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Geotechnical Engineering Questions and Answers by Topic

  • 7 topics
  • 16 questions
  • 16 from past papers
  • asked in 7 exams

Asked in these exams

ExamQuestionsYears
MH SET 6 2023-2025
SET Exam 6 2023-2025
RSMSSB 5 2025
RSMSSB JE 5 2025
DSSSB JE 4 2023
RSMSSB JE Agriculture 2 2025
GATE AG 1 2016

Recent past-paper questions

MH SET 2025 and 1 more

A building is to be founded on a clay stratum. The water table is practically at ground surface level. Laboratory test show that the unit weight of the soil is r = 20 kN/m3 and classify the soil as a firm clay. The uniaxial compressive strength is qu = 150 kN/m2.

Suppose that the footing will be square, 2 m wide, and the foundation depth will be 2 m below the ground surface, calculate the ultimate bearing capacity.

  1. A. 502.6 kPa
  2. B. 743.2 kPa
  3. C. 801.6 kPa
  4. D. 304.0 kPa
Answer and explanation

Answer: A

The ultimate bearing capacity is 502.6 kPa.

For a saturated clay loaded quickly, the undrained (\(\phi = 0\)) analysis applies. The undrained cohesion is half the unconfined compressive strength, \(c_{u} = 150/2 = 75\) kPa. With \(N_{c} = 5.14\), a shape factor of 1.2 for a square footing and the overburden \(\gamma D\) (\(N_{q} = 1\)):

\[ \begin{aligned} q_{u} &= 1.2\, c_{u} N_{c} + \gamma D \\ &= 1.2(75)(5.14) + 20(2) \\ &= 462.6 + 40 = 502.6\ \text{kPa} \end{aligned} \]

Because the analysis is in total stress, the water table at the surface does not alter it. A common slip is to use \(q_{u}\) itself as the cohesion, which roughly doubles the first term; the other options do not follow from this formula.

More Bearing Capacity of Soils questions

RSMSSB 2025 and 1 more

The size of bearing plate, which is used in plate load test varies from :

  1. A. 200 to 275 mm
  2. B. 300 to 750 mm
  3. C. 25 to 100 mm
  4. D. 100 to 200 mm
Answer and explanation

Answer: B

The bearing plate size varies from 300 to 750 mm.

IS 1888, the code for the plate load test, uses square steel plates of 300 mm to 750 mm side (or circular plates of equivalent area), at least 25 mm thick. A 300 mm plate is common for granular soils and larger plates for clayey soils.

Plates of 25 to 275 mm are too small: a small plate stresses only a shallow zone of soil and gives results that do not scale reliably to a real footing.

More Bearing Capacity of Soils questions

RSMSSB 2025 and 1 more

In standard penetration test, correction due to dilatancy/ submergence is given by:
(Symbols have their usual meanings)

  1. A. \(N_{e} = 15 + (N_{0} - 15) \times \frac{1}{2}\)
  2. B. \(N_{e} = 20 + (N_{0} - 18)\)
  3. C. \(N_{e} = 25 + (N_{0} - 16)\)
  4. D. \(N_{e} = 30 + (N_{0} - 12)\)
Answer and explanation

Answer: A

The dilatancy correction is \(N_{e} = 15 + (N_{0} - 15) \times \frac{1}{2}\).

In very fine or silty sands below the water table, rapid driving of the sampler causes dilatancy and a temporary build-up of pore pressure, which raises the observed blow count. Terzaghi and Peck recommended that, when the observed value \(N_0\) exceeds 15, the corrected value be taken as:

\[ N_e = 15 + \frac{1}{2}(N_0 - 15) \]

So only half the excess over 15 is counted. The other formulas, with base values of 20, 25 or 30, are not the standard correction.

More Bearing Capacity of Soils questions

RSMSSB 2025 and 2 more

Coulomb's theory of earth pressure is based on ________.

  1. A. The theory of elasticity
  2. B. The theory of plasticity
  3. C. Wedge theory
  4. D. Empirical rules
Answer and explanation

Answer: C

Coulomb's theory is based on the Wedge theory.

Coulomb (1776) treated the soil behind a retaining wall as a rigid wedge that slides on a plane failure surface. By considering the equilibrium of this wedge, including wall friction, and finding the most critical plane, he obtained the active or passive earth pressure.

Rankine's theory is the one based on the plastic equilibrium of the soil mass, not elasticity. Coulomb's method is a rational analysis, not a set of empirical rules.

More Earth Pressure Theories questions

RSMSSB 2025 and 1 more

In a compaction test if the compacting effort is increased, it will result in :

  1. A. no change in maximum dry density but decrease in OMC.
  2. B. increase in maximum dry density and OMC.
  3. C. increase in maximum dry density but OMC remains unchanged.
  4. D. increase in maximum dry density and decrease in OMC.
Answer and explanation

Answer: D

Greater compactive effort gives an increase in maximum dry density and decrease in OMC.

Heavier compaction, as in the modified Proctor test compared with the standard test, pushes the particles closer together, so the soil reaches a higher maximum dry density. Because the extra energy does part of the work that water does as a lubricant, that peak is reached at a lower water content. The compaction curve moves upward and to the left, along the line of optimums.

So the maximum dry density does change, ruling out option A, and the optimum moisture content falls, ruling out B and C.

More Compaction of Soils questions

Difficulty

  • Easy 3
  • Medium 7
  • Hard 4
  • Very Hard 2

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