546 Dr. G. Chree : Applications of 



upper surface being freely exposed to the liquid, the terminal 

 conditions are 



y = when z— — h/2, for, say, a?=?/=0, 

 zz = — (p —gp'hl'2) when z— + h/2. 

 The solution will be found to be 



xx=yy=-(p-gp'z), ~) 



^ h \ (35) 



zz= -. p -g( p - p >)- +gp Z ;^ 



^=/3/g=-(l/E)^(l-2 v ) P ^ V g(p--^)^gz{ V p-(l- v )p / }], 



(z + h/2) 



7= - E 



[(1-2^+^ (3,-2(1+^'} -^(p-2V)l 



J. (36) 



-£(*>+^){w-(1-i0p'}. 



1 2E 



The pressure per unit area exerted at the base of the 

 cylinder or prism, i. e. the value of — zz when z-=—h/ 2, is 

 gph + (p-gp'h/2). 



The total pressure thus equals, as it ought, the weight of 

 the solid plus the total pressure exerted by the liquid on the 

 upper flat surface. 



Our assumption that the vertical displacement y vanishes 

 when x=g = over the upper surface z = h/2 in case (i.), and 

 over the lower surface z= —h/2 in case (ii.) is really arbitrary. 

 The making a displacement vanish at a particular point is 

 merely equivalent to introducing a constant into the expres- 

 sion, and this is without effect on the strains or stresses. 

 Case (i.) includes several sub-cases. Thus if p'=p it applies 

 to a cylinder or prism floating wholly immersed in a liquid 



of equal density. The stress zz at the upper face then be- 

 comes a pressure and equals, as it ought, the pressure 

 p—gp'h\2 at the same depth in the liquid. If p exceed p' 

 we have the case of a cylinder supported from above, with 

 the whole, or a portion h, of its length immersed. Supposing 

 the portion /* alone immersed, the atmospheric pressure on 



the liquid surface being II, the value of zz over the surface 

 z— -\-hj2 becomes 



#(p-p')h-n. 



It is thus a pull or a push according as 



g{p-p f )h> or <II; 



