SAMPLE PROBLEMS ON PILES ENGR. NINO MAR H. CLARITO, RCE, SO2, RMP, RES CE 2020 A prestressed concrete pile 400 mm x 400mm in cross section and 20 m long is driven on a clayey soil with unconfined compressive strength of 110 KPa. Determine the skin friction, in KPa using the adhesion factor πΌ = 0.75 Solution ππ πππ = πΌ CPL C = πΆπ’ = ½ ππ’ πΆπ’ = ½(110) πΆπ’ = 55 KPa ππ πππ = 0.75 55 4 0.40 20 πΈππππ = 1,320 KN CE 2021 Friction pile, 400mm x 400mm in cross section is embedded 20m, into a layer of plastic clay to support a pile cap factored load 12,000 KN. The unconfined compressive strength of the clay was obtained at 120 KPa. Evaluate the number of piles required to support on the assumption that the pile group has 80% efficiency. Solution πππππ’π = ππππππ£πππ’ππ (πΈπππππππππ¦) ππππππ£πππ’ππ = ππ’ππ‘ππππ‘π (n) ππ’ππ‘ππππ‘π = ππ‘ππ + ππ πππ ππ‘ππ = π ππ π΄π‘ππ C = πΆπ’ = ½ ππ’ πΆπ’ = ½(120) πΆπ’ = 60 KPa Solution ππ‘ππ = π ππ π΄π‘ππ ππ‘ππ = (60)(9)(0.40π₯0.40) πΈπππ = 86.4 KN ππ πππ = πΌ CPL ππ πππ = 1 60 4 0.40 20 πΈππππ = 1,920 KN Solution ππ’ππ‘ππππ‘π = ππ‘ππ + ππ πππ ππ’ππ‘ππππ‘π = 86.4 + 1,920 ππ’ππ‘ππππ‘π = 2,006.40 KN/pile ππππππ£πππ’ππ = ππ’ππ‘ππππ‘π (n) ππππππ£πππ’ππ = 2,006.4(n) πππππ’π = ππππππ£πππ’ππ (πΈπππππππππ¦) = Pu 12,000 = 2,006.4(n) (0.80) N = 7.48 say 8 piles CE 2022 Evaluate the resisting capacity against axial load due to skin friction of a round wooden pile embedded into a layer of plastic clay given the following conditions: Size of Pile = 0.35m (diameter) Depth of penetration into the clay layer = 20 m Unconfined compressive strength of the clay = 110 KPa Solution ππ πππ = πΌ CPL C = πΆπ’ = ½ ππ’ πΆπ’ = ½(110) πΆπ’ = 55 KPa ππ πππ = 1.0 55 π 0.35 20 πΈππππ = 1,209.51 KN Problem A 0.36 m square prestressed concrete pile is driven in a clayey soil having an unconfined compressive strength of 110 KPa. Unit weight of clay s 18 KN/m^3. design capacity of pile is 360 KN. Factor of safety of 2 ο΅ Compute the length of the pile using πΌ πππ‘βππ ππ πΌ = 0.76 ο΅ Compute the length of the pile using λ πππ‘βππ ππ λ = 0.14 ο΅ Compute the length of the pile using β πππ‘βππ ππππ = 300 and the clay has an over consolidated ratio of 3. Solution ππ‘ππ = π ππ π΄π‘ππ C = πΆπ’ = ½ ππ’ πΆπ’ = ½(110) πΆπ’ = 55 KPa ππ‘ππ = (55)(9)(0.36 π₯ 0.36) ππ‘ππ = 64.15 KN Solution ππ’ππ‘ππππ‘π ππππππ€ππππ = πΉ. π. ππ’ππ‘ππππ‘π = ππ‘ππ + ππ πππ ππ‘ππ + ππ πππ ππππππ€ππππ = πΉ. π. 360 = 64.15 + ππ πππ 2 ππ πππ = 655.85 KN ππ πππ = πΌ CPL 655.85 = 0.76 55 4 0.36 πΏ π³ = 10.90 