794
Appendix: All Solutions to the Problems
∂
∂ x
u x
1 + u 2
x + u 2
y
+
∂
∂ y
u y
1 + u 2
x + u 2
y
=
u x x
1 + u 2
x + u 2
y − u x
u x u xx +u y u yx
1+u 2
x +u 2
y
1 + u 2
x + u 2
y
+
u yy
1 + u 2
x + u 2
y − u y
u x u xy +u y u yy
1+u 2
x +u 2
y
1 + u 2
x + u 2
y
=
u x x (1 + u 2
x + u 2
y ) − u x (u x u x x + u y u yx ) + u yy (1 + u 2
x + u 2
y ) − u y (u x u xy + u y u yy )
(1 + u 2
x + u 2
y ) 3
=
u x x (1 + u 2
y ) + u yy (1 + u 2
x ) − 2u x u y u xy
(1 + u 2
x + u 2
y ) 3
= 0
Because the denominator of the above equation is not zero, there is
u xx (1 + u
2
y ) + u yy (1 + u
2
x ) − 2u x u y u xy = 0
2.77 In plane stress state, the strain energy of linear elastic body is
J =
¨
D
E
2(1 − μ 2 )
(ε
2
x + ε
2
y + 2με x ε y ) +
E
4(1 + μ)
γ
2
xy
dxdy
where, E, μ are the modulus of elasticity and Poisson’s ratio for the material respectively, ε x =
∂u
∂ x
, ε y =
∂v
∂ y
, γ xy =
∂v
∂ x
+
∂u
∂ y
are the normal strains and shear strain
respectively, u, v are the displacements along the x, y direction respectively. Prove
u +
1 + μ
1 − μ
(u xx + v xy ) = 0, ,v +
1 + μ
1 − μ
(u xy + v yy ) = 0
Proof Substituting ε x =
∂u
∂ x
, ε y =
∂v
∂ y
, γ xy =
∂v
∂ x
+
∂u
∂ y
into the expression of strain
energy, we obtain
J =
E
2(1 + μ)
¨
D
1
1 − μ
(u
2
x + v
2
y + 2μu x v y ) +
1
2
(v x + u y )
2
dxdy
The Ostrogradsky equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
2
1 − μ
(u xx + 2μv xy ) + v xy + u yy = 0
2
1 − μ
(v yy + 2μu xy ) + v xx + u xy = 0
The first Ostrogradsky equation is transformed into