98
3 Continuum Mechanics and Nonlinear Elasticity
Solution
The first Piola-Kirchhoff stress vector is simply
T
0
x =
df x
dA 0
x
=
df x
dA 0
x
e x .
To compute the Cauchy stress vector, we need the deformed area dA x . The
prescribed deformation stretches and shears the element into a parallelogram while
maintaining its vertical dimension constant. Direct geometric analysis gives dy =
dY
√
1 + k 2 and, therefore, dA x = dA 0
x
√
1 + k 2 (see Fig. 3.15b). With this result,
the Cauchy stress vector is
T x =
df x
dA x
=
df x
dA 0
x
e x
√
1 + k 2
=
T 0
x
√
1 + k 2
.
Finally, the second Piola-Kirchhoff stress vector requires d ˜ f, as provided by
Eq. (3.102). With (X 1 , X 2 ) = (X, Y ) and (x 1 , x 2 ) = (x, y), the deformation
gradient tensor is given by
F =
F ij
=
∂x i
∂X j
=
λ k
0 1
= λ e x e x + k e x e y + e y e y ,
and Eq. (2.25) yields
F
−1
=
λ −1 −kλ −1
0
1
= λ
−1 e x e x − kλ
−1 e x e y + e y e y ,
which can be verified by computing F · F −1 = e x e x + e y e y = I. Substitution into
Eq. (3.102) gives
d ˜ f = F
−1
· df
= (λ
−1 e x e x − kλ
−1 e x e y + e y e y ) · (df x e x )
= df x λ
−1 e x ,
and (3.103) gives
˜
T x =
d ˜ f
dA 0
x
=
df x
dA 0
x
e x
λ
=
T 0
x
λ
.
It is instructive to note that dA x also can be computed using Eq. (3.87), i.e.,
dA x = J dA
0
x · F
−1 .
3 Continuum Mechanics and Nonlinear Elasticity
Solution
The first Piola-Kirchhoff stress vector is simply
T
0
x =
df x
dA 0
x
=
df x
dA 0
x
e x .
To compute the Cauchy stress vector, we need the deformed area dA x . The
prescribed deformation stretches and shears the element into a parallelogram while
maintaining its vertical dimension constant. Direct geometric analysis gives dy =
dY
√
1 + k 2 and, therefore, dA x = dA 0
x
√
1 + k 2 (see Fig. 3.15b). With this result,
the Cauchy stress vector is
T x =
df x
dA x
=
df x
dA 0
x
e x
√
1 + k 2
=
T 0
x
√
1 + k 2
.
Finally, the second Piola-Kirchhoff stress vector requires d ˜ f, as provided by
Eq. (3.102). With (X 1 , X 2 ) = (X, Y ) and (x 1 , x 2 ) = (x, y), the deformation
gradient tensor is given by
F =
F ij
=
∂x i
∂X j
=
λ k
0 1
= λ e x e x + k e x e y + e y e y ,
and Eq. (2.25) yields
F
−1
=
λ −1 −kλ −1
0
1
= λ
−1 e x e x − kλ
−1 e x e y + e y e y ,
which can be verified by computing F · F −1 = e x e x + e y e y = I. Substitution into
Eq. (3.102) gives
d ˜ f = F
−1
· df
= (λ
−1 e x e x − kλ
−1 e x e y + e y e y ) · (df x e x )
= df x λ
−1 e x ,
and (3.103) gives
˜
T x =
d ˜ f
dA 0
x
=
df x
dA 0
x
e x
λ
=
T 0
x
λ
.
It is instructive to note that dA x also can be computed using Eq. (3.87), i.e.,
dA x = J dA
0
x · F
−1 .