Section 4: Soil and Water Conservation Engineering
Soil Mechanics
Subtopic: Stability of slopes
Question Statement:
An infinite embankment slope of dry cohesionless sand ($c = 0$) has an effective internal friction angle of $\phi' = 35^\circ$. The slope angle with the horizontal is $\beta = 25^\circ$. Taking $\tan(35^\circ) = 0.7002$ and $\tan(25^\circ) = 0.4663$, the factor of safety ($FOS$) of the dry slope against translational shear sliding is ________ (round off to two decimal places).
Answer Key:
1.50
Step-by-Step Solution & Explanation
1. For an infinite cohesionless slope in completely dry conditions:
$FOS = \frac{\text{Resisting Shear Stress}}{\text{Mobilized Shear Stress}} = \frac{\sigma' \tan\phi'}{\tau}$
2. Stresses on a plane at depth $z$ parallel to the slope face:
$\sigma' = \gamma z \cos^2\beta$
$\tau = \gamma z \cos\beta \sin\beta$
$FOS = \frac{\gamma z \cos^2\beta \tan\phi'}{\gamma z \cos\beta \sin\beta} = \frac{\tan\phi'}{\tan\beta}$
3. Calculating $FOS$:
$FOS = \frac{\tan 35^\circ}{\tan 25^\circ} = \frac{0.7002}{0.4663} = 1.5016 \approx 1.50$