Nonlinear Problems in Machine Design 0-8493-2037-1

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Nonlinear Problems in

MACHINE

DESIGN

Nonlinear Problems in

MACHINE DESIGN Eliahu Zahavi David Barlam

CRC Press Boca Raton London New York Washington, D.C.

Library of Congress Cataloging-in-Publication Data Zahavi, Eliahu. Nonlinear problems in machine design / Eliahu Zahavi and David Barlam. p. cm. Includes bibliographical references and index. ISBN 0-8493-2037-1 (alk. paper) 1. Machine design. 2. Computer-aided design. I. Barlam, David. II. Title. TJ233.Z255 2000 00-063033 621.8′15—dc21 CIP

This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage or retrieval system, without prior permission in writing from the publisher. The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from CRC Press LLC for such copying. Direct all inquiries to CRC Press LLC, 2000 N.W. Corporate Blvd., Boca Raton, Florida 33431. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation, without intent to infringe.

© 2001 by CRC Press LLC No claim to original U.S. Government works International Standard Book Number 0-8493-2037-2 Library of Congress Card Number 00-063033 Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper

Preface Designing machines, by its very nature, is a creative process that requires—in addition to inner resources such as intuition, flair, and understanding—a comprehensive knowledge of engineering science. The task of designing modern machinery, dependable in performance and durability, challenges one with complex problems such as fatigue, friction, excessive deformation, and others that comprise nonlinear aspects of engineering analysis. Any attempt to provide simple answers can only be at the expense of the machine’s performance. Today, with advanced numerical programs that are capable of analyzing nonlinear processes, achieving the optimal solutions is within reality. But, as with all other automatic outputs, the need for professional oversight of computer generated results cannot be overestimated. The authors’ desire to share their accumulated knowledge in applying engineering theories from which the computer programs were derived, coupled with their practical experience in verifying the accuracy and applicability of the automatic results, has prompted this work. The book’s ultimate aim is to acquaint readers with the modern analytical methods of machine design, enabling them to use them in daily applications. To assure foolproof results, readers are presented with the tools to deal with the most prevalent problems. The emphasis in the book, therefore, is split between the theoretical analyses and the practical processes of problem solving. In the former, the authors give prominence to the appropriate theories, while the latter comprises a detailed presentation of the application process of the finite element method (FEM), accompanied by a number of representative cases. The illustrative problems are solved using two independent programs: ANSYS and MSC.NASTRAN finite element programs. The purpose of this double bind is to demonstrate the reliability of the results. A number of people provided valuable reviews and constructive advice that proved indispensable to us. To all of them (and among these, we would like to single out Simcha Barnefi), the authors are grateful. While the book was in progress, Frank Marx of Ansys, Inc. provided help with the computer programs, for which we thank him. We thank David Salton of MSI, Ltd., Israel, for his invaluable professional advice. Toni Barlam, a talented illustrator, deserves special thanks for his devotion to our project and contribution of most of the drawings in the book. Bob Stern, of CRC Press, deserves a very special mention: we are grateful to him for his patience, understanding, and guidance in completing this project.

About the Authors Dr. Eliahu Zahavi is professor emeritus at the Ben-Gurion University in Israel, where for the past 30 years he has been active both as a teacher and a researcher. He joined Ben-Gurion University after obtaining a DSc. He was responsible for developing the machine design group at Ben-Gurion. Prior to that, he was involved in the aerospace industry as a design engineer. His special interest is in the application of computers in machine design. He is the author of the books Finite Element Method in Machine Design and Fatigue Design, and he has written numerous papers on the topics of engineering science and engineering education. Dr. David Barlam is a leading Stress Engineer and a Senior Researcher at the Israel Aircraft Industry, specializing in stress and vibration in machinery—the field in which he has accumulated 30 years of experience, in academia as well as in industry. He is an adjunct professor at the Ben-Gurion University. Dr. Barlam’s current industrial experience since 1991, when he immigrated to Israel, includes dealing with diversified problems in aerospace and shipbuilding. Prior to that, he worked as a stress analyst in the heavy engine industry (in Russia). David Barlam received his doctoral degree in the finite element analysis at the Shipbuilding Institute of Leningrad (today, St. Petersburg).

Nomenclature vectors (Appendix A) half length of crack (Chapter 7) current slave node location (Chapter 6) crack length at failure (Chapter 7) factor of number of cycles (Chapter 7) initial crack length (Chapter 7) coefficients of element shape functions in constant-strain triangle (Chapter 2) ak unknown parameters in Galerkin and Rayleigh-Ritz approximation (Chapter 2) amn, bmn, cmn components of matrices (Appendix B) as, bs, a s ′ , b s ′ components of vectors (Appendix A) a0 previous slave node location (Chapter 6) a1, a2, a3 Cartesian coordinates in initial configuration (Chapter 5) 1 2 3 a,a,a Cartesian coordinates in initial configuration (Chapter 5) b width of plate (Chapter 7) b0, b1, b2 coefficients in stress-strain correlation equation (Chapter 5) c hardening modulus (Chapter 4) c fatigue ductility exponent (Chapter 7) di feasible direction (Chapter 3) ds principal values of a deviator (Appendix A) det … determinant of matrix or tensor (Appendix A) e vector of principal direction (Appendix A) e engineering strain (Chapter 4) eh pointwise error for displacements (Chapter 2) ek components of principal direction vector (Chapter 1) emnp permutation symbol (Appendix A) ex, ey, ez normal components of strain deviator (Chapter 1) eε pointwise error for strain (Chapter 2) eσ pointwise error for stress (Chapter 2) e1, e2, e3 principal values of strain deviator (Chapter 1) ek principal direction vectors (Appendix A) a, b, c a a af ai ai ai, bi,ci

i

ek f f, f′ , f″ f f

components of principal direction vectors (Appendix A) vector of contact forces (Chapter 6) function and its derivatives (Chapter 3) function of yield criterion (Chapter 4) Paris crack propagation function (Chapter 7)

fI fk(x) fij fn

compliance function (Chapter 7) coordinate functions (Chapter 1) function of stress distribution in the vicinity of crack (Chapter 7) normal forces in contact zone (Chapter 6)

trial

trial tangential force (Chapter 6)

n+1 t

h h0 h0

corrected tangential force (Chapter 6) correction of tangential force (Chapter 6) vector of penetration (Chapter 6) covariant base vectors (Appendix A) base vectors in initial configuration (Chapter 4) distance between the bodies in contact (Chapter 6) components of metric tensor in initial configuration contravariant (reciprocal) base vectors (Appendix A) reciprocal base vectors in initial configuration (Chapter 5) contravariant components of metric tensor in initial configuration (Chapter 5) height of cylinder in current configuration (Chapter 5) height of cylinder in initial configuration (Chapter 5) initial vertical position of a moving end of bar (Chapter 3)

ik, i k ′ k k l l l l lc l0 l0 l1, l3 m n n n´ ni p pn px , py , pz ∆p q qk r

unit direction vectors of Cartesian coordinate system (Appendix A) critical value of material (Chapter 4) spring rate (Chapter 9) length of bar (Chapters 3, 5) final length of specimen (Chapter 4) length of contact element (Chapter 6) half length of a leaf (Chapter 9) length of a bar in the current state (Chapter 3) length of bar in initial configuration (Chapters 3, 5) original length of specimen (Chapter 4) length of rod (Chapter 4) number of leaves (Chapter 9) unit normal vector (Chapter 1) hardening exponent (Chapter 4) cyclic strain hardening exponent (Chapter 7) number of cycles (Chapter 7) pressure (Chapter 11) vector of traction (Chapter 1) components of tractions (Chapter 1) pressure increment (Chapter 11) lateral distributed load (Chapter 2) specific loads (Chapter 1) position vector in undeformed (initial) configuration (Chapter 5)

ft

f ∆ft g gk gk gn gsk gk gk gsk

r r r rk rn r0 rk s sk sn

radius of yield surface projection on the deviatoric plane (Chapter 4) radius of cylinder in current configuration (Chapter 5) radius of a leaf (Chapter 9) residual (Chapter 1) covariant components of vector r (Appendix A) radius of yield surface at time step tn (Chapter 4) radius of cylinder in initial configuration (Chapter 5) contravariant components of vector r (Appendix A) deviatoric stress tensor (Chapter 4) nodal stresses (Chapter 2) deviatoric stress at time step tn (Chapter 4)

sn + 1 sx , sy , sz s1, s2, s3 s12, s23 t t t t, ti tn t(s) t–1, t–2, t–3 u u u, v, w ue uh ui ui un un us, s = 1,2,3 ut u1, u2 utot ∆ ui ∆u, ∆ui v v v v vn

trial value of deviatoric stress (Chapter 4) normal components of stress deviator (Chapter 1) principal deviatoric stress components deflections of leaves (Chapter 9) unit tangential vector (Chapter 6) traction (Chapter 5) time (Chapter 4) leaf thickness (Chapter 9) time step (Chapter 4) vector formed by components of tensor (Appendix A) tractions on the faces of elemental tetrahedron (Chapter 5) displacement vector (Chapter 1) exact solution (Chapter 2) components of displacement vector (Chapter 1) element nodal displacements (Chapter 2) finite element solution (Chapter 2) displacements of the nodes in contact zone (Chapter 6) displacement vector at iteration i (Chapter 3) vector of nodal displacements (Chapter 2) normal displacements in contact zone (Chapter 6) components of displacement vector (Chapter 5) tangential displacements in contact zone (Chapter 6) displacements of master nodes (Chapter 6) vector of total displacements (Chapter 6) correction vector in Newton–Raphson procedure (Chapter 3) increment of displacement (Chapter 3) virtual displacements field (Chapter 2) displacement vector (Chapter 5) virtual displacement field (Chapter 2) Coulomb law of friction (Chapter 6) virtual nodal displacements (Chapter 2)

r

v0 v0 vol w w w x x, y, z xe xi xS xS0 x1, x2 x1, x2, x3 x1, x2, x3 yA1, yA2, yB1, yB2 A A A [A] A, B, C A A Ai Ak A0 AT A–1 ∆Ac dA1, dA2, dA3 B B B Be BL BN BHN [C] C C

displacement vector of the end of bar (Chapter 5) vertical displacement of the end of bar (Chapter 5) volume of parallelepiped based on base vectors (Appendix A) displacement of beam (Chapter 2) constraint function (Chapter 6) width of leaf (Chapter 9) vector of coordinates within an element (Chapter 2) components of position vector vector of nodal coordinates (Chapter 2) vector of location of the nodes in contact zone after loading (Chapter 6) location of slave node after loading (Chapter 6) location of slave node before loading (Chapter 6) locations of master nodes (Chapter 6) Cartesian components in current configuration (Chapter 5) Cartesian components in current configuration (Chapter 5) deflections of bearing points of leaves (Chapter 9) matrix that defines a nonlinear strain vector (Chapter 5) matrix of contact surface (Chapter 6) antisymmetric part of any tensor (Appendix A) matrix of transformation (Appendix A) matrices (Appendix B) cross-section area (Chapters 1, 4) area of triangle element (Chapter 2) matrix of contact surface at iteration i (Chapter 6) cross-section area of rod-element (Chapter 2) cross-section area in initial configuration (Chapter 5) transpose matrix (Appendix B) inverse matrix (Appendix B) small area in current configuration (Chapter 5) areas of faces of elemental tetrahedtron (Chapter 5) strain-displacement transformation matrix (Chapter 2) matrix derivative of contact matrix A (Chapter 6) minimum specimen stiffness (Chapter 7) element strain-displacement transformation matrix (Chapter 2) linear strain-displacement transformation matrix (Chapter 5) nonlinear strain-displacement transformation matrix (Chapter 5) Brinell hardness matrix (Chapter 2) empirical constant in Miner equation (Chapter 7) constant in crack propagation equation (Chapter 7)

Cmn , m, n = 1,..., 6 C1, C2 D D D D Di Dprsq Ds DA DC E Ee Eep E1, E2, E3 EA EC EH C

material constants in general elastic stress-strain correlation (Chapter 1) material constituents of Mooney–Rivlin material (Chapter 5) deformation tensor (Chapter 5) differentiation matrix operator (Chapter 2) fourth-order tensor of material properties (Appendix A) deviatoric part of a tensor (Appendix A) damaging effect (Chapter 7) fourth-order tensor of material properties (Appendix A) deviator of stress tensor (Chapter 1) Almansy deformation tensor (Chapter 5) Cauchy–Green deformation tensor (Chapter 5) Young modulus (Chapter 1) elastic stiffness tensor (Chapter 4) elasto-plastic stiffness tensor (Chapter 4) principal values of Cauchy-Green or Almansy strain tensor (Chapter 5) Almansy strain tensor (Chapter 5) Cauchy–Green strain tensor (Chapter 5) Hencky strain tensor (Chapter 5)

C

E xx , E yy , C

,…, E zz F F F Fe Fn Fn Ftot r

Fi F∆ ∆F, ∆Fi G G GI Gk Gsk Gk Gsk

components of Cauchy–Green strain tensor in Cartesian coordinate system (Chapter 5) applied load (Chapter 3) yield surface (Chapter 4) load (Chapter 3) element nodal forces (Chapter 2) vector of nodal forces (Chapter 2) vector of nodal forces (Chapter 5) vector of total loads (Chapter 6) resistance of structure (Chapter 3) vector of hierarchical nodal forces (Chapter 2) increment of force (Chapter 3) shear modulus (Chapter 1) energy release rate (Chapter 7) energy release rate of mode I (Chapter 7) base vectors in current configuration (Chapter 5) covariant components of metric tensor in current configuration (Chapter 5) reciprocal base vectors in current configuration (Chapter 5) contravariant components of metric tensor in current configuration (Chapter 5)

H H Hi I I I I1, I2, I3 J Ji , Jc K K K K Ke Kb Kc K´ Kic Kks KL , KR , Kσ Kmax , Kmin Kt Kth Kε Kσ

tensor of simple shear (Chapter 5) slope of hardening function (Chapter 4) Hermit polynomials (Chapter 2) identity tensor (Chapter 5) bending moment of inertia (Chapter 2) moment of inertia of a single leaf (Chapter 9) invariants of tensor (Appendix A) Jacobian matrix (Chapter 2) Jacobians of transformation (Chapter 5) stiffness matrix (Chapter 4) stress intensity factor (Chapter 7) stress intensity factor of mode I (Chapter 7) correction factor of spring leaves deflection (Chapter 9) element stiffness matrix (Chapter 2) stiffness matrix of bodies in contact (Chapter 6) stiffness matrix of whole contact surface (Chapter 6) cyclic strength coefficient (Chapter 7) fracture toughness (Chapter 7) stiffness factors (Chapter 1) constituents of stiffness matrix in nonlinear problems (Chapter 5) maximum and minimum stress intensity factors (Chapter 7) theoretical stress concentration factor (Chapter 7) threshold stress intensity factor (Chapter 7) strain concentration factor (Chapter 7) stress concentration factor (Chapter 7)

e

normal stiffness matrix of contact element (Chapter 6)

e c, t

tangential stiffness matrix of contact element (Chapter 6)

e c

stiffness matrix of contact element (Chapter 6) tangent stiffness matrix (Chapter 5) tangent stiffness (Chapter 3)

K c, n K K Kt Kt

uu

u∆

Ke , Ke , ∆u

∆∆

Ke , Ke ∆K L N N Ni Nf Nk NS N0

constituents of hierarchical stiffness matrix (Chapter 2) range of stress intensity (Chapter 7) matrix of directional cosines (Chapter 2) matrix of shape functions (Chapter 2) axial force (Chapter 3) number of cycles (Chapter 7) number of load fluctuations up to failure (Chapter 7) interpolation functions (Chapter 2) matrix of normals (Chapter 6) matrix of normals (Chapter 6)

N1, N2, N3 N1, N2, N3

normal vectors in current configuration (Chapter 5) axial loads in rods (Chapter 4)

f

f

internal loads due to fictitious external load (Chapter 4)

r

r

internal residual loads (Chapter 4) orthogonal rotation tensor (Chapter 5) polynom of principal values (Appendix A) contact forces between leaves (Chapter 9) limit load (Chapter 4) limit value of tensile force of Signorini material in simple tension (Chapter 5) fictitious load (Chapter 4) Legendre polynomial of an order p (Chapter 2) load that corresponds to onset of yield (Chapter 4) concentrated force acting upon a small area (Chapter 1) tensor (Appendix A) concentrated load (Chapter 1) plastic potential (Chapter 4) vector of body forces (Chapter 2) position vector in current configuration (Chapter 5) residual vector of unbalanced forces (Chapter 5) vector-derivative of potential over displacements vector (Chapter 6) residual force (Chapter 3) stress ratio (Chapter 7) vector-derivative of potential over displacements vector at iteration i (Chapter 6) residual vector (Chapter 3) residual (Chapter 3) stress tensor (Chapter 5) symmetric part of any tensor (Appendix A) nominal stress (Chapter 7) stiffening factor (Chapter 9) maximum bending stresses in leaves (Chapter 9) maximum and minimum cyclic stress (Chapter 7) tensile strength (Chapter 9)

N 1, N 2 N 1, N 2 O P Pk Plim Plim Pf Pp Pyp ∆Pn Q Q Q R R R R R R Ri Ri Ri S S S SF Sk,max Smax , Smin Su Sxx , Syy , Szz , Sxy , Syz , Szx Syp S1, S2, S3 SC SP Sp T

components of stress tensor (Chapter 5) yield point (Chapter 9) components of stress tensor (Chapters 7, 8) Cauchy stress tensor (Chapter 5) first Piola stress tensor (Chapter 5) second Piola stress tensor (Chapter 5) tensor (Appendix A)

symbol of transposition of vector, tensor, or matrix components of tensor (Appendix A) transpose tensor (Appendix A) inverse tensor (Appendix A) tensor of third order tensor of fourth order strain tensor (Chapter 1) stress tensor (Chapter 1)

T Tsk , T sk ′ TT T–1 T(3) T(4) Tε Tσ 0

matrix of tangential vectors (Chapter 6)

TS 1 s

2 s

T , T , T U U

3 s

constituents of strain tensor (Chapter 1) left stretch tensor (Chapter 5) internal energy (Chapter 3)

Ui Ua U0 V V1, V2, V3 W W We Wi

stored elastic energy (Chapter 1) released strain energy due to crack propagation (Chapter 7) stored elastic strain energy (Chapter 7) right stretch tensor (Chapter 5) principal stretches (Chapter 5) work or strain energy section modulus of a single leaf (Chapter 9) external work internal work

Wi Wγ Wyp

specific internal work required destruction work (Chapter 7) critical value of distortion energy (Chapter 4)

v

internal dilatation (volumetric) specific work (Chapter 1)

d i

internal distortion specific work (Chapter 1) volumetric forces (Chapter 1) vector of location of the nodes in contact zone before loading (Chapter 6) global Cartesian coordinates (Chapter 5) function of hardening (Chapter 4) yield constant in perfectly plastic models (Chapter 4) slip surface (Chapter 6) tensor of backstress (Chapter 4) angle of rotation (Chapter 5) directional angles of unit normal (Chapter 1) coefficients of transformation of tensor components (Appendix A) angle between rods (Chapter 4) material constant of Mooney–Rivlin material (Chapter 5) parameter of simple shear (Chapter 5) elastic specific surface energy (Chapter 7)

Wi

W X, Y, Z Xi X1, X2, X3 Y Y0 Y(f) α α α, β, γ αmn β β γ γe

γp γxy , γyz, γxz γ rθ, γ θz, γ zr δ δ1, δ2 δmn ε ε εN ε εe εk εm ε r , ε θ, ε z εx, εy , εz ε1, ε2, ε3 εe εp p

εn + 1 δεi δε i

plastic specific surface energy (Chapter 7) shear strains (Chapter 1) shear strains in cylindrical coordinate system (Chapter 1) symbol of variation or increment of any quantity elongations (Chapter 5) Kronecker symbol (components of identity tensor) tensor of small deformation (Chapter 1) vector of strains (Chapter 2) nonlinear strain vector (Chapter 5) full (logarithmic) strain (Chapter 4) effective strain (Chapter 4) axial strain of rod-element (Chapter 2) mean strain normal strains in cylindrical coordinate system (Chapter 1) normal strains (Chapter 1) principal strains (Chapter 1) elastic component of total strain (Chapter 4) plastic component of total strain (Chapter 4) corrected plastic strain (Chapter 4) strain increment (Chapter 7)

e

elastic strain increment (Chapter 7)

p

plastic strain increment (Chapter 7) cyclic strain amplitude (Chapter 7) elastic strain amplitude (Chapter 7) plastic strain amplitude (Chapter 7) effective strain amplitude (Chapter 7) strain fracture limit (Chapter 7) polar angle (Chapter 1) angle of simple shear (Chapter 5) slopes of a plate (Chapter 2) slopes at the ends of beam (Chapter 2) curvilinear coordinates (Chapter 5) volumetric change (Chapter 1) Lame constants (Chapter 1) proportional factor in deformation plasticity (Chapter 4) line-search parameter (Chapter 3) principal values of tensor (Appendix A) Lagrange multipliers (Chapter 6) plastic flow scalar (Chapter 4) static coefficient of friction (Chapter 6) dynamic coefficient of friction (Chapter 6)

δε i ∆ε ∆εe ∆εp ∆εeff εf′ θ θ θx , θy θ1 , θ2 θ1 , θ2 , θ3 ϑ λ, µ λ λi λk λn, λt dλ µs µd

dµ κ κ κn κt ν ν ρ σ σˆ σ σe σh σm (σm)eff σn σ r , σ θ, σ z σx , σy , σz σyp σ1, σ2, σ3 σN ∆σ σf′ τmax τn τoct τ rθ, τ θz, τ zr τxy , τyz, τzx τ1, τ2, τ3 φk ξk χ0, χ1, χ2 πc η ∆ ∆ ∆ζ ∆e ∆k Ω Π

scalar in Ziegler’s stress correlation (Chapter 4) matrix of penalties (Chapter 6) hardening parameter (Chapter 4) normal penalty (Chapter 6) tangential penalty (Chapter 6) Poisson’s ratio lateral reduction (Chapter 5) radius of curvature (Chapter 9) vector of stresses (Chapter 2) interpolant of stress field (Chapter 2) axial stress (Chapter 1) effective stress vector of stresses obtained by FE solution (Chapter 2) mean stress effective mean stress (Chapter 7) normal component of traction (Chapter 1) normal stress components in cylindrical coordinate system (Chapter 1) normal stresses (Chapter 1) yield point (Chapter 4) principal stresses (Chapter 1) equivalent stress amplitude (Chapter 7) stress amplitude (Chapter 7) stress fracture limit (Chapter 7) critical value of shear stress (Chapter 4) tangential component of traction (Chapter 1) octahedral shear stress (Chapter 1) shear stresses in cylindrical coordinate system (Chapter 1) shear stresses (Chapter 1) maximum shear stresses (Chapter 1) integrals of Legendre polynomials (internal shape functions) (Chapter 2) coordinates of nodes (Chapter 2) coefficients in stress-strain correlation potential energy of contact element (Chapter 6) relative error (Chapter 2) determinant of a second order tensor (Appendix A) symbol of increment of any quantity scalar magnitude of correction term of tangential force (Chapter 6) vector of deviations (Chapter 2) coefficients of hierarchical approximation (Chapter 2) tensor of small rotation (Chapter 5) total potential energy

Πb Πc δΠc Φ Λ Λi ∇i

potential of bodies in contact (Chapter 6) potential of contact zone (Chapter 6) virtual work of contact forces (Chapter 6) minimized function (Chapter 3) vector of Lagrange multipliers (Chapter 6) vector of Lagrange multipliers at iteration i gradient in initial configuration

∇c

gradient in current configuration

Brackets: [...] (...) {...}

matrix brackets (Chapter 2) tensor brackets (Appendix A) vector brackets (Chapter 2)

Contents Part I: Theoretical Fundamentals Chapter 1 Basics of Solid Mechanics.......................................................................................1 1.1 Stress ................................................................................................................1 1.2 Linear Strain...................................................................................................17 1.3 Stress–Strain Relationship .............................................................................24 1.4 Variational Principles .....................................................................................34 1.5 Solution of the Boundary Value Problem......................................................38 Chapter 2 Finite Element Method .........................................................................................49 2.1 Introduction to Finite Element Theory..........................................................49 2.2 Isoparametric Elements..................................................................................60 2.3 Hierarchical Functions ...................................................................................70 2.4 Bending Elements: Beams and Plates ...........................................................76 2.5 Accuracy of FE Solution ...............................................................................82 Chapter 3 Nonlinear Problems ...............................................................................................89 3.1 Introduction ....................................................................................................89 3.2 Example: Two-Spar Frame ............................................................................89 3.3 Iterative Methods ...........................................................................................94 Chapter 4 Plasticity................................................................................................................107 4.1 One-Dimensional Theory.............................................................................107 4.2 Yield Criteria for Multi-axial Stresses ........................................................117 4.3 Constitutive Theories of Plasticity...............................................................122 4.4 Finite Element Implementation ...................................................................137 Chapter 5 Large Displacements ...........................................................................................141 5.1 Tensor Analysis of a Deformed Body .........................................................141 5.2 Deformation and Strain................................................................................148 5.3 Stress ............................................................................................................167 5.4 Constitutive Equations .................................................................................172

5.5

Finite Element Implementation ...................................................................180

Chapter 6 Contact Problems.................................................................................................185 6.1 Introduction ..................................................................................................185 6.2 Penalty Method ............................................................................................190 6.3 Lagrange Multiplier Method........................................................................202 6.4 Critical Review.............................................................................................209 Chapter 7 Fatigue Failure Prediction Methods ..................................................................211 7.1 Strain Method...............................................................................................211 7.2 Cumulative Damage.....................................................................................227 7.3 Fracture Mechanics ......................................................................................230

Part II: Design Cases Chapter 8 Design of Machine Parts .....................................................................................247 8.1 Nonlinear Behavior of Machine Parts .........................................................247 8.2 Failure of Machine Parts under Static Load ...............................................249 8.3 Fatigue of Machine Parts under Fluctuating Load......................................253 Chapter 9 Leaf Springs .........................................................................................................265 9.1 Introduction ..................................................................................................265 9.2 Design Fundamentals...................................................................................266 9.3 FE Analysis of Leaf Springs........................................................................273 9.4 Conclusions ..................................................................................................285 Chapter 10 Threaded Fasteners .............................................................................................289 10.1 Introduction ..................................................................................................289 10.2 Forces in Bolt Connection ...........................................................................292 10.3 Stresses .........................................................................................................297 10.4 Nonlinear Analysis Using FE Method ........................................................302 10.5 Conclusions ..................................................................................................330 Chapter 11 Flange Connection ...............................................................................................333 11.1 Introduction ..................................................................................................333 11.2 One-Dimensional Analysis ..........................................................................334 11.3 FE analysis ...................................................................................................339

11.4 Conclusions ..................................................................................................347 Chapter 12 Fretting Fatigue in an Axle.................................................................................351 12.1 Introduction ..................................................................................................351 12.2 Case Study: Axle Failure Due to Fretting ...................................................352 12.3 Design Improvement....................................................................................361 12.4 Conclusions ..................................................................................................364 Appendix A............................................................................................................367 Appendix B............................................................................................................387 Appendix C............................................................................................................391 Index ......................................................................................................................401

Part I Theoretical Fundamentals

1

Basics of Solid Mechanics

Solid mechanics is a part of a continuum mechanics. This chapter forms the foundation in machine design for problem solving within the linear theory of elasticity,1,2 discussing stresses, displacements, and strains. To stay within the linear theory, the discussion is limited to small displacements. The topic of nonlinearity, including large displacements, plastic materials, and contact, will follow in Chapters 3 through 6. Throughout the presentation, to facilitate infinitesimal analysis, bodies are assumed to be ideal, having such properties as flexibility, continuity, and isotropy. We disregard the microstructure of the material.

1.1 STRESS Stress is defined as an internal force per unit area of a section of a body under loading. One of the objectives of solid mechanics is to assess the internal state of a flexible body under load and define it in terms of stresses, a discussion of which follows.

1.1.1

DEFINITION

OF

STRESSES

The stresses in a body in state of equilibrium are analytically derived, using the method of sectioning. This method assumes the body is split into parts that are interconnected by internal forces to preserve it as one entity. Assume the body to be divided into two parts, I and II, with the internal forces acting on the cut surfaces (see Figure 1.1). Limiting the examination to section I, Figure 1.1b, consider an arbitrary infinitesimal area ∆Α on this cut surface of the section, with an outward normal vector n (Figure 1.2). It is assumed that forces acting upon ∆Α are reduced to concentrated force ∆Pn, applied at point Q. Note that the direction of ∆Pn does not necessarily coincide with a normal n. The average intensity of internal forces acting upon ∆A equals ∆P p n = ---------n ∆A

(1.1)

1

2

Nonlinear Problems in Machine Design

FIGURE 1.1 Imaginary sectioning of a body under load.

FIGURE 1.2 Forces acting on a cut surface.

Following the above pattern, the vector of traction at point Q is defined as ∆P dP p n = lim ---------n = ---------n dA ∆A → 0 ∆A

(1.2)

Area ∆Α, the direction of which is defined by normal n, has a vector of traction pn. As illustrated in Figure 1.3, area ∆Α through the same point Q, but in a different direction, will have different traction pn. Let us consider the body located within a Cartesian coordinate system (x,y,z). Assume the cut surface to be directed so that its normal n is parallel to z-axis. We divide traction pn into three Cartesian components: stress σz normal to area ∆Α, and two shear stresses, τzx and τzy, tangent to area ∆Α. See Figure 1.4. Next, let us cut out a section within the body in the form of an infinitesimal cube (see Figure 1.5). The edges of the cube are parallel to coordinates (x,y,z). Examining the Cartesian stresses acting upon the facets of the cube, we find three normal stresses, σx, σy, and σz, and six shear stresses, τxy, τyz, τzx, τxz, τzy, and τyz. The nine stresses form the stress tensor described by Equation (1.3).

3

Basics of Solid Mechanics

FIGURE 1.3 Vector of traction as a function of the orientation of the cut surface.

FIGURE 1.4 Partition of vector of traction into stress components.

σ x τ xy τ xz T σ = τ yx σ y τ yz

(1.3)

τ zx τ zy σ z It will be shown in Section 1.1.3 that τ xy = τ yx,

τ yz = τ zy,

τ zx = τ xz

(1.4)

See Equation (1.19). Thus, the number of stress components is reduced to three normal stresses, σx, σy, and σz, and three shear stresses, τxy, τyz and τzx—all forming a symmetric tensor,

4

Nonlinear Problems in Machine Design

FIGURE 1.5 Stress components acting upon an elemental cube.

σx sym T σ = τ yx σ y

(1.5)

τ zx τ zy σ z The above tensor, comprising the six stress components, presents a mathematical definition of the internal stress state at a specific point of the body.

1.1.2

PRINCIPAL STRESSES

To define principal stresses, consider a general relation between stress tensor Tσ, traction pn, and normal unit vector n. From the equilibrium of an infinitesimal tetrahedron, Figure 1.6, it follows that pn = Tσ ⋅ n

(1.6)

Consider a particular case in which traction pn is collinear with normal n, i.e., pn = σ n

(1.7)

Such collinear traction is known as the principal stress, and its direction n is the principal direction. When pn, is the principal stress, Equation (1.6) obtains the form σ n = Tσ ⋅ n

(1.8)

5

Basics of Solid Mechanics

FIGURE 1.6 Stress components acting upon a tetrahedron.

The latter is equivalent to the set of equations in Cartesian coordinates, σ cosα = σ x cos α + τ xy cosβ + τ xz cosγ σ cosβ = σ yx cos α + σ y cosβ + τ yz cosγ σ cosγ = σ zx cos α + τ zy cosβ + τ z cosγ

(1.9)

where cosα, cosβ, and cosγ are directional angles of normal n with respect to the coordinate axes. To solve the set of Equations (1.9), the following determinant must equal zero: σx – σ

τ xy

τ xz

τ yx

σy – σ

τ yz

τ zx

τ zy

σz – σ

= 0

(1.10)

See Appendix A. Equation (1.10) has three roots that are principal stresses: σ1, σ2, and σ3. In terms of the principal stresses, the stress tensor can be expressed as σ1 0 0 Tσ =

0 σ2 0 0 0 σ3

The invariants of the stress tensor are

(1.11)

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Nonlinear Problems in Machine Design

I 1 ( σ ) = σ1 + σ2 + σ3 = σ x + σ y + σz 2

2

2

I 2 ( σ ) = σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 = σ x σ y + σ y σ z + σ z σ x – τ xy – τ yz – τ xz 2

2

2

I 3 ( σ ) = σ 1 σ 2 σ 3 = σ x σ y σ z + 2τ xy τ yz τ zx – σ x τ yz – σ y τ zx – σ z τ xy

(1.12)

The invariants are used in the theory of plasticity, and they are essential to failure analysis in machine design. Plane Stress Condition In certain two-dimensional cases, the principal stresses have a simpler form. Consider the case of two-dimensional body in the form of a thin plate loaded within its plane, with stresses in z-direction equaling zero, i.e., σ z = τ yz = τ zx = 0

(1.13)

Here Equation (1.10), because of the two-dimensional stress condition, takes the form 2

2

σ – ( σ x + σ y )σ + ( σ x σ y – τ xy ) = 0

(1.14)

Solving the latter, one obtains two principal stresses.

1.1.3

EQUATIONS

OF

σ x – σ y 2 σx + σy 2 - + τ xy σ 1 = ---------------- +  --------------- 2  2

(1.15)

σ x – σ y 2 σx + σy 2 - + τ xy σ 2 = ---------------- –  --------------- 2  2

(1.16)

EQUILIBRIUM

To derive the equations of equilibrium, consider a solid body in static equilibrium that is subjected to applied forces and has geometric constraints. The applied forces comprise specific volumetric forces (X, Y, Z). The geometric constraints mean that some points of the body have prescribed displacements. Consider an infinitesimal cube within this body (see Figure 1.7). Because of the state of equilibrium, the total of the forces and moments acting upon the cube must equal zero. Let us relate first to the x components of stresses and forces, shown in the figure. Summing up the results, after multiplying the stresses by the respective areas and the force by the volume of the cube, we get the following equation:

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Basics of Solid Mechanics

FIGURE 1.7 Stress components acting upon an elemental cube: projection on xy-plane.

∂τ xy  ∂σ x   σ + --------dx – τ xy dzdx – σ x dydz +  τ xy + ---------dy   x ∂x  ∂y  ∂τ +  τ zx + --------zx-dz – τ zx dxdy + Xdxdydz = 0  ∂z 

(1.17)

Performing the same operation with y and z components, we obtain the following three differential equations ∂σ x ∂τ xy ∂τ zx -------- + --------- + --------- + X = 0 ∂z ∂y ∂x ∂τ xy ∂σ y ∂τ yz + -------- + --------- + Y = 0 --------∂z ∂y ∂x ∂τ zx ∂τ yz ∂σ z --------- + --------- + -------- + Z = 0 ∂z ∂y ∂x

(1.18)

In addition, the condition that the total of the moments acting on the cube is zero leads to the following equations: τ xy = τ yx ,

τ yz = τ zy ,

τ zx = τ xz

(1.19)

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Nonlinear Problems in Machine Design

On the boundary surface, we use separate equations because of different loading conditions. This is expressed by traction (px , py , pz) (see Figure 1.8). In consequence, here the equilibrium equations are σ x cosα + τ xy cos β + τ zx cos γ = p x τ xy cosα + σ y cos β + τ yz cos γ = p y τ zx cosα + τ yz cos β + σ z cos γ = p z

(1.20)

where α, β, and γ are the directional angles of the normal to the surface in respect to coordinate axes, see Figure 1.8.

1.1.4

MOHR REPRESENTATION

OF

STRESSES

The Mohr diagrams show how stress components change with the direction of the surface of a section. To describe the Mohr method, let us first analyze a twodimensional case, followed by a three-dimensional one. Two-Dimensional Stress Condition Consider a two-dimensional body, loaded within its plane, with the stresses in zdirection assumed to be zero. A section, in the form of an infinitesimal triangle of rectangular form OAB, is presented in Figure 1.9. Its perpendicular sides OA and OB coincide with coordinate directions x and y, while the inclined side AB forms an angle ϕ with OB; AB represents the surface of the section, the direction of which varies. The sides of the triangle relate to each other as follows: OA OB -------- = cos ϕ , -------- = sin ϕ AB AB

FIGURE 1.8 Traction acting on the boundary.

(1.21)

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Basics of Solid Mechanics

FIGURE 1.9 Stress components acting upon a plane triangle.

Stresses σx , σy , and τxy act upon sides OA and OB, while stresses σϕ and τϕ act upon side AB. The equilibrium is defined by the following set of two equations: σ ϕ cos ϕ – τ ϕ sin ϕ = σ x cos ϕ + τ xy sin ϕ σ ϕ sin ϕ + τ ϕ cos ϕ = σ x sin ϕ + τ xy cos ϕ

(1.22)

Solving the set of equations, one obtains σx + σy σx – σy - + ----------------- cos 2ϕ + τ xy sin 2ϕ σ ϕ = ---------------2 2 σx – σy - sin ϕ + τ xy cos 2ϕ τ ϕ = ---------------2

(1.23)

The latter two equations can be combined into one. 2 σ x – σ y 2 2 2 x + σ y σ – σ - + τ xy ---------------- + τ ϕ =  ---------------ϕ   2  2 

(1.24)

In Mohr diagram, Figure 1.10, Equation (1.24) is plotted as a circle in coordinate system (σϕ,τϕ). The diagram presents stresses σϕ and τϕ as functions of the inclination angle ϕ. Points A and B on the circle denote stresses acting upon sides OA and OB of the triangle, while C and D represent stresses acting on inclined side AB. Shear stress τϕ reaches a maximum at G that equals 2 x – σ y σ 1 – σ2 ---------------- + τ xy = σ --------------- 2  2 2

τ max =

The above maximum shear stress is called principal shear stress.

(1.25)

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Nonlinear Problems in Machine Design

FIGURE 1.10 Mohr circle for plane stress condition.

Three-Dimensional Stress Condition Within a three-dimensional body under load, consider a section in the form of an infinitesimal tetrahedron, formed by mutually perpendicular edges OA, OB, and OC (see Figure 1.11a). The perpendicular edges coincide with the principal directions 1, 2, and 3. Oblique plane ABC has a normal n with directional angles α, β, and γ. Stresses σ, σ2, and σ3 act upon the perpendicular facets, and traction pn acts upon the oblique plane. Traction pn has two stress components on plane ABC: normal σn and tangential τn. Consider now Figure 1.11b. It shows a sphere with the center coinciding with vertex O of the tetrahedron. Oblique plane ABC of the tetrahedron is tangent to the sphere, with point P denoting the contact. Radius OP of the sphere is normal to plane ABC. If contact point P is moved to a different location, the inclination of

FIGURE 1.11 Three-dimensional stress condition: (a) stresses acting upon a tetrahedron in principle coordinates, and (b) stress mapping on a unit sphere.

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Basics of Solid Mechanics

plane ABC will change. The new contact point will be defined by new directional angles α, β, and γ. It will have a different normal and tangential stress components σn and τn. Point P on the surface of the sphere is subject to the following condition: 2

2

2

cos α + cos β + cos γ = 1

(1.26)

The equilibrium of the tetrahedron is defined by the equations 2

2

2

σ n = σ 1 cos α + σ 2 cos β + σ 3 cos γ

(1.27)

and 2

2

2

2

2

2

2

2

2

2

2

2

2

τ n = p n – σ n = p x + p y + p z – σ n = σ 1 cos α + σ 2 cos β + σ 23 cos γ – σ n

(1.28)

Solving Equations (1.26) through (1.28) for cos2α, cos2β and cos2γ, we obtain 2

( σ2 – σn ) ( σ3 – σn ) + τn 2 cos α = ------------------------------------------------------( σ2 – σ1 ) ( σ3 – σ1 )

(1.29)

2

( σ3 – σn ) ( σ1 – σn ) + τn 2 cos β = ------------------------------------------------------( σ3 – σ2 ) ( σ1 – σ2 )

(1.30)

2

( σ1 – σn ) ( σ2 – σn ) + τn 2 cos γ = ------------------------------------------------------( σ1 – σ3 ) ( σ2 – σ3 )

(1.31)

Equations (1.29) through (1.31) form the basis for Mohr diagram for the internal state of the three-dimensional body. The diagram, shown in Figure 1.12, maps stresses σn and τn as functions of angles α, β, and γ. The diagram is formed by three Mohr half-circles in plane (σn,τn), with centers at points O', O" and O"'. The area bounded by three arcs corresponds to spherical triangle P1P2P3 (see Figure 1.11b). The area represents the domain of stresses acting on oblique plane ABC. To elucidate the method, consider the stresses at point P of Figure 1.11. Point P is located on arc AB, defined by the condition γ = const. From Equation (1.31), one derives σ 1 – σ 2 2 σ 1 + σ 2 2 2 2  --------------- σ – ----------------- = const (1.32) + τ cos γ ( σ – σ ) ( σ – σ ) = + n 1 3 2 3  n  2  2  Using the latter expression, constant angle γ is mapped in Figure 1.12 as circular arc AB, with center at point O'. The magnitude of angle γ is indicated by the inclinations of lines O"A and O"'B. In a similar manner, one can draw two more

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Nonlinear Problems in Machine Design

FIGURE 1.12 Mohr circles for three-dimensional stress condition.

arcs: arc CD, with its center at point O", which is defined by the conditions β = const; and arc EF, with its center at O"', defined by α = const. See Figure 1.13. The cross point of arcs AB, CD, and EF represents stresses that act at point P on the inclined surface of the tetrahedron, shown in Figure 1.11a. Also in Figure 1.12, one finds three maximum shear stresses at points Q1, Q2 and Q3 that equal 2 – σ 3 ,τ = σ 3 – σ 1 ,τ = σ 1 – σ 2 τ1 = σ ---------------- 2 ----------------- 3 ----------------2 2 2

These are the three principal shear stresses.

FIGURE 1.13 Determination of three-dimensional stresses using Mohr diagram.

(1.33)

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Basics of Solid Mechanics

Octahedral Stresses Consider a tetrahedron in space (σ1,σ2,σ3) as shown in Figure 1.14a. Assume the oblique plane to have equal directional angles, α = β = γ. According to Equation (1.27), the directional cosinuses are 1 cos α = cos β = cos γ = ------3

(1.34)

and the normal stress in the oblique plane is σ1 + σ2 + σ3 σ oct = ----------------------------3

(1.35)

According to Equation (1.28), the shear stress in the plane will be 1 2 2 2 τ oct = --- ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) 3

(1.36)

There are eight planes in space (σ1,σ2,σ3), with stresses as per Equations (1.35) and (1.36). The planes have normal cosines equal to ±1 ⁄ 3 and form a symmetric

FIGURE 1.14 Octahedral stresses: (a) octahedral stresses in space (σ1,σ2,σ3) and (b) eight planes forming an octahedron.

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Nonlinear Problems in Machine Design

octahedron (see Figure 1.14b). Stresses σoct and τoct acting on these planes are called octagonal stresses.

