Nonlinear Problems in
MACHINE
DESIGN
Nonlinear Problems in
MACHINE DESIGN Eliahu Zahavi David Barlam
CRC Press Boca Raton London New York Washington, D.C.
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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 621.8′15—dc21 00-063033 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
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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.
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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).
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Nomenclature a, b, c a a af ai ai ai, bi,ci
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) unknown parameters in Galerkin and Rayleigh-Ritz approximation ak (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) 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)
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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
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r 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)
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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)
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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)
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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)
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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)
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T Tsk , T sk ′ TT T–1 T(3) T(4) Tε Tσ
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)
0
TS 1 s
matrix of tangential vectors (Chapter 6) 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
Wi
d i
W X, Y, Z Xi X1, X2, X3 Y Y0 Y(f) α α α, β, γ αmn β β γ γe
internal dilatation (volumetric) specific work (Chapter 1) 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)
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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
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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
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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)
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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
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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
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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
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Part I Theoretical Fundamentals
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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
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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).
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Basics of Solid Mechanics
3
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,
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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)
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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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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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FIGURE 1.7 Stress components acting upon an elemental cube: projection on xy-plane.
x xy σ + ∂σ – σ dydz + τ + ∂τ – τ dzdx --------dx ---------dy x xy x ∂x xy ∂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 ∂τ xy ∂τ zx --------x + --------+ --------- + X = 0 ∂x ∂y ∂z ∂τ xy ∂σ y ∂τ yz --------- + -------- + --------- + Y = 0 ∂x ∂y ∂z ∂τ zx ∂τ yz ∂σ z --------- + --------- + -------- + Z = 0 ∂x ∂y ∂z
(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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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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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 σ ϕ = ---------------- + ----------------- cos 2ϕ + τ xy sin 2ϕ 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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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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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 2 σ 1 – σ 2 2 2 2 1 + σ 2 σ – 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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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 σ 2 – σ 3 ,τ = σ 3 – σ 1 ,τ = σ 1 – σ 2 τ 1 = ---------------- 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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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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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)
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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
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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
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s = =
17
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.
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R + dR = r + dr + u + du
(1.53)
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
u = [u v w] ,
r = [ x y z]
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)
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1 3 T ε = --- 2
19
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. 1 2 3 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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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 γ -----zx- -----zy2 2
γ xz -----2 γ yz -----2 εz
(1.63)
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21
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 xy γ ( ε x – ε ) cos α + -----cos β + -----xz- cos γ = 0 2 2 γ yx γ ------ 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 zx γ ----------zy2 2
γ -----xz2 γ yz -----2
= 0
εz – ε
FIGURE 1.18 Deformation of an infinitesimal cube in principal directions.
(1.65)
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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 V –V ( 1 + ε1 ) ( 1 + ε2 ) ( 1 + ε3 ) – 1 ϑ = ----------------0 = ----------------------------------------------------------------V0 1 = 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.
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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 zx γ zy ------ -----2 2
γ xz -----2 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)
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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)
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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 0 0 0 2(1 + ν) γ xy 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.
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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)
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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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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)
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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.
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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 2 1 2 2 2 --- σ e = --- [ ( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) ] 3 6
(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:
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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 0 – νε e – --------------------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)
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1 3 2 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.
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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 3 1 W i = --- σ m ε m = --- ( σ 1 + σ 2 + σ 3 ) ( ε 1 + ε 2 + ε 3 ) 2 6
(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 ) 2 6
(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)
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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)
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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 ∂U ---------i δε x + -------- δε + ---------i δε z ∂ε y y ∂ε z ∂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
∂δu ∂δv ∂δw ∂δv ∂δw σ x --------- + σ y --------- + σ z ---------- + τ yz --------- + ---------- ∂z ∂x ∂y ∂z ∂y ∂δw ∂δu ∂δu ∂δv +τ zx ---------- + --------- + τ xy --------- + --------- ∂x ∂y ∂z ∂x
dx dy 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
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δU i = – ∫ ∫ ∫
∂σ ∂τ xy ∂τ zx ∂τ xy ∂σ y ∂τ yz δu --------x + --------+ --------- + δν --------+ -------- + --------- + ∂x ∂x ∂y ∂z ∂y ∂z ∂τ ∂σ ∂τ + δ w --------zx- + --------yz- + --------z ∂x ∂y ∂z
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
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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 ∂x ∂y
∂ δε y = -----δv ∂y
∂ ∂ δγ yz = -----δw + -----δv ∂z ∂y
∂ δε z = -----δw ∂z
∂ ∂ δγ zx = -----δu + ------ δw ∂z ∂x
(1.129)
The virtual work performed by stresses along the virtual strains is ∂W io ∂W io ∂W io - δε y + ----------- δε ------------ δε x + ----------∂ε x ∂ε y ∂ε z z δW i = ∫ ∫ ∫ ∂W ∂W io ∂W io V io - δγ xy + ----------- δγ yz + ----------- δγ + ---------- ∂γ xy ∂γ yz ∂γ zx zx
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)
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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.
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du ε ( x ) = -----dx
(1.137)
d du ------ 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:
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d du ------ 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 d
du
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
∫
d du -----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
d f k( x) df s ( x ) - ∑ a s ------------- dx + ∫ q ( x ) f k ( x ) dx = 0 – ∫ EA --------------dx dx
(1.147)
Taking into consideration boundary conditions, as per Equation (1.139), Equation (1.147) becomes
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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.
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x2 – x f 1 ( x ) = --------------at x 1 ≤ x ≤ x 2 x2 – x1 x–x x3 – x f 2 ( x ) = ---------------1 at x 1 ≤ x 2 ; f 2 ( x ) = --------------at x 2 ≤ x ≤ x 3 x2 – x1 x3 – x2 ………… ………… ………… 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)
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Basics of Solid Mechanics
43
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 k ∑ ∂a k
= 0
(1.157)
K=1
The condition of stationarity now becomes ∂Π -------- = 0 ∂a k
k = 1, 2, … n
(1.158)
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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 d du ------ 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
1 du 2 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.
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45
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 = ∑ ------∂u k k
k=1 l
l
n
∫ EA ( x ) 0
d f k( x) ∑ uk ---------------dx
k=1
n
d f r( x)
-δu r ∑ --------------dx
n
– ∫ q ( x ) ∑ [ f k ( x )δu k ] dx – Qδu n = 0 0
dx
r=1
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.
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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 ∂x ∂y ∂z ∂τ xy ∂σ y ∂τ yz --------+ -------- + --------- + Y = 0 ∂x ∂y ∂z ∂τ zx ∂τ yz ∂σ z --------- + --------- + -------- + Z = 0 ∂x ∂y ∂z
(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∫ ------- ∂x ∂y ∂z
(1.173)
= 0
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∫ -----------------∂x ∂y ∂z from which we obtain
=
∫ S∫ p x v x dS 1
(1.175)
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47
∂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 ------∂y ∂z ∂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 ∂x ∂y ∂v ∂v ∂v ∂v ∂v ∂v + τ xy -------x + -------y + τ yx -------y + -------z + τ zx -------z + -------x ∂y ∂x ∂z ∂y ∂x ∂z
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).
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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).
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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
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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 x ∂τ xy -------- + --------- + X = 0 ∂x ∂y ∂τ xy ∂σ y --------- + -------- + Y = 0 ∂x ∂y
(2.1)
and the strain–displacement relation are
FIGURE 2.2 Common elements: (a) quadrilaterals and triangles, (b) hexahedrons and tetrahedrons, and (c) prisms.
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51
∂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 τ E 13 E 23 E 33 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 ----∂x ∂y 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
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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)
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53
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)
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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
v dx dy = 0
(2.20)
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 ∂x ∂y ∂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)
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55
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
D N = B
(2.24)
T
D v = Bv n
(2.25)
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)
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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)
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Finite Element Method
57
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).
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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)
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59
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)
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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,
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Finite Element Method
61
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
y ( ξ,η ) =
∑ yi N i ( ξ,η )
i=1
(2.53)
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.
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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
-x --------- y ∑ -------∂ξ k ∑ ∂ξ k
k = 1, 2, …n
(2.63)
It follows that
∂N k --------∂x ∂N k --------∂y
k ∂N ------- – 1 ∂ξ = J ∂N k - ------- ∂η
(2.64)
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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 ∂x ∂y ∂x ∂x ∂y ∂y d A e = ------dη ⋅ ------dη + ------ dξ – ------dη ------dη – ------ dξ ∂ξ ∂η ∂η ∂η ∂η ∂ξ
∂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)
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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, Ku n = F n
(2.71)
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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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)
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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.
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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.
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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
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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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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.
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U ∆ p = d--------------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 1 d 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
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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
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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)
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∆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 uu
∫e Be EBe d Ae
u∆
∫e Be EBe d Ae
Ke =
Ke = ∆∆
Ke
=
T
T
∆u
= Ke
∫e Be EBe d Ae 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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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∆ =
∫ ∫e Ne Re det J dξ dη + ∫–1 Ne pe dλ T
(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)
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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:
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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 1 x 3 2 3 N 1 ( x ) = H 0 -- = ---3 ( 2x – 3lx + l ) l l 1 3 1 x 2 2 N 2 ( x ) = l ⋅ H 1 -- = ---2 ( x – 2lx + l x ) l l 1 1 l – x 3 2 N 3 ( x ) = H 0 ---------- = – ---3 ( 2x – 3lx ) l l 1 3 1 l – x 2 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)
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79
where,
∑g ∫0 Be EI Be dx l
K =
T
(2.127)
2
d B e = --------2 N dx
(2.128)
wg = w0 θ0 w1 θ1 … L
F =
∫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
∂ M ∂ Mx ∂ My ----------- + 2 -------------xy- + ----------- + q ( x,y ) = 0 2 2 ∂x∂y ∂x ∂y 2
2
(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
3
2
2
Et ∂ w ∂ w - --------2- + v --------2- M x = -----------------------2 12 ( 1 – v ) ∂x ∂y 2
Et ∂ w ∂ w - --------2- + v --------2- M y = -----------------------2 ∂x 12 ( 1 – v ) ∂y 3
2
Gt ∂ w M xy = -------- -----------12 ∂x∂y
(2.133)
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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
2
∂ w ∂ w 12 ( 1 – v ) ∂ w - q ( x,y ) = 0 + --------4- + ------------------------------- + 2 ---------------2 2 3 4 ∂x Et ∂y ∂ x ∂y 4
4
(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 w = N we
(2.135)
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)
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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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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 x -z; σ x = -----------3 t
12M y -z; σ y = -----------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.
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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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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)
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85
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
eΠ
2
=
∑ ∫ eε eσ d( elem ) = T
1 e
n
∑ ∫ 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)
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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.
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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.
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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
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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
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91
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 l c = l o 1 + 2 ------- + ----22 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 - + -------2 l c ≈ l 0 1 + ------2 l 0 2l 0
(3.8)
h0 u u2 ε = ------- + -------2 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)
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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.
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93 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.
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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.
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3.3.1
95
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)
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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 ) + ---------------du 2 du 2
(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 ( ui ) = Ri 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)
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97 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.
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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.
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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.
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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
1 ∂ f + --- ∑ ∑ --------------- ( ∆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
1 ∂ Π + --- ∑ ∑ ----------------- ∆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
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101
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 + --2- ∂u 2 ∂u 1 i ∂u 2 ∂u 2 i … … … … … ∆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 2 i ∂u 2 ∂u n i i
…
…
2
2
∂ Π --------------- ∂u n ∂u 1 i
…
(3.40)
… 2
∂ Π ∂ Π ---------------- … ----------------- ∂u n ∂u 2 i ∂u n ∂u n 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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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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(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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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
Ri di = 0
(3.59)
which indicates an orthogonality condition. Usually, optimum parameters λi vary between the values 0.05 < λ2 <1.0. The line-search method is illustrated in Figure 3.8.
3.3.5
REGULA-FALSI METHOD (METHOD
OF
FALSE POSITION)
A simplified version of scalar parameter λi is obtained by the regula-falsi method as follows. In the quadratic expansion of potential Π, Equation (3.37), the Hessian matrix is approximated using gradients of two successive iterations. The expression of the potential takes the form ∂Π ( u i ) ∂Π ( u i – 1 ) ----------------- – ----------------------∂Π ( u i ) T 1 ∂u ∂u T Π ( u ) ≅ Π i + ----------------- ( u – u i ) + --- ( u – u i ) ----------------------------------------------- ( u – u i ) ∂u 2 ui – ui – 1 (3.60)
FIGURE 3.8 Concerning the line-search method.
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which is equivalent to 1 T T ( Ri – Ri – 1 ) - ( u – ui ) Π ( u ) ≅ Π i + R i ( u – u i ) + --- ( u – u i ) -------------------------( ui – ui – 1 ) 2
(3.61)
The minimum condition leads to the following equation: T ui – ui – 1 u = u i – R i ---------------------Ri – Ri – 1
(3.62)
We can combine the method of false position with the line-search method, since the latter expression represents a linear interpolation of u. Consider a descent direction, di = ui – ui – 1
(3.63)
Following the line-search procedure, we assume u = u i – 1 + λd i = – ( 1 – λ )d i
(3.64)
After combining Equation (3.64) with Equation (3.61), we obtain potential Π as function of a single parameter, λ. 1 T 2 T Π ( λ ) = Π ( 1 ) – ( 1 – λ )R i d i + --- ( 1 – λ ) d i ( R i – R i – 1 ) 2
(3.65)
Upon deriving the gradient [∂Π/∂u] and equating it to zero, we obtain the search parameter for local minimum, T
Ri di λ = 1 – ---------------------------------T ( Ri – Ri – 1 ) di
3.3.6
(3.66)
QUASI-NEWTONIAN METHODS
For problems with extended data, the inverse of a Hessian matrix may be problematic because of memory limits. In such cases, the quasi-Newtonian methods may be used. One introduces an approximation of the inverse Hessian matrix using information gathered from the minimization process. Let us consider a truncated Taylor series of the gradient of potential R ( u ) ≅ Ri + Ki ( u – ui )
(3.67)
The data obtained for two points ui and ui–1 are used to obtain an approximation of the inverse of stiffness matrix,
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∆R i = K i ∆u i or ∆u i = K i ∆R i
(3.68)
∆R i = R i – R i – 1
(3.69)
∆u i = u i – u i – 1
(3.70)
where
The recurrence equation of the quasi-Newtonian method takes the form
–1
Ki + 1
–1 –1 T = K i + α i ( ∆u i ) ( ∆u i ) = K i + α i
∆u 1 ∆u 2 ∆u 1 ∆u 2 … ∆u n (3.71) … ∆u n
In the above equations, scalar αi defines the magnitude of iterative step; αi and T ( ∆u i ) ( ∆u i ) form a matrix by which matrix Ki is updated. Additional recurrence scheme in use is the Broyden–Fletcher–Goldfarb–Shano (BFGS) algorithm,3 which may be presented in the following form: –1
T
K i + 1 = C i K i C i + α i ( ∆u i ) ( ∆u i )
T
(3.72)
where C i = I – α ( ∆R i ) ( ∆u i )
(3.73)
1 α i = -----------------------------T ( ∆u i ) ( ∆R i )
(3.74)
REFERENCES 1. Oden, J.T., Finite Elements of Nonlinear Continua, McGraw-Hill, New York, 1972. 2. Luenberger, D.G., Linear and Nonlinear Programming, Addison-Wesley, Reading, Mass, 1989. 3. Bazaraa, M.S., Sherali, L.D., and Shetty, C.M., Nonlinear Programming, Theory and Algorithms, John Wiley, New York, 1993. 4. Schweizerhof, K.H., and Wriggers, P., Consistent linearization of path following methods in nonlinear FE analysis. Comp. Meth. Appl. Mech. & Eng., 59, 261–279, 1986. 5. Crisfield, M.A., Non-linear Finite Element Analysis of Solids and Structures. Vol. 1, Essentials, John Wiley, Chichester, England, 1994. 6. Bathe, K.J., Finite Element Procedures. Prentice Hall, Englewood Cliffs, New Jersey, 1996.
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4
Plasticity
Plastic deformation is an irreversible process characterized by sustained changes in the original geometry. Consider monotonic loading of a body made of ductile metal. The body undergoes deformation that is proportional to the magnitude of the applied load. Up to a certain magnitude, the deformation is elastic and is reversible (see Chapter 1). After a critical point, parts of the body are subjected to plastic deformation, leaving a permanent distortion. In the following, we present the fundamentals of the theory of plasticity. We start with one-dimensional empirical correlations and extend the description to include the basics of the three-dimensional plasticity theory.1,2,3
4.1 ONE-DIMENSIONAL THEORY 4.1.1
STRESS-STRAIN
RELATIONSHIP
Tensional Test Tensional tests are performed on standard specimen to derive basic mechanical properties of materials. The deformation is measured as function of the applied load during axial tension of the specimen (see Figure 4.1). The standard data comprise diagrams of stresses versus strains. A typical stress-strain diagram from a specimen made of low-carbon or midcarbon steel is shown in the Figure 4.2a. One can observe the following features on this curve. At the start, until point A is reached, there is a linear relation between the stress and strain, which follows the Hooke’s law. At point A, the stress slightly drops, up to point A′ . From here to point B, the stress is practically constant, with small oscillations. Point A′ indicates a stress at which the specimen undergoes a marked elongation without an increase in load. This point is called yield strength or yield point and is designated as Syp. A rising curve up to point C indicates material’s hardening with an increase of the load. This point is called ultimate strength and is designated as Su. Beyond this point, the curve drops, ending abruptly, which indicates material’s final failure. High-carbon steels and nonferrous metals do not have a marked yield point. They exhibit a gradual transition from linear elastic to nonlinear plastic behavior (see Figure 4.2b). For such cases, the yield point is defined by drawing a parallel line to the elastic line at an offset of 0.2%. The cross point of the offset line with the stress-strain line is assumed to be the yield point Syp. 107
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FIGURE 4.1 Tension test.
In both cases, if the specimen is unloaded before reaching final failure, the stressstrain line follows path DE, which is nearly parallel with the elastic path (see Fig. 4.2). Point E at zero load indicates a permanent deformation of the specimen. Upon reloading from point E, the specimen follows Hooke’s law along the elastic path ED. Stress and Strain Definitions Let us consider a typical tensional-test diagram. To describe this test, one uses the engineering stress and strain, which are defined as follows: the engineering stress is the stress relative to the original cross-section area A0, P S = -----A0
(4.1)
while the engineering strain is the ratio of elongation versus original length l0, l–l e = -----------0 l0
(4.2)
See Figure 4.1. The engineering stress and strain defined by Equations (4.1) and (4.2) are applicable to small deformations only. When large plastic deformation takes place, a more functional definition becomes necessary, referred to as true stress and true stain. The true stress is defined as the load divided by the current (or deformed) cross-sectional area, P σ = --A
(4.3)
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FIGURE 4.2 Stress-strain curves: (a) low-carbon steel and (b) high-carbon steel.
The true (or logarithmic) strain is defined by the differential, dl dε = ----l
(4.4)
where l is the current length. Then, the full strain will be the integral, ε =
l
dl
∫l ---l0
l = ln --l0
(4.5)
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Comparing with Equation (4.2), we find ε = ln ( l + e )
(4.6)
Furthermore, assuming the material is incompressible, i.e., A 0 l 0 ≈ Al
(4.7)
P l P σ = --- = ------ --- = S ( l + e ) A0 l0 A
(4.8)
we obtain the relation
Figure 4.3 shows the difference between the engineering and the true stress-strain curves for a high-carbon steel. Within the elastic domain, up to the yield point σyp, the difference between cross section A and A0 is negligible, and both curves merge into a single line. True Stress-Strain Relationship Consider the behavior of true stress and strain while testing a high-carbon steel specimen, comprising loading along line OAB and unloading along BC. See Figure 4.4. Let us disregard 0.2% offset. Up to yield point A, the stress rises in direct proportion to strain, following the Hooke’s law. σ = Eε
FIGURE 4.3 Comparison of true and engineering stress-strain curves.
(4.9)
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FIGURE 4.4 Loading path beyond the yield point.
In the plastic domain, the true stress-strain relationship may be expressed by an empirical power function of Hollomon, σ = Kε
n
(4.10)
At point A, the total strain comprises two components, an elastic strain and a plastic one, e
ε = ε +ε
p
(4.11)
p
where the plastic strain ε corresponds to a permanent plastic deformation that remains after the specimen is unloaded along BC. The definition of elastic strain follows from Equation (4.9) as σ e ε = --E
(4.12)
The plastic strain is expressed by a revised correlation of Hollomon, σ ε = ---- K p
1 --n
(4.13)
Thus, the total strain equals 1 ---
σ n σ ε = --- + ---- K E
(4.14)
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Concerning stress-strain behavior in the plastic domain, to simplify the numerical solution, idealized models may be employed. A choice of such models depends on the practical application. If the structure is made of a material that undergoes plastic deformation without a marked increase in the load, a perfectly nonhardening plastic model can be used as shown in Figure 4.5a. In cases of increased resistance in the plastic region, so-called strain hardening, the stress-strain curve can be approximated by a linear correlation (see Figure 4.5b). For comparison, Figure 4.5c shows a strainhardening curve as per Equation (4.14). Bauschinger Effect Consider the following loading sequence. First, the specimen is subjected to tensional load with plastic deformation. Then, the load is removed and reloaded in compression with plastic deformation. It has been observed that the yield point in reloading is less then the yield point in the original loading (see Figure 4.6). This phenomenon is known as the Bauschinger effect. It was observed in polycrystalline metals and
FIGURE 4.5 Idealized stress-strain curves: (a) elastic-perfectly plastic model, (b) elastoplastic with linear hardening, and (b) elasto-plastic with nonlinear hardening.
FIGURE 4.6 Bauschinger effect.
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in single crystals. The effect is usually modelled by a simplified model, where it is assumed that the reduction of yield point in reloading equals the stress increment by which the specimen has been loaded beyond the initial yield point in the original direction.
4.1.2
RESIDUAL STRESSES: EXAMPLE
The Three-Truss Problem Prestressing of parts is a topic of practical consideration. It pertains to creating locally permanent plastic strain to obtain a favorable stress distribution at repeated loading. To understand the response of a structure that undergoes plastic deformation, consider a statically undetermined three-truss system as shown in Figure 4.7. This simple example will illustrate the elasto-plastic response of a structure typical for more general cases. The objective here is to determine the forces in the trusses. The considered structure is symmetric; all of the trusses are made of the same material and have the same cross-sectional area. Material behavior is elastic-perfectly plastic as in Figure 4.5a. Consider first the equilibrium condition of point A. N1 = N2 N 1 cos β + N 3 + N 2 cos β = P
FIGURE 4.7 Three-truss structure.
(4.15) (4.16)
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To determine the unknown internal forces Ni , one requires an additional equation, since the structure is statically undetermined. This will be provided by the condition of compatibility relating to the elongation of trusses, ∆l 1 = ∆l 3 cos β
(4.17)
We assume the displacements to be small; therefore, the changes of angle β are negligible. Elastic Deformation At first, consider deformation in the elastic range. Expressing truss elongations in terms of axial loads, Equation (4.17) becomes N 3 l3 N 1 l1 - cos β --------- = --------EA EA
(4.18)
l 3 = l 1 cos β
(4.19)
2
(4.20)
and since
one obtains the third equation, N 1 = N 3 cos β
From Equations (4.15), (4.16), and (4.20), we determine the unknown axial forces. 2 P–N Pcos β N 1 = N 2 = ----------------3 = ------------------------3 2 cos β 1 + 2cos β
(4.21)
P N 3 = ------------------------3 1 + 2cos β
(4.22)
One notes that N 3 > 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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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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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 – σ yp A P c cos β 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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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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Distortion energy for the multi-axial stress, in terms of the principal stresses, is 1+v d 2 2 2 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 1+v 2 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: 1 2 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)
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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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FIGURE 4.10 Von Mises yield surface in stress space.
FIGURE 4.11 Tresca yield surface in stress space.
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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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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)
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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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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 2 2 2 1 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)
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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). 1 3 p p 2 p 2 p 2 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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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)
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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 dκ ∂σ
(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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∂f e ------··E ··dεε σ ∂σ dλ = ----------------------------------------------------dY ∂f ∂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 ∂f ∂σ e e σ E ··dεε – E ·· ------ ----------------------------------------------------- = dσ σ dY ∂f ∂σ ∂f e ∂f ------- ------ + ------··E ·· -----σ σ dκ ∂σ ∂σ σ ∂σ
(4.87)
from which follows the required stress-strain correlation for isotropic hardening. e ∂f ∂f e E ·· ------ ------··E e σ ∂σ σ ∂σ σ = E – ----------------------------------------------------- ··dεε dσ dY ∂f ∂f e ∂f ------- ------ + ------··E ·· ------ σ σ ∂σ dκ ∂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 1 2 F = --- ( s··s ) – --- σ yp = 0 2 3
(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)
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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
ep
e
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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1 2 p p p p p dε e = ------------ ( dε 1 – dε 3 ) = --- ( dε 1 – dε 3 ) 1+v 3
(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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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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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 ∂σ ∂α
(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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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
4.3.4
ep
E ·· ( s – α ) ( s – α )··E e = E – --------------------------------------------------------------------------------------------e c ( s – α )·· ( s – α ) + ( s – α )··E ·· ( s – α ) e
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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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 + ------··E ·· -----c ------·· ------ + ------- -----σ σ ∂σ σ dκ ∂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 ∂f dY ∂f ∂f e c ------·· ------ + ------- ∂F ·· ------ - + ------··E 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)
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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 ∂f ∂f e ------··E e E ·· -----σ ∂σ σ ∂σ σ = E – --------------------------------- dεε dσ ∂f e ∂f ------··E ·· ------ σ σ ∂σ ∂σ
(4.135)
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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 ε = ε e + ε p = ---σ σ + --- ----e- s E 2 σe where σ denotes the stress tensor.
(4.142)
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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
ep
e
T
e
E ss E e = E – -----------------------------------------------------------4 2 dσ yp T T e 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.
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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)
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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)
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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.
