# Pavement Design Questions and Answers – Highway Materials – Subgrade Soil Tests – 2

«
»

This set of Pavement Design Questions and Answers for Freshers focuses on “Highway Materials – Subgrade Soil Tests – 2”.

1. ______ method is used to determine water content as well as the specific gravity of soil.
b) Sand bath
c) Oven drying
d) Pycnometer

Explanation: Pycnometer is a glass bottle that can be used to determine both water content and the specific gravity of soil. Measuring flask method is used to determine the specific gravity. Sand bath and oven drying methods are used to determine the water content of the soil sample.

2. Determine the deviator stress for a soil sample subjected to triaxial test having the following details:
Cohesion = 15 kN/m2
Angle of shearing stress = 30°
Cell pressure = 260 kN/m2
a) 1091.96 kN/m2
b) 571.96 kN/mm2
c) 1091.96 kN/mm2
d) 571.96 kN/m2

Explanation: Deviator stress is the difference between cell pressure and normal stress. The equation to find the normal stress is given by:
σ1 = σ3Nϕ + 2c$$\sqrt{N_{\phi}}$$
Substituting the values of σ3=260 kN/m2, c=15kN/m2 and
$${N_{\phi}}=(tan(45+\frac{\phi }{2}))^2= (tan(45+\frac{\phi }{2}))^2=3$$
σ1=260×3+ 2×15×$$\sqrt{3}$$=831.96 kN/m2
Deviator stress, σd13=831.96-260=571.96 kN/m2

3. Which test setup does the figure represent?

a) Direct shear
b) Plate bearing
c) CBR
d) Unconfined compression

Explanation: The setup in the figure is for plate bearing test. There are a series of stacked plates at the bottom, a loading device and a reaction frame. The setup for direct shear test consists of a box divided horizontally into two halves. The CBR test setup consists of dial gauges, plunger, proving ring and specimen mould. For unconfined compression test, the setup essentially consists of the triaxial cell, dial gauges and piston.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. What is the CBR value of the soil if the CBR value at 2.5 mm penetration is found to be 4.5% and that at 5 mm penetration is found to be 4.3%?
a) 4.5%
b) 4.4%
c) 4.6%
d) 4.3%

Explanation: The highest CBR value obtained is reported as the CBR value of the soil. Normally, the CBR at 2.5 mm penetration is found to be higher than the 5 mm penetration and hence is reported as the CBR value of the soil. If 5 mm CBR value is higher, the test is repeated and if the same results are obtained, then it is taken as the CBR value of the soil.

5. The cone penetration test is also called ______ test.
a) Swiss cone
b) Dutch cone
c) Dynamic cone penetration
d) Static cone penetration

Explanation: The cone penetration test was first used in the 1950s in the Duct lab for soil mechanics to determine the geotechnical properties of soil. Hence, the test is also called the Dutch cone test. Static and dynamic cone penetration tests are types of the cone penetration test.

6. What is the cohesion for a soil sample that fails under axial stress of 140 kN/m2 when subjected to unconfined compression test and has an angle of internal friction of 17°?
a) 38.33 kN/m2
b) 0 kN/m2
c) 38 kN/m2
d) 40 kN/m2

Explanation: Cohesion for the soil can be found using the equation for normal stress.
$$\sigma_1=\sigma_3 N\phi + 2c\sqrt{N\phi }$$
In an unconfined compression test, cell pressure σ3=0. It is a quick test and doesn’t involve the cell pressure. So, the equation can be rewritten as:
$$\sigma_1=2c\sqrt{(N\phi)}$$
Now substituting the known values, c can be found as:
$$c=\frac{\sigma_1}{2\sqrt{N\phi}} = \frac{\sigma_1}{2×\sqrt{(tan(45+\phi/2))^2}}$$ $$c=\frac{140}{2×\sqrt{(tan(45+17/2))^2}}=38.33 \,kN/m^2$$

7. Which of the below facts about the particle size distribution graph is false?
a) Uniformity coefficient can be obtained from the graph
b) Particle size is plotted on the X-axis
c) Helps in understanding the physical properties of soil
d) It is also called a semi-log graph

Explanation: The particles size distribution graph is plotted on a semi-log graph and it is not another name for it. It is sometimes called the gradation curve. Uniformity coefficient Cu is obtained as the ratio between particle size corresponding to 60% finer and 10% finer. The % finer is represented on the Y-axis and particle size on the X-axis. Gradation gives an idea of the physical property of soil.

