—by the two-terms formula , which takes into account the axial forces and bending moments (the first and second terms of (8.46). In the case of arched systems, the correct choice of directions in which independent mass movements occur is of great importance. In the case of a symmetric system it is advisable to consider separately the symmetric and antisymmetric vibration. Detailed example of the steady-state vibration of symmetrical uniform arch is presented in book (Karnovsky 2012). If the structure contains two different lumped masses m1 and m2, then it is convenient to make the transformation m1 ¼ m and m2 ¼ αm, where α is a certain positive number. In the case of three and more degrees of freedom the analytical transformations turn out to be cumbersome and hardly visible.
17.2.3 Impulsive Excitation Design diagram of massless “conditional dynamic structure” with two degrees of freedom is shown in Fig. 17.18. The term “conditional structure” means that restrictions on the type of structure (beams, frames, trusses, etc.), its supports, and its degree of redundancy are not imposed. The masses m1 and m2 simultaneously at the moment of time t ¼ 0 are subjected to instantaneous impulses S1 and S2. The initial velocity of each mass become υ1(0) ¼ S1/m1 and υ2(0) ¼ S2/m2, while initial displacements of each mass are equal to zero.
a
S1
S2 m2
m1 y1 J1
b
y2 J2 m2
m1 J1
c
m1 m2 J2 Fig.17.18 (a) Impulsive excitation of a “conditional” structure with two degrees of freedom; (b, c) Loading diagram of a structure according first and second mode of vibration
For dynamic analysis of structure the following steps should be performed (Kiselev 1980): 1. Select the generalized coordinates, construct unit states, plot corresponding bending moment diagrams, and determine unit displacements δik, i, k ¼ 1, 2. 2. Compile the frequency equation in the form of the force method (16.5a) and find the frequencies of free vibration ω1 and ω2.
D¼ Verification :
m1 δ11 1=ω2
m2 δ12
m1 δ21
m2 δ22 1=ω2
¼ 0 ! ω21 ,
1 1 þ 2 ¼ m1 δ11 þ m2 δ22 2 ω 1 ω2
3. Determine the vibration forms corresponding to frequencies ω1 and ω2.
ω22
ð17:33Þ
646
17
Dynamics of Elastic Systems: Forced Vibration
m1 δ11 ω21 1 m1 δ21 ω21 y2 ¼ ¼ , 2 y1 ω1 m2 δ12 ω1 m2 δ22 ω21 1 m1 δ11 ω22 1 m1 δ21 ω22 y ¼ 2 ¼ ¼ 2 y1 ω2 m2 δ12 ω2 m2 δ22 ω22 1
ρ21 ¼ ρ22
ð17:34Þ
where ρ21, ρ22 are coefficients of amplitude distribution. Verification :
ρ21 ρ22 ¼ m1 =m2
4. Compile the equation of free vibrations of the system y1 ðt Þ ¼ A sin ðω1 t þ φ1 Þ þ B sin ðω2 t þ φ2 Þ y2 ðt Þ ¼ ρ21 A sin ðω1 t þ φ1 Þ þ ρ22 B sin ðω2 t þ φ2 Þ
ð17:35Þ
To determine the unknowns A, B and phase angles φ1, φ2, the expressions for velocities y_ 1 ðt Þ ¼ dy1 =dt and y_ 2 ðt Þ ¼ dy2 =dt should be compiled, and then the expressions for displacements and velocities should be subjected to initial conditions. y1 ð0Þ ¼ 0,
y2 ð0Þ ¼ 0,
and
υ1 ð0Þ ¼ S1 =m1 ,
υ2 ð0Þ ¼ S2 =m2
Omitting the transformations, the final results are S S ρ22 1 2 ρ22 υ1 ð0Þ υ2 ð0Þ m1 m2 A¼ ¼ , ω1 ðρ22 ρ21 Þ ω1 ðρ22 ρ21 Þ S S ρ21 1 2 ρ21 υ1 ð0Þ υ2 ð0Þ m1 m2 ¼ , B¼ ω2 ðρ21 ρ22 Þ ω2 ðρ21 ρ22 Þ
φ1 ¼ φ2 ¼ 0,
ð17:36Þ
5. Determine the ratios between the forces that correspond to the frequencies ω1 and ω2 J 1 ¼ 1,
J2 ¼
m2 ρ for ω1 , and J 1 ¼ 1, m1 21
J2 ¼
m2 ρ for ω2 m1 22
These forces correspond to the first and second mode of vibration (Fig. 17.18b – vibration with one half wave) and (Fig. 17.18c, two half waves), respectively. Special Case Arbitrary deformable structure with two degrees of freedom is subjected to impulsive excitation S1 which is applied to mass m1 in the direction of the first generalized coordinate y1. In this case υ2(0) ¼ 0, and the displacements y1 and y2 become
ρ22 ρ21 S1 sin ω1 t þ sin ω2 t m1 ω1 ðρ22 ρ21 Þ ω2 ðρ21 ρ22 Þ
ρ22 ρ21 S y2 ðt Þ ¼ 1 ρ21 sin ω1 t þ ρ22 sin ω2 t m1 ω1 ðρ22 ρ21 Þ ω2 ðρ21 ρ22 Þ y1 ð t Þ ¼
These accurate analytical expressions allow plotting graphs of displacement, find the dynamic coefficients, and the distribution of internal forces in the system with two degrees of freedom subjected to the impulsive excitation.
17.2.4 General Case of Excitation Let us consider the massless conditional dynamic system (beam, truss, frame, etc.) with two degrees of freedom. Two masses m1 and m2 of structure are subjected to disturbing forces F1(t) and F2(t) which are arbitrary function of time t (Fig. 17.19).
17.2
Structures with Finite Number of Degrees of Freedom: The Force Method
647
F1(t)
a
F2(t) m2
m1 y1
b
y2
F (t) F1(t) F2(t)
t Fig. 17.19 Arbitrary excitation of conditional structure with two degrees of freedom. (a) Design diagram; (b) Excitation functions F1 and F2 as a function of time
Dynamic analysis of forced vibration begins with determining the frequencies of free vibration ω1,ω2 and coefficients of amplitude distribution ρ21, ρ22 (17.34). The displacements of mass m1 and m2 are y1(t) and y2(t). For their computation the following formulas should be used (Kiselev 1980) y1 ðt Þ ¼ y11 þ y12
1 ¼ ω1
ðt 0
F 1 ðτÞ þ ρ21 F 2 ðτÞ 1 sin ω1 ðt τÞdτ þ ω2 m1 þ m2 ρ221
ðt 0
F 1 ðτÞ þ ρ22 F 2 ðτÞ sin ω2 ðt τÞdτ, m1 þ m2 ρ222
ð17:37Þ
y2 ðt Þ ¼ ρ21 y11 þ ρ22 y12 Example 17.8 The massless uniform simply supported beam with overhang contains two lumped mass m1 and m2. Both of these masses are subjected to suddenly applied two-stepped loads F1 and F2 which then remain on the beam constantly (Fig. 17.20a). F1
a
m1
d
m2
y1
ω1
J2=1·ρ21
y2 ω1
b
J1=1
F2
y2=1·ρ21
J1=1
e
J2=1·ρ22 ω2
y1=1 ω2
c y1=1
y2=1·ρ22
Fig. 17.20 (a) Design diagram of weightless uniform beam with two degrees of freedom; (b, c) first and second mode of vibration; (d, e) forces which defines the first and second forms of vibrations
The response of each mass may be determined in the following order. 1. Choose the generalized coordinates (y1, y2), construct unit states (unit loads are applied to masses in the direction of generalized coordinates), plot corresponding unit bending moment diagrams M 1 , M 2 , and calculate unit displacements EIδik ¼ M i M k , i, k ¼ 1, 2. Vereshchagin’s procedure for computation of δik is most effective. In case of statically indeterminate two-span and three-span beam for construction of unit bending moment diagrams may be used Tables A.10 and A.11, respectively. In case of such beams with overhang the foci method may be applied. 2. Compile the frequency equation (17.28) and find frequencies of free vibration ω1 and ω2.
648
17
D¼
m1 δ11 1=ω2 m1 δ21
m2 δ12 m2 δ22 1=ω2
Dynamics of Elastic Systems: Forced Vibration
¼ 0 ! ω21 ,
ω22
ðaÞ
In case of different m1 and m2 should assume m1 ¼ m and m2 ¼ α1m, where α1 is a given positive number. 3. Determine the mode of vibration (the coefficients of amplitudes distribution) corresponding to frequencies ω1 and ω2 m1 δ21 ω21 y2 ρ21 ¼ ¼ , y1 ω 1 m2 δ22 ω21 1 m1 δ21 ω22 y ρ22 ¼ 2 ¼ y1 ω2 m2 δ22 ω22 1
ðbÞ
Verification of frequencies vibration and the coefficients of amplitudes distribution m 1 1 þ 2 ¼ m1 δ11 þ m2 δ22 ; ρ21 ρ22 ¼ 1 ; 2 m2 ω1 ω2 The expected forms vibration are shown in Fig. 17.20b, c. 4. Expression for response of both mass F 1 þ ρ21 F 2 F þ ρ22 F 2 ð1 cos ω1 t Þ þ 2 1 ð1 cos ω2 t Þ, ω21 m1 þ m2 ρ221 ω2 m1 þ m2 ρ222 y2 ðt Þ ¼ ρ21 y11 þ ρ22 y12 y1 ðt Þ ¼ y11 þ y12 ¼
ðcÞ
In case of different F1 and F2 should assume F1 ¼ F and F2 ¼ α2F, where α2 is a given positive number. 5. Determine the relationship between the inertial forces, which characterized forms of vibration with frequencies ω1 and ω2. J 1 ¼ 1,
J2 ¼
m2 ρ for ω1 , and J 1 ¼ 1, m1 21
J2 ¼
m2 ρ for ω2 m1 22
ðdÞ
Loading scheme that defines the first and second forms of vibrations is shown in Fig. 17.20d, e. 6. Determining the reactions and internal forces. The given structure is loaded by inertial forces as shown in Fig. 17.20d, e. The loading schemes (d) and (e) correspond to vibrations with a frequency ω1 and ω2, respectively. Corresponding bending in moment diagrams for ω1 and ω2 are M in 1 and M 2 . To test the determination of inertial forces, it is recommended to use the in in orthogonality property M 1 M 2 ¼ 0 . For multiplication of bending moment diagrams Vereshchagin rule is recommended. The determination of reactions and internal forces caused by inertial forces does not cause problems. Special Case Arbitrary deformable structure with two degrees of freedom is suddenly subjected to one-stepped force P1 which is applied to mass m1 in the direction of the first generalized coordinate y1 and then remains there. In this case the displacements y1 and y2 become
F 1 ð1 cos ω1 t Þ ω21 ð1 cos ω2 t Þ þ , ω21 m1 þ m2 ρ221 ω22 m1 þ m2 ρ222 y2 ðt Þ ¼ ρ21 y11 þ ρ22 y12
y1 ðt Þ ¼ y11 þ y12 ¼
ð17:38Þ
These exact analytical expressions allow plotting displacement graph, finding dynamic coefficients, and the distribution of internal forces in the system with the adopted stepped excitation to the mass m1.
17.3
Structures with Finite Number of Degrees of Freedom: Initial Parameters Method
This paragraph contains the development of the initial parameters method, previously studied in problems of statics (Ch. 8) and stability of deformable systems (Chap. 15), as applied to problems of dynamic analysis of structures. The application areas of the method are twofold. On the one hand, it is effective for the dynamic analysis of steady-state vibration of beams. On the other hand, it allows creating a table of reference data that is widely used in the displacement method for analysis of steadystate vibration of the framed structures with elements of different stiffness and containing lumped masses, bodies.
17.3
Structures with Finite Number of Degrees of Freedom: Initial Parameters Method
649
Two limiting cases related to taking into account the inertial properties of separate rod as the element of the rod structure will be considered in this chapter. This is a massless rod on which there are lumped inertia bodies having mass and moment of inertia, and a rod with uniformly distributed mass in the absence of an inertial body on it. The first case for a separate rod leads to a system with a finite number of degrees of freedom; this case is considered in current paragraph. The second case for a single rod leads to a system with an infinite number of degrees of freedom; this case will be considered in paragraph 17.5. For both cases the method of initial parameters can be effectively applied.
17.3.1 Resolving Equations The object under study is the massless uniform beam of flexural stiffness EI; the boundary conditions for beam are not specified. In general case, there are lumped bodies of mass m and moment of inertia J. Beam is subjected to force and kinematical excitations. The forces excitations include lumped forces and external couples, P(t) and Me(t). Kinematic excitations include linear and angular displacements of the supports, Δ(t) and φ(t). All excitations change by harmonic law with frequency excitation θ, i.e., P(t) ¼ P sin θt, etc. The number of lumped masses, force excitations and their location is not specified. Assume that the dissipative forces are neglected and superposition principle is hold. The problem is to determine the reactions, support moments, internal forces, and displacements. The peculiarity of the problem is necessity to take into account the forces of inertia and the moments of inertia of the bodies. Design diagram of beam with one lumped body and notations are shown in Fig. 17.21a. The origin of coordinates is placed on the left end of the beam, the x and y axes are directed to the right and upward, respectively. The position of the inertial body on the beam is determined by the coordinate am, the position P(t) and Me(t) by coordinates aP and aM. Fragment of the design diagram in the vicinity of the lumped mass is shown in Fig. 17.21b. A solid line (1) means the position of the axis of the rod before vibrations occur, while dotted line 2 means the position of the axis of the rod at an arbitrary time. Corresponding body position is determined by its vertical displacement ym sin θ t and angle of rotation φm sin θt.
a
y
P(t)=Psinθt
M0(t)
b Me(t)=Msinθt
m, J
φm
θ2mym θ2Jφm
x
Dynamic elastic curve (2) Tangent
0 Q0(t)
am
(1)
aP
R(t) aR
m, J
ym
aM x Fig. 17.21 Notation for initial parameters method. (a) General design diagram; kinematical initial parameters Δ0(t) φ0(t) are not shown; (b) Amplitude values of inertial force θ2mym and inertial couple θ2Jφm
The inertial force and the moment of inertia of the body are θ2mymsinθt and θ2Jφmsin θt. Since all excitations change in accordance with the harmonic law with the frequency of excitation θ, then all active, inertial, and reactive factors (reactions, internal forces, deflections) contain a multiplier sinθ t. Therefore, this factor will be omitted, meaning only the amplitude values of the excitations and factors determining the stress-strain state of the beam (displacements y, rotation angles, bending moments M, shear forces Q, and reactions). The equations of the initial parameters method represent expressions for the amplitude values of the required dynamic factors (Kiselev 1980). In general case, these equations are written as:
ð x aP Þ 3 ð x aR Þ 3 ð x aM Þ 2 ðx am Þ3 ð x am Þ 2 1 x2 x3 θ2 y ð x Þ ¼ y 0 þ φ0 x þ M 0 þ Q0 þ P þR þM þ mym þ Jφm EI 2! 3! 3! 3! 2! EI 3! 2!
650
17
Dynamics of Elastic Systems: Forced Vibration
ðx aP Þ2 ð x aR Þ 2 ð x am Þ 2 1 x2 θ2 M 0 x þ Q0 þ P þR þ M ð x aM Þ þ mym þ Jφm ðx am Þ EI 2 2 2 EI 2 M ðxÞ ¼ M 0 þ Q0 x þ ½M þ Pðx aP Þ þ Rðx aR Þ þ θ2 ½mym ðx am Þ þ Jφm ð17:39Þ QðxÞ ¼ Q0 þ ðP þ RÞ þ θ2 mym φð x Þ ¼ φ 0 þ
The initial parameters are supplied with the index “0.” Positive signs of each term in brackets (17.39) means that the corresponding factor (M0, Q0, P, .. . .) creates in section x positive bending moment. In equations (17.39), only those terms should be preserved, for which the difference x a is positive. In other words, in the expression for fixed x, only those loaded factors should be taken into account that are located to the left of the section x. Thus, equations (17.39) for 0 x am includes only M0, Q0, while for am x aP the additional terms, such as the inertia force θ2mym and the moment of inertia θ2Jφm, should be taken into account. In the case of an arbitrary number of lumped masses, reactions, concentrated forces, and couples, it is necessary, using the principle of superposition, to enter the corresponding terms in each equation. The principal difference of the system (17.39) from its static analogs (8.8, 8.9, Sect. 8.3.1) is that in addition to the active forces and reactions of the supports, dynamic equations of initial parameters contain inertial forces. They contain a multiplier θ2 and are represented by the last term in each equation. my ðx am Þ3 Jφ ðx am Þ2 and θ2 m . In their form, they coincide with The first equation contains two additional terms θ2 m EI 3! EI 2! the terms related to forces (Q0, P, R) and couples (M0, M), respectively. Each of this term represents the displacement of the beam at an arbitrary point x (if am x) caused by the amplitude of inertial force θ2mym and amplitude of inertial couple θ2Jφm, which are applied to beam at point x ¼ am; expressions for amplitude values contain linear ym and angular φm displacement of the beam in the section where the inertial body is located. As in the case of applying the method of initial parameters to static problems, in dynamic problems at the origin (i.e., at the left end of a beam), two conditions turn out to be zero, and two other initial parameters are unknown. At the right end of the beam, two conditions are known (in particular, one or both are zero). To determine the unknown initial parameters, the equations should be subjected to known conditions at the right end. It means, from the four equations (17.39) we need to choose two equations for parameters, the values of which are known at the right end. For example, at the free right end, the bending moment and shear force are zero; thus, we need to make equations for M(l) and Q(l). The method of initial parameters can be applied to the problems of determining the frequencies of free vibrations. To do this, all active actions should be discarded and the frequency θ should be replaced by the frequency ω of free vibration. Subject of the functions to boundary conditions at the right end is made as before. In the case of determining the frequencies of free vibrations, we obtain a system of linear homogeneous algebraic equations. Additional equations are compiled for the section am where the concentrated body is located. The determinant composed of coefficients for unknowns represents the frequency equation. The fundamental distinction between the equations of the initial parameters method (16.28, Sect. 16.4.4) and (17.39) is in the difference of the beams design diagrams: in the first case, design diagram of beams takes into account distributed mass, while in the second case the massless beams with a lumped masses are considered. Other difference is as follows: Eq. (16.28) are limited because they do not allow taking into account the effect of external influences. Example 17.9 The massless uniform clamped-pinned beam of length l ¼ a + b and bending stiffness EI contains a lumped mass m (Fig. 17.22). Determine the frequency of free vibration. Solution At the right end of the beam (at x ¼ l), the displacement and bending moment are zero. Therefore, from the system of equations (17.39), we choose the equations for y (x) and M (x). Such a choice will lead to a system of homogeneous algebraic equations Initial parameters y0=0, φ0=0
Boundary condition at x=l yl=0, Ml=0
m a
b
Fig. 17.22 Design diagram of clamped-pinned uniform beam with lumped muss
17.3
Structures with Finite Number of Degrees of Freedom: Initial Parameters Method
y ð x Þ ¼ y 0 þ φ0 x þ
ðx a m Þ3 1 x2 x3 ω2 M 0 þ Q0 þ mym EI 2! 3! EI 3!
651
ðaÞ
M ðxÞ ¼ M 0 þ Q0 x þ ω mym ðx am Þ 2
Here ω is the required frequency of free vibration, ym represents the displacement of the mass m, am ¼ a. Since initial parameters are y0 ¼ 0, φ0 ¼ 0, then system (а) for point х ¼ l after reducing by EI becomes ð l am Þ 3 l2 l3 þ Q0 þ ω2 mym ¼ 0, 2 6 6 2 M ðlÞ ¼ M 0 þ Q0 l þ ω mym ðl am Þ ¼ 0: yðlÞ ¼ M 0
ðbÞ
These equations contains unknown displacement ym. To determine it, we apply the first equation of system (a); with x ¼ am ¼ a we have 1 a2 a3 y ð am Þ ¼ M 0 þ Q0 EI 2 6 Substitute this expression in system (b) and take into account l am ¼ b l2 l3 1 a2 a3 b3 2 yð l Þ ¼ M 0 þ Q 0 þ ω m M 0 þ Q0 ¼0 EI 2 6 2 6 6 2 3 1 a a M 0 þ Q0 b¼0 M ðlÞ ¼ M 0 þ Q0 l þ ω2 m EI 2 6 We will reduce this system to homogeneous algebraic equations for unknown initial parameters M0 Q0 2 3 l ω2 m a 2 b 3 l ω2 m a 3 b 3 M0 þ þ Q0 þ ¼0 EI 2 6 EI 6 6 2 6 ω2 m a 2 ω2 m 1 a 3 b þ Q0 l þ b ¼0 M0 1 þ EI 2 EI EI 6 Nontrivial solution occurs, if 2 3 2 2 3 2 3 3 l þ ω ma b ; l þ ω ma b EI 12 EI 36 ¼ 0 6 D¼ 2 2 2 ω2 m a3 b 1 þ ω ma b lþ EI 2 EI 6 3 2 2 2 3 2 3 2 l ω mab ω mab ω m a2 b l ω2 m a 3 b 3 1þ þ lþ þ ¼0 EI 12 EI 6 EI 2 EI 36 2 6 The root of this equation represents the frequency of free vibration. After simple transformations we get sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12l3 EI ω¼ ma3 b2 ð3a þ 4bÞ rffiffiffiffiffiffiffiffirffiffiffiffiffiffiffi 768 EI Special case. If a ¼ b¼ l/2, then we get ω ¼ . It is obvious that for this simplest case we can use well-known 7 ml3 rffiffiffiffiffiffiffi 3 g 7 Pl P formula ω ¼ , where ystat ¼ , g¼ . ystat 768 EI m If the beam contains an intermediate support, then the number of unknowns includes nonzero initial conditions as well as the reaction of support. To compile an additional condition, the equation of the elastic line y(x) should be subordinated to the condition y(x ¼ aR) ¼ 0. The frequency equation represents the condition of nontriviality of homogeneous linear algebraic equations for unknowns.
652
17
Dynamics of Elastic Systems: Forced Vibration
17.3.2 Steady-State Vibrations This paragraph demonstrates the procedure of initial parameter method for the case of the harmonic excitation of the kinematic type. It is assumed that the massless beam of constant stiffness contains one lumped mass. Also, the reciprocal reactions theorem for dynamical analysis of steady-state vibration is shown. Example 17.10 Clamped-clamped beam of length l and stiffness EI contains one lumped mass m. The structure is subjected to angular harmonic excitation φ(t) ¼ φ0 sin θt of the left support. Calculate amplitude values of reactions of the left support (Fig. 17.23).
φ(t) m
M0 Q0 y
ym a
x
J=θ2mym b=l - a
Fig. 17.23 Design diagram of a structure
Solution The initial kinematical parameters are φ(t) ¼ φ0 sin θ t and y0(t) ¼ 0. The unknown initial forced parameters are M0(t) ¼ M0 sin θt and Q0(t) ¼ Q0 sin θt. Boundary conditions at the right clamped support are y(l ) ¼ 0 and φ(l) ¼ 0. Assume that the moment of inertia of mass m is neglected, while the inertial force J of mass m is taken into account. Since y0 ¼ 0, then equations of initial parameter method for amplitude values y(x) and φ(x) should be written as follows
ð x am Þ 3 1 x2 x3 M 0 þ Q0 θ2 mym EI 2! 3! 3!
2 ðx am Þ2 1 x 2 φð x Þ ¼ φ0 M 0 x þ Q0 θ mym , am ¼ a EI 2 2
y ð x Þ ¼ φ0 x
ðaÞ
In these equations, a negative sign in front of the bracket means the direction of the y axis downwards, while in front of the term, which contains θ2 means that the positive inertial force θ2mym creates a negative moment in the cross section x > a. Displacement and the force of inertia of mass m are
1 a2 a3 M 0 þ Q0 y m ð x ¼ a Þ ¼ φ0 a EI 2 6 2 3 Q a M a θ2 mym ¼ θ2 m φ0 a 0 0 EI 2 EI 6
ðbÞ
Substitution of the expression for the inertial force into (a) leads to the following formulas for linear and angular displacements
3 M 0 x2 Q0 x3 θ2 m M 0 a2 Q 0 a3 ð x aÞ yð xÞ ¼ φ 0 x φ0 a þ EI EI 2 EI 6 EI 2 EI 6 6
2 Q x2 θ 2 m M M a 2 Q a3 ð x aÞ φðxÞ ¼ φ0 0 x 0 þ φ0 a 0 0 EI EI EI 2 EI 2 EI 6 2 Considering boundary conditions at the right end (if х ¼ l then y ¼ 0, φ ¼ 0), then relationships (c) become
ðcÞ
17.3
Structures with Finite Number of Degrees of Freedom: Initial Parameters Method
3 M 0 l2 Q 0 l3 θ 2 m M a2 Q a3 ð l aÞ þ φ0 a 0 0 ¼0 EI 2 EI 6 EI EI 2 EI 6 6
2 M 0 l Q 0 x2 θ 2 m M 0 a2 Q 0 a3 ð l a Þ φð l Þ ¼ φ0 φ0 a þ ¼0 EI EI EI 2 EI 2 EI 6 2
653
y ð l Þ ¼ φ0 l
In the formulas (d) we will isolate members with unknowns M0 and Q0.
3 3 ð l aÞ 3 Q l3 M 0 l2 a2 ð l aÞ a3 ð l aÞ þ θ2 m þ 0 þ θ2 m ¼ φ0 l þ θ2 ma EI 2 2 6EI EI 6 6 6EI 6EI
2
2 ðl aÞ2 3 ð l aÞ 2 ð l aÞ 2 Q0 l M0 2 a 2 a 2 lþθ m þ þθ m ¼ φ0 1 þ θ ma EI 2 2EI EI 2 6 2EI 2EI
ðdÞ
ðeÞ
Now we need to introduce the dimensionless parameter θ/ω. For this purpose the second term in each bracket should be transformed. Unit displacement and frequency of free vibration for clamped-clamped beam are δ11
rffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffi a b 1 3l3 EI 3l3 EI !m¼ 2 3 3 ¼ 3 !ω¼ ¼ 3 3 mδ11 ma b ω ab 3l EI 3 3
The second term in the first bracket (e) becomes θ2 m
3 a2 ðl aÞ a2 b3 θ 2 l3 ¼ θ2 m ¼ 2 2 6EI 2 6EI ω 4a
Equations (e) in the equivalent form take the form
Q 0 l3 θ 2 l3 M 0 l2 θ2 l3 θ 2 l3 þ þ þ ¼ φ0 l þ 2 2 EI 2 ω2 4a EI 6 ω2 12 ω 2a
2 3 2 2 3 Q0 l M0 θ 3l θ l θ2 3l3 lþ 2 þ þ ¼ φ0 1 þ 2 2 EI EI 2 ω2 4b ω 4ab ω 2a b
ðeÞ
Solution of this set of equations is (subscript 0 at φ is omitted) θ2 3b 1 F ¼ 1 2 1þ 4a ω 1 θ2 =ω2
2 6EI θ 3ab þ b2 1 Q0 ¼ φ 2 G, G ¼ 1 2 1 þ 2 ω 2a l 1 θ2 =ω2 4EI M0 ¼ φ F, l
ðfÞ
Special case: If a ¼ b ¼ l/2, then the bending moment and shear force at the left clamped supports are M ð 0Þ ¼
4EI 1 7θ2 =4ω2 φ sin θt, l 1 θ2 =ω2
Qð0Þ ¼
6EI 1 3θ2 =ω2 φ sin θt: l2 1 θ2 =ω2
ðgÞ
These formulas present reference data for analysis of the dynamical structures, and are shown in Table A.27, row 1. Let us substitute these expressions in the first expression (b). Displacement at the middle of the beam caused by a static angular displacement φ becomes 6EI l3 l 1 x2 x3 l 1 4EI l2 yðxÞ ¼ φx M 0 þ Q0 ¼φ 2 φ¼ φ 8 EI 2 EI l 4 2! 8 3! 2! 3! x¼l=2 l
ðhÞ
According to (g), in case of θ/ω ¼ 1 resonance occurs. Using the found values of the initial parameters and any parameter θ/ω it is easy to construct a corresponding curve y(x) of the amplitude values of the dynamic displacements. In the case of a harmonic force excitation F(t) ¼ F sin θt or M(t) ¼ M sin θt, the corresponding terms should be introduced into the equations of the initial parameters method (17.39). Here, F and M are the amplitude values of the harmonic force F(t) and M(t) the harmonic couple, respectively.
654
17
Dynamics of Elastic Systems: Forced Vibration
Example 17.11 Uniform clamped-clamped beam AB with lumped mass m is subjected to unit harmonic rotation φ(t) ¼ φ sin θ t, φ ¼ 1 of the left support A (Fig. 17.24a). Determine the reactions at the right support B, inertial force and bending moment under the lumped mass. Assume a ¼ b ¼ l/2. Use formulas (g, h) from previous example. State 1 (Table A.27, row 1)
a MA1=4iF1μ
A
φ(t)
State 2 (Table A27, row 2) φ(t) B A
c
B
MB1=?
MB2=?
MA2=2iF2 μ a=l/2
RA1= - (6i/l)F5 μ
a=l/2
b RB1=?
b
b=a
RA2=- (6i/l)F6 μ
RB2=?
