Finite Element Analysis - Thermomechanics of Solids Part 6 ppt

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Finite Element Analysis - Thermomechanics of Solids Part 6 ppt

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95 Stress-Strain Relation and the Tangent-Modulus Tensor 6.1 STRESS-STRAIN BEHAVIOR: CLASSICAL LINEAR ELASTICITY Under the assumption of linear strain, the distinction between the Cauchy and Piola- Kirchhoff stresses vanishes. The stress is assumed to be given as a linear function of linear strain by the relation (6.1) in which c ijkl are constants and are the entries of a 3 × 3 × 3 × 3 fourth-order tensor, C . If T and E L were not symmetrical, C might have as many as 81 distinct entries. However, due to the symmetry of T and E L there are no more than 36 distinct entries. Thermodynamic arguments in subsequent sections will provide a rationale for the Maxwell relations: (6.2) It follows that c ijkl = c klij , which implies that there are, at most, 21 distinct coeffi- cients. There are no further arguments from general principles for fewer coefficients. Instead, the number of distinct coefficients is specific to a material, and reflects the degree of symmetry in the material. The smallest number of distinct coefficients is achieved in the case of isotropy, which can be explained physically as follows. Suppose a thin plate of elastic material is tested such that thin strips are removed at several angles and then subjected to uniaxial tension. If the measured stress-strain curves are the same and independent of the orientation at which they are cut, the material is isotropic. Otherwise, it exhibits anisotropy, but may still exhibit limited types of symmetry, such as transverse isotropy or orthotropy. The notion of isotropy is illustrated in Figure 6.1. In isotropic, linear-elastic materials (which implies linear strain), the number of distinct coefficients can be reduced to two, m and l , as illustrated by Lame’s equation, (6.3) 6 TcE ij ijkl kl L = () , ∂ ∂ = ∂ ∂ T E T E ij kl kl ij . TEE ij ij L kk L ij =+2 µλδ () () . 0749_Frame_C06 Page 95 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC 96 Finite Element Analysis: Thermomechanics of Solids This can be inverted to furnish (6.4) The classical elastic modulus E 0 and Poisson’s ratio n represent response under uniaxial tension only , provided that T 11 = T , T ij = 0. Otherwise, (6.5) It is readily verified that (6.6) from which it is immediate that (6.7) Leaving the case of uniaxial tension for the normal (diagonal) stresses and strains, we can write (6.8) (6.9) FIGURE 6.1 Illustration of isotropy. 1 2 3 4 T 21 3 4 E ET T ij L ij kk ij () .=− +       1 223 µ λ µλ δ E ==−=− T E E E E E L L L L L 11 11 22 11 33 11 () () () () () . ν 11 2 1 23 1 22 3E E =− +       = + µ λ µλ ν µ λ µλ , 1 2 1 µ ν = + E , ETTT TTT L 11 11 22 33 11 22 33 1 2 1 23 1 22 3 1 () () [ ( )], =− +       − + + =−+ µ λ µλ µ λ µλ ν E ETTT L 22 22 33 11 1 () [ ( )],=−+ E ν 0749_Frame_C06 Page 96 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC Stress-Strain Relation and the Tangent-Modulus Tensor 97 and (6.10) and the off-diagonal terms satisfy (6.11) 6.2 ISOTHERMAL TANGENT-MODULUS TENSOR 6.2.1 C LASSICAL E LASTICITY Under