Nepal Engineering Council · Civil Engineering · Chapter 2
Soil Mechanics and Foundation Engineering
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183 questions in 6 syllabus topics.
2.1 Soil properties and laboratory tests
32 questions · ACiE0201
1. The void ratio of a soil is defined as the ratio of
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Compare with porosity, which uses the total volume.
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Answer: C. volume of voids to volume of solids
e = Vv/Vs, whereas the ratio Vv/V is porosity and Vw/Vv is degree of saturation.
2. For a completely dry soil sample, the degree of saturation is
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Think about how much water fills the voids.
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Answer: D. 0
S = Vw/Vv and Vw = 0 for dry soil, so S = 0.
3. In the Casagrande apparatus, the liquid limit is the water content at which the standard groove closes over a length of 12 mm after
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A round number near the middle of the flow curve.
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Answer: A. 25 blows
By definition (IS 2720 Part 5) the liquid limit corresponds to 25 blows in the Casagrande device.
4. The plastic limit of a soil is the water content at which the soil thread, when rolled by hand, just begins to crumble at a diameter of
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About the thickness of a pencil lead, not a finger.
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Answer: B. 3 mm
The standard plastic-limit test rolls a thread of 3 mm diameter until it crumbles.
5. The shrinkage limit of a soil is the water content below which
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It marks the change from semi-solid to solid state.
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Answer: D. further loss of water does not cause any decrease in volume
Below the shrinkage limit the soil volume stays constant on drying, although the soil continues to lose water.
6. The plasticity index of a soil is
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It is a difference, not a ratio.
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Answer: C. liquid limit minus plastic limit
Ip = wL − wP, the range of water content over which the soil is plastic.
7. According to the Unified Soil Classification System, a soil is called fine-grained if more than 50% of it passes the
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Silt and clay are finer than 75 µm.
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Answer: D. 75 µm sieve
USCS (and IS 1498) put the coarse/fine boundary at the 75 µm sieve.
8. In the Unified Soil Classification System, the group symbol CH represents
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C is clay; the second letter is plasticity.
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Answer: A. inorganic clay of high plasticity
C = clay, H = high plasticity (LL above the limit, plotting above the A-line).
9. On the plasticity chart, the A-line is represented by
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Remember the constant 0.73.
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Answer: A. Ip = 0.73 (wL − 20)
Casagrande's A-line is Ip = 0.73(wL − 20); clays plot above and silts/organic soils below it.
10. In the Indian Standard (IS 1498) classification, fine-grained soils of medium compressibility have liquid limit
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Three ranges split at 35 and 50.
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Answer: D. between 35% and 50%
IS 1498: L < 35% low, 35–50% intermediate (medium), > 50% high compressibility.
11. In the MIT classification, the particle size range of silt is
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Silt lies between clay and sand sizes.
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Answer: B. 0.002 mm to 0.06 mm
MIT: clay < 0.002 mm, silt 0.002–0.06 mm, sand 0.06–2 mm, gravel > 2 mm.
12. Textural classification of soils is made using
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Three components, three corners.
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Answer: B. percentages of sand, silt and clay on a triangular chart
The textural (triangular) chart plots the sand, silt and clay fractions to name the soil.
13. A well-graded soil is indicated when the coefficient of curvature Cc lies between
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The curve must be smooth and concave.
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Answer: C. 1 and 3
Well-graded soils need Cc between 1 and 3 together with Cu > 4 (gravel) or > 6 (sand).
14. The hydrometer method of particle size analysis is based on
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Think of falling spherical grains.
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Answer: C. Stokes' law of settlement of spheres in a fluid
Particle settling velocity v = γw(G−1)D²/(18η) is Stokes' law, used to find D for fine grains.
15. Which laboratory permeability test is more suitable for clean sands and gravels?
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Flow must be measurable at steady head.
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Answer: B. Constant head test
Coarse soils pass enough flow for a steady constant head measurement; the falling head test suits fine soils.
16. The unconfined compression test is suitable mainly for determining the strength of
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Sample must stand without a mould.
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Answer: B. saturated cohesive soils
With no confining pressure, only cohesive soils that stand unsupported can be tested; qu = 2c for φu = 0.
17. The one-dimensional consolidation (oedometer) test is performed to determine the
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Settlement parameters.
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Answer: D. compressibility characteristics of soil
It gives e–log σ' curve, Cc, mv and cv for settlement computations.
18. According to the SPT N-value classification of Terzaghi and Peck, a sand with N = 35 is
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Between 30 and 50 on the SPT scale.
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Answer: A. dense
Sand: 0–4 very loose, 4–10 loose, 10–30 medium, 30–50 dense, > 50 very dense.
19. In a boring log, the SPT N-value is the number of blows required for the split spoon sampler to penetrate
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Seating drive is not counted.
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Answer: C. 300 mm after the initial seating of 150 mm
N is the blow count over the last 300 mm (second plus third 150 mm) of the 450 mm test drive.
20. The sensitivity of a clay is the ratio of its unconfined compressive strength in the
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Disturbance reduces the strength.
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Answer: C. undisturbed state to that in the remoulded state
St = qu(undisturbed)/qu(remoulded); a high St means large strength loss on disturbance.
21. A soil has void ratio 0.80 and specific gravity of solids 2.70. Taking γw = 9.81 kN/m³, its dry unit weight is
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Dry unit weight = G γw/(1+e).
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Answer: D. 14.71 kN/m³
γd = Gγw/(1+e) = 2.70×9.81/1.80 = 14.71 kN/m³.
22. A fully saturated clay has water content 20% and G = 2.70. Taking γw = 9.81 kN/m³, its saturated unit weight is
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Find e from S e = w G with S = 1 first.
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Answer: B. 20.64 kN/m³
e = wG = 0.54; γsat = (G+e)γw/(1+e) = 3.24×9.81/1.54 = 20.64 kN/m³.
23. A soil sample of mass 200 g is oven-dried to a constant mass of 170 g. Its water content is
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Water content uses the dry mass in the denominator.
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Answer: A. 17.6%
w = (Mw/Md)×100 = 30/170 = 17.6%, based on dry mass.
24. The maximum and minimum void ratios of a sand are 0.90 and 0.50. At a void ratio of 0.60 the relative density is
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Dr = (emax − e)/(emax − emin).
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Answer: D. 75%
Dr = (emax − e)/(emax − emin) = (0.90−0.60)/0.40 = 0.75.
