Bishop's Simplified Method

Bishop's Simplified Method is a widely used limit equilibrium technique for analyzing slope stability, especially suitable for circular slip surfaces. It improves on the Ordinary Method of Slices by including interslice normal forces and satisfies both moment and vertical force equilibrium. The key assumptions are:

  • Circular slip surface
  • Interslice side forces are horizontal and the moments they create are negligible
  • Satisfies:
    • Moment equilibrium
    • Vertical force equilibrium
  • Does not satisfy horizontal force equilibrium

With these assumptions, the forces acting on the slice are as follows:

bishop_slice.png

Where:

\(W\) = weight of the slice
\(\alpha\) = base inclination angle of the slice
\(\Delta \ell\) = length of the base
\(c', \phi'\) = effective cohesion and friction angle
\(u\) = pore water pressure
\(N\) = normal force on the base of the slice
\(S\) = shear force at the base

Recall that:

\(N = N' + u \Delta \ell\)

\(N' = N - u \Delta \ell\)

Where \(N'\) is the effective normal stress on the base of the slice.

Summing forces in the vertical direction:

\(\sum F_y = 0\)

\(N \cos \alpha + S \sin \alpha - W = 0\)

\((N' + u \Delta \ell) \cos \alpha + S \sin \alpha - W = 0\)

\(N' \cos \alpha + u \Delta \ell \cos \alpha + S \sin \alpha - W = 0 \qquad (1)\)

Where \(S\) is the mobilized shear force at the base of the slice. The shear force is given by:

\(S = \dfrac{1}{F} \left[c \Delta \ell + N' \tan \phi' \right] \qquad (2)\)

Substituting (2) into (1):

\(N' \cos \alpha + u \Delta \ell \cos \alpha + \dfrac{1}{F} \left[ c \Delta \ell + N' \tan \phi' \right] \sin \alpha - W = 0\)

Now solve for the effective normal force \(N'\). First, we rearrange the equation:

\(N' \cos \alpha + u \Delta \ell \cos \alpha + \dfrac{c \Delta \ell}{F} \sin \alpha + \dfrac{N' \tan \phi'}{F} \sin \alpha - W = 0\)

Next, we isolate all terms involving \(N'\):

\(N' \cos \alpha + \dfrac{N' \tan \phi'}{F} \sin \alpha = W - u \Delta \ell \cos \alpha - \dfrac{c \Delta \ell}{F} \sin \alpha\)

\(N' \left( \cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F} \right) = W - u \Delta \ell \cos \alpha - \dfrac{c \Delta \ell}{F} \sin \alpha\)

Finally, we can solve for \(N'\):

\(N' = \dfrac{W - u \Delta \ell \cos \alpha - \dfrac{c \Delta \ell}{F} \sin \alpha}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \qquad (3)\)

Next we use the general equation for the factor of safety based on moment equilibrium (resisting moments divided by driving moments):

\(F = \dfrac{\sum (c + \sigma' tan \phi') \Delta \ell}{\sum W sin \alpha}\)

\(N' = \sigma' \Delta \ell\)

thus:

\(F = \dfrac{\sum (c \Delta \ell + N' tan \phi')}{\sum W sin \alpha} \qquad (4)\)

Next, we substitute (3) into (4):

\(F = \dfrac{\sum \left[ c \Delta \ell + \left( \dfrac{W - u \Delta \ell \cos \alpha - \dfrac{c \Delta \ell}{F} \sin \alpha}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right) \tan \phi' \right]}{\sum W \sin \alpha}\)

To simplify the numerator, we multiply \(c \Delta \ell\) by \(\dfrac{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}}\) to combine like terms:

\(F = \dfrac{\sum \left[ \dfrac{c \Delta \ell (\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}) + (W - u \Delta \ell \cos \alpha - \dfrac{c \Delta \ell}{F} \sin \alpha) \tan \phi'}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right]}{\sum W \sin \alpha}\)

Now we rearrange the numerator:

\(F = \dfrac{\sum \left[ \dfrac{c \Delta \ell \cos \alpha + \dfrac{c \Delta \ell}{F} \sin \alpha \tan \phi' + (W - u \Delta \ell \cos \alpha) \tan \phi' - \dfrac{c \Delta \ell}{F} \sin \alpha \tan \phi'}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right]}{\sum W \sin \alpha}\)

Finally, the \(\dfrac{c \Delta \ell}{F} \sin \alpha \tan \phi'\) terms cancel out, leading to:

\(F = \dfrac{\sum \left[ \dfrac{c \Delta \ell \cos \alpha + (W - u \Delta \ell \cos \alpha) \tan \phi'}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right]}{\sum W \sin \alpha} \qquad (5)\)

Which is the standard equation for the factor of safety for Bishop's method.

The factor of safety \(F\) appears on both sides of the equation, so it must be solved iteratively.

Once \(F\) is determined, \(N'\) can be computed using equation (3) above.

