Principles of Foundation Engineering, SI Edition
8th Edition
ISBN: 9781305723351
Author: Braja M. Das
Publisher: Cengage Learning US
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Chapter 3, Problem 3.21P
To determine
Find the variation of penetration number
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A direct shear test, when conducted on a remolded sample of sand, gave the following observations at the time of failure: Normal load = 288 N shear load
= 173 N. The cross sectional area of the sample = 36 cm.sq.
1. Determine the angle of internal friction. (Select]
2. The magnitude of the major principal stress in the zone of failure. [Select]
3. Determine the magnitude of the deviator stress if a sample of the same sand with the same void ratio as given above was tested in a tri-axial apparatus
with a confining pressure of 60 kPa. ( Select ]
Solve this problem graphically and then analytically. A CU triaxial
test was performed on a dense sand specimen at a confining pressure 03=40 kPa. The
consolidated undrained friction angle of the sand is =39°, and the effective friction angle
is d'=34°. Calculate: (a) the major principal stress at failure, o1, (b) the minor and the
major effective principal stresses at failure, o'3f and oʻır, and (c) the excess pore water
pressure at failure, (Aua)f.
Q # 4. The following are the results of four drained direct shear tests on undisturbed normally
consolidated clay samples having a diameter of 50 mm. and height of 25 mm.
Test No.
Normal Force (N)
Shear Force at Failure (N)
1
67
23.3
133
46.6
213
44.6
4
369
132.3
Draw a graph for shear stress at failure against the normal stress and determine the drained angle of
friction from the graph. (Note: Use excel for calculation and drawing the graph).
Chapter 3 Solutions
Principles of Foundation Engineering, SI Edition
Ch. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Refer to Figure P3.3. Use Eqs. (3.10) and (3.11)...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Prob. 3.9PCh. 3 - Prob. 3.10P
Ch. 3 - Prob. 3.11PCh. 3 - Following are the standard penetration numbers...Ch. 3 - Prob. 3.13PCh. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - Prob. 3.23PCh. 3 - Prob. 3.24PCh. 3 - Prob. 3.25PCh. 3 - Prob. 3.26PCh. 3 - Prob. 3.27P
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- 3. A consolidated-undrained test was conducted on a normally consolidated clay sample. The results are as follows: Undrained angle of friction = 32.21°, Drained angle of friction = 63.54°, Deviator stress at failure = 511 lb/ft^2. Determine the pore water stress at failure in lb/ft^2. Round off to three decimal places.arrow_forward49. 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_forwardA consolidated-undrained test was conducted on a normally consolidated clay sample. The results are as follows: Undrained angle of friction = 38.65°, Drained angle of friction = 62.59°, Deviator stress at failure = 509 lb/ft^2. Determine the pore water stress at failure in lb/ft^2.arrow_forward
- 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 3.15arrow_forward3. 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 frictionarrow_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 99% CI for the population mean μ. Write down your interpretation.b. Explain why the size of CI changed between parts a - c.arrow_forward
- Table 1 gives data from a standard shearbox(direct shear) test on a sample of 125g of dry sand. The initial dimensions of the sample were 60mmx60mm on plan x20mm in height. The test was camried out at a constant normal effective stress of 50kPa. Take the specific gravity of the soil grains Gs=2.65. Plot graphs of (a) shear stress r against shear strain y; (b) volumetric strain evol against shear strain y; (c) specific volume v against shear strain y. (d) Comment on these graphs and estimate the peak and critical state effective angles of friction of the soil. Table 1: Shearbox Data Relative horizontal displacement x (mm) of shearbox lid y (mm) Upward vertical movement Shear stress t (kPa) 0.00 0.000 0.02 0.002 19 0.04 0.008 34 0.06 0.016 43- 0.08 0.026 47 0.20 0.064 56 0.32 0.128 51 0.48 0.192 46 0.64 0.256 41 0.80 0.288 37 0.96 0.320 34 1.12 0.321 33arrow_forwardHomeworkarrow_forwardQuestion Table 1 gives data from a standard shearbox(direct shear) test on a sample of 125g of dry sand. The initial dimensions of the sample were 60mmx60mm on plan x20mm in height. The test was caried out at a constant normal effective stress of 50k:Pa. Take the specific gravity of the soil grains Gs=2.65. Plot graphs of (a) shear stress t against shear strain y; (b) volumetric strain evol against shear strain y; (c) specific vohume v against shear strain y. (d) Comment on these graphs and estimate the peak and critical state effective angles of friction of the soil. Table 1: Shearbox Data Relative horizontal Upward vertical movement Shear stress t (kPa) displacement x (mm) of shearbox lid y (mm) 0.00 0.000 0.02 0.002 19 0.04 0.008 34 0.06 0.016 43- 0.08 0.026 47 0.20 0.064 56 0.32 0.128 51 0.48 0.192 46 0.64 0.256 41 0.80 0.288 37 0.96 0.320 34 1.12 0.321 33arrow_forward
- ZAIN IQ I. Homework 4... During an oedometer test on a specimen of saturated clay, the thickness of the specimen decreased from 19.931 mm to 19.720 mm under an increment of stress from 200 to 400 kPa which was maintained for 24 hours. The stress was then removed from the specimen, its thickness was measured water content determined as 26.8%. 19.842 mm, and its Taking Gs of the particles to be 2.70, calculate the void ratio before and after the application of the stress increment, and the coefficient of volume decrease (m,) for this stress range. (Ans: e,= 0.731, e2= 0.713, m, = 0.052 m/MN)arrow_forwardThe following results were obtained from a direct shear test on a dry uncemented carbonate sand: normal stress = 96.6 kPa, shear stress at failure = 67.7 kPa. By means of a Mohr's circle of stresses, find the magnitude and directions of the principal stresses acting on a soil element within the zone of failure. Assume uniformity of the sample. Show step-by-step your answer with the formulated equations and final results.arrow_forwardA direct shear test, when conducted on a remolded sample of sand, gave the following observations at the time of failure: Normal load = 288 N; shear load = 173 N. The cross-sectional area of the sample = 36 cm2. Determine the minor principal stress in kPa.arrow_forward
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