SPHERES AND CYLINDERS 121 



Since by Hooke's law the unit stress is proportional to the unit 

 deformation within the elastic limit, if p e denotes the unit stress on 

 the outside fiber of the thin ring, and p. on the inside fiber, then 



Pe *e * 



or Pir?=p e r?. 



Hence, if p h denotes the hoop stress on any element of the thin ring 

 and r the radius of this element, then 



(168) p h r z = constant, say C. 



It should be noted that this relation applies only to a thin ring and 

 not to the thick cylinder as a whole. It may be used, however, to 

 find the change in the hoop stress corresponding to a small change 

 in the radius, that is to say, the difference in the hoop stress on two 

 adjacent fibers, as explained in what follows. 



Now again consider a thin ring of the material and let its inter- 

 nal radius be r and its thickness Ar. Also, let p h denote the hoop 

 stress in this thin ring, p r the radial stress acting on its inner sur- 

 face, and p r + &p r the radial stress acting on its outer surface. Then 

 the difference in pressure on the inside and outside of the ring 

 must be equal to the total force holding the ring together ; that is, 



O r + Ap r ) 2 (r + Ar) - 2 rp r = 2p h kr ; 



but, since Ajt? r Ar is infinitesimal in comparison with the other terms, 

 this reduces to 



(169) 





If the ends of the cylinder are free from restraint, or if the cyl- 

 inder is subjected to a uniform longitudinal stress, the longitudinal 

 deformation must be constant throughout the cylinder. But the 

 lateral action of p r and p h produce longitudinal deformation in 



accordance with Poisson's law (article 8). Thus, if - - denotes 

 Poisson's ratio, the longitudinal deformation due to the action of 

 p r and p h is -&. + JE*_, O r (j> r +p h ). Therefore, in order that 



