576 THE DESIGN OF STEEL DETAILS. CHAP. XVII. 



Many other angles might be chosen but in no case could an angle smaller than 3" X 3" be 

 used, for the requirement for l/r would not be satisfied. Larger angles will give lighter sections 

 and be more rigid. Any angle 3%" X 3^" has a radius of gyration, r, of about 0.69 (Table 23), 

 giving an l/r of about 104, and an allowable unit stress of about 8,700 Ib. per sq. in. and requiring 

 an area of 2.47 sq. in., which would be provided by one angle 3^" X 3^" X %". The minimum 

 angle satisfying the l/r requirement is found as a guide in the selection of sections but is rarely a 

 satisfactory section, except for long members with low stresses such as lateral bracing. Table 41, 

 Part II, gives the safe loads for single angle struts fastened by both legs. 



See also 26, p. 203; 45, p. 203; "Fastening Angles," p. 207; 20, p. 209. 



If the angle is fastened by one leg only as in Fig. 3, the load is eccentric and the problem is 

 more difficult. An approximate solution is to consider only the area of the attached leg as effect- 

 ive. As before the least radius of gyration must be not less than 0.58 in., which corresponds to an 

 allowable unit stress of 7,300 Ib. per sq. in., requiring the area of the attached leg to be at least 2.95 

 sq. in. The requirement for radius of gyration would be satisfied by any 3%" X 3" angle, but 

 to provide 2.95 sq. in. of area if attached by the 3,V in. leg the thickness would have to be 2.95 

 -r- 3.50 = 0.85 in. requiring a 3^" X 3" X %" angle, which is a very poor section and would 

 be much heavier than a section with longer legs to satisfy the same conditions, and much less 

 rigid. The least radius of gyrations of any 5" X 3/^" angle is about 0.76 in. (Table 24), and the 

 allowable unit stress will be 



72 

 f e = 16,000 70 l/r = 16,000 70 X - J ? = 9,370 Ib. per sq. in., 



requiring an area of the attached leg of 



P 21,500 



A = -r = -- = 2.30 sq. m. 

 / 9,370 



2.-IQ 



which would be provided by a 5" X 3/^" angle of thickness equal to - = .46 in. An angle 



5" X 3^2" X 1 A" could be used with the 5 in. leg attached. 



Double Angle Strut. The member a-b Fig. 5 is to consist of two angles back to back sepa- 

 rated by % in. connection plates at the ends and washers % in. thick in the body of the member. 

 Design for a compressive stress of 50,000 Ib. 



References. 34, p. 57; 84, p. 60; 93, p. 61 ; 100, p. 61 ; 38, p. 141 ; 60, p. 142; 45, 

 p. 206; 16, p. 209; 20, p. 209; 231, p. 363; 10, p. 379. 



Solution. Using f e = 16,000 70 l/r Ib. per sq. in. as the allowable unit stress, and 125 as 

 the maximum value for the ratio l/r, the minimum value for r is found as follows 



/ 8 X 12 



Ir = 125, or r = = = 0.77 in. 



125 125 



The lengths about axes X-X and YY are equal, so that for a well designed member the radii 

 of gyration about the two axes should be as nearly equal as practicable. This condition is satis- 

 fied by using angles with unequal legs, short legs turned out. 



A member composed of two 2^" X 2" angles, % in- back to back, with short legs turned 

 out will have a least radius of gyration of about 0.78 in. (Table 40), the value for axis X-X being 

 about 0.78 in. and Y-Y about 0.95 in. The allowable unit stress is then / = 16,000 70 l/r 



8 X 12 



= 16,000 70 X - = 7,39O Ib. per sq. in., requiring an area of 

 0.7" 



P 50,000 



A = -j- = ~ = 6.76 sq. in. 

 / 7,390 



This area cannot be supplied by two 2^" X 2" angles, but even though it could, larger 

 angles would be more economical as well as more rigid. The minimum angle satisfying the l/r 



