Solutions for certain rectangular slabs continuous over flexible ...

26 ILLINOIS ENGINEERING EXPERIMENT STATION
1 (1 + A) P 7V
M() | - log, coth --
J x=./2 47r a
-- Pv 1+ (1-p)1) v coth-- 2Vv csch--2vv
2a 2a a a
M ] (1 + p) P Vh
=W - Irxr~ onih _ ~/O ~\
J x=a/2 47r " a
Pv P p- (1-)rv-- coth- 2iry
csch
2iy
-- ,
2a 2a a a
M() = 0.
xy I -a/2
The resultant bending moment My in the slab over the beam
becomes, when u = a/2 in (32),
] Pv
sinh -
a
sin --
a
M] = -_ - - -(36)
i_0 a 2yrv
cosh - + cosa
2irx
a
7V
rx
Since the corrective twisting moment in the slab over the beam is
zero, according to (35), the resultant twisting moment is given completely
by the effect of a concentrated load on the infinitely long slab
without the cross beam. The equation for the twisting moment in
the slab over the beam is, therefore,
cosh - cos -
U ] = 1 (1- ) Pv a a
M \ =M (0 \ =
(37)
,=0o J= 2a 2-v 2x (37)
cosh - + cos -
a
a
Westergaard* gives numerical values and curves obtained from an
*H. M. Westergaard, Computation of Stresses in Bridge Slabs Due to Wheel Loads, Public Roads,
V. 11, No. 1, March, 1930, p. 12-15. See Westergaard's Equation 75, Figures 12 and 14, and Table 5.