Earth Pressure — Rankine & Coulomb Theories
At-rest, active and passive earth pressure, Ka and Kp formulas, wall friction, pressure diagrams for layered soils and surcharge, and GATE-level worked examples
Last Updated: March 2026
Key Takeaways
- Three states of lateral stress: at-rest (K0), active (Ka), and passive (Kp); active requires small wall movement away from soil, passive requires large movement into soil.
- Rankine Ka = (1−sinφ)/(1+sinφ) = tan²(45°−φ/2); Kp = (1+sinφ)/(1−sinφ) = tan²(45°+φ/2); Ka × Kp = 1 always.
- At-rest pressure coefficient: K0 = 1−sinφ (Jaky’s formula for NC soil); K0,OC = K0,NC × OCR0.5.
- Active pressure intensity: σa = Kaγz − 2c√Ka; tension crack depth zc = 2c/(Kaγ√Ka) = 2c/(γ√Ka) for Ka cohesive soils.
- Coulomb’s theory accounts for wall friction (δ) and backfill inclination (β); gives higher passive pressure and slightly different active pressure than Rankine.
- Total active force Pa = ½KaγH² (cohesionless soil, level backfill) acting at H/3 from base.
- For c-φ soil, tension zone exists near the top; in practice the tension crack is assumed filled with water, adding hydrostatic pressure to the net active force.
1. States of Lateral Earth Pressure
A retaining wall or any embedded structure experiences lateral (horizontal) pressure from the soil it supports. The magnitude of this pressure depends on how much the wall moves relative to the soil. Three limiting states are defined:
| State | Wall Movement | Pressure Coefficient | Relative Magnitude |
|---|---|---|---|
| At-rest (K0) | No movement | K0 = 1 − sinφ | Ka < K0 < Kp |
| Active (Ka) | Wall moves away from soil (outward); soil expands | Ka = (1−sinφ)/(1+sinφ) | Minimum lateral pressure |
| Passive (Kp) | Wall moves into soil (inward); soil compressed | Kp = (1+sinφ)/(1−sinφ) | Maximum lateral pressure |
The wall displacement required to mobilise these states is markedly different: active pressure develops with very small outward wall movement (0.1–0.5 % of wall height for dense sands; 1–2 % for loose sands). Passive pressure requires ten to twenty times more displacement (1–5 % of wall height for dense; up to 10 % for loose sands). This asymmetry is crucial for design — passive resistance should only be relied upon where large displacements are acceptable.
2. At-Rest Earth Pressure (K0)
At-rest conditions apply when the wall or structure does not move — e.g., basement walls of rigid buildings, bridge abutments that are braced, and buried culverts.
2.1 Jaky’s Formula (Normally Consolidated Soil)
K0 = 1 − sinφ′
Valid for normally consolidated (NC) cohesionless soils
For φ′ = 30°: K0 = 0.50
For φ′ = 35°: K0 = 0.426
2.2 Overconsolidated Soil
K0,OC = K0,NC × OCR0.5 (Mayne & Kulhawy, 1982)
K0,OC can exceed 1.0 for heavily OC clays (e.g., OCR = 10 gives K0 ≈ 1.58 for φ′ = 30°)
2.3 At-Rest Pressure Intensity and Total Force
σ0 = K0 γ z (at depth z below surface)
Total at-rest force per unit length: P0 = ½ K0 γ H²
Acts at H/3 from base (triangular distribution)
3. Rankine’s Theory
Rankine (1857) derived lateral earth pressure for a cohesionless soil with a level backfill, assuming the wall is frictionless (smooth wall) and the failure occurs along planes inclined to the horizontal. Rankine’s theory is based on the state of plastic equilibrium throughout the soil mass.
3.1 Assumptions
- Soil is semi-infinite, isotropic, and homogeneous.
- The failure surface is a plane (not curved).
- The wall is vertical, smooth (no wall friction, δ = 0).
- Backfill surface is level (horizontal) — unless inclined backfill formulae are used.
- The soil is cohesionless in the basic formulation (c = 0, φ > 0).
