Section 4.4

4.4 Isotropic Hyperelasticity

Attention is now restricted to the case of isotropic hyperelastic materials.

4.4.1 Constitutive Equations in Material Form

Consider the general hyperelastic constitutive law 4.2.5:

∂W (C) S = 2 (4.4.1) ∂C

From §4.3.2, the strain energy for an isotropic hyperelastic material must be a function of the principal invariants of C:

W = W (IC (C),IIC (C),IIIC (C)) (4.4.2)

Then, using the relations 1.11.33 for the derivatives of the invariants (taking into account that C is symmetric),

∂W ∂W ∂I ∂W ∂II ∂W ∂III = C + C + C ∂C ∂IC ∂C ∂IIC ∂C ∂IIIC ∂C (4.4.3) ⎛ ∂W ∂W ⎞ ∂W ∂W ⎜ ⎟ −1 = ⎜ + IC ⎟I − C + IIICC ⎝ ∂IC ∂IIC ⎠ ∂IIC ∂IIIC

Further, the C−1 term can be replaced by a C2 term using the Cayley-Hamilton theorem. In summary then, expressing S in the form 4.3.9 (and 4.3.12),

2 ⎧ 2α 0I + 2α1C + 2α 2C ∂W (C) ⎪ S = 2 = ⎨ ∂C ⎪ −1 ⎩2β 0I + 2β1C + 2β −1C

∂W ∂W ∂W ∂W ∂W α 0 = + IC + IIC β 0 = + IC ∂IC ∂IIC ∂IIIC ∂IC ∂IIC ∂W ∂W ∂W α1 = − − IC β1 = − ∂IIC ∂IIIC ∂IIC ∂W ∂W α 2 = β −1 = IIIC ∂IIIC ∂IIIC Constitutive Equation in material description (4.4.4)

Solid Mechanics Part III 372 Kelly Section 4.4

4.4.2 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, W = W (Ib (b),IIb (b),IIIb (b)) . Then

2 ⎧ α 0I + α1b + α 2b ∂W (b) ⎪ = ⎨ (4.4.5) ∂b ⎪ −1 ⎩β 0I + β1b + β −1b

and the coefficients α i ,β i are the same as in 4.4.4

Taking the expression for ∂W / ∂C in 4.4.4, pre-contracting with F, post-contracting with F T , using the definitions C = F T F, b = FF T , and (pre- or post-) contracting the above similar expression for ∂W / ∂b in terms of the invariants of b with b, one finds that

∂W ∂W ∂W b = b = F F T (4.4.6) ∂b ∂b ∂C

Therefore, using S = JF −1σF −T , the constitutive equation in the material formulation, S = 2∂W (C) / ∂C , can be transformed into a spatial constitutive equation:

∂W (b) ∂W (b) σ = 2J −1 b = 2J −1b (4.4.7) ∂b ∂b

It should be emphasised that this expression, unlike the relation S = 2∂W (C) / ∂C , 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 2J −1b then leads to

2 ⎧ α 0I + α1b + α 2b −1 ∂W (b) ⎪ σ = 2J b = ⎨ ∂b ⎪ −1 ⎩β 0I + β1b + β −1b

−1 ⎡ ∂W ⎤ −1 ⎡ ∂W ∂W ⎤ α 0 = 2J ⎢ IIIb ⎥ β 0 = 2J ⎢ IIb + IIIb ⎥ ⎣∂IIIb ⎦ ⎣∂IIb ∂IIIb ⎦

−1 ⎡∂W ∂W ⎤ −1 ⎡∂W ⎤ α1 = 2J ⎢ + Ib ⎥ β1 = 2J ⎢ ⎥ ⎣ ∂Ib ∂IIb ⎦ ⎣ ∂Ib ⎦

−1 ⎡ ∂W ⎤ −1 ⎡ ∂W ⎤ α 2 = 2J ⎢− ⎥ β −1 = 2J ⎢− IIIb ⎥ ⎣ ∂IIb ⎦ ⎣ ∂IIb ⎦ Constitutive Equation in spatial description (4.4.8)