25. For a soil with LL = 60%, PL = 25% and natural water content 40%, the consistency index is
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Ic = (LL − w)/Ip.
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Answer: C. 0.57
Ip = 60−25 = 35; Ic = (wL − w)/Ip = 20/35 = 0.57.
26. A soil has wL = 65% and Ip = 40%. Its Unified classification is most likely
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Compare Ip with the A-line value at that LL.
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Answer: B. CH
A-line at wL = 65: 0.73×45 = 32.9 < 40, so the soil plots above the A-line with wL > 50: CH.
27. In a sieve analysis a 500 g dry sample leaves a cumulative 380 g retained on and above the 75 µm sieve. The percentage passing the 75 µm sieve is
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Passing = 100 − cumulative % retained.
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Answer: A. 24%
Passing = (500−380)/500 = 24%.
28. From the grain-size curve D10 = 0.15 mm and D60 = 0.45 mm. The uniformity coefficient is
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Cu = D60/D10.
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Answer: C. 3.0
Cu = D60/D10 = 0.45/0.15 = 3.0, so the soil is uniformly graded.
29. A sand has D10 = 0.10 mm, D30 = 0.40 mm and D60 = 0.80 mm. Its coefficient of curvature is
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Cc = D30²/(D10 D60).
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Answer: A. 2.0
Cc = D30²/(D10·D60) = 0.16/0.08 = 2.0.
30. A partially saturated soil has S = 80%, w = 25% and G = 2.70. Its void ratio is
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Use S e = w G.
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Answer: A. 0.84
e = wG/S = 0.25×2.70/0.80 = 0.84.
31. In a constant head test a sand specimen 15 cm long and 50 cm² in area passes 300 cm³ of water in 120 s under a head of 30 cm. The coefficient of permeability is
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k = Q L/(A h t).
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Answer: B. 0.025 cm/s
k = QL/(A h t) = 300×15/(50×30×120) = 0.025 cm/s.
32. In a falling head test (specimen length 10 cm, area 50 cm², standpipe area 1 cm²) the head falls from 100 cm to 50 cm in 600 s. The coefficient of permeability is
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k = 2.303 aL/(At) log(h1/h2).
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Answer: D. 2.31 × 10⁻⁴ cm/s
k = 2.303 aL/(A t)·log10(h1/h2) = 2.303×1×10/(50×600)×0.301 = 2.31×10⁻⁴ cm/s.
2.2 Stresses on soil and seepage
31 questions · ACiE0202
33. According to Terzaghi's principle, the effective stress in a saturated soil is
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Pore water carries part of the total stress.
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Answer: D. total stress minus pore water pressure
σ' = σ − u; the effective stress governs compression and shear strength of the soil skeleton.
34. If the water table rises to the ground surface in a uniform sand deposit, the effective stress at a given depth
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Buoyancy reduces the effective weight.
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Answer: D. decreases
Total stress rises slightly (γsat > γ) but pore pressure rises more, so σ' = z(γsat − γw) is lower than before.
35. Capillary water held above the water table, in a saturated capillary zone, affects the effective stress by
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Tension in water means a negative u.
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Answer: D. increasing it, because pore pressure there is negative
Pore pressure above the water table is −γw·h (suction), so σ' = σ + γw·h is larger.
36. Capillary rise in soils is greatest in
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Smaller pores, narrower tubes.
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Answer: A. fine-grained soils such as silt
hc = C/(e·D10) – the rise varies inversely with effective grain size, so fine soils show greater rise.
37. Quick sand condition occurs in a soil when
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Effective stress falls to zero.
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Answer: D. the upward hydraulic gradient equals the critical hydraulic gradient
At i = ic the upward seepage force balances the submerged weight, so σ' = 0 and sand loses strength.
38. Quick sand is
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It is a condition, not a material.
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Answer: C. a flow condition of cohesionless soil under upward seepage, not a type of soil
Any cohesionless soil, typically fine sand or silt, can become quick when σ' = 0 under upward flow.
39. In a flow net, the flow lines and equipotential lines intersect each other at
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Orthogonal families.
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Answer: A. right angles
For isotropic soil the two families of Laplace solution are orthogonal, forming curvilinear squares.
40. Which of the following cannot be obtained directly from a flow net under a concrete dam?
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Strength is a soil property, not a seepage quantity.
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Answer: C. Shear strength of the foundation soil
A flow net gives heads, discharge, gradients and uplift, not strength parameters.
41. The governing equation of steady two-dimensional seepage through homogeneous isotropic soil is
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Same equation as potential flow.
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Answer: D. Laplace's equation, ∂²h/∂x² + ∂²h/∂z² = 0
Continuity plus Darcy's law for incompressible steady flow gives Laplace's equation.
42. If v is the discharge velocity and n is the porosity, the seepage velocity in soil is
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Seepage velocity exceeds discharge velocity.
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Answer: B. v/n
Water flows only through the void area fraction n, so vs = v/n, always larger than v.
43. For the same clay, the swelling (recompression) index Cs compared with the compression index Cc is
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Rebound curve is flatter.
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Answer: A. much smaller
The unloading–reloading curve is flatter than the virgin curve; Cs is roughly 1/5 to 1/10 of Cc.
44. A clay with an overconsolidation ratio (OCR) greater than 1 is one in which
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Past maximum stress exceeds present stress.
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Answer: B. the present effective stress is less than the preconsolidation pressure
OCR = σ'c/σ'0; OCR > 1 means the soil was once subjected to a higher effective stress.
45. In the IS light (standard Proctor) compaction test, the soil is compacted in a mould of 1000 cm³ in three layers, each given 25 blows of a rammer weighing
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Light compaction uses the lighter rammer.
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Answer: A. 2.6 kg dropping 310 mm
IS 2720 Part 7 light compaction: 2.6 kg rammer, 310 mm fall; the heavy test uses 4.9 kg, 450 mm and 5 layers.
46. When the compactive effort applied to a soil is increased, the
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The peak moves up and left.
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Answer: A. maximum dry density increases and optimum moisture content decreases
Higher energy packs grains tighter at lower water content, shifting the curve up and to the left.
47. A clay compacted on the wet side of optimum moisture content generally has a
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Dry side is flocculated.
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Answer: A. dispersed structure
Wet of optimum, the particles orient parallel (dispersed), giving lower strength but higher flexibility and lower permeability.
48. The most suitable roller for compacting cohesive soils in the field is the
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Feet knead sticky soils.