3.2 Active and Passive Coefficients (Cohesionless, Level Backfill)
Ka = (1 − sinφ) / (1 + sinφ) = tan²(45° − φ/2)
Kp = (1 + sinφ) / (1 − sinφ) = tan²(45° + φ/2)
Ka × Kp = 1 (always, for Rankine level backfill)
3.3 Pressure Distribution (Cohesionless Soil, Level Backfill)
Active pressure intensity at depth z: σa = Ka γ z
Distribution: triangular (zero at top, maximum at base)
Total active force: Pa = ½ Ka γ H²
Acts at H/3 from base
Total passive force: Pp = ½ Kp γ H²
Acts at H/3 from base
3.4 Rankine Ka Values for Common φ
| φ (°) | Ka | Kp | Kp/Ka |
|---|---|---|---|
| 20 | 0.490 | 2.04 | 4.16 |
| 25 | 0.406 | 2.46 | 6.06 |
| 30 | 0.333 | 3.00 | 9.00 |
| 35 | 0.271 | 3.69 | 13.6 |
| 40 | 0.217 | 4.60 | 21.2 |
3.5 Inclined Backfill (Rankine)
For backfill inclined at angle β to horizontal (level backfill special case β = 0):
Ka = cosβ × [cosβ − √(cos²β − cos²φ)] / [cosβ + √(cos²β − cos²φ)]
Active pressure acts parallel to backfill surface (not horizontal)
Valid only when β ≤ φ
4. Active Pressure for c-φ Soils (Rankine)
4.1 Pressure Intensity
σa = Ka γ z − 2c √Ka
The −2c√Ka term represents cohesion reducing lateral pressure.
Passive: σp = Kp γ z + 2c √Kp
4.2 Tension Zone (Tension Crack)
Near the top of a cohesive backfill, σa becomes negative (tensile). Soil cannot sustain tension, so a tension crack forms to the depth zc where σa = 0:
Ka γ zc = 2c √Ka
zc = 2c / (γ √Ka)
For purely cohesive soil (φ = 0, Ka = 1): zc = 2cu / γ
4.3 Net Active Force Accounting for Tension Crack
Below the tension crack depth zc, the pressure is positive. The net active force acts over the height (H − zc):
Pa,net = ½ (KaγH − 2c√Ka)(H − zc)
Acts at (H − zc)/3 from base
With tension crack filled with water (critical condition):
Add hydrostatic force: Pwater = ½ γw zc²
This acts at zc/3 from base of tension crack
Total horizontal force = Pa,net + Pwater
4.4 Critical Height of Unsupported Cut (Hc)
Height at which net active force = 0 (no wall needed theoretically):
Hc = 4c / (γ √Ka) = 2 zc
For φ = 0: Hc = 4cu / γ
In practice, safe cut height = Hc/FOS (FOS = 1.5–2.0)
5. Passive Earth Pressure
Passive pressure is mobilised when the wall moves into the soil, compressing it. It is always much larger than active pressure for the same soil and wall height.
Passive pressure intensity: σp = Kp γ z + 2c √Kp
Total passive force: Pp = ½ Kp γ H² + 2c H √Kp
Acts at H/3 from base (for triangular distribution; resultant position shifts with cohesion)
Important caution: Passive pressure should be used in design with a factor of safety of 2–3 because: (1) large displacements are required to mobilise full passive resistance; (2) Coulomb’s passive pressure is an overestimate for walls with friction; (3) soil in the passive zone may be disturbed by construction. In stability analysis (e.g., sheet pile design), the full passive resistance is divided by an FOS before use.
6. Coulomb’s Wedge Theory
Coulomb (1776) proposed a different approach: assume a planar failure wedge sliding between the wall and the backfill, and find the inclination of the failure plane that gives the maximum active force (or minimum passive force) on the wall.
6.1 Assumptions
- The failure surface is a plane (Rankine allows general plastic state).
- The wall has friction: wall friction angle δ between soil and wall (0 ≤ δ ≤ φ).
- Backfill may be inclined at angle β to horizontal.
- Wall back may be inclined at angle α to vertical.
6.2 Coulomb’s Active Earth Pressure Formula
Pa = ½ Ka,C γ H²
Ka,C = sin²(α+φ) / {sin²α × sin(α−δ) × [1 + √(sin(φ+δ)sin(φ−β) / (sin(α−δ)sin(α+β)))]²}
For vertical wall (α = 90°) and level backfill (β = 0) and smooth wall (δ = 0): reduces to Rankine’s Ka
6.3 Effect of Wall Friction (δ)
| Wall Type | Typical δ |
|---|---|
| Smooth concrete (trowelled) | 10°–15° |
| Rough concrete (formed) | 15°–20° |
| Masonry / rough wall | 20°–25° |
| Common design value | δ = 2φ/3 |
Wall friction reduces active pressure (as wedge rotation is partially resisted) and significantly affects passive pressure. For passive pressure, Coulomb’s formula with wall friction overestimates because the assumed planar failure surface is a poor approximation when δ > 0 — a curved surface (Rankine or log-spiral) gives more accurate passive pressures.
