concrete beams gives for p A varying by .001, the values of bd' k, j, Cs, Cc, and r, which may be used in the preceding formulas. The economic percentage of steel for any given working stresses, Fs Fs and Fc, may be determined by computing re and finding Fc in the table a value of p that corresponds, or nearly corresponds, to re in the column headed r= EXAMPLE. fs Find the economic ratio of steel for n=15, 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, FS=12,500, Fc=600, M=500,000 in.-lb., d=22 in., and p=.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, Fs, and Fc, which is .01. As this is greater than the given p=.006; formula 7 must be employed. Substituting given values and noting in the table that for p=.006, Cs = .00531, it is obtained 500,000.00531XbX222X12,500. Whence, b= 15.6 in. (b) A = pbd = .006 X 15.6X22=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 b=18 in. and d=27 in. Also, let M=800,000 in.-lb., Fs =15,000, Fc=550, and n = 12. Required, A. SOLUTION.-Solving formulas 7 and 8 for Cs and Cc, respectively, and substituting known values, On referring to the table for n=12, it is found that for Cs =.00406, p=.0045; also, that for Cc.111, p=.0032. The former value of p being the greater, it must be used; therefore, Apbd= .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 Fs and Fc, or to find fs and fe for a given M. EXAMPLE 1.-Let b= 15 in., d=30 in., and p=.008. Find M for n=15, Fs=13,500, and Fc=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 p=.008, C=.168; therefore, substituting this value in formula 8, M = .168X15X302X500=1,134,000 in.-lb. EXAMPLE 2.-Let b=18 in., d=30 in., p=.012, n=12, and M=2,000,000 in.-lb. Find fs and fc. SOLUTION. In the table for n = 12, it is found that for p = .012, C1 = .0104 and Cc.178. Solving formula 6 for fs and fe and substituting known values, Values of R for Special Constants.-For the values n = 12 and n = 15 and certain unit stresses, F, and Fc, the calculations in the design of reinforced-concrete beams may be effected by formula 9, in which R has the value given in the following 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,= 600, n = 12, and b=20 in. Find: (a) d and (b) A. |