I need detailed help solving the problem 12.13 about skin friction resistance, please. TABLE 12.10 Variation of A with PileEmbedment Length, LEmbedmentlength, L (m)05101520253035405060708090TABLE 12.12 Variation of a with cu/Cu12345678910Open-ended pile0.50.40.3550.330.310.290.280.260.2550.25αλ0.50.3360.2450.2000.1730.1500.1360.1320.1270.1180.1130.1100.1100.110Closed-ended pile0.50.440.410.3950.380.3650.350.330.320.311.21.00.80.60.40.200.1TABLE 12.11 Variation of a (Interpo-lated Values Based onTerzaghi et al., 1996)CuPa≤0.10.20.30.40.60.81.01.21.41.61.82.02.42.8Note: Pa = atmospheric pressure≈100 kN/m²PI = 554030201512PI < 1010.3110.51.0α1.000.920.820.740.620.540.480.420.400.380.360.350.340.34FIGURE 12.26 Variation of awith c/ for the NGI-99 method[Eqs. (12.59a) and (12.59b)] [12.13] A concrete pile 406 mm × 406 mm in cross section is shown in Figure P12.13. Calculate the ultimate skin frictionresistance by using thea) a method [use Eq. (12.61) and Table 12.11]Qs = [fpAL = Σ acupALb) λ methodc) ẞ methodUse O'R = 20° for all clays, which are normally consolidated.6.1 m12.2 m406 mmFIGURE P12.13Qs Efp AL = Σacup ALtable=GroundwaterSilty clay=YsatCu = 35 kN/m²Plasticity index, PI = 1518.55 kN/m³Silty clay=YsatCu = 75 kN/m²Plasticity index, PI = 2019.24 kN/m³(12.61)

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[12.13] A concrete pile 406 mm × 406 mm in cross section is shown in Figure P12.13. Calculate the ultimate skin friction resistance by using the a) a method [use Eq. (12.61) and Table 12.11] Qs = [fpAL = [ac.PAL b) λ method c) ẞ method Use O'R = 20° for all clays, which are normally consolidated. 6.1 m 12.2 m 406 mm FIGURE P12.13 Q, = Σfp ΔL = Σacp ΔL Groundwater table Silty clay Ysat 18.55 kN/m³ Cu = 35 kN/m² Plasticity index, PI = 15 Silty clay Ysat = 19.24 kN/m³ Cu = 75 kN/m² Plasticity index, PI = 20 (12.61)

[12.13] A concrete pile 406 mm × 406 mm in cross section is shown in Figure P12.13. Calculate the ultimate skin friction resistance by using the a) a method [use Eq. (12.61) and Table 12.11] Qs = [fpAL = [ac.PAL b) λ method c) ẞ method Use O'R = 20° for all clays, which are normally consolidated. 6.1 m 12.2 m 406 mm FIGURE P12.13 Q, = Σfp ΔL = Σacp ΔL Groundwater table Silty clay Ysat 18.55 kN/m³ Cu = 35 kN/m² Plasticity index, PI = 15 Silty clay Ysat = 19.24 kN/m³ Cu = 75 kN/m² Plasticity index, PI = 20 (12.61)

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter12: Pile Foundations

Section: Chapter Questions

Problem 12.13P: A concrete pile 16 in. 16 in. in cross section is shown in Figure P12.13. Calculate the ultimate...

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