CONCRETE DESIGN 297 



the girder is limited to that shown. The steel in the bottom 

 should be kept up 2 in. to protect it from fire, so that the 

 effective depth, or d, is only 18 in. In the first place, the 



12 20 

 beam itself weighs 150X X- X 1 =250 Ib. per ft. The 



total load to be carried is therefore 450 + 250 = 700 Ib. per f t. , 

 or 700X30 = 21,000 Ib. in all. The span is here taken, for 

 convenience, as the clear distance between supports and 

 not the distance from center to center of support, which 



W/ 



is more correct. The maximum bending moment is 



8 



= 21,000 X = 78>750 ft _ lb f or 78,750X12 = 945,000 in.-lb. 



8 

 First find A by the formula given for rectangular beams, 



F s = - . Assume that j=l and that F s = 16,000. Then, 

 Ajd 



945,000 



If the beam were reinforced at the bottom only, the 

 allowable steel would be about .0075 X&c?=. 0075 X 12 X 18 

 = 1.62 sq. in. The excess of steel is then 3.75-1.62 = 2.13 

 sq. in. It is for this reason that reinforcement must be put 

 at the top of the beam; if it were not put there, the stress 

 in the concrete at that place would be too great. The 

 amount of steel required at the top is usually about two 

 and one-quarter times the excess at the bottom, or, in the 

 problem at hand, 2^X2.13 = 4.79, say 4f, sq. in. Assume 

 that this steel is placed 2 in. from the top. 



This completes the approximate design of the beam; that 

 is, its size and the amount and location of the steel at the 

 top and bottom have been assumed. It now remains to see 

 whether or not the values assumed will be safe. First k 

 must be found by the formula given, which is 



' 



The values of p. p' d. and d' are: p = '- -- =.01736 



12X18 



