CHAP 3 FEA for Nonlinear Elastic Problems Introduction

CHAP 3

FEA for Nonlinear Elastic Problems

Nam-Ho Kim

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

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

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

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?

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

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

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

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.

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

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

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. Rigid-body translation ! FeEB ii i i1 28 Generalized Lagrangian Strain • G-L strain is a special case of general form of Lagrangian strain tensors (Hill, 1968) 1 EU12m m 2m

Example – Polar Decomposition

• Simple shear problem X2, x2 2 I xXkX k L 11 2 3 JxX22 L KxX 33 X1, x1 FV1k • Deformation gradient F GW HX01 FV2 FV1k 1 • Deformation tensor CFFT GW GW3 HXkk2 1 GW2 7 HX3 3

X2 • Find eigenvalues and eigenvectors of C E1 E2 60o 123, 1 3 X1 EE1133, 1222 22 30 Example – Polar Decomposition cont. FV + ) 30 •In E1 –E2 coordinates C GW HX013 FV 12 32 • Principal Direction Matrix []EE12 GW HXGW32 12

• Deformation tensor in principal directions ) 44T C • Stretch tensor FV30 ) GW HXGW013

FV32 12 U 4 ) 4T GW HXGW12 52 3 31

Example – Polar Decomposition cont. •How U deforms a square? IY10IY32 IY I12 YX2, x2 UU4JZ J Z, 4 JZ J Z K[01K[12 K[ K 52 3 [

• Rotational Tensor 30o FV X1, x1 32 12 QFU41 GW HXGW12 32 X , x IY32 IY11.15 I12 Y I Y2 2 QQ4JZJZJ, 4 ZJZ K[12K[01 K 52 3 [ K [ –30o clockwise rotation 30o FV536 12 X1, x1 VFQ4T GW HXGW12 32 32 Example – Polar Decomposition cont. / • A straight lineXXtan21 will deform to Xxkx,Xx11 2 22 ; / X2, x2 x(xkx)tan212 ;xkx1 12tan / • Consider a diagonal line: / = 45o 25o X1, x1 x2 1 tan 24.9 x1k1 X2, x2 • Consider a circle XXr222 12 X1, x1 222 (x12 kx ) x 2 r 2222 x2kxx(1k)xr1122 Equation of ellipse 33

Deformation of a Volume • Infinitesimal volume by three vectors 4123 123 – Undeformed: dV0rstrst dXXX (d d ) e dX dX dX 4123 123 –Deformed: dVxijkijk dxxx (d d ) e dx dx dx

123 dVxijkijk e dx dx dx 2 CS CSx CS dX dx2 xi 123j xk edXdXdXDTDTDT 1 ijkEUXX rDT sEU X t dX dx3 rsEU t dX3 dx1 x xj x edXdXdXi k 123 ijk r s t XXXrst 123 erst J dX r dX s dX t JdV0 From Continuum Mechanics JdetF 123 eaaaijk ir js kt e rst deta 34 Deformation of a Volume cont. • Volume change dVx0 J dV

• Volumetric Strain

dV dV x0J1 dV0 • Incompressible condition: J = 1 • Transformation of integral domain OOOfd OOO fJd x0

Example - Uniaxial Deformation of a Beam

• Initial dimension of L0×h0×h0 deforms to L×h×h

xX L/L 111 1 0 L0 xX222 2 h/h 0 xX333 3 h/h 0 h0

• Deformation gradient h0

FV 00 Jdet F GW1 123 2 L F GW002 LhCS LA GW DT h HX00 3 LhEU LA 00 00 h

•Constant volume

LL J1; hh 00 AA 00LL 36 Deformation of an Area

• Relationship between dS0 and dSx 12 12 NXXdS0i0ijkjk d d NdS e dX dX 12 12 nxxdSxrxrstst d d n dS e dx dx Xj X NdS ek dx12 dx i0 ijkst xxst X XXXj X i iiNdS ek dx12 dx i0 ijkst xr xxxxrrst F(X)

N n dX2 dx2 X dS x dS 0 S x S dX1 0 dx1 x

Undeformed Deformed 37

Deformation of an Area cont.. • Results from Continuum Mechanics

xxx eeF rst ijk rst XXXijk X Xj X eeF 1 i k . rst ijk xxxrst

ndSrx • Use the second relation: XXXj X 1 iiNdS ek dx12 dx eF dx 12 dx i0 ijkstrstst xxxxrrst

T 4 nFNdS4 JT dS T 4;FN x0nF N n FNT 4

T dSx0 JFx ( ) NX ( ) dS 38 Stress Measures • Stress and strain (tensor) depend on the configuration • Cauchy (True) Stress: Force acts on the deformed config. 3f – Stress vector at : tnlim 5 x 3 3 S0x Sx Cauchy Stress, sym

– Cauchy stress refers to the current deformed configuration as a reference for both area and force (true stress) Undeformed configuration Deformedconfiguration f N S 0 P S x P n

Stress Measures cont. • The same force, but different area (undeformed area)

3f TPNlim T 3 3 S00 S0 First Piola-Kirchhoff Stress Not symmetric

– P refers to the force in the deformed configuration and the area in the undeformed configuration • Make both force and area to refer to undeformed config.

5 T 4T ddSdSfnx0 PN nFNdSx0 J dS

5 TT d(JdS)dSfFNPN00

PF J 15 : Relation between 5 and P

40 Stress Measures cont. • Unsymmetric property of P makes it difficult to use – Remember we used the symmetric property of stress & strain several times in linear problems •Make P symmetric by multiplying with F-T

1 SPF4T1T J F 445 F 5 44FSFT J Second Piola-Kirchhoff Stress, symmetric – Just convenient mathematical quantities • Further simplification is possible by handling J differently

6 44J5 FSFT

Kirchhoff Stress, symmetric

Stress Measures cont. •Example 5 5 6 OOO:dx00 OOO :Jd OOO :d x0 0 Integration can be done in 0 •Observation – For linear problems (small deformation): Ee – For linear problems (small deformation): 5 6 PS – S and E are conjugate in energy – S and E are invariant in rigid-body motion

42 Example – Uniaxial Tension • Cauchy (true) stress:5F , 5 = 5 = 5 = 5 = 5 = 0 11 A 22 33 12 23 13 • Deformation gradient:

L0 FV 1 00 GW1 11 F GW00,J12 GW 1 h0 HX00 3 h0 • First P-K stress

F1 FA F L P(JF 15 11 11 AAAA100 h h • Second P-K stress F

2 F1 FA FA F S(J44FF1T5 ) No clear physical 11 11 222 AA100AA A01 meaning

Summary • Nonlinear elastic problems use different measures of stress and strain due to changes in the reference frame • Lagrangian strain is independent of rigid-body rotation, but engineering strain is not • Any deformation can be uniquely decomposed into rigid- body rotation and stretch • The determinant of deformation gradient is related to the volume change, while the deformation gradient and surface normal are related to the area change • Four different stress measures are defined based on the reference frame. • All stress and strain measures are identical when the deformation is infinitesimal

