328 REINFORCED CONCRETE 



SOLUTION. (a) As p is not specified, the economic ratio of 

 steel will be used. This is given in Italic in the table for n= 12 

 as .00582. The corresponding value of R is 83.47. Then, 

 substituting in formula 9, and solving for d. 



,_ /^ooo j ooo = 



\83.47X20 



183.47X20 



(6) A = pbd=. 00582X20X34.6 = 4.03 sq. in. 



EXAMPLE 2. Let M = 800,000 in.-lb., 6 = 18, d = 27, F s 

 = 16,000, F c = 500, and n = 12. Find A. 



SOLUTION, Substituting given values in focmula 9 and 

 solving for R, ^ = ^^ = 60 97 



18X272. 



From the table the cor responding value of p is .0042. Then. 

 A = . 0042X18X27 = 2.04 sq. in. 



EXAMPLE 3. Find the safe value of M when 6 = 14, d = 30, 

 * = .006, =15, F s = 16,000, and F c =700. 



SOLUTION. From the table for the given constants, R = 85.00. 

 Therefore, M = 85X14X302= 1.071,000 in.-lb. 



Web Stresses. Two general methods are used for prevent- 

 ing failure of a beam by diagonal tension. These are: (1) 

 by bending up diagonally part of the horizontal reinforce- 

 ment,- and (2) by the use of special shear members, or 

 stirrjips. 



The following formulas may be employed for the purpose of 

 designing stirrups: 



For rectangular beams reinforced at the bottom, 



V = I~ (1) 



bjd 



For vertical stirrups, 



"* 



For stirrups inclined at 45, 



P-.T (3) 



In these formulas V is the total external vertical shear, in 

 pounds; v, the unit shear, in pounds per square inch; P, the 

 total stress in one stirrup, in pounds; and c, the horizontal 



