Solutions for certain rectangular slabs continuous over flexible ...

ILLINOIS ENGINEERING EXPERIMENT STATION
aY 2 + a -=O= 0, (68)
9) Y 9x2I J =o
(V 2 ) + (1- ) -- = 0, (69)
from the requirements that the moment and reaction must vanish
on the free edge.
Let the deflection again be made up of two parts,
w = wO + W2, (70)
the first due to the concentrated load on the infinitely long slab and
the second due to the corrective moment and reaction which must
be added along the line y = 0 to make it a free edge. Then wo is
represented by (9) and w 2 may be represented by an equation of
the form*
Pa 2 1
w 2 = , (a. + bay) e-"a sin au sin ax, (71)
2r3N 1.,.-. n'
where a, and b. are unknown parameters to be determined by the
boundary conditions.
The application of the condition equations, (68) and (69), leads
to two equations in a. and b. whose simultaneous solution gives the
results
1 - i 4 (1 + u)
a. = (- -- )2 + (1 + av) e-,
3 + p (1 -p)
2
1--•
b = - (1 + 2av) e-"".
3 +p
Thus
1 - Pa 2 1 4 ((1 + )
3 + p 2i7rN ... n (1 - ) 2 73)
+ (1 + av) + (1 + 2av) ay] e " -
(v + y ) sin au sin ax.
*A. Ngdai, Die elastischen Platten, 1925, p. 69.