8. What is the liquidity index of soil if the plastic limit is 24%, the liquid limit is 49% and the natural water content is 35%.
a) -54%
b) -44%
c) 44%
d) 54%

Explanation: The liquidity index Il can be calculated using the formula,
$$I_l=\frac{w-w_p}{I_P}×100%$$
Where Ip=wl-wp
w,wl,wp are natural water content, liquid limit and plastic limit respectively.
$$I_l=\frac{35-24}{49-24}×100%$$ = 44%

9. The diameter of the standard plate used to find the value of modulus of subgrade is ______
a) 7.5 cm
b) 75 mm
c) 75 cm
d) 750 cm

Explanation: Plate bearing test is used to determine the modulus of subgrade used in the design of highway pavements. The standard plate size used is 750 mm or 75 cm. It represents the diameter of the plate being used. Plates of diameter 750, 600, 450 and 300 mm are also used in the test.

10. A vane of 80 mm diameter and 160 mm height was pushed into the soil using a torque of 80 Nm. What will be the undrained shear strength of the soil?
a) 4.263×10-5 kN/m2
b) 4.263×105 kN/m2
c) 42.63 kN/m2
d) 4.263 kN/m2

Explanation: The equation to find out undrained shear strength of soil in a vane test is as follows:
$$C_u=\frac{T}{\pi(\frac{d^2 H}{2}+\frac{d^3}{6})}$$
Where T is the torque in kNmm,
d, H are the diameter and height of the vane used respectively in mm.
Substituting the known values of T = 80 Nm$$\approx$$ 80 kNmm, d = 80 mm and H = 160 mm,
$$C_u=\frac{80}{π((80^2×160)/2+80^3/6)} =4.263×10^{-5} kN/mm^2≈42.63 kN/m^2$$

11. Which of the below soil samples has better shear strength at the plastic limit?

 CONSISTENCY LIMITS SOIL A SOIL B SOIL C SOIL D Wl 38% 60% 33% 53% Wp 21% 21% 18% 26% If 4 12 8 14%

a) Soil A
b) Soil B
c) Soil C
d) Soil D

Explanation: The shear strength at plastic limit can be evaluated if the values of toughness index of soil samples are known. It is calculated as the ratio between the plasticity index and the flow index. The values of the plasticity index and toughness index of each soil sample can be tabulated in the tabular form as below:

 CONSISTENCY LIMITS SOIL A SOIL B SOIL C SOIL D Wl 38% 60% 33% 53% Wp 21% 21% 18% 26% If 4 12 8 14% Ip=Wl-Wp 17% 39% 15% 27% It=Ip/If 4.25 3.25 1.88 1.93

As per the values obtained in the table, the toughness index for soil A is the highest, indicating that it has better shear strength at the plastic limit.

12. In a CBR test, the load values corresponding to different deflection values are noted.
a) True
b) False

Explanation: CBR test is a type of penetration test, so the readings are taken for corresponding values of penetration. The load values corresponding to 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 5, 7, 10 and 12.5 mm penetration of plunger into the soil sample are taken.

13. Determine the sensitivity of the soil if the undisturbed undrained shear strength is 52.78 kN/m2 and the disturbed undrained shear strength is 23.54 kN/m2.
a) Less sensitive
b) Sensitive
c) Highly sensitive
d) Quick

Explanation: The sensitivity of a soil is defined as the ratio between undisturbed undrained shear strength and disturbed undrained shear strength. So, it can be found out as $$\frac{52.78}{23.54}$$=2.24. The soils having sensitivity in the range of 2-4 and it indicates that the soil is less sensitive. The range 4-8 indicates sensitive, 8-16 indicates highly sensitive and >16 indicates quick soil.

14. The test results from the dynamic cone penetration test can be related to CBR.
a) True
b) False

Explanation: Both dynamic cone penetration and CBR are penetration type tests used to determine the strength of the soil. The result can be correlated to the CBR value using an equation. The equation is given as:
$$CBR=\frac{292}{(DPI^{1.12})}$$
Where DPI represents the DCP Penetration Index.

15. In a drained test, the drainage is permitted during the application of ______
a) No stress
b) Deviator stress
c) Cell pressure
d) Cell pressure and deviator stress

Explanation: In a drained test, drainage of the specimen is permitted during the application of both cell pressure as well as deviator stress. It is also called slow test. Drainage is under the cell pressure in case of a consolidated-undrained test. In an undrained test, no drainage is allowed at any stage.

Sanfoundry Global Education & Learning Series – Pavement Design.

To practice all areas of Pavement Design for Freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.