Tiêu đề HPGE 11 Solutions.pdf ✅

Ap = 1 2 (117.9 kPa)(6 m) + (117.9 kPa)(9 m − 6 m) Ap = 707.4 kN m Compute for the skin friction: Qf = kpApP tan α Qf = (3) (707.4 kN m ) (4 × 0.3 m)(0.45) Qf = 1145.988 kN ▣ 2. Capacity due to end bearing. [SOLUTION] Formula for end bearing: Qb = qNqAtip Qb = (117.9 kPa)(80)(0.3 m) 2 Qb = 848.88 kN ▣ 3. Total allowable axial capacity if the factor of safety is 2. [SOLUTION] Compute the total ultimate bearing capacity of the piles: Qu = Qf + Qb Qu = 1145.988 kN + 848.88 kN Qu = 1994.868 kN Qa = Qu FS Qa = 1994.868 kN 2 Qa = 997.435 kN SITUATION 2. A nine-pile group composed of 0.3 m diameter pipe piles is embedded 17.5 m deep in soil conditions indicated by the sketch. Nc = 8.
▣ 4. Determine the capacity if the piles act individually. [SOLUTION] Compute for the capacity: Q = ∑(AtipcNc + ∑cLαP) Q = 9 [ π 4 (0.3 m) 2 (96 kPa)(8) + π(0.3 m)(7 m × 0.85 × 43 kPa + 7 m × 0.75 × 57.5 kPa + 3.5 m × 0.5 × 96 kPa)] Q = 6644.4 kN ▣ 5. Determine the total capacity if the piles act as a group. [SOLUTION] Formula for capacity of a pile group: Q = LBcNc ∑2(L + B)αcL Q = (2.1 m)(2.1 m)(96 kPa)(8) + 2(2.1 m + 2.1 m)[(0.85)(43 kPa)(7 m) + (0.75)(57.5 kPa)(7 m) + (0.5)(96 kPa)(3.5 m)] Q = 9482.97 kN ▣ 6. Determine the design load if FS = 3.