Rock slope stability in two dimensions¶
A rock slope is examined with one continuous fracture. It is assumed that rock matrix cannot fail and that the problem can be considered in two dimensions. This is the case when the azimuth of the slope and that of the fracture are on approximation the same, as shown in the following figure.
Assuming that no cohesion is present, to assess the stability of the slope, it is enough to know the inclination of the slope $\beta_s$, the inclination of the fracture $\beta$ and the friction angle $\phi$.
For sliding to take plece, two conditions need to be fulfilled:
- Sliding must be kinematically possible, i.e. possible from a geometrical point of view.
- Sliding must be statically possible, i.e. the driving forces need to be at least as large as the resisting forces.
The two criteria need to examined separately. Sliding takes place if both are satisfied.
Kinematic stability¶
A slope with a fracture is kinematically stable, when sliding is not possible from a geometric point of view. The concept is illustrated in the following figure.
- For the fracture set marked in blue, sliding is not kinematically possible. While the fractures are rather steeply inclined, movement is not allowed due to the geometry.
- For the red fractures sliding is kinematically possible. Rock blocks can move to the right under the influence of gravity.
- For the yellow fracture set sliding is also kinematically possible. Movement to the right is possible from a geometrical point of view, even though it is not likely due to the small inclination.
In short, the kinematic consideration concerns the geometry of the problem and is not concerned with the acting forces. Forces are the focus of the following section.
Static stability¶
Examination of the static stability focusses on the forces that are either acting or that can be mobilised. In the specific case the weight $W$ of the potentially sliding block is acting, along with the normal $N$ and tangential $T$ to the movement reaction, as shown in the figure below.
The weight can be decomposed into a component normal to the fracture $W_N$ and a component parallel to the fracture $W_T$. These can be calculated as
$W_N = W\cdot \cos(\beta)$ and $W_T = W\cdot \sin(\beta)$
In the direction normal to the fracture movement is not possible. As a result in this direction equilibrium needs to hold:
$ \Sigma F_N = N - W_N = 0 \Rightarrow N = W_N = W\cdot \cos(\beta)$
In the direction tangent to the fracture equilibrium is not necessary, as acceleration can be present, so that the reaction force $T$ cannot be directly evaluated. However, its maximum value is dictated by the shear resistance of the fracture:
$T_{max} = \tan(\phi) \cdot N $
If $T_max$ is larger than $W_T$ no sliding takes place, if $T_max$ is smaller than $W_T$ sliding takes place. The same can be expressed with the definition of the safety factor:
$F_s = \dfrac{T_{max}}{W_T}$
Sliding takes place if the safety factor is smaller than 1. Substituting yields for the specific case:
$F_s = \dfrac{\tan(\phi)}{\tan(\beta)}$
Note that the inclination of the slope and the weight of the block play no part in this calculation.
In the widget below $\beta_s$ is the inclination of the rock slope, $\beta$ is the inclination of the fracture and $\phi$ is the friction angle of the fracture. Note that for tangent of the slope inclination absolute values are used in the estimation of the safety factor.
