This question was previously asked in
DSSSB JE CE 2019 Official Paper Shift 1 (Held on 23 Oct 2019)
- Doubles
- Remains same
- Reduced to one third
- Reduces to half
Answer (Detailed Solution Below)
Option 4 : Reduces to half
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CT 1: Indian History
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Detailed Solution
Download Solution PDFExplanation:
Ultimate bearing capacity for shallow foundation is given by
qu= CNc+γDfNq+ 0.5BγNγ
Where, C = unit cohesion of soil, Df = Depth of foundation
γ = Unit weight of soil
Nc, Nq, Nγ= Bearing capacity factor which depends upon friction angle
For sand, C = 0, then for normal condition
qu1=γDfNq+ 0.5BγNγ
When the water table rises to the ground level then the soil becomes in submerged condition thenγsubused in the place ofγ
γsub= 0.5×γ
qu2=γsubDfNq+ 0.5BγsubNγ
qu2= 0.5× (γDfNq+ 0.5BγNγ)
qu2= 0.5× qu1
Hence, thewater table rises to the ground level of a footing resting on cohesionless soils, the bearing capacity approximatelyreduces to half.
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