Every retaining wall, basement wall, sheet pile, and braced excavation must resist lateral earth pressure — the horizontal force a soil mass exerts against a structure. Rankine’s theory (1857) provides the simplest closed-form equations to estimate active and passive pressures in cohesionless and cohesive soils. This guide covers the theory, derives the key formulas, includes pressure distribution diagrams, and works through two full numerical examples.
What Is Lateral Earth Pressure?
Lateral earth pressure is the horizontal stress exerted by the soil on a vertical or near-vertical structure. Its magnitude depends on three things:
- Soil properties — unit weight (γ), friction angle (φ), cohesion (c).
- Wall movement — whether the wall moves away from the soil (active), into the soil (passive), or stays fixed (at-rest).
- Surcharge and water — any surface loads and the water table position.
Three States of Lateral Pressure
| State | Wall movement | Coefficient | Pressure |
|---|---|---|---|
| At-rest (K₀) | No movement | K₀ ≈ 1 − sin φ | Intermediate |
| Active (Ka) | Wall moves away from soil | Ka = tan²(45° − φ/2) | Minimum (soil expands) |
| Passive (Kp) | Wall pushed into soil | Kp = tan²(45° + φ/2) | Maximum (soil compresses) |
The active state requires only 0.1–0.4% of wall height (H) in movement. The passive state needs much more — about 2–5% of H for dense sand — because the soil must be compressed to its full strength.
Rankine’s Assumptions
- The wall is vertical and smooth (no wall friction — the key simplification compared to Coulomb’s theory).
- The ground surface behind the wall is horizontal (Rankine extended this to inclined backfills, but the basic case assumes level ground).
- The soil is a semi-infinite, homogeneous mass.
- The failure surface is a plane inclined at (45° ± φ/2) to the horizontal.
- The soil obeys the Mohr-Coulomb failure criterion.
Rankine Active Pressure (Cohesionless Soil)
For a dry cohesionless soil (c = 0):
Ka = tan²(45° − φ/2) = (1 − sin φ) / (1 + sin φ)
Lateral pressure at depth z:
σh = Ka × γ × z
Total active thrust on a wall of height H:
Pa = ½ × Ka × γ × H²
This resultant acts at H/3 from the base (triangular pressure distribution).
Rankine Passive Pressure (Cohesionless Soil)
Kp = tan²(45° + φ/2) = (1 + sin φ) / (1 − sin φ)
Lateral pressure at depth z:
σh = Kp × γ × z
Total passive resistance:
Pp = ½ × Kp × γ × H²
Note: Kp = 1/Ka. For φ = 30°, Ka = 0.333 and Kp = 3.0 — passive resistance is 9 times the active pressure.
Rankine Pressure for Cohesive Soil (c–φ Soil)
When the soil has both cohesion and friction (c–φ soil):
Active: σh = Ka × γ × z − 2c√Ka
Passive: σh = Kp × γ × z + 2c√Kp
The cohesion term −2c√Ka creates a tension zone at the top of the wall in the active case. The depth of the tension crack is:
zc = 2c / (γ √Ka)
In design, the pressure in this tension zone is usually ignored (set to zero), and tension cracks are assumed to fill with water if present.
Effect of Surcharge
A uniform surcharge q on the ground surface adds a constant lateral pressure over the full height:
Δσh = Ka × q (active case)
This converts the triangular distribution to trapezoidal — the resultant shifts upward from H/3.
Effect of Water Table
Below the water table, use submerged unit weight γ’ = γsat − γw for the soil pressure component, and add hydrostatic water pressure γw × hw separately. Water pressure acts on the wall regardless of soil state — it is not multiplied by Ka or Kp.
Worked Example 1: Cohesionless Soil
Given: A 6 m high retaining wall retains dry sand with γ = 18 kN/m³ and φ = 30°. Calculate the active thrust and its point of application.
