BY FINITE DIFFERENCES OF PHYSICAL PROBLEMS, ETC. 



331 



where x a , z are the co-ordinates of D, and where /A is + 1 if AB is to the left of BD, 

 and 1 if they are reversed. Integrate by parts, then 



V D 



(x x a ) \zz a dx zz tt (dxf + (z z a ) \xx 9 dz I xx a (dz) 2 



Substitute the values of the stresses and 



= 0. 



X 



= -G-W2* - 



B 



where i is the length BD. Or, since by (13') and (14') (&) = (?f) -X, 



Z 



2/Ii 



(15'). 



a useful formula. 



When the boundary of the solid intersects AD, DB, as does the curly line in the 

 fig. 5, then draw horizontal and vertical lines to form a 

 figure enclosing the boundary. Apply the stresses to 

 FH instead of AD and to LK instead of BD, leaving 

 the rest of AFHJKLB unstressed. These stresses 

 must balance the surface forces as before, and now the 

 whole weight of the solid enclosed between AFJLB 

 and the boundary. On account of the change in level 



from AD to KJ or BL the term f/pz in zz is different, 

 and this term being integrated along FJ and LB has 

 the effect of subtracting the weight of AFJLBD from 

 that of the solid part. Consequently we may integrate 

 along AD, DB and keep everything else as before, 



f 1 ' 

 provided we replace +vgpv by gp (zz u )dx, where z follows the boundary curve. 



J A 



Similarly in (15') we must replace plgpv by gp (zz n ) (xx a )dx. 



JA 



4 "I'll. In order to simplify the arithmetic, the surface of the dam of the form 

 chosen has been represented by straight lines, either horizontal, vertical, or sloping at 

 45 degrees. It is one of the peculiar advantages of successive approximation methods 

 that a simple case like this comes in conveniently as the first stage of the solution for 

 any rather different boundary. 



Table of surface equations when the surfaces are horizontal, vertical, or at 45 degree* 

 (fig. 6), deduced from equations above : 



2 v 2 



Pi 



