Solutions for certain rectangular slabs continuous over flexible ...

ILLINOIS ENGINEERING EXPERIMENT STATION
due to loads distributed along the lines y = b/2 and y = -b/2,
respectively.
The boundary condition, from which k is to be determined, is
q=EEi = V'4 - V
d= L I =b/2+ Ldx] yb/2-e
= - 2N - (2w')
L
y-y b/2+e
where q is the load on the beam, positive downward, El and I, are the
modulus of elasticity and moment of inertia of the beam, respectively,
z is the downward deflection of either beam, and e is an infinitesimally
small positive distance. From this boundary condition one finds
ab
1 +--
2
k = (52)
(1+ab)e 2 + 1+- en
2
The deflection of each beam is
SPa
2
1 4
z = w = - - k- sin au sin ax, (53)
vy=b/2 21rN ,,...- n nnrH 1
and the bending moment in each beam is
d 2 z 2Pa 1
Mbeam = -EiI 1 - = - ,.- k sin au sin ax. (54)
dX 2 r 2 1,23- n2
If M( ) ) and M( 0) designate the bending moments due to the load P
on the infinitely long slab without cross beams, the resultant bending
moments under the load are
- a
Mj =MF -- -- (1+p)+(1-p)- sin 2 a,
M -u M -) u 7 1,2,3. n 2J
S(55)
Sh Pr ke r abe
M, =M -- - (1+U) -(l-M)- sin 2 au,
-u r-u i,- n 2
-0 vy-0
the terms given by the series being the effects of the cross beams.