Structural Analysis of Plane Frames: Solved Examples with Force and Displacement Methods (Springer Tracts in Civil Engineering) 3031352661, 9783031352669

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Springer Tracts in Civil Engineering

Isabella Giorgia Colombo · Matteo Colombo · Marco di Prisco · Anna Magri · Paolo Martinelli · Letizia Mazzoleni · Giulio Zani

Structural Analysis of Plane Frames Solved Examples with Force and Displacement Methods

Springer Tracts in Civil Engineering Series Editors Sheng-Hong Chen, School of Water Resources and Hydropower Engineering, Wuhan University, Wuhan, China Marco di Prisco, Politecnico di Milano, Milano, Italy Ioannis Vayas, Institute of Steel Structures, National Technical University of Athens, Athens, Greece

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Isabella Giorgia Colombo · Matteo Colombo · Marco di Prisco · Anna Magri · Paolo Martinelli · Letizia Mazzoleni · Giulio Zani

Structural Analysis of Plane Frames Solved Examples with Force and Displacement Methods

Isabella Giorgia Colombo DSC-Erba s.r.l. Erba, Italy Marco di Prisco DSC-Erba s.r.l. Erba, Italy Department of Civil and Environmental Engineering Politecnico di Milano Milan, Italy Paolo Martinelli Department of Civil and Environmental Engineering Politecnico di Milano Milan, Italy

Matteo Colombo Department of Civil and Environmental Engineering Politecnico di Milano Milan, Italy Anna Magri Leviat s.r.l. Bergamo, Italy Letizia Mazzoleni Cremonesi Workshop s.r.l. Brescia, Italy

Giulio Zani Department of Civil and Environmental Engineering Politecnico di Milano Milan, Italy

ISSN 2366-259X ISSN 2366-2603 (electronic) Springer Tracts in Civil Engineering ISBN 978-3-031-35266-9 ISBN 978-3-031-35267-6 (eBook) https://doi.org/10.1007/978-3-031-35267-6 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed 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

Preface

This book contains a wide range of worked examples concerning the solution of statically indeterminate frame structures. The book is aimed at providing a useful tool for anyone wishing to learn more about the application of practical methods for the evaluation of internal forces/moments in plane frames. It is intended to be both a valuable reference guide for structural engineering professionals and an ideal learning resource for students of civil engineering, building engineering and architecture programs. This book has been written in the cultural tradition of the Civil Engineering School and Building Engineering/Architecture School of the Politecnico di Milano and is dedicated to Structural Design courses. The course of Structural Design is placed as logical development after the course of Structural Mechanics. It is therefore assumed that the student has taken university courses in Statics and/or Structural Mechanics. This textbook presents applied examples of the main methods of structural analysis of statically indeterminate frame structures. The textbook begins with a brief description of kinematic analysis for plane frames (Chap. 1). The force method, the displacement method and the mixed method are applied for the solution of statically indeterminate plane structures. The textbook first deals with the solution of simple reference cases in which recurring structural situations such as inclined rods, extensional and rotational springs, thermal variations, symmetry and anti-symmetry (just to name a few) are treated individually (Chap. 2). The textbook then presents the complete solution of complex plane frames, in which the common structural situations individually analyzed in the previous chapter are combined (Chap. 3). The tabular coefficients used for the force and displacement methods are collected in Appendix A, while Appendix B summarizes the simple reference cases covered in each complex frame example. The book, due to its peculiar nature, is suitable for both sequential and subject-specific reading.

v

vi

Preface

We believe that most of the literature devoted to the structural analysis of statically indeterminate frame structures covers mainly theoretical aspects. We decided to write this book with the primary aim of providing practical and relevant worked examples for real structures, with a focus on static schemes that are readily understandable to students and designers alike. Milan, Italy

Isabella Giorgia Colombo Matteo Colombo Marco di Prisco Anna Magri Paolo Martinelli Letizia Mazzoleni Giulio Zani

Acknowledgement

The authors acknowledge the assistance of students Andrea Baldi, Matteo Gianoli and Giulia Lopez in preparing some of the figures in the textbook.

vii

Contents

1 Short Notes on Kinematic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Centre of Instantaneous Rotation (CIR) . . . . . . . . . . . . . . . . . . . . . . .

1 1 1

2 Solution of Simple Reference Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Problem A(I): Inclined Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Problem A(II): Inclined Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Problem B(I): Extensional Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Problem B(II): Extensional Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Problem C(I): Rotational Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Problem C(II): Rotational Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Problem D(I): Thermal Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Problem D(II): Thermal Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Problem E(I): Symmetry and Anti-symmetry . . . . . . . . . . . . . . . . . . 2.10 Problem E(II): Symmetry and Anti-symmetry . . . . . . . . . . . . . . . . . 2.11 Problem E(III): Symmetry and Anti-symmetry . . . . . . . . . . . . . . . . 2.12 Problem F(I): Imposed Displacement/Rotation . . . . . . . . . . . . . . . . . 2.13 Problem F(II): Imposed Displacement/Rotation . . . . . . . . . . . . . . . . 2.14 Problem G(I): Infinitely Rigid Bending Rod . . . . . . . . . . . . . . . . . . . 2.15 Problem G(II): Infinitely Rigid Bending Rod . . . . . . . . . . . . . . . . . . 2.16 Problem H: Concentrated Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17 Problem I: Statically Determined Portion . . . . . . . . . . . . . . . . . . . . . 2.18 Problem J: Unusual Restraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.19 Problem K(I): Distributed Loads/Moments . . . . . . . . . . . . . . . . . . . . 2.20 Problem K(II): Distributed Loads/Moments . . . . . . . . . . . . . . . . . . . 2.21 Problem L: Rod with Finite Axial Stiffness . . . . . . . . . . . . . . . . . . . .

7 7 19 27 35 42 53 59 65 71 75 82 92 98 103 106 111 116 122 127 132 137

3 Solution of Complex Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Worked Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Worked Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Worked Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Worked Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

143 143 158 164 172 ix

x

Contents

3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 3.30

Worked Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Worked Example 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

179 185 191 197 203 210 216 222 228 233 238 245 253 259 265 272 279 285 290 295 299 307 313 319 325 331

Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 Appendix B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

Chapter 1

Short Notes on Kinematic Analysis

1.1 Redundancy The force and displacement methods are used to solve statically indeterminate structures. Therefore, the first step to be performed is to count the degrees of freedom (DOF) and the number of restraints (NOR). Table 1.1 provides the number of restraint reactions for the most common restraint types.

1.2 Centre of Instantaneous Rotation (CIR) If the structure turns out to be statically determined it is solved by the equilibrium equations alone. If DOFs < NORs the structure is statically indeterminate. The kinematic analysis allows understanding which restraints are overabundant and have to be replaced by unknown reactions (statically indeterminate problem). A frame with displaceable joints (i.e. sway frame) is defined as a frame in which the joints, in addition to rotating, can also translate. If axial deformability is not neglected, every frame always results as sway frame. To examine if a frame is fixed-joint or sway, one must assess the degree of restraint of the structure by inserting a hinge at each inner and outer joint where rotation (absolute or relative) appears to be restrained, either it is perfectly restrained (rigid restraint) or only partially restrained (elastic restraint). The untied structure is called “associated truss structure”. If the associated truss structure is statically determinate or statically indeterminate, the frame has fixed joints. On the contrary, if the associated truss structure is hypostatic and allows for the possibility of rigid body motions, the frame is sway. The number of simple restraints to be introduced should be enough to lock any rigid body motion of the associated truss structure. To figure out which restraints to add, one must look for the centers of rotation of each member. It is necessary to downgrade all the flexural restraints, as shown in Table 1.2 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil Engineering, https://doi.org/10.1007/978-3-031-35267-6_1

1

2

1 Short Notes on Kinematic Analysis

Table 1.1 Restraint numbering

Some examples are given below. Example 1.1 Let’s consider the following structure:

After counting the number of restraints, it is concluded that the structure is statically indeterminate with redundancy grade equal to one. The downgrading of the restraints can be performed.

1.2 Centre of Instantaneous Rotation (CIR)

3

Table 1.2 Downgrading of restraints

The structure is free to move according to the kinematics shown in the figure below.

4

1 Short Notes on Kinematic Analysis

Rod number 3 is free to move horizontally, so the kinematic motion remains unchanged even considering only rods 1 and 2. The CIR of the entire structure can be found at the point of intersection of the action lines (dotted grey lines).

To prevent the kinematic motion, it is possible to block the vertical displacement of point B or, alternatively, the horizontal displacement of point D.

Example 1.2 Let’s consider the following structure:

The rod BC has an infinite flexural stiffness, so it transmits shear force and bending moment directly onto the slider without deforming. Therefore, the structure can be simplified as shown in the following figure.

1.2 Centre of Instantaneous Rotation (CIR)

5

Downgrading of the restraints is carried out. The structure turns out to be hypostatic and the rod rotates around the hinge A.

A roller is then introduced to prevent the vertical translation of point B.

Chapter 2

Solution of Simple Reference Cases

2.1 Problem A(I): Inclined Rod

• E I = constant. • EA → /(∞( l 2 • l∗ = + l2 = 2

√

5 l 2

Solution The structure is statically indeterminate, having 6 restraints and 3 degrees of freedom (NOR = 6 > DOF = 3). The kinematic analysis of the structure is then carried out; to do this, it is necessary to release the bending restraints. The member AB can be replaced with a roller with vertical axis, while member CD can be replaced with a roller with axis parallel to the member CD itself. In this way the member BC has a center of instantaneous rotation (CIR) indicated in the figure as CIR. Nodes B and C rotate with respect to the CIR and consequently the structure is a sway frame.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil Engineering, https://doi.org/10.1007/978-3-031-35267-6_2

7

8

2 Solution of Simple Reference Cases

The displacement method is used to solve the structure, adopting as hyperstatic unknowns the rotations ϕ1 and ϕ2 and the horizontal translation δ3 . An additional rotational restraint is introduced at nodes B and C (rigid block = ground spring with infinite rotational stiffness) and at point B an additional translational restraint is also added (roller with horizontal axis) such as to prevent the activation of the kinematics identified by the kinematic analysis. For each of the unknowns it is possible to write an equilibrium equation. In total, there will therefore be 2 equilibrium equations to the rotation and an equilibrium equation to the horizontal translation: ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0

2.1 Problem A(I): Inclined Rod

Kinematics • ϕ1 = 1

9

10

2 Solution of Simple Reference Cases

4E I 6E I 4E I + = l 2l l EI 2E I = m 21 = 2l l 21E I r31 = − 4l2 m 11 =

The reaction r31 is calculated by writing a rotational equilibrium equation of the whole structure with respect to the CIR. The member CD has no flexural deformation and consequently the shear force at the base D and the bending moment in D are zero. The axial forces in A and D have a null lever arm with respect to the CIR. In writing the rotational equilibrium, all the restraints reactions are considered, including those on the rigid blocks (moments), which represent an on-ground restraint. 4E I 4E I 2E I 6E I 2E I − − − =0 5l + r31 4l − 2 l l l 2l 2l 21E I r31 = − 4l2 • ϕ2 = 1

EI l ( √ ) √ 4E I 10 + 8 5 E I 4E I 8 5 EI + ∗ = 2+ = m 22 = 2l l 5 l 5 l √ 15 − 48 5 E I r32 = 20 l2 m 12 =

2.1 Problem A(I): Inclined Rod

11

• δ3 = 1

As a result of the global kinematics, node C undergoes a translation equal to δ. Using the trigonometric relationship: cos α =

l

√ 5 l 2

2 =√ 5

we have that: √ 1 5 δH = 2 = δ= √ cos α 2 5

Node C undergoes a vertical translation equal to δV . Considering that tan α =

1 l/2 = l 2

the following relationship is obtained: 1 δH = 2 6E I = δV 4l2 6E I = δV 4l2

δV = m 13 m 23

1 2 6E I 3E I 6E I = − 2 = 2 2 l 4l l √ 6E I 3 EI 12 5 − ∗2 δ = − l 4 l2 5 −

−21 E I 4 l2 √ EI 15 − 48 5 E I = l2 20 l2

For the calculation of the roller restraint reaction (r33 ), as before, the global rotational equilibrium equation of the whole structure is written:

12

2 Solution of Simple Reference Cases 12E I 12E I 6E I 6E I 6E I 6E I 5l + r33 4l − ∗3 δ5l∗ + 2 + ∗2 δ + 2 + ∗2 δ+ l3 l l l l l 6E I 6E I − 2 δV − δV = 0 4l 4l2 ( √ ) 495 + 192 5 E I r33 = 40 l3 −

It is worth noting that the displacements adopted are those referred to the orthogonal direction with respect to the direction of the member considered. • External load

1 m 10 = − pl2 3 1 2 m 20 = pl 3 1 r30 = pl 2 Calculation of unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 6E I 21 E I EI 1 − 4√l2 pl2 ϕ1 l l √ 3 ⎢ EI 10+8 5 E I 15−48 5 E I ⎥ ⎣ · ϕ2 ⎦ = ⎣ − 13 pl2 ⎦ ⎣ l 5 √ l 20 √ l2 ⎦ E I 15−48 5 E I 495+192 5 E I δ3 − 21 pl − 21 4 l2 20 l2 40 l3 ⎧ pl3 ⎪ ⎨ ϕ1 = 0.0448 E I 3 ⇒ ϕ2 = −0.0926 pl E I4 ⎪ ⎩ δ3 = −0.0300 pl EI

2.1 Problem A(I): Inclined Rod

13

The stiffness matrix of the solving system is symmetrical. This is due to the fact that we are using a pure solving method (the displacement method). Even in the case of a solution using the force method (also pure), a compatibility matrix characterized by symmetry should be obtained. On the contrary, in the case of the solution with the mixed method, the stiffness/compatibility matrix should be symmetrical in terms of absolute values, but antisymmetric if the signs are considered. Summary of Bending Moments Below is a summary of the bending moments acting on the structure, written with a positive or negative sign based on the convention defined in the figure. The superimposition of the effects is exploited for the writing of the internal forces.

Summary of Shear Forces Below is a summary of the shear forces acting on the structure, written with a positive or negative sign based on the convention defined in the figure.

14

2 Solution of Simple Reference Cases

Calculation of Shear Forces and Bending Moments • AB

2E I 6E I ϕ1 − 2 δ3 = 0.269 pl2 (internal stretched fibers) l l 4E I 6E I AB ϕ1 + 2 δ3 = −0.359 pl2 (internal stretched fibers) MB = − l l 6E I 12E I AB V A = − 2 ϕ1 − δ3 = −0.628 pl l l3 6E I 12E I VBAB = 2 ϕ1 − δ3 = −0.628 pl l l3 M AAB =

• BC

2.1 Problem A(I): Inclined Rod

15

4E I 2E I 6E I 4 pl2 ϕ1 + ϕ2 + = −0.359 pl2 (external stretched fibers) δV δ3 − 2 2l 2l 12 2l 2E I 4E I 6E I 4 pl2 ϕ1 − ϕ2 − = −0.170 pl2 (external stretched fibers) δV δ3 − MCBC = − 2 2l 2l 12 4l 6E I 6E I 12E I 2 pl ϕ2 − δV δ3 + V BBC = − 2 ϕ1 − = 1.094 pl 2 2l 2l2 2l3 6E I 6E I 12E I 2 pl = 0.906 pl ϕ2 − δV δ3 − VCBC = − 2 ϕ1 − 2 3 2 2l 2l 2l M BBC =

The abscissa x of maximum bending moment can be identified by equating to zero the derivative of the bending moment equation in the member BC :

M(x) = M BBC + VBBC · x − p · x ·

x px 2 = −0.359 pl2 + 1.094 plx − 2 2

d M(x) = 1.094 pl − px = 0 → x = 1.094l dx The abscissa x can also be determined by writing a proportion of similar triangles on the shear force diagram.

Being V (x) = d M(x) , the maximum of the function M(x) is obtained for x, such dx as V (x) = 0. Moreover, the following proportion applies: (|VB | + |VC |) : l = |VB | : x so it is possible to calculate x: x=

|VB | · 2l 1.094 pl · 2l = = 1.094l (|VB | + |VC |) (1.094 + 0.906) pl

16

2 Solution of Simple Reference Cases

For x = 1.094h, the value of bending moment is equal to: M(1.094l) = −0.359 pl2 + 1.094 pl(1.094l) −

p(1.094l)2 = 0.240 pl2 2

• CD

4E I 6E I ϕ2 − ∗2 δ · δ3 = −0.170 pl2 l∗ l 2E I 6E I CD M D = − ∗ ϕ2 + ∗2 δ · δ3 = 0.005 pl2 l l 6E I 12E I VCC D = − ∗2 ϕ2 + ∗3 δ · δ3 = 0.157 pl l l 6E I 12E I VDC D = − ∗2 ϕ2 + ∗3 δ · δ3 = 0.157 pl l l MCC D =

(external stretched fibers) (internal stretched fibers)

The axial force is calculated with translational equilibrium equations at the node. The shear forces in the rods are to be considered positive if they induce a clockwise rotation of the section considered. They are therefore positive if a configuration like the one reported in the figure is obtained:

2.1 Problem A(I): Inclined Rod

17

For the case under consideration, the following shear forces act on the rods:

The forces acting on the nodes have the opposite direction to those acting on the rods:

N AB = −VBBC = −1.094 pl N BC =

−VBAB 2VCBC

= −0.628 pl

NC D = √ + 5

(compression) (compression)

2 · 0.906 pl (−0.628 pl) N BC = = −1.091 pl (compression) + √ 2 2 5

18

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

2 Solution of Simple Reference Cases

2.2 Problem A(II): Inclined Rod

19

It is worth noting that the zero points of the bending moment correspond to the points of inflection (change of concavity) of the deformed shape.

