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Concept: Analyzes the stability of slopes where a block of soil/rock slides along a flat, planar surface.
Assumptions:
- 2D problem: We look at a slice of the slope.
- Forces are resolved parallel and normal to the sliding plane.
Factor of Safety (FS): Ratio of resisting forces to driving forces.
- $FS = \frac{\text{Resisting force}}{\text{Driving force}}$
- Resisting force: Cohesion ($c \cdot A$) + Friction ($\Sigma N \cdot \tan\phi$)
- Driving force: Sum of shear forces ($\Sigma S$) along the plane.
- $FS = \frac{cA + \Sigma N \tan\phi}{\Sigma S}$
General Formula for FS: $$FS = \frac{cA + (W \cos\psi_p - U - V \sin\psi_p) \tan\phi}{W \sin\psi_p + V \cos\psi_p}$$ Where:
- $c$: cohesion
- $A$: area of sliding plane
- $W$: weight of sliding block
- $\psi_p$: dip of the sliding plane
- $U$: total uplift force due to water pressure
- $V$: force from tension crack water pressure
- $\phi$: angle of internal friction

Tension crack: A crack that forms behind the slope crest due to tension.
Depth of crack ($z$): Influences the sliding block geometry.
Water in tension crack: Adds a force $V$ to the driving forces.
- $V = \frac{1}{2} \gamma_w z_w^2$ where $\gamma_w$ is unit weight of water, $z_w$ is water depth in crack.
Area $A$ of sliding plane:
- $A = (H + b \tan\psi_s - z) \csc\psi_p$
- $H$: slope height, $b$: distance from crest to crack, $\psi_s$: slope dip.

For tension crack in inclined upper slope surface: $$W = \gamma_r \left[ (1 - \cot\psi_f \tan\psi_p) (bH + \frac{1}{2}H^2 \cot\psi_f) + \frac{1}{2}b^2(\tan\psi_s - \tan\psi_p) \right]$$
For tension crack in the slope face: $$W = \frac{1}{2} \gamma_r H^2 \left[ (1 - \frac{z}{H})^2 \cot\psi_p (\cot\psi_p \tan\psi_f - 1) \right]$$ Where $\gamma_r$ is the unit weight of the rock/soil.

Uplift force ($U$): Water pressure within the sliding plane reduces effective normal stress, thus reducing friction.
Case 1: Ground water level above base of tension crack
- Water pressure acts on tension crack and sliding plane.
- Pressure decreases linearly from base of crack to zero at face.
Case 2: Surface water flows into crack
- If material is impermeable, $U$ can be zero or negligible.
Case 3: Ground water discharge blocked by freezing
- Uplift force $U$ can exceed normal pressure.
- For rectangular pressure distribution: $U = A \cdot p$, where $p = \gamma_w z_w$.
Case 4: Ground water level below base of tension crack
- Water pressure acts only on the sliding plane.
- Approximated by a triangular distribution.
- $U = \frac{1}{2 \sin\psi_p} z_w h_w \gamma_w$
- $h_w