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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
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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)
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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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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)
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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)
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g
sk
=
∂θ ∂θ
s
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,
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FIGURE 5.4
147
Deformation of an infinitesimal line due to deformation of a body.
p
q
dr 1 ⋅ dr 2 g pq dθ 1 dθ 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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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)
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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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The Cauchy–Green strain tensor is defined as 1 C T T 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 C 1 C T T 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)
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Using both expressions, one can write the Cauchy–Green strain tensor in the form 1 1 2 C 2 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 m ∂ s ∇ 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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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. ∂v 2 ∂w 2 ∂u 1 ∂u 2 C E xx = ------ + --- ------ + ------ + ------- ∂x ∂x ∂x 2 ∂x ∂v 2 ∂w 2 ∂v 1 ∂u 2 C E yy = ----- + --- ------ + ----- + ------- ∂y ∂y ∂y 2 ∂y ∂v 2 ∂w 2 ∂w 1 ∂u 2 C E zz = ------- + --- ------ + ----- + ------- ∂z ∂z ∂z 2 ∂z 1 ∂u ∂v ∂v ∂v ∂w ∂w 1 ∂u ∂u C C E xy = E yx = --- ------ + ------ + --- ------ ------ + ------ ----- + ------- ------- 2 ∂y ∂x 2 ∂x ∂y ∂x ∂y ∂x ∂y 1 ∂v ∂w ∂v ∂v ∂w ∂w 1 ∂u ∂u C C E yz = E zy = --- ----- + ------- + --- ------ ------ + ----- ----- + ------- ------- 2 ∂z ∂y 2 ∂y ∂z ∂y ∂z ∂y ∂z 1 ∂u ∂w ∂v ∂v ∂w ∂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)
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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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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
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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 ---c = ln 1 + ----------- ≈ ( 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 1 l 2 C T E = --- [ ∇ i R ⋅ ( ∇ i R ) – I ] = --- ---c – 1 2 2 li
(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 1 lc – li 1 C - = --- 1 + ----------- – 1 = ----------E = --- ------------ – 1 ≈ --- 1 + 2 ----------- (5.78) 2 2 2 li 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)
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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 ∂x ∂y ∂z
(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)
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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 l0 2 l0
(5.88)
Consequently, the stretch of the spar equals dL dL ------ = ------ = dl dx
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 <<1), the latter expression becomes 2 2 v0 h0 1 v0 dL 1 v 0 h 0 v 0 ------ ≈ 1 + --- 2 --------- + ---2- = 1 + --------- + --- ---22 2 2 l0 2 l0 dl l0 l0
(5.90)
which corresponds to the stretch derived in Chapter 3. Example 2: Cylinder Subject to Tension For illustration, consider now a cylindrical specimen that is subject to one-dimensional tension (see Figure 5.7). In the initial state, its height and radius are h0 and r0, respectively, while in the current configuration they equal h and r. The objective is to derive the Cauchy–Green and Almansi strain tensors. Let V1,V2, and V3, be the principal stretches. h r r V 1 = ----- ; V 2 = ---- ; V 3 = V 2 = ---h0 r0 r0
(5.91)
The position vector of a point in the initial configuration equals 1
2
3
r = a ii + a i2 + a i3
(5.92)
where a1, a2, and a3 are material coordinates. The current configuration is defined by the following equations: 1
1
x = V 1a ;
2
2
x = V 2a ;
3
3
x = V 2a ;
(5.93)
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FIGURE 5.7
Nonlinear Problems in Machine Design
Round bar subject to tension.
Thus, the position vector in the current configuration becomes 1
2
3
1
2
3
R = x i1 + x i2 + x i3 = V 1a i1 + V 2a i2 + V 2a i3
(5.94)
The above may be written in the following form: R = r ⋅V
(5.95)
where V is a right stretch tensor, V 0 0 1 V = 0 V2 0 0 0 V2
(5.96)
The base vectors in the initial configuration are ∂r g 1 = --------1 = i 1 ; g 2 = i 2 ; g 3 = i 3 ∂a
(5.97)
i2 × i3 1 2 3 g = ---------------- = i 1 = g1 ; g = g2 ; g = g3 (i1i2i3)
(5.98)
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The base vectors in the current configuration are ∂r ∂R G 1 = --------1 = --------1 ⋅ V = g 1 ⋅ V ; G 2 = g 2 ⋅ V ; G 3 = g 3 ⋅ V ∂a ∂a
(5.99)
The gradient of the current position vector is defined as k
g
∇i R = g G k = g gk ⋅ V = I ⋅ V = V
(5.100)
Furthermore, since ∇c r = ( ∇i R )
–1
(5.101)
one obtains the gradient of initial position vector ∇c r = V
–1
(5.102)
where V –1 denotes the tensor,
V
–1
=
1 ------ 0 0 V1 1 0 ------ 0 V2 1 0 0 -----V2
(5.103)
Consequently, the Cauchy–Green and Almansi deformation tensors, for the present case, are D
C
= ∇i R ⋅ ( ∇i R ) = V ⋅ V
T
D
A
= ∇c r ⋅ ( ∇c r ) = V
T
–1
T
= V
2
–1 T
(5.104) –1 2
⋅ (V ) = (V )
(5.105)
To derive the Cauchy–Green strain tensor, consider the length of an infinitesimal line in the initial configuration, 2
( dl ) = dr ⋅ dr = d r ⋅ I ⋅ dr
(5.106)
which, in the current state, becomes 2
C
( dL ) = dR ⋅ dR = dr ⋅ D ⋅ dr
(5.107)
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The change of the length of the infinitesimal line is 2
2
C
( dL ) – ( dl ) = dr ⋅ (D – I ) ⋅ dr 2
C
= dr ⋅ (V – I ) ⋅ dr = dr ⋅ 2E ⋅ dr
(5.108)
As a result of this, the Cauchy–Green strain tensor equals 1 2 C E = --- ( V – I ) 2
(5.109)
In terms of its components, it becomes 2 0 V1 – 1 0 1 C 2 E = --- 0 V2 – 1 0 2 2 0 0 V2 – 1
(5.110)
On the other hand, to derive the Almansi strain tensor, let us refer to the current configuration, where we have 2
( dL ) = dR ⋅ dR = dR ⋅ I ⋅ dR 2
(5.111)
A
( dl ) = dr ⋅ dr = dR ⋅ D ⋅ d R
(5.112)
and 2
2
A
( dL ) – ( dl ) = dR ⋅ (I – D ) ⋅ dR –1 2
A
= dR ⋅ [I – ( V ) ] = dR ⋅ 2E ⋅ dR
(5.113)
Consequently, the Almansi strain tensor equals 1 A –1 2 E = --- [ I – ( V ) ] 2 In terms of its components, it becomes
(5.114)
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1 0 1 – ------2 0 V1 1 1 A E = --- 0 1 – ------2 0 2 V2 1 0 0 1 – ------2 V2
(5.115)
Let us now compare the Cauchy–Green and Almansi strain tensors for small deformations. One may assume that the principal stretches are equal. h o + ∆h ∆h h - = 1 + ------V 1 = ----- = ----------------h0 h0 h0
(5.116)
w 0 + ∆w w ∆w V 2 = ------ = ------------------- = 1 + -------w0 w0 w0
(5.117)
One may assume, furthermore, the respective squares to be ∆h 2 ∆h ∆h 2 ∆h 2 V 1 = 1 + ------- = 1 + 2 ------- + ------- ≈ 1 + 2 ------ h0 h0 h0 h0 (5.118) ∆r 2 V 2 ≈ 1 + 2 -----r0
(5.119)
and their reciprocals to equal 1 ∆r 1 ------2 = -----------------------2 ≈ 1 – 2 -----r0 ∆r V1 1 + ---- r0 1 ∆r ------2 ≈ 1 – 2 -----r0 V2
(5.120)
(5.121)
We find for small deformations, therefore, that the Cauchy–Green and Almansi strain tensors are both equal.
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C A E ≈E ≈
∆h ------- 0 h0 ∆r 0 -----r0 0
0
0 0 ∆r -----r0
(5.122)
Example 3: Simple Shear Our illustration here pertains to the deformation of a cube, subject to simple shear (see Figure 5.8). Surface EFGH is moved so that it retains a form of parallelepiped. Our objective in this example is to derive the applicable strain tensors. The simple shear is defined by the following equations: X = ( x + y tan θ ) ; Y = y ; Z = z
(5.123)
In vectorial form, they are expressed as R = r +H ⋅r = r + r ⋅ H
T
T
= r ⋅ (I + H )
(5.124)
where r and R are the position vectors in the initial and current configurations, respectively. Tensor H equals
FIGURE 5.8
Simple shear.
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0 γ 0 H = 000 000
(5.125)
with γ = tan θ
(5.126)
The gradient of current position vector equals s
s
T
T
∇i R = g G s = g gs ⋅ ( I + H ) = I ⋅ ( I + H ) = I + H
T
(5.127)
which, in terms of its components, becomes 100 ∇i R = γ 1 0 001
(5.128)
The determinant of the matrix of the gradient, shown above, equals det ∇ i R = 1
(5.129)
which shows that the volume of cube remains constant during the deformation. The gradient of initial position vector equals
∇c r = ( ∇i R )
–1
1 –γ 0 = 0 1 0 0 0 1
(5.130)
which may be expressed in terms of H as V cr = I – H
(5.131)
Consequently, for this case, the Cauchy–Green deformation tensor becomes D
C
T
T
T
T
T
= ∇i R ⋅ ( ∇i R ) = ( I + H ) ⋅ ( I + H ) ⋅ ( I + H ) = I + H + H + H ⋅ H (5.132)
which, in term of its components, equals
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D
1 γ 0 = γ 1 + γ2 0 0 0 1
C
(5.133)
On the other hand, to derive the Almansi deformation tensor, let us refer to the current configuration, where we have ( ∇i R )
–1
= ∇c r = I – H
T
(5.134)
Consequently, the Almansi deformation tensor equals D
A
T
T
T
= ∇c r ⋅ ( ∇c r ) = ( I – H ) ⋅ ( I – H ) = I – H – H + H H
(5.135)
which may be expressed as
D
A
1 –γ 0 = –γ 1 + γ 2 0 0 1 0
(5.136)
To derive the Cauchy–Green and Almansi strain tensor, consider the following. Gradient ∇iR is expressed as the product ∇i R = U ⋅ O = O ⋅ V
(5.137)
so that D
C
= U
2
(5.138)
as per Equations (5.33) to (5.35). The deformation of the cube is confined to XYplane; therefore, rigid rotation O can be expressed as an orthogonal rotation tensor. cos α sin α 0 O = – sin α cos α 0 0 0 1
(5.139)
where rotation angle α is unknown. The stretch tensors, U and V, can be written as follows:
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U U 0 x xy U = U xy U y 0 0 0 1
(5.140)
V V 0 x xy V = V xy V y 0 0 0 1
(5.141)
To derive the unknown components of the left stretch tensor U, we use Equations (5.128), (5.139), and (5.140) to obtain a system of algebraic equations, U cos α – U sin α U sin α + U cos α 0 100 xy x xy x U xy cos α – U y sin α U xy sin α + U y cos α 0 = γ 1 0 0 0 1 001
(5.142)
which is equivalent to the following system of four equations: U x cos α – U xy sin α = 1 U x sin α + U xy cos α = 0 U xy cos α – U y sin α = γ U xy sin α + U y cos α = 1
(5.143)
Solving the latter, we find cos α – sin α 0 U = – sin α cos α – γ sin α 0 0 0 1
(5.144)
1 tan α = – --- γ 2
(5.145)
with
Using similar procedure, we derive
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cos α – γ sin α ( – sin α ) 0 V = – sin α cos α 0 0 0 1
(5.146)
Consequently, the Cauchy–Green strain tensor becomes 0 γ 0 1 E = --- γ γ 2 0 2 0 0 0
(5.147)
0 γ 0 1 E = --- 0 – γ 2 0 2 0 0 0
(5.148)
C
while the Almansi tensor equals
A
Let us now derive the principal components of stretch tensor V. By definition, they are roots of the following equation:
det V – VI =
cos α – γ sin α – V – sin α 0 =0 – sin α cos α – V 0 0 0 1–V
(5.149)
From Equations (5.146) and (5.149), we obtain 2 2 ( 1 – V ) V – ------------V + 1 = 0 cos α
(5.150)
Consequently, the principal components are 1 – sin α π α V 1 = -------------------- = tan --- – --- 4 2 cos α
(5.151)
1 + sin α π α V 2 = --------------------- = cotan --- – --- 4 2 cos α
(5.152)
V3 = 1
(5.153)
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It follows that V 1V 2V 3 = 1
(5.154)
The above confirms that the volume of the cube in simple shear does not change.
5.3 STRESS Stress is defined as an internal force per unit area of a section in a body under loading. The topic was introduced in Chapter 1, where it was limited to bodies undergoing small deformation. Presently, we are concerned with the bodies where deformation caused by loading results in large deformations. Consequently, we are faced with different stress tensors according to either initial (unloaded) or current (loaded) configuration. We discuss two kinds of stress tensors: the Cauchy (or true) stress tensor and the Piola stress tensors. The former refers to the current configuration, while the latter refer to the initial configuration.
5.3.1
CAUCHY STRESS TENSOR
Consider a body in the current configuration. Applying the section method (Chapter 1), let us examine a part of the body V bounded by a cut surface A (see Figure 5.9). The remaining part of the body acts upon part V by means of vector field t distributed over surface A. Vector t is called the stress vector (or traction) and is defined by the equation t = lim
∆ Ac → 0
∆P --------∆ Ac
(5.155)
where ∆P denotes force, and area ∆Ac refers to the current configuration. Parts V acts with stress t(–n) on a conjugate area of the remaining part defined by the normal –n so that – t( – n ) = t
FIGURE 5.9
Concerning the definition of stress.
(5.156)
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See Figure 5.10. (We neglect any possible distributed moments in the interaction between the two parts of the body.) Consider an elemental tetrahedron in the current configuration formed by vectors R 1dθ1, R 2dθ2, and R 3dθ3. See Figure 5.11. The faces of the tetrahedron are dA1, dA2, dA3, and dAc and their directions are specified by normal vectors N 1, N 2, N 3, and N. The vectors representing the faces dA1, dA2, and dA3, are expressed by the equations 1
1G 3 2 G dθ × G 2 dθ N 1 d A 1 = – --- --------2 G1 3 2
1G 1 3 N 2 dA 2 = – --- --------G dθ × G 3 dθ 2 G2 1
FIGURE 5.10
Stresses acting upon elemental area.
FIGURE 5.11
Stresses acting upon elemental tetrahedron.
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1G 2 1 N 3 d A 3 = – --- --------G dθ × G 1 dθ 2 G3 2 i
(5.157)
i
which is based on Equations (5.11). Factors ( G / G ) define normal vectors N 1, N 2, and N 3. The vector of face dAc equals 1 1 2 1 3 1 N d A c = --- ( AB × AC ) = --- ( G 2 dθ – G 1 dθ ) × ( G 3 dθ – G 1 dθ ) 2 2 1 2 3 1 2 3 1 = --- ( G 2 dθ × G 3 dθ + G 1 dθ × G 2 dθ + G 3 dθ × G 1 dθ ) 2 = –( N 1 d A1 + N 2 d A2 + N 3 d A3 )
(5.158)
which, upon substituting Equations (5.155) into (5.156), becomes 1
2
3
1G 1G 1G 3 2 1 3 2 1 N d A c = --- --------G dθ × G 2 dθ + --- --------G dθ × G 3 dθ + --- --------G dθ × G 1 dθ 2 G1 3 2 G2 1 2 G3 2 (5.159) Using the latter equation, one obtains areas dA1, dA2, and dA3 expressed in the form 1 3 2 1 d A 1 = --- G 3 dθ × G 2 θ = N ⋅ G 1 G d A c 2 1 1 3 2 d A 2 = --- G 1 dθ × G 3 θ = N ⋅ G 2 G d A c 2 1 2 1 3 d A 3 = --- G 2 dθ × G 1 θ = N ⋅ G 3 G d A c 2
(5.160)
The condition of equilibrium of the tetrahedron is expressed by the equation td A c + t–1 d A 1 + t–2 d A 2 + t–3 d A 3 = 0
(5.161)
It follows that 3
td A c = N ⋅
∑ G m tm G
m=1
Consequently, stress vector t equals
m
m
d Ac ; G =
G
mm
(5.162)
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t = N⋅
∑ G m tm
G
mm
= N ⋅S
C
(5.163)
m=1
In the above equation, S C is the Cauchy stress tensor. It is defined by the equation 3
S
C
=
∑ G m tm
G
mm
(5.164)
m=1
The Cauchy stress tensor correlates the internal forces acting upon an area with the direction of the area. It follows from equilibrium of moments acting on the tetrahedron that the Cauchy stress tensor is symmetrical. S
5.3.2
C
C T
= (S )
(5.165)
PIOLA STRESS TENSORS
There are two Piola stress tensors, a non-symmetric and a symmetric one. Both correlate the vector of traction with area in the initial configuration. The application of these tensors is more convenient because, unlike with the Cauchy tensor, the initial configuration is known at the start. To derive the Piola stress tensors, let us introduce ti within the initial configuration, ∆P ti = lim --------∆A i → 0 ∆ A i
(5.166)
To express ∆Ai in initial configuration, let us refer back to ∆Ac of the current configuration, defined by two infinitesimal vectors as N d A c = dR ′ × dR ″
(5.167)
Combining the above with Equation (5.33), we arrive at N d Ac =
G T ---- n ⋅ ( ∇ c r ) d A i g
(5.168)
where G = det G mn and g = det g mn are determinants of matrices of metrictensor components, and n denotes a normal vector referring to dAi. (Normal vectors in the current configuration are denoted by N, while normal vectors in the initial configuration are denoted by n.) The force acting on ∆Ac equals C
dP = td A c = N ⋅ S d A c =
G T C P ------- n ⋅ ( ∇ c r ) ⋅ S d A i = n ⋅ S d A i (5.169) g
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from which we obtain the equation ti = n ⋅ S
P
(5.170)
S P is the first Piola stress tensor (out of two). The first Piola stress tensor is also called the nominal stress tensor. It relates to the Cauchy stress tensor as follows: G T C ------- ( ∇ c r ) ⋅ S g
P
S =
(5.171)
Since the first Piola stress tensor is non-symmetric, a second Piola stress tensor is introduced, which is symmetric. It equals
P
P
S = S ⋅ ∇c r =
G mn T C ----------- ( ∇c r ) ⋅ S ⋅ ∇c r g rs
(5.172)
Notwithstanding the inconvenience, when accuracy is required, the Cauchy stress tensor is preferable, because it relates to the current configuration and therefore presents the true stress.
5.3.3
EXAMPLE
Example 4: Cylinder Subject to Tension Refer to Example 2 in Section 5.2.4. The problem’s concern now is the derivation of stress tensors. The cylinder is subject to tension caused by force P, which is applied at its ends (see Figure 5.7). The objective is to derive the applicable Cauchy and Piola stress tensors. The Cauchy stress tensor for the one-dimensional tension has the form
S
C
C S1 0 0 = 0 00 0 00
(5.173)
where the single stress component equals P P P C = ----------S 1 = ----2 = -------------2 2 2 r πV 2 r 0 V 2 A0
(5.174)
Principal stretches V1 and V2 are defined by Equation (5.91), and V0 is the crosssection area in the initial configuration.
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The first Piola stress tensor equals G T C –1 T C 2 ---- ( ∇ c r ) ⋅ S = V 1 V 2 ( V ) ⋅ S g
P
S =
(5.175)
In terms of the components, it becomes 2 P S = V 1V 2
1 ------ 0 0 V1 1 0 ------ 0 ⋅ V2 1 0 0 -----V3
P ------2 0 0 V2 = 0 00 0 00
P ------ 0 0 A0 0 0 0 0 00
(5.176)
The second Piola stress tensor equals P ------------ 0 0 V A P S = S ⋅ ∇c r = 1 0 0 00 0 00 P
(5.177)
5.4 CONSTITUTIVE EQUATIONS In Section 5.2, we discussed the deformation, and in Section 5.3 we described the stresses in a body under loading. To correlate the state of stress and the deformation, we need to consider their interdependence, which is reflected in constitutive equations. The constitutive equations comprise the physical properties of the material. We limit our discussion to ideally elastic materials with properties that are independent of the history of deformation, for which the loading and unloading processes are reversible. There are two approaches to derive the constitutive equations for such materials: Cauchy and Green. The Cauchy method postulates the existence of a law that directly links the stress and strain tensors; e.g., Hooke’s law for linear elastic bodies described in Chapter 1.5 Because the deformation process under consideration is very slow here, the dynamic phenomena are neglected. The method assumes complete reversibility: after unloading, the body returns to its original shape. The Green method is based on energy consideration. It postulates the existence of a potential energy function, defining stresses as its derivatives. One assumes there are no energy losses during the loading-unloading path. The work done by the external forces is completely converted into strain energy and is fully restored after the unloading. A detailed discussion of the adopted functions of potential energy has been provided by Truesdell and Noll,6 and others. The materials, for which such potential energy functions exist, are called hyper-elastic.
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The elastic material in the following description is homogeneous and isotropic. We assume the deformation process to be time independent. Since the stresses and the deformation both may refer to either the initial or the current configuration, we do not stipulate the type of configuration; either Cauchy–Green or Almansi deformation and strain tensors, and either Piola or Cauchy stress tensors, can be used.
5.4.1
CAUCHY METHOD
In the Cauchy approach, the stress tensor is expressed as a function of either the deformation or the strain tensor. Let us present the correlation of the stress tensor, as follows: S = f (D )
(5.178)
where D denotes the deformation tensor. The above can be expanded in a polynomial series. 2
3
S = a0 I + a1 D + a2 D + a3 D + …
(5.179)
where coefficients ak are function of invariants of tensor D. According to the Cayley–Hamilton theorem, tensor D must satisfy its own characteristic equation (see Appendix A); i.e., 3
2
D – I 1 × D + I 2 × D – I 3 × I =0
(5.180)
Consequently, powers of tensor D that are higher than 2 can be expressed in terms of D and D 2. Thus, the stress tensor may be presented as a quadratic function of deformation tensor D and its invariants. S = b 0 ( I 1 ,I 2 ,I 3 )I + b 1 ( I 1 ,I 2 ,I 3 )D + b 2 ( I 1 ,I 2 ,I 3 )D
2
(5.181)
Stress tensor S is independent of rotation, since the material is isotropic and therefore independent of directions. We may skip the rotation term of the deformation gradient by applying a polar decomposition (see Section 5.2.3). Thus, Equation (5.181) can be presented in the form S = χ 0 ( V 1 ,V 2 ,V 3 )I + χ 1 ( V 1 ,V 2 ,V 3 )V + χ 2 ( V 1 ,V 2 ,V 3 )V
2
(5.182)
where V is the right stretch tensor, and V1, V2, and V3 are its principal values.
