Chap. 9] 



SEISMIC METHODS 



445 



Solving for the normal stress and treating all equations of (9-6) in the 

 same manner, 



X. = 



+ 



du 



e (1 + (r)(l - 2(t) e(l + a) dx 



Y = - 



e 



z. = 



t (1 + tr)(l 

 



+ 



dv 



2(r) c(l + ff) dy 



+ 



dw 



(9-8) 



£ (1 + a)(l - 2<r) ' e(l +a) dz 

 The two coefficients in these relations are known as the Lame coefficients: 



(tE , E 



^ = 



(1 + a)(l - 2a) 



and 



V = 



2(1 + a)- 



If Poisson's ratio is j, 3t = v = fE. The quantity y is also known as the 

 shear or rigidity modulus. 

 With these coefficients, eq. (9-8) become 



Y„ = 0^ + 2^ 



du 

 dx 



ay 



dy 



(9-9) 



Z. = 0^ + 2y^^. 



These are the fundamental equations expressing the normal stresses as 

 functions of the volume changes and of the specific strains in the same 

 directions. 



Eqs. (9-9) state that the strains produced by normal stresses depend 

 on both X and y whereas the tangential stresses, as shown in the following 

 paragraph, depend on the rigidity modulus y alone. Eqs. (9-9) indicate 

 further that the normal stresses become equal when the rigidity is zero. 

 In that case all tangential stresses are zero and the normal stress is the 

 hydrostatic pressure. If Xx = Yj, = Z^ = — P in eqs. (9-9), by adding 

 eqs. (9-9) and substituting P we obtain 



-3P = 303i + 2iji0; P = -OCX + fv). 

 If ^ + fv = A;, then 



P = -k@ or -0 = P/k = PK. 



(9-10) 



