Tiêu đề Eval 6 Nov 2024 Solutions.pdf ✅

SITUATION 2. Shown is the figure for the pressure diagram ▣ 3. The coefficient of active lateral pressure is ka = 1 − sinφ 1 + sinφ ka = 1 − sin 32∘ 1 + sin 32∘ ka = 0.307 ▣ 4. Solving for the stresses first then the resultant force in each portion of the pressure diagram, σ1 = kaγ1h1 σ1 = (0.30726) (16.5 kN m3 ) (3 m)(1 m) σ1 = 15.209 kN m F1 = 1 2 σ1h1
F1 = 1 2 (15.209 kN m ) (3 m) F1 = 22.814 kN σ2 = 15.209 kN m F2 = (15.209 kN m ) (4 m) F2 = 60.837 kN σ3 = ka (γsat − γw)h2 σ3 = (0.30726) (18 kN m3 − 9.81 kN m3 ) (4 m)(1m) σ3 = 10.066 kN m F3 = 1 2 σ3h2 F3 = 1 2 (10.066 kN m ) (4 m) F3 = 20.132 kN σ4 = γwh2 σ4 = (9.81 kN m3 ) (4 m)(1 m) σ4 = 39.24 kN m F4 = 1 2 σ4h2 F4 = 1 2 (39.24 kN m ) (4 m) F4 = 78.48 kN Therefore, the total active force is F = F1 + F2 + F3 + F4 F = 22.814 kN + 60.837 kN + 20.132 kN + 78.48 kN F = 182.26 kN