piaras Part IV

3514 Non-Linear Elasticity Non-linear elastic materials, in particular elas tic materials subjected to large strains, are examined in this chapter. Mo st of the chapter is concerne d with hyperelastic materials, that is, elastic materials for which the stress can be derived from an elastic strain-energy potential function. Some of the different types of elastic materi al models are reviewed in §4.1. Hyperelastic materials are studied in the sections which follow. General constitutive relations are derived for both the material and spatial descriptions and for both compressible and incompressible materials. It is shown how the principle of material frame-indifference restricts the form a constitutive equation may take. The more commonly used isotropic material models are discussed in §4.4 and material anisotropy is discussed in §4.5. A de-coupled formulation of hyperelasticity is presented, which is useful for materials which are nearly incompressible. Finally, some of the equations for isotropic materials are linearised, showing how they are consistent with the linear elastic model. 352 Section 4.1 Solid Mechanics Part III Kelly 3534.1 Elastic Solids In this section is give n an overview of the comm on elasticity models. 4.1.1 The Linear Elastic Solid The classical Linear Elastic model , or Hookean model , has the following linear relationship between stress and strain: mn ijmn ijCεσ= ⋅= ,:εσ C (4.1.1) where ε is the small strain tensor, §2.7. Strain Energy In this purely mechanical theory of elastic materials, there is no di ssipation of energy – all the energy of the loads is stored as elastic stra in energy in the material as it deforms, and can be recovered. For the linear elastic material, the rate of de formation is equivalent to the rate of small- strain, εd&=, so the strain-energy function can be written as εσd dW := (4.1.2) and the total energy stored per unit volume over the complete history of straining is ∫=εσd W : (4.1.3) nnd the stress can be written as εσ∂∂=W (4.1.4) Reduction in the number of I ndependent Elastic Constants Since the stress and strain are symmetric, ji ijσσ= and nm mnεε= , the fourth-order elasticity tensor of stiffness coefficien ts contains the minor symmetries 1.9.65, ijlk jikl ijkl C C C == (4.1.5) and so the 81 coefficients reduce to 36 independent coefficien ts. Further, since C is independent of the strains, e eεεεσ ::21: : C C=⋅==∫∫d d W (4.1.6) Section 4.1 Solid Mechanics Part III Kelly 354and so ee∂∂∂=WC (4.1.7) Now ij mn mn ijW W εεεε ∂∂∂=∂∂∂ (4.1.8) and it follows that C possesses the major symmetries : mnij ijmn C C= (4.1.9) This reduces the number of independent elastic constants from 36 to 21. Problems involving Elastic Materials The six constitutive equations 4.1.1, together with the equations of motion and the 6 kinematic relations relating the strain s to the displacements, Eqn. 2.7.2, ()2/, , ij ji ij u u+=ε , gives a set of 15 equations in the 15 unknowns: the six stress components, the six strain components and the three displacement components. To maintain a linear theory, the acceleration term in the equations of motion must be linear; this is achieved by supposing the displacement gradients to be small: 22 22 ,tu dtud tuvxu tu dtduvi i i k ki i i i∂∂≈∂∂≈∂∂+∂∂== (4.1.10) When this acceleration term is included, the problem is dynamic . When the equations of equilibrium are used, the problem is static . Compatibility An alternative method of solution is to rem ove the displacements from the above system and solve only for the stresses and strains. In this case the strain-displacement relations are replaced by three compa tibility equations, and there are then 12 equations in 12 unknowns. Once the system is solved, the disp lacements can be obtained from the strains by integration. The Isotropic Linear Elastic Solid When the material is isotropic, the constitu tive equation holds in any coordinate system, mn ijmn ijCεσ ′=′ (4.1.11) Section 4.1 Solid Mechanics Part III Kelly 355and so the tensor of elastic c onstants is isotropic. The most general fourth-order isotropic tensor takes the form 1.10.7, () ( )jm in jn im mnij ijmnC δδδδμδλδ μλ + += ++⊗= ,II C II (4.1.12) with the fourth-order unit tensors given by 1.9.60, n m j i jm inn m j ijn im e e e ee e e e ⊗⊗⊗ =⊗⊗⊗ = δδδδ II (4.1.13) which has only two independent ma terial constants . Since the strain is symmetric, one has ()() () ε IIε IIεσ : 2:: ⋅+⊗=⋅++⊗=⋅= IIIC μλμλ (4.1.14) and the constitutive equation reduces to { ▲Problem 1} ()ij ij kk ij e eμδλσ μ λ 2 ,2 tr += += ε Iεσ (4.1.15) and the two elastic constants μλ, are called Lamé’s constants . Problems in Isotropic Elasticity The 15 equations mentioned earlier can be re duced by eliminating the strains from the constitutive equation and the kinematic e quation, and then substituting the resultant expression for stress into the equations of motion, giving Navier’s equations () ( ) ()22 22 2 2, div gradtubxu xxu i ki k ik ∂∂=+∂∂+∂∂∂+ =+∇+ + ρρμ μλρρμ μλ u b u u && (4.1.16) This set of three partial di fferential equations is appropr iate for problems involving displacement boundary conditions. The Lame’s constants and the Young’s modulus and Poisson’s ratio are related through () ( ) ()νμνννλμλλνμλμλμ +=−+=+=++= 12,21 1) (2,)2 3( E EE (4.1.17) The linear elastic constitutive equations in terms of the engineer ing constants reads Section 4.1 Solid Mechanics Part III Kelly 356 ⎥⎦⎤ ⎢⎣⎡+−+=++ −= ij ij kk ijij ij kk ij EE E εδενν νσσνδσνε 21 11 (4.1.18) The Bulk Modulus The tensor of elastic constants can be written in the alternative forms () ⎥⎦⎤ ⎢⎣⎡⊗−+⊗=⎥⎦⎤ ⎢⎣⎡⊗−++=⎥⎦⎤ ⎢⎣⎡⊗−+=⊗+= II IIIIIIII 31221 12122 IIII C μ κνν νννμλμ E (4.1.19) where the new constant introduced is the bulk modulus κ. This last expression then leads to the alternative form of the constitutive relations, ε Iεσ dev2)tr( μ κ+= (4.1.20) This expression shows that the stress can be decomposed into a s pherical component and a deviatoric component. For a pure volume change, 0 dev=ε , and there are no shear stresses, ()Iσσ tr= ; the bulk modulus is thus a measure of the resistance of the material to volume changes. 4.1.2 Geometrically Non-Linear Elastic Materials When the strains (displacement gradients) are not small, the behaviour of the material will inevitably be non-linear. This is due to the geometric non-linearity of the kinematic strain-displacement relations, for example using the Green-Lagrange strains and the reference configuration, () ()() ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ∂∂ ∂∂+∂∂+∂∂=+ + = jk ik ij ji ijXU XU XU XUE21Grad Grad Grad Grad21 T TU U U U E (4.1.21) The Kirchhoff Material The Kirchhoff material is an extension of the linear elastic model to the large strain range; the constitutive relation is a linear tensor relation, but non-line arities enter through ()uE : Section 4.1 Solid Mechanics Part III Kelly 357 mn ijmn ij EC S= = ,:E S C (4.1.22) where S is the PK2 stress tensor and E is the Green-Lagrange strain. Since both S and E are symmetric, the fourth-order tensor C has the minor symmetries, jimn ijmn C C= and ijnm ijmn C C= , and so has 36 independent coefficients. Following the same arguments as before, one can define a strain energy function (per unit reference volume) ij ijdES dW d dW = = ,:ES (4.1.23) and the total energy stored per unit volume over the complete history of straining is E E ::21 21C= = =∫ mn ij ijmn ij ij EEC dES W (4.1.24) and the stress can be written as ijijEWSW ∂∂=∂∂= ,ES (4.1.25) Again, the existence of the strain energy function implies that the matrix of elastic coefficients only has 21 i ndependent coefficients. When the Kirchhoff material is isotropic, the constitutive relation reduces further to ()ij ij kk ij E E S μδλ μ λ 2 ,2 tr += += E IE S (4.1.26) As mentioned in §2.7.2, the linear elastic m odel can not be used when there are large rigid body rotations, even if the displacemen t gradients are not large. The Kirchhoff model can be used in these cases. 4.1.3 Materially Non-Linear Elastic Materials An elastic material might also exhibit material non-linearity through a non-linear constitutive equation, for example the Cauc hy stress might be some non-linear function of a strain measure, or of the deformation gradient: ))((tFfσ= (4.1.27) where f is some tensor function of the deformation gradient F. This constitutive equation is called the Cauchy Elastic material model. As can be seen, the stress is dependent on the current state only, and not on the path hist ory, a requirement of elasticity. However, the stress in the case of a Cauc hy elastic material cannot in ge neral be written in terms of a strain-energy function. In other words, the work done might be path-dependent. Section 4.1 Solid Mechanics Part III Kelly 358Objectivity Requirements The notion of objectivity was introduced in §2. 8. When formulating constitutive relations for materials, one must ensure that the principle of material objectivity (or the principle of material frame indifference ) be satisfied. This principle states that A constitutive law must be independent of the location of the observer (or frame of reference that is taken) This implies that two observers, even if in relative motion with respect to each other, observe the same stress in a given body. Cons ider the Cauchy elastic material 4.1.27. The Cauchy stress is an objective tensor. Re ferring to the example of Eqns. 2.8.47-50 in §2.8.4, objectivity requires that the constitutive relation be of the form ()TRURfσ= (4.1.28) The constitutive relation can also be written in terms of other stress measures. For example, using ()T T T TR PU RUP PFσ = ==J , one has ()1-J UURf P= (4.1.29) For the PK2 stress, one has T 1−−=σFF SJ , so that ()1 1det-UUfUU S−= (4.1.30) which does not depend on the rotation. This last relation can also be written in the form () )( det2/1 2/1 22/1 2/1Cg C Cf CC S ≡ =− − (4.1.31) This is clearly objective, since S and C are unaffected by an observer transformation, S S=* and C C=*. 4.1.4 Hypoelastic Materials A hypoelastic material is one whose constitutive law relates the rate of stress to the rate of deformation d. This can be written in terms of the Cauchy stress as ),(dσσf=& . Consider a simple one-dimensional linear model, dE=σ& (4.1.32) Since, in one-dimension, the stretch is dXdx/=λ , the rate of deformation is equivalent to the spatial velocity gradient l and the rate of change of a line element dx is dxldt dxd=/)( , dividing through by dX gives λλ/&=d (see Eqn. 2.5.10), so that λλλσσλλσ lnEdE d E ===→= ∫∫& & (4.1.33) Section 4.1 Solid Mechanics Part III Kelly 359 This shows that the stress is clearly pa th-independent, depending only on the current stretch. In fact, the stress can be written as the derivative of a strain energy function according to λ σ d dW/= , where λ/E W= . In the three dimensional case, however, the ra te of deformation can not in general be written as the rate of change of some simple function, dt d/)(•=d , and so the above calculation cannot be done, implying that the hypoelastic material cannot be written in terms of a potential function, and the work done is path-dep endent. The path-dependence is, however, small when the strains are small. 4.1.5 Hyperelastic Materials A hyperelastic material (or Green elastic material) is defined to be an elastic material for which a strain-energy function W exists, a scalar function of one of the strain or deformation tensors, whose derivative with re spect to a strain component determines the corresponding stress component. From the a bove, the linear elastic model, the Kirchhoff model and the one-dimensional hypoelastic model are all examples of hyperelastic materials. The hyperelastic material is a subset of the Cauchy-elastic material. Hyperelastic material models for components und er large strains will be the subject of the following sections. 4.1.6 Problems 1. Show that () () εεIεεII 2 :tr : =+=⊗ II Section 4.2 Solid Mechanics Part III Kelly 3604.2 Hyperelasticity The Hyperelastic material is examined in this section. 4.2.1 Constitutive Equations The rate of change of internal energy W per unit reference volume is given by the stress power, which can be expressed in a number of different ways (see §3.7.6): CSESFPdσ &&& & : : : : ====J W (4.2.1) The internal energy is regarded as a function of a deformation variable. For example, The change in energy is due to the de formation which takes place, so take W to be a function of, say, the deformation gradient ) (tF, )(FW . It is assumed that in the reference configuration the stra in energy is zero, 0 )( =IW , and that it grows with deformation, 0)(≥FW1. The chain rule gives FFF& & :)( ∂∂=WW (4.2.2) From 4.2.1, FFFFP & & :)(:∂∂=W (4.2.3) Since F and F& can take on any value independent of the other, one must have FFP∂∂=)(W (4.2.4) This is a constitutive equation relating the kine matic variables to the force variables. From 4.2.1, alternative forms are: CCSEES∂∂=∂∂=)(2 ,)( W W (4.2.5) The procedure used in Eqns. 4.2.2-4 cannot be used for the stress power relation dσ:J since there is no function whose derivative is the rate of de formation. Instead, use the relation 3.5.6, T 1PFσ−=J , 4.2.4, F P∂∂= /W , to get 1 and that it tends to infinity as the material is either compressed to a point, 0 det→= F J or expanded to an infinite range, ∞→= FdetJ . Section 4.2 Solid Mechanics Part III Kelly 361jm imij FFWJWJ∂∂=∂∂=− − 1 T 1,)(σ FFFσ (4.2.6) An alternative relation can be obtained by first deriving a re lationship between the partial derivatives of the strain en ergy function with respect to the deformation gradient, ijF W∂∂/ , and with respect to the right Cauchy-Green tensor, ijC W∂∂/ . Suppose first that the strain energy is a func tion of the deformation gradient: j i ijFW We eF⊗∂∂≡∂∂ (4.2.7) The chain rule for () )(FCW W= gives j i ijmn mnFC CW We eF⊗∂∂ ∂∂=∂∂ (4.2.8) With FFCT= , im nj in mj km njki kn mjki ijkn km kn ijkm ijmnF F F FFFF FFF FCδδδδδδ += + =∂∂+∂∂=∂∂ (4.2.9) so that (and using the fact that C is symmetric), n i mnim m i nminn i mnim m i mninj i mnim nj j i mnin mj j i ij CWFCWFCWFCWFCWFCWFFW e e e ee e e ee e e e e e ⊗∂∂+⊗∂∂=⊗∂∂+⊗∂∂=⊗∂∂+⊗∂∂=⊗∂∂δ δ (4.2.10) or kjik ij CWFFW W W ∂∂=∂∂ ∂∂=∂∂2 , 2CFF (4.2.11) Now 4.2.6 can be re-written as T 12 FCFσ∂∂=− WJ (4.2.12) Similarly, Eqn. 4.2.4 cab be re-written as CF P∂∂=W2 (4.2.13) Section 4.2 Solid Mechanics Part III Kelly 362 The Strain Energy and the Right Stretch Tensor The right stretch tensor can be expressed as UUC= , where U is the right stretch, or, since U is also symmetric, UUCT= . One can see the similarity between this relation and the relation FFCT= so, using the same arguments as given above, one has for () )(UCW W= , (see 1.11.36) CUU∂∂=∂∂ W W2 (4.2.14) Since U is symmetric, UCUC CUU U ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂=∂∂ W W W W W2 2 2T T T (4.2.15) showing that U and C∂∂/W are coaxial. 