Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 17, Problem 17.6P
To determine
Find the variation of relative density with depth using Meyerhof method.
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3. Following are the results of a standard penetration test in fine dry sand.
N60
Depth (m)
1.5
7
13
3.0
18
4.5
22
6.0
7.5
24
For, the sand deposit, assume the mean grain size, D50, to be 0.26 mm and the
unit weight of sand to be 15.5kN/m3. Estimate the variation of relative density
with depth using the correlation developed by Cubrinovski and Ishihara.
Assume pas100kN/m2.
denined friction
In a deposit of normally consolidated dry sand a cone penetration test was conducted. Following are the result:
Depth(m)
Point resistance of cone, qc (MN/m²)
1.5
2.06
3.0
4.23
4.5
6.01
6.0
8.18
7.5
9.97
9.0
12.42
Assuming the dry unit weight of sand to be 16kN/m³, estimale the average peak friction angle, ф’, fo the sand. Use ф’=tan ¯1((0.38+0.27log(qc/ σ’o))
A standard penetration test was carried out in a normallyconsolidated sand at 25 ft depth where the N60 was determinedto be 28. The unit weight of the sand is 110 lb/ft3, andthe grain-size distribution suggests that D50 5 1.2 mm andCu 5 3.2. The age of the soil since deposition is approximately5000 years. Determine the relative density using thedifferent correlations discussed in Section 3.15
Chapter 17 Solutions
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- A dilatometer test was conducted in a sand deposit ata depth of 6 m. The groundwater table was located at a depth of 2 m below the ground surface. Given, for the sand: 3.29 - 14.5 kN/m3 and Yat 19.8 kN/m3. The contact stress during the test was 260 kN/m2. Estimate the soil friction angle, ф".arrow_forwardA standard penetration test is carried out in sand where the efficiency of the hammer nH =70%. If the measured N-value at 30 ft depth is 24, find N60 and (N1)60. The unit weight of the sand is 115.0 lb/ft3. Assume nB = nS = nR =1.arrow_forwardQ11: 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.arrow_forward
- 3. A civil engineer is interested in assessing the variation of shear strength of clay layers in a site of interest for different depths. After completing CU triaxial tests, the sample mean for the clay shear strength 40 kPa and the standard deviation was 15.5 kPa are obtained for the depth of 3 m from ground surface. For the following conditions: a. Calculate 99% CI for the population mean μ. Write down your interpretation.b. Explain why the size of CI changed between parts a - c.arrow_forward3.15 In a deposit of normally consolidated dry sand, a cone pen- etration test was conducted. Following are the results: 316 Depth (m) 1.5 3.0 4.5 6.0 7.5 9.0 Point of resistance of cone, q. (MN/m²) 2.06 4.23 6.01 8.18 9.97 12.42 Assuming the dry unit weight of sand to be 16 kN/m³, estimate the average peak friction angle, d', of the sand. Use Eq. (3.56). Rofararrow_forward3. A civil engineer is interested in assessing the variation of shear strength of clay layers in a site of interest for different depths. After completing CU triaxial tests, the sample mean for the clay shear strength 40 kPa and the standard deviation was 15.5 kPa are obtained for the depth of 3 m from ground surface. For the following conditions: a. Calculate 90% confidence interval (CI) for the population mean μ. Write down your interpretationb. Calculate 95% CI for the population mean μ. Write down your interpretationarrow_forward
- 49. The following data are given for the laboratory sample. o=175 kPa; e = 1.1; +Ao = 300 kPa; e = 0.9 If thickness of the clay specimen is 25 mm, the value of coefficient of volume compressibility is x 10-4 m²/kN A. 3x10-³ m²/kN B. 5x10-6 m²/kN C. 12x10-4 m²/kN D. 7.61×10-4 m²/kNarrow_forward21 The results of a constant head permeability test for a fine sand are as follows: Diameter of the sample = 37 cm Length of sample = 92 cm Constant head difference = 78 cm Time of collection = 337 secs Weight of water collected = 375 grams Find the seepage velocity in cm/min. if the void ratio is 0.6. Round off to four decimal places.arrow_forwardFollowing are the results of a standard penetration test in sand. Determine the corrected standard penetration number, (N1)60, at various depths. Note that the water table was not observed within a depth of 10.5 m below the ground surface. Assume that the average unit weight of sand is 17.3 kN/m3. Depth (m) N60 1.5 8 3.0 7 4.6 12 6.0 14 7.5 13arrow_forward
- Three samples of loose sand were tested under consolidated undrained (CU) conditions. The failure stresses and excess pore water pressure for the soil sample is given below: Cell Pressure = 243 kPa Deviator Stress = 127 kPa Pore Water Pressure = 82 kPa Determine the drained angle of inclination of failure plane in degrees. Round off to two decimal places. Answer: 53.22arrow_forwardThe standard penetration test results of a sand deposit at a certain site are given below in tabular form. The groundwater table in located at a depth of 2 m below the ground surface. The dry and saturated unit weights of sand are 17 kN/m³ and 19.0 kN/m', respectively. For an expected 10.8 earthquake magnitude M = 6 and maximum acceleration amax = 0.1 g, will liquefaction occur? Depth (m) NF (blows/30 cm) 1.5 8 3.0 7 4.5 12 6.0 15 7.5 17 9.0 17arrow_forwardb) Table Q2 provides the result of a standard penetration test in the sandy soil. The water table was not encountered during the test. Assume that the average unit weight of sand is 17.3 kN/m³ and use Skempton's relationship provided, determine the corrected standard penetration numbers, (N₁)eo at various depths given (CO1, PO2) (C3) 2 Skempton's equation: C = = 1+0.010! Table Q2 Depth (m) Noo 1.5 3.0 4.5 6.0 7.5 |4712419|arrow_forward
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