CONCRETE DESIGN 293 



The first of these will be found to be the more convenient 

 to use in most cases, and it may be transposed to the fol- 

 lowing: M = F s Ajd. 



As an example, design a girder on a 20-ft. span to carry 

 a load of 700 Ib. per ft. This load includes the weight of the 

 girder. The total load is 20X700 = 14,000 Ib. The bend- 

 ing moment is = '- = 35,000 ft.-lb., or 35,000 X 12 



8 8 



= 420,000 in.-lb. Assume the effective depth of the girder 

 that is, the distance from the top to the center of the steel 

 reinforcement, or d to be 18 in. Substituting the correct 

 values in the equation M = F s Ajd, 420,000 = 1 6,000 X A X$ 

 X 18, and A = 1.667 sq. in. 



Since A=pbd, and since p is limited to .00769, 1.667 

 = .00769X6X18 and 6 = 12.04, say 12$, in. 



All that now remains is to decide how much concrete 

 to put below the steel. This does not affect directly the 

 strength of the beam. Its principal uses are to hold the steel 

 in place and protect it from fire and rust. The Joint Com- 

 mittee recommends a thickness of 2 in. for girders, 1 in. 

 for beams, and 1 in. for slabs. Therefore, the girder just 

 designed would be 12 in. broad and 20 in. in total depth, 

 and it would have 1.667 sq. in. of steel 18 in. from the top. 



EXAMPLE. Design a floor slab on a 10-ft. span to carry a 

 load of 250 Ib. per sq. ft. This load includes its own weight. 



SOLUTION. Consider a section of the slab 12 in. wide. 

 The total load is 250X10 = 2,500 Ib. The maximum mo- 



Wl 2,500X10 



ment is - = = 3,125 ft.-lb., or 37,500 in.-lb. 

 8 8 



A = .00769 bd, but 6 = 12 in., as a strip that wide is consid- 

 ered. Therefore, A = . 00769 X 12 d=. 09 d. Substituting the 

 values for M, A, F s and / in the equation for the moment, 

 37,500 = 16,000 X. 09 dX$Xd; d*=29. 76, and d = 5.5. There- 

 fore, A = .09X5.5 = .495, say , sq. in. The slab must 

 therefore be 6 in. thick and have ^ sq. in. of steel every foot, 

 which must be placed 5$ in. from the top. Of course, all the 

 steel must not be put in one rod, but in several rods spaced 

 at equal distances so that it will average i sq. in. per ft. 