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Answer: B. sheepsfoot roller
Sheepsfoot (tamping) rollers knead clay layers, giving good bonding between lifts; vibratory rollers suit granular soils.
49. The main difference between compaction and consolidation is that compaction
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Air versus water, fast versus slow.
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Answer: B. reduces air voids rapidly by mechanical energy, while consolidation expels pore water slowly under sustained load
Compaction densifies partly saturated soil by expelling air; consolidation is time-dependent drainage of water in saturated soil.
50. A 5 m thick saturated clay layer (γsat = 19 kN/m³) is submerged under water table at ground surface. Taking γw = 9.81 kN/m³, the effective stress at 5 m depth is
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σ' = z(γsat − γw).
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Answer: C. 45.95 kPa
σ = 95 kPa, u = 49.05 kPa, σ' = 45.95 kPa (= 5×(19−9.81)).
51. A sand deposit has γ = 17 kN/m³ above the water table, which lies 3 m below ground, and γsat = 20 kN/m³ below it. Taking γw = 9.81 kN/m³, the effective stress at 7 m depth is
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Subtract pore pressure acting only below the water table.
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Answer: B. 91.76 kPa
σ = 51 + 80 = 131 kPa; u = 4×9.81 = 39.24 kPa; σ' = 91.76 kPa.
52. In a saturated capillary zone, a point 1.0 m above the water table has a pore water pressure of (γw = 9.81 kN/m³)
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Sign is negative above the water table.
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Answer: D. −9.81 kPa
u = −γw·h = −9.81 kPa; water above the water table is in tension.
53. The critical hydraulic gradient for a sand with G = 2.70 and void ratio 0.80 is
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ic = (G − 1)/(1 + e).
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Answer: C. 0.94
ic = (G−1)/(1+e) = 1.70/1.80 = 0.94.
54. A sand has G = 2.65 and e = 0.65. If the upward exit hydraulic gradient is 0.60, the factor of safety against quick sand is
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Fs = ic/ie.
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Answer: B. 1.67
ic = 1.65/1.65 = 1.0; Fs = ic/i = 1.0/0.60 = 1.67.
55. A flow net under a weir has 4 flow channels and 12 equipotential drops. With k = 2×10⁻⁵ m/s and total head loss 6 m, the seepage per metre length is
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q = k·H·(Nf/Nd).
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Answer: C. 4 × 10⁻⁵ m³/s
q = k H Nf/Nd = 2×10⁻⁵ × 6 × 4/12 = 4×10⁻⁵ m³/s per metre.
56. Water flows through a soil with k = 1×10⁻³ cm/s under hydraulic gradient 0.4. If the porosity is 0.4, the seepage velocity is
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vs = k·i/n.
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Answer: B. 1.0 × 10⁻³ cm/s
v = ki = 4×10⁻⁴ cm/s; vs = v/n = 1.0×10⁻³ cm/s.
57. Two horizontal soil layers each 2 m thick have k1 = 4×10⁻⁴ cm/s and k2 = 1×10⁻⁴ cm/s. The equivalent permeability for flow perpendicular to the layers is
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Layers act like resistances in series.
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Answer: C. 1.6 × 10⁻⁴ cm/s
kv = H/Σ(Hi/ki) = 4/(2/4e-4 + 2/1e-4) = 1.6×10⁻⁴ cm/s; kh would be 2.5×10⁻⁴.
58. Using Terzaghi–Peck's empirical relation Cc = 0.009 (LL − 10), the compression index of a normally consolidated clay with LL = 50% is
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Subtract 10 from LL first.
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Answer: B. 0.36
Cc = 0.009×(50 − 10) = 0.36.
59. A clay shows void ratio 1.10 at 100 kPa and 0.90 at 200 kPa on the virgin curve. Its compression index is
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Use a logarithmic stress ratio.
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Answer: A. 0.66
Cc = Δe/log(σ2/σ1) = 0.20/log 2 = 0.66.
60. For the data e = 1.10 at 100 kPa and e = 0.90 at 200 kPa, the coefficient of compressibility av is
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av = −Δe/Δσ'.
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Answer: C. 0.002 m²/kN
av = Δe/Δσ' = 0.20/100 = 0.002 m²/kN.
61. A clay has a preconsolidation pressure of 200 kPa and a present effective overburden of 80 kPa. Its OCR is
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OCR = σ'c/σ'0.
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Answer: D. 2.5
OCR = 200/80 = 2.5 (overconsolidated).
62. The zero air voids dry unit weight of a soil with G = 2.70 at water content 15% is (γw = 9.81 kN/m³)
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γd,zav = Gγw/(1 + wG).
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Answer: C. 18.85 kN/m³
γd = Gγw/(1+wG) = 26.49/1.405 = 18.85 kN/m³.
63. In a field density test, the compacted fill has a dry unit weight of 17.1 kN/m³, while the laboratory maximum dry unit weight is 18.0 kN/m³. The relative compaction is
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Field over laboratory maximum.
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Answer: A. 95%
R = γd(field)/γd(max) = 17.1/18.0 = 95%.
2.3 Shear strength of soil and stability of slopes
30 questions · ACiE0203
64. According to the Mohr–Coulomb failure criterion, the shear strength of a soil in terms of effective stresses is
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Friction depends on normal effective stress.
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Answer: D. τf = c' + σ' tan φ'
Strength has a cohesion part c' and a frictional part proportional to the effective normal stress.
65. A principal plane is a plane on which
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Think of the Mohr circle intercepts on the σ axis.
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Answer: B. the shear stress is zero
Principal planes carry only normal (principal) stresses; shear stress vanishes on them.
66. At failure, the failure plane makes an angle with the major principal plane equal to
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For φ = 0 the plane is at 45°.
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Answer: C. 45° + φ/2
The Mohr circle touches the Mohr–Coulomb envelope at an angle 2θ = 90° + φ, so θ = 45° + φ/2.
67. The relationship between major and minor principal stresses at failure for a c–φ soil is
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Nφ = tan²(45°+φ/2) is the flow value.
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Answer: C. σ1 = σ3 tan²(45° + φ/2) + 2c tan(45° + φ/2)
Obtained from the geometry of the Mohr circle tangent to the strength envelope.
68. A drawback of the direct shear test compared with the triaxial test is that
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Think about the split box.
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Answer: D. the failure plane is forced to occur at a predetermined location
In a shear box the failure plane is fixed, drainage cannot be controlled and stress distribution is non-uniform.
69. In an unconsolidated undrained (UU) triaxial test on a fully saturated clay, the Mohr failure envelope in terms of total stress is
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Total stress strength does not rise with σ3.
