Question

Geomechanics

(b) Figure 1 shows a water retaining embankment with a slope of 1:2 (vertical kN/m'. The clay soil has an undrained cohesion of 30 kPa and a saturated unit weight of i18 following information is given: Area FBDE-42m2 (above water level) Centroidal distance dA-13m Area AED 144m2 (below water level) Centroidal distance, ds -4.5m Angle AOF-83 Radius of circular slip plane, R- 22.5m (i) Determine the total disturbing moment by considering the forces Wa Ws and Pw (Gfi) Determine the total resisting moment, assuming that -0 and c -30 kPa over the entire slip surface. Determine the Factor of Safety (FOS) of the slope. If the water level decreases to the toe of the slope (ie to point A), explain the new condition of the slope, by calculating the new FOS. (iii) (iv) (15 marks) 8.0 m 0 0.0 m Ye-12.22m 4.0 m water level 6.0 m 3.0 m Potential slip Surface Figure 1: Question Ss

Answer 1

Disturbing moments: - Due to W_A, about point ' o 'M_A = w_A * d_A Here w_A = r * A * 1 = 18 * 42 = 756 KN/m M_A = 756 * 13 = 9828 kN-m/m w_B = w_B * d_B = r' * A_B *d_B = (18 - 10) * 144 * 1 * d_B = 1152 kN/m * d_B m_g = 5184 kN-m/m m_w = p_w * d_c * 1 = 1/2(r-w * Z_c^2) * d_c * 1 Z_c = 2 cu/r = 2 * 30/18 = 3.33 m Now M_w = 0.5 * 10 * 3.33^2 * 12.22 = 677.53 kN-m/m therefore Total Disturbing moment M_D = M_A + M_b + M_w = 15689.53 kN-m/m \r\n (ii) Resisting moment L = R theta * 2 pi/360 degree = 22.5 * 13 degree * 2 pi/360 degree = 32.594 m Resisting force = Cu * L = 30 * 32.594 = 977.82kN/m Now Resisting moment M_R = 977.82 * R = 22000.95