146
3 Continuum Mechanics and Nonlinear Elasticity
point, and the governing equations simplify considerably. The deformation and
stress tensors can then be written as
F = diag [F 1 , F 2 , F 3 ],
E = diag [E 1 , E 2 , E 3 ]
σ = diag [σ 1 , σ 2 , σ 3 ],
P = diag [P 1 , P 2 , P 3 ],
S = diag [S 1 , S 2 , S 3 ]
(3.245)
where “diag ” indicates a diagonal matrix. In the following equations, the summation
convention is suspended, and the chain rule ∂W/∂λ i = (∂W/∂E i )(∂E i /∂λ i ) is used
in deriving Eqs. (3.249) below. Equations that involve the gradient operator depend
on the specific coordinate system and are not listed here.
Kinematic Relations
F i = λ i
E i =
1
2 (λ
2
i − 1)
(3.246)
In terms of displacements, the forms depend on the specific coordinate system
Incompressibility
J = λ 1 λ 2 λ 3 = 1
(3.247)
Stresses
σ i = J
−1 λ i P i = J
−1 λ
2
i S i
(3.248)
Equations of Motion
Equations of motion depend on the specific coordinate system.
Constitutive Relations
σ i =
λ 2
i
J
∂W
∂E i
− p =
λ i
J
∂W
∂λ i
− p
P i = λ i
∂W
∂E i
−
J
λ i
p =
∂W
∂λ i
−
J
λ i
p
S i =
∂W
∂E i
−
J
λ 2
i
p =
1
λ i
∂W
∂λ i
−
J
λ 2
i
p
(3.249)
Before solving a problem, it is often useful to count the number of scalar
equations and dependent variables to be sure they match. Suppose all external loads,
including body forces, are given. Then, if symmetric stress and strain tensors are
used, we have six strain-displacement relations, three equilibrium equations, and
six constitutive relations. These 15 equations are needed to determine 15 unknowns:
six strain components, six stress components, and three displacement components.
If the material is incompressible, the incompressibility condition provides one more
equation to solve for the additional unknown p.