6.4 Coulomb vs Rankine for Passive Pressure
Coulomb’s passive pressure significantly overestimates when wall friction is large (δ > φ/3). For design, use Rankine’s passive pressure (which is conservative — lower) or a curved failure surface analysis. For active pressure, Coulomb’s theory with friction gives results close to Rankine for most practical wall geometries.
7. Pressure Diagrams — Practical Cases
7.1 Surcharge on Backfill
A uniform surcharge load q (kPa) on the backfill surface adds a uniform pressure Kaq throughout the depth:
σa = Ka(γz + q)
Additional force due to surcharge: Psurcharge = Ka q H (acts at H/2)
Total: Pa,total = ½KaγH² + KaqH
7.2 Layered Soil
For a two-layer soil (Ka1, γ1, H1 over Ka2, γ2, H2):
Bottom of Layer 1 (just above interface): σa = Ka1γ1H1
Top of Layer 2 (just below interface): σa = Ka2γ1H1
(Vertical stress is continuous; horizontal stress has a step change at the boundary due to Ka change)
At depth z in Layer 2: σa = Ka2(γ1H1 + γ2(z−H1))
7.3 Water Table Within Backfill
Above water table: use moist γ with Ka. Below water table: use submerged γ′ for effective stress calculation, then add hydrostatic pressure separately:
At depth z below WT (zw = depth to WT):
Effective horizontal stress: σa′ = Ka(γzw + γ′(z−zw))
Pore water pressure: u = γw(z−zw)
Total horizontal pressure: σa,total = σa′ + u
Water pressure is typically much larger than the change in effective earth pressure — this is why drainage of retaining walls is so important. A clogged drain that allows water to build up behind a wall can dramatically increase the total lateral force.
8. Culmann’s Graphical Method
Culmann’s method is a graphical application of Coulomb’s wedge theory for finding active earth pressure when the backfill is irregular in shape (sloping, broken, with surcharges). It avoids the complex algebra of Coulomb’s formula.
The procedure involves drawing the wall geometry and backfill to scale, then constructing a series of trial wedges. For each wedge, the equilibrium polygon of three forces (weight W, reaction R on failure plane, and force P on wall) is drawn. The locus of the tips of the force polygons forms the Culmann line; the maximum ordinate of this line to the reference line gives Pa.
Culmann’s method is covered in detail in structural design of retaining walls with complex backfill — for GATE CE, the analytical Coulomb and Rankine formulas are sufficient.
9. Rankine vs Coulomb — When to Use Which
| Criterion | Rankine | Coulomb |
|---|---|---|
| Wall friction | Ignores (δ = 0); conservative for active | Accounts for δ; more realistic for active |
| Backfill inclination | Handles inclined backfill analytically | Handles via formula or Culmann graphically |
| Wall inclination | Only vertical walls (basic form) | Handles inclined wall backs |
| Active pressure | Slightly conservative (higher) than Coulomb with friction | More accurate with friction |
| Passive pressure | Conservative (safer) — use for design | Overestimates when δ > φ/3 — unsafe |
| Cohesive soils | Extended to c-φ soils with tension crack analysis | Primarily for cohesionless soils |
| GATE CE usage | Primary theory tested | Ka formula for inclined/rough walls |
10. Worked Examples
Example 1 — Active and Passive Pressure on a Retaining Wall (Cohesionless Soil)
Problem: A smooth vertical retaining wall 5 m high retains dry sand with φ = 30°, γ = 17 kN/m³. Find: (a) Ka and Kp, (b) total active force and its point of application, (c) total passive force.
(a) Pressure Coefficients
Kp = 1/Ka = 3.00
(b) Total Active Force
= ½ × 141.5 = 70.8 kN/m
Point of application: H/3 = 5/3 = 1.67 m from base
(c) Total Passive Force
Note Pp/Pa = Kp/Ka = 9.0 — passive is nine times the active force for φ = 30°.
Example 2 — Active Pressure with Surcharge and Water Table
Problem: A 4 m retaining wall has a backfill of sand (φ = 35°, γsat = 19 kN/m³, γdry = 17 kN/m³). Water table is at 1.5 m below the top of the wall. Uniform surcharge q = 20 kPa. Find the total horizontal force per unit length of wall.
