Solutions for certain rectangular slabs continuous over flexible ...
Solutions for certain rectangular slabs continuous over flexible ...
Solutions for certain rectangular slabs continuous over flexible ...
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SOLUTIONS FOR CERTAIN RECTANGULAR<br />
SLABS<br />
The total deflection of the slab is<br />
w = Wo + w •<br />
where w 0 is the deflection of the infinitely long slab, without the cross<br />
beam, loaded as shown in Fig. 11, and w, is a corrective deflection<br />
given by the equation<br />
Pa 2 1 (1 + av)<br />
wi = Pa 2 -- 1 (1 av) (1+a•ly)e- a ( + t yl) sin ausinax. (44)<br />
2 ,2,N "4 n,<br />
nrH<br />
1<br />
The deflection of the beam is<br />
Pa 2 1 (1 + av) e-v sin a sin ax.<br />
z = -- -- sin au sin ax.<br />
27rN ..... n n'rHi1<br />
1 --<br />
(45)<br />
The curvatures due to w, are<br />
8 2 w, P 1 (1 + av)<br />
-------- -- (1+a.y|)e--a(+ I~'Ysin au sin ax,<br />
Ox 2 27rN 1..,3, n 4<br />
1+--<br />
nirHi<br />
0 2 wl P 1 (1 + av)<br />
2--- ----- -ay)e-<br />
1y 2 2irN t,..2, n 4<br />
nTrH 1<br />
-a(v+lyl)sin au sin ax.<br />
Under the load, at x = u, y = v, these become<br />
4 2 w P 1 (1 + av) 2 e- 2 a<br />
X--- = - - sin 2 Cu,<br />
Ox 2 J,- 2rN ....- n 4<br />
Y-v 1 + --- __<br />
nrH 1<br />
( 2 w, 1 P 1 (1 - a 2 2 ) e- 2 v<br />
- = - - sin 2 au,<br />
Oy' .-u 2rN 1... .. n 4<br />
»-» 1 + -- _-<br />
ndrH 1<br />
which lead to the corrective moments under the load.
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