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Answer: A. horizontal, with φu = 0
The strength is independent of the cell pressure, so the envelope is horizontal at τ = cu.
70. The vane shear test is most suitable for determining the in-situ
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Soft clay, undrained.
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Answer: A. undrained shear strength of soft clays
The vane is rotated to shear a cylindrical surface in soft saturated clay, giving cu.
71. The maximum shear stress at a point where the principal stresses are σ1 and σ3 acts on planes inclined to the major principal plane at
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Maximise sin 2θ.
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Answer: A. 45°
τ = (σ1−σ3)/2 sin 2θ is maximum when 2θ = 90°.
72. In a saturated soil, the pore pressure parameter B is equal to
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Water is incompressible relative to the skeleton.
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Answer: B. 1.0
For a fully saturated soil, an increase in cell pressure is carried entirely by the pore water, so B = 1.
73. When a dense sand is sheared under drained conditions, its volume
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Dense grains are interlocked.
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Answer: C. tends to increase (dilation)
Interlocked dense grains must ride over one another, producing dilation; loose sands contract.
74. In slope stability analysis, the factor of safety against sliding is defined as the ratio of
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Strength over demand.
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Answer: D. shear strength available along the surface to shear stress mobilised along it
FS = τf/τ along the potential slip surface; FS < 1 means failure.
75. The factor of safety of an infinite slope in dry cohesionless soil having angle of internal friction φ and slope angle β is
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The steepest stable slope equals φ.
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Answer: C. tan φ / tan β
Resisting force W cosβ tanφ over driving force W sinβ gives tanφ/tanβ; the slope is stable only if β < φ.
76. Taylor's stability number is defined as
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It is dimensionless.
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Answer: C. Sn = c / (Fc γ H)
Sn = c/(F γ H) where c is the mobilised cohesion requirement, γ unit weight and H slope height.
77. The Swedish slip circle method of slope stability analysis assumes the failure surface to be
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Name hints at the shape.
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Answer: A. a circular arc
Fellenius's Swedish circle method divides the sliding mass into slices above a circular arc.
78. A base failure of a slope generally occurs when
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Weak soil below toe.
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Answer: B. the slope is flat and the soil is soft clay overlying a firm stratum at depth
Flat slopes in cohesive soil with weak ground below, with small depth factor, tend to fail below the toe.
79. For the upstream slope of an earth dam, the most critical condition is generally
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Water level falls faster than pore pressure dissipates.
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Answer: D. sudden drawdown
Rapid lowering leaves high pore pressure in the shell with no stabilising water load.
80. For saturated clay slopes, the short-term (end of construction) stability is analysed using
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Quick loading, no time to drain.
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Answer: C. undrained strength (φu = 0 analysis)
Excess pore pressures have not dissipated, so total stress analysis with cu applies.
81. The major and minor principal stresses at a point are 300 kPa and 100 kPa. The normal stress on a plane inclined at 30° to the major principal plane is
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σn = (σ1+σ3)/2 + (σ1−σ3)/2 · cos 2θ.
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Answer: D. 250 kPa
σn = 200 + 100 cos 60° = 250 kPa.
82. For the same stresses (σ1 = 300 kPa, σ3 = 100 kPa), the shear stress on a plane at 30° to the major principal plane is
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τ = (σ1−σ3)/2 · sin 2θ.
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Answer: D. 86.6 kPa
τ = 100 sin 60° = 86.6 kPa.
83. A dry sand specimen in triaxial compression fails at σ3 = 100 kPa and σ1 = 400 kPa. The angle of internal friction is nearly
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sin φ = (σ1−σ3)/(σ1+σ3) for c = 0.
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Answer: B. 37°
sin φ = (400−100)/(400+100) = 0.6, φ = 36.9°.
84. A soil has c = 10 kPa and φ = 25°. The shear strength on a plane with effective normal stress 150 kPa is
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τf = c + σ tan φ.
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Answer: B. 79.9 kPa
τf = 10 + 150 tan 25° = 79.9 kPa.
85. A clay has c = 25 kPa and φ = 20°. In a triaxial test with σ3 = 100 kPa, the major principal stress at failure is
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Use tan(45° + φ/2) = tan 55°.
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Answer: A. 275 kPa
σ1 = 100 tan²55° + 2×25 tan 55° = 204 + 71 = 275 kPa.
86. An unconfined compression test on a saturated clay gives qu = 120 kPa. The undrained cohesion is
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Mohr circle radius equals cu.
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Answer: B. 60 kPa
For φu = 0, cu = qu/2 = 60 kPa.
87. In a direct shear test on a sand, a normal stress of 200 kPa produces a failure shear stress of 115 kPa. The friction angle is nearly
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tan φ = τ/σ for c = 0.
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Answer: B. 30°
φ = tan⁻¹(115/200) = 29.9°.
88. A vane of diameter 50 mm and height 100 mm requires a torque of 9.2 N·m to shear a soft clay. Assuming shear on both ends and the cylindrical surface, cu is nearly
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Torque is divided by a volume-like term.
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Answer: B. 20081 kPa
cu = T/[π d²(h/2 + d/6)] = 9.2/(π×0.0025×0.05833) = 20080.8 kPa.
89. A saturated clay (B = 1) in a triaxial test has A = 0.5. If the cell pressure increases by 50 kPa and the deviator stress by 60 kPa, the pore pressure change is
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Skempton's equation.
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Answer: C. 80 kPa
Δu = B[Δσ3 + A(Δσ1 − Δσ3)] = 50 + 0.5×60 = 80 kPa.
90. A soil with c' = 10 kPa and φ' = 30° has total normal stress 200 kPa and pore pressure 80 kPa on the failure plane. Its shear strength is nearly
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Use the effective normal stress.
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Answer: A. 79 kPa
σ' = 120 kPa; τf = 10 + 120 tan 30° = 79.3 kPa.
91. An infinite slope of dry sand with φ = 35° is inclined at 25° to the horizontal. The factor of safety is
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FS = tan φ / tan β.
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Answer: A. 1.50
FS = tan 35°/tan 25° = 0.700/0.466 = 1.50.
92. A slope of height 10 m in clay (γ = 18 kN/m³, c = 30 kPa) has a Taylor stability number 0.10. The factor of safety with respect to cohesion is nearly
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F = c/(Sn γ H).
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Answer: A. 1.67
F = c/(Sn γ H) = 30/(0.10×18×10) = 1.67.
