CEE 340 · Advanced Foundation Engineering · Ch. 7

Lateral Earth Pressure

Every retaining wall, basement wall, and sheet pile lives or dies by one question: how hard is the soil behind it pushing? The answer isn't a fixed number — it depends on which way the wall moves. That single idea is this whole chapter.

At-Rest (K₀) Active (Ka) Passive (Kp) Rankine & Coulomb
01

Why a Wall's Pressure Isn't Fixed

Water pushes on a dam with one predictable pressure — hydrostatic, straight from depth. Soil doesn't. Soil is a pile of grains that can lock together and support themselves, so how hard it pushes on a wall depends entirely on what the wall does.

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Think of a wall as one person holding back a crowd. If you stand rigid and nobody moves, the crowd leans on you at some resting pressure. Step back even a centimeter, and the crowd relaxes into the gap — pressure on you drops. Instead, lean into the crowd, and they brace back hard — pressure spikes. Same crowd, wildly different pressure, just from a few centimeters of your own movement.

Soil behaves the same way. A wall that doesn't move at all holds the soil in its natural, undisturbed state — at-rest pressure. A wall that yields (tips or slides away, even slightly) lets the soil relax and mobilize its own shear strength to hold itself up — pressure falls to a minimum called active pressure. A wall that gets pushed into the soil compresses it, forcing it toward failure in the other direction — pressure rises to a maximum called passive pressure.

The Master Diagram: Pressure vs. Wall Movement

K₀ Kₖ Kₚ ← wall moves away (active) wall pushes in (passive) → K (pressure coefficient)

A tiny outward movement (~0.1–0.5% of wall height) is enough to crash the pressure down to Ka. Getting the full passive resistance Kp takes a movement 10–50× bigger, in the opposite direction. This asymmetry is why designers can trust active pressure calculations after almost no movement, but can't fully trust passive resistance unless the wall has actually shifted a lot.

Why it matters: Every retaining wall, sheet pile, and basement wall in Ch. 8 gets designed using these three pressures. Get Ka or Kp wrong and the whole downstream design — overturning, sliding, bearing capacity — is wrong too.
02

At-Rest Pressure — K₀

Intuition

The wall that never moves

Basement walls braced by a floor slab on both ends, or a wall between two buildings that literally can't move — these hold soil at rest. Nothing has relaxed and nothing has been compressed; the soil is exactly as confined as when it was deposited.

Mechanics

Confinement, not failure

No shear strength is mobilized at all — the soil isn't anywhere near failure. Horizontal stress is just some fraction of vertical stress, set by how the soil was confined as it formed. Loose, normally-consolidated soil settles into a fairly predictable fraction.

The Math

Jaky's equation

\[ K_0 = 1 - \sin\phi' \] \[ \sigma'_h = K_0\,\sigma'_v \]

For normally consolidated soil. (Overconsolidated soil has higher K₀ — it "remembers" a bigger past confinement.)

Core Idea

K₀ is not a strength calculation — it's a snapshot of however the soil happens to sit right now. It's always between Ka and Kp, closer to the middle.

03

Active Pressure — Ka (Rankine)

Intuition

The wall that gives way

Tip a cantilever retaining wall forward, even by a millimeter per meter of height, and a wedge of soil behind it slides down and pushes it — but only with the minimum force needed to keep that wedge from falling further. The soil is doing as little work on the wall as physically possible.

Mechanics

Failure wedge at 45°+φ'/2

The soil is at the brink of shear failure along a plane tilted at 45°+φ'/2 from horizontal. Rankine's theory assumes a frictionless, vertical wall and a horizontal backfill — simple, but it's the backbone every real design builds on.

The Math

Rankine active coefficient

\[ K_a = \tan^2\!\left(45^\circ - \frac{\phi'}{2}\right) \] \[ \sigma'_a = K_a\,\sigma'_v \]
45° + φ'/2 wall yields → Active: wedge slides down & out 45° − φ'/2 ← wall pushed in Passive: bigger wedge resists, pushed up & out
Core Idea

Active pressure is the minimum the soil will ever push with. It's what happens once the wall has already yielded enough to let the soil find its easiest, most relaxed equilibrium.

04

Passive Pressure — Kp (Rankine)

Intuition

The wall that shoves back

Push a wall into the soil — this is what happens at the toe of a footing, or the embedded tip of a sheet pile — and the soil fights back at its maximum possible resistance. This resistance is what keeps footings and sheet piles from sliding or kicking out.

Mechanics

Failure wedge at 45°−φ'/2

Same theory, opposite direction: the failure plane flattens to 45°−φ'/2, and the wedge that resists is much bigger and much stronger. That's why Kp is always far larger than Ka.

The Math

Rankine passive coefficient

\[ K_p = \tan^2\!\left(45^\circ + \frac{\phi'}{2}\right) \] \[ \sigma'_p = K_p\,\sigma'_v \]
Why it matters: At φ'=30°, Kp/Ka = 9. The ground can shove back nine times harder than it pulls away — that nine-fold cushion is exactly why footings get embedded below grade and sheet piles get driven deep: you're banking on passive resistance you can't get any other way.
05

Cohesion & the Tension Crack

Clay has cohesion (c') — grains stick to each other even under zero confinement. That stickiness fights the active pressure, subtracting from it. But cohesion can't be squeezed, only pulled apart — near the top of the wall, where confinement is lowest, the math predicts negative pressure. Soil can't actually pull on a wall, so instead it just cracks open.

