110 REINFORCED CONCRETE CONSTRUCTION 



In the case of a single concentrated load at the center of span, 

 the shear V, due to a given load W, is 1/2W, and the moment M 

 is 1/4WI. Hence, 



But from preceding formulas, we have V = v'bjd and M 8 =pf f jbd 2 , 

 in which t/ = allowable shearing stress and f s = working stress 

 in steel. Substituting, we have 



from which 



d v' 

 For a uniformly distributed load, a similar process gives the ratio 



d~ v' 

 For beams loaded with equal loads at the third points, 



d v> 



Taking for example, v'=40 Ib. per square inch, /, = 16,000, 

 f e = 650, n 15, and, using an average value of 7/8 for /, we have 



the following ratios for -. 



For concentrated load at center of span*. ............... -, 6.16 



For uniformly distributed load ........................ ^ = 12.32 



For equal loads at the third points .................... -,= 9.24 



It should be clear that the strength of beams of greater relative 

 length than obtained by the formulas will be determined by 

 their moment of resistance, while that of shorter beams by their 

 shearing resistance. 



In the case of continuous beams the above formulas will apply 

 if I is taken as the length between points of inflection. It is often 

 convenient to know the extreme limit in design. The Joint 

 Committee recommends 120 Ib. per square inch for the shearing 

 strength of concrete when adequately reinforced against diagonal 

 tension. This is a low figure but is adopted in order to prevent 

 any likelihood of cracks opening up in the concrete. Suppose 



then, it is required to know the minimum value of -3 for a given 