1.1.5

STRESS DEVIATOR

The stress tensor may be split into two parts. They are a spherical part and a deviatoric part.  σ 0 0  m Tσ =  0 σm 0   0 0 σm

  σ –σ τ xy τ xz m   x  +  τ yx σ y – σ m τ yz   τ zx τ zy σ z – σ m  

    

(1.37)

where σm is a mean (hydrostatic) stress, equal to octagonal stress σoct, 1 1 1 σ m = --- ( σ x + σ y + σ z ) = --- ( σ 1 + σ 2 + σ 3 ) = --- I 1 ( σ ) 3 3 3

(1.38)

See Appendix A. The first (spherical) tensor represents hydrostatic pressure. The second tensor is the stress deviator, which we denote by Ds. It shows deviation of the general stress state from the hydrostatic state at any point. Again, see Appendix A. Introducing the stress components, s x = σ x – σm ,

s y = σ y – σm ,

sz = σz – σm ,

(1.39)

the stress deviator obtains the form  s τ τ   x xy xz  D s =  τ yx s y τ yz     τ zx τ zy s z 

(1.40)

In terms of principal values, the stress deviator equals  s 0 0  1 Ds =  0 s2 0   0 0 s3 with principal deviatoric stress components

    

(1.41)

15

Basics of Solid Mechanics

2σ 1 – σ 2 – σ 3 s 1 = σ 1 – σ m = ------------------------------3 2σ 2 – σ 3 – σ 1 s 2 = σ 2 – σ m = ------------------------------3 2σ 3 – σ 1 – σ 2 s 3 = σ 3 – σ m = ------------------------------3

(1.42)

The principal directions of the stress tensor and the stress deviator coincide (see Appendix A). The invariants are of the stress deviator are I 1 ( s ) = s1 + s2 + s3 = 0 I 2 ( s ) = s1 s2 + s2 s3 + s3 s1 I 3 ( s ) = s1 s2 s3

(1.43)

(Again, see Appendix A). The second invariant I2(s), above, may be expressed in other forms I 2 ( s ) = s1 s2 + s2 s3 + s3 s1 1 2 2 2 1 2 2 2 = – --- ( s 1 + s 2 + s 3 ) = – --- [ ( s 1 – s 2 ) + ( s 2 – s 3 ) + ( s 3 – s 1 ) ] 2 6 1 2 2 2 = – --- [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] 6

(1.44)

which follow from Equations (1.43). The invariant will be used again in Chapter 4, as we apply it to the theory of plasticity. Stress Deviator Tensor in Stress Space For an alternate presentation of the stress deviator, consider a rectangular coordinate system, where the coordinate axes are principal stresses (σ1,σ2,σ3) (see Figure 1.15). Within this system, let us define a plane expressed by the equation σ1 + σ2 + σ3 = 0

(1.45)

The plane intersects origin point O at coordinates σ1 = σ2 = σ3 = 0. It forms equal angles with each axis, and its unit normal vector is

16

Nonlinear Problems in Machine Design

FIGURE 1.15 Deviatoric plane.

1 n = ------- ( i 1 + i 2 + i 3 ) 3

(1.46)

Consider point P outside the said plane (see Figure 1.15). Vector OP, connecting points O and P, is a stress vector (representing stress tensor Tσ described in Section 1.1.1). The stress vector is expressed as σ = σ1 i1 + σ2 i2 + σ3 i3

(1.47)

The projection of σ onto the normal to plane σ1 + σ2 + σ3 = 0 equals 1 σ ⋅ n = ------- ( σ 1 + σ 2 + σ 3 ) = 3

3σ m

(1.48)

while a projection of σ onto plane σ1 + σ2 + σ3 = 0 produces a vector s = s1 i1 + s2 i2 + s3 i3

(1.49)

2

(1.50)

Vector s has a magnitude s =

2

2

s1 + s2 + s3

which by virtue of Equations (1.47) equals

17

Basics of Solid Mechanics

s = =

2

( s1 + s2 + s3 ) – 2 ( s1 s2 + s2 s3 + s3 s1 ) 0 – 2 ( s1 s2 + s2 s3 + s3 s1 ) =

–2 ⋅ I 2 ( Ds )

(1.51)

Vector s represents the deviator tensor Ds in the stress space. Since the vector is situated in plane σ1 + σ2 + σ3 = 0, the plane is referred to as the deviatoric plane.

1.2 LINEAR STRAIN The shape and dimensions of a flexible body are altered by the external loading, causing deformation and displacement. As a consequence, the distances between the points of the body change, and the angles between directions become different. The change of geometry is measured in nondimensional terms by means of strain. In the approach that follows, a linear correlation between the strain and the applied load is assumed, referring to cases with small deformations only. For analyses of those with large deformations, a nonlinear approach is necessary and is presented in Chapters 3 and 5.

1.2.1

DEFINITION

OF

STRAIN

Consider two points in a loaded body, located infinitesimally close (see Figure 1.16). The points are A defined by vector r and B defined by vector r + dr. Due to deformation under load, point A moves to A1, while point B moves to B1. The displacement of point A is governed by the equation R = r+u

(1.52)

where vector u defines the displacement. The displacement of B is governed by the equation

FIGURE 1.16 Concerning the definition of strain.

18

Nonlinear Problems in Machine Design

(1.53)

R + dR = r + dr + u + du

where vector u + du defines the displacement. Within this vector, the differential du is a relative displacement. Using the linear approach, vector u is defined solely by vector r. Thereby, vector du is expressed as follows du du = ------ ⋅ dr dr

(1.54)

The derivative of u with respect to r in the above equation is a tensor. In Cartesian coordinates, vectors u and r are expressed by their components T

r = [ x y z]

u = [u v w] ,

T

(1.55)

and the tensor becomes     du ------ =  dr    

∂u -----∂x ∂v -----∂x ∂w ------∂x

∂u -----∂y ∂v ----∂y ∂w ------∂y

∂u -----∂z ∂v ----∂z ∂w ------∂z

        

(1.56)

Let us divide the above tensor into three partial tensors     1 Tε =     

   1 2 T ε = ---  2   

 ∂u ------ 0 0  ∂x   ∂v 0 ----- 0  ∂y  ∂w  0 0 -------  ∂z 

0 ∂v ∂u ------ + -----∂x ∂y ∂w ∂u ------- + -----∂x ∂z

∂u ∂v ∂u ∂w ------ + ------ ------ + ------∂y ∂x ∂z ∂x ∂v ∂w ----- + ------0 ∂z ∂y ∂w ∂v 0 ------- + ----∂y ∂z

(1.57)

        

(1.58)

19

Basics of Solid Mechanics

   1 3 T ε = ---  2   

0 ∂v ∂u ------ – -----∂x ∂y ∂w ∂u ------- – -----∂x ∂z

∂u ∂v ∂u ∂w  ------ – ------ ------ – -------  ∂y ∂x ∂z ∂x  ∂v ∂w  ----- – ------0 ∂z ∂y   ∂w ∂v  0 ------- – ----∂y ∂z 

(1.59)

whereby Equation (1.56) is condensed as follows: du 1 2 3 ------ = T ε + T ε + T ε dr

(1.60)

Now, let us illustrate graphically the three component tensors, as shown in Figure 1.17. The illustration shows a deformed rectangle—a projection of the deformed elemental cube in the xy–plane. Consider the deformation as if it were composed of three independent acts: a volumetric deformation, a deformation of the form, and a pure rotation. (Additional projections of the deformed cube can be drawn in the yz- and zx-planes.) The relationship of each tensor to a corresponding deviation in the figure becomes apparent, as expressed by Equations (1.57) through (1.59) above. 3 2 1 Tensors T ε and T ε refer to deformations of the body, and tensor T ε represents a pure rotation of an infinitesimal neighborhood of a point that acts as a rigid body. 3 For the purpose of analysis, we may disregard the last tensor, T ε , and unite the former two tensors. ∂u -----∂x 1

2

Tε = Tε + Tε =

1  ∂u ∂v 1  ∂u ∂w --- ------ + ------ --- ------ + ------2  ∂y ∂x 2  ∂z ∂x  ∂v ----∂y

1  ∂v ∂u --- ------ + -----2  ∂x ∂y

1  ∂v ∂w --- ----- + ------2  ∂z ∂y 

1  ∂w ∂u 1  ∂w ∂v --- ------- + ------ --- ------- + ----2  ∂x ∂z  2  ∂y ∂z

(1.61)

∂w ------∂z

Equation (1.61) is an expression of the strain tensor. As seen from the equation, the tensor is symmetric. To simplify the above, we introduce the following expressions: ∂u ε x = ------ , ∂x

∂v ε y = -----, ∂y

∂w ε z = ------∂z

∂u ∂v γ xy = ------ + ------ , ∂y ∂x

∂w ∂v γ zy = ------- + ----∂y ∂z

∂v ∂w γ yz = ----- + ------- , ∂z ∂y

∂w ∂v γ zy = ------- + ----∂y zx

∂w ∂u γ zx = ------- + ------ , ∂x ∂z

∂u ∂w γ xz = ------ + ------∂z ∂x

(1.62)

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Nonlinear Problems in Machine Design

FIGURE 1.17 Deformation of an elemental cube: (a) volumetric deformation, (b) deformation of the form, and (c) pure rotation.

whereby the expression of the strain tensor T ε becomes γ xy ε x -----2 γ yx T ε = -----ε 2 y γ zx γ zy ------ -----2 2

γ xz -----2 γ yz -----2 εz

(1.63)

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Basics of Solid Mechanics

1.2.2

PRINCIPAL STRAINS

Each point within a body under loading has its own three principal strain directions. If we consider an elemental cube cut out from such a body, whose edges coincide with the said directions, then we will note there is only a volumetric deformation, with the cube retaining its rectangular form. See Figure 1.18. Here, the shear strains equal zero, and the normal strains become principal strains. In Cartesian coordinates (x,y,z), the principal strains and principal directions are derived from the equations γ γ xy cos β + -----xz- cos γ = 0 ( ε x – ε ) cos α + -----2 2 γ γ yx ------ cos α + ( ε y – ε ) cos β + -----yz- cos γ = 0 2 2 γ γ zx ------ cos α + -----zy- cos β + ( ε z – ε ) cos γ = 0 2 2

(1.64)

with the determinant being equal to zero, γ xy ε x – ε -----2 γ yx ------ ε y – ε 2 γ zy γ zx ----------2 2

γ -----xz2 γ yz -----2

= 0

εz – ε

FIGURE 1.18 Deformation of an infinitesimal cube in principal directions.

(1.65)

22

Nonlinear Problems in Machine Design

The roots of Equation (1.65) are the three principal strains ε1, ε2, and ε3. Angles (α, β, γ) in Equations (1.64) define the principal directions of the strain tensor. (See Appendix A.) Based on Hooke’s law, Section 1.3.1, we find that the principal strain directions and principal stress directions coincide. The strain tensor invariants are I 1 ( ε ) = ε x + ε y + εz = ε1 + ε2 + ε3 1 2 2 2 I 2 ( ε ) = ε x ε y + ε y ε z + ε z ε x – --- ( γ xy + γ yz + γ zx ) 4 = ε1 ε2 + ε2 ε3 + ε3 ε1 1 1 2 2 2 I 3 ( ε ) = ε x ε y ε z + --- γ xy γ yz γ xz – --- ( ε x γ yz + ε y γ xz + ε z γ xy ) 4 4 = ε1 ε2 ε3

(1.66)

The first invariant equals approximately the volumetric change of a cube with a unit volume. The volumetric change equals ( 1 + ε1 ) ( 1 + ε2 ) ( 1 + ε3 ) – 1 V –V ϑ = ----------------0 = ----------------------------------------------------------------1 V0 = ε1 + ε2 + ε3 + ε1 ε2 + ε2 ε3 + ε1 ε3 + ε1 ε2 ε3

(1.67)

which, within the linear approximation, disregarding the higher order terms, becomes ϑ ≈ ε1 + ε2 + ε2

(1.68)

In terms of the principal strains, strain tensor T ε has the form ε1 0 0 Tε =

0 ε2 0

(1.69)

0 0 ε3

1.2.3

STRAIN DEVIATOR

Similarly to stress deviator, discussed above, we define a strain deviator. To derive it, the strain tensor is split into two parts.

23

Basics of Solid Mechanics

εm 0 0 0 εm 0 +

Tε =

0 0 εm

γ xy -----2

ε x – εm γ yx -----2 γ zx -----2

ε y – εm γ zy -----2

γ -----xz2 γ yz -----2

(1.70)

εz – εm

Strain εm is a mean strain, defined as 1 1 ε m = --- ( ε x + ε y + ε z ) = --- ( ε 1 + ε 2 + ε 3 ) 3 3

(1.71)

The two tensors on the right side of Equation (1.70) are as follows. The first is a spherical tensor, and the second is the strain deviator, which we denote Dε. By denoting e x = ( ε x – εm ) e y = ( ε y – εm ) ez = ( εz – εm )

(1.72)

the strain deviator becomes γ xy e x -----2 yx D ε = γ-----e 2 y γ γ -----zx- -----zy2 2

γ -----xz2 γ yz -----2

(1.73)

ez

The three invariants of strain deviator equal I 1 ( Dε ) = e1 + e2 + e3 = 0 I 2 ( Dε ) = e1 e2 + e2 e3 + e1 e3 I 3 ( Dε ) = e1 e2 e3 The second invariant may be expressed in other forms,

(1.74)

24

Nonlinear Problems in Machine Design

I 2 ( Dε ) = e1 e2 + e2 e3 + e3 e1 1 2 1 2 2 2 2 2 = – --- ( e 1 + e 2 + e 3 ) = – --- [ ( e 1 – e 2 ) + ( e 2 – e 3 ) + ( e 3 – e 1 ) ] 2 6 1 2 2 2 = – --- [ ( ε 1 – ε 2 ) + ( ε 2 – ε 3 ) + ( ε 3 – ε 1 ) ] 6

(1.75)

which follow from Equations (1.74).

1.3 STRESS–STRAIN RELATIONSHIP The correlation between the stresses and strains in a loaded body is expressed by means of constitutive equations. Again, here, the stress–strain relationship is assumed to be linear.

1.3.1

HOOKE’S

LAW

The correlation between the stresses and strains in a loaded body can be expressed, in general form, by a system of equations. σ x = C 11 ε x + C 12 γ xy + C 13 γ xz + C 14 ε y + C 15 γ yz + C 16 ε z σ y = C 21 ε x + C 22 γ xy + C 23 γ xz + C 24 ε y + C 25 γ yz + C 26 ε z … … τ xz = C 61 ε x + C 62 γ xy + C 63 γ xz + C 64 ε y + C 65 γ yz + C 66 ε z

(1.76)

First, we analyze a three-dimensional problem, continuing with a two-dimensional one. Three-Dimensional Problem The constitutive equations for a three-dimensional body, using the experimentally obtained data with isotropic and linearly elastic materials, may be written in the form 1 ε x = --- [ σ x – v ( σ y + σ z ) ], E

τ xy γ xy = ----G

1 ε y = --- [ σ y – v ( σ x + σ z ) ], E

τ yz γ yz = ----G

1 ε z = --- [ σ z – v ( σ y + σ x ) ], E

τ zx γ zx = ----G

(1.77)

25

Basics of Solid Mechanics

where E is the modulus of elasticity, known as Young modulus, G is the shear modulus, and ν is the Poisson’s ratio. The Young and the shear modula are correlated by means of the equation E G = -------------------2(1 + ν)

(1.78)

Equations (1.77) can be presented in a matrix form as follows:           

 εx  1 –ν –ν 0 0 0 εy  – ν 1 – ν 0 0 0  1 –ν –ν 1 εz  0 0 0  = --E γ xy  0 0 0 2(1 + ν) 0 0  0 0 0 0 2(1 + ν) 0 γ yz   0 0 0 0 0 2 ( 1 + ν) γ zx 

          

 σx  σy   σz   τ xy   τ yz  τ zx  

(1.79)

Here we express the stress and strain as vectors, i.e., σ = [ σ x σ y σ z τ xy τ yz τ zx ] T ε = [ ε x ε y ε z γ xy γ yz γ zx ] T

(1.80)

The reciprocal of the above matrix equation is 1–ν --------------1 – 2ν ν   -------------- σx  1 – 2ν  σ  ν  y  -------------- σz  E 1 – 2ν   = ----------- τ xy  1 + ν 0    τ yz   τ  0  zx  0

ν --------------1 – 2ν 1–ν --------------1 – 2ν ν --------------1 – 2ν 0 0 0

ν --------------- 0 0 0 1 – 2ν ν --------------- 0 0 0  ε x  1 – 2ν  ε y 1–ν --------------- 0 0 0  εz 1 – 2ν  1 γ 0 --- 0 0  xy 2  γ yz 1  γ 0 0 --- 0  zx 2 1 0 0 0 --2

          

(1.81)

The stress–strain correlation shown in Equation (1.81) presents the basis for the use in the finite element analysis (see Chapter 2). Equations (1.77), (1.79), and (1.81) are known as a generalized Hooke’s law of the linear theory of elasticity.

26

Nonlinear Problems in Machine Design

Two-Dimensional Problems The two-dimensional analysis includes cases concerning plane stress, plane strain, and axial symmetry. We refer to parts that are flat, with forces acting within the plane, or to those axisymmetrical forms with axisymmetrical forces. The analysis can be treated as two-dimensional, because some stress and strain components can be disregarded as negligible. Plane-Stress Problem Consider a thin plate under plane load. Because stresses in the perpendicular direction are negligible, they are disregarded, so the following condition is satisfied: σz = 0 ,

τ yz = 0 ,

τ zx = 0

(1.82)

In this case, the equations expressing Hooke’s law have the following short form: 1 ε x = --- ( σ x – νσ y ) E 1 ε y = --- ( σ y – νσ x ) E τ xy γ xy = ----G

(1.83)

In matrix form, it becomes  εx  0   1 1 –ν  ε y  = --– ν 1 0   E 0 0 2(1 + ν)  γ xy 

 σx     σy     τ xy 

(1.84)

Upon reversing the above equation, we obtain the stress–strain relation, 1 ν 0  σx    E ν 1 0  σ  = --------------2 1–ν   1–ν 0 0 ----------- τ xy  2

 εx     εy     γ xy 

(1.85)

Plane-Strain Problem Consider a thick plate under plane load which does not change in z-direction. Here, the strains in z-direction are negligible, and we assume that εz = 0

γ yz = 0

γ zx = 0

(1.86)

27

Basics of Solid Mechanics

It should be noted that the stresses in z-direction may differ from zero, as per Equations (1.77). We obtain the stress–strain relation by reducing Equation (1.81) to the following: 1–ν --------------1 – 2ν  σx    E ν - ------------- σ y  = 1----------+ ν 1 – 2ν    τ xy  0

ν --------------- 0 1 – 2ν  ε   x  1–ν --------------- 0  ε y  1 – 2ν   γ 1  xy  0 --2

(1.87)

Axisymmetric Problem Consider a body of axisymmetric form with axisymmetric loading. We analyze this body with the help of cylindrical coordinates (r, θ, z) instead of the Cartsian system. Here, the displacements in radial, circumferential, and axial directions are (u, v, w). The stress components are σr, σθ,σz,τrθ,τθz, and τzr, while the strain components are εr,εθ,εz,γrθ,γθz and γzr. See Figure 1.19. The following strain components are negligible and are disregarded: τ rθ = τ θz = 0,

γ rθ = γ θz = 0

FIGURE 1.19 Stress components in cylindrical coordinates.

(1.88)

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Nonlinear Problems in Machine Design

The remaining strain components, in radial and circumferential directions, upon neglecting secondary terms, equal4 ∂u ε r = ------ , ∂r u 1 ∂ν ε θ = --- + --- -----r r ∂θ

(1.89)

See Figure 1.20. The strain components in the axial direction are ∂w ε z = ------- , ∂z

∂w ∂u γ rz = ------- + -----∂r ∂z

(1.90)

Using the above definitions, the Hooke’s law is expressed as 1 ε r = --- [ σ r – ν ( σ θ + σ z ) ] E 1 ε θ = --- [ σ θ – ν ( σ z + σ r ) ] E 1 ε z = --- [ σ z – ν ( σ r + σ θ ) ] E τ rz γ rz = ----G

FIGURE 1.20 Deformation of an elemental area in an axisymmetric problem.

(1.91)

29

Basics of Solid Mechanics

The stress and strain vectors in the axisymmetric case are σ = [ σ r σ θ σ z τ rz ] T ε = [ ε r ε θ ε z γ rz ] T

(1.92)

Therefore, the stress–strain relation for axisymmetric case becomes 1–ν --------------1 – 2ν  σ  ν  r  -------------- σθ  E 1 – 2ν   = -----------ν  σ z  1 + ν -------------  1 – 2ν  τ rz  0

ν --------------1 – 2ν 1–ν --------------1 – 2ν ν --------------1 – 2ν 0

ν --------------- 0 1 – 2ν  ε  r ν  --------------- 0   ε 1 – 2ν θ   1–ν εz   --------------- 0 1 – 2ν  γ rz    1 0 --2

(1.93)

Lame Equations The Hooke’s Law may be expressed in a different form, known as Lame’s Equations. Equations (1.77), as applied to a three-dimensional case may be rewritten as σ x = λϑ + 2µε x

τ xy = µγ xy

σ y = λϑ + 2µε y

τ yz = µγ yz

σ z = λϑ + 2µε z

τ xz = µγ xz

(1.94)

ϑ denotes volumetric change defined by Equations (1.67) and (1.68). Coefficients λ and µ are Lame constants. They relate to the Young modulus and shear modulus, as follows:

1.3.2

EFFECTIVE STRESS

AND

Eν λ = -------------------------------------( 1 + ν ) ( 1 – 2ν )

(1.95)

µ = G

(1.96)

STRAIN

Working machine parts in general are subject to multiaxil loading, resulting in multiaxial stresses and strains. To utilize the data derived from one-dimensional experiments for design purposes, it is necessary to substitute the multiaxial stresses and strains with effective one-dimensional magnitudes. We denote the effective stresses and strains as σe and εe, respectively.

30

Nonlinear Problems in Machine Design

There are several known methods to derive effective stresses and strains. Those presented here are based on von Mises theory and were chosen for their applicability to machine design. Von Mises Effective Stress The tensor of effective stress σe equals  σ 00  e Tσ =  0 0 0   0 00

    

(1.97)

Following von Mises, we match the deviators of the effective and of multiaxial stresses, disregarding hydrostatic pressure (σ1 + σ2 + σ3)/3.  σ  σ e – -----e 3   Ds =  0    0 

0 σ – -----e 3 0

 0    0 =   σe  – ----3 

 2σ 1 – σ 2 – σ  -----------------------------0 0 3   2σ 2 – σ 1 – σ 1  -------------------------------0 0 3   2σ 3 – σ 1 – σ 2  -------------------------------0 0  3

        

(1.98)

By equating the second invariants of the respective deviators, and using Equation (1.44), we derive the following expression: 1 1 2 2 2 2 --- σ e = --- [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] 6 3

(1.99)

from which we obtain the effective stress

σe =

1 2 2 2 --- [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] 2

(1.100)

In a different form, in terms of deviatoric components Si, the effective stress equals σe =

3 2 2 2 --- ( s 1 + s 2 + s 3 ) 2

(1.101)

Effective Strain The effective strain has the following three principal components, which follow from Hooke’s Law:

31

Basics of Solid Mechanics

1 ε 1 = ---σ e = ε e E 1 ε 2 = ---νσ e = – νε e E 1 ε 3 = ---νσ e = – νε e E

(1.102)

Consequently, the tensor of effective strain equals  ε 0 0  e T ε =  0 – νε 0   0 0 – νε e

    

(1.103)

Repeating the procedure used for effective stress, we match the deviators of the effective and multiaxial strain.  ε e – ε e – 2νε e  -------------------------------0 0 3   ε e – 2νε e  0 – νε e – --------------------0 3   – νε e – ε e – 2νε e  ---------------------------------------0 0  3  2ε 1 – ε 2 – ε 3  ----------------------------0 0 3   2ε 2 – ε 3 – ε 1  ----------------------------0 0 3   2ε 3 – ε 1 – ε 2  ---------------------------0 0  3

     =    

        

(1.104)

Further equating the second invariants of the respective deviators, we obtain the expression of the effective strain, 1 2 2 2 ε e = ------------------------- ( ε 1 + ε 2 ) + ( ε 2 – ε 3 ) + ( ε 3 – ε 1 ) (1 + ν) 2 In terms of deviatoric strains, ei, the effective strain equals

(1.105)

32

Nonlinear Problems in Machine Design

3 2 1 2 2 ε e = ------------ --- ( e 1 + e 2 + e 3 ) 1+ν 2

(1.106)

See Equation (1.75). Based on Hooke’s law, we can derive a correlation between the effective stress and strain as follows. First, applying Equations (1.77) to deviatoric stresses and strains, 1+ν e i = ------------ s i , E

i = 1, 2,3

(1.107)

and then combining Equations (1.101), (1.106), and (1.107), we obtain σ e = Eε e

1.3.3

(1.108)

WORK

The load acting upon a body performs work, which is transformed and stored within the body in the form of elastic strain energy. Analyzing the process, consider an elemental cube within the body (see Figure 1.5). To simplify the mathematical expressions, assume that coordinate axes (x, y, z) coincide with the principal directions. As a result, the shear stresses are zero, and the normal stresses equal σ x = σ1 , σ y = σ2 , σz = σ3

(1.109)

See Figure 1.21. Consider first the deformation of the cube in direction 1. As stress σ1

FIGURE 1.21 Stress components acting upon an elemental cube in principal coordinates.

33

Basics of Solid Mechanics

rises from zero to its final value, σ1, it causes an elongation of the cube, which equals ∂u du = ------ dx = ε 1 dx ∂x

(1.110)

σ1 dydz is a force acting on the cube. The work performed by the force equals 1 1 --- σ 1 dy dz du = --- σ 1 ε 1 dx dy dz 2 2

(1.111)

Applying the same procedure in directions 2 and 3 and summing up the results, we obtain the total work, 1 dW i = W i dx dy dz = --- ( σ 1 ε 1 + σ 2 ε 2 + σ 3 ε 3 )dx dy dz 2

(1.112)

W i is the specific work per unit volume, which equals the stored elastic strain energy. 1 W i = --- ( σ 1 ε 1 + σ 2 ε 2 + σ 3 ε 3 ) ≡ U i 2

(1.113)

Distortion Work We view the total work as a sum of two parts: dilatation (volumetric) work and distortion work. ν

d

Wi = Wi + Wi

(1.114)

The former refers to volume, while the latter applies to the shape. The dilatation work is the work done by mean stresses in three principal directions, ν 1 3 W i = --- σ m ε m = --- ( σ 1 + σ 2 + σ 3 ) ( ε 1 + ε 2 + ε 3 ) 6 2

(1.115)

while the distortion work is done by the deviatoric stresses. To derive the final expression of the distortion work, let us present it as d 1 1 W i = --- ( σ 1 ε 1 + σ 2 ε 2 + σ 3 ε 3 ) – --- ( σ 1 + σ 2 + σ 3 ) ( ε 1 + ε 2 + ε 3 ) 6 2

(1.116)

which, upon applying Hooke’s law, Equations (1.81), becomes d 1+ν 2 2 2 W i = ------------ [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] 6E

(1.117)

34

Nonlinear Problems in Machine Design

Now, using Equations (1.100) and (1.105), the above equation can be reduced to d 1 W i = --- σ e ε e 2

(1.118)

The magnitude of the distortion work is used to measure the onset of plastic deformation.

1.4 VARIATIONAL PRINCIPLES The variational principles include principles of minimum of potential energy, minimum of complementary energy, the Helliger-Reissner principle, and others.5 These principles provide solutions to the boundary value problem by relating to minimum of energy. We single out the principle of potential energy as applicable to problems in machine design. We also describe in this section the principle of virtual work. Although it is not a minimum principle per se, it relates to the minimum energy. The principles of minimum energy are applicable to conservative systems only. The principle of virtual work has a wider application, since it also covers non-conservative systems.

1.4.1

THE PRINCIPLE

OF

MINIMUM POTENTIAL ENERGY

Consider the body shown in Figure 1.22, where external forces (X,Y,Z) act upon volume V, while external surface tractions (px ,py ,pz) act upon boundary surface S1. Furthermore, displacements u0,v0,w0 are prescribed on boundary surface S2 (kinematic boundary condition). The total potential energy of the system is defined as the sum of the two, Π = Ui + W e

FIGURE 1.22 Concerning virtual work.

(1.119)

35

Basics of Solid Mechanics

where the first term is the internal potential energy stored in the body, 1 U i = --- ∫ ∫ ∫ ( σ x ε x + σ y ε y + σ z ε z + τ xy γ xy + τ yz γ yz + τ xz γ xz )dxdydz 2 v

(1.120)

while the second term is the work of the external forces acting upon the body, We = –

∫ ∫ ∫v ( Xu + Yv + Zw ) dxdydz + ∫ S∫ ( p x u + p y v + pz w ) dS

(1.121)

1

The principle of minimum potential energy in a given body is stated as follows: The causes of potential energy Π to become stationary are those admissible displacements that lead to the equilibrium. To prove the principle, consider the first variation of potential energy Ui. It equals  ∂U ∂U  ---------i δε x + ∂U ---------i δε y + ---------i δε z ∂ε z ∂ε y  ∂ε x δU i = ∫ ∫ ∫   v ∂U i ∂U i ∂U i  + --------- δγ + ---------δγ + ---------δγ  ∂γ xy xy ∂γ yz yz ∂γ zx zx

    d x d y dz    

(1.122)

which, based of Equation (1.20), becomes

δU i =

∫ ∫ ∫v

∂δw ∂δv ∂δu ∂δv ∂δw σ x --------- + σ y --------- + σ z ---------- + τ yz  --------- + ----------  ∂z ∂z ∂y ∂x ∂y  ∂δu ∂δv ∂δw ∂δu +τ zx  ---------- + --------- + τ xy  --------- + ---------  ∂y  ∂x ∂x  ∂z 

d x d y dz

(1.123)

Consider the first term of the integrant of the above equation. Integrating it by parts, we obtain the following: ∂δu

-dxdydz ∫ ∫ V∫ σ x -------∂x =

∂

∂σ x

- ( σ δu )dxdydz – ∫ ∫ ∫ δu -------- dxdydz ∫ ∫ V∫ ----∂x ∂x x V

(1.124)

Equivalent expressions can be derived for the remaining terms of Equation (1.123). Summing up the results and applying the Gauss divergence theorem, we get

36

Nonlinear Problems in Machine Design

δU i = – ∫ ∫ ∫

∂σ ∂τ xy ∂τ zx ∂τ xy ∂σ y ∂τ yz δu  --------x + --------+ -------- + --------- + + --------- + δν  -------- ∂x  ∂x ∂z  ∂y ∂z  ∂y ∂σ ∂τ ∂τ + δ w  --------zx- + --------yz- + --------z  ∂x ∂z  ∂y

V

d x d y dz

δu ( σ x cos α + τ xy cos β + τ zx cos γ ) + ∫ ∫ +δν ( τ xy cos α + σ x cos β + τ yz cos γ ) dS S1

+δw ( τ zx cos α + τ yz cos β + σ x cos γ )

(1.125) Angles α, β, and γ in the above equation refer to surface S1 with traction (px ,py ,pz), and they define the surface normal as per Figure 1.8. Inserting the equilibrium conditions as per Equations (1.18) and (1.20) into the above, we obtain δU i =

∫ ∫ V∫ ( Xδu + Yδv + Zδw ) dx dy dz

+ ∫ ∫ ( p x δu + p y δν + p 2 δw ) dS V

(1.126)

Comparing Equations (1.121) and (1.126), we obtain δΠ = δU i + δW e = 0

(1.127)

i.e.,

∫ ∫ V∫ ( σ x δε x + σ y δε y + σz δεz + τ xy δγ xy + τ yz δγ yz + τzx δγ zx ) dx dy dz – ∫ ∫ ∫ ( Xδu + Yδν + Zδw ) dx dy dz – ∫ ∫ ( p x δu + p y δν + p z δw ) dS = 0 V

S1

(1.128) Thus, we get a mathematical proof of the principle of stationary potential energy. It can be shown, after deriving a second variation of potential Π , that the true solution provides a minimum of the potential energy.5

1.4.2

THE PRINCIPLE

OF

VIRTUAL WORK

As mentioned above, the minimum principles are applicable to conservative systems only. We now introduce the principle of virtual work, which has an advantage over

37

Basics of Solid Mechanics

other principles, because it is applicable to both conservative and non-conservative systems, the latter involving irreversible processes.6 To explain the principle, assume a body whose outer forces and inner stresses are in equilibrium. Imagine that each point of the body is displaced by an arbitrary infinitesimal variation, in addition to its regular displacement, while it is subject to the kinematic boundary conditions. The displacement variation is called virtual displacement, and its components are denoted by δu , δv , and δw . The corresponding virtual strains equal ∂ δε x = ------ δu ∂x

∂ ∂ δγ xy = ------ δv + -----δu ∂y ∂x

∂ δε y = -----δv ∂y

∂ ∂ δγ yz = -----δw + -----δv ∂z ∂y

∂ δε z = -----δw ∂z

∂ ∂ δγ zx = -----δu + ------ δw ∂x ∂z

(1.129)

The virtual work performed by stresses along the virtual strains is  ∂W io ∂W io ∂W io - δε - δε y + ---------- ------------ δε x + ----------∂ε z z ∂ε y ∂ε x δW i = ∫ ∫ ∫   ∂W ∂W io ∂W io V io - δγ - δγ yz + ----------- δγ xy + ----------+ ---------- ∂γ xy ∂γ zx zx ∂γ yz

   d x d y dz   

(1.130)

which equals

δW i =



∫ ∫ V∫ 

σ x δε x + σ y δε y + σ z δε z +τ xy δγ xy + τ yz δγ yz + τ zx δγ zx

  d x d y dz 

(1.131)

On the other hand, the virtual work of applied external forces equals

δW e = –

∫ ∫ V∫ ( F x δu + F y δν + F z δw ) dx dy dz

(1.132)

+ ∫ ∫ ( p x δu + p y δν + p z δw ) dS S1

where Fi denotes the applied body forces and pi the applied surface pressures. In equilibrium, the sum of Equations (1.131) and (1.132) must equal zero. δW i + δW e = 0

(1.133)

38

Nonlinear Problems in Machine Design

The above expresses the principle of virtual work, which is: The total sum of the virtual work, performed by the external forces and internal stresses, equals zero.

1.5 SOLUTION OF THE BOUNDARY VALUE PROBLEM The essence of the boundary value problem in a loaded body is to derive the relationship between the acting forces, on one hand, and the stresses and strains, on the other. Two problem-solving approaches are presented here: (1) by variational principles, and (2) using differential equations, both presented below.

1.5.1

GALERKIN METHOD

One of the most practical methods used in solving the design problems is the method of Galerkin,7 with a direct approach to the differential equations. It presents a solution in the form of a series of specific functions with unknown coefficients that are determined in the solution process. The example that follows illustrates the procedure. Consider a one-dimensional bar of variable cross section A(x) (see Figure 1.23). The bar is subjected to a distributed axial load q(x)—body force per unit length. A concentrated load Q is applied to its right end. One has to find the displacement distribution. The problem can be stated by the differential equation dN ------- + q ( x ) = 0 dx

(1.134)

where N is a variable internal axial force, which equals N ( x) = σ( x) A( x)

(1.135)

Introducing the stress and strain expressions, σ = Eε

(1.136)

FIGURE 1.23 One-dimensional bar subjected to a distributed axial load q(x) and a concentrated load Q.

39

Basics of Solid Mechanics

du ε ( x ) = -----dx

(1.137)

du d ------ EA ( x ) ------ + q ( x ) = 0 dx dx

(1.138)

and

Equation (1.134) becomes

with the boundary conditions u = 0 at x = 0 Q du ------ = -------------- at x = l EA ( l ) dx

(1.139)

Equation (1.138) states the problem with unknown function u in mathematical form. The Galerkin method offers two approaches: a generalized one and a modified version. The generalized Galerkin approach is a series expressed in the form n

u ( x ) = u0 ( x ) +

∑ ak f k ( x )

(1.140)

k=1

where fk(x) are specific functions, so-called coordinate functions that disappear on the boundary, while constant ak are the unknown parameters to be determined. The term u0(x) is a function chosen to fit the boundary conditions as per Equation (1.139). According to Equation (1.138), function u0(x) equals Qx u 0 ( x ) = -------------EA ( l )

(1.141)

The second approach, the modified Galerkin version, has the form5 n

u( x) =

∑ ak f k ( x )

(1.142)

k=1

where coordinate functions fk(x) must suit the boundary conditions per Equation (1.139), and constants ak are the unknown parameters. In both approaches, to determine the unknown parameters ak , we insert into Equation (1.134) the approximate expression u(x), getting the following differential equation:

40

Nonlinear Problems in Machine Design

du d ------ EA ( x ) ------ + q ( x ) = r ( x ) dx dx

(1.143)

In Equation (1.143), r(x) is a residual that represents an error. It results from the fact that u is an approximation. To minimize the error r(x), we introduce the following equations: l

∫ r ( x ) wk ( x )dx = 0

k = 1, 2, … n

(1.144)

0

where wk(x) are weighted functions. Equation (1.144) represents a condition of orthogonality.8 Replacing wk(x) by coordinate functions fk(x), we obtain the following Galerkin equations: l

∫ r ( x ) f k ( x )dx = 0

k = 1, 2, … n

(1.45)

0

Combining Equations (1.143) and (1.145), we obtain l

- [ EA ( x ) ] ------ + q ( x ) f k ( x ) dx = 0 ∫  --- dx dx du

d

k = 1, 2, … n

(1.146)

0

From Equation (1.146), we can compute the unknown parameters ak. However, the equation in its present form is inconvenient because it needs a second derivative of fk(x). To circumvent this requirement, we introduce a weaker version of Equation (1.146) that eliminates a second differentiation. This is accomplished by integration by parts, as follows: l

∫ 

du d -----EA ------ + q ( x ) f k ( x ) dx = 0  dx dx

k = 1, 2…n

0

du = EA ------ f k ( x ) dx

l

l

0

0

du d f k ( x ) – ∫ EA ------ ----------------dx + ∫ q ( x ) f k ( x ) dx dx dx 0 l

= EA ( l )u′ ( l ) f k ( l ) – EA ( 0 )u′ ( 0 ) f k ( 0 ) l

n

l

0

s=1

0

df s ( x ) d f k( x) - dx + ∫ q ( x ) f k ( x ) dx = 0 - ∑ a s ------------– ∫ EA --------------dx dx (1.147)

Taking into consideration boundary conditions, as per Equation (1.139), Equation (1.147) becomes

41

Basics of Solid Mechanics l

n

l

0

s=1

0

d f k( x) d f s( x) - ∑ a s --------------- dx + ∫ q ( x ) f k ( x )dx = 0 Q f k ( l ) – ∫ EA --------------dx dx

(1.148)

The latter represents a set of equations, with k = 1, 2, … n, which provides the unknown parameters ak. For an illustration of the method, consider a solution where function u(x) is approximated by a broken line between nodes xk, k = 1, 2 … n, see Figure 1.24. We shall use modified Galerkin version, Equation (1.142). It follows from the figure that parameters ak equal the nodal displacements. ak = uk ,

k = 2, 3, … n

(1.149)

with a1 = u1 = 0, see Figure 1.24a. The coordinate functions that fit the approximation are shown in Figure 1.24b. They are piece-wise linear functions defined as follows:

FIGURE 1.24 Concerning the method of Galerkin: (a) approximate displacement distribution and (b) coordinate functions.

42

Nonlinear Problems in Machine Design

x2 – x f 1 ( x ) = --------------at x 1 ≤ x ≤ x 2 x2 – x1 x3 – x x–x at x 2 ≤ x ≤ x 3 f 2 ( x ) = ---------------1 at x 1 ≤ x 2 ; f 2 ( x ) = --------------x3 – x2 x2 – x1 ………… ………… ………… x – xn – 2 xn – x - at ( x n – 2 ≤ x ≤ x n – 1 ); f n – 1 ( x ) = -------------------- at ( x n – 1 ≤ x ≤ x n ) f n – 1 ( x ) = -------------------------xn – 1 – xn – 2 xn – xn – 1 x – xn – 1 - at ( x n – 1 ≤ x ≤ x n ) f n ( x ) = -------------------xn – xn – 1 (1.150) For simplification, let us introduce stiffness factors Kks and loads q k , as follows: l

K ks =

d f k( x) d f s( x)

----------------dx ∫ EA ---------------dx dx

(1.151)

0 l

qk =

∫ q ( x ) f k ( x )dx

(1.152)

0

Functions fk(x) are “nearly orthogonal,” i.e., only integrals of neighboring terms ( f′ k – 1 f′ k ), ( f′ k f′ k ) and ( f′ k f′ k + 1 ) are non-zero, while integrals of all other terms ( f′ k f′ s ) equal zero. See Figure 1.24b. Thus, integrating over respective segment, we obtain a series of so-called stiffness factors, xk

K ( k – 1 ),k =

∫

df k – 1 ( x ) df k ( x ) - ---------------dx EA ( x ) -------------------dx dx

xk – 1 xk + 1

K k,k =

∫

df k ( x ) df k ( x ) ---------------dx EA ( x ) --------------dx dx

xk – 1 xk + 1

K k,k + 1 =

∫ xk

df k ( x ) df k + 1 ( x ) ---------------------dx EA ( x ) --------------dx dx (1.153)

43

Basics of Solid Mechanics

and a series of specific loads, xk

qk =

∫

q ( x ) f k ( x )dx, k = 2, 3, …n

(1.154)

xk – 1

Consequently, Equation (1.148) is replaced by a set of linear equations, K 22 u 2 + K 23 u 3 = q 2 K 23 u 2 + K 33 u 3 + K 34 u 4 = q 3 ………… ………… ………… K n – 2 ,n – 1 u n – 2 + K n – 1 ,n – 1 u n – 1 + K n – 1 ,n = q n – 1 K n,n – 1 u n – 1 + K n,n u n = q n + Q

(1.155)

The latter set of equations is used to obtain the unknown displacements uk .

1.5.2

THE METHOD

OF

RAYLEIGH–RITZ

Let us proceed with another known method, Rayleigh–Ritz,9 which finds the unknown displacement distribution using the same approximation form as Galerkin. However, contrary to Galerkin’s method, the method is based on minimization of potential energy. The approximate solution of the unknown function u has the form of a series, n

u( x) =

∑ ak f k ( x )

(1.156)

k=1

For the derivation of the method, we use the principle of minimum potential energy, Equation (1.127), N

δΠ =

∂Π

- δa ∑ ------∂a k k

= 0

(1.157)

K=1

The condition of stationarity now becomes ∂Π -------- = 0 ∂a k

k = 1, 2, … n

(1.158)

44

Nonlinear Problems in Machine Design

Because potential Π is a sum of quadratic terms of ak, Equation (1.158) is a linear one. It represents a system of n simultaneous linear equations such that, solving it, one finds the unknown parameters ak. We repeat the illustration from the preceding section using the one-dimensional bar shown in Figure 1.23, to find the displacement distribution u. The problem is stated by the differential equation du d ------ EA ( x ) ------ + q ( x ) = 0 dx dx

(1.159)

with the boundary conditions u = 0 at x = 0 Q du ------ = -------------- at x = l EA (l) dx

(1.160)

The stored strain energy of the loaded bar equals l

l

0

0

1 1 du 2 U i = --- ∫ A ( x ) σεdx = --- ∫ EA ( x )  ------ dx  dx 2 2

(1.161)

and the potential of outer forces is l

W e = – ∫ q ( x )u ( x ) dx – Qu ( l )

(1.162)

0

Hence, the total potential becomes l

l

0

0

du 2 1 Π = --- ∫ EA ( x )  ------ dx – ∫ q ( x ) u ( x )dx – Qu ( l )  dx 2

(1.163)

We assume an approximate solution in the form of a series n

u( x) =

∑ uk f k ( x )

(1.164)

k=1

where fk(x) are functions defined by Equations (1.150). Constants uk are unknown parameters to be determined from minimizing potential Π.

45

Basics of Solid Mechanics

With the above approximation, potential Π becomes L

1 Π = --- ∫ EA ( x ) 2 0

2

n

l

n

0

k=1

d f k( x) dx – ∫ q ( x ) ∑ u k f k ( x )dx – Qu n ∑ uk ---------------dx

k=1

(1.165)

From the condition of stationarity, it follows that n

δΠ =

∂Π -------- δu k = ∑ ∂u k

k=1 l

l

n

n

∫ EA ( x ) 0

d f k( x) ∑ uk ---------------dx

k=1

d f r( x)

-δu r ∑ --------------dx

dx

r=1

n

– ∫ q ( x ) ∑ [ f k ( x )δu k ] dx – Qδu n = 0 0

k=1

(1.166)

The above equation can be presented in a simplified form as n

n

∑ ∑ K kr ur – Rk

δu k = 0

(1.167)

k=1 r=1

where l

K kr =

d f k( x) d f r( x)

- ---------------- dx , k ∫ EA ( x ) --------------dx dx

= 1, 2, … n

(1.168)

0 l

Rk =

∫ q ( x ) f k ( x ) dx,

k = 1, 2, …n – 1

(1.169)

0 l

Rn =

∫ q ( x ) f n ( x ) dx + Q

(1.170)

0

Eliminating coefficients δuk, Equation (1.166) becomes a set of n linear equations, n

∑ K kr ur – Rk = 0

k = 1, 2, … n

(1.171)

r=1

Equations (1.171) present a solution to the problem under consideration, providing variables uk. The equations are identical with Equations (1.155) of the Galerkin method, showing that the two methods, Rayleigh–Ritz and Galerkin, produce identical results.

46

1.5.3

Nonlinear Problems in Machine Design

COMPARISON

OF

METHODS: GALERKIN

VS.