93. A cohesionless infinite slope (φ = 30°, γsat = 20 kN/m³) of inclination 20° has seepage parallel to the surface with the water table at the ground surface. Taking γw = 9.81 kN/m³, the factor of safety is nearly
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Submerged unit weight appears in the numerator.
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Answer: D. 0.81
FS = (γ'/γsat) tan φ / tan β = 0.51×0.577/0.364 = 0.81.
2.4 Soil exploration, earth pressure and retaining structures
30 questions · ACiE0204
94. As per general practice (IS 1892), the depth of exploration below a foundation should be taken at least up to the depth where the increase in vertical stress due to the foundation is about
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Pressure bulb isobar.
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Answer: C. 10% of the foundation contact pressure
The significant depth is where Δσ falls to roughly 10% of the applied contact pressure (the 0.1q isobar).
95. Which method of boring is most suitable for obtaining continuous rock cores?
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Core means intact cylinder of rock.
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Answer: C. Rotary drilling with a diamond core bit
Rotary core drilling with a diamond or tungsten bit retrieves intact rock cores.
96. Undisturbed samples of soft clay for laboratory strength tests are best obtained using a
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Thin wall means low disturbance.
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Answer: D. thin-walled (Shelby) tube sampler
Thin-walled tubes with a low area ratio cause the least disturbance; the split spoon yields disturbed samples.
97. Disturbed soil samples collected from a borehole are normally adequate for determining
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Which tests do not need natural structure?
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Answer: A. grain size distribution and Atterberg limits
Index properties do not depend on the natural structure, so disturbed samples suffice.
98. The Standard Penetration Test uses a split spoon sampler driven by a hammer of weight
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Standard hammer is about 65 kg.
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Answer: D. 63.5 kg falling through 750 mm
IS 2131: 63.5 kg (140 lb) hammer, 750 mm free fall; N is the blow count for 300 mm penetration.
99. In the static cone penetration test (Dutch cone), the standard cone has an apex angle of
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Apex 60°.
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Answer: A. 60° and base area of 10 cm²
The Dutch mechanical cone is 60° apex with a 35.7 mm dia, 10 cm² base.
100. A plate load test may give misleading information about the bearing capacity and settlement of a footing resting on thick compressible clay because
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Size effect.
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Answer: A. the stressed zone below the small plate is much shallower than that of the actual footing
The plate's pressure bulb extends only about 2B_plate, missing deeper clay layers stressed by the large footing.
101. A site investigation report should typically include
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Think what the designer needs from the ground.
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Answer: A. borehole logs, soil profile, water table level, test results and foundation recommendations
The report records field and lab data and recommends type, depth and allowable pressure of foundation.
102. The coefficient of earth pressure at rest for a normally consolidated sand may be estimated by Jaky's formula as
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Last option is the active coefficient.
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Answer: B. K0 = 1 − sin φ
Jaky (1944): K0 ≈ 1 − sin φ'; the last option is Ka.
103. The correct order of the earth pressure coefficients for a given soil is
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Wall moving away vs towards the soil.
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Answer: C. Ka < K0 < Kp
Active pressure is the minimum, passive the maximum, with at-rest in between.
104. Which of the following is an assumption of Rankine's earth pressure theory?
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Contrast with Coulomb's theory.
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Answer: D. The wall is smooth and vertical, so no wall friction acts
Rankine's theory neglects wall friction and assumes a vertical frictionless wall; Coulomb's includes wall friction.
105. The magnitude of wall movement needed to mobilise full passive resistance compared with that for active pressure is
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Pushing into soil takes more movement.
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Answer: A. much greater
Passive state needs a much larger wall displacement (about 2–4% of height in sand) than the active state (0.1–0.5%).
106. The presence of a water table behind a retaining wall, if not drained, leads to
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Water pressure adds up.
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Answer: C. a large increase in lateral thrust on the wall
Hydrostatic pressure adds to the reduced soil pressure (using γ'), giving a larger net thrust.
107. To avoid tension at the base, the resultant of forces on the base of a gravity retaining wall of width B should lie within
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e ≤ B/6.
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Answer: B. the middle third of the base
Eccentricity e ≤ B/6 gives compressive pressure throughout the base.
108. The minimum factor of safety generally adopted for sliding of a retaining wall is
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Typical value for rigid walls.
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Answer: B. 1.5
Codes (IS 14458, IRC 78) require FS ≥ 1.5 against sliding (and 2.0 against overturning).
109. Which of the following is a method to increase the stability of a retaining wall against sliding?
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Increase resistance at the base.
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Answer: D. Providing a shear key under the base slab
A base key mobilises passive resistance in the soil beneath the base.
110. Weep holes and a filter drain are provided behind a retaining wall to
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Remove the water.
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Answer: C. relieve hydrostatic pressure from the backfill
Drainage removes water, preventing pore water thrust and seepage forces on the wall.
111. A reinforced earth wall resists earth pressure mainly by
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Friction with strips.
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Answer: D. friction between the backfill and tensile strips or geosynthetic layers
Tensile reinforcement interacts with the backfill by friction, creating a composite gravity-like mass.
112. For a backfill with φ = 30°, the Rankine active and passive earth pressure coefficients are respectively
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Kp is the reciprocal of Ka.
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Answer: C. 0.333 and 3.0
Ka = (1−sin30°)/(1+sin30°) = 1/3 and Kp = 1/Ka = 3.
113. A smooth vertical wall 6 m high retains dry cohesionless backfill (γ = 18 kN/m³, φ = 30°) with a horizontal surface. The Rankine active thrust per metre length is
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Pa = ½ γ H² Ka.
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Answer: D. 108 kN/m
Pa = ½ γ H² Ka = 0.5×18×36×(1/3) = 108 kN/m, acting at H/3 = 2 m above the base.
114. A smooth vertical wall 3 m high retains backfill of γ = 17 kN/m³ and φ = 30°. The Rankine passive resistance per metre length is
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Pp = ½ γ H² Kp with Kp = 3.
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Answer: A. 229.5 kN/m
Pp = ½ γ H² Kp = 0.5×17×9×3 = 229.5 kN/m.
115. A cohesive backfill with c = 20 kPa, φ = 0 and γ = 18 kN/m³ is retained by a vertical smooth wall. The depth of the tension crack is
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zc = 2c/(γ √Ka).
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Answer: B. 2.22 m
zc = 2c/(γ√Ka) = 2×20/18 = 2.22 m since Ka = 1.
