Engineering Questions with Answers - Multiple Choice Questions
Shear Strength of Cohesive Soil
1 - Question
The un-drained test is carried out on sample of clay, silt, and peat to determine _____________
a) Shear Strength of natural ground and Sensitivity
b) Pore pressure
c) None of the mentioned
d) All of the mentioned
View Answer
Answer: a
Explanation: The un-drained test is carried out on undisturbed sample of clay, silt and peat to determine the strength of the natural ground and also carried out on remoulded samples of clay to measure it its sensitivity.
2 - Question
In an un-drained test on saturated clays, both σ1’ and σ3’ is independent of ____________
a) Pore pressure
b) Shear strength
c) Cell pressure
d) Effective pressure
View Answer
Answer: c
Explanation: In an un-drained test on saturated clays (B=1), both the major principal effective stress σ1’ and the minor principal effective stress σ3’ are independent of the magnitude of cell pressure applied.
3 - Question
The consolidated-un drained test can be performed in ___________ methods.
a) 3
b) 2
c) 4
d) 1
View Answer
Answer: b
Explanation: The consolidated-undrained tests are performed by two methods: i) the remoulded specimens are sheared under a cell pressure. ii) the moulded specimens are consolidated under the same cell pressure and sheared with different cell pressure.
4 - Question
Which of the following cannot be obtained by using un-drained test?
a) Effective stress failure envelope
b) Shear strength
c) sensitivity
d) All of the mentioned
View Answer
Answer: a
Explanation: Since only one Mohr circle in terms of effective stresses, is obtained from all un-drained tests, effective stress failure envelope cannot be obtained from this test.
5 - Question
The change in the pore pressure during an un-drained shear can be explained by ___________
a) Lateral pressure
b) Effective stress
c) Pore pressure parameter
d) Mohr’s circle
View Answer
Answer: c
Explanation: The change in the pore pressure due to change in the applied stress, during an un-drained, any be explained in terms of empirical coefficients called pore pressure parameters.
6 - Question
Factor affecting pore pressure parameters is ___________
a) Type of shear
b) Temperature
c) Nature of the fluid
d) All of the mentioned
View Answer
Answer: d
Explanation: Type of shear, sample disturbance, and environment during shear such as temperature and nature of the fluid are the factors that affect pore pressure parameter.
7 - Question
The value of pore pressure parameter, at failure for saturated clay is __________
a) 1.2 to 2.5
b) 2 to 3
c) 0.3 to 0.7
d) 0.7 to 1.3
View Answer
Answer: b
Explanation: The approximate value of pore pressure parameter at failure for very loose, fine saturated sand, saturated clays is 2 to 3.
8 - Question
Negative pore pressure in clay or sand is developed due to __________
a) Expansion on loading
b) Over loading
c) Loose structure
d) Compaction
View Answer
Answer: a
Explanation: A negative pore pressure is developed when we apply load on a sample of clay or sand because both sand and clay tends to expand on loading.
9 - Question
If the pore pressure is measured during un-drained stage of the test, the result can be expressed in terms of __________
a) C’ and φ
b) cu
c) None of the mentioned
d) All of the mentioned
View Answer
Answer: a
Explanation: Both effective parameters C’ and φ’ can be expressed when pore pressure is measured during un-drained stage of the test.
10 - Question
he equation for the unconsolidated un drainage strength of clay is __________
a) τ = c + σ tan φ
b) τf = ccu + σ tan φcu
c) τ = c + σ
d) τ = σ tan φ
View Answer
Answer: b
Explanation: The equation for consolidated undrained strength of the preconsolidated clay in terms of total stress can be approximately expressed as τf = ccu + σ tan φcu where, σ is the normal pressure.
11 - Question
What will be the shearing resistance of a sample of clay in an unconfined compression test, falls under a load of 150 N? Take change of cross-section Af=2181.7 mm2.
a) 68.75 kN/m2
b) 34.38 kN/m2
c) 11.35 kN/m2
d) 0.6875 kN/m2
View Answer
Answer: b
Explanation: Given,
Af=2181.7 mm2; Pf=150 N
qu = Pf/ Af = 150/2181.7
= 68.75 kN/m2
Shear resistance = qu/2 = 68.75/2
= 34.38 kN/m2
Therefore, Shearing resistance = 34.38 kN/m2.