Foundation Design: Principles and Practices (3rd Edition)
3rd Edition
ISBN: 9780133411898
Author: Donald P. Coduto, William A. Kitch, Man-chu Ronald Yeung
Publisher: PEARSON
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Chapter 4, Problem 4.9QPP
To determine
The angle of internal friction for each test
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A sand sample is subjected to direct shear testing. Two tests areperformed. For test 1, The sample shears at a stress of 2500 psf whenthe normal stress is 4000 psf.Test 2, The sample shears at a stress of 3500 psf when the normalstress is 6000 psf.
Determine the following:a) Angle of Internal frictionb) Value of cohesionc) Compute the shear stress at a depth of 12 ft. if the unit weight ofthe soil is 150 pcf
A silty sand of density index (ID or Dr) = 59% was subjected to standard penetration tests at a depth of 3 m. Groundwater level occurred at a depth of 1.5 m below the surface of the soil which was saturated throughout and had a unit weight of 19.3 kN/m3. The average N count was 15.
During calibration of the test equipment, the energy applied to the top of the driving rods was measured as 350 Joules. Determine the (N1)60 value for the soil.
Chapter 4 Solutions
Foundation Design: Principles and Practices (3rd Edition)
Ch. 4 - Describe a scenario that would require a very...Ch. 4 - How would you go about determining the location of...Ch. 4 - A five-story office building is to be built on a...Ch. 4 - A two-story reinforced concrete building is to be...Ch. 4 - Discuss the advantages of the CPT over the...Ch. 4 - A standard penetration test was performed in a 150...Ch. 4 - Using the CPT data in Figure 4.22, a unit weight...Ch. 4 - Prob. 4.8QPPCh. 4 - Prob. 4.9QPPCh. 4 - A series of vane shear tests have been performed...
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- 9. A consolidated drained test was carried out on a sandy clay under a cell pressure of 250 kPa. A constant back pressure of 120 kPa applied throughout the test. The dimensions of the sample were 76 mm x 38 mm. Addional test data recorded at failure were: Load transducer force = 368 N 3 Measured change in volume = 2.42 x 10 m³ Axial deformation = 2.05 mm Determine the major principal stress, o, at failure. (455 kPa)arrow_forwardA consolidated undrained triaxial test was conducted on a normally consolidated clay sample and the results are as follows: chamber confining pressure = 118 kpa, Deviator stress at failure 93 kpa, pore water pressure = 52 kpa. These results were used to determine the undrained friction angle of the soil. Compute the deviator stress (kPa) at failure when the drain test was conducted in the chamber confining pressure change to 168 kpaarrow_forwardQuestion 1 A consolidated drained (CD) triaxial test was carried out on a sand sample with the known effective shear strength parameters, C' = 0 and ø′ = 30°. In the test, prior to the failure, when the sample was undergoing axial compression under constant cell pressure, the drainage valve was accidentally closed. At the failure, 360kPa deviatoric stress was recorded along with 70kPa pore water pressure. If the test is repeated without such error, and no back pressure is applied in either of the tests, what is the deviatoric stress (in kPa, in integer) at the failure? A 11 B C 350 D 500 200arrow_forward
- 5. A consolidated, undrained triaxial test is being caried out on a normally consolidated clay where c = 0 and 41= 26'. The triaxial specimen was consolidated under a cell pressure of 300 kPa and backpressure of 80 kPa. Skempton's A parameter at failure is estimated to be 0.80. The drainage valve has since been dosed and the vertical deviator stress increased to failure. What would be the deviator stress and pore water pressure at failure?arrow_forward5. A consolidated, undrained triaxial test is being carried out on a normally consolidated clay where c - 0 and p'= 26'. The triaxial specimen was consolidated under a cell pressure of 300 kPa and backpressure of 80 kPa. Skempton's A parameter at failure is estimated to be 0.80. The drainage valve has since been dosed and the vertical deviator stress increased to failure. What would be the deviator stress and pore water pressure at failure?arrow_forwardA cu test was performed on a 3.5cm diameter sand sample. A confining pressure of 2.0 kg/cm2 was applied, and the breaking axial load was 45 kg, and the gap water pressure at this time was u=1.0 kg/cm2. 1) What is the effective internal friction angle of this specimen? 2) Evaluate whether the experimental sample is loose/dense. Why? 3) Apply a confining pressure of 1.0 kg/cm2 to this specimen and repeatedly do a cu test. determine the specimen's loose/dense and explain the reason why. (where pressure u = 0.0 kg/cm2) ans is 35 / loose / densearrow_forward
- A Standard Penetration Test was performed at a depth of 15 feet in a soil with a Unit Weight of 115 lb/cubic feet. The water table was at 20 feet and the the N value was found to be 19. What is the overburden pressure at the point that the soil sample was taken? 1,000 tons/square feet 1230 lbs/square feet 575.3 lbs/square feet 0.86 tons/square feetarrow_forwarda. Explain the Mohr's circles shown in Figure Q2 for a series of undrained tests and consolidated undrained test carried out on the same clay. What are the strength parameters which can determined from each test? D. (a) (b) Figure Q2 b. Which is the third type of test not shown in the figure? What would you expect the Mohr's circles for this test to look like?arrow_forwardA consolidated-undrained tri-axial test was conducted on a normally consolidated clay sample and the results are as follows: Chamber confining pressure = 112 kPa, Deviator stress at failure = 90 kPa, Pore Water Pressure = 67 kPa. These results were used to determine the drained friction angle of the soil. Compute the deviator stress (kPa) at failure when the drained test was conducted with the chamber confining pressure changed to 156 kPa. Round off to two decimal places.arrow_forward
- Table 2.A consolidation test data An oedometer test is conducted for a soil sample.The sample is consolidated under vertical stress of 50 kPa. The Time (min) Dial gauge reading (mm) 194 0.01 0.02 193 0.05 190.5 oedometer test finished at 144 min and 0.1 186.5 data are given in Table I. Average thickness of the sample is 20 mm. Predict the coefficient of 0.2 181 0.5 173 1 168 2 164.5 consolidation (c, in mm?/min) using Casagrande log t method and Taylor square root of t method. Please use the empty graphical papers. 5 161.5 10 160 20 158.5 50 156.9 100 155.7 144 155.1arrow_forwardAnswer any one asap 1.A triaxial test is performed on a normally consolidated clay. The sample is further consolidated by the test of all-around confining pressure prior to the application of the axial load. The axial force is applied very slowly. The equipment pore pressure lines remain open so that drainage of the soil pore water can occur during the test. The sample fails when minor principal stress is 49 kPa and the major principal stress is 91 kPa. Determine the angle of internal friction with no decimal places. 2.A dry sand sample was tested in direct shear and yielded a shear strength of 132 kPa under a normal stress of 215 kPa. Lab results indicate the soil's wet unit weight is 18.1 kN/m3 and the saturated unit weight is 20.2 kN/m3. Determine the shear strength, in kPa, within the native sand deposit at a depth of 2.5 m below the water table. The water table is 3 m below ground surface. Provide your answer with 2 decimal places, do not include the units and assume a unit…arrow_forwardShear Stress 1. Please provide proper discussion and illustration. Clear and complete solution please thank you.arrow_forward
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