Question

12.16 The results of two consolidated-drained triaxial tests on a clayey sand are given here. Chamber Deviator stress pressure, (o,-o Specimen kN 70 140 (kN/m 155 265 Calculate the shear strength parameters of the soil.

Answer 1

Given Ist specimen chamber pressure(sigma_3) = 70 kn/m^2 deviator stress (sigma_d) = 155 kn/m^2 sigma_1 = sigma_3 + sigma_d Given sigma_d = sigma_1-sigma_3 155 = sigma_1-70 therefore sigma_1 = 225 kn/m^2 Now plastic equilibrium condition sigma_1 = sigma_3 tan^2(45 + emptyset/2) + 2c tan(45 + emptyset/2) 225 = 70 tan^2(alpha_f) + 2c tan(alpha_f) (because alpha_f = 45 + emptyset/2) II specimen sigma_3 = 140 kn/m^2 sigma_2 = sigma_1-sigma_3 = 265 kn/m^2 therefore sigma_1 = 405 kn/m^2 \r\n 405 = 140 tan^2(alpha_f) + 2C tan(alpha_f) Comparing 225 = 70 tan^2 alpha_f + 2C tan alpha_f-x multiply with 2 405 = 140 tan^2 alpha_f + 2C tan alpha_f-x multiply with After multiplying Eq with we get 450 = 140 tan^2 alpha_f + 4C tan alpha_f 405 = 140 tan^2 alpha_f + 2C tan alpha + f 45 = 2C tan alpha_f therefore C tan alpha_f = 22.5 Substitute in Eqn 225 = 70 tan^2 alpha_f + 2(22.5) 180 = 70 tan^2 alpha_f Tan^2 alpha_f = 2.571