J* MA1
A
m
B MB1
Notation: i =
Mm=i(1+2δ) μ RA1
q2 EI 1 , d = 2, m= l w 1– q 2 w2
RB1
Fig. 17.24 (a, b) Design diagram of clamped-clamped beam and reaction of supports in the real (first) state; (c) Second state. The functions F are presented in Table A.27. Factors φsinθt are omitted
Solution In the case of a static rotation of a clamped support A, displacement in the middle of the beam according to Example l 17.10, (or Table A.9, row 1) equals y ¼ φ. Therefore the equation of motion of the mass m becomes 8 l 1 yðt Þ ¼ φ sin θt 8 1 θ2 =ω2 rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi 1 192EI , then the mass m may be presented in terms of frequency Since the frequency of the free vibration ω ¼ ¼ mδ11 ml3 192EI vibration m ¼ 2 3 . Therefore the expression for inertial force of mass m becomes ω l J ðt Þ ¼ mθ2 yðt Þ ¼
θ2 =ω2 192EI 2 l 1 24EI θ ¼ φ sin θt φ sin θt 8 ω 2 l3 1 θ2 =ω2 l2 1 θ2 =ω2
The bending moment and shear force at the left clamped support are M ð 0Þ ¼ Q ð 0Þ ¼
4EI J l 4EI 1 7θ2 =4ω2 φ sin θt ¼ φ sin θt l 8 l 1 θ2 =ω2
6EI J 6EI 1 3θ2 =ω2 sin θt: φ sin θt þ ¼ 2 φ 2 2 l l 1 θ2 =ω2
The expressions for M(0) and Q(0) coincide with (g), which had been obtained by initial parameters method. Bending moment under lumped mass m l l 4EI J l EI 1 þ 2θ2 =ω2 ¼ M ð0Þ þ Qð0Þ ¼ φ sin θt ¼ φ sin θt M 2 2 l 8 l 1 θ2 =ω2 Bending moment at the right clamped support B is M ðl Þ ¼
2EI Jl 2EI 1 þ θ2 =2ω2 φ sin θt ¼ φ sin θt l 8 l 1 θ2 =ω2
Reaction at the right clamped support B may be determined from equation
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
X
Y ¼ RA1 J ðt Þ þ RB1 ¼ 0 ! RB1 ¼
655
6EI 1 þ θ2 =ω2 φ sin θt l2 1 θ2 =ω2
The scheme of forces for beam under given excitation (first state) is presented in Fig. 17.24b. According to Table A.27, row 1, reactions at the right clamped support B are unknown; this is shown in Fig. 17.24a. For computation of these reactions we will apply the reciprocal reactions theorem. For this purpose we need to create additional state 2 for beam with same boundary condition (Fig. 17.24c). For this state there are tabulated reactions at the left support A and unknown reactions at the right support B. According to the reciprocal reactions theorem we have MB1 ¼ MA2; RB1 ¼ RA2; MB2 ¼ MA1; RB2 ¼ RA1. Reactions which correspond these relationships are shown in the blocks of the same type. First state
Second state Tabulated data
M A1 = M ( 0 )1 =
4 EI 1 − 7θ 2 4 ω 2 sin θ t ( clockwise ) ϕ l 1 − θ 2 ω2
RA1 = Q ( 0 )1 = −
6 EI l2
ϕ
1 − 3θ 2 ω 2 1 − θ 2 ω2
( )
sin θ t ↓
M A2 = M ( 0 )2 = R A2 = Q ( 0 )2 =
2 EI 1 + θ 2 2 ω 2 sin θ t ( clockwise ) ϕ l 1−θ 2 ω2
6 EI l2
ϕ
1+ θ 2 ω2 1−θ 2 ω2
( )
sin θ t ↓
Computed data M B1 =
2 EI 1 + θ 2 2 ω 2 ϕ sin θ t ( clockwise ) l 1−θ 2 ω2 RB1 =
6 EI l2
ϕ
1+θ 2 ω2 1−θ 2 ω
( )
sin θ t ↑ 2
M B2 = RB 2 =
4 EI 1 − 7θ 2 4 ω 2 sin θ t ( clockwise ) ϕ l 1 − θ 2 ω2
6 EI l
2
ϕ
1 − 3θ 2 ω 2 1 −θ 2 ω2
( )
sin θ t ↓
Verification: bending moment under mass m taking into account the right forces
l 2EI 1 θ2 6EI θ2 l sin θt EI 1 þ 2θ2 =ω2 1þ φ 1 þ ¼ φ sin θt M m ¼ M B1 þ RB1 ¼ þ 2 2 2 2 2 2 l 2ω l ω 2 1 θ =ω l 1 θ2 =ω2 Now we will demonstrate application of the reciprocal reactions and displacements theorem. Let us consider a beam shown in Fig. 17.24a subjected to harmonic kinematical excitation of the left clamped support. We need to determine vertical dynamical displacement y(l/2) where the mass m is located. For this purpose we need to create additional state, which presents same clamped-clamped massless beam with one lumped mass m attached at the middle of the span. Corresponding scheme is shown in Table A.27, row 8. According to the reciprocal theorem, the displacement y(l/2) caused by harmonic angular displacement of the left support is equal to the bending moment at the left clamped support caused by harmonic force which is applied at the point where displacement should be determined, i.e., l l sin θ y ¼ r ð0Þjrow8 ¼ : 2 row 1 8 1 θ2 =ω2 For both design diagrams, the amplitudes of the kinematical and forced excitations are equal to unity. The initial parameters method in form (17.39) allows us to create a library of design diagrams for weightless homogeneous beams with a lumped mass; these beams are subjected to different harmonic excitations, such as kinematical and force excitations. This set of different design diagrams of separate beams together with the displacement method is very effective for dynamic analysis of steady-state vibration of various rod systems, especially for continuous beams, and frames.
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
The general procedure for analyzing steady-state bending vibration of deformable structures with a finite number of degrees of freedom is considered. The specific assumptions are discussed and the detailed procedure of the displacement method is presented. It is noted that after determining the unknowns of displacement method for further detailed analysis (construction of dynamic bending moment diagram, shear, etc.) it is advisable to apply the method of initial parameters. The application of group unknowns to the dynamic analysis of symmetric structures is discussed.
656
17
Dynamics of Elastic Systems: Forced Vibration
17.4.1 The Steady-State Vibration Design diagram of a structure. The object under study are beams and frames with rectilinear elements. Flexural stiffness of elements of a structure can be different, but within each element the stiffness is considered constant. Each element of the structure can contain one lumped mass, the position of which within the span can be arbitrary. The distributed mass of deformable element itself is negligible compared to the lumped mass. The presence of element without lumped mass is possible; in this case only the rigidity of the massless element is taken into account. Excitation character. In contrast to static problems, the structures are subjected to various dynamic effects, in particular, force harmonic excitations P(t) ¼ P sin θ t and/or kinematical settlements of supports Δ(t) ¼ Δ0 sin θ t (kinematical excitation). The points of application of force and mass may not coincide. The amplitudes of the disturbing forces (displacements) may be different, but the frequency of external excitation of all factors is the same. We believe that the system is ideal (in which there is no energy dissipation) and linear (for which the superposition principle is valid). In this case, the steady-state vibration of the structure occurs at the excitation frequency, i.e., all the characteristics of the dynamic state vary according to the harmonic law with the frequency of excitation θ. The statement of the problem. The problem is to determine the amplitude values of the characteristics of its dynamic state— bending moments, reactions, displacements, etc. Note that we are dealing with steady-state vibration of the deformable structure. The feature of such problems is when the system vibrates, the presence of the lumped mass leads to the appearance of the concentrated inertia harmonic force. For the analysis of the listed type of structures under the accepted assumptions, it is recommended to apply the displacement method. This classical method for analyzing the dynamic state of beam and framed systems with a high degree of static indeterminacy, as in the case of static and stability analysis, turns out to be very effective. This is explained by the high degree of algorithmization and the presence of reference data for reactions of individual elements subjected to various kinematic and force excitations. The primary system of the displacement method is constructed in the same way as in static problems: additional constraints that prevent angular and linear displacements of joints are introduced. In contrast to the static problems, the displacements of the introduced constraints occur according to law Z1(t) ¼ Z1 sin θ t, Z2 (t) ¼ Z2 sin θ t, . Similarly, the reaction of any i-th introduced constraint due to the harmonic displacement of the k-th introduced constraint also changes by law Rik(t) ¼ Rik sin θ t . Since in the original system the introduced constraints are absent, then the total reaction of this constraint equals zero Ri ðt Þ ¼ Ri1 sin θ t þ Ri2 sin θ t þ þ Rin sin θ t þ R1P sin θ t ¼ 0
ð17:40Þ
The number of such equations coincides with the degree of kinematic indeterminacy, or, in other words, coincides with the number of introduced constraints. Now we impart a convenient form for each term in (17.40). For this we need to introduce the coefficient r ik ¼
Rik sin θ t Rik ¼ ! r ik Z k ¼ Rik Z k sin θ t Zk
ð17:41Þ
The expression rik ¼ Rik/Zk represents the amplitude of the reactions in the ith introduced constraint caused by harmonic vibrational displacement of the k-th introduced constraint with unit amplitude. After that, the total reaction of i-th introduced constraint in terms of unit reactions rik and amplitude values of the primary unknowns Zi may be presented as follows Ri ðt Þ ¼ r i1 Z 1 þ r i2 Z 2 þ þ r in Z n þ R1P ¼ 0
ð17:42Þ
In the expanded form, the canonical equations of the displacements method in problems of dynamics have the form r 11 Z 1 þ r 12 Z 2 þ þ r 1n Z n þ R1P ¼ 0 : : : : : : : : : : : : r n1 Z 1 þ r n2 Z 2 þ þ r nn Z n þ RnP ¼ 0
ð17:42aÞ
The primary unknowns Zi of the displacement method are the amplitudes of the vibrational displacements Z1(t), , Zn(t) of the introduced constraints. Free terms RiP is the amplitude of reaction of the i-th introduced constraint caused by given vibration excitation. The unit reactions and the loaded terms of canonical equations are harmonic functions; for typical design diagrams of separate rod they are presented in Table A.27. It is assumed that the beam contains one lumped mass, while the distributed mass of the beam itself is not taken into account. The rod is subjected to harmonic excitation of force or kinematic type. Expressions of dynamic reactions include the correction factors Fi that take into account the inertial force of the lumped mass;
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
657
these functions depend on the position of the mass in the span, types of excitations, boundary conditions, and detuning (θ/ω). In case of static loading (θ ¼ 0), the correction functions are equal to unity. In case of the other type of boundary conditions (for example, a slider or elastic support) and the presence of two or more lumped mass, it is necessary to enhance the above table. For this purpose it is recommended to apply the reciprocal theorems and initial parameters method. If structural element does not have a lumped mass, then the reactions caused by unit displacement of the introduced constraint are determined by Tables A.3–A.6 as in the case of static problem. After determining the amplitude values of the displacements Z of the introduced constraints, the amplitude of bending moment at any point A of a structure may be calculated by formula M A ¼ M 1A Z 1 þ M 2A Z 2 þ þ M nA Z n þ M 0PA where M iA and M iA Z i are the bending moments at point A of a structure due to unit displacement of introduced constraint Zi ¼ 1, and due to amplitude displacement Zi, respectively; M 0PA is bending moment at same point A of a structure due to the amplitude value of harmonic excitation. Example 17.12 Design diagram of the portal frame with absolutely rigid crossbar is shown in Fig. 17.25a. Crossbar has distributed mass per unit length m. The bending stiffness of the massless vertical members is EI. The frame is subjected to horizontal harmonic force P(t) ¼ Psinθt . Provide dynamic analysis of the steady-state vibration.
P(t)
EI=∞, m
Z1=1
c
b
a P(t)
1
1
r11
J h
EI, m=0
A 3i/h
l
3i/h
Fig. 17.25 (a, b) Design diagram and primary system, (c) Bending moment diagram due by Z1 ¼ 1, i ¼ EI/h
Solution This structure has one degree of freedom since the vertical bars are massless itself and without lumped mass, and there is only one mass belonging to the horizontal element. The kinematical indeterminacy of the structure equals one. Indeed, since the crossbar is absolutely rigid, the joints of frame do not rotate (even if the connection of the vertical and crossbar would be rigid), and thus only horizontal displacement of the crossbar is possible. Therefore the structure has only one generalized coordinate, mainly, horizontal displacement of the crossbar (the frame with sidesway). The primary system is shown in Fig. 17.25b; additional support 1 prevents horizontal displacement. Thus, the primary unknown is the horizontal displacement Z1 at introduced constraint 1. Crossbar is nondeformable element; therefore it mass may be treated as lumped mass M ¼ ml. Canonical equation of displacement method is r11Z1 + R1P ¼ 0. The peculiarity of the system is that when calculating the unit reactionr11, the horizontal inertial force J ¼ Mθ2 of the lumped mass M¼ml should be taken into account. r 11 ¼ 2
3i EI J ¼ 6 3 Mθ2 2 h h
ðaÞ
Here 3i/h2 is reaction of introduced constraint 1 caused by unit displacement of this constraint and due to the bending of the non-inertial column. Free term R1P ¼ P, so the canonical equation and primary unknown are 6
EI P 2 Mθ Z1 P ¼ 0 ! Z1 ¼ h3 6EI=h3 Mθ2
ðbÞ
Now let us take into account the detuning θ/ω. In case of free vibration a structure performs vibration with frequency ω; EI EI therefore, according to (а) we have r 11 ¼ 6 3 Mω2 ¼ 0. If we substitute the expression 6 3 ¼ Mω2 into (b), then for h h primary unknown we get Z1 ¼
P ω2 M 1 θ2 =ω2
658
17
Dynamics of Elastic Systems: Forced Vibration
The bending moment at the bottom point A of the vertical bar is M A ¼ M A Z 1 þ M 0A ðPÞ Bending moment at point A of the primary system caused by force P is M 0A ðPÞ ¼ 0; therefore M dyn A ¼
3EI P 3EI P þ0¼ 2 h2 ω2 M 1 θ2 =ω2 h ω2 M 1 θ2 =ω2
ðcÞ
Since in case of static loading M stat A ¼ Ph=2, then dynamic coefficient becomes μ¼
M dyn 3EI P 1 6EI 1 1 A ¼ ¼ ¼ 2 2 2 3 2 2 2 2 Ph=2 M stat h ω M 1 θ =ω Mh ω 1 θ =ω 1 θ2 =ω2 A
ðdÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi θ2 θ2 ¼ . The resonance occurs if θ ¼ 6EI=Mh3 . ω2 6EI=Mh3 1 is inherent for systems with one degree of freedom. Expression for dynamic coefficient μ ¼ 1 θ2 =ω2 For given structure the square of detuning is
Example 17.13 The frame in Fig. 17.26a consists of three uniform massless rods of length l and bending stiffness EI. Each element contains one mass m which is located at the middle of the member. Mass m at the span 1-B is subjected to disturbing harmonic force P(t)¼Psinθt. Determine the amplitude value of angular rotation of the rigid joint 1 in case of the steady-state vibration.
a
b
Psin 1
A
B
a 1 b
A
m
m
c
Psin
m
r11 1 Ma
B
Mc
m
m
Mb
m
c
d C
C
R1P 1 M0b-P
Fig. 17.26 (a, b) Design diagram of frame and primary system; (c, d) Calculation of unit reaction and free term of canonical equation frame
Solution This structure has three degrees of freedom. Its kinematical indeterminacy is equal to one (while this structure contains six unknowns of the force method). The primary system of displacement method is obtained from a given structure by introducing constraint 1 at the fixed joint 1 (Fig. 15.26b). This constraint prevents angular displacement at support 1. The primary unknown is the angular displacement Z1 at support 1. Canonical equation of displacement method is r 11 Z 1 þ R1P ¼ 0
ðaÞ
Unit reaction r11. Suppose that the introduced constraint 1 vibrates with the frequency θ and the unit amplitude, i.e., Z1(t) ¼ 1 sin θt. According to Table A.27, row 1, the bending moments at specified sections a, b, c in the primary system are Ma ¼ Mb ¼ Mc ¼
4EI 4EI 7 θ2 1 F1 ¼ 1 l l 4 ω2 1 θ2 =ω2
The dotted lines in Fig. 17.26c, d show location of extended fibers in the vicinity of joint 1. Loaded reaction R1P. Mass m is subjected to force P(t) ¼ P sin θt. According to Table A.27, row 8
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
M 0bP ¼ PF 15 ¼ P
659
l 1 8 1 θ2 =ω2
Negative sign means that in the vicinity of joint 1 the extended fibers are located above, as shown in Fig. 17.26d. Equilibrium condition of joint 1 in the unit state (Fig. 17.26c) and loaded state (Fig. 17.26d) lead to the following results: r 11
4EI 12EI 7 θ2 1 ¼3 , F ¼ 1 l 1 l 4 ω2 1 θ2 =ω2
R1P ¼ M 0bP ¼
Pl 1 8 1 θ2 =ω2
ðbÞ
It can be seen that the expression for the amplitude of a unit reaction r11 and a load reaction R1P includes a detuning θ/ω. So the primary unknown, which presents the required amplitude value of angular rotation of joint 1, becomes Z1 ¼
R1P Pl2 1 ¼ r 11 96EI 1 7θ2 =4ω2
ðcÞ
For further dynamic analysis of the structure, the method of initial parameters is the most effective. For this, each element of the frame should be considered separately. The origin of coordinates should be placed in joint 1 and the x-axes are directed to support points A, B, C. Elements A-1 and C-1 are subjected only to the kinematic excitation of Z1, and the element 1-B of the excitation of Z1 and the harmonic force P(t). The method allows determining not only the amplitude of the displacement of each lumped mass m but also the bending moments and shear forces in the characteristic sections of each separate element. A detailed analysis for elements 1-A and 1-C is presented in Examples 17.10 and 17.11. For element 1-B the additional force P(t) should be taken into account. In particular, according to the formula (g) of Example 17.10 we have M ð 0Þ ¼
4EI 1 7θ2 =4ω2 φ sin θt l 1 θ2 =ω2
So amplitude value of bending moments Ma and Mc at points а and c of the entire frame may be calculated, if expression (c) for Z1 is substituted into formula for M(0) instead of φ M a ¼ M c ¼ M 0aðcÞ Z 1 ¼
Pl 1 24 1 θ2 =ω2
For computation of amplitude value of Mb we need to take into account Table A.27, line 8 M b ¼ M b Z 1 þ M 0b ðPÞ ¼
Pl 1 Pl 1 Pl 1 ¼ 24 1 θ2 =ω2 8 1 θ2 =ω2 12 1 θ2 =ω2
Control: Ma + Mb + Mc ¼ 0, so the equilibrium condition for joint 1 is satisfied. Comments 1. For practical purpose the dynamic analysis should be supplemented by computation of the frequency of the free vibration ω. Only in this case we can specify the frequency of excitation under which will be performed required value of detuning θ/ω. 2. To determine the bending moments at characteristic points of the system, shear forces, inertial force, and reactions, the method of initial parameters should be applied to each element separately. The bending moment diagram within each element consists of two straight lines with kink under a lumped mass. This is explained by the presence of inertial force applied to the mass. 3. If the system is exposed to a variable arbitrary load P(t), then it should be represented in a form of a trigonometric series, the dynamic response on each harmonic excitation separately should be determined, and the superposition principle should be applied. A large number of dynamical problems of different deformable structures is presented in the book (Kiselev 1980).
660
17
Dynamics of Elastic Systems: Forced Vibration
Example 17.14 Design diagram of the uniform continuous massless two-span beam A-1-B with equal spans of length l and two equal lumped masses m1 ¼ m2 ¼ m is shown in Fig. 17.27a. Bending stiffness of each span is EI. This structure is subjected to disturbing harmonic force P(t) ¼ Psinθt. Provide total dynamic analysis of the steady-state vibration.
Fig. 17.27 (a, b) Design diagram of the beam and corresponding primary system; (c, d) Bending moment diagram caused by unit angular displacement Z1(t) of introduced constraint and harmonic load P(t) respectively; (e, f) Computation of r11 and R1P; (g, h) Dynamic bending moment diagram (factor Pl), reactions and inertial forces (factor P); (h) Dynamic shear force diagram (factor P)
Solution This structure has two degrees of freedom, while its kinematical indeterminacy is the first degree. The primary unknown is the angular displacement Z1 at support 1. The primary system are obtained from a given structure by introducing constraint 1 at middle support 1 (Fig. 17.27b); this constraint prevents angular displacement at support 1. Canonical equation of displacement method is r 11 Z 1 þ R1P ¼ 0
ðaÞ
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
661
The bending moment diagram in primary system due to angular harmonic rotation Z(t) ¼ 1sinθt of introduced constraint 1 is shown in Fig. 17.27c. The bending moment diagram in primary system due to given force P(t) ¼ Psinθt is shown in Fig. 17.27d. Notation M1A and Q1A means bending moment and shear infinitely close left to the point 1 (i.e., this point belongs to the element 1A), while M1B and Q1B infinitely close right to the point 1 (element 1B); (here superscript “0” for bending moments and shear forces in primary system is omitted). Table 17.4 contains expressions for bending moments, shear forces, and inertial forces at the characteristic points of the primary system caused by harmonic rotation of introduced constraint 1 with unit amplitude Z(t) ¼ 1sinθt. This table also contains the bending moments at points A, 1-A, and 1-B caused by the given load P(t) ¼ Psinθt. Table 17.4 Bending moments and shear at characteristic points of the beam; θ/ω ¼ 0.5
All expressions are composed on the basis of Table A.27, and in some cases, in combination with the reciprocity reaction theorem. Numerical values are calculated under the following assumption: the rods are uniform, detuning θ/ω ¼ 0.5 (ω is a frequency of free vibration of the entire structure), and each mass is located in the middle of the span. Unit reaction (Fig. 17.27e) and loaded term (Fig. 17.27f) are EI EI EI þ 1:7142 ¼ 4:7142 , l l l ¼ 0:25Pl
r 11 ¼ M 01A þ M 01B ¼ 3 R1P ¼ M 01BP
ðbÞ
where M 01BP means bending moment at section 1, which belongs to span 1B, caused by load P in the primary system. Finally, the primary unknown becomes Z1 ¼
R1P 0:25Pl Pl2 ¼ ðradÞ ¼ 0:05303 r 11 EI 4:7142EI=l
ðcÞ
662
17
Dynamics of Elastic Systems: Forced Vibration
Amplitude Values of the Inertia Forces For computation of the dynamic characteristics of the system, it is necessary to take into account the inertial forces that arise during vibration of the system. Their calculation for each span is presented in Table 17.5. General expression for inertia forces is J ¼ mθ2y. For each element of the primary system, it is easy to write the expression for the frequency of free vibrations. This expression allows representing the mass m through the square of the frequency of free vibration ω2. The substitution m ¼ m (ω2) in the expression for J leads to the expression which will contain the square of the detuning θ2/ω2. The frequency of free pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vibration of each span is determined by the formula ω ¼ 1=mδ11 , where δ11 is the unit displacement. The parameter y is a dynamic displacement of a mass. Table 17.5 Inertial forces due to unit Z1 ¼ 1 and actual primary unknown Z1 ¼ 0.05303Pl2/EI (θ/ω ¼ 0.5)
0.4242 0.3636(**)
Computation of final dynamic bending moments and shear forces at specified points are presented in Tables 17.6 and 17.7, respectively.
Table 17.6 Amplitudes of bending moments at the specified points due to harmonic disturbing force P(t)¼Psinθt (Z ¼ 0.05303Pl2/EI rad)
Span
Left A-1 Right 1-B
Point
MZ=1
MZ=1Z
MP0
M=MZ=1Z+ MP0
A m 1left 1right m
3.0 -2.0 -3.0 1.7142 0.8571 EI/l
0.1591 -0.1061 -0.1591 0.0909 0.04545 Pl
0 0 0 -0.25 0.2083 Pl
0.1591 -0.1061 -0.1591 -0.1591 0.2538 Pl
Factor
Table 17.7 Amplitudes of shear forces at the specified points due to harmonic disturbing force P(t) ¼ Psinθt (Z ¼ 0.05303Pl2/EI rad)
Span
Point
QZ=1
QZ=1Z
QP0
Q=QZ=1Z+ QP0
Left A-1
A 1left 1right B
-10.0 -2.0 -1.7142 -1.7142 EI/l2
-0.5303 -0.1061 -0.0909 -0.0909 P
0 0 0.9167 -0.4167 P
-0.5303 -0.1061 0.8258
Right 1-B Factor
-0.5076
P
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
663
Final amplitude bending moment and shear diagrams, reactions R and inertial forces J are shown in Fig. 17.27g, h. Control For total verification of obtained results we will use the different approaches, i.e., considering each span separately and the structure in a whole: Left span A-1. Shear forces within portion A-m and 1-m are QAm ¼
dM 0:106 0:159 ¼ Pl ¼ 0:530P, dx l=2
Q1m ¼
0:159 ð0:106Þ Pl ¼ 0:106P l=2
Inertial force of mass m1 is J 1 ¼ Qright Qleft ¼ Q1m QAm ¼ 0:106P ð0:530PÞ ¼ 0:424Pð"Þ Right span 1-B. Shear forces within portion 1-m and m-B are Q1m ¼
0:159 ð0:2538Þ Pl ¼ 0:8256P, l=2
QBm ¼ 0:5076P
Total force at mass m2 includes the given and inertial forces; they are P and J P þ J 2 ¼ QBm Q1m ¼ 0:5076P 0:8256P ¼ 1:3332P,
J 2 ¼ 0:3332Pð#Þ
Amplitude values of reactions for entire structure and the equilibrium condition for entire structure RA ¼ QA ¼ 0:5303 Pð#Þ ; R1 ¼ Qright Qleft 1 ¼ 0:8258 ð0:106Þ ¼ 0:9318Pð"Þ, 1 RB ¼ QB ¼ 0:5076Pð"Þ X Y ¼ 0:5303P þ 0:4242P þ 0:9318P P 0:3332 þ 0:5076P ¼ 1:8635P þ 1:8634P Bending moment at support 1, using the left and right forces, respectively l 1 M left ¼ 0:159 0:530 1 þ 0:424 Pl ¼ ð0:371 0:530ÞPl ¼ 0:159Pl ¼ M R l J A A 1 1 2 2 l l 1 M right ¼ RB l P J 2 ¼ 0:5076 1:3332 Pl ¼ 0:159Pl 1 2 2 2 Assume the beam is loaded by the static force P at the middle of the span 1-B. In this case, the bending moment at support 2 3 3 1 is M 1 ðPstat Þ ¼ Plυ 1 υ2 : If υ ¼ 0.5 , then M 1 ðPstat Þ ¼ Pl ¼ 0:107Pl , M A ðPstat Þ ¼ Pl (see problem 7 28 54 0:159 Pl ¼ 1:49. 10.4). Dynamic coefficient becomes μ ¼ 0:107 Pl Comments In this example, the relative frequency of excitation, i.e., a certain detuning θ/ω, had been accepted, while the frequency of free vibration ω and the frequency of excitation θ separately are not computed. Let us show procedure for computation of absolute value of frequency excitation θ. In order to form the frequency equation of free vibration we need to put R1P ¼ 0 in Eq. (a). Since Z1 6¼ 0, then the frequency equation is r11 ¼ 0. According to Fig. 17.27e, the frequency equation is r11 ¼ M1A + M1B ¼ 0. According to Table 17.4 (the first row for both spans), the frequency equation in extended form after simplification becomes " # " # 7 θ2 16 θ2 4 1 þ3 1 ¼0 ðcÞ 4 ω2f f 7 ω2f p |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} member A1
member 1B
664
17
Dynamics of Elastic Systems: Forced Vibration
The peculiarity of the equation is that it contains frequencies ωff, ωfp related to elements of various types, such as fixed192EI 768EI fixed (f-f) and fixed-pinned (f-p). For these elements ω2f f ¼ and ω2f p ¼ . As the base element we take ml3 7ml3 ω2f p 4 4 fixed-fixed beam; thus, 2 ¼ ! ω2f p ¼ ω2f f . 7 7 ω f f Substituting this ratio into (c) and replacing θ ¼ ω1, we get "
ω21 ω2f f
#
2
3 ω21
7 16 6 7 þ 341 5¼0 4 7 4 2 ω |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} 7 f f ffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl member A1 4 1
member 1B
This equation leads to the following equation for ω21 in terms of frequency ω2f f of fixed-fixed beam 7 19
ω21 ¼ 0, ω2f f
so the required square of frequency of free vibration of original structure becomes ω21 ¼
7 2 7 192EI EI ω ¼ ¼ 70:7368 3 19 f f 19 ml3 ml
7 Verification. As the base element we take fixed-pinned member. In this case ω2f f ¼ ω2f p. Similar procedure leads to the 4 following equation for ω21 in terms of frequency ω2f p of fixed-pinned beam, and required square of frequency 7
76 ω21 49 49 768EI EI ¼ 0 ! ω21 ¼ ω2f p ¼ ¼ 70:7368 3 7 ω2f p 76 76 7ml3 ml
Since θ/ω ¼ 0.5, then the absolute value of the frequency excitation becomes rffiffiffiffiffiffiffi rffiffiffiffiffiffiffi EI EI θ ¼ 0:5 8:4105 ¼ 4:205 3 ml ml3
17.4.2 Group Unknowns Method This method is used to analyze symmetric systems. The essence of the method is that a given load is presented by symmetric and antisymmetric groups of loads; the primary unknowns are represented as group of symmetric and antisymmetric unknowns, and the system is analyzed separately for each of the group loads. The advantage of this approach to the analysis of symmetric systems is that the original system of canonical equations splits into two subsystems (with symmetric and antisymmetric unknowns). This leads to a significant simplification of the analysis. Moreover, in some cases, it is required to consider only one of the deformation scheme—symmetric or antisymmetric one. Example 17.15 Symmetrical portal frame contains three massless rods of length l and flexural stiffness EI. At the middle of each element a lumped mass m is located. The frame is subjected to harmonic force P sinθ t which is applied as shown in Fig. 17.28a. Consider the symmetric steady-state vibration of the frame, and construct the bending moment and shear force diagrams. Solution This structure has four degrees of freedom, while kinematical indeterminacy of frame is three. The primary system of displacement method contains three introduces constraints 1, 2, 3 as shown in Fig. 17.28b. In case of symmetrical structure it is possible to apply a new type of the primary unknowns, mainly, the group unknowns.
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
665
Fig. 17.28 (a, b) Design diagram and primary system; (c) Bending moment diagram due to simultaneous rotation of two introduced constraints Z1 ¼ Z2 ¼ 1, i ¼ EI/l, EC is elastic curve; (d) Bending moment diagram in primary system due to given load P(t), factor Pl; (e, f) Dynamic bending moment diagram MP, factor Plsinθt, and dynamic shear diagram QP, inertial forces J and reactions H and R, factor Psinθt
In case of symmetrical vibration the group unknown presents the simultaneous angular harmonic displacements of introduced constraints 1 and 2 in opposite direction. In case of antisymmetrical vibration the group unknown presents the angular harmonic displacements of introduced constraints 1 and 2 in the same direction. Below we will consider only symmetrical vibration. Corresponding elastic curve and bending moment diagram are shown in Fig. 17.28c. Ordinates of bending moments and shear forces are determined according to Таble А.27, rows 1,2, special case a ¼ b ¼ l/2. Primary System, Unit State The unit state presents the simultaneous rotation of two introduced constraints Z1 ¼ Z2 ¼ Z ¼ 1. Below are presented the amplitude values of the bending moments at specified points of structure for dimensionless frequency parameter θ/ω ¼ 0.5, where ω is a frequency of free vibration.
