Tiêu đề Eval 6 Nov 2024 Solutions.pdf ✅

EVAL EXAM 6 SOLUTIONS SITUATION 1. Shown is the Mohr’s Circle for the figure. Take note that the cohesion is 0 since the soil is sandy. ▣ 1. • A. From exterior angles of triangle, and using the figure, 2θ = 90∘ + 32∘ θ = 61∘ • B. From the formula, θ = 45∘ + φ 2 θ = 45∘ + 16∘ 2 θ = 61∘ ▣ 2. From the right triangle formed, sin 32∘ = 200 σ3 + 200 σ3 = 177.42 kPa

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
▣ 5. The ultimate bearing capacity for a strip footing is given by: qu = cNc + qNq + 0.5γBNγ But since assuming local shear failure, set c ′ = 2 3 c = 2 3 (50 kPa) = 100 3 kPa qu = ( 100 3 kPa) (8) + (19.2 kN m3 × 4.5 m) (1.94) + 0.5 (19.2 kN m3 ) (0.25) qu = 437.28 kPa ▣ 6. The ultimate bearing capacity for a circular footing is given by: qu = 1.3cNc + qNq + 0.3γBNγ qu = 1.3 ( 100 3 kPa) (8) + (19.2 kN m3 × 5 m) (1.94) + 0.3 (19.2 kN m3 ) (1.5 m)(0.25) qu = 535.07 kPa ▣ 7. Note that for the capacity of one pile is composed of end bearing and skin friction. Q = Qe + Qf Qe = cNcAtip Qf = ∑cuPαL At the end, the properties are: cu = 95 kPa, α = 0.5, Nc = 9 Qe = (95 kPa)(9) [ π 4 (0.4 m) 2 ] Qe = 107.44 kN Solving for the skin friction, Qf = (20 kPa)[π(0.4 m)](1)(2 m) + (60 kPa)[π(0.4 m)](1)(6 m) + (95 kPa)[π(0.4 m)](0.5)(4 m) Qf = 741.42 kN The capacity of one pile is: Q = 107.44 kN + 741.42 kN Q = 848.86 kN For a single pile failure mode, multiply the capacity of one pile by the number of piles Qgroup = 848.86 kN × 12 Qgroup = 10186.30 kN