528 Dr. J. Prescott on the Torsion of 



curve at K. Then, i£ ds represents the element of length 

 KK', the shear force on the element of area between the 

 two boundaries and the two normals at the ends of ds is 



$tds= +nh%ds. 



The moment of this about is 



dQ = nx OH xdsBg 



= nxOHxKK'xd? 



= 2ndA8%, (2S) 



where dA represents the area of the triangle OKK'. 

 Then the total torque in the tube is 



Q=2n8£$dA 



= 2nAS%, (29) 



which is the same equation as (25) since Sf and C represent 

 the same thing. 



The result in (29) is merely the torque due to the stresses 

 between two shear lines whose f's differ by Sf, the outer 

 curve haying the smaller f. Then the equation applies 

 equally well to the torque due to the shear between any two 

 shear lines which are infinitely near together, whether these 

 two shear lines are the boundary curves of a section or not. 

 If the lines are infinitely near together we may take A as 

 the area enclosed by the inner shear line. Then the total 

 torque on any section whatever is 



Q = 2wJAdf (30) 



By eliminating 8% from (27) and (29) we get the torque 

 in a thin tube in terms of the stress and thickness at any 

 point of the tubec Thus 



Q = 2AS*, ...... (31) 



whence 



8 = 2 4 < 32 > 



At different points of the tube Q and A are constant 

 while S and t may be variable and such that their product 

 is constant. The mean shear stress across the thickness of 

 the tube at any point varies therefore inversely as the 

 thickness, whence it follows that the shear stress in a tube 

 of variable thickness is greatest where the thickness is 

 least. Moreover, in any section whatever, the intensity of 

 the shear stress at any point of any one shear line is indicated 

 by the closeness of the next line. 