666
17
Dynamics of Elastic Systems: Forced Vibration
Vertical member 1-A and 2-B (Fig. 17.28c) 4EI 7 θ2 1 EI 7 2 1 EI 1 1 0:5 ¼ 4 ¼3 2 2 l 4 ω 1 θ =ω2 θ=ω¼0:5 l 4 l 1 0:52 EI θ2 1 EI 2EI vert Q1 ¼ 6 2 1 3 2 ¼ 6 2 0:333 ¼ 2 ω 1 θ2 =ω2 l l l 2 EI θ 1 EI vert l vert vert ¼ 1þ2 2 M m ¼ M 1 þ Q1 ¼2 2 l l ω 1 θ2 =ω2 2 2EI 1θ 1 EI vert 1þ MA ¼ ¼ 3 l 2 ω2 1 θ2 =ω2 l vert
M1
¼
Horizontal member 1- 2, using superposition principle, 4EI 7 θ2 1 2EI 1 θ2 1 2EI θ2 1 1 1þ 14 2 ¼ ¼ ¼ 0:0 l 4 ω2 1 θ2 =ω2 l 2 ω2 1 θ2 =ω2 l ω 1 θ2 =ω2 EI θ2 1 EI θ2 1 EI θ2 1 8EI hor Q1 ¼ 6 2 1 3 2 þ 6 1 þ ¼ 24 ¼ 2 2 2 2 2 2 ω2 2 2 2 2 ω ω l 1 θ =ω l 1 θ =ω l 1 θ =ω l 2 EI θ 1 EI hor l hor hor ¼2 1þ2 2 M m ¼ M 1 þ Q1 ¼4 2 l l ω 1 θ2 =ω2 hor M1
hor
hor
M 2 ¼ M 1 ¼ 0:0 Based on the superposition principle, these expressions take into account the simultaneous unit rotation of the first and second introduced constraints (the first and second terms, respectively). Primary System: Loading State Only horizontal member 1-2 is subjected to bending deformation (Fig. 17.28d) M 01P ¼ M 02P ¼ M 0mP ¼
Pl 1 ¼ 0:1667Pl, 8 1 θ2 =ω2
Pl 1 ¼ 0:1667Pl, 8 1 θ2 =ω2
Unit reaction and free term of canonical equation (θ/ω ¼ 0.5) r 11 4EI 7 θ2 1 2EI θ2 1 vert hor 1 14 2 þ ¼ ¼ M1 þ M1 ¼ l 4 ω2 1 θ2 =ω2 l 2 ω 1 θ2 =ω2 EI θ2 1 EI 6 15 2 ¼3 l l ω 1 θ2 =ω2 R1P Pl 1 ¼ 0:1667Pl ¼ M 01P ¼ 8 1 θ2 =ω2 2 Primary unknown becomes Z¼
R1P Pl2 ¼ r 11 8EI
1 6 15
θ ω2 2
¼
0:1667Pl Pl2 ¼ 0:05556 ðradÞ 3i EI
Bending moment diagram may be constructed by formula M P ¼ MZ þ M 01P (Table 17.8) Shear can be calculated by formula Q ¼ dM=dx. For horizontal member shear forces within portion 1-m and m-2 are Q1m ¼
0:1667 ð0:3889Þ Pl ¼ 1:1112P, l=2
Qm2 ¼ 1:1112P
17.4
Structures with Finite Number of Degrees of Freedom: Displacement Method
667
Inertial force Jhor which acts at mass m of the crossbar may be calculated using the following equation: X Y ¼ 0 : Q1m P J hor Qm2 ¼ 0 ! J hor ¼ 2 1:1112P P ¼ 1:2224P Table 17.8 Computation of bending moment at specified points, Z ¼ 0:05556 Pl2 =EI
Member
Point
M
MZ
M P0
M P = M Z + M P0
Strut
1vert m A
3.0 2.0 -3.0
0.1667 0.1111 -0.1667
0.0 0.0 0.0
0.1667 0.1111 -0.1667
Crossbar
1hor m
0.0 -4.0
0.0 -0.2222
0.1667 -0.1667
0.1667 -0.3889
EI l
Pl
Pl
Pl
Factor
Same approaches should be applied for vertical element 0:1111 ð0:1667Þ 0:1667 0:1111 Q1m ¼ Pl ¼ 0:1112P, QmA ¼ Pl ¼ 0:5556P l=2 l=2 X X¼0: Q1m J ver QmA ¼ 0 ! J vert ¼ 0:5556P 0:1112P ¼ 0:4444P All forces which act on the frame are shown in Fig. 17.28f. Reaction of support are H A ¼ 0:5556P ð!Þ,
RA ¼ Qhor 1m ¼ 1:1112P
In fact, all reactions and internal forces contains factor sinθ t; therefore ordinates in Fig. 17.28e,f means the amplitude values. X Verification : Y ¼ RA þ RB P J ¼ ð1:1112 þ 1:1112 1:0 1:2224ÞP ¼ ð2:2224 2:2224ÞP ¼ 0 Table 17.9 contains the bending moments at characteristic sections caused by static action of the load P (Umansky 1973, Table 8.2.9] and corresponding dynamic coefficients. Table 17.9 Dynamic coefficients μ ¼ Mdyn/Mstat at different points of the frame
Point of the frame 1, 2 At force P A, B Factor
Mdyn
Mstat
m = M dyn M stat
0.1667 0.3889 0.1667
1/12 1/6 1/24
Pl
Pl
2.0 2.33 4.0 -
In this example, the most dangerous in terms of dynamic coefficients are clamped supports A and B.
668
17
Dynamics of Elastic Systems: Forced Vibration
Comments 1. In this example, the relative frequency of excitation, i.e., a certain detuning θ/ω, had been accepted, while separately the frequency of free vibration ω and the frequency of excitation θ are not computed. 2. The symmetry property and the concept of group unknowns allows using only one canonical equation of the displacement method when analyzing symmetric frame vibrations. 3. In the case of symmetric vibration, the horizontal displacement of the crossbar is absent, the diagram of bending moments and longitudinal forces (symmetric internal forces) turns out to be symmetrical, and the diagram of shear forces (antisymmetric internal force) is antisymmetric. 4. In the case of antisymmetric vibrations, the vertical displacement of the point of the horizontal member on the axis of symmetry is absent (frame with lateral shifts); the diagram of bending moments and longitudinal forces turn out to be antisymmetrical, and diagram of shear forces turns out to be symmetrical. Extensive reference data relating to various structures are presented in the fundamental Handbooks (Blevins 1979; Young et al. 2012; Umansky 1973), etc.
17.5
Structures with Distributed Parameters
In this paragraph we consider the forced harmonic steady-state bending vibrations of a structures taking into account uniformly distributed masses of the separate elements itself. As the result of this assumption, we obtain a system with an infinite number of degrees of freedom. As before, we assume that the system is ideal (in which there is no energy dissipation) and linear (for which the superposition principle is valid). The problem is to determine the dynamic response of a structure—the bending moments, reactions, and dynamic coefficients. The general approach of the solving such problems is based on the initial parameters method and the classical displacement method. Thus, we observe the evolution of the design diagram of a single element: in static problems (Chap. 10) there is an element with certain rigidity; in problems of stability—it is an element with known rigidity and longitudinal compressive force. In problems of dynamics we deal with an element of known rigidity and lumped mass. Finally, we will consider design diagram containing elements with distributed masses.
17.5.1 Initial Parameter Method This method allows analyzing vibration of the one-span beams with uniformly distributed mass, subjected to forced and kinematical harmonic excitation, and to prepare the necessary reference data for applying the classical displacement method for dynamic analysis of structures. The behavior of a rectilinear beam with a uniform mass m0 distribution along its length is described by a fourth-order differential equation 4
EI
2
∂ y ∂ y þ m0 2 ¼ Pðt Þ ∂x4 ∂t
ð17:43Þ
This equation involves partial derivatives because the transversal displacement y is a function of two variables, coordinate x and time t, i.e., y ¼ y(x, t). Here, P(t) ¼ P sin θt is harmonic perturbation force of frequency θ; it is obvious, vibration can be excited by harmonic couple, as well as by the kinematically. Steady-state vibration of a beam occurs with a frequency of excitation θ, i.e., the all parameters of dynamic state contains factor sinθ t. This factor will be omitted; this means that we are dealing with amplitude values of the corresponding function.
17.5
Structures with Distributed Parameters
669
A fragment of a rectilinear element with a distributed mass m0, the loadings, and symbols are shown in Fig. 17.29. Initial parameters at y ¼ 0 (left end of element) are y0, θ0, M0, Q0. Amplitude values of
P(t)=Psinθt
y
M(t)=Msinθt M0(t)
0 Q0(t)
m0
y x, φ x Mx, Qx x
rP = x-ap
aP aM x
rM = x-aM
Fig. 17.29 Notation for separate uniform bending member. Force initial parameters are M0(t) and Q0(t); kinematical initial parameters y0(t) and φ0(t) are not shown. Disturbance excitations are P(t) ¼ Psinθt and/or M(t) ¼ Msinθt; mass per unit length of beam is m0; flexural stiffness EI ¼ constant
Solution of equation (17.43) in terms of Krylov-Duncan functions S,T,U,V is (Bezykhov et al. 1987) h i 1 1 1 1 1 y ð x Þ ¼ y0 Sx þ φ 0 T x þ M 0 U x þ Q0 V x þ MU rM þ PV rP 2 3 2 k k EIk EIk EIkh i 1 1 1 1 þ Q0 U x MT rM þ PU rP þ φð x Þ ¼ y 0 V x k þ φ0 S x þ M 0 T x 2 EIk EIk k EIk h i 1 1 2 M ðxÞ ¼ y0 U x EIk þ φ0 V x EIk þ M 0 Sx þ Q0 T x þ MSrM þ PT rP k k QðxÞ ¼ y0 T x k þ φ0 U x EIk2 þ M 0 V x k þ Q0 Sx þ ½MV rM k þ PSrP ð17:44Þ rffiffiffiffiffiffiffiffiffiffi 2 4 m θ 0 [length1], so notation Sx means S(kx), VrP ¼ V[k(x aP)], UrM ¼ U[k(x aM)], etc. The general Parameter k ¼ EI expressions for functions S, T, U, V are presented in (16.24). In each equation, the first four terms take into account the initial parameters. The terms in brackets take into account the lumped harmonic couple M and force P (active force and the reactions of intermediate supports); their number may be arbitrary. At special case (harmonic excitations are absent), this system of equations allows exploring free vibration (Chap. 16, Sect. 16.4.4). Below is shown the application of the initial parameters method to the analysis of steady-state vibration of beams subjected to kinematic and force harmonic excitation. Example 17.16 The clamped-clamped beam of length l, bending stiffness EI, and mass per unit length m0 is subjected to harmonic kinematical angular excitation φ(t) ¼ 1 sin θ t of the right support (Fig. 17.30). Determine the response of the beam in case of steady-state vibration. y Initial parameters y0=0, M0 φ0=0, M0≠0, Q0≠0,
φ(t) x Q0
Fig. 17.30 Beam with fixed ends subjected to kinematical harmonic excitation φ(t) ¼ 1 sin θ t of the right support
Solution The origin is placed at the left end of the beam. The positive initial parameters M0 and Q0 are shown in Fig. 17.30. At the right support (x ¼ l) two parameters are known. They are the vertical y(l) and angular displacements φ(l). So from the set of Eq. (17.44) we will use the general expressions for functions y(x) and φ(x)
670
17
1 1 1 y ð x Þ ¼ y 0 Sx þ φ0 T x þ M 0 U x þ Q0 V x , k EIk 2 EIk 3 1 1 φðxÞ ¼ y0 V x k þ φ0 Sx þ M 0 T x , þ Q0 U x EIk EIk2
Dynamics of Elastic Systems: Forced Vibration
rffiffiffiffiffiffiffiffiffiffi 2 4 m θ 0 k¼ EI
ðaÞ
Initial parameters are y0 ¼ 0 and φ0 ¼ 0. At the right support y(l) ¼ yl ¼ 0 and φ(l) ¼ φl ¼ 1; the negative sign means that support is rotated clockwise direction. Therefore from (a) we get a linear algebraic equations with respect to unknown initial parameters M0 and Q0 1 1 þ Q0 V ¼ 0, EIk 2 EIk 3 1 1 þ Q0 U ¼ 1 φð l Þ ¼ M 0 T EIk EIk 2
yð l Þ ¼ M 0 U
ðbÞ
Here index l at the Krylov functions is omitted, i.e., Ux¼l ¼ U ¼ U(k l), Tx¼l ¼ T ¼ T(k l), . . . Solution of this set of equations is U U 2 VT V M ð0Þ ¼ EIk 2 U VT Qð0Þ ¼ EIk 2
ðcÞ
The negative sign for Q(0) means that this shear trends to rotate beam around the opposite end of the beam counterclockwise. Bending moment and shear force at the right support are 1 SV TU ¼ Ek 2 k U VT US V 2 QðlÞ ¼ M 0 Vk Q0 S ¼ Ek 2 2 U VT M ðlÞ ¼ M 0 S Q0 T
ðdÞ
Now all reactions may be presented in terms of trigonometric and hyperbolic functions of dimensionless parameter u ¼ kl. For example, M ð0Þ ¼ EIk
V ð uÞ V u2 2EI u sinh u sin u 2EI ¼ EI ¼ ¼ f ð uÞ 2 l 2 l 2 1 cosh u cos u l U ðuÞ V ðuÞT ðuÞ U VT 2
ðeÞ
Formulas (c–e) present the amplitude value of corresponding response functions. Dynamic response involve function sinθ t. Coefficient 2EI/l presents response in case of static rotation of right clamped support through the angle φ ¼ 1. The complex f(u) presents the correcting function which takes into account nonuniformly distributed inertial forces along entire span. It is easy to show that, if u ¼ 0 (static loading), then function f(0) ¼ 1.0. To determine the free vibration frequency of a clamped-clamped beam with a uniformly distributed mass, we need to consider a homogeneous system of equations (b). Nontrivial solution M0 6¼ 0 and Q0 6¼ 0 takes place if U V=k ¼ 0 ! U 2 TV ¼ 0 D¼ T U=k According to (16.26) this equation may be presented in trigonometric and hyperbolic functions as follows: 1 U 2 ðklÞ T ðklÞV ðklÞ ¼ ð1 cosh kl cos klÞ ¼ 0 2 Frequency equation for fixed-fixed uniform beam becomes coshkl cos kl ¼ 1. Table A.24 contains the frequency equations, eigenvalues, and nodal points of mode shape for uniform beams with different boundary conditions. Tables A.25 and A.28 contain the analytical expressions for dynamic reactions of uniform one-span beams with distributed masses. These beams are subjected to harmonic kinematic and forced excitation. Reactions in Table A.25 are presented in terms of Krylov-Duncan function, while reactions in Table A.28 are presented in terms of hyperbolic-trigonometric functions.
17.5
Structures with Distributed Parameters
671
Example 17.17 The simply supported beam of length l, bending stiffness EI, and mass per unit length m0 is subjected to harmonic disturbing force P(t) ¼ P sin θ t located at the middle point of the span (Fig. 17.31a). Provide dynamic analysis of steady-state vibration. P(t)
a
m0, EI
A
B
x
φ0 EC
a=l/2
Q0 b
P/2
y
Initial parameters A y0=0, M0=0, φ0≠0, Q0 Q0≠0,
m0
x
a=l/2
Fig. 17.31 Design diagram of a beam and corresponding equivalent left part
Solution For dynamic analysis of this beam we will apply the initial parameter method. The origin is placed at point A of the beam. Due to symmetry, consider the equivalent left part of the beam (Fig. 17.31b). At the point on the axis of symmetry the sliding support is located. This boundary condition provides same elastic curve as the original design diagram. In the section on the axis of symmetry the angle of rotation φ(a) ¼ 0 and shear force Q(a) ¼ P/2. Therefore, from the system of equations of the initial parameters method we use two equations, namely for φ(x) and Q(x). The general expressions for these functions are φð x Þ ¼ y 0 V x k þ φ0 S x þ M 0 T x
1 1 þ Q0 U x , EIk EIk 2
QðxÞ ¼ y0 T x EIk þ φ0 U x EIk þ M 0 V x k þ Q0 Sx , 3
2
rffiffiffiffiffiffiffiffiffiffi 2 4 m θ 0 k¼ EI
ðaÞ
Taking into account initial parameters y0 ¼ 0, and M0 ¼ 0, at section x ¼ a we have 1 ¼0 EIk 2 P QðaÞ ¼ φ0 U a EIk 2 þ Q0 Sa ¼ 2 φðaÞ ¼ φ0 Sa þ Q0 U a
ðbÞ
Here and further notation Sa, Ta, etc., means S(ka), T(ka), etc. Homogeneous set of Eq. (b) allows compiling the equation of the frequencies of free vibration 1 Sa Ua EIk 2 ¼ 0 D ¼ U EIk 2 Sa a In expanded form, the frequency equation and its solution l l π π S2a U 2a ¼ cosh ka cos ka ¼ 0 ! cos ka ¼ cos k ¼ 0 ! k ¼ ! k ¼ 2 2 2 l rffiffiffiffiffiffi rffiffiffiffiffiffi EI π 2 EI ¼ 2 Thus, the frequency of free vibration becomes ω ¼ k 2 m0 m0 l Solution of nonhomogeneous set of Eq. (b) is P Ua 2EIk2 S2a U 2a P Sa Q ð 0Þ ¼ 2 S2a U 2a
φð 0 Þ ¼
ðcÞ
672
17
Dynamics of Elastic Systems: Forced Vibration
The bending moment and displacement at section under harmonic force are 1 P Sa T a U a V a M ðaÞ ¼ φ0 V a kEI þ Q0 T a ¼ , k 2k S2a U 2a
1 1 P Sa V a T a U a y ð a Þ ¼ φ0 T a þ Q 0 V a ¼ 3 3 k EIk 2k EI S2a U 2a
ðdÞ
Formulas (c, d) present the amplitude values of corresponding response functions. Dynamic response involves function sinθ t. According to (16.26), formulas (c, d) can be easily presented in terms of hyperbolic-trigonometric functions. Dynamic coefficient for any parameter is determined as ratio of its amplitude dynamic value and static one; for example, dynamic coefficient for displacement at y(a ¼ l/2) becomes yðaÞ
μdyn ¼
y ð aÞ y ð aÞ 24 Sa V a T a U a ¼ : ¼ yðaÞstat Pl3 =48EI k3 l3 S2a U 2a
ðeÞ
Since the amplitude dynamic expression and static one for any function (displacement, reaction, etc.) linearly depend on P, the dynamic coefficient does not depend on P. Obviously, expression (e) is approximate, since the static displacement is defined taking into account only the force P. Static displacement taking into account P and uniformly distributed load q0 ¼ m0g (g ¼ 9.81 m/s2) should be taken in form y(a)stat ¼ Pl3/48EI + 5q0l4/384EI. Let us find out the effect of excitation frequency θ on dynamic coefficients. Two versions of computation may be recommended. Version 1. Calculate the parameter k in terms of the absolute frequency of the excitation θ sffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi 2 4 q θ q pffiffiffi 0 k¼ ¼ 4 0 θ gEI gEI E ¼ 21.5 1010 N/m2, I ¼ 2.14 105 m4, l ¼ 4.75 m. In this case parameter k ¼ k(θ) is rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 263 4 θ ¼ 0:04913 θ m1 : k¼ 10 5 9:81 21:5 10 2:14 10 pffiffiffiffiffiffiffiffi If θ ¼ 144s1, then parameter k becomes k ¼ 0:04913 144 ¼ 0:5896 ðm1 Þ. Argument of Krylov functions ka ¼ kl/2 ¼ 0.5896 4.75/2 ¼ 1.4. Krylov-Duncan functions of this argument are
Let q0 ¼ 263 N/m,
Sa ¼ 1:16043,
T a ¼ 1:44487,
U a ¼ 0:99046,
V a ¼ 0:45942
For dynamic coefficient we get the following value: 24 Sa V a T a U a ¼ k 3 l3 S2a U 2a 24 1:16043 0:45942 1:44487 0:99046 ¼ 2:685 0:36558 0:58963 4:753 yðaÞ
μdyn ¼
Dynamic coefficients for Q(0) and M(0) are 3.174 and 2.386, respectively. Note that the dynamic coefficients related to the various characteristics of the system state (reaction, bending moments, etc.) are different. Version 2. Calculate the parameter k in terms of the relative frequency of the excitation i.e., with respect to ratio θ/ω. rffiffiffiffiffiffi π 2 EI π 4 EI The frequency of free vibration of entire beam is ω ¼ 2 ! m0 ¼ 4 2, so parameters k and ka may be presented m0 l l ω as follows: rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffi rffiffiffiffi 2 4 m θ π 4 θ2 π θ l π θ 0 ¼ k¼ ¼ ! ka ¼ k ¼ l ω2 l ω 2 2 ω EI Finally the relative excitation frequency (detuning θ/ω) and the factor in the expression (e) may be presented in the form
17.5
Structures with Distributed Parameters
673
θ ¼ ω
2ka π
2 ,
24 24 3 ¼ ¼ k3 l3 k3 ð2aÞ3 ðkaÞ3 yðaÞ
The dynamic coefficient computation for displacement in the middle point of the beam μdyn as a function of the ka parameter are presented in Table 17.10. In this table, the second and third column contains the relative excitation frequency θ/ω and coefficient 3/(ka)3, respectively. Columns with values of Krylov-Duncan function omitted. Detailed tables of Krylov-Duncan functions may be found in (Karnovsky and Lebed 2001, 2004a, b). yðaÞ
Table 17.10 Computation of dynamic coefficient μdyn as a function of ka or θ/ω ¼ (2ka/π)2
Dynamic coefficients for all types of response, in particular, formula (e) for displacement in the middle of the beam, contains expression S2a U 2a in the denominator. The equation S2a U 2a ¼ S2 ðkaÞ U 2 ðkaÞ ¼ 0 , according to tables of Krylov functions, is satisfied for the following parameters ka : π/2, 3π/2, . . . Graph of dynamic coefficient is shown in Fig. 17.32.
P(t) m0 2.684 1.19
1.22
1.0 0.0
2.0 1.0 1.4 π/2
1
0.59
3.0 π
2
4.0
4.6 3π/2
5.6
ka
3
Fig. 17.32 Simply supported beam loaded in the middle of the span by force P(t) = P, sin θt . Dynamic coefficient μdyn(0.5l) = ydyn/ystat and forms of vibration (curves 1, 2, 3) on the intervals ka = [0 π/2], [π/2 3π/2], etc. Corresponding dimensionless frequency intervals are θ/ω1 = [0 1]; [1 9], etc. Parameter ka = π corresponds to θ/ω1 = 4.0.
674
17
Dynamics of Elastic Systems: Forced Vibration
The dynamic coefficient is plotted as a function of dimensionless parameter ka. To each value of ka there is corresponding dimensionless detuning θ/ω. The axis of parameters ka is divided into intervals 0 π/2 3π/2 . The boundaries of these intervals is the detuning θ/ω at which resonance occurs; they are ka ¼ π/2, 3π/2. The dimensionless interval ka ¼ [0 π/2] corresponds to the first frequency interval θ/ω1 ¼ [0.0 1.0]; parameter ka ¼ 1.4 in the example above corresponds to the first frequency interval (row 2 of Table 17.10). When the detuning is equal to zero, the dynamic coefficient is equal to one (static case). With increasing detuning θ/ω, the dynamic coefficient increases, and if the frequency excitation θ and free vibration ω coincide, the dynamic coefficient becomes infinity. Parameter ka ¼ π/2, in which θ ¼ ω correspond to resonance. On the frequency interval θ/ω1 ¼ [0 1.0], the symmetric vibration with one half wave is observed (curve 1 is shown by the dotted line). Passing through point π/2 a change in the nature of the vibration occurs, which in the interval [π/2 3π/2] becomes antisymmetric with two half waves with one nodal point (curve 2). When the detuning is equal to π, an antiresonance phenomenon is observed, which consists in a noticeable weakening of the dynamic coefficient. If detuning θ/ω passing through parameter ka ¼ 3π/2, a resonance phenomenon and a transition to a symmetrical vibration with two nodal points (curve 3) are observed. Comments 1. Original equation (17.43) relates to the Bernoulli-Euler’s classical theory of transversal vibration of thin rods. It is assumed that the height of such rods is small compared to its length, h lc
General solution of these equations is Fc x, Pl
ð19:2Þ
F ðl cÞ ðl xÞ, Pl
ð19:3Þ
y1 ¼ C 1 cos nx þ D1 sin nx y2 ¼ C2 cos nx þ D2 sin nx
where n2 ¼ P/EI is parameter of compressed load P. It can be seen that the desired functions y1 and y2 linearly depend on the transverse force F and nonlinearly depend on the axial force P, since the parameter n is an argument of trigonometric functions. At the points A (x ¼ 0) and B (x ¼ l) the displacement y is zero. Equations (19.2) and (19.3) lead to C1 ¼ 0 and C2 ¼ D2 tan nl. Therefore, expressions for displacements within the left and right portions are
718
19 Nonlinear Structural Analysis
y1 ¼ D1 sin nx
Fc x Pl
y2 ¼ D2 tan nl cos nx þ D2 sin nx
ð19:4Þ
F ð l cÞ ð l xÞ Pl
For calculation of unknown coefficients D1 and D2 we can use the following conditions at the point C: y1 ¼ y2
and
dy1 dy2 ¼ dx dx
In the expanded form these conditions are D1 sin nðl cÞ ¼ D2 ½ sin nðl cÞ tan nl cos nðl cÞ D1 n cos nðl cÞ ¼ D2 n½ cos nðl cÞ þ tan nl sin nðl cÞ þ
F P
ð19:5Þ
Solution of these equations is D1 ¼
F sin nc , Pn sin nl
D2 ¼
F sin nðl cÞ Pn tan nl
Displacements at any point of the left part of the beam y1 ¼
F sin nc Fc sin nx x Pn sin nl Pl
ð19:6Þ
Knowing (19.6) we may construct the equations for slope and internal forces. In particular, the bending moment for left part of the beam M ðxÞ ¼ EI
d 2 y Fn sin nc sin nx ¼ P sin nl dx2
ð19:7Þ
Limiting Case If a force F is placed at the middle of the span (c ¼ 0.5l), then for section x ¼ 0.5l from (19.6) and (19.7) we get the maximum deflection and bending moment rffiffiffiffiffi l Fl3 3ð tan u uÞ l P y , u ¼ ¼ 2 2 EI 48EI u3 l d 2 y Fl tan u ¼ EI 2 ¼ ð19:7aÞ M 2 4 u dx Thus, in the presence of transverse load F and compressive force P, the displacement and bending moment are linearly dependent on the transverse load F and nonlinearly dependent on the compressive force P. If a compressive force P ! 0 (u ! 0), then 3ð tan u uÞ lim ¼ 1, u3 l Fl3 so y ¼ and 2 48EI u!0
tan u ¼ 1, lim u l Fl ¼ M 2 4 u!0
Now we can evaluate numerically effect of the compressive force P. Since a critical force rfor ffiffiffiffiffiffi simply supported column is π P Pcr ¼ π 2 EI=l2 , then the dimensionless stability parameter u may be rewritten as u ¼ . For P ¼ 0.36Pcrit we get 2 Pcr u ¼ 0.94248. Corresponding correction factors are 3ð tan u uÞ ¼ 1:555, u3
tan u ¼ 1:4604 u
i.e., a small compressive load increases the deflection at force F by 55%, and maximum bending moment increases by 46%. Thus, compressive force has unfavorable effect on the state of the beam-column, and therefore P-Δ analysis should not be ignored.
19.2
Compressed Rods with Lateral Loading
719
Notes: 1. The displacement and lateral force F according to (19.6) are related by linear law. Since the axial force P appears in parameter n2 ¼ P/EI, then the displacement and axial compressive force P in the same equation are related by nonlinear law. It means that superposition principle is applicable partially, i.e., only for lateral loads. 2. Equation (19.7) may be treated as the expression for influence line for bending moment of a simple-supported compressed beam. For this purpose we need to consider the section x as a fixed one, while a location of the unit force F is defined by a variable parameter c. It is obvious that this influence line becomes curvilinear. 3. Note, in Table A.21 case 2 for simply supported beam-column the following expressions are presented rffiffiffiffiffi l Fl υ P M ¼ tan , υ ¼ l 2 2υ 2 EI l Fl3 υ υ y tan ¼ 2 2 2 2EIυ3 It is easy to verify that formulas in Table A.21 coincide; for this it should be taken into account rffiffiffiffiffi (19.7a) given in the text and rffiffiffiffiffi P l P that the table parameter υ ¼ l and the parameter u ¼ adopted in the text are related by formula υ ¼ 2u. EI 2 EI
19.2.2 Initial Parameters Method This method is effective for analysis of structures which contains elements subjected to axial compression and lateral loads. Figure 19.3 presents a straight uniform element loaded by the compressive force P, the lateral forces Fi, and uniformly distributed load q; dotted line shows the initial nondeformable position (INDP) of the element. Fi q x
M0
INDP
y0
P
q0 Q0 Q0
y
ai x
y Fig. 19.3 Loading of the beam-column and positive initial parameters
Initial parameters are y0, θ0, M0, Q0, (or Q0 ). The shear force Q0 is directed perpendicular to deformed axis of the beam, while Q0 is directed perpendicular to nondeformed axis of the beam. From equilibrium condition for initial point of elastic curve we have Q0 ¼ Q0 cos θ0 þ P sin θ0 Q0 þ Pθ0 ! Q0 ¼ Q0 Pθ0 Differential equation of elastic curve is EI
d2 y ¼ M ðxÞ, dx2
where the bending moment at any section x is M ðxÞ ¼ Pðy y0 Þ þ M 0 þ Q0 x Differential equation becomes
X
F i ðx ai Þ
qx2 2
720
19 Nonlinear Structural Analysis
X d2 y 1 qx2 2 M þ n y ¼ þ Q x Py F ð x a Þ , 0 i i 0 0 EI 2 dx2
n¼
rffiffiffiffiffi P , EI
ð19:8Þ
Solution of this equation is y ¼ C 1 cos nx þ C 2 sin nxþ
1 1 X q n2 x 2 þy0 2 cos nx M 0 þ Q0 x þ 3 F i ½nðx ai Þ sin nðx ai Þ 4 1 2 n EI n EI n EI
ð19:9Þ
The first and second terms of this expression are solution of homogeneous Eq. (19.8), while other terms are partial solution of nonhomogeneous equation. For calculation of unknowns C1 and C2 we can use the following boundary conditions: at x ¼ 0 the initial displacement and slope are y ¼ y0 and y0 ¼ θ0. These conditions lead to Q0 M0 1 θ þ C1 ¼ 2 and C 2 ¼ n 0 n2 EI n EI Substitution of these constants into expression for y and differentiation with respect to x leads to the following formulas for displacement, angle of rotation, bending moment, and shear (Umansky, 1973, v2): sin nx M 0 1 cos nx Q0 nx sin nx þ y n EI EI n2 n3 dy M sin nx Q0 1 cos nx ¼ θ0 cos nx 0 θðxÞ ¼ þ θ dx n EI EI n2 d2 y sin nx þ M M ðxÞ ¼ EI 2 ¼ θ0 EI n sin nx þ M 0 cos nx þ Q0 n dx dM ¼ θ0 EI n2 cos nx M 0 n sin nx þ Q0 cos nx þ Q QðxÞ ¼ dx QðxÞ ¼ Q0 Q y ð x Þ ¼ y0 þ θ 0
ð19:10Þ
Each formula of system (19.10) contains two parts. The first part takes into account the initial parameters of the structure. Of the four initial conditions, two initial conditions are known from boundary condition at the left end of the beam. The second part depends on the lateral load; these terms are denoted by symbol (); they are presented in Table 19.1. The lateral load is taken into account by terms y, θ,. . ., while the preceding terms do not change. Additional terms; they should be implemented into (19.10) only at x a > 0. It means, for given section x, only loads located to the left of section x should be taken into account. Table 19.1 Additional terms of (19.10); n ¼
pffiffiffiffiffiffiffiffiffiffiffi P=EI (Umansky, 1973, v.2)
F
q a
a x
y
θ M Q Q
( x − a )2
q 2
2
n EI q 2
n EI −
+
cos n(x − a ) − 1 n
2
(x − a ) − sin n(x − a ) n
q n
2
−
F n EI 2
M a x
x
(x − a ) − sin n(x − a ) n
F [1 − cos n(x − a )] n 2 EI
−
M n 2 EI
−
[1 − cos n(x − a )]
M sin n(x − a ) nEI
[1 − cosn(x − a )]
−
F sin n(x − a ) n
M cos n(x − a)
q sin n(x − a ) n -q(x-a)
− F cos n(x − a ) -F
Mn sin n(x − a ) 0
19.2
Compressed Rods with Lateral Loading
721
The reader is asked to compare the equations of the initial parameters method for three different states of homogeneous rod: bended (Eqs. 8.8, 8.9, and 8.12), compressed (Eq. 15.9), and compressed-bended (Eq. 19.10). Special attention should be paid to the original design diagram and the set of assumptions, corresponding differential equations, and the structure of the final initial parameters equations.
