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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Following is the variation of the field standard penetration number in a sand deposit:
The groundwater table is located at a depth of 6 . Given: the dry unit weight of sand from 0 to a depth of 6 is 18 , and the saturated unit weight of sand for depth 6 to 12 is 20.2 . Using the equation
determine the average relative density of sand.
(Enter your answer to three significant figures.)
Average =
The following table gives the variation of the field standard penetration number in a sand deposit:
The groundwater table is located at a depth of 12 . The dry unit weight of sand from 0 to a depth of 12 is 17.6 . Assume the mean grain size of the sand deposit to be about 0.8 . Estimate the variation of the relative density with depth for sand. Use the equation
(Enter your answers to three significant figures.)
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
Chapter 12 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
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- Determine the relative density at each depth using attached equation. Assume moderately compressible sand and hence Qc = 1.arrow_forwardDetermine the relative density at each depth using attached equation. Assume moderately compressible sand and hence Qc = 1.arrow_forwardRefer to Problem 3.5. Using Eq. (3.22), determine the averagerelative density of the sand.arrow_forward
- Following is the variation of the field standard penetration number in a sand deposit: The groundwater table is located at a depth of 6 . Given: the dry unit weight of sand from 0 to a depth of 6 is 16 , and the saturated unit weight of sand for depth 6 to 12 is 22.2 . Estimate an average peak soil friction angle. Use the equation (Enter your answer to three significant figures.)arrow_forwardFollowing is the variation of the field standard penetration number () in a sand deposit: 1.5 6 3 8 4.5 9 6 8 7.5 13 9 14 The groundwater table is located at a depth of 6 . Given: the dry unit weight of sand from 0 to a depth of 6 is 19 , and the saturated unit weight of sand for depth 6 to 12 is 21.2 . Use the relationship given in the equation to calculate the corrected penetration numbers. (Round your answers to the nearest whole number.) 1.5 6 3 8 4.5 9 6 8 7.5 13 9 14arrow_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_forward
- 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?arrow_forwardRefer to Problem 3.5. Using Eq. (3.28), determine the averagerelative density of the sand. Assume it is a fine sand.Use Eq. (3.13) to obtain (N1)60.arrow_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_forward
- Problem attachedarrow_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_forwardRepeat Problem 3.6 using Eq. (3.29).arrow_forward
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