CHAP 3 FEA for Nonlinear Elastic Problems Nam-Ho Kim 1 Introduction • Linear systems – Infinitesimal deformation: no significant difference between the deformed and undeformed shapes – Stress and strain are defined in the undeformed shape – The weak form is integrated over the undeformed shape • Large deformation problem – The difference between the deformed and undeformed shapes is large enough that they cannot be treated the same – The definitions of stress and strain should be modified from the assumption of small deformation – The relation between stress and strain becomes nonlinear as deformation increases • This chapter will focus on how to calculate the residual and tangent stiffness for a nonlinear elasticity model 2 Introduction • Frame of Reference – The weak form must be expressed based on a frame of reference – Often initial (undeformed) geometry or current (deformed) geometry are used for the frame of reference – proper definitions of stress and strain must be used according to the frame of reference • Total Lagrangian Formulation: initial (undeformed) geometry as a reference • Updated Lagrangian Formulation: current (deformed) geometry • Two formulations are theoretically identical to express the structural equilibrium, but numerically different because different stress and strain definitions are used 3 Table of Contents • 3.2. Stress and Strain Measures in Large Deformation • 3.3. Nonlinear Elastic Analysis • 3.4. Critical Load Analysis • 3.5. Hyperelastic Materials • 3.6. Finite Element Formulation for Nonlinear Elasticity • 3.7. MATLAB Code for Hyperelastic Material Model • 3.8. Nonlinear Elastic Analysis Using Commercial Finite Element Programs • 3.9. Fitting Hyperelastic Material Parameters from Test Data • 3.9. Summary • 3.10.Exercises 4 3.2 Stress and Strain Measures 5 Goals – Stress & Strain Measures • Definition of a nonlinear elastic problem • Understand the deformation gradient? • What are Lagrangian and Eulerian strains? • What is polar decomposition and how to do it? • How to express the deformation of an area and volume • What are Piola-Kirchhoff and Cauchy stresses? 6 Mild vs. Rough Nonlinearity • Mild Nonlinear Problems (Chap 3) – Continuous, history-independent nonlinear relations between stress and strain – Nonlinear elasticity, Geometric nonlinearity, and deformation- dependent loads • Rough Nonlinear Problems (Chap 4 & 5) – Equality and/or inequality constraints in constitutive relations – History-dependent nonlinear relations between stress and strain – Elastoplasticity and contact problems 7 What Is a Nonlinear Elastic Problem? • Elastic (same for linear and nonlinear problems) – Stress-strain relation is elastic – Deformation disappears when the applied load is removed – Deformation is history-independent – Potential energy exists (function of deformation) • Nonlinear – Stress-strain relation is nonlinear (D is not constant or do not exist) – Deformation is large •Examples – Rubber material – Bending of a long slender member (small strain, large displacement) 8 Reference Frame of Stress and Strain • Force and displacement (vector) are independent of the configuration frame in which they are defined (Reference Frame Indifference) • Stress and strain (tensor) depend on the configuration • Total Lagrangian or Material Stress/Strain: when the reference frame is undeformed configuration • Updated Lagrangian or Spatial Stress/Strain: when the reference frame is deformed configuration • Question: What is the reference frame in linear problems? 9 Deformation and Mapping • Initial domain 0 is deformed to x – We can think of this as a mapping from 0 to x • X: material point in 0 x: material point in x • Material point P in 0 is deformed to Q in x xXu xXXuX(,t) (,t) displacement x u 0 Q P X x ,:1 One-to-one mapping Continuously differentiable 10 Deformation Gradient • Infinitesimal length dX in 0 deforms to dx in x • Remember that the mapping is continuously differentiable x x Q' ddxX; dd xFX P' u x X 0 X d d Q P • Deformation gradient: x F i u . 1 [],ij ij F1 10 u Xj X . 0x, – gradient of mapping Xx – Second-order tensor, Depend on both 0 and x – Due to one-to-one mapping: detF J 0. ddXF 1 x – F includes both deformation and rigid-body rotation 11 Example – Uniform Extension • Uniform extension of a cube in all three directions xX,xX,xX111222333 • Continuity requirement: i 0 Why? • Deformation gradient: FV 1 00 GW F GW002 GW HX00 3 •123 : uniform expansion (dilatation) or contraction • Volume change – Initial volume: dV0123 dX dX dX – Deformed volume: dVx dx 1 dx 2 dx 3 123 dX 1 dX 2 dX 3 123 dV 0 12 Green-Lagrange Strain • Why different strains? 