Solutions for certain rectangular slabs continuous over flexible ...

20 ILLINOIS ENGINEERING EXPERIMENT STATION
Since the beam is rigid, the total deflection must vanish on the
line y = 0. Therefore
for every value of x.
gives
J L J y=0 ==0
w] = wo + wi = 0
Substitution of (9) and (19) into this equation
a, = -(1 + av) e-a". (20)
Therefore the part of the deflection which is due to the beam reaction
is
Pa 2 1
wi- = 2 -(l+av)(1+aly)e-(v'+ll) sin au sin ax
2rN 1,2.,.. n
Pa 2 1
= 27- Z [l+a(v+'yI)]e-'(+Iul) sin au sin ax (21)
2xr3N f..... n
Pay 1
--- E--a ,
2r2N .X,.. n 2
y[e-«(+i'i) sin au sin az.
As in the previous solution, one may define a function
P 1
01 = NVw = - E - e-"(•~+Iv) sin au sin ax
7 1,2,3,.. n
(22)
(
+ - e- '' + l y l sin au sin ax.
A set of equ,2,,ations,
A set of equations,
2N - = - y--_
a82
ay
2N = y ,
8 xay Ox
(23)
v