86 Mr. C. Chree on Bars and 



M a very large number in ordinary units (e. g. for steel about 

 21 x 10 8 C.G.S. units), this will not cause any appreciable 

 variation in radius unless for cylinders of very large cross 

 section. However, in a cylinder of very large cross section 

 formed of layers of widely different materials exposed to a 

 severe tension or pressure, this tendency to a separation at 

 the edges at the junction of successive layers would become 

 noticeable. 



We shall next consider the torsion of a right circular 

 cylinder. Considering it first as uniform and isotropic, 

 regard u and w as vanishing, and v as independent of 0. 

 Then, for equilibrium, (1) and (3) are identically satisfied, 

 and (2) becomes 



^£ + 1*4 + ^=0, .... (29) 



dr 2 r dr r 1 dz l ' v ' 



a solution of which is 



v = T>r+Erz, (30) 



where D and E are constants. 



This contributes no term to S, and the consequent stresses 

 are 



T = = U, and S = nBr. 



This gives over every cross section the couple 



J 



„ 9a 7 ?27rEa 4 

 27rr z bdr= — - — •> 



denoting by a the radius of the cross section. 



Thus if Gr denote the couple of torsion applied at the one 

 end, the other being held, we get 



tittEo 4 



(x -^— ( 31 ) 



From (11) we see that the above solution satisfies the con- 

 ditions over the cylindrical surface. 



Suppose, now, the cylinder composed, as before, of a series 

 of layers of different isotropic materials. The internal equa- 

 tions for the equilibrium of each layer will be satisfied by a 

 solution of the form (30). From the equations (10), applied 

 at the surface separating two adjacent materials, we get the 

 value of S the same for both ; and supposing the surfaces of 

 these materials not to slip over one another, the values of v 

 must also agree there. Employing the same notation as in 

 the previous case, putting Di = 0, as the one end is supposed 

 fixed, we get 



2Gr 



n 1 F n = n 2 E 2 =...= 4) (32) 



