Solutions for certain rectangular slabs continuous over flexible ...

ILLINOIS ENGINEERING EXPERIMENT STATION
bi = (fi - 1) cosh av - (fj - 1) av sinh av, (99)
2 (1 - ,)2
c1 = (fa - 1) cosh av - av sinh av. (100)
A
The quantities fl, fa and A are independent of the position of the load
and are listed in Appendix B.
The deflection of each of the edge beams is found to be
1 Pa 2 1
z = Ws = -- f5 sin au sin ax, (101)
3
Sy b 7 N 1,2,3, -
where the quantity f5 is given in Appendix B.
in the edge beam is then
The bending moment
PaH 2 1
Mbeam -- -f5 sin au sin ax. (102)
7• 1,2,3,a" n
As the pair of loads traverses the span, a, the moment in the edge
beam varies until it becomes a maximum at x = a/2 when u = a/2.
The value of this maximum moment is
PaH 2 f5
max. Mbeam = - - (103)
The curvatures due to w' are
a 2 w' P 1
-x- --- E -(b 1 cosh ay-c 1 ay sinh ay) sin au
' 1
sin ax,
,rN ,23,•.. n
ay 2
'9w' P 1
=-- -[(b- 2c 1 ) cosh ay-ci ay sinh ay] sin ausin ax.
7rNw• .. n
(104)
At y = 0 these become
0 2 w' - P 1.*
- = -- - bi sin au sin ax,
Ox2 Y=0 7rN . n
0 2 w' 1
2W - P - (bi - 2cj) sin au sin ax.
y 2 J =0 rN u1,2,3,'- n
(105)