Lateral Earth Pressure Theories and Retaining Walls

Why walls are designed for active pressure

A rigid retaining wall holds soil on the back side. The retained soil pushes the wall outwards; the wall yields very slightly (about 0.001 to 0.005 H for sand, up to 0.01 to 0.02 H for clay) and the soil reaches the active state, the minimum lateral pressure. This is the destabilising force that causes overturning and sliding, so the wall must be designed against it ($K_a = \tan^2(45^\circ - \phi/2)$).

Why passive pressure is neglected

Passive pressure acts on the toe side of the embedded part of the wall and resists sliding. It is usually ignored, or only a fraction (about 50%) is used, because:

1. Large movement is needed: full passive resistance needs a movement of about 0.01 to 0.05 H toward the soil, much more than the small wall movement allowed at working load. At working movement only a small part is mobilised.
2. Soil in front may be removed: by scour, erosion, future excavation, utility trenches or other construction.
3. Soil may soften or shrink: wetting by surface water, seasonal moisture change, frost action and cracks reduce its strength; the top 0.5-1 m is particularly unreliable.
4. Construction disturbance: the toe backfill is often loose or poorly compacted.
5. Compatibility of strains: active state forms at very small strain, passive state at large strain, so both limit states cannot occur together.
6. Safety: neglecting passive resistance gives a conservative design, because $K_p$ is large and an overestimate is dangerous (unsafe).

Where passive resistance is used (e.g. a shear key), it is taken only below the depth of possible disturbance and with a factor of safety of at least 2 on $P_p$.

Assumptions: smooth vertical wall, horizontal backfill, Rankine theory; sand above the water table is taken at the given unit weight; the clay is saturated, so the effective stress uses $\gamma' = \gamma_{sat} - \gamma_w = 19.00 - 9.81 = 9.19\ \text{kN/m}^3$.

Coefficients

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-30^\circ/2) = 0.3333 \\ K_{a2} &= \tan^2(45^\circ-30^\circ/2) = 0.3333,\quad 2c\sqrt{K_{a2}} = 2\times10\times0.5774 = 11.55\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Sand, top: $0$.
• Sand, bottom ($z=3$ m): $p = K_{a1}\gamma H_1 = 0.3333\times17.5\times3 = 17.50$.
• Clay, top: $\sigma_v' = 17.5\times3 = 52.50$, so $p = 0.3333\times52.50 - 11.55 = 5.95$.
• Clay, bottom ($z=6$ m): $\sigma_v' = 52.50 + 9.19\times3 = 80.07$, so $p = 0.3333\times80.07 - 11.55 = 15.14$.
• Water pressure at the base: $u = 9.81\times3 = 29.43$ (zero at the water table).

 0 --|
     |\
sand | \
     |  \
 WT -|---17.50
     |  eff 5.95
clay |   \
     |    \ eff 15.14
 base|-----> + water 29.43

Forces per metre run and lever arms about the base

$$\bar y = \frac{\sum M}{\sum F} = \frac{189.72}{102.04} = 1.86\ \text{m above the base}$$

Answer: total thrust $P = 102.0$ kN/m (water included), acting horizontally at 1.86 m above the base of the wall.

Assumptions: smooth vertical wall, horizontal backfill, Rankine theory; sand above the water table is taken at the given unit weight; the clay is saturated, so the effective stress uses $\gamma' = \gamma_{sat} - \gamma_w = 19.50 - 9.81 = 9.69\ \text{kN/m}^3$.

Coefficients

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-35^\circ/2) = 0.2710 \\ K_{a2} &= \tan^2(45^\circ-30^\circ/2) = 0.3333,\quad 2c\sqrt{K_{a2}} = 2\times16\times0.5774 = 18.48\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Sand, top: $0$.
• Sand, bottom ($z=4.5$ m): $p = K_{a1}\gamma H_1 = 0.2710\times18\times4.5 = 21.95$.
• Clay, top: $\sigma_v' = 18\times4.5 = 81.00$, so $p = 0.3333\times81.00 - 18.48 = 8.52$.
• Clay, bottom ($z=7.5$ m): $\sigma_v' = 81.00 + 9.69\times3 = 110.07$, so $p = 0.3333\times110.07 - 18.48 = 18.21$.
• Water pressure at the base: $u = 9.81\times3 = 29.43$ (zero at the water table).

 0 --|
     |\
sand | \
     |  \
 WT -|---21.95
     |  eff 8.52
clay |   \
     |    \ eff 18.21
 base|-----> + water 29.43

Forces per metre run and lever arms about the base

$$\bar y = \frac{\sum M}{\sum F} = \frac{319.29}{133.64} = 2.39\ \text{m above the base}$$

Answer: total thrust $P = 133.6$ kN/m (water included), acting horizontally at 2.39 m above the base of the wall.