44 3.3 Nonlinear Elastic Analysis

Goals • Understanding the principle of minimum potential energy – Understand the concept of variation • Understanding St. Venant-Kirchhoff material • How to obtain the governing equation for nonlinear elastic problem • What is the total Lagrangian formulation? • What is the updated Lagrangian formulation? • Understanding the linearization process

46 Numerical Methods for Nonlinear Elastic Problem • We will obtain the variational equation using the principle of minimum potential energy – Only possible for elastic materials (potential exists) • The N-R method will be used (need Jacobian matrix) • Total Lagrangian (material) formulation uses the undeformed configuration as a reference, while the updated Lagrangian (spatial) uses the current configuration as a reference • The total and updated Lagrangian formulations are mathematically equivalent but have different aspects in computation

Total Lagrangian Formulation • Using incremental force method and N-R method – Total No. of load steps (N), current load step (n)

n1fff3 n n

• Assume that the solution has converged up to tn

• Want to find the equilibrium state at tn+1 Lastconvergedconfiguration Undeformed configuration (known) Currentconfiguration (known) n (unknown)

0 nu nP n+1 0P u n+1P

X x Iteration

48 Total Lagrangian Formulation cont. • In TL, the undeformed configuration is the reference •2nd P-K stress (S) and G-L strain (E) are the natural choice • In elastic material, strain energy density W exists, such that W stress strain

• We need to express W in terms of E

Strain Energy Density and Stress Measures • By differentiating strain energy density with respect to proper strains, we can obtain stresses •When W(E) is given W(E ) S Second P-K stress E •When W(F) is given WWE W 44: FFSPT First P-K stress FEF E • It is difficult to have W() because depends on rigid- body rotation. Instead, we will use invariants in Section 3.5

50 St. Venant-Kirchhoff Material • Strain energy density for St. Venant-Kirchhoff material W(EEDE ) 1 : : ab:ab 2 Contraction operator: ij ij • Fourth-order constitutive tensor (isotropic material) D11I ! 2 7 8EE – Lame’s constants: 7 (18 )(1 2 8 ) 2(1 8 )

nd – Identity tensor (2 order): 1 []ij th I( 1 ) – Identity tensor (4 order): ijkl2 ik jl il jk

Ia:9 a , 2nd-order sym. a 1a:tr()aaaa a ii 11 22 33

! – Tensor product: aaaij a kl (4th-order) 51

St. Venant-Kirchhoff Material cont. • Stress calculation – differentiate strain energy density W(E ) SDEE1E 7:tr()2 E – Limited to small strain but large rotation

EFF1UQQu1U111()(TTT2 )() 1 22 2

– Rigid-body rotation is removed and only the stretch tensor contributes to the strain WW – Can show S 2 EC

Deformation tensor

52 Example Y Deformed element • E = 30,000 and 8 = 0.3 2.0 • G-L strain: FV 0.389 0 E GW1.0 HX00.255 Undeformed element X • Lame’s constants: 0.7 1.5 8EE 17,308 7 11,538 (18 )(1 2 8 ) 2(1 8 )

•2nd P-K Stress:

FVF1 0 .389 0 V SE1E tr( ) 2 7 (.389 .255)GWG 2 7 W HXH01 0 .255 X FV11,296 0 GW HX0 3,565

1 FV1,872 0 5 FSFT GW J HX 021,516 53

Example – Simple Shear Problem

• Deformation map X2, x2 xXkX,xX,xX11 22233 FV1k 11FV0k F GWEFF1()T GW HX01 22HXkk2 X1, x1 • Material properties 8EE 40MPa 7 40MPa (18 )(1 2 8 ) 2(1 8 ) 20 • 2nd P-K stress Cauchy 10 FVk2k2 SE1E tr( ) 2 7 20GW MPa 2nd P-K HXGW2k 3k2 0 Shear stress 1 FV5k24 3k 2k 3k 3 -10 5 FSFT 20GW MPa J HXGW2k 3k32 3k -20 -0.4 -0.2 0.0 0.2 0.4 Shear parameter k 54 Boundary Conditions • Boundary Conditions

ug:,onh Essential (displacement) boundary Ts tPN:,on Natural (traction) boundary You can’t use S

• Solution space (set)

+13 ;uu[H ( )] , u:h g<

• Kinematically admissible space +13 ;uu[H ( )] , u:h 0<

Variational Formulation • We want to minimize the potential energy (equilibrium) >int: stored internal energy >ext: potential energy of applied loads >>>()uuuint () ext () :Tb T OO:W(Eufut )d OO d O s d 00o

• Want to find u % that minimizes the potential energy –Perturb u in the direction of % proportional to 6

uuu6 6

–If u minimizes the potential, >(u) must be smaller than >(u6) for all possible

56 Variational Formulation cont. • Variation of Potential Energy (Directional Derivative) >>6d (,uu )6 ( u u ) We will use “over-bar” for variation d 60 – > depends on u only, but > depends on both u and – Minimum potential energy happens when its variation becomes zero for every possible – One-dimensional example

>(u)

At minimum, all directional derivatives are zero u

Example – Linear Spring

k u f

• Potential energy: >44(u)1 k u2 f u 2 • Perturbation: >64646(u u)1 k (u u)2 f (u u) 2 d • Differentiation: HXFV>64(u u) k (u 644 u) u f u d6 • Evaluate at original state:

d FV>6 444 6 HX(u u) k u u f u 0 d 60

Variation is similar to differentiation !!! 58 Variational Formulation cont. • Variational Equation >W(E ) Tb : T (,uu )OO: : E d OO u f d O s u t d 0 00oE for all % – From the definition of stress Tb T : OO:SE:d OO uf d O s ut d 00o

Variational equation in TL formulation

– Note: load term is similar to linear problems – Nonlinearity in the strain energy term • Need to write LHS in terms of u and

Variational Formulation cont. • How to express strain variation Eu()11 ( C 1 ) . u . uTT . u . u 2200 00 6d Euu(, )6 Eu ( u ) d 60 1 .uu .TT . uuuu . . T . 2 00 0000 ....1 ()1uuTT u1u () 2 00 0 0 ..1 FuTT uF 2 00 .T Euu(, ) sym(0 u F )