Ka
Pressure at Key Depths
At z = 1.5 m (WT): σa′ = Ka(q + γdry×1.5) = 0.271×(20+17×1.5) = 0.271×45.5 = 12.33 kPa
u = 0 (at WT)
At z = 4.0 m (base): effective vertical stress = q + γdry×1.5 + γ′×2.5
γ′ = 19−9.81 = 9.19 kN/m³
σv′ = 20 + 17×1.5 + 9.19×2.5 = 20 + 25.5 + 22.975 = 68.48 kPa
σa′ = Ka×68.48 = 0.271×68.48 = 18.56 kPa
u = γw×2.5 = 9.81×2.5 = 24.53 kPa
Total horizontal pressure at base = 18.56 + 24.53 = 43.09 kPa
Total Horizontal Force (area of pressure diagram)
P2 (triangle 0 to 1.5m, dry soil): ½×Kaγdry×1.5² = ½×0.271×17×2.25 = 5.19 kN/m (acts at 4−0.5 = 3.5m from base)
P3 (rectangle 1.5m to 4m, from dry soil at WT): Kaγdry×1.5×2.5 = 0.271×17×1.5×2.5 = 17.32 kN/m (acts at 1.25m)
P4 (triangle 1.5m to 4m, submerged soil): ½×Kaγ′×2.5² = ½×0.271×9.19×6.25 = 7.79 kN/m (acts at 2.5/3 = 0.833m)
P5 (hydrostatic triangle): ½×γw×2.5² = ½×9.81×6.25 = 30.66 kN/m (acts at 0.833m)
Total P = 21.68 + 5.19 + 17.32 + 7.79 + 30.66 = 82.64 kN/m
Example 3 — Tension Crack in Cohesive Backfill
Problem: A clay backfill has c = 20 kPa, φ = 15°, γ = 18 kN/m³. The wall height H = 6 m. Find: (a) Ka, (b) tension crack depth zc, (c) net active force without and with water in the crack.
(a) Ka
√Ka = 0.767
(b) Tension Crack Depth
(c) Net Active Force — Without Water
= 63.50 − 30.68 = 32.82 kPa
Active zone height = H − zc = 6 − 2.90 = 3.10 m
Pa = ½ × 32.82 × 3.10 = 50.9 kN/m
Acts at 3.10/3 = 1.03 m from base
(d) Net Active Force — With Water in Crack
Acts at zc/3 = 0.97 m above base of crack = 0.97 + 3.10 = 4.07 m from wall base
Total = 50.9 + 41.25 = 92.15 kN/m (water nearly doubles the force!)
Example 4 — GATE-Style: At-Rest Pressure on Basement Wall
Problem (GATE CE type): A rigid basement wall 3.5 m high retains a saturated clay (K0 = 0.55, γsat = 19 kN/m³). Water table is at the top of the wall. Find the total lateral force per unit length of wall on the basement wall.
Effective at-rest pressure at base:
σ0′ = K0 γ′ H = 0.55 × 9.19 × 3.5 = 17.69 kPa
Total effective force: P0′ = ½ × 17.69 × 3.5 = 30.96 kN/m (at H/3 = 1.17m)
Hydrostatic force: Pw = ½ × 9.81 × 3.5² = ½ × 9.81 × 12.25 = 60.09 kN/m (at H/3 = 1.17m)
Total lateral force = 30.96 + 60.09 = 91.05 kN/m
11. Common Mistakes
Mistake 1: Applying Rankine’s Formula to Walls with Friction
What happens: Rankine’s Ka = (1−sinφ)/(1+sinφ) is used for a rough wall (with wall friction δ > 0). Rankine assumes a smooth wall (δ = 0); using it for a rough wall overestimates active pressure by 10–20 % (conservative for active, but incorrect theoretically).
Fix: For smooth walls (δ = 0) → Rankine. For rough walls (δ > 0) or inclined walls → Coulomb. For GATE problems, if δ is given, use Coulomb’s formula.
Mistake 2: Using Full Passive Resistance in Sliding Stability Checks
What happens: The full Coulomb passive resistance is used without applying a factor of safety, leading to an unconservative sliding check for retaining walls and sheet piles.
Fix: For sliding stability, use Pp/FOS (FOS = 2–3 for passive resistance) because: (1) full passive requires large displacement; (2) Coulomb overestimates passive with friction; (3) passive zone soil may be disturbed. For sheet pile design, net passive resistance per IS 4651 is used with appropriate mobilisation factors.