4.2.2 Objectivity of the Constitutive Equations An observer transformation (see §2.8) results in QF F=* and *CC= , so C C C C ∂∂=∂∂ /)( /)(* *W W , and, so, from 4.2.12, T T T 1 *2 QQσ QFCQF σ =∂∂=− WJ (4.2.16) which is the objectivity requirement for a sp atial tensor and so this constitutive law satisfies the requirement of material frame-indifference, when W is a function of C. This evidently holds true also when W is a function of E. It does not, however, hold true in general when W is a function of F, as in the constitutive law F F P ∂∂= /)(W . However, with W a function of C, the constitutive equation 4.2.13 can be seen to be objective. The objective constitutive equations in th is section are indicated by a box around them. 4.2.3 Elasticity Tensors The total differential of the PK2 stress can be written as C CCCSCCCSS d d d d21:21:)(2 :)(C≡⎟ ⎠⎞⎜ ⎝⎛⎟ ⎠⎞⎜ ⎝⎛ ∂∂=∂∂= (4.2.17) where C is the fourth-order tensor Section 4.2 Solid Mechanics Part III Kelly 363EES CCS ∂∂=∂∂=)( )(2C (4.2.18) and is a measure of the change in stress which occurs due to a change in strain. It is called the elasticity tensor (in the material description) or the tangent modulus . Since S and E are symmetric, C possesses the minor symmetries, ijlk jikl ijkl C C C == , and so has 36 independent components. However, if hyperelastic conditions hold, so that C C E E S ∂∂=∂∂= /)( 2 /)( W W , then EEE CCC ∂∂∂=∂∂∂=)( )(42 2W WC (4.2.19) and so C possesses the major symmetries, klij ijkl C C= , and only only 21 independent components. The rate form of the above equations is EEEECCCC CCS & & & :)(:)(2)(22 2 ∂∂∂=∂∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂=W W W dtd (4.2.20) or E C S &&& : :21C C== (4.2.21) 4.2.4 A Note on the Strain Energy Function Some confusion can arise in expressions such as ijijEWSW ∂∂=∂∂= ,ES (4.2.22) As mentioned at the end of §1.11.5, one can either • use the energy function ) ,,,,,(* *33 22 21 13 12 11 EEEEEE W W= , a function of 6 independent variables, in which case K,* 21,* 21,* 1313 1212 1111EWSEWSEWS∂∂=∂∂=∂∂= (4.2.23) • use the energy function ) ,,,,,,,,(33 32 31 23 22 21 13 12 11 EEEEEEEEEW W= , a function of 9 variables, not all of them independent, in which case K K , , , 2121 1212 1111EWSEWSEWS∂∂=∂∂=∂∂= (4.2.24) Section 4.2 Solid Mechanics Part III Kelly 364 and the symmetry of W with respect to the strains, which must be assumed here, implies that the stress is also symmetric. 4.2.5 Hyperelasticity with Constraints Consider a Hyperelastic material which is subject to the scalar constraint () 0=Fϕ or 0 :=∂∂= FF&&ϕϕ (4.2.25) Without the constraint, one has FPFFF&& & : :)(=∂∂=WW (4.2.26) The constitutive equation F P∂∂= /W can be derived from 4.2.26 when F& is arbitrary. However, for the material with the constraint, the F& cannot be “cancelled” out from each side; one has the conditions 0 : ,0 : =∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂− FFFFP & &φ W (4.2.27) In general then, F FP∂∂=∂∂−φαW (4.2.28) where α is some arbitrary scalar. Th e stress is therefore given by F FP∂∂+∂∂=φαW (4.2.29) and the rate of change of internal energy is φαφα &&& & +=∂∂+∂∂WWFFFFF: :)( (4.2.30) The strain energy is )( )( F Fαφ+ W (4.2.31) The second term here is evidently zero and so does not contribute to the strain energy. However, the stress is the derivative of the strain energy with respect to a kinematic quantity, and the derivative of this last term, the second term in 4.2.29, is not be zero. The scalar α is, or determines the magnitude of, a workless reaction stress. It ensures that the constraint is satisfied; it is not set a value in the constitutive equation, rather, it is Section 4.2 Solid Mechanics Part III Kelly 365determined by considering particular problems, through equilibrium and boundary conditions. For N constraints N i i K1,0)(==Fφ , 4.2.29 generalises to ∑ =∂∂+∂∂=N ii iW 1 F FPφα (4.2.32) Incompressible Materials An incompressible material is one whose vol ume remains constant throughout a motion, and so has the following constraint: 1 det== F J or 0=J& (4.2.33) From 1.11.34, T) (det /) (det−=∂∂ AA A A , one has T/−=F FJ ddJ , which equals T−F at 1=J . Thus the PK1 stresses can be written as T2−−∂∂= FCF P pW (4.2.34) and the strain energy f unction is of the form ())1 ( )( −− F F Jp W (4.2.35) with α−=p . Using 3.5.9, PFS1−= , one also has 1 )(2−−∂∂= CCCS pW (4.2.36) Using 2.5.20 and 2.5.18b, ()() C C E C EC FEF d & & & & & : : tr tr tr1 21 1 1 1 T − − − −−= = = == J J J J JJ (4.2.37) and the stress power is Jp W pWpW && & & && & & −= ⎪ ⎩⎪ ⎨⎧ −∂∂=−∂∂= −− F F FFFFPC C CCCCS : :)(:: :)(: T1 21 21 (4.2.38) consistent with 4.2.30. Section 4.2 Solid Mechanics Part III Kelly 366The strain energy of 4.2.35 is expressed in terms of F. This can be re-expressed in terms of C. Since C C F III det det ===J , the incompressibility constraint can be expressed as () 01 III=−=C Cϕ (4.2.39) and the strain energy can be written as () )1 III(21)( − − C CC p W (4.2.40) The factor of 1/2 has be en included so that the p in 4.2.35 is the same as the p in 4.2.40, since, from 1.11.33, 1 III−=∂∂CCC (4.2.41) leading to the same expression for th e PK2 stress as befoe, Eqn.4.2.36. Similarly, in terms of the Cauchy stress, one has{ ▲Problem 1} I FCCFσ pW−∂∂=T)(2 (4.2.42) Because of its role in this equation, the scalar p is called the hydrostatic pressure . Inextensible Constraint Consider a material which is inextensible in a direction defined by a unit vector Aˆ in the reference configuration. Th e constraint for this materi al is given by Eqn. 2.2.60, ()() 01ˆ ˆ =− = AFCA Fφ (4.2.43) Then, from 4.2.11, A AFACCAFCFF ˆ ˆ2ˆ ˆ22 ⊗=∂∂=∂∂=∂∂ φφ (4.2.44) The stress is then A AFCF P ˆ ˆ 2 2 ⊗+∂∂= αW (4.2.45) Section 4.2 Solid Mechanics Part III Kelly 367Post-contracting with TF (with 1=J ) then gives AFAF FCFσ ˆ ˆ 2 2T⊗+∂∂= αW (4.2.46) For inextensibility in two directions, () () ()() 01 ˆ ˆ ,01ˆ ˆ 2 2 2 1 1 1 =− = =− = AFCA F AFCA F φ φ (4.2.47) one has the stress 2 2 2 1 1 1T ˆ ˆ ˆ ˆ 2 AF AF AF AF FCFσ ⊗+⊗+∂∂= α αW (4.2.48) 4.2.6 Problems 1. Use the equation T )(−−∂∂= FFFP pW to show that the constitutive equation for an incompressible hyperelastic material can be written in terms of the Cauchy stress as I FCCFσ pW−∂∂=T)(2 Section 4.3 Solid Mechanics Part III Kelly 3684.3 Isotropic Elastic Materials 4.3.1 Constitutive Equations for Isotropic Elastic Materials Cauchy Stress and Isotropy A material is said to be isotropic if a rotation of a particle (in the undeformed state) has no influence on the stress tensor. From §2.8.6, the condition of isotropy is then )( ) ( )( )(TFσ FQσ Fσ Fσ = →=◊ (4.3.1) where the superscript ◊ refers to deformations relative to the rotated reference configuration, Fig. 2.8.6. (This is often expressed in the equivalent form )( )( FQσ Fσ= .) A Cauchy elastic material automatically satis fies the isotropy condition when the Cauchy stress is an arbitrary te nsor-valued function of the left Cauchy-Green tensor: )(bσσ= (4.3.2) To see this, note that ) ( ))((TFFσ Fbσ= , so ()( )()()())( ) )( ( ) (T T T TT T TFbσ FFσ QF FQσ FQ FQσ FQbσ == = = (4.3.3) Note that the condition of is otropy 4.3.1 must be satisfied for the particular rotation RQ=, so a condition to be satisf ied by isotropic materials is )( ) ( )( ) ( )(2/1 Tbσ bσ vσ FRσ Fσ === = (4.3.4) where v is the left stretch tensor, and again one has 4.3.2. In addition to satisfying the condition of isot ropy, the stress must sa tisfy the condition of objectivity: ) ( * *bσσ= , where the superscript * refers to objectivity transformations, §2.8.3. This requirement is automatically satisfied since both the Cauchy stress and the left Cauchy-Green tensors are objective spatial tensors: ) ( )( T TQbQσ QbQσ= Isotropy Condition for the (objective) Cauchy Stress (4.3.5) PK2 Stress and Isotropy From 3.5.7 and 2.8.58, a rigid body rotation of the reference configur ation alters the PK2 stress according to ()()TTT1T T 1QSQ FQσ FQ FσF S = = =− −−◊◊−◊◊◊J J (4.3.6) Section 4.3 Solid Mechanics Part III Kelly 369On the other hand, S is objective when written as a function of the material tensor C. Then, since TQCQ C=◊ (see Eqns. 2.8.58), ) ( )(T TQCQS QCQS = Isotropy Condition for the (objective) PK2 Stress (4.3.7) Isotropic Tensor Functions The constitutive relations 4.3.5 for the Cauchy stress and 4.3.7 for the PK2 stress are very similar. In general, the s econd-order tensor-valued function T of the second-order tensor variable B is an isotropic tensor functions if ()()T TQBQT QBQT = Isotropic Tensor Function (4.3.8) for all orthogonal tensors Q. Thus, σ is an isotropic tensor func tion of the tensor variable b and S is an isotropic tensor f unction of the tensor variable C. Isotropic functions are discussed in the Appendix to this Chapter, §4. A, and there it is show n that, for symmetric T and symmetric B, T must take t he form ()2 2 1 0 B B I BT ααα ++= Form for a symmetric isotropic tensor function of a symmetric tensor (4.3.9) where 2 1 0 ,,ααα are scalar functions of the principal scalar invariants of B, 1.9.38, 1.9.46: { }B B B III,II,Ii iαα= (4.3.10) Since the set of principal scalar invariants, the set { }3 2tr,tr,tr S SS and the set of eigenvalues {}3 2 1,,λλλ uniquely determine one another, the coefficients iα can be taken to be functions of any one of these three sets. Equation 4.3.9 can be rewritten in various alternative forms using the Cayley-Hamilton theorem, 1.9.45: 0I B B BB B B =−+− III II I2 3, (4.3.11) for example ()1 1 1 0− −++= B B I BT βββ (4.3.12) where 2 1 2 1 1 2 0 0 III , I , II αβααβααβB B B = += −=− . (4.3.13) Section 4.3 Solid Mechanics Part III Kelly 370The Cauchy-Elastic Solid Since for an isotropic Cauchy-elastic solid, the Cauchy stress is an isotropic tensor function of the left Cauchy-Green strain, ) (bσσ= , Eqn. 4.3.5 (a consequence of isotropy and objectivity) and since σ and b are symmetric, it follows from 4.3.9 that the Cauchy stress takes the form () ( )( )2 2 1 0 III,II,I III,II,I III,II,I b b I σb b b b b b b b b α α α + + = (4.3.14) or, alternatively, from 4.3.12, the form () ( )( )1 1 1 0 III,II,I III,II,I III,II,I− −+ + = b b I σb b b b b b b b b β β β (4.3.15) and these scalar functions are related through 4.3.13. Similar forms hold for the PK2 stress as a f unction of the right Cauchy-Green strain. 4.3.2 Strain Energy and Isotropy Consider the strain energy )(FW W= . From §2.8.4, objectivity requires that (see Eqn. 2.8.46), )( )( U F W W= (4.3.16) Also, the right stretch tensor is the square–r oot of the right Cauchy-Green tensor, so one can write )(CW W= , which is clearly objective, since ) ( )(*C C W W= . In addition to the objectivity requirement, is otropy requires that ) ( )(◊= C C W W , or, 2.8.58b ) ( )(TQCQ C W W= Isotropy Condition for the (objective) Strain Energy (4.3.17) Isotropic Scalar Functions The strain energy of Eqn. 4.3.17 is also an isotropic function; in general, the scalar function φ of the second-order tensor variable B is an isotropic scalar functions if ()()B QBQφ φ =T Isotropic Scalar Function (4.3.18) for all orthogonal tensors Q. Thus, W is an isotropic scalar func tion of the tensor variable C. It is shown in the Appendix to th is Chapter, §4.A, that, for symmetric B, φ must take the form (){ }()B B B B III,II,Iφφ= Form for an isotropic scalar function of a symmetric tensor (4.3.19) Section 4.3 Solid Mechanics Part III Kelly 371Thus the strain energy for an isotropic elastic material must be a function only of the three principal scalar invariants of the right Cauchy-Green strain (or of the three principal values iλ). Since the invariants for the right- and left-Cauchy-Green strain tensors are the same (see Eqn. 2.2.15), the strain energy can also be expressed as a function of the invariants of b. Section 4.4 Solid Mechanics Part III Kelly 3724.4 Isotropic Hyperelasticity Attention is now restricted to the cas e of isotropic hyperelastic materials. 4.4.1 Constitutive Equations in Material Form Consider the general hyperelastic constitutive law 4.2.5: CCS∂∂=)(2W (4.4.1) From §4.3.2, the strain energy for an isotropic hyperelastic material must be a function of the principal invariants of C: ( ))(III),(II),(I C C CC C CW W= (4.4.2) Then, using the relations 1.11.33 for the derivati ves of the invariants (taking into account that C is symmetric), 1IIIIII IIIII IIII IIIII III I − ∂∂+∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂ ∂∂+∂∂ ∂∂+∂∂ ∂∂=∂∂ C C IC C C C C C CC C CC CC CC C W W W WW W W W (4.4.3) Further, the 1−C term can be replaced by a 2C term using the Cayley-Hamilton theorem. In summary then, expressing S in the form 4.3.9 (and 4.3.12), C C CCC C CC C CC CC C CC C IC C I CCS IIIIII IIIIIIIII IIIII IIIIIIIII I2 2 22 2 2 )(2 1 21 10 01 1 1 02 2 1 0 ∂∂=∂∂=∂∂−=∂∂−∂∂−=∂∂+∂∂=∂∂+∂∂+∂∂=⎪⎩⎪⎨⎧ ++++ =∂∂= −− − W WW W WW W W W WW β αβ αβ αβββααα Constitutive Equation in material description (4.4.4) Section 4.4 Solid Mechanics Part III Kelly 3734.4.2 Constitutive Equations in Spatial Form As mentioned at the end of §4.3.2, the invariants of C and b are the same and the strain energy can be expressed in terms of the invariants of b, ( ))(III),(II),(I b b bb b bW W= . Then ⎪⎩⎪⎨⎧ ++++ =∂∂ − −1 1 1 02 2 1 0)( b b Ib b I bb βββααα W (4.4.5) and the coefficients i iβα, are the same as in 4.4.4 Taking the expression for C∂∂/W in 4.4.4, pre-contracting with F, post-contracting with TF, using the definitions T T, FFbFFC = = , and (pre- or post-) contracting the above similar expression for b∂∂/W in terms of the invariants of b with b, one finds that TFCFbbbb ∂∂=∂∂=∂∂ W W W (4.4.6) Therefore, using T 1−−=σFF SJ , the constitutive equation in the material formulation, C C S ∂∂= /)( 2W , can be transformed into a spatial constitutive equation: bbb bbbσ∂∂=∂∂=− − )(2)(21 1 WJWJ (4.4.7) It should be emphasised that this expression, unlike the relation C C S ∂∂= /)( 2W , holds only for isotropic materials. Differentiating the strain energy with respect to b (exactly as in Eqn. 4.4.3) and pre- or post-contracting the result with b 12−J then leads to ⎥⎦⎤ ⎢⎣⎡ ∂∂−= ⎥⎦⎤ ⎢⎣⎡ ∂∂−=⎥⎦⎤ ⎢⎣⎡ ∂∂= ⎥⎦⎤ ⎢⎣⎡ ∂∂+∂∂=⎥⎦⎤ ⎢⎣⎡ ∂∂+∂∂= ⎥⎦⎤ ⎢⎣⎡ ∂∂=⎪⎩⎪⎨⎧ ++++ =∂∂= − −−− −− −− −− b b bbb b bb bb bb bb b Ib b I bbbσ IIIII2II2I2 III I2IIIIIIIIII2 IIIIII2)(2 1 11 21 11 11 01 01 1 1 02 2 1 0 1 WJWJWJW WJW WJWJWJ β αβ αβ αβββααα Constitutive Equation in spatial description (4.4.8) Section 4.4 Solid Mechanics Part III Kelly 374Principal Directions of the Cauchy Stress and Left Cauchy-Green Tensor Recall that the eigenvalues of the right stretch tensor U are the principal stretches iλ, and that the eigenvalues of th e right Cauchy-Green tensor C and the left Cauchy-Green tensor b are the squares of the principal stretches, 2 iλ (the eigenvectors of b are those of C rotated with the rotation tensor R). Taking nˆ to be an eigenvector of b and 2λ the corresponding eigenvalue, and us ing the constitutive equation 2 2 1 0 b b Iσ ααα ++= , one has ( )n nb nb nI nσ ˆ ˆ ˆ ˆ ˆ4 22 1 02 2 1 0 λαλααααα ++=++= (4.4.9) Thus n nσ ˆ ˆσ= where σ is an eigenvalue of σ, the principal stress, given by term inside the brackets, and so the eigenve ctors, or principal directi ons, of the Cauchy stress and b coincide. If one takes a deformation for which 0 23 13==b b , it follows from 4.4.8 that, for arbitrary coefficient 1−β, one also has 023 13==σσ . The tensors b and σ are coaxial, that is, σb bσ= . From this relation, one finds that one must have 1222 11 1222 11 bb b−=− σσσ (4.4.10) This relation holds for all compressible isotr opic hyperelastic materials under the stress and deformation conditions stated, regardless of the particular constitutive relation. 