Solid Mechanics Part III 373 Kelly Section 4.4

Principal Directions of the Cauchy Stress and Left Cauchy-Green Tensor

Recall that the eigenvalues of the right stretch tensor U are the principal stretches λi , and that the eigenvalues of the right Cauchy-Green tensor C and the left Cauchy-Green tensor 2 b are the squares of the principal stretches, λi (the eigenvectors of b are those of C rotated with the rotation tensor R). Taking nˆ to be an eigenvector of b and λ2 the 2 corresponding eigenvalue, and using the constitutive equation σ = α 0I +α1b +α 2b , one has

2 2 4 σnˆ = α 0Inˆ + α1bnˆ + α 2b nˆ = (α 0 + α1λ + α 2λ )nˆ (4.4.9)

Thus σnˆ = σ nˆ where σ is an eigenvalue of σ , the principal stress, given by term inside the brackets, and so the eigenvectors, or principal directions, of the Cauchy stress and b coincide.

If one takes a deformation for which b13 = b23 = 0, it follows from 4.4.8 that, for arbitrary coefficient β −1 , one also has σ 13 = σ 23 = 0 . The tensors b and σ are coaxial, that is, bσ = σb . From this relation, one finds that one must have

σ −σ b − b 11 22 = 11 22 (4.4.10) σ 12 b12

This relation holds for all compressible isotropic 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 stretch with respect to b are given in §2.3.3. Introducing the 2 principal stresses, and noting that bnˆ i = λi nˆ i (no sum over the i) it follows then that (no sum over the i in what follows)

−1 ∂W (b) −1 ∂W (λ) ∂λk σ i nˆ i = σnˆ i = 2J b nˆ i = 2J b nˆ i ∂b ∂λk ∂b

−1 ∂W (λ) 1 −1 ∂W (λ) 1 = 2J b (nˆ k ⊗ nˆ k )nˆ i = 2J bnˆ i (4.4.11) ∂λk 2λk ∂λi 2λi

−1 ∂W (λ) = J λi nˆ i ∂λi or

−1 ∂W (λ) −1 ∂W (λ) −1 ∂W (λ) σ1 = J λ1 , σ 2 = J λ2 , σ 3 = J λ3 (4.4.12) ∂λ1 ∂λ2 ∂λ3

Solid Mechanics Part III 374 Kelly Section 4.4

Similarly, for the Piola-Kirchhoff stresses, one has

∂W 1 ∂W Pi = ,Si = (4.4.13) ∂λi λi ∂λi

Example (Uniaxial Stretch)

Consider a specimen of isotropic hyperelastic material under a uniaxial tension σ 11 , with

⎡λ1 0 0 ⎤ F = ⎢ 0 λ 0 ⎥ (4.4.14) ⎢ 2 ⎥ ⎣⎢ 0 0 λ3 ⎦⎥

Since σ and b are coaxial (see 4.4.9), the principal stresses are, from 4.4.8,

2 −2 ⎡σ 11 0 0⎤ ⎡1 0 0⎤ ⎡λ1 0 0 ⎤ ⎡λ1 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0⎥ = β ⎢0 1 0⎥ + β 0 λ2 0 + β 0 λ−2 0 (4.4.15) ⎢ ⎥ 0 ⎢ ⎥ 1 ⎢ 2 ⎥ −1 ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2 ⎥ ⎢ −2 ⎥ ⎣ 0 0 0⎦ ⎣0 0 1⎦ ⎣ 0 0 λ3 ⎦ ⎣ 0 0 λ3 ⎦

Subtracting the second of these equations from the first leads to

⎛ 1 ⎞ σ = λ2 − λ2 ⎜ β − β ⎟ (4.4.16) 11 ()1 2 ⎜ 1 −1 2 2 ⎟ ⎝ λ1 λ2 ⎠

Subtracting the third from the second leads to

⎛ 1 ⎞ 0 = λ2 − λ2 ⎜ β − β ⎟ (4.4.17) ()2 3 ⎜ 1 −1 2 2 ⎟ ⎝ λ2λ3 ⎠

which implies that λ2 = λ3 (the terms inside the second bracket can also be zero, but not if β1 > 0, β −1 ≤ 0 , as is usually the case).

The material parameters β 0 , β1 , β −1 can be obtained from such a test by measuring

σ 11 , λ1 and λ2 . On the other hand, if the constitutive relation, that is, the parameters, are known, then one can obtain the stretches by specifying the uniaxial stress. ■

4.4.4 Incompressible Materials

When the material is incompressible, then

J = det F = IIIC = IIIb = 1 (4.4.18)

Solid Mechanics Part III 375 Kelly Section 4.4

and the strain energy function is only dependent on two invariants:

W = W (IC ,IIC ) = W (Ib ,IIb ), (4.4.19) and the constitutive equations are the same as those given earlier for the compressible −1 material, with IIIb = 1, J = 1. However, the derivative ∂W / ∂IIIC is an unknown and in particular it is unknown at IIIC = 1.

