MEMBERS IN COMPRESSION. 



The net area through the pin hole (section m-m) must be 25 per cent in excess of the net 

 area of the body of the member according to a common specification. It will probably be neces- 

 sary to deduct the area of the pin hole and two rivet holes on each side, the rivet holes being so 

 n< -.ir the section m-m, see Fig. 4. The gross area through the pin hole is, web plates 2 X 18 X H 

 = 18.00 sq. in., angles 4 X 3.25 = 13.00 sq. in., fill plate 2X11 X *A Il.oo sq. in., outside 

 plate 2 X 17 X H ~ 17.00 sq. in. making a total gross area of 59.00 sq. in. The net area is 

 59.00 2X5. 5X1. 5 4X1 X iH =" 36.5 sq. in. The required net area through the pin 

 h.)K is 1.25 X 25.00 = 31.3 sq. in. 



Secthnm-m 



L 



Section n-n 



FIG. 4. RIVETED TENSION MEMBER. 



The net area back of the pin hole parallel with the axis of the member (section o-o) must not 

 be less than the net area in the body of the member (section n-n) = 25.0 sq. in. The total 

 thickness of the metal at this section is 1.50 in. for each side. Therefore the net length back of the 

 pin must be 25.00 -5- 2 X 1.50 = 8.33 in. Assuming that not over three rivets will come in this 

 section, the total length back of the pin hole must be at least 8.33 + 3.00 = 11.33 n. 



The number of rivets required and the size of pin plates is considered under " Riveted Connec- 

 tions and Joints." 



Unriveted Pipe. Design an unriveted iron pipe 12 in. in diameter to carry an internal 

 pressure of 400 Ib. per sq. in. 



From Structural Mechanics, Chap. XVI (Formula I2a), / = wD 2t; and / = w D 4- 2/, 

 where / is the thickness of metal, w = unit internal pressure, D = diameter and / the allowable 

 tensile stress which will be taken as 12,000 Ib. per sq. in. 



_ w-D _ 400 X 12 



* T ~~ "- ~ 0.20 in. 



2/ 2 X I2.OOO 



MEMBERS IN COMPRESSION. The design of compression members will be shown by 

 several examples. 



Single Angle Strut. Select an angle to carry a compressive stress of 21,500 Ib. The length 

 center to center of connections is 6' o", and both legs are to be fastened at the ends, Fig. 2. 



References. Specifications 34, p. 57; 39, p. 57; 84, p. 60; 85, p. 60; 93, p. 61; 

 38, p. 141; 43, p. 141; 60, p. 142; 100, p. 61; 45, p. 206; p. 207; 16, p. 209; 20, p. 209; 

 p. 223; 231, p. 363; 10, p. 379. 



Solution. Using / c = 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 as follows: 



l/r = 125, or r = - = $J*_!? = 0.58 in. 

 125 125 



Any 3" X 3" angle will satisfy the requirement for l/r (Table 23). The allowable unit stress 



72 



will then be 16,000 70 X -^ = 7,300 Ib. per sq. in. The area required will be 



5 



2.95 sq. in. 



A 2I '5 

 ~ 7 ~ 7.300 



The area of one angle 3" X 3" X 9/l6" is 3.06 sq. in., which is sufficient. 



