ASRB NET — Farm Machinery & Power / Land & Water Management Engineering / Agricultural Structures & Process Engineering · India · Agricultural Scientists Recruitment Board (ICAR)
Original practice questions written to the published syllabus · not past papersSoil as an engineering material: its phases, flow of water through it, compaction, consolidation, shear strength and stability.
In the exams: GATE AG Section 4 (soil mechanics asked in 6 of 8 papers 2019–26); ICAR AIEEA-PG Unit IV; NCEES PE soils and foundations
ICAR AICE-JRF/SRF (PhD) and ASRB NET — Soil and Water Conservation / Land and Water Management Engineering
Deeper treatment of seepage through and under earth dams (phreatic line, filters, piping), consolidation theory and its assumptions, drained versus undrained analysis, stability cases for earth dams including rapid drawdown, and methods of slices.
How our questions for ASRB NET spread across the units of Soil Mechanics. Difficulty: 0 easy, 13 medium, 23 hard.
| Seepage and flow nets | 7 | |
| Slope stability, including infinite slopes | 7 | |
| Shear strength and the Mohr-Coulomb criterion | 5 | |
| Terzaghi one-dimensional consolidation | 4 | |
| Permeability | 3 | |
| Effective stress | 3 | |
| Bearing capacity basics | 3 | |
| Compaction and the Proctor test | 2 | |
| Earth pressure: active and passive (Rankine) | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A deposit consists of two layers of equal thickness with permeabilities of 1 × 10⁻³ cm/s and 1 × 10⁻⁵ cm/s. What is the ratio of equivalent horizontal to equivalent vertical permeability?
A. 25.5
kh = (k1 + k2)/2 = 5.050e-04 cm/s. kv = 2/(1/k1 + 1/k2) = 2/(1000 + 100000) = 1.980e-05 cm/s. kh/kv = 25.5. 100 is simply k1/k2, which is not the ratio of the equivalent values. 51.0 forgets the 2 in the harmonic mean, taking kv = 1/(1/k1 + 1/k2) = 9.90e-06 cm/s.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
Downstream of a weir, the last square of a flow net has a length of 1.2 m along the flow line. The total head loss of 6 m is spread over 12 equal equipotential drops. The foundation sand has G = 2.68 and e = 0.72. What is the factor of safety against boiling at the exit?
A. 2.34
Head drop per equipotential = 6/12 = 0.5 m. Exit gradient = 0.5/1.2 = 0.417. ic = (G − 1)/(1 + e) = 0.977. FS = ic/iexit = 2.34. Using the whole head over one square (5.0) gives 0.20; 0.43 is the ratio upside down. A factor near 2.3 is below the 3 to 4 usually wanted, so the exit needs a loaded filter. 3.74 computes ic as G/(1 + e) = 1.558, forgetting buoyancy (1.558/0.417).
Further reading: Craig's Soil Mechanics
A soil (G = 2.70) has a standard Proctor maximum dry unit weight of 17.5 kN/m³ at an optimum moisture content of 16%. With γw = 9.81 kN/m³, what is the percentage air voids at optimum?
C. 5.4%
e = Gγw/γd − 1 = 2.70 × 9.81/17.5 − 1 = 0.5135. S = wG/e = 0.16 × 2.70/0.5135 = 0.841. n = e/(1 + e) = 0.3393. na = n(1 − S) = 5.4%, within the few-per-cent range typical at optimum. 15.9% is air content (fraction of voids), not air voids (fraction of total volume). 8.2% uses e instead of n (0.5135 × 0.159); 84.1% is the degree of saturation.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A clay has k = 1 × 10⁻⁹ m/s and mv = 0.25 m²/MN. With γw = 9.81 kN/m³, what is its coefficient of consolidation in m²/year?
C. 12.9 m²/year
mv = 0.25 m²/MN = 2.5 × 10⁻⁴ m²/kN. cv = k/(mv·γw) = 10⁻⁹/(2.5 × 10⁻⁴ × 9.81) = 4.077e-07 m²/s × 3.156 × 10⁷ s/year = 12.9 m²/year. Using 0.25 as if in m²/kN gives 0.0129; omitting γw gives 126. 1.29 is a factor-of-ten slip in the unit conversion.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A clay layer 3 m thick (e0 = 0.90) has σ′0 = 80 kPa and a preconsolidation pressure of 150 kPa at mid-depth. Cc = 0.30 and Cr = 0.05. What settlement results from a stress increase of 50 kPa at mid-depth?
