RECTANGULAR BEAMS 57 



being 0.96 sq. in.? E 8 = 30,000,000. # c = 2,500,000. / c = 500 Ib. per 

 square inch. f a = 16,000 Ib. per square inch. 



= E 8 30,000,000 _as_ 0.96 



E c 2,500,000 ~ P ~bd~ (8H12) " 



From equation (5) 



fc = V(2)(0.01)(12)+(0.01) 2 (12) 2 -(0.01)(12) 

 = 0.384 



From equation (6) 



M c = l/2(500)(0.384)(0.872)(8)(12) 2 

 = 96,500 in.-lb. 



From equation (7) 



M s = (0.01) (16,000) (0.872) (8) (12) 2 

 = 160,700 in.-lb. 



M c is the lesser of the two resisting moments and hence controls in the 

 design =96,500 in.-lb. 



Illustrative Problem. Suppose that the beam of the preceding problem 

 is 14 in. deep and is subjected to a bending moment of 130,000 in.-lb. 

 Compute the greatest unit stresses in the steel and concrete. 



From equation (5) 



k = \/(2) (0.0086) (12) +(0.0086) 2 (12) 2 - (0.0086) (12) 

 = 0.363 



From equation (6) 



130,000= (fj\ (0.363) (0.879) (8) (14) 2 



/ c = 520 Ib. per square inch 

 From equation (7) 



130,000 = (0.0086) (/.) (0.879) (8) (14) 2 

 f a = 11,000 Ib. per square inch 



Illustrative Problem. A beam is to be figured to withstand a bending 

 moment of 300,000 in.-lb. A 1:2:4 concrete will be used with E c = 

 2,000,000 and / c = 600 Ib. per square inch. The pull in the steel is to be 

 limited to 14,000 Ib- per square inch. Its modulus of elasticity E s is 

 30,000,000. 



From equation (4) 



p = 0.0084 

 With this value of p, Eq. (5) gives 



A; = 0.391 



/ = 0.870 



Either equation (6) or (7) may now be used in determining b and d since 