5.4.2
GREEN METHOD
Green’s approach postulates the existence of potential strain energy of the deformed body. Let W be the strain energy W = W(E), where E denotes a strain tensor, and
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the strain energy W is measured per unit volume of the body. Then, the stress tensor equals ∂W S = -------∂E
(5.183)
Because, for homogeneous isotropic materials, the specific strain energy does not depend on position r, the strain energy is a function of the invariants of strain tensor only. W = W ( I 1 ,I 2 ,I 3 )
(5.184)
where Ii = Ii(E). Therefore, we can state (S)
mn
∂W ∂I 1 ∂W ∂I 2 ∂W ∂I 3 - + -------- ------------ + -------- -----------= -------- ----------∂I 1 ∂E mn ∂I 2 ∂E mn ∂I 3 ∂E mn
(5.185)
Of special interest are incompressible rubber-like materials, for which the volume remains constant during the deformation, and the third strain invariant equals I 3(E ) = 1
(5.186)
Hence, the strain energy is a function of two invariants only. W = W ( I 1 ,I 2 )
(5.187)
A typical example is the Mooney material,7 whose simplest strain-energy relationship, without taking into consideration small volume changes, has the form W = C1( I 1 – 3) + C2( I 2 – 3)
(5.188)
where C1 and C2 are material constants. A more comprehensive expression for Mooney–Rivlin material has the form8 2
–2
W = C1(T 1 – 3) + C2(T 2 – 3) + β(T 3 – T 3 )
(5.189)
where –1 ⁄ 3
T 1 = 11 ( D ) × I 3
–2 ⁄ 3
( D );T 2 = I 2 ( D ) × I 3
1⁄2
( D ); T 3 = I 3 ( D ) (5.190)
are functions of invariants of the deformation tensor, and 1 + v C1 + C2 β = ------------ ⋅ -----------------1–2 24
(5.191)
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5.4.3
175
EXAMPLES
Example 5: Cylinder Subject to Tension Refer to Section 5.2.4, Examples 4. The cylinder is subject to deformation caused by force P, which is applied at its ends (see Figure 5.7). The Almansi strain tensor of the cylinder equals 1 0 1 – ------2 0 V1 1 1 A E = --- 0 1 – ------2 0 2 V 2 1 0 0 1 – ------2 V2
(5.192)
see Equation (5.115). The problem’s objective now is to derive a constitutive equation relating the stress and strain tensors and to define the correlation between the principal stretches V1 and V2. h r V 1 = ----- ; V 2 = V 3 = ---h0 r0
(5.193)
Assume that the cylinder’s material is Signorini material.1 Its properties are governed by a law correlating Cauchy stress tensor and Almansi strain tensor—a quadratic function with three material constants. In a simplified form, considered here, the function is limited to two Lame constants, S
C
1 2 A A A A = λI 1 ( E ) + --- ( λ + µ )I 1 ( E ) I + 2 [ µ – ( λ + µ )I 1 ( E ) ]E 2
(5.194)
The Lame constants equal Ev λ = ------------------------------------( 1 + v ) ( 1 – 2v )
(5.195)
E µ = -------------------2(1 + v)
(5.196)
E denotes the modulus of Young, and v is the Poisson ratio (see Chapter 1, Section 1.3.1). Invariant I1(E A) is the first invariant of Almansi strain tensor. 1 1 1 A I 1 ( E ) = --- 1 – ------2 + 2 1 – ------2 2 V1 V 2
(5.197)
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From Equation (5.194), we obtain the expressions for the principal components of the Cauchy stress tensor, S 1 v 2 2 ------1 = --------------- I 1 + ----------------------- I 1 + 1 – --------------- I 1 E 1 4 ( 1 – 2v ) 1 – 2v 1 – 2v 2µ
(5.198)
S S 1 v 2 2 ------2 = ------3 = --------------- I 1 + ----------------------- I 1 + 1 – --------------- I 1 E 2 = 0 4 ( 1 – 2v ) 1 – 2v 2µ 2µ 1 – 2v
(5.199)
and
In the above equations, E1 and E2 designate the principal components of the Almansi strain tensor. 1 1 E 1 = --- 1 – ------2 2 V 1
(5.200)
1 1 E 2 = --- 1 – ------2 2 V 2
(5.201)
A
(For simplicity Ei is substituted for E i .) The first invariant of Almansi strain tensor, in terms of the principal components E1 and E2, equals I 1 = E 1 + 2E 2
(5.202)
Equations (5.199) and (5.202) provide the following relation between the strain components: 1 2 2 E 2 – E 2 – vE 1 – --- E 1 = 0 4
(5.203)
Solving the above equation for E2, we obtain 1 2 E 2 = --- ( 1 – 1 + E 1 + 4vE 1 ) 2
(5.204)
The above solution suits Poisson ratios 0 ≤ v ≤ 1 ⁄ 2 , where 2
1 + E 1 + 4vE 1 > 0
(5.205)
To derive the correlation between the principal stretches V1 and V2, let us introduce the relative elongations as follows:
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h–h r–r δ 1 = --------------0 = V 1 – 1 ; δ 2 = δ 3 = -------------0 = V 2 – 1 h0 r0
(5.206)
which, in terms of E1 and E2, equal δ 1 = 1 – ( 1 – 2E 1 ) δ 2 = 1 – ( 1 – 2E 2 )
–1 ⁄ 2
–1 ⁄ 2
(5.207) = 1–∆
–1 ⁄ 4
(5.208)
where 2
∆ = 1 + E 1 + 4vE 1
(5.209)
Let us now express the correlation between relative elongations δ1 and δ2. δ v˜ = – -----2 δ1
(5.210)
–1 ⁄ 4 V2 – 1 ∆ –1 v˜ = – -------------- = – --------------------------------------–1 ⁄ 2 V1 – 1 ( 1 – 2E 1 ) –1
(5.211)
It follows that
Equation (5.211) presents a correlation between principal stretches V1 and V2 as a function of axial strain E1. Expanding the above equation into a truncated Taylor series of the first order, we obtain a linear approximation of lateral reduction factor. 1 2 v˜ ≈ v + --- E 1 ( 1 – 6v – 10v ) 4
(5.212)
We derive the constitutive equation, relating the stress and strain tensors, by expressing Signorini quadratic function in terms of axial strain E1. Then, the stressstrain correlation for cylinder in tension becomes µ 2 1⁄2 S 1 = --------------- [ 1 + 2v + E 1 – E 1 – ( 1 + 2v – E 1 )∆ ] 1 – 2v
(5.213)
From the above, one may also derive a correlation between the tensile load and axial strain. The cross-sectional area of the deformed cylinder equals 2
2
A = V 2 A0 = ( 1 + δ2 ) A0 = A0 ∆
–1 ⁄ 2
(5.214)
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where A0 denotes the undeformed cross section. Hence, the tensile force is µ A0 2 –1 ⁄ 2 - [ E – 1 – 2v + ( 1 + 2v + E 1 – E 1 )∆ ] P = S 1 A = -------------1 – 2v 1
(5.215)
Inserting Lame constant µ and expanding P into a truncated Taylor series of the second order in terms of E1, we obtain the force-strain correlation. 2
3 ( 1 + 2v ) 2 P = E A 0 E 1 – --- ---------------------- E 1 4 1+v
(5.216)
(Note: E denotes Young’s modulus.) The linear term EA0E1 corresponds to the forcestrain correlation of linear theory. At an infinite elongation, we have 1 δ 1 → ∞ ; E 1 → --2
(5.217)
whereby we arrive at a limit value of the tensile force. 1
--µ A0 2 - ( 5 + 8v ) – ( 1 – 4v ) P lim = ---------------------2 ( 1 – 2v )
(5.218)
The above equation defines the breaking point of the specimen as predicted by Signorini theory.1 Example 6: Simple Shear Refer to Section 5.2.4, Example 3. The cube is subject to simple shear, defined by the equation γ = tan θ
(5.219)
See Figure 5.8. The objective is to derive the applicable stress-strain relations. The Almansi strain tensor for simple shear of the cube equals 0 γ 0 1 E = --- γ – γ 2 0 2 0 0 0
(5.220)
1 2 A I 1 ( E ) = – --- γ 2
(5.221)
A
where the first invariant is
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See Equation (5.148). The cube is made of Signorini material as described in Example 5 above. Combining the expression of Almansi strain tensor, Equation (5.220), and the Signorini following constitutive equation as defined by Equation (5.194), we obtain the following stress-strain relations: 2
µγ 1 2 S x = S y = – ----------------------- 2v – --- γ 2 ( 1 – 2v ) 4 2
µγ 3 2 S z = – --------------- 1 – v + --- γ 1 – 2v 8 3
µγ S xy = S yx = µγ + ----------------------2 ( 1 – 2v ) S zx = S xz = S yz = S zy = 0
(5.222)
where v denotes λ v = --------------------2(λ + µ)
(5.223)
The above results lead to the following conclusions: 1. Normal forces have to be applied to all faces of the cube to perform a simple shear. The mean normal stress is not zero, i.e., 1 1 --- I 1 ( S ) = --- ( S x + S y + S z )≠0 3 3
(5.224)
2. The nonlinear solution, Equations (5.222), differs considerably from a linear one where the stresses are defined by the single equation, S xy = µγ
(5.225a)
The normal stresses of the nonlinear solution are of a second order in terms of γ, which is not shown in the linear theory. 3. The difference of normal stresses Sx – Sz is not equal to zero. 3
µγ S x – S z = µγ + ----------------------2 ( 1 – 2v )
(5.225b)
The above equation applies to elastic solids. In elastic liquids, where Sxy = 0, the difference becomes zero.
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5.5 FINITE ELEMENT IMPLEMENTATION In this section, we use the finite element nomenclature as described in Chapter 2. The finite element equations are derived from the virtual work principle,
∫ δεε ⋅ σ dV i – ∫S δu ⋅ F dS i
= 0
(5.226)
V
The virtual work is considered within a Lagrangian coordinate system. Here, ε = E c denotes the Cauchy–Green strain tensor, σ is a Piola stress tensor referring to initial state, and F is a vector of external forces acting on boundary S. (The superscript c is omitted for convenience.) Both strain and stress tensors can be expressed as six-component vectors, ε = [ ε x ,ε y ,ε z ,ε xy ,ε yz ,ε zx ] T
(5.227)
σ = [ σ x ,σ y ,σ z ,σ xy ,σ yz ,σ zx ] T
(5.228)
and
Within a specific element, the variation of the strain tensor is defined as δεε = Bδδ u
(5.229)
where u is the vector of nodal displacements, and matrix B represents a straindisplacement transformation of the element. The strain vector can be expressed as a sum of linear and quadratic terms, depending upon displacements u. ε = εL + εN
(5.230)
Also, matrix B is expressed as a sum, B = B L + BN
(5.231)
Matrix B L represents a linear function as defined in Equation (2.24). Matrix B N denotes a nonlinear function so that εN = B N u
(5.232)
To derive B N, consider the Cartesian components of nonlinear strain vector εN.
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181
1 ∂u 2 ∂v 2 ∂w 2 ( ε x ) N = --- ------ + ------ + ------- ∂x ∂x 2 ∂x 1 ∂u ∂u ∂v ∂v ∂w ∂w ( ε xy ) N = --- ------ ⋅ ------ + ------ ⋅ ----- + ------- ⋅ ------- 2 ∂x ∂y ∂x ∂y ∂x ∂y etc.
(5.233)
The above equations lead to the expression of εN in the form T
ax 0 0 T
0 ay 0
ax T 1 0 0 az 1 ε N = -- a y = --2- Aa 2 T T 0 az a y az T T az 0 a x T
(5.234)
T
ay ax 0 where A denotes the matrix and a is a vector whose components equal ax =
∂u -----∂x ∂v -----∂x ∂w ------∂x
∂u ------ ∂y ∂v a y = ----- ∂y ∂w ------- ∂y
az =
∂u -----∂z ∂v ----∂z ∂w ------∂z
(5.235)
Refer to Zienkiewicz.9 Using shape function N, Equation (2.17), the latter vectors become
ax =
∂N 1 --------∂x
0
0
∂N 2 --------∂x
0
0 … 0
0
∂N 1 --------∂x
0
0
∂N ---------2 ∂x
0 … 0
0
0
∂N 1 --------∂x
0
0
∂N ∂N ---------2 … ---------n ∂x ∂x
∂N 1 ∂N 2 ∂N N ---------I … ----------I = ---------I u ∂x ∂x ∂x and so on.
u
(5.236)
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Let us introduce a variation of Equation (5.234) in the form 1 1 dεε N = --- dAa + --- A da = AG da 2 2
(5.237)
Then, matrix B N can be expressed as B N = AG
5.5.1
(5.238)
STIFFNESS MATRIX
Equation (5.226) can be written in the form δu
T
∫B V
T
σ dV – δu T Fn = 0
(5.239)
where Fn denotes the vector of applied nodal forces. From the latter equation, one obtains the equilibrium condition, r
n
R (u ) = F – F = 0
(5.240)
where Fr is the structure’s resistance, represented by a vector of internal forces, r
F =
∫B V
T
σ dV
(5.241)
(See Section 3.3.1.) R(u) is a residual vector of unbalanced nodal forces, which appear during the computational process, and should be minimized. In the presence of geometric nonlinearity and nonlinear elastic stress-strain relations, Equation (5.240) becomes nonlinear and is solved by an iterative procedure. We solve the nonlinear problem, using the following recurrence equation: t
R (u i + 1 ) = R (u i ) + K ( u i )∆u i
(5.242)
See Chapter 3. Kt(u i) is a nonlinear tangent stiffness matrix that varies during the iterative process. At a specific iteration, Kt(u i) is obtained from Equation (5.240) through differentiation. r
dR (u i ) dF t K ( u i ) = ----------------- = --------i du i du i Thus,
(5.243)
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Large Displacements
183 r
d dF t T T K ( u ) = -------- = ------ ∫ dB σ dV + ∫ B dσ dV du V du V
(5.244)
(Index i is being omitted for simplicity.) Now, dB = d ( B L + B N ) = dB N
(5.245)
σ = D dεε = DB du dσ
(5.246)
Thus, t
K (u) =
T T N dB --------( B + B ) D ( B + B )dV + L N ∫ L N ∫ du σ dV V V
(5.247)
The above can be expressed as t
K (u ) = KL + KR + Kσ
(5.248)
where KL =
∫ B L DB L dV V
KR =
∫ ( B L DB N + B N DB L + B N DB N ) dV V
Kσ =
dB N - σ dV ∫ --------du
T
T
(5.249)
T
T
(5.250)
T
(5.251)
V
KL is the linear small-displacement stiffness matrix. It includes the elasticity matrix D, which is constant for linear elastic material. KR is so-called initial displacement matrix, which defines stiffness due to large rotations. The last term, Kσ, is known as the initial stress or geometric matrix.9
REFERENCES 1. Lurie, A.I., Nonlinear Theory of Elasticity (translated from Russian), North Holland, Amsterdam,1990. 2. Green, A.E., and Zerna, W., Theoretical Elasticity, Oxford University Press, Oxford, 1968. 3. Green, A.E., and Adkins, J.E., Large Elastic Deformations, Oxford University Press, Oxford, 1970.
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Nonlinear Problems in Machine Design 4. Nadai, A., Theory of Flow and Fracture of Solids, McGraw-Hill, New York, 1950. 5. Oden, J.T., Finite Elements of Nonlinear Continua, McGraw-Hill, New York, 1972. 6. Truesdell, C., and Noll, W., The nonlinear field theories of mechanics, Encyclopedia of Physics, Vol. III/3, Springer Verlag, New York, 1965. 7. Mooney, M., A theory of large elastic deformations, J. Appl. Mech., 11, 582–592, 1940. 8. Rivlin, R.S., Forty years of nonlinear continuum mechanics, Proceedings of the IX International Congress on Rheology, 1–29, Mexico, 1984. 9. Zienkiewicz, O.C., The Finite Element Method, McGraw-Hill, London, 1977.
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6
Contact Problems
6.1 INTRODUCTION In a working machine, the transfer of motion and forces between two parts takes place by means of contact. Prior to performing a stress analysis of the involved parts, it is necessary to consider the distribution of forces and displacements due to deformation and friction during contact. The interaction occurs along a contact surface, which acts as a geometrical constraint confining the bodies’ displacement. The contact surface, in most cases, is a variable and is a function of the applied loading. We classify the contact according to the effect of loading on the size of the surface as advancing, stationary, or receding. Only in the stationary contact does the surface remain constant. See Figure 6.1. The region encompassing the contact surface in all its variations, under the applied loading, is defined as the contact zone (see Figure 6.2). Within this domain, the rules governing the forces and displacements depend on their direction in relation to the contact surface: the normal direction and the tangential one. A major factor
FIGURE 6.1 Types of contact: (a) advancing, (b) stationary, and (c) receding.
FIGURE 6.2 Contact surface and contact zone. 185
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in considering the tangential direction is the frictional phenomena. The magnitude of forces in tangential direction depends on the surface characteristics of the parts and varies extensively, ranging from a mere 3 percent of the normal forces to as high as 40 percent. The behavior of leaf springs, for instance, is seen in the higher range (see Chapter 9), while threaded fasteners show up mostly in the lower range (see Chapter 10). In some instances, where the tangential forces are at a lower end and their magnitude is negligible, the friction is ignored. The solution to contact problems generally requires comprehensive numerical techniques involving advanced FE procedures.
6.1.1
MECHANICS
OF THE
CONTACT ZONE
Consider the theoretical aspects of contact condition. The vectors of forces and displacements in the contact zone comprise normal and tangential components. Because the correlations between the normal components are different from those existing between the tangential components, it is necessary to view each set independently. Moreover, the normal forces are determined from equilibrium conditions and are limited to compressive forces; the normal displacements are dictated by geometric constraints within the contact zone. The analysis of normal forces, therefore, follows the principles of classical mechanics, and consideration of constraint conditions suffices to establish the basis for analysis when the contact is free from friction. The tangential forces and displacements, on the other hand, depart from the rules of classical mechanics due to frictional phenomena that cause energy dissipation. Analysis of tangential components, therefore, is more complex and requires additional scrutiny. Normal Forces and Displacements The normal forces are confined to compression, i.e., fn ≤ 0
(6.1)
The displacements are constrained by the geometric inequality of the contact zone. (2)
(1)
un – un
= gn ≥ 0
(6.2)
where gn denotes the distance between the bodies. See Figure 6.3a. Expressions (6.1) and (6.2) are mutually dependent: as condition (6.1) turns to equality, condition (6.2) turns to inequality, and vice versa. This complementary relationship may be expressed as gn f n = 0
(6.3)
which is known in mathematical programming as the Kuhn–Tucker condition. All of the above provide sufficient mathematical definitions and form the basis for analysis of the contact condition when friction is not a factor.
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FIGURE 6.3 Forces and displacements in the contact zone: (a) normal and (b) tangential.
Tangential Forces and Displacements Analysis of the correlations between the tangential forces and displacements is more complicated because of frictional resistance to motion in the contact zone. Refer now to Figure 6.3b, with tangential forces and displacements present. There are several known approaches that consider the frictional effects in engineering problems. The most prevalent is Coulomb theory, presented below. The theory introduces coefficient of friction µ to define the magnitude of frictional resistance, µ f n . The frictional resistance represents a limiting value, which must not be greater than a tangential force. When a magnitude of tangential force is less than frictional resistance, a cohesion within the contact zone takes place. When a magnitude of tangential force reaches the value of frictional resistance, a sliding takes place. The former process is defined as sticking contact, while the latter is called sliding contact. The correlation between forces and displacements is shown in Figure 6.4. The figure presents schematically the two frictional modes, showing tangential force ft versus relative tangential displacement gt of one body against the other. Relative displacement gt is negligible in the sticking mode but becomes more significant in the sliding mode. The separation point between the sticking and sliding modes identified by singular point A in Figure 6.4 is an approximation used in
FIGURE 6.4 Tangential force vs. relative displacement.
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Coulomb theory. The present analysis considers the approximation. It must be noted, though, that in fact a transition zone, and not just a point, exists between the modes, as seen in the figure. See Kragelski et al.1 The two modes, according to Coulomb theory, are defined as outlined below. Sticking Mode of Friction Due to cohesion, there is no relative motion of bodies, i.e., the tangential displacements of both bodies at the contact surface are assumed to be equal. (2)
ut
(1)
= ut
(6.4)
The tangential forces acting on the contact surfaces do not reach the magnitude of frictional resistance. f t < µs f n
(6.5)
where µs is a static coefficient of friction. Sliding Mode of Friction One body slides over the other, i.e., their tangential displacements at the contact surface are unequal. (2)
(1)
ut – ut
= gt
(6.6)
The tangential force equals the frictional resistance and is directed against it, i.e., f t = – sgn ( g t ) µ d f n
(6.7)
where µd is the dynamic coefficient of friction. Note: Whenever sliding occurs, a dynamic process takes place. Therefore, dis(i) (i) placements u t and g t in Equation (6.6) should be replaced by velocities u˙ t and g˙ t , respectively. Similarly, in Equation (6.7), forces ft and fn should be replaced by rates f˙t and f˙n . In practice, when computing contact problems with friction, we apply incremental loading so that correlations (6.6) and (6.7) can represent a quasistatic condition that is equivalent to a dynamic process. More about incremental loading is presented below.
6.1.2
FINITE ELEMENT METHOD
The finite element method, in application to the contact problem, requires determination of the potential energy of the entire system as well as the contact zone. The contact is perceived as an engagement of surface nodes of one body with the facing elements of another body. Figure 6.5 shows a two-dimensional contact zone formed by the interacting nodes and the facing elements.
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189
FIGURE 6.5 Finite element representation of the contact zone.
Let us denote the potential energy of the contact surface by Πc. The potential energy of the whole system Π therefore comprises the potential energy Πb of the bodies in contact (see Chapter 3), supplemented by the new term Πc. 1 T T Π = Π b + Π c = --- ∫ ε σ dV – ∫ u FdS + Π c 2V S
(6.8)
The solution pertains to minimization of potential Π: one has to solve Equation (6.9) for unknown node displacements u, subject to constraints (6.1) and (6.2), ∂ ( Πb + Πc ) -δu = 0 δΠ = δΠ b + δΠ c = -------------------------∂u
(6.9)
Complying with the finite element theory, Equation (6.9) is based on an assumption that all contact forces are derivable from potential Πc. The main difficulty in solving Equation (6.9) pertains to determining δΠc, the first variation of potential energy, which is equivalent to virtual work of forces acting on the contact surface. The said virtual work is of tangential forces and thus requires a special treatment. To reach a solution applying the FE method, one assumes the tangential forces to be monogenic, i.e., derivable from potential energy, ignoring the dissipative nature of friction. Therefore, the solution of a contact problem becomes path dependent; to correct the effect of ignored energy dissipation, an incremental loading path must be used. Equation (6.9) is solved numerically, as conditions (6.1) and (6.2) are taken into consideration. The step requires the reliance on constrained optimization of mathematical programming.2,3 Presented here are two basic theories that, although different in their approaches, offer the desired solutions. Their ultimate expressions are easily adaptable to the FE method. One of the theories is known as the penalty function method, and the other as the Lagrange multipliers method. The main difference between the two is their approach to the potential energy term Πc. The penalty function method, due to its
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Nonlinear Problems in Machine Design
economy, has received a wider acceptance, albeit with inherent numerical inaccuracies. The method is acknowledged for its efficiency in solving contact problems with friction, while the Lagrange method, based on multipliers, is known for its accuracy. However, in applying the latter method, convergence toward solution is problematic at times. The discussion below refers to other methods that are in fact extensions of the Lagrange method, such as the augmented Lagrange method, which works with penalty function. Also included are alternate methods based on regularization, such as the perturbed Lagrangian method and the constraint function method.
6.2 PENALTY METHOD The penalty method of constrained optimization pertains to adding a penalty term to the optimized function to enhance the optimization process. In application to the contact problem, the penalty term includes a stiffness matrix of the contact surface based on a concept of an imaginary penetration of one body into another.4,5 The stiffness matrix of the contact surface is added to the stiffness matrix of the bodies under consideration so that, at any load increment, the equilibrium equation has the form [ K b + K c ]u = F
(6.10)
The magnitude of the contact surface is unknown; therefore, its stiffness matrix Kc is a nonlinear term. The total load and displacement values are F
tot
=
∑ ∆F
(6.11)
u
tot
=
∑ ∆u
(6.12)
To derive the stiffness matrix, the contact zone (encompassing the contact surface) is divided into a series of specially designed contact elements. The element represents an interaction between the surface node of one body with the respective element face of the other body. Figure 6.6 shows a contact element in a twodimensional application. It comprises a slave node (point S) and a master line, connecting nodes 1 and 2. S0 marks the slave node before the application of the load increment, and S marks the node after loading. The location of the three nodes i = S, 1, 2 is defined by the equation xi = Xi + ui
(6.13)
where Xi denotes the location before loading, and ui is the incremental displacement resulting from loading. The vector of all node displacements of the contact element equals
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191
FIGURE 6.6 Contact element.
u =
us u1 = u2
uS vS u1 v1 u2 v2
(6.14)
We introduce a local nondimensional Cartesian coordinate system (t,n) within the element, with origin at point 1 and direction t coinciding with master line 1–2, so that t = 0 at point 1 and t = 1.0 at point 2. The direction of the local coordinate system with respect to the global coordinate system is defined by angle β (Figure 6.6). This local coordinate system rotates along with the contact element, forming so-called co-rotational frame. The derivation of the element stiffness is based on the assumption that the contact is represented by an imaginary penetration of the slave node into the facing element. The penetration is resisted by the stiffness of the contact element. Thus, the following condition exists: gn < 0
(6.15)
The resulted expression, however, contradicts the fundamental concept of contact mechanics, Equation (6.1), and the Kuhn–Tucker law, Equation (6.3), which explain an unavoidable numerical error. To keep the penetration error small, the stiffness has to be large. The following is a derivation of the stiffness matrices of the contact element. The analysis will refer to the sticking and sliding modes of friction, presented separately.
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6.2.1
STICKING MODE
OF
FRICTION
A contact element in the sticking mode is illustrated in Figure 6.6. Point S marks the slave node in penetrated position after loading. The vector of the penetration equals g g = n gt
(6.16)
Normal component gn acts against the elastic stiffness of the element, while tangential component gt acts against frictional stiffness. The contact force that acts upon the slave node equals f = –κ g
(6.17)
where κ is a penalty parameter, which is expressed by the two-dimensional matrix
κ =
κn 0
(6.18)
0 κt
The potential energy of contact surface Πc is a sum of potential energies of the contact elements. Πc =
∑ πc
(6.19)
For a given penetration g, the potential energy of the contact element equals 1 T π c = --- g κ g 2
(6.20)
With the help of the above equation, we can derive the stiffness matrix. According to the basic finite element theory, the potential energy of the element with given displacements u is expressed as 1 T e π c = --- u K c u 2
(6.21)
e
with K c representing the stiffness matrix of the element. With potential energy πc known, we can obtain the stiffness matrix from the variational equation, T
e
δ ( δπ c ) = δu K c δu
(6.22)
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193
(See Chapter 2.) The first variation of the potential energy equals T δπ c = δg κ g = κ n g n δg n + κ t g t δg t
(6.23)
and the second variation is T δ ( δπ c ) = δg κ δg = κ n [ δg n δg n + g n δ ( δg n ) ] + κ t [ δg t δg t + g t δ ( δg t ) ]
(6.24)
To derive the expression of stiffness matrix, Equation (6.24) must be expressed in terms of displacements u as shown below. Variational Analysis of Penetration Components of Penetration According to Figure 6.6, the two components of g equal T
(6.25)
T
(6.26)
gn = ( xS – x1 ) n and g t = ( x S – x S0 ) t
(Note: The location of the slave node x S0 before loading is the same as x S0 .) Equation (6.26) can be approximated more conveniently as g t = ( a – a 0 )l
(6.27)
where l is the current (loaded) length of the master line, defined as T
l = ( x2 – x1 ) t =
( x2 – x1 ) T ( x2 – x1 )
(6.28)
The variable a is the current slave node’s coordinate location, 1 T a = --- ( x S – x 1 ) t l
(6.29)
while a0 is the previous (unloaded) location, 1 T a 0 = --- ( x S0 – x 1 ) t l
(6.30)
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First Variations The first variations of gn and gt are the components of virtual displacement δg and are derived from Equations (6.25) and (6.26) as follows: T
T
δg n = δ ( x S – x 1 ) n + ( x S – x 1 ) δn T
T
δg t = δ ( x S – x S0 ) t + ( x S – x S0 ) δt
(6.31) (6.32)
To further develop the above expressions, we introduce variation δβ, which represents the angle between current and previous t-directions (see Figure 6.7). Taking into consideration that angle δβ is very small, we can write 1 T δβ ≈ sin ( δβ ) = --- n ( δu 2 – δu 1 ) l
(6.33)
Consequently, variations δt and δn become 1 T δt = --- n [ n ( δu 2 – δu 1 ) ] l
(6.34)
1 T δn = – --- t [ n ( δu 2 – δu 1 ) ] l
(6.35)
The following variations are equal: δ ( xS – x1 ) = δ ( uS – u1 )
(6.36)
FIGURE 6.7 Variation of angle β: a denotes the original position of a contact element, and b denotes the rotated position of a contact element.
