4

4 i'a fh , mpx npy qmn — ^ / q(x,y) sin-sin—— dx dy (2.54)

From Equations 2.45 and 2.52 to 2.54 we can obtain the expression for deflection as

If the applied load is uniformly distributed with intensity q0, we have q(x y) = q0

and from Equation 2.54 we obtain

If the simply supported rectangular plate is subjected to any kind of loading given by q = q(x y) (2.52)

the function q(x, y) should be represented in the form of a double trigonometric series as oo oo q(x, y) = \ } qm„ sin-sin^ (2.53)

where qmn is given by

1 oo oo

where m and n are odd integers. qmn = 0 if m or n or both are even numbers. Finally, the deflection of a simply supported plate subjected to uniformly distributed load can be expressed as

16q0 ^^^^ sin(mpx/a) sin(npy/b) W = PD^ n^f mn((m2/a2) + (n2/b2))2 (2'57)

The maximum deflection occurs at the center. Its magnitude can be evaluated by substituting x = a/2 and y = b/2 in Equation 2.57 as

Equation 2.58 is a rapid converging series. A satisfactory approximation can be obtained by taking only the first term of the series; for example, in the case of a square plate,

4q0a4 q0a4

Assuming v = 0.3, the maximum deflection can be calculated as

q0a4

The expressions for bending and twisting moments can be obtained by substituting Equation 2.57 into Equations 2.36. Figure 2.33 shows some loading cases and the corresponding loading functions.

If the opposite edges at x = 0 and x = a of a rectangular plate are simply supported, the solution taking the deflection function as

a m=l can be adopted. Equation 2.59 satisfies the boundary conditions w = 0 and 02w/0x2 = 0 on the two simply supported edges. Ym should be determined such that it satisfies the boundary conditions along the edges y =±(b/2) of the plate shown in Figure 2.34 and also the equation of the deflection surface

04w 04 w 04w q0

q0 being the intensity of uniformly distributed load. The solution for Equation 2.60 can be taken in the form w = w1 + w2 (2-61)

for a uniformly loaded simply supported plate. wl can be taken in the form w1 = (x4 - 2ax3 + a3x) (2.62)

representing the deflection of a uniformly loaded strip parallel to the x axis. It satisfies Equation 2.60 and also the boundary conditions along x = 0 and x = a. The expression w2 has to satisfy the equation q 4 w2 q 4 w2 q 4 w2

Expansion coefficients qmn

0 0

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