Foundation Engineering Questions and Answers Part-25

1. Which of the following pose a limitation to plate load test?
a) Effect of size of foundation and Test on cohesive soil
b) Load increment
c) None of the mentioned
d) All of the mentioned

Answer: a
Explanation: A limitation of plate load test is
i) Effect of the size of foundation. For clayey soils the ultimate pressure for a large foundation is the same as that of the test plate. But in dense sandy soils, the bearing capacity increases, with the size of the foundation
ii) Plate load test is essentially a short duration test, and hence the test does not give the ultimate settlement, particularly in the case of cohesive soil.

2. For a c-φ soil, Housel suggested which of the following expression?
a) Q = A .q + P .s
b) Q = A .q
c) q f = M + N( BF/B p )
d) All of the mentioned

Answer: a
Explanation: In 1929, Housel suggested the following expression for a c-φ soil:
Q = A .q + P .s
Where, Q = total load on bearing area;
P = perimeter of footing;
S = perimeter shear;
A = contact area of footing or plate.

3. For clay soil the value of n can be taken as _____________ in the absence of test data.
a) 0. 4 to 0.5
b) 0.20 to 0.25
c) 0.003 to 0.05
d) 0.08 to 0.10

Answer: c
Explanation: The value of index n can be determined by carrying out two or more plate load tests on different size plate. In absence of test data, the following values of n can be adopted:
Dense sand : 0.4 to 0.5
Loose sand : 0.20 to 0.25
Clay : 0.003 to 0.05
Sand clay : 0.08 to 0.10.

4. The two commonly used penetration tests are ____________
a) Standard penetration test
b) Cone penetration test
c) All of the mentioned
d) None of the mentioned

Answer: c
Explanation: According to Indian standard, the two commonly used penetration tests are static cone penetration test and standard penetration test.

5. The values derived from penetration tests can be used for finding ____________
a) Depth of hard stratum and Strength of soil
b) Soil saturation
c) None of the mentioned
d) All of the mentioned

Answer: a
Explanation: The penetration test are useful for general exploration of erratic soil profiles, for finding depth to bed rock or hard stratum, and to have an approximation indication of the strength and other properties of soils.

6. The observed value of N in static cone penetration test is corrected by _________
a) Overburden and Dilatancy /submergence
b) Effective pressure
c) None of the mentioned
d) All of the mentioned

Answer: a
Explanation: The N-value for cohesion less soil shall be corrected for overburden and the values N0 obtained after overburden correction is corrected further for dilatancy.

7. Correction for increasing effective overburden pressure have been proposed by ___________
a) Gibbs and Holtz
b) Peck
c) Thornburn
d) All of the mentioned

Answer: d
Explanation: For a constant density index, the N value increase with increasing effective overburden pressure for which correction have been proposed by Gibbs and Holtz, peck, Thornburn, Whitman and others

8. The split tube used in static cone penetration test, is commonly known as ______________
a) Split spoon sampler
b) Split tube sampler
c) Tube sampler
d) All of the mentioned

Answer: a
Explanation: The split tube sampler, used in a test for standard penetration is commonly known as split spoon sampler resting on the bottom of the bore hole which is allowed to sink under its own weight.

9. The total blow required for the second and third 15 cm of penetration in standard penetration test is taken as ____________
a) Seating drive
b) Penetration resistance
c) Overburden pressure
d) Dilatancy/submergence

Answer: a
Explanation: In standard penetration test, the first 15 cm of drive may be considered to be a seating drive. The total blows required for the second and third 15 cm of penetration is termed as the penetration resistance N.

10. The expression for C n as given by Lio and Whitman is _____________
a) C n = √(σ’)
b) C n = √(100/σ’)
c) C n = 0.77log10(2000/σ’)
d) None of the mentioned

Answer: b
Explanation: In 1986, Lio and Whitman gave the following expression for
Normalizing factor C n:
C n = √ (100/σ’).