2.2 Problem A(II): Inclined Rod

• E I = constant • EA → ∞ • l∗∗ = 23 l /( ( l 2 • l∗ = + (2l)2 = 2

√

17 l 2

Solution The structure is statically indeterminate. The kinematic analysis is carried out, releasing the flexural constraints and tracing the structure back to a rod with two rollers at the ends. The analysis shows that the structure is a sway frame. The displacement method is used to solve the structure by adopting the rotations ϕ1 , ϕ2 and the horizontal translation δ3 as kinematic unknowns. The center of instantaneous rotation is placed at infinity, so horizontal translation is allowed.

20

2 Solution of Simple Reference Cases

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

2.2 Problem A(II): Inclined Rod

m 11 m 21

21

√ 8E I 24 + 8 17 E I EI = √ + = = 4.607 √ 17 3l l l 3 17 l 2 2E I 4E I EI = √ =√ = 0.970 17 l 17l l 2 4E I

r31 = −

6E I 24E I =− ∗∗2 l 9l2

The roller restraint reaction (r31 ) is obtained from the equilibrium to the global horizontal translation of the whole structure. It is worth noting that the rod CD has no flexural deformation, so shear force and moment in D are zero. • ϕ2 = 1

m 12 =

2E I

m 22 =

4E I

√

17 l 2

√

r32 = −

17 l 2

6E I l2

= 0.970 +

EI l

EI 4E I = 5.940 l l

22

2 Solution of Simple Reference Cases

• δ3 = 1

24E I 9l2 6E I m 23 = − 2 l 96E I 12E I EI r33 = + = 15.56 3 3 3 27l l l m 13 = −

• External load

The resultant P of the distributed load p can be decomposed into the two components Q and Q' . Adopting the following trigonometric relations: 2l

cos α =

√ 17 l 2

sin α =

√ 17 l 2

4 =√ 17

l/ 2 = √1 17

2.2 Problem A(II): Inclined Rod

23

we have that: • Q = P cos α = • Q ' = P sin α =

√4 P 17 √1 P 17

The actions q and q' are derived by distributing the concentrated loads Q and Q' over the length l∗ : √4 P 16 8 p · 2l 17 √ = p = 17 17 l 17 l 2 √1 P 4 Q' 2 p · 2l = p = √17 = 17 l∗ 17 l 17 l 2

Q q= ∗ = l q' =

From the equilibrium to horizontal translation of the entire structure it results that r30 = 0, being the load p vertical and columns without bending deformation and, therefore, no shear.

m 10 = − m 20 =

∗2

ql =− 12

16 17

p·

(√

)2

17 l 2

12

=−

4 2 1 pl = − pl2 12 3

1 2 pl 3

r30 = 0 The moments m10 and m20 can finally be derived through the relations:

24

2 Solution of Simple Reference Cases

p · (2l)2 1 4 = − pl2 = − pl2 12 12 3 4 pl2 1 2 = = pl 12 3

m 10 = − m 20

This procedure, in which the projection of the length in the direction perpendicular to the load is used as the span of the beam, is valid only for the determination of the bending moments; the shear forces acting at the ends of the rod (essential for the calculation of the internal forces and moments) will have to be sought through the writing of equilibrium equations. Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎡ ⎤ ⎡ 1 2 ⎤ EI EI 24 E I ⎤ ⎡ 4.607 l 0.970 l − 9 l2 pl ϕ1 3 ⎣ 0.970 E I 5.940 E I − 6E2 I ⎦ · ⎣ ϕ2 ⎦ = ⎣ − 1 pl2 ⎦ l l l 3 6E I EI EI δ 0 − 24 − 15.56 3 2 2 3 l l ⎧9 l 3 pl ⎪ ⎨ ϕ1 = 0.07905 E I 3 ⇒ ϕ2 = −0.09065 pl E I4 ⎪ ⎩ δ3 = −0.02141 pl EI Summary of Bending Moments

2.2 Problem A(II): Inclined Rod

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

25

26

Qualitative Deformed Shape

2 Solution of Simple Reference Cases

2.3 Problem B(I): Extensional Spring

27

2.3 Problem B(I): Extensional Spring

• E I = constant • EA → ∞ EI • k = 10l 3 Solution In order to control the shortening/elongation of the translational spring placed at node D, it is necessary to introduce at that point a roller with a vertical reaction parallel to the direction of the spring.

Therefore, the kinematic analysis of the structure is carried out considering the presence of such a roller.

28

2 Solution of Simple Reference Cases

The structure turns out to be a sway frame. The allowable horizontal displacement is prevented by the introduction of the roller, with horizontal reaction applied at node C. The displacement method is used to solve the structure, adopting as kinematic unknowns the rotations ϕ 1 and ϕ 2 and the translations δ3 and δ4 (δ4 corresponds to the vertical translation of the node D constrained to the ground by the translational spring).

⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0 ⎪ r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0 ⎪ ⎩ r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0

2.3 Problem B(I): Extensional Spring

29

Kinematics • ϕ1 = 1

2E I 6E I 4E I + = l l l EI m 21 = l 6E I r31 = 2 l 3E I r41 = − 2 2l m 11 =

The reaction r31 is obtained by writing the horizontal translational equilibrium of the whole structure knowing the base shear force at node A and knowing that the base shear force at node D is zero, since member CD has no flexural deformation. The reaction r41 is obtained by the rotational equilibrium of the whole structure with respect to node A. In writing this equilibrium all the restraining reactions must be considered, including those arising on the additional restraints (moments on the rigid blocks and forces on the rollers): 4E I 4E I 2E I 2E I + + + − r31 l + r41 2l = 0 l l 2l 2l 6E I 9E I − 2 l + r41 2l = 0 l l 3E I r41 = − 2 2l

30

2 Solution of Simple Reference Cases

• ϕ2 = 1

EI l 3E I 5E I 2E I + = m 22 = l l l 3E I r32 = 2 l 3E I r42 = − 2 2l m 12 =

• δ3 = 1

2.3 Problem B(I): Extensional Spring

31

6E I l2 3E I m 23 = 2 l 15E I r33 = l3 r43 = 0 m 13 =

• δ4 = 1

3E I 2l2 3E I m 24 = − 2 2l r34 = 0 m 14 = −

For problems falling in the elasticity field, it is permissible to apply the principle of superposition to subdivide the reaction r44 into the two contributions due to the imposed settlement δ4 (r44 ' ) and to the presence of a spring with stiffness k (r44 '' ).

32

2 Solution of Simple Reference Cases

Exploiting the relationship of the elastic reaction F = k·x, with x the displacement vector, for δ4 = 1 we have that: FM = k · x = k · 1 = k The compressed spring tends to move the roller upward, giving rise to a restraining reaction with direction concordant to the displacement δ4 . In general terms, the restraining reaction is assumed to be positive if it has concordant direction to the one chosen for the related displacement. 3E I 2l3 =k

' r44 = '' r44

' '' + r44 = r44 = r44

3E I 3E I EI 8E I +k = + = 2l3 2l3 10l3 5l3

• External load

pl2 3 pl2 m 20 = 3 r30 = 0 m 10 = −

r40 = − pl

2.3 Problem B(I): Extensional Spring

Calculation of Unknowns ⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0 ⎪ r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0 ⎪ ⎩ r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0 ⎤ ⎡ ⎤ ⎤ ⎡ ⎡ ϕ1 m 10 m 11 m 12 m 13 m 14 ⎢ m 20 ⎥ ⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎣ r31 r32 r33 r34 ⎦ · ⎣ δ3 ⎦ = −⎣ r30 ⎦ r41 r42 r43 r44 δ4 r40 ⎡ ⎤ ⎡ 6E I ⎡ ⎤ ⎤ pl2 EI 6E I 3E I − − ϕ 2 2 1 l l l 2l ⎢ pl32 ⎥ 5E I 3E I I ⎥ ⎢ ⎢ EI − 3E ϕ ⎥ ⎥ l l2 2l2 ⎥ · ⎢ 2 ⎥ = −⎢ ⎢ l ⎢ 3 ⎥ ⎣ 6E2 I 3E2 I 15E3 I 0 ⎦ ⎣ δ3 ⎦ ⎣ 0 ⎦ l l l 3E I 3E I 8E I δ4 − 2l2 − 2l2 0 − pl 5l3 ⎧ pl3 ⎪ ϕ1 = 1.19749 E I ⎪ ⎪ 3 ⎨ ϕ2 = 0.84431 pl EI 4 ⇒ pl ⎪ ⎪ ⎪ δ3 = −0.64786plE4 I ⎩ δ4 = 2.53919 E I Summary of Bending Moments

33

34

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

2 Solution of Simple Reference Cases

2.4 Problem B(II): Extensional Spring

Qualitative Deformed Shape

2.4 Problem B(II): Extensional Spring

35

36

2 Solution of Simple Reference Cases

• E I = constant • EA → ∞ EI • k = 10l 3 Solution The structure is a sway frame.

The displacement method is used to solve the problem by adopting as kinematic unknowns the rotation ϕ1 of node B, the rotation ϕ2 of node C, the horizontal translations δ3 of node C, and the horizontal translations δ4 of node D. The horizontal translation δ4 is introduced in order to control the deformations of the extensional spring.

The solving system, consisting of 4 equilibrium equations, is the following: ⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0 ⎪ r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0 ⎪ ⎩ r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0

2.4 Problem B(II): Extensional Spring

37

Kinematics • ϕ1 = 1

2E I 6E I 4E I + = l l l EI m 21 = l 6E I r31 = 2 l r41 = 0 m 11 =

The last equation (r41 = 0) is obtained from local equilibrium at node D, not having shear force on member CD. • ϕ2 = 1

38

2 Solution of Simple Reference Cases

EI l 2E I 3E I 5E I m 22 = + = l l l 3E I r32 = 2 l 3E I r42 = 2 l m 12 =

• δ3 = 1

6E I l2 3E I m 23 = 2 l 15E I r33 = l3 3E I r43 = 3 l m 13 =

2.4 Problem B(II): Extensional Spring

39

• δ4 = 1

m 14 = 0 3E I m 24 = 2 l 3E I r34 = 3 l 3E I 3E I EI 31E I r44 = 3 + k = 3 + = 3 l l 10l 10l3 • External load

pl2 3 pl2 m 20 = 3 r30 = 0 m 10 = −

r40 = 0

40

2 Solution of Simple Reference Cases

Calculation of Unknowns ⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0 ⎪ r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0 ⎪ ⎩ r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0 ⎤ ⎡ ⎤ ⎤ ⎡ ⎡ ϕ1 m 10 m 11 m 12 m 13 m 14 ⎢ m 20 ⎥ ⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎣ r31 r32 r33 r34 ⎦ · ⎣ δ3 ⎦ = −⎣ r30 ⎦ r41 r42 r43 r44 δ4 r40 ⎡ ⎤ ⎡ 6E I E I 6E I ⎤ ⎡ ⎤ pl2 0 − ϕ 2 1 l l l ⎢ pl32 ⎥ ⎢ E I 5E I 3E2 I 3E2 I ⎥ ⎢ ϕ2 ⎥ ⎥ l ⎢ l l l ⎥ · ⎢ ⎥ = −⎢ ⎢ 3 ⎥ ⎣ 6E2 I 3E2 I 15E3 I 3E3 I ⎦ ⎣ δ3 ⎦ ⎣ 0 ⎦ l l l l 3E I 3E I 31E I δ4 0 l2 l3 10l3 0 ⎧ pl3 ⎪ ϕ1 = 0.1870 E I ⎪ ⎪ 3 ⎨ ϕ2 = −0.2440 pl E I4 ⇒ pl ⎪ ⎪ ⎪ δ3 = −0.0908plE4 I ⎩ δ4 = 0.3240 E I Summary of Bending Moments

2.4 Problem B(II): Extensional Spring

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

41

42

Qualitative Deformed Shape

2.5 Problem C(I): Rotational Spring

• E I = constant • EA → ∞ • k = E2lI

2 Solution of Simple Reference Cases

2.5 Problem C(I): Rotational Spring

43

Solution The structure turns out to be a sway frame.

The displacement method is used to solve the structure, adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C, ϕ3 of node C and the horizontal translation δ4 of joint B. The two unknowns ϕ2 and ϕ3 at the ends of the spring are introduced to know uniquely, in each of the cases considered, the rotation of the spring and, consequently, the moment exchanged between the spring and the structure.

⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0 ⎪ m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0 ⎪ ⎩ 31 1 r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0

44

2 Solution of Simple Reference Cases

Kinematics • ϕ1 = 1

2E I 6E I 4E I + = l l l EI m 21 = l m 31 = 0 6E I r41 = − 2 l m 11 =

• ϕ2 = 1

2.5 Problem C(I): Rotational Spring

45

EI l 5E I 2E I +k = m 22 = l 2l EI m 32 = −k = − 2l r42 = 0 m 12 =

The term k represents the moment exerted by the spring as a result of unit rotation; for rotational type springs we have that M = k · ϕ, where ϕ represents the change in angular opening. Clockwise (positive) unit rotation of rigid block 2 gives rise to a clockwise moment of amplitude k within the spring, of direction concordant with rotation ϕ2 . In compliance with the internal equilibrium equations, the spring will exert a counterclockwise, hence negative, moment of amplitude k on block 3. • ϕ3 = 1

m 13 = 0 m 23 = −k = −

9E I 4E I +k = l 2l 6E I =− 2 l

m 33 = r43

EI 2l

46

2 Solution of Simple Reference Cases

• δ4 = 1

m 14 = − m 24 = 0

6E I l2

6E I l2 24E I = l3

m 34 = − r44 • External load

pl2 3 pl2 m 20 = 3 m 30 = 0 r40 = 0 m 10 = −

2.5 Problem C(I): Rotational Spring

47

Calculation of Unknowns ⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0 ⎪ m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0 ⎪ ⎩ 31 1 r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0 ⎤ ⎡ ⎤ ⎤ ⎡ ⎡ ϕ1 m 10 m 11 m 12 m 13 m 14 ⎢ m 20 ⎥ ⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎣ m 31 m 32 m 33 m 34 ⎦ · ⎣ ϕ3 ⎦ = −⎣ m 30 ⎦ r41 r42 r43 r44 δ4 r40 ⎡ 2 ⎡ 6E I E I ⎡ ⎤ ⎤ 6E I 0 − − pl3 ϕ 2 1 l l l ⎢ pl2 ⎢ E I 5E I − E I ⎢ ⎥ 0 ⎥ 2l 2l ⎢ l ⎥ · ⎢ ϕ2 ⎥ = −⎢ ⎢ 3 E I 6E I 9E I ⎣ 0 − − l2 ⎦ ⎣ ϕ3 ⎦ ⎣ 0 2l 2l 6E I 6E I 24E I δ4 − l2 0 − l2 0 l3 ⎧ pl3 ⎪ ϕ1 = 0.1242 E I ⎪ ⎪ 3 ⎨ ϕ2 = −0.1765 pl E3 I ⇒ pl ⎪ ⎪ E I4 ⎪ ϕ3 = 0.0327 pl ⎩ δ4 = 0.0392 E I Summary of Bending Moments

⎤ ⎥ ⎥ ⎥ ⎦

48

2 Solution of Simple Reference Cases

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

Remarks: i. The bending moment at point C has a value included between zero (Case a) and the moment MC ’ that would occur in the case of a continuous frame (Case B). The bending continuity restraint can in fact be seen as a special case of rotational spring with k → ∞, while the hinge as a special case of rotational spring with k → 0.

ii. It is possible to verify the correctness of the calculated moment at point C by computing the moment in the spring (Mspring ): Mspring = k(ϕ3 − ϕ2 ) =

EI pl3 · 0.2092 = 0.105 pl2 2l EI

2.5 Problem C(I): Rotational Spring

49

Qualitative Deformed Shape

Mixed Method The mixed method allows the number of unknowns in the problem to be reduced to 3. The structure can be solved by adopting as kinematic unknowns the rotation ϕ1 of node B, the horizontal translation δ2 of node B and as static unknown the moment X3 of node C. In this case, the unknown X3 directly represents the moment exerted on the spring and, consequently, exchanged with the structure.