4.4.3 Constitutive Equations in terms of the Stretches The derivatives of the st retch with respect to b are given in §2.3.3. Introducing the principal stresses, and noting that ii i n nb ˆ ˆ2λ= (no sum over the i) it follows then that (no sum over the i in what follows) i iii i ii k k k kik ki i ii WJWJWJWJWJ nnb nn n bnbb nbbb nσ n ˆ)(ˆ 21)(2 ˆ)ˆ ˆ(21)(2ˆ)(2 ˆ)(2 ˆ ˆ 11 11 1 λλλλλλ λλλλ λλσ ∂∂=∂∂=⊗∂∂=∂∂ ∂∂=∂∂== −− −− − (4.4.11) or 331 3 221 2 111 1)(,)(,)( λλλλλλλλλ∂∂=∂∂=∂∂=− − − WJσWJσWJσ (4.4.12) Section 4.4 Solid Mechanics Part III Kelly 375Similarly, for the Piola-Kirchhoff stresses, one has i ii iiWSWPλλ λ ∂∂=∂∂=1, (4.4.13) Example (Uniaxial Stretch) Consider a specimen of isotropic hyperel astic material under a uniaxial tension 11σ, with ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 321 0 00 00 0 λλλ F (4.4.14) Since σ and b are coaxial (see 4.4.9), the princi pal stresses are, from 4.4.8, ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡−−− − 2 32 22 1 1 2 32 22 1 1 011 0 00 00 0 0 00 00 0 100010001 00000000 λλλ β λλλ β βσ (4.4.15) Subtracting the second of these e quations from the first leads to ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−−=− 2 22 11 12 22 1 111 λλββλλσ (4.4.16) Subtracting the third from the second leads to ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−−=− 2 32 21 12 32 210λλββλλ (4.4.17) which implies that 3 2λλ= (the terms inside the second br acket can also be zero, but not if 0 ,01 1 ≤>−ββ , as is usually the case). The material parameters 1 1 0 ,,−βββ can be obtained from such a test by measuring 1 11,λσ and 2λ. On the other hand, if the constitutive relation, that is, the parameters, are known, then one can obtain the stretc hes by specifying the uniaxial stress. ■ 4.4.4 Incompressible Materials When the material is incompressible, then 1 III III det ====b C F J (4.4.18) Section 4.4 Solid Mechanics Part III Kelly 376 and the strain energy function is only dependent on two invariants: ()()b b C C II,I II,I W W W = = , (4.4.19) and the constitutive equations are the same as those given earlier for the compressible material, with 1 ,1 III1==−Jb . However, the derivative CIII/∂∂W is an unknown and in particular it is unknown at 1 III=C . Returning to the more general constitutive equations for an incompressible material, 4.2.36, 4.2.39, one has I FCFσ CCSC C C CpWpW−∂∂= −∂∂=− T 1 )II,I(2 ,)II,I(2 (4.4.20) Carrying out the differentiation as before, one now has, directly from these equations, CC C CC C I S II2 III I21 01 1 0 ∂∂−=⎥⎦⎤ ⎢⎣⎡ ∂∂+∂∂=−+=− W W Wp β βββ Constitutive Equation for an Isotropic Incompressible Hyperelastic Material (Material Form) (4.4.21) and b bI b bσ II2I21 11 1 1 ∂∂−=∂∂=−+= −− − W Wp α ααα Constitutive Equation for an Isotropic Incompressible Hyperelastic Material (Spatial Form) (4.4.22) Comparing these equations with the more general compressible equations: in the S equation, 4.4.4, the hydrostatic pressure is equivalent to CIII/ 2∂∂−W (with ) 1 III=C ; in the σ equation, 4.4.8, it is [ ]b b b II/ II III/ 2 ∂∂+∂∂−= W W p . Similarly, the equations can be written in te rms of the principal stretches and principal stresses (the incompressibility constraint is now 1321=λλλ ): pWSpWPpW i i ii i ii ii i 21 1,1,λλλ λλ λλσ −∂∂= −∂∂=−∂∂= (4.4.23) Section 4.4 Solid Mechanics Part III Kelly 377Example (Strip Biaxial Tension) Consider an incompressible isotropic hyperela stic material which is stretched in the 1e direction but its dimensions in the 2e direction is held constant, with ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = λλ /1000 100 0 F (4.4.24) With 03=σ ,one finds that 2 2II21 I2 λλb b∂∂−∂∂=W Wp (4.4.25) and the stresses are 2,1 ,1 II1 I222 22= ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−∂∂+⎟ ⎠⎞⎜ ⎝⎛−∂∂= iW Wσ ii iλλλλ b b (4.4.26) ■ Example (Biaxial Stretch) Consider the biaxial stretch of a thin sh eet of incompressible isotropic hyperelastic material, with ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 2121 /10 00 00 0 λλλλ F (4.4.27) From 4.4.18: ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−=∂∂= =−+=−− − b b II2 ,I2 ,3,2,1 ,1 12 12 1W Wip σi i i α α λαλα (4.4.28) Taking the stretches to be in-plane, with the larger faces stress-free, one has 03=σ so that, with 21 3 /1λλλ= , one can solve for p, 2 22 11 2 22 111λλαλλα−+ =p and the stresses in the sheet are Section 4.4 Solid Mechanics Part III Kelly 378() ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −=− − 2 22 12 22 11 1 2 2 22 12 12 21 1 11,1 λλλλααλλλλαα σ σ (4.4.29) If the strain energy is given in terms of the stretches, one would use 4.4.23 rather than 4.4.22. The PK1 stresses, T−=σF PJ , are, directly from these equations, () ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −==⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −==− − 3 22 122 11 1 22 2 2 23 112 21 1 11 11,1 λλλλααλ λλλλααλσPσP (4.4.30) The total forces acting on the late ral surfaces are then, from 3.5.4, () ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −==⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −==− − 3 22 122 11 1 0 02 2 2 23 112 21 1 0 01 11,1 λλλλααλλλλαα A AP F A AP F (4.4.31) where 0A is the area in the reference configuration. The material parameters 1 1,−αα can be determined from such a test by measuring the applied forces. This is helped by controlling the stretches 1λ and 2λ in such a way that the invariant bI is held constant, or the invariant bII is held constant. When the two applied forces are equal, 2 1F F= , one finds that ()() () 0 1 12 2 1 11 21 11 3 23 1 2 1 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛++−− −λλααλλααλλλλ (4.4.32) The term inside the first brackets gives the expected symmetric solution 2 1λλ=. However, the term inside the second brackets can also be zero for 2 1λλ≠, but this will only occur for fairly large stre tches for typical values of th e material parameters. The symmetric equal-stretch solution holds up until this happens. For example, for the typical value 1.0 /1 1=−−αα , the second term becomes zero when 17.32 1≈=λλ . The symmetric solution is unstable, in the sense that the forces must be precisely equal for it to hold. If there is even the slightest imbalance between 1F and 2F, then 2 1λλ≈ up to about a value of 3.17 but then the near-square will begin to deform into a rectangle, with the response now keeping clos e to the asymmetric soluti on. The long side of the rectangle will be along the direction of the slig htly larger force and will keep lengthening until the stretch in the other direction returns back to 1=λ . This large strain problem differs from the problems of linear elasticity, which have unique solutions. ■ Section 4.4 Solid Mechanics Part III Kelly 379Example (Simple Shear) The case of simple shear was discussed in detail in §2.2.6. In the equations following 2.2.40, let k2 tan12tan, cos 2sin, sin 2cos == = =θβθβθβ (4.4.33) Then the principal material directions can be expressed as 2 1 2 2 1 1 sin cos ˆ, cos sin ˆ E E N E E N ββ ββ +−= += (4.4.34) With RNn= , the principal spatial directions are (see figure 4.4.1 below) 2 1 2 2 1 1 cos sin ˆ, sin cos ˆ E E n E E n ββ ββ +−= + = (4.4.35) The principal stretches are 1 and2sin2cos1,2sin2cos1 2 1ββλββλ−=+= (4.4.36) and the principal stresses are, from 4.4.18, p σW Wp σp σ −+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−=∂∂= −+=−+= −− −− 1 1 31 1 2 212 21 22 112 11 1 II2 ,I2121 ααα αλαλαλαλα b b (4.4.37) Figure 4.4.1: principal directions for pure shear The strain energy here is a f unction of the invariants of b, which are ()1 III 3 II, 3 I2 2= += +=b b b k k (4.4.38) In the original Cartesian c oordinates, the stresses are 1e2e β β1ˆn2ˆn1ˆN 2ˆN Section 4.4 Solid Mechanics Part III Kelly 380 ββσσσβσβσσβσβσσ cos sin) (cos sinsin cos 2 1 122 22 1 222 22 1 11 −=+ =+ = (4.4.39) It follows that (this is a universal relati on, independent of the constitutive relation) ()12 12 22 11 2tan/2 σσβ σσ k= =− (4.4.40) The fact that and 22 11σσ≠ is called the Poynting effect . The stress are now () () () p σk σp k σp k σ −+=−=−++=−++= −−−− 1 1 331 1 122 1 1 2212 1 11 11 ααααααα α (4.4.41) The proportionality factor be tween the shear stress and the shear strain measure k is called the shear modulus: ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= b b II I2W Wμ (4.4.42) In the case of plane stress, the hydrostatic pressure p can be evaluated, and one finds that kσ kσ kσ μ α α = = =− 122 1 222 1 11 , , (4.4.43) The stresses acting on the deformed materi al are shown in Fig. 4.4.2. The unit normal e to the sloping surface (on the right hand side) is ()2 1211e e e k k− += (4.4.44) and, from Cauchy’s law, the tract ion acting on that surface is ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− += 011 22 1212 11 2σσσσ kk kt (4.4.45) The normal and shear traction are then (using the Poynting effect) () ()[]() () ()[]12 2 22 12 12 11 212 2 22 12 22 11 22 12 12 11 2 11 111 12 11 σ σσσσσ σσ σσσσσ kk k kktkk kk kk k kkt SN +=−+−+=+−=++−=−−−+= Section 4.4 Solid Mechanics Part III Kelly 381(4.4.46) and tend to 11σ and 12σ as 0→k . Figure 4.4.2: stresses acting on the material in simple shear It is clear then that the pure shear of mate rial discussed here can only occur when normal stresses are applied to the faces. Ther e is of course no volume change for an incompressible material; for a compressible mate rial, however, there will in general be a volume change unless normal stresses are applied, a phenomenon known as the Kelvin effect . ■ See also the shear problem using convected co ordinates discussed in the Appendix to this Chapter, §4.A.3. 4.4.5 Series Expansion of the Strain Energy Function It is sometimes helpful when seeking specifi c constitutive laws for isotropic hyperelastic materials to expand the strain energy function in to an infinite power series of the form () ( ) ( )∑∞ =−−− = = 0,,1 III3 II3I )III,II,I( rqpr q p pqr W WC C C C C C α (4.4.47) In the undeformed configuration, 1 III,3 II I ===C C C , and the strain energy is zero. For incompressible materials, this reads as () ( )∑∞ =−− = 0,3 II3I qpq p pq WC Cα (4.4.48) The series can be written in terms of the principal stretches by noting that k 12σ22σ 12σ 22σe Section 4.4 Solid Mechanics Part III Kelly 3822 32 22 12 12 32 32 22 22 12 32 22 1 IIIIII λλλλλλλλλλλλ =++=++= CCC (4.4.49) Next are considered some particular forms of the strain energy function. 4.4.6 Incompressible Material Models First, models of isotropic hyperelastic incompressible materials are examined. The Ogden Model (Incompressible) The strain energy for the Ogden model is defined to be () {}∑ =−++ =n i ii i i iW 13 2 1 3 2 1 3 ,,αααλλλαμλλλ (4.4.50) where the i iαμ, are experimentally determined material constants. Note that this model does not include terms involving the products of different stre tches, which appear in the most general power series 4.4.48. However, the Ogden model allows for fractional powers of iα. Only three terms in the series are usually needed to correlate the model with experimental data of incompressi ble rubber. Typical values are kPa10 0.2kPa2.1 0.5kPa 6300 3.1 3 22 11 1 −=−== == = μ αμ αμ α The Mooney-Rivlin Model (Incompressible) The Mooney-Rivlin model, applicable to incompressible materials, is ()()3 II 3I2 1 −+−=C C c c W (4.4.51) And, from 4.4.22, the stresses are then I b bσ p c c −−=−1 2 12 2 (4.4.52) From the simple shear example discussed in §4.4.4, the shear modulus for the Mooney- Rivlin material is ()2 12 cc+ , implying that the shear stress to shear strain measure k is linear. The Mooney-Rivlin model is often used to model rubber-like materials and the numerical value of 2c is usually much smaller than 1c. Section 4.4 Solid Mechanics Part III Kelly 383Note that the Mooney-Rivlin model can be vi ewed as a special case of the Ogden model, since setting 2 ,2 ,22 1 −=== αα n in that 4.4.50, and using the incompressibility constraint 12 32 22 1=λλλ gives {} {} {} {} 3 II23I23232 2 12 32 22 12 2 32 22 11 −−−=−++−−++=−−− C Cμ μλλλμλλλμW (4.4.53) The Neo-Hookean Model (Incompressible) The Neo-Hookean is a special case of the M ooney-Rivlin model, wherein 02=c , giving ()3I1−=Cc W (4.4.54) It is useful as a model of non-linear elas ticity within the small-strain range. The Varga Model (Incompressible) The Varga model is another special case of the Ogden model: )3 (3 2 1 1 −++= λλλc W (4.4.55) The Yeoh Model (Incompressible) The Yeoh model was motivated by the problem of predicting the behaviour of rubbers containing reinforcing fillers such as carbon black, for which the Mooney-Rivlin model does not succeed too well. The st rain energy here takes the form 3 32 2 1 )3I( )3I( )3I( −+−+−=C C C c c c W (4.4.56) As usual, 0 )(=IW in the reference configuration and the strain energy is a convex function, increasing monotonically with CI (and attains a minimum in the reference configuration, so that 0 I/=∂∂C W must contain no real root s, which places restrictions on the values that the constants 3 2 1,,ccc can take). The Arruda-Boyce Model (Incompressible) The Arruda-Boyce model is based on a pr oposed molecular network structure to a material, and the strain energy is () () ()⎥⎦⎤ ⎢⎣⎡+− +−+− = L 27 I1050119I2011I213 22 C C Cn nWμ (4.4.57) Section 4.4 Solid Mechanics Part III Kelly 384The two parameters are μ, the shear modulus, and n, the number of segments in a polymer chain. 4.4.7 Compressible Material Models The Mooney-Rivlin Model (Compressible) The compressible Mooney-R ivlin model reads as ()()3 II 3I ln) (2)1(2 1 2 12−+−++−−=b b c cJ cc Jc W (4.4.58) The stresses are then, from 4.4.8, with bIII=J , ()[] ()[] {}2 2 2 1 2 1 tr ) (4 III III III2b bb I σb b bc cc cc c − +++−− = (4.4.59) The Neo-Hookean Model (Compressible) Neo-Hookean models are characterised by th eir dependence on th e first (as in the incompressible case) and third invariants, but not on the second invariant. There are quite a number of Neo-Hookean models available, for example (the last term here is the incompressible Neo-Hookean term) () 3I21ln ) (ln22−+− =bμμλJ J W (4.4.60) With corresponding stresses [] ) ( lnI2 IIIIII2 11 1 Ib Ib I σ bb b −+ =∂∂+∂∂= −− − μλJ JWJWJ (4.4.61) The μλ, of this compressible Neo-Hookean mode l are the classical Lamé constants, which can be seen by particularising the constitutive equation to the small strain range. Another Neo-Hookean model is ()()3I21ln213/2 2− + =− b J J W μ κ (4.4.62) with corresponding stresses Section 4.4 Solid Mechanics Part III Kelly 385() bIb b σ dev ln)(tr31ln 3/2 13/2 1 − −− − + =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⎟ ⎠⎞⎜ ⎝⎛− + = J J JJ J J μκμκ (4.4.63) The Blatz and Ko Model (Compressible) Blatz and Ko proposed a compressible model based on a combination of theoretical arguments and experimental wo rk on foamed materials: ()() () ()⎥ ⎦⎤ ⎢⎣⎡− +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −+⎥⎦⎤ ⎢⎣⎡− +− =+ −1 III13IIIII 21 1 III13I2β β βμ βμ b bb b b f f W (4.4.64) where ννβ21−= (4.4.65) and νμ, denote the shear modulus and Poisson’s ratio, and ]1,0[∈f is an interpolation parameter. When the material is incompressible, 1 III=b , and the model reduces to the Mooney-Rivlin model. Also, by letting 1 =f , and replacing bIII with 2J, the Blatz and Ko model reduces to the Neo-Hookean model () ()11 23I22− +−=−β βμ μJ Wb (4.4.66) 4.4.8 Problems 1. Evaluate the PK2 stress for the Moon ey-Rivlin material (incompressible) Section 4.5 Solid Mechanics Part III Kelly 3864.5 Material Anisotropy 4.5.1 Material Symmetry The isotropic material was defined as one whose material response was unaffected by rigid body rotations of the reference confi guration. Other materi al symmetries are possible; to generalise the notion, instead of considering orthogonal transformations, consider an arbitrary deformation 0F of the reference configuration 0S bringing it to a new configuration ◊S, Fig. 4.5.1 (compare with th e isotropic case, Fig. 2.8.6). Figure 4.5.1: a deformation of the reference configuration Considering the Cauchy-elastic material, if the deformation 0F has no effect on the response of the material, then )( ) ( )( )(1 0 Fσ FFσ Fσ Fσ = →=− ◊ (4.5.1) When Q F=0 , one has the isotropic material. Setting 1 0−=FG , 4.5.1 can be cast in the most usual form: )( )( FGσ Fσ= (4.5.2) Note that the restriction 1 det±=G is assumed, since otherwise arbitrary dilatations could occur with no change in materi al response, which seems physically unreasonable. Note that the set of all tensors G which satisfy 4.5.2 forms a group (see the Appendix to this Chapter, §4.A.2) and hence is called the symmetry group of the material (with respect to the configuration 0S). Apart from isotropy, the two most important practical cases of material symmetry are transverse isotropy and orthotropy . Xreference configuration S◊S ◊X◊F Fx0SS0F Section 4.5 Solid Mechanics Part III Kelly 387 4.5.2 Transverse Isotropy Consider first the transversely isotropic material. Such a material has a single preferred direction, defined by a unit vector 0a in the reference configuration. Such a vector is illustrated in Fig. 4. 