19.2.2.1
Free-Clamped Beam: Precise Solution
Let us illustrate the initial parameters method for free-clamped beam subjected to axial compressed force P and lateral force F (Fig. 19.4). It is necessary to provide analysis of this structure on the basis of its deformable design diagram (compute the vertical and angular displacements at the free end and bending moment at the fixed support). F B
0 y0
P
P
x
x l
RB
y Fig. 19.4 Free-clamped compressed-bended beam
According to Table 19.1 the terms, which take into account a lateral load, are sin nðx aÞ F F sin nx F y ¼ 2 ¼ 3 ðnx sin nxÞ, ¼ 2 ð x aÞ x n n n EI n n EI EI a¼0 F F θ ¼ 2 ½1 cos nðx aÞa¼0 ¼ 2 ð1 cos nxÞ, n EI n EI F F M ¼ sin nðx aÞ ¼ sin nx n n Since Q0 ¼ 0,
ð19:11Þ
M0 ¼ 0, then (19.10) becomes sin nx F þ 3 ðnx sin nxÞ; n n EI F θðxÞ ¼ θ0 cos nx þ 2 ð1 cos nxÞ n EI
yð xÞ ¼ y0 þ θ 0
ð19:12Þ
These equations contain two unknown parameters. They are θ0 and y0. Boundary conditions are: 1. At x ¼ l (support B) the slope of elastic curve θ ¼ 0, so θðlÞ ¼ θ0 cos nl þ
F ð1 cos nlÞ ¼ 0, n2 EI
which leads immediately to the slope at the free end F 1 cos nl Fl2 2ð1 cos υÞ θ0 ¼ 2 ¼ , 2EI υ2 cos υ n EI cos nl
rffiffiffiffiffi P , υ ¼ nl ¼ l EI
2. At x ¼ l the vertical displacement y ¼ 0, so y ð l Þ ¼ y0 þ θ 0
sin nl F þ 3 ðnl sin nlÞ ¼ 0 n n EI
Taking into account (19.13), the vertical displacement at the free end becomes
ð19:13Þ
722
19 Nonlinear Structural Analysis
Fl2 1 cos υ sin υ F 3 ðυ sin υÞ ¼ n n EI υ2 EI cos υ h i Fl3 1 cos υ Fl3 3ð tan υ υÞ Fl3 sin υ ðυ sin υÞ ¼ φ ¼ 3 cos υ 3EI 3EI y υ EI υ3 y0 ¼
ð19:14Þ
If a beam is subjected to lateral force F only, then a transversal displacement at the free end equals Fl3/3EI. However, if additional axial force P acts, then the factor φy ¼
3ð tan υ υÞ υ3
must be included. 3. The moment at clamped support equals M ðlÞ ¼ θ0 EI n sin nl þ M ¼ ¼ Fl
tan υ ¼ FlφM υ
Fl2 1 cos υ F EI n sin υ sin υ ¼ 2 cos υ n υ EI
ð19:15Þ
Let the compressive force P and critical force Pcr be related as follows: P ¼ kPcr ¼ k
π 2 EI , 4l2
where k is any positive number, k 1. In this case the dimensionless parameter is rffiffiffiffiffi P π pffiffiffi υ¼l ¼ k EI 2 Coefficients φy and φM for different parameter k are presented in Table 19.2. Table 19.2 Coefficients φy and φM for different parameters k ¼ P/Pcr
Parameter k 0.0
0.2
0.4
0.6
0.8
1.0
jy
1.0
1.2467
1.6576
2.4792
4.9434
∞
jM
1.0
1.2051
1.5453
2.2234
4.2562
∞
Bolded data corresponds to case P ¼ 0; it means the beam is subjected to the lateral force only. Even if k ¼ P/Pcr ¼ 0.2, the transversal displacement at free end and bending moment at the fixed support are 24.6% and 20.5% higher than for same beam without compressive load. Thus refuse of the Navier assumptions about small deformation leads to the significant increasing of the displacement and bending moment. This fact can not be ignored choosing the design diagram and method for its analysis. Some important cases pffiffiffi in the expanded form are presented in Table A.21; note that this table contains formulas in terms of parameter υ ¼ nl ¼ l P=EI. Fundamental formulas presented in Table A.21 have the following features: 1. This table contains information for uniform one-span beams subjected to the axial compressive and lateral load, and allows performing the analysis for limiting case if compressive load is absent. For example, for pinned-pinned beam subjected to lateral concentrated load F only at middle span we get (case 2) lim M
υ!0
tan υ=2 Fl tan υ=2 Fl 1 l Fl Fl ¼ lim ¼ lim ¼ 1¼ 2 2 υ!0 2 υ!0 2 υ=2 2 2 4 υ
19.2
Compressed Rods with Lateral Loading
723
2. Superposition principle for different lateral loads is applicable for the same compressive load. For example, in case of simply supported beam subjected as shown in cases 1 and 2 in Table A.21, the bending moment at the middle span becomes M
l ql2 υ Fl υ ¼ 2 sec 1 þ tan 2 2 2υ 2 υ
3. If in two states of any elastic system the axial compressive forces are equal, then Maxwell theorem about reciprocity of the unit displacements is hold. For example, case 2 of Table A.21 (loading by force F) and case 3 (loading by couple M0) we will consider as the first and second conditions. Unit displacement caused by the force F ¼ 1 of the first state in the direction of the moment M0 of the second state is the angle of rotation on the left support θ ð 0Þ ¼ F
l2 υ sec 1 ¼ Fδ21 2 2 2EIυ
Unit displacement caused by the moment М0 ¼ 1 of the second state in the direction of the force F of the first state is the vertical displacement at the middle of the beam l l2 υ ¼ M0 1 ¼ M 0 δ12 sec y 2 2 2EIυ2 It can be seen that δ12 ¼ δ21, so the Maxwell relationship δik ¼ δki is hold. Summary Beam-column analysis allows determining the real deflections and internal forces, which are much higher than those which would be determined without taking into account compressive load. This is an essence of the P-Δ analysis. Neglecting of the compressive forces may lead to the failure of a structure; therefore, for high and multileveled structures, analysis on the basis of deformable design diagram is necessary. For P-Δ analysis we can use two different approaches. In the first approach, we need to complete and solve the secondorder differential equation of the beam for each specified loading. In the second approach (initial parameters method) we use the once derived formulas for displacements, slope, bending moments, and shear. These formulas can be used for any loads and integration procedure is not required.
19.2.3 P-Δ Analysis Analysis of a structure on the basis of deformable design diagram (P-Δ analysis) contains two steps. The first step presents the classical analysis of structure on the basis of nondeformable scheme. If a frame is subjected to arbitrary set of external forces, including compressive forces at the joints, then on the first stage of analysis these compressive forces should be omitted. This classical analysis is performed by any appropriate methods, which were discussed previously. For each member of the frame, the axial forces must be calculated. The second step begins from calculation of axial forces withptaking ffiffiffiffiffiffiffiffiffiffiffi into account external axial compressive forces. For each member, dimensionless parameter of compressive load υ ¼ l P=EI should be calculated. After that, the internal force diagrams of the first step should be reconstructed taking into account compressive load. The displacement method on this stage is more preferable. Unit reactions depend on parameter υ; the unit reactions are presented in Table A.20. The free terms of canonical equations depend on lateral loads. The following example demonstrates application of a beam-column theory for detailed analysis of two-span sidesway frame. Example 19.1 The frame in Fig. 19.5a is subjected to horizontal force F at the level of the crossbar and compressed forces P. The crossbar of the frame is connected with vertical members by means of the hinges. The stiffnesses of the columns and crossbar are EI and 2EI, respectively. Construct the internal force diagrams. Cross section for vertical members is W150 37 (I ¼ 22.2 106 mm4), the modulus of elasticity E ¼ 2 105(N/mm2).
724
19 Nonlinear Structural Analysis
P=180kN
a
b
P=180kN 1
F=30kN
i=0.333EI h=5m
EI B
2
i=0.2EI
C
l=6m
c
Z2
P
1
F 2EI
A
Z1
P
6m Z1=1
3i=1EI
r11 2
1
1
r21
3i=1EI
3i=1EI
r11 3i=1EI 1
M1
r21
Z2=1
r12
d
2
1
r22
2
M2
MB
MA r12
e
0.024EI
0.01435EI
MC Z2=1 1
P=180kN
f
2
0.01435EI
r22
P=180kN
F MP
63.64(50.0)
68.31(50.0)
63.64(50.0)
Fig. 19.5 (a, b) Design diagram of the frame and primary system; (c) Unit bending moment diagram due to Z1 ¼ 1 and calculation of unit reactions; (d, e) Unit bending moment diagram due to Z2 ¼ 1 and free body diagram for computation of the unit reactions; (f) Final bending moment diagram
Solution The primary system of the displacement method is shown in Fig. 19.5b. The introduced constraints are 1 and 2. The primary unknowns are angular displacement Z1 of constraint 1 and linear displacement Z2 of constraint 2. The bending stiffness per unit length for members are ivert ¼ EI/5 ¼ 0.2EI and ihor ¼ 2EI/6 ¼ 0.333EI. Stage 1. Analysis of the frame on the basis of nondeformable design diagram. In this step, we need to provide analysis without the axial compressive forces N. This analysis is presented in Chap. 10, Example 10.2. The internal forces are the following: the bending moments at the fixed supports are M A ¼ M B ¼ M C ¼ 1:667 F ¼ 50 kNm, where F is a horizontal force, which acts along the crossbar.
19.2
Compressed Rods with Lateral Loading
725
The axial forces at the columns are NA ¼ NB ¼ NC ¼ 0 and axial forces at the left and right crossbar are F ¼ 20 kN, N right ¼ 10 kN N left ¼ ðF QÞ ¼ F 3 Stage 2. Analysis of the frame on the basis of deformable design diagram. This step should be performed taking into account axial forces N. Axial forces which arise in each member are P1 ¼ N0 + P, where N0 is axial force in specified member as result of the stage 1, while P is external compressive load. Since for allp columns ffiffiffiffiffiffiffiffiffiffiffi N0¼0, then for left and right columns the axial force equals to given load P. Parameters of compressive load υ ¼ l P=EI for columns are sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 18 104 ðNÞ
υA ¼ υC ¼ 5000 ðmmÞ ¼ 1:007; 6 N 4 2 105 mm 2 22:2 10 ðmm Þ
υB ¼ 0:
Parameters of compressive load for crossbar are υleft
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 103 ðNÞ
¼ 6000ðmmÞ ¼ 0:0028; 5 N 2 10 mm2 2 22:2 106 ðmm4 Þ
υright ¼ 0:0020
For columns and for left and right parts of crossbar we will adopt the following parameters υ υA ¼ υC ¼ 1:0,
υB ¼ 0,
υleft ¼ υright ¼ 0:0:
Canonical equations of the displacement method are: r 11 Z 1 þ r 12 Z 2 þ R1P ¼ 0 r 21 Z 1 þ r 22 Z 2 þ R2P ¼ 0
ðaÞ
The bending moment diagram M 1 in the primary system due to the rotation of introduced constraint 1 is shown in Fig. 19.5c. Since parameters υ for crossbars are zero, then this member should be considered without effect of compressive load, so the bending moment diagram is bounded by straight lines. It is obvious that r 11 ¼ 2EI
kNm rad
and r 21 ¼ 0:
Figure 19.5d presents the bending moment diagram M 2 in the primary system due to the linear displacement of introduced constraint 2. For columns A and C we need to take into account parameter υ because these members are subjected to axial forces. Therefore bending moment diagrams along these members are curvilinear. The corrected functions according to Table A.23 are φ1(υ) ¼ φ1(1.0) ¼ 0.9313, η1(υ) ¼ η1(1.0) ¼ 0.5980. Specified ordinates of the bending moment diagram are 3i 3 0:2EI φ ðυÞ ¼ 0:9313 ¼ 0:1118EI l 1 5 3i 3 0:2EI ¼ 0:12EI MB ¼ ¼ l 5 MA ¼ MC ¼
Shear forces at specified sections are 3i 3 0:2EI η1 ðυÞ ¼ 0:5980 ¼ 0:01435EI l2 52 3i 3 0:2EI QB ¼ 2 ¼ ¼ 0:024EI l 52 QA ¼ QC ¼
It is obvious that r12 ¼ 0. The free-body diagram for the crossbar is presented in Fig. 19.5e. The equilibrium equation ∑X ¼ 0 for the crossbar becomes r 22 ¼ 2 0:01435EI þ 0:024EI ¼ 0:0527EI ðkN=mÞ It is obvious that the free terms of canonical equations are
726
19 Nonlinear Structural Analysis
R1P ¼ 0, and R2P ¼ F ¼ 30kN The canonical equations become 2EI Z 1 þ 0 Z 2 ¼ 0
ðbÞ
0 Z 1 þ 0:0527EI Z 2 F ¼ 0 The roots of these equations are Z 1 ¼ 0, Z 2 ¼
F 569:25 ¼ 0:0527EI EI
In case of linear classic analysis (nondeformable design diagram, Example 10.2) Z 2 ¼
F 416:67 ¼ . 0:072EI EI
The bending moment diagram can be constructed using the principle of superposition: M P ¼ M 1 Z 1 þ M 2 Z 2 þ M 0P
ðcÞ
Since Z1¼0 and acting load P does not cause bending of the members, then formula (c) becomes M P ¼ M 2 Z 2 . The bending moments at the clamped supports are M A ¼ M C ¼ 0:1118EI M B ¼ 0:12EI
569:25 ¼ 63:64 ðkNmÞ EI
569:25 ¼ 68:31 ðkNmÞ EI
The resulting bending moment diagram is presented in Fig. 19.5f. The numbers in parenthesis are bending moments calculated on the basis of the nondeformable design diagram (Example 10.2). We can see that P-Δ effect is significant, increasing of the horizontal displacement Z2 is 36.6%.
19.3
Static Nonlinearity
This section is devoted to the analysis of transverse vibration of a homogeneous beam with such boundary conditions under which the “restrained vibration” occur. This fact leads to a static nonlinearity of the dynamic problem. The essence of static nonlinearity is considered below.
19.3.1 Features of the Problem A uniform beam rests on two pivotally fixed supports, which prevent mutual horizontal displacement. The length of beam is l, moment of inertia and area of cross section are I and A, the mass of the beam per unit length m. We assume that the beam is brought from the state of rest, then left to itself, and, as a result, makes free vibrations. It is required to determine the frequency of free vibration. The peculiarity of the problem is in the fact that when beam performs vibration then the horizontal reactions H arise; these reactions by convention can be treated as a “thrust.” In the process of vibration the magnitude of the thrust changes. The value of initial thrust depends on the initial deflection of beam ymax (Fig. 19.6). E, A, I, m
H
l/2
y
H ymax l
Fig. 19.6 Static nonlinearity. Axial force H is a result of the free vibration of a beam
x
19.3
Static Nonlinearity
727
Let us consider the following fundamental relationships. Axial reaction H ¼ ε EA ¼
Sl EA, l
where ε is the relative elongation of the axial line of the beam. The length S of the elastic curve of a beam is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 dy 1þ dx S¼ dx ðl
ð19:16Þ
0
pffiffiffiffiffiffiffiffiffiffiffi 1 1 þ x ffi 1 þ x . . . to 2 Eq. (19.16). Therefore, the length of the elastic curve S and the corresponding axial elongation S l may be presented in form sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 # ðl 2 ðl " ðl dy 1 dy 1 dy 1þ dx ffi S¼ 1þ dx dx ¼ l þ dx 2 dx 2 dx Since the slope dy/dx is small compared with the unity, it is possible to apply the decomposition
0
0
ðl 2 1 dy Sl¼ dx 2 dx
0
ð19:17Þ
0
Thus, the axial reaction and bending moment can be calculated by formulas EA H¼ 2l
ðl 2 dy dx dx
ð19:18Þ
0
M ¼ M 0 Hy
ð19:19Þ
where M0 is the bending moment due to the transverse inertial force only and term Hy takes into account thrust H. It is seen that expressions (19.18) and (19.19) are nonlinear.
19.3.2 Transversal Vibration of a Rod Now we will derive the differential equation of transverse vibration. Moment-curvature equation is M d2 y ¼ EI dx2 Expression (19.19) for bending moment we substitute into the moment-curvature equation and twice differentiate it with respect to x
d4 y 1 00 M 0 Hy00 ¼ EI dx4
ð19:20Þ
Take into account the differential dependencies in bending and d’Alembert principle M 000 ¼ Q0 ¼ q ¼ m€y Here Q and q are shear force and distributed transversal load. Now we need to introduce expressions for M 000 ¼ m ∂2y/∂t2 and trust H from into (19.20), and take into account that required deflection of the beam is a function of two variables, i.e., y ¼ y(x, t). Thus, we get the governing differential equation (Bondar, 1971)
728
19 Nonlinear Structural Analysis 4
2
∂ y EA ∂ y EI 4 2l ∂x2 ∂x
ðl 2 2 ∂y ∂ y dx þ m 2 ¼ 0 ∂x ∂t
ð19:21Þ
0
This is partial nonlinear differential homogeneous equation of fourth order. This equation describes nonlinear Ð vibration of uniform rod with hinged fixed supports in which horizontal reaction occurs due to free vibrations. The factor (y0)2dx in the second term is a marker of static nonlinearity. This axial tension induced by the deflection of a beam is the source of the nonlinearity in the problem. If one of the supports is rolled, then static nonlinearity does not occur and Eq. (19.21) without second term becomes classical Bernoulli-Euler equation. Approximate Solution Let the transverse deflection of the beam be yðx, t Þ ¼ ymax T ðt Þ sin
πx l
πx determines the first form of vibration with amplitude l 00 ymax. This expression satisfies the boundary conditions y(0) ¼ y (0) ¼ y(l) ¼ y00 (l ) ¼ 0. To define an unknown function T(t) we apply the Bubnov-Galerkin procedure Here T(t) is unknown function of a time, while a factor ymax sin
ðl Lðx, t Þ sin
πx dx ¼ 0, l
ð19:22Þ
0
where operator L(x, t) is modified left part of the Eq. (19.21). 4
2
∂ y ∂ y Lðx, t Þ ¼ 4 a 2 ∂x ∂x
ðl 2 2 ∂y ∂ y dx þ b 2 ¼ 0, ∂x ∂t
A 1 a¼ , ¼ 2l I 2lr 2
rffiffiffi I r¼ , A
b¼
m EI
ð19:23Þ
0
Here r is radius of inertia of the cross section. Expression (19.22) means that the governing differential Eq. (19.21) and base πx are orthogonal. In expanded form, the procedure (19.22) is written as function sin l 2 3 l 2 ðl 4 2 ð 2 ∂ y ∂ y ∂y ∂ y πx 4 dx ¼ 0 ð19:24Þ a 2 dx þ b 2 5 sin l ∂x4 ∂x ∂x ∂t 0
0
This procedure allows proceeding to the ordinary differential equation for unknown function T(t). For this purpose we will compose the derivatives 2 2 ∂y π πx ∂ y π πx ¼ ymax T ðt Þ cos ; ¼ y T ðt Þ sin , max l l l l ∂x ∂x2 4 2 4 ∂ y π πx ∂ y πx ¼ ymax T ðt Þ sin ; ¼ ymax T€ ðt Þ sin dx l l l ∂x4 ∂t 2 Integrals in (19.24) take the form ðl 0
ðl 4 4 4 ∂ y πx π πx π l sin T ð t Þ sin 2 dx ¼ ymax T ðt Þ ; dx ¼ y max 4 l l l l 2 ∂x 0
19.3
Static Nonlinearity
729
ðl 0
2
∂ y πx sin dx l ∂x2
ðl 2 ðl 2 ∂y π l π 2 πx dx ¼ ymax T ðt Þ Tymax cos 2 dx ¼ l 2 l l ∂x 0
0
2 2 π l π l ymax ; T ðt Þ T 2 y2max l 2 l 2 ðl 2 ðl ∂ y πx πx l € sin dx ¼ ymax T ðt Þ sin 2 dx ¼ ymax T€ ðt Þ l l 2 ∂t 2 0
ð19:25Þ
0
The underlined components are explained below. Substitute these expressions in the formula (19.24) 4 2 2 π l π l π l l 2 2 ymax T ðt Þ a ymax T ðt Þ ymax T ðt Þ þ bymax T€ ðt Þ ¼ 0 l 2 l 2 l 2 2 After elementary transformations, we get 2
y T€ þ ω2 T þ ω2 max T3 ¼ 0 4r 2
ð19:26Þ
We received rffiffiffiffiffi an ordinary nonlinear second-order Duffing’s differential equation. The frequency of linear vibration of beam y2 π 2 EI . Nonlinear term ω2 max is ω ¼ 2 T 3 ¼ kT 3 presents the Duffing characteristic, so its harmonic linearization is m 4r 2 l 3 2 y2max ω T. Equation (19.26) after linearization becomes 4 4r 2 3 y2max 2 € T þω 1þ T¼0 ð19:27Þ 4 4r 2 The first frequency of nonlinear vibration of beam rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 3 ymax 2 ωnl ¼ ω 1 þ 4 2r
ð19:28Þ
Approximate frequency π2 ωffi 2 l
rffiffiffiffiffi rffiffiffiffiffi EI 1 3 ymax 2 π 2 EI 3 ymax 2 1þ 1þ ¼ 2 m 2 4 2r m 8 2r l
ð19:29Þ
It can be seen that with an increase in the initial deflection of the beam from the position of static equilibrium, the frequency of nonlinear vibration increases. Comments Insignificant, at the first glance, change in the design diagram, namely, the replacement of only one pivotally movable support by the pivotally fixed, resulted in a qualitatively new behavior of a structure. 1. The term “static nonlinearity” in a dynamic problem should not confuse the reader. This term refers to the static side of the problem, which is as follows. Force H tends to return the rod to its original rectilinear state and therefore may be considered Ðl ∂y 2 as a restoring force. In other words, it is one of the forces characterizing the vibrational process. Complex dx, 0 ∂x according to (19.18) determines the value of the restoring force. The underlined components in the relation (19.25) indicate the cause and the place of analysis where nonlinearity occurs. It can be seen that force H depends nonlinearly on the beam deformation ymax. It is this type of nonlinearity, H ¼ f(ymax), which is called the static one. 2. It can be seen that static nonlinearity has leaked into the resolving Eq. (19.21) through Eq. (19.17), which relate the elongations and deformations. We also note the second way in which static nonlinearity can manifest itself in the resolving equation, namely, through the equilibrium equation (Perelmuter & Slivker, 2003). The simplest example of static
730
19 Nonlinear Structural Analysis
nonlinearity (vibration of a lumped mass located on a thread) is given in book (Timoshenko, 1972). A more complex case (Mises's truss) may be found in the book (Karnovsky & Lebed, 2016). 3. Static nonlinearity leads to the fact that the frequency of free vibration of a nonlinear system, in contrast to the frequency of free vibration of a linear system, according to (19.28) depends on the initial conditions, that is, the amplitude of vibration ymax. The form of vibration does not depend on static nonlinearity (Bondar, 1971).
19.4
Physical Nonlinearity
This section is devoted to the analysis of transverse vibration of a uniform beam. It is assumed that one of the most important assumptions of the classical structural analysis, namely the stress-strain relationship, is not satisfied by Hooke’s law.
19.4.1 Features of the Problem Earlier in Chap. 14 we considered the analysis of a structure with elements the material of which does not obey Hooke’s law, namely, we dealt with the Prandtl diagram. This diagram consists of two parts: the linear part “stress-strain” and the yield plateau (Table 14.1). The nonlinearity of this type leads to a change in the design diagram in the process of increasing load. When the yield point is reached in one of the elements, it is excluded from the design diagram and the degree of static indeterminacy decreases by one. In contrast to this type of nonlinearity, we assume that the relationship between normal stress σ and strain ε is weakly nonlinear in the entire loading range (Bondar, 1971; Kauderer, 1958) σ ¼ Eε þ βε3
ð19:30Þ
Here β is the parameter of nonlinearity. Such nonlinearity does not lead to a sudden change in the design diagram. In other words, the design diagram in the deformation process remains unchanged. The term “weakly nonlinear” means that the parameter of nonlinearity is small, but how to estimate the “smallness” is not discussed.