50

2 Solution of Simple Reference Cases

In this case, the solving system is represented by the following equations (an equilibrium equation at rotation referring to node B, an equilibrium equation at horizontal translation, again referring to node B, and a compatibility equation, referring to node C): ⎧ ⎨ m 11 ϕ1 + m 12 δ2 + m 13 X 3 + m 10 = 0 r ϕ + r22 δ2 + r23 X 3 + r20 = 0 ⎩ 21 1 ϕ31 ϕ1 + ϕ32 δ2 + ϕ33 X 3 + ϕ30 = 0 Kinematics • ϕ1 = 1

3E I 11E I 4E I + = l 2l 2l 6E I r21 = − 2 l ) ( 1 1 = (positive rotation if clockwise) ϕ31 = ϕright − ϕleft = 0 − − 2 2 m 11 =

2.5 Problem C(I): Rotational Spring

• δ2 = 1

6E I l2 12E I 3E I 15E I r22 = + 3 = 3 l l l3 3 3 ϕ32 = ϕright − ϕleft = −0= 2l 2l m 12 = −

• X3 = 1

51

52

2 Solution of Simple Reference Cases

1 2 3 =− 2l

m 13 = − r23

ϕ33 = ϕright − ϕleft

( ) 2l 3l 2l 11l 1 l − − + = + = = 4E I 4E I k 4E I EI 4E I

• External load

m 10 = −

1 4 pl2 = − pl2 8 2

r20 = 0 pl3 6E I

ϕ30 =

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 δ2 + m 13 X 3 + m 10 = 0 r ϕ + r22 δ2 + r23 X 3 + r20 = 0 ⎩ 21 1 ϕ31 ϕ1 + ϕ32 δ2 + ϕ33 X 3 + ϕ30 = 0 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ r21 r22 r23 ⎦ · ⎣ δ2 ⎦ = −⎣ r20 ⎦ ⎡

ϕ31 ϕ32 ϕ33

11E I 2l ⎣ − 6E2 I l 1 2

I − 6E l2 15E I l3 3 2l

− 21 3 − 2l 11l 4E I

X3 ϕ30 ⎤⎡ ⎤ ⎡ 1 2 ⎤ pl ϕ1 2 ⎥ ⎦⎣ δ2 ⎦ = ⎢ ⎣ 0 ⎦ 3 pl X3 − 6E I

2.6 Problem C(II): Rotational Spring

53

Note that the stiffness/flexibility matrix of the solving system is antisymmetric: terms arising from the displacement method have opposite signs to those arising from the force method. ⎧ pl3 ⎪ ⎨ ϕ1 = 0.1242 E I 4 ⇒ δ2 = 0.0392 pl EI ⎪ ⎩ X = −0.1046 pl2 3

2.6 Problem C(II): Rotational Spring

• E I = constant • EA → ∞ • k = E2lI Solution The structure is statically indeterminate and results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C, ϕ3 of node D and the horizontal translation δ4 of node B. The rotation unknown ϕ3 allows to uniquely control the rotation of the spring and consequently to uniquely identify the moment that the spring exchanges with the structure.

54

2 Solution of Simple Reference Cases

⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0 ⎪ m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0 ⎪ ⎩ 31 1 r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0 Kinematics • ϕ1 = 1

2E I 6E I 4E I + = l l l EI m 21 = l m 31 = 0 6E I r41 = − 2 (global equilibrium to horizontal translation) l m 11 =

2.6 Problem C(II): Rotational Spring

55

• ϕ2 = 1

EI l 4E I 6E I 2E I + = m 22 = l l l 2E I m 32 = l 6E I r42 = − 2 l m 12 =

• ϕ3 = 1

m 13 = 0 2E I m 23 = l 9E I 4E I +k = m 33 = l 2l 6E I r43 = − 2 l In the calculation of m33 , the term k represents the moment exerted by the spring as a result of unit rotation; for rotational type springs we have that M = k · ϑ, where

56

2 Solution of Simple Reference Cases

ϑ represents the change in angular opening. The clockwise (positive) unit rotation of block 3 gives rise to a clockwise moment of amplitude k within the spring, of direction concordant with rotation ϕ3 . • δ4 = 1

6E I l2 6E I m 24 = − 2 l 6E I m 34 = − 2 l 24E I r44 = l3 m 14 = −

• External load

2.6 Problem C(II): Rotational Spring

57

pl2 3 pl2 m 20 = 3 m 30 = 0 r40 = 0 m 10 = −

Calculation of Unknowns ⎧ m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0 ⎪ m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0 ⎪ ⎩ 31 1 r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 m 14 ⎢ m 20 ⎥ ⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎣ m 31 m 32 m 33 m 34 ⎦ · ⎣ ϕ3 ⎦ = −⎣ m 30 ⎦ r41 r42 r43 r44 δ4 r40 ⎡ 2 ⎡ 6E I ⎤ ⎤ ⎡ pl 6E I EI 0 − ϕ 2 1 l l l ⎢ 3pl2 6E I ⎥ ⎢ 6E I 2E I ⎢ EI ⎥ − l2 ⎥ ⎢ ϕ2 ⎥ ⎢ − 3 l l ⎢ l =⎢ 2E I I ⎦·⎣ 9E I ⎣ 0 − 6E ϕ3 ⎦ ⎣ 0 l 2l l2 I I I 24E I δ4 − 6E − 6E − 6E 0 l2 l2 l2 l3 ⎧ 3 pl ⎪ ϕ = 0.084175 E I ⎪ ⎪ 1 3 ⎨ ϕ2 = −0.07071 pl E 3I ⇒ ⎪ ϕ3 = 0.053872 pl ⎪ E I4 ⎪ ⎩ δ4 = 0.016835 pl EI Summary of Bending Moments

⎤ ⎥ ⎥ ⎥ ⎦

58

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

2 Solution of Simple Reference Cases

2.7 Problem D(I): Thermal Variation

Qualitative deformed shape

2.7 Problem D(I): Thermal Variation

• E I = constant • EA → ∞ pl2 • αΔT = 5E t I l • t = 10 Solution The structure is statically indeterminate and results to be a sway frame.

59

60

2 Solution of Simple Reference Cases

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B.

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

4E I 6E I 4E I + = l 2l l EI m 21 = l 6E I r31 = − 2 l m 11 =

2.7 Problem D(I): Thermal Variation

61

• ϕ2 = 1

EI l 4E I 6E I 4E I + = m 22 = 2l l l 6E I r32 = − 2 l m 12 =

• δ3 = 1

6E I l2 6E I m 23 = − 2 l 24E I r33 = l3 m 13 = −

• Thermal variation The trapezoidal thermal distribution can be decomposed into two contributions: one constant and one linearly variable along the thickness.

62

2 Solution of Simple Reference Cases

Therefore, the principle of superposition can be used to study the contribution of thermal variation by analyzing the structure subjected to each of the two contributions separately.

m 10 = 0 6E I 3 pl4 9 · = − pl2 l2 50E I 25 18 12E I 3 pl4 = pl =+ 3 · l 50E I 25

m 20 = − r30

m 10 = E I ·

4 pl2 1 4 pl2 = = pl2 20E I 20 5

1 m 20 = − pl2 5 r30 = 0

2.7 Problem D(I): Thermal Variation

63

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎧ ⎡ 6E I ⎡ ⎡ 1 2⎤ ⎤ ⎤ pl3 EI I ⎪ − 6E ϕ1 − 5 pl ⎨ ϕ1 = −0.076 E I l l l2 3 I ⎦ ⎣ 6E I ⎣ EI · ϕ2 ⎦ = ⎣ 14 − 6E pl2 ⎦ ⇒ ϕ2 = 0.076 pl l l l2 25 EI 4 ⎪ I I 24E I ⎩ δ3 − 18 pl − 6E − 6E δ3 = −0.030 pl 25 l2 l2 l3 EI

Summary of Bending Moments

Summary of Shear Forces

64

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

2 Solution of Simple Reference Cases

2.8 Problem D(II): Thermal Variation

65

Remark The displacement δ3 of node B is negative and, therefore, directed to the left. Since the structure is symmetrically loaded, the displacement of node C is equal to δ3 and directed to the right. The total elongation of rod BC, due to the effect of constant thermal variation, results to be equal to δT . The latter value results to be equal to two times δ3 .

2.8 Problem D(II): Thermal Variation

• E I = constant • EA → ∞ pl2 • αΔT = 5E t I l • t = 10 Solution The structure is statically indeterminate and results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B.

66

2 Solution of Simple Reference Cases

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

4E I 6E I 4E I + = l 2l l EI m 21 = l 6E I r31 = − 2 l m 11 =

2.8 Problem D(II): Thermal Variation

67

• ϕ2 = 1

EI l 4E I 6E I 4E I + = m 22 = 2l l l 6E I r32 = − 2 l m 12 =

• δ3 = 1

6E I l2 6E I m 23 = − 2 l 24E I r33 = l3 m 13 = −

68

2 Solution of Simple Reference Cases

• Thermal variation

3 3 pl4 6E I 9 6E I m 10 = − αΔT l · =− =− · pl2 2 2 4l 2 50E I 4l2 200 9 pl2 m 20 = − 200 r30 = 0

m 10 = − m 20 = 0 r30 = 0

pl2 αΔT · E I =− t 5

2.8 Problem D(II): Thermal Variation

69

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎧ ⎡ 6E I ⎡ ⎡ 49 2 ⎤ ⎤ ⎤ pl3 EI I ⎪ − 6E pl ϕ1 ⎨ ϕ1 = 0.0562 E I l l l2 200 3 I ⎦ ⎣ 6E I 9 ⎣ EI · ϕ2 ⎦ = ⎣ 200 − 6E pl2 ⎦ ⇒ ϕ2 = 0.0162 pl l l l2 E I4 ⎪ I I 24E I ⎩ δ3 0 − 6E − 6E δ3 = 0.0181 pl l2 l2 l3 EI

Summary of Bending Moments

Summary of Shear Forces

70

2 Solution of Simple Reference Cases

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

Note that, in rod AB, there is a bending (mechanical) moment stretching the right fibers; the resulting mechanical curvature is compensated for by the thermal curvature of that rod; globally, it results in a deformation of that rod as in the figure, with the thermal curvature prevailing over the mechanical curvature.

2.9 Problem E(I): Symmetry and Anti-symmetry

71

2.9 Problem E(I): Symmetry and Anti-symmetry

• E I = constant • EA → ∞ Solution The structure is statically indeterminate and results to be a sway frame.

Since the frame is symmetrically and symmetrically loaded, it is possible to analyze half structure by introducing, at the axis of symmetry, an appropriate constraint that takes into account the actual conditions of the overall structure. Such a constraint is represented by a slider with a sliding plane parallel to the axis of symmetry. The constraint should be applied at node C located on the axis of symmetry. Since this node is located at the top of an axially inextensible rod, it is possible to further simplify the structure by transforming the slider and the vertical connecting rod in a fixed support (see image below). The frame is solved by the displacement method by adopting the rotation ϕ1 as the only unknown.

72

2 Solution of Simple Reference Cases

m 11 ϕ1 + m 10 = 0 Kinematics • ϕ1 = 1

2.9 Problem E(I): Symmetry and Anti-symmetry

m 11 =

73

4E I 8E I 4E I + = l l l

• External load

m 10 = −

pl2 12

Calculation of Unknowns m 11 ϕ1 + m 10 = 0 8E I pl2 · ϕ1 − =0 l 12 l pl3 pl2 ϕ1 = · = 0.0104 12 8E I EI Summary of Bending Moments

74

2 Solution of Simple Reference Cases

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams The solution of the system allows to plot the diagrams of the internal forces and moments and to plot the deformation of half structure. In order to represent the drawings of the entire structure, the following observations—valid for any symmetrically and symmetrically loaded structure—can be drawn: (a) the deformation of the structure results to be symmetrical; (b) the moment and axial force diagrams are symmetrical; (c) the shear force diagram is anti-symmetrical, meaning that a clockwise shear at a generic point of the structure results to be counterclockwise at its symmetrical point and vice-versa.

2.10 Problem E(II): Symmetry and Anti-symmetry

Qualitative Deformed Shape

2.10 Problem E(II): Symmetry and Anti-symmetry

75

76

2 Solution of Simple Reference Cases

• E I = constant • EA → ∞ Solution The frame is symmetrical and anti-symmetrically loaded. Due to the symmetry of the structure and the anti-symmetry of the loads, it is possible to simplify the problem by analyzing half of the structure and introducing, at the node located on the axis of symmetry, a roller with an axis parallel to the axis of symmetry in order to keep the loading condition of the whole structure in consideration. In this case, the introduction of the roller is more formal than substantial, given the presence of the rod CF with infinite axial stiffness.

It should be noted that, in the study of the simplified problem, the bending stiffness of the rod CF must be halved, as the CF rod lies exactly on the axis of symmetry: (E I )C F =

EI 2

The structure results to be a sway frame, so we add, at node C, a roller with horizontal reaction and associated displacement unknown δ3 .

The displacement method is used to solve the structure, adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C, and the horizontal translation δ3 of node C.

2.10 Problem E(II): Symmetry and Anti-symmetry

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

4E I 8E I 4E I + = l l l 2E I m 21 = l 6E I r31 = 2 l m 11 =

77

78

2 Solution of Simple Reference Cases

• ϕ2 = 1

2E I l 4E I 6E I 4E I + = m 22 = l 2l l 3E I r32 = 2 l m 12 =

• δ3 = 1

2.10 Problem E(II): Symmetry and Anti-symmetry

6E I l2 3E I m 23 = 2 l 18E I r33 = l3 m 13 =

• External load

pl2 12 =0

m 10 = m 20

r30 = pl −

pl pl = (global equilibrium of horizontal translation) 2 2

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎧ ⎡ 8E I 2E I 6E I ⎤ ⎡ ⎤ ⎡ pl2 ⎤ pl3 ⎪ ϕ − ⎨ ϕ1 = 0.011719 E I 2 1 l l l 12 ⎥ 3 ⎢ ⎣ 2E I 6E I 3E2 I ⎦ · ⎣ ϕ2 ⎦ = ⎣ 0 ⎦ ⇒ ϕ2 = 0.013021 pl l l l EI 4 ⎪ 6E I 3E I 18E I ⎩ δ3 − pl δ3 = −0.033854 pl l2 l2 l3 2 EI

79

80

2 Solution of Simple Reference Cases

Summary of Bending Moments

Summary of Shear Forces

Solving the system of equilibrium equations makes it possible to evaluate internal forces and moments and displacements in the analyzed half structure. In order to represent the diagrams of internal forces and moments and deformation for the entire structure, the following considerations, valid for all symmetrical structures loaded with anti-symmetrical loads, are taken into account: (a) the deformation results to be anti-symmetrical; (b) the shear force diagram results to be symmetrical (same sign on both sides of the axis of symmetry); (c) the bending moment and axial force diagrams result to be anti-symmetrical (opposite signs at two symmetrical points). Therefore, on the central column CF placed on the axis of symmetry we have: (i) the moment of the right side that sums to that of the left side; (ii) the shear of the right part that sums to that of the left part; (iii) the axial force of the right side being compensated by that of the left side (globally zero axial force).

2.10 Problem E(II): Symmetry and Anti-symmetry

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

81

82

2 Solution of Simple Reference Cases

2.11 Problem E(III): Symmetry and Anti-symmetry

• E I = constant • EA → ∞ Solution In the situation of symmetrical structure and generic loads, it is possible to decompose the loads into a symmetrical part and an anti-symmetrical part. In this way, it is possible to exploit the superposition of effects by analyzing the structure under consideration as the sum of a symmetrical structure symmetrically loaded and a symmetrical structure with anti-symmetrical loads.

2.11 Problem E(III): Symmetry and Anti-symmetry

83

Solution of Symmetrically Loaded Frame

The frame is symmetrically and symmetrically loaded; by introducing an appropriate restraint to consider this condition, only half frame can be analyzed. The displacement method is used to solve the structure.

m 11 ϕ1 + m 10 = 0

84

2 Solution of Simple Reference Cases

Kinematics • ϕ1 = 1

m 11 =

EI 5E I 4E I + = l l l

• External load

m 10 =

pl2 15 p l2 5 · − = − pl2 = − pl2 4 12 3 48 16

Calculation of Unknowns m 11 ϕ1 + m 10 = 0 5E I 15 2 · ϕ1 − pl = 0 l 48 15 2 l 1 pl3 ϕ1 = pl · = · 48 5E I 16 E I

2.11 Problem E(III): Symmetry and Anti-symmetry

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

85

86

Qualitative Deformed Shape

Solution of Anti-symmetrically Loaded Frame

2 Solution of Simple Reference Cases

2.11 Problem E(III): Symmetry and Anti-symmetry

87

The frame is symmetrically and anti-symmetrically loaded; by introducing an appropriate restraint to consider this condition, only half frame can be analyzed. It is solved by the displacement method.

⎧

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

4E I 7E I 3E I + = l l l 6E I = 2 l

m 11 = r21

88

2 Solution of Simple Reference Cases

• δ2 = 1

6E I l2 12E I = l3

m 12 = r22 • External load

3 pl2 4 12 3 pl = 4 2

m 10 = r20 Calculation of Unknowns ⎧

m 11 ϕ1 + m 12 δ2 + m 10 = 0 m 21 ϕ1 + m 22 δ2 + m 20 = 0 pl3 EI pl3 δ2 = −0.04688 EI

ϕ1 = 0.03125

2.11 Problem E(III): Symmetry and Anti-symmetry

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

89

90

2 Solution of Simple Reference Cases

Qualitative Deformed Shape

Moment, Shear and Axial Force Diagrams The diagrams of the internal forces and moments and the deformation of the initial structure—characterized by a symmetrical structure in geometry but generic loads— are obtained by simply summing the results obtained from the analysis of the symmetrically loaded structure with those obtained from the study of the anti-symmetrical loaded structure.

2.11 Problem E(III): Symmetry and Anti-symmetry

Qualitative Deformed Shape

91

92

2 Solution of Simple Reference Cases

2.12 Problem F(I): Imposed Displacement/Rotation

• E I = constant • EA → ∞ 1 pl4 • δ = 100 EI Solution The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B as kinematic unknowns.