5.2, showing also the unit vectors 3 2ˆ,ˆnn completing an orthonormal set. The symmetry group of the tran sversely isotropic material is the set of orthogonal tensors Q which transform the set {}3 2 0,,nna into the new orthonormal set {}3 2 0,,nna′′± . In particular, 0 0 a Qa±= (4.5.3) In order to ensure that the sense of 0Qa is immaterial, it is best to introduce the structural tensor 0 0a a⊗ , which transforms as the axes change according to 0 0 0 0 a a Qa Qa ±⊗±=⊗ (4.5.4) or ()0 0T 0 0 a a Qa aQ ⊗=⊗ (4.5.5) Figure 4.5.2: an orthonormal set of vectors The strain energy can now be taken to be a function of C, as in the isotropic case, and 0 0a a⊗ , which characterises the st ructure of the material: ) ,(0 0a aC⊗ =W W (4.5.6) Allowing for transformations of the undeformed configuration, ) , ( ) ,(T 0 0T 0 0 Qa Qa QCQS a aC ⊗ =⊗ W (4.5.7) with Q here restricted to the symmet ry group defined by 4.5.3. Then W is an isotropic scalar function of two symmetric tensors a nd so, from Table 4.A.1, takes the form 2n Q3n 0a3 2orn n′′ 0a0a− 3 2orn n′′ Section 4.5 Solid Mechanics Part III Kelly 388() ()() ( () () () () )2 0 02 2 0 0 0 02 0 03 0 02 0 0 0 03 2 tr, tr, tr, trtr, tr, tr,tr,tr,tr a aC a aC a aC a aCa a a a a a C C C ⊗ ⊗ ⊗ ⊗⊗ ⊗ ⊗ =W W (4.5.8) Since {▲Problem 1} () ()() ()() () ()3 22 2tr, tr,1 tr aa aaaaaaC aaC aCa aaC aa ⊗=⊗=⊗=⊗ =⊗ =⊗ (4.5.9) one arrives at the representation ()()()()() ()0 5 0 4 3 2 1 , ,, , , , aC aC C C C I I I I IW W= (4.5.10) where the fourth and fifth scalar (pseudo-) invariants 5 4,II are defined by aaC aCa2 5 4 ,= = I I (4.5.11) Note also that, from the definiti on of the stretch, Eqn. 2.2.17, 2 0 0 4 a Caaλ==I (4.5.12) where aλ is the stretch of the unit line element 0a. If the preferred direction is 3e, then the fourth and fifth invariants in terms of components are 2 332 232 13 02 0 533 0 0 4 C C C IC I ++==== aCaCaa (4.5.13) in which case the five invariants can be taken as { }2 232 13 33 3 2 1 ,,,, C CCIII + . Using the relations { ▲Problem 2} 0 0 0 05 0 04, aCa Ca aCa aC⊗+⊗=∂∂⊗=∂∂ I I (4.5.14) the PK2 stresses for a hyperelastic material are then () ⎥⎦⎤⊗+⊗∂∂+⊗∂∂+⎢ ⎣⎡ ∂∂+∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂ ∂∂= −=∑ 0 0 0 0 50 0 41 33 2 21 15 10 2),(2 aCa Ca a a aC C ICaCS IW IWIWIIW IWIIWI IW ii i (4.5.15) Section 4.5 Solid Mechanics Part III Kelly 389 Let a be a unit vector in the current configuration, in the direction of 0Fa, that is, 0Faaa=λ (4.5.16) Then, using Eqn. 3.5.7, T 1FSFσ−=J , with b FIF=T, 2 Tb FCF= (see Eqn.2.2.14), I F FC=− T1 and noting that C and b have the same principal invariants, 4.5.13 becomes () ⎥⎦⎤⊗+⊗∂∂+⊗∂∂+⎢ ⎣⎡ ∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂=− a ab baa aab b I σ 54 442 2 21 1 3312 IWIIWIIW IWIIW IWIJ (4.5.17) Using the Cayley-Hamilton theorem allows one to re-write the Cauchy stress as () ⎥⎦⎤⊗+⊗∂∂+⊗∂∂+⎢⎢ ⎣⎡ ∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=− − a ab baa aab b I σ 54 441 23 1 33 2212 IWIIWIIWIIW IWIIWI J (4.5.18) Expressing the scalar invariants in terms of b rather than C, the coefficients of the tensors in 4.5.17-18 are functions of the set { }aba aa b bba a2 2 3 2, ,tr,tr,tr λλ⊗ (4.5.19) Transversely Isotropic Materials with Constraints For an incompressible material, 13=I , and, analogous to 4.2.40, the strain energy takes the form () () ( ) ( )() )1 (21, ,, , ,3 0 5 0 4 2 1 −− Ip I I I IW aC aC C C (4.5.20) For a material which is inextensible in the direction of 0a, from 4.5.13, 04=I , and the strain energy takes the form () () () ( )() )1 (21, , , ,4 0 5 3 2 1 −− Iq I I I IW aC C C C (4.5.21) Section 4.5 Solid Mechanics Part III Kelly 3904.5.3 Orthotropy Consider now a material which is depe ndent on two characteristic directions, 0a and 0b; again the sense of these directions is immaterial. The strain energy is now of the form ) , ,(0 0 0 0 b ba aC ⊗⊗ =W W (4.5.22) As isotropic scalar function of three sy mmetric tensors depends on the following traces (see Table 4.A.1) () () () () () () () () () () () () () () () () () () () () () () () ()0 0 0 02 0 02 0 02 0 0 0 00 02 0 0 0 0 0 02 0 02 2 0 0 0 02 0 02 0 02 2 0 0 0 02 0 03 0 02 0 0 0 03 0 02 0 0 0 03 2 trtr, tr, tr, trtr, tr, tr, trtr, tr, tr, trtr, tr, tr, tr, tr, trtr,tr,tr b ba aCb b a a b ba ab b a a b ba ab bC b bC b bC b bCa aC a aC a aC a aCb b b b b b a a a a a aC C C ⊗⊗⊗ ⊗ ⊗⊗⊗ ⊗ ⊗⊗⊗ ⊗ ⊗ ⊗⊗ ⊗ ⊗ ⊗⊗ ⊗ ⊗ ⊗ ⊗ ⊗ (4.5.23) Using 4.5.9 (see Eqn. 1.9.10e) this redu ces to the set of nine invariants () ()0 0 0 02 0 0 02 0 0 0 02 0 0 03 2 , , , , ,tr,tr,tr Cbaba babCb CbbaCa CaaC C C ⋅ ⋅ (4.5.24) with 0 0 0 0 Cab Cba= . The term 0 0ba⋅ is the cosine of th e angle between the two characteristic directions; this does not cha nge during the deformation and so this term can be omitted, leaving ()0 0 0 0 02 0 0 0 02 0 0 03 2, , , , ,tr,tr,tr CbababCb CbbaCa CaaC C C ⋅ (4.5.25) An orthotropic material is one for which 0a and 0b are perpendicular, 00 0=⋅ba , making the last term here zero. This also then defines a third preferred direction, 0c, orthogonal to both 0a and 0b, which introduces extra terms 002 0 0 0 , cCc Ccc . But 02 0 02 0 02 020 0 0 0 0 0 trtr cCc bCb aCa CCcc Cbb CaaC ++=++= (4.5.26) so that 02 0 0 0 , cCc Ccc are redundant. Finally, the strain energy is of the form ( )02 0 0 0 02 0 0 03 2, , , ,tr,tr,tr bCb CbbaCa CaaC C CW W= (4.5.27) Section 4.5 Solid Mechanics Part III Kelly 391As before, the stresses in a hyperelas tic material can now be obtained by differentiation. 4.5.4 Problems 1. Show that, for unit vector a, (i) () aa aa ⊗=⊗2, (ii) ()1 tr=⊗aa (iii) ()() aCa aaC=⊗ tr (iv) ()() aaC aaC2 2tr =⊗ 2. Show that 0 0 0 05 0 04, aCa Ca aCa aC⊗+⊗=∂∂⊗=∂∂ I I 3. Show that (i) () aa Fa aFa⊗=⊗2 T 0 0 λ ( i i ) () baa F Ca aFa⊗= ⊗2 T 0 0 λ For (ii), it might help to note th e following relations (for vector b and second-order tensors A, B): ()() () () () () b AA AbABbAbABbBABAb T T T=⋅=≠ Section 4.A Solid Mechanics Part III Kelly 4064.A Appendix to Chapter 4 4.A.1 Isotropic Functions The scalar-, vector- and tensor-valued functions a,φ and T of the scalar variable φ, vector variable v and second-order tensor variable B are isotropic functions if ()()()() () () ( ) () () () () () ( ) () () ()T T T TTT QBQT QBQT QvQT QvT Q QT TBQa QBQa vQa Qva Qa aB QBQ v Qv = = == = == = φφφφφ φ φφ Isotropic Functions (4.A.1) for all orthogonal tensors Q. Isotropic functions are also called isotropic invariants . Here follow some examples. Examples (of Isotropic Functions) 1. The scalar-valued function of a second order tensor T T det)(=φ is an isotropic function since ()() T QTQ QTQ det detT T= = φ 2. The scalar-valued function of two second order tensors ()AB BA tr),(=φ is an isotropic function of its tw o tensor variables since ( )( )()()AB QABQ QBQ QAQ QBQ QAQ tr tr tr ,T T T T T= = = φ More generally, the function ()m mBAtr , m an integer, is isotropic 3. The vector-valued function a of a vector v and a second order tensor T, Av Tva=),( is an isotropic function since ( ) ()TvQa QTv QvQ QT Q QTQva , ,T T== =m m Indeed the function vT Tvam=),(, m an integer is an isotropic function. 4. The tensor-valued function of a second order tensor: 2)( T TA= is an isotropic function since () () ()() ()T T 2 T T2T TQTQA QQT QTQ QTQ QTQ QTQA = = = = Indeed the function mT TA=)(, m an integer is an isotropic function. ■ Section 4.A Solid Mechanics Part III Kelly 407 Restrictions on the form th at isotropic functions can take is next examined. Isotropic Scalar-valued Functions Consider first an isotropic scalar-valued function of a vector u, ()uφ , so that () ( ) Qu uφφ= . Since only the magnitude of u is invariant under an orthogonal tensor transformation, it follows that φ depends on u only through uuu⋅= , so ()uu⋅≡φφ . Here, uu⋅ is called the integrity basis of φ. Similarly, an isotropic scalar-valued f unction of two arguments is defined through ()()QvQu vu , ,φφ= (4.A.2) for every orthogonal Q, and its integrity basis consists of the three scalar invariants vvvuuu ⋅⋅⋅ , , (4.A.3) since only the lengths of the two vectors a nd the angle between them are preserved under a rotation. Consider next a scalar-v alued isotropic function φ of a symmetric second-order tensor S. Since S is symnmeytric, it has the sp ectral decomposition representation ∑⊗=i ii s s Sλ , where {}3 2 1,,λλλ are the eigenvalues and {}3 2 1ˆ,ˆ,ˆ nnn are the eigenvectors of S. Since S is isotropic, ()()( ) ( )( ) ∑ ∑ ⊗ =⊗ = =i i i i ii nQ nQ Qn n Q QSQ S ˆ ˆ ˆ ˆT Tλφ λφ φφ (4.A.4) Thus φ is independent of the orientati on of the principal directions of S and so must depend only on the three principal values, ()3 2 1,, )( λλλφ f=S (4.A.5) Note also that f must be a symmetric function of the eigenvalues. For example, take Q to be a positive rotation about 3ˆn. Then 2 1ˆ ˆ n nQ=, 1 2ˆ ˆ n nQ−= and3 3ˆ ˆ n nQ= , so () ()3 1 2 3 33 2 21 1 12T,, )ˆ ˆ ˆ ˆ ˆ ˆ( )( λλλ λ λ λφ φφ f=⊗+⊗+⊗= = n n n n n n QSQ S (4.A.6) and, similarly, the subscripts on any pair of eigenvalues in 4.A.5 can be interchanged. Since the set { }3 2tr,tr,tr S SS , the set of three principal scalar invariants {}S S S III,II,I and the set of eigenvalues {}3 2 1,,λλλ uniquely determine one anothe r, any of these sets can be regarded as the integrity basis of )(Sφ . Section 4.A Solid Mechanics Part III Kelly 408 Some important isotropic scala r-valued functions and their in tegrity bases are listed in Table 4.A.1 below. The integrity basis consis ts of that entry together with appropriate entries from higher up in the Table, for example the integrity basis for a tensor A and two vectors u and v is vuA uAvvvA vAvuuA uAuA A A vuvvuu 2 2 23 2 , , , , ,tr,tr,tr, , , ⋅⋅⋅ Isotropic Function Integrity Basis ()uφ ()Quφ= uu⋅ ()vu,φ ()QvQu,φ= vu⋅ ()Aφ ()TQAQφ= 3 2tr,tr,tr A A A ()Au,φ ()T,QAQQuφ= uuA uAu2, ()BA,φ ( )T T,QBQ QAQφ= 2 2 2 2tr, tr, tr, tr BA AB BA AB () vAu,,φ ( )Qv QAQQu , ,Tφ= vuA uAv2, () CBA,,φ ABCtr Scalar- valued functions CBA,, are symmetric tensors Four or more tensors redundant Table 4.A.1: Isotropic Scalar Functions and Integrity Bases Isotropic Vector-valued Functions Next, consider a vector-v alued isotropic function a of a vector v, so () ( ) QvavQa= . To find the dependence of a on v, consider the scal ar-valued function φ given by (note that φ here is linear in its first argument, u): ()()vauvu⋅=,φ (4.A.7) It follows that () ()()()()vu vauvQaQu Qva Qu QvQu , , φ φ =⋅=⋅=⋅= (4.A.8) and so φ is an isotropic function of its two vector arguments and must depend only on the three invariants 4.A.3, and so takes the general form () ()vvv uvvvuuu ⋅⋅=⋅⋅⋅ α φ , , (4.A.9) Finally, a must take the form () v vaα= (4.A.10) where the coefficient α is a function of the scalar invariant of v, i.e. vv⋅. Section 4.A Solid Mechanics Part III Kelly 409Note that the only isot ropic vector function a of a tensor B is the null vector oa=. Another important isotropic vector-valued func tions is that of a vector and symmetric tensor. This and their integrity base s are listed in Table 4.A.2 below. Isotropic Function Integrity Basis ()va ()()vQa Qva= v ()Ta ()()TQa QTQa=T o Vector-valued functions S is a symmetric tensor ()Sva, ()()SvQa QSQQva , ,T= vSSv2, Table 4.A.2: Isotropic Vector Functions and Integrity Bases Isotropic Tensor-valued Functions Consider next a second-ord er tensor-valued function T of a tensor B. To find how T depends on B, this time consider th e scalar-valued function φ given by (again, note that by definition φ is linear in its first argument, A) ()()[]BAT BA tr ,=φ (4.A.11) It follows that ( ) () [ ] ()[] ()[] ()[] ()BABATQB QATQBQT QAQQBQT QAQ QBQ QAQ ,trtrtrtr , TT TT T T T φφ ===== (4.A.12) Thus φ is an isotropic function of its two tensor arguments and so, if A and B are symmetric, is a function of the ten invari ants listed in Table 4.A.1. Since φ is linear in A, it can only depend on six of these ten invariants, namely 3 2tr,tr,tr,tr B B B A , 2tr, tr AB AB , and so takes the form ()[] ( ) [ ]2 2 1 0 tr tr B B I A BAT ααα φ ++ = = (4.A.13) and so T takes the form ()2 2 1 0 B B I BT ααα ++= Form for a symmetric isotropic tensor function of a symmetric tensor (4.A.14) where 2 1 0 ,,ααα are scalar functions of the invariants of B. Equation 4.A.13 can be rewritten in various alternative forms us ing the Cayley-Hamilton theorem, 1.9.45. Section 4.A Solid Mechanics Part III Kelly 410Some important symmetric isot ropic tensor-valued functions are listed in Table 4.A.3 below. Isotropic Function Integrity Basis ()vT ()()TQvQT QvT= vvI⊗, ()AT ()()T TQAQT QAQT = 2,, AAI ()vuT, () ()T, , QvuQT QvQuT = uvvu⊗+⊗ ()AuT, ()()T T, , QAuQT QAQQuT = Au Auu Au Auu ⊗⊗+⊗ , () vSuT ,, ( ) ()TT ,,, , QvAuQTQv QAQQuT = v Au Auvu Av Avu ⊗+⊗⊗+⊗ Tensor- valued functions BAT,, are symmetric tensors ()BAT, BAB ABA BA AB , ,+ Table 4.A.3: Isotropic (Symmetric) Te nsor Functions and Integrity Bases Some Results for Isotropic Functions Here follow some other important re sults regarding isotropic functions. 1. The principal values of an isotropic tensor function T of a tensor B are scalar invariants of B. To show this, let ()Bit be the principal values of )(BT and let ()TQBQit be the principal values of ) (TQBQT . Then ()() 0 )( det =− IB BTit , ( )0 ) ( ) ( detT T= − I QBQ QBQTit Because of the isotropy, and using the relation 1.9.13a, B A AB det det) det(= , the second of these can be written as ( )( ) () I QBQ BTQIQ QBQ QBQT I QBQ QBQT ) ( )( det) ( )( det ) ( )( det TT T T T T ii i tt t − =− = − This holds for all orthogonal Q and hence ) ( )(TQBQ Bi i t t= (4.A.15) which is the definition of an isotropic scalar invariant of B. 2. An isotropic tensor function T of a tensor B is coaxial with B. This follows directly from 4.A.14, since nB has the same principal directions as B. Section 4.A Solid Mechanics Part III Kelly 4113. Let T be a symmetric isotropic tensor function of the symmetric tensor B; if in addition the function T is a linear function of B, then it has the representation ()() B IB BT β α+ =tr (4.A.16) where βα, are arbitrary consta nts (independent of B). This follows directly from 4.A.14, noting that only the first invariant, Btr, is linear in B. It will be noted that this is the form of the (isotropic) linear elastic material model, 4.1.15. 4. Let C be a fourth-order isot ropic function, that is mnpq lq kp jn im ijkl CQQQQ C= (4.A.17) with the minor symmetries 1.9.65, ijlk jikl ijkl C C C == . Then it has the representation ( )jk il jl ik klij ijklC δδδδμδλδ + += (4.A.18) In terms of the identity tensors of §1.9.16 (compare with Eqn. 1.10.7), ()II C ++⊗= μλ II (4.A.19) To show this, consider a sy mmetric second-order tensor S and define S A :C= . Then the index notation for ) (SA is kl ijklSC , and A is clearly symmetric. Then () jn kl mnkl immn pqmn jq ipmn pqrs sn rm jq ipnl mn km pqrs ls kr jq ip nl mn km ijkl QS CQS CQQS C QQQSQ CQQQQ QSQ C :)(:) ( TT QSQAQSQA === δδ (4.A.20) from which it can be seen that A is a symmetric isotropi c tensor function of the tensor variable S. Further, A is linear in S, and for S symmetric, it follows that A takes the representation 4.A.16, ()() S IS SA μ λ 2 tr+= (4.A.21) In component form, this is () ()kl jk il jl ik kl iklijji ij kkijij kkij ij S SS S SS S A δδδδμδλδμλδμ λδ + + =++=+= 2 (4.A.22) Section 4.A Solid Mechanics Part III Kelly 412from which 4.A.18 follows. 