(After Huang, 1980)
Figure 14.9 shows a 2.5:1 slope, 20 m (65.6 ft) high, composed of three different soil layers. The soil data including $c^{′},\phi^{′}$ , and γ, as well as the location of the groundwater table, are given. Assuming a seismic coefficient of 0.1, determine both the static and the seismic factors of safety.

Question

(After Huang, 1980)
Figure 14.9 shows a 2.5:1 slope, 20 m (65.6 ft) high, composed of three different soil layers. The soil data including [latex]c^{′},\phi^{′}[/latex], and γ, as well as the location of the groundwater table, are given. Assuming a seismic coefficient of 0.1, determine both the static and the seismic factors of safety.

Accepted Answer

1 First locate the most dangerous failure surface as shown in Figure 14.9(b) and determine the distance YH with Figure 14.9(c)

[latex]YH = 5.5 m,[/latex]

[latex]Y=\frac{5.58 m}{20 m} =0.275[/latex]

2 Determine [latex]N_{s}, N_{f}[/latex], and [latex]N_{e }[/latex]from Figure 14.8 with S = 2.5 (given) and Y = 0.275:

[latex]N_{s} = 7.0, N_{f} = 2.0, N_{e } = 2.8[/latex]

3 Determine the average unit weight of soil:

[latex]\gamma =\frac{\left(131 ight) \left(18 ight)+\left(221 ight)\left(19 ight) +\left(534 ight) \left(20 ight) }{131+221+534}=19.5 kN/m^{3} (124.2 pcf)[/latex]

4 Determine the average effective cohesion, [latex]c^{′}[/latex]:
Measure the length of the failure arc through soils 1, 2, and 3 and note the lengths to be 40 m (12.2 ft), 17.6 m (5.4 ft), and 24 m (7.3 ft), respectively. Then the average effective cohesion, [latex]c^{′}[/latex] is given by

[latex]c^{′} =\frac{\left(40 ight) \left(5 ight)+\left(176 ight)\left(7.5 ight) +\left(24 ight) \left(10 ight) }{40+176+24}=7.0 kPa (1.0 psi)[/latex]

5 Determine the average coefficient of friction, tan[latex] \phi^{′}[/latex]:
Friction is developed from the component of weight of overlying soil that is normal to the failure surface. The necessary cos [latex] heta[/latex] values are given in Figure 14.9(c). A weight may be computed for each layer as

[latex]W_{1}=\left(131\cdot 18+187\cdot 19+293\cdot 20 ight) \cdot 0.95=11,182 kN/m[/latex]

[latex]W_{2}=\left(2\cdot 17\cdot 19+110\cdot 20 ight)\cdot 0.75=2,135 kN/m[/latex]

[latex]W_{3}=\left(131\cdot 20 ight)\cdot 0.46=1,205 kN/m[/latex]

The average value for tan[latex] \phi^{′}[/latex] may then be given as

[latex] an \phi ^{′} =\frac{\left(11,182 ight) \left( an 25 ight)+\left(2135 ight)\left( an 30 ight) +\left(1205 ight) \left( an 35 ight) }{11,182+2135+1205}=0.502[/latex]

6 Determine the average pore pressure, [latex] r_{u}[/latex] :

[latex]r_{u}=\frac{A_{sw}\gamma _{w} }{A_{t} \gamma }[/latex]

where [latex]A_{sw} ext{ and } A_{t} =[/latex] the area of sliding mass under water and total area of sliding mass, respectively, and [latex]\gamma_{w} ext{ and } \gamma =[/latex] the unit weight of the water and the average unit weight of the soil, respectively. If [latex]A_{sw}[/latex] was measured to be [latex]527 m^{2}[/latex], then [latex] r_{u}[/latex] is given by

[latex]r_{u}=\frac{527\cdot 9.8}{886\cdot 19.5} =0.299[/latex]

7 With all of the above values now calculated, the factor of safety, F, may be computed from Equation (14.14)

[latex]F=\frac{c ^{′}/\gamma H +\left(1- r_{u} ight) an \phi ^{′}/N_{f} }{1/N_{s} +C_{s}N_{e} } [/latex] (14.14)

(a) Static factor of safety

[latex]F=\frac{\left(7/19.5\cdot 20 ight)+\left(1- 0.299 ight)\cdot 0.502)/2.0 }{\left(1/7 ight)+\left(0/2.8 ight) } =1.36[/latex]

(b) Seismic factor of safety

[latex]F=\frac{\left(7/19.5\cdot 20 ight)+\left(1- 0.299 ight)\cdot 0.502/2.0 )}{\left(1/7 ight)+\left(0/2.8 ight) } =1.09[/latex]

As such, the influence of seismic activity is to reduce the factor of safety

Question 14.3: (After Huang, 1980) Figure 14.9 shows a 2.5:1 slope, 20 m (6...

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