RIVETING. 155 



As has been shown, if the plate and rivet be given the same 

 values, it would only be necessary to so arrange the diameter 

 and pitch of rivet that the area of the rivets should equal that 

 of the net section of plate to secure a perfect joint, but the 

 ultimate value of plate steel is about 60,000 Ibs., and that of 

 rivet metal 50,000 Ibs. per sq. in., and practice has further 

 increased the difference between the metals by allowing only 

 about 40,000 Ibs. ultimate strength to rivet-rods under shear. 



The area of the rivet-hole represents the true section of 

 the rivet when driven, and therefore the area of the rivet-hole, 

 multiplied by the shearing-value of the metal, gives the 

 strength of the rivet. 



The pitch of the rivet, representing a section of plate, 

 multiplied by its thickness and the tensile strength of the 

 metal, gives the strength of the solid plate, while the pitch 

 of the rivet, or length of section, less one-/^//"the diameter of 

 the rivet-hole at each end of the section, or for both ends, 

 the diameter of the rivet-hole, multiplied by the thickness of 

 the plate and its ultimate tensile strength, will give the 

 strength of the net section of plate. The relation of these 

 values expressing the "efficiency" of the joint in per cent, is 

 therefore found by dividing the greater value by the least. 



Pitch of Rivets. The pitch of the rivet is found by the 

 formula 



+ A where 



P= Pitch of rivet, 



A = Area of rivet-hole in decimal of an inch, 

 6" = Shearing- value of rivet, 

 T= Thickness of plate, 

 Q = Tensile strength of plate, 

 D= Diameter of rivet-hole in inches. 



Where rivet is in more than single pitch, multiply by 

 number of rivets in row. 



