Solutions for certain rectangular slabs continuous over flexible ...

70 ILLINOIS ENGINEERING EXPERIMENT STATION
beam. The resulting expression, w = w 1 + w 2 , is applicable when
y > 0 in the slab shown in Fig. 29.
The boundary condition, for determining k 3 , is
d * z 1 9 1
- 2N (V2W!) (168)
q = Eil- =
dx 4 L ay V ,=0
where z, is the deflection of the center beam. From (168) one finds
1 - bs
k3 = (1 - b3) k 2 (169)
4
_^
+ f i
nvrHi
where k2 is given by (163) and where f 1 and b3 are given in Appendix B.
The beam deflections are
] 16pa 5 1
I = W = -- k sin ax (170)
for the center beam, and
J Y= =0 "6E l/ 1,3,,.. n 6
1| 4pa4 _.1
z2 = w = -- - (d - kaf) sin ax (171)
Sy=" TrN l,.. nf
for the edge beams, where d3 and f 2 are given in Appendix B and where
k, is given by (169).
The bending moments in the beams are
for the center beam, and
for the edge beams.
16pal 3 k3
Mbeaml = 6pa--- sin ax (172)
7 4 1,, n 4
4pa'H2 1
Mbeam2 = - (d3 - k3f2) sin ax (173)
for the edge beams.-