small deformation, the fourth-order tangent-modulus tensor D in linear elas- ticity is defined by (6.12) In linear isotropic elasticity, the stress-strain relations are written in the Lame’s form as (6.13) Using Kronecker Product notation from Chapter 2, this can be rewritten as (6.14) from which we conclude that (6.15) 6.2.2 C OMPRESSIBLE H YPERELASTIC M ATERIALS In isotropic hyperelasticity, which is descriptive of compressible rubber elasticity, the 2 nd Piola-Kirchhoff stress is taken to be derivable from a strain-energy function that depends on the principal invariants I 1 , I 2 , I 3 of the Right Cauchy-Green strain tensor: (6.16) ETTT L 33 33 11 22 1 () [ ( )],=−+ E ν ETETET LLL 12 12 23 23 31 31 111 () () () .= + = + = + ννν E E E ddTDE= L . TE E=+2 µλ LL Itr(). VEC VEC()TE=⊗+[](),2 µλ II ii T L D =⊗+ITEN22 2(). µλ II ii T S EC s e sSeEcC == == === d d d d d d d d (a) (b) ww ww VEC VEC VEC 22 T c ((())). 0749_Frame_C06 Page 97 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC 98 Finite Element Analysis: Thermomechanics of Solids Now, (6.17) From Chapter 2, (6.18) The tangent-modulus tensor D o , referred to the undeformed configuration, is given by (6.19) and now (6.20) Finally, (6.21) In deriving A 3 we have taken advantage of the Cayley-Hamilton theorem (see Chapter 2). sn n T == ∂ ∂ = ∂ ∂ 2 φφ ii i i c ,, . w i i I I nnin 1 121 3 3 ==−= − ic CI VEC I(). dSDE s D E== oo TENdd )d22( , TEN o 22( ) d d D c =+ =44 φφ ij i j i i i i nn A A n T ,. A A A 1 22 1 9 3 1 3 21 2 2 1 == = =− =− = =−+ =−+⊕ =−−++⊕ − d d (a) d d d d [ (b) d d d d (c) i 0 i ii I c i in i ii I i i T T TT T 9 TT c c c c C c cC cCC cc CC I I VEC I I I VEC I ] [()] [()] [][ ] 0749_Frame_C06 Page 98 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC Stress-Strain Relation and the Tangent-Modulus Tensor 99 6.3 INCOMPRESSIBLE AND NEAR-INCOMPRESSIBLE HYPERELASTIC MATERIALS Polymeric materials, such as natural rubber, are often nearly incompressible. For some applications, they can be idealized as incompressible. However, for applica- tions involving confinement, such as in the corners of seal wells, it may be necessary to accommodate the small degree of incompressibility to achieve high accuracy. Incompressibility and near-incompressibility represent internal constraints . The principal (e.g., Lagrangian) strains are not independent, and the stresses are not determined completely by the strains. Instead, differences in the principal stresses are determined by differences in principal strains (Oden, 1972). An additional field must be introduced to enforce the internal constraint, and we will see that this internal field can be taken as the hydrostatic pressure (referred to the current configuration). 6.3.1 I NCOMPRESSIBILITY The constraint of incompressibility is expressed by the relation J = 1. Now, (6.22) and consequently, (6.23) The constraint of incompressibility can be enforced using a Lagrange multiplier (see Oden, 1972), denoted here as p . The multiplier depends on X and is, in fact, the additional field just mentioned. Oden (1972) proposed introducing an augmented strain-energy function, w ′ , similar to (6.24) in which w is interpreted as the conventional strain-energy function, but with