Note: E(u) is nonlinear, but Euu(, ) is linear

60 Variational Formulation cont. • Variational Equation

Tb T : for all % OO:SE:d OO uf d O s ut d 00o

a(uu , ) ()u

Energy form Load form

a(uu , )9+ ( u ), u

• Linear in terms of strain if St. Venant-Kirchhoff material is used • Also linear in terms of • Nonlinear in terms of u because displacement-strain relation is nonlinear 61

Linearization (Increment) • Linearization process is similar to variation and/or differentiation – First-order Taylor series expansion – Essential part of Newton-Raphson method •Let f(xk+1) = f(xk + 3uk), where we know xk and want to calculate 3uk

df(x ) f(xxk1 )43 f( k ) u k H.O.T. dx • The first-order derivative is indeed linearization of f(x) df ?343Linearization L[f]? f(xu ) u d ?0 x

df 6 4 Variation ff6 f(xu ) u d 60 x 62 Linearization of Residual • We are still in continuum domain (not discretized yet) •ResidualR(uuuu ) a( , ) ( ) • We want to linearize R(u) in the direction of 3u – First, assume that u is perturbed in the direction of 3u using a variable 6. Then linearization becomes

63FV T 3R(uu ) R L[R(uu )] 6GW 60 HXu

–R(u) is nonlinear w.r.t. u, but L[R(u)] is linear w.r.t. 3u – Iteration k did not converged, and we want to make the residual at iteration k+1 zero T FVk R(u ) R(uuuk1 )3GW k R( k ) 0 HXu 63

Newton-Raphson Iteration by Linearization • This is N-R method (see Chapter 2)

T FV k R(u ) kk GW3 uuR( ) f HXu x f(xk)

f(xk+1) 3 k • Update state uuuk13 k k u xk+1 xk x xXuk1 k1 FVR(uk ) k GW • We know how to calculate R(u ), but how about HXu ? [R(uuuu )] [a( , ) ( )] uu – Only linearization of energy form will be required – We will address displacement-dependent load later

64 Linearization cont. • Linearization of energy form

33FV L[a(uu , )] LHXOO00 S : E d OO [ S : E S : E ] d

– Note that the domain is fixed (undeformed reference) – Need to express in terms of displacement increment 3u • Stress increment (St. Venant-Kirchhoff material) S 3SEDE:: 3 3 E • Strain increment (Green-Lagrange strain)

3EFFFF1 () 3TT 3 2 CS3xXuu C() S 33FuDT 3 D T .3 EUXXX E U 0

Linearization cont.

• Strain increment 3EFFFF1 () 3TT 3 2 .3.31 ()uFTT F u 2 00 .3sym(uFT ) 0 !!! Linear w.r.t. 3u

• Inc. strain variation 33 .T EuF[sym(0 )] .3T sym(0uF ) ..3T sym(00uu ) !!! Linear w.r.t. 3u • Linearized energy form

333* L[a(,)]uuOO0 [ E : D : E S : E ]d a(; u uu ,)

– Implicitly depends on u, but bilinear w.r.t. 3u and – First term: tangent stiffness – Second term: initial stiffness 66 Linearization cont. • N-R Iteration with Incremental Force

–Let tn be the current load step and (k+1) be the current iteration – Then, the N-R iteration can be done by

a(*nuuu k ;3 k , ) ( u ) a( n uu k , ), 9+ u

– Update the total displacement nk1uuu3 nk k • In discrete form Tnk3 k T nk {}[dKdT ]{ } {}{ dR }

nk nk • What are []KT and{}R ?

Example – Uniaxial Bar du du • Kinematics u,22 u dX dX F =100N 2 1 2 du 1CS du 1 2 x Eu(u)11DT 2 2 dX 2EU dX 2 L0=1m • Strain variation du du du Eu(1u) 11dX dX dX 2 2 • Strain energy density and stress W1CS W(E)41 E(E)2 SEEEu(u)4DT2 112 11 11 11 2 2 E211 EU • Energy and load forms

L0 a(u,u)O S E AdX S AL (1 u )u (u) u F 0 11 11 11 0 2 2 2

9 • Variational equation RuSAL(1u)F2110 2 0,u 2 68 Example – Uniaxial Bar • Linearization 33 3 33 SEEE(1u)u11 11 2 2 Euu11 2 2 L a(u;u,u)* 3O 0 E 44343 E E S E AdX 0 11 11 11 11 332 EAL022211022 (1 u ) u u S AL u u • N-R iteration k2 k 3 k k k [E(1 u211021120 ) S ]AL u F S (1 u )AL k1 3 k k uuu2s2

Example – Uniaxial Bar

(a) with initial stiffness Iteration u Strain Stress conv 0 0.0000 0.0000 0.0000 9.999E01 1 0.5000 0.6250 125.00 7.655E01 2 0.3478 0.4083 81.664 1.014E02 3 0.3252 0.3781 75.616 4.236E06

(b) without initial stiffness Iteration u Strain Stress conv 0 0.0000 0.0000 0.0000 9.999E01 1 0.5000 0.6250 125.00 7.655E01 2 0.3056 0.3252 70.448 6.442E03 3 0.3291 0.3833 76.651 3.524E04 4 0.3238 0.3762 75.242 1.568E05 5 0.3250 0.3770 75.541 7.314E07

70 Updated Lagrangian Formulation • The current configuration is the reference frame – Remember it is unknown until we solve the problem – How are we going to integrate if we don’t know integral domain? • What stress and strain should be used? – For stress, we can use Cauchy stress (5) – For strain, engineering strain is a pair of Cauchy stress – But, it must be defined in the current configuration

CST 1 uu .DTsym(u ) DT x 2 EUxx

Variational Equation in UL • Instead of deriving a new variational equation, we will convert from TL equation CST 1 T 1 uu 5 44FSF EFFDTT J 2 DTXX EU ;SFJ 1T 445 F CST 1 TTDTuu 1 FFDT F F 2 EUXX CSTT 1 Xu uX FFT DT 2 DTxX Xx Similarly EU T T CS 3EF 434 F 1 T DTuu FFDT CS3T 3 2 EUxx 3 1 DTuu DT 44T 2 EUxx FF 72 Variational Equation in UL cont. • Energy Form

1T5 T a(uu , )OO S : E d OO (J F F ) : ( F F ) d 00 115 55 FFFFik kl jl mi mn nj mk nl kl mn mn mn

