Question

Problem 1: (20 points) Given A square footing (5 ft by 5 ft) is subjected to an inclined load as shown in g. Required The factor of safety against bearing capacity failure. 30 a 40 kips 5 ft 5 ft 1.0 a 0.8 0.6 0.4 0.2 Cohesive Soil Cohesive Soil 0.8 D/B 1 Granular Soil 0.6 Granular Soil 0.4 De-o- 0.2 020 40 60 80 90 2 40 60 80 90 Inclination a of Load to Vertical inclination α. Of Load to Vertical Inclination of Foundation to Horizontal

Answer 1

Since phi is not given (angle of internal friction), the shearing stress is stiff clays will be dominated by cohesion over internal friction. Because phi = 0 The general bearing capacity equation is: q_u = C N_c S_c d_c i_c + q N_q S_q d_q i_q + 0.5 r BN_r S_r d_r i_r Where: qult = ultimate bearing capacity c = q_u/2 = 3600/2 = 1800 lb/ft^2 N_c, N_q, N_r = Bearing Capacity factor For phi = 0, N_c = 5.14, N_q = 1, N_r = 0 S_c, S_q, S_r = shape factors S_c = 1 + B/L (N_q/N_c) = 1 + 5/5 times 1/5.14 S_c = 1.1946 S_q = 1 + B/L Tan phi S_q = 1 d_c, d_q, d_r = depth factors For phi = 0, d_q = d_c = 1 + 0.4 D_f/B d_c = 1 + 0.4 times 5/5 = 1.4 i_c, i_q, i_r = inclination factors: i_c = i_q = (1 - alpha/90)^2 i_c = i_q = (1 - 30/90)^2 = 0.444
factor of safety against bearing capacity failure is 5 hece safe