284 Mr Chree, On Long Rotating Circular Cylinders. [Feb. 8, 



length of the beam attains to something like ten times its greatest 

 diameter. Unless 77 = the present solution ought to be simi- 

 larly restricted, and it should not be applied to the portions of the 

 cylinder immediately adjacent to its ends. The solution con- 

 siders solely the action of the "centrifugal force", taking no 

 account of gravity or of the action of any forces applied at the 

 ends by the bearings. 



§ 2. The solution satisfies exactly the internal equations — 

 viz. (2) and (3) I.e. p. 203 — , and also the conditions at the 

 curved surface or surfaces. The only condition it fails to satisfy 

 exactly is that Hz vanish at every point of the flat ends. In- 

 stead of this we have to be content with satisfying 



J a 



2irrz7dr = 0, 



i.e. instead of making the normal stress over every element zero 

 we make the resultant normal stress zero. The solution is thus 

 based on the principle of statically equivalent load systems — 

 referred to in my previous paper, pp. 206-7 — and so can be 

 regarded as satisfactory only when the dimension 2a is small 

 compared to the dimension 21. 



The solution is found by starting with the expressions (16), 

 p. 205, for the displacements, which satisfy the internal equations. 

 We have then to determine the arbitrary constants by means of 

 the surface conditions, viz., 



^ = = rr, when r = a and when r = a', 



rz = = I 2irr zz dr when z = ± I. 



§ 3. It is unnecessary to reproduce the algebraical work, as 

 the solution may easily be verified. In terms of Young's modulus 

 E and Poisson's ratio rj it is as follows : 



2/2 *2\ 3 — 5?; 2 3 (1 - 277) (1 + 77) 



u = <o* P (a» + a>)r HE{l * v) -co* P r^ ^((-1) 



L 2 aV i! ( l+77)(3-2r;) 



w = - a ?p(a* + a'*)z 1 £ B (2), 



A „W. ^ (l-^)(3.-q) . ,(l- 2iy)(l+q ) 



A = a>p(a+a) ^ {1 _ v) -»P» 2E(1- V ) ' m " {6) * 



^ = a>>(a 2 -r 2 )(l-a7r 2 ) 8 -^|^ (4), 