depen- dence on I 3 ( = 1) removed. are called the deviatoric invariants. For reasons to be explained in a later chapter presenting variational principles, this form serves J I = = = = = det det ( ) det( )det( ) det ( ) , F F FF FF 2 2 3 T T I 3 1= . ′ = ′′ −− ′ = ′ =wwII pI I I I I I I () () , // 12 3 1 1 2 2 1 2 1 3 13 3 23 ,,, ′′ II 12 and 0749_Frame_C06 Page 99 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC 100 Finite Element Analysis: Thermomechanics of Solids to enforce incompressibility, with S now given by (6.25) To convert to deformed coordinates, recall that S = J F − 1 T F − T . It is left to the reader in an exercise to derive in which (6.26) in which It follows that p = − tr ( T ) / 3 since (6.27) Evidently, the Lagrange multiplier enforcing incompressibility is the “true” hydrostatic pressure. Finally, the tangent-modulus tensor is somewhat more complicated because d S depends on d E and d p . We will see in subsequent chapters that it should be defined as D ∗ using (6.28) Example: Uniaxial Tension Consider the Neo-Hookean elastomer satisfying (6.29) s e cc = ∂ ′ ∂ = ′′ + ′′ − ′ = ∂ ∂ ′ ′ = ∂ ∂ ′ ′ = ∂ ′ ∂       ′ = ∂ ′ ∂       w p w I w I II 22 12233 1 1 2 2 1 1 2 2 φφ φφ 1 nnn nn TT I . ′′ ψψ 12 and tmm== ′′ + ′′ −VEC p()T 22 11 2 2 ψψ i im im TT ′ = ′ = 12 00 and . tr p p p () . T = = ′′ + ′′ − =− =− it im im ii ii T T 1 TT T 22 3 122 ψψ TEN p p 22( )D s e s s ∗ = −                   d d d d d d T 0 d d d d s e s = ′′ + ′′ + ′′ ′ + ′′′ + ′′ ′ + ′′ ′ =− 4 1 1 2 2 11 1 1 12 1 2 21 2 1 22 2 2 33 [()() () ()] . φφ φ φ φφ A A nn nn nn nn TT TT p In wI=− [] = α I and subject to 13 31. 0749_Frame_C06 Page 100 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC Stress-Strain Relation and the Tangent-Modulus Tensor 101 We seek the relation between s 1 and e 1 , which will be obtained twice: once by enforcing the incompressibility constraint a priori , and the second by enforcing the constraint a posteriori . a priori : Assume for the sake of brevity that e 2 = e 3 . I 3 = 1 implies that . The strain-energy function now is The stress, s 1 , is now found as (6.30) a posteriori : Use the augmented function (6.31) Now (6.32) Thus, it follows that We conclude that (6.33) cc 21 1= / wc c =+− α []. 1 1 2 3 s w cc 1 11 32 221 1 ==−       d d α / . ′ =−− −wI p I α [][]. 13 3 2 1 s w c pc s w c pc s w c pc w p I 1 1 1 2 2 2 3 3 3 3 22 022 022 001 = ′ =− == ′ =− == ′ =− = ′ =→ = d d d d d d d d α α α / / / . cc c pc 23 1 2 12== =/ and / α . spc c c pc c c c 11 2 1 2 2 1 1 32 2 2 21 21 1 =− =− =−       =−       α α α α / / . / 0749_Frame_C06 Page 101 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC 102 Finite Element Analysis: Thermomechanics of Solids a posteriori with deviatoric invariants : Consider the augmented function with deviatoric invariants: (6.34) Hence, c 2 = c 3 . Furthermore, (6.34e) Equation 6.34 now implies that (6.35) and substitution into Equation 6.34b furnishes (6.36) as in the a priori case and in the first a posteriori case. Now, the Lagrange multiplier p can be interpreted as the pressure referred to current coordinates. 