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 12, Problem 12.5P
To determine
Find the corrected penetration number
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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
A soil profile is shown in Figure P3.2 along with the standard penetration numbers in the clay layer. Use Eqs. (3.8) and (3.9) to determine the variation of cu and OCR with depth. What is the average value of cu and OCR?
A 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.
Chapter 12 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
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- 21 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 is the variation of the field standard penetration number (N60) in a sand deposit: The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 18 kN/m3, and the saturated unit weight of sand for depth 6 to 12 m is 20.2 kN/m3. Use the relationship given in Eq. (3.13) to calculate the corrected penetration numbers.arrow_forwardThe results of a constant head permeability test for a fine sand are as follows: Diameter of the sample = 31 cm Length of sample = 98 cm Constant head difference = 52 cm Time of collection = 364 secs Weight of water collected = 448 grams Find the seepage velocity in cm/min. if the void ratio is 0.33. Round off to four decimal places. Answer: 0.3943arrow_forward
- Q11: 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_forwardGeotechnical Engineering The following are the results of a standard penetration test in sand. Estimate the soil friction angle at the 3.0 meter depths given N60 =7 . Note that the water table was not observed within a depth of 12 m below the ground surface. Assume that the average unit weight of sand is 15.5 kN/m3 and Pa = 100 kN/m2. a. 44.6 b. 37.6 c. 38.9 d. 34.4arrow_forwardA 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 3.15arrow_forward
- The 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_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_forwardIn 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))arrow_forward
- P-1 A soil profile is shown in Figure along with the standard penetration numbers in the clay layer. Use provided equations to Dry sand y = 16.5 kN/m³ 1.5 m Groundwater table Sand 1.5 m N60 Ysat = 19 kN/m³ determine the variation of 1.5 m • 8 Cu and OCR and preconsolidation pressure o', with depth. What is the average value of c, and OCR? Clay 1.5 m Ysat = 16.8 kN/m³ * A•8 1.5 m • 9 1.5 m 10 Sand 0.689 0.29N0.72 Pa Си N60 OCR = 0.193 o = 0.47NPa %3D %3D Shahid Chamran University of Abvaz 1 ר ר ררפarrow_forwardQuestion 40 The results of a constant head permeability test for a fine sand are as follows: Diameter of the sample = 25 cm Length of sample = 80 cm Constant head difference = 77 cm Time of collection = 388 secs Weight of water collected = 340 grams Find the seepage velocity in cm/min. if the void ratio is 0.49. Round off to four decimal places.arrow_forwardThe results of a constant-head permeability test for a fine sand sample having a diameter of 70 mm and a length of 140 mm are as follows (refer to Figure 7.5):• Constant-head difference = 550 mm• Water collected in 7 min = 450 cm3• Void ratio of sand = 0.8Determine:a. Hydraulic conductivity, k (cm/sec)b. Seepage velocityarrow_forward
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