116. A cohesive backfill (c = 20 kPa, φ = 0, γ = 18 kN/m³) is 5 m deep behind a smooth wall. The active pressure at the base is
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pa = γHKa − 2c√Ka.
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Answer: D. 50 kPa
pa = γ H Ka − 2c√Ka = 90 − 40 = 50 kPa.
117. A uniform surcharge of 20 kPa acts on the horizontal surface of a cohesionless backfill (φ = 30°) behind a 5 m high smooth wall. The additional thrust due to the surcharge is
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Surcharge pressure is uniform with depth.
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Answer: B. 33.3 kN/m
ΔP = q Ka H = 20×(1/3)×5 = 33.3 kN/m, acting at H/2.
118. A gravity wall weighing 300 kN/m acts at 1.5 m from the toe. The active thrust is 108 kN/m at 2.0 m above the base. Neglecting passive pressure, the factor of safety against overturning about the toe is
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Resisting moment / overturning moment.
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Answer: B. 2.08
FS = Mr/Mo = (300×1.5)/(108×2.0) = 450/216 = 2.08.
119. A wall weighing 250 kN/m rests on soil with base friction coefficient 0.55. If the horizontal active thrust is 108 kN/m, the factor of safety against sliding (neglecting passive resistance) is
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FS = μW/ΣH.
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Answer: B. 1.27
FS = μW/Pa = 0.55×250/108 = 1.27.
120. A retaining wall base of width 3.0 m has the resultant of all forces cutting the base at 1.2 m from the toe. The eccentricity of the resultant is
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e = B/2 − x̄.
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Answer: B. 0.30 m
e = B/2 − x = 1.5 − 1.2 = 0.30 m, which is less than B/6 = 0.50 m.
121. In an SPT in fine saturated sand below the water table, the observed N is 35. After the dilatancy correction, N is
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Correct only the excess over 15.
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Answer: A. 25
N' = 15 + ½(N − 15) = 15 + 10 = 25 for N > 15 (Terzaghi–Peck).
122. A sampling tube has inside diameter 70 mm (cutting edge) and outside diameter 76 mm. Its area ratio is
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Ar = (D2² − D1²)/D1².
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Answer: C. 17.9%
Ar = (D2² − D1²)/D1² = (5776−4900)/4900 = 17.9%.
123. The Rankine coefficient of earth pressure at rest for a sand with φ = 35°, estimated by Jaky's formula, is nearly
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K0 = 1 − sin φ.
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Answer: A. 0.43
K0 = 1 − sin35° = 0.43.
2.5 Fundamentals of foundation
30 questions · ACiE0205
124. A foundation is best defined as
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It is below the superstructure.
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Answer: B. the lowest part of a structure that transmits the load to the underlying soil or rock
The foundation transfers superstructure loads safely to the ground without excessive settlement or shear failure.
125. According to Terzaghi, a foundation is classified as shallow when the ratio of depth of foundation to its width is
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Depth not exceeding width.
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Answer: A. less than or equal to 1
Terzaghi: Df/B ≤ 1 for shallow foundations; larger ratios behave as deep foundations.
126. Which of the following is a deep foundation?
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Think of long slender members.
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Answer: D. Pile foundation
Piles, piers and well (caisson) foundations transfer loads to deeper strata; footings and rafts are shallow.
127. A combined footing is generally provided when
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Overlapping bases.
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Answer: C. two or more columns are so close that their individual footings overlap
Overlapping isolated footings are merged into a single combined footing to carry the columns.
128. A strap (cantilever) footing consists of
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Two footings and a connecting beam.
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Answer: C. two isolated footings connected by a beam
The strap beam transfers the moment from the eccentric exterior footing to the interior footing, useful near property lines.
129. A mat (raft) foundation is generally preferred when
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Large coverage of the plan.
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Answer: C. the footings would cover more than about half of the building plan area
When individual footings would cover over about 50% of the plan, a mat is more economical and reduces differential settlement.
130. A floating (compensated) foundation is one in which
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Net load is zero.
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Answer: D. the weight of excavated soil equals the weight of the building, so net pressure on the soil is nearly zero
Excavating soil equal to the building load removes in-situ stress equal to the added load, so settlement is small.
131. The main functions of a foundation include all of the following EXCEPT
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Which is not a function?
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Answer: D. increasing the self weight of the superstructure
A foundation need not increase the superstructure weight; the others are real functions.
132. Which of the following is the primary factor influencing the choice of foundation type?
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Think engineering, not aesthetics.
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Answer: C. Nature of the soil and magnitude of structural loads
Soil strength/compressibility, load, water table and nearby structures decide the type.
133. Pile foundations that derive most of their capacity from the shaft resistance along the sides are called
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Skin resistance.
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Answer: B. friction piles
Friction (floating) piles transfer load mainly through skin friction in deep soft soils.
134. Pile foundations are particularly useful when
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Weak top, strong bottom.
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Answer: B. the upper soil layers are weak and compressible while a firm stratum lies deeper
Piles bypass weak surface layers to reach stronger strata or mobilise skin friction.
135. For buildings on expansive (black cotton) soils in India, a commonly used foundation is
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Below the seasonally active zone.
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Answer: D. under-reamed piles
Bulbs in the pile anchor the structure below the active zone against swelling and shrinkage movements.
136. The depth of foundation in cohesive soils is generally kept below the zone of
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Soil moisture changes with seasons.
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Answer: A. seasonal moisture variation and organic top soil
Seasonal swell–shrink and organic topsoil are unsuitable to found upon.
137. The net safe bearing capacity is obtained from the net ultimate bearing capacity by
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Safe means reduced by a factor.
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Answer: A. dividing by a factor of safety
qns = qnu/F, where qnu = qu − γDf.
138. Foundation contact pressure on a rigid footing resting on cohesionless soil is usually
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Sand loses confinement at edges.
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Answer: A. greater at the centre than at the edges
In sand the edges can deform laterally under low confinement, so the pressure is concentrated at the centre; in clay it is higher at the edges.
139. A cellular raft is one that
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Cells mean voids.
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Answer: B. has box-like voids formed by slabs and walls, providing great stiffness
Cellular rafts form a rigid box, useful for heavy loads and differential settlement control; basements may be built in the voids.
140. Which statement about a mat foundation is correct?
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A raft is a shallow foundation.
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Answer: B. It reduces differential settlements by acting as a stiff plate over a wide area
A mat spreads column loads over the entire plan and bridges local soil variations.
