STEAM BOILERS 



345 



of the plate SV = 45,000, and resistance to crushing S c = 90,000. 

 Assume t = 7/16 inch, d = i inch, and p = 2 l / 2 inches. 



A strip of the joint equal in width to the pitch is sufficient to be 

 considered. 



1. Tearing between the two rivets. In this case there is a strip 

 to be torn in two, equal in width to the distance between the rivets 

 less the diameter of the rivet, i.e., p d, and it has a thickness 

 equal to t, i.e., the strip has a cross-section of an area (p d)t\ this 

 cross-section in square inches times the tensile strength will give 

 the pull required to fracture the joint: 



(p d) tSt = (2*/ 2 i ) X 7/16 X 55,000 = 36,095. 



2. Shearing one rivet. Since there is only one rivet in each 

 2^-inch strip, we have to consider the shearing of it only. 



FIG. 238 SINGLE-RIVETED 

 LAP JOINT 



FIG. 



239 DOUBLE-RIVETED 

 LAP JOINT 



The area to be sheared is the area of a cross-section of the rivet, or 

 3.1416 d* 



The pull which it will take to shear this rivet is the area times the 

 shearing strength : 



3. 1416 d* 



_ 3.1416 



S s = * -- X 45,000 = 35,343- 



3. Crushing. In this case it is common to consider that the area 

 to be crushed is the diameter of the rivet times the thickness, hence 



dtS c = i X 7/i6 X 90,000 = 39,375- 



The number of pounds it will take to fracture a strip of plate 

 -2^2 inches wide and 7/16 inch thick by tension is 

 2 J A X 7/ 16 X 55,000 = 60,155- 



