Question 14.2: An embankment has a height of 30 ft (9.1 m). The soil proper...
An embankment has a height of 30 ft (9.1 m). The soil properties are cohesion equals 800 psf (38.3 kN/m²), friction angle equals 25°, and the unit weight of soil is 122 pcf (19.2 kN/m³). Find the slope angle, which corresponds to a factor of safety of 2.0.
The given height, H = 30 ft. For a factor of safety = 2.0, the critical height is given by Equation (14.2):
F_{s} =\frac{\tan \phi }{\tan \phi _{c} } \text{ and } \frac{H_{c}}{H} (14.2)
H_{c} = F · H = 2.0 · 30 ft = 60 ft.The stability number, from Equation (14.9), is given by:
H_{c} =N_{s}\frac{c}{\gamma } (14.9)
N_{s}=\frac{\gamma H_{c} }{c} =\frac{122 lb/ft^{3} 60 ft }{800 lb /ft^{2} } =9.15\approx 9Then, by consulting Figure 14.6 with N_s = 9 \text{ and } \phi = 25° , a slope angle of approximately 70° or a 1:3 horizontal to vertical slope is obtained. Note that lowering the required factor of safety results in steeper slopes up to a maximum 90° vertical cut.