m Solution ππ πππ = PLλ (ππ£ + 2C) σπ£ = πΎπ h πΏ σπ£ = (18)(2) σπ£ = 9L 655.85 = (4)(0.36)(L)(0.14) [9πΏ + 2(55)] 3253.22 = L(9L + 110) 9πΏ2 + 110L – 3253.22 = 0 L = 13.86 m Solution ππ πππ = PL β ππ ππΆπ σπ£ = πΎπ h πΏ σπ£ = (18)(2) σπ£ = 9L β = (1- sin ππ )(tan ππ ) β = (1- sin300 )(tan 300 ) β = 0.289 655.85 = (4)(0.36)(L)(0.289)(9L) 3 L = 10.05 m Problem In a certain construction site a proposed warehouse building is to be constructed. During the unsupported excavation, a cave in occurred in a clay soil on a vertical trench when the trench was 10.2m deep. The unit weight of soil was 18.7KN/m^3. ο΅ Estimate the average unconfined compressive strength of clay. ο΅ Pile foundation was decided so a 0.30m x 0.30 m pile was driven on the same site with an allowable load of 200 KN. Compute the ultimate frictional resistance of the pile use if it has a factor of safety of 2. ο΅ Determine the length of the pile to be used using β πππ‘βππ ππ β = 0.30 Solution Note: for unsupported excavation, the active lateral force is zero π1 − π2 = 0 ½ πΎπ β2 − (2πΆ)(β) = 0 4πΆ h= πΎπ 10.2 = 4πΆ 18.7 C = 47.69 KPa ππ’ = 2πΆ ππ’ = 2 47.69 ππ = ππ. ππ π²π·π Solution ππ‘ππ = π ππ π΄π‘ππ ππ‘ππ = (47.69)(9)(0.30π₯0.30) ππ‘ππ = 38.63 KN ππ‘ππ + ππ πππ ππππππ€ππππ = πΉ. π. 200 = 38.63+ ππ πππ 2 πΈππππ = 361.37 KN Solution ππ πππ = PL β ππ = PL πππ£π σπ£ = πΎπ h πΏ σπ£ = (18.7)( ) 2 σπ£ = 9.35L 361.37 = (4)(0.30)(L)(0.3)(9.35L) L = 10.36 m Problem A square concrete pile 0.3m x 0.3m is required to support a load of 150 KN with a factor of safety of 3. the soil stratification consists of a 5m soft gray normally consolidated clay with a cohesion of 25 KPa on top of a deep over consolidated clay having an over consolidated ratio of 3.6 and a cohesion of 80 KPa. Use β = 0.25 ο΅ Compute the end bearing capacity of pile ο΅ Compute the frictional resistance capacity of pile ο΅ Compute the length of pile. Solution ππ‘ππ = π ππ π΄π‘ππ ππ‘ππ = (80)(9)(0.30 π₯ 0.30) πΈπππ = 64.80 KN Solution ππ‘ππ + ππ πππ ππππππ€ππππ = πΉ. π. 64.8+ ππ πππ 150 = 3 πΈππππ = 385.2 KN Solution ππ πππ = ππΏβ ππ = ππΏ πππ£π For the 5m length σπ£ = πΎπ h σπ£ = 17.8(2) + (18-9.81)(0.50) σπ£ = 39.695 KPa For the “h” m. length σπ£ = πΎπ h σπ£ = 17.8(2) + (18-9.81)(3) + (18.5-9.81)(h/2) σπ£ = 60.17 + 4.345h Solution ππ πππ = ππΏβ ππ 385.2 = (4)(0.30)(5)(0.25)(39.695) + (4)(0.30)(h)(0.25)(60.17 + 4.345h)( 3.6) 325.66 = 0.569h(60.17 + 4.345h) 572.34 = 60.17h + 4.345β2 h = 6.48 m Total length of pile L = 6.48 + 5 L= 11.48 m Problem