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195
Using Equations (6.29) and (6.36), Equation (6.31) becomes T
δg n = [ δu S – ( 1 – a )δu 1 – aδu 2 ] n
(6.37)
Referring to δgt , Equation (6.32) can be expressed in the form T
T
δg t = δ ( x S – x 1 ) t + ( x S – x 1 ) δt – a 0 δl
(6.38)
Now, recalling Equation (6.25), we have g T T ( x S – x 1 ) δt = -----n [ n ( δu 2 – δu 1 ) ] l
(6.39)
Furthermore, δl =
1 T T T ( x 2 – x 1 ) ( x 2 – x 1 ) = --- ( x 2 – x 1 ) δ ( x 2 – x 1 ) = δ ( u 2 – u 1 ) t (6.40) l
Consequently, we obtain g T T δg t = [ δu S – ( 1 – a 0 )δu 1 – a 0 δu 2 ] t + -----n ( δu 2 – δu 1 ) n l
(6.41)
The final expressions of the virtual components δgn and δg are T
δg n = ( N S ) δu
(6.42)
T g 0 δg t = T S + -----n N 0 δu l
(6.43)
and
where T
T
T
T
( N S ) = [ n , – ( 1 – a )n , – an ] T
T
( N 0 ) T = [ 0, – n , n ] 0 T
T
(6.44) (6.45)
T
T
( T S ) = [ t , – ( 1 – a 0 )t , – a 0 t ]
(6.46)
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Nonlinear Problems in Machine Design
The squares of the respective virtual components equal δg n δg n = δu
T
NS NS
T
δu
(6.47)
and T g g T 0 0 δg t δg t = δu T S + -----n N 0 T S + -----n N 0 δu l l
g g 0 T 0 o T T 0 T T = δu T S ( T S ) + -----n T S ( N 0 ) + N 0 ( T S ) + -----n N 0 ( N 0 ) δu l l
(6.48)
Second Variations The second variation of gn equals T
T
δ ( δg n ) = δ [ ( N S ) δu ] = δ ( N S ) δu
(6.49)
where T
T
T
T
δ ( N S ) = δ [ n , – ( 1 – a )n , – an ] T
T
T
T
T
= [ δn , – ( 1 – a )δn , – aδn ] + [ 0,δan , – δan ] 1 T T T T T T = – --- [ t , – ( 1 – a )t , – at ] [ n ( δu 2 – δu 1 ) ] + [ 0,n , – n ]δa l (6.50) Variation δa equals 1 1 1 T T T δa = --- δ ( x S – x 1 ) t + --- ( x S – x 1 ) δt – ---2 ( x S – x 1 ) tδl l l l 1 1 1 T T T T T = --- ( δu S – δu 1 ) t + ---2 ( x S – x 1 ) n [ N 0 δu ] – ---2 ( x S – x 1 ) t [ t ( δu 2 – δu 1 ) ] l l l g T 1 T = --- [ t, – ( 1 – a )t, – at ] δu + ----2-n N 0 δu l l (6.51) By placing T
T
T
T
( T S ) = [ t , – ( 1 – a )t , – at ]
(6.52)
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197
we get T g 1 1 T δN S = – --- T S [ ( N 0 ) δu ] – --- N 0 T S + -----n N 0 δu l l l
(6.53)
Consequently, by matrix processing, the second variation of gn becomes g 1 T T T T δ ( δg n ) = – --- δu T S ( N 0 ) + N 0 ( T S ) + -----n [ N 0 ( N 0 ) ] δu l l
(6.54)
The second variation of gt equals T
g 0 δ ( δg t ) = δ T S + -----n N 0 δu l
(6.55)
δg g g 0 T T T = δ ( T S ) + -----n δ ( N 0 ) + --------n ( N 0 ) – ----2-n ( N 0 ) δl l l l
(6.56)
where g 0 δ T S + -----n N 0 l
T
Now, 0 T
T
T
T
δ ( T S ) = [ δt , – ( 1 – a 0 )δt , – a 0 δt ] 1 T 1 0 T T T T T = --- [ n , – ( 1 – a 0 )n , – a 0 n ] [ n ( δu 2 – δu 1 ) ] = --- ( N S ) [ ( N 0 ) δu ] l l (6.57) where, for brevity, we write 0 T
T
T
T
( N S ) = [ n , – ( 1 – a 0 )n , – a 0 n ]
(6.58)
Furthermore, using expression (6.35), we have 1 1 T T T T T T δ ( N 0 ) = – --- [ 0, – t ,t ] [ n ( δu 2 – δu 1 ) ] = – --- ( T 0 ) [ ( N 0 ) δu ] l l
(6.59)
where we introduce the expression, T
T
T
( T 0 ) = [ 0, – t , t ]
(6.60)
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Nonlinear Problems in Machine Design
Also, T T δl = ( 0 – δu 1 + δu 2 ) t = ( T 0 ) δ u
(6.61)
Consequently, using Equations (6.56) to (6.61), the second variation of gt becomes g 1 T 0 T T T T δ ( δg t ) = --- δu N S ( N 0 ) + N 0 ( N S ) – -----n [ T 0 ( N 0 ) + N 0 ( T 0 ) ] δu (6.62) l l Stiffness Matrix in the Sticking Mode To define the stiffness matrix of the contact element in the sticking mode, we return to Equations (6.21) and (6.22). Combining both equations, we obtain δu K c δ u = κ n [ δg n δg n + g n δ ( δg n ) ] + κ t [ δg t δg t + g t δ ( δg t ) ] T
e
(6.63)
e
It follows that the stiffness matrix K c has two components, normal and tangential, so that Equation (6.63) can be separated into the following two equations: T
e
δu K c,n δu = κ n [ δg n δg n + g n δ ( δg n ) ] T
e
δu K c,t δu = κ t [ δg t δg t + g t δ ( δg t ) ]
(6.64) (6.65)
Inserting the variational expressions, we obtain the normal and tangential stiffness matrices, g g T T T T e K c,n = κ n N S ( N S ) – -----n T S ( N 0 ) + N 0 ( T S ) + -----n N 0 ( N 0 ) l l
(6.66)
and
e
K c,t
= κt
(6.67) gt 0 gn T T T T + ---- N S ( N 0 ) + N 0 ( N S ) – ----- ( T 0 ( N 0 ) + N 0 ( T 0 ) ) l l g g 0 0 T 0 0 T T T T S ( T S ) + -----n T S ( N 0 ) + N 0 ( T S ) + -----n N 0 ( N 0 ) l l
The stiffness matrix of the whole contact surface is the sum of the element matrices, i.e.,
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Kc =
∑ ( Kc,n + Kc,t )i e
e
(6.68)
i=1
where i denotes the engaged contact elements.
6.2.2
SLIDING MODE
OF
FRICTION
The derivation of the stiffness matrix of the contact element in sliding mode follows a similar path with the one in sticking mode. The basic difference stems from the fact that here tangential force ft equals in magnitude to the frictional resistance µ f n as per Equation (6.7). Normal force fn equals f n = κn gn
(6.69)
f t = – sgn ( g t ) µ d κ n g n = sgn ( g t ) µ d κ n g n
(6.70)
so Equation (6.7) becomes
where gn is negative, because it denotes penetration. Consequently, the first variation of the potential energy of a contact element equals δπ c = f n δg n + f t δg t = κ n g n δg n + sgn ( g t ) µ d κ n g n δg t
(6.71)
Stiffness Matrix in the Sliding Mode As before, the element stiffness is derived from the variational equation, T
e
δ ( δπ c ) = δu K c δu
(6.72)
where the second-variation term equals δ ( δπ c ) = κ n [ δg n δg n + g n δ ( δg n ) ] + sgn ( g t ) µ d κ n [ δg n δg t + g n δ ( δg t ) ] (6.73) The stiffness matrix contains normal and tangential components, e
e
e
K c = K c,n + K c,t
(6.74)
e
It follows that the normal stiffness matrix K c,n [see Equation (6.66)] remains the same while the tangential stiffness matrix becomes
K
e c,t
0 T T 0 0 g n 2N 0 ( N S ) + N S ( N 0 ) = sgn ( g t ) µ d κ n T S ( N S ) + ----- g l – -----n ( T ( N ) T + N ( T T )) 0 0 l 0 0
(6.75)
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Slip-Surface Correction To perceive the process of conversion from sticking to sliding mode and to describe the sliding mode in mathematical terms, it is convenient to visualize Equations (6.6) and (6.7) in geometrical form as a slip surface. To define the slip surface, let us express Equation (6.7) by means of function Y (f ) = f t – µ f n = 0
(6.76)
For better understanding, consider a three-dimensional contact. The tangential force acting on the contact surface becomes f ( f tl, f t2 ) , and function Y(f) takes the form Y (f ) =
2
2
f tl + f t2 – µ f n = 0
(6.77)
The function sets the limits of sticking contact. It defines slip conditions when the contact starts to slide. For illustration, function Y(f) is presented as slip surface in force-space ( f tl , f t2 ) . See Figure 6.8. (We assume the dynamic and static friction coefficients to be equal µ.) The function resembles yield function Y(σ) of plasticity theory, defining the start of plastic flow.6 A load step involving sliding mode at a first iteration in a numerical solution usually produces a result that exceeds the limits defined by slip surface Y(f). A correction becomes necessary so that it fits the slip surface. A return-mapping correction method is described as follows.7,8 Let us return to two-dimensional force space, Figure 6.9, as per Equation (6.76). Let points A, B, and C respectively denote tangential forces before the load step (n), after the step (trial), and after the correction (n + 1). The correct tangential force at the slip surface is
FIGURE 6.8 Slip surface Y = 0 in force space: a denotes force path in the sticking mode, and b denotes force path in the sliding mode.
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FIGURE 6.9 Return mapping in two-dimensional force space.
n+1
ft
trial
= ft
– ∆ft
(6.78)
In terms of displacements, it equals n+1
ft
trial
= κt ( gt
– ∆g t )
(6.79)
trial
where g t is the trial displacement and ∆gt is a correction term. To determine the value of the correction term, we introduce the scalar magnitude ∆ζ defined by the equation trial
∆g t = sgn ( f t ) ∆ς
(6.80)
Using ∆ζ, the correct tangential force at the slip surface is expressed by the equation n+1
ft
trial
= ft
trial
– κ t sgn ( f t ) ∆ς
(6.81)
During the computational process, ∆ζ is determined from the following conditions: ∆ζ = 0
if Y ( f
trial
trial
Y (f ) ∆ζ = ----------------κt
if
trial
) = ft Y (f
trial
trial
–µ f n
trial
) = ft
≤0
(6.82) trial
– µ fn
>0
(6.83)
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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)
Λ T = [ λ 1n , λ 2n , …, λ kn ]
(6.87)
T
where
1
2
k T
g = [ g n , g n , …, g n ]
(6.88)
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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 1 2 k 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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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
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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 Λ Ti + 1 A i + 1 = ( Λ Ti + ∆ i + 1 Λ T ) 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
∂A ∂ g 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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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 T g + Λ T δg + g T κ δg Λ ) = δΠ b ( u ) + δΛ 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)
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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
Y (f ) ∆ζ = ----------------κt
6.3.3
if
trial
) = ft
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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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 ) - – ----------------------+ εn w ( g n , f n ) = --------------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 π εt
(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)
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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.
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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
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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.
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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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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.
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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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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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By analogy with Equation (4.12) for monotonic loading (Chapter 4), the elastic strain amplitude in cyclic loading can be expressed as e
∆σ ∆ε -------- = ------2 2E
(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' ∆ε ∆σ ------ = -------- + -------- 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 2 2E
(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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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′ ∆σ ∆ε b -------- = ------- = -------f ( 2N f ) E 2 2E
(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
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219 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.
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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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∆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' ∆ε ∆σ ------ = ------- + ε′ f ----------- 2σ′ f 2 2E
(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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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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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 ∆σ ∆ε b - ( 2N f ) -------- = ------- = ------------------E 2 2E
(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 b c - ( 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)
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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
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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 x 1 e p -------- = --- ∑ ( δε x + δε x ) 2 cycle 2
∆γ xy 1 e p ---------= --- ∑ ( δγ xy + δγ xy ) 4 4 cycle
∆ε y 1 e p -------- = --- ∑ ( δε y + δε y ) 2 cycle 2
∆γ yz 1 e p ---------- = --- ∑ ( δγ yz + δγ yz ) 4 4 cycle
∆ε z 1 e p -------- = --- ∑ ( δε z + δε z ) 2 cycle 2
∆γ zx 1 e p ---------- = --- ∑ ( δγ zx + δγ zx ) 4 4 cycle
(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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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 2 ∆ε ∆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
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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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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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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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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.
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(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.
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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
∂Π 2πaσ ------- = ---------------- – 4γ e = 0 ∂a E
(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-
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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<
(7.58)
The displacement components, according to the same linear elastic solution, equal K r u i = ------- ------φ i ( θ ) 2G 2π
(7.59)
where K is a constant, called the stress intensity factor. (Note: Here G denotes the shear modulus.) Figure 7.22 shows three different loading modes: mode I, the opening mode, where the crack’s sides move perpendicularly apart; mode II, the sliding mode, where the crack’s sides slide in a parallel fashion, within the plane of the crack; and
FIGURE 7.21 Coordinate system in the vicinity of a crack.
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FIGURE 7.22 Three modes of loading in a crack.
mode III, the tearing mode, where the crack’s sides are torn apart, and the surfaces move out of the plane. Referring to Figure 7.21, the stresses and the displacements for mode I are as follows. • Stresses in xy-plane: KI θ θ 3θ - cos --- 1 – sin --- sin ------ σ x = -----------2 2 2 2πr KI θ 3θ θ - cos --- 1 + sin --- sin ------ σ y = ----------- 2 2 2 2πr KI θ 3θ θ - sin --- cos --- cos -----σ xy = -----------2 2 2 2πr
(7.60)
• Stresses in z-direction applicable to plane strain problems: σz = v ( σ x + σ y ) τ zx = τ yz = 0
(7.61)
• Stresses in z-direction applicable to plane stress problems: σ z = τ zx = τ yz = 0
(7.62)
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• Displacements in xy-plane: K r θ 2θ u = ------I- ------ cos --- κ – 1 + 2sin --- 2G 2π 2 2 K r θ 2θ v = ------I- ------ sin --- κ – 1 + 2cos --- 2 2G 2π 2
(7.63)
(where G denotes the shear modulus). For plane strain problems, κ equals κ = 3 – 4v
(7.64)
while for plane stress problems, κ becomes 3–v κ = -----------1+v
(7.65)
For modes II and III shown in Figure 7.22, the trigonometric functions fij and φ differ.18 Stress Intensity Factor Stress intensity factor KI signifies the relationship among the following factors: the geometry of the plate, the loading, and the length of the crack. It is defined as follows: K I = f I σ πa
(7.66)
In the above, fI is a compliance function that describes the geometry of the part, and σ is a stress in a remote distance, which designates the loading. For the specimen, shown in Figure 7.23, the compliance functions are presented below.19 Center Crack Loaded in Tension
fI =
πa sec -----2b
(7.67)
Edge Crack Loaded in Tension a a 2 a 3 a 4 f I = 1.12 – 0.231 --- + 10.55 --- – 21.72 --- + 30.39 --- b b b b
(7.68)
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FIGURE 7.23 Concerning mode I: (a) center-cracked plate, (b) edge-cracked plate, and (c) double-edge cracked plate.
Double Edge Cracks Loaded in Tension a a 2 a 3 f I = 1.12 + 0.203 --- – 1.197 --- + 1.93 --- b b b
(7.69)
For small cracks ( a/b<<1 ) , one can simplify the above equations by approximations, expressing the compliance functions as follows: K I = ασ πa
(7.70)
where
α = 1.0 for a center crack loaded in tension, α = 1.12 for an edge crack (or double cracks) loaded in tension Relationship between Energy Release Rate and the Stress Intensity Factor To derive this relationship, consider mode I. Assume that the strain energy released during the crack growth equals the work required to force its closure.8 See Figure 7.24. We shall evaluate the work along the segment 0 < x < δa on 0-x axis between the two instances, based on Equations (7.60) through (7.65). In the open condition, in Figure 7.24a, we assume that r = δ ( a – x ) and θ = π , whereby the displacements along the segment equal K δa – x v = ------I- --------------- ( κ + 1 ) 2G 2π
(7.71)
where G denotes the shear modulus. In the closed position, r = x and θ = 0 , Figure 7.24b, and the stresses along the same segment are
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FIGURE 7.24 Closure of a crack.
KI σ y = ------------2πx
(7.72)
Consequently, the strain energy that corresponds to the closure of crack equals 1 δa δU = – 2 ⋅ --- ∫ σv dx 2 0 2
δa δa – x K1 - (κ + 1)∫ = – -----------------------4Gπ x 0
(7.73)
Using the expression 2
x = δasin ϕ
(7.74)
the right-hand integral in Equation (7.73) becomes δa
∫0
π/2 δa – x πδa --------------- dx = δa ∫ ( 1 + cos 2ϕ ) dϕ = --------x 2 0
(7.75)
It follows that the energy release rate GI is a function of the stress intensity factor KI in the form 11 + v δU 2 G I = – ------- = --- ------------ ( κ + 1 )K I 4 E δa
(7.76)
For plane stress condition, the energy release rate, by virtue of Equation (7.65), can be expressed as 2
K G I = ------I E
(7.77)
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while, for the plane strain condition, based on Equation (7.64), the energy release rate equals 2
2 K G I = ( 1 – v ) ------I E
(7.78)
Equivalent relations can be derived for modes II and III. Fracture Toughness When the energy release rate reaches a critical value, an instability occurs, and the final fracture takes place. The instance of instability can be expressed by either the critical energy release rate Gc or the critical stress intensity factor Kc, the latter being called the fracture toughness. The fracture toughness is determined experimentally from tests with predetermined crack size a. Figure 7.25 presents a typical standard test specimen that is used to measure the fracture toughness of metals. Experimenting with metals for fracture toughness, one finds that the results are influenced by the thickness of the specimen. Fracture toughness KIc in the plane stress condition is larger than the one in the plane strain condition; KIc decreases assymptotically as the specimen’s thickness increases. See Figure 7.26. In test procedures for fracture toughness, for purposes of obtaining the condition of the plane strain, a certain thickness must be maintained. Only then can a limit value of fracture toughness be derived for use in standard test procedures. It was determined empirically that, for a plane strain condition, the minimum specimen thickness B is K 2 B ≥ 2.5 ------Ic- σ yp
FIGURE 7.25 Standard specimen for fracture toughness tests.
(7.79)
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FIGURE 7.26 Effect of specimen thickness on fracture.
7.3.3
CRACK PROPAGATION
UNDER
FATIGUE LOAD
Up to now, the discussion of crack propagation has been concerned with monotonic loading, resulting in steady crack growth with a final failure. Experience shows that, in metal parts subject to fluctuating loading, the process includes a preparatory phase, during which crack growth is subcritical, followed by the critical rupture. This is indicated in Figure 7.19, where surface of the crack, up to the rupture zone, is different from the remainder of the surface. This confirms the fact that the stress intensity factor has reached the critical value. Figure 7.27 presents the history of crack growth under a fluctuating load. Attention is given to the fact that the crack grows exponentially, slowly at first, with the rate of propagation accelerating continuously. Point C in the figure indicates the final failure, which occurs when the crack size reaches ac. Here, the stress intensity factor reaches the critical value Kc. In the figure, the fatigue histories of two specimens are compared. Both are subject to different stress ranges. One should note that the specimen subject to higher stresses reaches the critical point earlier and fails from a crack of a smaller size. In the following discussion, let us consider crack growth under fluctuating loading as a function of stress intensity factor K. Assume a load that fluctuates at a constant amplitude, where the stresses vary between constant limits Smax and Smin. The range of the stress intensity can be expressed as ∆K = K max – K min = α ( S max – S min ) πa
(7.80)
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FIGURE 7.27 Fatigue crack propagation at constant stress range ∆σ. Note: ∆σ1 > ∆σ2.
To define a mathematical formulation of crack propagation, we refer to typical test data20 shown in Figure 7.28. The figure presents the crack growth rate per cycle plotted versus ∆K. A careful study of the results shows that the data fit into a narrow band that can be approximated by a sigmoidal curve as shown in Figure 7.29. The curve can be divided into three distinct sections that represent different regions of crack growth: region I, crack formation; region II, moderate crack propagation; and region III, accelerated crack growth and fracture. Point ∆Kth denotes a threshold value, the onset of the crack growth. Point ∆K c = K c – K min reflects the final failure. Let us analyze region II. According to Paris et al.,8 the crack growth can be expressed in the form of a function da ------- = f ( ∆K ) dN
(7.81)
Indeed, the test data in region II fit an exponential correlation, da n ------- = C ( ∆K ) dN
(7.82)
where C is a constant factor and n an exponent taken from tests. See Appendix C. The above equation represents the fatigue crack propagation law. To compute the life expectancy of a cracked part, we use a two-step approach. Equation (7.83) is inverted, da dN = -------------------n C ( ∆K ) and the latter is integrated to get the expression (valid for n ≠ 2),
(7.83)
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FIGURE 7.28 Crack growth rate vs. stress intensity factor for aluminum 7075-T76, Smin = 0. Data from Hudson.19
FIGURE 7.29 Three regions of crack growth rate.
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Nonlinear Problems in Machine Design af 1 a f da 1 da N f = ---- ∫ ---------------n = -----------------------------------------------------n ∫ ------n/2 a C i ( ∆K ) C [ α ( S max – S min ) π ] ai a
2 1 1 = ----------------------------------------------------------------------n ---------------- – ---------------( n – 2 )/2 ( n – 2 )/2 af ( n – 2 )C [ α ( S max – S min ) π ] a i
(7.84)
Variable ai is the initial crack length computed from the equation ∆K th 1 a i = --- --------------------------------π α ( S max – S min )
2
(7.85)
using ∆K th = K th – K min
(7.86)
Variable af is the crack length at failure computed from Equation (5.88) on the basis of ∆K c 1 a f = --- --------------------------------π α ( S max – S min )
2
∆K c = K c – K min
(7.87)
(7.88)
Where the life expectancy is computed with the size of a crack given, Equation (7.85) becomes superfluous. Experimentally obtained values of Kth, and Kc for different metals are listed in the Appendix. Effect of Smin Let us now examine how the life expectancy analysis is influenced by changing the minimum stress. Consider three cases with Smin values that are positive, negative, or equal to zero. See Figure 7.30. Case (a) Smin = 0, Figure 7.30a Here, the range of stress intensity is ∆K = K max = αS max πa
(7.89)
Case (b) Smin < 0, Figure 7.30b The condition of compression at the minimum stress level reflects the fact that the crack stops growing; the propagation occurs at tensional stresses only. Therefore, we may ignore Smin and apply here the same Equation (7.89).
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FIGURE 7.30 Three cases of fatigue loading at constant amplitude.
Case (c) Smin > 0, Figure 7.30c The applied minimum stress can have a significant effect on crack growth. The resultant crack-growth rate can be expressed by revising the basic correlation, Equation (7.81), to become da ------- = f ( ∆K ,R ) dN
(7.90)
where R is the stress ratio defined by S min R = -------S max
(7.91)
Figure 7.31 shows the influence of stress ratio R on crack growth. There are a number of empirical correlations to fit the data. One, developed by Forman, Kearney, and Engle,21 has the form
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FIGURE 7.31 The influence of stress ratio on crack-growth rate for aluminum 7075-T76, Smin = 0. Data from Hudson.19 n
da A ( ∆K ) ------- = ------------------------------------------dN ( 1 – R ) ( K c – ∆K )
(7.92)
where A and n are material properties. Variable Amplitude Loading One of most difficult problems is to predict life expectancy in parts, subject to spectrum loading. It represents a condition where the loading cycles vary and produce different magnitudes of peaks and valleys. See Figure 7.16. It is known that fatigue damage is caused mostly by extreme events, even if few in number. The modern methods to arrive at life expectancy require selective recording of the more significant events, to be utilized in the analysis. Several approaches are possible to select the significant peaks and valleys. Figure 7.32 shows the overall range method.2 It pertains to defining then selecting the highest peaks and lowest valleys in a sequence. (The overall range is defined as the algebraic difference between successive selected valleys and peaks.) Out of numerous methods to compute the life expectancy of a cracked part under spectrum loading, we chose to present here two methods: the block method and the statistical method.
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FIGURE 7.32 Overall range method of counting peaks and valleys: (a) original history and (b) screened history.
The block method is based on arranging the random history of load fluctuations in terms of blocks.2 We present the crack growth in terms of blocks as follows: k
∆N B =
∆a
i ∑ -----------------------f ( ∆K i ,R i )
(7.93)
k=1
Summation of Equation (5.93) provides the final life in terms of number of fluctuations to failure, N =
∑ ∆N B
(7.94)
The other method, the statistical method, assumes that under random loading the rate of crack growth is practically the same as under constant amplitude loading.2 One introduces a statistical root-mean-square intensity factor based on the equation k
∑ ( ∆K i )
K rms =
i=1
2
-----------------------k
(7.95)
where k is the number of stress conditions ( ∆S,S m ) , per block. The equation allows us to compute the average fatigue crack growth rate using the correlation da n ------- = C ( ∆K rms ) dN
(7.96)
From here, the computation of life expectancy continues in the manner shown previously.
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REFERENCES 1. Woehler, A. 1858. ”Uber die Festigkeitsversuche mit Eisen und Stahl,” Z. Bauwesen. 8, 641. 2. SAE Fatigue Design Handbook, 3rd edition, Society of Automotive Engineers, Warrendale, PA, 1997. 3. Manson, S.S., Behavior of materials under conditions of thermal stress, NACA Tech. Note 2933, 1954. 4. Coffin, L.F., A study of cyclic-thermal stresses in a ductile material, Trans. ASME. 76, 931–950, 1954. 5. Morrow, J. Cyclic plastic strain energy and fatigue of metals, Internal Friction, Damping and Cyclic Plasticity, ASTM STP 378, 45–87, 1965. 6. Socie, D.F., Fatigue life prediction using local stress-strain concepts, Experimental Mechanics, 17, No, 2, 1977. 7. Smith, R.W., M.H. Hirschberg, and S.S. Manson, Fatigue behavior of materials under strain cycling in low and intermediate life range, NASA Tech. Note, D-1574, 1963. 8. Neuber, H., Theory of stress concentration for shear-strained prismatic bodies, Trans. ASME J. Appl. Mech., 28, 544, 1961. 9. Manson, S.S., Effect of mean stress and strain on cyclic Life, Machine Design, August 4, 129–135, 1960. 10. Morrow, J., Fatigue properties of metals, Fatigue Design Handbook, pp. 21–30, Society of Automotive Engineers, Warrendale, PA., 1968. 11. Bannantine, J.A., Comer, J.J., and Handrock, J.L., Fundamentals of Metal Fatigue Analysis, Prentice Hall, Englewood Cliffs, New Jersey, 1990. 12. Sines, G. Behaviour of metals under complex static and alternating stresses, Metal Fatigue, ed. Sines, G. and G.L. Waisman, McGraw-Hill, New York, 1959. 13. Fuchs, H.O., and Stephens, R.I., Metal Fatigue in Engineering, Wiley, New York, 1980. 14. Miner, M.A., Cumulative damage in fatigue. Trans. ASME, J. Appl. Mech., 67, A159, 1945. 15. Griffith, A.A., The phenomena of rupture and fracture in solids, Phil. Trans. Roy. Soc. A 221, 163–197, 1920. 16. Irwin, G.R., Analysis of Stresses and Strains Near the End of a Crack Traversing a Plate, Trans. ASME, Jour. Appl. Mechanics, 24, 361–364, 1957. 17. Paris, P.G., Gomez, M.P., and Anderson, W.E., A Rational Analytic Theory of Fatigue,” The Trend in Engineering, 13, 9–14, 1961. 18. Westergaard, H.M., Bearing pressures and cracks, Trans. ASME, J. Appl. Mech., 6, 49–53, 1939. 19. Sih, G.C., Handbook of Stress Intensity Factors for Researchers and Engineers, Institute of Fracture and Solid Mechanics, Bethlehem, PA: Lehigh University, 1973. 20. Hudson, C.M., NASA Tech. Note D-5390, 1969. 21. Forman, P.G., Kearney, V.E., and Engle, R.M., Numerical Analysis of Crack Propagation in Cyclic-Loaded Structures, Trans. ASME J. Basic. Eng., 89, 459–464, 1967.
