300 CONCRETE DESIGN 



Find F s by the formula M 



and find F c by the formula 



_ Mkd, 

 c ~ 



These values of F s and F c must be less than the allowable 

 values. 



These formulas neglect the compression in the stem. 

 For approximate results, the formulas for rectangular beams 

 may be used. 



As an example, design a beam to carry 5,184 Ib. per ft. 

 on a span of 10 ft. The load given includes the weight of 

 the beam, and the floor slab is 5 in. thick. The bending 



5 184 X 10 X 10 

 moment is - = 64,800 ft.-lb., or 64,800X12 



O 



= 777,600 in.-lb. Assume that ^ = 16,000 Ib., F c = 650 Ib., 

 and = 15, and that d = 16 in. and 6 1 = 16 in; b is governed 

 by the preceding rules. It must not exceed one-fourth the 

 span, or 30 in., and its overhang must be less than four times 

 the thickness of the slab, which would limit its width to 

 16 + 2X4X5 = 56 in. Therefore b is taken at 30 in. Solving 

 for A in the formula, M = A (d-$t)F s , 777,600 = A (16-2$) 



16 ' 000 - , 2X15X16X3.6 + 30X25 



Therefore, A =3.6 sq. in.; kd= 



2X15X3.6 + 2X30X5 



3.6X14.1 

 which is safe, and 



777,600X6.073 



777 600 



=15,319 Ib. per sq. in., 



which is also safe. 



Shear and Bond. One method of failure of beams that is 

 common is shown in Fig. 2. Such cracks are caused by 

 shear or diagonal tension. They are prevented by bend- 

 ing up some of the main reinforcing rods to make truss 

 rods and by the use of U-shaped stirrups. The size and the 



