THE EQUILIBRIUM OF AN ISOTROPIC ELASTIC PLATE. ISS) 
eliminate all but the symmetrical part of |, and to eliminate 0, > altogether. The 
solution is therefore, 
when p> p1, 
1 : 
+ TED = cosh xz, cosh xzJ.ykp,G,/«p , (k a pos. imag. root of sinh «h) } (i) 
Za | S , , . 
Za) 2 sinh xz, sinh xzJq'xp,G,y Kp, (« a pos. imag, root of cosh kh) 
when p</p,, p and p, have simply to be interchanged. 
The only stress across a cylinder p is 
p= w( =) 
a= = = — }. 
; dp p 
Hence if 2, , 2, be the transverse components of the traction on p=a, p=), and v, the 
transverse displacement at (p,, ©, 2) produced by this traction, Betti’s Theorem gives, 
as in (97), for the permanent part of v,, 
a ix i ee if i (2 einpaa 7H as eke | | 0, 2.a8,. 
Qr}o + 1 Barph ae war “~ 8rph, Py 
Also, since the distributions 2,, 02, have equal moments about Oz, 
af [8,40 | [%28,=0. 
In the case of a symmetrical deformation, we have therefore 
Ou wet ie (2 
S| —— = Q d. + Leal ae 
a es 2 Sta : 
The displacement v,=p, being merely a rigid body rotation about Oz, the permanent 
solution is practically 
v DE [ 0O,dz= _ o i QO dz 11) 
Si tian) =a 4uhp,) -r “ ; ; ; ; a 
This might have been got at once by omitting the term v= p,/87uhp from the source 
solution, in accordance with the method explained at the end of the last article. 
For a uniform system of couple about the edge-normal | (2,dz vanishes, and the 
permanent displacement is rigorously null. 
For the transitory part of the solution we will, to simplify the algebra, suppose the 
eylinder solid. 
Further, this not being a case where the separation into odd and even parts in z 
is of much consequence, we may shorten the formule by combining the two « series 
into one. 
Thus cosh« (h+z) cosh« (h+2,) being obviously equal to cosh xz cosh «z, when 