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Part II Design Cases
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8
Design of Machine Parts
This chapter and the four following it form the second part of the book—the part that deals with the application of the theories presented in the first part of the book, which is composed of Chapters 1 through 7. The design cases described in this part of the book were chosen by virtue of the relevance of the machine design problems they represent. This chapter serves as an introduction to the problems, the solution of which is found in the following chapters.
8.1 NONLINEAR BEHAVIOR OF MACHINE PARTS The ideal of modern machine design is to strive for a machine with maximum power and minimum weight—a machine with optimally sized parts and high stress concentration, safeguarded at critical locations. To achieve these design goals, one faces problems that require nonlinear analysis.
8.1.1
SOURCES
OF
NONLINEARITY
The nonlinear problems arise from several sources inherent in the behavior of machine parts under loading. Let us examine those that are caused by irreversible plastic deformation, formidable geometrical changes, and the parts’ interaction through contact. 1. Plastic deformation. Many parts designed for minimum weight have intricate forms that may include fillets, grooves, and holes. Such complex geometries, in combination with high loads, produce stress concentrations at critical locations. In parts made of metals, the stresses are often above the yield point, causing irreversible plastic deformation. To analyze the deformed state, plastic analysis becomes necessary, using nonlinear methods. Such analysis is especially important in the design of parts that are subject to fatigue loading. The theoretical background for solving plasticity and fatigue problems is discussed in detail in Chapters 4 and 7. 2. Large displacements. Parts designed for minimum weight are thin and flexible. When subjected to high loads, they undergo major deformations with large displacements and strains. Consequently, their geometries change considerably as illustrated by the example in Figure 8.1 The figure shows a leaf spring subject to bending under vertical load and resulting changes of the form. To analyze this deformed state, nonlinear methods become necessary. See Chapter 5. 247
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FIGURE 8.1 Leaf spring subject to bending under vertical load.
3. Contact with friction. Perhaps the most common inaccuracy in the stress analysis of machine parts is caused by misrepresentation of the transfer of forces between interacting members in a working machine. This is due to a common practice in linear analysis that disregards the contact phenomena and assumes the pattern of forces as given. This often leads to an inaccurate computational results. In fact, the transfer of forces is an unknown variable; the true loading of individual parts takes place through contact with friction, where the force distribution and the extent of contact zone are functions of the applied load. See Chapter 6. As an example, Figure 8.2 shows a section of internal combustion engine. Highpressure combustion gases start a chain of acting forces beginning with the piston, continuing through the connecting rod and the bearings, and ending at the crankshaft. At each connecting point, the transfer of forces takes place by means of contact. To properly analyze the stress distribution at each point, it is necessary to derive the true size of a contact surface and to determine the acting traction.
8.1.2
STATIC
VS.
FLUCTUATING LOAD
Historically, the tendency in diagnosing machine part failures has been based on the examination of stresses. The main concern was, as it is now, to know the stress
FIGURE 8.2 Section of an internal combustion engine.
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magnitude, but with one difference—in the past, no consideration was given to the fluctuating loads and their crucial effects. With time, objective observations proved that a machine part can withstand static stresses of high magnitude but will fail if these stresses fluctuate. In the following, we shall discuss the stress limits used in design against static load and fluctuating load, separating the two.
8.2 FAILURE OF MACHINE PARTS UNDER STATIC LOAD To safeguard machine parts against failure, standard static stress limits inherent in metals are used: either the yield point or the ultimate strength. The stress limits are determined from one-dimensional tests performed on standard specimen.
8.2.1
ONE-DIMENSIONAL TEST
The usual data recorded from a test are in a form of diagrams of stresses versus strains in axial tension. Two typical stress-strain diagrams from specimens made of steel are shown in Figure 8.3. Part a of the figure represents tests with low-carbon steel. At the beginning, there is a linear relation between the stress and strain, which follows Hooke’s law up to point A. After that, a small drop is observed, followed by a strain at a constant stress level with oscillations, which continues up to point B. The stress at which the specimen undergoes a marked elongation without an increase in load is called yield point. Further increase of the load is assisted by material hardening with a rising curve, which proceeds up to point C, the ultimate tensile strength. From this point on, the material softens, ultimately failing. High-carbon steels and nonferrous metals do not have a sharp yield point. A typical stress-strain curve of a specimen made of high-carbon steel is shown in Figure 8.3b. At the beginning, there is a linear relation between the stress and strain, based on the Hooke’s law, which continues up to point P, proportional limit. After this point, the curve exhibits a gradual transition from the linear elastic to the non-
FIGURE 8.3 Stress-strain curves: (a) low-carbon steel and (b) high-carbon steel.
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linear plastic behavior. Point B, located at a specified distance from P, is the yield point. It is usually defined by drawing a parallel line to the elastic line at an offset of 0.2%.
8.2.2
MULTIAXIAL ANALYSIS
Machine parts, by their very nature, have complicated forms that, in many instances are subject to combined loading, such as tension or compression, combined with torsion, or bending combined with torsion, or other forces. As a typical example, consider a bolt subjected to tension as shown in Figure 8.4. The intricate geometry of such a part, even when the applied load is one-dimensional, causes multidimensional stresses in critical locations. The stresses in the thread include three-dimensional normal and shear components. In advanced design cases, where multiaxial stresses are concerned, one uses finite element analysis. Failure Theories Over the history of machine design, several methods to derive failure criteria, known under a common name as failure theories, evolved. In designing a machine part subject to multiaxial stresses, one refers to stress limits such as yield point or ultimate stress, that are obtained from one-dimensional tests. To overcome the discrepancy between the multidimensional stresses and the one-dimensional test data, it is an accepted practice to combine the component stresses into a single equivalent stress and check it against the one-dimensional stress limits. Such correlations are called failure criteria. There are several hypotheses with different criteria of failure. A description of the most widely used failure theories follows.1,2 The one-dimensional limits used in failure theories depend on the kind of material. Generally, for ductile metals, the yield point of one-dimensional tests is assumed as the failure criterion while, for brittle materials such as cast iron, the ultimate strength is assumed.
FIGURE 8.4 A bolt subjected to tension.
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Maximum Normal Stress Theory The earliest known theory concerning a multiaxial stress state was the maximum normal stress theory of Rankine (circa 1850). It states that failure occurs when the maximum normal stress becomes equal to or exceeds the limiting value of the onedimensional stress test. To derive the conditions of failure, let us arrange the principal stresses in the following sequence: S1 ≥ S2 ≥ S3
(8.1)
If the first principal stress is a tensile stress, S1 > 0 then the condition of a failure according to this theory is S 1 ≥ S t,lim
(8.2)
If the third principal stress is a compressive stress, S3 < 0
(8.3)
S 3 ≤ S c,lim
(8.4)
then the condition of a failure is
where St,lim and Sc,lim are tensile and compressive limits, respectively. The two failure criteria, Equations (8.2) and (8.4), supplement each other. In the stress space, where coordinates are the principal stresses, the limiting surface is a cube. See Figure 8.5, which illustrates the equations. For a two-dimensional stress state, the limiting surface would become an outline of a square. It should be noted that Rankine theory gives inaccurate results for ductile materials where the yield point is used as a failure criterion rather than practical data. However, the theory confirms the limits of brittle materials where the ultimate strength is used as the failure criterion with St,lim and Sc,lim of different magnitudes. Maximum Shear Stress Theory The maximum shear stress theory applicable to ductile metals was initiated by Coulomb and revised by St. Venant and Tresca. It is also known as the Tresca criterion. According to the theory, a failure will occur when the maximum shear stress equals or exceeds the limit value of the shear stress, in an one-dimensional test. In a simple one-dimensional tension test the maximum shear stress, at time of failure, is S yp τ max = -----2
(8.5)
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FIGURE 8.5 Limit surface of the maximum normal stress theory.
A yielding of a ductile metal occurs when τ i ≥ τ max ,
i = 1, 2, 3
(8.6)
where τ1 = S1 – S2 ,
τ2 = S2 – S3 ,
τ3 = S1 – S3
(8.7)
In the stress space, the limiting surface forms a hexagonal cylinder such as the one shown in Figure 4.11, in the section dealing with Tresca theory. For a two-dimensional stress state (S1, S2), the corresponding limiting lines are shown in the Figure 8.6. In contrast with the maximum normal stress theory, the maximum shear theory experimentally confirms the observed behavior of ductile materials. Maximum Distortion Energy Theory The maximum distortion energy theory was originally proposed by Huber in 1904. Its main equations were derived from different hypotheses by von Mises (1913) and Henky (1925). The theory is also known as the von Mises criterion (Section 4.2.1). The maximum distortion energy theory states that failure will occur when the distortion energy per unit volume reaches or exceed a critical value obtained from the one-dimensional test. As shown in Chapter 1, the distortion energy per unit volume equals d 1+v 2 2 2 W i = ------------ [ ( S 1 – S 2 ) + ( S 2 – S 3 ) + ( S 3 – S 1 ) ] 6E
(8.8)
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FIGURE 8.6 Limit plot of the maximum shear stress theory for a two-dimensional stress condition.
See Equation (1.117). Comparing the above with the energy at the time of failure in a one-dimensional test, we obtain the limit condition, 1+v 1+v 2 2 2 2 ------------ [ ( S 1 – S 2 ) + ( S 2 – S 3 ) + ( S 3 – S 1 ) ] = ------------ ⋅ 2S yp 6E 6E
(8.9)
It follows that a failure in a multiaxial stress state occurs when 1 2 2 2 --- [ ( S 1 – S 2 ) + ( S 2 – S 3 ) + ( S 3 – S 1 ) ] ≥ S yp 2
(8.10)
In Figure 8.7, stress limits are plotted for a two-dimensional stress state (S1, S2) in accordance with the maximum distortion energy theory, comparing it with the other theories as well. The maximum distortion energy theory correlates well with the experimental data. Here, the results are closer to experiments than those of the maximum shear theory.
8.3 FATIGUE OF MACHINE PARTS UNDER FLUCTUATING LOAD In the majority cases, machine parts are subject to varying loads. Mostly, these are fluctuating loads, as shown in Figure 8.8, that cause failure due to fatigue. Such failures account for about 80% of all cases of destroyed machine parts, so fatigue analysis in machine design deserves a special attention.
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FIGURE 8.7 Limit plot of the maximum distortion energy theory for a two-dimensional stress condition.
FIGURE 8.8 Fluctuating load in a tractor axle.
It was determined that basing the design on static stress limits did not prevent fatigue failures under fluctuating loading. This finding prompted the use of a new property called the fatigue limit. Compared with the heretofore known limits applicable to static loading, this limit is lower than the yield point and the ultimate strength. Today, the fatigue limit is an unavoidable tool in machine design.
8.3.1
ONE-DIMENSIONAL MODEL
Consider first a simplified analysis: a one-dimensional part subject to a cycling load where the stress fluctuates at a constant amplitude between Smax and Smin (see Figure 8.9). The amplitude of the fluctuating stress is
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FIGURE 8.9 Stress pattern in a specimen subjected to variable loading: (a) mean stress, Sm = 0 and (b) general with Sm ≠ 0.
S max – S min S a = -----------------------2
(8.11)
S max + S min S m = -----------------------2
(8.12)
and the mean value equals
Figure 8.9 shows two simplified histories of cycling stresses: (a) reciprocating stresses where Sa ≠ 0 and Sm = 0, and (b) cycling stresses with Sa ≠ 0 and Sm ≠ 0. The failure data for reciprocating stresses (Sa ≠ 0 and Sm = 0) usually are provided by standard fatigue tests. The test results are presented in the form of S-N diagrams (see Figure 8.10). Test results with steel specimen indicate that after N ≈ 106, the S-N line runs parallel to the N-coordinate line. This indicates that, when the stresses in the specimen are below Sf , the specimen will be safe. Stress Sf is the fatigue limit, sometimes referred to as the endurance limit. For metals other than steel, the failure line declines steadily at all N values without a marked fatigue limit, Sf . The diagram shown in Figure 8.10 is known as Woehler’s diagram, derived from Woehler’s experiments (1848).3 Following the Woehler’s diagram, one may design a machine part either for an unlimited fatigue life (infinite life) or for a restricted life (finite life) with a predicted
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FIGURE 8.10 S-N diagram of fatigue test data.
life expectancy. There are different reasons for choosing either of the goals. The latter, for instance, provides lower weight, but the required analysis precision may be more costly. Infinite Life Designing a part with known load and with stresses within appropriate stress limits may assure an infinite life. To derive the relevant limits, consider a one-dimensional part subjected to stress history as shown in Figure 8.9b. Let us visualize stress history as a superposition of static stresses Sm over reciprocating cyclic stresses with amplitude Sa. There are two known stress limits: the ultimate tensile strength Su of the material for static stresses, and the fatigue limit Sf for the reciprocating stresses. (For the present, we ignore the yield-point limit Syp.) Both limits, Su and Sf, are shown as points A and B within the coordinate system (Sm, Sa) in Figure 8.11. Let us plot a line connecting points A and B. Line AB represents a border of a safe domain OAB; the stresses within the domain warrant an infinite fatigue life. The diagram in Figure 8.11 presents the limits of infinite life and is called a limit diagram. Line AB is defined by the equation Sm Sa ----- + ----- = 1 Su S f
(8.13)
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FIGURE 8.11 Stress-limit diagram.
Stresses within safe domain OAB are subject to the following condition: Sm Sa ----- + ----- ≤ 1 Su S f
(8.14)
From the above, one derives the following equation that is used in machine design: S S 1 ------- = -----m + ----aFS Su S f
(8.15)
where FS is called the factor of safety. In design for infinite life, we must meet the condition FS > 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.
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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.
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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.
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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)
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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 S K f Sa 1 ------- = -----m + ----------FS Su Sf
(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.
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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 σ' f – σ m n' ∆ε b c - ( 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).
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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, S m = S 1,m + S 2,m + S 3,m
(8.31)
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)
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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.
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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
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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 = --- -------------------- = -------------32 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
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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.
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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 k = --- = ---------------⋅ SF 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.
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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 M 1 ------ = --- = --------------------------- ≅ --------2 EI ρ 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.
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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 3EI 2
(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.
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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
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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.
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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
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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)
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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.
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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.
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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.
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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.
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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
63 × 6.3 63 × 6.7 63.60 2 × 20700 × --------------------- + 3 --------------------- + -------------- 12 12 12 × 1.15 = 34.19 N/mm k = 2 × --------------------------------------------------------------------------------------------------------------------3 570.0 3
3
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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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.
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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 570.0 × 67 4150 S = -----------------------------------------------------------------------------------------------------× ------------ = 595.71 MPa 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
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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.
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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
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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
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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
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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
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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.
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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
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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.
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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)
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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)
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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.,
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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)
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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 kb Ab E b κ b = --------------= ----------------------------kb + kc Ab E b + Ac E c
(10.18)
kc Ac E c κ c = --------------= ----------------------------- = 1 – κb kb + kc Ab E b + Ac E c
(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,
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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
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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)
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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)
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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
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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
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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
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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 2 π 2 2 A s = --- d 1 = --- 20.752406 = 338.241 mm 4 4 while the stressed area in a spiral thread is
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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.)
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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.
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FIGURE 10.16 Contact between threads.
FIGURE 10.17 FE model of bolt and nut.
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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.
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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
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FIGURE 10.20 Contact force distribution.
FIGURE 10.21 Von Mises stress distribution in bolt computed with ANSYS program.
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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.
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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.
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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.
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FIGURE 10.28 Location of critical cross sections in bolt.
FIGURE 10.29 Von Mises and principal stress distribution in cross section aa.
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FIGURE 10.30 Shear stress distribution in cross section bb.
FIGURE 10.31 Location of critical cross sections in the nut.
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FIGURE 10.32 Von Mises and principal stress distribution in cross section cc.
FIGURE 10.33 Shear stress distribution in cross section dd.
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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
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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
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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
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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 ∆F 27000 ∆S aver = ------- = ---------------- = 77.4 MPa As 348.91
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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
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FIGURE 10.39 Distribution of the first principal stress in the vicinity of thread: (a) stress distribution and (b) location of the highest stresses.
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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
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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 S 1,max 1555.2
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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 1 ∆K th 1 2.5 a th = --- ----------= --- ------------------------------ = 0.0000089 m π α∆S π 1.12 × 421.0
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
da A ( ∆K ) ------- = -------------------------------------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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FIGURE 10.41 Crack propagation rate in steels SAE 4340 and 4140.
–9
2.25
da 5.155 × 10 × ( ∆K ) ------- = -------------------------------------------------------dN ( 1 – 0.73 ) × 110 – ∆K 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),
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1 1 1 N f = -------------------------------------------------- ---------– ---------n n --- – 1 n --- – 1 n 2 --- – 1 C ( α ∆S π ) a i2 af 2
(10.38)
The length of a crack at failure equals 2 2 1 Kc 1 110.0 - = --- ------------------------------ = 0.0173m a f = --- --------- π α∆S π 1.12 × 421.0
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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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.
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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.
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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
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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.
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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 R = ------------ = ---------------- = 0.86 S 1,max 1423.5 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 1 ∆K th 1 2.5 a th = --- ----------= --- ------------------------------ = 0.0000419 m π α∆S π 1.12 × 194.4 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.
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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.
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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
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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. l d –1 k fg = ------------ + ------------ A f E f A g E g
(11.2)
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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)
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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 ∂F kb --------b- = ----------------∂F p k b + k fg
(11.6)
k fg ∂F ---------c = ----------------∂F p k b + k fg
(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)
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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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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 k fg = f ( p ) ; --------∂p It follows that
(11.11)
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kb ∂ F ∂ - >0 ----------2-c = ------ ---------------- ∂p k + k fg b ∂p
(11.12)
and 2
∂ Fc ∂ k fg -<0 ----------2- = ------ ----------------∂p k b + k fg ∂p
(11.13)
Based on the above equations, a working diagram is provided (see Figure 11.8). The diagram plots forces Fb and Fc as functions of the applied pressure. (For convenience, pressure p is used as an argument instead of pressure force Fp.) Forces Fb and Fc differ from their linear counterparts by a nonlinear increment, ∆F n = f ( p )
(11.14)
In effect, ∆Fb is a pressure force that disengages the flanges. Figure 11.9 presents an equivalent force-deflection diagram corresponding to Figure 11.8. See Junker7 and Bickford.8
11.3 FE ANALYSIS For the purpose of analysis, we choose a flange connection with the following properties. The two flanges are ASA standard 600 lb, nominal size 2 in., made of
FIGURE 11.8 Forces in flange connection vs. pressure load, considering flange nonlinearity.
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FIGURE 11.9 Load deflection diagram, considering flange nonlinearity.
cast steel. The dimensions are shown in Figure 11.10. The gasket is made of soft aluminum, 5 mm thick. The flanges and gasket are joined by eight bolts 5/8 in. dia., grade 5. The bolts are tightened at assembly to an initial load of 42,000 N. Subsequently, the flange connection is subjected to a rising pressure, starting from zero and ending at a critical point, when the connection starts to leak. The computations were expanded to use different tightening loads to check the influence of bolt tightening on the leakage.
FIGURE 11.10 Dimensions of flange.
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11.3.1 THE INCREMENTAL PROCEDURE The flange connection is symmetric, so the FE model could be limited to 1/32th part of the total volume, producing the form of a “pie”—a considerable economy. The model is shown in Figure 11.11. The two sides of the pie and the bottom xyplane serve as surfaces of symmetry and define the boundary conditions. The flange and gasket are built from hexahedron elements. The bolt is simulated by 13 spar elements connecting the upper edge of the bolt hole with the bottom xy-plane. See Figure 11.11b. The spar elements are provided with an initial strain to simulate the tightening. The contact between flange and gasket is simulated by contact elements (not seen in the figure). We assume a friction factor µ = 0.14. The pressure load acting on the flange is conceived as a pressure by the inside surface of the model, while the top of the model endures a tension (see Figure 11.11a. As the pressure load rises, additional pressure is applied to receded contact surface. For the solution of receding contact problem, we used MSC.NASTRAN program. A step-by-step approach is chosen to simulate the process of the continuously diminishing contact area. In incremental form, the equilibrium condition within the solid components can be expressed as K ( p )∆u = ∆F + ∆F′ ( p )
(11.15)
FIGURE 11.11 FE model of flange connection: (1) flange, (2) gasket, and (3) rod elements representing bolt.
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Stiffness matrix K includes contact elements. Since the contact elements become disengaged with rising pressure, K is a function of pressure. ∆F is the vector of acting pressure forces everywhere except within the contact domain. ∆F′ is the unknown vector of pressure forces acting at the receded contact surface. The incremental procedure comprises the following steps: Step 1. Solve Equation (11.15), assuming ∆F′ equals zero, i.e., with closed contact elements. Step 2. Check the contact elements for those that opened up. Step 3. Revise matrix K through elimination of opened contact elements, define vector ∆F′ representing pressure on the receded contact surface. Step 4. Solve Equation (11.15) with the revised K and ∆F′. Steps 2 through 4 are repeated until all contact elements open up—the point of the onset of the leakage. The process is illustrated in Figure 11.12, where the upper gasket surface in contact with the flange is shown. The solution of Equation (11.15) involves the use of a quasi-Newton subroutine.
11.3.2 FE COMPUTATIONAL RESULTS The flange connection was investigated for a bolt tightening that equals 42,000 N per bolt. The computational results are shown in Figures 11.13 through 11.20. The behavior of contact domain under rising pressure, during the period when pressure load is applied from the start and up to the critical pressure, is shown in Figure 11.13. The figure shows the contact pressure distribution along the surface of the gasket at different pressure loads. At bolt tightening, curve a, the contact pressure along the gasket surface is not uniform. At the inner border, it equals zero, rising at the outer border to 125 MPa, while the part of the gasket surface close to the inner border becomes exposed. As the pressure load rises, the contact pressure decreases, opening the contact surface wider (see curves b, c, and d) as explained in Figure 11.7. A plot of the tensional force in the bolt versus pressure load is shown in Figure 11.14. One notes that the line is almost parallel to the abscissa axis, indicating that the flange and gasket are practically rigid. The plot displays a sharp upturn at its end, marking the prying action of pressure force ∆F′ at the receding contact. Figure
FIGURE 11.12 FE model of contact area.
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FIGURE 11.13 Contact pressure distribution: (a) at bolt tightening and (b, c, and d) at the consecutive pressure loads.
FIGURE 11.14 Bold force vs. pressure load.
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FIGURE 11.15 Bolt force during loading and unloading.
FIGURE 11.16 Von Mises stresses in flange computed with MSC.NASTRAN program.
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FIGURE 11.17 Deformation of flange at leakage pressure (not to scale).
FIGURE 11.18 Maximum contact pressure (on gasket) vs. bolt tightening force.
345
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FIGURE 11.19 Leakage pressure vs. bolt tightening force.
FIGURE 11.20 Bolt force behavior for different tightening loads.
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11.15 presents a plot of the bolt force during loading and unloading. The figure displays a hysteresis, caused by an irreversible delay of the gasket displacement due to friction. Such hysteresis was observed in tests (see Sawa et al.9). Figure 11.16 presents von Mises stresses in the flange at maximum working pressure (56.242 MPa). The highest stresses occur at the upper edges of bolt holes, caused by the bearing pressure of the nut, as one would expect. More crucial are the stresses at the upper end of the flange connection, due to the changing geometry of the flange. In the present case, these stresses are within acceptable limits, about 300 MPa. Figure 11.17 shows the flange deformation (in a distorted form), indicative of a sliding contact between the flange and gasket. The behavior of the flange connection at different tightening loads is described in the following figures. Figure 11.18 shows the maximum gasket pressure versus the tightening force in the bolt. The leakage pressure, as function of the tightening force, is plotted in Figure 11.19. Figure 11.20 shows bolt force behavior at different tightening loads. Comparison with a Rigid Flange Let us compare the above results with a simplified linear computation. Consider the rigid flange behavior shown in Figure 11.6. At the time of leakage, the critical pressure force is F crit = F b = F i = 42, 000 N See Equation (11.10). Also at the time of leakage, the receded area, subject to pressure, is extending up to the outer diameter of contact surface. The receded area is π 2 2 A = --- 92.075 = 6658.453 mm 4 See Figure 11.10. Taking into consideration eight bolts, we obtain the critical pressure, 42000 p crit = 8 × ---------------------- = 50.462 MPa 6658.453 as compared to 56.242 MPa—the result of the nonlinear analysis. The error of about 10 percent is evident.
11.4 CONCLUSIONS The design of a high-pressure flange connection involving geometrical considerations and frictional phenomena faces problems that require nonlinear numerical solutions. The geometrical nonlinearity initially is caused by the unknown contact area and again by the changing form of the pressurized flange.
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The FE analysis finds the following: Receding contact. The problem of a receding contact is manifested mainly by the growing pressure that penetrates between the gasket and flange. The main difficulty encountered is the impossibility to predict, by direct analysis, the values of receding contact. The solution process is a series of increments in which the growing pressure is applied to the successive segments until the leakage is registered. Flange stiffness. The bolt force versus pressure load is practically constant, indicating rigid flange and gasket. A softening of the flange is shown in the plot of force versus pressure as a sharp upturn at the end, marking the prying action of the pressure at the receding contact. The stiffness decreases at the end, demonstrating a geometrical nonlinearity. Hysteresis. The plot of the bolt force during loading and unloading displays a hysteresis caused by an irreversible delay of the gasket displacement due to friction. The FE solutions to complex problems that deal with receding contact and sliding friction are not infallible. The accuracy of 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) The topic in item 3 above was discussed in Chapter 6, explaining that evading the errors may produce an unstable solution, which, by failing to converge, renders the problem insoluble.
REFERENCES 1. Roetscher, F., Die Maschinenelemente, Vol. 1, Springer Verlag, Berlin, 1927 (in German). 2. Osgood, C.C., Fatigue Design, Wiley, New York, 1970. 3. Osman, M.O.M., Mansour, W.M., and Dukkipasti, R.V., On the design of bolted connections with gaskets subjected to fatigue loading, ASME Paper no. 76-DET-57, 1976. 4. Motosh, I.N., Determination of joint stiffness in bolted connections, J. of Engineering for Industry, 98, pp. 858–861, 1976. 5. Shigley, J.E., and Mischke, C.R., Mechanical Engineering Design, 5th ed., McGrawHill, New York, 1988. 6. Juvinall, R.C., and Marshek, K.M., Fundamentals of Machine Component Design, John Wiley, New York, 1991. 7. Junker, G., Systematic computation of highly stressed bolted joints, VDI-Richtline 2230, Sept. 1972 (in German).
