Loading study handbook…

Soil Mechanics

4% of exam

Lateral earth pressure, consolidation and compaction, bearing capacity, settlement, and slope stability for water-resources structures.

5 concepts

A. Lateral earth pressure

Lateral Earth Pressure: Active, At-Rest and Passive

Earth-pressure coefficients (active, at-rest, passive) by Rankine and Coulomb, the pressure triangle and resultant thrust, plus surcharge, water, and wall friction.

Every retaining structure on the PE Civil exam — cantilever wall, sheet pile, braced cut, basement wall — is sized against the horizontal push of the soil behind it. That push is not a single number: it depends entirely on how much the wall is allowed to move. Let the wall yield away from the backfill and the soil relaxes to its minimum, the active state; push the wall into the soil and the resistance climbs to its maximum, the passive state; hold it perfectly still and you get the in-between at-rest state. The whole topic, covered in the NCEES PE Civil Reference Handbook §3.1 Lateral Earth Pressures, is fluency in three coefficients and the discipline to keep effective stress, water, and surcharge as separate bookkeeping lines.

At-rest pressure and K0

When a wall does not move — a rigid basement wall, a culvert, a tank wall braced top and bottom — the soil stays in its original at-rest condition and the horizontal effective stress is the at-rest coefficient

K_0

times the vertical effective stress. For a normally consolidated soil Jaky's relation

K_0 = 1-\sin\phi'

is the workhorse; overconsolidation raises it because the soil 'remembers' a higher past stress, so

K_0

is scaled by

\text{OCR}^{\Omega}

with

\Omega \approx \sin\phi'

. Because

K_0

sits between

K_a

and

K_p

, a non-yielding wall always carries more lateral load than a flexible one designed for the active case.

K_0 = 1-\sin\phi' \quad(\text{NC}), \qquad K_0 = (1-\sin\phi')\,\text{OCR}^{\,\Omega}\quad(\text{OC})

Rankine active and passive coefficients

Rankine theory assumes a smooth (frictionless) vertical wall, a horizontal cohesionless backfill, and a soil at the verge of shear failure. The active coefficient is the ratio of horizontal to vertical effective stress when the soil has expanded to failure; the passive coefficient is that ratio when the soil has been compressed to failure. They are exact reciprocals —

K_p = 1/K_a

— and for a typical

\phi' = 32^\circ

that gives

K_a \approx 0.31

against

K_p \approx 3.25

, a tenfold spread that explains why passive resistance is so valuable and why you mobilize it cautiously (it needs large movement).

K_a = \tan^2\!\left(45^\circ-\tfrac{\phi'}{2}\right) = \frac{1-\sin\phi'}{1+\sin\phi'}, \qquad K_p = \tan^2\!\left(45^\circ+\tfrac{\phi'}{2}\right) = \frac{1+\sin\phi'}{1-\sin\phi'}

Pressure distribution and resultant thrust

For a dry or moist cohesionless backfill the lateral pressure grows linearly with depth,

p_a = K_a\gamma z

, so the diagram is a triangle and the resultant thrust per unit length of wall is the area of that triangle,

P_a = \tfrac12 K_a\gamma H^2

, acting at one-third of the height above the base. The passive resultant has the identical form with

K_p

. Locating the resultant at

H/3

is what lets you take moments for overturning and sliding checks; getting that lever arm wrong is a classic stability-calculation error.

P_a = \tfrac12 K_a\gamma H^2 \ \text{at}\ \tfrac{H}{3}, \qquad P_p = \tfrac12 K_p\gamma H^2 \ \text{at}\ \tfrac{H}{3}

Surcharge and the water table

A uniform surcharge

q

on the backfill surface adds a constant horizontal pressure

\Delta\sigma_h = K\,q

over the full wall height — a rectangle stacked on the triangle, with its resultant at mid-height. Water is the bigger trap: below the water table you must switch to the buoyant (effective) unit weight

\gamma' = \gamma_{sat}-\gamma_w

for the soil pressure and then add the full hydrostatic water pressure

u = \gamma_w z_w

as its own triangle. Soil pressure uses

K

; water does not —

K_a

never multiplies the water term. Forgetting the water pressure routinely under-predicts the thrust by half.

\Delta\sigma_h = K\,q, \qquad u = \gamma_w z_w, \qquad \sigma'_h = K\,\sigma'_v

Cohesion: tension cracks and 2c'√K

For a

c'

-

\phi'

soil, cohesion reduces active pressure and boosts passive pressure by the term

2c'\sqrt{K}

. In the active case the pressure is negative (tension) near the surface down to the crack depth

z_c = 2c'/(\gamma\sqrt{K_a})

; soil cannot sustain tension, so that zone is ignored (or assumed water-filled if cracks open). On the passive side the same term is additive,

p_p = K_p\gamma z + 2c'\sqrt{K_p}

, giving cohesive soils meaningful resistance even at the surface.

p_a = K_a\gamma z - 2c'\sqrt{K_a}, \qquad p_p = K_p\gamma z + 2c'\sqrt{K_p}

Coulomb theory and wall friction

Rankine ignores friction between wall and soil; Coulomb does not. Coulomb's wedge analysis admits wall friction

\delta

, a battered wall face

\theta

, and a sloping backfill

\beta

, and is the basis for the design charts engineers actually use. Wall friction lowers the active coefficient (the friction helps hold the wedge) and the resultant tilts at angle

\delta

to the wall normal, so a fraction of the thrust acts downward — stabilizing against overturning. For a vertical wall with horizontal backfill and

\delta = 0

, Coulomb collapses exactly back to Rankine.

