An infinite soil slope with an inclination of 35- is subjected to seepage parallel to its surface- The soil has C- - 100 kN-m2 and - - - 30- Using the concept of mobilized cohesion and friction- at a factor of safety of 1-5 with respect to shear strength- the mobilized friction angle is

Factor of safety = Resisting shear forceDriving component of weight of soil F = C'+σ' _ntanϕ 'τ τ = C'F+σ '_ntanϕ 'F C_m = C'FOS' ; tanϕ'_m=tanϕ 'FOS τ =C_ ...

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Foundation Engineering

Infinite Slopes - Cohesive Soils

An infinite s...

Question

An infinite soil slope with an inclination of $35^{\circ}$ is subjected to seepage parallel to its surface. The soil has C' = $100 k N / m^{2}$ and $ϕ^{'} = 30^{\circ}$ . Using the concept of mobilized cohesion and friction, at a factor of safety of 1.5 with respect to shear strength, the mobilized friction angle is

{20.02}^{\circ}

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{21.05}^{\circ}

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{23.33}^{\circ}

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{30.0}^{\circ}

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Solution

The correct option is B ${21.05}^{\circ}$
$Factor of safety = \frac{Resisting shear force}{Driving component of weight of soil}$

$F = \frac{C^{'} + σ_{n}^{'} t a n ϕ^{'}}{τ}$

$τ = \frac{C^{'}}{F} + σ_{n}^{'} \frac{t a n ϕ^{'}}{F}$

$C_{m} = \frac{C^{'}}{F O S^{'}}; t a n ϕ_{m}^{'} = \frac{t a n ϕ^{'}}{F O S}$

$τ = C_{m} + σ_{n}^{'} t a n ϕ_{m}^{'}$

where,

$t a n ϕ_{m}^{'} = t a n^{- 1} (\frac{t a n ϕ^{'}}{F})$

$= t a n^{- 1} (\frac{t a n 30^{\circ}}{1.5})$

$Mobilised friction angle = {21.05}^{\circ}$

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An infinite soil slope with an inclination of 35∘ is subjected to seepage parallel to its surface. The soil has C' = 100kN/m2 and ϕ′=30∘. Using the concept of mobilized cohesion and friction, at a factor of safety of 1.5 with respect to shear strength, the mobilized friction angle is