2.12 Problem F(I): Imposed Displacement/Rotation

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

4E I 6E I 4E I + = 2l l l EI m 21 = l 6E I r31 = − 2 l m 11 =

• ϕ2 = 1

93

94

2 Solution of Simple Reference Cases

EI l 4E I 6E I 4E I + = m 22 = 2l l l 6E I r32 = − 2 l m 12 =

• δ3 = 1

6E I l2 6E I m 23 = − 2 l 24E I r33 = l3 m 13 = −

• External load

2.12 Problem F(I): Imposed Displacement/Rotation

95

4 pl2 12 2 4 pl m '20 = 12 ' r30 =0 m '10 = −

• Imposed displacement

6E I 4 pl2 209 2 − pl δ¯ = − 12 4l2 600 6E I 4 pl2 191 2 '' − pl m 20 = δ¯ = 2 12 4l 600 '' r30 = 0 ''

m 10 = −

Due to the superposition of effects, it is possible to calculate m 10 , m 20 and r30 . 6E I 4 pl2 209 2 − pl δ¯ = − 2 12 4l 600 6E I 4 pl2 191 2 ' '' − pl m 20 = m 20 + m 20 = δ¯ = 12 4l2 600 ' '' r30 = r30 + r30 = 0 '

''

m 10 = m 10 + m 10 = −

96

2 Solution of Simple Reference Cases

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎧ ⎡ 6E I ⎡ ⎡ 209 ⎤ ⎤ ⎤ pl3 EI I ⎪ − 6E ϕ1 ⎨ ϕ1 = 0.0704 E I l l l2 600 3 I ⎦ ⎣ 6E I 191 ⎦ ⎣ EI · ϕ2 ⎦ = ⎣ − 600 pl2 ⇒ ϕ2 = −0.0629 pl − 6E l l l2 E4 I ⎪ I I 24E I ⎩ δ3 0 − 6E − 6E δ3 = 0.0019 pl l2 l2 l3 EI Summary of Bending Moments

Summary of Shear Forces

2.12 Problem F(I): Imposed Displacement/Rotation

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

97

98

2 Solution of Simple Reference Cases

2.13 Problem F(II): Imposed Displacement/Rotation

• E I = constant • EA → ∞ 1 pl3 • θ = 12 EI Solution The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B. It is worth noting that the load p is introduced as a dimensionless variable and for formal consistency with the other examples.

2.13 Problem F(II): Imposed Displacement/Rotation

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

4E I 6E I 4E I + = l 2l l EI m 21 = l 6E I r31 = − 2 l m 11 =

• ϕ2 = 1

99

100

2 Solution of Simple Reference Cases

EI l 4E I 6E I 4E I + = m 22 = 2l l l 6E I r32 = − 2 l m 12 =

• δ3 = 1

6E I l2 6E I m 23 = − 2 l 24E I r33 = l3 m 13 = −

• Imposed rotation

2.13 Problem F(II): Imposed Displacement/Rotation

m 10 = −

101

pl2 2E I θ =− l 6

m 20 = 0 6E I pl r30 = 2 θ = l 2 Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ ⎡

r31 r32 r33

6E I l ⎣ EI l I − 6E l2

EI l 6E I l I − 6E l2

I − 6E l2 I − 6E l2 24E I l3

δ3 ⎤ ⎡

Summary of Bending Moments

Summary of Shear Forces

r30 ⎡

⎧ ⎤ pl3 pl2 ⎪ ϕ1 ⎨ ϕ1 = 0.0063 E I 6 3 ⎥ ⎦ · ⎣ ϕ2 ⎦ = ⎢ ⎣ 0 ⎦ ⇒ ϕ2 = −0.0271 pl E4I ⎪ ⎩ δ3 − pl δ3 = −0.026 pl 2 ⎤

EI

102

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

2 Solution of Simple Reference Cases

2.14 Problem G(I): Infinitely Rigid Bending Rod

103

2.14 Problem G(I): Infinitely Rigid Bending Rod

E I BC → ∞ E I AB = E IC D = E I EA → ∞ Solution The structure results to be a sway frame.

The presence of the infinitely rigid top member, which can only do rigid rototranslations, together with the beam-column continuity and the kinematics of the system, ensures that the two columns at the top do not experience any rotation. Consequently, no rotational unknowns are introduced at the intersection of beam and columns. Furthermore, it would not have been possible to introduce a rotation unknown at each of the two nodes B and C, since these rotations could never have been activated in the cases ϕ(B) = 1 and ϕ(C) = 1. The displacement method is used to solve the structure by adopting the horizontal translation δ1 of node B as the only kinematic unknown.

104

2 Solution of Simple Reference Cases

r11 δ1 + r10 = 0 Kinematics • δ1 = 1

r11 =

12E I 12E I 24E I + = l3 l3 l3

• External load

r10 = −

pl 2

2.14 Problem G(I): Infinitely Rigid Bending Rod

Calculation of Unknowns r11 δ1 + r10 = 0 24E I pl pl4 δ1 − = 0 ⇒ δ1 = 3 l 2 48E I Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

105

106

2 Solution of Simple Reference Cases

Qualitative Deformed Shape

2.15 Problem G(II): Infinitely Rigid Bending Rod

2.15 Problem G(II): Infinitely Rigid Bending Rod

107

• E I AB → ∞ • E I BC = E IC D = E I • EA → ∞ Solution The rod AB has infinite flexural stiffness and, as such, can only and exclusively undergo rigid roto-translations. The presence of a clamping at node A imposes that the translations and rotations of that point are zero and consequently that the translations and rotations of the whole rod AB are zero. Therefore, node B cannot neither rotate nor translate and, in accordance with the kinematic analysis, if node B does not move, the whole rod BC also cannot make the horizontal translation granted by the kinematic analysis. Therefore, the structure results to be fixed-joint frame.

Because of what was noted above, node B cannot undergo any rotation, so no rotation unknown is introduced at node B. The only unknown used is therefore the rotation ϕ1 of node C.

m 11 ϕ1 + m 10 = 0

108

2 Solution of Simple Reference Cases

Kinematics • ϕ1 = 1

m 11 =

4E I 6E I 4E I + = 2l l l

• External load

m 10 =

4 pl2 12

2.15 Problem G(II): Infinitely Rigid Bending Rod

109

Calculation of Unknowns m 11 ϕ1 + m 10 = 0 6E I pl2 pl3 ϕ1 + = 0 ⇒ ϕ1 = − l 3 18E I Summary of Bending Moments

Summary of Shear Forces

The values of bending moment and shear force on rod AB not explicitly shown in the previous figures are calculated by writing the equilibrium equations.

110

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

2 Solution of Simple Reference Cases

2.16 Problem H: Concentrated Load

111

2.16 Problem H: Concentrated Load

• EA → ∞ • P = pl Solution The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B.

112

2 Solution of Simple Reference Cases

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

4(2E I ) 10E I 4E I + = 2l l l EI m 21 = l 12E I r31 = − 2 l m 11 =

• ϕ2 = 1

2.16 Problem H: Concentrated Load

113

EI l 4E I 6E I 4E I + = m 22 = 2l l l 6E I r32 = − 2 l m 12 =

• δ3 = 1

12E I l2 6E I m 23 = − 2 l 36E I r33 = l3 m 13 = −

• External load

m 10 = 0 m 20 = 0 r30 = −P

114

2 Solution of Simple Reference Cases

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ ⎡

r31 r32 r33

10E I l ⎣ EI l I − 12E l2

EI l 6E I l I − 6E l2

δ3 12E I ⎤ ⎡

r30

⎧ 3 ⎡ ⎤ ⎪ ϕ1 = 0.063218 pl − l2 ϕ1 0 ⎨ E I3 I ⎦ ⎣ · ϕ2 ⎦ = ⎣ 0 ⎦ ⇒ ϕ2 = 0.045977 pl − 6E l2 E I4 ⎪ 36E I ⎩ δ3 P δ3 = 0.056513 pl l3

Summary of Bending Moments

Summary of Shear Forces

⎤

EI

2.16 Problem H: Concentrated Load

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

115

116

2 Solution of Simple Reference Cases

2.17 Problem I: Statically Determined Portion

• E I = constant • EA → ∞ Solution To the left of node A and to the right of node B can be identified two statically determined parts (appendages), which can be eliminated in the solution of the system as long as they are replaced by the reactions they exchange with the remaining part of the structure. Specifically, the cantilever on the left will be replaced by a shear force and moment concentrated at node A, while the triangle BDC will be replaced by two forces (one horizontal and one vertical) at both nodes B and C. For the determination of these reactions, the structures related to statically determined appendages should be solved; therefore, the cantilever and the triangle BDC are analyzed in the present case.

2.17 Problem I: Statically Determined Portion

117

The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node A, ϕ2 of node B and the horizontal translation δ3 of node A.

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0

118

2 Solution of Simple Reference Cases

Kinematics • ϕ1 = 1

4E I 2E I 6E I + = l l l EI m 21 = l 6E I r31 = − 2 l m 11 =

• ϕ2 = 1

EI l 4E I 6E I 4E I + = m 22 = 2l l l 6E I r32 = − 2 l m 12 =

2.17 Problem I: Statically Determined Portion

119

• δ3 = 1

6E I l2 6E I m 23 = − 2 l 24E I r33 = l3 m 13 = −

• External load

pl2 2 9 pl2 m 20 = − 32 9 pl 9 pl 9 pl 9 pl − − =− r30 = 4 4 8 8 m 10 =

It is emphasized that the moment pl2 /2 applied to node A, with counterclockwise sign, is precisely a nodal moment. The convention used in this text always considers the moments applied as a reaction moment on the rigid block, which in this case results to be of opposite sign and therefore clockwise. For this reason, the sign of the moment m 10 is positive.

120

2 Solution of Simple Reference Cases

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎧ ⎤ ⎡ ⎡ 6E I ⎡ ⎤ ⎤ 2 pl3 EI I ⎪ − 6E ϕ1 − pl2 ⎨ ϕ1 = −0.0352 E I l l l2 3 2 ⎥ ⎢ I ⎦ ⎣ 6E I ⎣ EI · ϕ2 ⎦ = ⎣ 9 pl ⎦ ⇒ − 6E ϕ2 = 0.121 pl l l l2 E I4 32 ⎪ I I 24E I ⎩ 9 pl δ3 − 6E − 6E δ3 = 0.0684 pl l2 l2 l3 8 EI Summary of Bending Moments

Summary of Shear Forces

2.17 Problem I: Statically Determined Portion

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

121

122

2 Solution of Simple Reference Cases

2.18 Problem J: Unusual Restraints

• E I = constant • EA → ∞ Solution The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B.

2.18 Problem J: Unusual Restraints

123

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

6E I l EI m 21 = l 6E I r31 = − 2 l m 11 =

• ϕ2 = 1

EI l 3E I m 22 = l r32 = 0 m 12 =

124

2 Solution of Simple Reference Cases

• δ3 = 1

m 13 = −

6E I l2

m 23 = 0 12E I r33 = l3 • External Load

pl2 4 pl2 =− 12 3 pl2 m 20 = 3 r30 = 0 m 10 = −

2.18 Problem J: Unusual Restraints

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎧ ⎡ 2 ⎤ ⎡ 6E I E I ⎡ ⎤ ⎤ pl3 pl I ⎪ − 6E ϕ1 ⎨ ϕ1 = 0.1666 E I l l l2 3 2 ⎥ ⎢ ⎣ E I 3E I 0 ⎦ · ⎣ ϕ2 ⎦ = ⎣ − pl ⎦ ⇒ ϕ2 = −0.1666 pl3 l l E4 I 3 ⎪ I I ⎩ δ3 0 12E − 6E 0 δ3 = 0.0833 pl l2 l3 EI Summary of Bending Moments

Summary of Shear Forces

125

126

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

2 Solution of Simple Reference Cases

2.19 Problem K(I): Distributed Loads/Moments

127

2.19 Problem K(I): Distributed Loads/Moments

• EA → ∞ Solution The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B.

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0

128

2 Solution of Simple Reference Cases

Kinematics • ϕ1 = 1

8E I 10E I 4E I + = 2l l l EI 2E I = m 21 = 2l l 12E I r31 = − 2 l m 11 =

• ϕ2 = 1

EI l 4E I 6E I 4E I + = m 22 = 2l l l 6E I r32 = − 2 l m 12 =

2.19 Problem K(I): Distributed Loads/Moments

• δ3 = 1

12E I l2 6E I m 23 = − 2 l 24E I 12E I 36E I r33 = + = l3 l3 l3 m 13 = −

• External load

m 10 = 0 m 20 = 0 r30 = − pl

129

130

2 Solution of Simple Reference Cases

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ϕ1 m 10 m 11 m 12 m 13 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎧ ⎡ 10E I ⎡ ⎡ ⎤ ⎤ ⎤ pl3 EI I ⎪ − 12E ϕ1 0 ⎨ ϕ1 = 0.06322 E I l l l2 3 I ⎦ ⎣ 6E I ⎣ EI · ϕ2 ⎦ = ⎣ 0 ⎦ ⇒ ϕ2 = 0.04598 pl − 6E l l l2 E I4 ⎪ I I 36E I ⎩ δ3 pl − 12E − 6E δ3 = 0.05651 pl l2 l2 l3 EI

Summary of Bending Moments

Summary of Shear Forces

2.19 Problem K(I): Distributed Loads/Moments

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

131

132

2 Solution of Simple Reference Cases

2.20 Problem K(II): Distributed Loads/Moments

• E I = constant • EA → ∞ Solution The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node C and the horizontal translation δ2 of node B.

⎧

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

2.20 Problem K(II): Distributed Loads/Moments

Kinematics • ϕ1 = 1

3E I 11E I 4E I + = l 2l 2l 6E I =− 2 l

m 11 = r21 • δ2 = 1

6E I l2 3E I 12E I 15E I = 3 + = 3 l l l3

m 12 = − r22

133

134

2 Solution of Simple Reference Cases

• External load

m 10 = 0 The following is a demonstration of using the force method, assuming the moment X at node C as static unknown, to solve the case at hand.

2.20 Problem K(II): Distributed Loads/Moments

Writing the compatibility equation at node C we obtain: ϕ11 X + ϕ10 = 0 → X = 0 r20 = 0 Calculation of Unknowns ⎧ m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0 ] [ ] [ ] [ ϕ1 m 10 m 11 m 12 · =− r21 r22 δ2 r20 [ 11E I ] [ ] [ ] ⎧ I − 6E ϕ1 0 ϕ1 = 0 2l l2 · = − ⇒ I 15E I − 6E δ δ2 = 0 0 2 2 3 l l Summary of Bending Moments

Summary of Shear Forces

135

136

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

2 Solution of Simple Reference Cases

2.21 Problem L: Rod with Finite Axial Stiffness

137

2.21 Problem L: Rod with Finite Axial Stiffness

• • • •

E I = constant E A ABC D → ∞ EI E A B D = 10l ∗2 √ ∗ l =l 2

Solution The structure results to be a fixed-joint frame.

The rod with finite axial stiffness can be transformed into a translational spring parallel to the axis of the rod. The stiffness of the spring results to be equal to k = El∗A where l∗ is the length of the rod with finite axial stiffness.

138

2 Solution of Simple Reference Cases

⎧

m 11 ϕ1 + m 12 η2 + m 10 = 0 r21 ϕ1 + r22 η2 + r20 = 0

Kinematics • ϕ1 = 1

3E I 7E I 4E I + = l l l 6E I √ = 2 2 l

m 11 = r21 • η2 = 1

m 12 =

6E I δ l2

2.21 Problem L: Rod with Finite Axial Stiffness

139

√ The force in the spring (Fspring ) along the direction of η2 (l∗ = l 2) is equal to: Fspring ∑

√ 2 EI EA EI EI = kη2 = k · 1 = ∗ = = = √ ∗3 l 10l 40 l3 10( 2l)3

FH = 0 √ √ 2 3E I √ 2 12E I √ r22 − 3 2− =0 2 − Fspring 3 2 l l 2 √ 1200 + 2 E I r22 = 40 l3 • External load

m 10 =

pl2 8

r20 = 0 Calculation of Unknowns ⎧ m 11 ϕ1 + m 12 η2 + m 10 = 0 r21 ϕ1 + r22 η2 + r20 = 0 ] [ ] [ ] [ ϕ1 m 10 m 11 m 12 · =− r21 r22 η2 r20 [ ] ⎧ √ ] [ pl2 ] [ 3 7E I 6 2E I ϕ1 = −0.02716 pl ϕ − 2 1 l√ E I 8 √l = · ⇒ 4 6 2E I 1200+ 2 E I η2 0 η2 = 0.007672 pl EI l2 40 l3

140

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

2 Solution of Simple Reference Cases

2.21 Problem L: Rod with Finite Axial Stiffness

Qualitative Deformed Shape

141

Chapter 3

Solution of Complex Frames

3.1 Worked Example 1

• • • • •

E I BC = E I D E → ∞ E I AB = E IC D = E I E F = E I EA → ∞ k = E2lI αΔ T t

=

pl2 100E I

Solution The structure is symmetrically loaded, so it can be simplified as shown in the figures below. The structure results to be a sway frame. Both the displacement method and the mixed method can be used to solve the frame. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil Engineering, https://doi.org/10.1007/978-3-031-35267-6_3

143

144

3 Solution of Complex Frames

Displacement Method The displacements method is used by adopting as kinematic unknowns the rotation ϕ1 and the horizontal translation δ2 . No rigid blocks are inserted at nodes B and C because, since the flexural stiffness of rod BC is infinite, it will only be able to perform rigid roto-translations dictated by the kinematics of the system. The rod BC will not be able to assume any flexural deformation. In the case where the horizontal displacement of node B is prevented (δ2 = 0), rod BC will not undergo any roto-translation; consequently, due to the flexural continuity between rods AB and BC, rod AB will also not be able to perform any rotation at endpoint B. As a consequence, in case δ2 is zero, the static scheme of the rod AB corresponds to that of a rod clamped at both ends.

3.1 Worked Example 1

145



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

m 11 =

4E I EI 9E I 4E I +k = + = l l 2l 2l r21 = −

8E I l2

It should be noted that at node B, since there is no rotational unknown, the actions shown in the figure all represent internal forces and moments and therefore the equilibrium at the rotation of the node must be respected; consequently, on the rod BC at node B a moment equal in modulus and opposite in direction to that acting at the top of the rod AB will act. A purely static representation of this equilibrium is given below.