Assumptions: smooth vertical wall, horizontal backfill, Rankine theory; sand above the water table is taken at the given unit weight; the clay is saturated, so the effective stress uses $\gamma' = \gamma_{sat} - \gamma_w = 19.50 - 9.81 = 9.69\ \text{kN/m}^3$.

Coefficients

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-35^\circ/2) = 0.2710 \\ K_{a2} &= \tan^2(45^\circ-35^\circ/2) = 0.2710,\quad 2c\sqrt{K_{a2}} = 2\times16\times0.5206 = 16.66\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Sand, top: $0$.
• Sand, bottom ($z=3$ m): $p = K_{a1}\gamma H_1 = 0.2710\times18\times3 = 14.63$.
• Clay, top: $\sigma_v' = 18\times3 = 54.00$, so $p = 0.2710\times54.00 - 16.66 = -2.02$.
• Clay, bottom ($z=7.5$ m): $\sigma_v' = 54.00 + 9.69\times4.5 = 97.60$, so $p = 0.2710\times97.60 - 16.66 = 9.79$.
• Water pressure at the base: $u = 9.81\times4.5 = 44.15$ (zero at the water table).

The effective pressure at the top of the clay is negative (-2.02), so a tension crack is assumed; the negative part is ignored. The pressure becomes zero at $z_0 = 4.5\times2.02/(9.79+2.02) = 0.77$ m below the water table (interface) and the clay diagram is a triangle from there.

 0 --|
     |\
sand | \
     |  \
 WT -|---14.63
     |  eff -2.02
clay |   \
     |    \ eff 9.79
 base|-----> + water 44.15

Forces per metre run and lever arms about the base

$$\bar y = \frac{\sum M}{\sum F} = \frac{292.41}{139.53} = 2.10\ \text{m above the base}$$

Answer: total thrust $P = 139.5$ kN/m (water included), acting horizontally at 2.10 m above the base of the wall.

Assumptions: smooth vertical wall, horizontal backfill, Rankine theory; sand above the water table is taken at the given unit weight; the clay is saturated, so the effective stress uses $\gamma' = \gamma_{sat} - \gamma_w = 19.50 - 9.81 = 9.69\ \text{kN/m}^3$.

Coefficients

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-35^\circ/2) = 0.2710 \\ K_{a2} &= \tan^2(45^\circ-35^\circ/2) = 0.2710,\quad 2c\sqrt{K_{a2}} = 2\times16\times0.5206 = 16.66\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Sand, top: $0$.
• Sand, bottom ($z=5$ m): $p = K_{a1}\gamma H_1 = 0.2710\times18\times5 = 24.39$.
• Clay, top: $\sigma_v' = 18\times5 = 90.00$, so $p = 0.2710\times90.00 - 16.66 = 7.73$.
• Clay, bottom ($z=7.5$ m): $\sigma_v' = 90.00 + 9.69\times2.5 = 114.22$, so $p = 0.2710\times114.22 - 16.66 = 14.30$.
• Water pressure at the base: $u = 9.81\times2.5 = 24.53$ (zero at the water table).

 0 --|
     |\
sand | \
     |  \
 WT -|---24.39
     |  eff 7.73
clay |   \
     |    \ eff 14.30
 base|-----> + water 24.53

Forces per metre run and lever arms about the base

$$\bar y = \frac{\sum M}{\sum F} = \frac{310.60}{119.16} = 2.61\ \text{m above the base}$$

Answer: total thrust $P = 119.2$ kN/m (water included), acting horizontally at 2.61 m above the base of the wall.

Assumptions

Rankine's theory:

• The soil is homogeneous, isotropic, semi-infinite and the wall back is vertical and smooth (no wall friction).
• The backfill surface is horizontal or a plane slope and the wall moves enough to reach the plastic state.
• Failure occurs by a state of plastic equilibrium in the whole soil mass; pressure acts parallel to the backfill slope.

Coulomb's theory:

• Soil is dry, homogeneous, isotropic and cohesionless.
• The failure surface is a plane through the heel of the wall; the failing wedge slides as a rigid body.
• Wall friction ($\delta$) acts on the back face; the wall back can be inclined and the backfill can slope.
• The resultant is found by wedge force equilibrium.

Four basic differences

Tentative dimensions are assumed from the wall height $H$ (including the footing depth) from past experience, then checked for stability and revised.

Gravity (masonry/plain concrete) wall

• Top width: 0.3 m minimum (about $H/12$ to $H/6$ up to 0.6 m).
• Base width: $B = 0.5H$ to $0.7H$.
• Thickness of base slab: about 0.3 m or $H/8$ to $H/6$.
• Front face vertical or battered (1 in 10 to 1 in 5).

