A 1.5:1 cut slope is to be made in a shale with an apparent dip of 16° as shown in Figure 14.17. Compute the factor of safety against failure along the lowermost daylighted bedding plane using a unit weight of 20.1 kN/m^3 and along bedding strength parameters of c' = 15 kPa and Φ' = 20°.

Question

A 1.5:1 cut slope is to be made in a shale with an apparent dip of 16° as shown in Figure 14.17. Compute the factor of safety against failure along the lowermost daylighted bedding plane using a unit weight of [latex]20.1 kN/m^{3}[/latex] and along bedding strength parameters of [latex]c^{\prime }= 15 kPa[/latex] and [latex]Φ^{\prime } = 20°[/latex].

Accepted Answer

[latex]W/b=\frac{1}{2} \left(85.0 m ight) \left(11.3 m ight) \left(20.1 kN/m^{3} ight) =9650 kN/m[/latex]

[latex]l=\frac{85.0 m}{\cos 16°} =88.4 m[/latex]

Compute the pore water pressure based on a visual estimate of the average [latex]z_{w}[/latex]. (ranges from 0 to 3.2 m):

[latex]u = γ_{w}z_{w} = (9.8 kN/m ^{3})(3.0 m) = 29 kPa [/latex]

[latex]F=\frac{c^{\prime }l+\left[\left(W/b ight) \cos\alpha - ul ight] an\Phi }{\left(W/b ight) \sin\alpha } =\frac{\left(15 kPa ight) \left(88.4 m ight) +\left[\left(9650 kN/m^{3} ight) \cos 16° - \left(29 kPa ight) \left(88.4 m ight) ight] an 20°}{\left(9650 kN/m^{3} ight) \sin 16°}= [/latex] 1.42

The computed factor of safety of 1.42 is slightly less than the minimum acceptable value of 1.50. Therefore, this design is probably not acceptable.

Question 14.1: A 1.5:1 cut slope is to be made in a shale with an apparent ...

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