RIVETING. 153 



and thus increase the efficiency of the joint. If the inner welt 

 or covering-strip had been of the same width as the outer one, 

 the net section of the plate would have been greatly reduced, 

 and the difference of strength between net section of plates 

 and rivets would have been greater, thus reducing the effi- 

 ciency of joint. The problem would be as follows : 



6-5 X 0.375 X 55,ooo = 134,062 = strength of solid plate. 



(6.5 0.8125x2) x 0-375 X55oo = 100,546 = strength 

 of net section of plate. 



0.5185 X 4 X 70,300= 145,802 = strength of 4 rivets in 

 double shear. Net section of plate is the weakest; therefore, 

 100,546 -:- 134,062 = only 75 per cent, efficiency of joint. 



Again, it may be suggested: Why not dispense with one 

 row of rivets in double shear, and extend the inner welt or 

 covering-strip so that the outer row of rivets in double pitch 

 and single shear could be used, thus increasing net section of 

 plate as in the original problem, but reducing at the same 

 time the shearing-resistance of the rivets? 



The solution of this problem would be as follows: 



6.5 X 0.375 X 55.000 = 134,062 = strength of solid 

 plate. 



(6.5 - 0.8125) X 0.475 X 55,ooo = 117,304 = strength of 

 net section. 



0.5185 X 2 X 70,300= 72,901 = strength of 2 rivets in 

 double shear. 



0-5i85 X 38,000= 19,703 = strength of I rivet in single 

 shear. 



This last result must be added to the result of 2 rivets in 

 double shear. 72,901 + 11,703 = 92,604 = strength of all 

 the rivets. 



The total strength of the rivets is the weakest ; therefore, 

 92,604 -^- 134,062 = 69 per cent, efficiency of joint. 



It may be further suggested that a rivet of smaller diame- 

 ter could be used. I will say that I have also considered such 