VIRTUAL WORK

A comparison of the methods of virtual work and Galerkin shows that identical results are reached when applying either of the methods. Consider a three-dimensional problem: a body subjected to volumetric forces (X,Y,Z). Surface tractions (px ,py ,pz) act upon part S1 of the boundary surface, while displacements u0 ,v0 ,w0 are prescribed over part S2—the kinematic boundary condition. See Figure 1.22. We begin with the Galerkin approach. The equilibrium of the body is defined by a set of differential equations, ∂σ x ∂τ xy ∂τ zx -------- + --------- + --------- + X = 0 ∂z ∂y ∂x ∂τ xy ∂σ y ∂τ yz + -------- + --------- + Y = 0 --------∂z ∂y ∂x ∂τ zx ∂τ yz ∂σ z --------- + --------- + -------- + Z = 0 ∂z ∂y ∂x

(1.172)

See Section 1.1.3. Let vector (vx , vy ,vz) represent weight functions in the form of an arbitrary displacement field, which satisfies kinematic boundary conditions. In accordance with Galerkin methods, we multiply the equilibrium equations by vx , vy , and vz. Consider the x-direction first. Accordingly, ∂σ x

∂τ xy

∂τ zx

+ --------- + --------- + X v x dx dy dz ∫ ∫ V∫  ------- ∂z ∂y ∂x

= 0

(1.173)

which can be presented in the form ∂( σxvx )

∂ ( τ xy v x )

∂ ( τ zx v x )

+ ------------------- + ------------------- + X v x dx dy dz ∫ ∫ V∫  ----------------- ∂x ∂y ∂z ∂v ∂v ∂v – ∫ ∫ ∫  σ x -------x + τ xy -------x + τ zx -------x dx dy dz = 0  ∂x ∂z  ∂y

(1.174)

V

According to Gauss’ divergence formula, the integral of the first three terms above can be expressed as follows ∂( σxvx )

∂ ( τ xy v x )

∂ ( τ zx v x )

+ ------------------- + ------------------- dx dy dz ∫ ∫ V∫  -----------------∂z  ∂y ∂x from which we obtain

=

∫ S∫ p x v x dS 1

(1.175)

47

Basics of Solid Mechanics

∂v x

∂v x

∂v x

+ τ xy ------- + τ zx ------- dx dy dz ∫ S∫ p x v x dS + ∫ ∫ V∫ X v x dx dy dz – ∫ ∫ V∫  σ x ------∂z  ∂y ∂x

= 0

(1.176)

1

After deriving similar expressions in y- and z-directions and summing up the results, we obtain the following equation:

∫ ∫ V∫ =

∂v ∂v ∂v σ x -------x + σ y -------y + σ z -------z ∂z ∂y ∂x ∂v ∂v ∂v ∂v ∂v ∂v + τ xy  -------x + -------y + τ yx  -------y + -------z + τ zx  -------z + -------x  ∂x ∂z   ∂z ∂y   ∂y ∂x 

d x d y dz

∫ ∫ ( p x v x + p y v y + pz vz ) dS + ∫ ∫ V∫ ( X v x + Y v y + Z vz ) dx dy dz

(1.177)

which represents the modified form of Galerkin method. Replacing weight functions (vx ,vy ,vz) with virtual displacements, δu = v x ,

δv = v y ,

δw = v z

(1.178)

we find the above equation to be the same as that reached by the method of virtual work, Equation (1.129). Thus, the modified form of Galerkin method is identical with the principle of virtual work: the weighted functions are treated as admissible virtual displacements, satisfying the kinematic boundary conditions. It should be noted that the Galerkin solution is derived independently of any variational formulation of the problem.

REFERENCES 1. Biezeno, C.B., and Grammel, R., Technishche Dynamik, Springer, Berlin, 1939 (in German). 2. Timoshenko, S.P., and Goodier, J.N., Theory of Elasticity, McGraw-Hill, New York, l970. 3. Truesdell, C., A First Course in Rational Continuum Mechanics, Academic Press, New York, l976. 4. Fung, Y.C., Foundations of Solid Mechanics, Prentice-Hall, Englewood Cliffs, New Jersey, l968. 5. Washizu, K., Variational Methods in Elasticity and Plasticity, Pergamon Press, Oxford, l982. 6. Lanczos, C., The Variational Principles of Mechanics. Toronto University Press, Toronto, l964. 7. Galerkin, B.G., Series solutions of some problems of elastic equilibrium of rods and plates. Vestnik Inzhinerov, l9, 897–908, l9l5 (in Russian).

48

Nonlinear Problems in Machine Design 8. Kantorovich, L.V., and Krylov, V.I., Approximate Methods of Higher Analysis, Interscience Publishers, New York, l964. 9. Ritz, W., Ueber eine neue Methode zur Loesung Variationsprobleme, J. Reine Angew. Math., l35, 1–61, l908 (in German).

2

Finite Element Method

Much like the methods described in Section 1.4, the finite element method (FEM) provides an approximate solution to the boundary value problem, which determines the relationship between stresses and strains and the acting forces in a loaded body. There are a number of ways to derive the main equations of the FEM that depend on its usage: in application to conservative systems, one may use either the principle of minimum of potential energy, or the principle of virtual work, in nonconservative systems, e.g., involving inelastic materials or friction, one must use the principle of virtual work or its equivalent—the Galerkin method.1

2.1 INTRODUCTION TO FINITE ELEMENT THEORY In the following derivation of FEM equations, we use the Galerkin method in a weak form (see Section 1.5.1). The body that is subjected to analysis is divided into subdomains, called elements (see Figure 2.1). The elements usually have simple geometric forms such as triangles, quadrilaterals, hexahedrons, tetrahedrons, prisms,

FIGURE 2.1 Body subject to analysis, divided into elements. 49

50

Nonlinear Problems in Machine Design

etc. (Figure 2.2). The elements are interconnected at so-called nodes. The FEM model must meet two fundamental requirements. 1. The equilibrium within and among the elements 2. A compatibility—continuos displacements within and across element boundaries

2.1.1

THE BOUNDARY VALUE PROBLEM

To simplify the presentation, we consider a two-dimensional body of unity thickness. See Figure 1.7. Body forces (X,Y) act upon the volume, while surface tractions (px ,py) are applied to part S1 of the boundary surface. Displacements (u0,v0) are applied to part S2 of the boundary surface. The problem pertains to a derivation of approximate expressions of strains and stresses in an explicit form. The pertinent equilibrium equations are as follows ∂σ ∂τ xy +X = 0 --------x + --------∂y ∂x ∂τ xy ∂σ y + -------- + Y = 0 --------∂y ∂x

(2.1)

and the strain–displacement relation are

FIGURE 2.2 Common elements: (a) quadrilaterals and triangles, (b) hexahedrons and tetrahedrons, and (c) prisms.

51

Finite Element Method

∂u ε x = --------x , ∂x

∂u ε y = --------y , ∂y

∂u ∂u γ xy = --------x + --------y ∂y ∂x

(2.2)

The stress–strain relation is obtained from the constitutive laws. In the case of linear elasticity, we apply the Hooke’s law, which, for two-dimensional problems, can be written as follows:  σx  E 11 E 12 E 13   =  σy  E 12 E 22 E 23   E 13 E 23 E 33 τ  xy 

 εx     εy     γ xy 

(2.3)

In a vectorial form, it equals σ = Eεε

(2.4)

where E is the elasticity matrix. Let us introduce the differentiation matrix operator, ∂ ∂ ------ 0 ----∂y ∂x D = ∂ ∂ 0 ----- -----∂y ∂x

(2.5)

whereby the equilibrium equations become σ+R = 0 Dσ

(2.6)

  R =  X ( x,y )   Y ( x,y ) 

(2.7)

where R equals

Let us express the strain–displacement relations, Equations (2.2), similarly, in the vectorial form ε = DT u

(2.8)

σ = E DT u

(2.9)

whereby the stress vector equals

52

Nonlinear Problems in Machine Design

Consequently, the problem under consideration is defined by the second-order differential equation T

D(E D u) + R = 0

(2.10)

with the boundary conditions u = u0

at S 2

σ = L E D T u = p at S 1 Lσ

(2.11) (2.12)

L denotes the following matrix:

L =

cos α 0 sin α 0 sin α cos α

(2.13)

which defines the direction of the outer normal vector to the boundary surface at S1 (Figure 2.3). Vector p is the vector of boundary traction. Galerkin Approach The two-dimensional body is conceived as a mesh made of elements interconnected at nodes (Figure 2.4). We replace displacement u in Equation (2.10) by an approximation, assuming

u =

N 1 ( x,y )u 1 + N 2 ( x,y )u 2 + …N n ( x,y )u n N 1 ( x,y )v 1 + N 2 ( x,y )v 2 + …N n ( x,y )v n

FIGURE 2.3 Direction of the outer normal to the boundary surface.

(2.14)

53

Finite Element Method

FIGURE 2.4 Shape function in space (x,y,N).

where ui and vi are displacements at the nodes i = 1,2,…n. N i ( x, y ) are prescribed interpolation functions, analogous to the coordinate functions of Galerkin, Section 1.4.2, and are shown in Figure 2.4 as surfaces in space ( x, y, N ) . N i reach the following values at the respective nodes: N i = 1 at node i N i = 0 at all other nodes

(2.15)

Equation (2.14) in vectorial form becomes      N1 0 N2 0 … Nn 0  u =  0 N1 0 N2 … 0 Nn      

 u1  v 1  u2   v2   …  u n  vn  

(2.16)

Matrix N above is known as the shape function.

N =

N1 0 N2 0 … Nn 0 0 N1 0 N2 … 0 Nn

(2.17)

Hence, Equation (2.16) can be written as follows: u = N un

(2.18)

54

Nonlinear Problems in Machine Design

Here, the aim is to determine displacements u, whose components (ui , vi) are called degrees of freedom. The number of degrees of freedom specifies the complexity of the problem. Let us introduce an arbitrary vector v defined as v = N vn

(2.19)

We assume that v represents a virtual displacement field that satisfies the kinematic boundary condition, Equation (2.11). Following the Galerkin method, we derive the following integral from Equation (2.6):

∫ D∫ ( Dσσ + R )

T

(2.20)

v dx dy = 0

where domain D pertains to the volume of the body. The first term in Equation (2.20) can be expressed as follows:

∫ D∫ ( Dσ )

T

v dx dy =

∂ ( σ x v x ) ∂ ( τ xy v x ) ∂ ( σ xy v y ) ∂ ( τ xy v y ) + ------------------- + -------------------- + ------------------- dx dy -----------------∂x ∂y ∂y ∂x

∫ D∫

∂v ∂v ∂v ∂v – ∫ ∫ σ x -------x + σ y -------y + τ xy  -------x + -------y dx dy  ∂y ∂y ∂x  ∂x D (2.21) We consider the right-hand first term of the equation. Applying the Gauss divergence theorem and Equation (2.12), we obtain the expression

∫ D∫

∂ ( σ x v x ) ∂ ( τ xy v x ) ∂ ( σ y v y ) ∂ ( τ xy v y ) -----------------+ ------------------- + ------------------ + ------------------- dx dy ∂x ∂y ∂y ∂x =

∫S [ ( σ x cos α + τ xy sin α )v x + ( σ y sin α + τ xy cos α )v y ] dS

=

∫S v

T

σ dS = Lσ

∫S v

T

p dS (2.22)

Then Equation (2.20) becomes

∫ D∫ ( D

T

T

v ) σ dx dy =

∫ D∫ v

T

R dx dy + ∫ v p dS T

S1

(2.23)

55

Finite Element Method

Note that Equation (2.23) presents the principle of virtual work in a general form, independent of any specific stress–strain relation. Therefore, it can be applied to inelastic materials as well. Let us introduce the following expressions referring to the shape function: T

(2.24)

D N = B T

(2.25)

D v = Bv n Then, Equation (2.23) becomes T T T T T T v n ∫ ∫ B σ dx dy = v n ∫ ∫ N R dx dy + v n ∫ N p dS D

D

(2.26)

S1

Vector vn is constant and therefore could be placed outside the integrals. Using Equation (2.9) and introducing T

D u = Bu n

(2.27)

we finally obtain

∫ D∫ B

T

EBu n dx dy =

∫ D∫ N

T

R dx dy + ∫ N p dS T

(2.28)

S1

Vector un is constant and can be placed outside the integral as well. The remaining integral itself is known as the stiffness matrix,

∫ D∫ B

K =

T

(2.29)

EB dx dy

Let us replace the body forces and surface tractions by the nodal forces Fn =

∫ D∫ N

T

R dx dy + ∫ N p dS T

(2.30)

S1

Using Equations (2.24) through (2.30), we obtain the fundamental expression of equilibrium of finite element theory, Ku n = F n

(2.31)

56

Nonlinear Problems in Machine Design

Solving it, one obtains nodal displacements un and, subsequently, stresses σ and strains ε, as functions of un. The process is illustrated in the following section.

2.1.2

SIMPLEX ELEMENTS

Explaining the computational process, we shall present a two-dimensional problem using the so-called simplex elements. The elements comprise three-node triangles and four-node tetrahedra. See Figure 2.2. Constant-Strain Triangle Let us divide the body into triangular elements as shown in Fig. 2.5. The elements are interconnected at nodes 1, 2,... n. In the following, we shall derive the shape function and then derive the corresponding stiffness, the stresses, and the strains. The shape function defined by Equation (2.13) equals the sum of the individual elements’ functions.

N =

N1 0 N2 0 … Nn 0 0 N1 0 N2 … 0 Nn

=

∑e Ne

(2.32)

Consider element e with nodes 1,2,3. Its shape function equals

Ne =

N e1 0 N e2 0 … N e3 0 0 N e1 0 N e2 … 0 N e3

and the vector of displacements within element e is

FIGURE 2.5 Two-dimensional body divided into three-node elements.

(2.33)

57

Finite Element Method

    N e1 0 N e2 0 … N e3 0  u = Ne ue =  0 N e1 0 N e2 … 0 N e3     

 u1  v1   u2   v2   u3  v3  

(2.34)

Let us assume a linear shape function, as follows: N ei = a i + b i x + c i y

i = 1, 2, 3

(2.35)

Coefficients ai ,bi , and ci are derived from the following equations: N e1 x 1 + N e2 x 2 + N e3 x 3 = x N e1 y 1 + N e2 y 2 + N e3 y 3 = y N e1 + N e2 + N e3 = 1

(2.36)

Thus, for node 1, one obtains x2 y3 – x3 y2 -; a 1 = ------------------------2A

y2 – y3 x3 – x2 b 1 = ---------------; c 1 = --------------2A 2A

(2.37)

The corresponding expressions for nodes 2 and 3 are obtained by cyclic permutation of the indices. Figure 2.6 shows shape-function components in (x,y,N) space. Based on the above, the derivative matrix of shape function, Equation (2.23), for element e becomes

FIGURE 2.6 Linear shape function components of a three-node element in space (x,y,N).

58

Nonlinear Problems in Machine Design

∂ ------ 0 ∂x T ∂ N e1 0 N e2 0 N e3 0 B e = D N e = 0 ----∂y 0 N e1 0 N e2 0 N e3 ∂ ∂ ----- -----∂y ∂x b1 0 b2 0 b3 0 =

0 c1 0 c2 0 c3 c1 b1 c2 b2 c3 b3

(2.38)

where bi and ci are constants. It follows that matrix Be is constant within the element. This allows us to replace the integral in Equation (2.29) by a sum of constant components, and the stiffness matrix of the body becomes K =

∑e Ke

=

∑e Be

T

EB e A e

(2.39)

where Ae denotes the area of an element. (It is assumed that the element’s thickness is constant and equals unity.) In accordance with the finite element method, the outer forces are perceived as acting upon the nodes only. The total vector of forces is presented as a sum of nodal forces in individual elements, Fn =

∑e Fe

(2.40)

As follows from Equation (2.30), they equal Fe =

∫ Ae∫ N

T

R d A e + ∫ N e p dL e T

(2.41)

Le

Le denotes the element side with acting traction pe. After defining the vector of forces and the stiffness matrix of the whole body, we can express the equilibrium equation as follows: Ku n = F n

(2.42)

Solving the above, we obtain the total vector of nodal displacements un, un =

∑e ue

(2.43)

59

Finite Element Method

Referring to Equations (2.8) and (2.9) for the individual elements and using nodal displacements u as per Equation (2.34), we obtain the vector of strains per element, ε = DT u = Be ue

(2.44)

and the vector of stresses per element σ = EB e u e

(2.45)

In the present solution, since matrix Be is constant, the computed strains and stresses within an element also become constant. The derived solution proves that the FEM model presented here fulfills the requirements of equilibrium and compatibility; the former by virtue of Equation (2.42), and the latter (because the displacements are common to both elements) by virtue of Equation (2.6). Tetrahedron Another simplex element is a four-node tetrahedron (see Figure 2.7). The element is used in three-dimensional problems. The displacement vector within tetrahedron has three components (u,v,w) and equals  u    u =  v  = Ne ue    w 

(2.46)

The nodal displacements vector ue has 12 components. ue = [ u1 v1 w1 u2 v2 w2 u3 v3 w3 u4 v4 w4 ]

FIGURE 2.7 Four-node tetrahedron.

T

(2.47)

60

Nonlinear Problems in Machine Design

The shape function of the tetrahedron is a 12 × 3 matrix. N e1 0 Ne =

0 N e2 0

0 N e1 0 0

0 N e3 0

0 N e2 0

0 N e1 0

0 N e4 0

0 N e3 0

0 N e2 0

0

0 N e4 0

0 N e3 0

(2.48)

0 N e4

The mathematical expressions of Ni are derived from the following equations: N e1 x 1 + N e2 x 2 + N e3 x 3 + N e4 x 4 = x N e1 y 1 + N e2 y 2 + N e3 y 3 + N e4 y 4 = y N e1 z 1 + N e2 z 2 + N e3 z 3 + N e4 z 4 = y N e1 + N e2 + N e3 + N e4 = 1

(2.49)

The differential matrix Be of the tetrahedron equals ∂ ------ 0 0 ∂x ∂ 0 ----- 0 ∂y ∂ 0 0 ----∂z Be = ∂ ∂ ----- ------ 0 ∂y ∂x ∂ ∂ 0 ----- ----∂z ∂y ∂ ∂ ----- 0 -----∂x ∂z

N e1 0

0 N e2 0

0 N e1 0 0

0 … N e4 0

0

0 N e2 0 … 0 N e4 0

0 N e1 0

0 N e2 … 0

(2.50)

0 N e4

The derivation of stiffness matrix K, as well as of the strain and stress vectors ε and σ, follows the same steps as in a three-node triangle, above.

2.2 ISOPARAMETRIC ELEMENTS The stresses and strains in the simplex elements, presented above, are constant within elements. As such, these elements have limited use. For more complicated problems, requiring a more accurate computation of stress and strain distributions, we use an isoparametric technique.2 The basic idea of this technique is to normalize the elements by expressing them, in addition to Cartesian coordinates, also in natural coordinates. The relationship of the elements, in the respective coordinate systems,

61

Finite Element Method

is defined by one-to-one mapping. The isoparametric elements include quadrilaterals and hexahedrons: the former are used in two-dimensional problems, whereas the latter are used in three-dimensional ones. The computational procedure, using the isoparametric approach, includes the same steps as in the procedure of simplex elements. 1. 2. 3. 4. 5.

definition of shape function N construction of strain matrix Be construction of stiffness matrix K computation of nodal displacements un computation of strain and stress vectors ε and σ

The results fulfill the two requirements: compatibility and equilibrium. The compatibility is met, because the displacements, along the borders of neighboring elements, depend only on nodes common to both elements; the equilibrium is reached by virtue of Galerkin approximation.

2.2.1

FOUR-NODE QUADRILATERAL

Definition of Shape Function To derive the shape function N, let us introduce a natural coordinate system, defined by one-to-one mapping. x = x ( ξ,η ) , y = y ( ξ,η )

(2.51)

Consider a four-node quadrilateral element (Figure 2.8). In a Cartesian coordinate system (x,y), the element has an arbitrary four-sided form. In the natural coordinate system (ξ,η), the element is mapped into a square.

FIGURE 2.8 Four-node quadrilateral element: (a) in a Cartesian coordinate system and (b) in a natural coordinate system.

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Nonlinear Problems in Machine Design

–1 ≤ ξ ≤ 1 , –1 ≤ η ≤ 1

(2.52)

Cartesian coordinates within the quadrilateral element are defined by the following interpolation: 4

∑ xi N i ( ξ,η )

x ( ξ,η ) =

i=1 4

∑ yi N i ( ξ,η )

y ( ξ,η ) =

(2.53)

i=1

where xi and yi are coordinates of the nodal displacements. The coefficients Ni equal 1 N 1 = --- ( 1 – ξ ) ( 1 – η ) ; 4

1 N 2 = --- ( 1 + ξ ) ( 1 – η ) 4

1 N 3 = --- ( 1 + ξ ) ( 1 + η ) ; 4

1 N 4 = --- ( 1 – ξ ) ( 1 + η ) 4

(2.54)

Equations (2.54), in terms of vectors, become x ( ξ,η ) = Nx e

(2.55)

where xe is the vector of nodal coordinates in the global Cartesian coordinate system, x e = [ x 1 y 1 …x 4 y 4 ]

T

(2.56)

while matrix N is the shape function.*

N =

N1 0 N2 0 N3 0 N4 0

(2.57)

0 N1 0 N2 0 N3 0 N4 According to the isoparametric approach, the displacements in natural coordinates (ξ, η)are correlated by the same shape function, i.e., 4

u =

∑ ui N i ( ξ,η ) i=1 4

v =

∑ vi N i ( ξ,η ) i=1

(2.58)

* To simplify, the presentation of N without the subscript e denotes the shape function of an element.

63

Finite Element Method

where ui and vi are the nodal displacements. Equation (2.58) in terms of vectors becomes u ( ξ,η ) = Nu e

(2.59)

where ue denotes the vector of nodal displacements, ue = [ u1 v1 u2 v2 u3 v3 u4 v4 ]

T

(2.60)

Construction of Strain Matrix The next step involves the computation of strain matrix Be. The matrix is defined by the equation ∂ ------ 0 ∂x T ∂ N1 0 N2 0 N3 0 N4 0 B e = D N = 0 ----∂y 0 N 1 0 N 2 0 N 3 0 N 4 ∂ ∂ ----- -----∂y ∂x

(2.61)

(See Section 2.1.) Now the Cartesian derivatives in the above equation are correlated to the natural coordinate system by the following transformation:       

∂ -----∂ξ ∂ -----∂η

   ∂    ----  ∂x   = J  ∂    ----  ∂y    

(2.62)

where J is a Jacobian matrix, ∂x -----∂ξ J = ∂x -----∂η

∂y -----∂ξ = ∂y -----∂η

∂N k

∂N k

∂N k -x ∑ -------∂η k

∂N k -y ∑ -------∂η k

--------- y -x ∑ -------∂ξ k ∑ ∂ξ k

k = 1, 2, …n

(2.63)

It follows that       

∂N ---------k ∂x ∂N k --------∂y

   ∂N k   -------  – 1  ∂ξ    = J   ∂N k   -   -------  ∂η  

(2.64)

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Nonlinear Problems in Machine Design

Hence the strain matrix Be becomes 1 0 0 0 –1 Be = 0 0 0 1 J 0 –1 0110 0 J ∂N ---------1 ∂ξ ∂N ---------1 × ∂η 0 0

0 0

∂N ---------2 ∂ξ ∂N ---------2 ∂η

∂N ---------1 ∂ξ ∂N 1 --------∂η

0 0

0 0

∂N ---------3 ∂ξ ∂N 3 --------∂η

∂N ---------2 ∂ξ ∂N 2 --------∂η

0 0

0 0 ∂N ---------3 ∂ξ ∂N 3 --------∂η

∂N ---------4 ∂ξ ∂N ---------4 ∂η 0 0

0 0 ∂N ---------4 ∂ξ ∂N 4 --------∂η

(2.65)

where 0 denotes a 2 × 2 zero-matrix. Construction of Stiffness Matrix The stiffness matrix of an element (see Section 2.1) equals Ke =

∫e Be EBe d Ae T

(2.66)

In the isoparametric approach, area dAe is defined in terms of natural coordinates. The area is shown in Figure 2.9a. Disregarding changes of higher order, the area becomes a parallelogram as shown in Figure 2.9b. It equals area ABCD = ABXV = ATDW + TUZY – 2 ( AUB )

(2.67)

In terms of (ξ, η) the latter expression equals ∂y ∂y ∂x ∂x ∂y ∂x d A e = ------dη ⋅ ------dη +  ------ dξ – ------dη  ------dη – ------ dξ  ∂ξ ∂ξ  ∂η   ∂η ∂η ∂η

∂x ∂y ∂x ∂y =  ------ ------ – ------ ------ dξdη =  ∂ξ ∂η ∂η ∂ξ

Consequently, Equation (2.67) becomes

∂x -----∂ξ ∂x -----∂η

∂y -----∂ξ dξdη = det Jdξdη ∂y -----∂η

(2.68)

65

Finite Element Method

FIGURE 2.9 Elemental area in natural coordinates: (a) dimensions and (b) geometric size relations.

Ke =

+1 +1

∫–1 ∫–1 Be EBe det J dξ dη T

(2.69)

The stiffness matrix of the whole body equals

K =

∑e Ke

=

+1

+1

∑e ∫–1 ∫–1

T

B e EB e det J dξ dη

(2.70)

Computation of Nodal Displacements Nodal displacements un are derived from the equilibrium equation, (2.71)

Ku n = F n See Section 2.1. Fn is the vector of nodal forces acting on the whole body, Fn =

∑e Fe

(2.72)

Fe is the vector acting on a single element. Fe =

+1

∫ ∫e Ne Re det J dξ dη + ∫–1 T

T

N e p e dλ

(2.73)

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Nonlinear Problems in Machine Design

One obtains the nodal displacements by solving the equation un = Fn K

–1

(2.74)

where un is the total vector of nodal displacements: a sum of nodal displacements of individual elements, un =

∑e ue

(2.75)

Computation of Strain and Stress Vectors At the last step, the strain and stress vectors are computed using the following equations:

2.2.2

ε = DT u = Be ue

(2.76)

σ = EB e u e

(2.77)

ISOPARAMETRIC QUADRILATERALS

OF

HIGHER ORDER

In the four-node quadrilaterals described above, components of shape function N are quadratic polynomials of natural coordinates (ξ,η); hence, matrix Be becomes a linear function of (x,y). It follows that the computed strains and stresses are also linear functions of (x,y). One can obtain more accurate expressions for strains and stresses by introducing higher order polynomials for the shape functions as described below. An added advantage of such elements is that they allow better approximation of curved boundaries of a region. Elements of higher-than-second degree are seldom used. Let us increase the order of the shape function by adding mid-side nodes as shown in Figure 2.10. The resulting elements are called second-order quadrilaterals. Their shape functions are third-degree polynomials as follows: For the corner nodes, 1 N 1 = – --- ( 1 – ξ ) ( 1 – η ) ( ξ + η + 1 ) 4 1 N 2 = --- ( 1 + ξ ) ( 1 – η ) ( ξ – η – 1 ) 4 1 N 3 = --- ( 1 + ξ ) ( 1 + η ) ( ξ + η – 1 ) 4 1 N 4 = – --- ( 1 – ξ ) ( 1 + η ) ( ξ – η + 1 ) 4

(2.78)

67

Finite Element Method

FIGURE 2.10 Eight-node quadrilateral: (a) with straight sides and (b) with curved sides.

And for the mid-side nodes, 1 2 N 5 = --- ( 1 – ξ ) ( 1 – η ) 2 1 2 N 6 = --- ( 1 – η ) ( 1 + ξ ) 2 1 2 N 7 = --- ( 1 – ξ ) ( 1 + η ) 2 1 2 N 8 = --- ( 1 – η ) ( 1 – ξ ) 2

(2.79)

The differential matrix Be for the second-order quadrilaterals equals ∂ ------ 0 ∂x ∂ N1 0 N2 0 N3 … N8 0 B e = 0 ----∂y 0 N 1 0 N 2 0 … 0 N 8 ∂ ∂ ----- -----∂y ∂x

(2.80)

The derivation of stiffness matrix K and strain and stress vectors ε and σ follows the same steps, as outlined above.

68

2.2.3

Nonlinear Problems in Machine Design

HEXAHEDRON

Hexahedron elements are shown in Figure 2.11. The figure depicts two kinds: the 8-node hexahedron and the second-order 20-node hexahedron. As with quadrilaterals, hexahedrons of higher degree are seldom used. Eight-Node Hexahedron This element is an extension of the four-node quadrilateral. Locating the origin of the natural coordinate system at the center of the element, we obtain the general form of its shape function, 1 N i = --- ( 1 + ξξ i ) ( 1 + ηη i ) ( 1 + ςς i ) 8

(2.81)

where ξi , ηi , and ζi are coordinates of the corner nodes i = 1,2,...,8 in the natural coordinate system. For node i = 1, the natural coordinates are ξ1 = –1 ; η1 = –1 ; ζ1 = –1

(2.82)

whereby we obtain

FIGURE 2.11 Hexahedron: (a) 8-node, (b) 20-node with straight sides, and (c) 20-node with curved sides.

69

Finite Element Method

1 N 1 = --- ( 1 – ξ ) ( 1 – η ) ( 1 – ς ) 8

(2.83)

For the other nodes, the expressions change accordingly. The Jacobian necessary for computation of stiffness is a three-dimensional matrix, ∂x -----∂ξ ∂x J = -----∂η ∂x -----∂ζ

∂y -----∂ξ ∂y -----∂η ∂y -----∂ζ

 ∂  ∂z ----- ------  ∂ξ  ∂ξ   ∂  ∂z ------ =  ------ [ x y z ] ∂η  ∂η   ∂  ∂z  ------  -----∂ζ  ∂ζ 

(2.84)

The differential matrix Be for the hexahedron is ∂ ------ 0 0 ∂x ∂ 0 ----- 0 ∂y ∂ 0 0 ----- N 1 0 0 N 2 0 0 … N 8 0 0 ∂z Be = 0 N1 0 0 N2 0 … 0 N8 0 ∂ ∂ ----- ------ 0 0 0 N 1 0 0 N 2 … 0 0 N 8 ∂y ∂x ∂ ∂ 0 ----- ----∂z ∂y ∂ ∂ ----- 0 -----∂x ∂z

(2.85)

The derivation of stiffness matrix K, as well as strain and stress vectors ε and σ, follows the same steps as before. Twenty-Node Hexahedron A 20-node hexahedron element comprises 8 corner nodes and 12 mid-edge nodes (see Figures 2.11b and c). The shape function of corner nodes (i = 1,2,...8) equals 1 N i = --- ( 1 + ξξ i ) ( 1 + ηη i ) ( 1 + ςς i ) ( ξξ i + ηη i + ςς i – 2 ) 8

(2.86)

The shape function for mid-edge nodes (i = 9,10,...20) depends on the edges. For nodes i = 10,12,14,16, the natural coordinates are

70

Nonlinear Problems in Machine Design

ξi = 0 ; ηi = ±1 ; ςi = ±1

(2.87)

1 2 N i = --- ( 1 – ξ ) ( 1 + ηη i ) ( 1 + ςς i ) 4

(2.88)

and the shape function is

For other mid-edge nodes, one obtains the shape function by interchanging the coordinates in Equation (2.88).

2.3 HIERARCHICAL FUNCTIONS While the method described in the preceding section raises the order of a shape function by adding mid-side nodes within an element, a different approach is described below. A higher-order shape function is created by adding high-order terms. The resulting shape function is called the hierarchical function. The process resembles a Taylor expansion of a function: the greater the number of terms in expansion, the better the approximation. A significant advantage of the hierarchical function is the fact that shape-function components of the lower order are not affected by an introduction of terms of higher order (contrary to the shape function, in the previously described original isoparametric approach).

2.3.1

ONE-DIMENSIONAL PROBLEM

As an introduction, consider a one-dimensional structure shown in Figure 2.12. Here, to begin with, we choose a simple linear approximation of displacements u = u1 N 1 + u2 N 2

(2.89)

where the shape function components equal 1 1 N 1 = --- ( 1 – ξ ), N 2 = --- ( 1 + ξ ) 2 2

FIGURE 2.12 One-dimensional structure under axial load.

(2.90)

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Finite Element Method

To increase the accuracy of the approximation, we introduce new variables ∆3, ∆4, etc. and add higher order terms Ni. u = u 1 N 1 + u 2 N 2 + ∆ 3 N 3 + ∆ 4 N 4 + …∆ p N p

(2.91)

where we assume that 2

2

N 3 = 1 – ξ , N 4 = ξ ( 1 – ξ ), etc.

(2.92)

Variables ∆3, ∆4, etc. are newly introduced degrees of freedom, shown in Figure 2.13. Term ∆3 denotes the deviation from the linear approximation, while term ∆4 is the deviation of the slope, obtained from the slope of previous approximation. In general, deviation ∆p is the derivative

FIGURE 2.13 Displacement approximations in one-dimensional structure as a function of hierarchical functions: (a) linear approximation, (b) including deviation from the linear approximation, and (c) including deviation of the slope.

72

Nonlinear Problems in Machine Design p–1

d U ∆ p = --------------p–1 dξ

; p>2

(2.93)

ξ=0

The shape function components of different orders define the curvature of displacement distribution, as illustrated in Figure 2.13. N1 and N2 are known as external components, while N3, N4, … Np are internal components. The latter equal zero at the ends of element (ξ = ±1), in accordance with the definition of shape function. Equation (2.91) is created by superposition of higher-order terms upon the lowerorder ones. Such polynomials are called hierarchical shape functions.3 The hierarchical shape functions possess an important feature in that they form a nested family, i.e., an advanced set of terms includes all lower-order terms as well. This fact simplifies the computational process, since the stiffness matrix and load vectors that refer to lower-order polynomials are computed at previous steps and may be stored in memory, as they remain unchanged. The method is known as p–adaptivity approach, where p denotes the order of approximation of polynomials. It leads to higher level of accuracy without increasing the number of elements. Legendre Polynomials The above shown hierarchical functions are not unique, and different polynomials may be used.3,4 One of the methods to derive hierarchical functions is based on Legendre polynomials. The Legendre polynomial of an order p is defined as p

p 1 d 1 2 -------- [ ( ξ – 1 ) ] P p ( ξ ) = ------------------- ---------( p – 1 )! 2 p – 1 dξ p

(2.94)

The required shape functions are obtained by integrating Legendre polynomials, φk ( ξ ) =

2k – 1 ξ --------------- ∫ P k – 1 ( ς )dς; k ≥ 2 2 –1

(2.95)

whereby the shape function equals N k + 1 = φk ( ξ )

(2.96)

The first two terms (k = 0,1) are the external shape functions N1 and N2, defined by Equations (2.90). The next terms (k = 2,3…) are the internal shape functions 3 2 N 3 ( ξ ) = ---------- ( ξ – 1 ) 2 2 5 2 N 4 ( ξ ) = ---------- ξ ( ξ – 1 ) 2 2

73

Finite Element Method

7 4 2 N 5 ( ξ ) = ---------- 5ξ – ( 6ξ + 1 ) 8 2

(2.97)

and so on. The derivation of stiffness matrix and stress and strain vectors, using hierarchical shape functions, is explained in the next section, dealing with a two-dimensional problem.

2.3.2

TWO-DIMENSIONAL PROBLEM

Hierarchical shape functions of a two-dimensional problem are created by adding high-order terms following the pattern of one-dimensional problem. As an explanation, consider the four-node quadrilateral element shown in Figure 2.8. The displacements within the element are expressed by the following polynomial approximations: u

u

u

v

v

v

u = u 1 N 1 + u 2 N 2 + u 3 N 3 + u 4 N 4 + ∆ 5 N 5 + ∆ 6 N 6 + …∆ p N p v = v 1 N 1 + v 2 N 2 + v 3 N 3 + v 4 N 4 + ∆ 5 N 5 + ∆ 6 N 6 + …∆ p N p u

(2.98)

v

where ui, vi, and ∆ i , ∆ i are degrees of freedom in the respective directions, and Ni are shape functions. The external shape functions are the same as defined by Equations (2.53). 1 N 1 = --- ( 1 – ξ ) ( 1 – η ) 4 1 N 2 = --- ( 1 – ξ ) ( 1 – η ) 4 1 N 3 = --- ( 1 – ξ ) ( 1 – η ) 4 1 N 4 = --- ( 1 – ξ ) ( 1 – η ) 4

(2.99)

The internal shape functions may be defined as multiplications of the one-dimensional functions of different orders in each direction. (int)

Nk

= φ i ( ξ )φ j ( η )

i = 2, 3…, p

j = 2, 3, …, p

(2.100)

Since the hierarchical functions equal zero at corner nodes, the shape functions along the element sides may be constructed as multiplications of the expressions defined by Equations (2.90) and (2.96). Consider a one-dimensional hierarchical function

74

Nonlinear Problems in Machine Design

of one local variables, e.g., ξ, and a linear function of the other variable, as for instance η; then, the approximation along an element side will be a multiplication of both functions. The resulting shape function components equal (1 – 2)

1 ( ξ,η ) = --- ( 1 – η )φ k ( ξ ) 2

at side 1–2, k = 2, 3, …, p

(3 – 4)

1 ( ξ,η ) = --- ( 1 + η )φ k ( ξ ) 2

at side 3–4

(2 – 3)

1 ( ξ,η ) = --- ( 1 + ξ )φ k ( η ) 2

at side 2–3

(4 – 1)

1 ( ξ,η ) = --- ( 1 – ξ )φ k ( η ) 2

at side 4–1

Nk

Nk

Nk

Nk

(2.101)

Equation (2.98) can be expressed in a matrix form as  u  u = [N N] e   ∆e 

(2.102)

where the external and internal shape functions are described by the following equations, respectively,

N =

N1 0 N2 0 N3 0 N4 0 0 N1 0 N2 0 N3 0 N4 (2.103)

N =

N5 0 N6 0 … 0 N p 0 0 N5 0 N6 … N3 0 N p

(2.104)

The vector of nodal displacements in the element equals ue = [ u1 v1 u2 v2 u3 v3 u4 v4 ]

T

(2.105)

while the vector of deviations is ∆ e = [ ∆ u5 ∆ v5 ∆ u6 ∆ v6 …∆ up ∆ vp ]

T

The derivation matrix of displacement u becomes a sum of two matrices,5,6

(2.106)

75

Finite Element Method

∆e D u = Bu e + B∆

(2.107)

∂ ------ 0 ∂x T ∂ N1 0 N2 0 N3 0 N4 0 B e = D N = 0 ----∂y 0 N 1 0 N 2 0 N 3 0 N 4 ∂ ∂ ----- -----∂y ∂x

(2.108)

T

where

∂ ------ 0 ∂x T ∂ N5 0 N6 0 … N p 0 B e = D N = 0 ----∂y 0 N 5 0 N 6 … 0 N p ∂ ∂ ----- -----∂y ∂x

(2.109)

The resulting element stiffness matrix contains four matrices uu

u∆

∆u

∆∆

Ke Ke

Ke =

(2.110)

Ke Ke where the component matrices equal Ke =

uu

∫e Be EBe d Ae

Ke =

u∆

∫e Be EBe d Ae

∆∆

∫e Be EBe d Ae

Ke

=

T

T

∆u

= Ke

T

(2.111)

The equilibrium equation of the element equals5,6 uu u∆  Fe  Ke Ke  ue     =  ∆u ∆∆  F∆  Ke Ke  ∆e 

(2.112)

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Nonlinear Problems in Machine Design

Force Fe is the force acting on a single element, the same as was derived for a simple four-node quadrilateral [see Equation (2.73)], Fe =

+1

T

+1

T

∫ ∫e Ne Re det J dξ dη + ∫–1 Ne pe dλ T

(2.113)

The supplementary term F∆ equals F∆ =

T

∫ ∫e Ne Re det J dξ dη + ∫–1 Ne pe dλ

(2.114)

The resulting strain and stress vectors are  u  ε = [ Be Be ]  e   ∆e 

(2.115)

 u  σ = E [ Be Be ]  e   ∆e 

(2.116)

As mentioned before, the hierarchical functions form a nested family; i.e., an advanced set of shape functions includes all lower-order terms as well, which simplifies the computational process. Note: due to the character of deviations ∆e, special care has to be taken to preserve intercontinuity between the neighboring elements. An adjustment of hierarchical magnitudes, associated with the element edges and those of a neighboring element, provides the required continuity of the field variables along this edge.7 Thus, an FEM model containing quadrilateral elements based on hierarchical functions fulfills the conditions of compatibility.

2.4 BENDING ELEMENTS: BEAMS AND PLATES Bending elements for modeling of beams and plates are unorthodox elements where the degrees of freedom include displacements and rotations. In some problems they may considerably simplify the analysis and, therefore, are of importance in machine design.

2.4.1

BEAM ELEMENT

Consider a beam subject to bending (Figure 2.14). The relevant equilibrium equation has the form 2

d M ---------= q( x) 2 dx

(2.117)

77

Finite Element Method

FIGURE 2.14 Beam subject to bending: (a) the beam and its FE model and (b) distribution of bending stresses.

where M is the local bending moment and q(x) is a distributed lateral load. Based upon the theory of Timoshenko,8 the correlation between the local bending moment and the displacement is 2

d w M = EI --------2dx

(2.118)

where E is the Young modulus and I is the moment of inertia of the cross section. Based on outer loading, as per Figure 2.14, we shall derive the local displacement distribution w(x). The differential equation to be solved equals 4

d w q( x) --------4- – ----------- = 0 EI dx

(2.119)

with the boundary conditions

w = w0 ,

dw ------- = θ 0 at dx

x = 0

w = wL ,

dw ------- = θ L at dx

x = L (2.120)

In accordance with the Galerkin procedure, we assume an arbitrary virtual displacement field v(x), which satisfies the kinematic boundary conditions. Multiplying Equation (2.119) by v(x) and performing integration by parts twice, one obtains the following equation:

78

Nonlinear Problems in Machine Design

w″′ ( L )v ( L ) – w″′ ( 0 )v ( 0 ) + w″ ( L )v′ ( L ) – w″ ( 0 )v′ ( 0 ) 2

2

L d wd v L q( x) + ∫ --------2- -------2- dx – ∫ ----------- v ( x ) dx = 0 0 dx dx 0 EI

(2.122)

A smooth curve of the deformed beam requires continuous displacements and slopes along the beam. This requirement limits the shape functions of a beam element to unbroken curves at interelement borders. [It requires elements with C1 continuity, i.e., continuity of up to the first derivative and integrability of at least the second derivative (see Section 2.5.1 below)]. After dividing the beam into elements, we assume the following approximation for displacements within an element (1–2): w ( x ) = w1 N 1 ( x ) + θ1 N 2 ( x ) + w2 N 3 ( x ) + θ2 N 4 ( x )

(2.123)

The interpolation polynomials are Hermite functions of the first order, 1 3 2 3 1 x N 1 ( x ) = H 0  -- = ---3 ( 2x – 3lx + l )  l l 1 3 2 2 1 x N 2 ( x ) = l ⋅ H 1  -- = ---2 ( x – 2lx + l x )  l l 1 3 2 1 l – x N 3 ( x ) = H 0  ---------- = – ---3 ( 2x – 3lx )  l  l 1 3 2 1 l – x N 4 ( x ) = l ⋅ H 1  ---------- = ---2 ( x – lx )  l  l

(2.124)

where 0 ≤ x ≤ l. The Hermite polynomials meet the continuity requirements, as follows: at node 1:

dN dN dN dN N 1 = 1 ,N 2 = N 3 = N 4 = 0, ----------2 = 1, ----------1 = ----------3 = ----------4 = 0 dx dx dx dx

at node 2:

dN dN dN dN N 3 = 1 ,N 1 = N 2 = N 4 = 0, ----------4 = 1, ----------1 = ----------2 = ----------3 = 0 dx dx dx dx (2.125)

Following a standard procedure, one obtains from Equations (2.122) through (2.124) the equilibrium equation, Kw g = F

(2.126)

79

Finite Element Method

where, K =

∑g  ∫0 Be EI Be dx l

T

(2.127)

2

d B e = --------2 N dx

(2.128)

wg = w0 θ0 w1 θ1 … F =

L

∫0 N

T

T

q ( x ) dx

(2.129) (2.130)

After obtaining the displacements wg from Equation (2.126), the stresses in the beam are derived using the equation My σ = -------I

2.4.2

(2.131)

PLATE ELEMENT

In stress analysis of plate element, subject to bending, one has to ensure that displacement w(x,y) and the slopes θ x and θ y are unique and continuous across element boundaries, in order to meet the C1 continuity conditions. Kirchhoff Plate Model Consider a thin plate in pure bending. The behavior of the plate is fully described by the classical theory of Kirchhoff–Love.9 For an isotropic plate of a constant thickness in a Cartesian coordinate system, the equilibrium equation is as follows: 2

2

2

∂ M ∂ My ∂ Mx - + q ( x,y ) = 0 ----------- + 2 -------------xy- + ----------2 2 ∂x∂y ∂y ∂x

(2.132)

q(x,y) is a lateral distributed load over the plate. Mx and My are bending moments and Mxy is a twisting moment expressed by the curvatures 3

2

2

3

2

2

Et ∂ w ∂ w -  --------2- + v --------2- M x = -----------------------2  12 ( 1 – v ) ∂x ∂y  Et ∂ w ∂ w -  --------2- + v --------2- M y = -----------------------2  12 ( 1 – v ) ∂y ∂x  3

2

Gt ∂ w M xy = -------- -----------12 ∂x∂y

(2.133)

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Nonlinear Problems in Machine Design

t is the plate thickness, E is Young modulus, and ν is Poisson’s ratio. See Figure 2.15. Based on the above, the differential equation to be solved is expressed as 4

4

2

4

∂ w ∂ w ∂ w 12 ( 1 – v ) - q ( x,y ) = 0 -------- + 2 ---------------+ --------4- + -----------------------2 2 3 4 ∂x ∂y Et ∂ x ∂y

(2.134)

which is to be supplemented by appropriate boundary conditions, regarding loading and mounting. We use Galerkin procedure to obtain a finite element solution of the plate. Figure 2.16 shows a rectangular element in xy-coordinates. (The analysis may be generalized to cover arbitrary quadrilaterals using natural coordinates.) The polynomial approximation of displacements within the element takes the form (2.135)

w = N we where w = w θx θy ∂w θ x = ------- ; ∂y

T

∂w θ y = – ------∂x

w e = w 1 θ x1 θ y1 … w 4 θ x4 θ y4

FIGURE 2.15 Plate subject to bending.