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8. Bickford, J.H., Introduction to the Design and Behavior of Bolted Joints, Marcel Dekker, New York, 1981. 9. Sawa, T., Higurashi, N., and Akagawa, H., A stress analysis of pipe flange connections, J. of Pressure Vessel Techn., 113, pp. 497–503, 1991.
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12
Fretting Fatigue in an Axle
12.1 INTRODUCTION This chapter introduces a method to deal with a phenomenon of fretting fatigue that occurs in the contact zone. It takes place in press-fitted assemblies where components are subjected to an oscillating relative motion of low amplitude, because the press fitting causes a high stress concentration. Press-fitted connections that are subject to fretting fatigue include press-fitted shafts, axles, splines, keyways, bolted and riveted connections, and others. The fretting process may be followed by chemical reactions, sometimes combined with electro-erosion processes, and it causes fretting corrosion. Let us analyze a connection comprising two press-fitted parts and describe the damaging conditions under fatigue loading. Because the parts are subjected to different deformations, a cyclic movement takes place between the mating surfaces. At a high stress level, this relative movement causes microcracks that lead to fatigue failure. Usually, the cracks are located in one of the surfaces, at the ends of a pressfitted assembly. See References 1–4. According to experimental evidence, the surface damage caused by fretting grows as the amplitude of the relative movement increases. The initial damage begins at very low contact pressures. Rising contact pressure, up to a limit, causes higher stress concentrations, increasing the damage. It was observed that the influence of pressure practically ends at about 50 MPa. A further pressure increase does not produce significant changes. The Process of Fretting Fatigue The occurrence of microcracks in the surfaces of a press-fitted connection represents the first stage of fretting fatigue. Subsequently, in the following stage, the microcracks grow under tensional stresses, independently of the fretting process. The initiation of microcracks resembles the condition occurring at the standard fatigue limit, and, thus, can be investigated by experiments. The second stage, the stage of the crack growth, defines the final failure of a machine part and may be analyzed using the methods of crack propagation. 351
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The process of fretting fatigue may be marked by a reduction factor that defines the reduction of the fatigue limit of a specimen subject to a fretting fatigue, as compared to one that is free of fretting, similar to a stress concentration factor. S S′ f = -----fKf
(12.1)
See Chapter 8. Such reduction factor depends on several factors: the materials of the mating parts, conditions of surfaces (e.g., roughness, hardness, etc.), the presence of lubrication, and others. The fretting reduction factor may reach Kf ≈ 2.0 to 3.0. See Horger.3 Preventive Measures There are several ways to limit the extend of the damage caused by fretting. The preventive measures include 1. 2. 3. 4.
suitable materials for the press-fitted parts surface treatments such as shot-peening, surface rolling, and nitriding use of lubricants use of special soft coatings
The most effective solution to limit the extent of fretting fatigue is a proper design, aimed at reduced stress concentration in the contact zone. For instance, in a pressfitted axle assembly, stress relief can be obtained by introducing a circular groove (Figure 12.1). The reduction may be achieved up to 50%. See Peterson.5
12.2 CASE STUDY: AXLE FAILURE DUE TO FRETTING Let us consider a part that belongs to a vehicle system shown in Figure 12.2. The axle shown in the figure, being subject to a cyclic bending moment, has broken
FIGURE 12.1 Press fit: (a) simple design and (b) improved design.
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FIGURE 12.2 Assembly of axle and wheel.
down. The maximum bending moment was about Mb = 20,000,000 mmN. The breakage occurred after about 275,000,000 rotations. Deep grooves were noted in the surface of the axle upon inspection, indicating fretting fatigue. The purpose of the following analysis is to provide a diagnosis of the failure and to suggest design improvements.
12.2.1 AXLE DESIGN The design of the axle is shown in Figure 12.2. The failure of the axle occurred next to a press-fitted wheel. The design called for a press fit H7/s6, as per the International Standard System of Limits and Fits ISO, with an interference between δ = 0.044 to 0.101 mm. The actual interference was δ = 0.100 mm, according to measurements taken after the failure. The axle was made of hot rolled steel SAE 4130, with the following properties: Brinell hardness
BHN = 250
Ultimate strength
Su = 810 MPa
Yield point
Syp = 710 MPa
The wheel was made of cast steel.
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The axle was subject to outer loading, which includes contact pressure, due to a press fit and a bending moment. The nominal contact pressure of the wheel upon the axle is according to Lame equation, Eδ d 2 p = ------- 1 – ---- D 2d
(12.2)
which equals 200, 000 ⋅ 0.100 100 2 p = --------------------------------------- 1 – --------- = 60.938 MPa 160 2 ⋅ 100 The bending moment varies randomly, with the highest value being Mb = 20,000,000 mmN. When the bending moment is at his highest, the amplitude of the cyclic stresses on the surface of the axle equals Mb 32 × 20, 000, 000 S a = ----------= ----------------------------------------- = 203.718 MPa 3 π 3 π × 100 ------ d 32
(12.3)
Because of the existence of multiaxial stresses, we first derive the equivalent stresses as described in Chapter 8. For the reciprocal cycling stress, one uses 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
(12.4)
For the mean stress, we apply the Sines correlation, S m = S 1,m + S 2,m + S 3,m
(12.5)
See Section 8.3.1. The cyclic stress is one-dimensional, so S a = S 1,a = 203.718 MPa The mean stress is multiaxial. The equivalent mean stress is a sum of radial and tangential components (the axial component equals zero) S m = S 2,m + S 3,m = – p – p = 121.876 MPa An equivalent one-dimensional cycling stress history is shown in Figure 12.3. One notes the difference in magnitude between the tension and compression stresses.
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FIGURE 12.3 One-dimensional equivalent stress history.
To be able to check the cyclic stresses against the stress limits, we construct a stress limit diagram as shown in Figure 12.4. The ultimate strength and yield point of the axle material are listed above. The fatigue limit approximately equals S 810 S f ≈ -----u = --------- = 405 MPa 2 2
(12.6)
Without considering fretting, it appears that the axle has an infinite life expectancy. However, this conclusion is proven wrong when the fretting reduction factor Kf = 2 is used. The resulting solution shows that the working stresses are beyond the limits and indicates a possibility of failure.
12.2.2 FINITE ELEMENT ANALYSIS To obtain a more accurate distribution of stresses in the axle, we perform a finite element analysis of the axle-wheel assembly. The finite element model used in the
FIGURE 12.4 Stress limit diagram in fretting fatigue.
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analysis is shown in Figure 12.5. Due to symmetry, the model is limited to one fourth of the axle-wheel assembly, with z = 0 and y = 0 being the symmetry planes. It represents section ABCD of Figure 12.2. The wheel is constructed of hexahedron elements; the axle is made of both hexahedron and prism elements. The press fit is simulated using contact elements at the interface between the wheel and the axle. The interference of δ = 0.100 mm is achieved by prescribing an elevated temperature for the axle, with a thermal expansion. The coefficient of friction in the contact zone is assumed to be µ = 0.2. The FE solution was performed using the MSC.NASTRAN program. It was based on the incremental method in combination with a quasi-Newton method. There were 10 incremental steps with a total of 40 iterations. The results are presented graphically in Figures 12.6 through 12.9. Figure 12.6 shows the distribution of radial stresses σr prior to application of the bending moment, which equals the pressure distribution caused by the press fit. The pressure of the wheel upon the axle is not constant; there is a slight gradient, contrary to the constant pressure of 60.9 MPa that was predicted based on Equation (12.2). Furthermore, contrary to our expectation, there is no noticeable stress concentration at the wheel edge. The effect of the bending moment, in addition to press fit, is shown in Figures 12.7 through 12.9. One notes the difference in stress behavior on the tension versus compression sides of the rotating axle: there is marked stress concentration on the compression side, while it is negligible on the tension side. The stress concentration is caused by the wheel’s edge digging into the axle, during rotation. Axial stress Sz shown in Figure 12.7, cycles between Smax = 240 MPa and Smin = −295 MPa. This
FIGURE 12.5 FE model of axle with press-fitted wheel.
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FIGURE 12.6 Stress distribution caused by press fit on the outer surface of the axle.
FIGURE 12.7 Distribution of axial stresses Sz caused by bending and press fit on the outer surface of the axle.
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FIGURE 12.8 Distribution of radial stresses Sr caused by bending and press fit on the outer surface of the axle.
FIGURE 12.9 Distribution of shear stresses τzr caused by bending and press fit on the outer surface of the axle.
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is different from the stresses computed with Equations (12.3) and (12.5). The difference could be attributed to the stiffening effect of the press-fitted wheel. The pressure of the wheel upon the axle varies considerably along the axis, as depicted by the distribution of radial stresses on the surface in Figure 12.8. One notes the pressure diminishing toward the edge on the tension side, and it is rising on the compression side. Here, we see a stress concentration on the compression side while there is none on the tension side. Figure 12.9 presents the shear stresses along the surface, which are attributed to frictional effects.
12.2.3 CRACK PROPAGATION Let us examine the stress distribution shown in Figures 12.7 through 12.9 to explain the failure of the axle. An arbitrary point on the surface of the axle, next to the edge of the wheel, is subjected to stresses that cycle between compression and tension, in coordination with the bending moment and axle rotation. During the cycling, the compression stresses caused by the bending moment are augmented by a concentrated pressure caused by press fit at the edge of the wheel (see point B in Figure 12.8). In addition, they are supplemented by concentrated shear stresses caused by friction (point C in Figure 12.9). The oscillating concentrated compression and shear stresses could cause microcracks, as explained before. The following oscillating tension, with maximum stresses marked by point A in Figure 12.7, acts toward crack propagation whenever a groove whose size reaches threshold limits is created. As noted above, deep grooves were seen in the surface upon inspection of the broken axle. For analysis, assume an arbitrary groove 0.2 mm deep, which is at the lower limit of a visibility with the naked eye. a i ≈ 0.2 mm = 0.0002 m The maximum tensional stress on the axle next to the edge of the wheel is the maximum principal stress S1. Figure 12.10 shows the distribution of principal stresses on the tension side of the axle. At the critical point, S 1 = S max = 235 MPa Using the expression of stress intensity (Chapter 7), ∆K I = αS max πa i
(12.7)
we obtain ∆K 1 = 1.12 × 235.0 × π × 0.0002 = 6.597 MPa m The above value is within the threshold limits of SAE 4130 steel (see Appendix C). Therefore, we can anticipate crack propagation.
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FIGURE 12.10 Distribution of maximal principal stresses S1 caused by bending and press fit on the tension side of the axle.
Let us now estimate the number of cycles to failure. As shown in Chapter 7, the crack propagation law states da n ------- = C ( ∆K ) dN
(12.8)
(See Section 7.3.3.). For martensitic steels, the above equation becomes da – 10 2.25 ------- = 1.35 × 10 × ( ∆K ) m/cycle dN See Appendix C. From Equation (12.8), one derives the expression of life expectancy,
2 1 1 N f = --------------------------------------------------n ------------– ------------(n – 2) (n – 2) ------------------------------( n – 2 )C ( αS max π ) 2 2 ai af
(12.9)
In the above equation, af is computed as follows: 2 1 ∆K Ic 2 1 100.0 a f = --- ------------= --- ------------------------------ = 0.04595m π αS max π 1.12 × 235.0
(12.10)
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It follows that, when crack of the size ai = 0.0002 m appears, the axle may fail after about 2 1 1 - --------------------N f = ------------------------------------------------------------------------------ – --------------------------- = 83,793 cycles 2.25 0.125 0.125 – 10 0.002 0.04595 0.25 × 10 ( 1.12 × 235.0 π ) The above confirms that fretting already could cause a fatigue failure at 84,000 cycles, with the computation based on a groove of 0.2 mm.
12.3 DESIGN IMPROVEMENT The axle failure was due to fretting fatigue, which was caused by concentrated compression and shear stresses in the axle. To eliminate fretting, a reduction of stress concentration of compression and shear stresses becomes necessary. This can be achieved by a change in the geometry of the wheel. The goal is to make the wheel more flexible in the vicinity of the edge. The recommended design of the wheel is shown in Figure 12.11. To make it more flexible, the wheel is provided with a large circular groove in the wall of the hub. This reduces the wall thickness at the edge to 5 mm instead of what it was in the original design.
FINITE ELEMENT ANALYSIS We shall present now the stress analysis of the improved design using the finite element method. The new FE model is shown in Figure 12.12. The computed stresses are plotted in Figures 12.13 through 12.16. Figure 12.13 shows stresses along the outer surface of the axle after press fitting and before the application of the bending moment. The magnitude and distribution of the stresses is substantially the same as in the old design (see Figure 12.6). However, upon application of the bending moment, the picture becomes different. See Figures 12.14 to 12.16. While stresses on the tension side—axial, radial, and shear—are practically the same as before, the stress distribution on the compression
FIGURE 12.11 Improved design of wheel: circular groove in the wall of the hub.
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FIGURE 12.12 FE model of axle with press-fitted wheel, improved design.
FIGURE 12.13 Stress distribution caused by press fit on the outer side surface of the improved design.
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FIGURE 12.14 Distribution of axial stresses Sz caused by bending and press fit on the outer axle surface of the improved design.
FIGURE 12.15 Distribution of radial stresses Sr caused by bending and press fit on the outer axle surface of the improved design.
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FIGURE 12.16 Distribution of shear stresses τzr caused by bending and press fit on the outer axle surface of the improved design.
side changes considerably. One notes the considerably reduced concentration stresses: the tip of radial stresses is reduced by about 33%, while that of shear stresses by 50%. Also, the tips of stress concentrations move inward on the wheel. This was confirmed in practice when the introduced design was built. The revised axle-wheel assembly did not show any fretting, and its life expectancy became practically infinite.
12.4 CONCLUSIONS Due to its complexity, fretting cannot be analyzed by standard analytical methods. The initial stage of fretting depends on the distribution of stresses due to contact. By changing the geometry of interacting parts, one may achieve more favorable conditions. By optimizing the conditions, one can avoid the occurrence of fretting. The FE analysis proved that a change in the wheel geometry results in a decrease of about 50% in the magnitude of shear stresses. This, in turn, slows down the development of microcracks. By changing the geometry, one also produces lower axial stresses, by approximately 33%, which prevents crack propagation. Decreasing of the level of both stresses leads to longer life expectancy. The problem considered here is characterized by nonlinearity, due to a frictional contact between the wheel and the axle, that could be solved only by using the FE method. However, the solutions to complex problems that deal with receding contact
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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) The topic of item 3 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. Waterhouse, R.B., Fretting Fatigue, Applied Science Publishers, London, 1981. 2. Collins, J.A., Failure of Materials in Mechanical Design, John Wiley, New York, 1993. 3. Horger, O.J., Ed., ASME Handbook—Metals Engineering Design, McGraw-Hill, New York, 1965. 4. Heywood, R.B., Designing Against Fatigue, Chapman and Hall, London, 1962. 5. Peterson, R.E., Stress Concentration Factors, John Wiley, New York, 1974.
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Appendix A Page 367 Tuesday, November 7, 2000 4:50 PM
Appendix A Basics of Tensor Calculus
A.1 BASIC DEFINITIONS To analyze physical processes, we need to define magnitudes and quantities to describe them. The designation of a numerical magnitude is arbitrary, since one needs to choose a reference frame, i.e., a coordinate system. But the relations between magnitudes that describe the physical processes are independent of the coordinate systems. Tensor calculus supplies a means to formulate relations between the physical magnitudes in a form that does not require any specification of a reference frame; i.e., the relations may be written in an invariant form.
A.1.1 SCALARS
AND
VECTORS
The simplest entity is a scalar. It is invariant by definition. To prescribe it, one needs one numerical value only that does not depend upon any coordinate system, e.g., work, energy, or temperature. The next, more complicated, entity is the vector. Its definition requires prescribing the numerical value and direction. It may be defined by its components in any arbitrary coordinate system. If we assume a Cartesian coordinate system, the vector is defined as follows: 3
a =
∑ ak i k
= ak i k
(A.1)
k=1
Here, ak are the components of the vector. See Figure A.1. Vectors ik are unit vectors that define the directions of coordinate axes. In the above equation, we apply the summation convention whereby the symbol Σ is dropped, and the summation is performed over the repeated indices. The length of the vector in terms of its Cartesian components is 2
2
2 1⁄2
a = ( a1 + a2 + a3 )
(A.2) 367
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FIGURE A.1
Nonlinear Problems in Machine Design
Vector and its components.
Scalar Product of Two Vectors The scalar product of two vectors equals a ⋅ b = ab cos ( a, b )
(A.3)
where cos(a, b ) denotes the angle between vectors a and b. The product is a scalar, i.e., independent of coordinate system. The latter equation is equivalent to a ⋅ b = am i m ⋅ bn i n = as bs
(A.4)
where the scalar product of two unit vectors equals i m ⋅ i n = δ mn
(A.5)
δ nm = 1 if m=n 0 if m≠n
(A.6)
δmn is the Kronecker delta,
The scalar product is subject to commutative law a⋅b = b ⋅a
(A.7)
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369
Vector Product of Two Vectors The vector product of two vectors equals c = a × b = am i m × bn i n
(A.8)
In this equation, c is a vector perpendicular to the plane of vectors a and b. Its direction is defined by the right-hand rule (see Figure A.2). The length of vector c is defined as follows: c = ab sin ( a, b )
(A.9)
The vector product of two vectors may be written as follows: c = a m i m × b n i n = e mnp a m b n i p
(A.10)
where the vector product of two unit vectors equals i m × i n = e mnp i p
(A.11)
and permutation symbol emnp is defined as follows: • emnp = +1 if each unit vector im, in, ip is different and m, n, p is an odd permutation • emnp = 0 if some or all unit vectors i m, in, ip are equal • emnp = –1 if each unit vector im, in, ip is different and m, n, p is an even permutation (A.12) The vector product is antisymmetric: a × b = –b × a
FIGURE A.2
Vector product of two vectors.
(A.13)
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Transformation of Coordinates The vector is invariant, i.e., it is independent from any coordinate system. Its components, however, depend on a selected reference frame. Consider two Cartesian coordinate systems (x1, x2, x3) and (x1′, x2′, x3′) where the second system is derived by rotating the first one. See Figure A.3. The unit vectors of the two systems are ik and ik′, respectively. Let us define i′ s ⋅ i k = cos ( i′ s ,i k ) = α sk
(A.14)
Consider decomposition of vector a in both coordinate systems. a = a s i s = a s ′i′ s
(A.15)
The components of a in the two systems equal, in terms of unit vectors, as = a ⋅ i s
(A.16)
a′ s = a ⋅ i′ s
(A.17)
Upon multiplying parts of Equation (A.13) by is and i′ s , respectively, one obtains a′ s = a ⋅ i′ s = a k i k ⋅ i′ s = α sk a k a s = a ⋅ i s = a′ k i k ′ ⋅ i′ s = α ks a′ k
FIGURE A.3
Rotation of Cartesian coordinate system.
(A.18) (A.19)
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371
Hence, we can state: An entity is called a vector or tensor of the first order if it has three components in any arbitrary system, and the components in one system are related to the components in another system by the characteristic law [Equations (A.18) and (A.19)].
A.1.2 TENSOR
OF
SECOND ORDER
Let us correlate the components of two vectors a(a1, a2, a3) and b(b1, b2, b3) in Cartesian coordinates as follows: a 1 = T 11 b 1 + T 12 b 2 + T 13 b 3 a 2 = T 21 b 1 + T 22 b 2 + T 23 b 3 a 3 = T 31 b 1 + T 32 b 2 + T 33 b 3
(A.20)
The latter system of equations can be written in the form a s = T sk b k
s = 1, 2, 3
(A.21)
The set of components Tmn defines a tensor of second-order T if, for an arbitrary non-zero vector b(b1, b2, b3), the multiplication produces a non-zero vector a(a1, a2, a3). Tensor T equals T T T 11 12 13 T = T 21 T 22 T 23 T 31 T 32 T 33
(A.22)
Equation (A.19) in vectorial form becomes a = T ⋅b
(A.23)
which presents a multiplication of a tensor and a vector. Equation (A.23) is one of several definitions of a tensor of the second order. Transformation of Coordinates Tensor T, defined by Equation (A.23), is independent of any coordinate system. Its components, however, are system dependent. The correlation between the respective components between two arbitrary Cartesian coordinate systems is derived as follows. From Equation (A.19), we obtain a′ s = α sk a k = α sk T km b m = α sk α lm T km b′ l
(A.24)
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Thus, a′ s = T sl ′b′ l
(A.25)
where T sl ′ represents the components in the new coordinate system T sl ′ = α sk α lm T km
(A.26)
Hence, we can state: An entity is called a tensor of the second order if it has nine components in any arbitrary coordinate system, and the components in one system are related to the components in another system by the characteristic law [Equation (A.26)]. Dyadic (Tensor) Product of Two Vectors Consider again Equations (A.20). The right-hand side of each equation is a scalar product of vector b and vector t(s). Vector t(s) is formed by components of tensor T, t( s ) = T sk i k
(A.27)
so that Equations (A.20) can be written as follows: a s = t( s ) ⋅ b
(A.28)
Multiplying both sides of (A.28) by unit vectors is and deriving the sum over s, one obtains a s i s = a = i s t( s ) ⋅ b = T sk i s i k ⋅ b
(A.29)
Hence, tensor T takes the following form: T = T sk i s i k = T 11 i 1 i 1 + T 12 i 1 i 2 + T 13 i 1 i 3 + T 21 i 2 i 1 + …
(A.30)
where ikis denotes a dyadic product of two unit vectors, 100 i1i1 = 0 0 0 000
010 i1i2 = 0 0 0 000
etc. (A.31)
The dyadic multiplication of two arbitrary vectors a and b defines a second-order tensor with the following matrix of components:
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373
a b a b a b 1 1 1 2 1 3 ab = a 2 b 1 a 2 b 2 a 2 b 3 a3 b1 a3 b2 a3 b3
(A.32)
Transpose Tensor, Symmetric, and Antisymmetric Tensors The transpose tensor is defined as follows: T T T 11 12 13 T T = T 21 T 22 T 23 T 31 T 32 T 33
T
T T T 11 21 31 = T 12 T 22 T 32 T 13 T 23 T 33
(A.33)
or T
T
T = ( T mn i m i n ) = T nm i m i n = T mn i n i m
(A.34)
A tensor is symmetric if the following condition is satisfied: T
, T mn = T nm
(A.35)
T
, T mn = – T nm
(A.36)
T=T A tensor is antisymmetric if T = –T
A.1.3 BASIC OPERATIONS
WITH
TENSORS
The basic operations include tensor addition and tensor multiplication. Tensor Addition A sum of two tensors of the same order is a tensor whose components are sums of corresponding components. T + Q = T mn i m i n + Q mn i m i n = ( T mn + Q mn )i m i n
(A.37)
Any tensor may be split into a sum of a symmetric tensor and an antisymmetric one. For explanation, consider a second-order tensor Q and divide it into two parts as follows: 1 1 T T Q = --- ( Q + Q ) + --- ( Q – Q ) 2 2
(A.38)
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The first part, on the right side of the equation, denotes a symmetric tensor, 1 T S = --- ( Q mn + Q nm )i m i n = S 2
(A.39)
while the second part is an antisymmetric tensor, 1 T A = --- ( Q mn – Q nm )i m i n = – A 2
(A.40)
Tensor Multiplication 1. The right and left scalar products of a tensor and a vector are vectors. T ⋅ a = T mn i m i n ⋅ a k i k = T mn a k i m ( i n ⋅ i k ) = T mn a k i m δ nk = T mn a n i m = b m i m
(A.41)
a ⋅ T = a k i k ⋅ T mn i m i n = a k T mn ( i k ⋅ i m )i n = a k T mn δ k i n = T mn a m i n = b′ n i n
(A.42)
It should be noted that a ⋅ T is different from T ⋅ a , except when tensor T is symmetric. 2. The scalar product of two tensors is a tensor. T ⋅ Q = T mn i m i n ⋅ Q kl i k i l = T mn Q kl i m δ nk i l = T mn Q nl i m i l = R ml i m i l (A.43) Generally, the scalar product of two arbitrary tensors is nonsymmetric. T ⋅ Q ≠Q ⋅ T
(A.44)
3. The double-scalar product (or convolute) of two second-order tensors is a scalar. T··Q = T mn i m i n ··Q kl i k i l = T mn Q kl δ nk δ ml = T lk Q kl
(A.45)
Identity Tensor We define the following tensor as an identity tensor: 100 I = ikik = 0 1 0 001
(A.46)
A scalar multiplication of an identity tensor with a vector does not change the vector, i.e.,
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375
a ⋅ I = as i s = a
(A.47)
The identity tensor is symmetric, i.e., a⋅I = I⋅a
(A.48)
Inverse Tensor An inverse tensor T –1 of the second-order tensor T is defined by the following equation: T ⋅T
–1
= I
(A.49)
It can be shown that the components of an inverse tensor are 1 –1 T ts = ----- e trn e sqm T qr T mn 6t
(A.50)
where t is the determinant of the matrix of tensor components, t = det T mn
(A.51)
and factor etrn is defined by Equations (A.12). Tensors of Higher Order The tensor (dyadic) products of tensors are tensors of a higher order. Thus, a tensor product of a second-order tensor and a vector is to a tensor of the third order. T
(3)
(3)
= Ta = T mn i m i n a s i s = T mn a s i m i n i s = T mns i m i n i s
(A.52)
The tensor (dyadic) product of two second-order tensors is to a tensor of the fourth order. T
(4)
= T prsq i p i r i s i q
(A.53)
etc. Examples from the Theory of Elasticity The stress and strain tensors in the Cartesian coordinate system equal σ = σ kl i k i l
(A.54)
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and ε = ε mn i m i n
(A.55)
Hooke’s law in the tensor form can be written as follows: σ = D ··εε
(A.56)
where D is a fourth-order tensor of material properties D = D prsq i p i r i s i q
(A.57)
Consequently, the stress tensor in terms of the strain and material properties tensors equals σ = D prsq i p i r i s i q ··ε kl i k i l = D prsq ε kl i p i r ( i q ⋅ i k ) ( i s ⋅ i l ) = D prsq ε kl i p i r δ qk δ sl = D prlk ε kl i p i r
(A.58)
and the stress components equal σ pr = D prlk ε kl
(A.59)
The strain energy is the convolute of stress and strain tensors, σ ··εε = σ kl i k i l ··ε nm i m i n = σ mn ε nm
(A.60)
which is a scalar.