4.A.2 The Symmetry Group The nonempty set G with a binary operation, that is, to each pair of elements Gba∈, there is assigned an element G ab∈, is called a group if the following axioms hold: 1. associative law : ) ( )( bcacab= for any Gcba∈,, 2. identity element : there exists an element Ge∈, called the identity element, such that a ea ae== 3. inverse : for each Ga∈, there exists an element G a∈−1, called the inverse of a, such that eaa aa ==−− 1 1 Consider the set of tensors G of 4.3.2. Since for two tensors 1G and 2G in G, )( ) ( ) (1 2 1 Fσ FGσ GFGσ = = (4.A.23) G∈2 1GG . The associative law clearly holds, the identity element is I and the inverse of G is 1−G. Thus the set of tensors G forms a group. 4.A.3 Shear of an Isotropic Square Block Consider a combined stretch and simple shear of an isotropic hyperela stic material, Fig. 4.A.1. Relative to the Ca rtesian coordinate system 3 3 3 2 2 2 2 2 1 1 1 , , X x X x Xk X x λ λ λλ = = += (4.A.24) Then ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 321 322 1 0 00 00 0 1000100 1 0 00 00 λλλ λλλλ k k F (4.A.25) and so can be considered to be a homogeneous stretch followed by a simple shear. The left Cauchy-Green strain and inverse are ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + −− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡+ =− 2 32 12 2 22 12 12 1 1 2 32 22 22 22 22 2 1 /1 0 00 / /1 /0 / /1 , 0 000 λλλλλ λ λλλλλλ k kk kk k b b (4.A.26) The compressible and incompressible isot ropic relations are (4.4.8 and 4.4.22 respectively) Section 4.A Solid Mechanics Part III Kelly 4131 1 1)(1 1 1 0)( − −− − ++−=++= b b Iσb b Iσ ααβββ pic (4.A.27) Substituting in the Cauchy-Green strains, one finds that 023 13==σσ and ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−=− − 2 112 21)( 12 2 112 21)( 121,1 λαλασλβλβσ k ki c (4.A.28) Using this relation, it can then be seen that 12 2 22 22 2 22 1 22 11 σλλλλσσkk+−=− (4.A.29) which holds for both compressible and incomp ressible materials, and is the universal relation analogous to 4.4.40. Here, however, th e stretches can be chosen so as to make the normal stress-difference zero. Figure 4.A.1: block under st retch and simple shear Introduce now base vectors 2 1,gg along the edges of the deformed block, with corresponding contravariant base vectors 1g and 2g, Fig. 4.A.1, so that 33 22 2 113 3 2 1 2 1 1 , ,, , e g e g e e ge g ee g e g = = −==+== kk (4.A.30) The metric coefficients are 2λk 2Sσ2Nσ e 2λ 1λxy 1g2g2g 1g1Nσ1Sσ oB A Section 4.A Solid Mechanics Part III Kelly 4141 , 10 0010 1 , 1 0 00 10 12 2= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−+ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ +=g kk k g k kk gij ij (4.A.31) From 3.9.2, the unit normals to th e block surfaces are (see 4.4.44) 2222 2 2 22 1 111 1 1ˆ , 1ˆ egg ne e gg n === +−=== g kk g (4.A.32) The stress components with respect to the cu rvilinear system can be obtained from the transformation rule in §1.13.1: [] [] [ ] [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −=⋅= =⋅ ⋅ ⋅ 10001001 , k A A Aj ij ij nmnTi mijge σ σ (4.A.33) leading to [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− +− = 3322 22 2122 12 222 12 11 0 000 2 σσ σσσσσσσ σ kk k k ij (4.A.34) The normal and shear stresses acting on the surfa ces of the block are (see Fig. 4.A.1) are 12 111 1 12 2 22 2 , , , σσσσσσσσ = = = =S N S N (4.A.35) In order that the normal stresses acting on the block are zero, then, one requires 0 2 ,012 11 22 =− = σσ σ k (4.A.36) From 4.A.29, this means that ()2 22 2 11λ λ k+= (4.A.37) A physical interpretation of this results is that the lengths of the sides of the deformed block are equal, oB oA= in Fig. 4.A.1. In this case, 4.A.34 reduces to, using 4.A.28, [] []() ()1 2 2 1 2 1 11 22 1 111 2 2 1 )( 12)( 12 )()( 1 1g g g gg g g g ⊗+⊗⎥ ⎦⎤ ⎢ ⎣⎡ ⎭⎬⎫ ⎩⎨⎧−+⎭⎬⎫ ⎩⎨⎧=⊗+⊗ ⎭⎬⎫ ⎩⎨⎧= ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ −− λαβλ αβσσ σσ kkic iijcij (4.A.38) Section 4.A Solid Mechanics Part III Kelly 415Thus a state of pure shear is achieved, with only shear stresses acting on the faces, and a square block deforms into a rhombic block. Consider now the (incompressible) Ne o-Hookean model, Eqn. 4.4.54, for which b Iσ12c p+−= (4.A.39) The stress components are then ij ij ijbc pg12+−=σ (4.A.39) The metric components ijg are given by 4.A.31. The contravariant components of the left Cauchy-Green strain can be obtained from coordinate transformation equations similar to 4.A.33 (with ij ijb b= in the Cartesian system), leading to [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ − ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡+ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− =2 32 22 1 2 32 22 22 22 22 2 1 0 00 00 0 10001001 0 000 1000100 1 λλλ λλλλλλ k kk k k bij (4.A.40) with 1321=λλλ . Then, with the stress taking the representation 4.A.38, with 1 12c=α , 01=−α , () () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−+ −= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ++2 32 22 1 12 2 2 112 2 11 0 00 00 0 2 10 0010 1 0 0 00 0 1/ 20 1/ 2 0 λλλ λλ c kk k p k kck kc (4.A.41) Solving leads to () () ()6/12 3 23/12 13/12 1 1 , 112 −− +== +=+= k kk c p λλ λ (4.A.42) The solution shows that 11>λ and 12<λ and so the block deforms as in Fig. 4.A.2. Section 4.A Solid Mechanics Part III Kelly 416 Figure 4.A.2: simple shear of a Neo-Hookean block Note that, in contrast to the decomposition 2λk 2λ 1λxy 12σ12σ Section 5.2 Solid Mechanics Part III Kelly 3605.2 Plasticity I: Hypoel astic-Plastic Models The two main types of classical plasticity model for large strains are the hypoelastic- plastic model and the hyperelastic-plastic model. The first of these is discussed in this section. 5.2.1 Hypoelasticity In the hypoelastic-plastic models of plasti city, the elastic response is assumed to be hypoelastic. As discussed in §4.1.4, hypoelastic materials are characterised by constitutive relations of the form ),(dσσf=& (5.2.1) and, in mutiaxial problems, this implies that the response cannot be expressed in terms of an elastic strain energy function. Thus the response is path-dependent and dissipation may occur even though the material is supposed to be elastic. However, the idea with the hypoelastic-plastic models is that the elastic strains are assumed to be relatively small so that any error in the conservation of energy is very small, and so can be neglected. The stress-rate σ& in the hypoelastic equation 5.2.1 must be objective. Thus the material derivative of the Cauchy stress can not be used, but any of the many objective rates, for example the Jaumann, Green-Naghdi, Truesdell, etc., can be used. A large class of hypoelastic materials is encompassed in the linear relation between objective stress-rate and the rate of deformation: d Cσ :∇∇= (5.2.2) where ∇σ is an objective stress rate and ∇C is the corresponding fourth order tensor of elastic moduli, which may itself depend on the stress, in which case it must be an objective function of the stress. For a given finitely deformed state, the (small) increments in stress and strain are linearly related and are recovered upon unloading. However, for finite deformations, the work done in a closed path may not be zero. The elastic modulus tensor ∇C is also called the tangent modulus . It possesses the minor symmetries due to the symmetry of d and ∇σ. It is usually assumed to possess also the major symmetries. Example Consider the following hypoelas tic constitutive equation: d Cσ :J Jσ=∇ (5.2.3) where J∇σ is the Jaumann stress-rate, Eqn. 3.5.20, σw wσσ+−& . The Truesdell stress- rate T∇σ, defined by Eqn. 3.5.22, ()σdσl lσσ trT+−−& , is then Section 5.2 Solid Mechanics Part III Kelly 361 () () () dIσC Cσdσd dσσσdσl lσσw wσσσ : ˆtrtr TT ⊗+−=+−−=+−−−+= ∇∇ ∇ JJJ T σ (5.2.4) where σd dσ dC +=:ˆ . Thus 5.2.3 can be expressed as d Cσ :T Tσ=∇ with the Truesdell tangent modulus given by IσC C ⊗+−ˆ Jσ. It is interesting to note that, if JσC is constant, then T∇σ is not. Further, if JσC has the major symmetries, then T∇σ does not (since Iσ⊗ does not). For this reason one often uses the Kirchhoff stress στJ= rather than the Cauchy stress. For example, the Jaumann rate of the Kirchhoff stress is τw wτττ +−=∇&J. Then, with ) (trdJJ=& , ( ) () d CdC Cσd dστ σ ::ˆ 1T 1 TJJ T JJ J J στ ≡−=−− = −∇− ∇ (5.2.5) Now if TσC has the major symmetries, then so does JτC. 5.2.2 Hypoelastic – Plastic Model Additive Decomposition of the Rate of deformation The rate of deformation is now decomposed additively into elastic and plastic parts according to p ed dd+= (5.2.6) The elastic response is then ()p e dd Cσ −=∇:σ (5.2.7) The yield condition is ()0 ,=ασf (5.2.8) where α represents any other variable(s) besides σ upon which f depends. The flow rule is ()ασG d ,λ&=p (5.2.9) Section 5.2 Solid Mechanics Part III Kelly 362The tensor G is usually expressed in the form ()σασ G ∂∂= /,g , where g is the plastic potential; λ is the plastic multiplier. For associative flow-rules, gf=. The variables α are assumed to follow an evolution law of the form ()ασAα ,λ&&= (5.2.10) The loading and unloading conditions may be expressed as 0 ,0 ,0 =≤≥ f fλ λ & & (5.2.11) These are known as the Kuhn-Tucker conditions . The first of these states that the plastic multiplier rate is non-ne gative; for plastic loading, 0 >λ& , otherwise 0 =λ& . The second states that the stress must lie on or inside the yield surface. The last condition assures that the stress state remains on the yield surface during plastic flow. This last condition can also be expressed in rate form, 0=f& . This is known as the consistency condition : 0 : : =∂∂+∂∂= αασσ&&&f ff (5.2.12) Isotropic Materials From the objectivity requirement, the yield function ()ασ,f in Eqn. 5.2.8 must be an objective scalar function of σ. This implies that f must be a function only of the invariants of σ. Thus the form ()σf necessarily represents the yield function of an isotropic material. For anisotropic materials, one must use a different stress measure, for example the yield function could be expressed in term of the PK2 stress, () 0=Sf . Consider then an isotropic material, with ()3 2 1,,IIIff= , where iI are the invariants of σ. Then it can be shown that { ▲Problem 1} σσσσ∂∂=∂∂ f f (5.2.13) that is, the tensors σ and σ∂∂/f are coaxial. Consider now a formulation in terms of the Jaumann stress-rate. From 5.2.13, it follows that {▲Problem 2} J f f∇ ∂∂=∂∂σσσσ: :& (5.2.14) The consistency condition 5.2.12 can now be expressed in terms of the Jaumann stress rate: Section 5.2 Solid Mechanics Part III Kelly 3630 : : =∂∂+∂∂=∇αασσ& &f ffJ (5.2.15) The plastic multiplier can now be solved for, using the hypoelastic relation 5.2.7, the flow rule 5.2.9, the evolution equation 5.2.10 and 5.2.15 { ▲Problem 3}: G CσAαd Cσ : : :: : J eJ e f ff σσ λ ∂∂+∂∂−∂∂ =& (5.2.16) Substituting this expression into the flow rule 5.2.9 and using the hypoelastic relation 5.2.7 then leads to { ▲Problem 4} () G CσAαCσG C C C d Cσ : : :: : ,: J eJ eJ e J eJ J J f ff σσ σ σ σ σ ∂∂+∂∂−⎟ ⎠⎞⎜ ⎝⎛ ∂∂⊗ −= =∇ (5.2.17) The tensor JσC is the elasto-plastic tangent modulus . It consists of an elastic component J eσC and a component which results from plastic flow. When the flow is associative, so that σ G∂∂= /f , the elasto-plastic modulus possesses the major symmetries. The plasticity equations can also be based upon other stress-rates; for example, the Truesdell stress-rate. For th e same reasons as discussed earlier, although JσC has the major symmetries for associative flow rules, the Truesdell modulus will not. If, however, the equations 5.2.6-17 are formulated in terms of the Kirchhoff stress, then the Truesdell modulus will possess the major sy mmetries (in fact, all the relevant relations above are valid for the Kirchhoff stress, with σ simply being replaced by τ). Note that, if the elastic strains are small and plastic de formations are isochoric (volume preserving), then 1≈J and στ≈. Small Strains In the case of small strains, in the above relations one replaces d with ε& and decomposes the small strain rate according to p eεεε&&&+= . The stress rate is the time derivative of the Cauchy stress, since objective stress-rates are not now a consideration. The hypoelastic relation is ()pεεCσ &&&−=: and the remaining relations follow, for example the elasto- plastic tangent modulus is given by 5.2.17b with σ eC replaced with the small-strain elastic modulus tensor C. The small-strain formulation is valid for anisotropic elastic moduli. 5.2.3 J 2 Flow Theory In the J2 Flow Theory, one assumes the material is isotropic, plastic flow is independent of the hydrostatic pressure, plastic flow is incompressible and the yield surface used is the Section 5.2 Solid Mechanics Part III Kelly 364Von Mises yield surface. An effective stress is used to generalise the uniaxial behaviour to multiaxial stress states. First to be examined is the isotropic hardening model. Isotropic Hardening It is useful to express the equations 5.2.6-17 now in terms of the Kirchhoff stress and the Jaumann stress-rate. In that case, one again has Eqn. 5.2.6, p ed dd+= , with the hypoelastic Eqn. 5.2.7 now reading ()p J eJdd Cτ −=∇:τ (5.2.18) The Von Mises yield function can be expressed as () 0 : 3 ,23 2 =−=−= Y Y J f ss ατ (5.2.19) where 2J is the second invariant of the deviatoric Kirchhoff stress and τ sdev= . The term ss:23 acts as an effective stress σˆ. The (associative) flow rule 5.2.9 can be expressed as (see the Appendix to this section for details of the differentiation involved here) () () sss ττGτG d : 23, 23=∂∂= =fpλ& (5.2.20) The only variable α necessary in the model is the scalar accumulated effective plastic strain: p p p p p p pd d d dt d εε:32ˆ ,ˆ ˆ ˆ ,ˆ = == ≡ ∫∫ε εεεεα& (5.2.21) The proposed evolution law for α is simply (Eqn. 5.2.10 with the function 1=A ) ()pελα&&& ˆ== (5.2.22) With the definition of the plastic modulus being ()() pp p ddYHεεεˆˆˆ= (5.2.23) the consistency condition 5.2.12 is, from 5.2.19, () 0 ˆˆ : = −∂∂=p pHff εε&&&ττ (5.2.24) Section 5.2 Solid Mechanics Part III Kelly 365The plastic multiplier rate is now given by Eqn. 5.2.16, ()τCτd Cτ ∂∂ ∂∂+∂∂ =f fHf J epJ e : : ˆ: : ττ ελ& (5.2.25) Finally, the elasto-plastic tangent modulus, Eqn. 5.2.17, is ()τCτCττC C C d Cτ ∂∂ ∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂⊗⎟ ⎠⎞⎜ ⎝⎛ ∂∂ −= =∇ f fHf f J epJ eJ e J eJ J J : : ˆ: : ,: ττ τ τ τ τ ε (5.2.26) It is useful to decompose the response into volumetric and deviatoric components. First express the elastic moduli as in Eqns. 4.1.19, ⎥⎦⎤ ⎢⎣⎡⊗−+⊗= II II C312 IμκτJ e (5.2.27) where n m j ijn im e e e e ⊗⊗⊗ =δδ I . Note then that { ▲Problem 5} μ μτ τ3 : : , 2 : =∂∂ ∂∂ ∂∂=∂∂ τCτττCf f f fJ eJ e (5.2.28) so that the tangent modulus 5.2.26b can be expressed as () ττII II C∂∂⊗∂∂ +−⎥⎦⎤ ⎢⎣⎡⊗−+⊗=f f HJ μμμκτ 3/ 13/4 312 I (5.2.29) Kinematic Hardening To accommodate kinematic hardening, one introduces as another hardening parameter the back stress α. The yield condition 5.2.8 is now (in terms of the Kirchhoff stress) () ()()()0 ˆ : ˆ,,23=−−−=p pY f ε ε αsαs ατ (5.2.30) The flow rule is () ()() ()()αsαsαs τατGατG d −−−=∂∂= = : 23, ,, 23fpλ& (5.2.31) The evolution equation for the back-stress is, for example using a linear hardening rule and the Jaumann stress-rate, Section 5.2 Solid Mechanics Part III Kelly 366()G dα λ&c cp J==∇ (5.2.32) Note that the use of the Jaumann rate in the back-stress evolution law can lead to physically unreasonable stress oscillations in simple shear fo r large deformations (Nagtegaal and DeJong, 1981). However, this behaviour is not significant provided the elastic strains are not too large. A formulation based upon the Green-Naghdi strain can eliminate such stress oscillations (Johnson and Bammann, 1984). 