A pipe pile in clay is driven as shown in the figure. The pipe has a diameter of 400mm. ο΅ Compute the skin resistance using πΌ πππ‘βππ if πΌ =0.50 for the lower 6m length of pile. ο΅ Compute the skin resistance using λ πππ‘βππ ππ λ = 0.14 for the entire length of pile. ο΅ Compute the skin resistance using β πππ‘βππ ππππ = 300 and the top and bottom clay layer is normally consolidated. Solution ππ πππ = πΌ CPL ππ πππ = 1.0 30 π 0.40 4 + 0.50(60) π 0.40 6 πΈππππ = 376.99 KN Solution ππ πππ = ππΏ λ (ππ£ + 2C) = For the 4m length σπ£ = πΎπ h 4 σπ£ = (18)(2) σπ£ = 36 KPa For the 6m length σπ£ = πΎπ h σπ£ = 18 4 + (20 − 9.81)(3) σπ£ = 102.57 KPa ππΏπππ£π Solution ππ πππ = ππΏ λ (ππ£ + 2C) = ππΏπππ£π ππ πππ = (π)(0.40)(4)(0.14)[36 + 2(30)] + (π)(0.40)(6)(0.14)[102.57 + 2(60)] ππ πππ = 67.56 + 234.94 πΈππππ = 302.5 KN Solution ππ πππ = ππΏ β ππ = ππΏ πππ£π From 0 to 4m σπ£ = πΎπ h 4 σπ£ = (18)(2) σπ£ = 36 KPa From 4 to 10m σπ£ = πΎπ h σπ£ = 18 4 + (20 − 9.81)(3) σπ£ = 102.57 KPa Solution β = (1- sin ππ )(tan ππ ) β = (1- sin300 )(tan 300 ) β = 0.289 ππ πππ = ππΏ β ππ = ππΏ πππ£π ππ πππ = (π)(0.40)(4)(0.289)(36) + (π)(0.40)(6)(0.289)(102.57) ππ πππ = 52.30 + 223.50 πΈππππ = 275.80 KN CE 2018 A pile cap is shown. The column is 400mm x 400 mm and carries a service dead load of 900 KN and a service live load of 1,300 KN. The centroid of main reinforcing bars is located 85 mm from the bottom of the footing. Use fc’ = 21 MPa and fy = 415 MPa. Use NSCP 2001 determine the following: ο΅ Required thickness based on wide beam shear ο΅ Required thickness based on punching shear ο΅ Factored moment at the critical section Solution ο΅ Pu = 1.4 DL + 1.7 LL ο΅ Pu = 1.4(900) + 1.7(1,300) ο΅ Pu = 3,470 KN ο΅ Pu/pile = 3,470/9 ο΅ Pu/pile = 385.56 KN/pile Solution Wide beam shear Vu = 385.56(3) = 1,156.67 KN Vu = φVc = φ(1/6) ππ ′ ππ€ d 1,156,670 = 0.85(1/6) 21 (2,800) d d = 636.32 mm Therefore, t = d + d’ = 636.32 + 85 t = 721.32 mm Solution Punching shear Vu = 385.56(8) = 3,084.48 KN Vu = φVc = φ(1/3) ππ ′ ππ d 3,084,480 = 0.85(1/3) 21 (4)(400 + π) d d = 596.18 mm Therefore, t = d + d’ = 596.18 + 85 t = 681.18 mm Solution Mu = Vu(z) Mu = 1,156.67(0.80 – 0.20) Mu = 694.002 KN-m Problem ο΅ A pile group consists of 9 friction piles in cohesive soil. Each pile has a diameter of 0.3m and center to center spacing of 1.2m. The ultimate capacity of each pile is 300 KN. ο΅ Compute the design capacity of the pile group using a factor of safety of 2 if the soil is cohesionless. ο΅ Compute the pile group efficiency using Converse-Labarre equation. Solution ππ = π(ππ’ ) ππ = 9 300 ππ = 2,700 KN ππππππ€ = ππ πΉ.