JOINTS AND CONNECTIONS 



185 



by the pin must have enough bearing area to prevent crushing. 

 This being attained the pin is designed to resist shear and bending. 



Pin connections require a minimum of material in the members. 

 The cost of fabrication with pin connected trusses is not high. 

 Such trusses cost less than ri vetted trusses. The' joint is flexible, 

 if the pin does not rust, for all forces meet on the axis of the pin. 

 It is theoretically a perfect joint and for many years was favored 

 by American bridge engineers for the reasons given. European 

 engineers always favored the ri vetted joint because of its rigidity 

 and all joints were designed to take care of eccentric stresses. 

 Under heavy traffic it was found that the pin holes wore badly 

 and thus the trusses became too flexible. When pins rusted into 

 place eccentric stresses were set up and frequently the members 

 were too small to take care of them. The pin-connected joint 

 at the present time is not high in favor with bridge engineers, but 

 it is all right for roof trusses. 



Referring to Fig. 110 the bearing area is found by assuming 

 the whole load to rest on a strip having a length equal to the 

 combined thicknesses of the pieces 

 connected, with a width equal to 

 the diameter of the pin. Some- 

 times, for example in the case of 

 a built-up member, an extra thick- 

 ness of steel is rivetted to the 

 side of a member in order to ob- 

 tain increased bearing area at the 

 pin hole. This is often cheaper 

 than to increase the thickness of 

 the metal in the member through- 

 out the whole length. 



The shear on the pin seldom determines the thickness, the bear- 

 ing area and bending stresses being usually of greater importance. 

 The shear, however, should in all cases be investigated. The num- 

 ber of joints, that is the divisions between the pieces, will be one 

 less than the number of pieces. Divide the sum of the loads 

 on the pin by the number of joints to obtain the shear on each 

 joint, if the pieces are of equal thickness. If they are not equally 

 thick the shear on each joint will be equal to the sum of the re- 

 actions of the pieces on either side of the joint. If the pin is 

 found to be too small to carry the shear the diameter must be 



