Appendix: All Solutions to the Problems
961
∂ F
∂ V
=
∂ F
∂ W
dw
d V
− T =
dw
d V
− T ,
∂ F
∂ T = F · R T − V ,
∂ F
∂ u
= −F,
∂ F
∂ u∇
= T · R
The Euler equations of the functional are
dW
dV
− T
= 0 (constitutive equation, in V )
(1)
F · R
T
− V = 0 (compatibility equation, in V )
(2)
T
· R
· ∇ + ¯
F = 0 (equilibrium equation, in V
(3)
The integral of the stress boundary corresponding to the variation of
u∇ is
S σ
(T
· R · N) · δudS, the result combined it with −
S σ
δu · PdS is
S σ
(T
· R · N − P) · δudS = 0, namely
T
R · N − ¯
P
S σ
= 0(equilibrium condition on S σ )
(4)
Taking the variation to T
· R on the displacement boundary, we have
(u − u)| S u = 0 (compatibility equation, on S u )
(5)
10.49 The functional of the generalized variational principles (complementary
energy form) in finite elasticity using the Atluri stresses T
and left stretch tensor V
as conjugate variables is
Π
c
g J [T
, V ] =
V
{W c (T
) − T
: R
T
+ u · [(T
· R) · ∇ + F]}dV
−
S σ
u · (T
· R · N − P)dS −
S u
u · T
· R · NdS
Given that the right polar decomposition of deformation gradient F is F = R · U,
write out the Euler equations and boundary conditions of the functional.
Solution Putting F = W c (T
) − T
: R
T
+ u · [(T
· R) · ∇ + F], where
u · [(T
· R) · ∇] = ∇ · [u · (T
· R)] − u∇ : (T
· R) = ∇ · [u · (T
· R)] − u∇ · R
T : T
Finding the various partial derivatives
∂ F
∂ T
=
∂ F
∂ W c
dW c
dT
− R
T
− (u∇) · R
T
=
dW c
dT
− (I + u∇) · R
T
=
dW c
dT
− F · R
T
∂ F
∂ u
= (T
· R) · ∇ + F
The Euler equations of the functional are