920
Appendix: All Solutions to the Problems
y| x=
l
2
=
ql
4
8π 4 E I
∞
n=1
1
n 4 (1 − cos nπ) =
ql
4
8π 4 E I
π
4
48
=
ql
4
384E I
(18)
This is the same as the result obtained by directly using Eq. (12).
9.8 Let there be completely flexible thin film in the Ox y plane, the periphery of the
film is simply supported, and it bears the tension τ in per unit length, the surface
density of the film is ρ, and assuming that it is a constant everywhere in the film. If
a normal distributed pressure q(t, x, y) acts on the film, thus the film produces the
deflection w, let the film be a small deformation. Find the equation of motion for the
film and prove that the static deflection satisfies the equation
∂
2 w
∂ x 2 +
∂
2 w
∂ y 2 = −
q
τ
Hint: According to the theory of differential geometry, the surface area of the film
in the deformation condition is
A =
¨
S
1 + w 2
x + w 2
y dxdy
Solution Let the equation of the film be
w = w(x, y)
The kinetic energy of the film is
T =
¨
S
1
2
ρw
2
t dxdy
The area before the film bearing the load is
S =
¨
S
dxdy
The energy of deformation after the film stretched is proportional to the change
of the area of the film, the proportionality coefficient is τ (tension), at this point, the
energy of deformation of the film is
U = τ
¨
S
(
1 + w 2
x + w 2
y − 1)dxdy
In the case of small deformation, using the Taylor formula, there is
1 + w 2
x + w 2
y ≈ 1 +
1
2
w
2
x +
1
2
w
2
y