Active pressure with cohesion
\[ \sigma'_a = K_a\,\sigma'_v - 2c'\sqrt{K_a} \]
Depth of the tension crack
\[ z_c = \frac{2c'}{\gamma\sqrt{K_a}} \]

Below zc, pressure builds normally. Design practice ignores the cracked zone's pressure entirely — but that crack can fill with rainwater and push with full hydrostatic force, which is often the more dangerous case in real walls.

06

Live Calculator & Groundwater

One tool, three pressure states. Groundwater matters because below the water table the soil's effective weight drops (buoyancy), but full hydrostatic pressure gets added back on top — so the total pressure diagram kinks at the water table instead of staying a straight triangle.

07

Sloped Backfill & Coulomb's Theory

Rankine's theory needs a frictionless, vertical wall and a flat backfill — clean for learning the concept, but most real walls have batter, wall friction, and sloped fill. Two extensions handle that:

Generalized Rankine (sloped backfill, angle α)
\[ K_a = \cos\alpha\,\frac{\cos\alpha - \sqrt{\cos^2\alpha - \cos^2\phi'}}{\cos\alpha + \sqrt{\cos^2\alpha - \cos^2\phi'}} \]

Pressure now acts parallel to the sloped ground surface, not horizontally.

Coulomb's theory (wall friction δ)

Coulomb drops the frictionless-wall assumption and lets the wall drag on the sliding wedge, which lowers Ka and (for rough walls) meaningfully raises Kp. The trade-off: the failure surface is no longer a flat plane, so the closed-form expression is heavier. In practice, engineers pull Ka/Kp straight from published charts for the given δ, α, and wall batter β rather than deriving it by hand.

Why it matters: In the 2017 Rangamati–Chattogram hill-tract landslides, several valley-side retaining walls failed by overturning once their foundations lost support — a reminder that lateral earth pressure theory isn't academic. Undersized or misjudged pressure assumptions on a slope show up as real collapsed roads and walls.
08

Worked Examples

Example 1

At-Rest Force on a Braced Basement Wall

Problem: A 5 m tall basement wall retains dry, normally consolidated sand with γ = 17 kN/m³ and φ' = 32°. The wall is fully braced (no movement). Find the at-rest force per meter of wall and its location.

1
\( K_0 = 1-\sin32^\circ = 1-0.530 = 0.470 \)
2
Pressure at base: \( \sigma'_h = K_0\gamma H = 0.470(17)(5) = 39.9 \text{kPa} \) (triangle from 0 at top to 39.9 kPa at base)
3
Resultant force: \( P_0 = \tfrac12 K_0 \gamma H^2 = \tfrac12(0.470)(17)(5)^2 = 99.9 \text{kN/m} \)
4
Location: acts at \( H/3 = 1.67\text{m} \) above the base (centroid of the triangle).
Answer: P₀ ≈ 99.9 kN/m, acting 1.67 m above the base.
Example 2

Active Force with Cohesion & Tension Crack

Problem: A frictionless 6 m wall retains a clay backfill: γ = 18 kN/m³, φ' = 26°, c' = 12 kPa. Find the tension crack depth and the active force after the crack forms.

1
\( K_a = \tan^2(45-13) = \tan^2(32^\circ) = 0.390 \)
2
Tension crack depth: \( z_c = \dfrac{2c'}{\gamma\sqrt{K_a}} = \dfrac{2(12)}{18\sqrt{0.390}} = 2.14\text{m} \)
3
Pressure at base (z=6 m): \( \sigma'_a = K_a\gamma H - 2c'\sqrt{K_a} = 0.390(18)(6) - 2(12)\sqrt{0.390} = 42.1 - 15.0 = 27.1\text{kPa} \)
4
Net triangle runs from 0 at \( z_c=2.14\text{m} \) to 27.1 kPa at the base, over a height of \( 6-2.14=3.86\text{m} \): \( P_a = \tfrac12(27.1)(3.86) = 52.3\text{kN/m} \)
Answer: Crack extends 2.14 m deep; net active force below it ≈ 52.3 kN/m. (The cracked zone above is ignored in the force — unless it fills with water.)
09

Quick Reference & Quick Check

K₀1 − sin&phi'
Katan²(45° − φ'/2)
Kptan²(45° + φ'/2)
Resultant forceP = ½ K γ H² (triangle)
Location (no water/cohesion)H/3 above base
Failure plane (active)45° + φ'/2 from horizontal
Failure plane (passive)45° − φ'/2 from horizontal
Tension crack depthzc = 2c'/(γ√Ka)

1. A cantilever wall tips forward by 2 mm at the top. Which pressure state is now acting on it?

2. Why is Kp always much bigger than Ka for the same soil?

3. A tension crack forms in clay backfill because: