BY FINITE DIFFERENCES OF PHYSICAL PROBLEMS, ETC. 327 



regarded as a ridge upon the surface of a very large slab of rock which, with the 

 dam, forms one homogeneous isotropic elastic system ; together with the further 

 assumption that this substratum suffered from no internal stress before the dam was 

 built, except, of course, that due to its own weight. The consideration of the results 

 that follow from these assumptions will be continued in 4'1'9 below. 



4'1'2. Conventions as to Symbols. The Z axis vertically downwards. 



The X axis horizontally directed from the water towards the tail. 



6, the angle which a line makes with the Z axis, to be reckoned positive when the 

 rotation is from the Z axis to the X axis through 270 and so onwards. 



The stresses xx and zz positive when they pull adjacent portions of the material 



together, and therefore the shear xz positive when the lower portion is pulling the 

 upper towards the tail. 



q and s distances along the outwardly directed normal to the masonry and rock 

 and along the boundary drawn to the right of it. 



N and T the normal and tangential stresses on the surface related to q and .? in the 



same way that zz and xz are related to z and x. 



I and n the cosines of the angles which the outwardly drawn normal makes with 

 the X and Z axes respectively. 



4'1'3. Position of the Origin. PEARSON and POLLARD (p. 37) take this at the join 

 of the front and the top. It will be more convenient in what follows to take it 

 vertically below this point at the level of the surface of our hypothetical slab of bed 

 rock. This is done throughout. The water surface stands at z = p, and the 

 pressure due to it is accordingly gp' (+p + z), where p' is the density of the water. 



4'1'4. Specific Constants of the Masonry and Rock. Following PEARSON and 

 POLLARD I take the density p = 2'25 times that of water (p. 29) and POISSON'S ratio 

 f] as | (p. 33). 



4'1'5. Units. Distances are reckoned in metres. Forces in metric tons, each 

 equal to the weight of a cubic metre of water. Consequently gp' the weight of unit 

 volume of water is equal to unity. And gp*= 2 - 25. 



4'1'6. Equations to be Solved. 'The stresses may be expressed in terms of a single 

 scalar ^ which satisfies 



V^EEfJl+25-^ + |V)X = ....... (1) 



\dx* dx 2 dy 2 dz 4 / f 



(2), (3), (4). 



The surface equations then become 