B. 16.6 mm
Final stress 80 + 50 = 130 kPa is below σ′p = 150 kPa, so only recompression occurs: S = [Cr·H/(1 + e0)]·log(130/80) = (0.05 × 3/1.90) × 0.2109 = 16.6 mm. Using Cc treats the clay as normally consolidated and gives 99.9 mm, about six times too much; leaving out (1 + e0) gives 31.6 mm. OCR = 150/80 = 1.88. 38.3 mm uses the natural logarithm instead of log10 (0.0789 × ln 1.625).
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A soil has c′ = 10 kPa and φ′ = 25°. In a drained triaxial test at a cell pressure of 150 kPa, what major principal effective stress will cause failure?
B. 401.0 kPa
σ′1 = σ′3·tan²(45° + φ′/2) + 2c′·tan(45° + φ′/2). tan 57.5° = 1.5697, tan² = 2.4639. σ′1 = 150 × 2.4639 + 2 × 10 × 1.5697 = 369.6 + 31.4 = 401.0 kPa. Leaving out cohesion gives 369.6; the cohesion term uses tan(45° + φ′/2), not its square. 418.9 kPa squares the tangent in the cohesion term (369.6 + 2 × 10 × 2.4639); 385.3 kPa leaves out the 2 (369.6 + 10 × 1.5697).
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A saturated clay (B = 1.0) has a pore-pressure parameter A = 0.5. Under undrained loading, σ3 increases by 50 kPa and σ1 increases by 150 kPa. What is the change in pore pressure?
B. 100 kPa
Δu = B[Δσ3 + A(Δσ1 − Δσ3)] = 1.0 × [50 + 0.5 × (150 − 50)] = 100 kPa. A multiplies the change in deviator stress (Δσ1 − Δσ3), not Δσ1 itself. 75 kPa applies A to Δσ1 alone (0.5 × 150).
Further reading: Skempton; Craig's Soil Mechanics
A dry soil slope at 30° has c′ = 10 kPa, φ′ = 25° and γ = 18 kN/m³. What is the factor of safety for sliding on a plane parallel to the surface 3 m deep (infinite-slope analysis)?
A. 1.24
Normal stress σ = γz cos²β = 18 × 3 × 0.75 = 40.5 kPa; shear stress τ = γz sinβ cosβ = 23.38 kPa. Strength = 10 + 40.5 × tan25° = 28.89 kPa. F = 28.89/23.38 = 1.24. tanφ/tanβ = 0.81 ignores cohesion; using cosβ instead of cos²β gives 1.36. 1.07 takes τ = γz sinβ without the cosβ (28.89/27.0).
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A 6 m high cutting in saturated clay (cu = 30 kPa, φu = 0, γ = 18 kN/m³) has a slope for which Taylor's chart gives a stability number cu/(FγH) = 0.18. What is the factor of safety?
C. 1.54
Sn = cu/(FγH), so F = cu/(Sn·γ·H) = 30/(0.18 × 18 × 6) = 30/19.44 = 1.54. 0.65 is the reciprocal. Some texts define the stability number the other way up, so always check the chart's definition. 0.28 stops at cu/(γH) = 30/108 without the stability number; 9.26 leaves out H (30/(0.18 × 18)).
Further reading: Taylor; Das and Sobhan, Principles of Geotechnical Engineering
A slip circle through a dry embankment (c′ = 8 kPa, φ′ = 28°) is divided into four slices with weights 200, 400, 450 and 250 kN/m and base inclinations −5°, 15°, 35° and 55°. The total arc length is 18 m. By the ordinary (Fellenius) method of slices, what is the factor of safety?
B. 1.33
ΣW sinα = 200 × sin(-5°) + 400 × sin(15°) + 450 × sin(35°) + 250 × sin(55°) = 549.0 kN/m. ΣW cosα = 1097.6 kN/m. Resistance = c′L + ΣW cosα·tanφ′ = 8 × 18 + 1097.6 × 0.5317 = 727.6 kN/m. F = 727.6/549.0 = 1.33. The slice with negative α resists sliding, so its W sinα must stay negative; taking it as positive gives 1.25. Omitting cohesion gives 1.06. 0.75 is the ratio inverted (549.0/727.6).
Further reading: Craig's Soil Mechanics
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