Bearing Capacity of Soils Questions and Answers

Answer: B

The figure shows a compaction pile: a pile driven into loose granular soil to compact it. The horizontal arrows point outward from the pile along its whole length, showing the soil being pushed sideways and densified, which raises its bearing capacity. An end-bearing pile would show load passing to a firm stratum at the tip, a friction pile would show shear resistance along the shaft carrying the load, and a tension pile (uplift pile) resists upward pull. The arrows here do not show those actions.

Answer: A

The false statement is that a pile foundation is preferred when there are small fluctuations in the sub-soil water level. Piles are preferred when the water level fluctuates a great deal, because shallow footings would be disturbed by the changing moisture, so it is large fluctuations that favour piles. The other three are genuine conditions for choosing piles: an expansive top soil, a heavy superstructure load that must be carried to deeper, stronger strata or spread over a wider area, and a weak top soil of poor bearing capacity that cannot support the structure.

Answer: B

Meyerhof's theory takes into account the shear strength of the soil above the base of the footing, treating the soil there as a surcharge which supplies shear resistance along an extended failure surface. Terzaghi's theory replaces the soil above the footing base by a surcharge of equal weight, ignoring that shear strength. Vesic's and Hansen's methods are later modifications that add shape, depth and inclination factors to the general equation, and the shear above the base is not the distinguishing feature of those.

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.

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.

Answer: B

The key gives 3313 kPa.

Above the water table the sand is dry, so the overburden at the base of the footing is q = 20 × 2 = 40 kPa. For φ' = 35° the bearing capacity factors are N_q ≈ 33.3 and N_γ ≈ 48. For a 4 m by 8 m rectangle (B/L = 0.5) the shape factors are about 1 + 0.5 tan 35° = 1.35 for the overburden term and 1 - 0.4 × 0.5 = 0.8 for the self-weight term:

\[ \begin{aligned} q_{u} &= q N_{q} s_{q} + 0.5\gamma B N_{\gamma} s_{\gamma} \\ &= 40(33.3)(1.35) + 0.5(20)(4)(48)(0.8) \\ &\approx 1798 + 1536 \approx 3330\ \text{kPa} \end{aligned} \]

That agrees with the keyed 3313 kPa to within the rounding of the factors, which vary slightly between textbooks. 1105 kPa is roughly a third of the true value, and 5002 and 8017 kPa are far too high for a footing only 2 m deep.

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.