In order to obtain the second Piola“Kirchho¬ stress in terms of the

principal directions and stretches, recall Equation (5.23) and note that the

identity, the right Cauchy“Green tensor, and its inverse can be expresses as

(see Equations (2.30) and (3.30)),

3

I= (5.69a)

N± — N±

±=1

3

»2 N ± — N ±

C= (5.69b)

±

±=1

3

’1

»’2 N ± — N ±

= (5.69c)

C ±

±=1

Substituting these equations into Equation (5.23) gives S as,

3

(2ΨI + 4ΨII»2 + 2IIIC ΨIII »’2 )N ± — N ±

S= (5.70)

± ±

I=1

Given that the term in brackets is a scalar it is immediately apparent that for

an isotropic material the principal axes of stress coincide with the principal

axes of strain. The terms ΨI , ΨII, and ΨIII in Equation (5.70) refer to

the derivatives with respect to the invariants of C. Hence it is necessary

to transform these into derivatives with respect to the stretches. For this

purpose note that the squared stretches »2 are the eigenvalues of C, which

±

according to the general relationships (2.60a“c) are related to the invariants

of C as,

IC = »2 + »2 + »2 (5.71a)

1 2 3

IIC = »4 + »4 + »4 (5.71b)

1 2 3

IIIC = »2 »2 »2 (5.71c)

123

20 HYPERELASTICITY

Di¬erentiating these equations gives,

‚IC

1= (5.72a)

‚»2

±

‚IIC

2»2 = (5.72b)

±

‚»2±

‚IIIC

IIIC

= (5.72c)

»2 ‚»2

± ±

which upon substitution into Equation (5.70) and using the chain rule gives

the principal components of the second Piola“Kirchho¬ tensor as derivatives

of Ψ with respect to the principal stretches as,

3

‚Ψ

S= S±± N ± — N ± ; S ±± = 2 (5.73)

‚»2

±

±=1

5.6.2 SPATIAL DESCRIPTION

In order to obtain an equation analogous to (5.73) for the Cauchy stress,

substitute this equation into Equation (4.45b) to give,

3

2 ‚Ψ

T

’1

σ=J = (F N ± ) — (F N ± ) (5.74)

F SF

J ‚»2

±

±=1

Observing from Equation (3.44a) that F N ± = »± n± yields the principal

components of Cauchy stress tensor after simple algebra as,

3

»± ‚Ψ 1 ‚Ψ

σ= σ±± n± — n± ; σ±± = = (5.75)

J ‚»± J ‚ ln »±

±=1

The evaluation of the Cartesian components of the Cauchy stress can

be easily achieved by interpreting Equation (5.75) in a matrix form using

Equation (2.40d) for the components of the tensor product to give,

3

σ±± [n± ][n± ]T

[σ] = (5.76)

±=1

where [σ] denotes the matrix formed by the Cartesian components of σ

and [n± ] are the column vectors containing the Cartesian components of

n± . Alternatively, a similar evaluation can be performed in an indicial

manner by introducing T±j as the Cartesian components of n± , that is,

21

5.6 ISOTROPIC ELASTICITY IN PRINCIPAL DIRECTIONS

3

n± = j=1 T±j ej , and substituting into Equation (5.75) to give,

3 3

σ= σ±± T±j T±k ej — ek (5.77)

±=1

j,k=1

The expression in brackets in the above equation gives again the Cartesian

components of the Cauchy stress tensor.

5.6.3 MATERIAL ELASTICITY TENSOR

To construct the material elasticity tensor for a material given in terms of

the principal stretches it is again temporarily convenient to consider the

™ ™

time derivative Equation (5.12), that is, S = C : E. From Equation (3.123)

™

it transpires that E can be written in principal directions as,

3 3

1 d»2 1

™ ±

W±β »2 ’ »2 N ± — N β

E= N± — N± + (5.78)

± β

2 dt 2

±=1 ±,β=1

±=β

where W±β are the components of the spin tensor of the Lagrangian triad,

that is, N ± = 3 W±β N β . A similar expression for the time derivative

™

β=1

of S can be obtained by di¬erentiating Equation (5.73) to give,

3 3

2

‚ 2 Ψ d»β ‚Ψ ™

™ ™

S= 222 N± — N± + 2 2 (N ± — N ± + N ± — N ± )

‚»± ‚»β dt ‚»±

±=1

±,β=1

3 3

2

‚ 2 Ψ d»β

= 222 N± — N± + (S±± ’ Sββ )W±β N ± — N β

(5.79)

‚»± ‚»β dt

±,β=1 ±,β=1

±=β

Now observe from Equation (5.78) that the on-diagonal and o¬-diagonal

™

terms of E are,

d»2 ™

±

= 2E±± (5.80a)

dt

™

2E±β

W±β = ; (± = β) (5.80b)

»2 ’ »2

± β

22 HYPERELASTICITY

Substituting Equations (5.80a“b) into (5.79) and expressing the components

™ ™

™

of E as E±β = (N ± — N β ) : E yields,

3 3

‚2Ψ ™ S±± ’ Sββ ™

™= 4 2 2 Eββ (N ± — N ± ) + 2 E±β N ± — N β

S

»2 ’ »2

‚»± ‚»β ± β

±=1 ±,β=1

±=β

3

4‚ 2 Ψ

= N± — N± — Nβ — Nβ

‚»2 ‚»2

± β

±,β=1

3

S±± ’ Sββ ™

+ 2 N± — Nβ — N± — Nβ : E (5.81)

2 ’ »2

»± β

±,β=1

±=β

™ ™

Comparing this expression with the rate equation S = C : E, the material

or Lagrangian elasticity tensor emerges as,

3

‚2Ψ

C= 4 2 2 N± — N± — Nβ — Nβ

‚»± ‚»β

±,β=1

3

S±± ’ Sββ

+ 2 (5.82)

N± — Nβ — N± — Nβ

»2 ’ »2

± β

±,β=1

±=β

Remark 5.6.1. In the particular case when »± = »β isotropy implies

that S±± = Sββ , and the quotient (S±± ’ Sββ )/(»2 ’ »2 ) in Equation (5.82)

± β

must be evaluated using L™Hospital™s rule to give,

‚2Ψ ‚2Ψ

S±± ’ Sββ

lim 2 2 =4 (5.83)

’

»± ’ »2 ‚»2 ‚»2 ‚»2 ‚»2

»β ’»± ±

β β β β

5.6.4 SPATIAL ELASTICITY TENSOR

The spatial elasticity tensor is obtained by pushing the Lagrangian tensor

forward to the current con¬guration using Equation (5.14), which involves

23

5.6 ISOTROPIC ELASTICITY IN PRINCIPAL DIRECTIONS

the product by F four times as,