19.4.2 Transversal Free Vibration of Uniform Beam Consider the free vibration of the simply supported uniform beam of length l and uniformly distributed mass m. The beam material is subject to the law (19.30). According to Bernoulli-Euler’s plane sections hypothesis, the relative elongation ε of the fiber at a distance z from the neutral axis is ε¼
z ¼ kz ρ
ð19:31Þ
Here ρ is the radius of curvature of the elastic curve, k is the curvature of the neutral layer. This ratio is approximate and is valid for the case of moderately large displacements when the longitudinal displacements are small compared to the transversal displacements. Also, we believe that in the case of plane bending, the longitudinal fibers do not press on each other. Given these assumptions and according to (19.31), normal stresses (19.30) are calculated as follows: β σ ¼ Ekz 1 þ k2 z2 E Equilibrium equations and static equivalence conditions are
ð19:32Þ
19.4
Physical Nonlinearity
731
X
ð X ¼ 0 ! σdA ¼ 0;
X
ð M y ¼ 0 ! σzdA ¼ M
A
ð19:33Þ
A
Since the function (19.32) is odd, the first condition (19.33) means that the neutral axis passes through the center of gravity of the cross section. Substitution (19.32) into the second formula (19.33) and execute the integration
k EI 2 þ β k2 I 4 ¼ M
ð19:34Þ
Here the moment of inertia of the n-th order is ð I n ¼ zn dA,
ðn ¼ 2, 4Þ;
A
For rectangular cross section I 2 ¼ bh3 =12, I 4 ¼ bh5 =80; for circular cross section I 2 ¼ πd4 =64, I 4 ¼ πd 6 =512. Assume that the transversal displacements of a beam are small. We take an approximate relation k ¼ y00 ; substitution of expression for curvature k of the neutral layer in (19.34) leads to the differential equation of the curved axis of a physically nonlinear elastic beam h i 2 y00 EI 2 þ β I 4 ðy00 Þ ¼ M
ð19:35Þ
We differentiate this equation twice taking into account the bending differential relations and the d’Alembert principle, i.e. M 00 ¼ Q0 ¼ q ¼ m€y. Finally, we get the governing nonlinear partial differential equation, which describes free transversal vibration of beam the material of which does not obey Hooke’s law; the stress-strain relationship is presented in form (19.30) (Bondar, 1971): Lðx, t Þ ¼ EI 2
2 2 4 4 2 3 2 2 ∂ y ∂ y ∂ y ∂ y ∂ y ∂ y þ 6βI þ 3βI þm 2 ¼0 4 4 4 2 3 2 4 ∂x ∂x ∂x ∂x ∂x ∂t
ð19:36Þ
It can be seen that the physical nonlinearity (19.30) is taken into account by the second and third terms. If we put a parameter β ¼ 0, we obtain the classical equation of free vibrations of a homogeneous beam. We will assume that with free vibration, the displacement of the beam occurs according to the law yðx, t Þ ¼ f1 ðt Þ sin
πx , l
ð19:37Þ
where f1(t) is the unknown function of time. To define it, we apply the Bubnov-Galerkin procedure. According to this procedure, the differential operator of the resolving equation and the function defining the form of vibration (in our case sinπx/l ) must be orthogonal: ðl Lðx, t Þ sin
πx dx ¼ 0 l
0
Here L(x, t) is the left side of the homogeneous Eq. (19.36). Expanded orthogonality condition is written as ðl " 0
# 2 2 4 4 2 3 2 2 ∂ y ∂ y ∂ y ∂ y ∂ y ∂ y πx EI 2 4 þ 6βI 4 2 þ 3βI 4 þ m 2 sin dx ¼ 0 l ∂x ∂x ∂x3 ∂x2 ∂x4 ∂t
Let us form the derivatives 2 2 ∂ y π πx ¼ f ðt Þ sin ; l 1 l ∂x2
3 3 ∂ y π πx ¼ f ðt Þ cos ; l 1 l ∂x3
ð19:38Þ
732
19 Nonlinear Structural Analysis
4 4 ∂ y π πx ¼ f ðt Þ sin ; 4 l 1 l ∂x
2
∂ y € πx ¼ f 1 ðt Þ sin l ∂t 2
Integrals included in (19.22) are as follows ðl
ðl 0
2
0 3
∂ y ∂ y ∂x2 ∂x3
ðl 0
ðl 4 4 4 ∂ y πx π πx π l dx ¼ sin f ð t Þ sin 2 dx ¼ f ðt Þ 1 4 l l l l 1 2 ∂x
2
∂ y ∂x2
2
2
0
2 ðl 2 πx π πx π 3 πx πx l π 8 3 sin dx ¼ f1 ðt Þ sin f1 ðt Þ cos sin dx ¼ f ðt Þ l l l l l l 8 l 1 0
2 ðl 2 8 4 ∂ y πx π πx π 4 πx πx π 3 3 dx ¼ sin dx ¼ sin f ð t Þ sin f ð t Þ sin f ðt Þ l 1 1 4 l l l l l l l 1 8 ∂x 0
In expanded form, the expression (19.38) takes the form EI 2
4 8 8 π l π 3l π 3 3 l f1 ðt Þ 6βI 4 f1 þ 3βI 4 f ðt Þ l þ m €f 1 ðt Þ ¼ 0 l 2 l 8 l 1 8 2
ð19:39Þ
After simple transformations we get 4 3 βI 4 π 4 3 €f 1 ðt Þ þ EI 2 π f 1 ðt Þ þ f ¼0 4 EI 2 l 1 m l
ð19:40Þ
This nonlinear homogeneous ordinary differential equation of the second order with constant coefficients and cubic term f13 (Duffing equation, 1918) describes in the first approximation transverse free vibration of uniform simply supported beam taking into account the physical nonlinearity (19.30) of beam material. rffiffiffiffiffiffiffi π 2 EI 2 . It is obvious that the frequency vibration in linear formulation is ω ¼ 2 m l 3 For linearization of the nonlinear Duffing term, which contains f13 , we apply linearization f13 ffi y2max f1 . The frequency 4 vibration in the nonlinear formulation is equal to π2 ω1 ¼ 2 l
rffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffi EI 2 9 β I4 π 4 2 π 2 EI 2 9 β I 4 π4 2 ymax ffi 2 y 1þ 1þ 16 EI 2 l 32 EI 2 l4 max m m l
ð19:41Þ
This result shows that, in contrast to a linear conservative system, the frequency of free vibration of physically nonlinear structure, as in the case of static nonlinearity, depends not only on the parameters of the system but also on the initial conditions. Consider the problem in the second approximation. We assume that the displacement of the beam is described by the equation yðx, t Þ ¼ f1 ðt Þ sin
πx 2πx þ f2 ðt Þ sin l l
ð19:42Þ
To determine the function of time f1(t) and f2(t), we again apply Bubnov-Galerkin procedures ðl
πx Lðx, t Þ sin dx ¼ 0, l
0
where L(x, t) is, as before, the differential operator (19.20). The set of nonlinear differential equations is
ðl Lðx, t Þ sin 0
2πx dx ¼ 0, l
ð19:43Þ
19.5
Geometrical Nonlinearity
733
4 4 π 3 β I4 π 4 3 2 €f 1 ðt Þ þ 1 π EI 2 f1 ðt Þ þ 24β I 4 f f þ f ¼0 m l l 12 4 m l 1 4 4 4 π 2 π 3 €f 2 ðt Þ þ 1 π 16EI 2 f2 ðt Þ þ 24β I 4 f f þ 192β I 4 f ¼0 m l l 1 2 l 2
ð19:44Þ
It is seen that the functions f1(t) and f2(t) are connected. Therefore, it should be expected that, in contrast to static nonlinearity, physical nonlinearity affects the mode of vibration If we break the connection between the equations, we get 4 3 β I4 π 4 3 €f 1 ðt Þ þ 1 π EI 2 f1 ðt Þ þ f ¼0 m l 4 m l 1 4 4 π 3 €f 2 ðt Þ þ 1 π 16EI 2 f2 ðt Þ þ 192β I 4 f ¼0 m l l 2
ð19:45Þ
With this assumption, the nonlinear frequency of the second form of vibration becomes 4π 2 ω2 ¼ 2 l
rffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffi EI 2 192 β I 4 π 4 2 4π 2 EI 2 β I4 π 4 2 ymax ffi 2 ymax 1þ 1þ6 16 EI 2 l m m EI 2 l l
ð19:46Þ
The numerical integration of systems (19.44) allows establishing the influence of the interaction of vibration shapes on the vibration frequencies and on the dynamic process This example shows the consequences of the computational character and new feature of behavior of a beam caused by the rejection of the assumption that the material obeys Hooke’s law.
19.5
Geometrical Nonlinearity
This section is devoted to the stability analysis of a simply supported uniform flexible rod with large deflections. It is assumed that the compressing force retains orientation in the process of deforming the rod. A detailed mathematical analysis is presented.
19.5.1 General In the nonlinear theory of deformations it is assumed that the displacements of points of the system are comparable with its dimensions and the angles of rotation of the sections cannot be considered small compared with unity (Filin, 1981, vol.1). This means the rejection of the assumption of small deformations and, as a result, the transition to the analysis of the structure in a deformed scheme. Similar problems arise in the analysis of structures with significant flexibility. A feature of such structures is that displacements are essentially nonlinearly dependent on the load, although the material works elastically. In contrast to the case of small displacements, in the problem of large displacements it is impossible to determine the support reactions using the principle of solidification. All reactions of support are significantly dependent on large displacements during bending. Therefore, all geometrically nonlinear problems, in principle, are statically indeterminate. This provision is easily illustrated by the following example. A cantilever beam of length l is loaded with force P at the free end. For large deformations, the free end is also displaced horizontally by an amount u. Therefore, the bending moment in clamped support will be M0 ¼ P(l u), unlike in the linear case M0 ¼ Pl. In the case of large displacements, the principle of superposition of solutions turns out to be unfair. For example, if in the system (Fig. 19.7a) the force P is increased n times, then the displacement will not increase n times. It is especially important to take into account, when there are several forces acting on the structure. Therefore, all methods and techniques discussed in the classical course of the theory of structures based on the principle of superposition, in the theory of nonlinear displacements cannot be claimed.
734
19 Nonlinear Structural Analysis
a
c
P
l
P
u
b
P
Fig. 19.7 The various cases of load behavior with the large deformations of the rod
In geometrically nonlinear problems, in addition to the assumption of large displacements, one more assumption should be made. We are talking about the behavior of force in the process of growing deformations. It is possible the following cases of load behavior: 1. The load retains its position at any level of deformation (Fig. 19.7a). 2. The load is invariably directed perpendicular to the tangent to the elastic line at the point of application of force (Fig. 19.7b). 3. The force is invariably directed along the tangent to the elastic line at the point of application of force (Fig. 19.7c). Each of these cases of load behavior leaves an imprint on the singularities of solving geometrically nonlinear problems. It should be noted the markers of the three classes of problems of mechanics of rods with large displacements (Popov, 1986). 1. The markers of the main class of geometrically nonlinear problems are the following: the initial curvature of the rod is constant (in particular, zero), the flexural rigidity of the rod is constant, and bending occurs under the action of concentrated forces and moments applied at the ends of the rod. 2. The problems which can be reduced to the main class include all those cases when the elastic line can be divided into a finite number of portions of a finite length, so that everyone finds itself in the conditions of the problem of the main class. These conditions satisfy, in particular, a rod of a stepwise variable stiffness. 3. The problems which cannot be reduced to the main class include all those problems in which at least one of the above conditions is not observed. Among them are problems with distributed loads, problems with arbitrarily varying initial curvature and cross-sectional area. The detailed theory of plane bending of rods for above three classes is presented in the book (Popov, 1986). For solution of geometrically nonlinear problems a rather high level of mathematical apparatus is applied. The problem of rod stability in a geometrically nonlinear formulation has been the subject of research by many authors. We limit ourselves to detailed classical solution to the Euler problem.
19.5.2 Stability of a Flexible Rod Consider a uniform centrally compressed rod with hinges at the ends. Length of the rod is l, flexural stiffness is EI (Fig. 19.8a). Assume the rod is flexible enough, but even with large displacements the stress does not exceed the limit of proportionality, and the material obeys Hooke’s law. In addition, we will assume that the internal axial force in any cross section of the rod is directed horizontally (Fig. 19.8b). The origin of coordinates 0 is placed at the left support, the curvilinear coordinate s is measured along the elastic curve s ¼ OA.
a
y
b
y(x)
P
P x
l
M y
x
ds A ξ
P
Fig. 19.8 (a) Design diagram of simply supported beam and (b) fragment of a deflected shape
0
* dx
P dy
x
19.5
Geometrical Nonlinearity
735 00
Instead of an approximate differential equation EIy ¼ Py, we take the exact equation of the curved axis of the beam rffiffiffiffiffi 1 y00 1 P 2 ¼ þ n y ¼ 0, n ¼ ð19:47Þ i3=2 , ρ h ρ EI 2 1 þ ðy0 Þ Here, the curvature 1/ρ is represented in Cartesian coordinates and an exact expression is accepted for it. It can be seen that Eq. (19.47) is nonlinear, and nonlinearity is of a geometric type, since it is caused by large deformations. In the case of large displacements, the elastic line of the beam is convenient to write in curvilinear coordinates; in this case 1 dξ the curvature should be determined by the formula ¼ , where ξ is angle of inclination of the tangent to the elastic line ρ ds (Fig. 19.8b). Therefore (19.47) dξ ¼ n2 y ds
ð19:48Þ
Differentiating (19.48) by s and taking into account that dy/ds ¼ sin ξ, as shown in Fig. 19.8b, we get d2 ξ dy ¼ n2 ¼ n2 sin ξ ds ds2
ð19:49Þ
This ordinary differential second-order equation is nonlinear, since the variable ξ on the right-hand side is found as an argument of a circular function, i.e., sinξ. For all diverse problems of basic type, the exact differential equation of an elastic line has the following single compact form l2
d2 ξ ¼ β2 sin ξ ds2
or
d2 ξ ¼ n2 sin ξ, ds2
β2 ¼
Pl2 EI
ð19:50Þ
The Eq. (19.49) of the centrally compressed flexible rod coincides with the general Eq. (19.50) for the main class of problems of flexible rods (Popov, 1986); this equation formally coincides with differential equation of large vibration of mathematical pendulum (Lojtsyansky & Lurie, 1983). If using this analogy, then we might write the change in the angle ξ along the elastic line. However, this is not enough to determine the deformed state of the rod. For determining the displacement of the system, deep research will be required (Feodosiev, 2016). The authors consider it appropriate to present below a derivation of the basic relations (therewith intermediate mathematical transformations are omitted), so that the reader can evaluate the possible mathematical difficulties that await him, if he abandons the assumptions of the linear classical theory of structures. Let us find the first integral (19.50), i.e., the curvature of the rod. For this purpose (19.50) should be presented in form d2 ξ ξ ξ dξ ξ 2 cos , multiply both sides of (19.50) by the corresponding parts of the obvious identity 2 ds ¼ 4d , ¼ 2n sin 2 2 ds 2 ds2 and integrate by parts both sides of the obtained relationship ð 2
ð d2 ξ dξ ξ ξ 2 ds ¼ 4n dξ sin cos 2 2 ds2 ds
ð19:51Þ
ξ
After cumbersome transformations, we obtain 2 dξ ξ ¼ 4n2 m2 sin 2 , ds 2
ð19:52Þ
where m2—arbitrary constant. The problem under consideration is reduced to an elliptic integral of the first kind in the Legendre form. Indeed, if we introduce new variable ψ such that sin then Eq. (19.52) becomes (Feodosiev, 2016)
ξ ¼ m sin ψ, 2
ð19:53Þ
736
19 Nonlinear Structural Analysis
dξ ¼ 2n ds
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m2 m2 sin 2 ψ ¼ 2n m cos ψ ! dξ ¼ 2n m cos ψ ds
ð19:54Þ
The meaning of introducing new variable ψ (19.53) is the following: as a result of their substitution in the equation of the slope of elastic curve (19.52), we get rid of the squares and do not introduce the double sign inherent to (19.52) when extracting the root. Expression (19.54) gives the curvature value of the elastic line at an arbitrary point, and as a result, the physical meaning of cosψ becomes clear. d ξ d Let us differentiate relationship (19.53) by s, i.e., sin ¼ ðm sin ψ Þ. This procedure leads to the following ds 2 ds result 1 ξ dξ dψ dξ cos ψ dψ cos ¼ m cos ψ ! ¼ 2m 2 2 ds ds ds cos ξ=2 ds
ξ Since cos ¼ 2
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ξ 1 sin 2 ¼ 1 m2 sin 2 ψ , then finally we get 2
cos ψ dξ ¼ 2m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dψ 1 m2 sin 2 ψ
ð19:55Þ
Equate the right-hand sides (19.54) and (19.55) and then integrate both parts dψ nds ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 sin 2 ψ ðψ dψ ns ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 sin 2 ψ
ð19:56Þ ð19:57Þ
ψ0
Here ψ 0 represents the value of the function ψ at the origin. Since at s ¼ 0 the bending moment is zero, so the curvature dξ/ds ¼ 0. Therefore, according to (19.54) cosψ 0 ¼ 0 ! ψ 0 ¼ π/2, and the expression (19.57) becomes π=2 ð
ns ¼ 0
dψ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 sin 2 ψ
ðψ 0
dψ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 sin 2 ψ
ð19:58Þ
These expressions are called elliptic integral of the first kind. The first integral with limits of integration 0 π/2 is called complete elliptic integral of the first kind in the form of Legendre π=2 ð
K¼ 0
dψ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 sin 2 ψ
In elementary functions, these integrals cannot be represented. The value of complete elliptic integral of the first kind of K with parameter m (m < 1) is found using a numerical sequence π=2 ð
K¼ 0
" # 2 2 2 dψ π 1 1 3 1 3 5 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1þ m2 þ m4 þ m6 þ 2 24 246 1 m2 sin 2 ψ 2
The values of the complete elliptic integrals depending on the modulus of the integral m are presented in mathematical tables (Abramovich & Stegun, 1964). Due to the symmetry of the elastic curve, at the midpoint (s ¼ l/2) the angle of inclination of the tangent ξ ¼ 0. According to equality sin(ξ/2) ¼ m sin ψ, the function ψ at this point vanishes and, therefore, for a point s ¼ l/2 the expression (19.58) takes the form
19.5
Geometrical Nonlinearity
737
l n ¼ 2
π=2 ð
0
dψ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ K, 1 m2 sin 2 ψ
rffiffiffiffiffi P n¼ EI
ð19:59Þ
The minimum value of this integral is achieved when m ¼ 0 ; therefore nl/2 π/2. Corresponding first critical load Pcr ¼ π 2EI/l2. It means that in order the equilibrium form with a curved axis can exist, it is necessary the fulfilment of Pl2 π 2 ! P > Pcr. Thus, the paradox of critical force (Sect. condition P > Pcr (!). Indeed, from condition nl/2 π/2 we get EI 15.3.3) is easily explained when considering a problem in a nonlinear formulation. Within the framework of the linear theory of stability of rods, it was established that a loss of stability is possible along two or more half waves. Suppose that the loss of stability occurs in two half-waves. Then the angle ξ and, as the result, the function ψ vanish when s ¼ l/4. In this case, Eq. (19.58) takes the form l n ¼ 4
π=2 ð
0
dψ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 sin 2 ψ
ð19:60Þ
In order for this equality to be fulfilled, the parameter n must satisfy the condition nl/4 π/2. Hence immediately follows the expression for the second critical force, known from the linear theoryPcr2 ¼ 4π 2EI/l2. Similarly, other critical forces are determined. These results should be interpreted as follows. If force P < Pcr1 ¼ π 2EI/l2, then one form of equilibrium of the rod with a straight axis is possible. If a forceP > Pcr1 , then along with the rectilinear axis, one more form of equilibrium of the rod with the curvilinear axis is possible. If forceP > Pcr2, then three forms of equilibrium are possible. This is a rectilinear axis, a form of equilibrium of a rod with an axis curved along one half-wave and along two half-waves. Thus, if the force exceeds the j-th critical force Pcr. j ¼ j2π 2EI/l2, then the equilibrium forms of the rod are one straight and j curvilinear forms. After the equilibrium forms are determined, the question should be answered: which of the indicated forms of equilibrium are stable and which are unstable? In the framework of the linear theory to answer this question is impossible. Careful analysis in the framework of the nonlinear theory allows making the following conclusions: If the force P < Pcr1, then the original rectilinear form is the only stable form of equilibrium. If the force P > Pcr1 , a stable form is only one—with an axial line curved along one half-wave. All other forms of equilibrium (over two or more halfwaves) are unstable. That is why for practice only the first form of loss of stability and, accordingly, the first critical force are important. Now we will determine displacements of a centrally compressed rod at P > Pcr1. According to Fig 19.8b we have dx/ds ¼ cosξ and dy/ds ¼ sinξ; therefore 2ξ 2ξ dx ¼ 1 2 sin ds ¼ 2 1 sin ds ds, 2 2 ξ ξ dy ¼ 2 sin cos ds 2 2 These expressions using the relations (19.53) and (19.56) may be brought to the form qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
dψ 2 dx ¼ 2 1 m2 sin 2 ψ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ds ¼ 1 m2 sin 2 ψ dψ ds, n n 1 m2 sin 2 ψ ! qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dψ m 2 2 dy ¼ 2m sin ψ 1 m sin ψ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 sin ψdψ 2 n 2 n 1 m sin ψ The coordinates of the point of the elastic line of a curved rod 2 x¼ n
ðψ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 sin 2 ψ dψ s; ψ0
m y¼2 n
ðψ sin ψdψ ψ0
Since ψ 0 ¼ π/2, the final expressions for the coordinates of the points of the elastic line of the rod are
738
19 Nonlinear Structural Analysis
2 26 x¼ 4 n
π=2 ð
3 ðψ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7 1 m2 sin 2 ψ dψ 1 m2 sin 2 ψ dψ 5 s;
y¼2
m cos ψ n
ð19:61Þ
0
0
The first relationship (19.61) contains the elliptic integrals of the second kind. Numerical values of complete elliptic integrals are presented in Table A.30. Relationship P-ymах. The greatest transverse displacement happens at s ¼ l/2 where ψ ¼ 0. Therefore ymax ¼ 2m/n. If it is necessary to determine the displacement caused by a particular force P > Pcr1, then the following procedure for numerical analysis may be recommended: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi 1. Compute critical force, the ratio P/Pcr, and parameter n: Since Pcr ¼ π 2 EI=l2 , then n ¼ P=EI ¼ ðπ=lÞ P=Pcr . pffiffiffiffiffiffiffiffiffiffiffiffi 2. Calculate the complete elliptic integral K. According to (19.59) K ¼ nl=2 ¼ ðπ=2Þ P=Pcr . 3. Using Table A.30 find the corresponding module of the elliptic integral m. 4. Calculate the absolute value of displacement ymax ¼ 2m/n and dimensionless displacement ymax 2m 2m ¼ pffiffiffiffiffiffiffiffiffiffiffiffi ¼ nl l π P=Pcr If it is necessary to construct a graph ymax-P, then it is advisable to conduct calculations in tabular form for different values of the modulus m of the elliptic integral (or corresponding α ) as shown in Table 19.3. Data for K are taken from Table A.30. Table 19.3 Dimensionless displacement ymax / l for various compressive forces P > Pcr
Parameter of elliptic integral m (a °), a = arcsin m 0.087 5°
0.174 10°
0.259 15°
0.342 20°
0.423 25°
0.500 30°
0.574 35°
0.643 40°
0.707 45°
K
m=0.0 a ° = 0.0° 1.5708
1.5738
1.5828
1.5981
1.6200
1.6490
1.6858
1.7312
1.7868
1.8541
P Pcr
1.0000
1.0038
1.0153
1.0351
1.0636
1.1020
1.1518
1.2147
1.2939
2.3932
ymax l
0.0
0.0553
0.1099
0.1621
0.2111
0.2565
0.2966
0.3316
0.3599
0.3813
The second row presents dimensionless force P/Pcr ¼ n2l2/π 2. According to the formula (19.59), parameter n ¼ 2K/l. Therefore P/Pcr ¼ 4K2/π 2. The third row presents the required dimensionless transverse displacement ymax/l ¼ 2m/nl ¼ m/K of the rod. Table 19.3 establishes a correspondence between a given force and displacement of the rod in the supercritical zone. It can be seen that at P ¼ Pcr the deflection of the rod is absent, but already with a very small increase in the force, transverse displacements appear which then increase sharply with increasing P. Assume that P ¼ 1.0153Pcr; in this case, the maximum transverse displacement of the rod is ymax ¼ 0.1099l. Exceeding the compressive force over the critical by only 1.5% leads to transverse displacement of the rod, which is 11% of the total length of the rod. Thus, the behavior of the system in the supercritical zone can be investigated only on the basis of a nonlinear theory. This is how the paradox of critical force that the reader confronted in Sect. 15.3.3 can be explained. As for the magnitude of the critical force, the linear and nonlinear theories lead to the same result. The exact analytic solution (19.59) may be represented in equivalent form (Nikolai, 1955, page 441]. rffiffiffiffiffi π=2 ð EI dψ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l¼2 P 2 1 f ðP=4EI Þ sin ψ 0
Problems
739
This equation establishes the relationship between the length of the rod l and its flexural rigidity on the one hand and, on the other hand, the acting force and the transverse displacement f of the rod midpoint. For practical purposes, this equation is not convenient. The appropriate solution in the series is ) rffiffiffiffiffi( 2 2 2 3 2 2 EI 1 f P 1 3 2 f 2P 135 f P l¼π 1þ þ þ þ P 2 4EI 24 4EI 246 4EI Thus, the nonlinear theory allowed us to establish the analytical relationship “force-maximum displacement” in the supercritical zone, find out which forms of equilibrium are stable, and explained the paradox in the linear problem (Sect. 15.3.3). Some nonlinear problems of structural mechanics which might be interesting for reader (beams in magnetic field, beams on elastic foundation, pipeline under moving liquid, behavior of structures in the wind flow) is presented by Bondar (1971), Kazakevych & Kulyabko (1994) as well as may be found in Karnovsky & Lebed, (2001, 2004b). Comments 1. Labor-consuming computational procedures are one of the reasons for the emergence of various approximate methods. These methods are based on linearization of differential equations, approximation of elliptic integrals, etc. The above exact analytical solution allows considering it as a test problem for assessing the accuracy of alternative methods. 2. Significant difficulties in solving geometrically nonlinear problems explain the need for introducing into the classical course of structural analysis a soft, controversial and questionable, at first glance, assumption about analyzing the structure using a nondeformed scheme. The payment for this assumption is that in the framework of linear theory there is no fundamental possibility of analyzing the structure in the area of large deformations. 3. This section provides the stability analysis of the simplest structure based on an exact nonlinear differential bending equation. The laborious and mathematically complex analysis is the result of rejection of the fundamental Navier hypothesis about the analysis of a structure on the base of its undeformed design diagram. This example in the classic Structural Analysis course has the following purposes: a) get an exact solution, reveal new effects and explain the contradictory result of linear theory. b) show the significance of the Navier hypothesis in particular (and all the assumptions in general), adopted in the classical course of analysis. It is the classical assumptions are the basis on which the rigorous classical theory of structures is developed, which has enormous applications and generalizations and is brought to an elegant simplicity and perfection. c) draw the attention of the reader that the choice of assumptions and their justification must inevitably be present in any analysis of an arbitrary design scheme of a structure.
Problems In problems 19.1–19.4 it is necessary rffiffiffiffiffiffiffiffi to analyze the structure on the basis of deformable design diagram. In all cases rffiffiffiffiffi P kPcr parameter υ ¼ nl ¼ l ¼l . EI EI 19.1. The uniform free-clamped beam is subjected to axial compressive force P and lateral distributed load q (Fig. P19.1). Calculate the vertical and angular displacement at free end, and the bending moment at fixed support. Estimate effect of axial load for different parameter k, P ¼ kPcr, Pcr ¼ π 2 EI=ð4l2 Þ.
Fig. P19.1
740
19 Nonlinear Structural Analysis
19.2. The uniform beam AB is subjected to axial compressive force P and lateral uniform distributed load q (Fig. P19.2). Calculate slope at the support, the deflection and bending moment at the middle of the beam. Estimate effect of axial load for different parameters k, P ¼ kPcr, Pcr ¼ π 2 EI=l2 .
Fig. P19.2
19.3. The uniform beam AB is subjected to axial compressive force P and external moment M0 at support A (Fig. P19.3). Calculate the slope at support A and reactions of supports A and B. Estimate effect of axial load for different parameters k, P ¼ kPcr, Pcr ¼ π 2 EI=ð0:7lÞ2 .
Ans.
M0 P
B
A l
Fig. P19.3
19.4. The uniform beam AB is subjected to axial compressive force P and transversal force F at the middle point of the beam. Calculate slope at support A, and linear displacement and bending moment at the point of force F.
F P
B
A l/2
l/2
Fig. P19.4
19.5. What is the necessity to consider nonlinear problems of structural analysis? 19.6. Explain the partial implementation of the superposition principle in the case when rod is subjected to simultaneous action of compressive force and displacement of the support. The reciprocity of unit displacements theorem is valid in case of this loading—true or false? 19.7. What is the essence of the various types of nonlinearity—static, physical, and geometrical? Give an example of constructive nonlinearity and mixed nonlinearity. 19.8. Explain how each of static, physical, and geometrical nonlinearities appears in the resolving differential equation of the system. 19.9. Describe some physical features of nonlinear systems.
Chapter 20
Conclusion: Once More About Modeling of Structures
Conclusion contains some remarks about modeling of engineering structures. Concepts of the transition from the constructive scheme of the structure to its design diagram, as well as the difference between the problem of structural analysis and the problem of straight structural design, are briefly discussed.
20.1
Some Problems of Structural Modeling
Throughout the course, we dealt with a ready-made design diagram of a structure. However, how it was adopted, and what considerations were taken into account, had not been discussed. After the reader has mastered the fundamental provisions of the course, it makes sense to briefly discuss some issues (Blekhman et al. 2005; Feodos’ev 1975). An attempt to create a design model of a structure that absolutely reflects the work of a real structure creates a situation of uncertainty. Uncertainty is generated by different reasons. They are the inaccessibility of certain types of necessary information, incompleteness of data, and ambiguity of interpretation of the same factors (Perelmuter and Slivker 2003). As an example of unavailability (or rather incomplete) information, a ceiling truss can be given, for which its own weight is unknown, since sections can only be determined after computation of the internal forces in the truss elements. There is a closed circle. There are classical approaches to solving problems in conditions of uncertainty. These are decision-making based on subjective experience of experts (expert assessments), objective previous experience (application of probability theory), and making the best decision from among those achievable (minimax estimate). These approaches aim to assess the plausibility of the design diagram and, if possible, indicate ways to increasing of the plausibility. The inaccessibility and incompleteness of certain types of information are fundamental for any object. This means that it is impossible to compile a design model that takes into account all the features of the structural scheme of the structure. Other factors should be noted that determine the approximation character of the design diagram. Among them are the following: 1. The use of simplified model of the structure, when simplifications are introduced (consciously or unconsciously) into the design diagram of the structure. Example of a conscious simplification of the design diagram is the representation of a through column of the truss type as the uniform rack of constant cross section. An example of an “unconscious” simplification of flat slab which is supported by regular grid of columns is the representation of the load on one of the columns. 2. Practically all of the model parameters associated with tolerances on the measured values are approximate. 3. Possible inaccuracies and under-determinations associated with the lack of clear boundaries in the concepts used. The above factors may lead to distortions of significance of types of simplifications, errors, and contradictions. Note that in the case of unique structures, the experiment is a mean of developing appropriate computational models. Experiment, on the one hand, allows checking the feasibility of accepted assumptions. On the other hand, the experiment allows discovering new phenomena or new facts. In this case, the system of accepted assumptions should be revised. For complex systems, not all hypotheses, assumptions, and simplifications are verifiable on a natural object or on a model. However, in many cases, a complex structure can be divided into a number of simple subsystems. It is on such subsystems that individual assumptions could be verified.
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8_20
741
742
20 Conclusion: Once More About Modeling of Structures
It should be remembered that for a specific engineering structure, in addition to general assumptions, a number of specific assumptions should be taken into account. For example, in a static analysis of a statically determinate structure, it is necessary to take into account only the type of supports, its geometry, the method of connecting the elements, and the given load. In the case of a statically indeterminate structure, the stiffness of individual elements should also be taken into account. In the case of stability analysis, the behavior of the force in the process of loss of stability of the structure should be taken into account. For example, will the force maintain its original position unchanged or will it somehow change? In the case of the analysis of free vibrations, the law of mass distribution for each element should also be taken into account. In the simplest case, the masses can be considered as lumped. To obtain more accurate results, distributed masses should be taken into account; in this case, a significant complication of the analysis is expected. Therefore, a number of serious fundamental questions arise. Among them are the following: 1. Is it worth to provide a complex analysis to increase the accuracy of the results? 2. Is it possible to achieve a significant increase in accuracy via proceeding to a more detailed calculation diagram? 3. Is the object so important and unique that a detailed design diagram and high accuracy of the analysis are necessary, and the associated difficulties of analysis are justifiable? 4. Is the importance and significance of all assumptions relating to the model commensurate? 5. Is high accuracy achieved with high details of some structure characteristics and low details of others? Answers to these questions can only be based on a deep understanding of the laws of structural mechanics.
20.2
Common Modeling Problems
Among the general principles of modeling structures note the following: 1. Initial hypotheses may be conditional, but they must put the structure in less favorable conditions than those in which it really is. 2. It is necessary to have a system of structural models, each of which has its own limits of application. 3. Each model should not only correctly and fully reflect the behavior of a real object, but also be so simple that the structural analysis does not become unnecessarily complicated. 4. It is necessary to understand and justify the goals of analysis, for the achievement of which it is necessary to complicate the design diagram which leads to the inevitable complication of computational procedures. A special role in the construction of the design diagram of structure is devoted to the models of individual elements (rod, plate, etc.), the choice of the standardized idealized material (elastic, ideally plastic, etc.), idealized loads, supports, and ways of connecting elements. Analysis of the structures consisting of one-dimensional elements (as the simplest ones) whose material obeys Hooke’s law represents the central part of the basic structural mechanics course. Note that it is necessary to speak not so much about the loads as about the exposures on the structure. Their brief classification is presented in Table 20.1. Table 20.1 Classification of exposures External exposures Internal exposures
Force exposures Loads Controlled preliminary load
Kinematical exposures Displacements of supports Temperature displacements
The duality of the force and kinematic description of ideal connections greatly facilitates the analysis and its verification. This is “the impossible mutual displacement means the appearance of force” and vice versa “possible mutual displacement means the absence of force.” Between these two extreme cases are cases of elastic supports and elastic connections of elements. Ways of connecting elements impose a serious imprint on the behavior of the structure. It is often difficult to find a seeming similarity between real and idealized connections. This is especially noticeable in trusses; the real connection of the truss elements is carried out on bolts or on welding; the support joints and intermediate joint are shown in Fig. 20.1.