π. 2,700 ππππππ€ = 2 ππππππ = 1,350 KN Solution Ε = 1 − π[ π −1 π+π(π−1) ] 90ππ π tan π = π 0.3 tan π = 1.2 π = 14.040 Ε = 1 − 14.04[ Ε = 0.792 Ε = 79.2% 3 −1 3+3(3−1) ] 90(3)(3) Problem A pile group consists of four friction piles in cohesive soils. Each pile has a diameter of 0.30 m and a spacing of 0.90m center to center. Unit weight of clay is 19.8 KN/m^3 with an unconfined compressive strength of 190 KN/m^2. length of pile is 10m. Determine the allowable group capacity based on individual pile failure. Use factor of safety of 2 and πΌ =0.56 . Use ConverseLabarre equation for pile group efficiency. Solution ππ‘ππ = π ππ π΄π‘ππ π ππ‘ππ = (190/2)(9)( )(0.30)^2 4 πΈπππ = 60.44 KN ππ πππ = πΌ CPL ππ πππ = 0.56 190/2 π 0.30 10 πΈππππ = 501.40 KN Solution ππ‘ππ + ππ πππ ππππππ€ππππ = πΉ. π. 60.44 + 501.40 ππππππ€ππππ = 2 ππππππ€ππππ = 280.92 KN for an individual pile Solution π −1 π+π(π−1) Ε = 1 − π[ ] 90ππ π tan π = π 0.3 tan π = 0.9 π = 18.430 2 −1 2+2(2−1) Ε = 1 − 18.43[ ] 90(2)(2) Ε = 0.795 Ε = 79.5% Solution πππππ’π πππππ€ = ππππππ€ππππ (n)(Eff) πππππ’π πππππ€ = (280.92)(4)(0.795) πΈπππππ πππππ = 893.33 KN (group piles) Problem A nine pile group consists of 0.30 m diameter friction concrete piles 12m long. The piles are driven into clay having an unconfined compressive strength of 180 KPa and the unit weight of clay is 18 KN/m^3. the spacing of piles is 0.75m center to center. ο΅ Find the allowable group pile capacity based on individual pile failure using factor of safety of 3. ο΅ Find the block capacity of pile group using factor of safety of 3. ο΅ Find the minimum pile spacing center to center to achieve 100% efficiency. Solution ππ’ππ‘ππππ‘π ππππππ€ππππ = πΉ. π. ππ’ππ‘ππππ‘π = ππ‘ππ + ππ πππ ππ‘ππ + ππ πππ ππππππ€ππππ = πΉ. π. ππ‘ππ = π ππ π΄π‘ππ C = πΆπ’ = ½ ππ’ πΆπ’ = ½(180) πΆπ’ = 90 KPa Solution ππ‘ππ = π ππ π΄π‘ππ π ππ‘ππ = (90)(9)( )(0.30)^2 4 πΈπππ = 57.26 KN ππ πππ = πΌ CPL ππ πππ = 1 90 π 0.30 12 πΈππππ = 1,017.88 KN Solution ππ‘ππ + ππ πππ ππππππ€ππππ = πΉ. π. 57.26 + 1,017.88 ππππππ€ππππ = 3 ππππππ€ππππ = 358.38 KN Solution πππππ’π πππππ€ = ππππππ€ππππ (n)(Eff) 2 π + π − 2 π + 4π· Ε= ππππ· m = (no. of columns) n = (no. of rows) 2 3 + 3 − 2 (0.75) + 4(0.30) Ε= (3)(3)(π)(0.30) Ε = 0.849 πππππ’π πππππ€ = (358.58)(9)(0.849) πΈπππππ πππππ = 2,739.91 KN Solution 2 π + π − 2 π + 4π· Ε= ππππ· 2 3 + 3 − 2 (π) + 4(0.30) 1= (3)(3)(π)(0.30) S = 0.91m