5.2.4 Drucker – Prager Model The yield criterion for a Drucker-Prager material is a modification of the Von Mises function so as to incorporate a plastic response due to a hydrostatic pressure: () 0 3 ,1 2 =−+= YI J f α ασ (5.2.33) where Iσσ : tr1==I . The associated flow rule is then () () I sss σσGσG d α λ + =∂∂= = : 23, 23fp& (5.2.34) where σ sdev= . A suitable non-associative flow-rule might be () ()2 233 , : 23, J ggp= =∂∂= = sss σσGσG dλ& (5.2.35) 5.2.5 Elastic Moduli and Objectivity It was mentioned above that objectivity re quires that a yield function of the form ()σf necessarily represents an isotropic material. Objectivity also places restrictions on the elastic moduli. First, consider a constant modulus tensor J eτC. Objectivity of a constitutive equation requires that ()()* *:e J eJd Cττ=∇ (5.2.36) or ()T T: QQd C Q Qτe J eJ τ=∇ (5.2.37) so, using the index notation, () () [ ]()()kle pqrsJ e ls rk qj pi ijJQQQQ d C ττ=∇ (5.2.38) In order that objectivity be satisfied, one must have Section 5.2 Solid Mechanics Part III Kelly 367() ()pqrsJ e ls rk qj pi ijklJ e QQQQτ τC C= (5.2.39) which shows that, if the modulus is constant, it must be isotropic . If one requires an anisotropic elastic modulus, one must use a formulation based on a configuration other than the sp atial configuration, as discus sed in the next sub-section. 5.2.6 Corotational Stress Formulation The models discussed thus far have two drawbacks: 1. the yield function must be an isotropic function of the stress 2. if the elastic moduli are constant, then the moduli must be isotropic To overcome these restrictions to isotropic ma terials, one can formulate a plasticity model in terms of, for example, the corotational stress. The corotational stress is defined by Eqn. 3.5.12, σRRσT U= (5.2.40) Using a formulation based on the Kirchhoff st ress, the corotational Kirchhoff stress is τRRτT U= (5.2.41) Recall from §3.5.3 that the PK2 stress S is the pull-back of the Kirchhoff stress τ, ()T 1 # 1 *−− −= = τFFτ Sχ ; the corotational stress Uτ is the pull-back of τ but with respect to R and not F: ()T 1 # 1 * U−− =−= = τR Rτ τRFχ . Define now the corotational rate of deformation: dRR dT U= (5.2.42) with the decomposition p e U U U d d d+= (5.2.43) Note that the corotational stress and rate of deformation are insensitive to rigid body rotations Q to the current configuration: ()()()()() ()()()()()UT T *T * UUT T *T * U d QR QdQ QR dRR dτ QR QQτ QRτRRτ = = == = = (5.2.44) The corotational stress-rate is Section 5.2 Solid Mechanics Part III Kelly 368RτRRτRτRRτRRτ & & & &T T T T U ++==⋅ (5.2.45) A rigid body rotation Q to the current configuration also results in { ▲Problem 6} ()U* Uττ&&= (5.2.46) With the corotational stress-rate objective, the elastic response given by e e U U U :d Cττ=& (5.2.47) Note that rigid body rotations to the current configuration leave Uτ& and e Ud unchanged and so τ eUC remains unchanged also. Thus, unlike the isotropic restriction 5.2.39, there is no objectivity restriction here on the form of τ eUC and so τ eUC can represent in general an anisotropic response. Another way of looking at this is as follows: for an anisotropic material and using a fixed coordinate system, the components of an elastic modulus tensor C will in general change as the material rotates. With the corotational formulation, however, the axes rotate with the material, and so material rotation has no effect on τ eUC. The remainder of the formulation follows as before, for example the yield condition would be () 0 ,U U=ατf and the flow rule ()U U U ,ατG dλ&=e. The evolution law for the Uα variable(s) is ()U U U ,α Aατλ&&= . Again, the scalar yield function may now represent in general anisotropic material behaviour . The loading and unloading conditions are again given by 5.2.11. The plastic multiplier rate is now (see 5.2.16) G CτAαd Cτ : : :: : U U UU U U ττ λ ee f ff ∂∂+∂∂−∂∂ =& (5.2.48) and () G CτAαCτG C C C d Cτ : : :: : ,: Ue U UUe UUe Ue U U U ττ τ τ τ τ ∂∂+∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⊗ −= =∇ f ff J (5.2.49) 5.2.7 Problems 1. The derivatives of the invariants of the stress tensor are (see Eqn. 1.11.33) Section 5.2 Solid Mechanics Part III Kelly 3691 3 2 12 3 12 1I I II, II,I−=+−=∂∂−=∂∂=∂∂σ IσσσσIσIσ Use these relations to show that, for the yield function ()3 2 1,,IIIff= , the tensors σ and σ∂∂/f are coaxial. 2. Use the fact that σ and σ∂∂/f are coaxial, and Eqn. 1.9.3h, ()()()B AC CAB BCA : : :T T= = , to show that J f f∇ ∂∂=∂∂σσσσ: :& where σw wσσσ +−=∇&J. 3. Derive Eqn. 5.2.16 4. Use the index notation to verify that () ()()dCA GC GdCAC :: : :::: ⊗= where GdA,, are second order tensors and C is a fourth-order tensor. Hence derive the elasto-plastic modulus 5.2.17b. 5. Use the relations 1.9.62 and 1.9.64 to show that, for an arbitrary tensor A, () () A A A A CAA IA A C dev: dev2 tr : :dev2 tr : 2μ κμ κ ττ + =+= J eJ e with J eτC given by 5.2.27. Hence, using 1.9.31, show that, for a deviatoric tensor A, AA A CAA A C : 2 : :2 : μμ ττ == J eJ e Finally, use 5.2.20, sss τ : 23 23=∂∂f to derive Eqns. 5.2.28, μτ3 : :=∂∂ ∂∂ τCτf fJ e 6. The application of a rigid body rotation Q to the current configuration results in a change to the corotational stress rate 5.2.45, () ()() () () () ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛=+ + = ⋅ ⋅ ⋅ QR QQτ QR QR QQτ QR QR QQτ QRRτR RτR RτRτ T T T T TT* * *T * * *T * * *T * U& & && Use the relation 0QQQQQQ =+=⋅ & & T T T to show that ()UT * U ττRRτ & & ==⋅ . 5.2.8 Appendix to §5.2 Differentiation of the Von Mises Yield Function The Von Mises yield criterion is 02=−= k J f where ijijss J21 2= . Using the product rule of differentiation, in index notation: Section 5.2 Solid Mechanics Part III Kelly 370() () () () 231313131212121 2 2 222222121 1 21 Js ssssss ssssssssssss ssss ssss J ij pq pqijpq pqmmij ijmnij nj mi pq pqmnmnkjki nj mi pq pqmnijmn kk mn pq pqmnijmn mn pq pqijmn mn pq pq ijmn mn ij = =−=− =− =∂−∂=∂∂=∂∂=∂∂=∂∂ δδδδδδδδδδσδσσσσ σ σ (5.2.50) In tensor notation: () ()() () () 2313131212121 2 2 :2:2tr: :2tr: :2: : 21: 21 :1 21 : JJ s sssssIs sII sssσIσσ sssσss ssσss ss σss σ ==−=⊗− =∂−∂=∂∂=∂∂=∂∂=∂∂ I (5.2.51) Section 5.3 Solid Mechanics Part III Kelly 3715.3 Plasticity II: The Multip licative Decomposition of F Whereas the hypoelastic-plastic formulation involved an additive decomposition (of the deformation rate), the hyperelastic-plastic formulation involves a multiplicative decomposition (of the deformation gradient). 5.3.1 The Multiplicativ e Decomposition of F The key feature of the hyperelastic-plastic formulation is the multiplicative decomposition of the deformation gradient according to p eFFF= (5.3.1) where eF and pF are, respectively, the elastic part of the deformation gradient and the plastic part of the deformation gradient . pF, brings line elements Xd from the undeformed configuration to an intermediate configuration XFx d dp= and then eF brings them to their current location, Fig. 5.3.1. XFFxFx d d dp e e== . (5.3.2) Figure 5.3.1: Decomposition of the deformation gradient The idea now is to relate the current state of stress to the elastic part of the deformation: ()eFfσ= (5.3.3) where 0)(=If (5.3.4) Xd xdpF FeFintermediate configuration xd Section 5.3 Solid Mechanics Part III Kelly 372In other words, the intermediate configuration is stress-free. At any stage of the deformation, one can imagine unloading the body instantaneously, arriving at the stress- free intermediate configuration. Thus it appe ars that the intermediate configuration has a clear physical interpretation – a configuration which encompasses any permanent deformation after elastic unloading. Unfortunate ly, there are difficulties associated with the intermediate configuration which makes it a more abstract concept than this. For example, consider a stressed element in the current configuration. Unload the element so that it is stress free. According to 5.3.3, this is accomplished through ()1−eF back to the intermediate configuration. Since each elem ent in the current configuration can undergo this process independently, there is no reason that the individual elements will be compatible in the intermediate configuration, and so the intermediate configuration is in general an incompatible configuration. Thus, although mappings of line elements between the various configurations are given by Eqn. 5.3.2, the eF and pF in 5.3.2 cannot necessarily be expressed as gradients of position vectors; they can be expressed as gradients only in the special case where the deformation does lead to a compatible intermediate configuration. When the material is unloaded from the continuous current configuration (as in a thought experiment), the body moves from a Euclidean space to the non-Euclidean space of the intermediate configuration. The elastic pa rt of the deformation gradient restores compatibility, bringing the body back into the Euclidean space of the current configuration. It is thus best to consider the decomposition 5.3.1 as merely a mathematical means by which the constitutive response of a material can be conveniently described. It is not to say that a material actually does undergo a “plastic” deformation to the intermediate configuration and then subsequently undergoe s an “elastic” deformation to reach the current configuration. Rather, the intermed iate configuration is a useful abstract configuration. Issues: 1. Conceptual de-straining 2. no unique intermediate configuration 3. if the yield surface has shifted, unloading to zero stress might necessitate plastic flow The decomposition 5.3.1 was introduced into finite-strain elastoplasticity by Lee and Liu in (1967). Since then has followed theory development, explanation and criticism. See, for example, Lee (1969), Hahn (1974), Ne mat-Nasser (1979), Lee and McMeeking (1980), Lee (1981), Menat-Nasser (1982), Simon and Ortiz (1985), 5.3.2 Strain Tensors In terms of the elastic and plastic parts of the deformation gradient, the Right Cauchy- Green strain C and Left Cauchy-Green strain b are Section 5.3 Solid Mechanics Part III Kelly 373()() () ()T TTT TT e p p ep e e p F FFF FFbFF F F FFC ==== (5.3.5) The Euler-Lagrange and Euler-Almansi strains are ()() ()( ) () () () () ()()1 1 T T1T T 21 2121 21 −− − −−−=−=− =−= e p p ep e e p F F F FI bI eI FF F F IC E (5.3.6) Define now elastic and plastic Right Cauchy-Green strains and elastic and plastic Left Cauchy-Green strains: () () () ()T TT T ,, p p p p p pe e e e e e FF b F F CFF b F F C = == = (5.3.7) and elastic and plastic Green-Lagrange strains and elastic and plastic Euler-Almansi strains: () ()() () ()()11 21,2121,21 −− −= −=−= −= p p p pe e e e bI e I C EbI e I C E (5.3.8) The overbars in Eqns. 5.3.7-8 are used to signi fy that the tensor is associated with the intermediate configuration. The action of these twelve tensors on line elements in the reference, intermediate and current configurations can be summarised as { ▲Problem 1} )2( )1( )2( )1()2( )1( )2( )1()2( )1( )2( )1( xCx x xXCX x xXCX x x d d d dd d d dd d d d ep =⋅=⋅=⋅ , () ())2(1)1( )2( )1()2(1)1( )2( )1()2( 1 )1( )2( )1( x bx x xx bx X Xxbx X X d d d dd d d dd d d d ep −−− =⋅=⋅=⋅ (5.3.9) and {▲Problem 2} () () () xEx x xX XE X xX XE X x d d d dd d d dd d d d ep =−=−=− 2 22 22 2 212121 , () () () xxe x xxex X xxxe X x d d d dd d d ddd d d ep =−=−=− 2 22 22 2 212121 (5.3.10) They are illustrated in Fig. 5.3.2. Note that p e p eEEECCC ≠ ≠ ,. Section 5.3 Solid Mechanics Part III Kelly 374 Figure 5.3.2: Strain Tensors and associated configurations Define now a new strain measure E, which can be called the intermediate strain , associated with the intermediate configuration, through ()()1 T − −=p pFE F E (5.3.11) This is the push-forward for covariant components of the Green-Lagrange strain from the reference configuration to the intermed iate configuration (see Eqn. 2.10.18a): ()b irE E→=*χ (5.3.12) Using 5.3.7-8, this strain can be decomposed into an elastic component and a plastic component according to { ▲Problem 3} p ee EE+= (5.3.13) This compares with the push-forward of the Green-Lagrange strain from the reference configuration to the current configuration, which equals the Euler-Almansi strain, ()1 T *−− →= = EFF E eb crχ . The push-forward of the strain E to the current configuration is the Euler-Almansi strain (compare with 5.3.11): ()()()1 T *− − →= =e e ci FE F E eχ (5.3.14) which can be verified using Eqns. 5.3.11, 5.3.1 and the relation 1 T−−= EFFe . Finally, introduce the tensors Xd xdintermediate configuration p pCE, p pbe, e eCE,e ebe, be, CE, Section 5.3 Solid Mechanics Part III Kelly 375() ()()11 2121 −−− =−=−=−= b b ee eCC EE E e e pp p e (5.3.15) In each configuration, there is now a set of el astic and plastic strain tensors, with their sum equaling a total strain. These sets are related by push-forward/pull-back operations {▲Problem 4}: ()()() ()() () ()() ()1 T *1 T *1 T * − − →− − →− − → = == == = p p pbp irpp e pbe irep p b ir FE F E eFE F E EFE F E E χχχ , ()() ()() () ()p p pbp irpp e pbe irep p b ir Fe F e EFE F E EFE F E E T1 *T1 *T1 * = == == = − ←− ←− ← χχχ (5.3.16a) () () ()1 T *1 T *1 T * − − →− − →−− → = == == = FEF E eFEF E eEFF E e pbp crpebe creb cr χχχ , () () () FeF e EFeF e EeFF e E pbp crpebe creb cr T 1 *T 1 *T 1 * = == == = − ←− ←− ← χχχ (5.3.16b) ()()() ()() () () () ()1 T *1 T *1 T * − − →− − →− − → = == == = e p ebp cipe e ebe ciee e b ci Fe F e eFE F E eFE F E e χχχ , ()() () () () ()e p ebp cipe e ebe ciee e b ci Fe F e eFe F e EeF F e E T1 *T1 *T1 * = == == = − ←− ←− ← χχχ (5.3.16c) These nine strain tensors and their rela tionships through push-forward (pull-back) operations are illustrated in Fig. 5.3.3. Figure 5.3.3: Push-Forward operations on strain tensors The action of the three new tensors introduced above, p eeEE ,, , analogous to 5.3.10, is given by intermediate configuration p eE EE+=p ee EE+= p eeee+=()1 T− −•F F()()()1 T − −•p pF F ()()()1 T − −•e eF F Section 5.3 Solid Mechanics Part III Kelly 376() () () xEx X xx xe X xX XE x x d d d dd d d dd d d d pe =−=−=− 2 22 22 2 212121 (5.3.17) and xxe xEx X XEx xe xex X XExxe xEx XXE d d d d d dd d d d d ddd d d d d e e ep p p = == === (5.3.18) The various tensors are summarised in Table 5.3.1. Configuration Green-Lagrange Euler-Almansi Other Reference pEE, p eEE E−= Intermediate eE pe p ee EE+= Current eee, e pee e−= Table 5.3.1: Strain Tensors for the Multiplicative Decomposition 5.3.3 Strain-Rate Tensors The velocity gradient l is 1−=FFl& (5.3.19) In terms of the elastic and plastic deformation gradients { ▲Problem 5}, ()()()1 1 1 −− −+ =e p p e e eF FFF FFl & & (5.3.20) The rate of deformation and spin are () ()T T 21skew ,21sym ll l w ll l d −== +== (5.3.21) One can now define elastic and plastic velo city gradients, and decompose the total velocity gradient according to p elll+= (5.3.22) There are, however, a number of different possible definitions of el and pl. The two most prevalent definitions are discussed in what follows. Section 5.3 Solid Mechanics Part III Kelly 377Elastic and Plastic Velocity Gradients 1 The first term in Eqn. 5.3.20 is analogous to Eqn. 5.3.19, 1−=FFl& . It only involves elastic deformations and hence can be called the elastic part of the velocity gradient, el. The second term is denoted by pl and is called the plastic part of the velocity gradient (although it involves both elastic and plastic deformations). Thus one has () () ()1 11 −−− == e p p e pe e e F FFF lFFl && (5.3.23) The rate of deformation and spin are then ()[] ()() [ ] ()[] () ()[ ]1 1 11 1 1 skew skewsym sym −− −−− − + =+ = e p p e e ee p p e e e F FFF FF wF FFF FF d & && & (5.3.24) With the corresponding elastic and plastic parts: ()() ()() ()() ()()T TT T 21,2121,21 p p p p p pe e e e e e l l w l l dl l w l l d −= +=−= += (5.3.25) Elastic and