Complete Formulation

For a complete implementation of Bishop's Simplified Method, we need to consider additional forces acting on the slice. The full set of forces are shown in the following figure:

bishop_complete.png

Where:

\(D\) = distributed load resultant force
\(\beta\) = inclination of the distributed load (perpendicular to slope)
\(kW\) = seismic force for pseudo-static seismic analysis
\(c.g.\) = center of gravity of the slice
\(P\) = reinforcement force on base of slice
\(T\) = tension crack water force
\(H\) = pile/pier force at point \(e\) on the failure surface
\(\theta_p\) = angle of pile force from horizontal (positive = counterclockwise/upward)

Each of these forces is described in detail in the Ordinary Method of Slices (OMS) section. The forces \(D\), \(kW\), \(P\), \(T\), and \(H\) are included in the Bishop's method factor of safety equation as follows:

Vertical Force Equilibrium

First we first need to consider how these forces affect the vertical force equilibrium. The vertical force equilibrium equation becomes:

\(N \cos \alpha + S \sin \alpha + P \sin \alpha + H \sin \theta_p - W - D \cos \beta = 0\)

\((N' + u \Delta \ell) \cos \alpha + S \sin \alpha + P \sin \alpha + H \sin \theta_p - W - D \cos \beta = 0\)

\(N' \cos \alpha + u \Delta \ell \cos \alpha + S \sin \alpha + P \sin \alpha + H \sin \theta_p - W - D \cos \beta = 0 \qquad (6)\)

The shear force on the base of the slice remains the same as before:

\(S = \dfrac{1}{F} \left[c \Delta \ell + N' \tan \phi' \right] \qquad (7)\)

Substituting (7) into (6) and solving for N':

\(N' \cos \alpha + u \Delta \ell \cos \alpha + \dfrac{1}{F} \left[c \Delta \ell + N' \tan \phi' \right] \sin \alpha + P \sin \alpha + H \sin \theta_p - W - D \cos \beta = 0\)

\(N' \cos \alpha + u \Delta \ell \cos \alpha + \dfrac{1}{F} c \Delta \ell \sin \alpha + \dfrac{1}{F} N' \tan \phi' \sin \alpha + P \sin \alpha + H \sin \theta_p - W - D \cos \beta = 0\)

\(N' \cos \alpha + \dfrac{1}{F} N' \tan \phi' \sin \alpha = W + D \cos \beta - P \sin \alpha - H \sin \theta_p - u \Delta \ell \cos \alpha - \dfrac{1}{F} c \Delta \ell \sin \alpha\)

\(N' \left( \cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F} \right) = W + D \cos \beta - P \sin \alpha - H \sin \theta_p - u \Delta \ell \cos \alpha - \dfrac{1}{F} c \Delta \ell \sin \alpha\)

Finally, we can solve for \(N'\):

\(N' = \dfrac{W + D \cos \beta - P \sin \alpha - H \sin \theta_p - u \Delta \ell \cos \alpha - \dfrac{c \Delta \ell}{F} \sin \alpha}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \qquad (8)\)

Moment Equilibrium

The moment equilibrium equation about the center of the slip circle must also be revised to include the moments from the additional forces. The mobilized shear force is \(S_{mob} = S/F\), where \(S = c \Delta \ell + N' \tan \phi'\) is the full shear strength. The reinforcement force \(P\) and the pile force \(H\) are known applied forces and are not factored by \(F\). Taking moments about the center of the circle:

\(R \sum \dfrac{S}{F} + R \sum P + \sum D \sin \beta \, a_{dy} + \sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right] = R \sum W \sin \alpha + \sum D \cos \beta \, a_{dx} + k\sum W \, a_s + T \, a_t \qquad (9)\)

Where:

\(a_{dx}\) = horizontal distance from center of circle to point \(d\)
\(a_{dy}\) = vertical distance from center of circle to point \(d\)
\(a_s\) = vertical distance from center of circle to center of gravity of the slice
\(a_t\) = vertical distance from center of circle to point \(c\)
\(a_{ey}\) = vertical distance from center of circle to point \(e\): \(Y_o - y_e\)
\(a_{ex}\) = horizontal distance from center of circle to point \(e\): \(x_e - X_o\)

The pile force \(H\) is decomposed into horizontal (\(H \cos \theta_p\)) and vertical (\(H \sin \theta_p\)) components, each with its own moment arm about the circle center.