Mistake 3: Forgetting Hydrostatic Pressure When Water Table is Present
What happens: Kaγsatz is used for the horizontal pressure below the water table, without separately adding the hydrostatic pressure γw(z−zw). This significantly underestimates total lateral pressure (water pressure is often larger than the effective earth pressure).
Fix: Below the water table: total horizontal pressure = effective earth pressure (Kaγ′z component) + pore water pressure (γw×depth below WT). Calculate separately and add.
Mistake 4: Placing the Resultant Active Force at H/3 When Surcharge is Present
What happens: When a surcharge adds a uniform rectangular pressure block to the triangular earth pressure diagram, the combined pressure diagram is trapezoidal, not triangular. The resultant does not act at H/3 but at a higher point.
Fix: Calculate the resultant of each component separately (rectangle and triangle) and find the combined moment arm about the base. The resultant acts at: ȳ = (Prectangle×H/2 + Ptriangle×H/3) / (Prectangle + Ptriangle).
Mistake 5: Ignoring the Tension Zone in Active Pressure of Cohesive Soils
What happens: The full triangular pressure diagram (including the negative/tensile region near the top) is used to calculate the total active force. Integrating the negative portion reduces the apparent total force, leading to an unconservative design.
Fix: The tension zone is ignored (soil cannot pull on the wall). Total active force is calculated only from the positive pressure region below zc. For the most critical (largest) force, assume the tension crack is filled with water and add the hydrostatic force from the crack.
12. Frequently Asked Questions
Q1. Why is passive pressure so much larger than active pressure for the same soil and wall height?
Active pressure develops when the soil is allowed to expand — the soil grains move apart, interlocking decreases, and the horizontal stress drops to its minimum stable value. Only the residual friction and cohesion prevent further expansion. Passive pressure develops when the soil is compressed — grains are pushed together, interlocking increases dramatically, and you are literally trying to shear a wedge of soil upward and outward against both its own weight and its friction. The mechanics are fundamentally different. For φ = 30°, Kp/Ka = 3.0/0.333 = 9. For φ = 40°, this ratio grows to about 21. The ratio Kp/Ka = tan⁴(45°+φ/2) increases rapidly with φ, explaining why passive pressure becomes increasingly dominant in dense soils.
Q2. What is the physical meaning of the tension zone in cohesive soils?
In a cohesive backfill, cohesion provides tensile strength — the soil can resist being pulled apart by a small tensile stress. Near the top of a retaining wall, the active pressure formula gives a negative value (tension), meaning the soil would need to pull on the wall to maintain equilibrium. In reality, the soil separates from the wall (or a tension crack forms in the soil) to a depth zc where the tensile stress reaches the cohesive tensile strength. Below zc, compression takes over and the wall is pushed outward. The tension crack is critical for stability because: (1) it reduces the effective length of the wall-soil contact, concentrating forces on the lower portion; (2) it can fill with rainwater, dramatically increasing the total lateral force — a common trigger for retaining wall failures after heavy rainfall.
Q3. How is Coulomb’s theory different from Rankine’s at a fundamental level?
Rankine’s theory starts with the state of stress in the soil mass and derives pressure from the condition of incipient plastic failure throughout the entire soil body — it is a stress field approach. The failure planes emerge naturally from the Mohr-Coulomb criterion. Coulomb’s theory starts with a force equilibrium of a single rigid triangular wedge of soil, assuming a planar failure surface, and finds the inclination that maximises the force on the wall — it is a limit equilibrium approach. Coulomb’s method is more versatile (handles wall friction, inclined walls, irregular backfills) but less rigorous for passive pressure. Rankine’s method is more theoretically elegant but limited to smooth vertical walls and level or simply inclined backfills.
Q4. Why must lateral earth pressure always be accompanied by a water pressure analysis for submerged or partially submerged walls?
Lateral earth pressure theories (Rankine and Coulomb) compute the effective horizontal stress from the soil skeleton — the force transmitted through grain-to-grain contacts. They do not include water pressure, which acts independently and is isotropic (equal in all directions). For a wall with a submerged backfill, the total lateral force on the wall is the sum of the effective earth pressure (from Kaγ′) and the hydrostatic pressure (γw×depth below WT). In many practical situations, the hydrostatic pressure exceeds the effective earth pressure — making drainage the single most cost-effective method of reducing lateral wall pressure. A well-drained backfill with KaγH² can be a small fraction of the force on a poorly drained wall carrying Kaγ′H² + ½γwH².