146

3 Solution of Complex Frames

 

M B2+3 = 0 → NCC ' =

2E I l2

FH1+2+3 = 0 → r21 = −

8E I l2

• δ2 = 1 The figure shows a superposition of effects between a unit horizontal translation and a counterclockwise rotation of magnitude α of node B of rod AB. As a result of the translation of node B, the infinitely rigid rod BC makes a rigid rotation of amplitude α with respect to the CIR. All points of rod BC are rotating by α and, due to the flexural continuity at point B between rods AB and BC, the top of the column (i.e. rod AB) will also undergo a counterclockwise rotation α.

3.1 Worked Example 1

147

m 12 = − r22 =

8E I l2

30E I l3

Similarly to what has been described for node B, also at node C the continuity between rods BC and CC’ generates, within rod CC’, a flexural deformation related to the rotation α of end C. In particular, it is possible to write an equilibrium equation at rotation.  

• External load

M B2+3 = 0 → NCC ' =

FH = 0 → r22 =

12E I l3

18E I 12E I 30E I + = 3 3 l l l3

148

3 Solution of Complex Frames

m '10 = 0 ' r20 =

7 pl 12

• Constant thermal variation

m ''10 = 0 '' r20 =0

• Linear thermal variation

3.1 Worked Example 1

149

m ''' 10 = −

pl2 E I αΔ t =− t 100

''' r20 =−

pl 100

The reaction r''' 20 is calculated by writing static equilibrium equations, on the structure made statically determined by downgrading the fixed end restraints and the flexural continuity points. 

M B2+3 = 0 → NCC ' = −



''' =− FH = 0 → r20

pl 100

pl 100

By applying the superposition of effects it is possible to calculate m 10 and r20 . m 10 = −  r20 =

pl2 100

 1 43 688 7 − pl = pl pl = 12 100 1200 75

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       m 11 m 12 ϕ1 m 10 · =− r21 r22 δ2 r20  9 EI      1 −8 El2I pl2 ϕ1 2 l 100 · = −8 El2I 30 El3I δ2 − 43 pl 75 ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = −0.06038 EI ⇒ 4 ⎪ pl ⎪ ⎩ δ2 = −0.03521 EI

Summary of Bending Moments A summary of the bending moments acting on the structure is given in the figure, written with concordant or discordant sign, with respect to the moments defined in the convention.

150

3 Solution of Complex Frames

Summary of Shear Forces A summary of the shear forces acting on the structure is given in the figure, written with concordant or discordant sign, with respect to the forces defined in the convention.

3.1 Worked Example 1

Moment, Shear and Axial Force Diagrams

151

152

3 Solution of Complex Frames

Qualitative Deformed Shape

Mixed Method The mixed method is used by adopting as static unknowns the moments X1 , X2 and X3 and as kinematic unknown the horizontal translation δ4 .

3.1 Worked Example 1

153

⎧ ϕ11 X 1 + ϕ12 X 2 + ϕ13 X 3 + ϕ14 δ4 + X 10 = 0 ⎪ ⎪ ⎨ ϕ21 X 1 + ϕ22 X 2 + ϕ23 X 3 + ϕ24 δ4 + X 20 = 0 ⎪ ϕ X + ϕ32 X 2 + ϕ33 X 3 + ϕ34 δ4 + X 30 = 0 ⎪ ⎩ 31 1 r41 X 1 + r42 X 2 + r43 X 3 + r44 δ4 + δ40 = 0 Kinematics The rod BC, due to its infinite flexural stiffness, following the application of the moment at node C has no bending deformation. • X1 = 1

ϕ11 =

l 2E I

(concordant with the direction of X1 ) ϕ21 = ϕ31 = 0

154

3 Solution of Complex Frames

r41 = − 1l (discordant with the direction of δ4 ) • X2 = 1

ϕ12 = 0 ϕ22 = 3El I (concordant with the direction of X2 ) ϕ32 = 6El I (concordant with the direction of X3 ) r42 = 2l (concordant with the direction of δ4 ) • X3 = 1

ϕ13 = 0 ϕ23 = 6El I (concordant with the direction of X2 ) ϕ33 = 3El I + k1 = 3El I + E2lI = 73 ElI (concordant with the direction of X3 )

3.1 Worked Example 1

155

r43 = − 1l (discordant with the direction of δ4 ) • δ4 = 1

ϕ14 =

1 l

2 1 1 ϕ24 = − − = − l l l ϕ34 =

1 l

r44 = 0 (all the restraining reactions are zero because the rods rotate and translate without flexural deformation) • External load

156

3 Solution of Complex Frames

' ϕ10 =

pl3 24E I

' ϕ20 =0 ' ϕ30 =0

' r40 =

pl 2

• Constant thermal variation

'' ϕ10 =0 '' ϕ20 =0 '' ϕ30 =0 '' r40 =0

3.1 Worked Example 1

157

• Linear thermal variation

''' ϕ10 =0

''' ϕ20 =

pl3 200E I

''' ϕ30 =

pl3 200E I

''' r40 =0

Due to the superposition of effects, it is possible to calculate ϕ10 , ϕ20 , ϕ30 , r40 . ϕ10 =

pl3 24E I

ϕ20 =

pl3 200E I

ϕ30 =

pl3 200E I

r40 =

pl 2

158

3 Solution of Complex Frames

Calculation of Unknowns ⎧ ϕ11 X 1 + ϕ12 X 2 + ϕ13 X 3 + ϕ14 δ4 + X 10 = 0 ⎪ ⎪ ⎨ ϕ21 X 1 + ϕ22 X 2 + ϕ23 X 3 + ϕ24 δ4 + X 20 = 0 ⎪ ϕ X + ϕ32 X 2 + ϕ33 X 3 + ϕ34 δ4 + X 30 = 0 ⎪ ⎩ 31 1 r41 X 1 + r42 X 2 + r43 X 3 + r44 δ4 + δ40 = 0 ⎡ ⎡ ⎤ ⎡ ⎤ ⎤ X1 ϕ10 ϕ11 ϕ12 ϕ13 ϕ14 ⎢ ϕ20 ⎥ ⎢ ϕ21 ϕ22 ϕ23 ϕ24 ⎥ ⎢ X 2 ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎥ ⎣ ϕ31 ϕ32 ϕ33 ϕ34 ⎦ · ⎣ X 3 ⎦ = −⎣ ϕ30 ⎦ r41 r42 r43 r44 δ4 r40 ⎡ pl3 ⎤ ⎡ l ⎤ ⎡ ⎤ 0 0 1l X1 I 2E I ⎢ 24E pl3 ⎥ ⎢ 0 l l − 2 ⎥ ⎢ X2 ⎥ ⎥ 3E I 6E I l ⎥·⎢ ⎢ ⎥ = −⎢ 200E ⎢ pl3 I ⎥ ⎣ 0 l 7l 1 ⎦ ⎣ X 3 ⎦ ⎣ ⎦ 6E I 3E I l 200E I pl δ4 − 1l 2l − 1l 0 EI

By adopting the mixed method, the stiffness/flexibility matrix results to be symmetric in absolute value.

⇒

⎧ ⎪ X 1 = −0.013 pl2 ⎪ ⎪ ⎪ ⎪ 2 ⎪ ⎨ X 2 = −0.241 pl X 3 = 0.030 pl2 ⎪ ⎪ ⎪ ⎪ ⎪ pl4 ⎪ ⎩ δ4 = −0.035 EI

For diagrams of internal forces and moments, refer to the graphs obtained with the displacement method.

3.2 Worked Example 2

3.2 Worked Example 2

159

• • • •

E I = constant EA → ∞ αΔ T l∗ = δ ∗ t = l5 /  2 / • l∗ = l2 + 2l = 45 l = • cosα =

l l∗

=

√2 l 5l

=

√

√

5 l 2

2 5 5

Solution The structure results to be a sway frame. A horizontal roller should be added at node A; however, the frame is not subject to external loads (but only to thermal variation and imposed displacement), so the vertical rod AB is merely subjected to axial force. To solve the problem, it is therefore sufficient to consider rotation ϕ1 as the unique displacement unknown.

Displacement Method

m 11 ϕ1 + m 10 = 0

160

3 Solution of Complex Frames

Kinematics • ϕ1 = 1

m 11

/ 4 EI 4E I 8√ E I = ∗ =4 5 = l 5 l 5 l

• Imposed displacement

m '10 =

√ 6E I 24 E I 12 5 E I 6E I δ δ δ = = = · · · δ  √ 2 cos α 5 l2 cos α 5 l2 (l∗ )2 cos α 5 l 2

3.2 Worked Example 2

161

• Thermal variation

m ''10 = −

m ''' 10 =

12 E I 6E I αΔ T l∗ =− · δ 2 ∗ 2 5 l2 (l )

10E I 2E I αΔ T 8E I = ∗ 2 ·δ = 2 ·δ t l (l )

Due to the superposition of effects, it is possible to calculate m 10 .

162

m 10 =

3 Solution of Complex Frames

m '10

+

m ''10

+

m ''' 10

√ √ 12 5 E I 12 E I 8E I 12 5 + 28 E I = δ− δ+ 2 δ= δ 5 l2 5 l2 l 5 l2

Calculation of Unknowns m 11 ϕ1 + m 10 = 0 

  √  12 5 + 28 E I 8E I δ · [ϕ1 ] = − √ 5 l2 5l ⇒ ϕ1 = −3.065

Summary of Bending Moments

Summary of Shear Forces

δ l

3.2 Worked Example 2

Moment, Shear and Axial Force Diagrams

163

164

3 Solution of Complex Frames

Qualitative Deformed Shape

3.3 Worked Example 3

• • • • • •

E I = constant EA → ∞ k = 2El I P = 2 pl pl2 αΔ T = 10E t= I t = 3l

pl3 30E I

Solution The structure results to be a fixed-joint frame.

3.3 Worked Example 3

165

Force Method The force method is used to solve the structure by adopting as static unknowns the moment X 1 and the moment transmitted by the internal rotational spring X2 .



ϕ11 X 1 + ϕ12 X 2 + ϕ10 = 0 ϕ21 X 1 + ϕ22 X 2 + ϕ20 = 0

Kinematics • X1 = 1

ϕ11 =

3l 25l 4l + = 3E I 4E I 12E I

166

3 Solution of Complex Frames

ϕ21 =

4l 2l = 6E I 3E I

ϕ12 =

2l 4l = 6E I 3E I

• X2 = 1

ϕ22 =

5l 1 7l 4l + + = 3E I 3E I k 2E I

• External load

' ϕ10 =

p(3l)3 9 pl3 2 pl3 41 pl3 P(4l)2 + = + = 16E I 48E I 16E I EI 16E I ' ϕ20 =

2 pl3 p(4l)2 = 16E I EI

3.3 Worked Example 3

167

• Linear thermal variation

'' ϕ10 =0 '' ϕ20 =0

• Constant thermal variation

αΔ T = δV =

pl3 30E I

3 αΔ T · 4l = 6αΔ T l 2

δV =

pl4 6 pl4 = 30E I 5E I

δH =

3 3 pl4 δV = 4 20E I

δ=

5 pl4 δV = 4 4E I

168

3 Solution of Complex Frames

γ =

α=

1 pl4 pl3 δ = · = 5l 5l 4E I 20E I

β=

δV 1 pl4 pl3 = · = 4l 4l 5E I 20E I

1 3 pl4 3 pl3 δH δH = · = = 2l 2l 20E I 40E I (2/3)3l

''' ϕ10 = −β + γ = −

3 pl3 pl3 pl3 + = 20E I 40 E I 40E I

''' ϕ20 =β −α =

pl3 pl3 − =0 20E I 20E I

Due to the superposition of effects, it is possible to calculate ϕ10 and ϕ20 . ' '' ''' ϕ10 = ϕ10 + ϕ10 + ϕ10 =



1 41 +0+ 16 40



' '' ''' ϕ20 = ϕ20 + ϕ20 + ϕ20 = (2 + 0 + 0)

pl3 207 pl3 = EI 80 E I

pl3 pl3 =2 EI EI

Calculation of Unknowns 

ϕ11 X 1 + ϕ12 X 2 + ϕ10 = 0 ϕ21 X 1 + ϕ22 X 2 + ϕ20 = 0       ϕ11 ϕ12 X1 ϕ10 · =− ϕ21 ϕ22 X2 ϕ20    25l 2l    207 pl3 X 1 12E I 3E I · = − 80 plE3 I 2l 7l X2 2 EI 3E I 2E I  X 1 = −1.517 pl2 ⇒ X 2 = 0.860 pl2

3.3 Worked Example 3

169

Summary of Bending Moments

Calculation of Shear Forces by Writing the Equilibrium Equations

V (x) = R A − p · x = 1.1 p · l − p · x = 0 The shear force is zero at a distance x from point A equal to: x = 1.1l M(x) = R A · 1.1l − 0.316 pl2 − p 

(1.1l)2 = 0.289 pl2 2

M BAB = 0 → R A · 3l + 1.517 pl2 − 0.316 pl2 − p · 3l · 1.5l = 0 → R A = 1.1 pl



F→ = 0 → R A + R B A − p · 3l = 0 → R B A = 1.9 pl

170



3 Solution of Complex Frames

MCBC = 0 → R BC · 4l − 1.517 pl2 − 0.86 pl2 − 2 pl · 2l = 0 → R BC = 1.59 pl



F↑ = 0 → RC B + R BC − 2 pl = 0 → RC B = 0.41 pl

 

 

MCC D = 0 → R D =

M AAC = 0 → RC A =

1 pl 30

0.86 pl = 0.172 pl 5

F→ = 0 → 1.9 pl − 0.033 pl − N AC cosα + 0.172 plsinα = 0 → N AC = 2.46 pl F↑ = 0 → 0.41 pl + NC D + 2.46 plsinα + 0.172cosα = 0 → NC D = −2.024 pl

3.3 Worked Example 3

Moment, Shear and Axial Force Diagrams

171

172

3 Solution of Complex Frames

Qualitative Deformed Shape

3.4 Worked Example 4

• EA → ∞ pl3 • θ = 12E I • k = El3I pl2 • αΔ T = 24E t I Solution The structure is symmetric and symmetrically loaded and can be simplified as shown in the figure. In simplifying the structure, the stiffness of the spring is doubled, since the spring in the whole structure would be subjected to twice the displacement evaluated in the simplified structure.

3.4 Worked Example 4

173

The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the horizontal translation δ2 of node B.

Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

174

3 Solution of Complex Frames

Kinematics • ϕ1 = 1

m 11 =

8E I 12E I 4E I + = l l l r21 =

12E I l2

m 12 =

12E I l2

• δ2 = 1

r22 =

24E I 26E I + 2k = l3 l3

3.4 Worked Example 4

175

• External load

m '10 =

pl2 12

' r20 =0

• Imposed rotation

m ''10 = − '' r20 =−

4E I pl2 ·θ =− l 3

12E I · θ = − pl l2

176

3 Solution of Complex Frames

• Linear thermal variation

m ''' 10 =

pl2 4E I αΔ T = t 6 ''' r20 =0

Due to the superposition of effects, it is possible to calculate m 10 and r20 . m 10 = m '10 + m ''10 + m ''' 10 =

pl2 pl2 pl2 pl2 − + =− 12 3 6 12

' '' ''' r20 = r20 + r20 + r20 = − pl

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       ϕ m m 11 m 12 · 1 = − 10 r21 r22 δ2 r20    12E I 12E I    pl2 ϕ − 2 1 l l 12 · =− 12E I 26E I δ2 − pl l2 l3 ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = −5.853 · 10−2 EI ⇒ 4 ⎪ pl ⎪ ⎩ δ2 = 6.548 · 10−2 EI

3.4 Worked Example 4

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

177

178

Qualitative Deformed Shape

3 Solution of Complex Frames

3.5 Worked Example 5

179

3.5 Worked Example 5

• • • •

E I = constant E A → 3∞ pl θ = 24E I I k = 6E l3

• l∗ =

√

5l 2

Solution The structure is symmetric and symmetrically loaded; it can be simplified as shown in the figure below.

The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the horizontal translation δ2 of node B.

180

3 Solution of Complex Frames

Displacement Method



Kinematics • ϕ1 = 1

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

3.5 Worked Example 5

181

m 11 =

3E I 4E I EI + ∗ = 6.683 l l l

Writing the rotational equilibrium at point O, we obtain: r21 ·

3E I 6E I 3 l 4E I 2E I + + ∗ + − 2 l=0 2 l l l l 2 r21 =

EI 6E I 3E I 2 − ∗ · = 0.633 2 2 l l l l

• δ2 = 1

2 1 = l/2 l √ √ 5 2 l= 5 αl∗ = · l 2 α=

m 12 = r22 =

6E I 3E I √ EI − ∗ 2 · 5 = 0.633 2 2 l (l ) l

EI 12E I 3E I √ 2 + 2k + ∗ 2 · 5 · = 34.733 3 3 l (l ) l l

182

3 Solution of Complex Frames

• External load

pl = p ∗ l∗ → p ∗ =

2 pl 2l = p √ = √ p = 0.894 p l∗ 5l 5

pl2 8  2  pl pl2 3 pl 2 − = · =− 8 2 l 4 m '10 = −

' r20

• Imposed rotation

m ''10 = −

2E I pl2 ·θ =− l 12

3.5 Worked Example 5

183 '' r20 =−

6E I pl ·θ =− 2 l 4

Due to the superposition of effects, it is possible to calculate m 10 and r20 . m 10 = −

pl2 5 pl2 pl2 − =− 8 12 24

r20 = − Calculation of Unknowns 

pl 3 pl − = − pl 4 4

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       m 11 m 12 ϕ m · 1 = − 10 r21 r22 δ2 r20       6.683 ElI 0.633 El2I ϕ1 0.208 pl2 · = 0.633 El2I 34.733 El3I δ2 pl ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = 2.849 · 10−2 EI ⇒ 4 ⎪ pl ⎪ ⎩ δ2 = 2.827 · 10−2 EI

Summary of Bending Moments

184

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.6 Worked Example 6

Qualitative Deformed Shape

3.6 Worked Example 6

185

186

3 Solution of Complex Frames

• E I = constant • EA → ∞ 1 pl3 • αΔ T = 104 EI / • l1 = 45 l / • l2 = 13 l 4 Solution The structure is symmetric and symmetrically loaded; it can be simplified as shown in the figure below.

