A drilled shaft is shown in Figure P10.1. Determine the net allowable point bearing capacity. Given Ye = 15.6 kN/m³ Ys = 17.6 kN/m³ 6' = 35° D, = 2 m D = 1.2 m Lj = 6 m L = 3 m Factor of safety C, = 35 kN/m² 3 L=L, Q,-Σα"c,p ΔL (10.37) L=0 atmospheric pressure = 100 kN/m² (~2000 lb/ft?). So, conservatively, we may assume that where Pa = a* = 0.4 (10.39)

Principles of Foundation Engineering (MindTap Course List)
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ISBN:9781305081550
Author:Braja M. Das
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Chapter10: Drilled-shaft Foundations
Section: Chapter Questions
Problem 10.1P
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For the drilled shaft described in Problem 10.1, what skin resistance would develop in the top 6 m, which are in clay ?

A drilled shaft is shown in Figure P10.1. Determine the net allowable point bearing
capacity. Given
Ye = 15.6 kN/m³
Ys = 17.6 kN/m³
6' = 35°
D, = 2 m
D = 1.2 m
Lj = 6 m
L = 3 m
Factor of safety
C, = 35 kN/m²
3
Transcribed Image Text:A drilled shaft is shown in Figure P10.1. Determine the net allowable point bearing capacity. Given Ye = 15.6 kN/m³ Ys = 17.6 kN/m³ 6' = 35° D, = 2 m D = 1.2 m Lj = 6 m L = 3 m Factor of safety C, = 35 kN/m² 3
L=L,
Q,-Σα"c,p ΔL
(10.37)
L=0
atmospheric pressure = 100 kN/m² (~2000 lb/ft?).
So, conservatively, we may assume that
where Pa =
a* = 0.4
(10.39)
Transcribed Image Text:L=L, Q,-Σα"c,p ΔL (10.37) L=0 atmospheric pressure = 100 kN/m² (~2000 lb/ft?). So, conservatively, we may assume that where Pa = a* = 0.4 (10.39)
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