(2.136)

(2.137) T

(2.138)

81

Finite Element Method

FIGURE 2.16 Rectangular plate element.

The resulting FE equilibrium equation is Kw g = F

(2.139)

with index g denoting the vector of displacements of the whole plate. The stiffness matrix of the plate equals K = t ∫ B EB d A T

(2.140)

A

where 2

B =

∂ – --------2 ∂x

0

0

∂ – --------2 ∂y

0

0

0

∂ – 2 -----------∂x∂y

0 2

N

(2.141)

2

1 v 0 3 Et - v 1 0 E = -----------------------2 12 ( 1 – v ) 1–v 0 0 ----------2

(2.142)

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Nonlinear Problems in Machine Design

For a plate loaded by lateral distributed load q(x,y), the load vector equals F =

∫A N

T

q ( x,y ) d A

(2.143)

Following the computation of displacements wg, the stresses in the plate are computed using the equations 12M y -z; σ y = -----------3 t

12M x -z; σ x = -----------3 t

12M xy -z τ xy = -------------3 t

(2.144)

Supplementary Degrees of Freedom In the above described plate element, the shape function contains 12 components corresponding to the number of degrees of freedom. To provide a continuity of displacements and their derivatives across element borders, one needs 4 supplementary degrees of freedom and a shape function containing 16. To simplify it, one may use the plate element of Bogner et al.,10 which has an additional displacement component in the form 2

∂ w θ xy = -----------∂x∂y

(2.145)

whereby the displacement vector becomes w = w θ x θ y θ xy

T

(2.146)

The shape function of the element is made of Hermite polynomials as follows: N = N1 + N2 + N3 + N4

(2.147)

N 1 = H 10  --x- H 10  --y- bH 10  --x- H 11  --y- aH 11  --x- H 11  --y- abH 11  --x- H 11  --y-  a  b  a  b  a  b  a  b (2.148) with similar expressions for N2, N3, and N4. See Equations (2.124) for the Hermite polynomials that relate to the beam element.

2.5 ACCURACY OF FE SOLUTION After an FE solution is obtained, a question arises concerning its accuracy. Two factors determine the accuracy of a solution.

Finite Element Method

83

1. continuity of variables across element borders 2. magnitude of numerical error The accuracy depends on the number of degrees of freedom. The methods to increase the accuracy are as follows: 1. The mesh is locally refined by reducing the size of elements, while their shape functions, i.e., the interpolating polynomials, stay unchanged (h–method). The number of elements and nodes is increased, enlarging the number of degrees of freedom. See Figure 2.17a. Property h denotes a characteristic dimension of the element. 2. The number of degrees of freedom is increased by adding mid-side nodes without changing the element size. The order of shape functions is increased. 3. The number of degrees of freedom is increased by adding displacement derivatives at the nodes while number of nodes and the shape and size of elements remain the same (p–method). See Figure 2.17b. The order of shape functions is increased, so derived shape functions become hierarchical functions.

2.5.1

CP–1-CONTINUITY

The continuity of displacements and their derivatives across the borders of adjacent elements is specified by property Cp–1, where superscript p ≥ 1 defines its order. In particular, C0 continuity means that only the displacements are continuous across element borders; C1 continuity implies a continuity of both displacements and slopes across the borders; while Cp–1 continuity (p ≥ 3) defines a continuous curvature of

FIGURE 2.17 FE meshes: (a) h-adaptivity method and (b) p-adaptivity method.

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Nonlinear Problems in Machine Design

higher order. Thus, a finite element system based on linear shape functions has C0 continuity (see Figure 2.17a). An element system based on mid-side nodes will have C1. An element system based on hierarchical functions (Figure 2.17b) will have Cp–1 continuity with p ≥ 2. As noted in Section 2.3.2 concerning hierarchical functions, special care is taken to preserve continuity of field variables across the borders of adjacent elements. This is achieved by adjusting the hierarchical magnitudes of the bordering elements. The order of continuity can be derived from the governing differential equation of the boundary value problem. Thus, referring to the Galerkin solution of the boundary value problem (Section 1.5), the minimum order of continuity is defined by the highest order of derivatives in the weak formulation of the Galerkin method.

2.5.2

ERROR MEASURE

OF

FE SOLUTION

The reliability of the finite element solution is determined using an error measure. The simplest way to define the error of FE solution is to compare it with an existing theoretical one. The main difficulty in the above approach is that theoretical solutions exist only for a restricted number of problems. In most cases, a decision concerning the quality of the FE solution can be made only after the problem is solved. Therefore, to accept the solution, one needs a measure of error that is based on the obtained numerical solution only. Different criteria concerning errors have been proposed. Here, we consider the error measures of Zienkiewicz and Zhu.11 We define a local displacement error as follows: e u = u ( x ) – uˆ ( x )

(2.149)

where u is a theoretical value of the displacement and uˆ denotes the numerical solution, uˆ ( x ) = N ( x )u n

(2.150)

The strain and stress errors are defined accordingly. T T e ε = D u – D uˆ = ε – εˆ

(2.151)

e σ = σ – σˆ = E ( ε – εˆ )

(2.152)

where σˆ and εˆ are stresses and strains obtained from the FE solution. Because of the difficulty of local error estimation, integrated forms are used. Thus, we define the displacement error measure to equal the norm, eu =

∫ e u e u dD D T

(2.153)

85

Finite Element Method

where D is the domain subject to numerical solution. Another widely used error measure, based on potential energy, that fits the FEM is the norm

∫ e ε e σ dD D T

eΠ =

∫ eσ D D T

=

–1

e σ dD =

∫ eε Deε dD D T

(2.154)

which is the so-called energy error. It is often replaced by the stress error norm, eσ =

∫ e σ e σ dD D T

(2.155)

In the finite element application, the error measure of the whole domain is computed from the individual element contributions. Thus, for instance, the energy error measure becomes n

n

eΠ

2

=

∑ ∫ eε eσ d( elem ) = T

1 e

∑ ∫ eσ D T

–1

e σ d( elem )

(2.156)

1 e

Often, in practice, a relative error measure is used. eΠ η = --------Π

(2.157)

where Π is the full strain energy of the body computed from the FE solution. Because an analytical solution is usually not known, the theoretical values are replaced by a continuous field, constructed on the basis of stresses (or displacements) obtained from the FE solution. Thus, the stress error is computed by the equation e σ = σ – σˆ

(2.158)

where is a smooth stress distribution obtained by a suitable interpolation. Zienkiewicz and Zhu11,12 derive the smooth stress by nodal averaging of stresses. They introduce an interpolation formula, σ = Nσ σn

(2.159)

and assume the condition that unbalanced nodal forces caused by stress discontinuities are offset by forces resulting from the smoothed stresses,

∫N D

T

σˆ dD =

∫N D

T

σ dD

(2.160)

86

Nonlinear Problems in Machine Design

Combining Equations (2.159) and (2.160), one obtains

∫N D

T

σn = N dDσ

∫N D

T

T

ED N dDu n

(2.161)

from which follows –1 σ n =  ∫ N T N dD ∫ N T ED T N dDu n   D

(2.162)

D

If the error measure is not within an accepted interval, an improvement of initial assumptions has to be made by decreasing the size of elements or increasing the order of interpolation polynomials, or both. The process is repeated until an accepted error level is achieved. The process of improvement, based on consecutive solutions, is known as the adaptivity process. The adaptivity process is controlled by relative error measure [Equation (2.157)]. The error measure is applied to the whole domain as well as to the individual elements. One compares the contributions to the total error with those from the single elements. Those elements with errors that are beyond the user-defined limits are subject to change. The process is repeated until all element errors are within accepted limits. The adaptivity process may be performed using number of methods that include the h-method, p-method, and h/p-method. In the h/p-method, a mixed procedure may be performed where both the element size and the interpolating polynomials are changed. The mesh can be refined, and the order of interpolating (hierarchical) functions can be increased, or both at the same time.13,14,15

REFERENCES 1. Bathe, K.-J., Finite Element Procedures, Prentice Hall, Englewood Cliffs, New Jersey, 1996. 2. Zienkiewicz, O.C., The Finite Element Method, 3d edition, McGraw-Hill, London, 1977. 3. Sabo, B., and Babuska, I., Finite Element Analysis, Wiley, New York, 1991. 4. Zienkiewicz, O.C., and Morgan. K., Finite Elements and Approximation, Wiley, New York, 1983. 5. Taylor, R.L., Beresford, P.J., and Wilson, E.L., A non-conforming element for stress analysis, Int. J. Numer. Methods Eng., 10, 1211–1219, 1976. 6. Zienkiewicz, O.C., de S.R. Gago, J.P., and Kelly, D.W., The hierarchical concept in finite element analysis, Comp. & Struct., 1–4, 53–65, 1983. 7. Chih-Chao Chang, Quasi-Newtonian methods for the solution of hierarchical finite element equations, Comp. & Struct., 3, 423–431, 1993. 8. Timoshenko, S.P., Strength of Materials, Van Nostrand, Princeton, 1956. 9. Timoshenko, S.P., and Woinowsky-Krieger, S., Theory of Plates and Shells, McGrawHill, New York, 1959. 10. Bogner, F.K., Fox, R.L., and Schmit, L.A., The generation of interelement-compatible stiffness and mass matrices by the use of interpolation formlae, Proc. Conf. Matrix Methods in Struct. Mech., AF Inst. of Techn., Wright-Patterson AF Base, Ohio, 1965.

Finite Element Method

87

11. Zienkiewicz, O.C., and Zhu, J.Z., A simple error estimator and adaptive procedure for practical engineering analysis, Int. J. Numer. Methods Eng., 24, 337–357, 1987. 12. Zienkiewicz, O.C., and Zhu, J.Z., The three R’s of engineering analysis and error estimation and adaptivity, Comp. Methods in Appl. Mech. and Eng., 82, 95–113, 1990. 13. Demkowicz, L., Oden, J.T., Rachowicz, W., and Hardy, O., Toward a universal h-p adaptive finite element strategy, Part 1, Constrained approximation and data structure, Comp. Methods Appl. Mech. Eng., 77, 79–112, 1989. 14. Oden, J.T., Demkowicz, L., Rachowicz, W., and Westermann, T.A., Toward a universal h–p adaptive finite element strategy, Part 2. A posteriori error estimation, Comp. Methods Appl. Mech. Eng., 77, 113–180, 1989. 15. Rachowicz, W., Oden, J.T., and Demkowicz, L., Toward a universal h–p adaptive finite element strategy, Part 3. Design of h–p meshes, Comp. Methods Appl. Mech. Eng., 77, 181–212, 1989.

3

Nonlinear Problems

3.1 INTRODUCTION Nonlinearity is of practical concern in machine design, because displacements in machine parts present nonlinear functions of the applied loads. A simplified analysis may disregard nonlinearity, but more accurate designs, called for by a need for cost awareness and a concern for life expectancy, must take the nonlinear approach. The problems of nonlinearities stem from several sources. 1. Material nonlinearity exists in materials that do not follow the classical Hooke’s law; i.e., the relation of stress and strain is nonlinear, which is also true for the relations of forces and displacements. The materials referred to here have such properties as nonlinear elasticity, plasticity, and rubber-like qualities. In numerical handling particularly, plasticity due to energy dissipation presents difficulties. 2. Geometric nonlinearity exists when the loading causes large displacements, large rotations, large strains, or a combination of any of these factors to such an extent that the relation of forces and displacements becomes nonlinear. 3. Contact problem exists whenever contact is encountered, because, in most cases, the geometry of the contact zone is unknown. With the presence of friction, the contact problem becomes especially complicated because of energy dissipation. A numerical solution of the nonlinear problems classified above is based on expressing the equilibrium of a structure in the form

∫L K du

= F

(3.1)

where stiffness K is a function of displacements. L denotes the loading path. Stiffness K requires a special numerical treatment, illustrated below by a simple case.

3.2 EXAMPLE: TWO-SPAR FRAME The following problem serves as an introduction to nonlinear programming. Consider a shallow two-spar frame, Figure 3.1. The frame is symmetric, with one degree 89

90

Nonlinear Problems in Machine Design

FIGURE 3.1 Two-spar frame with one degree of freedom.

of freedom, subject to a vertical load F directed downward. The applied load causes deformation of the frame, whereby point C moves down by displacement u. The objective of the problem is to define the displacement as a function of the applied load. The load is resisted by forces N in the spars. At point C, the equilibrium of forces is expressed by the equation 2N sin α = F

(3.2)

It is evident that the two left-hand variables, force N and spar angle α, are functions of displacement u, thus representing a nonlinear problem. We shall discuss the theoretical aspects of the problem, followed by numerical solutions.

3.2.1

THEORY

According to the rule of stationarity, when a body is in static equilibrium, the first variation of its potential energy equals zero, i.e., δΠ = δU – δW e = 0

(3.3)

where U denotes the internal strain energy, and We is the external work. The internal strain energy of the two-spar system can be expressed as follows, if for simplification we neglect the cross-sectional changes of the spars, 1 2 U ( u ) = 2 ⋅ --- ∫ σε ⋅ dV = E A 0 l 0 ε 2V

(3.4)

E is the Young modulus of material, l0 denotes the initial length, A0 represents the cross-sectional area of the spars, and subscript 0 denotes the initial state. Strain ε is defined as

91

Nonlinear Problems

lc – l0 ∆l ε = ----- = -----------l0 l0

(3.5)

where subscript c denotes the current state. From the geometric analysis it follows that 2

2

2

2

lc = hc + ( l0 – h0 ) 2

2

2

2

= h 0 + 2h 0 u + u + l 0 – h 0

(3.6)

See Figure 3.1. Thus, the spar length in the deformed state equals h 0 u u 2 1 ⁄ 2 - + ----2l c = l o  1 + 2 ------2  l0  l0

(3.7)

We assume that α, hc , and u are very small. Consequently, expanding Equation (3.7) into a truncated Taylor series, we get 2

h0 u u  l c ≈ l 0  1 + ------- + -------2 2  l 0 2l 0

(3.8)

h0 u u2 - + -------2 ε = ------2 l 0 2l 0

(3.9)

It follows from the above that

whereby the internal strain energy of the of two-spar frame becomes 2

h0 u u  - + ------U ( u ) = E A 0 l 0  ------ l 2 2l 2 0 0 2 2

2

3

4

A 0 E 4h 0 u + 4h 0 u + u - ⋅ --------------------------------------------= --------3 4 l0

(3.10)

The first variation of the internal energy, representing the internal virtual work, equals A0 E ∂U 3 2 - ( 2h 0 u + 3h 0 u )δu δU = -------δu = --------3 ∂u l0 On the other hand, the first variation of external work equals

(3.11)

92

Nonlinear Problems in Machine Design

δW e = Fδu

(3.12)

Subsequently, the first derivative of potential energy becomes A0 E ∂Π 2 2 3 - ( 2h 0 u + 3h 0 u + u ) – F = R ------- = --------3 ∂u l0

(3.13)

which is a residual force R. In equilibrium, due to the condition of stationarity, the residual force vanishes, i.e., ∂Π ------- = R = 0 ∂u

(3.14)

The second derivative of the potential energy equals the stiffness of the structure 2 A0 E ∂ Π 2 2 - ( 2h 0 + 6h 0 u + 3u ) = K ( u ) ---------2- = --------3 l ∂u 0

(3.15)

Using the term K(u), the equilibrium defined by Equation (3.14) can be expressed in the form A0 E 2 2 3 --------- ( 2h 0 u + 3h 0 u + u ) = 3 l0

∫u K ( u ) ⋅ du

= F

(3.16)

which confirms the basic expression, Equation (3.1). In finite element nomenclature, stiffness K = Kt is called the tangent stiffness. It can be expressed as a sum of two parts. 2 A0 E 2 3 A0 E t 2 -h 0 + ------------ ( 2h 0 u + u ) K ( u ) = -----------3 3 l0 l0

(3.17)

The first term, on the right-hand side of the equation, is constant, while the second term is a function of displacement and is a nonlinear term. In cases where the displacements are very small, the second term is often disregarded, and the problem becomes linear. The nonlinear problems are solved by either incremental or iterative methods.

3.2.2

INCREMENTAL METHOD

The solution, based on this method, consists of a chain of consecutive incremental solutions along a loading path, as described below. We replace Equation (3.16) by an equivalent equation.

93

Nonlinear Problems t

K ( u i ) ⋅ ∆u i = ∆F i

(3.18)

where ∆Fi and ∆ui are the load and displacement increments, respectively. Subscript i denotes the consecutive increment. Assume that displacement ui, at the beginning of increment i, is given. Stiffness t t K ( u i ) = K i is constant within increment i, since it is a function of ui. Increment ∆u, is a linear function of ∆Fi in the form t –1

∆u i = [ K i ] ∆F i

(3.19)

We compute displacement ui+1 at the end of increment i by the recurrence formula, t –1

u i + 1 = u i + ∆u i = u i + [ K i ] ∆F i

(3.20)

To compute ui+2 at the next increment, the stiffness matrix Kti +1 is recalculated using displacement value ui+1. This procedure is repeated for each consecutive increment until the total of load increments reaches the external load F. Numerical Example For illustration, we use the two-spar frame with the following data: l0 = 500 mm A0 = 50 mm2 h0 = 25 mm E = 200,000 MPa F = 3,000 N Our computation of displacement u involves 5 increments. The steps are shown in Figure 3.2, while the corresponding numerical data are presented in Table 3.1. An assessment of accuracy of the obtained results is shown in Figure 3.2, where they are plotted next to a curve derived from Equation (3.13). The numerical result of the final displacement equals 16.3 mm, while the curve provides a different answer: 14.6 mm. The error of the incremental computation, therefore, is about 11 percent (u = 16.3 mm versus u = 14.6 mm). The incremental method has two main drawbacks: the inaccuracies and the extended computational time. The inaccuracies stem from the fact that the equilibrium at each increment is not fully satisfied and therefore the incremental path deviates considerably from the true path, see Figure 3.2. The process is time cont suming, because a new stiffness, K is computed and inverted at each increment. The remedy to improve the accuracy is to make the steps smaller. However, then the computation time, will be even greater. A more efficient methods are the iterative methods, which are presented below.

94

Nonlinear Problems in Machine Design

FIGURE 3.2 Solution of two-spar frame problem: (a) theoretical solution and (b) incremental solution.

TABLE 3.1 Computation of Displacements by Incremental Method Increment

∆Fi (N)

ui (mm)

1

100.0

0.0

100

1.0

1.0

2

200.0

1.0

112.24

1.782

2.782

3

400.0

2.782

135.24

2.958

5.740

4

800.0

5.740

176.79

4.525

10.265

5

1500.0

10.265

248.47

6.037

16.302

Kt(ui) (N/mm)

∆ui (mm)

ui +1 (mm)

3.3 ITERATIVE METHODS The iterative methods for deriving the displacements in a loaded body aim to find the final solution in directly, without following the loading path. To begin with, estimates are assigned to unknown values of the displacements. The process then proceeds to adjust the results, reducing the error with each iteration until a sufficiently accurate result is reached. This is decided when the calculated error is within the desired limits. Described below are two self-adjusting methods: the Newton–Raphson method and the Newton–Raphson modified method. Both of them provide high accuracy and rapid convergence.

95

Nonlinear Problems

3.3.1

NEWTON–RAPHSON METHOD

To explain the method, first consider the solution of a function of a single variable. f ( x) = 0

(3.21)

The solution is shown graphically in Figure 3.3, which presents the curve y = f(x): Cross point x0 between the curve and line 0x marks the exact solution of Equation (3.21). Let us assume that point xi on line 0x is an approximation, replacing point x0, while distance yi represents the error. If here we draw a tangent line to curve y = f(x), it will lead to point xi+1. The latter is a better choice, because the subsequent error y i+1 will be smaller. In mathematical form, the new approximation is defined by the equation f ( xi ) x i + 1 = x i – -----------f′ ( x i )

(3.22)

The Newton–Raphson method reaches a satisfactory result by repeating Equation (3.22) in recurring computations. In general, an acceptable result is obtained once function y = f(x) is either monotonously increasing or monotonously decreasing. In the following, we shall apply the Newton–Raphson method to a one-dimensional structure such as the spar-frame (Figure 3.1). Let us express displacement u of the frame in the form u = u i + ∆u

FIGURE 3.3 Newton–Raphson iteration method.

(3.23)

96

Nonlinear Problems in Machine Design

where ui is a rough estimate, and ∆u is the sought correction. Then the potential energy of the structure can be approximated by a truncated Taylor series, 2

dΠ ( u i ) 1 d Π ( ui ) 2 -∆u - ∆u + --- -----------------Π ( u ) ≅ Π ( u i ) + ---------------2 du 2 du

(3.24)

When the structure is at equilibrium, Π(u) reaches a minimum; i.e., the first derivative of Π(u) equals zero. 2

dΠ ( u i ) d Π ( u i ) dΠ ( u ) -∆u = 0 - + --------------------------------- = ---------------2 du du du

(3.25)

From here, combining Equations (3.23) and (3.25), we obtain the following expression of displacement u: 2

d Π ( ui ) u = u i – -----------------2 du

–1

dΠ ( u i ) ----------------du

(3.26)

The displacement so defined is in fact a new approximation, u = ui+1, since it is derived from a truncated Taylor series. Equation (3.26), therefore, can be regarded as a recurrence equation similar to Equation (3.22). In numerical application, we define the first derivative of Π at point ui as residual Ri, dΠ ( u i ) ----------------- = R ( u i ) = R i du

(3.27) t

and the second derivative of Π at point ui as stiffness K i , 2

d Π ( ui ) t t ------------------ = K ( ui ) = K i 2 du

(3.28)

Consequently, Equation (3.26) becomes the recurrence formula, t –1

ui + 1 = ui – Ri [ K i ]

(3.29)

Residual Ri represents the error due to approximation of displacement u. It may be construed as unbalanced force, which keeps the structure out of equilibrium, and is equal to the difference between the resistance of the structure and the applied load, r

Ri = F i – F

(3.30)

97

Nonlinear Problems r

where F is the applied load, while F i is the structures’ resistance. In the case of the two-spar frame, for instance, the latter equals A0 E r 2 2 3 r - ( 2h 0 u i + 3h 0 u i + u i ) F i = F ( u i ) = --------3 l0

(3.31)

as per Equation (3.16). The computational process is plotted in Figure 3.4, where a trial solution for u0 = 0 was used at the beginning. Numerical Example Consider again the two-spar frame. Applying the Newton–Raphson method, the process of computation involves five iterations. The obtained result is u = 14.637 mm. The steps are presented in Table 3.2. One finds a magnitude of an error by relating last correction ∆u to the ultimate result: 0.007/14.637 equals about 0.05 percent, which is less by a factor of 200 from the error of the incremental method of 11 percent.

3.3.2

OTHER ITERATIVE PROCEDURES

The two-spar frame described above is a strain-hardening system where the stiffness increases with loading, as shown in Figure 3.4. A different sequence of corrections follows in strain-softening systems, where the stiffness decreases with loading. In such systems, typical for nonlinear materials, the correction of displacements is in

FIGURE 3.4 Newton–Raphson iteration method applied to a strain-hardening system.

98

Nonlinear Problems in Machine Design

TABLE 3.2 Computation of Displacements by Newton–Raphson Method Increment

ui (mm)

Kt(ui) (N/mm)

1

0.0

100.00

2

30.0

3

Ri (N)

ui+1 –ui (mm)

ui +1 (mm)

–3000.0

30.0

30.0

676.00

7560.0

–11.183

18.817

18.817

410.774

1539.0

–3.747

15.070

4

15.070

336.952

–0.426

14.644

5

14.644

327.195

–0.007

14.637

143.55 2.309

the positive direction (Figure 3.5). This permits us to introduce a modified Newton–Raphson method where the initial stiffness matrix is applied throughout the process without repeating matrix computation and inversion. The process is shown in Figure 3.6. The recurrence formula for modified Newton–Raphson method is t –1

ui + 1 = ui – Ri [ K 0 ]

(3.32)

Even though the modified method has a slower convergence, the result is obtained faster and, because of a lesser computational effort, it is less costly. Often, in addition to data at the maximum value of the applied load, the information corresponding to the full loading path is required. This is especially important

FIGURE 3.5 Newton–Raphson iteration method applied to a strain-softening system.

Nonlinear Problems

99

FIGURE 3.6 Modified Newton–Raphson iteration method.

in problems where the loading process is irreversible, as for instance problems involving plastic materials, contacts with frictional forces, and other dissipation sources. In such problems, a mixed procedure can be applied, combining the incremental with one of the Newton–Raphson methods. To obtain the full loading path, the load is divided into a number of increments ∆Fi where, for each load increment, the described iterative correction procedure is applied. See Figure 3.7.

FIGURE 3.7 Mixed iteration method.

100

3.3.3

Nonlinear Problems in Machine Design

FINITE ELEMENT APPLICATION

Most nonlinear problems in machine design contain many degrees of freedom, presented in finite element form. Therefore, one uses a truncated Taylor series for a function of many variables. ∂f f ( x 1 ,x 2…, x n ) = [ f ( x 1 ,x 2…, x n ) ] 0 + ∑  ------- ∆x i   ∂x i 0 (i) 2

∂ f 1 + --- ∑ ∑  --------------- ( ∆x i ∆x j ) 2 i∆ ( j )  ∂x i ∂x j 0

(3.33)

In finite element application, the Newton–Raphson procedure refers to displacements combined into the displacement vector,    u =    

u 1  u1   …  u n 

(3.34)

u = u i + ∆u

(3.35)

Let us express vector u in the form

where i denotes the values of the last iteration, meaning that ui is known. The components of the vector are the degrees of freedom, u n = ( u n ) i + ∆u n

(3.36)

The potential function of the system is expressed by truncated Taylor series, ∂Π Π ( u 1 ,u 2 ,…u n ) = [ Π ( u 1 ,u 2 ,…u n ) ] i + ∑  ------- ∆u n  ∂u  i n 2

∂ Π 1 + --- ∑ ∑  ----------------- ∆u n ∆u p 2 n p  ∂u n ∂u p i

(3.37)

with subscript i denoting known values computed at a previous iteration. In matrix form, Equation (3.37) becomes

101

Nonlinear Problems Π ( u 1 ,u 2 ,…u n ) = [ Π ( u 1 ,u 2 ,…u n ) ] i +             

T

 ∂Π    -------   ∂u 1 i  ∂Π    ------ ∂u 2 i   …  ∂Π    ------ ∂u n i  

2

      

2

2

∂ Π   ∂ Π  ∂ Π  -------------------------------- …  -----------------      ∂u 1 ∂u n i  ∆u  ∂u 1 ∂u 1 i ∂u 1 ∂u 2 i  1    2 2 2 ∂ Π ∂ Π   ∂ Π   1  ∆u 2   -------------------------------- …  -----------------  ∂u 2 ∂u n i   ∂u 2 ∂u 1 i  ∂u 2 ∂u 2 i  + --2-   …   … … … …  ∆u   n    2 2 2 ∂ Π   ∂ Π  ∂ Π  -------------------------------- …  -----------------  ∂u n ∂u 1 i  ∂u n ∂u 2 i  ∂u n ∂u n i T

∆u 1 ∆u 2 … ∆u n

      

∆u 1  ∆u 2   …  ∆u n 

(3.38)

We substitute, in the above equation, the residual vector Ri for gradient of Π at point ui ,             

 ∂Π    -------   ∂u 1 i  ∂Π    ------ ∂u 2 i  = R i  …  ∂Π    ------ ∂u n i  

(3.39)

t

and replace Hessian matrix of Π at point ui by stiffness matrix K i , 2

2

2

2

2

∂ Π   ∂ Π  ∂ Π  -------------------------------- …  -----------------  ∂u 1 ∂u 1 i  ∂u 1 ∂u 2 i  ∂u 1 ∂u n i 2

∂ Π   --------------- ∂u 2 ∂u 1 i

∂ Π ∂ Π   ---------------- …  ----------------- = K t  ∂u 2 ∂u n i  ∂u 2 ∂u 2 i i

…

…

2

2

∂ Π   --------------- ∂u n ∂u 1 i

…

(3.40)

… 2

∂ Π ∂ Π   ---------------- …  -----------------  ∂u n ∂u n i  ∂u n ∂u 2 i

Consequently, we obtain the expression of potential energy Π(u), 1 T t T Π ( u ) = Π i + R i ∆u + --- ∆u K i ∆u 2

(3.41)

According to the requirement of stationarity at the minimum point, the gradient of Π(u) must be zero

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Nonlinear Problems in Machine Design

dΠ ( u ) t ---------------- = R i + K i ∆u = 0 du

(3.42)

From the above equation the Newton–Raphson recurrence formula is derived t –1

∆u i = – [ K i ] R i

(3.43)

u i + 1 = u i + ∆u i

(3.44)

Equations (3.43) and (3.44) represent a minimization procedure to reach a minimum value of ui. To posses the minimum, function Π(u) must be convex, and its second derivative must be positive semi-definite. With these conditions, vector ∆ui defines the direction toward the minimum. It means that T

t

di Ki di ≥ 0

(3.45)

where di = ∆ui. From here, we derive the expression T

( –Ri ) di ≥ 0

(3.46)

The above-described minimization procedure is performed in a direction defined by the Hessian matrix, and not in the direction defined by gradient of Π, as in the classical gradient methods. In the FE application, residual Ri represents the vector of unbalanced nodal forces acting during the iterative process. The residual equals the difference between r applied load F and the structure resistance F i r

Ri = F – Fi

(3.47)

r

Structure resistance F i is computed from the stresses at nodal displacements ui using the equation r

Fi =

∫B (vol)

T

σ ( u i ) d( vol )

(3.48)

See Chapter 2. With an acceptable solution, the residual Ri approaches zero. See Figure 3.3.

3.3.4

LINE SEARCH

Line search is used to make the Newton–Raphson procedure more efficient. It helps to overcome convergence difficulties when higher-order terms in the Taylor expansion of potential Π are not negligible, and the quadratic approximation [Equation

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Nonlinear Problems

(3.37)] becomes inaccurate. In general, line search is used to determine a minimum point on a given line. In the following application, Equation (3.43) is used to derive a feasible step direction, while the minimum point, i.e., the optimal length of step, is determined by the line search. The length of the step is defined by a scalar parameter λi. The application of the line-search method is described below. At a loading step with a given point ui, the new point is computed using the equation u i + 1 = u i + ∆u i = u i + λ i d i

(3.49)

The feasible direction toward the minimum value of potential is defined the same way as in the Newton–Raphson method. t –1

di = –[ Ki ] Ri

(3.50)

See Equation (3.43). To determine the optimal length of step λi, one must first solve the minimization problem for the following function: Φ ( λ ) = Π ( ui + λi di )

(3.51)

We introduce a trial vector, u

(i)

= ui + λi di

(3.52)

with ui and di given, to obtain a one-dimensional minimization problem. Let us replace the function by a quadratic polynomial, derived from the Taylor expansion and neglecting higher-order terms. In the vicinity of point λi. the polynomial equals 2

∂Φ ( λ ) 1 ∂ Φ ( λi ) 2 - ( λ – λi ) Φ ( λ ) = Φ ( λ i ) + ----------------i - ( λ – λ i ) + --- -----------------2 ∂λ 2 ∂λ

(3.53)

At the minimum point, the following condition must be satisfied: 2

∂Φ ( λ ) ∂ Φ ( λ i ) ∂Φ ( λ ) - ( λ – λi ) = 0 ---------------- = ----------------i - + -----------------2 ∂λ ∂λ ∂λ

(3.54)

which is equivalent to ∂Π ( u i + λ i d i ) T d ( u i + λ i d i ) --------------------------------- ---------------------------∂u dλ 2

d ( ui + λi di ) T ∂ Π ( ui + λi di ) d ( ui + λi di ) ----------------------------------- ---------------------------- ( λ – λ i ) = 0 + ---------------------------dλ ∂u∂u dλ

(3.55)

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Nonlinear Problems in Machine Design

or ∂Φ ( λ ) T T t ---------------- = R i d i + d i K i d i ( λ – λ i ) = 0 ∂λ

(3.57)

Consequently, at the (i + 1) iteration, the length of step λi+1 is determined from the equation T

Ri di λ i + 1 = λ i – ---------------T t di Ki di

(3.58)

For optimum λi+1, the following condition must be satisfied: T

(3.59)

Ri di = 0

which indicates an orthogonality condition. Usually, optimum parameters λi vary between the values 0.05 < λ2 N 1 . Plastic Deformation Now consider the external load to be increasing. Before the stresses in the trusses reach yield point σyp, the axial forces rise in proportion to external load P, as per Equations (4.21) and (4.22). Since the axial force in the central truss is greater than in the side trusses, the central truss will be the first to yield. After the central truss yields, its axial load remains constant and equal. N 3 = σ yp A

(4.23)

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Plasticity

If we assume that there are no side trusses, the central truss will grow without limit because of the perfectly plastic behavior of the material. De facto there are side trusses, which are still in elastic domain, restricting the growth of the central truss in the structure. Since N3 is known, the structure becomes statically determined, and the equilibrium condition is sufficient to determine forces N1, and N2. From Equation (4.16), it follows that P – σ yp A N 1 = N 2 = -------------------2 cos β

(4.24)

The external load that corresponds to the onset of yield in the central truss is determined from Equation (4.22). 3

P yp = ( 1 + 2cos β )σ yp A

(4.25)

A further increase of external load is sustained by the side trusses until they reach the yield point, i.e., N 1 = N 2 = σ yp A

(4.26)

The corresponding external load is determined from Equation (4.16) as follows: P lim = σ yp A ( 1 + 2cosβ )

(4.27)

This is the limit load of the structure. After this limit is reached, all three trusses will grow until final failure. Residual Stresses Consider the sequence of loading and unloading of the structure, as per Figure 4.8. After reaching the value P c ( P yp ≤ P c ≤ P lim ) , a load is reduced to zero. The behavior of stresses is shown in Figure 4.9. Stresses in the side trusses follow the loading path OA and return along the unloading path AO. The stress in the central truss, because of plastic deformation, will follow the loading path OBC and unloading path CD. Due to plastic deformation of the central truss (indicated by plastic strain in the figure, shown as εp), the structure is subject to residual forces and stresses. The objective is to determine the internal (residual) forces in the trusses after unloading. The sequence of the loading and unloading process is presented as a sum of two stages, the first prompted by load Pc and the second by adding a fictitious load P f = P c . Consequently, the internal, i.e., residual, loads after unloading are r

r

r

c

c

f

N1 = N2 = N1 + N1 f

N3 = N3 + N3

(4.28) (4.29)

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Nonlinear Problems in Machine Design

FIGURE 4.8 Loading and unloading of three-truss structure.

σ

FIGURE 4.9 Stress-strain curves for three-truss structure. c

f

where N i are caused by the force Pc while N i are caused by Pf as per Equations (4.21) and (4.22). It follows that 2

P c cos β f f N 1 = N 2 = – ------------------------3 1 + 2cos β

(4.30)

–Pc f N 3 = ------------------------3 1 + 2cos β

(4.31)

whereby the residual forces equal 2

P c cos β P c – σ yp A Pc σ yp A r r N 1 = N 2 = ---------------------- – ------------------------- = ---------------------------------------------- – -------------3 3 2 cos β 1 + 2cos β 2 cos β ( 1 + 2cos β ) 2 cos β

(4.32)

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Plasticity

Pc r f N 3 = N 3 + N 3 = σ yp A – ------------------------3 1 + 2cos β

(4.33)

Consider loading to limit load P c = P lim with subsequent unloading to zero. We find σ yp A σ yp A ( 1 + 2 cos β ) r - – -------------N 1 = ---------------------------------------------3 2 cos β ( 1 + 2cos β ) 2 cos β 2

sin β ->0 = σ yp A ------------------------3 1 + 2cos β

(4.34)

σ y A ( 1 + 2 cos β ) r N 3 = σ y A – --------------------------------------3 1 + 2cos β 2

2sin β cos β -< 0 = – σ yp A --------------------------3 1 + 2cos β

(4.35)

The residual internal forces are self-balanced. Such a phenomenon occurs in any structure that first has been loaded to reach plastic deformation, after which the external load has been removed. The obtained results emphasize that the residual stresses and deformations are dependent on the stress levels reached at a loading step, i.e., they are path dependent (history dependent).

4.2 YIELD CRITERIA FOR MULTI-AXIAL STRESSES In the case of a one-dimensional (uniaxial) tension as described above, yielding begins when the following condition occurs: σ – σ yp = 0

(4.36)

which represents a yield criterion. For multi-axial stresses in a three-dimensional body, the yield criterion must be expressed by a combination of the stress components. Below, we present two criteria for multi-axial stresses, von Mises and Tresca criteria, which are used in machine design.

4.2.1

VON MISES CRITERION

The von Mises criterion (also known as either the Huber or Hencky criterion) is based on the assumption that the onset of plasticity occurs when the distortion strain energy Wd reaches a critical value, d

W = W yp

(4.37)

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Nonlinear Problems in Machine Design

Distortion energy for the multi-axial stress, in terms of the principal stresses, is 1+v 2 2 2 d W = ------------ [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] 6E

(4.38)

See Equation (1.117). To derive the critical value Wyp, consider first the uniaxial tension test where σ 2 = σ 3 = 0 . Here, the von Mises criterion equals 1+v 2 ------------ σ 1 = W yp 3E

(4.39)

Extending the above equation to multi-axial stresses, the von Mises yield criterion becomes 1+v 2 1+v 2 2 2 ------------ [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] = ------------ σ yp 3E 6E

(4.40)

1 2 2 2 2 --- [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] = σ yp 2

(4.41)

or

In terms of the second invariant of stress deviator, the above equation becomes 1 2 I 2 ( s ) + --- σ yp = 0 3

(4.42)

The bold s denotes a stress deviator. See Equation (1.44). Referring to Section 1.1.4, we find that the von Mises criterion can also be expressed in terms of octahedral shear stress as follows: 2 1 2 2 2 τ oct = --- ( σ 1 – σ 2 ) + ( σ 2 – σ 1 ) + ( σ 3 – σ 1 ) = ------- σ yp 3 3

(4.43)

See Equation (1.36).

4.2.2

TRESCA CRITERION

The Tresca theory, as opposed to the von Mises theory, is based on shear stress and not on distortion energy. Here, it is assumed that yield will occur when the maximum shear stress reaches the critical value k of a material; i.e., τ max = k

(4.44)

119

Plasticity

In terms of principal stresses, the yielding condition may be written as 1  1 1 max  --- σ 1 – σ 2 , --- σ 2 – σ 3 , --- σ 1 – σ 3  = k 2 2 2  

(4.45)

See Section 1.1.4. Material constant k is derived from uniaxial tension tests where σ 1 = σ yp , σ 2 = σ 3 = 0

(4.46)

so that, from Equation (4.45), one obtains 1 k = --- σ yp 2

(4.47)

Alternatively, performing a pure shear test, one obtains k = τ max = τ yp

(4.48)

σ yp τ max = -----2

(4.49)

Hence, it follows that

The latter equation, which follows from Tresca theory, differs from the von Mises expression, Equation (4.43). Experiments show that the von Mises theory predicts yielding more accurately than Tresca’s.

4.2.3

YIELD SURFACE

A yield criterion can be visualized as an yield surface in a three-dimensional stress space with coordinate axes ( σ 1 ,σ 2 ,σ 3 ) . See Figures 4.10 and 4.11. In a generic form, the yield criterion may be expressed as F (σ) = f (σ) – Y = 0

(4.50)

where f(σ) is taken from the left sides of Equations (4.41) and (4.45), while Y expresses the respective right sides, σyp and k, of the same equations. The latter σ) < Y, the considered part is in represent a measure of plastic deformation. When f(σ σ) = Y, the stresses are located on the yield surface, which an elastic state. When f(σ indicates a plastic deformation. For materials with perfectly plastic behavior, the yield surface is fixed in the stress space. For strain-hardening materials, the yield surface moves as the plastic deformation progresses.

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Nonlinear Problems in Machine Design

FIGURE 4.10 Von Mises yield surface in stress space.

FIGURE 4.11 Tresca yield surface in stress space.

121

Plasticity

Let us present the yield surfaces according to von Mises and Tresca. Both of them relate to a central axis expressed by the equation σ1 = σ2 = σ3

(4.51)

which represents the hydrostatic stress (uniform tension or compression). The hydrostatic-stress line is normal to the deviatoric plane (see Section 1.1.5). Von Mises Yield Surface The von Mises yield surface is a cylinder whose axis coincides with the hydrostaticstress line, which follows from Equation (4.41). See Figure 4.10. The projection of yield surface on the deviatoric plane is a circle whose radius equals 1 2 2 2 r = ------- ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) 3 =

2

2

3

s1 + s2 + s2 =

2 --- σ yp 3

(4.52)

Tresca Yield Surface The yield surface fulfills a condition expressed by three equations derived from Equation (4.45). σ 1 – σ 2 = ±σ yp

σ 2 – σ 3 = ±σ yp

σ 3 – σ 1 = ±σ yp

(4.53)

The surface consists of three sets of pairs of parallel planes, forming a symmetric hexagonal prism. See Figure 4.11. The axis of symmetry coincides with the hydrostatic-stress line. The projection of yield surface on the deviatoric plane is a regular hexagon, Figure 4.12. According to both theories (von Mises and Tresca), the loci must coincide for one-dimensional loadings along the following principal axes: σ 1 = σ yp and σ 2 = σ 3 = 0 σ 2 = σ yp and σ 3 = σ 1 = 0 σ 3 = σ yp and σ 1 = σ 2 = 0

(4.54)

It follows that edges of the hexagonal prism must be located on the cylindrical surface or, in other words, the Tresca surface is inscribed in the von Mises surface. See Figure 4.12.

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Nonlinear Problems in Machine Design

FIGURE 4.12 Von Mises and Tresca yield loci on deviatoric plane.