Cantilever wall (reinforced concrete)

• Base width: $B = 0.5H$ to $0.7H$ (0.4H to 0.5H if no surcharge and good soil).
• Base slab thickness: $H/12$ to $H/10$ (not less than 300 mm).
• Stem thickness at top: 200-300 mm minimum; at the base $H/12$ to $H/10$ (about equal to base slab thickness).
• Toe projection: $B/3$ (about 0.25B to 0.3B); the remainder is the heel.
• Shear key (if sliding governs): depth about $0.1H$ to $0.15H$ below the base, under the stem.
• Depth of foundation $D_f$: not less than 1 m or below frost and scour depth.

        <--0.3m-->
        |  stem  |  backfill
        |        |
     H  |        |
        |  H/12  |
   _____|________|_______________
   |toe |    heel                |
   |B/3 |                        |   base slab H/12..H/10
   |<-------- B = 0.5-0.7H ----->|

Counterfort/buttress wall ($H > 6$ m to 8 m)

• Base width: $0.5H$ to $0.6H$; slab and stem thickness $H/20$ to $H/15$ (200-300 mm min).
• Counterfort spacing 0.3H to 0.6H (about 2.5-3.5 m), counterfort thickness 250-400 mm.
• Counterforts on the backfill side (tied to stem and heel); buttresses on the front side (rarely used).

Trial section is then checked for overturning ($FS \ge 2$), sliding ($FS \ge 1.5$), bearing and tension at the base.

Culmann's method is a graphical wedge method that finds the active thrust on a wall for any backfill surface, including inclined backfill and surcharge.

Procedure (cohesionless backfill with inclined surface and line load)

1. Draw the wall $AB$ and the backfill surface to scale. Mark the heel $B$.
2. Draw the $\phi$-line $BD$ from $B$ at angle $\phi$ to the horizontal.
3. Draw the pressure line $BE$ from $B$ at angle $\psi = \theta-\delta$ above $BD$, where $\theta$ is the angle the wall back makes with the horizontal (on the fill side) and $\delta$ is wall friction.
4. Draw several trial failure planes $BC_1, BC_2, \dots$ from $B$ to points on the ground surface.
5. Compute the weight $W_1, W_2,\dots$ of each trial wedge $ABC_i$ per metre. If a line load $Q$ lies inside a wedge, add $Q$ to that wedge's weight; for a uniform surcharge $q$, add $q\times$(horizontal width of the wedge) to the weight.
6. Choose a scale of force and mark $W_i$ along $BD$: $b_1, b_2, \dots$
7. From each $b_i$ draw a line parallel to $BE$ to meet the corresponding trial plane $BC_i$ at $c_i$.
8. Join $c_1 c_2 \dots$ to get the Culmann curve. The step in the curve at the line load position shows the effect of $Q$ (jump when the plane passes beyond $Q$).
9. Draw a tangent to the curve parallel to $BD$. The ordinate from the point of tangency $c_m$ to $BD$, parallel to $BE$, scaled by the force scale, is the active thrust $P_a$ (maximum).
10. The critical failure plane is $BC_m$; $P_a$ acts at $H/3$ above the base (uniform fill) at angle $\delta$ to the normal to the wall.

        Q (line load)
   A    v      ground surface
   |\______________________ C1 C2 Cm C3
   |                    ..
   |             .. ...    Culmann
   |        ..      curve
   |   ..    /
 B +--------/---> b1 b2 bm b3 (weights on BD)
   BD at phi, BE at psi from BD

Effect of surcharge on active pressure

• A uniform surcharge $q$ is equivalent to an extra soil height $h_e = q/\gamma$. It adds a uniform (rectangular) pressure $\Delta p_a = K_a q$ over the whole height, so the total thrust increases by $K_a q H$.
• The resultant thrust moves higher than $H/3$ because the pressure distribution changes from triangular to trapezoidal, increasing the overturning moment.
• A line load adds a localised pressure on the wall (a discontinuity in the Culmann curve) and increases the thrust only when the load is within the sliding wedge; the closer to the wall the larger the effect.
• Hence surcharges (traffic, buildings) must always be included in wall design.

A retaining wall is checked for the following modes; $\sum V$ = total vertical load, $\sum H$ = total horizontal thrust.

1. Overturning about the toe

$$FS_{ot} = \frac{\sum M_R}{\sum M_O} \ge 2 \ (1.5 \text{ minimum in some codes})$$

$M_R$ = resisting moment of wall weight, soil above heel, and passive pressure (if reliable); $M_O$ = overturning moment of lateral thrust about the toe.

