DOUBLE ANGLE STRUT. 577 



requirement is found so as to guide in the selection of angles but is rarely a satisfactory section, 

 c\< cpt for a long member with low stresses, such as lateral bracing. 



Try two angles 4" X 3" with the short legs turned out, Y% in. back to back. From Table 

 40 it is seen that for any thickness the least radius of gyration will be about the axis X-X, and 



8 X 12 

 will be about 1.26 in., giving an allowable unit stress of / = 16,000 70 X z~ 10,670 



Ib. per sq. in., which requires an area of 50,000 + 10,670 4.68 sq. in. The area of 2 angles 

 4" X 3" X i" 4.96 sq. in., which will satisfy the conditions. If the estimated radius of gyra- 

 tion does not agree closely enough with the actual radius of gyration, another calculation should 

 be made, but this is not often necessary. 



The spacing of the washers should be such that the //r of one angle between the washers is not 



8 X 12 

 greater than the l/r for the whole member, or l/r - -r- = 76.2, / = 76.2 X .64 = 48.7 in., 



0.64 being the least radius of gyration of one angle 4" X 3" X W (Table 24). One washer in 

 the center will be sufficient. 



\Y 

 bd 



, . _ *i 



d V a 



FIG. 5. DOUBLE ANGLE STRUT. 



If lengths about the two axes are different, as is often the case in roof trusses and portals, the 

 greatest value for l/r should be used, the corresponding length and radius of gyration being taken; 

 for example in designing the member b-d, Fig. 5, as a strut the length corresponding to the axis 

 Y-Y is 12' o", and to the axis X-X is 6' o". To make an efficient member the long legs should 

 be turned out and r v should be equal to 2 X r,. 



The minimum allowable values of r x and r v are found as follows, 



/ r 6 X 12 

 l/r = 125, r x = -__=o.58in.; 



125 125 



in 



From Table 39 it is seen that any 2%" X 2" angle with long legs turned out and % in. back 

 to back is the smallest angle which will satisfy the requirements for l/r, r x = 0.58 in. and r y = 1.26 

 in. (approx.). The values for l/r are 124 and 114, respectively, 124 being the greater. The 

 allowable unit stress is then 



f c = 16,000 70 X 124 = 7,320 Ib. per sq. in. 

 If the stress in b-c is the same as that in c-d, 19,000 Ib. compression, the required area is, 



P 19,000 



A = -f = 2.60 sq. in. 

 / 7.320 



which will be taken by 2 angles 2%" X 2" X 5/16", having r x = 0.58 in., and r y = 1.26 in. 

 (Table 39). If the stresses in b-c and c-d are not equal proceed as above and design for the 

 maximum. The spacing of the washers should not be greater than, / = 124 X 0.42 = 52.1 in., 

 0.42 in. being the least radius of gyration of one angle 2^" X 2" X 5/l6". 

 38 