5 5 OOSE:d OO :Jd OO :d 00 x

– We just showed that material and spatial forms are mathematically equivalent • Although they are equivalent, we use different notation: 5 a(uu , )OO : d Is this linear or nonlinear? x • Variational Equation

a(uu , )9+ ( u ), u What happens to load form? 73

Linearization of UL

• Linearization of a(,x uu ) will be challenging because we don’t know the current configuration (it is function of u) • Similar to the energy form, we can convert the linearized energy form of TL *03 33 • Remember a(;uuu , )OO0 [ ED : : ES : E ]d • Initial stiffness term CS33TT 31T5 1 DTuu uu SE:J(): FF DT 2 EUXX XX CS 1 33uu uu 5JF11 F DTmm mm ik kl jl DT 2XXEUij XX ij CS33 1 uumm uu mm 5J DT @3 kl kl(,)uu 2xEUkl x x kl x 74 Linearization of UL cont. • Initial stiffness term 35 @ 3 @ 3 ..3T SE:J:(,) uu (,)sym(uuxx u u )

• Tangent stiffness term (:ED :3 E ) ( FTT 44 F )::( D F 434 F ) 3 FFDFki kl lj ijmn pm pq F qn FV1 JFFDFFGW 3 klHX ki lj ijmn pm qn pq J 4th-order spatial constitutive tensor ED::3 E J:: c 3 1 where cFFFFD ijklJ ir js km ln rsmn

Spatial Constitutive Tensor • For St. Venant-Kirchhoff material ! 7 7 D11I()2 Drsmn rs mn ( rm sn rn sm ) • It is possible to show 1 FV cbb(bbbb). HX 7 ijklJ ij kl ik jl il jk • Observation – D (material) is constant, but c (spatial) is not – SDE:,5 c :

76 Linearization of UL cont. • From equivalence, the energy form is linearized in TL and converted to UL 3 5 @ L[a(uu , )]OO [ : c : : ]J d 0 * 3 3 5 @ a(;uuu , )OO [ : c : : ]d x • N-R Iteration a(*nkuuu ;3 k , ) ( u ) a( nk uu , ), 9+ u • Observations – Two formulations are theoretically identical with different expression – Numerical implementation will be different – Different constitutive relation

Example – Uniaxial Bar

• Kinematics F =100N 1 2 duuu22 du , x dx 1 u22 dx 1 u L0=1m dx • Deformation gradient: F1u,J1u 11dX 2 2

11 • Cauchy stress:5FSF E(u u)(12 u) 11J2 11 11 11 2 2 2

u (u) FT1 E F 2 • Strain variation: 11 11 11 11 1u2

L • Energy & load forms: a(u,u)5O (u)Adx 5 Au (u) u F 0 11 11 11 2 2

59 •Residual:Ru211 AF 0,u 2

78 Example – Uniaxial Bar 1 • Spatial constitutive relation: c FFFFE (1 u)E3 1111J 11 11 11 11 2

L • Linearization: O 33(u)c ( u)Adx EA(1 u )2 u u 0 11 1111 11 2 2 2 L 5 A O 5@(u,u)Adx 3 11 u 3 u 0 11 11 2 2 1u2 L a* (u;3 u,u)O (u)c 35@3 ( u) ( u,u) Adx 0 11 1111 11 11 5 EA(1 u )2 u 3 u11 Au 3 u 222 22 1u2

Iteration u Strain Stress conv 0 0.0000 0.0000 0.000 9.999E01 1 0.5000 0.3333 187.500 7.655E01 2 0.3478 0.2581 110.068 1.014E02 3 0.3252 0.2454 100.206 4.236E06

Section 3.5 Hyperelastic Material Model

80 Goals • Understand the definition of hyperelastic material • Understand strain energy density function and how to use it to obtain stress • Understand the role of invariants in hyperelasticity • Understand how to impose incompressibility • Understand mixed formulation and perturbed Lagrangian formulation • Understand linearization process when strain energy density is written in terms of invariants

What Is Hyperelasticity? • Hyperelastic material - stress-strain relationship derives from a strain energy density function – Stress is a function of total strain (independent of history) – Depending on strain energy density, different names are used, such as Mooney-Rivlin, Ogden, Yeoh, or polynomial model • Generally comes with incompressibility (J = 1) – The volume preserves during large deformation – Mixed formulation – completely incompressible hyperelasticity – Penalty formulation - nearly incompressible hyperelasticity • Example: rubber, biological tissues – nonlinear elastic, isotropic, incompressible and generally independent of strain rate • Hypoelastic material: relation is given in terms of stress and strain rates 82 Strain Energy Density • We are interested in isotropic materials – Material frame indifference: no matter what coordinate system is chosen, the response of the material is identical – The components of a deformation tensor depends on coord. system – Three invariants of C are independent of coord. system • Invariants of C

222 Itr()CC1112233123C C No deformation I1 = 3 I(tr)tr() 1 FVCC2 2 22 22 22 I = 3 22 HX 122331 2 I3 = 1 222 Idet3123C

– In order to be material frame indifferent, material properties must be expressed using invariants

– For incompressibility, I3 = 1 83

Strain Energy Density cont. • Strain Energy Density Function

– Must be zero when C = 1, i.e., 1 = 2 = 3 = 1 mnk W(I,I,I)123B A mnk1 (I 3)(I 2 3)(I 3 1) mnk1

– For incompressible material mn W(I12 ,I )B A mn1 (I 3) (I 2 3) mn1

– Ex: Neo-Hookean model 7 W(I ) A (I 3) A 1101 10 2 – Mooney-Rivlin model W(I12 ,I ) A 101 (I 3) A 012 (I 3)

84 Strain Energy Density cont. • Strain Energy Density Function – Yeoh model 23 W(I)A(I3)A(I3)A(I3)11 101 201 301 –Ogden model Initial shear modulus N N 7 1 W( , , )B i iii 3 7B 7 1123 123 ii i1 i 2 i1

–When N = 1 and a1 = 1, Neo-Hookean material –When N = 2, 1 = 2, and 2 = 2, Mooney-Rivlin material

Example – Neo-Hookean Model • Uniaxial tension with incompressibility 1231/ • Energy density 2 W A(I3)A(222 3)A( 2 3) 10 1 10 1 2 3 10 • Nominal stress CSCS W1 7 1 P2A110 DTDT EU 22EU(1 ) 50 Linear elastic

-50

-100

Neo-Hookean Nominal stress -150

-200

-250 -0.8 -0.4 0 0.4 0.8 Nominal strain 86 Example – St. Venant Kirchhoff Material • Show that St. Venant-Kirchhoff material has the following strain energy density

2 W(EEE )7HXFV tr( ) tr(2 ) 2 W(EEE ) tr( ) tr(2 ) SE 7tr( ) EEE •First term tr(E ) tr(E1E ) : 1 E tr(E ) !tr(E11E11E ) ( : ) ( ) : E • Second term EEij ji EE EE 2E ik jl ji ij jk il lk lk lk Ekl 87

Example – St. Venant Kirchhoff Material cont. • Therefore

tr(EE ) tr(2 ) SE tr( ) 7 EE ():211E ! 7 E HXFV()2:11 ! 7 I E

88 Nearly Incompressible Hyperelasticity • Incompressible material – Cannot calculate stress from strain. Why? • Nearly incompressible material – Many material show nearly incompressible behavior – We can use the bulk modulus to model it

•Using I1 and I2 enough for incompressibility?

