324 REINFORCED CONCRETE 



concrete beams gives for p = , varying by .001, the values of 

 bd 



k,j, C s , C c , and r, which may be used in the preceding formulas. 

 The economic percentage of steel for any given working stresses, 



F s and F c , may be determined by computing r e = and finding 



F c 

 in the table a value of p that corresponds, or nearly corresponds, 



to r e in the column headed r . 

 Jc 



EXAMPLE. Find the economic ratio of steel for n=15, 

 F s = 16,000, and F c = 500. 



F s 16,000 



SOLUTION. r e = = =32 



FC 500 



On referring to the table for n= 15, it is found that the nearest 

 corresponding value of r is 31.95, for which p is .005, which is 

 the economic ratio of steel. 



To Design a Beam. The following practical examples will 

 serve to show the way in which the table may be used in 

 designing a beam: 



EXAMPLE 1. Let the following values be given: n=15, 

 F s = 12,500, F c <= 600, M = 500,000 in.-lb., d = 22in., and = .006. 

 Required: (a) the value of b and (b) that of A. 



SOLUTION. (a) By the preceding method, find from the 

 table the economic steel ratio for the given n, F s , and F Ct 

 which is .01. As this is greater than the given /> = .006; for- 

 mula 7 must be employed. Substituting given values and 

 noting in the table that for p = .006, C s = .00531, it is obtained 

 500,000 = .00531X6X222X12,500. Whence, &=15.6 in. 



(b) A=pbd = . 006X15.6X22 = 2.06 sq. in. 



NOTE. If, in the preceding example, the given steel ratio p 

 were greater than the economic steel ratio, formula 8 would 

 have to be used. If the economic steel ratio were used, either 

 formula 7 or 8 would give the same result. 



EXAMPLE 2. Let the dimensions of the beam be fixed, as 

 &-18 in. and d = 27 in. Also, let M = 800,000 in.-lb., F s 

 -15,000, F C = 55Q, and n = 12. Required, A. 



SOLUTION. Solving formulas 7 and 8 for C s and C c , respect- 

 ively, and substituting known values, 



