EBK PRINCIPLES OF FOUNDATION ENGINEERIN
8th Edition
ISBN: 8220100547058
Author: Das
Publisher: CENGAGE L
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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.
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)
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Chapter 10 Solutions
EBK PRINCIPLES OF FOUNDATION ENGINEERIN
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- STRENGTH OF MATERIALS answer: 17.68 kN and 11.8 MN/m2 compressive *what is the solution to this problem?arrow_forwardEarth 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 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
- 3. The A-36 steel drill shaft of an oil well extends 12000 ft into the ground. Assuming that the pipe used to drill the well is suspended freely from the derrick at A, determine the maximum average normal stress in each pipe string and the elongation of its end D with respect to the fixed end at A. The shaft consists of three different sizes of pipe, AB, BC, and CD, each having the length, weight per unit length, and cross-sectional area indicated. AAB= 2.50 in.² WAB= 3.2 lb/ft ABC= 1.75 in? WBC= 2.8 lb/ft ACD= 1.25 in.² WCD= 2.0 lb/ft B C D 5000 ft 5000 ft 2000 ftarrow_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_forwardThe 50-kg stone has a speed of vA= 8 m/s when it reaches point A. Determine the normal force it exerts on the incline when it reaches point B. Neglect friction and the stone’s size.arrow_forward
- A 20 mm diameter hole is drilled on the centerline of a steel strap 50 mm wide by 5 mm thick, subjected to an axial pull P = 10 kN. Determine the approximate maximum unit stress adjacent to the hole. Hint: This is a stress concentration question. The first step is to calculate the K value. Select one: O a. 0.1 kN/mm? b. 0.0733 kN/mm? c. 0.146 kN/mm2arrow_forwardThe same torque is applied to two tubes with identical metal sheets, determine the ratio of square/ circle of the shear stresses. Given: r = 16 mm, %3D S = 17 mm, and t = 2.80 mm. %3D トarrow_forwardA hollow circular tube is subjected to a tension force of 30 KN, a tensile stress of 20 MPa was recorded.lf the outer diameter of the tube is 100 mm, the inner diameter will be eqaual to. a. 93.95 mm O b. 89.95 mm C. 85.95 mm o d. 91.95 mmarrow_forward
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