572 THE DESIGN OF STEEL DETAILS. 



In selecting bars in tension the area is determined by the formula: 



CHAP. XVII. 



where A is the required area, P the total tension in the bar and ft the allowable unit tensile stress. 

 The following problems are given to illustrate the use of the tables in selecting the details for 

 bars, etc. 



Loop Bar. Select a loop bar to carry a tensile stress of 48,000 lb., one end passing around a 

 3 in. pin and the other end around a 3^ in. pin, the center to center distance between pins being 

 30' o". 



References. Specification 8, p. 55; 33, p. 57; 84, p. 60; 91, p. 61; 104, p. 61; 108, 

 p. 62; 116, p. 62; 37, p. 141; 49, p. 142; 61, p. 142; 14, p. 206; 36, p. 206; 15, p. 209; 

 36, p. 210; 230, p. 363; 8, p. 379; 42, p. 381; 28, p. 385. 



Solution. Using an allowable unit stress of ft = 16,000 lb. per sq. in., the area required is, 



P 48,000 



A = -r = -^ - = 3-OO sq. in. 

 /t 16,000 



A bar i% in. square has an area of 3.06 sq. in. (Table 6), and a 2 in. round bar has an area of 3.14 

 sq. in. (Table 6). Either bar could be used. Using the i% in. square bar the additional length 

 required to pass around a 3 in. pin is i' n" (Table 92), and for a 3^ in. pin is 2' i", making it 

 necessary to add 4' o" to the center to center distance of pins to obtain the total length of bar. 



If a turnbuckle is used the upset required on a i % in. square bar is 2^ in. in diameter and 5^ 

 in. long (Table 89), requiring 4^ in. extra material to make each upset, or 9 in. for the two up- 

 sets. The weight of a turnbuckle for a 2^ in. screw is 25 lb. (Table 94). The clearance between 

 the ends of the screws for all turnbuckles is 5 in. (Diagram at top of Table 92). 



The total length and weight of the i% in. square bar is therefore: 



c. to c. of pins, less 5 in., = 29' 7 

 Material for 2 loops = 4' o 

 Material for 2 upsets = o' 9 

 One Turnbuckle 

 Total Length 



of i% in. square bar, @ 10.41 lb. per ft. (Table 6) = 308.0 lb. 

 of i% in. square bar, @ 10.41 lb. per ft. (Table 6) = 41.6 lb. 

 of i% in. square bar, @ 10.41 lb. per ft. (Table 6) = 7.8 lb. 

 @ 25 lb. (Table 94) = 25.0 lb. 



Total Weight = 382.4 lb. 



= 34' 4" 



If a sleeve nut is used, instead of a turnbuckle, its weight for a 2 J^ in. screw, is 19 lb. (Table 

 94). The clearance between the ends of the screws is 3 in. for all sleeve nuts (Diagram at the top 

 of Table 92). 



in. square bar when a sleeve nut is used is therefore: 



The total length and weight of I ^ 

 c. to c. of pins, less 3 in., = 29' 9" of I 

 Material for 2 loops = 4' o" of I 

 Material for 2 upsets = o' 9" of I 

 One sleeve nut 



Total Length 



= 34' 6" 



in. square bar, @ 10.41 lb. per ft. (Table 6) = 309.8 lb. 



in. square bar, @ 10.41 lb. per ft. (Table 6) = 41.6 lb. 



in. square bar, @ 10.41 lb. per ft. (Table 6) = 7.8 lb. 



@ 19 lb. (Table 94) = 19.0 lb. 



Total Weight = 378.2 lb. 



Bar with Clevises. Select a bar to carry a tensile stress of 48,000 lb., the ends to be held 

 by clevises, the distance center of pins being 12' o". 



References. Same as for loop bar, also 41, p. 58; 39, and 41, p. 141; 17, 18, and 19, 

 p. 209. 



Solution. Using an allowable unit stress of ft = 16,000 lb. per sq. in., the area required is, 



_ P _ 48,000 _ 



ft ~ 16,000 



A bar i% in. square has an area of 3.06 sq. in. (Table 6), and a 2 in. round bar has an area of 3.14 

 sq. in. (Table 6). Either bar could be used. Using the i% in. square bar a No. 6 clevis is 

 required (Table 93). 



