580 THE DESIGN OF STEEL DETAILS. CHAP. XVII. 



References 33, p. 57; 4 2 P- 58; 45. P- 58; 14. P-'^H; 39, p. 141; 50, p. 142; 55, 

 p. 142; 17, p. 209; 29, 30, p. 210. Properties of Carnegie I-Beams are given in Tables 7 to 

 13 inclusive. Properties of Bethlehem Girder and I-Beams are given in Tables 151 to 160, 

 inclusive. 



Solution. The bending moment is 



M = y%w-P = y% X looo X I6 2 = 32,000 ft.-lb. = 32,000 X 12 in.-lb. = 384,000 in.-lb. 

 From applied mechanics, 



c 

 The section modulus required is then, 



'' I M 384,000 . , 



S =- = =?-2 - = 24.0 in. 8 

 c f 16,000 



The section modulus of a 9 in. I @ 35 lb. is 24.8 in. 3 , and of a 10 in. / @ 25 lb. is 24.4 in. 3 (Taole 

 7), either of which will carry the load, but the IO in. I @ 25 lb. being lighter is the more economical, 

 and being the minimum section is more easily obtained. 



The allowable bending moments in ft.-lb. for I-Beams, using a fiber stress of 16,000 lb. per 

 sq. in., are given in Table 7. The I-Beam could have been selected directly from the moment 

 making use of these values. The allowable bending moments for other unit stresses are propor- 

 tional. 



The safe uniform load, in tons, for I-Beams are given in Table 12, using a fiber stress of 

 16,000 lb. per sq. in. The I-Beam could have been selected directly from the load by using 

 this table. Safe loads for other unit stresses are proportional. 



If the I-Beam is not supported to prevent lateral deflection the allowable fiber stress must be 

 reduced by the compression formula as shown in Table 120. 



Design an I-Beam 14' o" long to carry a concentrated load of P = 20,000 lb. at the center 

 of the beam. The maximum moment is at the center, and is, M = %P-l = M X 20,000 X 14 

 = 70,000 ft.-lb. = 840,000 in.-lb. 



The required section modulus is, S = M/f = 840,000 -j- 16,000 = 52.5. In Table 7, the 

 lightest beam that will carry the load is a 15 in. I @ 42 lb., which has a value of 5 = 58.9 in. 3 , 

 and a bending moment of 79,000 ft.-lb. A 12 in. / @ 55 lb. will also carry the load, but is not an 

 economical section. A concentrated load, P, at the center will give the same maximum stresses 

 as a uniformly distributed load of 2P. From Table 12, a 15 in. / @ 42 lb. will carry a uniformly 

 distributed load of 22 tons, which is sufficient. 



Two I-Beams with Separators. Design a girder consisting of two I-Beams fastened together 

 by means of separators, the girder having a span of 16' o" and carrying a uniform load of 2,000 

 lb. per linear ft. 



References 33, p. 57; . 19, p. 105; 39, p. 141; 17, p. 209; 30, p. 210. 



Solution. The bending moment is 



M = | w.l 2 = | X 2000 X i6 2 = 64,000 ft.-lb. = 798,000 in.-lb. 

 From mechanics, 



The section modulus required is, 



I M 798,000 . . 



S = - = -r = ~ - - 48.0 in. 3 

 c f 16,000 



Each I-Beam must have a section modulus of f X 48.0 = 24.0 in. 3 The section modulus 

 of one 9 in. I @ 36 lb., is 24.8 in. 3 and of one 10 in. / @ 25 lb., is 24.4 in. 3 , either of which will 

 carry one-half the load, but the 10 in. / @ 25 lb. being lighter is the more economical, and being 

 the minimum section is more easily obtained. 



The allowable bending moments, in ft.-lb. for I-Beams, using a fiber stress of 16,000 lb. per 



