574 



THE DESIGN OF STEEL DETAILS. 



CHAP. XVII. 



Solution. If fastened by both legs as in Fig. 2 the load may be considered as axial and the 

 required net area, using an allowable unit stress of / = 16,000 Ib. per sq. in., is 



_ P _ 40,000 _ 

 ft 16,000 



2.50 sq. m. 



Try one angle 4" X 4" X W- Gross area = 2.86 sq. in. (Table 23 or Table 25). Net 

 area, deducting one % in. hole for a % in. rivet = 2.86 .33 = 2.53 sq. in. (Table 116). This 

 angle will satisfy the conditions. This result can be obtained directly from Table 29. 



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

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

 ive. The solution would then be, as before 



P 40,000 



A -r = -4 = 2.50 sq. in. 



ft 16,000 



1 



FIG. 2. ANGLE CONNECTED BY BOTH LEGS. 



FIG. 3. ANGLE CONNECTED BY ONE LEG. 



Try one angle 6" X 4" X /^" with 6 in. leg attached. Gross area of 6 in. leg = 6 X J^ 

 = 3.00 sq. in., net area = 3.00 .44 = 2.56 sq. in., which will satisfy the conditions. 



Built-up Tension Member. Design a built-up member to carry a tensile stress of 390,000 

 Ib., using % in. rivets. 



References 33, p. 57; 83, p. 60; 84, p. 60; 89, p. 61; 90, p. 61; 101, p. 61; 37, 

 p. 141; 44, p. 141; 61, p. 142; 75, p. 143; 14 and 26, p. 206; 28, p. 210; 38, p. 210; 

 52, p. 211; 82, p. 213; p. 219; ii, p. 382. 



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



A 39>229 _ 

 /( 16,000 



24.4 sq. in. 



Try 4 angles 3^" X 3^" X W and 2 plates 18 in. X K in., as shown in Fig. 4. Gross area 

 = 18.00 + I3-OO = 31.00 sq. in. Referring to Fig. 4, it will be seen that the section n-n is the 

 least section in the body of the member and that four rivet holes should be deducted from each 

 side to obtain the net section, giving a net area of 31.00 4.00 2.00 = 25.00 sq. in., 4.00 sq. 

 in. being the area of holes in the plates and 2.00 sq. in. being the area of holes in the angles, de- 

 ducting I in. holes for ^ in. rivets. This section has sufficient area, 24.4 sq. in. being required. 



If the ends of the members are to be riveted they should be designed as outlined under 

 "Riveted Connections and Joints" in this chapter. 



If the ends are to be pin-connected they may be designed as follows. Assume that 5^ in. 

 pins are to be used at each end. The bearing area required allowing a unit stress of 24,000 Ib. 

 per sq. in., is 390,000 -f- 24,000 = 16.2 sq. in. This requires a total thickness of plates of 16.2 -5- 

 5.5 = 2.95 in., or 1.48 in. on each side. The web plates are J^ in., the fill plates must be at least 

 Yi in., the thickness of the angles being K in., and using % in. outside plates the total thickness of 

 plates is 1.50 in., which satisfies the conditions, 1.48 in. being required. 



