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Solutions for certain rectangular slabs continuous over flexible ...

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SOLUTIONS FOR CERTAIN RECTANGULAR<br />

SLABS<br />

The solution represented by (217) is satisfactory to use when the<br />

ratio b/a is not too small. When the ratio b/a is small, in particular<br />

when it is so small that the effect of the supports at y = ± a/2<br />

becomes negligible, a bending moment obtained from (217) is in<br />

the <strong>for</strong>m of a series whose convergence is very slow, and whose<br />

terms may be made up of small differences of large numbers. This<br />

difficulty may be avoided by putting wi in a special <strong>for</strong>m in the<br />

following manner:<br />

The deflection, wi, may be written as the sum of two parts, wo<br />

and wor, where w 0 is the deflection of the <strong>rectangular</strong> slab, b by a,<br />

simply supported on four edges, and loaded by a concentrated <strong>for</strong>ce P<br />

at the center. The remaining part, wcor, is a correction to be added<br />

to w 0 to give the effect of the continuity of the slab across the beams<br />

at x = +b/2. From Equation (109), with the modifications resulting<br />

from the position of the load and from the orientation of the<br />

axes, one finds<br />

Pa 2 1<br />

wo = - - (f cosh ax - sinh ax<br />

27rN ..... " n3 (219)<br />

+ ax cosh ax - f/ ax sinh ax) cos ay<br />

where<br />

sinh ab - ab sinh ab<br />

f l =<br />

,T f = -<br />

( 220)<br />

1 + cosh ab 1 + cosh ab<br />

The corrective deflection function is found as the difference<br />

between wi and w 0 . Thus the corrective deflection function is<br />

Wcor W- 1 WO<br />

Pab 1 ab ab (2 2 1 )<br />

=- - k - tanh - cosh ax - ax sinh ax cos ay<br />

2Tr 2 N ni,... \2 2<br />

where<br />

ab<br />

tanh - 2<br />

k 5 = 1 + b + sinh b + cosh b (222)<br />

1 + ab + sinh ab + cosh ab

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