Solutions for certain rectangular slabs continuous over flexible ...

SOLUTIONS FOR CERTAIN RECTANGULAR
SLABS
10. The Infinitely Long Slab Supporting a Concentrated Load Midway
Between Two Flexible Cross Beams.-Because of symmetry about
the x axis as shown in Fig. 12, it is sufficient to obtain the deflection
Fia. 12
functions for y > 0. Letting wi and w 2 represent the total deflection
of the slab for b/2 > y > 0 and for y > b/2 respectively, one may
write
Pa 2 1
wo -- (1 + ay) e - " sin au sin a«x,
27rN ..... n
(47)
Pa 2 k ab\ a b
wl= wo-- - 1+ -)cosh ay - ay sinh ay e
T sin au sin ax (48)
rN,,.,... n3LR 2 /
and
Pa 2 k ab ab ab sinsin(49)
w2=w- - E - (1+ay) cosh---sinh- e-sin ausin ax(49)
rN 1.,.... n' 2 2 2 ]
where k is to be determined from the boundary condition. In Equations
(48) and (49) the correction to w 0 has been expressed as the sum
of the deflections of the infinitely long slab,
and
Pa 2 k
w' = - 2 -- (1 + aly - b/21)e-aly-b/21 sin au sin ax (50)
27rN ,.,,... n'
Pa 2 k
w" = - - (1 + aly + b/2)e-aly+b/2' sin au sin ax, (51)
2r3N 1.,.3... nf