A cut slope is to be made in a soft saturated clay with its sides rising at an angle of 60° to the horizontal (Figure 15.17).
Given: $c_u$ = 40 kN/m² and γ = 17.5 kN/m³.
a. Determine the maximum depth up to which the excavation can be carried out.
b. Find the radius, r, of the critical circle when the factor of safety is equal to 1 (Part a).
c. Find the distance $\bar{BC}$ .

Question

A cut slope is to be made in a soft saturated clay with its sides rising at an angle of 60° to the horizontal (Figure 15.17).
Given: [latex]c_u[/latex] = 40 kN/m² and γ = 17.5 kN/m³.
a. Determine the maximum depth up to which the excavation can be carried out.
b. Find the radius, r, of the critical circle when the factor of safety is equal to 1 (Part a).
c. Find the distance [latex]\bar{BC}[/latex].

Accepted Answer

Part a
Since the slope angle β = 60° > 53°, the critical circle is a toe circle. From Figure 15.13, for β = 60° , the stability number = 0.195.
[latex]H_{cr}=\frac{c_u}{\gamma m}=\frac{40}{17.5 imes 0.195}=[/latex] 11.72 m
Part b
From Figure 15.17,
[latex]r=\frac{\bar{DC}}{\sin \frac{ heta}{2}}[/latex]
But
[latex]\bar{DC}=\frac{\bar{AC}}{2}=\frac{\left(\frac{H_{cr}}{\sin \alpha} ight)}{2}[/latex]
so,
[latex]r=\frac{H_{cr}}{2\sin \alpha \sin \frac{ heta}{2}}[/latex]
From Figure 15.14, for β = 60°, α = 35° and θ = 72.5°. Substituting these values into the equation for r we get
[latex]r=\frac{H_{cr}}{2\sin \alpha \sin \frac{ heta}{2}}\=\frac{11.72}{2(\sin 35)(\sin 36.25)}=[/latex] 17.28 m
Part c
[latex]\bar{BC}=\bar{EF}=\bar{AF}-\bar{AE}\=H_{cr}\left(\cot \alpha - \cot 75^\circ ight)\=11.72 \left(\cot 35 - \cot 60 ight)=[/latex] 9.97 m

Question 15.3: A cut slope is to be made in a soft saturated clay with its ...

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