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. (w - PL) / PI
- B. (LL - w) / PI
- C. (w - LL) / PL
- D. (PI - w) / LL
Explanation: Liquidity index compares the natural water content with the plasticity range, so it is (w - PL) divided by PI.
Correct answer: (w - PL) / PI- A. Percentage of sand fraction
- B. Percentage of clay fraction
- C. Percentage of silt fraction
- D. Percentage of gravel fraction
Explanation: Clay activity is the plasticity index divided by the percentage of particles finer than 0.002 mm.
Correct answer: Percentage of clay fraction- A. Arithmetic mean of layer permeabilities
- B. Geometric mean of layer permeabilities
- C. Harmonic mean of layer permeabilities
- D. Maximum permeability of the layers
Explanation: For flow perpendicular to soil layers, the equivalent permeability is obtained from a thickness-weighted harmonic mean.
Correct answer: Harmonic mean of layer permeabilities- A. Elastic distortion of soil grains
- B. Sudden collapse of air voids
- C. Immediate expulsion of pore water
- D. Creep and rearrangement of soil structure
Explanation: Secondary consolidation is the gradual deformation that continues after primary consolidation and is mainly caused by creep and particle…
Correct answer: Creep and rearrangement of soil structure- A. Neither consolidated nor drained before shearing
- B. Consolidated but not drained before shearing
- C. Consolidated and drained before shearing
- D. Drained but not consolidated before shearing
Explanation: In the unconsolidated undrained test, drainage is prevented during both application of cell pressure and shearing.
Correct answer: Neither consolidated nor drained before shearing- A. Cohesion and footing embedment
- B. Soil unit weight and footing width
- C. Water content and plasticity index
- D. Pile length and shaft roughness
Explanation: The term 0.5 gamma B N-gamma is the soil unit weight contribution in the bearing capacity equation.
Correct answer: Soil unit weight and footing width- A. A flexible raft above the ground
- B. Separate piles with no interaction
- C. A single enlarged pile shaft
- D. A frictionless column in the soil
Explanation: In block failure, the piles and the enclosed soil move together as one composite block.
Correct answer: A single enlarged pile shaft- A. Convert a shallow footing into a pile
- B. Increase the soil plasticity index
- C. Eliminate the need for reinforcement
- D. Reach a zone with smaller moisture variation
Explanation: Moisture changes near the ground surface cause swelling and shrinkage in expansive soils.
Correct answer: Reach a zone with smaller moisture variation- A. Soft saturated clay
- B. Dense gravel deposits
- C. Clean coarse sand
- D. Sound weathered rock
Explanation: The field vane test directly measures the undrained shear strength of soft, saturated cohesive soil.
Correct answer: Soft saturated clay50. The air content of a partially saturated soil is defined as the ratio of the volume of air to the:
- A. total volume of soil
- B. volume of voids
- C. volume of soil solids
- D. volume of water
Explanation: Air content is the proportion of the void space occupied by air, so it is Va divided by Vv.
Correct answer: volume of voids51. In soil classification, the effective grain size D10 represents the particle size corresponding to:
- A. 50 percent finer by weight
- B. 10 percent coarser by weight
- C. 10 percent finer by weight
- D. 90 percent finer by weight
Explanation: D10 is read from the particle-size distribution curve at 10 percent finer.
Correct answer: 10 percent finer by weight- A. hydraulic gradient
- B. coefficient of permeability
- C. specific gravity
- D. porosity
Explanation: Darcy discharge velocity is based on the total cross-sectional area, whereas actual water movement occurs through the voids.
Correct answer: porosity- A. specific gravity and void ratio
- B. coefficient of permeability and grain size
- C. cohesion and plasticity index
- D. liquid limit and water content
Explanation: The critical gradient is approximately (Gs minus 1) divided by (1 plus e).
Correct answer: specific gravity and void ratio- A. low dry density after compaction
- B. excessive exit hydraulic gradient
- C. high plasticity of clay
- D. large angle of internal friction
Explanation: Piping occurs when seepage forces remove soil particles, usually near the downstream exit where the hydraulic gradient is high.
Correct answer: excessive exit hydraulic gradient- A. constant air content of five percent
- B. maximum dry density at optimum water content
- C. complete saturation at each water content
- D. complete removal of water from the soil
Explanation: The zero-air-voids line is the theoretical limit for a soil with no air in its voids, meaning complete saturation.
Correct answer: complete saturation at each water content- A. gamma w divided by k mv
- B. mv divided by k gamma w
- C. k mv gamma w
- D. k divided by mv gamma w
Explanation: The usual relationship is Cv = k divided by mv gamma w. It combines permeability with compressibility and the unit weight of water…
Correct answer: k divided by mv gamma w- A. pore water is allowed to escape
- B. no volume change is permitted
- C. the specimen remains completely undrained
- D. only total stress is considered
Explanation: Drainage valves remain open in a drained test, allowing pore water to escape and permitting effective stress changes during shearing.
Correct answer: pore water is allowed to escape- A. 1 plus sin phi
- B. 1 minus sin phi
- C. tan squared of 45 degrees plus phi over 2
- D. tan squared of 45 degrees minus phi over 2
Explanation: Jaky's expression for normally consolidated soil is K0 = 1 minus sin phi.
Correct answer: 1 minus sin phi- A. remains unchanged for every slope angle
- B. increases because confinement is greater
- C. decreases because the failure zone is reduced
- D. becomes equal to the pile capacity
Explanation: A nearby slope reduces the lateral confinement and available soil mass supporting the footing, so bearing capacity decreases.
Correct answer: decreases because the failure zone is reduced- A. γwGs(1 + e)
- B. γwGs/(1 + e)
- C. γw(1 + e)/Gs
- D. γw(Gs + e)/(1 + e)
Explanation: For saturated soil, all voids contain water, so the total weight includes the solids and water occupying the voids.
Correct answer: γw(Gs + e)/(1 + e)