Solid Cylinders of Elliptic Section. 73 



in number. There are three internal equations to be satisfied 

 at every point within- the solid, viz. : — 



m 



dx \dx 2 dy 2 dz 2 ) dx 



(3) 



dy 2 ' dz 2 ) ' dx 



J 



where 



Y = co 2 p(x 2 +y 2 )/2. 



Of the six surface conditions, three apply at every point of 

 the plane faces z— ±1, viz. :-— 



£ = 0, (4) 



7so = 0, (5) 



Tz = (6) 



The remaining three apply at every point of the cylindrical 

 surface (1), usually denoted here the rim. Denoting the 

 direction-cosines of the normal to (1) by X and /jl, these 

 conditions are : 



F = \xx + fjixy = 0, (7) 



G=>uy + /W = 0, (8) 



K=\2v + fjuyz=0 (9) 



My previous solution satisfied exactly the internal equations 

 (3), and also all the surface conditions except (7) and (8). 

 Instead of these it gave 



F = C* 2 , G=C , 3 S 



where C and C are independent of z. Thus over the length 

 21 of a generator, there remained a resultant force, whose 

 components parallel to the axes of x and y were 



r +i o i +i 9 



F^=|C/ 3 , Gdz=p>P (10) 



Professor Pearson's objection to this solution as applied to 

 a thin disk is not that it fails exactly to satisfy (7) and (8), 

 but that it leads to an unequilibrated resultant, given by 

 (10), for the stresses over a generator. To remove this 

 objection all that is necessary is to determine two of the arbi- 

 trary constants of my general solution from the equations 



