10.6 Variational Problems of Functionals with Parallel-Type Inner …
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Since K is a symmetric tensor, there is
u∇ · K : F = K
T
· ∇u : F
T
= K · ∇u : (I + ∇u)
(4)
Because of F = I + u∇, taking the variation to F in −F : u∇ · K , there is
−δ F : u∇ · K = −u∇ · K : δ F = −u∇ · K : δu∇ = −K · ∇u : ∇δu (5)
Taking the partial derivative to the various terms with K, we obtain
∂ F
∂ K
=
∂ F
∂Σ c1 (K )
dΣ c1 (K )
d K
−
1
2
[∇u + u∇ + (∇u) · (u∇)] =
dΣ c1 (K )
d K
−
1
2
(∇u + u∇ + ∇u · u∇)
(6)
Taking the partial derivative to the various terms with u, we obtain
∂ F
∂ u
= (F · K ) · ∇ + ρ 0 b,
∂ F
∂ u∇
= u∇ · K − u∇ · K = 0
(7)
Of cause, if taking the partial derivative of F with respect to the left gradient of
u, then the result is zero too, that is
∂ F
∂∇u
= K · ∇u − K · ∇u = 0.
The system of Euler equations of the functional can be written as
(F · K ) · ∇ + ρ 0 b = 0 (in V )
(8)
dΣ
c1
(K )
d K
−
1
2
(∇u + u∇ + ∇u · u∇) = 0 (in V )
(9)
Combine the variation
A [u · δ(F · K )] · ndA of the surface integral corresponding to u · (F · K ) · ∇ term with the variation
−
A u
u 1 · δ(F · K ) · ndA of the surface integral on the displacement boundary into
A u
(u − u 1 ) · δ(F · K ) · ndA, the variation of the surface integral on the stress
boundary is
A t
(F · K · N − P 1N ) · δudA, therefore, the boundary conditions of
the functional are
u = u 1 (on A u )
(10)
F · K · N = P 1N (on A P )
(11)
Example 10.6.5 The functional of Hu Haichang-Washizu generalized variational
principle (that is the generalized total potential stationary value principle) is
Π
∗
[u, E, K ] =
V
Σ(E) +
1
2
(u∇ + ∇u + ∇u · u∇) − E
: K − u · ρ 0 b
dV