–No, I1 and I2 actually vary under hydrostatic deformation

– We will use reduced invariants: J1, J2, and J3

1/3 2/3 1/2 JIIJII113223 JJI 3 3

• Will J1 and J2 be constant under dilatation?

Locking • What is locking – Elements do not want to deform even if forces are applied – Locking is one of the most common modes of failure in NL analysis – It is very difficult to find and solutions show strange behaviors • Types of locking – Shear locking: shell or beam elements under transverse loading – Volumetric locking: large elastic and plastic deformation • Why does locking occur? – Incompressible sphere under hydrostatic pressure

No unique pressure for given displ. p sphere Pressure

Volumetric strain 90 How to solve locking problems? • Mixed formulation (incompressibility) – Can’t interpolate pressure from displacements – Pressure should be considered as an independent variable – Becomes the Lagrange multiplier method – The stiffness matrix becomes positive semi-definite

Displacement

Pressure

4x1 formulation

Penalty Method

• Instead of incompressibility, the material is assumed to be nearly incompressible • This is closer to actual observation • Use a large bulk modulus (penalty parameter) so that a small volume change causes a large pressure change • Large penalty term makes the stiffness matrix ill-conditioned • Ill-conditioned matrix often yields excessive deformation • Temporarily reduce the penalty term in the stiffness calculation • Stress calculation use the penalty term as it is

Unique pressure FV1 for given displ. GW 7 GW10

Pressure [K] GW1 GW HX Volumetric strain 1 92 Example – Hydrostatic Tension (Dilatation)

I xX FV 00 FV 2 00 L 11 GW GW 2 JxX22 F GW00 C GW00 L GW GW 2 KxX33 HX00 HX00

• Invariants 246 I3123 I3 I I1 and I2 are not constant • Reduced invariants

1/3 JII3113 J1 and J2 are constant 2/3 JII3223 1/2 3 JI33

Strain Energy Density • Using reduced invariants W(J,J,J)123 W(J,J) D12 W(J) H3

–WD(J1, J2): Distortional strain energy density

–WH(J3): Dilatational strain energy density • The second terms is related to nearly incompressible behavior K W(J) (J 1)2 H32 3 – K: bulk modulus 2 7 for linear elastic material 3 1 Abaqus: W(J) (J 1)2 H32D 3

94 Mooney-Rivlin Material • Most popular model – (not because accuracy, but because convenience) W(J,J,J)123 W(J,J) D12 W(J) H3 K A (J 3) A (J 3) (J 1)2 10 1 01 22 3

–Initial shear modulus ~ 2(A10 + A01)

– Initial Young’s modulus ~ 6(A10 + A01) (3D) or 8(A10 + A01) (2D) –Bulk modulus = K • Hydrostatic pressure W W pK(J1)H 3 JJ33 – Numerical instability for large K (volumetric locking) – Penalty method with K as a penalty parameter 95

Mooney-Rivlin Material cont. • Second P-K stress

WWJJ W WJ S 12 3 a EEEEJJ123 J a ,E E S AJ10 1,EE AJ 01 2, K(J 3 1)J 3, E

– Use chain rule of differentiation 1/3 JII113 1/31 4/3 2/3 J(I)II(I)I JII223 1,EE 3 1,3 1 3 3, E 1/2 J(I)II(I)I2/32 5/3 JI33 2,EE 3 2,3 2 3 3, E J(I)I 1 1/2 3,EE2 3 3,

I21,E 1 I21,E 1 I4(1tr)42,E E1 E I2(I)2,E 11C I(24tr)4[eeEE]E1 E 9 1 3,E 4 imn jrs mr ns I2I3,E 3C 96 Example

1 •Show I1,EE 2,I11CC 2, 2(I 1 ),I2I 3, E 3 I tr(CCCCCC ), I11 tr( ), I tr( ) •Let 1223 3 •Then I I,I11 I23 I,II I II 11 226 1 2 33 1 12 • Derivatives

III 12 ,C,CC 3 ij ji jk ki CCCij ij ij

II I 12 ,IC,IC 3 1 ij 1 ij ji 3 ji CCij ij C ij and 2 CE

Mixed Formulation • Using bulk modulus often causes instability – Selectively reduced integration (Full integration for deviatoric part, reduced integration for dilatation part) • Mixed formulation: Independent treatment of pressure W(J,p)H3 p(J 3 1) – Pressure p is additional unknown (pure incompressible material) – Advantage: No numerical instability – Disadvantage: system matrix is not positive definite • Perturbed Lagrangian formulation 1 W(J,p) p(J 1) p2 H3 3 2K

– Second term make the material nearly incompressible and the system matrix positive definite 98 Variational Equation (Perturbed Lagrangian) • Stress calculation 1 W(J,J,J) A(J 3) A(J 3) p(J 1) p2 123 101 012 3 2K S AJ10 1,EEE AJ 01 2, pJ 3, • Variation of strain energy density

WW,,pEE Wp p SE:(J1)p 3 K • Introduce a vector of unknowns: ru (,p) FV a(rr , )OO HX S : E pH d 0 p HJ 1 Volumetric strain 3 K

Example – Simple Shear • Calculate 2nd P-K stress for the simple shear deformation

– material properties (A10, A01, K) X2, x2 FV110 FV 110 GWT GW FCFFGW010 GW 120 HXGW001 HXGW 001 45o

X1, x1 I4,I4,I1123

I21,E 1 FV620 GW I2(I)2402,E 11C GW HXGW006 FV420 GW 1 I2I2203,E 3C GW HXGW002

100 Example – Simple Shear cont.

FV54 0 JII1/3 4 113 42 GW JI1,EE 1, I 3, EGW 410 33GW HX00 1 JII2/3 4 223 FV750 GW 8 2 JI2,EE 2, I 3, EGW 520 JI1/2 1 3 3 33 HXGW001

S AJ10 1,EE AJ 01 2, K(J 3 1)J 3, E

FV 5A10 7A 01 4A 10 5A 01 0 2 GW GW4A10 5A 01 A 10 2A 01 0 3 GW HX00AA10 01

Note: S11, S22 and S33 are not zero

101

Stress Calculation Algorithm

T •Given: {E} = {E11, E22, E33, E12, E23, E13} , {p}, (A10, A01) {}1CE1 {11 1 0 0 0}T { } 2{} {} ICCC1123 ICCCCCCCCCCCC2 121323445566 I(CCCC)C(CCCC)C(CCCC)C3 12 443 46 155 45 266 {I}1,E 2{1110} {I2,233112456E } 2{C C C C C C C C C } 222 {I3,E } 2{C 2 C 3 C 5 C 3 C 1 C 6 C 1 C 2 C 4 CC56 CC 34 CC 64 CC 15 CC 45 CC} 26