6.3.2 N EAR -I NCOMPRESSIBILITY As will be seen in Chapter 18, the augmented strain-energy function (6.37) serves to enforce the constraint (6.38) ′ =− [] −− = ′ =−       − == ′ =−       − = wII p I s w cI I I I c p I c s w cI I I I c p I c α α α 13 13 3 1 13 13 1 3 43 3 1 3 1 2 23 13 1 3 43 3 2 3 2 3 2 1 22 11 3 022 11 3 0 / [ ] (a) d d (b) d d (c) ss w cI I I I c p I c 3 33 13 1 3 43 3 3 3 3 22 11 3 = ′ =−       − d d (d) α d d ′ =→ = w p I01 3 . 2 3 21 2 2 α cc c p c −       = , s c 1 1 32 21 1 =−       α / , ′′ = ′′ −−−wwII pI p (, ) [ ] 12 3 2 1 2 1 2 κ pI=− − [] κ 3 1. 0749_Frame_C06 Page 102 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC Stress-Strain Relation and the Tangent-Modulus Tensor 103 Here, k is the bulk modulus, and it is assumed to be quite large compared to, for example, the small strain-shear modulus. The tangent-modulus tensor is now (6.39) 6.4 NONLINEAR MATERIALS AT LARGE DEFORMATION Suppose that the constitutive relations are measured at a constant temperature in the current configuration as (6.40) in which the fourth-order tangent-modulus tensor D can, in general, be a function of stress, strain, temperature, and internal-state variables (discussed in subsequent chapters). This form is attractive since and D are both objective. Conversion to undeformed coordinates is realized by (6.41) If s = VEC ( S ) and e = VEC ( E), then (6.42) Recalling Chapter 2, it follows that (6.43) in which (6.44) is the tangent-modulus tensor D o referred to undeformed coordinates. TEN p p 22 1 D s e s s * . () = −                   d d d d d d T κ T o = DD, T o ˙ ˙ . SD DE = = −− −− −− J J FDF FF FF T TT 1 11 ˙ ( ˙ ) () ( ˙ ) ()()() ˙ () ˙ () ˙ . s =⊗ =⊗ ⊗ =⊗ ⊗⊗ =⊗ ⊗ = −−−− −− − − −− − − −− − − J J J() J IF F FF IFF I F F FF F IIF FF F Fe e TT T TT TT 11 11 1 11 11 22 22 22 22 VEC TEN VEC TEN VEC TEN TEN o DE D DE D D E ˙ ˙ ,SDE= o DD o ITEN TEN=⊗ ⊗ −− − − 22 22 11 (())JFF F F TT 0749_Frame_C06 Page 103 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC 104 Finite Element Analysis: Thermomechanics of Solids Suppose instead that the Jaumann stress flux is used and that (6.45) Now (6.46) For this flux, there does not appear to be any way that can be written in the form of Equation 6.42, i.e., is determined by . To see this, consider (6.47) In the second term in Equation 6.47a, VEC(L + W) cannot be eliminated in favor of VEC(D), and hence in favor of . Note that (6.48) I + U 9 is singular, as seen in the following argument. Recall that U 9 is symmetric and thus has real eigenvalues. However, and thus the eigenvalues of U 9 are either 1 or −1. Some of the eigenvalues must be −1. Otherwise, U 9 would be the identity matrix, in which case it would not, in general, have the permutation property identified in Chapter 2. Thus, some of the eigenvalues of I + U 9 vanish. Instead, we write (6.49) and we see that if the Jaumann stress flux is used, is determined by and the spin W. Recall that the spin does not vanish under rigid-body rotation. 6.5 EXERCISES 1. In classical linear elasticity, introduce the isotropic stress and isotropic linear strain as TD ∆ = D. ˙ ()( ) ˙ ( ˙ )( ) ( ) . STDTT DE T E TT =+−+−−       =+−+−− [] −− −−− −− − J() J FLWLWF FFF FF LW LWF TT TT T T T ∆ tr tr 1 11 1 ˙ S ˙ E VEC TEN VEC TEN VEC TEN TEN VEC VEC TEN ( ˙ () ( ˙ () ( ) ()[ ( ) ( ) () ( )] () [ [ ] SDED DDT DTT )) J J] = ′ − ′′ + ′ =⊗ + ′′ =⊗ ⊗−⊗ −−− − − 22 22 22 22 22 2 12 9 LW IF F F FF IF II U TTT TT (a) (b) (c) VEC() ˙ E VEC VEC VEC()(DLLIUL T )( ()).