RECTANGULAR BEAMS 119 



M _ yP _ (5300)ao)(10)(12) _ ^^ ^ 



o o 



795,000 



"(0.0077) (16,000) (0.874) 

 Assume 6 = 14 in. 



3 in. 



We shall take 6 = 14 in., and d = 23 in. 



v <> = TJ = i = 82 lb - P er square inch. (Allowable 35.) 



od 



Thus web reinforcement is needed. Area of cross-section, 6d = (14)(23) = 

 322 sq. in. 



a s = (322) (0.0077) =2.48 sq. in. 



We shall select ten 9/16-in. round rods =2.485 sq. in. If all the rods 

 extended straight to end of beam, the maximum bond would be 



(26,500) 

 " = (1.77)(10)(7/8)(23) =74 - 5 lb ' per Square mch " 



The spacing of the rods is shown in Fig. 56. Weight of designed beam per 

 linear foot is 



(26)(14)(150) ,, (26)(14)(145) 



~~ -379lb,oi _ 



. . (5280) (5) (1.5) 

 Minimum distance from end of beam to edge of support is ^-T 



6.3 in., allowing 50 per cent for unequal distribution of reactions over bearing 

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

 Reviewing the beam we have designed, 



(5280)(10)(10)(12) 

 ^ = (8 ) (2.485)(0.874) ( 23 L ) = 



(2) (15,800) (0.0077) 



fc = n o7o = 64 7 lb. per square inch. 

 U.o/o 



which is perfectly satisfactory. 



Web reinforcement is unnecessary at a distance from support 



_10 (40) (14) (0.874) (23) 

 Xi ~ 2 ~ 5280 



= 5-2.13 = 2.87 ft. or 34.4 in. 



If inclined rods are employed, the total stress to be taken is represented by 

 triangle ABC, Fig. 57. BC represents two-thirds of the horizontal shear 

 at the support per 1 in. length of beam. 



2 Vb (5278) (5) 



= 3' bjd f 0.874 (23) =87? lb " 

 r = (0.7)(AD) = (0.7)(34.4)=24.1 in. 



Hence total stress to be taken by the rods 

 = ^(r) =^(24.1) = 10,600 lb. 



But the area of one rod multiplied by 16,000 gives its tensile value, or tensile 

 value of one rod = (0.2485) (16,000) =3980 lb. Thus, only three rods are 