The structure results to be a fixed-joint frame and it is solved with the force method by adopting as static unknown the moment X 1 applied at node B.

3.6 Worked Example 6

187

Force Method

ϕ11 X 1 + ϕ10 = 0 Kinematics • X1 = 1

ϕ11 =

l1 1 l l + = (l + l1 ) = 0.706 3E I 3E I 3E I EI

188

3 Solution of Complex Frames

• External load

' ϕ10 =

• Constant thermal variation

p ∗ (l1 )3 24E I

3.6 Worked Example 6

189

ω=

αΔ tl2 1 δc = l cosβ l

3 l 3 l → cosβ = /2 = 0.832 2 13 l 4 /  / pl3 13 13 2 1 pl3 l = ω= 104E I 4 4 3 l 48E I

l2 cosβ =

'' ϕ10 = −ω −

3 pl3 ω =− ω=− 2 2 32E I

' '' ϕ10 = ϕ10 + ϕ10 = 0.0153

pl3 48E I

Calculation of Unknowns ϕ11 X 1 + ϕ10 = 0 0.706

l pl3 X 1 + 0.0153 =0 EI EI X 1 = −0.022 pl2

Summary of Bending Moments

190

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.7 Worked Example 7

Qualitative Deformed Shape

3.7 Worked Example 7

• EA → ∞ pl2 ) • α(Δ T = 6E t I

191

192

3 Solution of Complex Frames

• l = 9t • k = 2El I Solution The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B and ϕ2 of node C.

Displacement Method



m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0

3.7 Worked Example 7

193

Kinematics • ϕ1 = 1

m 11 =

12E I (2E I ) 3(2E I ) 4E I + + = l l l l m 21 =

2E I l

m 12 =

2E I l

• ϕ2 = 1

m 22 = k +

6E I 4E I = l l

194

3 Solution of Complex Frames

• External load, linear and constant thermal variation

m 10 = − =−

αΔ T 3 pl2 αΔ T 6E I − − 2 (2E I ) + (2E I ) 12 t 2 t l



 3 3(2E I ) αΔ T l + (3αΔ T l) = 2 l2

pl2 pl2 pl2 pl2 + + = 12 6 6 4

m 20 =

pl2 pl2 pl2 − =− 12 6 12

3.7 Worked Example 7

195

Calculation of Unknowns 

m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0       m 11 m 12 ϕ1 m 10 · =− m 21 m 22 ϕ2 m 20  2   12E I 2E I    pl ϕ1 l l 4 2 · = − 2E I 6E I pl ϕ − 2 l l 12 ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = −2.451 · 10−2 EI ⇒ 3 ⎪ pl ⎪ −2 ⎩ ϕ2 = 2.206 · 10 EI

Summary of Bending Moments

Summary of Shear Forces

196

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.8 Worked Example 8

Qualitative Deformed Shape

3.8 Worked Example 8

• • • •

E I = constant EA → ∞ 3 pl αΔ T = 100E I 2E I k= l

Solution The structure results to be a sway frame.

197

198

3 Solution of Complex Frames

The mixed method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the horizontal translation δ3 of node A and as static unknown the moment X2 at node D. Mixed Method

⎧ ⎨ m 11 ϕ1 + m 12 X 2 + m 13 δ3 + m 10 = 0 ϕ ϕ + ϕ22 X 2 + ϕ23 δ3 + ϕ20 = 0 ⎩ 21 1 r31 ϕ1 + r32 X 2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

3.8 Worked Example 8

199

m 11 =

3E I 6E I 13E I 4E I + + = l l l l ϕ21 = r31 = −

1 2 6E I l2

• X2 = 1

m 12 = − ϕ22 =

l 1 7 l l + + = 3E I 6E I k 8 EI r32 = −

• δ3 = 1

1 2

3 l

200

3 Solution of Complex Frames

6E I l2  / 2 l 2 3 24E I = = l3 2E I l m 13 = −

ϕ23 r33 =

12E I 24E I 36E I + = l3 l3 l3

• External load and constant thermal variation

m 10 = − ϕ20 =

3 pl2 3 2 9 pl2 pl + =− 32 100 800

3 pl3 7 pl3 pl3 − = 48E I 200E I 1200E I r30 = 0

3.8 Worked Example 8

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 X 2 + m 13 δ3 + m 10 = 0 ϕ ϕ + ϕ22 X 2 + ϕ23 δ3 + ϕ20 = 0 ⎩ 21 1 r31 ϕ1 + r32 X 2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ m 11 m 12 m 13 ϕ1 m 10 ⎣ ϕ21 ϕ22 ϕ23 ⎦ · ⎣ X 2 ⎦ = −⎣ ϕ20 ⎦ r31 r32 r33 δ3 r30 ⎤ ⎡ ⎡ 13E I ⎤ 3 I ⎤ ⎡ − 800 pl2 − 21 − 6E ϕ1 l l2 ⎥ 7 pl3 7l 3 ⎣ 1 ⎦ · ⎣ X 2 ⎦ = −⎢ ⎦ ⎣ 1200E 2 8E I l I I 3 36E I δ − 6E − 0 3 2 3 l l l ⎧ pl3 ⎪ ⎪ ⎪ ϕ1 = −0.00011 ⎪ ⎪ EI ⎨ ⇒ X 2 = −0.0051 pl2 ⎪ ⎪ ⎪ ⎪ pl4 ⎪ ⎩ δ3 = −0.00044 EI Summary of Bending Moments

201

202

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.9 Worked Example 9

Qualitative Deformed Shape

3.9 Worked Example 9

• E I = constant • EA → ∞ EI • k = 2l 3

203

204

3 Solution of Complex Frames

Solution Initially, a roller is introduced at node C to control the displacement at the translational spring. We then proceed with the kinematic analysis of the structure.

The structure results to be a sway frame and it is solved by the displacement method. The introduction of a roller in B with a vertical constraint reaction allows to prevent any movement of the structure. The rotation ϕ1 of joint C, the translations η2 of joint C and η3 of joint B are adopted as kinematic unknowns. Displacement Method

⎧ ⎨ m 11 ϕ1 + m 12 η2 + m 13 η3 + m 10 = 0 r ϕ + r22 η2 + r23 η3 + r20 = 0 ⎩ 21 1 r31 ϕ1 + r32 η2 + r33 η3 + r30 = 0

3.9 Worked Example 9

205

Kinematics • ϕ1 = 1

m 11 = 

12E I 27 E I 3E I + = l 5l 5 l

  3 3E I 3E I 3E I 3 +  5l  − 2 l + l = 0 M BBC D = 0 → r21 · l + 4 l l 4 4 r21 = − 

F↑ = 0 → r31 + r21 − r31 =

• η2 = 1

1 EI 5 l2

16 E I 5 l2

3E I =0 l2

206

3 Solution of Complex Frames

m 12 = 

3E I 80 E I 1 EI − =− 2 2 l 25 l 5 l2

3E I 7 3E I 3 3E I 5 ∗ · l − 3 l + 2 −  2 = 0 M BBC D = 0 → r22 5l 4 l 4 l 3 4

' r22

109 E I = 15 l3

' '' ' r22 = r22 + r22 = r22 +k =

r32 = − • η3 = 1

64 E I 15 l3

233 E I 30 l3

3.9 Worked Example 9

207

m 13 =

16 E I 5 l2

r23 = − r33 =

64 E I 15 l3

109 E I 15 l3

Remark: the relative displacement between the two ends of rod BC, in the direction orthogonal to the rod is 5/3. • External load

m 10 = −

7 pl2 128

r20 = −

35 pl 32

r30 = −

21 pl 32

Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 η2 + m 13 η3 + m 10 = 0 r ϕ + r22 η2 + r23 η3 + r20 = 0 ⎩ 21 1 r31 ϕ1 + r32 η2 + r33 η3 + r30 = 0

208

3 Solution of Complex Frames

⎡

⎤ ⎡ ⎤ ⎡ ⎤ m 11 m 12 m 13 ϕ1 m 10 ⎣ r21 r22 r23 ⎦ · ⎣ η2 ⎦ = −⎣ r20 ⎦ r31 r32 r33 η3 r30 ⎤ ⎡ ⎡ ⎡ 27 E I ⎤ ⎤ EI 7 − 15 El2I 16 pl2 ϕ1 − 128 5 l 5 l2 ⎣ − 1 E2I 233 E3I − 64 E3I ⎦ · ⎣ η2 ⎦ = −⎣ − 35 pl ⎦ 5 l 30 l 15 l 32 21 16 E I E I 109 E I η3 pl − 32 − 64 5 l2 15 l3 15 l3 ⎧ pl3 ⎪ ⎪ ϕ = −0.2044 ⎪ 1 ⎪ EI ⎪ ⎪ ⎨ pl4 ⇒ η2 = 0.3463 ⎪ EI ⎪ ⎪ ⎪ 4 ⎪ pl ⎪ ⎩ η3 = 0.3837 EI Summary of Bending Moments

Summary of Shear Forces

3.9 Worked Example 9

Moment, Shear and Axial Force Diagrams

209

210

Qualitative Deformed Shape

3.10 Worked Example 10

• E I = constant • EA → ∞ pl3 • αΔ T = 24E I

3 Solution of Complex Frames

3.10 Worked Example 10

211

Solution The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node C and the horizontal translation δ2 of node B. The addition of a horizontal roller in B prevents any movements of the structure. Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

212

3 Solution of Complex Frames

Kinematics • ϕ1 = 1

m 11 =

10E I l

r21 = − • δ2 = 1

3E I l2

3.10 Worked Example 10

213

3E I l2   3 EI = 18 + √ 2 l3 m 12 = −

r22

• External load and constant thermal variation

m 10 = − r20 =

pl2 8

pl 8

214

3 Solution of Complex Frames

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       m 11 m 12 ϕ1 m 10 · =− r21 r22 δ2 r20  10E I     pl2  3E I ϕ l  − l2  8 · 1 = I EI √3 18 + − 3E δ2 − pl 2 3 l 8 2 l  3 ϕ1 = 0.0111 pl EI 4 ⇒ δ2 = −0.0046 pl EI

Summary of Bending Moments

Summary of Shear Forces

3.10 Worked Example 10

Moment, Shear and Axial Force Diagrams

215

216

3 Solution of Complex Frames

Qualitative Deformed Shape

3.11 Worked Example 11

• • • •

E I = constant EA → ∞ 3 pl αΔ T = 8E I k = ElI

Solution It is observed that the right-hand portion of the structure (C-D-E) constitutes an isostatic appendage; the effect of thermal variation on the rod DE simply results in a rigid kinematic motion of the appendage itself, caused by the axial elongation of the oblique rod, and will therefore only be taken into account in the deformation tracking, since it does not generate any internal forces and moments in the structure under consideration. We then proceed to solve the isostatic appendage.

3.11 Worked Example 11

217

The structure can be simplified as:

We then proceed to analyze the simplified structure, which results to be a sway frame.

The mixed method is used to solve the structure by adopting as kinematic unknown the horizontal translation δ2 of node B and as static unknown the moment X1 transmitted by the spring.

218

3 Solution of Complex Frames

Mixed Method



ϕ11 X 1 + ϕ12 δ2 + ϕ10 = 0 r21 X 1 + r22 δ2 + r20 = 0

Kinematics • X1 = 1

ϕ11 =

l 1 7 l l 19 l l + + = + = 3E I 4E I k 12 E I EI 12 E I r21 =

3 2l

3.11 Worked Example 11

219

• δ2 = 1

  3 1 3E I l + 2 =− ϕ12 = − l l 6E I 2l r22 =

3E I 6E I 3E I + = l l l

• External load

ϕ10 =

pl3 pl2 l pl3 − = 24E I 8 6E I 48E I

220

3 Solution of Complex Frames

r20 = −

5 pl 7 pl pl − pl + =− 2 8 8

Calculation of Unknowns 

ϕ11 X 1 + ϕ12 δ2 + ϕ10 = 0 r21 X 1 + r22 δ2 + r20 = 0       ϕ11 ϕ12 X1 ϕ10 · =− r21 r22 δ2 r20    19 l    3 pl3 − 2l X1 − 48E 12 E I I · = 7 pl 3 6E I δ2 2l l3 8  X 1 = 0.101 pl2 ⇒ 4 δ2 = 0.1205 pl EI Summary of Bending Moments

3.11 Worked Example 11

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

221

222

Qualitative Deformed Shape

3.12 Worked Example 12

• EA → ∞ ql4 • δ = 24E I ql3 • αΔ T = 12E I l • h = 10

3 Solution of Complex Frames

3.12 Worked Example 12

223

Solution The structure results to be a fixed-joint frame and it is solved with the displacement method, by adopting as kinematic unknown the rotation ϕ at node B.

Displacement Method

m 11 ϕ1 + m 10 = 0 Kinematics • ϕ1 = 1

224

3 Solution of Complex Frames

m 11 =

13E I l

• External load

m '10 = −

ql2 8

• Imposed displacement

m ''10 =

3E I ql2 δ = l2 8

3.12 Worked Example 12

225

• Constant thermal variation

m ''' 10 = −

3 9E I αΔ T = − ql2 l 4

• Linear thermal variation

m '''' 10 = −

5 2E I αΔ T = − ql2 h 2 6

Due to the superposition of effects, it is possible to calculate m 10 .   1 1 3 5 19 '''' + − − ql2 = − ql2 m 10 = m '10 + m ''10 + m ''' + m = − 10 10 8 8 4 6 12

226

3 Solution of Complex Frames

Calculation of Unknowns m 11 ϕ1 + m 10 = 0   l ql3 m 10 19 ql2 = 0.1217 ⇒ ϕ1 = − =− − m 11 12 E I 13E I EI Summary of Bending Moments

Summary of Shear Forces

3.12 Worked Example 12

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

227

228

3 Solution of Complex Frames

3.13 Worked Example 13

• E I = constant • EA → ∞ pl2 ) • 2α(Δ T = 24E h I Solution The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the horizontal translation δ2 of node B.

3.13 Worked Example 13

229

The geometric quantities l∗ and l1 are calculated: / l2 +

l2 = 1.118l 4

9l2 +

9l2 = 3.354l 4

∗

l = / l1 = Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

m 11 =

4E I 4E I EI + = 7.578 l∗ l l

230

3 Solution of Complex Frames



6E I 6E I 6E I − ∗ + ∗ 2 · l1 = 0 l l (l )   1 6E I 6E I 6E I EI = + − l1 = −2.367 l∗ l 1.25l 2l l

M O = 0 → r21 · 2l −

r21 • δ2 = 1

√ EI 5 3E I 6E I = −2.367 − · l2 l (l∗ )2 2   √  5 6E I M O = 0 → r22 · 2l + 2 · + (l∗ )2 2 √   12E I 5 6E I 1 − ∗ 3 · · l1 = 0 · −2 l2 2 2 (l ) m 12 = +

r22 = • External load

12E I 12E I EI + 2 ∗ = 13.73 3 3 4l l l l

3.13 Worked Example 13

231

• Linear thermal variation

m 10

  2α(Δ T ) pl2 pl2 − EI =− =− 12 h 8 r20 = −

Calculation of Unknowns 

pl 4

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       m 11 m 12 ϕ1 m 10 · =− r21 r22 δ2 r20    7.578E I    2 I − 2.367E ϕ1 − pl8 l l2 · =− I 13.73E I − 2.367E δ2 − pl l2 l3 4 ⎧ 3 pl ⎪ ⎪ ⎨ ϕ1 = 0.0229 EI ⇒ 4 ⎪ ⎪ ⎩ δ2 = 0.0222 pl EI

Summary of Bending Moments

232

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.14 Worked Example 14

233

Qualitative Deformed Shape

3.14 Worked Example 14

• • • •

E I AB = E I E I BC → ∞ EA → ∞ I k = 4E l3

Solution The structure results to be a sway frame. The mixed method is used to solve the structure by adopting as static unknown the moment X1 of node B and as kinematic unknowns the vertical translations δ2 of node A and δ3 of node B.