20.2
Common Modeling Problems
743
b
a
c 1-1 Gusset plate
Gusset plate
1 1 Immovable support Abutment
Lower chord
2 angles back to back
Movable support
Fig. 20.1 Real connection of the truss elements. (a, c) Support joints; (b) intermediate joint
Looking on real connections of elements of truss (Fig. 20.1) the obvious question arises: Why such connections are presented on design diagram as hinged? Based on easily realizable assumptions and experimental data, a justification was given to this question (Chap. 4). It is possible to consider this problem in terms of the flexibility of individual elements. In a truss with rigid joints the rods experience deformations of bending and elongation, and in the hinged scheme—only elongation (or shortening). The stiffness of the rod in bending has order EI/l3, and in tension (compression) order EA/l. For thin rod EI=l3 r 2 1 ¼ ¼ 1/3
5
M 1 3 2 2 3M 1 2 3M RB 1 2l
u
MA
q
MA
RB
11 ql 40
t1 t 2 2h 3EI t1 t 2 2hl
ql 2 2 u 2 u2 8 qa u qlk1 , R A 2
ql 2 k1 , qa RB
u
k1
u
k1
0.2 0.4 0.5
0.0162 0.0512 0.0703
0.6 0.8 1.0
0.0882 0.1152 0.1250
Appendix
753
EI,
A
Table A.4 Reactions of fixed-fixed beams
h
B
l No
Loading conditions
Reactions and bending moment diagrams MB
MA
A=1
1
RA
RB
A=1
MB
MA MA 2
MB
MA
C
MB RA
3 ul
l
RB
ul
l
MA
(u+ =1)
6 EI
MB
l2 12 EI
RA
RB
MA
u
MC
2u 2
l3
MB
MA
P
MA
RB
RA
=1
Expressions for bending moments and reactions 4 EI 2 EI MA , MB l l 6 EI R A RB l2
C
MB
2
RA
2
RB
2
Pl, 2
Pl
1 2u P
u 1 2 P
u
RA
0 .5
MC
MA
M
1/3 2/3 MA
(u+ =1)
MB
u > 2/3 MB
MA
M l R Aul M A
MA
RA
5 MA
M CA
M
ql 2 12
MB
RA MB
M 4
3M 2 l M M CB 2
MB
RB
MB
RB
MA
q
Pl 8
6u
0.5 : M A
RA
P 2
RB
CB
ul
u 2 Pl
MB
RB l 2
1 ql 2 ql 2 24 (continued)
754
Appendix
Table A.4 (continued)
MA 6
MB
q
RA
RB
MA
7
Temperature gradient t1
MA
MB
MA RA
t2
RA
MB
RB
RA
EI RB
MB RB
ql 2 30 3 ql 20
t1 t2 h 0
MB
MA
q
8
MA
ql 2 , 20 7 ql , 20
MA
RA
RB
MA
ql 2 k1
MB
ql 2 k 2
MB
ql 2 2 u u 3
3 2 u 4
u
k1
k1
0.2 0.4 0.5 0.6 0.8 1.0
0.0151 0.0437 0.0573 0.0684 0.0811 0.0833
0.0023 0.0149 0.0260 0.0396 0.0683 0.0833
MB
ul MA
MA
ql 2 2 u 3 4u 6
MB
3 2 u 2
Appendix
755
EI,
A
Table A.5 Reactions of fixed-sliding beams
h
B
l No
Loading conditions
Reactions and bending moment diagrams MB
MA
A=1
1
RA
A=1
MA
MB
MA
MB
2(1) MA
C
RA ul
ul
l
RA
MB
3 MA
MB
MC RA
MC MA
MB
MA
ul
l (u+ =1)
MA
MB
q
MB RA
MA MB
6
(1)
MB
MA RA
t2
MA
0
ql 2 3 ql 2 6 ql
MA
RA
Temperature gradient t1
uM
RA
MB
MA 5
M
MB
RA
6 EI l2 12 EI l3
Plu 2 u 2 2 Plu 2 MB P
MB
C
4
QB
MA
l C
(u+ =1) M
MB
MB
MA
P
MA
QB
RA =1
Expressions for bending moments and reactions EI M A MB l RA 0
MB
l 2
ql 2 24
MB
EI h
M
MA RA
0
t
t1
QB is the force that should be applied to obtain vertical displacement =1
t2
t
756
Appendix
Table A.6 Reactions of uniform beams with elastic supports (k and krot are the stiffnesses of the supports) No
Loading conditions A=1
Reactions and bending moment diagrams MA
EI
1
RB
RA
k
A=1
l
MA
krot A=1
MB
MA
EI
2
MB
MA
RA
MA
MA 3
RB
RA
k
1
MA
RB
RA l
Expressions for bending moments and reactions 3EI 1 MA 3EI l 1 k l3 3EI 1 R A RB 2 3EI l 1 k l3
RA
MA
k l 1 rot 3EI 3EI l 1 krotl 4 EI 6 EI l2
3 1 2 4
3EI 1 2 3EI l 1 kl 3 3EI 1 RB 3 3EI l 1 kl 3
P C
4
RA ul
MA k
ul
MA
l (u+ =1)
MA
l
MC
MB
q MA 5
k
RA
u2 1 3 u 3EI 2 1 kl 3
Pl u
RB
C
RB
RB
MA
MA RB
krotl EI krotl EI
Pu 2 1 3 u 3EI 2 1 kl 3
ql 2 2
1
3 1 4 1 3EI kl 3
3 1 ql 3EI 8 1 kl 3
Appendix
757
Table A.7. Reactions of stepped one span beams
Dimensionless parameters are
EI1 EI 2
1; a 1
; b 1
2
3
; c 1
;
f
4
1
;
Fixed-pinned beam Design diagram
q
EI2
EI1
B
A
A
B
A
B =1
l
l
RB
3f ql 8c
RB
3EI1
RB
RB
l 3c
Fixed-fixed beam q EI1 A l
EI2 l
A
=1
B
B
=1 MB
MB
RB
MB RB
MB
RB
RB 9bf 8c 2 ql 2 2 12 4ac 3b 2bc 3af ql 2 4ac 3b 2
B
A
MB RB
EI 1
6b
4ac 3b 2 l 2 EI 1 12 a 4ac 3b
2
l
3
MB RB
4c EI1 2 4ac 3b l 6b EI1 4ac 3b 2 l 2
758
Appendix
Table A.8. Continuous uniform beams with arbitrary loading within the overhang P 1
0 a
3
2
4
l
l
Beam is loaded by force P at the free end; bending moments and reactions of the supports, a*=a/l. The number of span
Bending moments (Factor Pa)
Reactions of supports (Factor P)
2
3
4
M0 M1 M2 M3 M4
-1.0 0.2500 -
-1.0 0.2667 -0.0667 -
-1.0 0.2678 -0.0714 0.0179 -
R0 R1 R2 R3 R4
1+1.25a* -1.5a* 0.25a* -
1+1.267a* -1.60a* 0.40a* -0.067a* -
1+1.268a* -1.607a* 0.429a* -0.107a* 0.018 a*
Example. If a three-span beam is loaded by uniformly distributed load q within overhang, then M0 R0
qa 2 qa 2 qa 2 , M1 0.2667 , M2 0.0667 , 2 2 2 a a a 1 1.267 1.60 qa, R2 0.4 qa, R3 qa, R1 l l l
0.067
a qa. l
Appendix
759
Table A.9 Elastic curves of one span uniform beams caused by the unit angular and linear displacements 1; 1 ( x #
l-x
x
l 2
y x
Expression for elastic curve
y 1 l 8
1
f1
x2 l
x 2
f1(x) f2(x)
2
3
f3(x)
1 l 8 1 2
f4
3
x3 l2
x2
1 3
1 2
4
l2
x2 l
f2
f3
x3
x3
2
l2
x2
2
l2
l3
x3 l3
f4(x)
5
3 l 16
=1
f5
x 3
x2 l
x3 2l 2
f5(x)
f6(x)
6
7 f7(x)
f8(x)
8
5 16
5 16
1 2
1 2
9 f9(x)
f6
1 3
f7
3
x2
x3
l2
2l 3
x2
x3
2l 2
2l 3
f8 x
f9 x
x l
1
x l
0.375ql 1.250ql -0.125ql2 0.438ql -0.063ql2 -0.063ql
RA RB MB RA MB RA
l
q l=2a
P
-0.094P
0.313P 1.375P -0.188Pl 0.406P -0.094Pl
a
Y
RA =0.094P
A
0:
l
l/2
0.094P
RB
RB =0.688P
B
MB =0.094Pl
b
-0.167P
0.667P 2.667P -0.333Pl 0.833P -0.167Pl
l=3b
P b P b c c
0
RB
0.688P
c
l=4c
P
B
Pc
-0.234P
1.031P 3.938P -0.469Pl 1.266P -0.234Pl
P
l
RC=0.406P (RC for row 3 and RA for row 2are equal) M=0.203Pl
C
P 0.406P
P
Example For uniform beam shown below to find the reaction of supports.
Location of the load
A
Type of load in the loaded span
Table A10. Reactions and bending moments due to different types of loads
Two span beam with equal spans.
Tables A10, A11. Continuous uniform beams with equal spans and with hinged supports
l
l
q
-0.039ql
0.172ql 0.656ql -0.078ql2 0.211ql -0.039 ql2
l/2
C
760 Appendix
0.400ql 1.100ql -0.100ql2 0.450ql -0.050ql2 -0.050ql -0.050ql2 1.200ql -0.117ql2 -0.033ql2 0.017ql2 -0.067l2
RA RB MB RA MB RA MB RB MB MC MB MC
l
q l=2a
P
1.300P -0.175Pl -0.050Pl 0.025Pl -0.100Pl
-0.075P -0.075Pl
0.350P 1.150P -0.150Pl 0.425P -0.075Pl
a b
A
q
l
B
l/2
l
P1
C
l/3
P2
l
l/3
P2 D
MC
MB
0.033ql 2
0.117ql 2
l
2.533P -0.311Pl -0.089Pl 0.044Pl -0.178Pl
-0.133P -0.133Pl
0.733P 2.267P -0.267Pl 0.867P -0.133Pl
l=3b
P b P b
0.075P1l 0.178P2l
0.075P1l 0.044P2l
Example For beam shown below to find the bending moments at supports B and C.
Location of the load
A
B
c
Type of load in the loaded span
Table A11 Reactions and bending moments due to different types of loads
Three span beam with equal spans
3.750P -0.438Pl -0.125Pl 0.063Pl -0.250Pl
-0.188P -0.188Pl
1.125P 3.375P -0.375Pl 1.313P -0.188Pl
l=4c
Pc P c Pc
l
C
l
l
q
0.626ql -0.073ql2 -0.022ql2 0.011ql2 -0.042ql2
-0.032ql -0.032ql2
0.188ql 0.563ql -0.063ql2 0.219ql -0.032 ql2
l/2
D
Appendix 761
762
Appendix
Two span continuous uniform beam with different spans q1
The bending moment at support B and maximum bending moment at the first and second spans may be calculated by formula M kql 2 . Parameters k are presented in Table A.12
q2
B
l1
A
l2
Table A12. Bending moments due to uniformly distributed load
l1:l2 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.2 2.4 2.6 2.8 3.0 Factor
Load q1 is applied in the first span only MB M1 max -0.063 -0.079 -0.098 -0.119 -0.143 -0.169 -0.197 -0.227 -0.260 -0.296 -0.333 -0.416 -0.508 -0.610 -0.722 -0.844 q1l 22
0.095 0.114 0.134 0.155 0.178 0.203 0.228 0.256 0.285 0.315 0.347 0.415 0.488 0.570 0.655 0.743 q1l 22
Load q2 is applied in the second span only MB M2 max
Load q1= q2=q is applied both in the first and second span MB M1 max M2 max
-0.063 -0.060 -0.057 -0.054 -0.052 -0.050 -0.048 -0.046 -0.045 -0.043 -0.042 -0.039 -0.037 -0.035 -0.033 -0.031
0.095 0.096 0.097 0.098 0.099 0.100 0.101 0.102 0.103 0.103 0.104 0.106 0.107 0.108 0.109 0.110
-0.125 -0.139 -0.153 -0.174 -0.195 -0.219 -0.245 -0.274 -0.305 -0.339 -0.375 -0.455 -0.545 -0.645 -0.755 -0.875
0.070 0.090 0.111 0.133 0.157 0.183 0.209 0.237 0.267 0.298 0.330 0.398 0.473 0.553 0.639 0.730
0.070 0.065 0.059 0.053 0.047 0.040 0.033 0.026 0.019 0.013 0.008 0.001 * * * *
q2 l22
q2 l22
ql22
ql22
ql22
*)Within the second span the bending moments are negative.
C
Appendix
763
Uniform continuous beam with three different spans and hinged supports The load q is applied at the first span. The bending moments at supports B and C are MB k1ql12 , M C k 2 ql12 . Parameters k1 and k2 are presented in Table A.13a.
q1 A
B
l1
C
D
l3
l2
Table A.13a Bending moments at supports B and C (factor ql12 ) l3:l2
l1 l2
0.3 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2
0.3
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
0.034 0.013 0.041 0.016 0.053 0.021 0.062 0.024 0.069 0.027 0.075 0.029 0.079 0.031 0.083 0.032 0.086 0.033 0.089 0.034
0.033 0.012 0.041 0.015 0.053 0.019 0.062 0.022 0.069 0.024 0.074 0.027 0.079 0.028 0.083 0.029 0.086 0.031 0.089 0.032
0.033 0.010 0.040 0.012 0.052 0.016 0.061 0.019 0.068 0.021 0.074 0.023 0.078 0.024 0.082 0.026 0.085 0.027 0.088 0.028
0.032 0.009 0.040 0.011 0.051 0.014 0.060 0.017 0.067 0.019 0.073 0.020 0.078 0.021 0.081 0.023 0.085 0.023 0.087 0.024
0.032 0.008 0.039 0.010 0.051 0.013 0.060 0.015 0.067 0.017 0.072 0.018 0.077 0.019 0.081 0.020 0.084 0.021 0.087 0.022
0.032 0.007 0.039 0.008 0.050 0.011 0.059 0.013 0.066 0.015 0.072 0.016 0.077 0.017 0.080 0.018 0.084 0.019 0.087 0.020
0.031 0.007 0.038 0.008 0.050 0.010 0.059 0.012 0.066 0.014 0.072 0.015 0.076 0.016 0.080 0.017 0.084 0.017 0.086 0.018
0.031 0.006 0.038 0.007 0.050 0.010 0.059 0.011 0.066 0.013 0.071 0.014 0.076 0.015 0.080 0.015 0.083 0.016 0.086 0.017
0.031 0.006 0.038 0.007 0.050 0.009 0.058 0.010 0.065 0.012 0.071 0.013 0.076 0.013 0.080 0.014 0.083 0.015 0.086 0.015
0.031 0.005 0.038 0.006 0.049 0.008 0.058 0.010 0.065 0.011 0.071 0.012 0.076 0.013 0.079 0.013 0.083 0.014 0.086 0.014
Example Calculate the bending moment at the supports B and C if l1 distributed load q=2kN/m is applied at the first span.
8m, l2
10m, l3
6m and uniformly
Solution: Relationships l1 l2 0.8, l3 l2 0.6 . For this case k1 0.061, k2 0.019 . Bending moments at supports B and C are: MB k1ql12 0.061 2 64 7.808 kNm , M C 0.019 2 64 2.432 kNm .
764
Appendix
The load q is applied at the second span.
q2
The bending moments at supports B and C are MB k1ql22 , MC k2ql22 . Parameters k1 and k2 are presented in Table A.13b.
A
B
l1
D
C
l2
l3
Table A.13b Bending moments at supports B and C (factor ql22 ) l3:l2
l1 l2
0.3 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2
0.3
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
0.070 0.070 0.064 0.072 0.055 0.077 0.048 0.078 0.043 0.080 0.038 0.082 0.035 0.082 0.032 0.084 0.030 0.085 0.027 0.086
0.072 0.064 0.066 0.066 0.057 0.069 0.050 0.071 0.044 0.074 0.040 0.075 0.036 0.077 0.033 0.078 0.031 0.078 0.029 0.079
0.075 0.055 0.069 0.056 0.060 0.060 0.052 0.062 0.047 0.064 0.042 0.065 0.038 0.066 0.035 0.068 0.033 0.067 0.030 0.069
0.078 0.048 0.072 0.050 0.062 0.052 0.054 0.054 0.048 0.056 0.044 0.057 0.040 0.058 0.037 0.059 0.034 0.060 0.032 0.061
0.080 0.043 0.073 0.044 0.063 0.047 0.056 0.049 0.050 0.050 0.045 0.051 0.041 0.052 0.038 0.053 0.035 0.054 0.033 0.055
0.081 0.038 0.075 0.040 0.064 0.042 0.057 0.044 0.051 0.045 0.046 0.046 0.042 0.047 0.038 0.048 0.036 0.049 0.034 0.049
0.083 0.035 0.076 0.036 0.066 0.038 0.058 0.040 0.052 0.041 0.047 0.042 0.043 0.043 0.039 0.044 0.036 0.044 0.034 0.045
0.083 0.031 0.077 0.033 0.067 0.035 0.059 0.037 0.053 0.038 0.048 0.039 0.044 0.040 0.040 0.040 0.037 0.041 0.035 0.041
0.085 0.030 0.079 0.031 0.068 0.033 0.060 0.034 0.054 0.035 0.049 0.036 0.044 0.037 0.041 0.037 0.038 0.038 0.035 0.038
0.086 0.027 0.079 0.028 0.069 0.030 0.061 0.032 0.054 0.033 0.049 0.033 0.045 0.034 0.041 0.035 0.038 0.035 0.036 0.036
Appendix
765
The load q is applied at the third span q3 The bending moments at supports B and C are M B k1ql32 , MC k2ql32 . Parameters k1 and k2 are presented in Table A.13c.
B
A
l1
D
C
l2
l3
Table A.13c Bending moments at supports B and C (factor ql32 ) l3:l2
l1 l2
0.3 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2 k1 k2
0.3
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
0.013 0.034 0.012 0.033 0.010 0.033 0.009 0.032 0.008 0.032 0.008 0.032 0.007 0.031 0.006 0.031 0.006 0.031 0.005 0.031
0.016 0.041 0.015 0.041 0.012 0.040 0.011 0.040 0.010 0.039 0.008 0.039 0.008 0.038 0.007 0.038 0.007 0.038 0.006 0.038
0.021 0.053 0.019 0.053 0.016 0.052 0.014 0.051 0.013 0.051 0.011 0.050 0.010 0.050 0.010 0.050 0.009 0.050 0.008 0.049
0.024 0.062 0.022 0.062 0.019 0.061 0.017 0.060 0.015 0.060 0.013 0.059 0.012 0.059 0.011 0.059 0.010 0.058 0.010 0.058
0.027 0.069 0.024 0.069 0.021 0.068 0.019 0.067 0.017 0.067 0.015 0.066 0.014 0.066 0.013 0.066 0.012 0.065 0.011 0.065
0.029 0.075 0.027 0.074 0.023 0.074 0.020 0.073 0.018 0.072 0.016 0.072 0.015 0.072 0.014 0.071 0.013 0.071 0.012 0.071
0.031 0.079 0.028 0.079 0.024 0.078 0.021 0.078 0.019 0.077 0.017 0.077 0.016 0.076 0.015 0.076 0.013 0.076 0.013 0.076
0.032 0.083 0.029 0.083 0.026 0.082 0.023 0.081 0.020 0.081 0.018 0.080 0.017 0.080 0.015 0.080 0.014 0.080 0.013 0.079
0.033 0.086 0.031 0.086 0.027 0.085 0.023 0.085 0.021 0.084 0.019 0.084 0.017 0.084 0.016 0.083 0.015 0.083 0.014 0.083
0.034 0.089 0.032 0.089 0.028 0.088 0.024 0.087 0.022 0.087 0.020 0.087 0.018 0.086 0.017 0.086 0.015 0.086 0.014 0.086
766
Appendix
Settlements of supports of uniform beams with equal spans and hinged supports Bending moments at supports are M
k
EI l2
, where
is a vertical settlement of support directed
downward. Coefficient k are presented in Table A.14. Table A.14 Bending moments due to the vertical settlement of supports (factor EI l 2 ) Settlement of support Design diagram of the beam A
B
A
A l
C
D
C l
D
C
B
B
Supports moments
l
E l
A
B
C
D
E
MB
-1.500
3.000
-1.500
-
-
MB MC
-1.600 0.400
3.600 -2.400
-2.400 3.600
0.400 -1.600
-
MB MC MD
-1.6072 0.4286 -0.1072
3.6429 -2.5714 0.6429
-2.5714 4.2857 -2.5714
0.6429 -2.5714 3.6429
-0.1072 0.4286 -1.6072
Example. Three-span uniform beam ABCD with equal spans has settlement B of support B. Calculate the bending moment at all supports. Solution. EI 1. Bending moment at support B is M B 3.6 2 (extended fibers below of the neutral line). l 2.
Bending moment at support C is
MC
2.4
EI l2
(extended fibers above of the neutral line).
A
l
P=1
B
l
+
A
3
l2 6 EI
+
B
u u3
l2 6 EI
factor l 2/6EI .
IL( B)
IL( A) factor l 2/6EI .
B MA
MA
A *
ul
l
P=1
B
l
3
l 2
+
B
u2 1 u
l2 4 EI
factor l 2/4EI
IL( B)
IL(MA) factor l/2.
B
Clamped-pinned beam
1. Positive sign for angle of rotation: for left support clockwise, for right support counterclockwise. 2. Sign * means the inflection points of elastic curve. 3. Each beam is divided by 10 equal segments.
Note:
A
ul
0.171 0.288 0.357 0.384 0.375 0.336 0.273 0.192 0.099
0.099 0.192 0.273 0.336 0.375 0.384 0.357 0.288 0.171
0.171 0.288 0.357 0.384 0.375 0.336 0.273 0.192 0.099 0.009 0.032 0.063 0.096 0.125 0.144 0.147 0.128 0.081
Pinned-pinned beam
MA
A *
ul
l
P=1 l
*
Clamped-clamped beam
0.081 0.128 0.147 0.144 0.125 0.096 0.063 0.032 0.009
MA
u
2
l
MB
u2
0.009 0.032 0.063 0.096 0.125 0.144 0.147 0.128 0.081
Table A15 One-span beams. Influence lines for boundary effects.
l
factor l.
IL(MB)
factor l.
IL(MA)
MB
B
Appendix 767
768
Appendix
Influence lines for continuous uniform beam with equal spans and hinged supports (Odd sections are not shown) 0
2
4
6
8
10
l
12
0
2
l
4
6
8
l
10
12
14
16
18
l
l
Table A.16a Two span beam. Influence lines for bending moments and shear forces Position of the load P=1 0 1 2 3 4 5 6 7 8 9 10 11 12
Ordinates of influence lines of bending moments at sections (factor l) 1
2
3
4
5
6
Ordinates of influence line Q0
0.0 0.1323 0.0988 0.0677 0.0402 0.0172 0.0 -0.0106 -0.0154 -0.0156 -0.0123 -0.0068 0.0
0.0 0.0976 0.1976 0.1354 0.0803 0.0343 0.0 -0.0212 -0.0309 -0.0313 -0.0247 -0.0135 0.0
0.0 0.0632 0.1298 0.2031 0.1205 0.0516 0.0 -0.0318 -0.0463 -0.0469 -0.0370 -0.0203 0.0
0.0 0.0285 0.0619 0.1041 0.1606 0.0687 0.0 -0.0424 -0.0617 -0.0626 -0.0494 -0.0270 0.0
0.0 -0.0060 -0.0061 +0.0051 +0.0340 +0.0860 0.0 -0.0530 -0.0772 -0.0782 -0.0617 -0.0338 0.0
0.0 -0.0405 -0.0740 -0.0938 -0.0926 -0.0636 0.0 -0.0636 -0.0926 -0.0938 -0.0740 -0.0405 0.0
1.0000 0.7928 0.5927 0.4062 0.2407 0.1031 0.0 -0.0636 -0.0926 -0.0938 -0.0740 -0.0405 0.0
Example. Force P is located at section 8. Calculate the bending moment at specified points and construct the bending moment diagram. P 0.0926Pl
C
A B
RA=0.0926P
M
0.1605Pl
RB=0.8519P
RC=0.2407P
Solution. 1. Bending moment at the section 6 (support B) is M 6 0.0926 Pl . 2. Reaction of support A is RA= 0.0926P kN and directed downward. 3. Reaction at support C l P RC M B 0 : RC l P 0.0926 P l 0 RC 0.0926 P 0.2407 P kN 3 3 4. Reaction at support B: RB Y 0 : P 0.0926 P 0.2407 P RB 0 RB 0.8519 P kN 5. Bending moments at section 8 is 2 2 M 8 RC l 0.2407 P l 0.1605 Pl 3 3 Since structure is symmetrical and points 4 and 8 are symmetrically located, then bending moments M8 (if load P is located at point 8) and M4 (if load P is located at point 4) are equals. Last bending moment may be taken immediately from Table A.16a.
0.0 0.0618 0.1273 0.2000 0.1174 0.0495 0.0 -0.0285 -0.0395 -0.0375 -0.0271 -0.0131 0.0 0.0085 0.0123 0.0125 0.0099 0.0054 0.00
0.0 0.0267 0.0585 0.1000 0.1565 0.0659 0.0 -0.0379 -0.0526 -0.0500 -0.0362 -0.0175 0.0 0.0113 0.0165 0.0167 0.0132 0.0072 0.0
Influence lines for shear force at section 6 are shown below.
0.0 0.0967 0.1960 0.1333 0.0782 0.0329 0.0 -0.0190 -0.0263 -0.0250 -0.0181 -0.0088 0.0 0.0057 0.0082 0.0083 0.0066 0.0036 0.0
0.0 -0.0083 -0.0102 0.0 0.0289 0.0826 0.0 -0.0474 -0.0658 -0.0625 -0.0452 -0.0219 0.0 0.0141 0.0206 0.0208 0.0165 0.0090 0.0 0
0.0 -0.0432 -0.0790 -0.1000 -0.0987 -0.0677 0.0 -0.0569 -0.0789 -0.0750 -0.0543 -0.0263 0.0 0.0169 0.0247 0.0250 0.0197 0.0108 0.0 2
l
4
0.0 -0.0342 -0.0625 -0.0792 -0.0782 -0.0536 0.0 0.0872 0.0364 0.0083 -0.0028 -0.0036 0.0 0.0028 0.0041 0.0042 0.0033 0.0018 0.0
0.0987
0.0 0.1318 0.0980 0.0667 0.0391 0.0165 0.0 -0.0095 -0.0132 -0.0125 -0.0090 -0.0044 0.0 0.0028 0.0041 0.0042 0.0033 0.0018 0.0 6
8
0.0 -0.0252 -0.0461 -0.0583 -0.0576 -0.0395 0.0 0.0644 0.1516 0.0917 0.0487 0.0191 0.0 -0.0113 -0.0165 -0.0167 -0.0132 -0.0072 0.0
0.1234 0.0000 1.0000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
l
10
12
0.0 -0.0162 -0.0296 -0.0375 -0.0370 -0.0254 0.0 0.0418 0.1002 0.1750 0.1002 0.0418 0.0 -0.0254 -0.0370 -0.0375 -0.0296 -0.0162 0.0
Ordinates of influence lines of bending moments at sections (factor l) 3 4 5 6 7 8 9
0.6913
2
0.3087
1
14
l
16
1.0000 0.7901 0.5877 0.4000 0.2346 0.0990 0.0 -0.0569 -0.0789 -0.0750 -0.0543 -0.0263 0.0 0.0169 0.0247 0.0250 0.0197 0.0108 0.0 18
(continued)
right
IL Q6
0.0 0.0540 0.0987 0.1250 0.1234 0.0846 0.00/1.00* 0.8639 0.6913 0.5000 0.3087 0.1361 0.0 -0.0846 -0.1234 -0.1250 -0.0987 -0.0540 0.0
Ordinates of Inf. Lines Q0 Q6right
-0.1234
Position of the load P=1
-0.0987
Table A.16b Three span beam. Influence lines for bending moments and shear forces
Appendix 769
MC
RD l
Bending moment at point 12 (support C) is M 12 located at section 10. Final bending moment diagram is presented below.
5.
Reaction of support D: RD
0.6913 P , then reaction of support B is
4.
RB
RA 2l
P
A
RA=0.0789P
RB l
P
0.1515Pl RB=0.7702P
B
0.0789Pl
RC=0.363P
C
0.0543Pl M
RD=0.0543P
D
2 l 0 RD 0.0543 P 3 0.0543 Pl . The same result may be taken immediately from Table A.16b for section 6, if load P is
RB=RA+ 0.6913P=0.0789P+ 0.6913P=0.7702P.
RA
Since Q6right
3.
Solution 1. Bending moment at point 6 (support B) is M 6 0.0789 Pl . 2. Ordinate of influence line Q0 0.0789 , so reaction of support A is RA= 0.0789P and directed downward.
Example Continuous uniform three-span beam with equal span is loaded by force O at section 8. Construct the bending moment diagram. Use the influence line method (Table A.16b)
Table A.16b (Continued)
770 Appendix
f l 1 4
*)
7*)
6
5
4
3
2
1
t
t t
q
1 2u
0
0
l
2
6 EI
ql u 2 2u 2 u3 2
13 ql 32
P1 u
2
P 2
RA
is a thermal expansion coefficient
=1
P
q
=1
ul
l/2
ul
l/2
P
Loading
45 EI 4 f2
45 EI 4 lf 2
15 EI 2 lf
tt
ql 3 u 10 15u 6u 2 8f
2
ql 2 16 f
15 Pl 2 u 1 u2 4 f
15 Pl 64 f
H
Pl 32
MA
HA
15 EI 2 f
15 EI 2 lf
9 EI l
tt
ql 2 u 1 3u 3u 2 u 3 2
2
ql 2 64
A
RA
MA
Pl u 1 u 2 2 5u 2
Table A17 Reactions and bending moments due to different types of loads.
