Solutions for certain rectangular slabs continuous over flexible ...

ILLINOIS ENGINEERING EXPERIMENT STATION
we is the deflection of an infinitely long slab due to a concentrated
load at the point x = u, y = -v. The part of the deflection w'
is given by the equation
Pa 2 1
w/ = l--- -- [1 + a (y + v)] e - " (+') sin au sin ax. (64)
2rN 21,2,..- n
A particular problem is solved, then, by taking the difference
between two solutions. Westergaard's numerical values and curves,
mentioned previously, may be used in each of these solutions.
13. The Finite Edge Fixed.-When the finite edge, coinciding with
I
FIG. 15
the x axis, is fixed as indicated in Fig. 15, one may write the expression
for the deflection of the slab as the sum of three parts:
w = wo + w' + 2w, (65)
where wo, wo and wi are given by Equations (9), (64) and (21)
respectively. In (21) one may replace ly\ by y since the slab no longer
exists for y < 0.
Since the moments, shears and reactions are obtainable in finite
form for each part of (65), they are obtainable for the entire solution.
The results obtained in previous sections may be applied to this
solution.
It is of interest to observe that the part of the deflection produced
by the fixing of the edge, obtained as the difference between (65)
and (63), namely,
Pay 1
wf = 2 (wi ± W 1 ) = -__N _ - ay e - *(v+) sin au sin ax, (66)