Pavement Design Questions and Answers – Highway Materials – Subgrade Soil Tests – 1

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This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “Highway Materials – Subgrade Soil Tests – 1”.

1. The tests to evaluate the strength of subgrade soil can be broadly classified as ______
a) Shear and penetration tests
b) Shear and bearing tests
c) Bearing and penetration tests
d) Shear, bearing and penetration tests
View Answer

Answer: d
Explanation: There are three broad classifications of test types to evaluate subgrade soil strength. They are shear, bearing and penetration tests. Each contains several other tests under them to evaluate the strength.
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2. The ______ test can be conducted in-situ as well as in the laboratory.
a) CBR
b) Unconfined compression
c) Triaxial
d) Direct shear
View Answer

Answer: a
Explanation: CBR is a type of penetration test and there have been procedures developed to allow conducting the test both in-situ and in the laboratory. Unconfined compression test, triaxial test and direct shear test are all types of shear tests. These are usually conducted in the laboratory only.

3. The figure below represents the Mohr’s envelope. What does the symbol “σd” represent in the figure?
The symbol “σd” represent in the figure is Deviator stress in the Mohr’s envelope
a) Differential stress
b) Direct stress
c) Derived stress
d) Deviator stress
View Answer

Answer: d
Explanation: The stresses obtained in the triaxial test are plotted in the Mohr’s envelope. The lateral pressure is represented as σ3 and the vertical pressure is represented as σ1. “σd” represents the deviator stress and is obtained as the difference between σ1 and σ3.
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4. What does CBR stand for?
a) California Binding Ratio
b) Combined Binding Ratio
c) California Bearing Ratio
d) Combined Bearing Ratio
View Answer

Answer: c
Explanation: CBR stands for California Bearing Ratio and it is a test used to determine the strength of subgrade soil. It was developed by the California State Highway Department and hence the name.

5. Modulus of subgrade reaction is defined as pressure sustained per unit volume of subgrade.
a) True
b) False
View Answer

Answer: b
Explanation: Modulus of subgrade reaction is represented by the letter k and it is defined as the pressure sustained per unit deformation of the subgrade. It is obtained after conducting the plate load test.
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6. What is the angle of internal friction for a saturated soil sample that fails under axial stress of 100 kN/m2 in an unconfined compression test and the failure plane makes an angle of 50° with the horizontal?
a) 10°
b) 20°
c) 15°
d) 25°
View Answer

Answer: a
Explanation: The angle of internal friction, Φ is related to the failure angle, θ by the equation:
Φ=2θ-90
Substituting θ=50° from the question,
Φ=2×50°-90=10°

7. Determine the CBR value of the specimen at 2.5 mm penetration if the dial reading obtained from the graph is 35 divisions and 100 division of load dial represents 190 kg load in the calibration chart of proving ring.
a) 4.8%
b) 4.9%
c) 4.8
d) 4.9
View Answer

Answer: b
Explanation: The equation for computing CBR value of the soil is as below:
CBR at 2.5mm penetration in %
= \(\frac{(Load \,sustained \,by \,specimen \,at \,2.5mm \,penetration)}{(Load \,sustained \,by \,standard \,aggregate \,at \,2.5mm \,penetration)}×100%\)
Load sustained by specimen=\(35×\frac{190}{100}=66.5 kg\)
CBR at 2.5mm penetration=\(\frac{66.5}{1370}×100=4.85\approx \)4.9%
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8. The test to obtain values of consistency limits are carried out as per which IS code?
a) IS 2720 Part V
b) IS 2720 Part VII
c) IS 2720 Part IV
d) IS 2720 Part III
View Answer

Answer: a
Explanation: Consistency limits include liquid limit, plastic limit and shrinkage limit. These are tested using the Casagrande apparatus and the guidelines to be followed for performing the test are outlined in IS 2720 Part V. IS 2720 Part VII gives guidelines on tests for water content determination, IS 2720 Part IV gives guidelines on the grain size distribution and IS 2720 Part III gives guidelines on specific gravity tests.

