Topics Covered: Soil Classification, Soil Compaction and Consolidation, Shear Strength and Foundation
Geotechnical Engineering Topic-wise Questions for JE Exam
Topic: Soil Properties & Classification
1. The water content of a soil sample is 25%. If the mass of solids is 360g, the mass of water is:
Answer: A
Water content (w) = (Mass of water)/(Mass of solids). Given w=25% and Mₛ=360g, M_w = w × Mₛ = 0.25 × 360 = 90g. Total mass M = Mₛ + M_w = 360 + 90 = 450g.
2. The void ratio of a soil sample is 0.60 and its specific gravity is 2.70. The dry density is:
Answer: B
Dry density (ρₐ) = Gₛρ_w/(1+e) = (2.70 × 1)/(1+0.60) = 1.6875 g/cm³ ≈ 1.69 g/cm³. Where Gₛ=specific gravity, ρ_w=density of water (1g/cm³), e=void ratio.
3. The soil classification symbol 'SM' represents:
Answer: A
As per Unified Soil Classification System (USCS), 'SM' stands for Silty Sand - sand with silt (35% or less passing #200 sieve). 'SC'=Clayey Sand, 'ML'=Sandy Silt, 'CL'=Sandy Clay.
4. The plasticity index of a soil is the difference between:
Answer: A
Plasticity Index (PI) = Liquid Limit (LL) - Plastic Limit (PL). It indicates the range of water content over which soil remains plastic. High PI (>17) indicates clayey soils, low PI (7-17) silty soils, PI<7 sandy soils.
5. The standard Proctor test is conducted with:
Answer: B
Standard Proctor test (IS 2720 Part VII) uses 4.5kg hammer with 450mm free fall, compacted in 3 layers with 25 blows per layer in 1000cm³ mold. Modified Proctor uses 4.5kg hammer with 450mm drop, 5 layers, 25 blows/layer.
6. The coefficient of permeability of clay is typically in the range:
Answer: C
Typical permeability ranges: Gravel (10⁰-10⁻² cm/s), Sand (10⁻²-10⁻⁴ cm/s), Silt (10⁻⁴-10⁻⁶ cm/s), Clay (10⁻⁶-10⁻⁸ cm/s). Permeability decreases with decreasing particle size and increasing plasticity.
7. The sensitivity of clay is defined as the ratio of:
Answer: B
Sensitivity (Sₜ) = (Undisturbed shear strength)/(Remolded shear strength). Classification: Sₜ=1 (insensitive), 1-2 (low), 2-4 (medium), 4-8 (sensitive), 8-16 (extra sensitive), >16 (quick). Marine clays often have high sensitivity.
8. The effective size (D₁₀) of a soil is the particle size corresponding to:
Answer: A
D₁₀ (effective size) is the diameter where 10% of particles are finer (90% coarser). It controls permeability - Hazen's formula k = CD₁₀² (cm/s) where C=0.01-0.015. Uniformity coefficient Cᵤ = D₆₀/D₁₀.
9. The activity of clay is defined as the ratio of:
Answer: B
Activity (A) = PI/(% clay fraction <2μ). Classification: A<0.75 (inactive), 0.75-1.25 (normal), >1.25 (active). Montmorillonite has high activity (1-7), kaolinite low (0.3-0.5).
10. The relative density of a sandy soil is 70%. This soil is classified as:
Answer: D
Relative density (Dᵣ) classification: 0-15% (very loose), 15-35% (loose), 35-65% (medium dense), 65-85% (dense), 85-100% (very dense). Dᵣ = (eₘₐₓ - e)/(eₘₐₓ - eₘᵢₙ) × 100%.
Topic: Soil Compaction & Consolidation
11. The zero air void line in compaction test represents:
Answer: A
Zero air void line (S=100%) is calculated as ρᵈ = Gₛρ_w/(1+wGₛ). It's the theoretical maximum density at full saturation. Actual compaction curve always lies below it as 100% saturation is unattainable in standard compaction.
12. The coefficient of consolidation (cᵥ) is calculated from:
Answer: C
cᵥ can be determined from both Casagrande's log-time method (cᵥ = 0.197H²/t₅₀) and Taylor's √time method (cᵥ = 0.848H²/t₉₀), where H=drainage path length, t₅₀/t₉₀=time for 50%/90% consolidation.
13. The preconsolidation pressure of a soil can be determined by:
Answer: C
Preconsolidation pressure (σₚ') is determined from e-logσ' curve in consolidation test using Casagrande's graphical method. It represents maximum past effective stress experienced by the soil.
14. The compression index (Cₑ) is the slope of:
Answer: A
Compression index Cₑ = Δe/Δlogσ' for normally consolidated clays. Typical values: 0.1-0.3 (silty clays), 0.3-0.5 (soft clays), >0.5 (highly compressible clays). For remolded clays, Cₑ ≈ 0.007(LL-10%).
15. The time factor (Tᵥ) for 50% consolidation is:
Answer: B
Time factor Tᵥ = cᵥt/H². Values: 0% (Tᵥ=0), 50% (0.197), 90% (0.848), 100% (∞). For double drainage, H=half thickness; single drainage, H=full thickness.
