BY FINITE DIFFERENCES OF PHYSICAL PROBLEMS, ETC. 335 



(iii.) The stress function i/> 3 in a weightless medium, when on the surface z = 0, 

 xz = = zz from x = to x = + oo, and zz = p from x = to x = oo. 



The stretch s z is made to vanish at x = + by making 3xx = zz = p in that region. 

 At x = oo the stresses due to (ii.) will vanish, and therefore so will also the corre- 

 sponding s x . Consequently, if we add together any multiples of the (%)'s specified in (i.), 

 (ii.), and (iii.), we have s x = at x = + oo when the rock has its proper density p. 

 (iii.) is to be so placed that its origin is at the meet of the front and surface, (ii.) is 

 to have the point of application of the forces placed where the resultant of the water 

 pressure and weight of the dam cuts the plane 2 = 0. 



It will be necessary to consider distributions ^ and i/; :i in some detail. We are 

 indebted to Mr. J. H. MICHELL for both of them. (See ' Proc. Lond. Math. Soc.,' 

 1901.) 



First, \l> 2 . For a point force of unit magnitude acting OH the straight boundary 

 at the origin of the polar co-ordinates r, 6'. The stress function ^ is given by 

 \}{ 2 = Tr~ l r8' sin 0', and the force is in the positive direction of the line from which 6' 

 is measured. The stresses are as follows : 



_1 3^ a !9i/>_ _ 2 cos^ 00=^ = V rf = _ 1/1 iA = (}l) ) /-> 0) /oD 

 ~ r 2 30" 7- 3r ~ TIT ' ar 2 dr\r W) 



Next, i// 3 . The Stress Function due to the Lake. MICHELL shows that if unit 

 normal pressure be applied along a finite length of a straight boundary of an otherwise 

 unlimited plate in which the sideways stretch vanishes, then the form of the stress 

 function is 



r ......... (22), 



where r, < and r', <' are polar co-ordinates centred A and B, and AB is the initial 

 line. He further proves that the axes of principal stress at any point P are the 

 bisectors of the angle APB, and that if this angle is equal to a, the magnitudes of the 

 principal stresses are 



(a + sin a)/7r along the internal bisector ...... (23), 



(a sin )/7T along the external bisector ...... (24). 



We want the limit of ^ 3 ' in the neighbourhood of one end B when the other end A 

 is removed to an infinite distance. Let the origin of our co-ordinates x, z coincide 

 with B. Then let A be at a very great distance x = t, r'<$> +z, r 1 = t + x, 



\li a ' =-- {r 2 (f>(t + x)z}. Now remove the infinite but stressless tz and we have 



' 



$3 = {i*<t> xz}/2ir, (f> being equal to &+^TT in the notation of 4'1'2. On the upper 

 surface 9 2 /3a; 2 = tf/dr 2 . Therefore the traction zz is equal to + <j)/2ir. 



