Solutions for certain rectangular slabs continuous over flexible ...

ILLINOIS ENGINEERING EXPERIMENT STATION
The bending moment in the beam is
d 2 z 2Pa 1 (1 + av) e-""
Mbeam = - EII-- = -- sin au sin ax (46)
dx 2 .... np 2 4
1+--
nrHi
which becomes a maximum when u = x = a/2 and v = 0.
these conditions
2Pa 1 1
max. Mbeam --
7 2 ,3.. n 2 , 1 l---.-
4
Under
n-rH 1
Pa 2Pa 1 1
4 72' . n 2 nrHi
1+-
4
In the slab over the beam the curvatures due to w 1 are
2
l
2 w1 = P 1 (1 + av) e-"v
-=-- - --- sin au sin ax,
x 2 J=( a 27N 7... n 4
1+n~rH,
giving corrective moments in the slab over the beam
m)1 _m M(1+ )P 1 (1+av)ēa
M ] M = ( r 12-, -3 na
=0 =0 27 1,2, .. n l÷1"----- 4
nrH 1
sin au sin ax.
When the load is over the beam, v = 0, and the resultant moments
in the slab over the beam become
1 = ] (1 + A) -
AM. = MyA -- = -- Mbeam
v-0 v-0
(1 + ±) P 1 1
= - E- sin au sin ax.
2r ,,,.. n nrH,
1+ 4 4