Step 1 — Active earth pressure coefficient:
Ka = tan²(45° − 30°/2) = tan²(30°) = (0.5774)² = 0.333
Step 2 — Pressure at base (z = 6 m):
σh = Ka × γ × H = 0.333 × 18 × 6 = 36 kPa
Step 3 — Total active thrust:
Pa = ½ × 0.333 × 18 × 6² = ½ × 0.333 × 18 × 36 = 108 kN/m
Step 4 — Point of application:
y = H/3 = 6/3 = 2 m from the base
Result: The wall must resist an active thrust of 108 kN per metre run, acting 2 m above the base.
Worked Example 2: Cohesive Soil with Surcharge
Given: A 5 m high wall retains a c–φ soil with γ = 19 kN/m³, φ = 25°, and c = 10 kPa. A uniform surcharge q = 15 kPa acts on the backfill surface.
Step 1 — Coefficients:
Ka = tan²(45° − 25°/2) = tan²(32.5°) = (0.6379)² = 0.407
√Ka = 0.638
Step 2 — Pressure at z = 0 (top):
σh = Ka × q − 2c√Ka = 0.407 × 15 − 2 × 10 × 0.638 = 6.1 − 12.76 = −6.66 kPa (tension — set to 0)
Step 3 — Depth where pressure becomes zero:
Ka(γz + q) = 2c√Ka
z = (2c√Ka − Kaq) / (Kaγ) = (12.76 − 6.1) / (0.407 × 19) = 6.66 / 7.73 = 0.86 m
Step 4 — Pressure at base (z = 5 m):
σh = 0.407 × (19 × 5 + 15) − 2 × 10 × 0.638 = 0.407 × 110 − 12.76 = 44.77 − 12.76 = 32.01 kPa
Step 5 — Active thrust (from z = 0.86 m to 5 m):
Net height = 5 − 0.86 = 4.14 m
Pa = ½ × 32.01 × 4.14 = 66.3 kN/m
Acting at 4.14/3 = 1.38 m above the base.
Rankine vs Coulomb: When to Use Which?
| Criterion | Rankine | Coulomb |
|---|---|---|
| Wall friction (δ) | Ignored (δ = 0) | Included |
| Wall face | Must be vertical | Can be inclined |
| Backfill surface | Horizontal (basic case) | Can be inclined |
| Failure surface | Plane | Plane (can overestimate Kp) |
| Complexity | Simpler | Slightly more complex |
| Typical use | Preliminary design, sheet piles, smooth walls | Gravity walls, rough concrete walls |
For most exam problems and preliminary designs, Rankine is preferred for its simplicity. For final design of gravity retaining walls where wall friction is significant, Coulomb or log-spiral methods are more accurate.
Failure Plane Inclination
In the Rankine active case, the failure plane makes an angle of (45° + φ/2) with the horizontal. In the passive case, the failure plane is at (45° − φ/2).
For φ = 30°: the active failure plane is at 60° from horizontal, and the passive failure plane is at 30° from horizontal.
Frequently Asked Questions
What is the difference between active and passive earth pressure?
Active pressure develops when the wall moves away from the soil — the soil expands, reaches its minimum strength state, and pushes against the wall with the least force. Passive pressure develops when the wall is pushed into the soil — the soil is compressed, mobilizes its maximum resistance, and pushes back hard. Passive pressure is always much larger than active pressure.
Why is Rankine’s theory called a “lower bound” for passive pressure?
Because Rankine assumes a plane failure surface. In reality, the passive failure surface curves (a log-spiral). By assuming a plane, Rankine overestimates the passive earth pressure coefficient — it is not a lower bound in the traditional sense. For accurate passive pressure, the log-spiral method or Caquot-Kerisel tables should be used.
How much wall movement triggers the active state?
Very little. For dense sand, about 0.1% of the wall height (0.6 mm for a 6 m wall). For loose sand, about 0.4% (2.4 mm for a 6 m wall). This is why active pressure is the default design condition — the wall almost always moves enough to reach it.
Does Rankine theory apply to clayey soils?
Yes, with modifications. The cohesion term (−2c√Ka or +2c√Kp) accounts for cohesion. However, for undrained conditions in saturated clay, the total stress analysis uses φu = 0 and the lateral pressure simplifies to σh = γz − 2cu (active) or σh = γz + 2cu (passive).
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