(After Koppula, 1984)
Consider a cohesive slope of height, H inclined at 60° to the horizontal. Let the shear strength of the soil be given by $a_{0}/ γ = 0.02$ and $c_{0} / γ H = 0.3$ . Determine the factor of safety, $F_{s}$ , when the seismic coefficient A = 0 (no earthquake) and A = 0.4 g (strong earthquake), where g = the acceleration due to gravity (9.81 m/s²).

Question

(After Koppula, 1984)
Consider a cohesive slope of height, H inclined at 60° to the horizontal. Let the shear strength of the soil be given by [latex] a_{0}/ γ = 0.02 [/latex] and [latex]c_{0} / γ H = 0.3[/latex]. Determine the factor of safety, [latex]F_{s}[/latex], when the seismic coefficient A = 0 (no earthquake) and A = 0.4 g (strong earthquake), where g = the acceleration due to gravity (9.81 m/s²).

Accepted Answer

When [latex]\beta =60°[/latex] and the seismic coefficient A=0, from Figure 14.12 we obtain the stability factor [latex] N_{1} =3.2[/latex], and from Figure 14.13 we obtain the stability factor [latex] N_{2} =5.3 [/latex]. The factor of safety F, can be computed from Equation (14.29) as

[latex]F = N_{1}\frac{\alpha _{0} }{\gamma }+N_{2} \frac{c_{0} }{\gamma H}[/latex] (14.29)

[latex]F=\left(3.2\right) \left(0.02\right) +\left(5.3\right) \left(0.3\right) =1.65[/latex]

When the seismic activity is increased to 0.4 g, the values of [latex] N_{1} [/latex]and [latex] N_{2}[/latex] are found to be 2.0 and 2.75, respectively, and F is calculated as

[latex]F=\left(2.0\right) \left(0.02\right) +\left(2.75\right) \left(0.3\right) =0.87[/latex]

Question 14.4: (After Koppula, 1984) Consider a cohesive slope of height, H...

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