RECTANGULAR BEAMS 



55 



Substituting 



which reduces to 



EJcd E,d(l~k) 

 l-k 



(2) 



The stress-deformation diagram of the concrete being straight, 

 the center of action of the compressive stresses is at a point 2/3 

 of the height of OA from 0. The lever arm for the resisting 



Deformation 



Diagram 



Stress 

 Diagram 



FIG. 28. 



Cross 

 section 



moment of the summation of the compressive stresses with 

 respect to the neutral axis, is therefore represented by2/3/bd. 



The center of action of the tensile stresses is at a point distant 

 OC from 0. The lever arm for the resisting moment of the sum- 

 mation of the tensile stresses with respect to the neutral axis, 

 is therefore represented by d(l k). 



The total resisting moment of the beam is the sum of the 

 moments of the total compressive stresses and of the total tensile 

 stresses about the neutral axis. Therefore, we have 

 M= 2/3 kd(l/2f c kbd) + d(l-k)a s f s 



= 1 /3 f c k*bd* +a s f s d (1 -k) (3) 



Eliminating k between equations (1) and (2), the following 

 formula for steel ratio results 



P = 



(4) 



fc \nfc 



This formula shows that for a given concrete and ratio of 

 working stresses, p has the same value for all sizes of beams. 



Introducing the value of f a from equation (2) into equation 

 (1), we have 



l/2k 2 bd-a s n(l-k)=0 

 or 



l/2k 2 b-pbn(l-k)=0 



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