Returning to the more general constitutive equations for an incompressible material, 4.2.36, 4.2.39, one has

∂W (I ,II ) ∂W (I ,II ) S = 2 C C − pC−1 , σ = 2F C C F T − pI (4.4.20) ∂C ∂C

Carrying out the differentiation as before, one now has, directly from these equations,

−1 S = β 0I + β1C − pC

⎡∂W ∂W ⎤ ∂W β 0 = 2⎢ + IC ⎥ β1 = −2 ⎣∂IC ∂IIC ⎦ ∂IIC Constitutive Equation for an Isotropic Incompressible Hyperelastic Material (Material Form) (4.4.21) and

−1 σ = α1b + α −1b − pI

∂W ∂W α1 = 2 α −1 = −2 ∂Ib ∂IIb 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 − 2∂W / ∂IIIC (with IIIC = 1 ) ; in the σ equation, 4.4.8, it is p = −2[∂W / ∂IIIb + IIb ∂W / ∂IIb ].

Similarly, the equations can be written in terms of the principal stretches and principal stresses (the incompressibility constraint is now λ1λ2λ3 = 1):

∂W ∂W 1 1 ∂W 1 σ i = λi − p, Pi = − p, Si = − 2 p (4.4.23) ∂λi ∂λi λi λi ∂λi λi

Solid Mechanics Part III 376 Kelly Section 4.4

Example (Strip Biaxial Tension)

Consider an incompressible isotropic hyperelastic material which is stretched in the e1 direction but its dimensions in the e 2 direction is held constant, with

⎡λ 0 0 ⎤ ⎢ ⎥ F = ⎢0 1 0 ⎥ (4.4.24) ⎣⎢0 0 1/ λ⎦⎥

With σ 3 = 0 ,one finds that

∂W 1 ∂W 2 p = 2 2 − 2 λ (4.4.25) ∂Ib λ ∂IIb and the stresses are

⎡∂W ⎛ 1 ⎞ ∂W ⎛ 1 ⎞⎤ σ = 2 λ2 − + ⎜λ2 − ⎟ , i = 1, 2 (4.4.26) i ⎢ ⎜ i 2 ⎟ ⎜ 2 ⎟⎥ ⎣⎢ ∂Ib ⎝ λ ⎠ ∂IIb ⎝ λi ⎠⎦⎥ ■

Example (Biaxial Stretch)

Consider the biaxial stretch of a thin sheet of incompressible isotropic hyperelastic material, with

⎡λ1 0 0 ⎤ F = ⎢ 0 λ 0 ⎥ (4.4.27) ⎢ 2 ⎥ ⎣⎢ 0 0 1/ λ1λ2 ⎦⎥

From 4.4.18:

⎛ ∂W ∂W ⎞ 2 −2 ⎜ ⎟ σ i = α1λi + α −1λi − p, i = 1, 2, 3, ⎜α1 = 2 ,α −1 = −2 ⎟ (4.4.28) ⎝ ∂Ib ∂IIb ⎠

Taking the stretches to be in-plane, with the larger faces stress-free, one has σ 3 = 0 so that, with λ3 = 1/ λ1λ2 , one can solve for p,

1 2 2 p = α1 2 2 + α −1λ1 λ2 λ1 λ2 and the stresses in the sheet are

Solid Mechanics Part III 377 Kelly Section 4.4

⎛ 1 ⎞ ⎛ 1 ⎞ σ = α −α λ2 ⎜λ2 − ⎟, σ = α −α λ2 ⎜λ2 − ⎟ (4.4.29) 1 ()1 −1 2 ⎜ 1 2 2 ⎟ 2 ()1 −1 1 ⎜ 2 2 2 ⎟ ⎝ λ1 λ2 ⎠ ⎝ λ1 λ2 ⎠

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, P = JσF −T , are, directly from these equations,

