Solutions for certain rectangular slabs continuous over flexible ...

ILLINOIS ENGINEERING EXPERIMENT STATION
The resultant bending moments over the beam are
M] = Mo )+M ] , M]0 = M)~+M ± .
J -=0 I= J
J y=0- L Jy=0
One finds
] Pv 1 1 \ V1
My = - sinh -- ,
J=o 4a B A / a
(32)
where A and B are given by (27). There is also a twisting moment
in the slab over the beam which is given by the formula
M,] = Mi>) + M( = Mi]
J y=0 L J V=0 J L=0
sin -----) sin ---
(1- u)Pv a a
8a A B
*(33)
The maximum bending moment in the slab over the beam is, therefore,
not normal to the beam at every point along its length.
The upward reaction of the cross beam upon the slab, or the
downward load on the beam, is
q = 2V (' ) = -2N - (v2wl) = -2-
J I 1=L ay J ayJ
where e is an infinitesimally small positive distance. Therefore
P 1 \ rv rv V rv/1 1 2rv \
q=- --- l sinh---cosh-+- +- cosh--1 1
2a\B AL a a a 2a\B A A a
where A and B are given by (27).