Solutions for certain rectangular slabs continuous over flexible ...

SOLUTIONS FOR CERTAIN RECTANGULAR
SLABS
corresponding to (15), may be verified by differentiation of (21) and
(22) and substitution into (23).
From (22) one may write, by analogy with (10), (11) and (14),
the finite form of 4i, namely,
P A 1 Pv( 1 1 \ 7-(v + |yl)
S= - log- + - - - - sinh (24)
4vr Bi 4a B A! a
where
A c (v+| y\) r(x ± u)
= cosh - cos (25)
B 1 J a a
To verify this solution, note the boundary condition
0 2 w1 = a 2 wo a 2 w i
ax 2 2
jo L +x -\, ax 2
which, from (15) and (23), becomes
Substitution of 40 into this equation gives
] P A Pv (1 1 rvy
€o = -- log + - -- - sinh-- (26)
,J=o 4r B 4a B A a
where
A ry ir(x ± u)
= cosh-- - cos a (27)
B a a
It is easily verified, then, that Equation (24) for 01 has the following
properties:
(1) It satisfies V 2 1i = 0 at all points.
(2) It vanishes at x = 0, x = a and y = co.
(3) It satisfies (26) at y = 0.
Equation (24), satisfying both the differential equation and the
boundary conditions, is therefore the correct solution.