Plastic Velocity Gradients 2 The rate of deformation is the Lie derivative of the Euler-Almansi strain, () ()[] () d FEF E e e = = =⎟ ⎠⎞⎜ ⎝⎛=−− →− ← →1 T *1 * * L & &b crb b cr crb vdtdχ χχ (5.3.26) In the same way, define the elastic and pl astic rates of deformation to be the Lie derivatives of the tensors ee and pe respectively: () ()[]() () ()[] ()1 T *1 * *1 T *1 * * LL − − →− ← →− − →− ← → = =⎟ ⎠⎞⎜ ⎝⎛=≡= =⎟ ⎠⎞⎜ ⎝⎛=≡ FEF E e e dFEF E e e d pbp crb bp cr crp b vpebe crb be cr cre b ve dtddtd & && & χ χχχ χχ (5.3.27) One can also define rates of deformation in the intermediate configuration by taking Lie derivatives of the strain tensors in the intermediate configuration: () ()[]()() ()1 T *1 * * L− − →− ← → = =⎟ ⎠⎞⎜ ⎝⎛=≡p pb irb b ir irb vdtdFE F E E E d & &χ χχ (5.3.28) and Section 5.3 Solid Mechanics Part III Kelly 378 () ()[]()() () () ()[] ()() ()1 T *1 * *1 T *1 * * LL − − →− ← →− − →− ← → = =⎟ ⎠⎞⎜ ⎝⎛=≡= =⎟ ⎠⎞⎜ ⎝⎛=≡ p p pbp crb bp ir irp b vpp e pbe crb be ir ire b ve dtddtd FE F E e e dFE F E E E d & && & χ χχχ χχ (5.3.29) Note that p ed dd+= (5.3.30) These rates of deformations as Lie derivatives are illustrated in Fig. 5.3.4. Figure 5.3.4: Rate of Deformation and Lie Derivatives Note also that the rates of deformations in the intermediate and current configurations are related through the push-forward/pull-back operations { ▲Problem 6} ()()() () () () ()() ()1 T *1 T *1 T * − − →− − →− − → = == == = e p ebp cipe e ebe ciee e b ci Fd F d dFd F d dFd F d d χχχ (5.3.31) Finally, note that direct relationships between d and e&, and between d and E&. From 2.5.18, elelde −−=T& (5.3.32) intermediate configuration () () ()p b vpe b veb v E dE dE d LLL === p eddd+=()()()1 T − −•p pF F ()1 T− −•F F p eeee+=p ee EE+= p eE EE+= p eE EE&&&+=p eddd+= () () ()p b vpe b veb v e de de d LLL === Section 5.3 Solid Mechanics Part III Kelly 379Similarly, analogous to 5.3.19, introduce a plastic velocity gradient, ()1−=p p pFF l& (5.3.33) so that {▲Problem 7} ()p plEEldE −−=T & (5.3.34) Similarly { ▲Problem 8}, le el d eleel d e p p p pe e e e −−=−−= TT &&, () ()pp p p p ppe e p e e le el d elE El d E −−=− −= TT && (5.3.35) These can all be expressed as Cotter-Rivlin rates of the strains: le el e dleele deleled p p p pe e e e ++=++=++= TTT &&& , () () ()pp p p p ppe e p e ep p le el e dlE El E dlEEl Ed ++=+ +=++= TTT &&& (5.3.36) Push forward now the plastic velocity gr adient to the current configuration: () () ( ) ()()()1 T 1 1 T *− − − − − → = =p e p e e p p ebp ci FFF F F FF F l & & χ (5.3.33) Section 5.3 Solid Mechanics Part III Kelly 380 Alternatively, introduce a velocity gradient l which is the pull-back for covariant components of the spatial velocity gradient. For covariant components, ()()e e b ci lF F l lT1 *= =− ←χ (5.3.26) so that () xxl xlFxFxlF Fx xlx dd d d d d dde e e e= = =T (5.3.27) Here, so the rate of change of a line element xd in the intermediate configuration is () xl x d dp=. Section 5.3 Solid Mechanics Part III Kelly 381 The velocity gradients p elll,, are defined on the current configuration. These can be pulled back to the intermediate configurati on. The pull-backs to be used are for mixed components ()()i jpi je i j l ll⋅ ⋅⋅ , , (see the Appendix to this section, §5.3.3, which gives the component forms of the tensors used in this section). These pull-backs are (see Eqn. 2.10.20c): ()() () ( ) () ()e p e p cipe e e e ciee e ci Fl F l lFl F l llF F l l 1 \1 *1 \1 *1 \ 1 * −− ←−− ←−− ← = == == = χχχ (5.3.20) so, using 5.3.19, p elll+= (5.3.21) where () ()11 −− == p p pe e e FF lF F l && (5.3.22) the second of these being 5.3.14. The push-forwards for the velocity gradients defined on the intermediate configuration, the inverse of the operations 5.3.20, are ()() ()() ()()1 \ *1 \ *1 \ * − →− →− → = == == = e pe p cipe ee e ciee e ci FlF l lFlF l lFlF l l χχχ (5.3.23) and 5.3.18-19 are recovered. These velocity gradients are summarised in Fig. 5.3.5. Section 5.3 Solid Mechanics Part III Kelly 382 Figure 5.3.5: Velocity Gradients Rate of Deformation and Spin The rate of deformation and spin are given by the symmetric and skew-symmetric parts of the velocity gradient. From 5.3.17, ()[]()() [ ] ()[] () ()[]ae p p e ae ese p p e se e 1 1 11 1 1 −− −− − − + =+ = F FFF FF wF FFF FFd & && & (5.3.?) The velocity gradients can be decomposed into their symmetric and skew-symmetric parts, that is, the rate of deformation and spin: ()() ()() ()() ()()T TT T 21,2121,21 p p p p p pe e e e e e l l w l l dl l w l l d −= +=−= += (5.3.24) If the velocity gradients have the mixed components ()()i jpi je i j l ll⋅ ⋅⋅ , , then, in order that the rate of deformation and spin tensors have covariant com ponents, one should include the metric tensor Ig= in 5.3.24 as follows (see the Appendix to this section, §5.3.3): ()() ()() ()() ()() gl gl w gl gl dgl gl w gl gl d T TT T 21,2121,21 p p p p p pe e e e e e −= +=−= += (5.3.25) For two tensors ba,, the push forward and pull-back (for covariant components) of their contraction can be expressed as intermediate configuration p elll+=()()1−•e eF Fp elll+= Section 5.3 Solid Mechanics Part III Kelly 383 () ()()()() () ( ) () () () () ( ) () () () () () ()\ 1 *1 *1 T1 */ 1 *T T T T 1 *\ * *1 1 T*/ *1 T T T 1 T * b a bFFaFFb a bFF aFF abFF abb a FbF aFFb a bFF aFF abFF ab − − −− − − −− −−−− − − − = == === == = = χχχχ χχχχχ χ bb bbb b (5.3.26) Using the relations () () () I gFF gI gFF gC gFF g = ===== − −− −− T T / 1 *1 \ 1 *T 1 * χχχb (5.3.27) and Eqns. 5.3.20, the pull-back of the rate of deformation and spin tensors of 5.3.25 are () () () () ()() () () ()() () ()() ()e ee e e ee e e eb ci cib cib cib ci CllCC Fl FlCC Fl FlCg l l gglgl d d TTT1TTT1 */T1 *\ 1 *1 *T 1 *1 * 2121212121 +=⎟⎠⎞⎜⎝⎛+=+=+ =+ = = −−− ←− − ←− ←− ←− ← χχ χχχ χ (5.3.28) Similar calculations lead to, in summary, () () ()() ()() ()() ()()e p pe p e p pe pe e ee e e e ee ee e e e Cl lC w Cl lC dCl lC w Cl lC dCllC w CllC d T TT TT T 21,2121,2121,21 −= +=−= +=−= += (5.3.29) The tensor eC plays the role of the metric tensor on the intermediate configuration and appears here so as to ensure that the mixe d components of the velocity gradients become covariant components, consistent with the covariant components of d and w. Note that p e p ew w w d dd += += , (5.3.30) Section 5.3 Solid Mechanics Part III Kelly 3845.3.4 Stress Tensors The pull-back of the Kirchhoff stress τ is the PK2 stress S: ()T 1 # 1 *−− − ←= = τFFτ Scrχ (5.3.) Introduce a new stress tensor S to be the analogous pull-back of τ to the intermediate configuration(see Eqn. 2.10.20b), ()()()T 1 # 1 *− −− ←= =e e ci Fτ Fτ Sχ (5.3.) or, equivalently, the push-forward of S to the intermediate configuration (see Eqn. 2.10.18b), ()()()T # *p p ir FSF S S = =→χ (5.3.) Stress Rates Define S& to be the material time derivative of S: S Sdtd=& (5.3.) This is equivalent to the pull-back of the Kirchhoff stress to the intermediate configuration: () ()[] () ()T# *# # 1 * *T# ,L e ecici cie e civ dtd FSFSτlττlττ &&& ==⎟ ⎠⎞⎜ ⎝⎛=−−= →− ← →← χχχ (5.3.) Another useful Lie derivative is that of S, where the pull back is to the reference configuration: () ()[] () ()T# *# #1 * *T# ,L p pirir irp p irv dtd FSFSSlSSlSS &&& ==⎟ ⎠⎞⎜ ⎝⎛=−−= →− ← →← χχχ (5.3.) Section 5.3 Solid Mechanics Part III Kelly 385These are related through ()T # ,## , * L L FSFτ S &==→ ← → crv irv ciχ (5.3.) 5.3.5 Problems 1. Use Eqns. 5.3.2 to show that the action of the Right Cauchy-Green strain tensors e pCCC ,, and the Left Cauchy-Green strain tensors e pbbb ,, on line elements in the reference, intermediate and current configurations are given by Eqns. 5.3.9. 2. Show that the action of the Green-Lagrange strain tensors e pEEE ,, and the Euler- Almansi strain tensors e peee ,, on line elements in the reference, intermediate and current configurations are given by Eqns. 5.3.10. 3. Derive the decomposition 5.3.13, p ee EE+= . 4. Verify all nine push-forward/pull-back operations, 5.3.16a-c. 5. Use Eqn. 2.5.4, 1−=FFL& , and 5.3.1, to show that ()( ) ( )1 1 1 − − −+ =e p p e e eF FFF FFl & & 6. Use relations 5.3.26-29 to verify relations 5.3.31. 7. Use the relation 5.3.11, ()()1 T − −=p pFE F E , and Eqns. 2.5.5, i.e. ()()()( ) () ( )T T. T 1. 1,− − − −−= −=p p p p p pF l F l F F , to show that ()p plEEldE −−=T &. 5.3.6 Appendix to §5.3: Convected Coordinates Deformation Gradients Using a convected coordinate formulation, the line elements can be expressed as ii ii iid d d d d d g x g x G X Θ= Θ= Θ= , , (5.3.5) The deformation gradients are i ie i ip i i g g F G g F G gF ⊗= ⊗= ⊗= , , (5.3.6) and the inverse deformation gradients are Section 5.3 Solid Mechanics Part III Kelly 386() ()i ie i ip i i g g F g G F g G F ⊗= ⊗= ⊗=− −−1 11, , (5.3.6) Strain Tensors From these, the Cauchy-Green tensors are j iij e j iij p j iijj i ije j i ijp j i ij g G Gg g g g g b g g b g g bg g C G G C G G C ⊗= ⊗= ⊗=⊗= ⊗= ⊗= , ,, , (5.3.6) and their inverses are () () () ()j i ije j i ijp j i ijj iij e j iij p j iij g G Gg g g g g b g g b g g bg g C G G C G G C ⊗= ⊗= ⊗=⊗= ⊗= ⊗= − −−− −− 1 111 11 , ,, , (5.3.6) with the metric coefficients given by j i ij j i ij j i ijj i ij j i ij j i ij G g gG g g GG gg ggGG gg gg ⋅= ⋅= ⋅=⋅= ⋅= ⋅= , ,, , (5.3.6) The Green-Lagrange and Euler-Almansi strains are () () () () () ()j i ij ije j i ij ijp j i ij ijj i ij ije j i ij ijp j i ij ij g g G g G gg g G g G g g g e g g e g g eg g E G G E G G E ⊗−= ⊗−= ⊗−=⊗−= ⊗−= ⊗−= 21,21,2121,21,21 (5.3.6) Also, ()j i ij ijp eG g g g e EE ⊗−=+=21 (5.3.6) Strain-Rate Tensors The time rates of change of the Green-Lagrange strain and the plastic Green-Lagrange strain are j i ijp j i ij g g G G E G G E ⊗= ⊗= && && 21,21 (5.3.6) The spatial velocity gradient is j i ji j ii jxvl g g g g l ⊗∂∂=⊗=⋅ (5.3.6) The rate of deformation is the symmetric part of l: Section 5.3 Solid Mechanics Part III Kelly 387 ()()i ji ji j l l d⋅ ⋅ ⋅+= +=21,21Tll d (5.3.6) For covariant components, ( )()ji ijm j imm j imm j im ij l l lg lg dg d +=+ ==⋅ ⋅ ⋅21 21 (5.3.6) When l is expressed in terms of mixed components, as in ???, one ends up with mixed components for the rate of deformation, ???. One can obtain the covariant components for d directly, by expressing d in the form () () ( ) ( ) ()() ()j i ji iji i mj m jj mm j ii l ll l g gg gg g g g g g glgl d ⊗+=⊗⊗+⊗ ⊗=+=⋅ ⋅ 2121 21T (5.3.6) where g is the metric tensor. Note that the rate of deformation and the plastic rate of deformation are the push-forwards of the corresponding Green-Lagrange strains, so () ()j i ijbp j i ijbg g g g E d g g E d ⊗= = ⊗== & & && 21,21 * * χ χ (5.3.6) or 1 T−−= FEFd& and ()()1 T − −=p p pFE F d & . Section 2.2 Solid Mechanics Part IV Kelly 572.2 Thermoelasticity In this section, thermoelasticity is considered. By definiti on, the constitutive relations for such a material depend only on the set of field variables { }θθGrad,,F . This general case will be considered toward the end of th is section. Simpler cases, isothermal {}F, small- strain {}ε, one-dimensional {}ε, will be considered first. In what follows, upper case letters will be used to denote the energy potentials per unit volume . As before, lower case denotes per unit mass so, for example, u Uρ= for internal energy. 2.2.1 Isothermal Small Strain Elasticity For isothermal deformations, the basic ther modynamic statement can be written as (see Eqn. 1.6.24) Φ+Ψ=+= &&&ρφψρεσ: (2.2.1) where ψρ&&=Ψ (rate of change of the free energy per unit volume) 0≥=Φρφ (dissipation per unit volume) In isothermal elasticity1, Φ= 0, and so one need only specify a free energy function. One-Dimensional Analysis Consider a simple one-dimensional extension of material. The free energy is a function of kinematic state variables. The only state variable is the single strain ε and the only stress is σ and hence )(εεσΨ=&& (2.2.2) i.e. εεεεσ & & dd )(Ψ= (2.2.3) so that, since ε& is arbitrary, εεσdd )(Ψ= (2.2.4) 1 this is discussed more rigorously in §2.2.2 below Section 2.2 Solid Mechanics Part IV Kelly 58which agrees with the general relation 1.6.33d. This is the stress-strain (constitutive) equation. In the context of elasticity th eory, the Helmholtz free energy function is the elastic strain energy functi on (Eqn. 2.2.4 is the same as Part I, Eqn. 5.4.9). The classical linear elastic material corre sponds to choosing a simple quadratic free energy function: ()2 21εε E=Ψ (2.2.5) so εσE= , εσ&&E= (2.2.6) A nonlinear example is () () εεεεεσσεεε εσεεεεε && & 2 3 2 13 32 2 1 04 33 22 1 0 3 2/)(4 3 2)( E E EE E E EE E E E ++=∂∂=+++=+++=Ψ (2.2.7) One should ensure that (at least (1)) (1) 0 )0(=σ (zero strain at zero stress) which implies that 00=E (2) () 0 /0≠=εεσdd (so one has a non-zero initial modulus) (3) () 0 /02 2==εεσd d (so that, with (2), the origin is a point of inflection, ensuring similar tensile and compressive strain behaviour) Note that Ψhas the dimensions of stress, so that the Ei also have the dimensions of stress. In the rate formulation : εσ&&D= , D is the instantaneous elastic modulus (or stiffness ) ded/σ or 2 2/εd dΨ . D is a constant in the linear model, but is, in general, a function of strain, as in the above example, where 2 3 2 1 3 2 εε E E ED ++= . Three-dimensional Analysis Generalising the above to three di mensions, the free energy function is )(ijεΨ )(εΨ (2.2.8) so that the basic thermomechanical identity becomes ij ijij ijij εεεεσ & && ∂Ψ∂=Ψ= )( εεεεσ & && : )( :ddΨ=Ψ= (2.2.9) Section 2.2 Solid Mechanics Part IV Kelly 59Since this again holds for all strain rates one can deduce that ijijεσ∂Ψ∂= εσddΨ= (2.2.10) Again, for the linear elastic material, the free energy function is assumed to be a quadratic function of the strain: klij ijklDεε21=Ψ εDε ::21=Ψ (2.2.11) so that kl mnklij ijmn kl mnklmnkl ij ijkl kl mnij ijklmnmn DD DD D εε εεεε εεεεσ =+ =∂∂+∂∂=∂Ψ∂= 21 2121 21 (2.2.12) provided the fourth or der stiffness tensor D possesses the major symmetry: klij ijkl D D= (2.2.13) (D satisfies the minor symmetries jikl ijkl D D= and ijlk ijkl D D= by virtue of the symmetry of the stress and strain.) Eqn. 2.1.12 is the stress-strain equation for an anisotropic linear elastic material. The rate form is formally kl ijkl ijDεσ & &= (2.2.14) For a nonlinear material Ψis no longer quadratic, and the instantaneous modulus tensor is a function of the strains. If one starts by proposing the ra te form of the elastic law, with a given modulus function ()εD , then in general one cannot integrate E qn. 2.2.14 to find a free energy function (although one can achieve this in one-dim ension). Such a model is termed hypoelastic . A model in which Ψdoes exist is called hyperelastic . Hyperelastic models automatically satisfy the laws of th ermomechanics, hypoelastic ones do not. Note that the strain energy must be positive, 0>Ψ . From 2.2.11, this implies that 0>klij ijklDεε 0 ::>εDε (2.2.15) for all strain tensors, that is, the stiffness tensor must be positive definite. Section 2.2 Solid Mechanics Part IV Kelly 60Isotropic Elasticity The theory of isotropic elasticity with sma ll strains is a special case of that for large strains discussed in Part III, §4.4. The theo ry discussed there applies here, only now with the left Cauchy Green strain replaced by the small strain tensor. Thus, the free energy function must be a function of a set of three invariants of the strain tensor ε, ),,(3 2 1 EEEΨ , which here are taken to be 3 32 21 trtrtr εεε = ===== kijkijijijii EEE εεεεεε (2.2.16) The stress-strain relati on is hence given by: ij ij ijijij E EE EE E ε ε εεσ ∂∂ ∂Ψ∂+∂∂ ∂Ψ∂+∂∂ ∂Ψ∂=∂Ψ∂= 3 32 21 1 (2.2.17) It is readily verified that { ▲Problem 1} kjik ijij ijij ij EEE εεεεεδε 3 /2 // 321 =∂∂=∂∂=∂∂ 2 321 3 /2 /I / εεεεε =∂∂=∂∂=∂∂ EEE (2.2.18) so that 2 3 2 13 2 ε ε IσE E E ∂Ψ∂+∂Ψ∂+∂Ψ∂= (2.2.19) which is a quadratic