Principles of Foundation Engineering, SI Edition
8th Edition
ISBN: 9781305446298
Author: Braja M. Das
Publisher: Cengage Learning US
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Textbook Question
Chapter 10, Problem 10.15P
A free-headed drilled shaft is shown in Figure P13.10. Let Qg = 260 kN, Mg = 0, γ = 17.5 kN/m3, ϕ′ = 35°, c' = 0, and Ep = 22 × 106 kN/m2. Determine
- a. The ground line deflection, xo
- b. The maximum bending moment in the drilled shaft
- c. The maximum tensile stress in the shaft
- d. The minimum penetration of the shaft needed for this analysis
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A free-headed drilled shaft, shown in Figure 4, has an elastic modulus, Ep = 20,000 MPa.
M, = 880 kN m
Q = 245 kN,
Sand
at = 19 kN/m3
O' = 34°
1.2 m
Figure 4
(a) Determine the ground line deflection, x.
Refer to Figure 11.26b. For the drilled shaft with bell, given:Thickness of active zone, Z = 9 mDead load = 1500 kN Live load = 300 kNDiameter of the shaft, Ds = 1 mZero swell pressure for the clay in the active zone = 600 kN/m2Average angle of plinth-soil friction, Φ'ps = 20°Average undrained cohesion of the clay around the bell = 150 kN/m2. Determine the diameter of the bell, Db. A factor of safety of 3 against uplift is required with the assumption that dead load plus live load is equal to zero.
Problem II. The figure below shows the cross-section of a thin-walled shaft subjected to a torque MT = 100 kN-m
If a = 200 mm, t = 20 mm, and G = 77.2 GPa,
t
t
t
2a
2a
a) Determine the shear stress in semi-circle portion of the profile.
b) Compute the maximum torque that can be applied if the allowable shear stress is 200 MPa
Chapter 10 Solutions
Principles of Foundation Engineering, SI Edition
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- A compound shaft is shown. Segment (1) has an outside radius of 100 mm and thickness of wall is 10 mm. Segment (2) has an outside diameter dimension of 150mm and a wall thickness of 10 mm. The compound shaft is subjected to torques TB = 42 kN-m and TC = 18 kN-m. Evaluate the maximum shear stress magnitude in each shaft segment. Answer in MPaarrow_forwardFor the drilled shaft described in Problem 19.7, estimate the total elastic settlement at working load. Use Eqs. (18.45), (18.47), and (18.48). Assume that Ep = 20 106 kN/m2, s = 0.3, Es = 12 103 kN/m2, = 0.65 and Cp = 0.03. Assume 80% mobilization of skin resistance at working load. (See Part c of Problem 19.7) 19.7 Figure 19.16 shows a drilled shaft without a bell. Here, L1 = 6 m, L2 = 7 m, Ds = 1.5 m, cu(1) = 50 kN/m2, and cu(2) = 75 kN/m2. Find these values: a. The net ultimate point bearing capacity. Use Eqs. (19.23) and (19.24) b. The ultimate skin resistance. Use Eqs. (19.26) and (19.28) c. The working load, Qw (FS = 3) FIG. 19.16arrow_forwardExample : Figure shows bellow a drilled shaft without a bell Assume the folowing values : L1 =6 m Cu(1) = 50 KN/m L2 = 7 m Cu2) = 75 KN/m? Ds = 1.5m Determine : The net ultimate point bearing capacity by use general equation. b. The ultimate skin friction by use general equation. The working load, Qw, factor of safety = 3 a. C.arrow_forward
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