102
3 Continuum Mechanics and Nonlinear Elasticity
This equation relates all three stress tensors through the deformation gradient tensor.
Solving for the pseudo-stress tensors gives
P = J F
−1
· σ = S · F
T
S = J F
−1
· σ · F
−T
= P · F
−T .
(3.120)
Any of the three stress tensors can be used to formulate and solve a given
problem. Each has both advantages and disadvantages that can depend on the
particular problem. A major consideration is symmetry. For example, we will show
later that σ is a symmetric tensor. Hence, only six of the nine components of the
Cauchy stress tensor are independent, thereby reducing the number of unknown
stresses from nine to six in the most general case. In contrast, since F is generally
not symmetric, Eq. (3.120) 1 shows that P is not always symmetric and may not be
the best choice for some problems. On the other hand, since σ = σ T , Eq. (3.120) 2
gives
S
T
= J (F
−1
· σ · F
−T )
T
= J (F
−T )
T
· (F
−1
· σ )
T
= J F
−1
· (σ
T
· F
−T ) = J F
−1
· σ · F
−T
= S.
Thus, S also is a symmetric tensor.
Regardless of which stress tensor is used to solve a problem, it is important to
remember that the Cauchy stress tensor provides the most physically meaningful
results. If P or S is chosen, σ can always be computed afterwards using Eq. (3.119).
Example 3.16 Let σ = σ ij e i e j , P = P ij e i e j , and S = S ij e i e j . Derive the
component form of Eq. (3.119).
Solution
In terms of the first Piola-Kirchhoff stresses, Eq. (3.119) yields
σ = J
−1 F · P
σ ij e i e j = J
−1
F ij e i e j
· (P kl e k e l )
= J
−1 F ij P kl e i (e j · e k )e l = J
−1 F ij P kl δ jk e i e l
= J
−1 F ik P kl e i e l = J
−1 F ik P kj e i e j .
To obtain the σ ij in terms of the second Piola-Kirchhoff stresses, we first modify the
above relation to obtain
F · S = F ik S kj e i e j .
3 Continuum Mechanics and Nonlinear Elasticity
This equation relates all three stress tensors through the deformation gradient tensor.
Solving for the pseudo-stress tensors gives
P = J F
−1
· σ = S · F
T
S = J F
−1
· σ · F
−T
= P · F
−T .
(3.120)
Any of the three stress tensors can be used to formulate and solve a given
problem. Each has both advantages and disadvantages that can depend on the
particular problem. A major consideration is symmetry. For example, we will show
later that σ is a symmetric tensor. Hence, only six of the nine components of the
Cauchy stress tensor are independent, thereby reducing the number of unknown
stresses from nine to six in the most general case. In contrast, since F is generally
not symmetric, Eq. (3.120) 1 shows that P is not always symmetric and may not be
the best choice for some problems. On the other hand, since σ = σ T , Eq. (3.120) 2
gives
S
T
= J (F
−1
· σ · F
−T )
T
= J (F
−T )
T
· (F
−1
· σ )
T
= J F
−1
· (σ
T
· F
−T ) = J F
−1
· σ · F
−T
= S.
Thus, S also is a symmetric tensor.
Regardless of which stress tensor is used to solve a problem, it is important to
remember that the Cauchy stress tensor provides the most physically meaningful
results. If P or S is chosen, σ can always be computed afterwards using Eq. (3.119).
Example 3.16 Let σ = σ ij e i e j , P = P ij e i e j , and S = S ij e i e j . Derive the
component form of Eq. (3.119).
Solution
In terms of the first Piola-Kirchhoff stresses, Eq. (3.119) yields
σ = J
−1 F · P
σ ij e i e j = J
−1
F ij e i e j
· (P kl e k e l )
= J
−1 F ij P kl e i (e j · e k )e l = J
−1 F ij P kl δ jk e i e l
= J
−1 F ik P kl e i e l = J
−1 F ik P kj e i e j .
To obtain the σ ij in terms of the second Piola-Kirchhoff stresses, we first modify the
above relation to obtain
F · S = F ik S kj e i e j .