Solutions for certain rectangular slabs continuous over flexible ...

SOLUTIONS
FOR CERTAIN RECTANGULAR SLABS
q, = E,I - = -2N - (V1) . (148)
dx 4 Vy ,=o
From this condition, one finds
b 1 + (1 + av) e- *
(
ki = (149)
4
-H +fl
nTrH1
The deflection of the center beam may then be expressed by the
equation
Pa 2 1 4
z = - --- ki sin au sin ax. (150)
S rWN .2.. n 3 nirH1
The deflection of each of the outer beams is found to be
1 Pa 2 1
Z2 = W2 =- E (fA - k 1 f 2 ) sin au sin ax. (151)
where the quantities f2 and fr are given in Appendix B.
The deflection of the slab for y > 0 has been expressed by Equations
(145) and (146) as the sum of two parts: [1] the deflection of
the infinitely long slab of span a due to the pair of concentrated loads
shown in Fig. 26, and [2] a correction Wcor = w' + w. Thus:
Pa 2 1
Weor -- 7 (bi cosh ay - ci ay sinh ay) sin au sin ax
73N n. na
Pa 2 kci (152)
- - (fi cosh ay - sinh ay + ay cosh ay
- f3ay sinh ay) sin au sin ax.
The quantities bi, ci, ki, fi and fs are stated in Appendix B.
25. Concentrated Load Over the Center Stringer.-By letting v = 0
in the preceding solution one obtains the solution for the special case