RIVETING. 147 



slightly. In these experiments the steel plate was 55,000 

 Ibs. tensile strength per sq. in. 



Assuming 38,000 Ibs. as the safe estimate, we must de- 

 cide upon the thickness of plate, diameter of rivet-hole, and 

 pitch of rivets. In deciding upon these elements in the prob- 

 lem, we must so adjust the size and pitch of rivets as to make 

 the shearing-resistance of the rivets as near the strength of 

 net section as possible. I will assume the elements of the 

 problem to be as follows : 



Steel plate, tensile strength per square inch of section, 

 55,000 Ibs. 



Thickness of plate T 5 g- in. = decimal 0.3125. 



Diameter of rivet-hole \\ in. = decimal 0.8125. 



Area of rivet-hole = decimal 0.5185. 



Pitch of rivets i-J- ins. ^decimal 1.875. 



Shearing-resistance of iron rivets per square inch = 38,000 

 Ibs. 



Then 1.875 X 0.3125 X 55,000=132,226 Ibs. = strength 

 of solid plate. 



(1.875 0.8125) X 0.3125 X 55,000 = 18,262 = strength 

 net section of plate. 



0.5185 X 38,000:= 19,703 Ibs. = strength one rivet in 

 single shear. 



Net section of plate is the weakest, therefore 18,262 -~ 

 32,226 = 56.6 per cent, efficiency of joint. 



Double-riveted Joints (Fig. 41). In double-riveted joints 

 we find an accession of strength over single-riveted joints of 

 nearly 20 per cent. This arises from the wider lap and the 

 better distribution of the material. The rivets are pitched 

 wider, and there is more rivet-area to be sheared, together 

 with a larger percentage of net section of plate to be broken. 



Steel plate, tensile strength per square inch of section, 

 55,000 Ibs. 



