248 REINFORCED CONCRETE CONSTRUCTION 



The moment of the stresses about the gravity axis, eliminating 

 /', and/ s as before, is 



or, if the quantity within the brackets is designated by L, then 



, M (11) 



M=f c bt*L, or 70=-^ 



The position of the neutral axis must be determined before 

 equation (11) can be used. Since Wx = M we may multiply 

 equation (9) by x and equate it to equation (10). Proceeding 

 in this manner the following equation results 



x 4 



By solving this formula for- ' using the values n 15 and 2r = -* 



t -o 



we have 



t 

 Diagram 14 based on equation (13) gives values of k for various 



values of p and ~- L in equation (1 1) , if solved for n = 15, and 



t 



4 

 2r=, is as follows: 



,12p. * (14) 



-~ + (3 



Diagram 15 is based on this equation and gives values of L for 

 various values of k and p . 



The method of procedure in solving problems under Case II is 

 as follows: (1) determine k from Diagram 14; (2) find L from 

 Diagram 15; (3) solve equation (11) for/ c ; (4) find unit stresses 

 in the steel irom formulas (7) and (8) . 



Illustrative Problem. A beam is 9 in. wide and 20 in. deep. The rein- 

 forcement both above and below consists of 1 steel rod 1 in. in diameter 

 embedded at a depth of 2 in. At a certain section, the normal component 

 of the resultant force is 60,0001b., acting at a distance of 3.4 in. from the 

 gravity axis. Assume n = 15. Compute the maximum unit compressive 

 stress in the concrete. 