Cross section of the arch uniform or slightly changing through the span.
Hingeless parabolic arches, f
3 5u
15 EI 2 f
15 EI 2 lf
3EI l
tt
ql 3 u 1 2u u 2 2
2
ql 2 64
Pl 2 u 1 u 2
Pl 32
MB
l
C
3 Pl 64
Mc
HB
u 0.5
15 EI tt 4 f
15 EI 4 lf
3 EI 2 l
ql 3 u 2 5u 2u 2 8
2
0
u 0.5 Pl 2 u 3 10u 5u 2 4
RB
B
MB
Appendix 771
IC , cos
IC 45 , k 4 AC f 2 1
1
*)
P
t t
q
Increase of temperature
=1
=1
l/2
ul
l/2
P
P
0
0
l
2
6 EI C
13 ql 32
1 2u 1 u
2
P 2
RA
45 EI C 4 f2
45 EI C 4 lf 2
15 EI C 2 lf
ql 2 k 16 f
15 Pl 2 u 4 f
15 Pl k 64 f
H
tt
2
k
5k u 1 2
15 EI C 2 lf
9 EI C l
t tk
ql 2 11 8k 192
2
15 EI C f 2
Plu
MA
Pl 5 k 1 8 4
**)
HA
MB
15 EI C f 2
15 EI C 2 lf
3EI C l
l
f
0, k
1
C
t tk
ql 2 8k 5 192
5k 2
Pl 5 k 1 8 4
Plu 2
RA
A
Ix
*) I C , AC are moment of inertia and area of cross section at the crown C. If axial forces are neglected, then **) The axial forces are taken into account
6
5
4
3
2
1
Loading
Table A.18 Reactions and bending moments due to different types of loads
Ix
Hingeless non-uniform parabolic arches
MA
1
1
t tk
15 EI C 4 lf
3 EI C 2 l
15 EI C 4 f
HB
5k 2 2
ql 2 1 k 48
Pl 2 u 2
u 0.5
Pl 5 1 k 8 8
Mc
RB
B
MB
772 Appendix
Appendix
773
l
Q 2 x dx for analysis of gentile inextensible cables;
Table A.19 Load characteristic D 0
Q(x) is the shear of the reference beam, L is the total unstressed length of the cable, l and f are the span and a sag of the cable. Load characteristic D
Loading P
a
P 2l 1
1 l
q
P 2l for a 4
l 2
L l
1
q 2l 3 ; 12
2 l
l0
q l
4
q
q 2l 3 ; l0 80
l/2
l/2
q 2l 3 1 12 12
P
a1 5
q
q 2l 3 1 12
w
6
q l a1
b/2 w
7
q b l a1 8
w q b l
1
q 2l 3 1 12
q 2l 3 1 12 12 a1 , 1 l
2
1
1
1
3
1
3 2
4 3 5 16
L l
12 b , l
2 1
,
1 2 1
1
b , l
,
q 2l 3 4 for b 12
2 2
12
P ql
l 2
for a1
2
l ; 2
2
a1 , l
1
6 4
for b
2
18 f 1 5 l
12
1
w q l, w
2 1
q
2
,
w q
q 2l 3 4 for b 12
b/2
w
1 1
q 2l 3 1 3 12
l b
2
8 f 3 l
q 2l 3 45
3
q
a l
,
q w 2l3 1 12 1 12 12 2 12 a1 b w , , 2 1 l l q w
l, w
2 2 2
q
12
1
q 2l 3 4 for b 0, w q 12
12
2 1
2
2
,
774
Appendix
Table A.20 Reactions of beams subjected to compressed load and unit settlement of support,
Pinned-clamped beam
Clamped-clamped beam
1
l
i l 3i
R M
R
1 1
P 1 l
i l2 i 3 l
R
3
M
1
l
R 6
2
P 1
P
1
4
MB
i l
i 2 l2 i 6 4 l
R 12
M
4
M P
i l 4i
6
3
MA
P
R
R M
2i
M
2
P l
3
MB
EC
P
P
1 R
1
MA
M
EC
P
l
M P
P
P
1
1 P
3
P
l
l
M
i tan
i
M
sin
Pinned-pinned beam
P
1
P 1
l
R
P
P
i
M 2
l
l2 M
Expression for corrected functions
tan
Clamped-free beam
P
4 R
i
1 4
and
1 2
i tan
are presented in Table A.22.
P
P . EI
Appendix
775
Members subjected to axial compression and bending (beam-columns) Table A.21 Beam-columns. Reactions and deflections,
4
q
P
q
P
P
ql 2
l 2
ql 4
l 2
y
sec
2
EI
1 2
EI
2
tan
2
1
0
2
2
Fl 3 2 EI Fl
0
3
M l
tan
3
2
2
sec
2
R l
1
2
P l/2
l 2 0 l
M 0l 2 2 EI M 0l EI 2 M 0l 2
1 cos cos
2
2
1 2
2
sin
1
sin
cot 1
1
cos
P
R l
2
2
P l/2
sec
2
M l 0
1
,K
M0
l/2
M0 sec 2 2
EI
F K sin
6
P
y
Fl K sin
2
2
2 EI
l 2
l/2
Fl K sin
l 2
M
M0
M
1 cos P
l/2
Fl tan 2 2
l 2
y
cos
2
F
l/2
l 2
2 cos
P
P l/
sin
5
P
2 cos cos
3
2 sin
F
M
EI
8 1 2
l/2
sin ql 2 2 2 sin 2 ql
M l 2
ql 3
0
1
2
sec
4
P
l/2
l M
(Umansky 1973, vol.2)
Pinned-clamped beam-column
Simply-supported beam-column 1
P EI
l
l/2
1 cos M0 l sin cos sin M0 sin cos M 0l 2 sin 2 cos sin cos EI
sin 2
1 2
cos
776
Appendix
Table A.21 (continued) Beam-columns. Reactions and deflections; Free-clamped beam-column
7
Clamped-sliding beam-column
10
q P
q P
P
P
l
y 0
l
ql 2 cos
M l
2
ql 4
cos 4
EI
ql 3
0
EI
8
sin cos
1
sin cos
1
M 2
y
l 2
9
tan
Fl 3
tan
EI
Fl 2 EI
2
1 cos
y
1
2 EI
tan
3
cot
4
2 4
l 2
l/2
Fl tan 2 4 Fl tan 2 4 3 Fl tan 4 EI 3
4
Positive signs for reactions and displacements P l
0
l 2
M 0
1
P
y 0
ql 4
2
1
2
P
M
2
1
2
l/2
M0
M l
ql 2
csc
F
l
Fl
2
P
P
0
2
11
P
y 0
ql 2
M 0
sin cos
3
l 2
2
F
M l
P EI
l
M0 cos M 0l 2 2
EI M 0l tan EI
1 cos
1
1. Positive reactions are directed upward. 2. Positive moments extend the lower fibers. 3. Positive displacements are directed downward. 4. Positive angle of rotation at support coincides with positive direction of the moment at support.
Appendix
777
Table A.22 Special functions for stability analysis Functions
Form 1
Form 2
2
1
2
1 3 sin
tan
1 sin 4 2 2 cos
8 tan
tan
2
4 sin
tan
1
4
2
1
2 1
2 1
2
15 2
1
sin 1 2 2 2 cos sin
1
tan
tan
2
30 2
13 4 25200
60
3
2
sin cos
1 3 sin
3
1
cos cos
3 1 cos 1 12 2 sin cos
sin
sin
tan
cos sin
tan
525 11 4 25200
4
60
1
1
84000 2
2
sin cos
Numerical values of these functions in terms of dimensionless parameter
1
1
4
5
525
2
4
10
8400
2
sin
4
2
2
3 tan
2
1
cos sin
1 6 2 sin
3
2
sin cos
2
sin 3
2
tan 3 tan
Maclaurin series
7 4 360
6 2
4
3
45 4
0
2
3 are presented in Table A.23
778
Appendix
Table A.23 Special functions for stability analysis by displacement method 1(
0.0 0.2 0.4 0.6 0.8 1.0 1.1 1.2 1.3 1.4 1.5 /2 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0 5.2 5.4 5.6 5.8 6.0 6.2 2
)
1.0000 0.9973 0.9895 0.9856 0.9566 0.9313 0.9164 0.8998 0.8814 0.8613 0.8393 0.8225 0.8153 0.7891 0.7609 0.7297 0.6961 0.6597 0.6202 0.5772 0.5304 0.4793 0.4234 0.3621 0.2944 0.2195 0.1361 0.0424 0.0000 -0.0635 -0.1847 -0.3248 -0.4894 -0.6862 -0.9270 -1.2303 -1.6268 -2.1726 -2.9806 -4.3155 -6.9949 -15.330 227.80 14.669 7.8185 5.4020 4.1463 3.3615 2.3986 1.7884 1.3265 0.9302 0.5551 0.1700 0.0000
2(
)
1.0000 0.9980 0.9945 0.9881 0.9787 0.9662 0.9590 0.9511 0.9424 0.9329 0.9226 0.9149 0.9116 0.8998 0.8871 0.8735 0.8590 0.8437 0.8273 0.8099 0.7915 0.7720 0.7513 0.7294 0.7064 0.6819 0.6560 0.6287 0.6168 0.5997 0.5691 0.5366 0.5021 0.4656 0.4265 0.3850 0.3407 0.2933 0.2424 0.1877 0.1288 0.0648 -0.0048 -0.0808 -0.1646 -0.2572 -0.3612 -0.4772 -0.7630 -1.1563 -1.7481 -2.7777 -5.1589 -18.591 -
3(
)
1.0000 1.0009 1.0026 1.0061 1.0111 1.0172 1.0209 1.0251 1.0298 1.0348 1.0403 1.0445 1.0463 1.0529 1.0600 1.0676 1.0760 1.0850 1.0946 1.1050 1.1164 1.1286 1.1417 1.1559 1.1712 1.1878 1.2057 1.2252 1.2336 1.2463 1.2691 1.2940 1.3212 1.3508 1.3834 1.4191 1.4584 1.5018 1.5501 1.6036 1.6637 1.7310 1.8070 1.8933 1.9919 2.1056 2.2377 2.3924 2.7961 3.3989 4.3794 6.2140 10.727 37.308 +
4(
)
1.0000 0.9992 0.9973 0.9941 0.9895 0.9832 0.9798 0.9751 0.9715 0.9669 0.9619 0.9620 0.9566 0.9509 0.9448 0.9382 0.9313 0.9240 0.9164 0.9083 0.8998 0.8909 0.8814 0.8716 0.8613 0.8506 0.8393 0.8275 0.8225 0.8153 0.8024 0.7891 0.7751 0.7609 0.7457 0.7297 0.7133 0.6961 0.6783 0.6597 0.6404 0.6202 0.5991 0.5772 0.5543 0.5304 0.5054 0.4793 0.4234 0.3621 0.2944 0.2195 0.1361 0.0424 0.0000
1(
)
1.0000 0.9840 0.9362 0.8557 0.7432 0.5980 0.5131 0.4198 0.3181 0.2080 0.0893 0.0000 -0.0380 -0.1742 -0.3191 -0.4736 -0.6372 -0.8103 -0.9931 -1.1861 -1.3895 -1.6040 -1.8299 -2.0679 -2.3189 -2.5838 -2.8639 -3.1609 -3.2898 -3.4768 -3.8147 -4.1781 -4.5727 -5.0062 -5.4903 -6.0436 -6.6968 -7.5058 -8.5836 -10.196 -13.158 -27.781 +221.05 7.6160 0.4553 -2.2777 -3.8570 -4.9718 -6.6147 -7.9316 -9.1268 -10.283 -11.445 -12.643 -13.033
2(
)
1.0000 0.9959 0.9840 0.9641 0.9362 0.8999 0.8789 0.8557 0.8307 0.8035 0.7743 0.7525 0.7432 0.7100 0.6747 0.6374 0.5980 0.5565 0.5131 0.4675 0.4198 0.3701 0.3181 0.2641 0.2080 0.1498 0.0893 0.0207 0.0000 -0.0380 -0.1051 -0.1742 -0.2457 -0.3191 -0.3951 -0.4736 -0.5542 -0.6372 -0.7225 -0.8103 -0.9004 -0.9931 -1.0884 -1.1861 -1.2865 -1.3895 -1.4954 -1.6040 -1.8299 -2.0679 -2.3189 -2.5939 -2.8639 -3.1609 -3.2898
Appendix
779
Table A.24 Frequency equation, eigenvalue and nodal points for one-span uniform beams, k 4 Type of beam
Frequency equation
Eigenvalue
n
m 2 EI
Nodal points = x / l of mode shape X
#
n
1
Pinned-pinned
sin knl
1 2 3
3.14159265 6.28318531 9.42477796
0; 1.0 0; 0.5; 1.0 0; 0.333; 0.667; 1.0
2
Clamped-clamped
cos knl cosh knl
1 2 3
4.73004074 7.85320462 10.9956079
0; 0; 0;
1.0 0.5; 1.0 0.359; 0.641;
1.0
1 2 3
3.92660231 7.06858275 10.21017612
0; 0; 0;
1.0 0.440; 1.0 0.308; 0.616;
1.0
1 2 3
1.87510407 4.69409113 7.85475744
0 0; 0;
0.774 0.5001;
1, 2 3 4
0 4.73004074 7.85320462
Rigid-body modes 0.224; 0.776 0.132; 0.500; 0.868
1 2 3
0 3.92660231 7.06858275
Rigid-body mode 0; 0.736 0; 0.446; 0.853
3
4
Pinned-clamped
Clamped-free
0
1
tan knl tanh knl
0
cos knl cosh knl
1
5
Free-free
cos knl cosh knl
6
Pinned-free
tan knl tanh knl
1
0
0.868
kl,
Table A.25 Amplitude values of dynamical reactions of uniform beams with distributed mass m0 ;
t
sin t
t
sin t M(l)
M0 m0, EI l Q0
Bending moment M(0)=M0
Bending moment M(l)
4 EI f3 l
2 EI f5 l
(t)
f3 2
TU 4U 6 EI
(t)
l f4
1 (Kiselev 1980)
Shear Force Q(l)
Shear force Q(0)
2
SV TV
f5
l
SU T 4U
2
2
f8
TV
(t)
3EI
(t)
l
2
12 EI
6 U
l
2
Psin t a
UV 3 TU
3
3
U
f7
TV
3EI l
2
TV
l
ST UV
f13
P U klVka VklU ka k U kl2 TklVkl
P
3
U
12 U
2
TV
2
ST SV
f12
l3 S2 SV
T
3 TU
3EI
f13
U klU ka
TV
3EI f12 l2
U2 3 TU
U kl2
f9
f11
2
ST SV
2
12 EI f9 l3
UV 3 TU
3
U
6 U
f7
2
f11
3EI
P U klVka VklU ka k U kl2 TklVkl
2
2
f6
12 U 2 TV
f11
2
f11
4U
f6
l2
SU T 2
f4
f8
2
2
6 EI
f4
2
l
V 2 U 2 TV 6 EI
f4
2
6 EI
3EI f10 l T2 V2 f10 3 TU SV
3
5
1,
Q(l)
1
4
are amplitudes of angular and linear displacements,
,
f14 2
f14
TklVka
P
TklVkl
SV
S
3 TU
U klU ka U kl2
SV
TklVka TklVkl
Notes: 1. In all cases the span of a beam and mass per unit length are l and m0, respectively 4
2
2.
In all cases the corrective functions fi, has argument
3. 4.
All Krulov-Duncan functions S,T,U,V contain index , which is for short is omitted ; Each reaction contain factor sin t ; since this factor is omitted, then tabulated data should be considered as
kl, k
m0
EI
their amplitude values. More detail Table may be found in books (Kiselev 1980; Karnovsky and Lebed 2004a). Example Clamped-clamped beam AB of length l is subjected to angular harmonic excitation
t
of support B. Mass per unit length of a beam is m0. Determine the amplitude reaction of support. Solution. According case 1 of Table A.25and reciprocal reaction theorem we get: 2EI 6EI 4EI 6EI MA f 5 , QA f6 , M B f 3 , QB f4 2 l l l l2 B
A MA
MB
(t) QA
QB
sin t
Appendix
781
Table A.26 One span uniform beams subjected to harmonic excitation Psin M0
Design diagram Psin t
M0
Q0
at middle point of a span (a=l/2) Q0 ,
0
M0
P U klVka VklU ka k U kl2 TklVkl
Q0
P
U klU ka TklVka
M0
P Vkl Tka TklVka k S klVkl TklU kl
Q0
P
S klVka U klTka S klVkl TklU kl
Q0
P
S klVka TklU ka S klVkl TklU kl
U kl2 TklVkl
a Psin t
M0
Q0
a
Psin t
a 0
0
Q0 a
Psin t
Q0
P U klVka VklU ka EIk 2 S klVkl TklU kl
S ka P 2 2 2 S ka U ka
0
Q0
U ka P 1 2 2 2 2 EIk S ka U ka
Notes: 1. In all cases the span of a beam and mass per unit length are l and m0, respectively 2. Characteristic parameter k
4
m0
2
EI
3. Index ka and kl for all functions S,T,U,V means argument of corresponding function; 4. Each response (M0, Q0,
0
contain factor sin t ; since this factor is omitted, then tabulated data should be
considered as their amplitude values. More detailed Tables may be found in (Bezykhov et al. 1987; Kiselev 1980).
782
Appendix
Table A.27 Amplitude values of dynamical reactions of massless uniform beams with one lumped mass m.
EI
const
1 1
2
2
;
t
sin t ,
t
sin t ,
are amplitudes of angular and linear
,
frequency of excitation (kinematical -cases 1-6, forced cases frequency of free vibration, displacements; 8,9). General scheme shows positive reactions (Kiselev 1980) M0
M(l)
m a
b=l-a
Q0
Bending moment M(0)=M0
Shear force Q(0)=Q0
Functions
Q(l)
1
2
6 EI
4 EI F1 l
(t)
(t)
2
l
2 EI F2 l
2
6EI l2
F1
1
F5
1
2
l2
4
F3
12EI l3
F2
1
F6
1
2
F7
6 EI l2
F4
12EI l3
1
6
4a
F4
1
3EI l2
2
1
F9
1
F12
2
F12
1
F10
1
l2
F10
l3
2
2
F13
7
2
F13
1
2 2
(t) 3EI l2
F11
3EI l3
F11
1
2
F14
2
F14
1
2
8 Psin t
PF15
PF16
9
Psin t
PF17
PF18
F18
1
1 3
F9
1
2
3l 3 EI ma 3b 3
2
16 7
2
2b 4b
F10
1
40 7
3a b a 3 3a 4b
F13
1
256 7
4l
3
2
2l b 3a 4b
8 7
F14
48 7
3
6l ab 3a 4b
ab 2
2
ab b 1 2l l 3a 2
a3
2
3
2l
2l
F18
and for clamped pinned element
2
12l 3 EI mb2 a 3 3a 4b
2
2
2 2
2
2
1 1
2
2
1
2
1
2
2
1
F17
2
1
1 21
F16
3
2
2
l 81
F15
l2 l
1
1
2
2
F11 1
(*) The lumped mass m and parameters a and b of it location on the beam appear in the formulas for frequency of free vibration. For clamped clamped element
2
2l 2 3a a 2 3a
2
2
2 2
b 2 3a b
1 2
F8
2
40 7
2
2
1
1
1
1
F17
F4
2
F12
2
1
1 5
2b 4b
1
F16
F7
2l 2 3a a 2 3a
2
1
F5 2
3
3l 4ab
1
F15
F3 2
4l a 3a 4b
2
(*)
2
2
2
2
3EI
1
2
1 2
2
2
(t) 3EI
F6
l 2b
2
F8
2
3EI F9 l
1
2
2
b 3a b
1
2
F8
(t)
F2
F5
F7
1 3
l
2 2a
2
2 2
F5
2
1 2 (*) 2 2
2
(t)
5
2a
2
7 4
F1 1
3ab b 2
1
2
F6
F3 6 EI
3b 4a
1
2
F5
(t)
3
Special case a=b=l/2
2
2
3l 16
11 16
Appendix
783
Table A.28 The exact dynamic reactions of uniform beams with distributed mass due by kinematic harmonic excitation of the ends supports t 1 sin t, t 1 sin t, and forced excitation Psin . The length of a beam i=EI/l,
is l, distributed mass per unit length m0, the bending stiffness is EI=const, the frequency excitation 4
dimensional parameter u
2 4
m0
l
EI (Darkov 1989)
Design diagram 1
Amplitude values of moments
MA
MB
MA MB
1 QA
MA
MB 1
MB
MA
MA
1
6i l 6i l
u
5
u
6
u
12i
QA 6
l2 12i l2
u
QB
3i
1
u
QA QB
QA
4
5
QB
QB
QA
3
3
u u
2
6i l 6i l
QA
QB
MA
2
4i 2i
Amplitude values of shear
QB
MA MA
1
3i l
7
u
QA
QB
QA
3i l 3i l
10
u
11
u
4
u
7
u
l2 3i
9
u
l2
12
u
8
u
9
u
3i
QB
5
MA
MA
1
6i l
4
u
QA QB
QA
6
l2 3i l2
QB
Psin θt
MA l/2 QA
3i
QB
MB
MA
MB
Pl 8
1P
u
QA
QB
P 8
2P
u
784
Appendix
The correction functions are 2 sinh u sin u u ; 1 u 3 coshu sin u sinh u cosu 3
u
5
u
7
u
9
u
11
u
1P
u
u sinh u sin u ; 2 1 coshu cosu u2 sinh u sin u ; 6 1 coshu cosu u2 sinhu sin u ; 3 coshu sin u sinhu cosu
8
2
u
4
u
6
u
u
u coshu sin u sinh u cosu ; 4 1 coshu cosu u 2 coshu sin u sinh u cosu ; 2 coshu sin u sinh u cosu u 2 coshu cosu ; 6 1 coshu cosu u2 2 coshu cosu ; 3 coshu sin u sinhu cosu
cosh u cosu u3 ; 10 u 3 coshu sin u sinh u cosu u 3 sinhu sin u ; 12 u 12 1 coshu cosu u u cosh cos 2 2 2 ; 2P u u u u u cosh sin sinh cos 2 2 2 2 2
u 3 coshu sin u sinh u cosu ; 12 1 cosh u cosu 1 cosh u cosu u3 ; 3 cosh u sin u sinh u cosu u u 2 sinh sin 2 2 u u u u u cosh sin sinh cos 2 2 2 2
Numerical values of these function may be found in (Klein 1972).
Appendix
785
Table A.29 Approximate dynamical reactions of uniform beam due by unit displacements of the ends supports . The length of a beam is l, distributed mass per unit length m0, the bending stiffness is EI, the i=EI/l, k=m0 2l3 (Smirnov 1947; Bolotin 1964) frequency vibration
1
1
1 1
1
1
786
Appendix
Table A.30 Complete elliptic integrals of the first and second kinds Elliptic integrals of the first and second kinds, K and E, are given in two equivalent forms: as a function of the
arcsin m .
module m, as well as an angle function 2
First kind
2
d
K 1 sin
0
2
sin
K
2
2
1 sin 2
E
1 m2 sin 2
0
2
Second kind
d
sin 2
d
1 m 2 sin 2
E
0
d
0
Complete elliptic integrals of the first and second kinds Abramovich and Stegun (1964) 2
m
sin
K
2
0
0 5 10 15 20 25 30 35 40 45
0.0000 0.0872 0.1736 0.2589 0.3420 0.4226 0.5000 0.5735 0.6428 0.7071
2
d 1 m sin
1.5708 1.5738 1.5828 1.5981 1.6200 1.6490 1.6858 1.7312 1.7868 1.8541
2
1 m 2 sin 2
E 0
1.5708 1.5678 1.5689 1.5442 1.5238 1.4981 1.4675 1.4323 1.3931 1.3506
d
Bibliography
General Textbooks and Manuals Anokhin, N. N. (2000). Structural mechanics (part 2) Statically indeterminate systems. Moscow: ACB Publishers. Bezykhov, N. I., Lyzhin, O. V., & Kolkynov, N. V. (1987). Stability and structural dynamics. Moscow: Vysshaya Shkola. Bychkov, D. V. (Ed.). (1961). Manual to practical exercises in structural mechanics. Moscow: Gosstrojizdat. Craig, R. R., Jr. (2000). Mechanics of materials (2nd ed.). New York: Wiley. Darkov, A. V. (Ed.). (1989). Structural mechanics. Moscow: Mir Publishers. Feodosiev, V. I. (2016). Mechanics of materials (16th ed.). Moscow: Nauka. Ghali, A., & Neville, A. M. (2017). Structural analysis. A unified classical and matrix approach (7th ed.). Boca Raton: CRC Press. Karnovsky, I. A., & Lebed, O. I. (2010). Advanced methods of structural analysis (1st ed.). New York: Springer. Kassimali, A. (2014) Structural analysis (5th ed.). Thomson-Engineering. Kiselev, V. A. (1960). Structural mechanics. Moscow: Stroizdat. Kiselev, V. A. (1980). Structural mechanics, special course: Dynamics and stability (3rd ed.). Moscow: Stroizdat. Klein, G. K. (Ed.). (1980). Manual for structural mechanics. Statics of bar systems. Moscow: Vysshaya Shkola. Klein G.K., Rekach, V.G., Rozenblat, G.I. (1972). Manual for structural mechanic. Stability, structural dynamics, and spatial systems. (2nd ed.). Moscow: Vysshaya Shkola. Leet, K. M., Uang, C. M., & Gilbert, A. M. (2008). Fundamentals of structural analysis (3rd ed.). New York: McGraw-Hill. Nelson, J. K., & McCormac, J. C. (2003). Structural analysis: Using classical and matrix methods (3rd ed.). New York: Wiley. Parvanova, S. (2011). Structural analysis II. Sofia: University of Architecture, Civil Engineering and Geodesy. Rabinovich, I. M. (1960). A course of structural mechanics. Part 1 (1960), part 2 (1954). Moscow: Gosstrojizdat. Rzhanicyn, A. R. (1982). A structural mechanics. Moscow: Vysshaya Shkola. Smirnov, A. F. (1947). Static and dynamic stability of structures. Moscow: Transzheldorizdat. Smirnov, A. F., Aleksandrov, A. V., Lashchennikov, B. Y., & Shaposhnikov, N. N. (1984). Structural mechanics, dynamics and stability of structures. Moscow: Stroizdat. Timoshenko, S. P., & Gere, J. M. (1972). Mechanics of materials. New York: Van Nostrand Reinhold Company. Utku, S., Norris, C. H., & Wilbur, J. B. (1990). Elementary structural analysis. New York: McGraw-Hill College. Wissmann, J., & Sarnes, K. D. (2009). Finite Elemente in der Strukturmechanik. Berlin: Springer.
Handbooks Abramovich, M., & Stegun, I. (1964). Handbook of mathematical functions with formulas, graphs, and mathematical tables. Cambridge: Cambridge University Press. Birger, I. A., & Panovko, Y. G. (Eds.). (1968). Strength, stability, vibration. Handbook (Vol. 1–3). Moskow: Mashinostroenie. Blevins, R. D. (1979). Formulas for natural frequency and mode shape. New York: Van Nostrand Reinhold. Chelomey, V. N. (1978–1981). Vibration in technic handbook (Vol. 1–6). Moscow: Mashinostroenie. Elishakoff, I. (2020). Handbook on Timoshenko-Ehrenfest beam and Uflyand-Mindlin plate theories. Singapore: World Scientific. Harris, C. M., & Piersol, A. G. (2002). Harris’ shock and vibration handbook (5th ed.). New York: McGraw-Hill. Ivovich, V. A. (1981). Transfer matrices in the dynamic of elastic systems. Handbook. Moscow: Mashinostroenie. Karnovsky, I. A., & Lebed, O. I. (2001). Formulas for structural dynamics. New York: McGraw-Hill. Karnovsky, I. A., & Lebed, O. I. (2004a). Free vibrations of beams and frame. Eigenvalues and eigenfunctions. New York: McGraw-Hill Engineering Reference. Karnovsky, I. A., & Lebed, O. (2004b). Non-classical vibrations of arches and beams. Eigenvalues and eigenfunctions. New York: McGraw-Hill Engineering Reference. Young, W. C., Budynas, R. G., & Sadegh, A. M. (2012). Roark’s formulas for stress and strain (8th ed.). New York: McGraw-Hill. Umansky A. A. (Ed.). (1973). Handbook of designer (2nd ed. Vol. 1 (1972). Vol. 2 (1973)). Moscow: Strojizdat.
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Related List of Literature Babakov, V. M. (1965). Theory of vibration (2nd ed.). Moscow: Nauka. Blekhman, I. I., Myshkis, A. D., & Panovko, J. G. (2005). Applied mathematics. Subject, logic, features of approaches with examples from mechanics (3rd ed.). Moscow: URSS. Bokaian, A. (1988). Natural frequencies of beams under compressive axial loads. Journal of Sounds and Vibration, 126(1), 49–65. Bolotin, V. V. (1964). The dynamic stability of elastic systems. San Francisco: Holden–Day. Bondar, N. G. (1971). Nonlinear autonomous problems of the mechanical elastic systems. Kiev: Budivelnik. Nonlinear Autonomous Systems of the Structural Mechanics, Moscow: Strojizdat, 1972. Cook, R. D., Malkus, D. S., & Plesha, M. E. (1989). Concepts and applications of finite element analysis (3rd ed.). New York: Wiley. Clough, R. W., & Penzien, J. (1975). Dynamics of structures. New York: McCraw Hill. Feodos’ev, V. I. (1975). Ten lectures-conversations on the strength of materials (2nd ed.). Moscow: Nauka. Filin, A.P. Applied mechanics of rigid deformable body, (mechanics of materials, continuous media theory, and structural mechanics), vol. 1–3 (1975/1978/1981). Moscow: Nauka. Filippov, A. P. (1970). Vibration of deformable systems (3rd ed.). Moscow: Mashinostroenie. Galef, A. E. (1968). Bending frequencies of compressed beams. Journal of the Acoustical Society of America, 44(8), 643. Hayashi, C. (1964). Nonlinear oscillations in physical systems. New York: McGraw Hill Book Company. Karnovsky, I. A. (2012). Theory of arched structures. Strength, stability, vibration. Berlin: Springer. Karnovsky, I. A., & Lebed, E. (2016). Theory of vibration protection. Berlin: Springer. Kauderer, H. (1958). Nichtlineare Mechanic. Berlin: Springer. Kazakevych, M. I., Kulyabko, V. V. (1994). Complex Study of Dynamics and Aerodynamics of Long-span Pipelines and Bridges. East European Conference on Wind Engineering, Warsaw, 1994. Korenev, B. G., & Reznikov, L. M. (1993). Dynamic vibration absorbers-theory and technical applications. New York: Wiley. Lojtsyansky, L. G., Lurie A. I. (1983). Course of theoretical mechanics (Vol. 1, 2, Vol. 3 (1934)). Moscow: Nauka. Majid, K. I. (1974). Optimal design of structures. Hoboken: Wiley. Newland, D. E. (1989). Mechanical vibration. Analysis and computation. New York: Longman Scientific and Technical. Nowacki, W. (1963). Dynamics of elastic systems. New York: Wiley. Panovko, Y. G., & Gubanova, I. I. (1973). Stability and oscillations of elastic systems: Modern concepts, paradoxes and errors. New York: NASA Technical Report Server. Perelmuter, A. V., & Slivker, V. I. (2003). Numerical structural analysis. Models, methods and pitfalls. Berlin: Springer. Popov, E. P. (1986). Theory and analysis of flexible elastic rods. Moscow: Nauka. Popov, E. P. (1973). Applied theory of control processes in the nonlinear systems. Moscow: Nauka. Shabana, A. A. (1991). Theory of vibration. Discrete and continuous systems (Vol. II). Berlin: Springer. Thomson, W. Y. (1981). Theory of vibration with applications (2nd ed.). Upper Saddle River: Prentice Hall. Timoshenko, S., Young, D. H., & Weaver, W. (1974). Vibration problems in engineering (4th ed.). New York: Wiley. Timoshenko, S. P. (1972). In Grigoljuk (Ed.), A course of elasticity theory (2nd ed.). Kiev: Naukova Dymka.