141. Test pits are suitable for exploring subsoil only up to depth of about
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Hand excavation limit.
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Answer: D. 3 m, above the water table
Open pits are practicable for shallow depths (about 3 m) and allow visual inspection and undisturbed block sampling.
142. A column carries a service load of 900 kN. If the safe bearing capacity is 150 kPa, the side of a square footing required (ignoring footing weight) is nearly
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A = P/qs, then B = √A.
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Answer: C. 2.45 m
A = 900/150 = 6 m²; B = √6 = 2.45 m.
143. A footing at 1.5 m depth in soil of γ = 18 kN/m³ has an ultimate bearing capacity of 600 kPa. With a factor of safety of 3 on net capacity, the safe bearing capacity is
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qs = qnu/F + γDf.
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Answer: D. 218 kPa
qnu = 600 − 27 = 573; qns = 191; qs = qns + γDf = 191 + 27 = 218 kPa.
144. A 2 m × 2 m square footing carries a vertical load of 500 kN at an eccentricity of 0.2 m along one axis. The maximum contact pressure is
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qmax = (P/A)(1 + 6e/B).
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Answer: C. 200 kPa
qmax = (P/BL)(1 + 6e/B) = 125×1.6 = 200 kPa.
145. Using Rankine's formula, the minimum depth of foundation for a safe pressure of 120 kPa in soil of γ = 18 kN/m³ and φ = 30° is nearly
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Df = (q/γ) Ka².
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Answer: B. 0.74 m
Df = (q/γ)[(1−sinφ)/(1+sinφ)]² = 6.67×(1/3)² = 0.74 m.
146. A fully compensated (floating) raft is needed for a building that imposes a uniform gross pressure of 100 kPa on a soil of unit weight 18 kN/m³. The depth of excavation required is nearly
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Excavated weight equals applied load.
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Answer: A. 5.56 m
Df = q/γ = 100/18 = 5.56 m.
147. The four columns of a building each carry 1000 kN on an 8 m × 8 m raft. Neglecting self weight, the average contact pressure is
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Total load / raft area.
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Answer: B. 62.5 kPa
q = 4000/64 = 62.5 kPa.
148. A load-bearing wall transmits 150 kN per metre run to a strip footing. If the safe bearing capacity is 100 kPa, the required width of footing is
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B = Q/qs per metre.
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Answer: B. 1.5 m
B = Q/qs = 150/100 = 1.5 m for a 1 m length.
149. A bored pile of diameter 0.4 m and length 12 m in clay has an average adhesion of 40 kPa along the shaft. Neglecting end bearing and taking a factor of safety of 2.5, the safe load is nearly
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Shaft area = π d L.
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Answer: A. 241 kN
Qu = π d L α c = π×0.4×12×40 = 603 kN; Qs = 241 kN.
150. The total area of the building is 300 m². Individual footings would cover 160 m². The percentage of area covered is nearly
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Coverage = footing area / plan area.
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Answer: A. 53%
160/300 = 53% > 50%, so a raft is generally preferred.
151. The gross bearing pressure under a raft at depth 2 m is 140 kPa. If the soil has γ = 18 kN/m³, the net pressure causing settlement is
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Subtract the overburden pressure removed.
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Answer: A. 104 kPa
qnet = q − γDf = 140 − 36 = 104 kPa.
152. Two columns 4 m apart carry 600 kN and 900 kN. The resultant load acts at what distance from the 600 kN column?
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Take moments about the 600 kN column.
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Answer: D. 2.4 m
x = 900×4/1500 = 2.4 m, so the combined footing is made to be centred about this point.
153. A 2 m wide square footing carries 800 kN. Using the 2:1 load distribution method, the average vertical stress at 2 m below the footing base is
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Area at depth z is (B + z)².
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Answer: C. 50 kPa
The loaded area grows by z on each side: (B+z)² = 16 m²; Δσ = 800/16 = 50 kPa.
2.6 Bearing capacity and foundation settlements
30 questions · ACiE0206
154. The ultimate bearing capacity of a foundation is defined as
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It relates to shear failure, not settlement.
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Answer: C. the minimum gross pressure at the base at which the soil fails in shear
qu is the gross intensity of loading at which shear failure of the supporting soil occurs.
155. The net allowable bearing pressure of a foundation is governed by
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Two criteria, take the governing one.
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Answer: A. the lesser of the pressure based on shear failure and the pressure based on permissible settlement
Design must satisfy both safety against shear failure and tolerable settlement; the smaller controls.
156. General shear failure of a foundation is most likely in
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Strong soil, defined failure planes.
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Answer: B. dense sand or stiff clay
Dense/stiff soils show a clear failure surface with heave and sudden collapse; loose soils fail by punching or local shear.
157. Punching shear failure of a foundation occurs mainly in
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Foundation punches through.
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Answer: D. loose sand or very compressible soils
In compressible soils the footing sinks with little heave and no well-defined failure surface.
158. Which of the following is an assumption of Terzaghi's bearing capacity theory?
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Overburden acts as a uniform surcharge.
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Answer: A. The base of the footing is rough and the shear strength of soil above the base is neglected, replaced by a surcharge
Terzaghi: strip footing, rough base, Df ≤ B, soil above the base replaced by a surcharge γDf.
159. Terzaghi's equation for ultimate bearing capacity of a strip footing is
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Self weight term has a factor of ½.
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Answer: A. qu = cNc + γDf Nq + ½γB Nγ
The three terms represent cohesion, surcharge and self weight (width) contributions; the third option is for square footings.
160. For a square footing, Terzaghi's bearing capacity equation is
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Compare with the circular shape factors.
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Answer: D. qu = 1.3cNc + γDf Nq + 0.4γB Nγ
Shape factors are 1.3 and 0.4 for a square footing; 1.3 and 0.3 for a circular one.
161. For local shear failure, Terzaghi suggests using the reduced strength parameters
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Two-thirds.
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Answer: B. c' = (2/3)c and tan φ' = (2/3) tan φ
Reduced strength simulates the progressive failure of loose or soft soils.
162. For a footing on saturated clay (φ = 0) at the surface, Terzaghi's bearing capacity factors are
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Nγ vanishes when φ = 0.
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Answer: B. Nc = 5.7, Nq = 1, Nγ = 0
For φ = 0, Nq = 1 and Nγ = 0, giving qu = 5.7c + γDf for a strip.
163. If the water table rises from well below the foundation level to the ground surface, the ultimate bearing capacity of a footing in sand
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Both the Nq and Nγ terms become smaller.
