Appendix: All Solutions to the Problems
959
∂ F
∂U
=
∂ F
∂Σ(U)
dΣ(U)
dU
− S =
dΣ(U)
dU
− S,
∂ F
∂ S
= R
T
· F − U,
∂ F
∂ u
= −F,
∂ F
∂ u∇
= R · S
The Euler equations of the functional are
dΣ(U)
dU
− S = 0 (constitutive equation, in V )
(1)
R
T
· F − U = 0 (compatibility equation, in V )
(2)
(R · S) · ∇ + F = 0 (equilibrium equation, in V )
(3)
The integral of the stress boundary corresponding to the variation of
u∇ is
S σ
(R · S · N) · δudS, the result combined it with −
S σ
δu · PdSis
S σ
(R · S · N − P) · δudS = 0, that is
(R · S · N − P)
S σ
= 0 (equilibrium condition, on S σ )
(4)
Taking the variation to S on the displacement boundary, we have
(u − u)| S u = 0 (compatibility equation, on S u )
(5)
10.47 The functional of the nonlinear generalized variational principle
(complementary energy form) is
Π
J 2
G [S, u] =
V
{Σ
c
(S) − R
T
: S + u · [(R · S) · ∇ + F]}dV
−
S σ
u · (R · S · N − P)dS −
S u
u · R · S · NdS
Given that the right polar decomposition of deformation gradient F is F = R · U,
write the Euler equations and boundary conditions of the functional.
Solution The integrand is F = Σ
c
(S) − R
T
: S + u · [(R · S) · ∇ + F], where
u · [(R · S) · ∇] = [u · (R · S)] · ∇ − u∇ : (R · S) = ∇ · [u · (R · S)] − u∇ : (R · S)
u · [(R · S) · ∇] = ∇ · [u · (R · S)] − u∇ : (R · S) = ∇ · [u · (R · S)] − R T · u∇ : S
Since F = R · U, there is R
T
· F = R
T
· R · U = I · U = U. Finding the various
partial derivatives