REINFORCED CONCRETE 325 



80MOO 



18X272X15,000 



On referring to the table for =12, it is found that for C s 

 = .00406, = .0045; also, that for C c = .lll, = .0032. The 

 former value of p being the greater, it must be used; therefore, 

 A = 6d = . 0045X18X27 = 2.2 sq. in. 



To review a beam. To review a beam means to investigate 

 one that has already been built. In this case, b, d, p, and n will 

 be known, and it will be required either to determine M for 

 given F s and F c , or to find f s and f c for a given M. 



EXAMPLE 1. Let 6=15 in., d = 30 in., and = .008. Find 

 M for n= 15, F s = 13,500, and F c = 500. 



SOLUTION. By the method already given, it is found that 

 the economic steel ratio is .0066. As this is less than the given 

 value of p, formula 8 must be employed. From the table for 

 n= 15 and = .008, C c = .168; therefore, substituting this value 

 in formula 8, M = . 168X15X302X500 =1,134,000 in.-lb. 



EXAMPLE 2. Let 6 = 18 in., d = 30 in., = .012, = 12, and 

 M = 2.000,000 in.-lb. Find f s and f c . 



SOLUTION. In the table for = 12, it is found that for p 

 = .012, C s = .0104 and C c = .178. Solving formula 6 for f s and 

 fc and substituting known values, 

 2,000,000 



2,000,000 

 fc 



.178X18X302 



Values of R for Special Constants. For the values n = 12 

 and = 15 and certain unit stresses, F s and F c , the calculations 

 in the design of reinforced-concrete beams may be effected by 

 formula 9, in which R has the value given in the follow- 

 ing tables. The economic steel ratio for each set of units of 

 these tables is printed in Italic. The application of this 

 table will best be seen from the examples that follow: 



EXAMPLE 1. Let M = 2,000,000 in.-lb.,F,= 16,000, F c = 600, 

 n=12, and 6 = 20 in. Find: (a) d and (6) A. 



