198 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
The displacements ty , V, W2 and U3, V3, W; are similarly decomposable into equilibrat- 
ing systems U,, Vo, W,-and U,, V:, Weowath other pyebents cones A forces. 
and couples. The contributions to u(a’, y’, 2), v(x’, y’, 2’), w(x’, y’, 7) im (97) from 
the various systems conveying forces and couples amount on the whole merely to a neNg 
body displacement of the plate. If this be neglected, then the value of u(x’, y’, 2’), 
for instance, becomes simply the work difference from the system U,, V,, W,, the 
edge tractions due to which are equilibrating, and can be balanced by a free system 
Uy, Vy, Wy’. We then obtain from (97) 
ule’, y', 2) = | | \ X(U, + Uy) + ¥(V,+.Vy) +Z(W, + W,’) \ ds 
and similarly for v, w. 
37. Exact solutions of special problems for a circular plate. 
As already stated, we are not at present in a position to complete the solution of 
the problem of arbitrary edge tractions, even for the simplest form of edge. The 
method just indicated may be used, however, whatever be the form of the edge, to 
obtain approximately the boundary conditions which define the permanent part of the 
solution. But before entering on this important application, we shall consider a few 
special problems which admit of exact solution. All of these have reference to a plate 
bounded by a right cireular cylinder, with or without a concentric circular aperture, 
and to systems of displacement symmetrical about the axis. 
The radius of the external edge is a, of the internal edge b; and the axis of z 
coincides with the axis of the cylinder. uw, v, w are the displacements in the directions 
in which the coordinates p, , 2 increase. | 
Problem 1. Symmetrical transverse displacement. ; 
The displacement v, in the most general case, is given by a series involving cosines. 
and sines of multiples of . We can determine the symmetrical term of the series. 
This constitutes the whole solution when the plate is subjected only to symmetrical 
torsional force. 
For a transverse force Q, applied at the point (p,, ,, 2) we have seen in article 29 
that the solution is 
dy’ 
y= 
dp, 
(oe = OCLs ae ee 
py Ee, 4ap(a+ 1) 
1 dd’ 
weg p, da, 
The solution for a constant linear distribution of transverse force on the circle p=p,, 
z=2,, of intensity Q,/27p, per unit length, is found by integrating this with respect 
to w, from 0 to 27, and dividing by 27. The result of the integration is simply to: 