History Benvenuto, E. (1991). An introduction to the history of structural mechanics. New York: Springer-Verlag. Bernshtein, S. A. (1957). Essays in structural mechanics history. Moscow: Gosstrojizdat. Elishakoff, I. (2019). Who developed the so-called Timoshenko beam theory? Mathematics and Mechanics of Solids, 25(1), 108128651985693. Endzhievsky, L. V., & Tereshkova, A. V. (2013). History of crashes and catastrophe. Krasnoyarsk: Siberian Federal University. Nikolai, E. L. (1955). On Euler’s works on the theory of longitudinal bending. Proceedings on Mechanics. Moscow: Gostexteoretizdat. Oden, J. T. (1990). Historical comments on finite elements. In S. G. Nash (Ed.), A history of scientific computing (pp. 152–166). New York: ACM Press. Timoshenko, S. P. (1953). History of strength of materials. New York: McGraw Hill. Todhunter, I., & Pearson, K. (1960). A history of the theory of elasticity and of the strength of materials (Vol. I and II). New York: Dover Publications. (Originally published by the Cambridge University Press in 1886 and 1893).
Index
B Babakov, I.M., 595 Belyaev, N.M., 703 Bendixen, A., 367 Bernoulli, D., 746 Bernoulli, J., 36 Bertrand, J.L., 226 Bespalov, 211 Betti, E., 250 Blekhman, I.I., 744 Bokaian, A., 695–697 Bolotin, V.V., 620, 680–690, 706, 785 Bondar, N.G., 739 Boussinesq, J., 553 Bresse, J.A.C., 211, 292, 299, 589, 674 Bubnov, I.G., 732
C Castigliano, C.A., 187, 219, 223, 250 Cauchy, A.L., 117–118, 128 Chebyshev, P.L., 9, 10, 131 Chelomey, V.N., 703, 705–707, 715 Clapeyron, B.P.E., 187, 212, 285, 291, 747 Clebsch, A., 367 Cook, R.D., ix Coriolis, G.G., 701, 702 Cotterill, J.H., 223 Coulomb, C.A., 747 Cremona, L., 76 Culmann, K., 71, 299
D D’Alembert, J.R., 616, 727, 731 Den Hartog, J.P., 707 Dinnik, A.N., 559 Dirac, P.A.M., 712 Duffing, G., 729, 732 Duhamel, J.M.C., 615, 620, 621, 626, 631 Duncan, 590
E Ehrenfest, P., 691 Euler, L., 746, 748
F Feodosiev, V.I., 540, 715, 735 Filin, A.P., 223 Filippov, A.P, 702 © Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8
Föppl, L., 109 Fourier, J.B.J., 588–589 Frahm, H., 707
G Galef, A.E., 693, 696–697, 713 Galerkin, B.G., 732 Galileo, G., 745, 746 Gauthey, É., 747 Gerber, G.H., 45 Green, G., 621 Gvozdev, A.A., 253, 405
H Hamilton, W.R., 748 Helmholtz, H., 250, 748 Henneberg, L., 99 Heppel, 291 Hooke, R., 745 Howe, 65, 72 Huygens, C., 172
I Ince, 705 Inglis, C.E., 698
J Jourawski, D.I., 68, 71
K Kirpichev, V.L., 325 Krylov, A.N., 698 Krylov, N.M., 590
L Lagrange, J.L., 36, 747, 748 Laissle, F., 291 Lamè, G., 212, 291, 747 Legendre, A.M., 735, 736 Leibniz, W., 746 Leverrier, U.J.J., 571 Lévy, M., 299, 325 Leytes, S., 367 L’Hôpita, 601 Lukin, 322, 324, 325 Lyapunov, A.M., 748 789
790 M Mahistre, M., 211 Majid, K.I., 744 Malkus, D.S., ix Mandelshtam, L.I., 703 Manderla, H., 367 Mariotte, E., 746 Mathieu, É.L., 704, 705, 713 Maupertuis, P.L., 226 Maxwell, J.C., 76, 748 Menabrea, L.F., 226, 748 Mises, R., 730 Mohr, O.C., 211, 252, 748 Molinos, L., 291 Morgaevsky, A.B., 698 Müller-Breslau, H., 37, 211 Myshkis, A.D., 744
N Navier, C.L., 291, 367, 722, 746, 747 Newland, D.E., 621, 698, 702, 703, 705
O Ormondroyd, J., 707 Ostenfeld, A., 367 Ostrogradsky, M.V., 36
P Panovko, Ja.G., 744 Parvanova, S., 99 Perelmuter, A.V., 744 Plesha, M.E., ix Poisson, S.D., 747 Popov, E.P., 734 Prandtl, L., 730 Pronnier, C., 291
R Rabinovich, I.M., 299 Rayleigh, J.W., 250, 252, 253, 341, 589, 674, 748 Rippenbejn, Ya.M., 299 Ritter, A., 71, 73, 74, 78, 80, 81, 86, 87, 92, 93, 98, 101
Index S de Saint-Venant, Barre, 748 Salimbeni, L., 133 Schallenkamp, A., 698 Schübler, A., 291 Schwedler, J.W., 61, 71, 109, 125–128, 137 Semikolenov, G., 45, 209, 274, 299 Shchapov, N.P., 246 Simpson, T., 241–243, 334, 340 Slivker, V.I., 744 Smirnov, A.F., 605–607, 614 Stokes, G.G., 698 Strutt, J.W., see Rayleigh, J.W.
T Taylor, W.P., 76 Timoshenko, S.P., 291, 589, 691–693, 713
U Umansky, A.A., 231, 324
V Vereshchagin, A.K., 188, 240, 242–246, 254, 302, 327, 332, 647, 648, 748
W Warren, 65 Wichert, 70, 99, 101 Willis, R., 698 Winkler, E., 32, 33, 299, 367
Y Yasinsky, F.S., 102, 106 Young, T., 357, 746
Z Zhemochkin, B.N., 299
Index
A Absolute maximum of bending moment, 32–34, 43 Ancillary diagram displacement–load (Z-P), 453, 456–464, 471, 481 internal forces–deformation (S-e), 453, 457–458, 476, 491 joint- load (J-L), 453–456, 471, 473, 474, 478 Arches askew, 148–151 Boussinesq equation, 553 differential equations, 551, 552 geometry parameters, 133 hingeless, 327, 421–427, 551, 558, 579 influence lines, 35 nil points, 145–148 rational, 560 three-hinge, 131–157, 239, 551, 552, 559, 560 two-hinged, 131, 326, 327, 551, 558–560 Asssumptions Navier, 746 Saint-Venant, 748
B Beam-columns differential equation, 740 deflection, 721 Beams Bernoulli-Euler, 589, 693 continuous, 46, 282, 289, 291, 342, 349, 371, 373, 414–421, 432–438, 443, 458, 470–480, 540–551, 604, 605, 655, 675 with elastic supports, 196, 370, 542, 698 free vibration equation, 609–618, 621, 622, 625–630, 651, 653, 657, 659, 661 Gerber-Semikolenov, 45, 46, 55, 209 influence lines, 25, 51–53, 55, 56, 398, 434, 443, 447, 449 Timoshenko-Ehrenfest, 691 universal equation, 253 Betti theorem, 250–252 Bolotin functions, 682 Boundary condition, 15, 31, 162, 165, 168, 190–193, 195–197, 204–205, 207, 223, 253, 291, 400, 443, 524, 531–533, 539, 541, 542, 551, 552, 554, 556, 559, 563, 569, 587–588, 591, 593, 600, 604, 608, 609, 612, 615, 617, 624, 627, 649, 650, 652, 655, 657, 670, 671, 690, 691, 693–697, 699, 701, 703, 704, 712, 713, 716, 720, 721, 726, 728, 743 Boussinesq equation, 553 Bresse formula, 361 Bubnov-Galerkin procedure, 701, 728, 731 Buckling, 523–525, 532, 533, 551–553, 556, 696, 747
© Springer Nature Switzerland AG 2021 I. A. Karnovsky, O. Lebed, Advanced Methods of Structural Analysis, https://doi.org/10.1007/978-3-030-44394-8
C Cable arbitrary load, 160, 161 catenary, 160, 172, 178–181 differential equation, 167 direct problem, 160, 162, 172 inverse problem, 160, 162, 172 Castigliano theorem, 188, 216–223, 254 Cauchy-Clebsch condition, 193 Cauchy theorem, 117–118, 128 Change of temperature, 46, 62, 187, 227, 232–235, 254, 305, 347–352, 354, 356–358, 385, 386, 395, 451, 455, 459, 491 Chebyshev formula, 131 Clapeyron theorem, 284–291 Comparison of, 23 redundant and determinate structures, 358 Compressed force conservative, 559 critical, 524–532, 534–536, 538–552, 555–564, 693, 696, 702, 718, 722, 737, 738 nonconservative, 524 Concomitant vibration, 626 Conjugate beam method, 188, 204–211, 246, 249, 254, 258, 284 Connecting line, 34, 35, 56, 79–81, 85–88, 94, 97, 142–143, 146–148 Constraint absolutely necessary, 274, 275, 353 conventionally necessary, 357 redundant, 5, 6, 274–276, 299, 325, 356–358, 638 Constructive scheme, 65, 67, 741 Continuous beams change of temperature, 45, 62, 187, 451, 455, 459, 521 foci method, 361, 647 influence lines, 36, 37, 39, 40, 54 plastic analysis, 516–518 settlement of supports, 235 stability equation, 562 Convolution integral, 621 Coordinates global, 452, 453, 457, 468–470, 472, 475, 478, 483, 485, 491, 492 local, 451, 452, 461, 465–468, 470, 472, 474, 476, 482, 485, 487–489, 491, 492 Critical load (Load unfavorable position), 3, 29, 30, 32, 524–532, 534–536, 538–552, 555–564, 737, 738, 747 Critical load (stability) critical load parameter, 538, 542–548, 558, 563 Critical speed of moving load, 713 Cutting panel, 310
791
792 D Deflection computation beam-columns, 723 Castigliano theorem, 254 conjugate beam method, 204–210 double integration method, 189–192, 243, 253 elastic load method, 188, 232, 246–249, 254, 267 graph multiplication method, 187, 188, 240–243 initial parameters method, 253 Maxwell-Mohr integral, 228, 229, 232, 233, 235, 240, 241, 249, 253, 254, 262, 267, 319, 430, 748 statically indeterminate structures, 338–340 work-energy method, 215, 226, 253, 259 Deflection types angular, 17, 187 linear, 216, 232 mutual angular, 217, 229, 232, 233, 265, 266 mutual linear, 187, 217, 228, 233 Degrees of freedom Chebyshev formula, 9, 10, 131 for dynamical analysis, 569 for static analysis, 690 for stability analysis, 656, 703, 733, 742 Degree of indeterminacy static, 273, 296, 308, 332, 437, 439, 568, 656, 675, 712 kinematical, 367–371, 385, 389, 393, 431, 442–447, 451, 453, 454, 461, 469, 470, 474, 482, 491, 657, 658, 660, 664, 668 mixed, 405–411 Design diagrams deformable, 523 non-deformable, 54, 182–184 Dimensions of ordinates of influence lines, 19, 28, 32, 38–40, 52, 53, 78, 80, 98, 105, 144, 145, 153, 417, 421, 426, 434, 440, 442, 445, 447, 479, 480, 498, 719 unit displacements, 414, 430 unit reactions, 373 Dirac function, 712 Displacement method canonical equations, 371, 373, 374, 377, 378, 382, 383, 387, 389, 392, 431, 432, 482, 542–544, 557, 604, 607, 618, 657, 658, 660, 668, 677, 678, 682, 684, 723, 725 conception, 367 influence lines, 413–450 kinematical indeterminacy, 367, 368, 370, 453, 658, 660, 664 matrix form, 451 primary systems, 369–371, 382, 387–389, 392, 406, 432, 437, 439, 451, 453, 455, 456, 465, 467, 469–471, 478, 480, 491, 492, 540, 543, 544, 557, 579, 604, 605, 607, 618, 656, 658, 660, 664, 677, 680, 724 primary unknowns, 367, 371, 378, 400, 432, 438–440, 451, 472, 491, 492, 656, 677 settlements of supports, 367, 388, 453, 456 stability analysis, 540, 542–545, 549, 551, 553, 559, 656 Duffing equation, 732 Duhamel integral, 620–626, 631 Dummy load method actual state, 229–231 unit state, 228–231
Index E Elastic center, 309–315, 329, 330, 423 curves, 187–203, 207, 212, 216, 223, 224, 245, 253, 255–258, 268, 277, 284, 296, 297, 302, 304, 306, 308, 350, 360–363, 368, 370, 372, 374, 377, 378, 383, 389, 392, 394, 395, 398, 400–403, 415, 421, 442–445, 447, 450, 453, 524, 531–533, 541–543, 546, 550, 552, 569, 579, 581, 583, 587, 615, 622, 624, 626, 631, 634, 665, 671, 678, 680, 682, 694, 699, 700, 716, 717, 719, 721, 727, 730, 734, 736 joints, 68, 210, 232, 246, 247, 249, 267, 304, 306, 308, 315, 361, 368, 370, 374, 383, 392, 400, 403, 429, 430, 445, 523, 528, 529, 543, 550, 557, 583 load method, 188, 232, 246, 247, 249, 267, 429, 430, 445 supports, 68, 190, 196, 225, 226, 257, 258, 316, 356, 357, 359, 360, 365, 370, 400, 489, 524–539, 547–548, 551, 559, 561, 566, 594, 595, 597, 615, 617, 655, 704, 707, 742 Envelope diagrams, 31–32 Errors of fabrication, 187, 235–240, 273, 276, 279, 357, 358, 367, 385, 386, 388, 392–394 Euler critical load, 704 Excitation forced, 640, 675 kinematical, 346, 620, 631, 632, 634, 636, 640, 641, 649, 655, 656, 669, 707, 742
F Failure, 46, 523, 525, 723 Fictitious beams, 188, 204–210, 247–249, 253, 284–286, 428, 430 Finite element method change of temperature, 521 concept, 491 fixed loads, 491 influence lines, 495 settlement of supports, 494 Flexibility matrix, 385, 490, 577, 586 Focus, 292–294, 693 Föppl dome, 109 Forced vibrations, 346, 385, 566, 600, 615–690, 703, 704 Force method canonical equations, 278, 279, 282, 283, 287, 299, 302, 305, 310, 313, 315, 319, 320, 322, 324, 325, 331, 336, 341, 343, 347, 351–354, 357, 365, 413, 414, 423, 442 changes of temperature, 276, 279, 347–351, 367, 385, 451, 491, 503 conception, 134 degree of redundancy, 327 errors of fabrication, 273, 276, 353, 357, 367 matrix form, 278 primary system, 273–284, 287, 288, 291, 298–300, 303–312, 315, 317, 319, 320, 322, 323, 325, 327, 328, 330–333, 336–344, 347–350, 352, 354–357, 359, 360, 363, 365, 367, 369–392 primary unknowns, 371, 374, 387, 389, 392 settlements of supports, 389, 390 static indeterminacy, 273, 309 symmetrical unknowns, 423 Fourier method, 588–589 Frames stability, 540–551, 557, 558 statically determinate, 57, 302 statically indeterminate, 61, 281, 302, 304, 367, 368, 371, 373, 387, 388, 394, 405, 431, 439 symmetrical, 394 plastic analysis, 516–518
Index with sidesway, 542, 545 three-hinged, 267, 275 Frequency of free vibration displacement methods, 680, 688 force method, 617–618, 636–648
G Generalized coordinates, 216–218, 220, 223–225, 254, 525–531, 568, 580, 639, 641, 645–648, 657, 748 Generalized forces, 216–224, 226, 233, 235–237, 253, 254, 340, 345, 748 Graph multiplication method bending members, 460 Green integral, 621 Group unknowns method, 664–667 Gvozdev’s theorem, 253
H Homogeneous function, Euler’s theorem, 218 Henneberg method, 101–103, 106, 119, 124
I Indirect load application, 15, 34–35, 41, 55–57, 80, 139, 142, 146–148, 155, 448 Influence lines method analytical construction, 442, 449 application, 25–34, 40, 144–145 connecting line, 34, 79, 80, 85–87 indirect load application, 41, 55–57, 142 kinematical method, 35–40, 43, 54, 95–99, 413 models, 35, 38–40, 95, 413, 442–445 nil points, 145–148, 153 Initial parameters methods, 535–539, 541, 594–600, 613, 615, 648–655, 657, 668, 669, 671, 674, 686, 687, 690, 717, 719–723 Interaction schemes, 46, 56, 61, 63, 209, 447 Ince-Strutt diagram, 705–706, 713
K Kinematical analysis degrees of freedom, 453 geometrically changeable structures, 4, 7 geometrically unchangeable structures, 52 infinitesimally changeable structure, 4 infinitesimally rigid structure, 4 redundant constrains, 5–6, 10 Kinematical method for influence lines, 35–40, 43, 54, 95–99, 413, 442–447 Krulov-Dunkan functions, 780
L Least work theorem, 223–226 Lévy-Kirpichev theorem, 325 Load conservative (nonconservative), 524 critical, 29, 30, 32, 524–526, 528, 529, 531, 532, 534, 536, 538–540, 543–548, 550, 551, 555–560, 562, 563, 699, 702, 703, 713, 737, 747 dynamical, 16, 187, 211, 566, 623, 697, 698 fixed (static), 16, 48, 96, 211, 400, 472, 524, 526, 528, 529, 620–626, 635, 644, 667, 672, 698
793 moving, 15, 16, 19, 23, 25, 28, 29, 31–34, 36, 38, 40–42, 51, 78, 80, 83, 84, 95, 103, 151, 413, 414, 416, 420, 427, 431, 432, 434, 440, 442, 444, 445, 447, 450, 476–480, 566, 620, 697–702, 713 seismic, 566, 620 tracking, 551, 559, 560 unfavorable, 16, 28–33, 40, 718 Load path, 46–48, 51, 61, 82–83, 87, 89 Load unfavorable position, 29 Loss of stability (buckling), 523–525, 532, 533, 551–553, 555, 556, 696, 747 Lukin formula, 322
M Matrices ancillary, 451, 463 deformations, 461, 491 flexibility, 385, 490, 577, 586 modal, 572–575, 578, 581–586, 611, 612 static, 451, 461, 491 stiffnesses, 461, 466, 467, 470, 472, 474, 476, 482, 486, 487, 489–491, 569 Matrix stiffness methods, 451–494, 586 Maxwell-Cremona diagram, 76–78 Maxwell-Mohr integral, 430 Maxwell theorem, 723 Menabrea principle, 223–226, 261, 269 Method displacements, 16, 96, 97, 99, 188, 193–196, 200, 202, 204, 207, 210, 211, 216, 226–228, 232, 235, 240, 244, 246, 249, 253–255, 269, 274, 279, 280, 283, 302, 305, 313, 319, 341, 351, 353, 365, 367–371, 373, 374, 376–378, 380, 382–384, 386, 387, 389, 391, 394–396, 400, 405, 406, 408–410, 413, 429, 431–440, 445, 451, 453, 456, 468–470, 473, 480, 482–484, 490–492, 525–528, 540–551, 570, 579–581, 587, 604, 605, 607, 614, 615, 618, 647, 648, 655–657, 659, 660, 668, 675, 678, 680, 682, 684, 689, 721, 723, 725, 746 dummy load (Maxwell-Mohr integral), 430 elastic loads, 188, 232, 246, 247, 249, 255, 429, 430, 442, 445 forces, 11, 12, 15, 16, 37, 54, 71, 73, 74, 76, 78, 83, 95–99, 101, 103, 105, 118, 119, 122–125, 129, 157, 216, 226, 228, 232, 247, 254, 273, 274, 276, 278, 279, 282–284, 287, 289, 291, 294, 298, 299, 302, 305, 307, 310, 313, 315, 322, 324, 325, 327, 331, 336, 340, 341, 343, 347, 351–354, 357, 360, 365, 367, 376, 378, 380, 384, 386, 387, 389, 391, 394–396, 400, 405, 406, 408, 410, 413, 414, 423, 429, 430, 432, 435, 437, 438, 442, 450, 470, 490, 491, 525, 526, 528, 529, 538, 541, 542, 544, 570, 579, 581, 587, 605, 615, 630, 639, 645, 652, 653, 655–657, 659, 669, 671, 675, 689, 719, 721, 723, 742, 748 graph multiplication, 187, 188, 240, 242, 244, 254, 264, 280, 305, 416, 444 influence lines, 16–25, 35–40, 54, 63, 78, 83, 95–99, 103, 105, 145–148, 157, 298, 367, 384, 413, 414, 420, 431–440, 442, 445, 449–451, 630, 687, 698 initial parameters, 188, 193, 194, 196, 200, 202, 204, 249, 253, 256, 301, 376, 378, 536, 600, 605, 649, 650, 652, 655, 659, 669, 671, 689 mixed, 410, 411 strain energies, 253 Mises truss, 730 Missed frequencies, 608–609 Mixed method, 253, 405–410 Modal matrix, 571–579, 581–586, 611, 612 Mode shape vibration, 576, 688
794 Mohr integral, 188, 228, 229, 232, 233, 235, 240, 241, 243, 249, 254, 255, 262, 267, 319 Müller-Breslau’s principle, 37, 54, 442–447 Multispan redundant beams (continuous beams), 348, 389, 414, 470–476 Multispan statically determinate beams influence lines, 52, 53 structural presentation, 46
N Nil points method, 145–150, 153–155, 157 Nonlinearities constructive, 716 geometrical, 740 physical, 716, 740 statics, 716, 740
O Orthogonality conditions, 575, 579, 586, 731
P Paradox of critical load, 539–540 P-Δ effect, 724 Plastic analysis direct method, 504–506 kinematical theorem, 508 schemes of failure, 516 static theorem, 508 Plastic hinge, 716 Prandtl diagram, 730 Primary system displacement methods, 369–373, 376, 378, 382, 388, 389, 392, 397, 406, 437, 438, 441, 453–456, 465, 467, 469–471, 478, 540, 543, 546, 549, 605, 607, 618, 656, 664, 677, 680, 724 force method, 279, 287, 298, 299, 332, 340, 341, 347, 348, 385, 387, 396, 397, 406, 407, 409, 420, 437, 540 mixed method, 406–408 Primary unknown displacement method, 406, 408 force method, 274, 276, 279, 283, 302, 319, 320, 323, 325, 332, 336, 352, 413, 416, 427, 437–439, 442, 604, 617, 658 mixed method, 406–408 Principle Menabrea, 226, 748 Müller-Breslau, 37, 442, 445 Saint-Venant, 748 superposition, 215, 278, 371, 380, 383, 391, 650, 715, 726, 733 virtual displacements, 35, 236, 238, 239, 747, 748
Q Quasi-static loading, 698–701, 713
R Rayleigh theorem, 250, 252, 748 Reciprocal theorems displacements, 657 displacements and reactions, 249–253, 406, 641 reactions, 656 works, 250–251
Index S Sag, 161–164, 166, 169–172, 174, 175, 181–186 Schwedler dome, 107, 109, 125–128 Separation variable method, 589 Settlements of supports, 46, 62, 155, 187, 188, 200–202, 235–240, 273, 276, 291, 305, 340–346, 357, 358, 385, 386, 389–392, 395, 399, 451, 456, 459, 470, 473–476, 491, 494, 521, 656, 687, 766 Shchapov formula, 246 Simpson rule, 241, 340 Smirnov functions, 614 Space trusses attached, 113 meshwork, 107, 122 statics, 107 Stability methods double integration, 717–719 energy, 526 initial parameter, 535–539, 594 static, 524, 677 Statically determinate arches, 326 beams, 25, 45, 52, 205, 254, 274, 280, 281, 283, 587, 609 frames, 57, 60, 61 trusses, 68–70, 96, 103, 107, 114, 318–322 Statically indeterminate arches, 326, 327, 382, 386, 427 beams, 196, 225, 276, 281, 299, 387, 437, 451, 470, 511, 567, 609 frames, 61, 274, 275, 305, 367, 368, 371, 387, 405, 439, 610 trusses, 318–325, 352, 353, 414, 427–431, 442, 484, 497, 748 Steady-state vibration, 615, 620–630, 632, 633, 635–637, 639–641, 643, 645, 648, 652–658, 664, 668, 669, 671, 675, 677, 679, 680, 682, 686, 688, 690, 702, 710, 712 Stiffness matrix global coordinates, 468–470, 475, 478, 483, 491 local coordinates, 451, 461, 465–468, 470, 472, 482, 485, 488, 489, 491, 492 Strain energy methods, 211–226, 253 Structure geometrically changeable, 4, 7, 8, 10, 84, 97, 99, 126, 206, 274, 275, 358, 368, 461, 567 geometrically unchangeable, 4, 8, 9, 11, 12, 36, 37, 47, 48, 54, 58, 69, 75, 84, 90, 93, 95, 99, 101, 112–115, 117, 118, 120–122, 124, 131, 274, 358, 368, 567 infinitesimally changeable, 4, 6, 11, 21, 98, 115, 511 infinitesimally rigid, 4 statically determinate, 15, 26, 40, 41, 45, 47, 52–54, 58, 65, 68–70, 72, 75, 83, 84, 90, 93, 95, 99, 101–103, 107, 112–114, 117, 118, 120–122, 124, 131, 132, 148, 742 statically indeterminate, 48, 110, 187, 202, 206, 224, 232, 273–276, 279, 287, 290, 305, 309, 310, 319, 336, 338, 340, 341, 345, 347, 353, 356, 358, 367, 386, 414, 416, 420, 427, 441, 442, 447, 461, 463, 486, 568, 632, 742 Substitute bar method, 99–102 Superposition principle, 26, 100, 101, 124, 261, 276, 277, 280, 284, 298, 303, 339, 344, 352, 358, 385, 395, 397, 489, 570, 631, 633, 635–638, 640, 645, 649, 656, 659, 666, 668, 717, 719, 723, 740 Symmetrical frames, 314, 317, 394–397, 610, 612, 614
T Temperature changes, 232, 235, 273, 276, 279, 347–356, 367, 386, 453, 455 Three moment equations, 282, 284–291, 293, 349, 360
Index
795
Thrusts, 66, 90–95, 103, 105, 132, 134–136, 138–145, 149–153, 155, 156, 159–167, 169–176, 178, 179, 181–186, 239, 327, 329, 331, 332, 335, 424, 426, 726, 727 Timoshenko-Ehrenfest beam, 691–693, 713 Transfer matrices, 600–604, 613 Trapezoid rule, 241, 242, 244, 635 Tributary area, xxxi, xxxii, xxxiv, xxxvii Trusses general fundamental conception, 65 kinematical analysis, 6, 69, 82, 103, 118 statically indeterminacy, 319, 323 Trusses generation complex, 65, 66, 70, 99–103, 106, 107, 119, 123 compound, 65, 66, 69–70, 82, 83, 87, 107, 115, 122 simple, 66, 68, 69, 71–78, 81–85, 96, 98, 99, 103, 119, 120, 122 with subdivided panels, 65, 66, 81–90, 104 Trusses analysis classical methods, 71, 76, 78 force method, 318–322, 347–366 elastic load method, 232 influence lines, 78–81 Trusses specified types Howe truss, 65, 72 K-truss, 75, 104 Pratt truss, 81 three-hinged, 65, 66, 90–92, 99 Warren truss, 65 Wichert truss, 70, 99, 101
force method, 226, 232, 334 mixed method, 405–407 Unknown antisymmetrical, 310, 312, 330, 395, 396 displacement methods, 368, 376, 386–388, 392, 396, 406, 431, 432, 437–441, 453, 464, 472, 480, 491, 492, 543, 544, 548, 557, 604, 658, 660, 677 force method, 386, 406, 617 mixed method, 405, 406 symmetrical, 175, 310, 312, 314, 321, 330, 395, 396, 491, 665
U Umansky formula, 231, 324 Unit displacements computations, 280, 281, 583, 647 dimensions, 373, 385 Unit reactions computations, 583, 678 dimensions, 468 Unit state displacement methods, 579
W Winkler rule, 32–34 Work external forces, 211–215, 747 internal forces, 67, 211–215, 316, 747, 748
V Vereshchagin’s rule Simpson rule, 241 trapezoid rule, 241 Vibration methods displacement method, 565, 571, 579–587, 604–607, 614 Krylov-Duncan, 590–595, 597, 599, 601, 605, 612 Vibration types concomitant, 626 free, 565–614 forced, 346, 385, 566, 600, 615–690 steady state, 615, 620–630, 632, 633, 635–637, 639–641, 643, 645, 648, 652–658, 664, 668, 669, 671, 675, 677, 679, 680, 682, 686, 688, 690, 702, 710, 712 transient, 619, 624 Virtual displacement principle, 36, 95
Y Yasinsky’s approach, 102, 106