Active Earth Pressure Calculator: Rankine Method Explained Step by Step

Published 28 August 2026

Why you need this calculation

Every retaining wall, every excavation shoring, every buried wall eventually needs an answer to a simple question: how hard does the soil behind it push? That answer drives the wall's design — thickness, reinforcement, checks against overturning and sliding — and, despite appearances, it can be worked out by hand in a few minutes, with a single formula.

Below is the theory, a complete numerical example, and at the end a free active earth pressure calculator that does exactly this calculation, with a pressure diagram and an instant result.

What active earth pressure is

Soil left free behind a wall tends to spread laterally and push against it. If the wall can yield slightly — rotating or translating away from the soil — the soil mass reaches the active state: the state of minimum thrust, the one normally used to design ordinary retaining walls.

Rankine's theory relates horizontal pressure to vertical pressure through a coefficient, $K_a$:

$$K_a = \tan^2\left(45° - \frac{\varphi}{2}\right)$$

where $\varphi$ is the soil's internal friction angle. For a sand with $\varphi = 30°$, $K_a = 1/3$ — horizontally the soil pushes with a third of what it pushes vertically.

The horizontal pressure at any depth follows from the vertical effective stress $\sigma'_v$:

$$p_a = K_a \sigma'_v - 2c\sqrt{K_a} \ge 0$$

The term $-2c\sqrt{K_a}$ is the beneficial effect of cohesion: cohesive soil reduces the thrust, and near the surface the active pressure can even turn negative (tension). Since soil cannot carry tension, the pressure is capped at zero over that depth — which is why a crack often appears behind walls in clayey ground.

Worked example, step by step

Take a retaining wall with:

  • height $H = 5$ m
  • soil unit weight $\gamma = 18$ kN/m³
  • friction angle $\varphi = 30°$ (so $K_a = 1/3$)
  • cohesion $c = 0$ (sand)
  • surface surcharge $q = 10$ kN/m²
  • no groundwater

Step 1 — the coefficient. $K_a = \tan^2(45° - 15°) = \tan^2(30°) = 0{,}333$.

Step 2 — pressure at the surface (surcharge only, $\sigma'_v = 0$):

$$p_{top} = K_a \cdot q = 0{,}333 \times 10 = 3{,}33 \text{ kN/m}^2$$

Step 3 — pressure at the base of the wall ($\sigma'_v = \gamma H = 18 \times 5 = 90$ kN/m²):

$$p_{base} = K_a (\sigma'_v + q) = 0{,}333 \times (90 + 10) = 33{,}3 \text{ kN/m}^2$$

The pressure diagram is trapezoidal: a uniform band of $3{,}33$ kN/m² (from the surcharge) plus a triangle growing linearly to $30$ kN/m² (from the soil's own weight).

Step 4 — the resultant force, by summing the two components of the trapezoid: the uniform band $K_a q H$ and the triangle $\tfrac{1}{2} K_a \gamma H^2$:

$$E = K_a q H + \tfrac{1}{2} K_a \gamma H^2 = 16{,}7 + 75{,}0 = 91{,}7 \text{ kN/m}$$

Step 5 — the point of application, measured from the base of the wall, by weighting the two components by their lever arms ($H/2$ for the band, $H/3$ for the triangle):

$$z_R = \frac{16{,}7 \times 2{,}5 + 75{,}0 \times 1{,}67}{91{,}7} \approx 1{,}82 \text{ m}$$

That is it — five steps and one formula give you everything needed to size the wall: the peak pressure, the total force, and where it acts.

What changes with a water table

Take the same wall, but add a water table at $z_w = 2$ m above the base — 3 m of "dry" soil above the water, 2 m submerged, with a submerged unit weight $\gamma' = 10$ kN/m³. Here is the diagram the calculator produces for this case:

SOIL PROFILE ▼ water q=10.0 kN/m² z=0 H=5.0m zw=2.0m LATERAL PRESSURES SOIL FREE 3.33 48.00 kN/m² z=0 H=5.0m E=106.3 kN/m
The exact diagram produced by the calculator for the water-table example — soil profile (left) and resulting lateral pressures (right).

Below the water table two things happen at once: the soil's weight drops (submerged, $\gamma'$ instead of $\gamma$) — but on top of the active soil pressure, the hydrostatic water pressure is added separately, and it does not go through $K_a$ (water has no internal friction; it pushes directly, with coefficient 1).

At the surface ($z = 5$ m), nothing changes from the dry example: $p = 3{,}33$ kN/m² (surcharge only — the water is much lower).

At the water table ($z = 2$ m), still a soil effect, but now $\sigma'_v = \gamma(H - z_w) = 18 \times 3 = 54$ kN/m²:

$$p = K_a(\sigma'_v + q) = 0{,}333 \times (54 + 10) = 21{,}33 \text{ kN/m}^2$$

At the base of the wall ($z = 0$), the effective stress keeps growing — but with $\gamma'$, not $\gamma$ — and the hydrostatic pressure $u = \gamma_w z_w = 10 \times 2 = 20$ kN/m² is added:

$$\sigma'_v = \gamma(H-z_w) + \gamma' z_w = 54 + 10 \times 2 = 74 \text{ kN/m}^2$$

$$p = K_a(\sigma'_v + q) + u = 0{,}333 \times 84 + 20 = 48{,}00 \text{ kN/m}^2$$

The diagram is no longer a single trapezoid: it has a kink right at the water table, where the slope changes. Broken down into four simple pieces — the uniform band from the surcharge, the triangle of dry soil above the water, the trapezoid of submerged soil below it, and the triangle of hydrostatic pressure — the resultant force and its point of application work out as follows:

Component Force (kN/m) Lever arm from base (m)
Uniform band (surcharge)16.672.50
Dry soil triangle27.003.00
Submerged soil trapezoid42.670.95
Hydrostatic triangle20.000.67
Total106.33zR ≈ 1.66

Compared with the dry example above — same geometry, same surcharge, water added — the total force grows from 91.7 to 106.3 kN/m (+16%), and the point of application drops from 1.82 m to 1.66 m: water does not just increase the thrust, it also moves it closer to the base, where overturning moment matters less but shear and uplift matter more. That is why drainage behind a wall is not optional.

Assumptions and limitations to keep in mind

  • Rankine's theory assumes a vertical wall, horizontal backfill, and no wall friction. For a battered wall, sloped backfill, or wall friction, Coulomb's theory gives different results.
  • This covers active pressure only — the minimum. At-rest pressure ($K_0$, a rigid wall that cannot move) and passive pressure ($K_p$, a wall pushing into the soil) are separate calculations.
  • The analysis gives the loading; wall stability (overturning, sliding, bearing capacity) is checked separately.
  • Seismic conditions are not covered (Mononobe-Okabe dynamic thrust).

Calculate it online, no spreadsheet and no PLAXIS

For a quick check — or for a complete design brief with a diagram — you don't need a commercial FEM package like PLAXIS, nor a spreadsheet built from scratch. JointLib's active earth pressure calculator runs exactly the steps above, instantly, right in your browser: enter the wall height, soil parameters, surcharge, and water table (if any), and get the pressure diagram, the resultant force, and its point of application.

Related calculations

An unhandled error has occurred. Reload 🗙