ANALYTICAL DEVELOPMENT 



The problem is to develop expressions for the normal and tangential stresses for 

 the active and passive states of stress, with the angle of the investigation, a , being 

 one of the variables. Other variables in these expressions are: the known stress con- 

 dition ( CT a > T a ); the slope inclination, 9; and the strength characteristics of the soil, 

 c and <j>. There is no need to develop separate equations for ( a c >T c^act anc ^ ( CT c ,T c)pass 

 because the development results in a quadratic equation that yields solutions for both 

 the active and passive states of stress . 



The equation for a line parallel to the ground surface and passing through point 



t = a tanG + b , (14) 



(a ,t ), as shown in figure 13, is 

 a a 



where b is the intercept. 



Substituting the known stress condition (a ,t ) , given by equations 5 and 8, into equa- 

 tion 14 yields: 



Y.Z sinecosO + y _(Z-Z )sin6cos9 

 t w sat w 



(15) 



= Y.Z cos 2 9tan0 + y,(Z-Z )cos 2 0tan0 + b. 

 t w b w 



Solving for b from equation 15: 



b = (y + -Yj[( z " z )sin9cos9]. (16) 

 sat b L w J 



Recalling that y w = ^ Y sa t~ Y b'' ' ec ( uat ^ on ^ ma y ^ e written as: 



b = y sin6(Z-Z )cos0. (17) 

 w w J 



Equation 17 is recognized as the equation for the seepage force per unit area, S: 



Y iV = 

 w 



Y sine (Z- 

 w 



Z )cos( 

 w 



(18) 



Where : 



i = sine (the hydraulic gradient as shown in figure 5) ; 



V = (Z-Z w )cos6 (the volume of the column of soil between the water table and plane 

 A- A' shown in figure 2) . 



The intercept, b, is therefore equal to the seepage force per unit area, S. Substi- 

 tuting the expression for b, from equation 17, into equation 14, yields the equation 



for a line through point (a ,x ), with a slope equal to tan9: 



a a 



x = atan9 + y (Z-Z )sin9cos9 = atane + S. (19) 



12 



