56 REINFORCED CONCRETE CONSTRUCTION 



from which 



k=V2pn+ (pn) 2 -pn (5) 



Sustituting the value of a s f 8 from (1) into (3), we get 



M c = l/2f c k(l-l/3k)bd 2 (6) 



Substituting the value of f c from (1) into (3) and remembering 

 that a s =pbd, 



M 8 =pf 8 (l-l/3k)bd 2 (7) 



Solving equation (1) for f c , 



It will be noted that equation (6) gives the resisting moment 

 when the maximum allowable value of f c is introduced as the 

 limiting factor, and that equation (7) gives the resisting moment 

 when the maximum allowable value of / is the limiting factor. 

 The lesser of these two resisting moments, when proper working 

 values are assigned to/ c and/, is the allowable resisting moment 

 of the beam in question. 



The distance from C to T is in most formulas denoted by jd\ 

 namely, / denotes the ratio of this distance to the total depth 

 of the beam. Since jd = dl/3kd, then /=! l/3k and the 

 above formulas may be simplified by substituting j for the 

 quantity (I -I /3k.) 



The formulas which we shall need to use in the design of re- 

 inforced concrete beams are collected for convenience as follows : 





2 -pn (5) 



or M.= __ (6) 



M 

 or M'=. (7) 



(8) 



Illustrative Problem. What will be the resisting moment (M) for a 

 beam whose breadth (6) is 8 in. with a distance from the center of the rein- 

 forcement to the compression surface (d) of 12 in., the area of steel section 