{J } I1/3 {I }1 I I 4/3 {I } 1,EE 3 1,3 1 3 3, E {J } I2/3 {I }2 I I 5/3 {I } 2,EE 3 2,3 2 3 3, E {J } 1 I1/2 {I }, 3,EE2 3 3, For penalty method, use {}S A{J}10 1,EEE A{J} 01 2, p{J} 3, K(J3 – 1) instead of p 102 Linearization (Penalty Method) • Stress increment 3 3 3 SEDEW:,,EE : • Material stiffness

S D AJ AJ K(J 1)J KJ ! J E 10 1,EE 01 2, EE 3 3, EE 3, E 3, E • Linearized energy form

* 3FV 33 a(;uuu , )OO HX ED : : ES : E d 0

103

Linearization cont. • Second-order derivatives of reduced invariants

1 141474 J I I33 I (I !!! I I I ) II 3 I I II 3 I 1,EE 1, EE33393 1, E 3, E 3, E 1, E 1 3 3, E 3, E 1 3 3, EE 2 2102585 J I I33 I (I !! I I I ) II 3 I ! I II 3 I 2,EE 2, EE33393 2, E 3, E 3, E 2, E 2 3 3, E 3, E 2 3 3, EE 113 1 JIIIII22 ! 3,EE4233 3, E 3, E 3, EE

I1,EE 0 ! I42,EE 11I !11 11 I4I3,EE 3CC I 3 CIC

104 MATLAB Function Mooney • Calculates S and D for a given deformation gradient

% % 2nd PK stress and material stiffness for Mooney-Rivlin material % function [Stress D] = Mooney(F, A10, A01, K, ltan) % Inputs: % F = Deformation gradient [3x3] % A10, A01, K = Material constants % ltan = 0 Calculate stress alone; % 1 Calculate stress and material stiffness % Outputs: % Stress = 2nd PK stress [S11, S22, S33, S12, S23, S13]; % D = Material stiffness [6x6] %

105

Summary • Hyperelastic material: strain energy density exists with incompressible constraint • In order to be material frame indifferent, material properties must be expressed using invariants • Numerical instability (volumetric locking) can occur when large bulk modulus is used for incompressibility • Mixed formulation is used for purely incompressibility (additional pressure variable, non-PD tangent stiffness) • Perturbed Lagrangian formulation for nearly incompressibility (reduced integration for pressure term)

106 Section 3.6 Finite Element Formulation for Nonlinear Elasticity

107

Voigt Notation • We will use the Voigt notation because the tensor notation is not convenient for implementation –2nd-order tensor V vector –4th-order tensor V matrix • Stress and strain vectors (Voigt notation)

T {}S {SS11 22 S} 12

T {}E {E11 E 22 2E} 12

– Since stress and strain are symmetric, we don’t need 21 component

108 4-Node Quadrilateral Element in TL • We will use plane-strain, 4-node quadrilateral element to discuss implementation of nonlinear elastic FEA • We will use TL formulation • UL formulation will be discussed in Chapter 4

X 2 4 t 3 (–1,1) (1,1)

1 2 X1 (–1,–1) (1,–1) FiniteElementat ReferenceElement undeformeddomain

109

Interpolation and Isoparametric Mapping • Displacement interpolation

Nodal displacement vector (uI, vI) Ne usuB N()II I1 Interpolation function • Isoparametric mapping – The same interpolation function is used for geometry mapping

Nodal coordinate (XI, YI) Ne XsXB N()II I1

N1 (1s)(1t) 1 4 N(1s)(1t)1 Interpolation (shape) function 2 4 • Same for all elements N(1s)(1t)1 3 4 • Mapping depends of geometry N(1s)(1t)1 4 4 110 Displacement and Deformation Gradients • Displacement gradient

N N u e N()s e B I u B I uN()ui,j I,js Ii XXI1 I1

. T 0u {uuuu} 1,1 1,2 2,1 2,2

N()s – How to calculate I ? X

• Deformation gradient

TT {}F {FF11 12 F 21 F} 22 {1u 1,1 u 1,2 u 2,1 1 u 2,2 }

– Both displacement and deformation gradients are not symmetric

111

Green-Lagrange Strain • Green-Lagrange strain

IYE IYu(uuuu)1 LL11 LL1,12 1,1 1,1 2,1 2,1 1 {}E JZJ E22 u 2,2 (uu 1,2 2,1 u 2,2 u 2,2 ) Z LLL2 L 2E K[12 K[uuuuuu1,2 2,1 1,2 1,1 2,1 2,2

– Due to nonlinearity, {}EBd []{} – For St. Venant-Kirchhoff material, {}SDE []{}

FV 20 7 GW []D 7GW 2 0 HXGW007

112 Variation of G-R Strain

• Although E(u) is nonlinear, Euu (, ) is linear

.T Euu(, ) sym(0 u F ) {}EBd [N ]{}

FV GWFN11 1,1 FN 21 1,1 FN 11 2,1 FN 21 2,1 FN 11 4,1 FN 21 4,1 GW GW [B ] GW FN FN FN FN FN FN NGW 12 1,2 22 1,2 12 2,2 22 2,2 12 4,2 22 4,2 GW FN FN FN FN FN FN G 111,2211,2112,2212,2114,2214,2 W G W HXFN12 1,1 FN 22 1,1 FN 12 2,1 FN 22 2,1 FN 12 4,1 FN 22 4,1

Function of u Different from linear strain-displacement matrix

113

Variational Equation • Energy form a(uu , )OO S : E d 0 TT {}dBSOO [N ]{}d 0 {}{dFTint } • Load form

:Tb T ()uufutOO: d O S d 00

Ne :B usfstTb;

•Residual

Tint9+ Text {}{dFd ()}{}{ dF }, {} d h 114 Linearization – Tangent Stiffness

3 3 • Incremental strain {}[]{}EBdN • Linearization 3TTFV 3 OOED:: E d {} dGW OO [][][]d{} BNN DB d 00HX

3TTFVB 3 OOSE:d {} dGW OO [][][]d{} BGG B d 00HX

FVSS 0 0 GW11 12 SS 0 0 []B GW12 22 GW00SS GW11 12 HX00SS12 22

FVN0N0N0N0 GW1,1 2,1 3,1 4,1 N0N0N0N0 GW1,2 2,2 3,2 4,2 []BG GW GW0N1,1 0N 2,1 0N 3,1 0N 4,1 GW HX0N2,1 0N 2,2 0N 3,2 0N 4,2 115

Linearization – Tangent Stiffness • Tangent stiffness

FVTTB []KBDBBBTNNGG0OO HX [][][][][][]d 0

• Discrete incremental equation (N-R iteration)

TTextint3 9 + {}[dKTh ]{} d {}{ dF F }, {} d

–[KT] changes according to stress and strain – Solved iteratively until the residual term vanishes

116 Summary • For elastic material, the variational equation can be obtained from the principle of minimum potential energy • St. Venant-Kirchhoff material has linear relationship between 2nd P-K stress and G-L strain • In TL, nonlinearity comes from nonlinear strain- displacement relation • In UL, nonlinearity comes from constitutive relation and unknown current domain (Jacobian of deformation gradient) • TL and UL are mathematically equivalent, but have different reference frames • TL and UL have different interpretation of constitutive relation.

