of an Elastic Plate under Normal Pressure. 115 



Now it is clear that the mean shear force S is zero 

 because, in such a surface of revolution as we have assumed, 

 the circumferential shear force on the circular sections is 

 just as likely to act in one direction as in the other ; and 

 since it cannot act in both at once, it must be zero. That is, 



=n£- = 0, (56) 



'whence *=/(y) + F(*) '"(57) 



Moreover, the mean tensions P x and P 2 cannot be functions 

 of x. But 



p 2 =e|^=ef"0); .... (58) 



QX 



and since this cannot be a function of x it can only be a 

 constant or zero. Then let 



F"0)=~A. 

 It follows that 



F(.r)=iAa? 2 + B. .*..... (59) 



There can be no term containing the first power of x on 

 account of the symmetry about the #-axis. 



Since P 2 is constant, it has the same value at the edges of 

 the plate as it has at any other point of the plate. As we 

 are assuming that there are no actions on the circular edges 

 of the plate, the stress P 2 must be zero, and therefore 

 A must be zero. Moreover, the constant B can be merged 

 into /(?/), so that 



4>=f(j/) (60) 



The equation lor p now is 



2 Eft 3 dfiiDj 2Wid 2 $ 



3 1 — a* dx i a dy' 



(61) 



Also the equation connecting w^ and (/> is 

 d±(j> _ 1 d?w x 

 dy^ a dy 2 ' 



from which ^ 2 4=-(if 1 + C), {62) 



dy 2 a 



since symmetry requires no term involving the first power 

 of y. 



The constant C depends only on the position of the xy 

 plane, and since we left this position indefinite in that it had 

 only to be parallel to a certain tangent plane, we mav put 



12 



