Principles of Foundation Engineering (MindTap Course List)
9th Edition
ISBN: 9781337705028
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 13, Problem 13.6P
To determine
Find the maximum allowable load on the shaft.
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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.
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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)
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_forwardDetermine the ultimate load-carrying capacity of the drilled shaft shown in Figure P13.4, using the Reese and ONeill (1989) method.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
- [2] A 12 m long and 600 mm diameter drilled shaft installed in a sand soil. The nominal side friction capacity of the drilled shaft, Σfn As, is 800 kN and nominal toe-bearing capacity, qn'At, is 700 kN. The modulus of elasticity of the drilled shaft, E, is 30,000 MPa. Compute the settlement of the drilled shaft for a surface load of 600 kN.arrow_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_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_forward
- Figure 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 drilled shaft designed in accordance with the AASHTO code must support the following downward and uplift axial design loads: P = 850 k, Pup. = 270 k. The soil profile consists of: Undrained Shear Strength, s,, (lb/ft²) Depth (ft) Soil Description Unit Weight, y (lb/ft³) N60 0-15 Clayey silt 115 1200 15-35 Silty clay 112 1800 35-55 Sandy silt (nonplastic) 115 24 55-80 Silty sand 124 43 Practice Problems 597 The groundwater is at a depth of 50 ft. Using the AASHTO resistance factors, select a diameter and depth for a single drilled shaft to support these design loads. Use a load factor of 0.9 on the weight of the shaft. Note there are many different diameter-length combinations that would be satisfactory, but select one that you think would be most appropriate.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_forward
- A drilled shaft is constructed in a uniform clay layer of 40 ft. and tipped in a uniform, dense sand layer with N60 of 30. The drilled shaft has a diameter of 3 ft. and embedded length is 40 ft. Assume a total unit weight of 125 pcf for clay and 120 pcf for sand. The water table is at a depth of 15-ft. The unit weight of clay below water table is 125 pcf as well. The clay layer has undrained shear strength of 2 ksf. Assume depth of zone of seasonal moisture change to be 5-ft. Find the ultimate load capacity of the drilled shaft. Use alpha method for clay and beta method for sand. Clay y=Ysat=125 pcf c,=2 ksf 15 ft Water table 40 ft Dense sand: N=30; Ysat=120 pcfarrow_forwardThis is I'm uploading second time. If you don't know the solution please leave it but don't copy from other experts solution. Give me right solution with clear calculations.arrow_forwardA 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 marrow_forward
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