117

Section 3.7 MATLAB Code for Hyperelastic Material Model

118 HYPER3D.m • Building the tangent stiffness matrix, [K], and the residual force vector, {R}, for hyperelastic material • Input variables for HYPER3D.m

Variable Array size Meaning MID Integer Material Identification No. (3) (Not used) PROP (3,1) Material properties (A10, A01, K) UPDATE Logical variable If true, save stress values LTAN Logical variable If true, calculate the global stiffness matrix NE Integer Total number of elements NDOF Integer Dimension of problem (3) XYZ (3,NNODE) Coordinates of all nodes LE (8,NE) Element connectivity

119

function HYPER3D(MID, PROP, UPDATE, LTAN, NE, NDOF, XYZ, LE) %*********************************************************************** % MAIN PROGRAM COMPUTING GLOBAL STIFFNESS MATRIX AND RESIDUAL FORCE FOR % HYPERELASTIC MATERIAL MODELS %*********************************************************************** %% global DISPTD FORCE GKF SIGMA % % Integration points and weights XG=[-0.57735026918963D0, 0.57735026918963D0]; WGT=[1.00000000000000D0, 1.00000000000000D0]; % % Index for history variables (each integration pt) INTN=0; % %LOOP OVER ELEMENTS, THIS IS MAIN LOOP TO COMPUTE K AND F for IE=1:NE % Nodal coordinates and incremental displacements ELXY=XYZ(LE(IE,:),:); % Local to global mapping IDOF=zeros(1,24); for I=1:8 II=(I-1)*NDOF+1; IDOF(II:II+2)=(LE(IE,I)-1)*NDOF+1:(LE(IE,I)-1)*NDOF+3; end DSP=DISPTD(IDOF); DSP=reshape(DSP,NDOF,8); % %LOOP OVER INTEGRATION POINTS for LX=1:2, for LY=1:2, for LZ=1:2 E1=XG(LX); E2=XG(LY); E3=XG(LZ); INTN = INTN + 1; % % Determinant and shape function derivatives [~, SHPD, DET] = SHAPEL([E1 E2 E3], ELXY); FAC=WGT(LX)*WGT(LY)*WGT(LZ)*DET; 120 % Deformation gradient F=DSP*SHPD' + eye(3); % % Computer stress and tangent stiffness [STRESS DTAN] = Mooney(F, PROP(1), PROP(2), PROP(3), LTAN); % % Store stress into the global array if UPDATE SIGMA(:,INTN)=STRESS; continue; end % % Add residual force and tangent stiffness matrix BM=zeros(6,24); BG=zeros(9,24); for I=1:8 COL=(I-1)*3+1:(I-1)*3+3; BM(:,COL)=[SHPD(1,I)*F(1,1) SHPD(1,I)*F(2,1) SHPD(1,I)*F(3,1); SHPD(2,I)*F(1,2) SHPD(2,I)*F(2,2) SHPD(2,I)*F(3,2); SHPD(3,I)*F(1,3) SHPD(3,I)*F(2,3) SHPD(3,I)*F(3,3); SHPD(1,I)*F(1,2)+SHPD(2,I)*F(1,1) SHPD(1,I)*F(2,2)+SHPD(2,I)*F(2,1) SHPD(1,I)*F(3,2)+SHPD(2,I)*F(3,1); SHPD(2,I)*F(1,3)+SHPD(3,I)*F(1,2) SHPD(2,I)*F(2,3)+SHPD(3,I)*F(2,2) SHPD(2,I)*F(3,3)+SHPD(3,I)*F(3,2); SHPD(1,I)*F(1,3)+SHPD(3,I)*F(1,1) SHPD(1,I)*F(2,3)+SHPD(3,I)*F(2,1) SHPD(1,I)*F(3,3)+SHPD(3,I)*F(3,1)]; % BG(:,COL)=[SHPD(1,I) 0 0; SHPD(2,I) 0 0; SHPD(3,I) 0 0; 0 SHPD(1,I) 0; 0 SHPD(2,I) 0; 0 SHPD(3,I) 0; 0 0 SHPD(1,I); 0 0 SHPD(2,I); 0 0 SHPD(3,I)]; end 121

% % Residual forces FORCE(IDOF) = FORCE(IDOF) - FAC*BM'*STRESS; % % Tangent stiffness if LTAN SIG=[STRESS(1) STRESS(4) STRESS(6); STRESS(4) STRESS(2) STRESS(5); STRESS(6) STRESS(5) STRESS(3)]; SHEAD=zeros(9); SHEAD(1:3,1:3)=SIG; SHEAD(4:6,4:6)=SIG; SHEAD(7:9,7:9)=SIG; % EKF = BM'*DTAN*BM + BG'*SHEAD*BG; GKF(IDOF,IDOF)=GKF(IDOF,IDOF)+FAC*EKF; end end; end; end; end end

122 Example Extension of a Unit Cube • Face 4 is extended with a stretch ratio = 6.0

•BC: u1 = 0 at Face 6, u2 = 0 at Face 3, and u3 = 0 at Face 1 • Mooney-Rivlin: A = 80MPa, A = 20MPa, and K = 107 10 01 X % Nodal coordinates 2 XYZ=[0 0 0;1 0 0;1 1 0;0 1 0;0 0 1;1 0 1;1 1 1;0 1 1]; Face 1 % 4 3 % Element connectivity LE=[1 2 3 4 5 6 7 8]; 8 % 7 Face 4 % No external force EXTFORCE=[]; Face 6 2 % X1 % Prescribed displacements [Node, DOF, Value] 1 SDISPT=[1 1 0;4 1 0;5 1 0;8 1 0; % u1=0 for Face 6 5 6 1 2 0;2 2 0;5 2 0;6 2 0; % u2=0 for Face 3 Face 3 1 3 0;2 3 0;3 3 0;4 3 0; % u3=0 for Face 1 X3 2 1 5;3 1 5;6 1 5;7 1 5]; % u1=5 for Face 4 % % Load increments [Start End Increment InitialFactor FinalFactor] TIMS=[0.0 1.0 0.05 0.0 1.0]'; % % Material properties MID=-1; PROP=[80 20 1E7]; 123

