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 SLABS<br />
designated by qo, the deflection of the slab may be found from (113)<br />
by replacing P by (qo du) and then integrating with respect to u.<br />
One finds the deflection of the slab to be<br />
2 qoa 3 1<br />
w = -• E - (f2 cosh ay -<br />
f4 ay sinh ay) sin ax. (139)<br />
The deflection of either edge beam is<br />
2qoa 3 1<br />
] 2a d f, sin ax<br />
Z y=b 7 4 N n 4<br />
4qoa 4 1<br />
- EsI s 1 - (1 - fs) sin ax<br />
rSiE212 i,.,.- n 5<br />
(140)<br />
qox<br />
24 Ed 2<br />
(a 3 - 2ax 2 + x3) -<br />
4qoa 4<br />
75E212<br />
E<br />
1<br />
. n5<br />
f, sin ax<br />
where<br />
Af 8 = (3 + p) (1 - g) sinh 2ab - 2(1 - 0)2 ab.<br />
(141)<br />
The curvatures of the slab at the center are<br />
.w2 2q o a (- 1) 2<br />
4X 2 J=a/2 ..<br />
y=0<br />
82w _ -y2W] 2qoa 2a E (-1) (2-<br />
- 2 (f2 - 2f4).<br />
y2 Jx-a/2 7N .. n<br />
y=0<br />
2<br />
(142)<br />
The bending moment in either edge beam is<br />
2qoa 2 H 2 1<br />
Mbeam = -1 f 7 sin ax<br />
72 ,3.,- n 2<br />
qox 4qoa 2 1<br />
- (a - x)- - - sin ax.<br />
2 r 3 1,3.6,.. n<br />
(143)
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