FIGURE 2.33 Typical loading on plates and loading functions. Copyright 2005 by CRC Press a b/2

FIGURE 2.34 Rectangular plate.

x y and must be chosen such that Equation 2.6 1 satisfies all boundary conditions of the plate. Taking w2 in the form of series given in Equation 2.59 it can be shown that the deflection surface takes the form q0 / 4 -, 3 i 3 \

a ala

Observing that the deflection surface of the plate is symmetrical with respect to the x axis we keep in Equation 2.65 only the even function of y; therefore, Cm = Dm = 0. The deflection surface takes the form

Developing the expression in Equation 2.62 into a trigonometric series, the deflection surface in Equation 2.65 is written as q0a4'*1/ 4 , mpy mpy mny\ mpx

Substituting Equation 2.66 in the boundary conditions

one obtains the constants of integration Am and Bm and the expression for deflection may be written as

4qo«4 ^ 1 f, am tanh am + 2 2aimy w = —;— > —- 1--:-cos h —-L-

p5D f-r< mM 2 cos h am b m=1,3,5,... ^ m am 2y . , 2amv\ . mux . .

2 cos h am b b J a where am = (mub)/2a. Maximum deflection occurs at the middle of the plate, x = a/2, y = 0, and is given by

Solution of plates with arbitrary boundary conditions is complicated. It is possible to make some simplifying assumptions for plates with the same boundary conditions along two parallel edges to obtain the desired solution. Alternatively, energy method can be applied more efficiently to solve plates with complex boundary conditions. However, it should be noted that the accuracy of results depends on the deflection function chosen. These functions must be so chosen that they satisfy at least the kinematics boundary conditions.

Figure 2.35 gives the formulas for deflection and bending moments of rectangular plates with typical boundary and loading conditions.

2.5.4 Bending of Circular Plates

In the case of a symmetrically loaded circular plate, the loading is distributed symmetrically about the axis perpendicular to the plate through its center. In such cases, the deflection surface to which the middle plane of the plate is bent will also be symmetrical. The solution of circular plates can be conveniently carried out by using polar coordinates.

Stress resultants in a circular plate element are shown in Figure 2.36. The governing differential equation is expressed in polar coordinates as

where q is the intensity of loading.

In the case of a uniformly loaded circular plate Equation 2.70 can be integrated successively and the deflection at any point at a distance r from the center can be expressed as

where q0 is the intensity of loading and a is the radius of the plate. Q, C2, and C3 are constants of integration to be determined using the boundary conditions. For a plate with clamped edges under uniformly distributed load q0 the deflection surface reduces to w = JL (a2 - r2)2 (2.72)

The maximum deflection occurs at the center where r = 0 and is given by

q0 a4

64D v 7

Case no.

Structural system and static loading

Deflection and intensity

0 0

Post a comment