SLABS, CROSS-BEAMS, AND GIRDERS 149 



The percentage of steel is greater than 0.0077 and consequently the resulting 

 moment is determined by the concrete. 



M = 1/2(650) (0.392) (0.869) (12) (4.25) 2 



= 24,000 in. -Ib. 

 It is assumed to carry a load of 



Thus, this arrangement of the two reinforcing systems is satisfactory. 



The assumed and calculated weights are identical and the slab need not 

 be redesigned. 



In practice it is always more convenient to use the same sized rods and 

 to space them the same in both directions and, if the floor spans are nearly 

 square, this should be done. In the construction of the floor slab, the 

 spacing of the rods should vary as described for a square panel. It will 

 be found that web reinforcement is not necessary. 



Illustrative Problem. Design the center cross-section of the 12-ft. 

 supporting beam for the slab of the preceding problem. Assume a center 



span floor panel so that M = -^ may be used throughout. (Art. 54.) The 



beam receives its load from two floor panels. 



The triangular load on the beam due to the live load plus dead weight 

 of slab is equal to w. 2 bl (Art. 56) = (0.41)(325)(12)(13.2) =21,100 Ib. Assume 

 dead weight of stem of beam at 110 Ib. per linear foot. Then maximum 



21 000 

 shear occurs at either support and equals ^ -- f-(HO)(6) = 12,200 Ib. 



The only purpose of the concrete below the neutral axis is to bind together 

 the tension and compression flanges, and consequently its section is deter- 

 mined by the shearing stresses involved and space for the necessary rods. 

 The shearing stress v should not be greater than 120. The area b'd (unless 

 the value of / should turn out to be less than 7/8) should not be less than 

 12,200 



Some rough calculations show that if four rods are to be used and all in 

 one row, the breadth of stem necessary for the rods controls. A breadth 

 b' of 9 in. and a depth d of 16 in. (total depth 18 in.) will be tried. 



The bending moment due to the live load plus weight of slab is given 

 by the formula M = 1/12 w 2 bH (Art. 57) for a simply supported beam 

 carrying the load from one panel only. For a beam fully continuous and 

 carrying the load from two panels. 



M = l/12(0.41)(325)(144)(13.2)(2/3)(2)(12) 



= 340,000 in.-lb. 



The moment due to the dead weight of stem (113 Ib.) is 16,300 in.-lb. or 

 a total maximum moment of 356,300 in.-lb. 



The flange of the T-beam is 6 in. thick and its breadth is controlled 

 by one-fourth the span, or 36 in. Using approximate formula (b) of 

 Art. 59, 



__ M __ 356,300 



'" (/.)(<*- l/2fl " (16,000) (13.0) ~ 