A.1.4 PRINCIPAL VALUES
AND
PRINCIPAL DIRECTIONS
Consider a symmetric tensor of second order, T = T mn i m i n = T
T
(A.61)
Next consider a unit vector e whose direction is parallel to scalar product T ⋅ e , i.e., T ⋅ e = λe
(A.62)
where λ is a scalar. The direction of vector e, as expressed by Equation (A.62), is called principal direction of tensor T. To derive a principal direction and scalar λ, we shall write Equation (A.62) in the following form:
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377
( T – λI ) ⋅ e = 0
(A.63)
In Cartesian coordinates, it becomes ( T mn – λδ mn )e n = 0,
m = 1, 2, 3,
n = 1, 2, 3
(A.64)
where en are components of vector e. The latter represents a system of three algebraic equations. ( T 11 – λ )e 1 + T 12 e 2 + T 13 e 3 = 0 T 21 e 1 + ( T 22 – λ )e 2 + T 23 e 3 = 0 T 31 e 1 + T 32 e 2 + ( T 33 – λ ) e 3 = 0
(A.65)
To find principal direction e, Equations (A.65) must be solved for e1, e2, e3. The equations may be supplemented with the normality condition, 2
2
2
e1 + e2 + e3 = 1
(A.66)
which excludes a zero solution. To obtain a non-zero solution, the following determinant must equal zero: T 11 – λ
T 12
T 13
T 21
T 22 – λ
T 23
T 31
T 32
T 33 – λ
= 0
(A.67)
Equation (A.67) may be written in a polynomial form. 3
2
2
2
2
P ( λ ) = – λ + λ ( T 11 + T 22 + T 33 ) – λ ( T 11 T 22 – T 12 + T 22 T 33 – T 23 + T 33 T 11 – T 31 ) T 11 T 12 T 13 – T 21 T 22 T 23
= 0
T 31 T 32 T 33 (A.68) Equation (A.68) has three roots, λ 1 ,λ 2 ,λ 3 , called principal values of tensor T. Using roots λ 1 ,λ 2 ,λ 3 , Equation (A.68) may be written in an alternate form.
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P ( λ ) = ( λ1 – λ ) ( λ2 – λ ) ( λ3 – λ ) 3
2
= – λ + λ ( λ1 + λ2 + λ3 ) – λ ( λ1 λ2 + λ2 λ3 + λ1 λ3 ) + λ1 λ2 λ3 (A.69) Tensor T has three principal directions e i (i = 1, 2, 3), corresponding to three principal values λ i . The principal directions are derived by solving Equations (A.65). One obtains, for each principal value λi (i = 1, 2, 3), a set of three components. λ i T 12 T 13 1 e = --- λ i T 22 T 23 , ∆ λ i T 32 T 33
T 11 λ i T 13 1 e = --- T 21 λ i T 23 , ∆ T 31 λ i T 33
i 1
i 2
T 11 T 12 λ i 1 e = --- T 21 T 22 λ i ∆ T 31 T 32 λ i i 3
(A.70) where
T 11 T 12 T 13 ∆ =
(A.71)
T 21 T 22 T 23 T 31 T 32 T 33
The three principal directions e i are mutually orthogonal. If we assume a coordinate system where the coordinate axes coincide with principal directions, tensor T is expressed as λ 0 0 1 T = λ1 e e + λ2 e e + λ3 e e = 0 λ2 0 0 0 λ3 1 1
2 2
3 3
(A.72)
The inverse tensor then equals T
–1
–1 1 1
–1 2 2
–1 3 3
= λ1 e e + λ2 e e + λ3 e e
(A.73)
A specific case is when all principal values are equal, λ 1 = λ 2 = λ 3 = λ . Then, the tensor is called spherical, and it may be written as follows: T = λI
(A.74)
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379
Invariants of Symmetric Tensor of Second Order Since tensor T is independent of any coordinate system, the coefficients of polynom P(λ) in Equations (A.68) and (A.69) do not depend upon a coordinate system either. They equal I 1 ( T ) = λ 1 + λ 2 + λ 3 = T 11 + T 22 + T 33 I 2 ( T ) = λ1 λ2 + λ2 λ3 + λ1 λ3 2
2
2
= T 11 T 22 – T 12 + T 22 T 23 – T 23 + T 11 T 33 – T 13 T 11 T 12 T 13 I 3 ( T ) = λ1 λ2 λ3 =
T 21 T 22 T 23 T 31 T 32 T 33
(A.75)
I1(T), I2(T) and I3(T)are called invariants of tensor T. In terms of the invariants, Equation (A.68) can be written as 3
2
– λ + I 1 ( T )λ – I 2 ( T )λ + I 3 ( T )I = 0
(A.76)
According to the Cayley–Hamilton theorem,3 tensor T must satisfy the same equation, i.e., 3
2
– T + I 1 ( T )T – I 2 ( T )T + I 3 ( T )I = 0
(A.77)
The Cayley–Hamilton theorem allows us to express powers of tensor that are higher than 2 in terms of T and T 2, where 2 1 1
2
2 2 2
2 3 3
T = T ⋅ T = λ1 e e + λ2 e e + λ3 e e
(A.78)
A.1.5 DEVIATOR A tensor whose first invariant I1 equals zero is called deviator. A symmetric tensor of the second order can be split into spherical and deviatoric parts. The spherical part by definition equals 1 T 0 = --- I 1 ( T )I 3
(A.79)
1 D = T – T 0 = T – --- I 1 ( T )I 3
(A.80)
The deviatoric part (deviator) is
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It follows that the first invariant of deviator D equals zero. The principal directions of the deviator are the same as those of tensor T. The principal values of the deviator equal 1 d s = λ s – --- I 1 ( T ) , 3
s = 1, 2, 3
(A.81)
The invariants of the deviator are I 1(D ) = d1 + d2 + d3 = 0 1 2 I 2 ( D ) = d 1 d 2 + d 2 d 3 + d 1 d 3 = I 2 ( T ) – --- I 1 ( T ) 3 1 2 3 I 3 ( D ) = d 1 d 2 d 3 = I 3 ( T ) – --- I 1 ( T )I 2 ( T ) + ------ I ( T ) 3 27 1
(A.82)
In another form, the second invariant of the deviator equals 1 2 1 2 2 2 2 2 I 2 ( D ) = – --- ( d 1 + d 2 + d 3 ) = – --- [ ( λ 1 – λ 2 ) + ( λ 2 – λ 3 ) + ( λ 1 – λ 3 ) ] 2 2
(A.83)
A.1.6 GENERAL COORDINATE SYSTEMS The following description relates to general coordinate systems, including nonCartesian systems. Let us introduce three non-coplanar vectors g1, g2, g3. From the vectors, we form a coordinate system so that an arbitrary vector r can be expressed as a sum of components, 1
2
3
r = r g1 + r g2 + r g3
(A.84)
See Figure A.4. Vectors g1, g2, g3 are base vectors, which are not necessary unit vectors. Coefficients r1, r2, r3, defining the size of the components, are derived below. Reciprocal Base Systems To derive coefficient r1, we consider the fact that the two projections—of component r1g1 and of vector r— upon a normal to the plane created by pair of base vectors (g2, g3) must be equal. Since the normal is parallel to vector product (g2 × g3), 1
r g 1 ⋅ ( g 2 × g 3 ) = r ⋅ (g 2 × g 3 ) See Figure A.5. Thus, we obtain coefficient r1,
(A.85)
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381
FIGURE A.4
Non-Cartesian coordinate system.
FIGURE A.5
Concerning the derivation of r1.
r ⋅ (g 2 × g 3 ) 1 r = -----------------------------g1 ⋅ ( g2 × g3 )
(A.86)
r ⋅ (g 1 × g 1 ) r ⋅ (g 3 × g 1 ) 2 r = -----------------------------= -----------------------------g1 ⋅ ( g3 × g1 ) g1 ⋅ ( g2 × g3 )
(A.87)
r ⋅ (g 1 × g 2 ) r ⋅ (g 1 × g 2 ) 3 = -----------------------------r = -----------------------------g3 ⋅ ( g1 × g2 ) g1 ⋅ ( g2 × g3 )
(A.88)
In a similar way, we derive
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The mixed scalar-vector products in denominators of the three equations represent the volume of a parallelepiped whose sides are vectors g1, g2, g3. vol = g 1 ⋅ ( g 2 × g 3 ) = g 2 ⋅ ( g 3 × g 1 ) = g 3 ⋅ ( g 1 × g 2 )
(A.89)
See Figure A.6. From Equations (A.86) through (A.88), we derive three vectors, g1, g2, g3. ( g2 × g3 ) 1 g = -----------------------------g1 ⋅ ( g2 × g3 )
(A.90)
( g3 × g1 ) 2 g = -----------------------------g1 ⋅ ( g2 × g3 )
(A.91)
( g1 × g2 ) 3 g = -----------------------------g1 ⋅ ( g2 × g3 )
(A.92)
so that 1
1
2
2
3
r = r ⋅g ; r = r ⋅g ; r = r ⋅g
3
(A.93)
Vectors g1, g2, g3 are called reciprocal base vectors. They are respectively perpendicular to planes determined by pairs of base vectors (g2, g3), (g3, g1) and (g1, g2). Consider now the scalar product gs ⋅ gk . From Equations (A.90) through (A.92), we get
FIGURE A.6
g1 ⋅ ( g2 × g3 ) 1 g 1 ⋅ g = -----------------------------= 1 g1 ⋅ ( g2 × g3 )
(A.94)
g1 ⋅ ( g3 × g1 ) 2 = 0 g 1 ⋅ g = -----------------------------g1 ⋅ ( g2 × g3 )
(A.95)
Parallelepiped formed by base vectors.
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383
etc. Thus, the scalar product of a base vector and a reciprocal base vector equals Kronecker delta, k
k
k = 1, 2, 3,
gs ⋅ g = δs ,
s = 1, 2, 3
(A.96)
It follows that vector r can be referred to either the base system g1, g2, g3 or the reciprocal base system g1, g2, g3, 1
2
3
s
r = r g1 + r g2 + r g3 = r gs 1
2
3
r = r1g + r2g + r3g = rsg
(A.97)
s
(A.98)
where the respective coefficients equal s
s
r = r ⋅g ,
rs = r ⋅ g s
(A.99)
Let us introduce the following scalar products, s
k
g s ⋅ g k = g sk , g ⋅ g = g
sk
(A.100)
Then, the reciprocal relations of gs and gs can be expressed as s
s
k
sk
g = g ⋅ g gk = g gk k
g s = g s ⋅ g k g = g sk g
(A.101)
k
(A.102)
Consequently, vector r can be presented as follows: sk
s
r = r s g g k = r g sk g
k
(A.103)
i.e., its components with upper and lower indices are related to each other by the following equations: k
r = rsg
sk
k
(A.104)
, r s = r g ks
Identity Tensor Following Equations (A.97) and (A.98), Equation (A.84) can be written as 1
2
3
1
2
3
r = r ⋅ (g 1 g + g 2 g + g 3 g ) = r ⋅ (g g 1 + g g 2 + g g 3 )
(A.105)
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The dyadic products in parentheses represent an identity tensor, s
s
I = gs g = g gs
(A.106)
r = r ⋅I
(A.107)
so that
Transformation of Coordinates Consider two sets of coordinate systems, a primary set (gs, gs) and a secondary one s ( g s ′, g ′ ) . We shall derive a transformation law using the identity tensor in terms of both sets. From Equation (A.106), we obtain s
s
s
s
I = g s g = g g s , I′ = g s ′g′ = g′ g s ′
(A.108)
Although the identity tensor remains the same in both equations, we marked it with prime in the second equation for identification with the secondary set. Consider now vector r as expressed in Equation (A.97). Let us multiply the vector by the secondary identity tensor, s
k
k′
s
k
r ⋅ I ′ = r g s ⋅ g′ g′ k = r α s g k ′ = r′ g k ′
(A.109)
where k′
α s = g s ⋅ g′ k
s
k
(A.110)
k′
r′ = r α s
(A.111)
Using Equation (A.98), we also have s
s
k
k
r ⋅ I ′ = r s g ⋅ g k ′g′ = r s α k′ g′ = r′ k g′
k
(A.112)
where s
s
α k′ = g ⋅ g k ′
(A.113)
s
r k ′ = r s α k′
(A.114)
We also define the correlations between respective coordinates as follows: s
s
g k ′ = g k ′ ⋅ I = g k ′ ⋅ g g s = α k′ g s
(A.115)
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Appendix A
385 k
k
k
k′ s
s
g′ = g′ ⋅ I = g′ ⋅ g s g = α s g
(A.116)
Upon comparing Equations (A.114) and (A.115), we find that vector components with lower indices, rs and r k ′ , transform with the same laws as base vectors gs and g k ′ . We call rs and r k ′ covariant components and gs and g k ′ covariant base vectors. Also, comparing Equations (A.111) and (A.116), we see that vector components k with upper indices, rs and r′ , transform with the same laws as reciprocal base k k vectors gs and g′ , which are different from the above laws. rs and r′ are called k contravariant components and gs and g′ contravariant base vectors. k′ s Consider coefficients of transformation α s and α k′ . The matrices of these coefficients are mutually inverse. Therefore, given a scalar product of the secondary base vector and reciprocal base vector, we obtain from Equations (A.115), (A.116), and (A.96) the following equation: m
s
m′
l
l
s
m′
s
m′
g k ′ ⋅ g′ = α k′ α l g s ⋅ g = δ s α k′ α l = α k′ α s
(A.117)
Its right-hand side is the product of two matrices: s
m′
[ A ] = [ α k′ ] and [ A ] = [ a s ]
(A.118)
Since the left-hand side of Equation (A.117) represents the components of the identity tensor (whose matrix is identity matrix), we obtain [ A ] ⋅ [ A ] = [ I ] and [ A ] = [ A ]
–1
(A.118)
REFERENCES 1. Malvern, L.E., Introduction to the Mechanics in a Continuous Medium, Prentice-Hall, Englewood Cliffs, NJ, 1969. 2. Green, A.E., and Zerna, W., Theoretical Elasticity, Oxford University Press, Oxford, 1968. 3. Truesdell, C., and Noll, W., The nonlinear field theories of mechanics, Encyclopedia of Physics, Vol. III/3, Springer Verlag, New York, 1965.
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Appendix B Page 387 Tuesday, November 7, 2000 4:51 PM
Appendix B Basics of Matrix Calculus
A matrix is an m × n array of elements (components) arranged in rows and columns as follows a 11 a 12 … a 1n A =
a 21 a 22 … a 2n … … … … a ml a m2 … a mn
(B.1)
The matrices obey several laws, which are presented below. Two matrices are equal if their corresponding components are equal. This may be written as follows: A = B if a kl = b kl , k = 1, 2, …m, l = 1, 2, …n
(B.2)
The sum of two matrices is a matrix formed by adding corresponding components, i.e., C = (A + B ) where c kl = a kl + b kl , k = 1, 2, …m , l = 1, 2, …n (B.3) In the same manner, the difference of two matrices is defined as C = A – B where c kl
= a kl – b kl , k = 1, 2, …m , l = 1, 2, …n (B.4)
The product of two matrices whose dimensions are m × n and n × l is the matrix m × l, formed as follows: C m × l = A m × nBn × l
where
c ij = a ip b pj , i = 1.2, …m , j = 1, 2, …l p = 1, 2, …n
(B.5)
Note: In the above equation, we apply the summation convention whereby the summation is performed over the repeated indices. 387
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The identity matrix I is a square n × n matrix with elements aij = δij, i,j = 1,2,...,n, where δij denotes the Kronecker delta. That is, 1 I = 0 … 0
0 1 … 0
… … … …
0 0 … 1
(B.6)
The transpose of a matrix is the matrix formed by interchanging the rows and columns. If A is an m × n matrix, then the transpose A T is an n × m matrix, a 11 a 12 … a 1n A
T
=
T
a 21 a 22 … a 2n … … … … a m1 a m2 … a mn
a 11 a 21 … a m1 =
a 12 a 22 … a m2 … … … … a 1n a 2n … a mn
(B.7)
Its components are T
a kl = a lk
(B.8)
An n × n matrix (square matrix) is symmetric if it equals its transpose A = A
T
(B.9)
A square matrix has an associated determinant that is a single number. Its elements have the same arrangements as those of the matrix a 11 a 12 … a 1n det A =
a 21 a 22 … a 2n … … … … a m1 a m2 … a mn
= a
(B.10)
The determinant of a product of two matrices is equal to the product of their determinants. AB = A ⋅ B
(B.11)
An inverse matrix A –1 is a matrix that satisfies the condition, –1
A A = I
(B.12)
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Appendix B
389
Square matrix A has an inverse if it is nonsingular, i.e., its determinant is non-zero. A ≠0
(B.13)
REFERENCES 1. Ayres, F., Jr., Theory and Problems of Matrices, Schaum’s Outline Series, McGrawHill, New York, 1962. 2. Strang, G., Linear Algebra and Its Applications, Academic Press, New York, 1976.
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Appendix C Page 391 Tuesday, November 7, 2000 4:51 PM
Appendix C Tables
391
SAE Steel 1015 1020 1040 1045 1045 1045 1045 1045 1045 1144 1144 4130 4130 4140 4142 4142 4142 4142 4142 4142 4142
Process Normalized HR plate plate As forged Q&T Q&T Q&T Q&T Q&T Q&T CD strain rel. Drawn Q&T Q&T Q&T drawn Drawn Drawn Q&T Q&T deform. Q&T Q&T deform. Q&T deform.
BHN 80 108 225 225 410 390 450 500 595 265 305 258 365 310 310 335 380 400 450 475 450
228 262 345 634 1365 1276 1517 1689 1862 717 1020 779 1358 965 1048 1234 1379 1448 1586 1896 1862
Syp monotonic (MPa) 241 241 386 414 827 758 965 1276 1724 552 565 565 827 621 745 1248 827 896 1069 1103 1069
Syp cyclic (MPa) 825 895 1540 1225 1860 1585 1795 2275 2725 1000 1585 1275 1695 1825 1450 1250 1825 1895 2000 2070 2105
σ′f (MPa) 0.95 0.41 0.61 1.00 0.60 0.45 0.35 0.25 0.07 0.32 0.27 0.92 0.89 1.20 0.22 0.06 0.45 0.50 0.40 0.20 0.60
ε′f 0.22 0.18 0.18 0.18 0.15 0.17 0.15 0.12 0.13 0.15 0.18 0.13 0.12 0.14 0.18 0.14 0.17 0.16 0.16 0.15 0.16
n′ –0.11 –0.12 –0.14 –0.095 –0.073 –0.074 –0.07 –0.08 –0.081 –0.08 –0.09 –0.083 –0.081 –0.08 –0.10 –0.08 –0.08 –0.09 –0.08 –0.082 –0.09
b –0.64 –0.51 –0.57 –0.66 –0.70 –0.68 –0.69 –0.68 –0.60 –0.58 –0.53 –0.63 –0.69 –0.59 –0.51 –0.62 –0.75 –0.75 –0.73 –0.77 –0.76
c 68 62 60 65 51 59 55 51 41 33 25 67 55 60 29 28 48 47 42 20 37
%RA
207 203 200 200 200 207 207 207 207 197 199 221 200 201 200 199 207 200 207 200 200
E (GPa)
392
415 440 620 725 1450 1345 1585 1825 2240 930 1035 895 1425 1075 1060 1250 1415 1550 1750 2035 1930
Su (MPa)
TABLE C.1 Monotonic and Cyclic Stress–Strain Properties of Selected Metals
Appendix C Page 392 Tuesday, November 7, 2000 4:51 PM
Nonlinear Problems in Machine Design
155
Aluminum 2014–T6 Sol. tr. & artif. aging Sol. tr. & strain hard. Sol. tr. & artif. aging
Process Q&T Q&T HR & anneal. Q&T Q&T Q&T Annealed Q&T Q&T HR & anneal. CD HR & anneal.
97 379 469
510 469 579
Syp monotonic (MPa) 1724 1689 634 1372 1172 1531 455 786 1379 255 745 221
524
427
414
Syp cyclic (MPa) 1344 1724 455 827 758 1000 524 648 1048 717 876 345
1317
1103
848
σ′f (MPa) 2170 2655 1200 2000 1655 1931 1040 1220 1855 2413 2275 1655
0.19
0.22
0.42
ε′f 0.09 0.07 0.45 0.48 0.73 0.40 0.16 0.41 0.38 1.02 0.89 0.60
0.146
0.065
0.16
n′ 0.13 0.12 0.18 0.15 0.14 0.15 0.15 0.12 0.09 0.36 0.17 0.26
–0.52
–0.124
–0.106
b –0.081 –0.089 –0.095 –0.091 –0.076 –0.071 –0.071 –0.073 –0.057 –0.15 –0.12 –0.15
–0.52
–0.59
–0.65
c –0.61 –0.76 –0.54 –0.60 –0.62 –0.57 –0.47 –0.60 –0.65 –0.69 –0.77 –0.57
33
25
25
%RA 35 27 43 38 57 42 14 33 32 74 69 64
71
69
69
E (GPa) 207 207 193 200 193 193 207 193 200 186 172 193
Appendix C
Grain Direction for all presented materials is L. Source: SAE Handbook. 1989. Warrendale, PA: Society of Automotive Engineers.
BHN 475 560 243 409 350 430 260 280 410 160 327 145
SAE 4142 4142 4340 4340 4340 5160 9262 9262 9262 30304 30304 30310
Su (MPa) 1930 2240 825 1470 1241 1670 925 1000 1565 745 951 641
TABLE C.1 Monotonic and Cyclic Stress–Strain Properties of Selected Metals (continued)
Appendix C Page 393 Tuesday, November 7, 2000 4:51 PM
393
Appendix C Page 394 Tuesday, November 7, 2000 4:51 PM
394
Nonlinear Problems in Machine Design
TABLE C.2 Plane-Strain Fracture Toughness of Selected Materials Material
Properties
KIc (MPa m0.5)
Steel 4130
BHN = 390
66
4340
BHN = 380
110
BHN = 430
75
BHN = 550
53
4340 H13
HP 9-4-30
air melted
45
vacuum arc remelted
60
longitudinal direction: Rc = 40
78
Rc = 46
52
Rc = 50
33
Rc = 53
24
Rc = 44–48
99–115
Rc = 49–53
66–99
UTS = 1500 MPa
80–120
UTS = 1750 MPa
50–100
Tempered martensite
UTS = 2000 MPa
30–70
18Ni(200)
maraging, UTS = 1500 MPa
155–200
18Ni(250)
maraging, UTS = 1800 MPa
120
18Ni(300)
maraging, UTS = 2050 MPa
80
18Ni(350)
maraging, UTS = 2450 MPa
35–50
18Ni(cast)
maraging, UTS = 1750 MPa
105
21-6-9
BHN = 225
110
PH 13-8Mo
BHN = 400
110
15-5PH
BHN = 330
121
17-4PH
BHN = 300
66
AISI TYPE 301
BHN = 380
165
Aluminium 2024-T351
38.4
2024-T851
32.6
6061-T651
32.9
7075-T73
BHN = 140
33
7075-T6
BHN = 150
31
Sources: 1. Metals Handbook, 1990, Vol. 1, 10th ed., Metals Park, OH, American Society of Metals. 2. L. Schwarmann, Material Data of High-Strength Aluminium Alloys for Durability Evaluation of Structures, 2nd ed., 1988, Aluminium-Verlag, Dusseldorf.
Appendix C Page 395 Tuesday, November 7, 2000 4:51 PM
Appendix C
395
TABLE C.3 Threshold Data for Selected Materials Material
R = Kmin/Kmax
∆Kth (MPa m0.5)
Mild steel
0.13
6.6
0.35
5.2
0.49
4.3
Low alloy steel
0.64
3.2
0.0
6.6
0.32
5.2
0.5
4.4
0.62
3.3
0.73
2.5
9310
0.3
5.5
0.9
3.3
A508 Class 2
0.1
6.7
0.5
5.6
A517-F
A533 Grade B Class 1
18/8 Austenitic
0.7
3.1
0.21
5.8
0.4
4.5
0.88
3.3
0.1
7.7
0.3
5.9
0.5
4.9
0.8
3.1
0.0
6.0
0.33
5.9
0.62
4.6
0.74
4.1
Gray cast iron
0.0
7.0
0.5
4.5
Monel
0.0
7.0
0.5
5.2
Inconel
0.67
3.6
0.0
7.1
0.57
4.7
0.71
4.0
Sources: 1. Barsom, J.M, and S.T. Rolfe, 1977, Fracture and Fatigue Control in Structures, Englewood Cliffs, NJ: Prentice-Hall. 2. Pook, L.P., and Smith R.A. Theoretical background to elastic fracture mechanics, in Fracture Mechanics, Current Status, Future Prospects, ed. Smith, R.A., l979, New York: Pergamon Press.
Appendix C Page 396 Tuesday, November 7, 2000 4:51 PM
396
Nonlinear Problems in Machine Design
TABLE C.4 Crack Growth Rate of Selected Metals da n ------- = C ( ∆K ) dN Ferritic–pearlitic steels 3.0 da – 12 ------- ( m/cycle ) = 6.9 × 10 × ( ∆KMPa m ) dN 3.0 da – 10 ------- ( in/cycle ) = 3.6 × 10 × ( ∆Kksi in ) dN
Martensitic steels 2.25 da – 10 ------- ( m/cycle ) = 1.35 × 10 × ( ∆KMPa m ) dN 2.25 da –9 ------- ( in/cycle ) = 6.6 × 10 × ( ∆Kksi in ) dN
Austenitic steels 3.25 da – 12 ------- ( m/cycle ) = 5.6 × 10 × ( ∆KMPa m ) dN 3.25 da – 10 ------- ( in/cycle ) = 3.0 × 10 × ( ∆Kksi in ) dN
Source: Rolfe, S.T., and Barsom, J.M., Fracture and Fatigue Control in Structures, Prentice-Hall, Englewood Cliffs, New Jersey, l977.
Appendix C Page 397 Tuesday, November 7, 2000 4:51 PM
Appendix C
397
TABLE C.5 Basic Dimensions of Unified Threads
Size
Basic major dia., in.
Threads per in.
Basic pitch dia., in.
Minor dia. external threads, in.
Minor dia. internal threads, in.