2. Sliding along the base

$$FS_{s} = \frac{\sum V\tan\delta_b + c_a B + P_p}{\sum H} \ge 1.5$$

$\delta_b$ = friction angle between base and soil (about $\tfrac23\phi$), $c_a$ = adhesion. A shear key is added if $FS$ is low.

3. Bearing capacity and base pressure

Eccentricity $e = B/2 - \bar{x}$, with $\bar{x} = (\sum M_R - \sum M_O)/\sum V$; need $e \le B/6$ (no tension).

$$q_{max,min} = \frac{\sum V}{B}\left(1 \pm \frac{6e}{B}\right)$$

Check $q_{max} \le q_a$ (i.e. $FS = q_u/q_{max} \ge 3$) and $q_{min} \ge 0$.

4. Overall (deep-seated) slope stability

Check circular slip surfaces below and around the wall (Fellenius or Bishop method), $FS \ge 1.5$.

5. Settlement and structural checks

Estimate settlement of the base and check bending and shear of stem, heel and toe in RC walls.

Culmann's passive method finds the minimum passive resistance by trial wedges in front of the wall.

Procedure

1. Draw the wall $AB$, the backfill surface in front of the wall and the heel/toe $B$ to scale.
2. Draw the $\phi$-line $BD$ from $B$ at angle $\phi$ to the horizontal, but on the opposite side from the active case (below the horizontal, going into the soil), because the resultant of the failure plane tilts the other way in the passive state.
3. Draw the pressure line $BE$ from $B$ at angle $\psi = \theta+\delta$ to $BD$, where $\theta$ is the angle of the wall back with the horizontal and $\delta$ the wall friction (the thrust acts on the other side of the normal in the passive state).
4. Draw trial failure planes $BC_1, BC_2, BC_3,\dots$ from $B$ to the ground surface, in front of the wall.
5. Calculate the weight $W_i$ of each trial wedge $ABC_i$ (per metre), adding any surcharge on it.
6. Mark $W_i$ to a scale on $BD$ to get points $b_1, b_2, \dots$
7. Through each $b_i$ draw a line parallel to $BE$ to cut the trial plane $BC_i$ at $c_i$.
8. Join $c_1 c_2 c_3 \dots$ to obtain the Culmann curve.
9. Draw a tangent to the curve parallel to $BD$ on the side closest to $BD$. The minimum distance from the curve to $BD$, measured parallel to $BE$ and multiplied by the force scale, gives the passive resistance $P_p$. (Passive failure occurs along the plane giving the least resistance, so the minimum ordinate is taken, whereas it is the maximum for active.)
10. The critical plane is $BC_m$; $P_p$ acts at one-third of the height above the base.

   A  ____ ground surface ___ C3 C2 Cm C1
   |  \
   |   \_ trial planes
 B +----+-----> b1 b2 bm b3  (W on BD, below horizontal)
   BE at psi = theta + delta from BD
   P_p = minimum ordinate of Culmann curve

The earth pressure coefficient $K$ is the ratio of the horizontal effective stress to the vertical effective stress at a point in the soil:

$$K = \frac{\sigma_h'}{\sigma_v'}$$

Its value depends on the soil strength and on how much the wall moves.

Relation: $K_a < K_0 < K_p$ and $K_a K_p = 1$ (for horizontal backfill and smooth vertical wall). For example, for $\phi = 30^\circ$: $K_a = 0.333$, $K_0 = 0.5$, $K_p = 3$.

Geometry and assumptions: top width 1 m, base width 3 m, height 4 m; the soil face (back) is vertical and the front face is inclined, so the toe is the front bottom corner. Backfill slopes at $\beta = 20^\circ$, $\phi = 30^\circ$, $\gamma = 20$. The only $\delta$ given, $24^\circ$, is used both as wall friction (for Coulomb's $K_a$) and as base friction (sliding). Passive resistance is neglected.

1. Coulomb active thrust

$$K_a = \frac{\sin^2(\theta+\phi)}{\sin^2\theta\,\sin(\theta-\delta)\left[1+\sqrt{\dfrac{\sin(\phi+\delta)\sin(\phi-\beta)}{\sin(\theta-\delta)\sin(\theta+\beta)}}\right]^2} = 0.4162\quad(\theta = 90^\circ)$$ $$P_a = \tfrac12 K_a\gamma H^2 = 0.5\times0.4162\times20\times4^2 = 66.59\ \text{kN/m}$$

Acts at $H/3 = 1.333$ m above the base, at $24^\circ$ below the horizontal:

$$P_h = P_a\cos24^\circ = 60.83,\qquad P_v = P_a\sin24^\circ = 27.08\ \text{kN/m}$$

2. Weights and moments about the toe

Overturning moment: $M_o = P_h\times1.333 = 60.83\times1.333 = 81.11$ kN m/m.

