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Foundation Engineering

Bearing Capacity - Terzaghi Theory

The ultimate ...

Question

The ultimate bearing capacity of a soil is 300 kN/ $m^{2}$ . The depth of foundation is 1 m and unit weight of soil is 20 kN/ $m^{3}$ . Choosing a factor of safety of 2.5, the net safe bearing capacity is

100 kN/

m^{2}

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112 kN/

m^{3}

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80 kN/

m^{3}

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100.5 kN/

m^{3}

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Solution

The correct option is B 112 kN/ $m^{3}$

$γ = 20 k N / m^{3}$
FOS = 2.5
$q_{u} = 300 k N / m^{2}$

Net safe bearing capacity

$q_{n s} = \frac{q_{n u}}{F O S} = \frac{q_{u} - γ D_{f}}{F O S}$

$= \frac{300 - 20 \times 1}{2.5} = 112 k N / m^{2}$

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Similar questions

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The ultimate bearing capacity of a soil is 300 kN/m2. The depth of foundation is 1 m and unit weight of soil is 20 kN/m3. Choosing a factor of safety of 2.5, the net safe bearing capacity is

The correct option is B 112 kN/m3 γ=20 kN/m3 FOS = 2.5 qu=300 kN/m2 Net safe bearing capacity qns=qnuFOS=qu−γDfFOS =300−20×12.5=112 kN/m2

The ultimate bearing capacity of a soil is 300 kN/ $m^{2}$ . The depth of foundation is 1 m and unit weight of soil is 20 kN/ $m^{3}$ . Choosing a factor of safety of 2.5, the net safe bearing capacity is

The correct option is B 112 kN/ $m^{3}$

$γ = 20 k N / m^{3}$
FOS = 2.5
$q_{u} = 300 k N / m^{2}$

Net safe bearing capacity

$q_{n s} = \frac{q_{n u}}{F O S} = \frac{q_{u} - γ D_{f}}{F O S}$

$= \frac{300 - 20 \times 1}{2.5} = 112 k N / m^{2}$