152 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
When the solution of the general problem has thus been obtained in terms of series. 
of surface integrals, several questions present themselves for treatment, among which 
may be specially mentioned 
(i) For how many times in succession may these series be differentiated term by — 
term with respect to the coordinates a, y, 2? 
(ii) When the thickness of the plate is infinitesimal, but f(x, y) does not vary as h 
tends to zero, what are the orders of the various parts of the solution, and of the 
related physical quantities ? 
(iii) How are the answers to these two questions affected by discontinuity in the 
applied traction, or its x, y derivatives ? 
A perfectly general discussion of these questions would be tedious and difficult, 
and it will probably be more useful to consider the points suggested in the light of a 
special case, in which the integrations required can be performed, and the outstanding 
features of the solution can be grasped with comparative ease. 
10. Detailed solution of a special case. Term by term differentiations. 
The solution we propose to work out is to satisfy the following boundary con- 
ditions :— 
=~ 
Zz 
nN 
= 4rpJ,,(Bp) cosmo, on z=h 
= —ArpJ,,(Bp) cosmo, onz= -h 
= |) on z= +h, when p>a 
; when p<a 
— 
a — yO on zth. 
p, ®, 2 are the cylindrical coordinates of the point (a, y, z), so that x=p cosa, 
y=p sin, 
6 is any constant, and m is an integer. 
The solution, obtained from (20) by integration, is 
¢=¢,+¢,; 6=6,+6,, where 
= 3 aise 2 ) > 
Bi - Fal ev cane a 
3 1 9 . (29) 
= FF 1802 Dee 
0, ik ae 5? VWF — 2h2zy ) ron” F 
with F = | | x(B)Tm(Bp’) cos mu'p’do'de’ 
= cosh «kh sinh xz 
po = eaG Tose Ie—ih| | eB InBe cos mw’ p dp’ do’ 
y 
_ Sileos kh + 2h sinh xh) sinh xz ae | 
HCO ee et) 
(30) 
the integrals being taken over the circle of radius a. 
Consider in the first place the part of the solution defined by ¢,, 6,. 
The value of the surface integral in (30) takes different forms when p> and <a. 