Section at minor dia., in.2
Tensile stress area, in.2
Coarse thread series, UNC 1 (.073)
0.0730
64
0.0629
0.0538
0.0561
0.00218
0.00263
2 (.086)
0.0860
56
0.0744
0.0641
0.0667
0.00310
0.00370
3 (.099)
0.0990
48
0.0855
0.0734
0.0764
0.00406
0.00487
4 (.112)
0.l120
40
0.0958
0.0813
0.0849
0.00496
0.00604
5 (.125)
0.1250
40
0.1088
0.0943
0.0979
0.00672
0.00796
6 (.138)
0.1380
32
0.1177
0.0997
0.1042
0.00745
0.00909
8 (.164)
0.1640
32
0.1437
0.1257
0.1302
0.01196
0.0140
10(.190)
0.1900
24
0.1629
0.1389
0.1449
0.01450
0.0175
12(.210)
0.2160
24
0.1889
0.1649
0.1709
0.02060
0.0242
1/4
0.2500
20
0.2175
0.1887
0.1959
0.0269
0.0318
5/16
0.3125
18
0.2764
0.2443
0.2524
0.0454
0.0524
3/8
0.3750
16
0.3344
0.2983
0.3073
0.0678
0.0775
7/16
0.4375
14
0.3911
0.3499
0.3602
0.0933
0.1063
1/2
0.5000
13
0.4500
0.4056
0.4167
0.1257
0.1419
9/16
0.5625
12
0.5084
0.4603
0.4723
0.162
0.182
5/8
0.6250
11
0.5660
0.5135
0.5266
0.202
0.226
3/4
0.7500
10
0.6850
0.6273
0.6417
0.302
0.334
7/8
0.8750
9
0.8028
0.7387
0.7547
0.419
0.462
1
1.0000
8
0.9188
0.8466
0.8647
0.551
0.606
1-1/8
1.1230
7
1.0322
0.9497
0.9704
0.693
0.763
1-1/4
1.2500
7
1.1472
1.0747
1.0954
0.890
0.969
1-3/8
1.3750
6
1.2667
1.1705
1.1946
1.054
1.155
1-1/2
1.5000
6
1.3917
1.2955
1.3196
1.294
1.405
1-3/4
1.7500
5
1.6201
1.5046
1.5335
1.74
1.90
2
2.0000
4-1/2
1.8557
1.7274
1.7594
2.30
2.50
2-1/4
2.2500
4-1/2
2.1057
1.9774
2.0094
3.02
3.25
2-1/2
2.5000
4
2.3376
2.1933
2.2294
3.72
4.00
2-3/4
2.7500
4
2.5876
2.4433
2.4794
4.62
4.93
3
3.0000
4
2.8376
2.6933
2.7294
5.62
5.97
3-1/4
3.2500
4
3.0976
2.9433
2.9794
6.72
7.10
Appendix C Page 398 Tuesday, November 7, 2000 4:51 PM
398
Nonlinear Problems in Machine Design
TABLE C.5 Basic Dimensions of Unified Threads (continued)
Size
Basic major dia., in
Threads per in
Basic pitch dia., in
Minor dia. external threads, in
3-1/2
3.5000
4
3.3376
3.1933
3.2294
7.92
8.33
3-3/4
3.7500
4
3.5876
3.4433
3.4794
9.21
9.66
4
4.0000
4
3.8376
3.6933
3.7294
10.61
11.08
Minor dia. internal threads, in
Section at minor dia., in2
Tensile stress area, in2
Fine thread series, UNF 0 (.060)
0.0600
80
0.0519
0.0447
0.0465
0.00151
0.00180
1 (.073)
0.0730
72
0.0640
0.0560
0.0580
0.00237
0.00278
2 (.086)
0.0860
64
0.0759
0.0668
0.0691
0.00339
0.00394
3 (.099)
0.0990
56
0.0874
0.0771
0.0797
0.00451
0.00523
4 (.112)
0.1120
48
0.0985
0.0864
0.0894
0.00566
0.00661
5 (.125)
0.1250
44
0.1102
0.0971
0.1004
0.00716
0.00830
6 (.138)
0.1380
40
0.1218
0.1073
0.1109
0.00874
0.01015
8 (.164)
0.1640
36
0.1460
0.1299
0.1339
0.01285
0.01474
10(.190)
0.1900
32
0.1697
0.1517
0.1562
0.01750
0.0200
12(.210)
0.2160
28
0.1928
0.1722
0.1773
0.02260
0.0258
1/4
0.2500
28
0.2268
0.2062
0.2113
0.0326
0.0364
5/16
0.3125
24
0.2854
0.2614
0.2674
0.0524
0.0580
3/8
0.3750
24
0.3479
0.3239
0.3299
0.0809
0.0878
7/16
0.4375
20
0.4050
0.3762
0.3834
0.1090
0.1187
1/2
0.5000
20
0.4675
0.4387
0.4459
0.1486
0.1599
9/16
0.5625
18
0.5264
0.4943
0.5024
0.189
0.203
5/8
0.6250
18
0.5889
0.5560
0.5649
0.240
0.256
3/4
0.7500
16
0.7094
0.6733
0.6823
0.351
0.373
7/8
0.8750
14
0.8286
0.7874
0.7977
0.480
0.509
1
1.0000
12
0.9459
0.8978
0.9098
0.625
0.663
1-1/8
1.1250
12
1.0709
1.0228
1.0348
0.812
0.856
1-1/4
1.2500
12
1.1959
1.1478
1.1598
1.024
1.073
1-3/8
1.3750
12
1.3209
1.2728
1.2848
1.260
1.315
1-1/2
1.5000
12
1.4459
1.3978
1.4098
1.521
1.501
Appendix C Page 399 Tuesday, November 7, 2000 4:51 PM
Appendix C
399
TABLE C.6 Basic Dimensions of Metric Threads
Nominal dia.
Pitch, mm
Basic pitch dia., mm
Minor dia., extern. threads, mm
Section at minor dia., mm2
tensile stress area, mm2
0.5 0.6 0.7 0.75 0.8 1 1 1.25 1.5 1.75 2 2 2.5 2.5 2.5 3 3 3.5 3.5 4 4
2.675 3.110 3.545 4.013 4.480 5.350 6.350 7.188 9.026 10.863 12.701 14.701 16.376 18.376 20.376 22.051 25.051 27.727 30.727 33.403 36.402
2.387 2.764 3.141 3.580 4.019 4.773 5.773 6.466 8.160 9.853 11.546 13.546 14.933 16.933 18.933 20.319 23.319 25.706 28.706 31.093 34.093
4.47 6.00 7.75 10.1 12.7 17.9 26.2 32.8 52.3 76.2 105 144 175 225 282 324 427 519 647 759 912
5.03 6.78 8.78 11.3 14.2 20.1 28.9 36.6 58.0 84.3 115 157 192 245 303 353 459 561 694 817 976
1 1.25 1.25 1.5 1.5 1.5 1.5 1.5 1.5 2 2 2 2 3 3
7.350 9.188 11.188 11.026 13.026 15.026 17.026 19.026 21.026 22.701 25.701 28.701 31.701 34.051 37.051
6.773 8.466 10.466 10.160 12.160 14.160 16.160 18.160 20.160 21.546 24.546 27.546 30.546 32.319 35.319
36.0 56.3 86.0 81.1 116 157 205 259 319 364 473 596 732 820 979
Regular threads 3 (3.5) 4 (4.5) 5 6 (7) 8 10 12 (14) 16 (18) 20 (22) 24 (27) 30 (33) 36 39 Fine threads 8 10 12 (12) (14) 16 (18) 20 (22) 24 (27) 30 (33) 36 (39)
39.2 61.2 92.1 88.1 125 167 216 272 333 354 496 621 761 865 1030
Appendix C Page 400 Tuesday, November 7, 2000 4:51 PM
400
Nonlinear Problems in Machine Design
TABLE C.7 Mechanical Requirements for Bolts Metric threads Tensile strength (stress) min. MPa
Yield strength (stress) min. MPa
Tensile strength (stress) Elongation Reduction min. min., of area, MPa % %
Property class
Nominal dia.
Proof load (stress) MPa
8.8 9.8 10.9 12.9
M17 thru M36 M1.6 thru M16 M6 thru M36 M1.6 thru M36
600 650 830 970
830 900 1040 1220
660 – 940 1100
Nom. size dia. in
Proof load (stress) psi
Tensile strength (stress) psi
1/4 thru 1
85000
120000
92000
14
35
Over 1 to 1/2 No. 6 thru 5/8
74000 85000
105000 120000
81000 –
14 –
35 –
No. 6 thru 1/2 1/4 thru 1
85000
120000
92000
14
35
1/4 thru 1-1/2
105500
133000
115000
12
35
1/4 thru 1-1/2
120000
150000
130000
12
35
1/4 thru 1-1/2 1/4 thru 1
120000 120000
150000 150000
130000 130000
10 10
35 35
830 – 1040 1220
12 – 9 8
35 – 35 35
American threads
Grade on 5
5.1
5.2 7 8
8.1 8.2
Products Bolts Screws Studs Sems Bolts Screws Bolts Screws Bolts Screws Bolts Screws Studs Studs Bolt Screws
Yield strength Elongation Reduction (stress) min., of area, psi % %
2037-Index Page 401 Tuesday, November 7, 2000 4:52 PM
Index A Almansi strain tensor, 157, 160, 161, 164, 175–179 Analogy, 217, 334 Associated flow rule, 123, 124, 127, 134 Axial stress, 117, 118, 316, 318, 356, 364 Axisymmetric, 27, 29, 302, 304, 306
B Bar, 38, 44, 260 Base vector, 143-146, 156, 158, 159, 380–385 Basis, 11, 25, 85, 123, 186, 216, 222, 229, 232, 242 Beam, 76–79, 82, 266, 268, 269 Beam element, 76, 78, 82 Bending moment, 77, 269, 271, 352, 354, 356, 359, 361 Body force, 37, 50, 55 Bolt, 250, 289–331, 334-342, 347, 348 Boundary conditions, 37–40, 44–47, 52, 77, 80, 341 Boundary value problem, 34, 38, 49, 50, 84
C Cartesian coordinates, 5, 18, 21, 62, 151, 377 Cauchy stress, 167, 170, 171, 175 Cauchy stress tensor, 170 Cauchy-Green deformation tensor, 148, 150, 157, 163 Cayley-Hamilton theorem, 173, 379 Coefficient of friction, 187, 188, 356 Compatibility, 50, 59, 61, 76, 114, 293 Constitutive equations, 24, 172 Constitutive law, 51 Constitutive relations, 210 Contact element, 190–192, 198, 199, 276, 304, 342, 356 Contact force, 189, 192, 202, 270, 300, 306, 337
Contact mechanics, 191 Contact problem, 89, 186–190, 209, 210, 341 Contact surface, 185, 188–192, 198, 200, 202, 248, 277, 341, 342, 347 Contact zone, 89, 185–187, 190, 202, 248, 351, 352, 356 Continuity, 1, 76–79, 82, 83, 84 Continuum, 1, 47, 140, 184 Contravariant base vectors, 146, 385 Convergence, 98, 102, 190, 209, 302 Convolute, 135, 136, 374, 376 Coordinate axes, 15, 32, 119, 367, 378 Coordinate functions, 39, 40, 41, 53 379, 380, 384 Coordinates Cartesian, 5, 18, 21, 62, 151, 377 curvilinear, 142 cylindrical, 27 material, 141, 151, 155, 157 natural, 63 nodal, 62 Coulomb, 187, 188, 203, 208, 251 Covariant base vectors, 145, 146, 385 Crack formation, 232, 240 Crack growth, 236, 239, 240, 243, 245, 351 Crack propagation, 211, 233, 239, 240, 262–264, 322–325, 329, 330, 331, 359, 364 Curvilinear coordinates, 142 Cyclic, 57, 211–217, 222, 223, 246, 256, 258, 321, 322, 331, 351, 352, 354, 355 Cylindrical coordinates, 27
D Deformation gradient, 173 Deformation tensor, 148–150, 157, 159, 163, 164, 173, 174 Deformation theory, 122, 136, 137 Determinant, 5, 163, 375, 377, 388, 389 401
2037-Index Page 402 Tuesday, November 7, 2000 4:52 PM
402 Dimension, 83 Displacement large, 1, 89, 141, 247, 265, 277, 278 nodal, 56, 61–66, 74, 102, 180 normal, 186 small, 1, 155 tangential, 187, 188 virtual, 37, 47, 77 Displacement distribution, 38, 43, 44, 77 Displacement error, 84 Displacement vector, 59, 82, 100, 142, 156 Distortion, 33, 34, 107, 118, 252, 253 Divergence, 46, 54 Divergence theorem, 54 Dyadic, 372, 375, 384
E Effective plastic strain, 125, 136 Effective strain, 30, 31, 226, 263 Effective stress, 29–32, 125, 129, 136, 225 Elastic deformation, 114, 138, 184, 211 Elastic material, 122, 173, 183 Elastic strain, 32, 33, 111, 217, 218, 220, 223, 224, 231 Elasticity, 1, 25, 47, 51, 89, 137, 141, 183, 210, 211, 260, 265, 276, 302, 375, 385 Elasticity matrix, 51, 137, 183 Elasto-plastic matrix, 137 Energy dissipation, 89, 186, 189, 279, 287 Energy error, 85 Engineering strain, 108 Engineering stress, 108 Equations constitutive, 24, 172 equilibrium, 8, 46, 50, 51, 65, 75, 79, 81, 190 Galerkin, 40 linear, 43, 44, 45, 286 Equilibrium equation, 8, 46, 50, 51, 65, 75, 79, 81, 190 Error, 40, 83-87, 93, 94-97, 191, 209, 277, 303, 347
F Failure criteria, 250, 251 Fatigue failure, 211, 230, 254, 280, 289, 322, 331, 351, 361
Nonlinear Problems in Machine Design Final failure, 107, 108, 115, 239, 240, 351 Finite element, 49, 58, 80, 84–87, 92, 100, 137, 141, 180, 188–192, 202, 209, 210, 250, 263, 266, 273, 285, 302, 355, 361 Flange connection, 293, 297, 333-337, 339342, 347, 349 Flow rule, 123, 124, 127, 134 Flow theory, 122, 123 Fracture, 184, 211, 217, 229-233, 238, 240, 246, 302, 323 Fracture mechanics, 211, 230, 232 Fracture toughness, 238, 323 Fretting, 351, 352, 353, 355, 361, 364, 365 Frictionless contact, 202, 204, 206, 209 Functional, 108
G Galerkin, 38-41, 43, 45, 46, 47, 49, 52–54, 61, 77, 80, 84 Galerkin equation, 40 GalerkIn method, 38, 39, 45, 47, 49, 54, 84 Gasket, 333–342, 347, 348 Gauss, 35, 46, 54 Green strain tensor, 150–152, 157, 159, 160, 166, 180 Griffith, 231, 232, 246
H Hencky, 117, 153–155 Hencky strain tensor, 153, 154 Hexahedron, 68, 69, 341, 356 Hierarchical function, 70, 72, 73, 76, 83, 84 Hooke’s law, 22–26, 29, 32, 33, 51, 89, 107, 108, 123, 127, 172, 249, 376 Huber, 117, 252 Hydrostatic stress, 121
I Identity matrix, 385, 388 Identity tensor, 374, 375, 383–385 Initial strain, 276, 277, 333 Initial stress, 138, 183 Instability, 209, 238 Integral, 54, 55, 58, 109, 136, 237 Integration, 40, 210, 324
2037-Index Page 403 Tuesday, November 7, 2000 4:52 PM
Index Interpolation, 53, 62, 78, 85, 86, 105 Interpolation function, 53 Interpolation functions, 53 Invariant, 15, 22, 23, 118, 174–176, 226, 367, 370, 379, 380 Inverse matrix, 388 Isoparametric, 60, 61, 64, 66, 70 Isoparametric elements, 60, 61 Iteration, 94, 100, 104, 139, 200, 302 Iterative method, 92–94
J Jacobian, 63, 69
K Kronecker, 368, 383, 388 Kronecker delta, 368, 383, 388 Kuhn-tucker, 186, 191, 202, 203, 208
L Lagrange, 189, 190, 202–209 Lagrange method, 190, 202, 206, 209 Lagrange multiplier, 189, 202, 204, 207, 209 Lagrangian, 149, 180, 190, 202, 206–210 Lame, 29, 175, 178, 354 Lame constant, 29, 175, 178 Large deformation, 17, 155, 209 Large displacement, 1, 89, 141, 247, 265, 277, 278 Large strain, 89 LEFM, 232 Left stretch tensor, 154, 165 Legendre polynomials, 72 Linear elastic fracture mechanics, 232 Linear equation, 43, 44, 45, 286 Linear form, 297 Linear strain, 17, 131, 150, 151 Load vector, 72, 82
403 Material coordinates, 141, 151, 157 Matrix elasticity, 51, 137, 183 elasto-plastic, 137 Hessian, 101, 102, 104, 105 identity, 385, 388 inverse, 388 stiffness, 55, 58–61, 64–69, 72, 73, 81, 93, 98, 101, 105, 137, 182, 183, 190–193, 198, 199, 205, 209 transpose, 388 Mean strain, 23, 123 Mean stress, 33, 221, 222, 226, 227, 246, 263, 317, 321, 327, 329, 354 Metric tensor, 146, 147 Minimization, 43, 102, 103, 105, 189, 204 Minimum, 34, 35, 36, 43, 49, 84, 96, 102–105, 221, 242, 247 Minimum energy, 34 Modulus, 25, 29, 77, 80, 90, 132, 175, 178, 233, 235, 236, 266, 302 Mohr diagram, 8, 9, 11 Moment, 77, 79, 266, 269, 271, 352–354, 356, 359, 361 Mooney-Rivlin, 174 Multiaxial strain, 31, 224 Multiaxial stress, 29, 30, 250, 251, 253, 261, 354
N Natural coordinate system, 63 Natural coordinates, 60, 64, 66, 68, 69, 80 Newton-Raphson method, 94–99, 103, 277, 316, 327 Nodal coordinates, 62 Nodal displacements, 56, 61–66, 74, 102, 180 Nodal force, 55, 58, 85, 102, 182 Nondimensional, 208 Nonlinear analysis, 141, 297, 302, 333, 334, 338, 347 Normal displacement, 186 Normal strain, 21 Normal stress, 2, 3, 13, 32, 179, 251, 252
M
O
Mapping, 61, 200, 210 Master, 190, 191, 193 Material coordinate, 141, 151, 155, 157
Octahedral stress, 12 Orthogonal, 42, 153, 164, 378 Orthogonality, 40, 104
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P
R
Penalty, 189–192, 202, 206–209, 273, 281, 287, 331, 348, 365 Penalty function method, 189 Piola stress tensor, 167, 170–172, 180 Plane strain, 26, 234, 235, 238, 276 Plane stress, 6, 26, 234, 235, 237, 238 Plastic deformation, 34, 107, 108, 111–119, 122, 125, 127, 134, 136–138, 211, 220, 232, 247, 260, 262, 263, 301, 302, 307, 325, 330 Plastic strain, 111, 113, 115, 123–126, 129, 134, 136, 139, 217, 218, 223, 224, 246 Plasticity, 6, 15, 47, 89, 107, 122, 123, 136, 140, 200, 211, 246, 247, 316 Plate, 6, 26, 79–82, 231, 233, 235, 246, 260, 293, 304 Poisson, 25, 80, 125, 136, 175, 176, 226 Polar decomposition, 153, 173 Polynom, 379 Polynomial, 72, 73, 80, 103, 173, 377 Positive definite, 154 Potential, 34, 35, 36, 43–45, 49, 85, 90, 92, 96, 100-105, 123–129, 172, 188–193, 199, 202–208, 232, 322 Potential energy, 34–36, 43, 49, 85, 90, 92, 96, 101, 172, 188–193, 199, 202, 207, 208, 232 Prandtl-Reuss, 124, 125 Press fit, 351, 353, 354, 356, 359, 361 Pressure flange, 347 Pressure vessel, 293, 295, 302, 333 Principal strain, 21, 22 Principal stress, 4–6, 15, 22, 118, 119, 129, 251, 315, 316, 322, 330, 359 Prism, 121, 356
Radial return, 138, 139, 140 Rayleigh-Ritz, 43, 45 Reciprocal, 25, 144–146, 227, 354, 380, 382, 383, 385 Reciprocal base, 144, 145, 380, 382, 383, 385 Relaxation, 317, 327, 330 Reliability, 84 Residual, 40, 92, 96, 101, 102, 113–117, 182, 204, 206, 215, 266, 281–283, 287, 325 Resistance, 96, 97, 102, 112, 182, 187, 188, 199, 287 Right stretch tensor, 154, 158 Rigid body, 19, 131, 151, 153 Ritz, 43, 45, 48 Rubber, 89, 174
Q Quadratic, 44, 66, 103, 104, 146, 173, 175, 177, 180, 208 Quadrilateral, 61, 62, 68, 73, 76, 276, 280, 304 Quadrilateral element, 62, 73, 280
S Scalar product, 145, 146, 368, 372, 374, 376, 383, 385 Shape function, 53–57, 60, 61, 62, 66–78, 82–84, 181 Shear strain, 21 Shear stress, 2, 3, 9, 12, 13, 32, 118, 226, 251, 298–301, 315, 316, 359, 361, 364 Signorini, 175, 177–179 Simple shear, 162, 178, 179 Simplex element, 56, 59–61 Slave, 190–193 Sliding contact, 347 Sliding friction, 206, 207, 287, 337, 348, 365 Sliding mode, 187, 188, 199, 200, 233 Small deformation, 17, 108, 161, 167 Small displacement, 1, 155 Stationary contact, 185 Sticking contact, 200 Sticking friction, 206 Sticking mode, 187, 188, 192, 198, 199, 202 Sticking-friction contact, 202–204 Stiffness matrix, 55, 58, 60, 64–67, 69, 72–75, 81, 93, 101, 105, 137, 182, 183, 190, 192, 193, 198, 199, 205, 209, 342
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Index Strain effective, 30, 31, 226, 263 elastic, 32, 33, 111, 217, 218, 220, 223, 224, 231 engineering, 108 initial, 276, 277, 333 large, 89 linear, 17, 131, 150, 151 mean, 23, 123 multiaxial, 31, 224 normal, 21 plane, 26, 234, 235, 238, 276 plastic, 111, 113, 115, 123–126, 129, 134, 136, 139, 217, 218, 223, 224, 246 principal, 21, 22 shear, 21 total, 111, 135, 136, 219, 221 true, 217, 220 virtual, 37 Strain component, 26–28, 176, 224 Strain deviator, 22, 23, 123, 125, 136 Strain energy, 32, 33, 44, 85, 90, 91, 117, 172–174, 231, 232, 236, 237, 246, 376 Strain hardening, 112, 125, 131, 134, 212, 217, 302 Strain softening, 212 Strain tensor, 19, 20, 22, 127, 141, 148, 149, 150–154, 157, 159, 160–162, 164, 166, 172–180, 375, 376 Strain vector, 29, 73, 180 Stress axial, 117, 118, 316, 318, 356, 364 Cauchy, 167, 170, 171, 175 effective, 29–32, 125, 129, 136, 225 engineering, 108 hydrostatic, 121 initial, 138, 183 mean, 33, 221, 222, 226, 227, 246, 263, 317, 321, 327, 329, 354 multiaxial, 29, 30, 250, 251, 253, 261, 354 normal, 2, 3, 13, 32, 179, 251, 252 octahedral, 12 plane, 6, 26, 234, 235, 237, 238 principal, 4–6, 15, 22, 118, 119, 129, 251, 315, 316, 322, 330, 359
405 shear, 2, 3, 9, 12, 13, 32, 118, 226, 251, 298–301, 315, 316, 359, 361, 364 true, 108, 110, 111, 171, 217, 220 Stress component, 3, 4, 8, 10, 14, 117, 123, 137, 171, 233, 263, 376 Stress deviator, 14, 15, 118 Stress intensity factor, 233, 235–239 Stress relaxation, 327 Stress space, 15, 16, 119, 123, 127, 128, 131, 132, 134, 251, 252 Stress tensor, 4, 5, 14, 15, 167, 170–176, 180, 376 Stress vector, 15, 51, 60, 61, 66, 67, 69, 76, 167, 169 Stress-strain relation, 24, 26, 27, 29, 51, 55, 107, 110, 111, 122, 125, 135, 136, 140, 178, 179, 182 Stretch tensor, 153, 154, 158, 165, 166
T Tangent stiffness, 92, 182, 205, 210 Tangent stiffness matrix, 182, 205 Tangential displacement, 187, 188 Taylor series, 91, 96, 100, 105, 155, 177, 178, 205 Tensor Almansi strain, 157, 160, 161, 164, 175, 176, 178, 179 Cauchy stress, 167, 170, 171, 175 Cauchy–Green deformation, 148, 150, 157, 163 deformation, 148–150, 157, 159, 163, 164, 173, 174 Green strain, 150–152, 157, 159, 160, 166, 180 Hencky strain, 153, 54 identity, 374, 375, 383–385 left stretch,154–165 metric, 146, 147 Piola stress, 167, 170–172, 180 right stretch, 154,158 transpose, 373 Tetrahedron, 4, 10–12, 59, 60, 168–170 Threshold, 240, 322, 323, 330, 359 Threshold value, 240, 322 Torque, 283-285, 292, 333
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V
Torsion, 250 Total potential, 44, 203, 232 Total potential energy, 232 Total strain, 111, 135, 136, 219, 221 Traction, 2, 4, 8, 10, 36, 52, 58, 167, 170, 248 Transformation, 180, 370, 371, 384, 385 Transformation of coordinates, 370, 371, 384 Transpose, 154, 373, 388 Transpose tensor, 373 Tresca criterion, 118, 251 Triangle, 8, 9, 11, 56, 60, 266, 291 Triangular element, 56, 304 True strain, 217, 220 True stress, 108, 110, 111, 171, 217, 220 Truss, 113–115
Variational formulation, 47 Variational principle, 34, 38 Vector base, 143–146, 156, 158, 159, 380–385 displacement, 59, 82, 100, 142, 156 load, 72, 82 strain, 29, 73, 180 stress, 15, 51, 60, 61, 66–69, 76, 167, 169 unit, 4, 367–372, 376, 380 Vector product, 369, 382 Virtual displacement, 37, 47, 77 Virtual strain, 37 Virtual work, 34, 36–38, 46, 47, 49, 55, 91, 180, 189 Volumetric deformation, 21 Von Mises criterion, 117
U
Weak form, 49 Windup, 283, 284, 285
Ultimate strength, 107, 249, 251, 254, 353, 355 Uniqueness, 140 Unit vector, 4, 367–372, 376, 380
W
Y Yield surface, 119, 121–125, 127, 131–134, 138, 140