3. Base pressure

$$\bar x = \frac{M_r - M_o}{\sum V} = \frac{449.25-81.11}{219.08} = 1.680\ \text{m},\quad e = |-0.180|\ \text{m} < B/6 = 0.5$$ $$q = \frac{\sum V}{B}\left(1 \pm \frac{6e}{B}\right)$$

The resultant falls toward the heel, so $q_{max}$ is at the heel and $q_{min}$ at the toe:

$$q_{max} = 99.4\ \text{kN/m}^2,\qquad q_{min} = 46.7\ \text{kN/m}^2$$

(both compressive, so no tension).

4. Factors of safety

$$FS_{ot} = \frac{M_r}{M_o} = \frac{449.25}{81.11} = 5.54\ (>2,\ \text{safe})$$ $$FS_{s} = \frac{\sum V\tan24^\circ}{P_h} = \frac{219.08\times0.4452}{60.83} = 1.60\ (>1.5,\ \text{safe})$$

Answer: $q_{max} = 99.4$ and $q_{min} = 46.7$ kN/m$^2$; $FS_{ot} = 5.54$; $FS_{sliding} = 1.60$.

Geometry and assumptions: top width 1 m, base 3 m, height 4 m, vertical back face against the soil, inclined front face (toe at the front bottom corner). The back is vertical and no wall friction is given, so Rankine's theory for a surcharge angle $\beta = 20^\circ$ is used; passive resistance is ignored.

1. Active thrust (Rankine, sloping backfill)

$$K_a = \cos\beta\,\frac{\cos\beta-\sqrt{\cos^2\beta-\cos^2\phi}}{\cos\beta+\sqrt{\cos^2\beta-\cos^2\phi}} = 0.4142$$ $$P_a = \tfrac12 K_a\gamma H^2 = 0.5\times0.4142\times20\times4^2 = 66.27\ \text{kN/m}$$

It acts at $H/3 = 1.333$ m above the base, parallel to the backfill slope ($20^\circ$):

$$P_h = 62.28\ \text{kN/m},\qquad P_v = 22.67\ \text{kN/m}$$

2. Moments about the toe

$$M_o = 62.28\times1.333 = 83.03\ \text{kN m/m}$$

3. Base pressures

$$\bar x = (436.00-83.03)/214.67 = 1.644\ \text{m},\qquad e = |-0.144| < B/6 = 0.5$$ $$q_{max} = 92.2\ \text{kN/m}^2\ (\text{heel}),\qquad q_{min} = 50.9\ \text{kN/m}^2\ (\text{toe})$$

4. Factors of safety

$$FS_{ot} = \frac{M_r}{M_o} = \frac{436.00}{83.03} = 5.25\ (>2,\ \text{safe})$$

Sliding (for information) with $\mu = 0.45$: $FS_s = 214.67\times0.45/62.28 = 1.55$.

Answer: $q_{max} = 92.2$ kN/m$^2$, $q_{min} = 50.9$ kN/m$^2$, $FS_{overturning} = 5.25$.

Culmann's graphical method for active thrust on a wall with a cohesionless backfill and no surcharge:

1. Draw the wall $AB$ and the backfill surface to scale and mark the heel $B$.
2. Draw the $\phi$-line $BD$ from $B$ at angle $\phi$ to the horizontal.
3. Draw the pressure line $BE$ from $B$ making angle $\psi = \theta-\delta$ with $BD$ ($\theta$ = wall back angle with horizontal, $\delta$ = wall friction).
4. Draw trial failure planes $BC_1, BC_2, BC_3,\dots$ from $B$ to the ground surface.
5. Find the area of each wedge $ABC_i$ and its weight $W_i = \gamma\times\text{area}$ per metre run.
6. To a suitable scale, mark $W_1, W_2, W_3,\dots$ on $BD$ as $b_1, b_2, b_3,\dots$
7. From each $b_i$ draw a line parallel to $BE$ to meet $BC_i$ at $c_i$.
8. Join $c_1, c_2, c_3,\dots$ with a smooth curve (Culmann curve).
9. Draw a tangent to the curve parallel to $BD$; let it touch at $c_m$.
10. Draw $c_m b_m$ parallel to $BE$ to meet $BD$. Its length, converted by the force scale, is the active thrust $P_a$. $BC_m$ is the critical failure plane.
11. $P_a$ acts at $H/3$ above the base of the wall, inclined at $\delta$ to the wall normal.

    A |\                      ground
      |  \_____________________ C1 C2 Cm C3
      |     curve ..  .
      |        ..  /
    B +---b1--b2--bm--b3--> D   (phi from horizontal)
          BE at psi from BD, weights on BD

The plot shows lateral force $P$ on the wall against wall movement $\Delta$ (away from or toward the soil).

