THE EQUILIBRIUM OF AN ISOTROPIC ELASTIC PLATE. 205 
39. Problem 3. To determine the permanent modes, having given the 
symmetrical edge tractions zp and pp. 
Supposing the cylinder solid, we have only to make a slight modification in the 
process of (i), (ii1) in last article. 
Extensional mode, 
In (i) the balancing solution must now be taken as 
E@uly 28 =a)(a+1) 2 
ae, x T-a GC 
p 
—, W= 
= a? 
This, along with 62=@=log p, gives pp and pz =0 at p=a. 
The balanced solution gives at the edge 
8atl wwe AS sae le 
ea Stn T-a a 
Hence 
U (Py %)=py Ae A 
= 3} eS eet 
Mepa)=2 5.1% | 72 ea 
f (4aP, + a—3 2L,)dz « 
—h 
This gives the permanent mode exactly. The ordinary approximate theory omits the 
term in Z, from the integral. 
In Curer’s solution of the problem of a rotating disc, the stress z vanishes identi- 
h 
cally at the edge, while | _Padz=0. His solution is therefore exact, so far as the 
fundamental mode is concerned. 
As in last article, we infer from the form of the above solution that for any other 
than the permanent mode is (4p ppta—32%)=0. 
A -h 
This may be verified by actual integration. Further, in the case of a hollow 
eylinder the solution is of the form 
(py »%)=Ap, + B/p, 
(py » 2) = 2A(a — 3)z,/(a + 1) } 
and the coefficients A , B are found from the conditions that 
h ———— 
| (4p - ppt+a—3z- 2p) dz must, for p=a and p=b 
=i? 
have the same value for the assumed form and for the given tractions. 
Flexural mode. 
Referring to (iii) of last article, the solution balancing 
6==zlogp at p=a is now 
_ +l rly ot O-alatl) 
u : = 2 7 
T-a a’ =a, Wor i-a a 
The balanced solution gives at the edge 
+l 2 oe neg) te 
