Solutions for certain rectangular slabs continuous over flexible ...

ILLINOIS ENGINEERING EXPERIMENT STATION
E
The boundary conditions at y = 0 are
8 2 w 8 2 w o,
Sy 2 2 J =0 =
0,
j
d4 z 1 (1-- a3w ,),
q=E, i- = R, =-N -(V2w) +Adx
a 4 ] J
L y ax 2 ay 1 y,
(83)
where q is the load on the edge beam, positive when downward, and
where
z = w = Wo + W±
J »=o L J yi0
is the deflection of the beam. Part of the deflection of the slab, wo,
is given by (9), and the remainder is given by a correction
Pa 2 1
y
S= -- - (c. + day) e-- sin au sin ax (84)
27r3N ,•,,. n'
where cn and d, are to be determined by the boundary conditions (83).
From the simultaneous equations in c. and d. given by the substitution
of w and z into Equations (83), one finds
2nirH 1 (1 + av) -4(1 + A) - (1 - ut) 2 (1 + av)
c. = 2nrHj + (3 + A) (1 - )
e
2nirHi - (1 - g) 2 (1 + 2av) e-'
S= 2nrHi + (3 + p) (1 - j)
(85)
where
HI1 ----- --
Ei
aN
The deflection of the beam is, therefore,
] 2Pa 2 1
2 +(1-)p)av
z-w3 (1- e-' sin au sin ax. (86)
Ja 7r"NN.7-nl 2nrH,+(3+j)(1-A)