9. Which of the below indices is a measure of shearing strength of soil at the plastic limit?
a) Plasticity index
b) Consistency index
c) Flow index
d) Toughness index
View Answer

Answer: d
Explanation: Toughness index is obtained as the ratio between the plasticity index and flow index. It represents the shearing strength of soil at the plastic limit. Plasticity index gives a measure of clay content, consistency index gives a measure of firmness of soil and flow index gives a measure of shear strength.
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10. As per the Mohr-Coulomb theory, the point inside the Mohr’s circle is considered to be ______
a) Not possible
b) Failure
c) Stable
d) Unstable
View Answer

Answer: c
Explanation: If the point is inside the Mohr’s circle, then the soil is considered to be stable as the point lies below the failure envelope. If the failure envelope just touches the circle, then the soil is considered to have failed. It is not possible to have the point above the failure envelope.

11. Soil compacted at optimum moisture content reduces swelling of the soil.
a) True
b) False
View Answer

Answer: a
Explanation: When the soil is compacted, the soil particles come closer and the density increase. The water deficiency of the soil hence increases and due to this, the swelling of soil mass decreases.

12. According to the given data, which of the soil is the most plastic one?

CONSISTENCY LIMITS SOIL A SOIL B SOIL C SOIL D
Wl 34% 50% 48% 44%
Wp 15% 18% 12% 20%

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

Answer: c
Explanation: The plasticity index is considered to indicate how plastic the soil is. The more the plasticity index, the more plastic the soil is. The plasticity index for each soil can be computed as below.
Ip= wl-wp
For soil A, Ip= wl-wp=34-15=19%
For soil B, Ip= wl-wp=50-18=32%
For soil C, Ip= wl-wp=48-12=36%
For soil D, Ip= wl-wp=44-20=24%
Since soil C has the highest plasticity index of 36%, it is the most plastic soil.

13. Which of the below sieve sets for gravel has been arranged in sequential order?
a) 2, 1 mm, 600, 425, 212, 150, 75 microns
b) 600, 425, 212, 150, 75 microns
c) 80, 40, 20, 10, 4.75 mm
d) 40, 20, 10, 5, 4.75 mm
View Answer

Answer: c
Explanation: The gravel fractions are larger in size and hence the sieve sets are those in mm. The largest sieve size used is 80 mm, then 40, 20, 10 and 4.75 mm are used in decreasing order of particle size. For sand fractions, the size ranges from 4.75 mm to 75 microns. The sieve sets used are 2, 1 mm, 600, 425, 212, 150 and 75 microns.

14. In the vane shear test, the difference between the initial and final reading gives the ______
a) Angle of failure
b) Angle of repose
c) Angle of internal friction
d) Angle of torque
View Answer

Answer: d
Explanation: The vane shear test consists of equipment with four steel blades and it is inserted in the sample and rotated at 0.1° per second using torque. The reading before rotating is the initial reading and final reading is taken when the sample fails. The difference between these gives the angle of torque.

15. Find the angle of shearing resistance of sand subjected to drained triaxial test for the following data:
Deviator stress, σd = 475 kN/m2
Cell pressure, σ3 = 200 kN/m2
a) 23.88°
b) 32.88°
c) 33°
d) 32°
View Answer

Answer: b
Explanation: The equation for normal stress, σ1 is given by:
σ13d=200+475=675 kN/m2
The formula for the drained triaxial test is given by:
σ13NΦ + 2c\(\sqrt{N_{\phi}}\) Where \({N_{\phi}}\) is \((tan(45+\frac{\phi }{2}))^2\) Substituting values from the question and taking c = 0 (sand has no cohesion),
675 = 200× \((tan(45+\frac{\phi }{2}))^2\)+0
Φ=32.88°

Sanfoundry Global Education & Learning Series – Pavement Design.

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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