σ ⎛ 1 ⎞ σ ⎛ 1 ⎞ P = 1 = α −α λ2 ⎜λ − ⎟, P = 2 = α −α λ2 ⎜λ − ⎟ (4.4.30) 1 ()1 −1 2 ⎜ 1 3 2 ⎟ 2 ()1 −1 1 ⎜ 2 2 3 ⎟ λ1 ⎝ λ1 λ2 ⎠ λ2 ⎝ λ1 λ2 ⎠

The total forces acting on the lateral surfaces are then, from 3.5.4,

⎛ 1 ⎞ ⎛ 1 ⎞ F = P A = A α −α λ2 ⎜λ − ⎟, F = P A = A α −α λ2 ⎜λ − ⎟ 1 1 0 0 ()1 −1 2 ⎜ 1 3 2 ⎟ 2 2 0 0 ()1 −1 1 ⎜ 2 2 3 ⎟ ⎝ λ1 λ2 ⎠ ⎝ λ1 λ2 ⎠ (4.4.31) where A0 is the area in the reference configuration.

The material parameters α1 , α −1 can be determined from such a test by measuring the applied forces. This is helped by controlling the stretches λ1 and λ2 in such a way that the invariant Ib is held constant, or the invariant IIb is held constant.

When the two applied forces are equal, F1 = F2 , one finds that

⎛ ⎛ α ⎞ α ⎞ λ − λ ⎜ 1+ λ3λ3 ⎜1+ −1 λ λ ⎟ − −1 λ + λ 2 ⎟ = 0 (4.4.32) ()1 2 ⎜()1 2 ⎜ 1 2 ⎟ ()1 2 ⎟ ⎝ ⎝ α1 ⎠ α1 ⎠

The term inside the first brackets gives the expected symmetric solution λ1 = λ2 .

However, the term inside the second brackets can also be zero for λ1 ≠ λ2 , but this will only occur for fairly large stretches for typical values of the material parameters. The symmetric equal-stretch solution holds up until this happens. For example, for the typical value −α −1 /α1 = 0.1, the second term becomes zero when λ1 = λ2 ≈ 3.17 . 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 F1 and F2 , then λ1 ≈ λ2 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 close to the asymmetric solution. The long side of the rectangle will be along the direction of the slightly larger force and will keep lengthening until the stretch in the other direction returns back to λ = 1.

This large strain problem differs from the problems of linear elasticity, which have unique solutions. ■

Solid Mechanics Part III 378 Kelly Section 4.4

Example (Simple Shear)

The case of simple shear was discussed in detail in §2.2.6. In the equations following 2.2.40, let

1 2 cos 2β = sinθ, sin 2β = cosθ, tan 2β = = (4.4.33) tanθ k

Then the principal material directions can be expressed as

ˆ ˆ N1 = sin βE1 + cos βE2 , N 2 = −cos βE1 + sin βE2 (4.4.34)

With n = RN , the principal spatial directions are (see figure 4.4.1 below)

nˆ 1 = cos βE1 + sin βE 2 , nˆ 2 = −sin βE1 + cos βE 2 (4.4.35)

The principal stretches are

1+ cos 2β 1− cos 2β λ = , λ = and 1 (4.4.36) 1 sin 2β 2 sin 2β and the principal stresses are, from 4.4.18,

2 1 σ1 = α1λ1 + α −1 2 − p λ1 1 ⎛ ∂W ∂W ⎞ σ = 2α λ2 + α − p ⎜α = 2 ,α = −2 ⎟ (4.4.37) 2 1 2 −1 2 ⎜ 1 −1 ⎟ λ2 ⎝ ∂Ib ∂IIb ⎠

σ 3 = α1 + α −1 − p

e ˆ 2 ˆ n2 N1 ˆ N2

nˆ 1 β β e 1

Figure 4.4.1: principal directions for pure shear

The strain energy here is a function of the invariants of b, which are

2 2 Ib = 3 + k , IIb = 3 + k (IIIb = 1) (4.4.38)

In the original Cartesian coordinates, the stresses are

Solid Mechanics Part III 379 Kelly Section 4.4

2 2 σ 11 = σ 1 cos β + σ 2 sin β 2 2 σ 22 = σ 1 sin β + σ 2 cos β (4.4.39)

σ 12 = (σ 1 −σ 2 )sin β cos β

It follows that (this is a universal relation, independent of the constitutive relation)

σ 11 −σ 22 = (2 / tan 2β )σ 12 = kσ 12 (4.4.40)

