116 REINFORCED CONCRETE CONSTRUCTION 



weight, design the beam, compute the weight, and if necessary revise the 

 design. Suppose a beam is to span 10 ft. and is to support 600 Ib. per foot 

 (not including weight). We shall assume the weight of beam at 85 Ib. per 

 foot. 



p = 0.0077 



k = 0.378 

 = 0.874 



103,000 



(0.0077) (16,000) (0.874) ~ 

 Assume b = 7 in. 



QfiO 



^2 = ^ = 137, orrf=11.7in. 



We will take 6 = 7 in., and d = ll 1/2 in. Thus, web reinforcement will be 

 needed. 



V (685) (5) 



Vo = bd = (7)(11.5) =43 lb ' per Square mch ' 



Allowable average shear is 35 lb. per square inch. We shall provide the web 

 reinforcement by means of vertical stirrups. Area of cross-section, 6d= 

 (7) (11. 5)= 80.5 sq. in. 



a 8 = (80.5) (0.0077) =0.62 sq. in. 



We shall select three 1/2-in. square bars =(3) (0.250) =0.750 sq. in. No 

 difficulty will be encountered in the spacing of these bars with a 7-in. width. 

 Weight of designed beam per linear foot is (total depth 13 in.) 



The assumed and calculated weights are close enough so that the beam need 

 not be redesigned. We shall, however, consider the loading as 695 lb. per 

 foot in the following computations. The maximum 



(5) (695) 

 ^ = (3)(4) ( 0.5)(7/8T(ir5) = 58 lb. per square mch. 



The bond stress is satisfactory. 



Minimum distance from end of beam to edge of support is ^-/sc^r*?* =2.2 



U)v*oU) 



in., allowing 100 per cent for unequal distribution of reactions over support- 

 ing surface. A distance of 6 in. will be taken. 

 Reviewing the beam we have designed, 



A; = 0.406 

 7=0.865 



- (2) (14,000) (0.0093) 



/ c = ^^ ' L\ - ^ = 640 lb. per square mch. 

 0.40o 



which is satisfactory. 



t = (0.012)(11.5) =0.138 in. 

 We will use U-shaped stirrups bent at the upper end and 1/4-in. round rods 



