Solutions for certain rectangular slabs continuous over flexible ...

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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.-
SOLUTIONS FOR CERTAIN RECTANGULAR SLABS VII. EXAMPLES OF SPECIAL SOLUTIONS OBTAINABLE FROM THOSE GIVEN IN PREVIOUS CHAPTERS 29. Rectangular Slab Having Two Opposite Edges Simply Supported and Two Edges Supported on Beams Which Are Continuous Over Interior Supports; Concentrated Load on the Longitudinal Center Line.-The slab is shown in Fig. 30. If the concentrated interior -yl'Confnuous F-'xi/be Beam Ez 2 7 p1/ 7 b __________ a ----- / FIG. 30 reactions are designated by Q, one may write an equation for the deflection of the slab, for y > 0: Pa 2 1 w -- (f, cosh ay - sinh ay 27r'N .a-n.. + ay cosh ay - f3 ay sinh ay) sin au sin ax (1 74 ) Qa2 1 NQa - (f2 cosh ay - fh ay sinh ay) sin av sin ax. 7rN 12, n 3 This equation is simply the sum of two equations of the form of (109) and (113), respectively. The deflection of the continuous edge beam is S Pa 2 1 ] Z = W = -- 2 3 N f2 sin au sin ax y-b( 2*N n_. 3 » 7 2 nQa 1 [(175) - - f7 sin av sin ax. 7r1N 1,7,.. n 3
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H ILL I N 0 I S UNIVERSITY OF ILLIN
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SOLUTIONS FOR CERTAIN RECTANGULAR S
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