The fact that and σ 11 ≠ σ 22 is called the Poynting effect. The stress are now

2 σ11 = α1 (1+ k )+ α −1 − p σ = α + α 1+ k 2 − p 22 1 −1 () (4.4.41) σ12 = ()α1 −α −1 k

σ 33 = α1 + α −1 − p

The proportionality factor between the shear stress and the shear strain measure k is called the shear modulus:

⎛ ∂W ∂W ⎞ μ = 2⎜ + ⎟ (4.4.42) ⎝ ∂Ib ∂IIb ⎠

In the case of plane stress, the hydrostatic pressure p can be evaluated, and one finds that

2 2 σ11 = α1k ,σ 22 = α −1k , σ12 = μk (4.4.43)

The stresses acting on the deformed material are shown in Fig. 4.4.2. The unit normal e to the sloping surface (on the right hand side) is

1 e = ()e1 − ke 2 (4.4.44) 1+ k 2 and, from Cauchy’s law, the traction acting on that surface is

⎡σ 11 − kσ 12 ⎤ 1 ⎢ ⎥ t = σ 12 − kσ 22 (4.4.45) 2 ⎢ ⎥ 1+ k ⎣⎢ 0 ⎦⎥

The normal and shear traction are then (using the Poynting effect)

1 k(2 + k 2 ) k t N = 2 []()()σ 11 − kσ 12 − k σ 12 − kσ 22 = σ 11 − 2 σ 12 = σ 22 − 2 σ 12 1+ k 1+ k 1+ k 1 1 t = []k()()σ − kσ + σ − kσ = σ S 1+ k 2 11 12 12 22 1+ k 2 12

Solid Mechanics Part III 380 Kelly Section 4.4

(4.4.46)

and tend to σ 11 and σ 12 as k → 0 .

σ 22

k σ12

e

σ12

σ 22

Figure 4.4.2: stresses acting on the material in simple shear

It is clear then that the pure shear of material discussed here can only occur when normal stresses are applied to the faces. There is of course no volume change for an incompressible material; for a compressible material, 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 coordinates 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 specific constitutive laws for isotropic hyperelastic materials to expand the strain energy function into an infinite power series of the form

∞ p q r W = W (IC ,IIC ,IIIC ) = ∑α pqr ()()()IC − 3 IIC − 3 IIIC −1 (4.4.47) p,q,r=0

In the undeformed configuration, IC = IIC = 3, IIIC = 1, and the strain energy is zero. For incompressible materials, this reads as

∞ p q W = ∑α pq ()()IC − 3 IIC − 3 (4.4.48) p,q=0

The series can be written in terms of the principal stretches by noting that

Solid Mechanics Part III 381 Kelly Section 4.4

2 2 2 IC = λ1 + λ2 + λ3 2 2 2 2 2 2 IIC = λ1 λ2 + λ2λ3 + λ3λ1 (4.4.49) 2 2 2 IIIC = λ1 λ2λ3

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 μ i αi αi αi W ()λ1 ,λ2 ,λ3 = ∑ {}λ1 + λ2 + λ3 − 3 (4.4.50) i=1 α i

where the μi ,α i are experimentally determined material constants. Note that this model does not include terms involving the products of different stretches, which appear in the most general power series 4.4.48. However, the Ogden model allows for fractional powers of α i . Only three terms in the series are usually needed to correlate the model with experimental data of incompressible rubber. Typical values are

α1 = 1.3 μ1 = 6300 kPa

α1 = 5.0 μ 2 = 1.2 kPa

α 2 = −2.0 μ3 = −10 kPa

The Mooney-Rivlin Model (Incompressible)

The Mooney-Rivlin model, applicable to incompressible materials, is

W = c1 (IC − 3)+ c2 (IIC − 3) (4.4.51)

And, from 4.4.22, the stresses are then

−1 σ = 2c1b − 2c2b − pI (4.4.52)

From the simple shear example discussed in §4.4.4, the shear modulus for the Mooney-

Rivlin material is 2(c1 + c2 ), 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 c2 is usually much smaller than c1 .