4.3 CONSTITUTIVE THEORIES OF PLASTICITY There are two approaches to express the stress-strain relation in plastic deformation: one following the flow theory and the other following the deformation theory. The flow or incremental theory takes into consideration the irreversible work done during plastic deformation, whereby the final stress and strain become path dependent.1,2,3 Contrary to the above, the deformation theory assumes there exists a single-valued relation between stress and strain that is path-independent during the loading process, similar to the behavior of nonlinear elastic material.4

4.3.1

INCREMENTAL

OR

FLOW THEORY

As noted above, the final state of the material, due to the irreversible nature of plastic deformation, is path dependent: it depends on the order in which the loads were applied. The deformation, therefore, is considered a flow process, and its analysis is performed in terms of infinitesimal increments. The basic assumptions of the incremental or flow theory are as follows: 1. In accordance to experimental evidence in the uniaxial tests, the total infinitesimal strain can be presented as a sum of plastic and elastic parts. p

dε = dε + dε

e

(4.55)

123

Plasticity

2. The volume change is governed by Hooke’s law, i.e., dσ dε m = ( 1 – 2v ) ---------mE

(4.56)

where εm and σm are the mean strain and stress. From this assumption, it follows that p

dε m = 0

(4.57)

The above condition is interpreted as an indication that the material in the plastic state is incompressible. Associated Flow Rule Let us define a function of stress components in the stress space so that its gradient and the plastic strain increment coincide. ∂Q ( σ ) p dε = dλ ---------------σ ∂σ

(4.58)

where σ denotes a tensor, and εp denotes a plastic strain deviator. dλ is a scalar σ) represents a plastic potential. The plastic potential resemparameter. Function Q(σ bles elastic potential with a gradient e ∂U dε ---------i σ ∂σ

(4.59)

See Section 1.4.1. σ) forms the basis for the flow theory of plasticity. Plastic potential function Q(σ The function represents a surface in the stress space that is assumed to be identical to the yield surface, Q(σ) = F (σ)

(4.60)

where F(σ) is defined by Equation (4.50). From here, we can obtain the normality condition, ∂F ( σ ) p dε = dλ ---------------σ ∂σ

(4.61)

Equation (4.61) discloses that the vector of plastic strain increment dεp is normal to the yield surface in stress space.5,6 If we assume a direct relation between infinitesimal stress increment dσ and the plastic strain increment dεp, then, for elasto-

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Nonlinear Problems in Machine Design

plastic loading of a hardening material, the yield surface is convex, and stress dσ is directed toward the outside of the yield surface. See Figure 4.13. Because of the σ) with yield function F(σ σ), Equation (4.61) is known as association of potential Q(σ the associated flow rule. Prandtl–Reuss Relations Consider a potential based on the von Mises yield criterion, Equation (4.41). 1 2 F ( σ ) = I 2 ( s ) + --- σ yp = 0 3

(4.62)

1 1 2 2 2 f ( σ ) = – I 2 ( s ) = --- ( s 1 + s 2 + s 3 ) = --- ( s ⋅ ·s ) 2 2

(4.63)

∂F ( σ ) ∂f ( σ ) ---------------- = -------------- = s σ σ ∂σ ∂σ

(4.64)

It follows that

and

See Equation (1.44). Thus, Equation (4.61) becomes p dεε = dλ s

(4.65)

The corresponding convolute equation therefore will become p p 2 dεε ··dεε = ( dλ ) s··s

FIGURE 4.13 Normality rule of strain increment.

(4.66)

125

Plasticity

Let us introduce the effective stress derived from Equation (1.101), σe =

3 2 2 2 --- ( s 1 + s 2 + s 3 ) = 2

3 --- ( s··s ) 2

(4.67)

and the incremental effective plastic strain derived from Equation (1.106). 3 1 p 2 p 2 p 2 p dε e = ------------ --- [ ( dε 1 ) + ( dε 2 ) + ( dε 3 ) ] = 1+v 2

2 p p --- ( dεε ··dεε ) 3

(4.68)

The latter equation is based on the assumption that, due to incompressibility, the Poisson’s ratio is v = 0.5. Then, from Equation (4.66), we derive the scalar parameter, p

3 dε dλ = --- -------e2 σe

(4.69)

whereby the increment of plastic strain deviator equals p

3 dε p dεε = --- -------e- s 2 σe

(4.70)

In terms of Cartesian components, the latter equation becomes p

dε p , dγ xy = 3 -------e- τ xy σe

p

dε p , dγ yz = 3 -------e- τ yz σe

p

dε p , dγ zx = 3 -------e- τ zx σe

3 dε p dε x = --- -------e- s x 2 σe 3 dε p dε y = --- -------e- s y 2 σe 3 dε p dε z = --- -------e- s z 2 σe

p

p

p

(4.71)

Equations (4.71) are known as the Prandtl–Reuss stress-strain relations.

4.3.2

ISOTROPIC HARDENING

For materials that harden during plastic deformation, the yield surface under loading changes. This phenomenon is called strain hardening. The change of the yield surface follows a hardening rule. Here, we shall consider an isotropic hardening rule, proceeding with an incremental derivation of the stress-strain correlation. Isotropic hardening assumes that the yield surface expands uniformly while maintaining a constant center and shape. See Figure 4.14. The isotropic hardening model is defined by the yield function,

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Nonlinear Problems in Machine Design

FIGURE 4.14 Isotropic hardening of von Mises yield surface.

F ( σ ,κ ) = f ( σ ) – Y ( κ ) = 0

(4.72)

where f(σ) reflects a yield criterion (either von Mises or Tresca) and Y(κ) is a monotonously increasing function of hardening parameter κ. In its simplest form, κ denotes the accumulated incremental plastic strain, κ =

∫t

p dεε

(4.73)

where … denotes either the Euclidean norm x =

x··x

(4.74)

x = max x ij

(4.75)

or the infinite norm

p In the case of von Mises potential, dεε is the Euclidean norm,

p dεε =

p p dεε ··dεε =

3 p --- dε e 2

(4.76)

In an alternative form, κ is assumed to be the accumulated plastic work, p

κ = W =

∫t s··dεε p

=

∂f

------ dλ ∫t s·· ∂σ σ

(4.77)

127

Plasticity

Our present objective is to derive a stress-strain correlation in the form σ = E ep ··dεε dσ

(4.78)

where Eep, is an elasto-plastic tensor, and dε is a strain tensor increment that includes elastic and plastic parts, e p dεε = dεε + dεε

(4.79)

The above can be expressed by means of Hooke’s law, applying the associated flow rule. ∂f e –1 σ + dλ -----dεε = ( E ) ··dσ σ ∂σ

(4.80)

It follows that, to derive Eep, we first have to define dλ. The derivation is based on an assumption of consistency, whereby we assume that, during the process of plastic deformation, the reference point in the stress space must remain on the yield surface, satisfying yield function F ( σ ,κ ) = 0 . The yield function, upon differentiation, becomes ∂F ∂F σ + ------- dκ = 0 dF = ------- ··dσ σ ∂σ ∂κ

(4.81)

As a hardening parameter, let us choose κ defined by Equation (4.73). Then, from Equation (4.72), it follows that ∂f dY σ – ------- dεε p = 0 dF = ------··dσ σ ∂σ dκ

(4.82)

which, using Equation (4.61), becomes ∂f dY ∂f = 0 σ – ------- dλ -----------··dσ σ ∂σ dκ σ ∂σ

(4.83)

Combining the above with Equation (4.80) and multiplying both sides by the term σ ··E e ) , we get ( ∂f /∂σ ∂f ∂f ∂f ∂f e σ + dλ ------··E e ·· -----------··E ··dεε = ------··dσ σ σ σ σ ∂σ ∂σ ∂σ ∂σ σ ··σ σ) , We derive dλ from Equations (4.83) and (4.84), eliminating ( ∂f /∂σ

(4.84)

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Nonlinear Problems in Machine Design

∂f e ------··E ··dεε σ ∂σ dλ = ----------------------------------------------------∂f dY ∂f e ∂f ------- ------ + ------··E ·· -----σ σ ∂σ dκ ∂σ σ ∂σ

(4.85)

Consequently, Equation (4.80) becomes ∂f ∂f e ------ ------··E ··dεε σ ∂σ σ ∂σ σ + ----------------------------------------------------dε = ( E ) ··dσ ∂f dY ∂f e ∂f ------- ------ + ------··E ·· -----σ σ dκ ∂σ ∂σ σ ∂σ e –1

(4.86)

Multiplying both sides of Equation (4.86) by Ee and rearranging the terms, we obtain ∂f e ------··E ··dε σ ∂σ ∂f e e σ E ··dεε – E ·· ------ ----------------------------------------------------- = dσ σ dY ∂f ∂f ∂σ e ∂f ------- ------ + ------··E ·· -----σ σ dκ ∂σ ∂σ σ ∂σ

(4.87)

from which follows the required stress-strain correlation for isotropic hardening. e e ∂f ∂f   E ·· ------ ------··E  e  σ ∂σ σ ∂σ σ =  E – ----------------------------------------------------- ··dεε dσ dY ∂f ∂f e ∂f  ------- ------ + ------··E ·· ------  σ σ dκ ∂σ ∂σ σ ∂σ

(4.88)

The stress-strain correlation can be used in conjunction with the von Mises yield function or with that of Tresca. Using the von Mises Yield Function We consider von Mises potential, 1 2 1 F = --- ( s··s ) – --- σ yp = 0 3 2

(4.89)

with the gradient in stress space that equals ∂f ------ = s σ ∂σ

(4.90)

The norm of the gradient of the potential is ∂f ( σ ) = -------------σ ∂σ

∂f ( σ ) ∂f ( σ ) --------------·· -------------- = σ σ ∂σ ∂σ

s··s =

2 --- σ yp 3

(4.91)

129

Plasticity

The function of hardening parameter in Equation (4.89) is 1 2 Y ( κ ) = --- σ yp 3

(4.92)

The differential of the hardening parameter dκ is p dκ = dεε =

p p dεε ··dεε =

3 p --- dε 2

(4.93)

Hence, the derivative of Y with respect to κ will be dY ------- = dκ

2

dσ yp 1 2 dσ yp 2 2 2 dY - = --- --- σ yp ------------ --------p = --- --- ---------p p 3 3 dε e 3 3 3 dε e dε e

(4.94)

Inserting the relevant expressions into Equation (4.88), we obtain the elasto-plastic tensor, e

E

ep

e

E ··ss··E e = E – ----------------------------------------------4 2 dσ yp e --- σ yp ----------p- + s··E ··s 9 dε e

(4.95)

Note: If, instead of expressing the hardening parameter by Equation (4.73), we p assume the expression κ = W , the same results will be obtained.3 Using the Tresca Yield Function The Tresca potential, as derived from Equation (4.45), is σ yp 1 F ( σ ) = --- ( σ 1 – σ 3 ) – ------ = 0 2 2

(4.96)

where we assume σ1 and σ3 to be the respective highest and lowest principal stresses, 1  1 1 1 --- ( σ 1 – σ 3 ) = max  --- σ 1 – σ 2 , --- σ 2 – σ 3 , --- σ 1 – σ 3  2 2 2 2  

(4.97)

The effective stress and effective incremental plastic strain according to Tresca theory are σe = σ1 – σ3 = s1 – s3

(4.98)

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Nonlinear Problems in Machine Design

2 1 p p p p p dε e = ------------ ( dε 1 – dε 3 ) = --- ( dε 1 – dε 3 ) 3 1+v

(4.99)

σ equals It follows that gradient ∂f /∂σ    ∂f ( σ ) -------------- =  σ ∂σ   

1 --- 0 0 2 0 0 0 1 0 0 – --2

    = f˙   

(4.100)

To define the norm of the gradient, we use the infinite norm of Equation (4.75). ∂f ( σ ) = 1---------------2 ∂σ

(4.101)

The function of the hardening parameter in Equation (4.95) is 1 Y ( κ ) = --- σ yp 2

(4.102)

The differential dκ equals p p p p p p p dκ = dεε = max { dε 1 – dε 2 , dε 2 – dε 3 , ( dε 3 – dε 1 ) }

3 p p p = dε 1 – dε 3 = --- dε e 2

(4.103)

Hence, the derivative of Y with respect to κ is 1 dσ yp dY ------- = --- ---------3 dε ep dκ

(4.104)

Inserting the relevant expressions into Equation (4.86), we obtain the elastoplastic tensor.

E

ep

e e E ··f˙f˙··E e = E – --------------------------------------1 dσ yp ˙ e ˙ --- ----------p- + f ··E ··f 6 dε e

Note: In Equations (4.94) and (4.105), the following term,

(4.105)

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Plasticity

dσ yp ---------- = H p dε e p

is the slope of an empirical function Y versus ε e , obtained from one-dimensional elasto-plastic tests. If H is constant, then it represents the linear strain hardening.

4.3.3

KINEMATIC HARDENING

The kinematic hardening rule is based on the assumption that, during the process of plastic deformation, the yield surface moves in the stress space as a rigid body, with shape and size remaining unchanged. See Figure 4.15. Thus, the advancing yield surface during the plastic deformation is defined by the equation, F = f (σ – α) – Y = 0

(4.106)

Here, Y is a function of the yield strength that determines the size of the yield surface. Tensor α, called backstress, is a hardening parameter that defines the translation of the yield surface. The main weakness of the isotropic hardening model is that it fails to take into account the Bauschinger effect. Differentiating F at constant Y produces the equation, ∂f ∂f σ + ------- ··dα α = 0 dF = ------··dσ σ α ∂σ ∂α

(4.107)

The following correlation was proposed by Prager:3 α = cdεε p dα

FIGURE 4.15 Kinematic hardening of von Mises yield surface.

(4.108)

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Nonlinear Problems in Machine Design

whereby it is assumed that the yield surface moves in an outer normal direction. Scalar c is a hardening modulus obtained from tests. Prager’s assumption leads to the equations α = dσ σ dα

(4.109)

∂f ∂f ------ = – ------σ α ∂α ∂σ

(4.110)

and

Consequently, the increment of the yield function in the stress space can be expressed as ∂f ∂f ∂f σ – ------··cdλ ------ = 0 ------··dσ σ σ σ ∂σ ∂σ ∂σ

(4.111)

Using Equation (4.84) again as above, ∂f ∂f ∂f ∂f e σ + dλ ------··E e ·· -----------··E ··dεε = ------··dσ σ σ σ σ ∂σ ∂σ ∂σ ∂σ

(4.112)

We obtain the scalar parameter dλ, ∂f e ------··E ··dε σ ∂σ dλ = ------------------------------------------------------∂f ∂f ∂f e ∂f c ------·· ------ + ------··E ·· -----σ ∂σ σ σ σ ∂σ ∂σ ∂σ

(4.113)

∂f ∂f e ------ ------··E ··dεε σ ∂σ σ ∂σ σ + ------------------------------------------------------dεε = ( E ) ··dσ ∂f ∂f ∂f e ∂f c ------·· ------ + ------··E ·· -----σ ∂σ σ ∂σ σ σ ∂σ ∂σ

(4.114)

from which we obtain

e –1

Repeating the above-described procedure for isotropic hardening, we obtain stress-strain correlation for kinematic hardening. e ∂f ∂f e   E ·· ------ ------··E  e  σ ∂σ σ ∂σ σ =  E – ------------------------------------------------------- ··dεε dσ ∂f ∂f ∂f ∂f e  c ------·· ------ + ------··E ·· ------  σ ∂σ σ ∂σ σ σ ∂σ ∂σ

(4.115)

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Plasticity

Let us derive the expression of elasto-plastic tensor for kinematic hardening using the von Mises yield function. 1 1 2 f ( σ ) – Y = --- [ ( s – α )·· ( s – α ) ] – --- σ yp = 0 2 3

(4.116)

Here, the gradient of yield function equals ∂f ------ = ( s – α ) σ ∂σ

(4.117)

which leads to the elasto-plastic tensor, E ·· ( s – α ) ( s – α )··E e = E – --------------------------------------------------------------------------------------------e c ( s – α )·· ( s – α ) + ( s – α )··E ·· ( s – α ) e

E

4.3.4

ep

e

(4.118)

COMBINED HARDENING MODEL

Next let us derive the expression of the elasto-plastic tensor for materials with combined kinematic and isotropic hardening. The relevant yield function will be F = f (σ – α) – Y (κ) = 0

(4.119)

In the following presentation, assume the isotropic hardening parameter to be p κ = ε

(4.120)

and Prager’s correlation for kinematic hardening to equal α = cdεε p dα

(4.121)

From Equation (4.119), it follows that ∂f ∂f dY σ + ------- ··dα α – ------- dεε p = 0 dF = ------··dσ σ α ∂σ ∂α dκ

(4.122)

As a first approximation, assume the condition3 ∂f ∂f ------- = – -----σ α ∂σ ∂α

(4.123)

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Nonlinear Problems in Machine Design

See Equation (4.110). Then, the increment of yield function can be expressed as ∂f ∂f ∂f dY ∂f σ = cdλ ------·· ------ + ------- dλ -----------··dσ σ σ ∂σ σ dκ ∂σ ∂σ σ ∂σ

(4.124)

We shall use again Equation (4.81), whereby we obtain ∂f e ------··E ··dεε σ ∂σ dλ = --------------------------------------------------------------------------------∂f ∂f dY ∂f ∂f e ∂f c ------·· ------ + ------- -----+ ------··E ·· -----σ ∂σ σ dκ ∂σ σ σ ∂σ ∂σ ∂σ σ

(4.125)

Repeating the procedure described above, we derive the stress-strain correlation for the combined hardening model. e ∂f ∂f e   E ·· ------ ------··E  e  σ ∂σ σ ∂σ σ =  E – --------------------------------------------------------------------------------- ··dεε dσ ∂f dY ∂f ∂f ∂f ∂F + ------··E e ·· ------  c ------·· ------ + ------- ------ ∂σ  σ ∂σ σ dκ ∂σ σ σ ∂σ ∂σ σ

4.3.5

(4.126)

PERFECTLY PLASTIC MODEL

This is a special case where the yield surface is fixed in the stress space. The yield function here equals F (σ) = f (σ) – Y 0 = 0

(4.127)

where Y0 is constant. The stress increment during the plastic deformation is tangent to the yield surface (see Figure 4.13). Thus, in the plastic state, at each stress increment, the strain increment becomes infinite. In the case of an uniaxial tension loading, after the stress reached the yield point, the plastic flow will continue indefinitely. Subsequently, there is no one-to-one dependence between stress and strain as is true for the strain hardening models, which renders the perfectly plastic model inadequate. Since the plastic strain increment is in the normal direction to yield surface, while the stress increment is tangent, the normality condition of the associated flow rule becomes ∂f σ = 0 ------··dσ σ ∂σ

(4.128)

See Equation (4.61). For the same reason, the convolute of the stress increment and the plastic strain increment equals zero. σ ··dεε p = 0 dσ

(4.129)

135

Plasticity

The two expressions above are insufficient for the determination of a relation between the stress and strain increments, therefore there is a need for different rule. Let us repeat the expression of total strain increment. ∂f e p e –1 σ + dλ -----dεε = dεε + dεε = ( E ) ··dσ σ ∂σ

(4.130)

σ ··E ) and To derive dλ, we multiply both sides of the above equation by ( ∂f /∂σ introduce the convolute, e

∂f ∂f ∂f ∂f e σ + dλ ------··E e ·· -----------··E ··dεε = ------··dσ σ σ σ σ ∂σ ∂σ ∂σ ∂σ

(4.131)

∂f e ------··E ··dεε σ ∂σ dλ = --------------------------∂f e ∂f ------··E ·· -----σ σ ∂σ ∂σ

(4.132)

whereby we obtain

Hence, the total strain increment as per Equation (4.130) becomes ∂f ∂f e ------ ------··E ··dεε σ ∂σ σ ∂f ∂σ σ + ------ ---------------------------------dεε = ( E ) dσ σ ∂f ∂σ e ∂f ------··E ·· -----σ σ ∂σ ∂σ e –1

(4.133)

Multiplying both sides of Equation (4.133) by Ee and rearranging the terms, we obtain ∂f ∂f e ------ ------··E ··dεε σ ∂σ σ ∂σ σ E dεε – E ---------------------------------- = dσ ∂f e ∂f ------··E ·· -----σ σ ∂σ ∂σ e

e

(4.134)

from which follows the required stress-strain relation for the perfectly plastic model, e e ∂f ∂f  ------ ------··E   e E ·· ∂σ σ ∂σ σ σ =  E – --------------------------------- dεε dσ ∂f e ∂f  ------··E ·· ------   σ σ  ∂σ ∂σ

(4.135)

136

4.3.6

Nonlinear Problems in Machine Design

DEFORMATION THEORY

The deformation theory of plasticity expresses the stress-strain relation in plastic deformation differently from the incremental theory, presented above, by assuming that there exists a single-valued relation between stress and strain, and that the relation is path-independent.4 The respective mathematical expressions are presented in an integral form. The deformation theory assumes that the plastic strain deviator is proportional to the stress deviator. ε p = λs

(4.136)

The corresponding convolute equation will be ε p ··εε p = λ 2 s··s

(4.137)

As in the incremental approach, we shall refer to the effective stress and strain. The effective stress in tensorial form equals 3 --- ( s··s ) 2

σe =

(4.138)

The effective plastic strain based on deviatoric components is derived from Equation (1.106), assuming Poisson’s ratio κ = 0.5. It equals p

εe =

2 p p --- ( ε ··ε ) 3

(4.139)

Then, from Equation (4.137), we derive the scalar parameter p

3ε λ = --- ----e2 σe

(4.140)

Consequently, Equation (4.136) becomes p

3ε ε p = --- ----e-s 2 σ6

(4.141)

The total strain is the sum of elastic and plastic strains where the elastic part is governed by Hooke’s law. Thus, the total strain tensor equals p

1 3ε e p σ + --- ----e- s ε = ε + ε = ---σ E 2 σe where σ denotes the stress tensor.

(4.142)

137

Plasticity

The numerical analysis of engineering problems is predominantly based on the incremental theory. However, for proportional loading, where all stress components rise in the same ratio, the deformation theory provides the same results as application of the incremental theory.

4.4 FINITE ELEMENT IMPLEMENTATION As mentioned above, the numerical analysis of elasto-plastic engineering problems is predominantly based on the incremental theory. An essential feature of such problems is the material stiffness matrix, which changes continually in response to the reached stress level. Computation of the stiffness matrix, as described in Chapter 2, is based on Equation (2.29). We define the incremental loading procedure with the equilibrium at the consecutive load increments* K∆u = ∆F

(4.143)

where the stiffness matrix equals K =

∫B V

T

EBdV =

∑B V

e

T

E B + ∑B E B e

T

V

ep

(4.144)

p

The stiffness matrix in Equation (4.144) comprises elastic and plastic terms corresponding to the respective parts of the structure. Ee is the elasticity matrix of those elements which remain elastic. Eep is the elasto-plastic matrix applicable to those elements that are subject to plastic deformation. Elasto-plastic matrix Eep is derived in accordance with the hardening rule for the given material. Thus, for elements of combined isotropic and kinematic hardening material, subject to von Mises yield condition, the elasto-plastic matrix equals e

E

ep

T

e

E ss E e = E – -----------------------------------------------------------4 2 dσ yp T e T css + --- σ yp ----------p- + s E s 9 dε e

(4.145)

See Equation (4.126). Due to the nonlinear nature of plastic deformation, the computation at each load step uses an iterative solution method, either Newton–Raphson or another (see Chapter 3). Upon obtaining nodal displacements ∆u, the strains and stresses are computed using the following equations: ∆εε = B∆u

(4.146)

σ = E ep ∆εε ∆σ

(4.147)

See Chapter 2. * Note: In this section, the bold expressions are matrices and vectors, not tensors.

138

Nonlinear Problems in Machine Design

A load step involving plastic deformation first produces a result that exceeds the limits defined by the yield surface. A correction becomes necessary so that the solution fits the yield surface. The correction method is based on a radial return algorithm.7,8 To illustrate the radial return algorithm, assume von Mises yield condition with isotropic hardening. The consecutive steps are shown in Figure 4.16. At time t = tn , the initial stress, based on Equation (4.147), is located on the yield surface (point A) with radius r.

r n sn =

2 --- σ yp 3

(4.148)

A trial step is computed assuming an elastic deformation, E s n + 1 = s n + ------------ ∆εε n + 1 1+v

(4.149)

followed by a check of whether the new condition (point B) is located outside the yield surface. p

F = f ( s n + 1 ) – Y ( ε e,n ) > 0

(4.150)

We compute the unit normal to the yield surface, sn + 1 n n + 1 = -------------sn + 1

FIGURE 4.16 Radial return method at isotropic hardening.

(4.151)

139

Plasticity

and the new plastic strain ε np + 1 = ε np + ∆λ n + 1 s n + 1

(4.152)

Using an iteration method, either Newton–Raphson or another, we obtain the corrected stress at point C. p

s n + 1 = Y ( ε e,n + 1 )n n + 1

(4.153)

For kinematic hardening materials, the computational procedure is slightly different. Figure 4.17 presents a radial return for combined kinetic and isotropic hardening. Here, Equation (4.149) is replaced by E s n + 1 = s n + ------------ ∆εε n – α n 1+v

(4.154)

Furthermore, before locating point C, as per Equation (4.153), one has to update parameter α. α n + 1 = α n + c∆λ n + 1 s n + 1

FIGURE 4.17 Radial return method for combined hardening.

(4.155)

140

Nonlinear Problems in Machine Design

so that p s n + 1 = α n + 1 + Y ( ε e,n + 1 )n n + 1

(4.156)

The term radial return method is derived from the method’s main assumption that the stress vector is corrected by a radial return to the yield surface, proposed by Mendelson.9

REFERENCES 1. Hill, R., The Mathematical Theory of Plasticity, Clarendon Press, Oxford, 1950. 2. Martin, J., Plasticity: Fundamentals and General Results, MIT Press, Cambridge, Massachusetts, 1975. 3. Khan, A.S., and Huang, S., Continuum Theory of Plasticity, John Wiley, New York, 1995. 4. Kachanov, L.M., Foundations of Theory of Plasticity, North Holland, Amsterdam, 1971. 5. Drucker, D.C., A more fundamental approach to plastic stress-strain relations, Proc. First U.S. Nat. Congr. Appl. Mech., 487–, 1951. 6. Koiter, W.T., Stress-strain relations, uniqueness and variational theorems for elastic plastic materials with singular yield surface, Quart. J. Appl. Math., 11, 350–354, 1953. 7. Schreyer, H.L., Kulak, R.L., and Kramer, J.M., Accurate numerical solutions for elastic-plastic models. J. Pressure Vessel Techn., Trans. ASME, 101, 226–234, 1979. 8. Simo, J.C., and Taylor, R.L., Consistent tangent operators for rate-independent elastoplasticity. Comp. Meth. Appl. Mech. Eng., 48, 101–118, 1985. 9. Mendelson, A., Plasticity: Theory and Application, Macmillan, New York, 1968.

5

Large Displacements

The shape and dimensions of a flexible body are altered by the external loading, causing deformation and displacements. Depending on the flexibility of the body and the size of the load, the displacements may be small or large. The magnitude of displacements, either small or large, affects the character of the problem, linear or nonlinear. In contrast to the linear theory of elasticity, presented in Chapter 1, the focus here is on large displacements, where a distinction is made between the initial (unloaded) state and the current (loaded) state. A problem of this type requires a nonlinear analysis. The contents of this chapter are arranged into five subjects. 1. 2. 3. 4. 5.

Introduction to tensors in a deformed body Description of geometric changes within the body Derivation of stresses Correlation of stresses and deformation Finite element implementation

5.1 TENSOR ANALYSIS OF A DEFORMED BODY We introduce deformation and strain tensors in terms of derivatives of body-point coordinates. Consider changes caused by the loading between the initial (unloaded) and current (loaded) configurations, assuming the process of deformation to be continuous. The following discussion focuses on purely geometric character, without relating to the forces acting on the body. See Refs. 1, 2, and 3.

5.1.1

MATERIAL COORDINATES

Consider deformation of the body in a global Cartesian coordinate system (X1, X2, X3). As the body shifts from the initial configuration to the current one, point P of the body moves to P′. Point P and point P′ are defined by position vectors r and R, respectively. See Figure 5.1. The vectors are identified by their components as follows: 1

2

3

1

2

3

s

(5.1)

3

1

2

3

s

(5.2)

r ( a ,a ,a ) = a i 1 + a i 2 + a i 3 = a i s 1

2

R ( x ,x ,x ) = x i 1 + x i 2 + x i 3 = x i s

141

142

FIGURE 5.1

Nonlinear Problems in Machine Design

Displacement of a point in a body due to deformation.

where the summation rule for repeated indices is used. Consequently, the change from initial to current state is expressed by the displacement vector, s

s

u = R – r = ( x – a )i s

(5.3)

For convenience in future analysis, we introduce here a general system of curvilinear s s 1 2 3 coordinates θ = θ ( X ,X ,X ) . s

s

1

2

3

θ = θ ( X ,X ,X )

(5.4) 1

2

3

The initial position vector in terms of the curvilinear coordinates is r ( θ ,θ ,θ ) , 1 2 3 while the current position vector now becomes R ( θ ,θ θ ,t ) . Coordinates 1 2 3 ( θ ,θ , θ ) are called material coordinates.2,3 Cartesian components as and xs are assumed to be single-valued functions of 1 2 3 ( θ ,θ , θ ) , differentiable as many times as needed. For a continuous deformation to take place, the following Jacobians must be greater than zero: 1

1

1

2

2

2

∂a ∂a ∂a --------1 --------2 --------3 ∂θ ∂θ ∂θ ∂a = ∂a ∂a ∂a > 0 J i = --------------1 --------2 --------3 -k ∂θ ∂θ ∂θ ∂θ s

3

3

3

∂a ∂a ∂a --------1 --------2 --------3 ∂θ ∂θ ∂θ

(5.5)

143

Large Displacements

and 1

1

1

2

2

2

∂x ∂x ∂x --------1 --------2 --------3 ∂θ ∂θ ∂θ ∂x = ∂x ∂x ∂x > 0 J c = --------------1 --------2 --------3 -k ∂θ ∂θ ∂θ ∂θ s

3

3

(5.6)

3

∂x ∂x ∂x --------1 --------2 --------3 ∂θ ∂θ ∂θ

5.1.2

BASE VECTORS

We define base vectors in the initial configuration as ∂r ∂r ∂r g 1 = --------1 ; g 2 = --------2 ; g 3 = --------3 ∂θ ∂θ ∂θ 1

2

(5.7)

3

The base vectors are tangent to ( θ ,θ , θ ) curves (see Figure 5.2). Using base vectors, the differential of initial position vector r equals ∂r 1 ∂r 2 ∂r 3 ∂r s dr = --------1 dθ + --------2 dθ + --------3 dθ = --------s dθ ∂θ ∂θ ∂θ ∂θ 1

2

3

= ( g 1 dθ + g 2 dθ + g 3 dθ ) = g s dθ

FIGURE 5.2

Base vectors in curvilinear coordinates.

s

(5.8)

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Nonlinear Problems in Machine Design

Reciprocal base vectors in the initial configuration are derived as follows: g2 × g3 g3 × g1 g1 × g2 1 2 3 g = --------------------, g = --------------------, g = -------------------( g1 g2 g3 ) ( g1 g2 g3 ) ( g1 g2 g3 )

(5.9)

where the denominators are scalar triple products of the base vectors, ( g1 g2 g3 ) = g1 ⋅ ( g2 × g3 ) = g2 ⋅ ( g3 × g1 ) = g3 ⋅ ( g1 × g2 )

(5.10)

The direction of reciprocal base vector g1 is perpendicular to the directions of base vectors g2 and g3 (see Figure 5.3). The directions of reciprocal base vectors g2 and g3 are derived by equivalent relations. See Appendix A. In a similar way, we define base vectors in the current configuration (see Fig. 5.2), ∂R ∂R ∂R G 1 = --------1 ; G 2 = --------2 ; G 3 = --------3 ∂θ ∂θ ∂θ

(5.11)

whereby the differential of position vector R in the current configuration becomes 1

2

3

dR = G 1 dθ + G 2 dθ + G 3 dθ = G s dθ

s

The reciprocal base vectors for the current configuration equal

FIGURE 5.3

Reciprocal base vectors.

(5.12)

145

Large Displacements

G2 × G3 G3 × G1 G1 × G2 1 2 3 - , G = ------------------------- , G = ------------------------G = ------------------------(G1G2G3) (G1G2G3) (G1G2G3)

(5.13)

Vectors gs and Gs are called covariant base vectors; vectors gs and Gs are called contravariant base vectors. Scalar Products of Base Vectors Below, we present three kinds of scalar products concerning base and reciprocal base vectors. 1. Scalar products of base vectors of initial and current configurations equal the covariant components of metric (unit) tensors, g sk = g s ⋅ g k

(5.14)

G sk = G s ⋅ G k

(5.15)

In a Cartesian coordinate system (X1,X2,X3), the tensor components equal 3

∂a ∂a p n g sk = --------ps --------nk i ⋅ i = ∂θ ∂θ

G sk

∂α ∂a p

-p --------k ∑ -------s ∂θ ∂θ

(5.16)

p=1

3

∂x ∂x p n = --------ps --------nk i ⋅ i = ∂θ ∂θ

∂x p ∂x p

-s --------k ∑ ------∂θ ∂θ

(5.17)

p=1

where

p

n

i ⋅i = δ

pn

 =  1 if p=n  0 if p≠n

(5.18)

Note: there is no summation at repeated indices in Equation (5.18). 2. Scalar products of reciprocal base vectors of initial and current configurations equal the contravariant components of metric (unit) tensors, g

sk

= g ⋅g

s

G

sk

= G ⋅G

s

k

(5.19) k

which, in a Cartesian coordinate system (X1,X2,X3), become

(5.20)

146

Nonlinear Problems in Machine Design 3

g

sk

=

s

∂θ ∂θ

k

s

k

-------- -------∑ ∂a p ∂a p

(5.21)

p=1

3

G

sk

=

∂θ ∂θ

- -------∑ ------∂x p ∂x p

(5.22)

p=1

3. Scalar products of covariant base vectors with contravariant base vectors equal either 1 or zero, as explained below. From Equation (5.9), we obtain the relations for scalar products of base vectors and reciprocal base vectors as g1 ⋅ ( g2 × g3 ) g2 ⋅ ( g3 × g1 ) 1 2 g ⋅ g 1 = -----------------------------= 1, g ⋅ g 2 = -----------------------------= 1, ( g1 g2 g3 ) ( g1 g2 g3 ) g3 ⋅ ( g1 × g2 ) g 3 ⋅ g 3 = -----------------------------= 1 ( g1 g2 g3 )

(5.23)

It follows that, in the initial configuration,  s s g ⋅ g k = δ k =  1 if s=k  0 if s≠k

(5.24)

Similarly, in the current configuration  s s G ⋅ G k = δ k =  1 if s=k  0 if s≠k

(5.25)

Note: There is no summation at repeated indices in Equations (5.24) and (5.25). Metric Tensor The metric (unit) tensor defines metric properties of space: the lengths of infinitesimal lines and the angles between arbitrary directions. See Figure 5.4. In the initial configuration, the length of infinitesimal line dl is defined by the quadratic form, 2

s

k

s

( dl ) = dr ⋅ dr = g s dθ ⋅ g k dθ = g sk dθ dθ

k

(5.26)

where gsk denotes the components of metric tensor. Using gsk , the angle between two arbitrary directions dr 1and dr 2 in the initial configuration is defined by the quadratic form,

147

Large Displacements

FIGURE 5.4

Deformation of an infinitesimal line due to deformation of a body.

p

q

g pq dθ 1 dθ 2 dr 1 ⋅ dr 2 - = -----------------------------------------------------------cos α = ---------------------s k m n dr 1 dr 2 g sk dθ 1 dθ 2 g mn dθ 1 dθ 2

(5.27)

In the current configuration, the length of infinitesimal line dL is defined as 2

s

k

s

( dL ) = dR ⋅ dR = G k dθ ⋅ G k dθ = G sk dθ dθ

k

(5.28)

where Gsk denotes components of the metric tensor. The angle between two arbitrary directions dR 1 and dR 2 can be defined by an expression similar to Equation (5.27). Gradient Fields Due to the different metric properties caused by deformation, gradient fields in the initial and current configurations are also different. The gradient in the initial configuration equals s ∂ ∇ i = g --------s ∂θ

(5.29)

while, in the current configuration, it becomes s ∂ ∇ c = G --------s ∂θ

(5.30)

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Nonlinear Problems in Machine Design

5.2 DEFORMATION AND STRAIN The deformation and strain tensors define the body changes from the initial to the current configuration. The deformation tensor measures the change in terms of lengths and angles in the current configuration, while the strain tensor reflects the difference between the current and initial configurations.

5.2.1

DEFORMATION TENSORS

A deformation tensor is derived from the square of an infinitesimal line when the deformed body moves from the initial configuration to a current one (see Figure 5.4). Two kinds of deformation tensors are described: the Cauchy–Green tensor, which defines the infinitesimal line in the current configuration in terms of the initial configuration, and the Almansi tensor, which defines the same in terms of the current configuration. To derive the Cauchy–Green deformation tensor, consider an infinitesimal change of the current position vector. ∂R s s s k k s s dR = --------s dθ = G s dθ = δ k G s dθ = g k dθ ⋅ g G s = dr ⋅ g G s ∂θ

(5.31)

From Equation (5.29), we have s ∂R s ∇ i R = g --------s = g G s ∂θ

(5.32)

Comparing Equations (5.31) and (5.32), one obtains dR = dr ⋅ ∇ i R

(5.33)

from which follows the derivative of vector R with respect to r, dR T -------- = ( ∇ i R ) dr

(5.34)

As noted above, to measure the deformation, we use the square of the length of an infinitesimal line. According to Equation (5.28), 2

T

C

( dL ) = dR ⋅ dR = dr ⋅ ∇ i R ⋅ ( ∇ i R ) ⋅ dr = d r ⋅ D ⋅ dr

(5.35)

Tensor D C is the Cauchy–Green deformation tensor, defined as D

C

T

s

k

s k

= ∇ i R ⋅ ( ∇ i R ) = g G s ⋅ G k g = G sk g g

(5.36)

149

Large Displacements

Similarly, we derive the Almansi deformation tensor. One may express the gradient of position vector r in respect to the current configuration as s ∂r s ∇ c r = G --------s = G g s ∂θ

(5.37)

Thereby the length of an infinitesimal line in the initial configuration, Equation (5.26), becomes 2

T

A

( dl ) = dR ⋅ ∇ c r ⋅ ( ∇ c r ) ⋅ dR = dR ⋅ D ⋅ dR

(5.38)

Tensor D A is the Almansi deformation tensor, defined as D

A

T

s

k

s

= ∇ c r ⋅ ( ∇ c r ) = G g s ⋅ g k G = g sk G G

k

(5.39)

Based on Equations (5.26) and (5.35), one obtains the relation C

g m ⋅ D ⋅ g m  dL ------ =  -------------------------- dl  m  gm ⋅ gm 

1⁄2

=

G mm ---------g mm

(5.40)

which allows a geometric interpretation of the diagonal terms of the deformation tensor. The angle between two infinitesimal vectors (dr)m and (dr)n in the initial configuration equals g mn cos ( α mn ) i = ------------------g mm g nn

(5.41)

as derived from Equation (5.27). For the current configuration, one obtains the angle G mn cos ( α mm ) c = ---------------------G mm G nn

(5.42)

Equations (5.41) and (5.42) provide a geometric interpretation of non-diagonal terms of the deformation tensor.

5.2.2

STRAIN TENSORS

A strain tensor measures the change of an infinitesimal line and angles, when the body moves from the initial configuration to a current one, using the difference of squares of the infinitesimal line’s length in both configurations as a criterion (Figure 5.4). Defined are two kinds of strain tensors: the Cauchy–Green (also called Lagrangian) and the Almansi (also known as Eulerian), relating to the initial or the current configuration, respectively.

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Nonlinear Problems in Machine Design

The Cauchy–Green strain tensor is defined as 1 T T C E = --- [ ∇ i R ⋅ ( ∇ i R ) – ∇ i r ⋅ ( ∇ i r ) ] 2

(5.43)

1 C 1 C s k E = --- ( D – I ) = --- ( G sk – g sk )g g 2 2

(5.44)

which equals

The Almansi strain tensor is defined as 1 1 A A s k E = --- ( I – D ) = --- ( G sk – g sk )G G 2 2

(5.45)

Linear Strain Tensor and Rotational Tensor The Cauchy–Green strain tensor can be divided into two parts, linear and rotational, as follows. Consider the tensor in terms of displacements. The gradient of vector R in terms of displacement vector u equals s ∂ s s ∂u ∇ i R = g --------s ( r + u ) = g g s + g --------s = I + ∇ i u ∂θ ∂θ

(5.46)

Recalling the Cauchy–Green deformation tensor, we define it as D

C

T

T

= ( I + ∇i u ) ⋅ ( I + ∇i u ) – I + ∇i u + ( ∇i u ) + ∇i u ⋅ ( ∇i u )

T

(5.47)

from which follows the new expression of the Cauchy–Green strain tensor, 1 1 C T T C E = --- ( D – I ) = --- [ ∇ i u + ( ∇ i u ) + ∇ i u ⋅ ( ∇ i u ) ] 2 2

(5.48)

The linear part on the right-hand side of Equation (5.48) is a symmetric tensor. 1 ε = --- [ ∇ i u + ( ∇ i u ) T ] 2

(5.49)

An accompanying antisymmetric tensor will be 1 Ω = --- [ ( ∇ i u ) T – ∇ i u ] 2

(5.50)

151

Large Displacements

Using both expressions, one can write the Cauchy–Green strain tensor in the form 1 1 2 2 C E = ε + --- ( ε – Ω ) ⋅ ( ε + Ω ) = ε + --- ( ε + ε ⋅ Ω – ε ⋅ Ω + Ω ) 2 2

(5.51)

Tensor ε in the above expression is a linear strain tensor. Tensor Ω is a rotation tensor that defines the rotation of an infinitesimal proximity of the point assumed to be a rigid body. When ε is of lesser order of magnitude than Ω, the Cauchy–Green strain tensor may be expressed in the approximate form 1 2 C E ≈ ε + --- Ω 2

(5.52)

In appropriate cases, the last equation can be reduced to 1 2 C E ≈ --- Ω 2

(5.53)

Cauchy–Green Strain Tensor in Cartesian Coordinates For practical application, consider the Cauchy–Green strain tensor in a Cartesian coordinate system. Assume material coordinates (a1,a2,a3) in lieu of (θ1,θ2,θ3), whereby the displacement vector takes the form 1

2

3

u = u s ( a ,a ,a )i

s

(5.54)

The gradient of u, with respect to the initial configuration, equals ∂u m s s m ∂ ∇ i u = i --------- ( u s i ) = --------s- i i ∂a m ∂a m

(5.55)

∂u m s T ( ∇ i u ) = --------m- i i ∂a s

(5.56)

and its transpose is

Consequently, per Equation (5.48), the Cauchy–Green strain tensor becomes 1 ∂u ∂u C E = --- --------s- + --------m- + 2 ∂a m ∂a s

3

∂u ∂u j

---------j -------∑ ∂a m ∂a s

j=1

m s

i i

(5.57)

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Nonlinear Problems in Machine Design

Let us designate the position components as follows: x = a1 , y = a2 z = a3

(5.58)

and the displacement components as u = u1 , v = u2 w = u3

(5.59)

Equation (5.57) yields nine components of the Cauchy–Green strain tensor. Due to the symmetry, the components can be reduced to six. ∂w 2 ∂v 2 ∂u 1 ∂u 2 C E xx = ------ + ---  ------ +  ------ +  -------  ∂x   ∂x ∂x 2  ∂x ∂w 2 ∂v 2 ∂v 1 ∂u 2 C E yy = ----- + ---  ------ +  ----- +  -------  ∂y   ∂y ∂y 2  ∂y ∂w 2 ∂v 2 ∂w 1 ∂u 2 C E zz = ------- + ---  ------ +  ----- +  -------  ∂z   ∂z ∂z 2  ∂z  ∂w ∂w ∂v ∂v 1 ∂u ∂v 1 ∂u ∂u C C E xy = E yx = ---  ------ + ------ + ---  ------ ------  +  ------ ----- +  ------- ------- 2  ∂y ∂x 2  ∂x ∂y   ∂x ∂y  ∂x ∂y  1 ∂v ∂w ∂w ∂w ∂v ∂v 1 ∂u ∂u C C E yz = E zy = ---  ----- + ------- + ---  ------ ------  +  ----- ----- +  ------- ------- 2  ∂z ∂y  2  ∂y ∂z   ∂y ∂z  ∂y ∂z  ∂w ∂w ∂v ∂v 1 ∂u ∂w 1 ∂u ∂u C C E zx = E xz = ---  ------ + ------- + ---  ------ ------  +  ------ ----- +  ------- ------- 2  ∂z ∂x  2  ∂x ∂z   ∂x ∂z  ∂x ∂z 

(5.60)

Principal Directions; Strain Invariants Strain tensor E (either Cauchy–Green E C or Almansi E A) is defined by six components that depend on coordinate directions. The number can be further reduced to three components using coordinates in principal directions. The principal components are designated as (E1,E2,E3). They are the roots of the following equation: 3

2

λ – I 1 ( E )λ + I 2 ( E )λ – I 3 ( E ) = 0 where I1(E), I2(E) and I3(E) are the invariants,

(5.61)

153

Large Displacements

I 1 ( E ) = E xx + E yy + E zz = E 1 + E 2 + E 3 2

2

2

I 2 ( E ) = E xx E yy + E yy E zz + E xx E zz – E xy – E yz – E xz E xx E xy E xz I 3 ( E ) = E xy E yy E yz = E 1 E 2 E 3 E xz E yz E zz

(5.62)

which are independent of a coordinate system. See Appendix A.

