Pile_review_exercise-soln

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Mechanical Engineering

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Jan 9, 2024

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1. Consider a deep cohesive soil with the following properties: Top Layer: ɣ = 116 pcf Lower Layer: ɣ = 124 pcf e 0 = 0.48 e 0 = 0.86 C c = 0.35 C c = 0.52 q u = 2500 psf q u = 1800 psf Find: The factor of safety for the given design load and the expected settlement of the pile group under the design load if each pile is smooth steel 18 ” dia. Use the Coyle & Sulaiman method Solution: Part a) S/d = 3ft / 18” = 2 < 3 → use: Q g = 2 x D x (W + L) x f + 1.3 x c x N c x W x L where: W = L = 3 + 3 + 1.5 = 7.5 ft f 1 = c 1 x α 1 c 1 = q u1 / 2 = 2500 / 2 = 1250 psf = 1250 x 0.7 = 875 psf From graph: α 1 = 0.7 f 2 = c 2 x α 2 c 2 = q u1 / 2 = 1800 / 2 = 900 psf = 900 x 0.8 = 720 psf From graph: α 1 = 0.8 D 1 = 17 ft D 2 = 28 ft Q g = 2 x D 1 x (W + L) x f 1 + 2 x D 2 x (W + L) x f 2 + 1.3 x c 2 x N c x W x L = 2 x 17 x (7.5 + 7.5) ft x 875 lb/ft 2 + 2 x 28 x (7.5 + 7.5) ft x 720 lb/ft 2 + 1.3 x 900 lb/ft 2 x 5.14 x 7.5 ft x 7.5 ft = 446.25 + 604.8 + 338.3 = 1389.35 kips SF = Q g /Q D = 1389.35 / (450 x 2) = 1.54 Bottom of lower layer @ elev: 0’

Part b) Deep clay → imaginary footing at 2 / 3 pile depth = 2 / 3 x 45’ = 30 ft → @ elev: (100 – 30) = 70’ Clay layer extends from bottom of imaginary footing down to elev: 0’ → (70’ – 0’) = 70 ft Mid-point of clay layer is at 35 ft. S c = ((C c x H) / (1 + e 0 )) x log(p/p 0 ) Overburden pressure @ mid-point of clay layer: p 0 @ 35’ = 9 ft x 116 lb/ft 2 + 8 x (116 – 62.4) lb / ft 3 + (83 – 35) ft x (124 – 62.4) lb / ft 3 = 1044 + 428.8 + 2956.8 = 4429.6 lb / ft 2 Area of imaginary footing at mid-point of clay layer: W = L = 7.5 + 35 = 42.5 ft Area = 42.5 x 42.5 = 1806.25 ft 2 Δp @ 35’ = Load / Area = (450 tons x 2000 lb / ton) / 1806.25 ft 2 = 498.27 lb / ft 2 P @ 35’ = p 0 @ 35’ + Δp @ 35’ = 4429.6 + 498.27 = 4927.87 lb / ft 2 S c = ((C c x H) / (1 + e 0 )) x log(p/p 0 ) = ((0.52 x 70 ft) / (1 + 0.86)) x log (4927.87 / 4427.87) = 19.57 x log (1.1124) ft = 19.57 x log (1.1125) ft = 19.57 x 0.0463 ft x 12 in / ft = 10.87 in 2. Consider a dense sandy soil with the following properties: Top Layer: ɣ = 18.22 kN/m 3 ɸ = 38  K = 0.52 Lower Layer: ɣ = 19.48 kN/m 3 ɸ = 43  K = 0.39 Find: The factor of safety for the given design load and the expected settlement of the pile group under the design load given that a pile load test on a single 0.5 m diameter pile resulted in a settlement of 3” . Individual pile diameter is 0.5 m. Solution: Part a) Convert metric unit weight and pile Ø to imperial units to avoid more complicated conversions later. Pile Ø = 0.5 m x (3.28 ft / m) = 1.64 ft 1 kN/m 3 = 6.423 lb /ft 3 Unit weight δ top layer = 18.22 kN/m 3 x 6.423 lb /ft 3 = 117.0 lb /ft 3 Unit weight δ lowee layer = 19.48 kN/m 3 x 6.423 lb /ft 3 = 125.1 lb /ft 3 Dense sand → d c = 20 Ø = 20 x 1.64 ft = 32.8 ft Determine vertical pressures p v : p v@elev: 91 ’ = 9 ft x 117.0 lb /ft 3 = 1053 lb /ft 2 p v@elev: 83 ’ = 1053 + 8 ft x (117.0 – 62.4) lb /ft 3 = 1053 + 436.8 = 1489.8 lb /ft 2 p v@elev: 67.2 ’ = 1489.8 + 15.8 ft x (125.1-62.4) lb /ft 3 = 1489.8 + 990.7 = 2480.5 lb /ft 2

Q ult = Q friction + Q tip → since there are 2 layers of clay over the length of the pile, calculate Q friction for each layer separately Determine areas under the p v curve: For top layer: Area p v top = ½ x 1053 lb/ft 2 x 9 ft +1053 lb/ft 2 x 8 ft + ½ x (1489.8 – 1053) lb/ft 2 x 8 ft = 4738.5 + 8424.0 + 1747.2 = 14909.7 lb / ft For lower layer: Area p v lower = 1489.8 lb/ft 2 x 15.8 ft + ½ x (2480.5 – 1489.8) lb/ft 2 x 15.8 ft +2480.5 x 12.2 ft = 23538.8 + 7826.5 + 30262.1 = 61627.5 lb/ft Circumference of pile= π x D = π x 0.5 m x 3.28 ft / m = 5.15 ft tan δ = 0.20 (smooth steel) Q friction = Q friction top + Q friction lower = Area under pv curve x circumference x K x tan δ = Area p v top x circumference x K top x tan δ + Area p v lower x circumference x K lowerp x tan δ = 14909.7 lb / ft x 5.15 ft x 0.52 x 0.20 + 61627.5 lb/ft x 5.15 ft x 0.39 x 0.20 = 7985.6 + 24755.8 = 32741.4 lbs = 32.74 kips Q tip = Area of tip x p v @ tip x N q* Area of tip = π r 2 = π x (0.25m x 3.28 ft / m) 2 = 2.11 ft 2 For φ = 43° N q* = 180 Q tip = Area of tip x p v @ tip x N q* = 2.11 ft 2 x 2480.5 lb /ft 2 x 180 = 942093 lbs = 942.09 kips Q ult = Q friction + Q tip = 32.74 + 942.09 = 974.8 kips Q g = Q ult x # piles = 974.8 x 9 = 8774 kips Q design = (450 tons x 2 kips / ton) = 900 kips SF = Q g / Q design = 8774 kips / 900 kips = 9.75 Part b) Settlement of pile group = settlement of test pile x (smallest dimension of pile group / diameter of test pile ) S = S o x (B/B) ½ = 3 inches *(7.5 ft / (0.5m x 3.28 ft / m)) = 6.42 inches

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