  P
  ^
  |                                  ____ Pp (passive)
  |                              _.-'
  |                          _.-'
  |       P0 (at rest)    _.-'
  |       *            _.-'
  |        \       _.-'
  |         \___.-'  
  |   Pa ----'----   (active)
  +----+-----------+----------+---> movement
  <-- away from soil   0   toward soil -->
  (0.001-0.005H)            (0.01-0.05H)

• At zero movement the force is $P_0$ (at rest, $K_0$).
• Moving away from the soil, the force falls rapidly and reaches the minimum active value $P_a$ after a small movement (0.001-0.005 H in sand).
• Moving toward the soil, the force rises and reaches the maximum passive value $P_p$ only after a large movement (0.01-0.05 H in sand; more for clay).
• Order: $P_a < P_0 < P_p$.

In a purely cohesive soil ($\phi=0$) the active pressure at depth $z$ is $p_a = \gamma z - 2c$, which is negative (tension) near the surface.

• The tension zone extends to a depth $z_0 = \dfrac{2c}{\gamma}$ (zero pressure point). Soil cannot carry tension for long, so tension cracks form to this depth and the soil there gives no pressure on the wall.
• The pressure diagram is therefore a triangle from $z_0$ to the base, with maximum value $\gamma H - 2c$ at the bottom. Thus the active thrust is

$$P_a = \tfrac12(\gamma H-2c)(H-z_0) = \tfrac12\gamma H^2 - 2cH + \frac{2c^2}{\gamma}$$

and it acts at $(H-z_0)/3$ above the base.

• Neglecting the cracks (taking tension) would under-estimate the thrust. If the cracks fill with water, an extra hydrostatic thrust $\tfrac12\gamma_w z_0^2$ acts on the wall and increases pressure.

Rankine's theory is limited because it assumes a smooth, vertical wall back (no wall friction), a horizontal or uniformly sloping backfill and thrust parallel to the backfill surface. It cannot deal with an inclined wall back, a broken surface, or a rough wall; and it gives no direction for the thrust.

Coulomb's wedge theory removes these limits:

• It includes wall friction ($\delta$) and wall adhesion, so the result is more realistic for rough concrete/masonry walls.
• It allows an inclined wall back and sloping backfill.
• The thrust direction (at $\delta$ to the normal) is obtained, and loads such as surcharge and line loads can be included through the wedge weight (and the Culmann graphical method).
• It handles irregular ground surfaces. Hence Coulomb's theory is more general and gives the more accurate active thrust, a step ahead of Rankine.

General guidelines before checking stability

1. Choose the wall type from height: gravity wall for about 3-4 m, cantilever for 4-8 m, counterfort/buttressed for more than 8 m or with large surcharge.
2. Assume trial dimensions from the usual proportions (base width 0.5-0.7 H, stem and slab thickness H/12-H/10, toe about B/3).
3. Find the soil data: unit weight, $\phi$, $c$ of backfill and foundation soil, water table, bearing capacity.
4. List the loads: self-weight, soil weight over heel, active thrust, surcharge, water pressure, seismic force.
5. Select the earth pressure theory (Rankine for vertical back with smooth wall assumption; Coulomb for rough and inclined walls) and compute the active pressure; ignore passive pressure in the top 0.5-1 m, or take only a reduced part.
6. Check overturning, sliding, bearing, tension at base and overall stability.

Design considerations

• Backfill: preferably free-draining granular material (sand, gravel); avoid clay, which gives high and variable pressure.
• Drainage: weep holes (100 mm at 2-3 m spacing), a filter or geotextile and a longitudinal drain at the heel to remove water pressure.
• Stability requirements: FS against overturning $\ge 2$, sliding $\ge 1.5$, bearing $\ge 3$; resultant within the middle third.
• Foundation depth: below frost, scour and softened soil; use a shear key if sliding is a problem.
• Structural design: stem as a cantilever, heel and toe slabs in bending and shear; reinforcement on tension faces.
• Joints: expansion joints about 20-30 m and contraction joints about 10 m in concrete walls.
• Seismic effect and surcharge where relevant; allow for the effect of tension cracks in cohesive backfill.

A soil mass is in a state of plastic equilibrium when every point in it is on the verge of shear failure; the shear stress at each point equals the shear strength given by the Mohr-Coulomb criterion $\tau_f = c + \sigma\tan\phi$. The Mohr circle of stress at every point then just touches the failure envelope.