5.2.3

POLAR DECOMPOSITION; HENCKY STRAIN TENSOR

The deformation of an infinitesimal proximity at any point may by considered as a combination of a stretch deformation and a rotation of rigid body. The former refers to deformation of dimensions, while the latter pertains to angular motion. See Figure 5.5. Gradient ∇iR can be expressed as the product of an orthogonal rotational tensor and a stretch tensor. ∇ i R = U ⋅ O or ∇ i R = O ⋅ V

(5.63)

Tensor O is an orthogonal rotation tensor that meets the condition T

O⋅O = I

(5.64)

and U and V are symmetrical stretch tensors, i.e., U = U

FIGURE 5.5

T

; V =V

T

Stretch deformation and rotation of a body.

(5.65)

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Nonlinear Problems in Machine Design

Equations (5.63) are known as a polar form of tensor ∇iR. Tensor U is called a left stretch tensor, while tensor V is a right stretch tensor. Tensors U and V are positive definite and are related through the equations, T

(5.66)

V = O ⋅ U ⋅O

(5.67)

U = O ⋅ V ⋅O T

One can derive tensors U, V, and O from ∇iR, as follows. According to definition, the Cauchy–Green tensor equals D

C

T

T

= ∇i R ⋅ ( ∇i R ) = U ⋅ O ⋅ O ⋅ U

T

= U ⋅I ⋅U

T

= U

2

(5.68)

and its transpose is T

T

( ∇i R ) ⋅ ∇i R = V ⋅ O ⋅ O ⋅ V

T

= V ⋅I ⋅V

T

= V

2

(5.69)

From here, we obtain the expressions for the respective tensors. T 1⁄2

U = [ ∇i R ⋅ ( ∇i R ) ] T

V = [ ( ∇i R ) ⋅ ∇i R ] O = ( ∇i R ) ⋅ V

1⁄2

–1

(5.70) (5.71) (5.72)

Hencky Strain Tensor The Hencky strain tensor is derived from the right stretch tensor V. Let us express the latter tensor in the form V = V 1i1i1 + V 2i2i2 + V 3i3i3

(5.73)

where (V1,V2,V3) are the principal components. Then, the Hencky strain tensor is defined as E

H

= ln ( V i ) i i i i

(5.74)

For illustration, consider the Hencky tensor in the following application: tensional deformation of one-dimensional rod. The Hencky strain tensor in a onedimensional application equals

155

Large Displacements

E

H

T

= ln [ ( ∇ i R ) ⋅ ∇ i R ]

1⁄2

l = ln  ---c  li

(5.75)

where li is the initial length of the rod and lc is the current length. The logarithmic expression in Equation (5.75) is known as a natural (true) strain.4 For small displacements, one may use an approximation derived from truncated Taylor series. E

l l c – l i l c – l i lc – li - = ----------- ≈ ( ln 1 ) +  ----------= ln  ---c = ln  1 + ---------- li     li li l

H

(5.76)

Let us now compare, in a one-dimensional application, Equation (5.75) to the Cauchy–Green strain tensor, which can be expressed as 1 l 2 1 T C E = --- [ ∇ i R ⋅ ( ∇ i R ) – I ] = ---  ---c – 1 2  li 2

(5.77)

For small displacements, it is approximated by the truncated Taylor series. 2

2

l c – l i l c – l i 2 lc – li 1 lc – li 1 1 C - = ---  1 + ----------- – 1 = ----------E = --- ------------ – 1 ≈ --- 1 + 2  ----------- (5.78) 2     2 li 2 2 li li li One may conclude from the comparison that, for small displacements, Hencky and Cauchy–Green formulas, Equations (5.76) and (5.78), provide the same result.

5.2.4

EXAMPLES

Example 1: Deformation of Two-Spar Frame For illustration, consider the two-spar problem of Chapter 3, raising the acting force of the previous example so that it causes a large deformation. The frame in the initial and current configurations is shown in Figure 5.6. It is symmetrical, so only half of the frame is analyzed. The position of point A is fixed, while point B descends vertically, its displacement being represented by vector v0. The objective is to derive the deformation of the frame. During the deformation process, the spar remains unbent, and we assume that the cross-sectional area does not change. Let us analyze the spar within a Cartesian coordinate system (x,y,z), where the coordinate x coincides with the spar’s initial configuration. This system is designated as a material coordinate system. For a typical point of the spar, the respective position vectors in the initial and current configurations are r = xi x , 0 ≤ x ≤ l 0 R = r +v

(5.79) (5.80)

156

Nonlinear Problems in Machine Design

FIGURE 5.6

Deformation of two-spar frame.

The displacement vector for a point with the current position x in initial configuration equals x x v = ---v 0 = ---v 0 ( i x sin α + i y cos α ) l0 l0

(5.81)

so that the position vector in the current configuration becomes x R = xi x + ---v 0 ( i x sin α + i y cos α ) l0

(5.82)

The base vectors in the initial and current configurations, respectively, are ∂r ∂r ∂r g x = ------ = i x ; g y = ------ = 0 ; g z = ------ = 0 ∂z ∂y ∂x

(5.83)

v v˙ ∂R G x = -------- =  1 + ----0 sin α i x +  ----0 cos α i y ; G y = 0; G z = 0  l0    l0 ∂x

(5.84)

See Equations (5.7) and (5.11). The length of a line element in initial and current configurations is defined by the respective equations, 2

( dl ) = d x

2

2  v v 2 2 2 ( dL ) = dR ⋅ dR = G x ⋅ G x d x =  1 + 2 ----0 sin α + ---2-0 d x l0 l0   2  v 0 h 0 v 0 2 =  1 + 2 --------- + ---2- d x 2 l0  l0 

(5.85)

(5.86)

157

Large Displacements

Thus, the x-component of the Cauchy–Green deformation tensor equals 2

v0 h0 v0 C D xx = 1 + 2 --------- + ---22 l0 l0

(5.87)

and the x-component of Cauchy–Green strain tensor is 2

v0 h0 1 v0 1 C C E xx = --- ( D xx – 1 ) = --------- + --- ---22 2 2 l0 l0

(5.88)

Consequently, the stretch of the spar equals dL dL ------ = ------ = dx dl

D

C xx

2  v 0 h 0 v 0 =  1 + 2 --------+ --2 2 l0  l0 

1⁄2

(5.89)

For a shallow spar (assuming h0,v0 0

(6.83)

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Nonlinear Problems in Machine Design

6.3 LAGRANGE MULTIPLIER METHOD The Lagrange multiplier method of constrained optimization pertains to adding a linear term to the optimized function to enhance the optimization process. The linear term includes constraints multiplied by unknown variables—the Lagrangian multipliers. As applied to the contact condition, the added linear term is the potential energy of contact surface, Πc, derived from the Kuhn–Tucker condition. Along with the classical Lagrangian method, which is limited to the sticking mode of contact with friction, other methods are considered here that are in fact extensions of Lagrange method, such as the augmented Lagrange method, which works with penalty function. Also included are alternative methods based on regularization, such as the perturbed Lagrangian method and the constraint function method.

6.3.1

CLASSICAL LAGRANGE METHOD

The classical Lagrangian method uses the Kuhn–Tucker condition where the contact forces are expressed by Lagrange multipliers. In its plain form, the method is applicable to frictionless contact. However, with certain modifications, it can be applied to sticking-friction contact.9 The frictionless contact is presented first, followed by the sticking-friction contact. Frictionless Contact We begin with the Kuhn–Tucker condition, pertaining to normal forces in the contact zone, defined as gn f n = 0

(6.84)

Using the Kuhn–Tucker condition to represent the potential energy of the contact surface, the energy potential Π to be minimized equals Π ( u,λ n ) = Π b ( u ) + ∑ λ n g n k

k

(6.85)

k

where λn are Lagrange multipliers, and superscripts 1, 2,... k denote the consecutive segments in the contact zone in finite element representation (Figure 6.10). In vectorial form, Equation (6.85) becomes Λ ) = Πb ( u ) + Λ g Π ( u,Λ

(6.86)

T 1 2 k Λ = [ λ n , λ n , …, λ n ]

(6.87)

T

where

1

2

k T

g = [ g n , g n , …, g n ]

(6.88)

203

Contact Problems

FIGURE 6.10 Segment in contact zone in finite element representation.

Sticking-Friction Contact Consider the forces and displacements in tangential direction as per Coulomb law. Here we have f t ≠ 0 at g t = 0

(6.89)

Joining both, we obtain the supplementary Kuhn–Tucker condition, gt f t = 0

(6.90)

With Equation (6.90), the total potential Π now equals Π ( u,λ n ,λ t ) = Π b ( u ) + ∑ ( λ n g n + λ t g t ) k

k

k k

(6.91)

k

The latter can be expressed by the same vectorial Equation (6.86) as the frictionless form, with vectors ΛT and g now equal to  k   1   2   λn   λn   λn  Λ =  ,  2 , …,  k  1  λt   λt   λt        T

 1   2   k   gn   gn   gn  g=  ,  2 , …,  k  1  gt   gt   gt       

(6.92)

T

(6.93)

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Nonlinear Problems in Machine Design

Minimization of Potential Π Minimization of potential Π, enhanced by Lagrange multipliers, leads to the final FE equations. It is derived by equating the first variation of the potential to zero. The first variation of Π produces the residual, Λ ) = δΠ b ( u ) + δΛ Λ g + Λ δg δΠ ( u,Λ T

T

(6.94)

Let us introduce the matrix ∂g A = -----∂u

(6.95)

Using Equations (6.42) and (6.43), matrix A for frictionless contact equals A = [ ( NS )

1,T

, ( NS )

2,T

k,T T

, …, ( N S ) ]

(6.96)

while, for the sticking-friction contact, it becomes 1,T  ( NS )  A =  1,T g   T 0S + -----n N 0 l   

  ,  

2,T  ( NS )  2,T  g   T 0S + -----n N 0 l   

k,T   ( NS )   k, T , …,  g   T 0S + -----n N 0  l    

    

T

(6.97) Thus, Equation (6.94) is expressed as follows: Λ ) = Rδu + Λ Aδu c + δΛ Λ g δΠ ( u,Λ T

T

T T Λ = ( R + Λ A )δu + g δΛ

(6.98)

where R denotes the derivative, δΠ R = R ( u ) = ----------b δu

(6.99)

Minimization of potential Π, by equating Equation (6.98) to zero, leads to the system of equations, R(u) + Λ A = 0

(6.100)

g(u) = 0

(6.101)

T

205

Contact Problems

Linearization of FE Equations The linearization pertains to expanding Equations (6.100) and (6.101) into Taylor series and skipping the higher-order terms. In Equations (6.100) and (6.101), R, A, and g are functions of u and Λ. It is essential to express them in linear terms of Λ to facilitate incremental solution. increments ∆u and ∆Λ Assume displacement ui to be the solution at the i increment. At the i + 1 increment, the first term of Equation (6.100) equals ∂R R i + 1 = R i +  ------- ∆ i + 1 u = R i + K t ∆ i + 1 u – ∆ i F  ∂u 

(6.102)

where Kt is a tangent stiffness matrix defined by 2

∂ Πb K t = ----------2 ∂u

(6.103)

and ∆iF is the increment of external load as derived in Chapter 2, “Finite Element Method.” The second part of Equation (6.100), neglecting second-order terms, is ∂A T T T Λ i + 1 A i + 1 = ( Λ i + ∆ i + 1 Λ ) A i +  ------- ∆ i + 1 u  ∂u  i T T T = Λi Ai + Ai ∆i + 1 Λ + Λi Bi ∆i + 1 u

(6.104)

Matrix B is a derivative of A, 2

∂ g ∂A B = ------- = --------2 ∂u ∂u

(6.105)

which is derived from Equations (6.55) and (6.56). Equation (6.101) equals ∂g g i + 1 = g i +  ------ ∆ i + 1 u = g i + A i ∆ i + 1 u  ∂u

(6.106)

Thus, the system of equations at i + 1 increment equals ( K i + Λ i B i )∆ i + 1 u + A i ∆ i + 1 Λ = – R i – Λ i A i + ∆ i F T

T

Ai ∆i + 1 Λ = –gi which, in the matrix form, becomes

T

(6.107) (6.108)

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Nonlinear Problems in Machine Design

T T Ki + Λi Bi Ai  ∆i + 1 u  Ai 0  ∆i + 1 Λ

6.3.2

   – R i – Λ Ti A i + ∆ i F =     –gi 

    

(6.109)

AUGMENTED LAGRANGE METHOD

The augmented Lagrange method pertains to regularization of the classical Lagrange method. The adaptation is accomplished by adding a penalty function from the penalty method.10,11 The augmented Lagrange method, unlike the classical method, can be applied to all contact conditions: to sticking friction, sliding friction, and to a frictionless contact. By adding the penalty term, Equation (6.86) is revised, and the following augmented form is obtained: 1 Λ ) = Π b ( u ) + Λ T g + --- g T κ g Π ( u,Λ 2

(6.110)

The first variation of potential (6.110) produces the residual Λ ) = δΠ b ( u ) + δΛ Λ T g + Λ T δg + g T κ δg δΠ ( u,Λ

(6.111)

The condition that the residual must equal zero leads to the following system of equations: R(u) + A (Λ + κ g) = 0

(6.112)

g(u) = 0

(6.113)

T

T

Now, T T Ai + 1 κ gi + 1 = κ Ai + 1 gi + 1

∂A = κ A i +  ------- ∆ i + 1 u ( g i + A i ∆ i + 1 u )  ∂u  = κ ( Ai gi + Bi gi ∆i + 1 u + Ai Ai ∆i + 1 u )

(6.114)

Thus, for the augmented Lagrangian method, Equation (6.107) of the classical Lagrangian method is changed to become ( K t + Λ i B i + κ A i A i + κ B i g i )∆ i + 1 u + A i ∆ i + 1 Λ = – R ( u i ) – Λ i A i – κ A i g i + ∆ i F T

T

T

(6.115)

207

Contact Problems

Now, the corresponding matrix form is

T T ( Ki + Λi Bi + κ Ai Ai + κ Bi gi ) Ai  ∆i + 1 u    Ai 0  ∆i + 1 Λ 

 T  =  – Ri – Λi Ai – κ Ai gi + ∆i F  –gi 

     (6.116)

Slip-Surface Correction A load step during a numerical solution, when reaching a sliding friction mode, requires a correction as applied to the penalty method (as described in Section 6.2.2). The correction of the tangential force at the slip surface follows the same steps, as shown in Figure 6.9. Here, the corrected tangential force acquires an additional term, in accordance with Lagrangian form of the potential energy, becoming now n+1

ft

trial

= ft

trial

+ λ t – κ t sgn ( f t ) ∆ς

(6.117)

During the computational process, ∆ζ is determined from the following conditions: ∆ζ = 0

if

Y (f

trial

trial

) = ft

trial

Y (f ) ∆ζ = ----------------κt

6.3.3

if

Y (f

trial

trial

–µ f n

trial

) = ft

≤0

trial

–µ f n

(6.118)

≤0

(6.119)

OTHER METHODS

A brief description of other methods based on regularization of the Lagrange multiplier method follows. They include the perturbed Lagrangian method and the constraint function method. Both stem from the classical Lagrangian, with the adaptation accomplished by adding penalty terms. The penalty parameters applied here are ε,

ε =

which tend to zero, i.e., ε « 1 .

εn 0 0 εt

(6.120)

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Nonlinear Problems in Machine Design

Perturbed Lagrangian Method The added penalty term to the potential energy is a quadratic expression similar to the one of the augmented Lagrangian method.12,13 The potential to be minimized equals T

1g g Λ ) = Π b ( u ) + Λ T g + --- -------Π ( u,Λ 2 ε

(6.121)

The FE equations that are derived here follow closely the ones derived above. Constraint Function Method The constraint function method defines the Kuhn–Tucker condition and the Coulomb law in different forms.14 The Kuhn–Tucker condition, Equation (6.3), is approximated by a hyperbolic formula, gn f n = εn

(6.122)

which is converted to 2

gn – f n ( gn + f n ) w ( g n , f n ) = --------------- – ----------------------+ εn 2 2

(6.123)

g n and f n denote nondimensional forms of gn and fn. The Coulomb law of friction is approximated by g 2 v ( g t ,τ ) = τ – --- arc tan ----t εt π

(6.124)

ft τ = -------µfn

(6.125)

where

The potential energy Π to be minimized takes the form Π ( u,λ n ,λ t ) = Π b ( u ) + λ n w ( g n ,λ n ) + λ t v ( g t λ t ) where λn and λt are Lagrangian multipliers identical with fn and τ.

(6.126)

Contact Problems

209

6.4 CRITICAL REVIEW The Penalty Method Due to its effectiveness, coupled with the fact that it has much lesser memory requirements, this method has received a wider acceptance in spite of the inaccuracies found in the solutions. Its main weakness pertains to the fact that, to reduce errors, the assigned values of penalty parameters must be high.15 At the same time, with larger parameters, there is a danger of numerical instability that disrupts the solution process altogether. Lagrange Multiplier Method The advantage of the Lagrange multiplier method lies in its approach, since its concept corresponds to the constraints of the contact: no imaginary penetration needs to be assumed. An error-free solution is granted, as a result of the accord, providing it is used in sticking-friction and frictionless contact. To widen the range of applicability to include the sliding-friction mode, one has to introduce penalty parameters that lead to the augmented Lagrange and other methods. The method’s disadvantage is in excessive memory requirements due to the presence of many variables. Another disadvantage lies in the numerical processing difficulties due to a problematic form of the global stiffness matrix, which contains a zero term in the main diagonal. There are means to overcome the difficulties by renumbering variables or introducing penalty parameters, which again leads to augmented Lagrange and other methods. The Augmented Lagrange Method This method succeeds in merging the advantages of the penalty method with the pure Lagrangian method while moderating the disadvantages of both. Namely, there is no need for excessive penalty parameters to reach convergence; the danger of instability is thus avoided. The addition of a penalty term improves the form of the global stiffness matrix, converting it into a non-singular one. There is also no requirement to renumber variables. The other versions of Lagrange methods, such as the perturbed Lagrange method and constraint function method, have similar positive properties, as the augmented Lagrange method.

REFERENCES 1. Kragelski, I.V., Dobychin, M.N., and Kombalov, V.S., Friction and Wear—Calculation Methods (translated from Russian), Pergamon Press, Oxford, 1982. 2. Fletcher, R., Practical Methods of Optimization, Constrained Optimization, Vol. 2, John Wiley, New York, 1981. 3. Bertsekas, D.P., Constrained Optimization and Lagrange Multiplier Methods, Academic Press, New York, 1982. 4. Wriggers, P., Vu Van, T., and Stein, E., Finite element formulation of large deformation impact-contact problems with friction. Comp. & Struct., 37, 319–331, 1990.

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5. Wriggers, P., and Simo, J.C., A note on tangent stiffness for fully nonlinear contact problems. Commun. Appl. Numer. Meth., 1, 199–203, 1985. 6. Michalowski, R., and Mroz, Z., Associated and non-associated sliding rules in contact friction problems. Arch. Mech., 39, 259–276, 1978. 7. Tornstenfelt, B.A., An automatic incrementation technique for contact problems with friction. Comp. & Struct., 19, 393–400, 1984. 8. Giannokopoulos, A.G., The return mapping method for the integration of friction constitutive relations. Comp. & Struct., 32, 157–168, 1989. 9. Crisfield, M.A., Non-linear Finite Element Analysis of Solids and Structures, Vol. 2, Advanced Topics, John Wiley, Chichester, England. Chapter 23, Contact with friction, pp. 411–446, 1997. 10. Chaudaray, A.B., and Bathe, K.J., A solution method for static and dynamic analysis of contact problems with friction. Comp. & Struct., 24, 855–873, 1986. 11. Simo, J.C., and Laursen, T.A., An augmented Lagrangian treatment of contact problems involving friction. Comp. & Struct., 42, 97–116, 1992. 12. Simo, J.C., Wriggers, P., and Taylor, R.L., A perturbed Lagrangian formulation for the finite element solution of contact problems. Comp. Meth. in Appl. Mech. & Eng., 50, 163–180, 1985. 13. Kikuchi, N., and Oden, J.T., Contact Problems in Elasticity: A Study of Variational Inequalities and Finite Element Methods. SIAM, Philadelphia, 1988. 14. Bathe, K.J., and Bouzinov, P.A., On the constraint function method for contact problems. Comp. & Struct., 64, 1069-1085, 1997. 15. Barlam, D., and Zahavi, E., The reliability of solutions in contact problems. Comp. & Struct., 70, 35–45, 1999.

7

Fatigue Failure Prediction Methods

This chapter covers the phase of machine-part analysis that confronts the problems of fatigue failure. The concern with life expectancy of machine parts goes back to the beginning of industrial revolution (the middle of nineteenth century) and, in particular, with the advent of railroads in central Europe. The first known investigators concerned with fatigue phenomena were designers of axles for locomotives and wagons, whose objective was to analyze the machine parts operating under fluctuating load to prevent fatigue failure. Woehler’s experiments with axles were the first known laboratory tests to derive and quantitatively define the limits of fatigue.1 In Chapter 8, we present a simplified empirical procedure, based on Woehler’s work, known as the stress method. The fatigue process in a machine part comprises two stages. The first is an “invisible” process that culminates in a small crack. The second is continuous crack propagation, which ultimately leads to a failure. Therefore, two independent analytical methods that supplement each other are used for design against fatigue.2 The method called the strain method deals with “invisible” fatigue and relates to the plastic deformation caused by fluctuating loads. The other method refers to crack propagation and is known as the method of fracture mechanics. The two methods are described below.

7.1 STRAIN METHOD The strain method is based on the theories of elasticity and plasticity. Originally, it was initiated by Manson3 and Coffin.4 Research accomplishments of Morrow,5 Socie,6 and their followers further expanded and perfected the method. The theoretical foundation of the method addresses two attributes that are created by loading: an elastic deformation existing below the yield point of the material and a plastic deformation that appears above it. The following description begins with material properties that are characteristic to fatigue problems and continues with the analysis of fatigue life.

7.1.1

CYCLIC PROPERTIES

Material properties in cycling loading, such as yield point and the hardening behavior, differ from those in monotonic loading. Since special testing facilities are 211

212

Nonlinear Problems in Machine Design

necessary to obtain the cyclic properties, let us consider an electrohydraulic test system. See Figure 7.1. The system is fully automated: it comprises a computer (d) for test programming and control, a servo-controller (e) to monitor the applied load, a load cell (f), and an extensometer (g) to check the specimen’s changing geometry. The computer reads the load and strain signals from the sensors and responds with command signals to the controller. For each load fluctuation, there is a signal defining the load magnitude and direction in accordance with a programmed load history in the computer, which includes the consecutive maxima and minima of force, stress, displacement, and strain. The original signals are analog and are translated into digital data for computer input. Consider a stress-controlled test where the prescribed strain fluctuation data are stored in computer memory. Assume the metal specimen to have hypothetical properties that stay constant under load cycling. The specimen is subjected to completely reversed loading, and symmetric tension and compression beyond the yield point, with both occurring in succession. In real life, the metals respond to the cyclic loading with changes of either strain hardening or strain softening (known as cycle hardening or cycle softening). We choose using a metal specimen with a cyclesoftening response. The process is presented schematically in Figure 7.2. The loading

FIGURE 7.1 Electrohydraulic fatigue test system.

Fatigue Failure Prediction Methods

213

FIGURE 7.2 Cyclic stress-strain curve.

history begins with tension at point O. At first, the deformation is elastic and is represented by straight line OA. Beyond point A (the yield point in tension), the deformation is plastic, as reflected by curve AB. After this, beyond point B (the turning point) the unloading follows, as indicated by BO' . At the reverse loading (compression), due to the Bauschinger effect, the process is different, as reflected by path O' A'B' . The changed pattern continues with another unloading B'O″ , after which the specimen is reloaded again, as indicated by O″A″B″ . From here, the process continues in the same manner. The starting loading path, OAB, and the first reverse loading path, O' A'B' , are nonsymmetrical: yield point A' is below the yield point A. Due to continuous cycling, the Bauschinger effect diminishes. To show both responses to cyclic loading (cycle hardening and cycle softening), let us consider the behavior of two specimens of different metals. See Figure 7.3. For both specimens, the strain ε has a prescribed constant amplitude. Figure 7.3a reflects the behavior of a cycle-softening response, where the stress σ follows with an assymptotically decreasing amplitude. After many cycles, a stable state is achieved, as seen in the figure, by the stabilized hysteresis loop. Figure 7.3b shows a cycle-hardening response where the stress σ follows with an asymptotically increasing amplitude, with the final hysteresis loop reached at the end. The stabilized hysteresis loop shown in Figure 7.4 is the final result of the tests using both cycle-softening and cycle-hardening specimens. The size of the hysteresis is defined by the stress range ∆σ and the strain range ∆ε; both ranges are functions of the imposed fluctuating load. Cyclic Stress-Strain Correlation Let us examine the stress and strain amplitudes in the stable state, ∆σ/2 and ∆ε/2 , using again Figure 7.4. Out of several test programs available to arrive at a correlation

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Nonlinear Problems in Machine Design

FIGURE 7.3 Stress vs. strain in a strain-controlled test: (a) specimen made of cyclesoftening material and (b) specimen made of cycle-hardening material.

FIGURE 7.4 Stabilized hysteresis loop showing stress and strain amplitudes in the stable state.

Fatigue Failure Prediction Methods

215

of the two amplitudes, we shall focus on the multistep program (incremental step program). A specimen is subjected to a succession of load blocks comprising gradual increases and decreases of strain amplitudes. See Figure 7.5. A series of resulting hysteresis loops is plotted in a common σ – ε diagram. We get a stress-strain curve by connecting the tips of the respective loops (see Figure 7.6). The curve represents a correlation between the cyclic stress and strain amplitudes. The metal production process has a direct bearing on whether the material has cycle-softening or cycle-hardening characteristics: annealed metals, free of any prior treatment, stressed in plastic range and later released, tend to be cycle hardening. On the other hand, treated metals, subject to residual stresses released during the

FIGURE 7.5 Multistep (incremental step) test program.

FIGURE 7.6 Derivation of the cyclic stress-strain.

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Nonlinear Problems in Machine Design

cyclic loading, tend to be cycle softening. For illustration, we show cyclic stressstrain curves of both treated and untreated steels and compare the cyclic curves with monotonic ones (see Chapter 4). Figure 7.7 shows the curves of steel SAE 9262: Figure 7.7a, quenched and tempered and Figure 7.7b, annealed.2 The curves overlap in the elastic range and diverge in the plastic range. The plastic range of the cyclic curve in Fig. 7.7a is below the plastic range of the monotonic curve. This is due to the cycle-softening response of the given specimen. An opposite effect is noted in Fig. 7.7b, which is due to the cycle-hardening response of the specimen. With the help of the graphic presentation of the cyclic stress-strain curve, let us derive the stress-strain correlation in mathematical form as used in machine design. Let us compare the curves in Figures 4.4 and 7.7 and note that the cyclic and the monotonic curves have similar forms. The similarity in graphic form enables us to use the mathematical expression of the monotonic correlation as the basis for cyclic correlation, as shown below. By definition, the cyclic strain amplitude comprises two components, elastic and plastic. e

p

∆ε ∆ε ∆ε ------ = -------- + --------2 2 2

(7.1)

FIGURE 7.7 Cyclic and monotonic stress-strain curves for SAE 9262 steel: (a) cyclesoftening material, quenched and tempered, BHN = 410, and (b) cycle-hardening material, annealed, BHN = 260. Data from SAE Handbook.1

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Fatigue Failure Prediction Methods

By analogy with Equation (4.12) for monotonic loading (Chapter 4), the elastic strain amplitude in cyclic loading can be expressed as e

∆ε ∆σ -------- = ------2E 2

(7.2)

Note: Here and in the following equations, σ and ε denote true stress and true strain, respectively, as defined in Chapter 4. The plastic strain amplitude in cyclic loading, analogous to Equation (4.13) for monotonic loading, can be expressed as 1 ----

∆σ n' ∆ε --------- =  --------  2K' 2 p

(7.3)

Upon summation, Equations (7.2) and (7.3) provide the total cyclic strain amplitude. 1 ----

∆σ n' ∆σ ∆ε ------ = -------- +  -------- 2K'  2K' 2

(7.4)

which is similar to Equation (4.14) for monotonic loading. K' is the cyclic strength coefficient, and n' is the cyclic strain-hardening exponent. Equation (7.4) may be simplified with the help of data from cyclic tests. Let σ′ f and ε′ f denote the stress and strain that cause a specimen’s fracture at the first reversal of cyclic loading. The cyclic strength coefficient K′ can be expressed in terms of σ′ f and ε′ f . σ′ f K′ = ------------n' ( ε′ f )

(7.5)

Thereby Equation (7.4) becomes 1 ----

∆σ n' ∆σ ∆ε ------ = ------- + ε′ f  -----------  2σ′ f  2E 2

(7.6)

This is the final mathematical expression of the cyclic stress-strain correlation. For the reader’s convenience, cyclic properties K′,n′,σ′ f and ε′ f of selected metals are given in the Appendix.

7.1.2

FATIGUE LIFE

OF

ONE-DIMENSIONAL SPECIMENS

The theory of the strain method is based on one-dimensional experiments investigating fatigue failure. The cyclic tests with one-dimensional specimens provide a

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Nonlinear Problems in Machine Design

relation between the stress amplitude and the number of cycles to failure, defined at the onset of a crack. The empirical correlation is expressed as follows: ∆σ b ------- = σ′ f ( 2N f ) 2

(7.7)

Nf is the number of load fluctuations up to the instance of specimen failure, and 2Nf denotes the number of load reversals. The equation (known as Basquin equation) is confirmed by experimental data7 throughout the range Nf = 10 to 106, as shown in Figure 7.8a. Equation (7.7) permits us to redefine the elastic and plastic strains, Equations (7.2) and (7.3), in terms of the number of 2Nf . The elastic strain now becomes e σ′ ∆ε ∆σ b -------- = ------- = -------f ( 2N f ) E 2E 2

(7.8)

while the plastic strain is

FIGURE 7.8 Fatigue properties of annealed SAE 4340 steel: (a) stress amplitude vs. cyclic life and (b) plastic strain amplitude vs. cyclic life. From Smith et al.9

219

Fatigue Failure Prediction Methods 1 ----

∆σ n' ∆ε c --------- = ε′ f  ---------- = ε′ f ( 2N f )  2σ' f  2 p

(7.9)

where c is the fatigue ductility exponent defined by b c = ---n′

(7.10)

Equation (7.9), called the Manson-Coffin rule, was proven by tests as shown in Figure 7.8b. To arrive at the final correlation between the fatigue life and the total strain, we combine Equations (7.8) and (7.9) as follows: σ′ ∆ε b c ------ = -------f ( 2N f ) + ε′ f ( 2N f ) E 2

(7.11)

Figure 7.9 presents a graphic interpretation of Equation (7.11) using log-log coordinates. In the figure, the above final correlation is a curve formed by combined elastic and plastic lines. Stress Concentration Up to now, we have discussed the fatigue analysis of a smooth specimen. We now consider a specimen with a notch, which causes local high stresses and thereby presents a more complicated problem. See Figure 7.10. The analysis now has to consider the strains and stresses in the critical location within the notch which, in fact, are multiaxial. To relate to the problem within a one-dimensional approach, let us use a stress concentration factor, which already reflects the local high stresses.

FIGURE 7.9 Derivation of correlation strain vs. cyclic life.

220

Nonlinear Problems in Machine Design

FIGURE 7.10 Stress concentration in notched specimen.

When all stresses are within elastic range, one can state that σ = K tS

(7.12)

where Kt is the theoretical stress concentration factor, S = P/A is the nominal stress, and σ is the peak stress. An equivalent equation can be written for the elastic strains. ε = K te

(7.13)

When peak stress σ is higher than the yield point, we encounter a local plastic deformation. Here, the local true stress σ and true strain ε have different concentration factors. Therefore, instead of Equations (7.12) and (7.13), we now have σ = K σS

(7.14)

ε = K εe

(7.15)

and

assuming the nominal stress S to be elastic. Concentration factors Kσ and Kε are interdependent and can be expressed by means of an empirical relation. Out of the numerous available correlations, we use the most prevalent one, Neuber’s rule.8 2

K σK ε = K t

In application to fatigue analysis, Neuber’s rule can be expressed as

(7.16)

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Fatigue Failure Prediction Methods

∆σ ∆ε 2 ------- ------ = K t ∆S ∆e

(7.17)

∆σ ∆ε ∆S 2 1 ------- ------ =  K t ------- -- 2 E 2 2

(7.18)

which is equivalent to

The above represents a hyperbolic relationship of ∆ε/2 and ∆σ/2 . Since ∆ε/2 and ∆σ/2 are also connected by the equation 1 ----

∆σ n' ∆σ ∆ε ------ = ------- + ε′ f  -----------  2σ′ f  2E 2

(7.19)

the strain in the notched part, ∆ε/2 , can be determined from the two equations above. Figure 7.11 presents a graphical solution, where cross point A indicates the needed total strain, ∆ε/2 . After obtaining the total strain, the fatigue life is computed using Equation (7.11). Mean Stress Let us expand the analysis of fatigue cases to include those where the stresses vary between arbitrary maximum and minimum. To obtain the fatigue-life correlation, the relevant equations in the preceding sections need to consider the mean stress. The mean stress σm is illustrated in Figure 7.12. Presented below are two independent analytical procedures to derive the correlation.

FIGURE 7.11 Determination of strain at the root of a notch.

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Nonlinear Problems in Machine Design

FIGURE 7.12 Fluctuating stress between upper and lower bonds with a mean equal to the steady stress.

Manson’s Procedure9 To consider σm, the procedure introduces an equivalent stress amplitude σN. The equivalent stress amplitude is so defined that a cyclic process with amplitude σN and mean stress σm = 0 will have the same fatigue life Nf as a process with amplitude ∆σ/2 and mean stress σ m ≠ 0 . We derive σN on the basis of the diagram (∆σ/2 , σm) in Figure 7.13. The inclined lines, Nf = const, represent the constant fatigue life. According to the diagram, the equivalent stress amplitude σN equals ∆σ/2 σ N = σ′ f -------------------σ′ f – σ m

FIGURE 7.13 Lines of constant life in (∆σ/2, σm) coordinates.

(7.20)

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Fatigue Failure Prediction Methods

Inverting the above equation, we obtain the stress amplitude ∆σ/2 in the form σ′ f – σ m ∆σ ------- = σ N ------------------σ′ f 2

(7.21)

The correlation between equivalent stress amplitude σN and fatigue life, as it follows from Equation (7.7), is σ N = σ′ f ( 2N )

b

(7.22)

Using Equations (7.21) and (7.22), we get the relationship between elastic strain e amplitude ∆ε /2 and the fatigue life. e σ′ f – σ m ∆ε ∆σ b - ( 2N f ) -------- = ------- = ------------------E 2E 2

(7.23) p

To obtain the relationship between the plastic strain amplitude ∆ε /2 and fatigue life, we use Equation (7.3) in conjunction with Equations (7.21) and (7.22). 1 ----

1 ----

p σ′ f – σ m n' ∆σ n' ∆ε c - ( 2N f ) --------- = ε′ f  ----------- = ε′ f  ------------------   2σ′ f σ′ f  2

(7.24)

Combining Equations (7.23) and (7.24), we get the final correlation between strain amplitude ∆ε/2 and fatigue life as derived by Manson. 1 ----

σ′ f – σ m σ′ f – σ m n' ∆ε c b - ( 2N f ) + ε′ f  ------------------- ( 2N f ) ------ = ------------------ σ′ f  E 2

(7.25)

Morrow Procedure10 This procedure, used to derive the necessary fatigue-life correlation, introduces failure stress σ′ f . We assume σ′ f to be the maximum stress in cyclic loading where, at the amplitude ∆σ/2 = ( σ′ f – σ m ) , the fatigue life equals one, or Nf = 1, as follows: 1 ---

∆σ/2 b 2N f =  --------------------  σ′ f – σ m

(7.26)

Consequently, the correlation of stress versus fatigue life becomes ∆σ b ------- = ( σ′ f – σ m ) ( 2N f ) 2

(7.27)

224

Nonlinear Problems in Machine Design

and the elastic strain amplitude equals e σ′ f – σ m ∆ε b - ( 2N f ) -------- = ------------------E 2

(7.28)

p

Similarly, we assume that, at amplitude ∆ε /2 = ε′ f , the fatigue life equals one, or Nf = 1. Therefore, ∆σ/2 ∆ε --------- = ε′ f -------------------σ′ f – σ m 2 p

1 ---n'

(7.29)

It follows from Equations (7.27) and (7.29) that the correlation between plastic strain amplitude ∆εp/2 and the fatigue life is 1 b --n'

p ( σ′ f – σ m ) ( 2N f ) ∆ε --------- = ε′ f ----------------------------------------σ′ f – σ m 2

= ε′ f ( 2N f )

c

(7.30)

Summing up Equations (7.28) and (7.29), we obtain the fatigue life correlation, ( σ′ f – σ m ) ∆ε b c - ( 2N f ) + ε′ f ( 2N f ) ------ = -----------------------E 2

7.1.3

FATIGUE LIFE

UNDER

(7.31)

MULTIAXIAL LOADING

In general, working machine parts are subject to multiaxial loading, resulting in multiaxial stresses and strains. This section deals with the analysis of complicated geometries and loadings, where stress and strain components act simultaneously and in different directions. Realizing that most components in machine parts originate from the same source—the applied load—it is conservative to assume that they fluctuate in phase, reaching the peaks and valleys simultaneously. The fatigue analysis of multiaxial fluctuating loading involves three stages: first, derivation of fluctuating multiaxial strain components; second, replacement of the components by one-dimensional equivalents; and third, computation of fatigue life. Multiaxial Strain Components Let us consider the first stage of the analysis and derive the expression for fluctuating multiaxial strain components. As described in Chapter 4, stress and strain components in multiaxial loading may rise either in the same ratio (proportional loading) or in nonequal ratios (nonproportional loading. The following analysis is restricted to nonproportional loading; the resulting correlations can be easily adapted to proportional loading. As an illustration of nonproportional loading, consider Figure 7.14. The figure shows a typical case: a problem concerning a thick plate with a

225

Fatigue Failure Prediction Methods

FIGURE 7.14 Stresses in the notch in a thick plate subjected to nonproportional loading.

notch, where the strains in z-direction do not rise proportionately to the strains in the xy-plane. For the purpose of analysis, the applied load is viewed as a succession of small discrete steps with the resulting strain increments, e

p

δε i = δε i + δε i

(7.32)

where index i denotes the directions x, y, and z. Since, during a stress cycle, the principal directions of the body at the critical points may change, we shall use Cartesian strain components in lieu of principal ones. Consequently, during the fluctuating loading, the respective strain amplitudes equal ∆ε 1 e p --------x = --- ∑ ( δε x + δε x ) 2 cycle 2

∆γ xy 1 e p ---------= --- ∑ ( δγ xy + δγ xy ) 4 cycle 4

∆ε y 1 e p -------- = --- ∑ ( δε y + δε y ) 2 cycle 2

∆γ yz 1 e p ---------- = --- ∑ ( δγ yz + δγ yz ) 4 cycle 4

∆ε 1 e p --------z = --- ∑ ( δε z + δε z ) 2 cycle 2

∆γ 1 e p ---------zx- = --- ∑ ( δγ zx + δγ zx ) 4 cycle 4

(7.33)

See Figure 7.15, showing stress and strain increments and the resulting amplitudes in multiaxial loading. Equivalent Strains and Stresses The second stage of analysis, as mentioned before, deals with replacement of the components by one-dimensional equivalents. There are several known theories to derive equivalent stresses and strains, also known as effective stresses and strains.

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Nonlinear Problems in Machine Design

FIGURE 7.15 Hysteresis loops in multiaxial loading showing stress and strain amplitudes in the x-direction: (a) loading with zero mean stress and (b) loading with a mean stress other than zero.

The theories presented here were chosen for their special applicability to fatigue analysis. Von Mises Correlation To derive the effective strain amplitude from the above components, we use von Mises correlation.  ∆ε ∆ε 2 ∆ε ∆ε 2 ∆ε 2   --------x – --------y +  ∆ε --------y – --------z +  --------z – --------x  2  2   2 ∆ε eff 2  2  2  1 ----------- = ------------------------  2 2 2 2 (1 + v) 2 ∆γ ∆γ ∆γ xy + σ  ---------+  ---------yz- +  ---------zx-   4   4   4  

1

--2   (7.34)   

where v is a mean value of the Poisson ratio, corresponding to the elastic and plastic deformation stages. It can be determined using the equation11 e

p

∆ε eff ∆ε eff - + 0.5 ---------v ---------2 2 v = --------------------------------------e p ∆ε eff ∆ε eff ----------- + ----------2 2

(7.35)

Sines Correlation For multiaxial mean stresses, we use Sines12 empirical correlation, ( σ m ) eff = a ( σ x,m + σ y,m + σ z,m )

(7.36)

derived from tests. The term in the parentheses denotes hydrostatic pressure, which is an invariant independent of shear stresses, while a is an empirical factor. The

Fatigue Failure Prediction Methods

227

Sines equation has an advantage over other methods because of this. It recognizes the fact that a mean tension stress shortens fatigue life, whereas a mean compressive stress extends it. The other methods, such as von Mises or Tresca, disregard the influence of tension and compression on the fatigue life. Equation (7.36) has a disadvantage, since it requires the empirical factor a, which is not readily available. The expression can be used in a modified form, simplified by Fuchs and Stephens,13 which eliminates factor a, ( σ m ) eff = ( σ x,m + σ y,m + σ z,m )

(7.37)

Fatigue-Life Correlation Here, we come to the final stage of the analysis of the life expectancy, using two cases. First, where a machine part is subject to reciprocal multiaxial loading, its life expectancy is computed using the equation σ′ ∆ε eff b c ---------- = -------f ( 2N f ) + ε′ f ( 2N f ) E 2

(7.38)

In the second case, where there is an acting mean load, we use the effective mean stress ( σ m ) eff , whereby the life expectancy is computed using one of the following two equations. Manson Equation σ′ f – ( σ m ) eff σ′ f – ( σ m ) eff ∆ε eff b c - ( 2N f ) + ε′ f ---------------------------- ( 2N f ) ----------- = ---------------------------E σ' f 2

(7.39)

Morrow Equation σ′ f – ( σ m ) eff ∆ε eff b c - ( 2N f ) + ε′ f ( 2N f ) ---------- = ---------------------------E 2

(7.40)

7.2 CUMULATIVE DAMAGE Until now, our discussion was restricted to problems with constant maxima and minima throughout the fluctuating loading. However, one of most difficult problems is to predict life expectancy in parts with spectrum loading. Spectrum loading (irregular loading) represents a condition where the loading cycles vary and produce different magnitudes of peaks as well as valleys. See Figure 7.16. The fatigue life in spectrum loading is a function of accumulated effects on the part throughout its performance, which is summed up as the function of cumulative damage. There are a number of methods to approach the analysis of cumulative damage, among them a widely applicable method known as the Palmgren–Miner rule, described below.

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Nonlinear Problems in Machine Design

FIGURE 7.16 Spectrum loading: recording of fluctuating load in a tractor axle.

Palmgren–Miner Rule The Palmgren–Miner rule is based on a linear concept, assuming that the changes in the sequence of nonuniform loading cycles will not affect the fatigue life. The rule was originally introduced by Palmgren in the design of bearings and adapted by Miner14 to aircraft structures. To derive the Palmgren–Miner rule, let us replace the real sequence of cycles that takes place in irregular loading with an assumed sequence of groups of uniform cycles. See Figure 7.17. Each group, comprising ni number of uniform cycles, represents a corresponding load level i. At each load level, the theoretical life expectancy is Ni number of cycles. The damaging effect of a single cycle at this level is assumed to be

FIGURE 7.17 Referring to Palmgren–Miner rule: (a) spectrum loading and (b) blocks of uniform cycles.