16. The swelling index (Cₛ) is typically:
Answer: B
Swelling index (Cₛ) ≈ 1/5 to 1/10 of Cₑ for most clays. Cₛ represents slope of e-logσ' curve during unloading/reloading in overconsolidated range. Typical range: 0.02-0.1.
17. The compaction energy in standard Proctor test is:
Answer: A
Standard Proctor energy = (4.5kg × 9.81m/s² × 0.45m × 25 blows × 3 layers)/944cm³ ≈ 593 kJ/m³. Modified Proctor energy is 4.5× larger (2696 kJ/m³) due to 5 layers and heavier compaction.
18. The overconsolidation ratio (OCR) is defined as:
Answer: A
OCR = Preconsolidation pressure (σₚ')/Present effective overburden pressure (σ₀'). OCR=1 (normally consolidated), OCR>1 (overconsolidated), OCR<1 (underconsolidated). Higher OCR indicates more preloaded/stiffer soil.
19. The coefficient of volume compressibility (mᵥ) is given by:
Answer: B
mᵥ = Δe/[(1+e₀)Δσ'] = aᵥ/(1+e₀) where aᵥ=coefficient of compressibility. Units: m²/kN. Typical range: 10⁻¹-10⁻⁴ m²/kN (high for soft clays, low for stiff clays). Related to cᵥ by cᵥ = k/(mᵥγ_w).
20. The secondary consolidation is primarily due to:
Answer: C
Secondary consolidation occurs after primary consolidation (pore pressure dissipation) due to creep - plastic adjustment of soil particles. It's significant in organic soils/peats. Coefficient of secondary compression Cₐ = Δe/Δlogt.
Topic: Shear Strength & Foundations
21. The angle of internal friction (φ) for loose dry sand is typically:
Answer: B
Typical φ values: Loose dry sand (25°-30°), dense dry sand (35°-40°), silty sand (28°-34°), gravel (35°-50°). φ increases with density, angularity, and uniformity of particles.
22. The unconfined compressive strength (qᵤ) of a clay is related to its undrained cohesion (cᵤ) by:
Answer: B
For φ=0 condition (saturated clays), qᵤ = 2cᵤ where qᵤ=unconfined compressive strength. Undrained shear strength sᵤ = cᵤ = qᵤ/2. Typical range: 5-50 kPa (very soft to stiff clays).
23. The Terzaghi's bearing capacity factor Nᵥ for φ=30° is:
Answer: D
Terzaghi's bearing capacity factors for general shear failure: Nᵥ = 0.5(e^(3π/4-φ/2)tanφ)²/(cos²(45°+φ/2)). Values: φ=0° (Nᵥ=0), 20° (4.5), 30° (22.5), 40° (109). Nₑ and N_q also increase with φ.
24. The standard penetration test (SPT) 'N' value is corrected for:
Answer: C
SPT N-value is corrected for: (1) Overburden (Cₙ = (Pₐ/σ₀')⁰·⁵ ≤1.7, Pₐ=100kPa), and (2) Dilatancy (applied if N'>15 in fine sands/silts: N'' = 15 + 0.5(N'-15)). Hammer efficiency corrections may also apply.
25. The ultimate bearing capacity of a square footing is given by (Terzaghi's equation):
Answer: A
Terzaghi's bearing capacity for square footing: qᵤ = 1.3cNₑ + qN_q + 0.4γBNᵥ. For strip footing: qᵤ = cNₑ + qN_q + 0.5γBNᵥ. Shape factors: 1.3 (square), 1.2 (circular) for cohesion term.
26. The negative skin friction in piles occurs when:
Answer: A
Negative skin friction occurs when surrounding soil settles more than the pile, causing downward drag. Common in compressible soils (fill, soft clay) or when water table drops. It reduces pile capacity and increases settlement.
27. The group efficiency of pile foundations is typically:
Answer: A
Group efficiency (η) = (Group capacity)/(Sum of individual capacities). For friction piles, η<1 due to overlapping stress zones (typical 0.7-1.0). For end-bearing piles on hard stratum, η≈1.
28. The critical depth for calculating bearing capacity of shallow foundations in sand is:
Answer: B
Critical depth (Dₑ) = 1.0B for square/rectangular footings in sand, where B=width. Below this depth, qN_q term becomes constant. For continuous footings, Dₑ=2B. This accounts for limiting overburden effect.
29. The modulus of subgrade reaction is expressed in units of:
Answer: B
Modulus of subgrade reaction (kₛ) = p/δ (kN/m³), where p=pressure (kN/m²), δ=settlement (m). Typical values: 15,000-50,000 kN/m³ (dense sand), 5,000-15,000 kN/m³ (medium clay). Plate load test is used to determine kₛ.
30. The minimum factor of safety against bearing capacity failure for shallow foundations is:
Answer: C
Minimum FOS against bearing failure is 2.5-3 for dead load + live load, and 2.0 for dead load + live load + wind/earthquake load as per IS 6403. Higher FOS (3-4) is used when uncertainty in soil parameters is high.