22 • Length change: ddddddxXxxXXTT ddddXFFXTT X T X d(XFF1XTT )d Ratio of length change • Right Cauchy-Green Deformation Tensor CFF T • Green-Lagrange Strain Tensor dx dX 1 EC1() The effect of rotation is eliminated 2 To match with infinitesimal strain 13 Green-Lagrange Strain cont. • Properties: – E is symmetric: ET = E CSu u 1 DTi j – No deformation: F = 1, E = 0 ij DT 2XEUj Xi 1 CSuuTT uu E DT 2 EUXX XX Displacement gradient 1 .uu .TT . uu . 2 00 00 Higher-order term 1 –When .u 1 , Euu..T 0 2 00 – E = 0 for a rigid-body motion, but 0 14 Example – Rigid-Body Rotation • Rigid-body rotation xXcosXsin11 2 xXsinXcos21 2 xX33 • Approach 1: using deformation gradient FVcos sin 0 FV100 GW GWT F GWsin cos 0 FF GW010 HXGW001 HXGW001 EFF101 ()T 2 Green-Lagrange strain removes rigid-body rotation from deformation 15 Example – Rigid-Body Rotation cont. • Approach 2: using displacement gradient uxXX(cos1)Xsin1111 2 uxXXsinX(cos1)2221 2 uxX0333 FVcos 1 sin 0 .GW 0u GWsin cos 1 0 HXGW000 FV2(1 cos ) 0 0 GW ..T 00uuGW02(1cos)0 HXGW000 Euuuu0....1 ()TT 2 00 00 16 Example – Rigid-Body Rotation cont. • What happens to engineering strain? uxXX(cos1)Xsin1111 2 uxXXsinX(cos1)2221 2 uxX0333 FVcos 1 0 0 GW GW0cos10 HXGW000 Engineering strain is unable to take care of rigid-body rotation 17 Eulerian (Almansi) Strain Tensor 22 • Length change: ddddddxXxxXXTT ddxxTTT1 d xFFx d d(x1FFTT1 )d x d(x1bT1 )d x • Left Cauchy-Green Deformation Tensor bFF T b–1: Finger tensor • Eulerian (Almansi) Strain Tensor 1 e1b()1 2 Reference is deformed (current) configuration 18 Eulerian Strain Tensor cont. • Properties – Symmetric u – Approach engineering strain when 1 x – In terms of displacement gradient 1 CSuuTT uu DT e . 2 EUxx xx x x 1 ....uuTT uu Spatial gradient 2 xx xx • Relation between E and e EFeF T 19 Example – Lagrangian Strain •Calculate F and E for deformation in the figure • Mapping relation in 0 I 4 3 LXN(s,t)X(s1)B Y L II I1 4 Deformed element J 2.0 L 4 1 LYB NII (s,t)Y (t 1) K 2 1.0 I1 Undeformed element • Mapping relation in x X 0.7 1.5 I 4 Lx(s, t)B N (s, t)x 0.35(1 t) L II J I1 L 4 L y(s, t)B NII (s, t)y s 1 K I1 20 Example – Lagrangian Strain cont. • Deformation gradient Q' xxs x P' u x F 0 X d XsX d Q P FVFV0.354/30 GWGW HXHX10 02 x(,)st FV00.7 X(,)st GW HX4/3 0 Reference domain (s, t) • Green-Lagrange Strain 1 FV0.389 0 EFF1()T GWTension in X1 dir. 2 HX0 0.255 Compression in X2 dir. 21 Example – Lagrangian Strain cont. •AlmansiStrain FV0.49 0 bFF4T GW HX01.78 FV0.52 0 e1b1 1 GWCompression in x1 dir. 2 HX00.22 Tension in x2 dir. • Engineering Strain FV10.7 .uF1GW 0 HX1.33 1 FV10.32 ..1 uuT GWArtificial shear deform. 2 00 HX0.32 1 Inconsistent normal deform. Which strain is consistent with actual deformation? 22 Example – Uniaxial Tension ' '( • Uniaxial tension of incompressible material ( 1 = ) • From incompressibility xX111 ; 1/2 xX222 1231 2 3 xX • Deformation gradient and deformation tensor 333 FV FV 2 00 00 GW GW 1 F GW001/2 C GW00 GW GW 1 HX00 1/2 HX00 •G-L Strain FV 2 10 0 GW 1 E GW0101 2 GW1 HX00 1 23 Example – Uniaxial Tension •AlmansiStrain (b = C) FV 2 00 FV100 2 GW1 GW b 1 GW00e GW01 0 GW2 GW HX00 HX001 • Engineering Strain FV 10 0 GW 0101/2 10% GWstrain GW HX00 1/2 1 • Difference 1122 E(1)e(1)11 11 11 1 22 24 Polar Decomposition • Want to separate deformation from rigid-body rotation • Similar to principal directions of strain • Unique decomposition of deformation gradient FQUVQ – Q: orthogonal tensor (rigid-body rotation) – U, V: right- and left-stretch tensor (symmetric) • U and V have the same eigenvalues (principal stretches), but different eigenvectors 25 Polar Decomposition cont. ddxQUX44 FVQ 4 4 VQd X e e 3 1 e V Q 2 , e 3 3 E , e 3 1 1 E 1 E , E , e 2 3 3 2 2 U , E Q , E 2 2 1 1 FQU •Eigenvectors of U: E1, E2, E3 •Eigenvectors of V: e1, e2, e3 • Eigenvalues of U and V: 1, 2, 3 26 Polar Decomposition cont. • Relation between U and C UCUC2 – U and C have the same eigenvectors. – Eigenvalue of U is the square root of that of C •How to calculate U from C? • Let eigenvectors of C be []EEE123 •Then,) T C where FV 2 00 GW1 )GW00 2 Deformation tensor in 2 principal directions GW 2 HX00 3 27 Polar Decomposition cont. Useful formulas T •AndU ) 3 !2 CEEB ii i i1 FV 1 00 3 GW ! )00 UEEB ii i GW2 GW i1 HX00 3 3 ! QeEB ii i1 3 • General Deformation !2 beeB ii i ddx FXbQUXb d i1 3 1. Stretch in principal directions ! VeeB ii i 2. Rigid-body rotation i1 3 3.