Solutions for certain rectangular slabs continuous over flexible ...

48 ILLINOIS ENGINEERING EXPERIMENT STATION
where A 1 and Bi are given by (25). From (4) and (78)
1 + - P B 1
M1 = - M' = log, . (79)
3 + A 27r A 1
Again, previous results may be used to determine M" and M", since
(79) represents the moment sum, (M, + M,), in the infinitely long
slab due to a load of 2P/(3 + p) at the point x = u, y = -v.
The third part of the correction, w'", may be written at once in
finite form, since it differs from (66) only by a constant multiplier.
Thus
1 - p' Pvy B,
w'" = loge -- (80)
3 + u 4rN A 1
Moments may be obtained in finite form from (80) by differentiation.
The corrective bending moments due to w 2 , obtained as the sum
of the effects of w', w" and w"', may be summarized as follows:
(1 + ) (5 - p) P B 1
M(2) = - -log,--
(3 + IA) 81r A,
1- P ,1 1 r(y+v)
•
+- - [(1+3)v+(1- )y](--) sinh
3+A 8a B5i A-) a
Scos 7(v + y) 7r(x - u)
cosh cos - 1 '(81)
(1 - Prvy L)2 a a
3 + ± 4a 2 Bi
c r (v + y) "(x+ u)
cosh cos - 1
a
a