Standard penetration tests were carried out in normally consolidated fine sands data were collected, with average unit weights assumed for the entire depth. Determine the (N)60 values using the cor- rection factors proposed by (a) Liao and Whitman (1986) and (b) Skempton (1986). different locations. The following Average unit weight (kN/m³) Depth (m) 3.5 18.0 12 8.8 18.0 8 12.4 18.5 15 18.5 18.5 19 23.6 19.0 21 16.9 19.0 26

Standard penetration tests were carried out in normally consolidated fine sands data were collected, with average unit weights assumed for the entire depth. Determine the (N)60 values using the cor- rection factors proposed by (a) Liao and Whitman (1986) and (b) Skempton (1986). different locations. The following Average unit weight (kN/m³) Depth (m) 3.5 18.0 12 8.8 18.0 8 12.4 18.5 15 18.5 18.5 19 23.6 19.0 21 16.9 19.0 26

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter3: Natural Soil Deposits And Subsoil Exploration

Section: Chapter Questions

Problem 3.12P

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Following 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 13
= A standard penetration test is carried out in sand where the efficiency of the hammer n 50%. If the measured N-value at 7.5 m depth is 20, find Noo and (N₁)60. The unit weight of the sand is = 15 MR = 1 and 18.08 kN/m³. Assume ¹ = 0.95. If you know that: ns
b) 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|
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))
In a standard penetration test in sands, the blow count measured at 14 m depth was 26. An automatic hammer released by a trip with an efficiency of 70% was used in the test. The unit weight of sand is 17.5 kN/m³. Determine: (a) N60 (N1) 60 using Peck, Hanson, and Thornburn (1974) correlation with Pa = 100 kPa. The most likely value of the friction angle (not the range). The possible range of relative density per Peck et al. (1974) The Young's Modulus. The table below presents two correlations proposed for estimating Young's Modulus. kg Leonards (1986) E .cm² E (kPa): = 8N60 = aPa (kPa)N60 Kulhawy & Mayne (1990) In the table a = 5 for fine sands, 10 for clean normally consolidated sands, and 15 for clean over consolidated sands. If a specific method is not specified, you may use any possible correlations. While doing so, you must clearly indicate which correlation was used.
Following 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.
An SPT was performed at a depth of 7 m in sand of unit weight 20.40 kN/m3. The N-value was found to be 22. Compute the corrected N-value by the two methods presented in this chapter
The following are the results of a standard penetration test in sand. Determine the standard penetration numbers, (N₁)60(CORR), at the various depths given. Note that the water table is at depth 5.0 m. Assume that the average unit weight of sand is 15.5 kN/m³ above the water table and the saturated unit weight of sand is 17.81 kN/m³, while the unit weight of water = 9.81 kN/m³. Depth (m) 0.0 1.5 3.0 4.5 6.0 7.5 N60 0 10 15 19 24 28
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 17
2- Subsurface geotechnical investigations carried out in a deposit of sand provided the field standard penetration numbers N60 as given in the table below. During the geotechnical investigations, it is also observed that underground water table is encountered at a depth of 2.0 m below the ground surface. Laboratory tests have shown that physical properties of deposit of sand are as: Dry unit weight = 17.6 kN/m³ and saturated unit weight = 19.6 kN/m. Determine, for an earthquake magnitude of 8.25, if liquefaction will occur at the site. Assume that maximum peak ground acceleration at the site is: amax = 0.25g. DEPTH (M) N60 (BLOWS/30CM) 6. 7 9 11 12 14 18 21 Note: the gravitational acceleration could be taken as 9.81 m/s.
Q3/ A standard penetration test was carried out at a site. The soil profile is given in Fig. 3 with the penetration values. The average soil data are given for each layer. Compute the corrected values of N60, (N1)60 and o average for sand layers only. Assume: standard 10 sampler, united state safety hammer and diameter of borehole 150mm. Depth (m) 0 12 14 Sand 7-18.5 kN/m 18.5 kN/m Sand 19.81 kN/m 2m +4+E Im N-value
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 ררפ