234

3 Solution of Complex Frames

Mixed Method

⎧ ⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0 r X + r22 δ2 + r23 δ3 + r20 = 0 ⎩ 21 1 r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0 Kinematics • X1 = 1

ϕ11 =

l 3E I

r21 = − r31 = • δ2 = 1

1 l

1 l

3.14 Worked Example 14

235

ϕ12 = r22 = k =

1 l 4E I l3

r32 = 0 • δ3 = 1

ϕ13 = −

1 l

r23 = 0 r33 = 0 • External load

ϕ10 =

pl3 24E I

r20 = −

pl 2

236

3 Solution of Complex Frames

r30 = −

pl 2

Calculation of Unknowns ⎧ ⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0 r X + r22 δ2 + r23 δ3 + r20 = 0 ⎩ 21 1 r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0 ⎡ ⎡ ⎤ ⎡ ⎤ ⎤ X1 ϕ10 ϕ11 ϕ12 ϕ13 ⎣ r21 r22 r23 ⎦ · ⎣ δ2 ⎦ = −⎣ r20 ⎦ r31 r32 r33 δ3 r30 ⎤ ⎡ ⎡ l 1 ⎤ 1⎤ ⎡ pl3 − X 1 3E I l l 24E I ⎥ ⎣ − 1 4E3 I 0 ⎦ · ⎣ δ2 ⎦ = −⎢ ⎦ ⎣ − pl l l 2 pl 1 δ 0 0 − 3 l 2 ⎧ 2 ⎪ ⎨ X 1 = 0.5 pl4 pl ⇒ δ2 = 0.25 E I ⎪ ⎩ δ = 0.4583 pl4 3 EI Summary of Bending Moments

Summary of Shear Forces

3.14 Worked Example 14

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

237

238

3 Solution of Complex Frames

3.15 Worked Example 15

• E I = constant • EA → ∞ • k = EI l

Solution The structure is a sway frame. The statically determined part CD is eliminated and replaced by a resultant force and moment.

3.15 Worked Example 15

239

Displacement Method

⎧ ⎪ m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m ϕ + m ϕ + m ϕ + m δ + m = − pl2 21 1 22 2 23 3 24 4 20 pl2 ⎪ m ϕ + m ϕ + m ϕ + m δ + m 31 1 32 2 33 3 34 4 30 = 2 ⎪ ⎪ ⎩r ϕ + r ϕ + r ϕ + r δ + r = 0 41 1 42 2 43 3 44 4 40 Kinematics • ϕ1 = 1

m 11 =

4E I EI 7E I EI 4E I +k = + = = 2.33 3l 3l l 3l l m 21 =

EI 2E I = 0.67 3l l m 31 = 0

r41 = −

6 EI 2 EI EI =− = −0.67 2 9 l2 3 l2 l

240

3 Solution of Complex Frames

• ϕ2 = 1

m 12 = m 22 =

EI 2 EI = 0.67 3 l l

4 EI 32 E I EI 4 EI + = = 2.13 5 l 3 l 15 l l m 32 =

r42 = −

EI 2 EI = 0.4 5 l l

6 EI 2 EI EI =− = −0.67 2 9 l2 3 l2 l

• ϕ3 = 1

m 13 = 0 m 23 =

EI 2 EI = 0.4 5 l l

3.15 Worked Example 15

241

m 33 =

1 EI EI 4 EI + = 1.3 5 l 2 l l r43 = 0

• δ4 = 1

m 14 = −

6 EI EI = −0.67 2 9 l2 l

m 24 = −

6 EI EI = −0.67 2 9 l2 l m 34 = 0

r44 =

12 E I 4 EI EI = = 0.44 3 3 3 27 l 9 l l

• External load

m 10 = 0

242

3 Solution of Complex Frames

m 20 = − m 30 =

25 2 pl 12

25 2 pl 12

r40 = 0 Calculation of Unknowns ⎧ ⎪ m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0 ⎪ ⎪ ⎨ m ϕ + m ϕ + m ϕ + m δ + m = − pl2 21 1 22 2 23 3 24 4 20 2 ⎪ m 31 ϕ1 + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = pl2 ⎪ ⎪ ⎩r ϕ + r ϕ + r ϕ + r δ + r = 0 41 1 42 2 43 3 44 4 40 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ m 11 m 12 m 13 m 14 ϕ1 m 10 ⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥ ⎢ m 20 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎣ m 31 m 32 m 33 m 34 ⎦ · ⎣ ϕ3 ⎦ = −⎣ m 30 ⎦ r41 r42 r43 r44 δ4 r40 ⎡ ⎤ ⎡ ⎤ ⎡ 0 0 −0.67 El2I 2.33 ElI 0.67 ElI ϕ1 25 2 ⎢ 0.67 E I 2.13 E I 0.4 E I −0.67 E2I ⎥ ⎢ ϕ2 ⎥ ⎢ pl − pl2 ⎢ l l l l ⎥·⎢ ⎢ ⎥ = ⎢ 12 2 EI EI 25 2 ⎣ ⎦ ⎦ ⎣ 0 0.4 l 1.3 l 0 ϕ3 ⎣ − 12 pl + pl2 EI EI EI δ4 0 0.44 l3 −0.67 l2 −0.67 l2 0 ⎧ pl3 ⎪ ⎪ ⎪ ϕ1 = 0.4236 ⎪ ⎪ EI ⎪ ⎪ ⎪ 3 ⎪ pl ⎪ ⎪ ⎨ ϕ2 = 1.6943 EI ⇒ ⎪ pl3 ⎪ ⎪ ϕ = −1.7393 ⎪ 3 ⎪ ⎪ EI ⎪ ⎪ ⎪ 4 ⎪ pl ⎪ ⎩ δ4 = 3.1769 EI

⎤ ⎥ ⎥ ⎥ ⎦

3.15 Worked Example 15

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

243

244

Qualitative Deformed Shape

3 Solution of Complex Frames

3.16 Worked Example 16

245

3.16 Worked Example 16

• • • • •

E I = constant EA → ∞ 1 pl4 δ = 10 EI k = ElI P = pl

Solution The ABCD structure is the statically indeterminate frame, while the remaining portion of the structure constitutes an isostatic appendage.

The three-hinged arch FEC results loaded by a thermal variation that deforms the rod, but since FEC is statically determined this does not induce any thermal co-action.

246

3 Solution of Complex Frames

The statically determined portion BEFGC can then be easily analyzed by the node method; globally it transmits to the statically indeterminate portion ABCD the nodal actions highlighted in the following figure. We then proceed to solve this structure using the displacement method.

The structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node D and the horizontal translation δ3 of node C.

3.16 Worked Example 16

247

Displacement Method

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

  7E I 3 EI = m 11 = 2 + 2 l 2l m 21 = 0

248

3 Solution of Complex Frames

r31 = −

6E I 3E I =− 2 2l (2l)2

• ϕ2 = 1

m 12 = 0 m 22 =

4E I 3E I +k = l l

r32 = − • δ3 = 1

3E I l2

3.16 Worked Example 16

249

m 13 = −

6E I 3E I =− 2 2 2l (2l)

m 23 = −

3E I l2

r33 =

12E I 3E I 9E I + 3 = l 2l3 (2l)3

m 10 =

3 2 3E I 1 pl4 = pl 2 4l 10 E I 40

• Imposed displacement

m 20 = 0 r30 = 0 Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎡ ⎤ ⎡ ⎤ ⎤ m 11 m 12 m 13 ϕ1 m 10 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30

250

3 Solution of Complex Frames

⎤ ⎡ ⎤ ⎡ 7 3 ⎤ ⎡ 0 − 2l ϕ1 0.075 pl2 EI ⎣ 2 ⎦ 0 4 − 3l ⎦ · ⎣ ϕ2 ⎦ = −⎣ 0 l 9 3 3 δ3 0 − 2l − l 2l2 ⎧ 3 pl ⎪ ⎨ ϕ1 = −0.030 E I pl3 ⇒ ϕ2 = −0.015 E I ⎪ 4 ⎩ δ3 = −0.020 pl EI Summary of Bending Moments

Summary of Shear Forces

3.16 Worked Example 16

Moment, Shear and Axial Force Diagrams

251

252

Qualitative Deformed Shape

3 Solution of Complex Frames

3.17 Worked Example 17

253

3.17 Worked Example 17

• EA → ∞ pl2 h • αΔ T = 16E I l • h = 10 Solution The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B.

254

3 Solution of Complex Frames

Displacement Method

⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 Kinematics • ϕ1 = 1

m 11

  4 EI EI 8+ = 10.53 = l 1.581 l m 21 =

EI 2 EI = 1.265 1.581 l l

3.17 Worked Example 17

r31 = −

255

12E I 6 sin α E I EI + = −11.241 2 2 2 2 l 1.581 l l

• ϕ2 = 1

m 12 = 1.265

r32 • δ3 = 1

EI l 

 4 4 EI EI + = 5.060 m 22 = l 1.581 1.581 l   6 sin α 6 EI EI · cos α + = −1.581 2 = 2 − 2 l 1.581 1.581 l

256

3 Solution of Complex Frames

m 13 = −11.241 m 23 = −1.581 r33 =

EI l2

EI l2

EI EI (24 + 2.733 + 0.304) = 27.037 3 3 l l

• External load and linear thermal variation

m 10 = − m 20 =

3 2 1 2 1 pl + pl = − pl2 16 16 8

3 2 1 2 1 2 pl − pl − pl = 0 16 8 16

3.17 Worked Example 17

257

r30 = 0.375 pl Calculation of Unknowns ⎧ ⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0 m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0 ⎩ 21 1 r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ m 11 m 12 m 13 ϕ1 m 10 ⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦ r31 r32 r33 δ3 r30 ⎤ ⎡ ⎤ ⎡ 1 2 ⎤ ⎡ 10.53 1.265 − 11.241 ϕ1 − 8 pl l EI ⎣ ⎦ · ⎣ ϕ2 ⎦ = −⎣ ⎦ 1.265 5.060 − 1.518 0 l l 11.241 1.518 27.037 δ3 0.375 pl − l − l l2 ⎧ 3 ⎪ −3 pl ⎪ ⎪ ϕ1 = −4.888 · 10 ⎪ EI ⎪ ⎪ ⎨ 3 pl ⇒ ϕ2 = −3.610 · 10−3 ⎪ EI ⎪ ⎪ ⎪ 4 ⎪ pl ⎪ −2 ⎩ δ3 = −1.610 · 10 EI Summary of Bending Moments

258

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.18 Worked Example 18

Qualitative Deformed Shape

3.18 Worked Example 18

• E I = constant • EA → ∞

259

260

3 Solution of Complex Frames

Solution The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the horizontal translation δ2 of node B. Member CD represents a statically determined portion.

Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

3.18 Worked Example 18

261

Kinematics • ϕ1 = 1

m 11 = (3 + 4) r21 = (−6 + 3)

7E I EI = l l

3E I EI =− 2 l2 l

• δ2 = 1

m 12 = (−6 + 3) r22 = (12 + 3)

EI 3E I =− 2 2 l l

15E I EI = 3 l l3

262

3 Solution of Complex Frames

• External load

1 m '10 = − pl2 8 ' r20 =

3 pl 8

m ''10 =

1 2 pl 4

'' r20 =

3 pl 4

3.18 Worked Example 18

263

m ''' 10 = 0 ''' r20 = pl

Due to the superposition of effects, it is possible to calculate m 10 and r20 . 1 2 1 2 1 pl = pl2 m 10 = m '10 + m ''10 + m ''' 10 = − pl + 8 4 8 ' '' ''' r20 = r20 + r20 + r20 =

Calculation of Unknowns 

3 3 17 pl + pl = pl 8 4 8

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       m 11 m 12 ϕ m · 1 = − 10 r21 r22 δ2 r20       pl2 E I 7 − 3l ϕ1 · = − 178pl δ2 l − 3l 15 l2 8 ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = −8.5937 · 10−2 EI ⇒ 4 ⎪ pl ⎪ ⎩ δ2 = −1.5885 · 10−1 EI

Summary of Bending Moments

264

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.19 Worked Example 19

Qualitative Deformed Shape

3.19 Worked Example 19

• E I = constant • E A BC = E AC D → ∞ I • E A AC = 768E l2 pl2 αΔ T • h = 6E I • m = pl 2

265

266

3 Solution of Complex Frames

Solution Node C can move vertically due to the axial deformability of rod AC. The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node C and the horizontal translation δ2 of node A.

The BC rod, because of the particular loading condition, can be considered an isostatic appendage. It transfers to the frame ACD a vertical load equal to pl/2.

 

F↑ = 0 → VB = VC

M BBC = 0 → VC · l − m · l = 0 → VC = 0.5 pl VB = VC = 0.5 pl

3.19 Worked Example 19

267

Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

m 11 =

4E I 8E I 4E I + = l l l r21 =

6E I l2

268

3 Solution of Complex Frames

• δ2 = 1

m 12 = r22 = k +

6E I l2

12E I 780E I = 3 l l3

• External load

m '10 = −

pl2 12

' r20 =−

pl 2

3.19 Worked Example 19

269

m ''10 = 0 '' r20 =−

pl 2

• Linear thermal variation

m ''' 10 =

2E I αΔ T h

''' r20 =0

Due to the superposition of effects, it is possible to calculate m10 and r20 : m 10 = m '10 + m ''10 + m ''' 10 = −

2E I αΔ T pl2 pl2 +0+ = 12 h 12

' '' ''' r20 = r20 + r20 + r20 =−

pl pl − + 0 = − pl 2 2

270

3 Solution of Complex Frames

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       m 11 m 12 ϕ1 m 10 · =− r21 r22 δ2 r20  2      pl E I 8 6l ϕ1 12 · = − δ l 6l 780 − pl 2 2 l  3 ϕ1 = −1.1444 · 10−2 pl E I ⇒ 4 δ2 = 1.37 · 10−3 pl EI

Summary of Bending Moments

Summary of Shear Forces

3.19 Worked Example 19

Moment, Shear and Axial Force Diagrams

271

272

3 Solution of Complex Frames

Qualitative Deformed Shape

3.20 Worked Example 20

• E I = constant • EA → ∞ pl2 • αΔ T = 10E h I • l = 5h Solution Taking advantage of symmetry, the frame is simplified as shown in the following figure for the kinematic analysis.

3.20 Worked Example 20

273

The simplified structure results to be a fixed-joint frame and it is solved by the displacement method adopting as kinematic unknown the rotation ϕ1 of node B. The loads that the rod AC transfers to node C are initially calculated:

Displacement Method

m 11 ϕ1 + m 10 = 0

274

3 Solution of Complex Frames

Kinematics • ϕ1 = 1

m 11 =

8E I l

• External load, constant and linear thermal variation

3.20 Worked Example 20

275



 6E I 3E I E I αΔ T δT − − l2 l2 h

m 10 =

15 2 pl + 64

m 10 =

pl2 359 2 15 2 3E I 3 pl2 pl + 2 · − = pl 64 l 100 10 1600

Calculation of Unknowns m 11 ϕ1 + m 10 = 0 ϕ1 = −

l pl3 m 10 359 pl2 · = −0.02805 =− m 11 1600 8E I EI

276

Summary of Bending Moments

Summary of Shear Forces

3 Solution of Complex Frames

3.20 Worked Example 20

Moment, Shear and Axial Force Diagrams

277

278

Qualitative Deformed Shape

3 Solution of Complex Frames

3.21 Worked Example 21

279

3.21 Worked Example 21

• • • •

E I = constant EA → ∞ I k = 6E l3 Pl2 αΔ T = 12E I

Solution Taking advantage of symmetry, the frame is simplified as shown in the following figure. The simplified structure results to be a sway frame.

The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the vertical translation δ2 of node C.

280

3 Solution of Complex Frames

Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

m 11 =

8E I l

r21 = −

6E I l2

3.21 Worked Example 21

281

• δ2 = 1

m 12 = − r22 =

6E I l2

12E I 18E I +k = l3 l3

• External load

m '10 = −Pl ' r20 = −P

282

3 Solution of Complex Frames

• Imposed displacement

m ''10 = '' r20 =−

6E I δT = Pl l2

12E I δT = −2P l3

Due to the superposition of effects, it is possible to calculate m 10 and r20 . m 10 = m '10 + m ''10 = 0 ' '' r20 = r20 + r20 = −3P

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       E I 8 − 6l ϕ1 0 · =− δ2 −3P l − 6l 18 l2 ⎧ 2 ⎪ ⎪ ϕ1 = 1 Pl ⎨ 6 EI ⇒ 3 ⎪ 2 ⎪ ⎩ δ2 = Pl 9 EI

3.21 Worked Example 21

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

283

284

Qualitative Deformed Shape

3 Solution of Complex Frames

3.22 Worked Example 22

285

3.22 Worked Example 22

• E I = constant • EA → ∞ I • k = 10E l Solution The structure is symmetrically loaded, and taking advantage of symmetry, the frame is simplified as shown in the following figure. A fixed end constraint is inserted at node C as the continuity constraint of the rod BD is recreated, while the presence of the rod CF prevents any possible vertical displacement.

The kinematic analysis shows that the simplified structure is a fixed-joint frame— as it represents a three-hinged arch—and it is solved with the force method by adopting as static unknown the moment X1 of node B.

286

3 Solution of Complex Frames

Force Method

ϕ11 X 1 + ϕ10 = 0 Kinematics • X1 = 1

ϕ11 =

4l 1 9l l 47l 5l + + = + = 4E I 4E I k 4E I 10E I 20E I

3.22 Worked Example 22

287

• External load

ϕ10 =

5 pl3 13 pl3 2 p(4l)3 + = 48E I 3 EI 3 EI

Calculation of Unknowns ϕ11 X 1 + ϕ10 = 0 X1 = −

260 2 ϕ10 13 pl3 20 E I · =− pl =− ϕ11 3 E I 47 l 141 ⇒ X 1 = −1.8439 pl2

288

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.22 Worked Example 22

Qualitative Deformed Shape

289

290

3 Solution of Complex Frames

3.23 Worked Example 23

• • • • •

E I AB = E IC D = E I E I BC → ∞ EA → ∞ I k1 = 2E l3 3E I k2 = l

Solution The structure results to be a sway frame. The mixed method is used to solve the structure by adopting as static unknown the moment X1 of node B and as kinematic unknowns the vertical translations δ2 of node A and δ3 of node B.

