Appendix: All Solutions to the Problems
899
8.16 Given the circular thin plate as shown in figure of the problem 8.16, the radius
is R, the periphery is fixed, subject to the uniform distribution load q, let the deflection surface function be w = a
1 −
r
2
R 2
2
, using the Galerkin method to find the
deflection.
Solution The differential equation of circular thin plate is
2 w =
d
4 w
dr 4 +
2
r
d
3 w
dr 3 −
1
r 2
d
2 w
dr 2 +
1
r 3
dw
dr
=
q
D
(1)
The deflection surface function is
w = a
1 −
r
2
R 2
2
= a
1 − 2
r
2
R 2 +
r
4
R 4
(2)
The all-order derivatives are
d
4 w
dr 4 =
24a
R 4 ,
2
r
d
3 w
dr 3 =
48a
R 4 ,
1
r 2
d
2 w
dr 2 =
12a
R 4 −
4a
R 2 r 2 ,
1
r 3
dw
dr
=
4a
R 4 −
4a
R 2 r 2 (3)
Substituting Eqs. (3) into Eq. (1), we get
2 w =
64a
R 4 =
q
D
(4)
The Galerkin equation is
R
0
64a
R 4 −
q
D
1 −
r
2
R 2
2
dr = 0
( 5 )
The integral result of Eq. (5) is
8R
15
64a
R 4 −
q
D
= 0
( 6 )
or
a =
q R
4
64D
(7)
In fact, Eq. (7) can be obtained directly by Eq. (4). Substituting Eq. (7) into Eq.
(2), we get
w =
q R
4
64D
1 −
r
2
R 2
2
(8)