Free Soil Mechanics and Foundations MCQs with Answers
78 Soil Mechanics and Foundations MCQs from Civil Engineering, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Soil mechanics examines index properties, phase relationships, classification, permeability, seepage, compaction, consolidation, shear strength and earth pressure. Foundations apply these principles to bearing capacity, settlement and shallow or deep foundation selection, with clear distinction between compaction, which expels air, and consolidation, which expels water over time.
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- A. Mass of water to mass of dry soil
- B. Mass of water to mass of wet soil
- C. Mass of solids to mass of wet soil
- D. Mass of voids to mass of solids
Explanation: Water content is calculated as the mass of water divided by the mass of soil solids, usually expressed as a percentage.
Correct answer: Mass of water to mass of dry soil- A. 25.0 percent
- B. 33.3 percent
- C. 50.0 percent
- D. 66.7 percent
Explanation: Porosity n is related to void ratio e by n = e/(1 + e). Therefore, n = 0.50/1.50 = 0.333, or about 33.3 percent.
Correct answer: 33.3 percent- A. Liquid limit plus plastic limit
- B. Plastic limit minus shrinkage limit
- C. Liquid limit minus plastic limit
- D. Liquid limit minus shrinkage limit
Explanation: Plasticity index is PI = LL minus PL, and it represents the range of water content over which the soil remains plastic.
Correct answer: Liquid limit minus plastic limit- A. Degree of saturation only
- B. Specific gravity of solids
- C. Plasticity index
- D. Hydraulic gradient
Explanation: Darcy's law gives discharge per unit area as q = ki, where k is the coefficient of permeability and i is the hydraulic gradient.
Correct answer: Hydraulic gradient- A. Effective stress becomes nearly zero
- B. Total stress becomes zero
- C. Pore pressure becomes zero
- D. The soil reaches maximum dry density
Explanation: Quicksand occurs when upward seepage reduces the effective stress to nearly zero, so the soil loses its shear strength.
Correct answer: Effective stress becomes nearly zero- A. Soil minerals
- B. Pore air
- C. Pore water
- D. Organic matter
Explanation: Compaction rapidly densifies soil mainly by expelling air from its voids through mechanical energy.
Correct answer: Pore air7. For a normally consolidated clay layer, the preconsolidation pressure is approximately equal to the:
- A. Present pore water pressure
- B. Current total overburden pressure
- C. Current effective overburden pressure
- D. Shear strength at failure
Explanation: In normally consolidated clay, the current effective overburden pressure is the greatest effective stress the soil has experienced.
Correct answer: Current effective overburden pressure- A. Cohesive resistance
- B. Volumetric strain
- C. Pore water pressure
- D. Frictional resistance
Explanation: The Mohr-Coulomb equation is τ = c + σ tan φ, where c is cohesion and σ tan φ is frictional resistance.
Correct answer: Frictional resistance- A. 0.25
- B. 0.33
- C. 0.50
- D. 1.00
Explanation: Rankine's active coefficient is Ka = (1 − sin φ)/(1 + sin φ). For φ = 30 degrees, Ka = 0.5/1.5, which is approximately 0.33.
Correct answer: 0.33- A. A strong bearing layer lies near ground level
- B. The upper soil is weak and a deeper bearing layer is available
- C. The applied load is negligible
- D. The groundwater table is absent
Explanation: Piles are useful when weak or highly compressible near-surface soil cannot safely support the structure.
Correct answer: The upper soil is weak and a deeper bearing layer is available- A. Volume of solids to total volume
- B. Volume of air to volume of solids
- C. Volume of water to volume of voids
- D. Volume of water to total volume
Explanation: Degree of saturation is the percentage of the void space occupied by water, so it is Vw divided by Vv.
Correct answer: Volume of water to volume of voids- A. Solid state to semisolid state
- B. Semisolid state to plastic state
- C. Liquid state to gaseous state
- D. Plastic state to liquid state
Explanation: At the liquid limit, soil changes from a plastic consistency to a liquid consistency.
Correct answer: Plastic state to liquid state- A. 1.0 and 3.0
- B. 0.5 and 1.0
- C. 3.0 and 5.0
- D. 5.0 and 10.0
Explanation: For well-graded sand, Cu must exceed 6 and Cc must lie from 1 to 3. These limits indicate a broad and suitably distributed range of…
Correct answer: 1.0 and 3.0- A. Increases substantially
- B. Decreases substantially
- C. Remains constant
- D. Becomes uniformly graded only
Explanation: Capillary rise is greater in smaller pores, which are commonly associated with finer soil particles.
Correct answer: Decreases substantially- A. q = kH Nd/Nf
- B. q = HNf/(kNd)
- C. q = kH Nf/Nd
- D. q = kNd/(HNf)
Explanation: The standard flow-net expression is q = kH(Nf/Nd), where Nf is the number of flow channels and Nd is the number of potential drops.
Correct answer: q = kH Nf/Nd- A. Shrinkage limit
- B. Liquid limit
- C. Saturation moisture content
- D. Optimum moisture content
Explanation: The optimum moisture content is the water content at which a specified compactive effort produces maximum dry unit weight.
Correct answer: Optimum moisture content- A. Rate at which consolidation settlement occurs
- B. Ultimate consolidation settlement
- C. Shear strength at failure
- D. Initial specific gravity of solids
Explanation: The coefficient of consolidation combines the effects of permeability and compressibility and controls the time rate of excess pore-water…
Correct answer: Rate at which consolidation settlement occurs- A. 0 degrees
- B. 15 degrees
- C. 30 degrees
- D. 45 degrees
Explanation: In the total-stress analysis of saturated clay under undrained conditions, the apparent cohesion carries the shear resistance and phi is…
Correct answer: 0 degrees- A. 0.33
- B. 1.00
- C. 2.00
- D. 3.00
Explanation: The Rankine passive coefficient is Kp = (1 + sin phi)/(1 - sin phi). For phi equal to 30 degrees, this gives 1.5 divided by 0.5, or…
Correct answer: 3.00- A. Columns are widely spaced on strong rock
- B. The soil has high bearing capacity near the surface
- C. Only a single light column carries the structure
- D. Individual footings would cover most of the building area
Explanation: A raft supports many columns over a large area and is useful when individual footings would occupy most of the plan area.
Correct answer: Individual footings would cover most of the building areaSoil Mechanics and Foundations MCQs: common questions
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