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Fatigue Failure Prediction Methods

1 D i = ----Ni

(7.41)

The damage after ni cycles at this level equals n n i D i = -----i Ni

(7.42)

We can express now the accumulated damage from the entire loading as

∑ ni Di

=

n

-i ∑ ---Ni

n n n = -----1- + -----2- + -----3- + … N1 N2 N3

(7.43)

One hundred percent damage of the machine part (which will cause fracture) can be expressed as

∑ ni Di

= 1.0

(7.44)

Equation (7.44), known as the Miner equation, forms the basis of the analysis. For practical application, the Miner equation is written in the form

∑ ni Di

= C

(7.45)

where C is an empirical constant assume to vary in the range 0.7 ≤ C ≤ 2.2

(7.46)

The total number of loading cycles at time of failure is N =

∑ ni

= n1 + n2 + n3 + …

(7.47)

To facilitate the use of Equation (7.45), let us divide both sides by N, obtaining nD

i i ∑ --------N

=

n

i ∑ --------NiN

C = ---N

(7.48)

Now, let us introduce factor ai , expressed as n a i = ----i N

(7.49)

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Nonlinear Problems in Machine Design

It represents the number of cycles ni at load i relative to the total number N. Values of ai are obtained from processing data accumulated from the actual loading histories as illustrated in Figure 7.18. It follows that C N = ------------ai ∑ ---Ni

(7.50)

The above is used to compute life expectancies of parts subjected to spectrum loading. The cumulative damage should be affected by the sequence of loading cycles, but it appears that the Palmgren–Miner rule, even though it disregards the sequence, achieves realistic results.

7.3 FRACTURE MECHANICS As mentioned in this chapter’s introduction, the manifestation of fatigue failure in a machine part is the onset of a crack. The crack starts with microscopic imperfections and dislocations, which are followed by a visible split that leads to an ultimate fracture. The theory of crack propagation was originated by Griffith15 and applied to fatigue by Irwin,16 Paris,17 and others. Theoretically, we divide the process of crack propagation into three discrete periods: 1. initial damage in a microscopic scale 2. visible damage, cracks initiation, and growth 3. the final instantaneous fracture Fracture mechanics is concerned with the process covering the two latter stages. Both stages are visible upon examination of the damaged surface of a broken part

FIGURE 7.18 Cumulative load distribution in spectrum loading: creating a histogram.

Fatigue Failure Prediction Methods

231

(see Figure 7.19). The fatigue zone is relatively smooth, with the location of cracks’ origins showing (in general) quite clearly, while the surface in the rupture zone is rough.

7.3.1

GRIFFITH THEORY

OF

FRACTURE

Griffith theory, based on energy criteria, was originally developed for ideally brittle materials such as glass and ceramics. It was expanded to apply to elasto-plastic materials such as metals. To explain the fundamentals, consider a brittle plate of unit thickness subjected to uniform tension as shown in Figure 7.20a and b. After being stretched, the plate is fixed (see Figure 7.20c). The stored elastic strain energy of the plate is

FIGURE 7.19 Typical fatigue fracture surface.

FIGURE 7.20 Concerning the equilibrium in a stretched body with a crack.

232

Nonlinear Problems in Machine Design 2

σ U 0 = -----wl E

(7.51)

Assume that an incipient fracture took place, producing a crack of length 2a (see Figure 7.20d). Let Ua denote the released strain energy due to crack formation and Wγ the destruction work required to produce the crack. The total potential energy of the system is Π = U0 – Ua + W γ

(7.52)

According to Griffith,15 assuming the crack to be of an elliptic form, the released strain energy equals 2

2

πa σ U a = -------------E

(7.53)

During the formation of the crack, energy has to be supplied to destruct the material and create crack surfaces. The required destruction work equals W γ = 4aγ e

(7.54)

where γe denotes specific surface energy per unit area, assumed to be elastic. By minimizing total potential Π, the Griffith theory defines the equilibrium condition at crack formation as 2

2πaσ ∂Π ------- = ---------------- – 4γ e = 0 E ∂a

(7.55)

The following expression is known as the energy release rate, 2

πaσ G = ------------ = 2γ e E

(7.56)

The above forms the basis of crack propagation theory. Equation (7.56) was later modified to cover metals with plastic deformation. 2

πaσ G = ------------ = 2 ( γ e + γ p ) E

7.3.2

(7.57)

LINEAR ELASTIC FRACTURE MECHANICS

The analysis of crack propagation, based on the linear theory of elasticity, is called linear elastic fracture mechanics (LEFM). It approaches the process of crack prop-

Fatigue Failure Prediction Methods

233

agation with a fundamental assumption that the material is linearly elastic and the fracture is brittle. The basic theory of crack propagation relates to two-dimensional space. Let us start the analysis with defining the stresses and displacements in the area of crack. Figure 7.21 shows an infinite plate with an existing crack, subject to tension. Shown in the figure is a coordinate system of the crack. The general expression of the Cartesian stress components in the vicinity of the crack tip ( r 1. Figure 8.12 shows another diagram, which presents the same stress limits using different coordinates, Sm, Smax. It is called a Goodman diagram. Goodman, in his pioneering work,4 introduced the linear correlation AB in accordance with Equation 8.13. Therefore, line AB is called the Goodman line. It is an accepted practice to include yield point Syp , as a stress limit to further restrict the safe domain. The stress limit is shown as line CD in Figure 8.12. The stresses within new domain ODCB fulfill the additional condition S max = S m + S a ≤ S yp

(8.16)

so that the limit line now becomes BCD. Subsequently, we perceive the additional safety factor FS´ S yp ->1 FS′ = -------S max

(8.17)

Both safety factors, Equations (8.15) and (8.17), supplement each other in design analysis.

258

Nonlinear Problems in Machine Design

FIGURE 8.12 Goodman diagram.

The stress-limit diagrams shown in Figures 8.11 and 8.12 are applicable to cyclic stresses where Sa ≠ 0 and Sm ≥ 0. Experience shows that, for negative values, Sm < 0, the limit stresses are defined by yielding,5 as illustrated in Figure 8.13. The figure shows a complete stress-limit diagram in Sm,Sa coordinates, including both tension and compression.

FIGURE 8.13 A complete stress-limit diagram including tension and compression.

Design of Machine Parts

259

The magnitude of the safety factor depends on the machine part’s future functions and the uncertainties associated with the particular case. The following factors are usually taken into account when choosing a suitable factor of safety FS: • • • • •

the the the the the

uncertainty of the material properties uncertainty of the service loads weight of the machine part cost of the material influence of environmental conditions, e.g., corrosion

It is practical to assign, a priori, larger safety factors for heavier parts. For lowweight parts, one has to use low safety factors, incurring expensive analytical and experimental work, since obtaining a reliable low safety factor raises the cost of production considerably. The accepted range of safety factors in stationary machinery is at least, FS = 1.5 to 3.0, or even higher, with the latter figure being applicable to steady loading. In airborne equipment such as jet engines and helicopter drive mechanisms, where the weight must be kept down, the low safety factor is applicable. Here, as a rule, one uses safety factors of about FS = 1.1 to 1.2. In motorized vehicles, however, applicable numbers are FS = 1.5 to 2.0. Stress Concentration Consider a part with geometry that includes notches or holes. Under loading, the stresses concentrate, rising sharply to a peak, at the notch border. See Figure 8.14.

FIGURE 8.14 Stress concentration: (a) in a round bar with a circular notch, subjected to bending and (b) in a plate with a hole subjected to axial load.

260

Nonlinear Problems in Machine Design

This stress concentration is a function of the notch curvature: as the radius of curvature gets smaller, the stress concentration increases. The locus of stress concentration represents the danger point. The practical approach to derive the stresses is as follows. First, a nominal stress is computed disregarding notch curvature. Then the peak stress is computed when the nominal stresses are multiplied by a stress concentration factor. For example, in the round bar subject to bending, shown in the Fig. 8.14, the peak stress at the circular notch equals 32M S peak = K t S nom = K t ----------2πD

(8.18)

while the peak stress in the plate with a hole subject to axial load is F S peak = K t S nom = K t -------------------t(w – d )

(8.19)

The theoretical stress concentration factor Kt is evaluated, basing it on the elasticity theory or using photoelastic methods. It is defined by the equation theoretical peak stress in notched speciment K t = --------------------------------------------------------------------------------------------------------nominal stress in notched specimen

(8.20)

where the nominal stress ignores the stress concentration. The peak stresses derived by using Kt are usually higher then the peak stresses observed in practice. Therefore, an experimental stress concentration factor Kf is introduced, called the fatigue notch factor. To derive the fatigue notch factor from tests, we compare test data of a notched specimen to those of an unnotched specimen. Then, the fatigue notch factor is defined by the equation6 fatigue limit of unnotched speciment K f = ---------------------------------------------------------------------------------------fatigue limit of notched specimen

(8.21)

Experience shows that the fatigue notch factor for metals is less than or equal to the theoretical stress concentration factor, K f ≤ Kt

(8.22)

The latter inequality appears to be the result of localized plastic deformation taking place within the notch. An empirical correlation relating Kt and Kf has the form5 effective stress increase K f – 1 q = ------------------------------------------------------------- = -------------theoretical stress increase K t – 1

(8.23)

261

Design of Machine Parts

where q is called the notch sensitivity. For practical purposes, Equation (8.23) may be rewritten as K f = 1 + q(kt – 1)

(8.24)

When dealing with fatigue problems, it is customary to employ a corrected fatigue limit for stress concentration. S S' f = -----fKf

(8.25)

This is illustrated in Figure 8.15 in the form of a corrected S-N line. Subsequently, we can rewrite Equations (8.15) and (8.17) in the form K f Sa S 1 ------- = -----m + ----------Sf Su FS

(8.26)

S yp FS′ = ----------------------Sm + K f Sa

(8.27)

and

Equations (8.26) and (8.27) present a simplified approximate solution only. A more accurate solution of problems involving stress concentration requires a multiaxial stress analysis as described in Section 8.3.2. Finite Life A machine part with stresses beyond the limits shown in the limit diagrams of Figures 8.12 and 8.13 has a limited (finite) life. The analysis involves fatigue life

FIGURE 8.15 S-N diagram corrected for stress concentration.

262

Nonlinear Problems in Machine Design

prediction methods, as described in Chapter 7. It is based on plastic deformation without computing crack propagation. The correlation between fatigue life Nf and one-dimensional strain amplitude ∆ε/2 can be defined by either of the two expressions below. Manson Equation 1 ----

σ' f – σ m n' σ' f – σ m ∆ε c b - ( 2N f ) + ε' f  ------------------ ( 2N f ) ------ = ----------------- σ' f  E 2

(8.28)

Morrow’s Equation σ' f – σ m ∆ε b c - ( 2N f ) + ε' f ( 2N f ) ------ = -----------------E 2

(8.29)

See Section 7.1.2. A simplified procedure to define the fatigue life is to employ a constant life diagram based on the Woehler’s S-N line. The procedure is shown in Figure 8.16. Figure 8.16a presents a section of Woehler’s line with stresses above Sf/Kf. Figure 8.16b shows a derived constant life diagram in which the inclined lines are loca of Sa, Sm stresses that cause fatigue life N. The procedure is known as the stress method.

8.3.2

MULTIAXIAL ANALYSIS

As we know, machine parts that are subject to combined loading or have an intricate geometry require a multiaxial analysis. The most applicable analysis is based on the

FIGURE 8.16 Concerning the stress method: (a) Woehler’s line and (b) constant life diagram (not to scale).

263

Design of Machine Parts

finite element analysis. The analytical approach depends on life requirement of the part: infinite or finite. Infinite Life In the infinite-life analysis, we combine the component stress amplitudes into an equivalent stress amplitude and the mean component stresses into mean equivalent stress. For amplitudes, we use the von Mises equation, Sa =

1 2 2 2 --- [ ( S 1,a – S 2,a ) + ( S 2,a – S 3,a ) + ( S 3,a – S 1,a ) ] 2

(8.30)

For the mean stresses, we use the Sines correlation, (8.31)

S m = S 1,m + S 2,m + S 3,m

as derived in Section 7.1.3. (We assume that the stress components reach their peak values and valleys simultaneously.) One continues with one-dimensional analysis using Equations (8.15) through (8.17) and Figures 8.11 through 8.13. Finite Life As we know, multiaxial analysis of metal parts under fluctuating loading that have limited life is based on plastic deformation and crack propagation, including two stages. See Chapter 7. The number of cycles to failure in the first stage is determined using effective strain amplitude and effective mean stress. It can be computed using one of the two equations below. Manson Equation σ' f – ( σ m ) eff σ' f – ( σ m ) eff ∆ε eff b ---------- = ---------------------------( 2N f ) + ε' f ---------------------------E σ' f 2

1 ---n'

( 2N f )

c

(8.32)

Morrow Equation σ' f – ( σ m ) eff ∆ε eff b c ----------- = ---------------------------( 2N f ) + ε' f ( 2N f ) E 2

(8.33)

See Chapter 7, Section 7.1.3. After the onset of a crack, the remaining number of cycles relating to crack propagation is determined by integrating and solving the equation da dN = -------------------n C ( ∆K ) See Section 7.3.3.

(8.34)

264

Nonlinear Problems in Machine Design

Fail-Safe Designs In finite-life design, in addition to the analytical prediction of fatigue life, a practical approach has been devised to ensure safety when a failure of machine part happens. It pertains to damage-tolerant or “fail-safe” design where, initially, the design provides for load-carrying members that are assigned to take over the load temporarily in case of a component’s failure.8 The fail-safe design to ensure load redistribution when failure occurs requires a multiple load path so that applied loads will go through other members, avoiding the damaged area. A typical example of a fail-safe design is a bolted connection between two machine members comprising a series of bolts designed to withstand the applied load. A failure of a single bolt will not interfere with a function of the machine; however, it may shorten the fatigue life. The same could happen during crack propagation in machine component. After a crack has started to propagate, the component may continue to function for a predicted period.

REFERENCES 1. Juvinall, R.C., Engineering Consideration of Stress, Strain and Strength, McGrawHill, New York, 1967. 2. ASME Handbook, Metals Engineering Design, ed. Horger, O.J., McGraw-Hill, New York, 1965. 3. Woehler, A. 1858. “Uber die Festigkeitsversuche mit Eisen und Stahl,” Z. Bauwesen. 8, 641. 4. Goodman, J., 1899. Mechanics Applied to Engineering, Longmans, London, 1899. 5. Juvinall, R.C., and Marshek, K.M., Fundamentals of Machine Component Design, John Wiley, New York, 1991. 6. Peterson, R.E., Stress Concentration Factors, John Wiley, New York, 1974. 7. SAE Fatigue Design Handbook, 3rd edition, Society of Automotive Engineers, Warrendale, PA, 1997. 8. Osgood, C.C., Fatigue Design, Pergamon Press, Oxford, England, 1982.

9

Leaf Springs

9.1 INTRODUCTION The spring is a machine part used to absorb sudden loads and to accumulate elastic energy. There are different mechanical designs and forms of springs. The spring under consideration is called a leaf spring. This type of spring has an advantage over other kinds of springs because of its compact design and essential structural role. Its importance, first and foremost, comes from the part’s unique role, utilized in motor vehicles to provide the absorption of irregular loads caused by uneven roads. The leaf spring is also used in other machines such as heavy presses that operate under loads at large displacements. Since the displacements undergo intermittent absorptions and releases, a sturdy design of the part must be provided—the design that combines optimum strength with a needed elasticity. This is achieved by an assembly of narrow leaves acting in unison as bending beams. A typical leaf spring is shown in Figure 9.1. Figure 9.1a shows the leaves hold together by a center bolt and clamp. Figure 9.1b and c show different spring ends used in practice. The top leaf is designated as the main leaf. The leaves are bent with the ends facing upward. When a spring is designed to be used in a reversed

FIGURE 9.1 Leaf spring: (a) spring (1, center bolt; 2, clamp), (b) eye spring end, and (c) plain spring end.

265

266

Nonlinear Problems in Machine Design

position, the main leaf is at the bottom. The load is applied simultaneously at each end of the main leaf, while the reaction forces concentrate in the center of the spring, or vice versa. All leaves are subjected to significant deflections and change in curvature. The normal working load is a vertical load that engages the leaves, bending them in a direction that relieves the curvature. The change in geometry causes them to slide while the interleaf friction hinders the motion. This phenomenon causes a part of applied energy to be transformed into irreversible work and to dissipate. The magnitude of dissipated work depends on several factors such as the condition of leaf surfaces, applied load, and speed of sliding. The exposure to varying loads subjects leaf springs to hazardous stresses that result in fatigue. The best known deterrent against fatigue is the surface treatment of the metal, namely shot peening, done prior to assembly. Processing with shot peening produces residual compressive stresses in the surface layer. Consequently, the tensile stresses at the surface provide protection against fatigue.1,2 The following design analysis of a leaf spring presents two approaches. One uses a simplified theory, and the other, for more complex problems, uses the finite element method.

9.2 DESIGN FUNDAMENTALS 9.2.1

THEORETICAL STRESS DISTRIBUTION

Originally, the leaf spring was conceived as a beam of uniform strength.3 Our discussion begins with this concept to help lay down the basic theoretical principles. Consider Figure 9.2. Due to symmetry, only one-half of the leaf spring is analyzed, representing a cantilever beam. The beam has the form of a flat triangle loaded at its apex, where the maximum bending stresses are identical throughout. Dividing the triangle into parallel leaves and stacking them on top of each other, an ideal leaf spring is achieved (see Figure 9.2b). The maximum bending stress at the fixed end of the spring is determined from the correlation of a beam of a rectangular cross-section, Pl S = ---------mW

(9.1)

where m is the number of leaves, and W is the section modulus of a single leaf. The deflection of the free end of the spring is assumed to be the same as the deflection of a beam of the constant cross-section of a width that equals two-thirds of the base of the triangle. The deflection equals 3

3

1 Pl Pl f = --- -------------------- = -------------32 2EmI --- EmI 3 

(9.2)

where I is the moment of inertia of a single leaf. A different equation, presented by Wahl,3 considers a trapezoidal beam (Figure 9.3). Here, the deflection becomes

267

Leaf Springs

FIGURE 9.2 Triangular beam theory: (a) bending triangle and (b) leaf spring.

FIGURE 9.3 Trapezoidal beam.

3

Pl f = K -------------3EmI

(9.3)

K is a correction factor that is a function of the number of leaves m (Figure 9.4). Equations (9.1) to (9.3), in their simplified form, are presented here to help understand the more accurate derivatives included in the standard design formulae used in practice today.

268

Nonlinear Problems in Machine Design

FIGURE 9.4 Correction factor for deflection of leaf spring.

SAE Design Formulae The formulae that form standards for practical application are based on widely accumulated professional experience. The published source for designing leaf springs is the SAE Spring Design Manual, issued by the Spring Committee of the Society of Automotive Engineers.1 Introducing the varying leaf thickness, the simplified formulae above become as follows. The expression for bending stresses, derived from Equation (9.1), is lt S = ------------ ⋅ P 2∑ I

(9.4)

The deflection, derived from Equation (9.2), equals 3

Pl 1 f = ---------------- ⋅ ------2E ∑ I SF

(9.5)

SF denotes a stiffening factor, which is a function of leaf engagement as explained below. Derived from Equation (9.5) is the load rate, 2E ∑ I P ⋅ SF k = --- = ---------------3 f l

(9.6)

Leaf Engagement The above triangular beam conjecture implies that the leaves are engaged throughout their lengths and are bearing on each other. This, however, is contrary to reality.

269

Leaf Springs

Only a small area near the tip is engaged, while the rest of the leaf is free.1 To prove it analytically, consider two adjoining leaves of a leaf spring under load (Figure 9.5a). The leaves are subject to bending as follows: Face 1 of the upper leaf is compressed, while face 2 of the lower leaf is subject to tension. The curvatures of both faces are expressed by the differential equation of a beam. 2

d y --------2 2 d y dx 1 M ------ = --- = --------------------------- ≅ --------2 ρ EI dy 2 d x 1 +  ------  dx

(9.7)

Equation (9.7) is of a curve that has no turning points, i.e., curvature d2y/dx2 nowhere equals zero. Two curves of this kind with different radii ρ will cross at two points (see Figure 9.5b). Therefore, according to behavior of one-half of the leaf spring, any pair of leaves whose patterns are defined by bending moment M can have one bearing point only—point A in Figure 9.5. In practice, as stated above, the leaves are engaged in a small area near the tip. The size of the area depends on the form of leaf ends, which may be square, tapered, or trimmed. The area affects spring deflection f through the stiffening factor SF. For leaves with square ends, SF = 1.15 while, for tapered ends, SF = 1.10. 1 Example Let us derive more accurate stress distribution using the following example.4 Consider the three-leaf spring shown in Figure 9.6. For simplicity, assume the three

FIGURE 9.5 Concerning leaf engagement: (a) leaf separation under load and (b) engagement of two adjoining leaves.

FIGURE 9.6 Three-leaf spring.

270

Nonlinear Problems in Machine Design

leaves to be of the same thickness. The contact between the leaves takes place at bearing points A and B, while the rest become disengaged. The stress distribution is based on the assumption that the deflections of adjoining leaves at the bearing points are equal, i.e., y A1 = y A2 ; y B2 = y B3

(9.8)

The deflections are expressed as functions of contact forces P1 and P2 as follows (see Figure 9.7). The deflection of the main leaf 1 at point A is 3

l y A1 = --------- ( 14P – 8P 1 ) 3EI

(9.9)

The deflection of the middle leaf 2 at the same point is 3

l 5 y A2 = ---------  8P 1 – --- P 2 2  3EI 

(9.10)

The deflection of the middle leaf 2 at point B is 3

l 5 y B2 = ---------  --- P 1 – P 2  3EI  2

(9.11)

FIGURE 9.7 Load distribution in a three-leaf spring: (a) forces and (b) bending moments.

271

Leaf Springs

The deflection of the bottom leaf 3 equals 3

l y B3 = ---------P 2 3EI

(9.12)

It follows from Equations (9.8) that 32P 1 – 5P 2 = 28P 5P 1 – 4P 2 = 0

(9.13) (9.14)

from which one gets P 1 = 1.087P ; P 2 = 1.359P

(9.15)

The bending moments produced by forces P1, P2, and P are shown in Figure 9.7b. One finds the critical bending stresses to be at points A, B, and C. (Point C represents the support of the bottom leaf, as shown in Figure 9.6.) Comparing the resulting stresses, we get S 2, max = 1.087S 1,max ; S 3,max = 1.359S 1,max

(9.16)

where indices 1, 2, and 3 refer to the respective leaves. One notes that the smallest bending stress occurs in the main leaf, while the greatest is in the bottom leaf.

9.2.2

CURVED LEAVES

In practice, the leaves are designed in form of arcs, with each leaf having a different radius, scaled down accordingly.1 As a result of this geometry, there are spaces between the leaves. See Figure 9.8. At assembly, the center bolt and clamp act to pull the leaves together, changing the curvatures and causing mechanical prestress. Let us consider prestressing of leaves in the spring as shown in Figure 9.8. The leaves are of equal thickness, and stresses and deflection can be determined from Equations (9.1) and (9.2). Tying the leaves together is explained schematically in Figure 9.9. For simplicity, assume that the tying is accomplished by tightening the center bolt, while ignoring the center clamp. Figure 9.9a presents the condition prior to bolt tightening, and Figure 9.9b shows the local leaf deformation due to the tying. The local compression of the leaves affects only the vicinity of the center bolt and have negligible influence on the bending stresses in them. As the tightening force P rises, the leaves draw closer together. The distribution of forces between the leaves is shown in Figure 9.10. The forces necessary to bring the leaves together, ignoring the leaf curvatures, may be expressed as follows

272

Nonlinear Problems in Machine Design

FIGURE 9.8 Curved leaves in a leaf spring.

FIGURE 9.9 Tightening of leaves by a center bolt: (a) condition before tightening and (b) deformation caused by tightening.

FIGURE 9.10 Forces acting on the leaves during bolt tightening.

273

Leaf Springs 3

2 P′a P′l 2 s 12 = -----------2- ( 3l 2 – 4a 2 ) – -----------248EI 48EI

(9.17)

3

P′′a P′′l 2 2 s 23 = -----------3- + ------------2 ( 3l 2 – 4a 2 ) 48EI 48EI

(9.18)

To bring leaves 1 and 2 together, the tightening force must equal P = P′ ; for leaves 2 and 3, it must be P = P″ . The compressive stresses induced in the main leaf through tightening equal Eh P′l S = ---------- = s 12 -----2 mW l

(9.19)

See Equations (9.1) and (9.2). The external load applied upon the leaf spring induces bending stresses that are superimposed on the preliminary ones (i.e., created at assembly). In consequence, the maximum tensile stresses in the main leaf decrease, and those in the bottom leaf increase. Meanwhile, in the intermediate leaves, the stresses vary proportionally to the leaf location. Because the bottom leaves are subjected to a greater stresses, the spaces between stay small, thus lessening the influence of the preliminary stresses.

9.3 FE ANALYSIS OF LEAF SPRINGS The finite element method of analyzing displacements and stresses is known to give better results than any other approach. Furthermore, it allows analysis of frictional effects that cannot be considered by standard machine design equations. In the following analysis, the leaves are represented by a two-dimensional FE model, and the contact between them is simulated by contact elements using the penalty method. The friction factor is assumed to be µ = 0.45.

9.3.1

PROBLEM DEFINITION

The leaf spring under consideration is adapted from an example given by the SAE Manual on Leaf Springs.1 The spring has five leaves, the same as in the example. However, contrary to the SAE example, here the spring is assumed to be symmetric. The leaves are made from alloy steel SAE 9260 with the following properties: Brinell hardness

BHN = 400

Tensile strength

Su = 1560 MPa

Yield point

Syp = 1350 MPa

Two kinds of leaves are analyzed: leaves that were not processed by shot peening, and leaves that were. The relevant information is summarized in tables and figures. Figure 9.11 shows the leaves tied by a center clamp and the spring’s end supports. The load is applied

274

Nonlinear Problems in Machine Design

FIGURE 9.11 Design details of a leaf spring: (a) center clamp and (b) end support.

at the bottom of the center clamp and is directed upward. At assembly, the bolts are tightened, prestressing the leaves and affecting the form of arcs. The design data are as follows: Design load

4150 N

Maximum load

6432 N

Length of center clamp

100 mm M 8 × 1.25, class 4.8

Bolts Tightening force

10 kN

Table 9.1 presents the dimensions of free leaves, before tying. Table 9.2 shows their position in relation to each other. The nomenclature used in the tables is explained in Figure 9.12. TABLE 9.1 Spring Leaves–Main Dimensions Leaf no.

Half Length l (mm)

Thickness t (mm)

Width w (mm)

Radius r (mm)

1

570.0

6.7

63.0

1603.0

2

454.6

6.3

63.0

1481.0

3

352.4

6.3

63.0

1399.0

4

245.4

6.3

63.0

1359.0

5

135.7

6.0

63.0

1300.0

Equations used to derive the dimensions, listed in Table 9.2, are as follows: li 180 - --------α i = -------------------r i + 0.5 t i π

(9.20)

275

Leaf Springs

FIGURE 9.12 Leaf geometry nomenclature.

TABLE 9.2 Spring Leaves–Derived Dimensions Leaf No.

α (degrees)

h (mm)

x (mm)

1

20.330933

99.864506

556.949945

2

19.549885

68.933472

446.57488

134.447283

3

14.400052

43.952472

349.91838

90.9863898

4

10.337378

22.026573

243.50555

49.0688909

5

5.9670278

9.0435508

135.14296

63.6287013

x i = r i sin α i

(9.21)

h i = r i ( 1 – cos α i )

(9.22)

ci =

9.3.2

c (mm)

2

2

( r i = 1 + t i – 1 ) – xi + hi – r i

(9.23)

FE SOLUTION

Let us analyze the deformation of the spring and the stress distribution. Two FE models are developed: a coarse model for spring deformation and a precise model for the stress distribution. The stress distribution is derived first for leaves that were not processed by shot peening, and then for the processed leaves.

276

Nonlinear Problems in Machine Design

Coarse Model The leaf spring is represented by a plane model, using plane strain relations of elasticity. Because of symmetry, only half of the spring is being modeled. The leaves are made of quadrilateral elements with height that equals the leaf’s thickness. The contact between the leaves is simulated by contact elements. The center clamp is simulated by link (rod) elements. Figure 9.13 shows the leaves in the loose condition. The spring ends are supported by contacts with two quadrilateral elements representing a mounted body (see Figure 9.14). The upper quadrilateral restricts the upward displacement of the spring, while the lower one restricts the downward displacement. To achieve a tightening force of 10 kN in each of four bolts in the center clamp, it is necessary to specify an initial strain as part of the input data of the link elements. An initial strain of 0.2791 was derived by a preliminary FE solution (see below).

FIGURE 9.13 Finite element model of a leaf spring.

FIGURE 9.14 Finite element representation of spring end support.

277

Leaf Springs

The FE model is created and solved using the ANSYS program. Two nonlinearities are present: varying contact surfaces between the leaves and large displacements. A solution process based on the Newton–Raphson method is used, proceeding with small incremental loading steps. Preliminary Step: Tightening of Leaves The first computational step pertains to deformation of leaves before loading, due to tying. It is performed to evaluate the initial strain in the link element. The result confirms a tightening force of 10 kN in each of the clamp bolts. By trial and error, the initial strain is found to be 0.22791. The deformation of the model due to tightening is illustrated in Figure 9.15 and shows the leaves before and after tying. Because of tightening, the center clamp moves up 1.433 mm. Final Solution: Incremental Loading. The maximum normal load is 6432 N. A two-dimensional plane-strain model is used, 1 mm thick. The load per millimeter of spring width, applied to half-spring, is 6432 F/w = 0.5 × ------------ = 51.048 N/mm 63.0 The load acting at the bottom of the center clamp is applied as a pressure on two bottom elements, with a length equal to the length of half the clamp. The length of the two corresponding elements is 50 mm. The pressure upon the elements equals

FIGURE 9.15 Deformation of leaves due to tightening, computed by the ANSYS program.

278

Nonlinear Problems in Machine Design

51.048 p = ---------------- = 1.021 MPa 50.0 The loading process is divided into 40 equal increments of ∆p = 0.0255 MPa each. The computational results are shown in Figure 9.16. It follows that the spring deflection at maximum load equals 142.70 mm. In a similar way, we find the deflection with design load equaling 94.228 mm. The net deflection under the applied load is smaller by 1.433 mm, due to the shrinkage caused by initial tightening (see Figure 9.15). The load rate, based on the maximum load, equals 6432 k = ------------------------------------ = 45.531 N/mm 142.70 – 1.433 while the load rate that is based on the design load equals 4150 k = ------------------------------------ = 44.722 N/mm 94.228 – 1.433 The difference in rates is explained by nonlinearity due to friction and large displacements. Neglecting friction and bolt tightening, the load rate equals, respectively,

FIGURE 9.16 Deflection of the leaf spring computed using the ANSYS program: (a) at design load, F = 4150 N, and (b) at maximum load, F = 6432 N.

279

Leaf Springs

6432 k = ---------------- = 40.08 N/mm 160.47 4150 k = ---------------- = 39.06 N/mm 160.26 Comparing SAE design results, we note lower values, due to considering nonlinear effects, neglected by SAE formulae. For computation of the load rate, we use Equation (9.6) multiplied by a factor of two (the equation presented above is for a cantilever spring). The result is 3

3

3

63 × 6.3 63.60 63 × 6.7 2 × 20700 × --------------------- + 3  --------------------- + -------------- 12  12 12 k = 2 × --------------------------------------------------------------------------------------------------------------------× 1.15 = 34.19 N/mm 3 570.0 Hysteresis The loading history of the spring under consideration is presented in Figure 9.17. To determine the energy dissipation due to friction, the load is plotted against spring deflection, Figure 9.18. Energy dissipated during a loading cycle is expressed by the area within the hysteresis loop.

FIGURE 9.17 Loading history of the leaf spring.

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Nonlinear Problems in Machine Design

FIGURE 9.18 Hysteresis loop of a leaf spring computed using the ANSYS program.

Precise Model The most common failures in leaf springs are fatigue failures. To derive the life expectancy of a leaf spring by means of a fatigue analysis, an accurate stress distribution must be determined. For this purpose, a precise FE model of the leaf spring is prepared. The model comprises small plane-strain quadrilateral elements created by subdividing the elements of the coarse model (see Figure 9.19). To

FIGURE 9.19 Precise FE model of a leaf spring.

281

Leaf Springs

facilitate the stress analysis, elements bordering with the leaf surface are considered to be thinner than those inside the leaf. (They are made 0.8 mm thick, corresponding to the depth of the layer of residual stresses discussed below.) The applied load in our computation equals the designed load, 4150 N. The precise model of the leaf-spring is created and solved using the MSC.NASTRAN program. For contact simulation, it uses the penalty method. As before, the solution proceeds in steps applying incremental loading. The resulting bending stress distribution is presented in Figure 9.20. Figure 9.20a shows preliminary stresses created at assembly due to bolt tightening, while Figure 9.20b presents the final stresses caused by loading. One notes that maximum bending stresses differ from one leaf to another. The bottom leaf has the highest stresses, similar to the example discussed above, Equation (9.16). It is interesting to note how the FE results compare with those based on the SAE formula, Equation (9.4). According to SAE, the highest stress appears in the main leaf, where it equals 4150 570.0 × 67 × ------------ = 595.71 MPa S = -----------------------------------------------------------------------------------------------------3 3 3 2 63 × 6.3 63 × 6.0 63 × 6.7 2 × --------------------- + 3  --------------------- + -------------------- 12  12 12 Table 9.3 presents a comparison of stresses in all leaves computed by the FE method with those resulting from Equation (9.4). TABLE 9.3 Maximum Bending Stresses in Leaves Leaf no.

Max. bending stress, MPa FE method

Max, bending stress, MPa Eq. (9.4)

1

334.56

595.71

2

383.62

560.15

3

421.21

560.15

4

448.84

560.15

5

615.07

533.47

One notes the large difference of stress values in leaf no. 1. Part of the difference stems from preliminary stresses due to tying of the leaves, a factor disregarded in the SAE formula. Shot Peening and Stress Peening There is a special surface process, prior to assembly, used to increase the fatigue life of leaf springs. In its simpler application, it is referred to as shot peening. Shot peening induces compressive residual stresses near a surface of the leaf. Upon superposition with working stresses, the maximum bending stresses at the surface decrease, increasing the fatigue life.2,5 (See Chapter 7.) When shot peening is done while the leaf is subject to tensile stresses, the process is called the stress peening

282

Nonlinear Problems in Machine Design

FIGURE 9.20 Bending stresses in a leaf spring computed using MSC.NASTRAN program: (a) after assembly and bolt tightening and (b) final stresses after loading.

process. The magnitude of the residual stresses is then considerably larger. Stress peening is usually done on one side of the leaf, while shot peening is applied to both sides. Figure 9.21 shows residual stresses in a leaf caused by both processes. The depth of the compressed layer is assumed to be 0.8 mm. The residual stresses caused by stress peening reach 550 MPa, while those caused by shot peening equal 400 MPa.

Leaf Springs

283

FIGURE 9.21 Distribution of residual stresses in leaf no. 5 computed using MSC.NASTRAN program: (a) caused by stress peening on one side of the leaf and (b) caused by shot peening on both sides of the leaf.

To insert residual stresses into the FE model, we compute thermal stresses induced by an imaginary thermal loading of the leaves; the thermal stresses produce the same effect as shot peening. The distribution of stresses due to loading of a leaf spring processed by stress peening and shot peening is shown in Figure 9.22. The figure presents a superposition of working stresses upon the residual stresses in leaf no. 5, illustrating the beneficial effects of both processes. One notes a substantial decrease of tensile stresses on the surface of the leaf, which renders the desired prolongation of fatigue life. The introduction of residual stresses changes stress equilibrium in leaves. The stress peening process causes a nonsymmetrical deformation, resulting in a decreased curvature (see Figure 9.23a). The shot peening, when done on both sides of the leaf, causes a symmetrical deformation without any change in curvature (see Figure 9.23b). Table 9.4 lists new and old curvature radii in leaves processed by stress peening. Windup Torque and Longitudinal Load Leaf springs in vehicles are occasionally loaded by horizontal forces and torques in addition to standard vertical loading. Leaf springs designed for this purpose must

284

Nonlinear Problems in Machine Design

FIGURE 9.22 Superposition of working stresses upon residual stresses in leaf no. 5, computed using the MSC.NASTRAN program: (a) in a leaf with stress peening on one side of the leaf and (b) in a leaf with stress shot peening on both sides of the leaf.

be provided with eye ends, shown in Figure 9.1b, to accommodate the horizontal loading. Consider the deformation and stress distribution of a leaf spring subject to loading as follows: Vertical load

Fn = 4150 N

Horizontal load

Ft = 3780 N

Windup torque

T = 2,765,700 mmN

285

Leaf Springs

FIGURE 9.23 Deformation of a leaf caused by residual stresses computed using MSC.NASTRAN program: (a) in a leaf with stress peening on one side of the leaf and (b) in a leaf with stress shot peening on both sides of the leaf.

TABLE 9.4 Leaf Deformation Due to Stress Peening on One Side Leaf no.

Radius before Stress Peening (mm)

Radius after Stress Peening (mm)

1

1603.0

1830.5

2

1481.0

1782.5

3

1399.0

1659.0

4

1359.0

1571.5

5

1300.0

1471.5

The FE model is shown in Figure 9.24a. For the purpose of analysis, since there is no symmetry, the half-spring model is extended to cover the whole spring. The location of the left-end support is fixed, while the right-end support is sliding. (The FE model is without the eye-end since deformation, and stresses in the eye end were not computed.) The horizontal load and windup torque are applied at the center clamp. The resulting spring deformation is shown in Figure 9.24b. The computed stress distribution is presented in Figure 9.25.

9.4 CONCLUSIONS The design of a leaf spring includes consideration of geometric and frictional phenomena that present a nonlinear numerical problem. The problems are solved twice in the text, first using the finite element method, followed by the simplified design

286

Nonlinear Problems in Machine Design

FIGURE 9.24 Leaf spring subjected to vertical load, windup torque, and longitudinal load computed using MSC.NASTRAN program: (a) loading of leaf spring and (b) deformation of leaf spring.

FIGURE 9.25 Stress distribution in the leaves due to vertical load, windup torque, and longitudinal load computed using the MSC.NASTRAN program.

formulae. The comparison of the two distinctly different solutions confirms the anticipated discrepancies, explained by the inaccuracies of the linear equations. The FE method is therefore more reliable. The FE solution illuminated the following properties of spring leaf: • The stress distribution depends on the leaf’s location, as the bending stresses at the bottom leaves are higher than at the top leaves. The FE

Leaf Springs

287

analysis shows a very large difference between the stresses in the first and last leaves (about 100 percent). See Table 9.3. • Interleaf friction and tying of leaves, due to increased resistance, causes higher spring rates, contrary to the simplified formula, which provided a constant spring rate. • Shot peening produces lower tensile stresses at the surface, enhancing fatigue life. Stress peening (shot peening, while the leaf is subject to tensile stresses) produces better results; however, it causes a change of curvature of the leaf, since it is done on one side only. • A comparison of results obtained by simplified equations and FE analysis shows a difference of about 30% of spring deflection. The advantage of FE computation over that of simplified equations, such as SAE design formulae, is due to the ability to consider the effects of the following phenomena: 1. 2. 3. 4.

The leaves are engaged at the tips only. The friction between the leaves causes resistance to spring’s deflection. The applied load causes considerable geometric changes. Tying the leaves by center bolt and clamp makes the spring less flexible.

In addition, the FE method is able to analyze what, until recent developments, could have been evaluated only through tests, namely: (1) the energy dissipation and hysteresis caused by friction, and (2) the positive influence of shot peening and stress peening. However, the solutions to complex problems that deal with receding contact and sliding friction are not infallible. The accuracy of FE solutions must be critically examined with regard to the following considerations: 1. precision of the FE model 2. exactness of the iterative process 3. awareness of penalty parameters and the resulting numerical errors (as discussed in Chapter 6.) Topic 3, above, was discussed in Chapter 6, explaining that evading the errors may produce an unstable solution that, by failing to converge, renders the problem insoluble.

REFERENCES 1. Design and Application of Leaf Springs, Spring Design Manual, SAE AE-11, Part 1, Society of Automotive Engineers, Warrendale, PA, 1990. 2. Almen, J.O., and Black, P.H., Residual Stresses and Fatigue in Metals, McGraw-Hill, New York, 1963. 3. Wahl, A. M., Mechanical Springs, McGraw-Hill, New York, 1963.

288

Nonlinear Problems in Machine Design 4. Biderman, V.L., Design calculation of leaf springs. Stress Calculations in Machine Design, ed. Ponomaryev, S.D., Vol. I, Mashgiz, Moscow, 1956 (in Russian). 5. Zahavi, E., and Torbilo, V., Fatigue Design: Life Expectancy of Machine Parts, CRC Press, Boca Raton, Florida, 1996.

10

Threaded Fasteners

10.1 INTRODUCTION In the assembly of machines, threaded fasteners are immensely important. As the links of interacting parts, they are the ones that transmit forces, created by a load, to joined parts. Since the fasteners become loci for concentrated forces within the machine, we focus on threaded fasteners for a detailed study of stresses. As we know, stresses, and in particular fluctuating stresses that are caused by dynamics of forces, constitute one of the major reasons for fatigue failure of machine parts.

10.1.1 TYPES

OF

THREADED FASTENERS

The size of threaded fasteners depends mainly on availability of space for the parts. The forms of the fasteners are dictated by the constraints of the design, namely, ways of joining the parts. Commercially, three forms are available, as shown in Figure 10.1:

FIGURE 10.1 Threaded fasteners: (a) bolt with nut, (b) screw, and (c) stud with nut.

1. Fasteners, comprising a bolt and nut. The connected parts are clamped between the bolt’s head and the nut. 289

290

Nonlinear Problems in Machine Design

2. Fasteners that are screws in a form of a bolt, without a nut. The fastener is introduced into one of the parts, pulling the other part to create the connection. 3. Fasteners, comprising a headless bolt (a stud) and nut. The stud is introduced permanently into one of the parts, while a nut clamps the parts together. Fastener type 1, comprising a bolt and nut, is in use more than the others. This type of bolt comes in two forms: standard and one that is referred to as a high-strength bolt, illustrated in Figure 10.2. The latter is a bolt with a narrow shank, having a large filet radius at the first thread. Its flexibility and lower stress concentration considerably increase the life expectancy of the bolt. This type of a bolt is reserved for special uses because of its higher costs.

10.1.2 THREAD GEOMETRY To illustrate the thread geometry, let us refer to Figure 10.3. The spiral form of the thread provides an axial displacement of the nut, which, when tightened, creates an

FIGURE 10.2 Types of bolts: (a) standard and (b) high-strength.

FIGURE 10.3 Thread spiral.

291

Threaded Fasteners

axial force in the bolt. To ensure self-locking by friction between the nut and the bolt, and to avoid an opening of the nut, spiral angle α, shown in the figure, must be small enough. The following condition must met: p tan α = --------- < µ πd m

(10.1)

where p denotes the pitch of the thread, dm is the mean thread diameter, and µ is the friction coefficient. In standard threads, the value of tanα varies between 1/30 to 1/16 for coarse threads and between 1/40 to 1/25 for fine threads. The generally accepted value for the friction coefficient in machine design is about µ = 0.14. The standard thread profile for threaded fasteners is shown in Figure 10.4.1 The main thread parameters are d (the outer diameter of bolt), and p (the pitch). The standard thread has a triangular form with angle 2β = 60°. The theoretical thread height, assuming an exact triangle, equals p 1 H = --- ----------------- = 0.866025 p 2 tan 30°

(10.2)

The actual thread height, caused by a reduction of the corners of the triangle, is 5 h = --- H = 0.541266 p 8

FIGURE 10.4 Metric and unified thread standards.