=+=+ 1 2 1 2 9 U I 9 2 = VEC VEC VEC VEC() () ( ) ( ), ˙ ˙ SD D D D= ′ − ′′ − ′′ +ED W2 ˙ S ˙ E se L ==tr tr() ( ,SE) 0749_Frame_C06 Page 104 Wednesday, February 19, 2003 5:06 PM © 2003 by CRC CRC Press LLC [...]... (d) torsional flow: vr = 0, vθ = f ( z ), vz = 0 nd 6 In undeformed coordinates, the 2 Piola-Kirchhoff stress for an incompressible, hyperelastic material is given by s= © 2003 by CRC CRC Press LLC ∂w ′ = 2ϕ1n1 + 2ϕ 2 n′ − pI3n3 ′ ′ ′ 2 ∂e 0749_Frame_C 06 Page 1 06 Wednesday, February 19, 2003 5: 06 PM 1 06 Finite Element Analysis: Thermomechanics of Solids Find the corresponding expression in deformed...0749_Frame_C 06 Page 105 Wednesday, February 19, 2003 5: 06 PM Stress-Strain Relation and the Tangent-Modulus Tensor 105 and introduce the deviatoric stress and strain using 1 sd = s − si 3 1 ed = e − ei 3 Verify that iTsd = 0 iTed = 0 sd = 2 µed s = (2 µ + 3λ )e 2 Verify that Equation 6. 3 can be inverted to furnish e=  1  λ T t − (2 µ + 3λ ) (i t... 8 Obtain λ in terms of E and ν (Problem 10, Chapter 5.) (L 9 The bulk modulus K is defined by tkk = 3Κekk ) Obtain K as a function of E and ν (Problem 11, Chapter 5.) 10 The 2" × 2" × 2" shown in Figure 6. 2 is confined on its sides facing the ± x faces by rigid, frictionless walls The sides facing the ± z faces are free The top and bottom faces are subjected to a compressive force of 100 lbf Take E =... psi and ν = 1/3 Find all nonzero stresses and strains What is the volume change? What are the principal stresses and strains? What is the maximum shear stress? (Problem 12, Chapter 5.) v E x z FIGURE 6. 2 Strain in a constrained plate © 2003 by CRC CRC Press LLC . 0749_Frame_C 06 Page 97 Wednesday, February 19, 2003 5: 06 PM © 2003 by CRC CRC Press LLC 98 Finite Element Analysis: Thermomechanics of Solids Now, (6. 17) From Chapter 2, (6. 18) The tangent-modulus. 0749_Frame_C 06 Page 96 Wednesday, February 19, 2003 5: 06 PM © 2003 by CRC CRC Press LLC Stress-Strain Relation and the Tangent-Modulus Tensor 97 and (6. 10) and the off-diagonal terms satisfy (6. 11). v vvvfr vvfrv vvfzv rz rz rz rz = () == === () == () = == () = ,, ,, ,, ,, θ θ θ θ 00 00 00 00 s e nnn= ∂ ′ ∂ = ′′ + ′′ − w pI22 11 2 2 33 ϕϕ . 0749_Frame_C 06 Page 105 Wednesday, February 19, 2003 5: 06 PM © 2003 by CRC CRC Press LLC 1 06 Finite Element Analysis: Thermomechanics of Solids Find the corresponding expression

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  • Finite Element Analysis: Thermomechanics of Solids

    • Table of Contents

    • Chapter 6: Stress-Strain Relation and the Tangent-Modulus Tensor

      • 6.1 STRESS-STRAIN BEHAVIOR: CLASSICAL LINEAR ELASTICITY

      • 6.2 ISOTHERMAL TANGENT-MODULUS TENSOR

        • 6.2.1 C LASSICAL E LASTICITY

        • 6.2.2 C OMPRESSIBLE H YPERELASTIC M ATERIALS

      • 6.3 INCOMPRESSIBLE AND NEAR-INCOMPRESSIBLE HYPERELASTIC MATERIALS

        • 6.3.1 I NCOMPRESSIBILITY

        • 6.3.2 N EAR -I NCOMPRESSIBILITY

      • 6.4 NONLINEAR MATERIALS AT LARGE DEFORMATION

      • 6.5 EXERCISES

      • References

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