132 RESISTANCE OF MATERIALS 



APPLICATIONS 



216. The outside diameter of a pipe is 4 in., and thickness of wall \ in. Find 

 the safe internal fluid pressure by Clavarino's formula for a working stress in the 

 steel of 10,000 lb./in. 2 



Solution. The thickness ratio in this case is = i = 0.126 in. Also, D = 4 in., 

 d = 3 in., p = 10,000 lb./in. 2 , and consequently 



... _ 10 ( 16 - 9 ) 105 o 00 = 2869 lb /in 2 



13 x 16 + 4 x 9 



217. A cast-iron gear, 8 in. external diameter, 3 in. wide, and l|in. internal 

 diameter, is to be forced on a steel shaft. Find the stresses developed, the pressure 

 required to force the gear on the shaft, and the tangential thrust required to shear 

 the fit, that is, to produce relative motion between gear and shaft. 



2.D + i 



Solution. From the formula K = * the allowance is found to be .004 in., 



1000 



making the diameter of the shaft Z> 2 = 1.754 in. Also, since D^ = 1.75 in. and 

 D 3 = 8 in., we have H= 1.1005. Hence, assuming E l = 15,000,000 lb./in. 2 and 

 E 2 = 30,000,000 lb./in. 2 , we have 



p l = 23,550 lb./in. 2 , p 2 = 21,400 lb./in. 2 



To find the pressure required to force the gear on the shaft it is first necessary 

 to calculate the pressure between the surfaces in contact. From the relation 



p 9 = w this amounts to 



w = 21,400 lb./in. 2 



The coefficient of friction depends on the nature of the surfaces in contact. As- 

 suming it to be /* = .15 as an average value, and with a nominal area of contact 

 of TT x If x 3 = 16.497 in. 2 , the total pressure P required is 



P = 16.497 x 21,400 x .15 = 52,955 lb. = 26.5 tons. 

 To find the torsional resistance of the fit, we have, as above, 



Bearing area = 16.497 in. 2 , Unit pressure = 21,400 lb./in. 2 , 



p = .15, radius of shaft - .875 in. 

 Hence the torsional resistance is 



M t = 16.497 x 21,400 x .15 x .875 = 46,336 in.-lb. 

 Consequently the tangential thrust on the teeth of the gear necessary to shear the 



fit iS 4fi33fi 11 KQA1K r Q 



ajL^ajL = 11,004 ID. = o.o tons. 



218. The outside diameter of a steel pipe is 5 in., thickness of wall ^ in., and 

 internal fluid pressure 1500 lb./in. 2 Find by Clavarino's formula the maximum 

 fiber stress in the wall of the pipe. 



219. The outside diameter of a steel pipe is 8 in., the internal fluid pressure is 

 2000 lb./in. 2 , and the allowable stress in the steel is 15,000 lb./in. 2 Find the required 

 thickness of pipe wall. 



220. Solve problem 216 by the other four formulas listed in article 79 and com- 

 pare the results. 