Isolating the shear term on the left side:

\(R \sum \dfrac{S}{F} = R \sum W \sin \alpha + \sum D \cos \beta \, a_{dx} + k\sum W \, a_s + T \, a_t - R \sum P - \sum D \sin \beta \, a_{dy} - \sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]\)

Dividing by \(R\):

\(\dfrac{1}{F} \sum S = \sum W \sin \alpha + \frac{1}{R}\sum D \cos \beta \, a_{dx} + \frac{k}{R}\sum W \, a_s + \frac{1}{R} T \, a_t - \sum P - \frac{1}{R}\sum D \sin \beta \, a_{dy} - \frac{1}{R}\sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]\)

Solving for \(F\):

\(F = \dfrac{\sum S}{\sum W \sin \alpha + \frac{1}{R}\sum D \cos \beta \, a_{dx} + \frac{k}{R}\sum W \, a_s + \frac{1}{R} T \, a_t - \sum P - \frac{1}{R}\sum D \sin \beta \, a_{dy} - \frac{1}{R}\sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]}\)

Substituting \(S = c \Delta \ell + N' \tan \phi'\):

\(F = \dfrac{\sum \left( c \Delta \ell + N' \tan \phi' \right)}{\sum W \sin \alpha + \frac{1}{R}\sum D \cos \beta \, a_{dx} + \frac{k}{R}\sum W \, a_s + \frac{1}{R} T \, a_t - \sum P - \frac{1}{R}\sum D \sin \beta \, a_{dy} - \frac{1}{R}\sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]}\)

Substituting (8) for \(N'\):

\(F = \dfrac{\sum \left(c \Delta \ell + \left[ \dfrac{W + D \cos \beta - P \sin \alpha - H \sin \theta_p - u \Delta \ell \cos \alpha - \dfrac{c \Delta \ell}{F} \sin \alpha}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right] \tan \phi' \right)}{\sum W \sin \alpha + \frac{1}{R}\sum D \cos \beta \, a_{dx} + \frac{k}{R}\sum W \, a_s + \frac{1}{R} T \, a_t - \sum P - \frac{1}{R}\sum D \sin \beta \, a_{dy} - \frac{1}{R}\sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]}\)

To simplify the numerator, we multiply \(c \Delta \ell\) by \(\dfrac{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}}\):

\(F = \dfrac{\sum \left[ \dfrac{c \Delta \ell (\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}) + (W + D \cos \beta - P \sin \alpha - H \sin \theta_p - u \Delta \ell \cos \alpha) \tan \phi' - \dfrac{c \Delta \ell}{F} \sin \alpha \tan \phi'}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right]}{\sum W \sin \alpha + \frac{1}{R}\sum D \cos \beta \, a_{dx} + \frac{k}{R}\sum W \, a_s + \frac{1}{R} T \, a_t - \sum P - \frac{1}{R}\sum D \sin \beta \, a_{dy} - \frac{1}{R}\sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]}\)

Now, we can rearrange the numerator:

\(F = \dfrac{\sum \left[ \dfrac{c \Delta \ell \cos \alpha + \dfrac{c \Delta \ell}{F} \sin \alpha \tan \phi' + (W + D \cos \beta - P \sin \alpha - H \sin \theta_p - u \Delta \ell \cos \alpha) \tan \phi' - \dfrac{c \Delta \ell}{F} \sin \alpha \tan \phi'}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right]}{\sum W \sin \alpha + \frac{1}{R}\sum D \cos \beta \, a_{dx} + \frac{k}{R}\sum W \, a_s + \frac{1}{R} T \, a_t - \sum P - \frac{1}{R}\sum D \sin \beta \, a_{dy} - \frac{1}{R}\sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]}\)

Finally, the \(\dfrac{c\Delta\ell}{F} \sin \alpha \tan \phi'\) terms cancel out, leading to:

\(F = \dfrac{\sum \left[ \dfrac{c \Delta \ell \cos \alpha + (W + D \cos \beta - P \sin \alpha - H \sin \theta_p - u \Delta \ell \cos \alpha) \tan \phi'}{\cos \alpha + \dfrac{\sin \alpha \tan \phi'}{F}} \right]}{\sum W \sin \alpha + \frac{1}{R}\sum D \cos \beta \, a_{dx} + \frac{k}{R}\sum W \, a_s + \frac{1}{R} T \, a_t - \sum P - \frac{1}{R}\sum D \sin \beta \, a_{dy} - \frac{1}{R}\sum \left[ H \cos \theta_p \, a_{ey} + H \sin \theta_p \, a_{ex} \right]} \qquad (10)\)

This is the complete formulation for Bishop's Simplified Method. Note that:

  • The reinforcement force \(P\), the pile force \(H\), and the distributed load resisting moment \(D \sin \beta\, a_{dy}\) appear in the denominator because they are known forces that are not factored by the safety factor \(F\)
  • \(P\) and \(H\) affect the numerator indirectly through their effect on \(N'\)
  • The water force \(T\) only applies to the uppermost slice

The factor of safety \(F\) appears on both sides of the equation, so it must be solved iteratively, just like the basic formulation.

Summary

Assumes horizontal side forces
Satisfies moment and vertical force equilibrium
Applicable to circular slip surfaces
Requires iteration to solve for \(F\)
More accurate than OMS, especially for effective stress analysis with high pore pressures