Mixed Method

3.23 Worked Example 23

291

⎧ ⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0 r X + r22 δ2 + r23 δ3 + r20 = 0 ⎩ 21 1 r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0 Kinematics • X1 = 1

ϕ11 =

1 l 2 l + = 3E I k2 3 EI r21 = −

1 l

r31 =

2 l

ϕ12 =

1 l

• δ2 = 1

r22 = 0 + k1 = r32 = 0

2E I l3

292

3 Solution of Complex Frames

• δ3 = 1

  2 1 1 + =− ϕ13 = − l l l r23 = 0 r33 = 0 • External load

ϕ10 =

pl3 24E I

r20 = −

pl 2

r30 = −

pl 2

3.23 Worked Example 23

Calculation of Unknowns ⎧ ⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0 r X + r22 δ2 + r23 δ3 + r20 = 0 ⎩ 21 1 r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0 ⎡ 3 ⎤ ⎡2 l ⎤ 2 1 ⎤ ⎡ pl − X 1 3 EI l l 24E I ⎥ ⎣ 2 0 0 ⎦ · ⎣ δ2 ⎦ = −⎢ ⎦ ⎣ − pl l 2 1 2E I δ3 − l 0 l3 − pl 2 ⎧ ⎪ X 1 = 0.25 pl2 ⎪ ⎪ ⎪ ⎪ ⎨ pl4 ⇒ δ2 = 0.2917 E I ⎪ ⎪ ⎪ ⎪ pl4 ⎪ ⎩ δ3 = 0.375 EI Summary of Bending Moments

Summary of Shear Forces

293

294

Moment, Shear and Axial Force Diagrams

Qualitative Deformed Shape

3 Solution of Complex Frames

3.24 Worked Example 24

295

3.24 Worked Example 24

• E I = constant • EA → ∞ • k = 3El I Solution The structure results to be a fixed-joint frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotations ϕ1 and ϕ2 of node D.

Displacement Method

296

3 Solution of Complex Frames



m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0

Kinematics • ϕ1 = 1

m 11 =

6E I 3E I +k = l l

m 21 = −k = −

3E I l

m 12 = −k = −

3E I l

• ϕ2 = 1

m 22 =

3E I 9E I 3E I + +k = l l l

3.24 Worked Example 24

297

• External load

m 10 =

pl2 8

m 20 = 0 Calculation of Unknowns 

m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0  2     pl E I 6 −3 ϕ1 =− 8 · ϕ2 l −3 9 0 ⎧ 3 ⎪ ⎪ ϕ1 = −0.025 pl ⎨ EI ⇒ 3 ⎪ pl ⎪ ⎩ ϕ2 = −0.008 EI

Summary of Bending Moments

298

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.25 Worked Example 25

Qualitative Deformed Shape

3.25 Worked Example 25

299

300

3 Solution of Complex Frames

• E I = constant • EA → ∞ • k = 2El I pl2 • αΔ T = 10E τ I • l = 5τ Solution The structure results to be a sway frame. In particular, the vertical displacement of node D is due to the presence of the slider, while nodes B and C are fixed. Therefore, the frame is solved by the displacement method by adopting as kinematic unknowns the rotations ϕ1 and ϕ2 of node C.

Displacement Method



m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0

3.25 Worked Example 25

301

Kinematics • ϕ1 = 1

m 11 =

3E I 9E I 4E I + +k = l l l

m 21 = −k = − • ϕ2 = 1

2E I l

302

3 Solution of Complex Frames

m 12 = −k = − m 22 =

2E I l

3E I EI +k = l l

• External load, constant and linear thermal variation

3.25 Worked Example 25

m 10 = −

303

3E I 5 pl2 · l2 2 10E I m 20 = −

Calculation of Unknowns 

  l 3 l = − pl2 5 20 pl2 3

m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0       3 E I 9 −2 − 20 pl2 ϕ1 =− · 2 ϕ2 l −2 3 − pl3 ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = 0.0486 EI ⇒ 3 ⎪ pl ⎪ ⎩ ϕ2 = 0.1435 EI

304

Summary of Bending Moments

Summary of Shear Forces

3 Solution of Complex Frames

3.25 Worked Example 25

Moment, Shear and Axial Force Diagrams

305

306

Qualitative Deformed Shape

3 Solution of Complex Frames

3.26 Worked Example 26

307

3.26 Worked Example 26

• E I = constant • E A ABC D E → ∞ • E AC F = 5 El2I 1 pl4 • δ = 100 EI Solution The frame is studied by exploiting symmetry. The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the vertical translation δ2 of node C. The rod CF, with finite axial deformability, can be idealized as a translational spring of stiffness k: k=

5 EI 1 E AC F = 2 l 2 l3

308

3 Solution of Complex Frames

Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

m 11 =

4E I 8E I 4E I + = l l l r21 = −

6E I l2

3.26 Worked Example 26

309

• δ2 = 1

m 12 = − r22 =

6E I l2

12E I 12E I 5 EI EI +k = + = 14.5 3 3 3 3 l l 2 l l

• External load

m '10 = −

pl2 pl2 + 12 16

' r20 =−

pl 2

310

3 Solution of Complex Frames

• Imposed displacement

m ''10 = 0 '' r20 = −k · δ = −

5 pl 5 EI 1 pl4 =− · 3 2 l 100 E I 200

In the adopted scheme, the forces on the spring are positive in the case of spring shortening (see case δ2 = 1). For the case under study, the roller has no translation (δ2 = 0) and the constraint at the base of the spring undergoes a lowering causing '' the spring to elongate, with which a force r20 < 0 is then associated. Due to the superposition of effects, it is possible to calculate m 10 and r20 . m 10 = m '10 + m ''10 = − ' '' r20 = r20 + r20 =−

Calculation of Unknowns 

pl2 pl2 pl2 + =− 12 16 48

21 pl pl 5 pl − =− 2 200 40

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0       2 E I 8 − 6l ϕ1 − pl 48 · =− δ2 l − 6l 14.5 − 2140pl l2 ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = 0.043151 EI ⇒ 4 ⎪ pl ⎪ ⎩ δ2 = 0.0541 EI

3.26 Worked Example 26

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

311

312

Qualitative Deformed Shape

3 Solution of Complex Frames

3.27 Worked Example 27

313

3.27 Worked Example 27

• E I = constant • EA → ∞ • k = E5lI pl2 • αΔ T = 10E h I Solution The frame is studied by exploiting symmetry. The structure results to be a fixedjoint frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the rotation ϕ2 of node D.

314

3 Solution of Complex Frames

At point C, in order to consider symmetry, it would be necessary to introduce a slider with a sliding plane parallel to the symmetry axis. In that case, a kinematic analysis of the simplified structure shows that it results a fixed-joint frame, and consequently from the kinematic point of view point C can neither rotate nor translate and could be assimilated to a fixed end. Despite what has been said from the kinematic point of view, the slider is retained since from the static point of view no forces parallel to the axis of symmetry arise in that point. Displacement Method



m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0

3.27 Worked Example 27

315

Kinematics • ϕ1 = 1

m 11 =

4E I 4 · 2 EI EI 4E I + + √ = 11.58 l l l 5 l m 21 =

2E I l

Due to what was previously discussed, although there is a slider at point C, since point C—based on the kinematic analysis—cannot translate, the stiffness coefficients related to the fixed end—fixed end conditions are adopted for the rod BC. • ϕ2 = 1

m 12 = m 22 =

2E I l

4E I EI 21 E I 4E I +k = + = l l 5l 5 l

316

3 Solution of Complex Frames

• External load

m 10 = −

2E I αΔ T 2 pl2 17 1 2 1 2 − pl = − − pl = − pl2 h 12 10 12 60 m 20 = 0

Calculation of Unknowns 

m 11 ϕ1 + m 12 ϕ2 + m 10 = 0 m 21 ϕ1 + m 22 ϕ2 + m 20 = 0      17  E I 8 + √85 2 − 60 ϕ1 = − pl2 · 21 ϕ 0 l 2 2 5 ⎧ pl3 ⎪ ⎪ ⎨ ϕ1 = 2.67 · 10−2 EI ⇒ 3 ⎪ ⎪ ⎩ ϕ2 = −1.27 · 10−2 pl EI

3.27 Worked Example 27

Summary of Bending Moments

Summary of Shear Forces

317

318

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.28 Worked Example 28

319

Qualitative Deformed Shape

3.28 Worked Example 28

• E I = constant • EA → ∞ Solution Rods BEC represent a statically determined appendage (three-hinged arch), and therefore only the actions that the rods BEC transmit to the rods ABCD at nodes B and C can be carried over to the frame ABCD.

320

3 Solution of Complex Frames

The structure results to be a sway frame and it can be solved with the displacement method.

The structure is symmetric and anti-symmetrically loaded.

Taking advantage of the anti-symmetry, the structure can be divided in half, and a roller can be inserted at midspan of rod BC. At the roller location the moment is zero and a shear force is present. The rotation ϕ1 of node B and the horizontal displacement δ2 of joint B are adopted as kinematic unknowns.

3.28 Worked Example 28

321

Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

m 11 =

4E I 7E I 3E I + = l l l

322

3 Solution of Complex Frames

r21 = −

6E I l2

• δ2 = 1

m 12 = − r22 =

6E I l2

12E I l3

• External load

m 10 = r20 = −

pl2 12

pl pl − = − pl 2 2

3.28 Worked Example 28

323

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0  2      pl E I 7 − 6l ϕ1 · = − 12 6 12 δ2 l − l l2 − pl ⎧ 3 ⎪ ⎪ ϕ1 = 0.104 pl ⎨ EI ⇒ 4 ⎪ pl ⎪ ⎩ δ2 = 0.135 EI

Summary of Bending Moments

Summary of Shear Forces

324

Moment, Shear and Axial Force Diagrams

3 Solution of Complex Frames

3.29 Worked Example 29

325

Qualitative Deformed Shape

3.29 Worked Example 29

• • • •

E I AB → ∞ E I D BC = E I E A ABC → ∞ 1 EI E A B D = 10 l2

• δ=

1 pl4 200 E I

Solution The infinitely rigid rod AB is a statically determined appendage, so it can be simplified with a roller with a horizontal axis applied at node B and a concentrated vertical load also applied at node B.

326

3 Solution of Complex Frames

To consider the finite stiffness of the rod BD, a spring with stiffness k = E AlB D = equal to the axial stiffness of the rod is introduced. The structure is then simplified as follows: 1 EI 10 l3

The structure is solved with the displacement method by adopting the rotation ϕ1 of node B and the vertical displacement δ2 of node D as kinematic unknowns. Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

3.29 Worked Example 29

327

Kinematics • ϕ1 = 1

m 11 =

3E I 6E I 3E I + = l l l r21 =

3E I l2

m 12 =

3E I l2

• δ2 = 1

r22 =

3E I 31 E I +k = l3 10 l3

328

3 Solution of Complex Frames

• External load and imposed displacement

m 10 = − r20 = −

3E I 3 pl2 pl2 7 − 2 δ=− − pl2 = − pl2 8 l 8 200 50

3E I 3 41 13 13 pl − 3 δ = − pl − pl = − pl 8 l 8 200 25

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0     7 2  E I 6 3l ϕ1 − 50 pl · = − 31 δ2 − 41 pl l 3l 10l 2 25

3.29 Worked Example 29

329

 ⇒

3

ϕ1 = −0.467 pl EI 4 δ2 = 0.981 pl EI

Summary of Bending Moments

Summary of Shear Forces

Moment, Shear and Axial Force Diagrams

330

Qualitative Deformed Shape

3 Solution of Complex Frames

3.30 Worked Example 30

331

3.30 Worked Example 30

• E I = constant • EA → ∞ Solution The structure results to be a sway frame. The displacement method is used to solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the horizontal translation δ2 of node C.

332

3 Solution of Complex Frames

Displacement Method



m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0

Kinematics • ϕ1 = 1

m 11 =

3E I 7E I 4E I + = l l l r21 =

6E I l2

3.30 Worked Example 30

333

• δ2 = 1

m 12 = r22 =

6E I l2

12E I l3

• External load

m 10 =

pl2 8

m 20 = − pl

334

3 Solution of Complex Frames

Calculation of Unknowns 

m 11 ϕ1 + m 12 δ2 + m 10 = 0 r21 ϕ1 + r22 δ2 + r20 = 0      1 2 E I 7 6l pl ϕ1 =− 8 6 12 · δ2 − pl l l l2 ⎧ 3 ⎪ ⎪ ϕ1 = −0.1563 pl ⎨ EI ⇒ 4 ⎪ pl ⎪ ⎩ δ2 = 0.1615 EI

Summary of Bending Moments

Summary of Shear Forces

3.30 Worked Example 30

Moment, Shear and Axial Force Diagrams

335

336

Qualitative Deformed Shape

3 Solution of Complex Frames

Appendix A

Table A.1 provides the influence coefficients related to the force method. Rotations and displacements are referred to the extremities of the rods. Coefficients provided in Table A.1 are associated to unitary forces and moments. Table A.1 Influence coefficients for the force method Case ID

Rotation/displacement values

1

ϕ11 = ϕ21 =

2

3

4

l 3E I l 6E I

ϕ22 =

l 6E I l 3E I

ϕ11 =

l EI

δ11 =

l2 2E I

ϕ12 =

l 4E I m 21 = 21

ϕ11 =

(continued)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil Engineering, https://doi.org/10.1007/978-3-031-35267-6

337

338

Appendix A

Table A.1 (continued) Case ID

Rotation/displacement values

5

ϕ11 =

l2 2E I

δ11 =

l3 3E I

Pl2 16E I

6

ϕ10 = ϕ20 =

7

ϕ10 =

( ) Pb l2 −b2 6lE I

ϕ20 =

Pab(2l−b) 6lE I

8

ϕ10 = ϕ20 =

9

ϕ10 =

pl 48E I

m 20 =

pl 8

10

11

pl3 24E I

2

2

ϕ10 =

7 pl3 360E I

ϕ20 =

pl3 45E I

ϕ10 = ϕ20 =

Ml 24E I

(continued)

Appendix A

339

Table A.1 (continued) Case ID

Rotation/displacement values

12

ϕ10 =

pl3 24E I

δ10 =

pl4 30E I

ϕ10 =

pl3 6E I

δ10 =

pl4 8E I

13

14

ϕ10 = ϕ20 =

15

ϕ10 =

16

3δ 2l I m 20 = 3E l2

δ l

δ

ϕ10 = ϕ20 = αlΔT t (α = thermal expansion coefficient) (t = beam height)

Table A.2 provides the influence coefficients related to the displacement method. Forces and moments are referred to the extremities of the rods (reaction forces and moments on constraints). Coefficients provided in Table A.2 are associated to unitary displacements and rotations.

340

Appendix A

Table A.2 Influence coefficients for the displacement method CASE ID

Force/moment values

1

m 11 = m 21 =

4E I l 2E I l

r11 = r21 =

2

m 11 =

6E I l2

3E I l

m 21 = 0

r11 = r21 =

3

m 11 = m 21 = r11 = r21 =

4

3E I l2

m 11 =

6E I l2 12E I l3

3E I l2

m 21 = 0

r11 = r21 =

5

3E I l3

m 11 = m 21 = r11 = r21 = 0

EI l

(continued)

Appendix A

341

Table A.2 (continued) CASE ID

Force/moment values

6

m 11 =

12μ(E I )2 +4E I l l(l+4μE I )

m 21 =

6μ(E I )2 +2E I l (l+4μE I )(l+3μE I )

I ) +2lE I r11 = r21 = 3 6μ(E l2 (l+4μE I ) 2

l+2μE I l+3μE I

(μ = rotational spring stiffness)

7

m 11 =

6lE I +12μ(E I )2 l2 (l+4μE I )

m 21 =

6E I l(l+4μE I )

I l+μE I r11 = r21 = 12E l3 l+4μE I (μ = rotational spring stiffness)

8

m 10 = m 20 = r10 = r20 =

9

m 10 = r10 = r20 =

10

P 2

3Pl 16 11P 16 5P 16

m 10 = m 20 = r10 = r20 =

11

m 10 =

Pl 8

pl2 12

pl 2

pl2 8

m 20 = 0 r10 = r20 =

5 pl 8 3 pl 8

(continued)

342

Appendix A

Table A.2 (continued) CASE ID

Force/moment values

12

m 10 = m 20 = r10 = r20 =

13

m 10 =

M 8

r10 = r20 =

14

M 4 3M 2l

9M 8l

m 10 = m 20 =

2E I αΔT t

n 10 = n 20 = αΔT E A (A = beam cross-section) (t = beam height)

15

m 11 = m 21 = r10 = 0

16

m 10 =

pl2 3

m 20 =

pl2 6

EI l

r10 = pl

17

3E I αΔT t r20 = 3E ItlαΔT

m 10 = r10 =

Appendix B

Table B.1 summarizes, for each complex frame example (Chap. 3), the simple reference cases covered in Chap. 2 present in it. Table B.1 Simple reference cases covered in each complex frame example Simple reference cases Complex frame ID

A

1

•

2

•

3

•

4

B

C

D

E

•

•

•

• • •

5

•

6

•

9

•

10

•

• •

8

•

•

•

•

•

•

•

•

•

•

•

•

• •

•

• •

16

• •

L

•

15 •

K

•

•

•

18

J

•

•

17

I

•

12 14

H

•

•

11 13

G

•

•

7

F

• •

• •

•

• • •

19

•

20

•

•

•

•

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil Engineering, https://doi.org/10.1007/978-3-031-35267-6

(continued) 343

344

Appendix B

Table B.1 (continued) Simple reference cases Complex frame ID

A

21 22

B

• •

24

E

F

•

•

G

H

I

J

K

L

•

•

•

•

•

•

•

25

•

26

•

• •

28

• •

•

•

•

•

•

•

•

29 30

D

• •

23

27

C

• •

•

• •

•

•