Question 14.3: (After Huang, 1980) Figure 14.9 shows a 2.5:1 slope, 20 m (6...
(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.
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)
YH = 5.5 m,Y=\frac{5.58 m}{20 m} =0.275
2 Determine N_{s}, N_{f}, and N_{e }from Figure 14.8 with S = 2.5 (given) and Y = 0.275:
N_{s} = 7.0, N_{f} = 2.0, N_{e } = 2.83 Determine the average unit weight of soil:
\gamma =\frac{\left(131\right) \left(18\right)+\left(221\right)\left(19\right) +\left(534\right) \left(20\right) }{131+221+534}=19.5 kN/m^{3} (124.2 pcf)4 Determine the average effective cohesion, c^{′}:
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, c^{′} is given by
5 Determine the average coefficient of friction, tan \phi^{′}:
Friction is developed from the component of weight of overlying soil that is normal to the failure surface. The necessary cos \theta values are given in Figure 14.9(c). A weight may be computed for each layer as
W_{2}=\left(2\cdot 17\cdot 19+110\cdot 20\right)\cdot 0.75=2,135 kN/m
W_{3}=\left(131\cdot 20\right)\cdot 0.46=1,205 kN/m
The average value for tan \phi^{′} may then be given as
\tan \phi ^{′} =\frac{\left(11,182\right) \left(\tan 25\right)+\left(2135\right)\left(\tan 30\right) +\left(1205\right) \left(\tan 35\right) }{11,182+2135+1205}=0.5026 Determine the average pore pressure, r_{u} :
r_{u}=\frac{A_{sw}\gamma _{w} }{A_{t} \gamma }where A_{sw} \text{ and } A_{t} = the area of sliding mass under water and total area of sliding mass, respectively, and \gamma_{w} \text{ and } \gamma = the unit weight of the water and the average unit weight of the soil, respectively. If A_{sw} was measured to be 527 m^{2}, then r_{u} is given by
r_{u}=\frac{527\cdot 9.8}{886\cdot 19.5} =0.299
7 With all of the above values now calculated, the factor of safety, F, may be computed from Equation (14.14)
F=\frac{c ^{′}/\gamma H +\left(1- r_{u}\right)\tan \phi ^{′}/N_{f} }{1/N_{s} +C_{s}N_{e} } (14.14)
(a) Static factor of safety
F=\frac{\left(7/19.5\cdot 20\right)+\left(1- 0.299\right)\cdot 0.502)/2.0 }{\left(1/7\right)+\left(0/2.8\right) } =1.36(b) Seismic factor of safety
F=\frac{\left(7/19.5\cdot 20\right)+\left(1- 0.299\right)\cdot 0.502/2.0 )}{\left(1/7\right)+\left(0/2.8\right) } =1.09As such, the influence of seismic activity is to reduce the factor of safety
