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Principles of Foundation Engineering (MindTap Course List)
9th Edition
ISBN: 9781337705028
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Textbook Question
Chapter 13, Problem 13.4P
Determine the ultimate load-carrying capacity of the drilled shaft shown in Figure P13.4, using the Reese and O’Neill (1989) method.
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Figure P10.7 shows a drilled shaft without a bell. Assume the following values:L1 = 6 m cu(1) = 50 kN/m2L2 = 7 m cu(2) = 75 kN/m2Ds = 1.5 mDetermine:a. The net ultimate point bearing capacity [use Eqs. (10.33) and (10.34)]b. The ultimate skin friction [use Eqs. (10.37) and (10.39)]c. The working load Qw (factor of safety = 3)
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.
A drilled shaft constructed in medium sand is shown in the figure below. Given information is: y
= 18 kN/m', '= 38°. Sand is medium-density sand, and the average standard penetration number
(N60) within 2Ds below the drilled shaft is 19. Using the method proposed by Reese and O'Neill,
determine the following:
(a) The net allowable point resistance for a base movement of 25 mm.
(b) The shaft frictional resistance for a base movement of 25 mm.
(c) The total load that can be carried by the drilled shaft for a total base movement of 25 mm.
1 m
11 m
12 m
- 2 m
Chapter 13 Solutions
Principles of Foundation Engineering (MindTap Course List)
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Determine the ultimate load-carrying capacity of...Ch. 13 - For the same data given in Problem 13.4, determine...Ch. 13 - Prob. 13.6PCh. 13 - A 3 ft diameter straight drilled shaft is shown in...Ch. 13 - Prob. 13.8PCh. 13 - Figure P13.9 shows a drilled shaft extending into...Ch. 13 - A free-headed drilled shaft is shown in Figure...
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- A free-headed drilled shaft is shown in Figure P13.10. Let Qg = 260 kN, Mg = 0, = 17.5 kN/m3, = 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 analysisarrow_forwardFigure P13.9 shows a drilled shaft extending into clay shale. Given: qu (clay shale) = 1.81 MN/m2. Considering the socket to be rough, estimate the allowable load-carrying capacity of the drilled shaft. Use FS = 4. Use the Zhang and Einstein procedure.arrow_forwardA 3 ft diameter straight drilled shaft is shown in Figure P13.7. Determine the load-carrying capacity of the drilled shaft with FS = 3. Take / as 0.8 for the sand.arrow_forward
- For the same data given in Problem 13.4, determine the load-carrying capacity of the drilled shaft, limiting the settlement to 10.0 mm. 13.4 Determine the ultimate load-carrying capacity of the drilled shaft shown in Figure P13.4, using the Reese and ONeill (1989) method.arrow_forwardFor the drilled shaft described in Problem 19.7, determine these values: a. The ultimate load-carrying capacity b. The load-carrying capacity for a settlement of 25 mm Use the procedure outlined in Section 19.8. 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_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_forward
- Earth Sciences Downward force is 4000 kg and rotational force is 3000 kg. Contact surface area (cross sectional area) is 100 cm2. Also, the rock sample (diameter: 20 cm) is tested by a uniaxial compressive strength machine and the sample was cracked at 30000 kg. Answer the following questions:a. Find the resultant force acting on the rock formation.b. Find the bearing strength of the rock drilled.c. Can we drill under these circumstances?arrow_forwardFor the drilled shaft described in Problem 10.1, what skin resistance would develop in the top 6 m, which are in clay ?arrow_forwardRefer 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.arrow_forward
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