Title HPGE 12 Solutions.pdf ✅

σA = 0.65γHkab σA = 0.65 (18 kN m3 ) (13.25 m) ( 1 − sin 38° 1 + sin 38°) (3 m) σA = 110.633 kN m Isolate the segment from the top of the soil to point B. ∑MB = 0 −FA(3 m) + (110.633 kN m ) (4.25 m) ( 4.25 m 2 ) = 0 FA = 333.05 kN ▣ 2. Strut load at B. [SOLUTION] Compute for FB1: FA + FB1 = (110.633 kN m ) (4.25 m) FB1 = (110.633 kN m ) (4.25 m) − 333.05 kN FB1 = 137.14 kN Isolate segment BC: FB2 = σA(3 m) 2

Friction force along the failure plane is given by: Ff = μN The weight of the soil that will slide in the failure plane assuming unit width is given by: W = 1 2 γh 2 [ sin(β − θ) sin β sin θ ] W = 1 2 (16.5 kN m3 ) (5 m) 2 [ sin(45° − 30°) sin 45° sin 30°] W = 150.985 kN Compute for the friction force: Ff = μN Ff = (tanφ)(W cos θ) Ff = tan 25° (150.985 kN)(cos 30°) Ff = 60.973 kN ▣ 5. Cohesive force along the failure plane. [SOLUTION] Fc = cA Fc = c H sin θ b Fc = (7 kPa) ( 5 m sin 30°) (1 m) Fc = 70 kN ▣ 6. Factor of safety against sliding. [SOLUTION]