(10.3)

292

Nonlinear Problems in Machine Design

The outer bolt diameter d, the major diameter of external thread, refers to the actual thread height h. The minor diameter of internal thread, the inner nut diameter, equals d 1 = d – 2h = d – 1.25H

(10.4)

The pitch diameter is defined as the theoretical mean diameter, which is H d 2 =  d + 2 × ---- – H = d – 0.75H  8

(10.5)

10.2 FORCES IN BOLT CONNECTION The forces in a bolt are mainly axial forces. Subsequently, the bolt’s elongation is the dominant deformation. Because of the prevailing axial action, we shall first discuss a one-dimensional analysis.

10.2.1 BOLT LOADING Bolts installed in a machine undergo two-stage loading: preloading at the assembly and subsequent loading caused by acting forces in the working parts. Preloading The preloading force is caused by application of torque in tightening the nut. An estimate of this force can be derived by means of an empirical relation, T = CF i d

(10.6)

where T is the torque, Fi denotes the axial force in the bolt, and C is an empirical coefficient that can be assumed to be 0.2, based on experience.2 There are ways of getting more accurate measurement of T using, for instance, a special torque wrench or measuring the nut displacement. Let us follow the process of preloading, illustrating the working of a bolted joint comprising two plates and a bolt. See Figure 10.5. As noted above, we apply a onedimensional analysis, assuming that all forces and displacement act in the axial direction of the bolt. The preloading causes a bolt extension, δb = δb ′ + δb ″

(10.7)

δc = δc ′ + δc ″

(10.8)

and compression of the plates,

See Figure 10.5. Together they form a grip displacement that equals ∆ = δb + δc

(10.9)

293

Threaded Fasteners

FIGURE 10.5 Preloading of bolt and nut.

(Note: δb and δc denote the absolute values of the respective displacements, whether they denote tension or compression.) The grip displacement amounts to the difference between the dimensions of the unloaded bolt and plates. Assume the bolt and plate deflections to be linear functions of the force. Then, F F δ b = -----i and δ c = -----i kb kc

(10.10)

where kb and kc are the respective stiffnesses of the bolt and the plates. Consequently, the grip displacement equals 1 1 ∆ = F i  ---- + ----  k b k c

(10.11)

Subsequent Loading by Outer Forces Let us analyze the subsequent loading of the bolted joint as illustrated in Figure 10.6. Consider the bolted flange connection of a pressure vessel. We shall define the final bolt loading after initial tightening and an outer applied force Fp, caused by internal pressure in the vessel. For simplicity, assume that only part of the flange is participating, as indicated in Figure 10.6b. Let Fb be the tensional force in the bolt while applying pressure in the vessel, while Fc is the resulting compressive force acting on the flange. The condition of equilibrium states that Fb = F p + Fc

(10.12)

(Note: Fb and Fc denote the absolute values.) The compatibility condition requires that ∆, which is the difference between the dimensions of the unloaded bolt and flange, remain unchanged, i.e.,

294

Nonlinear Problems in Machine Design

FIGURE 10.6 Bolted joint design: (a) design of a flange and (b) schematic representation.

F F 1 1 ∆ = -----b + -----c = F i  ---- + ----  k b k c kb kc

(10.13)

Combining the latter two equations, we obtain the expressions for Fb and Fc as follows: kb F b = F i + ---------------F kb + kc p

(10.14)

kc -F F c = F i – -------------kc + kc p

(10.15)

Both equations are illustrated in graphic form by means of a Roetscher diagram,3 where the forces acting upon the bolted joint are plotted versus the deflection of the bolt and flange (see Figure 10.7). Bolt and Flange Stiffnesses Stiffnesses kb and kc are functions of the geometry and the elastic constants of the bolt and flange. Assuming, as before, a one-dimensional condition, the bolt stiffness is defined by the formula Ab E b k b = ----------lb

(10.16)

295

Threaded Fasteners

FIGURE 10.7 Roetscher diagram.

A one-dimensional model of the flange used in machine design is shown in Figure 10.8. We assume the flange to be replaced by two truncated cones, with their stiffness equal to that of the flange. The stiffness of the model is defined by the equation Ac E c k c = ----------lc

(10.17)

where Ac is a nominal cross-section, which is equal to mean cross-section of the two cones. One-dimensional tests of bolted joints4 confirm the validity of such assumption. According to Birger et al4 the conical angle α can be assumed to be in the range of 25–33°; Osgood 5suggests a smaller range of 22–26°. If we disregard the thickness of the washer lw in Figure 10.8, and assume l b ≈ l c , then the coefficients in Equations (10.14) and (10.15) take the form Ab E b kb = ----------------------------κ b = --------------Ab E b + Ac E c kb + kc

(10.18)

Ac E c kc = ----------------------------- = 1 – κb κ c = --------------Ab E b + Ac E c kb + kc

(10.19)

10.2.2 BOLT FATIGUE Bolts in joints subject to cycling loads (e.g., pressure vessels with pulsating pressure) are subject to fatigue. The magnitude of varying load acting on the bolt can be subdivided into a mean term and an alternating term, respectively,

296

Nonlinear Problems in Machine Design

FIGURE 10.8 One-dimensional model of a bolted joint.

Fb + Fi Fb – Fi F m = ---------------- ; F a = ---------------2 2

(10.20)

In another form, the alternating term equals ∆F 1 kb - F = ----------b F a = --- --------------2 kb + k f p 2

(10.21)

The impact of flange thickness on the fatigue of the bolt is a topic that warrants further examination. Consider the case where the thickness of the joint l increases while the bolt size d remains unchanged. As a result of Equations (10.14) to (10.19), the force acting on the bolt, ∆Fb, decreases. Figure 10.9 illustrates two extreme cases, comparing a very thick flange, where (κb < κc,), with a thinner flange, where (κb < κc,). In both cases, the preload Fi and the applied outer load Fp are the same. A thick flange is preferable to the thinner one, because the amplitude of the varying load acting upon the bolt is considerably smaller. In practice, in bolted joints, subject to fatigue coefficient, κb equals up to 0.1. In the above discussion, we assumed a linear behavior of bolt and flanges, using one-dimensional analysis. However, considering contact phenomena, the linear

297

Threaded Fasteners

FIGURE 10.9 Comparison of Roetscher diagrams for bolts and flanges of different stiffnesses: (a) k b < k c and (b) k b > k c .

assumption becomes inaccurate. A more accurate analysis of bolted flange connections is presented in Chapter 11.

10.3 STRESSES Standard stress formulas used in machine design for bolts assume a linear dependence between load and deformation, ignoring nonlinear phenomena. In this section, we analyze the stresses in a typical bolt using the standard linear formulas. A more accurate, nonlinear analysis of the same bolt is described in the next section using the FE method.

10.3.1 LINEAR STRESS ANALYSIS Consider the standard bolt of Figure 10.2a. The locations that we chose for the stress analysis in the bolt are those that are known in practice to fail. The bolts usually fail in one of three locations: A, the thread root next to the bottom face of the nut; B, the thread runout section; or C, the intersection of the head and the shank. See Figure 10.10. The predominant place where the bolts break is at location A; therefore, we consider it first. The load is transferred from the nut into the bolt through contact in the thread. Because of the elastic character of the bolted joint, the first engaged thread, closest to the bottom of the nut, gets the biggest portion of the load, while the lessening of the load in the remaining coils occurs consecutively. See Figure 10.11. Photoelastic tests and numerical computations6 show that, in standard bolts and nuts, 30 percent of the total load acts on the first coil. f 1 ≈ 0.3F

(10.22)

298

Nonlinear Problems in Machine Design

FIGURE 10.10 Locations of high stresses in a bolt.

FIGURE 10.11 Load distribution in the thread coils of a bolt.

There are two kinds of stresses at location A: tensional stresses at the root of the thread, and shear stresses at the base of the thread. See Figure 10.12. The maximum tensional stress is computed by the equation, F S z = K t ----As

(10.23)

299

Threaded Fasteners

FIGURE 10.12 Stresses in the thread of a bolt: (a) tensional stresses and (b) shear stresses.

where the cross-sectional area subject to the stresses is π 2 A s = --- ( d – 0.9743 p ) 4

(10.24)

The shear stresses are caused by force f1 acting upon the coil. One may assume a constant distribution across the base,7 whereby the stress becomes f1 τ = ----------πd 1 b

(10.25)

In Equation (10.25), the minor diameter of internal thread equals d 1 = d – 2h = d – 1.25H

(10.26)

H 2 b = p – ---- ⋅ ------- = p – 0.288675H 4 3

(10.27)

and the thread width is

At location B in Figure 10.10, there are tensional stresses caused only by the total load F = 270,000 N. The maximum stress is computed again by the equation, F S z = K t ----As

(10.28)

At location C in the same figure, there are tensional stresses with concentration at the intersection of the head and the shank. The highest stress at this point equals

300

Nonlinear Problems in Machine Design

4F S z = K t --------2 πd

(10.29)

where Kt is the stress concentration factor different from the factor above. Stress Analysis of a Sample Bolt Consider the stresses in a bolt, size M24 × 3, Class 12.9. The nut height is m = 24 mm, and the number of threads in the nut equals n = 8. The bolt is subject to a tensional force of F = 270,000 N caused by preloading at assembly. The bolt material specification according to Class 12.9 are as follows: Ultimate tensile strength Yield point Proof stress

Su = 120 MPa Syp = 1100 MPa Sproof = 970 MPa

See Appendix C. Let us compute the stresses at location A first. The contact force acting upon the thread coil at the bottom of the nut equals f 1 = 0.3 × 270, 000 = 81, 000 N The thread dimensions are as follows: d p H d1 b

= = = = =

24.0 mm 3.0 mm 0.866025 × 3.0 = 2.598075 mm 24.0 – 1.25 × 2.598075 = 20.752406 mm 3.0 – 0.288675 × 2.598075 = 2.250001 mm

At location A, the cross-section area is π 2 2 A s = --- ( 24.0 – 0.9743 × 3.0 ) = 348.909 mm 4 We assume a stress concentration factor Kt = 2.4.5 Hence, the tensional stress at the thread root, as per Equation (10.23), equals 270, 000 S z = 2.4 × --------------------- = 1857 MPa 348.909 The maximum shear stress at location A, as per Equation (10.25), is 81, 000 τ = --------------------------------------------------------------- = 552 MPa π × 20.752406 × 2.250001

301

Threaded Fasteners

At location B, the tensional stress at the thread root is the same as at location A, i.e., 270, 000 S z = 2.4 × --------------------- = 1857 MPa 348.909 At location C, the concentration factor is Kt = 2.5.5 Hence, the tensile stress equals 4 × 270, 000 - = 1492 MPa S z = 2.5 × ----------------------------2 π × 24.0 The above results indicate that stresses are above the limits, causing the material at the thread root and at the intersection of head and shank to fail. The above formulas, however, do not provide data concerning the central part of the bolt. Analysis Based on Average Stresses A more practical analysis refers to the average stress in the threaded part of the bolt.1 For the design criterion, one uses the average tensional stress, ignoring the stress concentration caused by thread geometry and the shear stresses. The average tensional stress is expressed as F S z = ----As

(10.30)

where As is determined from Equation (10.24). For the bolt under consideration, the average tensile stress in the threaded section of the bolt equals 270,000 S z = ------------------- = 773.8 MPa 348.909 The proof stress of bolt material is 970 MPa, indicating that the bolt is within limits.

10.3.2 INACCURACY

OF

LINEAR ANALYSIS

The results obtained from Equations (10.23), (10.28), and (10.29), which were based on elastic analysis, are inaccurate because of two following facts. 1. The assumption of linear behavior. In the example shown, the bolt of class 12.9 has a yield point of 1,100 MPa. The computed stresses at locations A, B, and C are well above the yield point, which means that the material is subject to plastic deformation. 2. Ignoring the deformation of threads in the bolt and the nut. Due to the deformation, the contact between the bolt and nut becomes nonuniform, which changes the stress distribution considerably. Based on the above numerical example, one notes the advantage of designing a bolt based on the average tensile stress only,1 Equation (10.30). Using the averages allows

302

Nonlinear Problems in Machine Design

us to eliminate the plastic deformation from the computation, since it is limited to a restricted range at the roots of the thread and at the intersection of head and shank. Experience shows that a bolt so designed may be used safely, provided that the computed average stress is sufficiently below the yield point. A more detailed analysis that takes into consideration the plastic behavior of the material and the nonuniform contact pressure is presented in the next section.

10.4 NONLINEAR ANALYSIS USING FE METHOD Here, we analyze stresses in a bolted joint, subjected to preloading and subsequent fatigue loading, using the finite element method. We apply nonlinear solutions, taking into consideration the elastic-plastic behavior of the material and frictional contact phenomena in the thread. The solution process utilizes theoretical correlations and iterative techniques presented in Chapters 3, 4, and 6. The bolted joint analyzed here refers to a pressure vessel as shown in Figure 10.6. The bolt is standard, size M24 × 3, class 12.9, with a proof stress of 970 MPa (the same bolt analyzed in Section 10.3.1). The bolt is subjected to a tensional force of F = 270,000 N at preloading, with a subsequent varying force increment ∆Fb = 0–27,000 N. The material of class 12.9 is a chrome-molybdenum-nickel steel SAE 4340, heat treated to BHN = 350. Our computation is based on the monotonic properties of the material, because the bolt is subject to tension only (see explanation below). The properties of SAE 4340 steel, BHN = 350, are as follows: Ultimate tensile strength

Su = 1241 MPa

Yield point (monotonic)

Syp = 1172 MPa

Modulus of elasticity

E = 193,000 MPa

True fracture strength

σf = 1655 MPa

True fracture ductility

εf = 0.89

Strain hardening exponent

n

= 0.066

See Appendix C. The stress-strain diagram of the material is shown in Figure 10.13.

10.4.1 FE MODEL To facilitate the convergence of the iteration process and to save computational time and memory space, an axisymmetric model of the bolt and nut is introduced. Accordingly, the thread in the bolt and nut is presented as a set of concentric rings instead of a spiral coil. See Figure 10.14. This fact introduces a computational error. The stressed area in our model equals π π 2 2 2 A s = --- d 1 = --- 20.752406 = 338.241 mm 4 4 while the stressed area in a spiral thread is

303

Threaded Fasteners

σ, MPa

ε FIGURE 10.13 Stress-strain curve of steel SAE 4340, heat treated to BHN = 350.

FIGURE 10.14 Representation of thread coils by concentric rings: (a) bolt with spiral thread and (b) spiral thread replaced by concentric rings.

π π 2 2 2 A s = --- ( d – 0.9743 × p ) = --- ( 24.0 – 0.9743 × 3.0 ) = 348.909 mm 4 4 It follows that the computed stresses in concentric rings will be somewhat higher than stresses in a spiral thread. (In a strictly elastic analysis, the error will be about 3%. In elastic-plastic analysis, it is considerably less.)

304

Nonlinear Problems in Machine Design

The geometrical model of the bolted joint is shown in Figure 10.15. It comprises three parts: bolt, nut, and flange. The bolt in the model is limited to a shank without the head. The head including part of the shank is the subject of a separate analysis that follows. Since the model is axisymmetric, all parts, including the nut, have cylindrical forms. The dimensions of the parts are as follows: Bolt, diameter, including shank Nut diameter Nut height Plate thickness

24 40 24 20

mm mm mm mm

The loading of the bolted joint is shown in Figure 15b. The load is applied at the bottom of the shank in the form of distributed tension. The load is transferred through contacts, first to the nut and then into the flange. The bottom of the flange is fixed in the perpendicular direction to the surface and moves freely in the tangential direction. The details of contact between the threads is shown in Figure 10.16: the bottom surfaces of the bolt’s thread are in contact with the top surfaces of the nut’s thread. Another contact, between the bottom of the nut and the top of the flange, is shown in Figure 10.15b. Figure 10.17 presents the FE model. The model is created as two-dimensional and axisymmetric. All the parts are constructed from quadrilateral and triangular elements. The contacts between the respective parts is represented by contact elements with a friction coefficient µ = 0.14.

FIGURE 10.15 Geometric model of the bolted joint: (a) axonometric view and (b) details of boundary conditions.

Threaded Fasteners

FIGURE 10.16 Contact between threads.

FIGURE 10.17 FE model of bolt and nut.

305

306

Nonlinear Problems in Machine Design

The FE mesh is drawn to accommodate the anticipated stress distribution. Wherever high stresses and steep stress gradients are expected, a denser mesh is provided. We expect the highest stresses in the thread at the lowest engaged coils (location A in Figure 10.15), while the lower stresses are in the central part of the bolt and at the outer part of the nut; there are large elements in the center of bolt and outer part of nut, and small elements in the threads. The densest mesh is at the lowest engaged threads, as shown in an enlarged view in Figure 10.18. Two versions of analysis are to be performed. One version utilizes the ANSYS program, which applies the two-dimensional axisymmetric model as shown in Figures 10.17 and 10.18. The second version, used for comparison, utilizes the MSC.NASTRAN program with a three-dimensional model, shown in Figure 10.19. The elements are hexahedrons and prisms. The model is created as a section (1/36) of full circle, due to the axial symmetry.

10.4.2 SOLUTION: FIRST STAGE, PRELOADING The solution process of the problem comprises two stages: computing stresses and strains, first at preloading and then at the subsequent fatigue loading. As stated in the problem’s data, at preloading the tensional force is F = 270,000 N. The static nonlinear solution comprises 10 incremental loading steps; at each step the quasiNewton method is used with 4 to 10 iterations. The results are presented in Figures 10.20 through 10.34. Let us refer to the figures. Figure 10.20 shows the computed distribution of contact forces between the respective threads. According to the data, the largest force occurs at bottom coil, where it equals f 1 = 2007 × 36 = 72, 252 N

FIGURE 10.18 FE model, thread details.

307

Threaded Fasteners

FIGURE 10.19 FE model of the bolt used in MSC.NASTRAN program.

This amounts to about 27% of the total load, i.e., f 1 = 0.2676F

(10.31)

as opposed to 30% obtained in the approximate analysis (see Section 10.3.1). Figure 10.21 shows the distribution of von Mises stresses in the bolt, obtained by the ANSYS program. The highest stresses occur at locations A and B, as predicted in Section 10.3.1. Figures 10.22 and 10.23 present stress distributions at those locations. The maximum von Mises stress appears at location A, at the internal corner, where it reaches Se = 1187.3 N (see Figure 10.22). At location B, the stress reaches Se = 1143 N (see Figure 10.23). The shadowed areas in both figures indicate plastic domains. (Although stress Se = 1143 N is below the yield point, it is above the proportional limit, marking the onset of plastic deformation.) A similar analysis performed with MSC.NASTRAN program produces slightly different results with maximum von Mises stress at location A, where it equals Se = 1214 N (a difference of 2%) (see Figure 10.24). Figure 10.25 shows von Mises stresses in the nut. The

308

Nonlinear Problems in Machine Design

FIGURE 10.20 Contact force distribution.

FIGURE 10.21 Von Mises stress distribution in bolt computed with ANSYS program.

Threaded Fasteners

309

FIGURE 10.22 Von Mises stress distribution at location A.

ANSYS 5.5.1 MAR 29 1999 08:01:21 NODAL SOLUTION STEP= 1 SUB= 119 TIME= 1 SEQV˚˚˚˚(AVG) DMX= .200766 SMN= 150.524 SMX= 1143 A= 221.42 B= 363.212 C= 505.004 D= 646.795 E= 788.587 F= 930.379 G= 1072

FIGURE 10.23 Von Mises stress distribution at location B.

310

Nonlinear Problems in Machine Design

FIGURE 10.24 Von Mises stress distribution at location A computed with MSC.NASTRAN program.

FIGURE 10.25 Von Mises stress distribution in the nut computed with ANSYS program.

Threaded Fasteners

FIGURE 10.26 Von Mises stress distribution at the bottom thread of the nut.

FIGURE 10.27 Designation of cross sections in bolt and nut.

311

312

Nonlinear Problems in Machine Design

FIGURE 10.28 Location of critical cross sections in bolt.

FIGURE 10.29 Von Mises and principal stress distribution in cross section aa.

Threaded Fasteners

FIGURE 10.30 Shear stress distribution in cross section bb.

FIGURE 10.31 Location of critical cross sections in the nut.

313

314

Nonlinear Problems in Machine Design

FIGURE 10.32 Von Mises and principal stress distribution in cross section cc.

FIGURE 10.33 Shear stress distribution in cross section dd.

315

Threaded Fasteners

FIGURE 10.34 Axial stress across the critical cross section as a function of loading computed with MSC.NASTRAN program.

maximum stress appears at the internal corner of the bottom coil, where it reaches Se = 1029 N. See Figure 10.26. Figures 10.27 through 10.30 show the stress distributions in critical sections, including von Mises, principal, and shear stresses. Figures 10.27 and 10.28 indicate the location of sections aa and bb in the bolt, while Figures 10.29 and 10.30 show the stress distributions in the stated sections. Likewise, Figures 10.27 and 10.31 indicate the location of sections cc and dd in the nut, while Figures 10.32 and 10.33 show the stress distributions in the stated sections. To summarize the results in the bolt, the highest stresses are Von Mises stress

Se = 1187.3 MPa

First principal stress

S1 = 1713.0 MPa

Maximum shear stress

τrz = 664.7 MPa

The value S1 = 1713.0 MPa is well above the yield point, which may be misleading, since von Mises stress at this location is next to the yield point. To summarize the results in the nut, the highest stresses are

316

Nonlinear Problems in Machine Design

Von Mises stress

Se = 1028.7 MPa

First principal stress

S1 = 1284.9 MPa

Maximum shear stress

τrz = 585.1MPa

Stress-Concentration Factor Based on computational results, one can check the accuracy of stress concentration factors that were obtained by photoelastic methods, see Peterson.8 We find that the stress concentration factor changes when the peak stress reaches a plastic domain (a fact that cannot be proved by photoelastic tests). Figure 10.34 presents the computed distribution across the section aa of axial stresses Sz for various levels of loading. The shown stress concentration factors are computed using the formula S z,peak K f = -----------S z,aver

(10.32)

Here, we use fatigue factor Kf in lieu of theoretical factor Kt, due to the occurring plasticity. The results are shown in Table 10.1. TABLE 10.1 Stress Concentration Factors Fraction of full load

Sz,peak (MPa)

Domain

Sz,aver (MPa)

Kf

0.2

731.1

elastic

159.2

4.592

0.4

1215.8

plastic

318.3

3.820

0.6

1367.4

plastic

477.5

2.864

0.7

1413.2

plastic

557.1

2.537

0.8

1460.5

plastic

636.6

2.294

0.9

1566.3

plastic

716.2

2.186

1.0

1650.0

plastic

795.8

2.073

One notes that, in the elastic domain, the stress concentration factor equals Kf = 4.6, which compares well with Kf = 4.8 quoted by Peterson.8 In the plastic domain, Kf varies considerably, depending on the magnitude of stresses. It appears that, with rising average stresses, the concentration factor in the plastic domain decreases.

10.4.3 SOLUTION: SECOND STAGE, FATIGUE LOADING According to the problem statement, in the second stage, the bolt is subjected to fatigue loading by load increments ∆F = 27,000 N. Figure 10.35 presents the loading history with the force varying between Fmin = 270,000 N and Fmax = 297,000 N. Our analysis was limited to four loading cycles. For the solution, we used the MSC.NASTRAN program. The nonlinear solution involved 5 loading steps, at each cycle, with 4 to 10 iterations at each step, applying the modified Newton–Raphson method. The computational results are shown in Figures 10.36 through 10.40. Figure

317

Threaded Fasteners

FIGURE 10.35 Loading history of the bolt.

10.36 presents plots of von Mises stresses and strains versus time at the critical location with the highest stresses, while Figure 10.37 shows a plot of stresses versus strains. The respective maxima and minima during the loading and unloading process are listed in Table 10.2. TABLE 10.2 Stress and Strain Cycling Cycle

Se,min (MPa)

εe,min

Se,max (MPa)

εe,max

Se,mean (MPa)

1

1223.3

0.0470584

1240.8

0.0569077

1232.1

2

895.3

0.0545450

1220.8

0.0567725

1058.2

3

887.0

0.0544879

1232.8

0.0568537

1059.9

4

888.2

0.0544964

1234.1

0.0568629

1061.2

One notes a large relaxation of stresses after the first cycle. The plot of the mean stress shows a drastic decrease from 1232 MPa to 1061 MPa, after which it appears to stabilize. Figures 10.36 to 10.38 show that the behavior becomes fully elastic after the stresses are stabilized. Stress-Concentration Factor The results for fluctuating load allow us to compute stress concentration factors at fatigue loading. The stress range of the alternating stress at the fourth cycle is

318

Nonlinear Problems in Machine Design

FIGURE 10.36 Von Mises stress and strain vs. time.

∆S = ∆S e,max – ∆S e,min = 345.9 MPa The corresponding change of load equals ∆F = F max – F min = 27000 N and the range of the average axial stress in the thread is 27000 ∆F ∆S aver = ------- = ---------------- = 77.4 MPa 348.91 As

319

Threaded Fasteners

FIGURE 10.37 Stress-strain diagram, von Mises stresses and strains.

FIGURE 10.38 First principal stress variation vs. time.

Thus, the stress concentration factor at fatigue loading becomes 345.9 ∆S K f = ------------- = ------------- = 4.47 77.4 ∆S aver

320

Nonlinear Problems in Machine Design

FIGURE 10.39 Distribution of the first principal stress in the vicinity of thread: (a) stress distribution and (b) location of the highest stresses.

321

Threaded Fasteners

FIGURE 10.40 Principal stress distribution in cross section aa.

We find that the stress concentration factor equals the stress concentration factor for static loading, in purely elastic state at low loading (see Table 10.1). Life Expectancy From the stress and strain data, one can compute the life expectancy of the bolt. In the following, we use the Morrow’s equation, σ f ′ – σm ∆ε b c - ( 2N f ) + ( 2N f ) ------ = ------------------E 2

(10.33)

See Chapter 7. Let us repeat the cyclic properties of steel SAE 4340, BHN = 350: Fatigue strength coefficient

σ f ′ = 1655 MPa

Fatigue strength exponent

b = –0.076

Fatigue ductility exponent

c = –0.62

The mean stress equals σm = 1061 MPa. The strain range is ∆ε = 0.0568629 – 0.0544964 = 0.0023665 With the above data, Equation (10.33) becomes 1655 – 1061 0.0023665 – 0.076 – 0.62 + ( 2N f ) ------------------------- = ------------------------------ ( 2N f ) 193, 000 2

322

Nonlinear Problems in Machine Design

Solving the above equation, we obtain life expectancy Nf ≈ 1,000,000 cycles. The number of cycle of that magnitude may be considered as infinite life expectancy. The above conclusion can be confirmed by referring to Figure 10.37, the plot of stresses versus strains. In the figure, the cyclic hysteresis becomes a single line, AB, indicating that there is no plastic work during the cyclic process, which means an infinite life expectancy. However, one should note that there may exist highly localized, tiny cracks that eventually may lead to a fatigue failure.9 See the following section. Crack Propagation in Thread Consider the impact of surface imperfections on the life expectancy of the bolt. When there is a surface indentation causing a stress intensity at greater than the threshold value, ∆K > ∆K th , a crack will start to propagate. See Chapter 7. The threshold equals ∆K th = α∆S πa th

(10.34)

ath is the size of an initial crack that starts to propagate under the existing stress. Stress range ∆S at the critical point equals ∆S = S 1,max – S 1,min

(10.35)

See Figure 10.38. The distribution of S1 in the vicinity of the thread is shown in Figure 10.39. Point A is the location of the highest stress value in the whole bolt, while point B is the location of the highest stress value on the surface. A cross-sectional plot in Figure 10.40 shows the stress behavior in the plastic zone. Table 10.3 presents the maxima and minima of S1 at point B, during cycling. TABLE 10.3 Cycling of First Principal Stress Cycle

S1,min (MPa)

S1,max (MPa)

1

1549.1

1570.7

2

1150.1

1541.4

3

1134.1

1553.3

4

1134.2

1555.2

Point B at the surface is a potential source for a crack. The stress range and stress ratio at that point, after the cycling process is stabilized, equal ∆S = 1555.2 – 1134.2 = 421.0 MPa S 1,min 1134.2 R = ------------ = ---------------- = 0.73 1555.2 S 1,max

323

Threaded Fasteners

The value of threshold stress intensity of steel SAE 4340, BHN 350, at R = 0.73 is 0.5

∆K th = 2.5 ( MPa ⋅ m ) See Appendix C. Consequently, the size of an indentation at point B that may cause crack propagation equals 2 2 2.5 1 1 ∆K th = ---  ------------------------------ = 0.0000089 m a th = ---  ----------π  1.12 × 421.0 π  α∆S 

Now, harmful surface irregularities stem mainly from manufacturing processes. Threads in bolts are made either by a cutting operation or rolling. Consider first threads made by rolling. The surface roughness in a rolling operation is in the range Ra = 0.08−0.63 µm. The expected surface irregularity will be of the size a i = 4.5 × 0.63 = 2.84µm = 0.00000284 m Since a i < a th here, the crack will not propagate, and we expect an infinite life. On the other hand, when the bolt is manufactured by cutting (turning), the surface roughness is in the range Ra = 0.32−3.2 µm (depending on the size of the bolt). Thus, a thread with a roughness of 3.2 µm may have a local surface indention of a size a i = 4.5 × 3.2 = 14.4µm = 0.0000144 m Here, since a i > a th , a crack may start to propagate. To compute the number of cycles to the final fracture, we use the crack propagation rates for steels SAE 4340 and 4140, as presented in Figure 10.41. In the figure, lines R = 0 and R = 0.1 are taken from SAE Fatigue Design Handbook,10 while the line R = 0.73 is extrapolated using the Forman equation, m

A ( ∆K ) da ------- = -------------------------------------dN ( 1 – R )K c – ∆K as follows (see Chapter 7). By extrapolation, we obtain constants A and m, A = 5.155 × 10

–9

and m = 2.25

Fracture toughness Kc for steel SAE 4340, BHN 350, equals 0.5

K c = 110.0 ( MPa ⋅ m ) After inserting A, m, Kc , and R = 0.73 into Equation (10.36), we obtain

(10.36)

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Nonlinear Problems in Machine Design

FIGURE 10.41 Crack propagation rate in steels SAE 4340 and 4140.

–9

2.25

5.155 × 10 × ( ∆K ) da ------- = -------------------------------------------------------( 1 – 0.73 ) × 110 – ∆K dN A plot of the resulting line with R = 0.73 is presented in Figure 10.41. The crack propagation rate with R = 0.73 can be expressed by Paris equation, n da n ------- = C ( ∆K ) = C ( α ∆S πa ) dN

(10.37)

where constants C an n are extrapolated from the line R = 0.73 as C = 1.6 × 10

– 10

and n = 2.45

The number of cycles to failure is obtained by integration of Equation (10.37),

325

Threaded Fasteners

  1 1  1 – ---------N f = --------------------------------------------------  ---------n  n --- – 1  n--- – 1 C ( α ∆S π ) n  --2- – 1 2 af  ai 2 

(10.38)

The length of a crack at failure equals 2 2 110.0 1 1 Kc  - = ---  ------------------------------ = 0.0173m a f = ---  ---------    π 1.12 × 421.0 π α∆S

Hence, the estimated number of cycles to failure is 1 1 1  --------------------------------- – ------------------------ N f = ----------------------------------------------------------------------------------------------2.45  0.225 0.225 – 10 0.0000144 0.0173  0.225 × 1.6 × 10 ( 1.12 × 421.0 π ) = 18, 860cycles Comparing the results for rolled and cut threads, we understand why, in practice, bolts with rolled threads are preferred over bolts with cut threads. (It should be noted that, in machine design, when high loads are present, only rolled threads are used.) Rolled threads have additional advantage: residual compressive stresses at the surface constitute another safeguard against crack propagation.

10.4.4 STRESSES

AT THE

BOLT HEAD

Let us analyze the conditions at the bolt head. We focus at the intersection of the head and the shank as a possible failure location. Conforming to our assumption of axial symmetry, we assume a cylindrical form for the bolt head. The FE model of the head and the shank is shown in Figure 10.42. The dimensions are as follows: Head diameter

40 mm

Head height

24 mm

Shank diameter

20 mm

Radius of fillet

1.2 mm

As before, at preloading the bolt is subjected to a load of F = 270,000 N, with a subsequent fatigue loading increment ∆Fb = 27,000 N. Preloading The results of computation are shown in Figures 10.43 and 10.44. Figure 10.43 shows von Mises stress distribution in the vicinity of fillet, which is critical. The highest stress is 1185 MPa, which practically equals the yield point (1172 MPa). This indicates that plastic deformation is practically negligible. Figure 10.44 presents plots of principal and Cartesian stresses in critical cross section across the shank.

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Nonlinear Problems in Machine Design

FIGURE 10.42 FE model of bolt head.

FIGURE 10.43 Von Mises stress distribution in critical location of bolt head, computed with MSC.NASTRAN program.

Threaded Fasteners

327

FIGURE 10.44 Stress distribution in critical cross section of bolt head.

Fatigue Loading Our analysis, which included four loading cycles, was based on using MSC.NASTRAN program and applying the modified Newton–Raphson method. The results of computation are shown in Figure 10.45 and Table 10.4, both of which present the history of von Mises stress and strain cycling. As before, in bolt-nut analysis, we encounter the phenomenon of stress relaxation: the mean stress decreases from 1116 to 1041 MPa after the fourth cycle.

328

Nonlinear Problems in Machine Design

FIGURE 10.45 Von Mises stress and strain variation vs. time in the critical location of the bolt head.

TABLE 10.4 Stress and Strain Cycling Cycle

Se,min (MPa)

εe,min

Se,max (MPa)

εe,max

Se,mean (MPa)

1

1107.6

0.0151878

1125.2

0.0177960

1116.4

2

958.8

0.0166684

1125.3

0.0178109

1042.1

3

956.5

0.0166675

1125.1

0.0178100

1040.8

4

956.3

0.0166669

1125.0

0.0178098

1040.7

329

Threaded Fasteners

To compute the life expectancy, we use the value of mean stress σm = 1041 MPa. The strain range is ∆ε = 0.0178098 – 0.0166669 = 0.0011429 Inserting the above data into Equation (10.33), we obtain 1655 – 1041 0.0011429 – 0.076 – 0.62 + ( 2N f ) ------------------------- = ------------------------------ ( 2N f ) 193, 000 2 9

Solving the above equation, we find the life expectancy to be N f > 1.0 × 10 cycles. Since the numerical result is more than one million cycles, we may assume an infinite life expectancy. Crack Propagation To check the likelihood of crack propagation, consider the cycling of first principal stress (Figure 10.46 and Table 10.5). As seen in the figure, the stresses relax after the first cycle. Upon relaxation, according to Table 10.6, the stress range equals ∆S = 1423.5 – 1229.1 = 194.4 MPa

FIGURE 10.46 First principal stress variation vs. time in the critical location of the bolt head.

330

Nonlinear Problems in Machine Design

TABLE 10.5 Cycling of First Principal Stress Cycle

S1,min (MPa)

S1,max (MPa)

1

1398.8

1423.4

2

1231.9

1423.9

3

1229.3

1423.6

4

1229.1

1423.5

The stress ratio is S 1,min 1229.1 - = ---------------- = 0.86 R = -----------1423.5 S 1,max For lack of exact data, we use a conservative approach and assume the value of threshold stress intensity to be R = 0.73. 0.5

∆K th = 2.5 ( MPa ⋅ m ) The corresponding surface indent, which may induce crack propagation, equals 2 2.5 1 1 ∆K th = ---  ------------------------------ = 0.0000419 m a th = ---  ----------π  1.12 × 194.4 π  α∆S  2

The surface irregularity in the rolling operation is a i = 4.5 × 0.63 = 2.84µm = 0.00000284 m whereas in the cutting operation it equals a i = 4.5 × 3.2 = 144µm = 0.0000144 m In both cases, a i < a th , which means there is no danger of crack propagation. However, even with the above results, if there is corrosion or mechanical damage of the surface, harmful irregularities may occur that could cause crack propagation.

10.5 CONCLUSIONS The case examined above combines two kinds of nonlinearities: frictional contact and plastic deformation. The FE analysis finds the following: 1. Life expectancy prediction. This is achieved while considering the existing plastic deformation in the roots of threads. No accurate prediction can be achieved without fully considering plastic deformation.

Threaded Fasteners

331

2. Stress concentration factor. The solution revealed that, under static load, the factor depends on the loading level; as the load increases, the plastic zones in the roots of the thread expand, causing lower stress concentration. 3. Surface irregularity. The results for life expectancy are not fully dependable because of extraneous circumstances in real life. The computed life expectancy of the given bolt in fatigue loading approaches infinity in spite of the high loading. In reality, however, a surface irregularity of a microscopic magnitude (0.01 µ) could easily initiate crack propagation, causing fatigue failure. Such surface irregularity could be caused either by improper manufacturing methods or corrosion. In accepting the above FE solution, one must take into consideration the following: 1. The accuracy of the computation depends on the precision of the FE model. 2. It depends on the precision of the iterative process to solve nonlinearity. 3. Computation of contact involves penalty parameters that may lead to numerical errors, as discussed in Chapter 6. Evading the errors may cause unstable solution processes that do not converge, thus making the problem insoluble.

REFERENCES 1. SAE Handbook, Society of Automotive Engineers, Warendale, PA, 1989. 2. Niemann, G., Maschinenelemente, Vol. 1, Springer Verlag, Berlin, 1975 (in German). 3. Roetscher, F., Die Maschinenelemente, vol. 1, Springer Verlag, Berlin, 1927 (in German). 4. Birger, I.A., and Yosilevich, G.B., Thread and Flange Connections, Machinostroyenye, Moscow, 1990 (in Russian). 5. Osgood, C.C., Fatigue, 2d edition, Pergamon Press, Oxford, 1980. 6. Miller, D.L., Marshek, K.M., and Naji, M.R., “Determination of load distribution in threaded connections,” Mechanism and Machine Theory, 18, pp. 421–430, 1983. 7. Black, P.H., and Adams, O.E., Jr., Machine Design, McGraw-Hill, New York, 1968. 8. Peterson, E.E., Stress Concentration Factors, John Wiley, New York, 1974. 9. Sandor, B.I., Fundamentals of Cyclic Stress and Strain, University of Wisconsin Press, Madison, Wisconsin, 1972. 10. Fatigue Design Handbook, AE-10, 3rd edition. Society of Automotive Engineers, Warendale, PA, 1997.

11

Flange Connection

11.1 INTRODUCTION In the assembly of high-pressure vessels, the design of flanges requires nonlinear analysis. Our analysis here focuses on the interaction of bolts and flanges and the forces and deformation occurring when the connection is subjected to pressure loading. Consider a flange connection, a system comprising two flanges with a gasket between them, held together by bolts (see Figure 11.1). The loading history is shown in Figure 11.2 in schematic form. The loading begins with the assembly, when the bolts and nuts are closed manually. This is followed by the application of torque to tighten bolts, thus introducing an initial strain in bolts and flanges. Final, the flange connection is subjected to an internal pressure load. As the system is pressurized, the parts become deformed, and the contact between the flanges and the gasket recedes. At a critical pressure, the flanges separate, and leakage takes place. The system’s behavior poses two interesting engineering problems, which are given a special attention below. One problem concerns flanges’ disengagement under pressure, while the other relates to the expansion of flanges at a rate different from that

FIGURE 11.1 Bolted flange connection. 333

334

Nonlinear Problems in Machine Design

FIGURE 11.2 Flange connection at three consecutive steps: (a) after assembly, (b) at an intermediate pressure load, and (c) at the leaking point.

of the gasket, which causes sliding with friction. The solution to both problems: the diminishing contact and the friction require a nonlinear analysis and complex numerical computations. The two objectives to be achieved include defining the bolt forces caused by loading and predicting the critical pressure that is present as the flange connection starts to leak.

11.2 ONE-DIMENSIONAL ANALYSIS The deformation and forces caused by loading are analyzed assuming that the two flanges, the gasket, and the bolts act in accordance with the spring analogy: they are replaced by one-dimensional springs with forces and deflections in one direction only.

11.2.1 LINEAR SOLUTION Consider a linear approximation according to Roetscher.1 We assume that the contact domain, including the area between the flanges and the gasket, is constant and without friction. The choice of such constant areas is discussed by Osgood,2 Osman,3 Motosh,4 and Shigley and Mischke.5 The one-dimensional model is shown in Figure 11.3. (For a review of the basics of the one-dimensional theory, refer to Chapter 10.) It is assumed that the load is applied at the outer surfaces of the flanges. Accordingly, the force-deflection functions of bolts, flanges, and gasket are represented by linear functions. Figure 11.4 presents a corresponding diagram. The force-deflection line of the bolt begins at point A and continues at a constant slope kb. Ab E b k b = ----------lb

(11.1)

The term kb is the bolt stiffness. The behavior of the flanges and gasket is expressed in a combined single function. The joint force-deflection line of the flanges and gasket begins at point B and continues at a constant slope kfg. d –1 l k fg =  ------------ + ------------  A f E f A g E g

(11.2)

335

Flange Connection

FIGURE 11.3 One-dimensional model of a flange connection.

FIGURE 11.4 Load deflection diagram according to Roetscher.

Af is a hypothetical flange area subjected to compression. It is assumed to be constant. Ag is the matching gasket area. (Index f denotes the flange, while index g denotes the gasket.) kfg denotes the stiffness of the flange–gasket system. In the linear approximation, stiffness kfg is constant. As the pressure load Fp rises, the bolt load Fb increases correspondingly. Accordingly, the correlation between bolt load Fb and pressure load Fp is kb -F F b = F i + ----------------k b + k fg p

(11.3)

336

Nonlinear Problems in Machine Design

On the other hand, the correlation between the load acting upon the flanges Fc and pressure load Fp equals k fg -F F c = – F i + ----------------k b + k fg p

(11.4)

Fc equals the compression force acting upon the gasket. When the pressure load Fp reaches a critical value, the compression force Fc is reduced to zero, the flange connection becomes disengaged, and leakage takes place. The critical pressure load Fp = Fcrit is defined by the equation k fg -F F i = ----------------k b + k fg crit

(11.5)

An alternate description of the bolt and flange behavior is derived by a graphical representation of the derivatives kb ∂F --------b- = ----------------k b + k fg ∂F p

(11.6)

k fg ∂F ---------c = ----------------k b + k fg ∂F p

(11.7)

and

See Figure 11.5. In lieu of force versus deflection, the figure depicts force versus the pressure load. Line ABC represents the tensional force in the bolt, while DE is the compression force between the flanges and the gasket. The two points B and E indicate the onset of leakage, which depends on the stiffnesses of bolt, flanges, and gasket. Of practical importance in the design of flange connections is the following case. The flanges and gasket are thick and rigid, so we can neglect bolt’s stiffness, i.e., k b /k fg ≈ 0 . The respective gradients here can be expressed as kb -----------------≈0 k b + k fg

(11.8)

k fg -----------------≈1 k b + k fg

(11.9)

According to Equations (11.3) and (11.4), we then have F b ≈ F i ; F c ≈ –F i + F p

(11.10)

Flange Connection

337

FIGURE 11.5 Forces in flange connection vs. pressure load, linear approximation.

At the critical pressure F b ≈ F p ≈ F crit , the flanges become disengaged, and leakage takes place. See Figure 11.6. After leakage occurs, the respective forces are F b ≈ F p and F c ≈ 0 . The assumption of a rigid flange and gasket may often help to predict the behavior of a flange connection. See Juvinall et al.6

11.2.2 NONLINEAR FLANGE BEHAVIOR Consider again the loading history of the flange connection shown in schematic form in Figure 11.2, presented now in detail in Figure 11.7. The contact area changes in three consecutive steps: Step 1. Step 2. Step 3.

Full contact between the flange and the gasket, in a manual assembly, marked by MN Diminished contact M′N after bolt tightening Further receding of contact M″N , due to pressure

At a critical pressure, area M″N is reduced to zero, and leakage takes place. The flange and the gasket expand at a different rate so that contact area M″N is subject to sliding friction. As the contact recedes, the newly created opening becomes accessible to pressure, which causes a further deformation of the flanges while the contact force between the flanges and gasket diminishes. In effect, stiffness decreases in the flanges and the gasket.

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Nonlinear Problems in Machine Design

FIGURE 11.6 Bolt force vs. pressure load, rigid flange.

FIGURE 11.7 Receding contact in flange connection.

Were we to present the above described phenomena in a one-dimensional form and apply Equations (11.3) and (11.4) to the nonlinear analysis, we would find that the stiffness of flange-gasket system is a function of the applied pressure. ∂k fg -0 ----------2- = ------  ---------------- ∂p k + k fg b ∂p

(11.12)

and 2

∂ Fc ∂ k fg -