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Shalnov [3]
3 years ago
11

A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla

ced 0.5 m into a soil with a cohesion of 25 kPa and an angle of internal friction of 5°. Most nearly, what is the value of the Terzaghi bearing capacity factor, N?(A) 0.5(B) 1.1(C) 1.6(D) 7.3
Engineering
1 answer:
creativ13 [48]3 years ago
7 0

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed =  0.5 m

cohesion = 25 kPa

internal friction angle =  5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for  a pure cohesive soil

N_{\gamma } = 0

and we know formula for N_{\gamma } is

N_{\gamma } = (Nq - 1 ) × tan(Ф)   ..................1

so here Ф is very less  N_{\gamma } should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

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Answer:

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Given:

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Density of water 1000 kg/m³.

Calculation:

Step1

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S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

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Attached below is the remaining  part of the detailed solution

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