STRENGTH OF MATERIALS 281 



X* = .5. The net area is 3.25 -.5 = 2.75. The carrying 

 capacity of the angle is therefore 2.75X16,000 = 44,000 Ib. 



2. Riveted joints also are examples of simple stress. In 

 the joint shown in Fig. 1, the rivet is in single shear, because 

 there is only one section e of the rivet subjected to a shearing 

 stress. The amount R that one rivet will carry being equal 

 to the area of the cross-section of the rivet multiplied by the 

 unit shearing stress, or R = sA, the number of rivets required 

 to transfer a stress T by single shear is 



In Fig. 2, the rivet is subjected to shear on two sections, d 

 and e, and it is said to be in double shear. The amount of stress 

 that one rivet can carry in double shear is twice that of one in 

 single shear, and, using the preceding notation, 



T 

 n = W 



The bearing value of a rivet is the compressive stress in- 

 duced by the rivet in bearing on the plate, and is also cal- 

 culated by the simple-stress formula, P = sA, P being the value 

 of a rivet in bearing, 5 the unit working stress in bearing, and 

 A the bearing area, which, as it is customary to assume, is the 

 thickness of the plate multiplied by the diameter of the rivet. 

 In calculating the required number of rivets, both the shearing 

 and the bearing value of one rivet are determined and the 

 critical value (the smaller) used. 



The following tables give the shearing and bearing values 

 of rivets, in pounds, for different values of the working stress. 



3. Strength of Cylindrical Shells and Pipes With Thin Walls. 

 When a cylinder is subjected to internal pressure, the tensile stress 

 developed in the walls or shell of the cylinder is called circumfer- 

 ential stress, or hoop tension. Let s be the intensity of this stress; 

 d, the internal diameter of the cylinder; p, the intensity of pressure 

 on the inner surface of the cylinder ; and t the thickness of the shell. 



pd 

 Then. < = 



Pd 

 and *- 