• Rankine's general state: this condition exists throughout the whole semi-infinite soil mass, which happens when the soil is stretched or compressed horizontally enough.
• Active state: the soil expands horizontally (e.g. wall moves away). The vertical stress $\sigma_v$ is the major principal stress and the horizontal stress falls to the minimum, $\sigma_h = K_a\sigma_v - 2c\sqrt{K_a}$.
• Passive state: the soil is compressed horizontally. The horizontal stress becomes the major principal stress, $\sigma_h = K_p\sigma_v + 2c\sqrt{K_p}$.
• Failure planes make angle $45^\circ \pm \phi/2$ with the major principal plane. It is the basis of Rankine's earth pressure theory.

Earthquake effect on earth pressure

Ground shaking adds inertia forces to the soil wedge behind the wall. In the pseudo-static (Mononobe-Okabe) method these are taken as horizontal $k_h W$ and vertical $k_v W$ forces, where $k_h$ and $k_v$ are seismic coefficients.

• The active thrust increases to $P_{ae} = \tfrac12\gamma H^2(1-k_v)K_{ae}$, with $K_{ae} > K_a$.
• The passive resistance decreases.
• The resultant acts higher (the dynamic increment about $0.6H$ above the base), increasing overturning moment.
• Water-saturated loose sand may liquefy, causing large extra pressure. Walls in seismic zones must be designed for these forces.

Horizontal strain needed to reach the active state

• (i) Coarse-grained soil (sand): very small, about 0.001 to 0.005 times $H$ (0.1 to 0.5%) of wall movement; dense sand needs less, loose sand more.
• (ii) Fine-grained soil (clay): much larger, about 0.01 to 0.04 times $H$ (1 to 4%).

For a vertical cut in clay with $\phi = 0$, cohesion $c$ and unit weight $\gamma$, Rankine's active pressure at depth $z$ is

$$p_a = \gamma z - 2c$$

since $K_a = 1$. The pressure is zero at $z_0 = 2c/\gamma$ and tensile above it.

For an unsupported (vertical) cut, the soil is stable up to the height at which the total active thrust over the full height is zero; i.e. the compressive force below balances the tension above, so no support is needed:

$$\begin{aligned} P_a &= \int_0^{H_c}(\gamma z - 2c)\,dz = \tfrac12\gamma H_c^2 - 2cH_c = 0 \\ H_c &= \frac{4c}{\gamma} \end{aligned}$$

Result: $H_c = 4c/\gamma$ (for $c = c_u$ in undrained condition, $H_c = 4c_u/\gamma$). Using a factor of safety, the safe height is $H = H_c/F$.

Note: because soil cannot carry tension, tension cracks of depth $2c/\gamma$ form; considering them, the practical safe height is smaller (the theoretical value with the cracks is $2.67c/\gamma$ by Terzaghi's analysis), so a factor of safety of 2 or more is used.

For a purely cohesive soil, $\phi=0$, $K_a = 1$, so Rankine's active pressure at depth $z$ is

$$p_a = \gamma_t z - 2c$$

Pressure diagram

• At the top ($z=0$): $p_a = -2c$ (tension).
• Zero at depth $z_0 = 2c/\gamma_t$.
• At the base ($z=H$): $p_a = \gamma_t H - 2c$.

 z=0  |<-2c            (tension, ignored)
      |
 z0   |----+ 0         z0 = 2c/gt
      |     \
      |      \
 H    |-------+ gt*H - 2c

Soil cannot carry tension, so a tension crack forms to depth $z_0$ and the tensile part is ignored. The pressure then acts only on a triangle of height $(H - z_0)$ with base value $\gamma_t H - 2c = \gamma_t(H - z_0)$.

Magnitude of active force

$$\begin{aligned} P_a &= \tfrac12(\gamma_t H - 2c)(H - z_0) = \tfrac12\gamma_t (H - z_0)^2 \\ &= \tfrac12\gamma_t H^2 - 2cH + \frac{2c^2}{\gamma_t} \end{aligned}$$

Line of action is at the centroid of the triangle, one-third of its height above the base:

$$\bar{y} = \frac{H - z_0}{3} = \frac{H - 2c/\gamma_t}{3} \ \text{(from the base)}$$

If the tension zone were not neglected (no crack), $P_a = \tfrac12\gamma_t H^2 - 2cH$, but this under-estimates the thrust on the wall, so the crack is allowed for in design.

Assumptions: smooth vertical wall, horizontal backfill, no groundwater (none is given), Rankine theory.