polynomial in the strain tensor. If the free energy function does not depend on the third strain invariant 3E, the tensor relation is actually linear, though it should be appreciated th at the coefficients kE∂Ψ∂/, k = 1,2,3, will in general depend in a nonlinear manner on the strain invariants. The classical linear model is recovered by choosing 22 1 21E Eμλ+=Ψ ()2 2 21tr tr εεμλ+ =Ψ (2.2.20) for then ij ij kk ij μεδλεσ 2+= ()ε Iεμ λ 2 trσ += (2.2.21) Section 2.2 Solid Mechanics Part IV Kelly 61The Gibbs Function and the Complementary Energy From section 1.3, the Helmholtz free energy and the Gibbs energy functions are related by a Legendre transformation. Formally, cons ider again the one-dimensional stress-strain relation 2.2.4: εεσdd )(Ψ= (2.2.22) To invert this expression a nd express the strain as a func tion of stress, first partially integrate the equation with respect to strain to obtain )( )(σεσε C+Ψ= (2.2.23) where )(σC is an arbitrary function of inte gration, dependent only on stress. Differentiating with respect to σ leads to σσεddC )(= (2.2.24) which is the inverse of the original equation. From Table 1.3.1, it is clear that the function C is the negative of the Gibbs energy functi on (per unit volume). In this context, C is the complementary energy discussed in Part I, §5.4. One says that the free energy is the dual of the complementary energy, and vice versa . For example, if 3εK=Ψ , 23εσ K= (2.2.25) then {▲Problem 2} 23)3/(2)( K K C σσ= . (2.2.26) This theory readily extends to the general three dimensional situation. The original stress strain equation is given by 2.2.10 so that )( )(ij ij ijij Cσεεσ +Ψ= )( )( : σεεσ C+Ψ= (2.2.27) and the strains are now ijij ijddC σσε)(= , σσεddC )(= (2.2.28) For the linear material, εDε ::21=Ψ , σ Dε :1−= and the dual Complementary function is C = σ Dσ : :1 21 −. The fact that D in the linear case is positive definite ensures that it is invertible. The inverse 1−D is called the elastic flexibility tensor C (see 2.2.11): Section 2.2 Solid Mechanics Part IV Kelly 62 kl ij ijklC C σσ21= σCσ ::21=C (2.2.29) 2.2.2 Large Strain Thermoelasticity In the general large strain thermoelastic case, the constitutive relations are of the form () () () () ()θθψψθθθ ∇=∇=∇=∇=∇= ,θ,,θ,,θ,,θ,,θ, FFFFqqFσσ ssuu (2.2.30) Here, for convenience, the material gradient of 2.1.13 has been repl aced with the spatial gradient θ∇ (through θ θ Grad1−=∇ F ). The dissipation inequality 1.6.20 is 01:1≥∇⋅−+−−= θρθρψθφ q dσ&&s (2.2.31) Using the relation Part III, 3.7.25, FσF dσ &: :T−= , this can be expressed as 01:T≤∇⋅+ −+−θθψρθρ q FσF& &&s (2.2.32) With ()⋅ ∇⋅∇∂∂+∂∂+∂∂= θθψθθψψψ && & FF: (2.2.33) one has ()01:T≤∇⋅+∇⋅∇∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂−−⎟ ⎠⎞⎜ ⎝⎛ ∂∂+⋅ −θθθθψρψρ θθψρ q FFσF & & s (2.2.34) For any given {}θθ∇,,F , this inequality must hold for any arbitrary values of θ&&,F and ⋅ ∇θ. Thus the coefficients must be zero: ()0 , ,T=∇∂∂ ∂∂=∂∂−=θψ ψρθψFFσ s (2.2.35) Section 2.2 Solid Mechanics Part IV Kelly 63The third of these implies that the free ener gy is independent of th e temperature gradient: ()θψψ ,F= (2.2.36) and so the entropy and stress are also ; further, the in ternal energy is θψθψ∂∂−=u (2.2.37) Stress-Strain Relation The stress-deformation relation 2.2.35b can be expressed in a number of different ways, for example, in terms of the PK1 stress T−=σF PJ and the PK2 stress T 1−−= FσF SJ (the first of these follows from the symmetry of the Cauchy stress) FF SFPFF FFσ ∂∂=∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂=∂∂= −ψρψρψρψρ 1 00T T (2.2.38) Introducing the free energy per unit reference volume, Ψ, these read as FFSFP FFσ∂Ψ∂=∂Ψ∂=∂Ψ∂=− − 1 T 1, , J (2.2.39) In the case of material symmetries, they take reduced forms as discussed in Part III, Chapter 4. For example, for isotropic materi als the Cauchy stress is a function of the left Cauchy-Green strain only: bbσ∂Ψ∂=−12J (2.2.40) Gibbs Relations One also has the useful Gibbs relations {▲Problem 3} Section 2.2 Solid Mechanics Part IV Kelly 64θρθρθρθρψ && &&& &&& &&& & s gs hs us − −=+ −=+ =− = PFPFFPFP :1:1:1:1 0000 (2.2.41) Isothermal Elasticity For isothermal materials, the fr ee energy becomes the strain energy W per unit reference volume, e.g. in terms of the PK1 stress, th e stress and stress power can be expressed as () ()FPFFF FFP && & : : , =∂∂=∂∂=WWW (2.2.42) Since there are no temperature gradients, the entropy inequality 2.2.34 becomes an equality and any process is completely reversible. For a reversible process, the dissipation is 0==s&θφ and so, as mentioned in §1.6.4, su ch a process is also isentropic. Adiabatic Elasticity For adiabatic conditions, the heat flux is zer o, so again the entropy inequality 2.2.34 becomes an equality and any process is reversible and isentropic. Working with the internal energy, from 2.2.41b, with its canonical variables, ),(s uu F= , one has FP FσF & & & :1:1 0T ρ ρ= =−u (2.2.43) The internal energy thus becomes the st rain energy and one again has 2.2.42. Energy Potentials and Canonical Variables Following on from §1.3, the thermodynamic potentials (per unit reference volume) in terms of canonical variables take the form () ()()()s H H GGs UU , ,, ,, ,, P P F F = = = Ψ=Ψ θ θ (2.2.44) The various relationships for these potentials are given in Table 2.2.1, which is a reproduction of Table 1.3.1 for finite strain reversible processes. The fourth column contains the Gibbs relations 2.2.41. Thermo- dynamic Symbol and Definition Rate-form relationship Maxwell relation Section 2.2 Solid Mechanics Part IV Kelly 65potential appropriate variables Internal energy ),(FSU S U &&& θ+= FP: FF P FFP ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂= SU SU ss θθ , Enthalpy ),(PSH FP:−=UH S H &&& θ+−= PF: PP F PPF ⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂∂= SH SH SS θθ , Helmholtz free energy ),(FθΨ S Uθ−=Ψ θ&&& S−=Ψ FP: FF P FFP ⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂=⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂−= θθ θθ Ss , Gibbs free energy ),(PθG FP:−−= S UGθ θ&&& S G −−= PF: PP F PPF ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂∂−= θθ θθ SG Gs , Table 2.2.1: Thermodynamic Pote ntials (per unit reference volume) The Gibbs Relation and Classical Thermodynamics In increment form, the second of the Gibbs relations 2.2.41 can be expressed as ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−=−FσF d du ds :1 1T ρθ (2.2.45) Consider an elastic (inviscid) fluid, for which by definition the stress is a function of F only through the density ρ: ()I σ θρ,p−= (2.2.46) where p is the hydrostatic pressure. Using { ▲Problem 4} F F:T−−=ρρ&, (2.2.47) relation 2.2.45 reduces to the classi cal thermodynamics Gibbs relation ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−= ρρθdpdu ds21 (2.2.48) Heat Flux Section 2.2 Solid Mechanics Part IV Kelly 66The first law is q dσQ FP div1:1Div1:1 0 0 ρρρ ρ −=− =& &u (2.2.49) Recall that, in this expression, the divergen ce of the (Cauchy) heat flux represents the heat entering a material element, Eqn. 1.6.4, dvdtQ v∫−= qdivδ (2.2.50) For a homogeneous material, ()() qdiv /1/ −=v dtQδ . Using 2.2.41b, one has the useful relation be tween the rate of entropy and the heat flux vector for a thermoelastic material: q Q div1Div1 0 ρ ρθ −= −=s& (2.2.51) Heat Conduction From 2.2.34, the dissipation inequality reads () 0 ,, ≤∇⋅∇θθθFq (2.2.52) This is Fourier’s inequality and merely states that heat fl ows from a hot region to a cold region. One usually assumes Fourier’s law of heat conduction : ()θθ∇−= ,FK q (2.2.53) where K is the thermal conductivity tensor. This constitutive law is sometimes referred to as Duhamel’s law of heat conduction , to distinguish it from the case of a thermally isotropic material, for which I K k=, which is then referred to as Fourier’s law. If one considers the left hand side of Eqn. 2.2.52 to be a function of θ∇, this function attains a maximum (of zero) when 0=∇θ . Thus () () 00Fq qqq = =+∇⎟ ⎠⎞⎜ ⎝⎛ ∂∇∂=∇⋅∂∇∂ =∇=∇,, 0T 0θ θθθθ θθ (2.2.54) This implies that, without a temperature gradie nt, there is no heat fl ow. Since it is a maximum, Section 2.2 Solid Mechanics Part IV Kelly 67()()θθθ θθ∂∇∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∇∂=∇⋅ ∇∂∂ =∇q qqT 022 (2.2.55) must be negative semi-definite. This implies that the symmetric part of the thermal conductivity tensor K in 2.2.53 must be positive semi-definite. Consider next the quantity 2 2/ : x K∂∂θ . Since ()()i j j i x x x x ∂∂∂∂=∂∂∂∂ / / / / θ θ , 2 2/x∂∂θ is symmetric. Further, the double contra ction of a symmetric tensor and a skew tensor is zero. Thus 2 2 2 2/ : sym / : x K x K ∂∂=∂∂ θ θ . The heat flux appears in the first law only as qdiv . Evaluating this give s, with Fourier’s law, j iij j iij jij i xxKxxK xKx ∂∂∂−∂∂ ∂∂−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=θ θ θ2 divq (2.2.56) For a homogeneous material, 0 /=∂∂i ijx K , and so 22 : sym divxK q∂∂−=θ (2.2.57) Thus, for homogeneous materials, it is only the symmetric part of the thermal conductivity tensor which enters the energy balance. In f act, it is almost universally accepted that K is symmetric, whether the material is homogeneous or not. Heat Capacity and Latent Heat The specific heat capacity is the heat requi red to raise the temperature of a body by unit temperature (per unit mass). From the first law, 2.2.49, when there is no deformation ( 0F=& ), all the heat absorbed is transferred to internal energy. The specific heat capacity at constant deformation is then by definition (see Eqn. 1.1.15) () θθ ∂∂=,F Fuc (2.2.58) The heat capacity can also be expr essed in terms of the free energy { ▲Problem 5}: ()() 22, , θθψθθθθ∂∂−=∂∂=F F Fsc (2.2.59) From 2.2.51, one then has the equation of heat conduction {▲Problem 6} Q FFDiv1: 02 2 ρ θψθθθθψθ −=∂∂∂−∂∂∂− & & (2.2.60) Some of the heat energy results in a temperat ure change, the first term, which is the heat capacity term. The second term represents a thermoelastic coupli ng effect whereby a Section 2.2 Solid Mechanics Part IV Kelly 68deformation induces a temperature change; this term is often neglected as it is usually small in applications. This second term can be interpreted as a latent heat term, whereby heat can be absorbed without a change in temperature. In terms of the enthalpy, the first law 2.2.49 can be expressed as Q PF Div1:1 0 0ρ ρ− =& &h (2.2.61) At constant stress, the heat energy now re sults in an increase in enthalpy. The specific heat capacity at constant stress is then by definition () θθ ∂∂=,P phc (2.2.62) The heat capacity can also be expressed in terms of the Gibbs ener gy (using the relations in Table 2.2.1): ()() 22, , θθθθθθ∂∂−=∂∂=P P Pg sc (2.2.63) For isotropic engineering materials, e.g. me tals, concrete, which undergo small strains, IFεc c= , with K kg/mN103≈εc and IPσc c= , with σc being slightly greater than εc (by a few percent). 2.2.3 Small Strain Thermoelasticity The Free Energy and Thermal Strains In the linear theory, the free en ergy is a function of the small-strain and the temperature, ()θψ,ε, and, as with the strains, temperature changes are assumed to be small. Assume that in the reference st ate the temperature is 0θ and the material is strain-free. Then, expanding the free energy in a Taylor series an d retaining only terms up to second order, () () () ()2 0 022 02 0022 0 0 00 21:: :21: , θθθψ θψθθψθθθψ ψψθψ −∂∂+∂∂∂−+∂∂+−∂∂+∂∂+= εεεεε εεε (2.2.64) where the notation 0 means that the quantity is evalua ted at the reference state. Note that the specific heat at constant deformation (strain) occurs in the last term here (see Eqn. 2.2.59). The stress is now Section 2.2 Solid Mechanics Part IV Kelly 69() 02 0 022 0:θψθθρψρψρψρ∂∂∂−+∂∂+∂∂=∂∂=εεεεεσ (2.2.65) Assuming the material is stress -free in the reference state, 0ε=, 0θθ=, then () 0ε=∂∂0/ψ , and 2.2.65 can be expressed as () ()0 0 :θθ θθβεσ −−= −−= βεDσij kl ijkl ijD (2.2.66) where the stiffness tensor is 22 2 εD∂∂=∂∂∂=ψρεεψρ kl ijijklD (2.2.67) and θψρθεψρβ∂∂∂−=∂∂∂−=εβ2 2 ijij (2.2.68) It was seen earlier that D is positive definite and so is al ways invertible. Define then the tensor α through βαD=: ( s o βCα := , where C is the flexibility tensor) so that ()[] () [ ]0 0 : θθ θθαε σ −−= −−= αεDσkl kl ijkl ijD (2.2.69) The quantity ()0θθ−α is called the thermal strain and α is the tensor of thermal expansion coefficients . For isotropic engi neering materials, Iαα= , with 1 5K10−−≈α . Note that, from 2.2.67-8, with the derivative s evaluated at the reference configuration, D and α are constant tensors. One can generali ze to a theory with temperature-dependent properties, by expressing 2.2.64 in the form () ( ) () ()()() ( )2 0 00 21:1:1:21)( , θθθ θρθθθρθθθ θψ − + −−+−+= EBA εβε Dε ε (2.2.70) The Gibbs Energy The temperature dependence of a thermoelastic material can be conveniently expressed using the Gibbs free energy, ()θ,σgg= : σε∂∂−=∂∂−=g g ijij ρσρε (2.2.71) Then, immediately, Section 2.2 Solid Mechanics Part IV Kelly 70 θ θασε &&&&&& ασCε+= += :ij kl ijkl ijC (2.2.72) where now θρθσραρσσρ ∂∂∂−=∂∂∂−=∂∂∂−=∂∂∂−= σασσC g gg gC ijijkl ijijkl 2 22 2 (2.2.73) To complete the set of second derivatives of the Gibbs function, one also has the heat capacity at constant stress (see Eqn. 2.2.63): θθθ∂∂∂−=gc2 σ (2.2.74) Heat Capacity Just as the stress, Eqn. 2.2.65, was expre ssed in terms of the free energy 2.2.64, the entropy can be expressed as the linear function () 022 0 02 0:θψθθθψ θψ θψ ∂∂−−∂∂∂−∂∂−=∂∂−= εεs (2.2.75) The first term on the right represents the refe rence entropy, at the re ference state, and can be omitted. The last term involves the specifi c heat at constant strain. The second term involves the tensor β (see Eqn. 2.2.68). Thus ε εDα c sθθθ ρ0::1 −+ = (2.2.76) For an isotropic material, the stiffness te nsor is given by Part III, Eqn. 4.1.12, () ( )jm in jn im mnij ijmnD δδδδμδλδ μλ + += ++⊗= ,I I II D (2.2.77) in which case ( )()kk kk ij kkij ij kl ijkl ij Kαε αεεαεDα ε μλμ λδδ 3 2 3 2 =+=+ = (2.2.78) where K is the bulk modulus. With ()()0 0 0 / / θθθθθθ −≈− , the entropy is now ()ε ε ε cKs 00tr3,θθθ ραθ−+ = (2.2.79) Section 2.2 Solid Mechanics Part IV Kelly 71 With Fourier’s law, θgradk−=q , the equation of heat conduction, 2.2.60, can be expressed as (with 0θθ≈) ()θρραθθ graddiv tr3 0k Kc = + εε & & (2.2.80) In order to express these relati ons in terms of stress, from 2.2.69, the isotropic stiffness tensor 2.2.77 and Part III, 4.1.20, ()I Iεε σ0 3)tr( dev2 θθα μ −−+= K K (2.2.81) Then, with the trace of a deviator zero, ()0 9)tr(3 tr θθα−−= K Kε σ (2.2.82) Thus 2.2.79 can be expressed as () ()ε σ σ cKs 00 029tr ,θθθθθρα ραθ−+− += (2.2.83) The specific heat capacity at constant stress is then, from 2.2.63, ()() θθθθθθ∂∂≈∂∂=, , 0σ σ σs sc (2.2.84) leading to the relation ραθKcc2 09=−εσ (2.2.85) In terms of stress, the equati on of heat conduction 2.2.80 reads ()θρραθθ graddiv tr0kc = +σσ & & (2.2.86) 2.2.4 Problems 1. Differentiate the invariants 2.2. 16 to derive relations 2.2.18. 2. Derive the complementary function 2.2.26. 3. Derive the Gibbs relations 2. 2.41 for an elastic material. Section 2.2 Solid Mechanics Part IV Kelly 724. Use the conservation of mass relation ρρ J=0 and the relation Part III, 2.5.19, F F& & :T−=JJ , to derive the relation 2.2.47, F F: /T−−=ρρ& . 5. Show that the heat capac ity is proportional to the s econd derivative of the free energy, Eqn. 2.2.59. 6. By taking the entropy s to be a function of the variables ()θ,F , use the relation 2.2.51, QDiv0−=s&θρ , and the relation between entropy and free energy, Eqn. 2.2.35a, to derive the equati on of heat conduction 2.2.60, Q FFDiv1: 02 2 ρ θψθθθθψθ −=∂∂∂−∂∂∂− & & 7. Derive the equation of heat conduction 2.2.86.