Page 152 - Aircraft Stuctures for Engineering Student

136 Bending of thin plates

This equation is valid for all values of x and y so that

or in alternative form

giving

A,=- 1 amn
dD [(m2/a2) + (n2/b2)I2

Hence

in which am, is obtained from Eq. (5.29). Equation (5.30) is the general solution for a
thin rectangular plate under a transverse load q(x, y).

Example 5.1
A thin rectangular plate a x b is simply supported along its edges and carries a uni-
formly distributed load of intensity 40. Determine the deflected form of the plate
and the distribution of bending moment.

Since q(x, y) = qo we find from Eq. (5.29) that
16%
mxx . my
am, = %r sin-sin- dxdy = -
aboo a b Gmn
where in and n are odd integers. For m or n even, am, = 0. Hence from Eq. (5.30)

16q0 2 2 sin(m.rrx/a) sin(nry/b)
w=- 0)
40 m=1,3,5 n=1,3,5 mn[(m2/a2> + (n / 11
2 b2 2
The maximum deflection occurs at the centre of the plate where x = a/2, y = b/2.
Thus

(ii)

This series is found to converge rapidly, the first few terms giving a satisfactory
answer. For a square plate, taking Y = 0.3, summation of the first four terms of the
series gives
a4
w,, = 0.0443% -
Et3