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Answer: B. reduces to about half
Submerged unit weight γ' ≈ γsat − γw ≈ ½γ lowers both the surcharge and self weight terms.
164. In a purely cohesive (φ = 0) soil, increasing the width of a footing placed at the surface
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Which term contains B?
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Answer: D. does not change the ultimate bearing capacity
Nγ = 0 for φ = 0, so qu = cNc does not depend on width (though settlement increases).
165. Consolidation of a saturated clay is the process of
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It is time dependent.
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Answer: A. gradual volume decrease with time due to expulsion of pore water under sustained load
Excess pore pressure dissipates and effective stress rises, compressing the soil skeleton with time.
166. Terzaghi's one-dimensional consolidation theory assumes
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One-dimensional flow only.
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Answer: B. the soil is homogeneous and saturated, with Darcy's law valid and drainage only in the vertical direction
The 1-D theory assumes saturated homogeneous soil, incompressible grains and water, and constant k and mv.
167. In a consolidating clay layer of thickness H drained at the top only (resting on an impermeable base), the drainage path length is
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Water leaves by one face only.
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Answer: D. H
For single drainage Hdr = H; for double drainage Hdr = H/2.
168. The three components of total settlement of a foundation on clay are
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Think of time sequence.
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Answer: D. immediate, primary consolidation and secondary compression settlement
Si + Sc + Ss; Sc usually dominates in soft clay, Ss in organic soils.
169. Secondary compression of a clay occurs due to
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Occurs after pore pressure dissipation.
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Answer: C. plastic readjustment of soil particles at constant effective stress after primary consolidation
It is creep of the soil skeleton, significant for organic soils and highly plastic clays.
170. Differential settlement of a structure is more critical than uniform settlement because it
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Think about relative movement.
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Answer: C. produces tilting, cracking and distress in members
Unequal settlement induces angular distortion and extra bending in the superstructure.
171. A strip footing is founded at the surface of a saturated clay with cu = 40 kPa. Using Terzaghi's theory, the ultimate bearing capacity is
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qu = 5.7c for a strip.
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Answer: A. 228 kPa
qu = 5.7c = 5.7×40 = 228 kPa for φ = 0 and Df = 0.
172. A square footing is placed at the surface of a saturated clay with cu = 30 kPa. Using Terzaghi's theory, qu is
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Apply the shape factor 1.3.
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Answer: A. 222.3 kPa
qu = 1.3 c Nc = 1.3×5.7×30 = 222.3 kPa.
173. A 2 m wide strip footing at 1 m depth rests on cohesionless soil (γ = 18 kN/m³) with Terzaghi factors Nq = 22.5, Nγ = 19.7 (φ = 30°). The ultimate bearing capacity is
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Use qu = γDf Nq + ½γBNγ.
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Answer: C. 760 kPa
qu = γDf Nq + ½γBNγ = 405 + 354.6 = 760 kPa.
174. A 2 m × 2 m square footing at 1.5 m depth in sand (γ = 18 kN/m³, φ = 30°, Nq = 22.5, Nγ = 19.7) has an ultimate bearing capacity of
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Square: 0.4 γ B Nγ.
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Answer: B. 891 kPa
qu = γDf Nq + 0.4γBNγ = 607.5 + 283.7 = 891 kPa.
175. For the strip footing (B = 2 m, Df = 1 m, φ = 30°, Nq = 22.5, Nγ = 19.7) in sand with γsat = 20 kN/m³ and the water table at ground level, qu (γw = 9.81 kN/m³) is nearly
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Replace γ by the submerged unit weight.
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Answer: B. 430 kPa
γ' = 10.19 kN/m³; qu = 10.19×(22.5 + 19.7) = 430 kPa.
176. A soil has φ = 30°. For local shear failure the reduced friction angle is nearly
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Reduce tan φ, not φ.
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Answer: A. 21°
tan φ' = (2/3) tan 30° = 0.385, φ' = 21.1°.
177. A 4 m thick normally consolidated clay (e0 = 0.90, Cc = 0.30) has mean effective overburden 100 kPa. If the stress increases by 100 kPa, the primary consolidation settlement is nearly
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Use log10 of the stress ratio.
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Answer: D. 190 mm
Sc = Cc H/(1+e0) log10[(σ0 + Δσ)/σ0] = 0.3×4/1.9×log 2 = 190 mm.
178. A 5 m thick clay layer has coefficient of volume compressibility mv = 2×10⁻⁴ m²/kN. A uniform stress increase of 80 kPa causes a settlement of
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Sc = mv·Δσ·H.
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Answer: B. 80 mm
Sc = mv Δσ H = 2×10⁻⁴ × 80 × 5 = 0.08 m.
179. A clay layer 3 m thick has initial void ratio 1.2, which reduces to 1.1 under load. The consolidation settlement is nearly
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S = HΔe/(1+e0).
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Answer: D. 136 mm
S = H Δe/(1+e0) = 3×0.1/2.2 = 136 mm.
180. A 5 m thick clay layer drained at top and bottom has cv = 1×10⁻³ cm²/s. Taking Tv = 0.197 for 50% consolidation, the time required is nearly
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t = Tv Hdr²/cv with Hdr = H/2.
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Answer: C. 143 days
Hdr = 2.5 m = 250 cm; t = Tv Hdr²/cv = 0.197×62500/10⁻³ = 1.23×10⁷ s = 143 days.
181. A 6 m clay layer drained only at the top takes 8 years to reach 50% consolidation. If a sand layer also exists at its bottom (double drainage), the time for the same degree is
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Time is proportional to path length squared.
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Answer: A. 2 years
t ∝ Hdr²; halving the drainage path reduces time to one quarter: 8/4 = 2 years.
182. A plate load test with a 0.3 m square plate on sand gives 15 mm settlement at the design pressure. The settlement of a 2 m square footing at the same pressure by Terzaghi–Peck's relation is nearly
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Settlement in sand is less than linear with width.
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Answer: C. 45 mm
Sf = Sp[Bf(Bp+0.3)/(Bp(Bf+0.3))]² = 15×(1.2/0.69)² = 45 mm.
183. A 5 m thick clay (e0 = 1.0) has Cα = 0.02. Secondary settlement between 1 year and 10 years after primary consolidation is
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Log of the time ratio is 1.
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Answer: C. 50 mm
Ss = Cα H/(1+e0) log(t2/t1) = 0.02×5/2×log 10 = 0.05 m.