RECTANGULAR BEAMS 113 



M , either bd~ = -j-r-. or bd 2 = 7 may be used to determine 



pfs] 



cross-section, when the p used is obtained from the formula 

 1/2 



7* I(L 



fc We 



SHEAR 



BOND 

 V 



u = 



2ojd 



VERTICAL STIRRUPS 



2 7s I v'bjd 



27 



(uniform loading only) 



i = 2A^r d (see table) 



Js 



INCLINED RODS 

 I' =7^ diameters (see table) x 2 = or <-M p^J 



(uniform loading only) 



3a a f s jd 2 0.77s , . r x . 



s = ' a = o (45 degrees inclination) 



Note. Throughout the following illustrative problems, those working 

 stresses will be used which have been recommended by the Joint Committee 

 for a 2000 Ib. concrete. Allowable stress in the steel will be taken at 16,000 

 Ib. per square inch. The beams are assumed to be simply supported and to 

 rest upon concrete supports. Slide rule will be used wherever possible. 



Illustrative Problem. Design a beam to span 40 ft. and to support 600 

 Ib. per foot (including weight of beam). The breadth and depth may be 

 varied at will as far as the conditions of design are concerned, the only con- 

 dition being that the beam must have the required strength. 



1/2 

 P = ift nnn / i a nnn T = - 0077 



16,000 / 16,000 \ 

 650 \ 15X650+ / 



.0077) (15) + (0.0077) 2 (15) 2 - (0.0077) (15) =0.378 

 7 = 1-1/3 (0.378) = 0.874 



> 



M 1,440.000 



pf* (0.0077) (16,000) (0.874) 