Example Extension of a Unit Cube

Time Time step Iter Residual 0.05000 5.000e-02 2 1.17493e+05 Not converged. Bisecting load increment 2

Time Time step Iter Residual 0.02500 2.500e-02 2 2.96114e+04

3 2.55611e+02 6000 4 1.84747e-02

5 1.51867e-10 5000

Time Time step Iter Residual 4000 0.05000 2.500e-02 2 2.48106e+04 Stress 3 1.69171e+02 3000 4 7.67766e-03

5 2.39898e-10 2000

Time Time step Iter Residual 1000 0.10000 5.000e-02 2 8.45251e+04 3 1.88898e+03 0 4 8.72537e-01 1 2 3 4 5 6 Stretch ratio 5 1.86783e-07 ... Time Time step Iter Residual 1.00000 5.000e-02 2 8.55549e+03 3 8.98726e+00 4 9.88176e-06 5 1.66042e-09 124 Hyperelastic Material Analysis Using ABAQUS • *ELEMENT,TYPE=C3D8RH,ELSET=ONE – 8-node linear brick, reduced integration with hourglass control, hybrid with constant pressure • *MATERIAL,NAME=MOONEY *HYPERELASTIC, MOONEY-RIVLIN 80., 20.,

– Mooney-Rivlin material with A10 = 80 and A01 = 20 • *STATIC,DIRECT – Fixed time step (no automatic time step control) y

z 125

Hyperelastic Material Analysis Using ABAQUS

*HEADING *MATERIAL,NAME=MOONEY - Incompressible hyperelasticity (Mooney- *HYPERELASTIC, MOONEY-RIVLIN Rivlin) Uniaxial tension 80., 20., *NODE,NSET=ALL *STEP,NLGEOM,INC=20 1, UNIAXIAL TENSION 2,1. *STATIC,DIRECT 3,1.,1., 1.,20. 4,0.,1., *BOUNDARY,OP=NEW 5,0.,0.,1. FACE1,3 6,1.,0.,1. FACE3,2 7,1.,1.,1. FACE6,1 8,0.,1.,1. FACE4,1,1,5. *NSET,NSET=FACE1 *EL PRINT,F=1 1,2,3,4 S, *NSET,NSET=FACE3 E, 1,2,5,6 *NODE PRINT,F=1 *NSET,NSET=FACE4 U,RF 2,3,6,7 *OUTPUT,FIELD,FREQ=1 *NSET,NSET=FACE6 *ELEMENT OUTPUT 4,1,8,5 S,E *ELEMENT,TYPE=C3D8RH,ELSET=ONE *OUTPUT,FIELD,FREQ=1 1,1,2,3,4,5,6,7,8 *NODE OUTPUT *SOLID SECTION, ELSET=ONE, U,RF MATERIAL= MOONEY *END STEP 126 Hyperelastic Material Analysis Using ABAQUS • Analytical solution procedure – Gradually increase the principal stretch from 1 to 6 – Deformation gradient FV 00 GW F GW01/ 0 GW HX001/

–Calculate J1,E and J2,E –Calculate 2nd P-K stress S AJ10 1,EE AJ 01 2,

– Calculate Cauchy stress 1 5 44FSFT J – Remove the hydrostatic component of stress 555 11 11 22 127

Hyperelastic Material Analysis Using ABAQUS • Comparison with analytical stress vs. numerical stress

128 Section 3.9 Fitting Hyperelastic Material Parameters from Test Data

129

Elastomer Test Procedures • Elastomer tests – simple tension, simple compression, equi-biaxial tension, simple shear, pure shear, and volumetric compression

70 uni-axial bi-axial 60 pure shear

30 Nominal stress 20

0 0 1 2 3 4 5 6 7 Nominal strain 130 Elastomer Tests • Data type: Nominal stress vs. principal stretch

F L F L

Simple tension test Pure shear test F F

F L

Volumetric compression test Equal biaxial test 131

Data Preparation • Need enough number of independent experimental data – No rank deficiency for curve fitting algorithm • All tests measure principal stress and principle stretch

Experiment Type Stretch Stress

E Uniaxial tension Stretch ratio '= L/L0 Nominal stress T = F/A0

E Equi-biaxial Stretch ratio = L/L0 in y- Nominal stress T = F/A0 tension direction in y-direction

E Pure shear test Stretch ratio '= L/L0 Nominal stress T = F/A0

E Volumetric test Compression ratio '= L/L0 Pressure T = F/A0

132 Data Preparation cont. • Uni-axial test 123,1/

U T 2(13 )(A A ) 10 01 IY T23 FV A10 T(A10 ,A 01 , ) {xb } { }HX 2( ) 2(1 ) JZ K[A01

2 • Equi-biaxial test 12,1/ 3 1U T2()(AA) 52 2 10 01 • Pure shear test 123,1,1/

U T2()(AA) 3 10 01

133

Data Preparation cont. • Data Preparation

Type111 44 4 123 i i1 NDT EEEE EE E TTTT123 TT i i1 T NDT

• For Mooney-Rivlin material model, nominal stress is a

linear function of material parameters (A10, A01)

134 Curve Fitting for Mooney-Rivlin Material

• Need to determine A10 and A01 by minimizing error between test data and model NDT 2 E minimizeB Tk1001k T(A ,A , ) A,A10 01 k1

• For Mooney-Rivlin, T(A10, A01, k) is linear function – Least-squares can be used IYA {}b JZ10 K[A01 IYTE IYT FVx() T LL1 LL1 GW1 LLTE LLT GWx() T {}TE JZ2 {TbXb }JZ2 1 {} []{} GW LL LLGW LLE LLK[T GW T K[TNDT NDT HXx()NDT

135

Curve Fitting cont. • Minimize error(square)

{}{}eeTETE { T TT }{ T } {}{}TXbTXbETE {}{}2{}[]{}{}[][]{}TTET E bXTbXXb T T E T T

• Minimization b Linear regression equation

[][]{}XXbXTTTE []{ }

136 Stability of Constitutive Model • Stable material: the slope in the stress-strain curve is always positive (Drucker stability) • Stability requirement (Mooney-Rivlin material)

d: D :d 0

• Stability check is normally performed at several specified deformations (principal directions) 55 dd11 dd 22 0 FVIYDD d ;