Coefficients

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-15^\circ) = 0.3333,\quad \sqrt{K_{a1}} = 0.5774 \\ K_{a2} &= \tan^2(45^\circ-17^\circ) = 0.2827 \end{aligned}$$

Layer 1 (0-4 m, $c=10$, $\phi=30^\circ$, $\gamma=18$): $p = K_{a1}\gamma z - 2c\sqrt{K_{a1}}$

• $z=0$: $p = -2\times10\times0.5774 = -11.55$ kN/m$^2$ (tension).
• Zero at $z_0 = 2c/(\gamma\sqrt{K_{a1}}) = 1.925$ m; the tension zone is ignored (tension crack).
• $z=4$ m (top layer): $p = 0.3333\times72 - 11.55 = 12.45$ kN/m$^2$.

Layer 2 (4-8 m, $c=0$, $\phi=34^\circ$, $\gamma=20$): $\sigma_v$ at 4 m $= 18\times4 = 72$ kN/m$^2$.

• Top: $p = 0.2827\times72 = 20.36$ kN/m$^2$.
• Base ($z=8$ m): $\sigma_v = 72 + 20\times4 = 152$, $p = 0.2827\times152 = 42.97$ kN/m$^2$.

 0  |  (tension zone, 0 to 1.92 m, ignored)
    |
1.92|\
    | \
    |  \
 4m |---12.45
    |-- 20.36
    |   \
    |    \
 8m |-----42.97

Forces per metre run (arm above base)

$$\bar y = 283.79/139.58 = 2.03\ \text{m above the base}$$

Answer: resultant active thrust $P_a = 139.6$ kN/m, horizontal, acting 2.03 m above the base (distribution as shown).

Assumptions: smooth vertical wall back, horizontal ground, no water. Rankine passive pressure $p_p = K_p\sigma_v + 2c\sqrt{K_p}$, with total height 10 m of soil (5 m + 5 m) acting on the back.

Coefficients

$$\begin{aligned} K_{p1} &= \tan^2(45^\circ+15^\circ) = 3.000 \\ K_{p2} &= \tan^2(45^\circ+7.5^\circ) = 1.6984,\quad \sqrt{K_{p2}} = 1.3032 \\ 2c\sqrt{K_{p2}} &= 2\times30\times1.3032 = 78.19\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Top layer: $z=0$: $p=0$; $z=5$ m: $\sigma_v=18\times5=90$, $p = 3.000\times90 = 270.0$.
• Bottom layer top: $p = 1.6984\times90 + 78.19 = 231.05$.
• Bottom layer base ($z=10$ m): $\sigma_v = 90+20\times5 = 190$, $p = 1.6984\times190 + 78.19 = 400.89$.

 0 |
   |\
   | \
 5 |--270.0
   |--231.05
   |     \
10 |------400.89

Forces (arm above the base of the 10 m wall)

$$\bar y = 8095.78/2254.85 = 3.59\ \text{m}$$

Answer: total passive thrust $P_p = 2255$ kN/m, acting horizontally at 3.59 m above the base (bottom of the wall).

Assumption: the wall height is not stated, so $H = 6$ m is assumed (the method is the same for any $H$). Rankine's theory for a sloping backfill ($\beta = 20^\circ$, $\phi = 30^\circ$) gives a pressure parallel to the slope.

Active coefficient

$$K_a = \cos\beta\,\frac{\cos\beta - \sqrt{\cos^2\beta - \cos^2\phi}}{\cos\beta + \sqrt{\cos^2\beta - \cos^2\phi}} = 0.4142$$

(a) Dry backfill

$$\begin{aligned} P_a &= \tfrac12 K_a\gamma H^2 = 0.5\times0.4142\times18\times6^2 = 134.2\ \text{kN/m} \end{aligned}$$

It acts at $H/3 = 2.00$ m above the base, inclined at $20^\circ$ to the horizontal (horizontal component 126.1 kN/m, vertical component 45.9 kN/m).

(b) Water table 2 m below the top

Top 2 m dry ($\gamma=18$), lower 4 m submerged ($\gamma'=12$).

• At 2 m: $p = 0.4142\times18\times2 = 14.91$ kN/m$^2$.
• At 6 m: $p = 0.4142\times(36 + 12\times4) = 34.79$ kN/m$^2$.
• Water pressure at the base: $9.81\times4 = 39.24$ kN/m$^2$.

Horizontal total $= 114.32\cos20^\circ + 78.48 = 185.9$ kN/m acting at

$$\bar y = \frac{107.43\times2.12 + 78.48\times1.333}{185.9} = 1.79\ \text{m}$$

and the vertical component of the soil thrust is 39.1 kN/m (resultant 190.0 kN/m).

Answer:

• Dry: $P_a = 134.2$ kN/m at 2.00 m above the base.
• With the water table: the effective soil thrust falls to 114.3 kN/m (point 2.12 m), but with water pressure (78.5 kN/m) the total horizontal thrust rises to 185.9 kN/m (+60 kN/m, about 47% more), and its point of application drops to 1.79 m above the base.