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 19.2P
To determine
Find the net allowable point bearing capacity.
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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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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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- 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_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
- A 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_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 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
- [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_forwardAn 8 in diameter drilled shaft will penetrate through an expansive stiff clay to a depth well below the active zone. It will carry a downward load of 5,200 lb. The undrained shear strength of the clay is 900 psf and the active zone extends to a depth of 10 ft. Determine the following: What is the uplift skin friction load? Is a tensile failure in the shaft possible (do not forget to consider the weight of the shaft)? assume: unit weight concrete = 150pcf, Concrete has no tensile strength. Assume full active zone in calculations.arrow_forwardAn 8 in diameter drilled shaft will penetrate through an expansive stiff clay to a depth well below the active zone. It will carry a downward load of 5,200 lb. The undrained shear strength of the clay is 900 psf and the active zone extends to a depth of 10 ft. Determine the following:a. What is the uplift skin friction load?b. Is a tensile failure in the shaft possible (do not forget to consider the weight of the shaft)?c. What is the required reinforcing steel to prevent a tensile failure (if necessary)? Use fy = 40 k/in2 and a load factor of 1.7.arrow_forward
- Question 2. Drilled shaft A 500-mm diameter, 15-m deep 4.0m drilled shaft foundation is to be constructed in the soil profile shown below. The design axial load is 700 kN. The factor of safety for downward loading is 3.0, and the factor of safety for upward loading is 6.0. Compute the allowable downward and upward loads capacity and determine if the design is acceptable. :15:0 m Medium Clay f = 30 kPa Groundwater table Stiff Clay f₁ = 60 kPa 9₁ '= 2000 kPaarrow_forwardA 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_forward
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