Solid Mechanics Part III 382 Kelly Section 4.4

Note that the Mooney-Rivlin model can be viewed as a special case of the Ogden model, since setting n = 2, α1 = 2, α 2 = −2 in that 4.4.50, and using the incompressibility 2 2 2 constraint λ1 λ2λ3 = 1 gives

μ1 2 2 2 μ 2 −2 −2 −2 W = {}λ1 + λ2 + λ3 − 3 − {}λ1 + λ2 + λ3 − 3 2 2 (4.4.53) μ μ = 1 {}I − 3 − 2 {}II − 3 2 C 2 C

The Neo-Hookean Model (Incompressible)

The Neo-Hookean is a special case of the Mooney-Rivlin model, wherein c2 = 0 , giving

W = c1 (IC − 3) (4.4.54)

It is useful as a model of non-linear elasticity within the small-strain range.

The Varga Model (Incompressible)

The Varga model is another special case of the Ogden model:

W = c1 (λ1 + λ2 + λ3 − 3) (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 strain energy here takes the form

2 3 W = c1 (IC − 3) + c2 (IC − 3) + c3 (IC − 3) (4.4.56)

As usual, W (I) = 0 in the reference configuration and the strain energy is a convex function, increasing monotonically with IC (and attains a minimum in the reference configuration, so that ∂W / ∂IC = 0 must contain no real roots, which places restrictions on the values that the constants c1 ,c2 ,c3 can take).

The Arruda-Boyce Model (Incompressible)

The Arruda-Boyce model is based on a proposed molecular network structure to a material, and the strain energy is

⎡1 1 2 11 3 ⎤ W = μ ()IC −1 + ()IC − 9 + 2 ()IC − 27 +L (4.4.57) ⎣⎢2 20n 1050n ⎦⎥

Solid Mechanics Part III 383 Kelly Section 4.4

The 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-Rivlin model reads as

2 W = c(J −1) − 2(c1 + c2 )ln J + c1 (Ib − 3)+ c2 (IIb − 3) (4.4.58)

The stresses are then, from 4.4.8, with J = IIIb ,

2 2 σ = {}[]c()IIIb − IIIb − 4(c1 + c2 ) I + []c1 + c2 ()trb b − c2b (4.4.59) IIIb

The Neo-Hookean Model (Compressible)

Neo-Hookean models are characterised by their dependence on the 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)

λ 1 W = (ln J ) 2 − μ ln J + μ()I − 3 (4.4.60) 2 2 b

With corresponding stresses

−1 ∂W −1 ∂W σ = 2J IIIbI + 2J b ∂IIIb ∂Ib (4.4.61) = J −1[]λ ln JI + μ(b − I)

The λ,μ of this compressible Neo-Hookean model are the classical Lamé constants, which can be seen by particularising the constitutive equation to the small strain range.

Another Neo-Hookean model is

1 1 W = κ ()ln J 2 + μ(J −2 / 3I − 3) (4.4.62) 2 2 b with corresponding stresses

Solid Mechanics Part III 384 Kelly Section 4.4

−1 ⎛ −2 / 3 ⎛ 1 ⎞⎞ σ = J ⎜κ ln J + μJ ⎜b − (trb)I⎟⎟ ⎝ ⎝ 3 ⎠⎠ (4.4.63) = J −1 ()κ ln J + μJ −2 / 3devb

The Blatz and Ko Model (Compressible)

Blatz and Ko proposed a compressible model based on a combination of theoretical arguments and experimental work on foamed materials:

μ ⎡ 1 ⎤ μ ⎡⎛ II ⎞ 1 ⎤ −β ⎜ b ⎟ +β W = f ⎢()Ib − 3 + ()IIIb −1 ⎥ + ()1− f ⎢⎜ − 3⎟ + ()IIIb −1 ⎥ (4.4.64) 2 ⎣ β ⎦ 2 ⎣⎝ IIIb ⎠ β ⎦ where

ν β = (4.4.65) 1− 2ν and μ,ν denote the shear modulus and Poisson’s ratio, and f ∈[0,1] is an interpolation parameter. When the material is incompressible, IIIb = 1, and the model reduces to the 2 Mooney-Rivlin model. Also, by letting f = 1, and replacing IIIb with J , the Blatz and Ko model reduces to the Neo-Hookean model

μ μ 1 W = ()I − 3 + (J −2β −1) (4.4.66) 2 b 2 β

4.4.8 Problems

1. Evaluate the PK2 stress for the Mooney-Rivlin material (incompressible)

Solid Mechanics Part III 385 Kelly