Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 19, Problem 19.6P
To determine
Find the maximum load that can be allowed on the drilled 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.
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.
Chapter 19 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Redo Problem 19.2. Use Eq. (19.4) and Es = 600 pa....Ch. 19 - For the drilled shaft described in Problem 19.2,...Ch. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - For the drilled shaft described in Problem 19.7,...Ch. 19 - For the drilled shaft described in Problem 19.7,...Ch. 19 - Prob. 19.10P
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- Show me all clear calculationsarrow_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_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_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
- For 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 10.1, what skin resistance would develop in the top 6 m, which are in clay ?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_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_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_forwardFor 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_forward
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