K_a = \dfrac{\cos^2(\phi'-\theta)}{\cos^2\theta\,\cos(\delta+\theta)\left[1+\sqrt{\dfrac{\sin(\phi'+\delta)\sin(\phi'-\beta)}{\cos(\delta+\theta)\cos(\theta-\beta)}}\,\right]^2}

Exam strategy

First decide the state from the wall's freedom to move: rigid/braced →

K_0

; free to yield →

K_a

on the driving side and

K_p

on the resisting side. Compute

K_a

and

K_p

from

\phi'

and remember they are reciprocals as a fast check. Build the pressure diagram in separate strips — moist soil, buoyant soil, surcharge rectangle (

Kq

), water triangle (

\gamma_w z_w

, no

K

) — find each resultant and its own lever arm, then sum. Use Rankine unless the problem explicitly gives wall friction

\delta

or a sloping wall/backfill, in which case it wants Coulomb (usually via a supplied chart). Keep

\gamma

vs

\gamma'

straight below the water table; that single substitution is the most common point-loser in the topic.

Key equations

At-rest coefficient (Jaky)

K_0 = 1-\sin\phi'

Normally consolidated soil, non-yielding wall. $\phi'$ = effective friction angle. For overconsolidated soil multiply by $\text{OCR}^{\,\Omega}$ , $\Omega\approx\sin\phi'$ .

Rankine active coefficient

K_a = \tan^2\!\left(45^\circ-\tfrac{\phi'}{2}\right) = \frac{1-\sin\phi'}{1+\sin\phi'}

Rankine passive coefficient

K_p = \tan^2\!\left(45^\circ+\tfrac{\phi'}{2}\right) = \frac{1+\sin\phi'}{1-\sin\phi'}

Active / passive pressure with cohesion

p_a = K_a\gamma z - 2c'\sqrt{K_a}, \quad p_p = K_p\gamma z + 2c'\sqrt{K_p}

Active / passive thrust

P_a = \tfrac12 K_a\gamma H^2, \qquad P_p = \tfrac12 K_p\gamma H^2

Tension crack depth

z_c = \dfrac{2c'}{\gamma\sqrt{K_a}}

Surcharge contribution

\Delta\sigma_h = K\,q

Constant added horizontal pressure (psf) from a uniform surface surcharge $q$ ; resultant $Kq\,H$

Hydrostatic water pressure

u = \gamma_w z_w

Acts on the wall in addition to the soil pressure; $K$ does NOT apply to water. $\gamma_w = 62.4\,\text{pcf}$

Coulomb active coefficient

K_a = \dfrac{\cos^2(\phi'-\theta)}{\cos^2\theta\,\cos(\delta+\theta)\left[1+\sqrt{\frac{\sin(\phi'+\delta)\sin(\phi'-\beta)}{\cos(\delta+\theta)\cos(\theta-\beta)}}\right]^2}

Worked examples

Active thrust on a smooth cantilever wall

Problem. A 18 ft high wall retains a dry cohesionless backfill with $\phi' = 32^\circ$ and $\gamma = 120\,\text{pcf}$ . Using Rankine theory, find the active earth-pressure coefficient, the pressure at the base, and the resultant thrust and its point of application.

Solution. Coefficient:

K_a = \tan^2(45^\circ-16^\circ) = \tan^2 29^\circ = 0.307

P_a = \tfrac12(0.307)(120)(18)^2 \approx 5{,}970\ \text{lb/ft}

Wall with surcharge and a water table

Problem. A 20 ft wall retains sand with $\phi' = 30^\circ$ . The top 8 ft is moist ( $\gamma = 120\,\text{pcf}$ ); below the water table (8 ft depth) the sand is saturated ( $\gamma_{sat} = 125\,\text{pcf}$

Passive resistance and at-rest pressure

Problem. Dense sand has $\phi' = 36^\circ$ , $\gamma = 125\,\text{pcf}$ , and is normally consolidated. For a 10 ft embedment, compare $K_a$

Coulomb versus Rankine with wall friction

Problem. A vertical wall ( $\theta = 0$ ) retains horizontal granular backfill ( $\beta = 0$ ) with $\phi' = 34^\circ$

Common pitfalls

•Applying the earth-pressure coefficient to the water pressure. Soil uses $K$ ; water is full hydrostatic $\gamma_w z_w$ with no $K$ . Keep them as separate diagrams.
•Using total unit weight below the water table. Soil pressure below the table uses the buoyant weight $\gamma' = \gamma_{sat}-\gamma_w$ ; using $\gamma_{sat}$
•Confusing at-rest with active. A braced or rigid (non-yielding) wall carries $K_0$ pressure, which is larger than $K_a$ . Designing such a wall for $K_a$
•Placing the resultant at mid-height. A triangular pressure diagram has its resultant at $H/3$ from the base; only a uniform (surcharge) strip resolves at mid-height.
•Forgetting that mobilizing passive resistance needs large movement. Full $K_p$ requires far more displacement than full $K_a$ ; many designs apply a factor of safety or a reduced $K_p$
•Mixing up $K_a$ and $K_p$ — they are reciprocals. Confirm $K_a<1<K_p$
•Ignoring the tension zone for cohesive backfill. Active pressure is negative down to $z_c = 2c'/(\gamma\sqrt{K_a})$

References

NCEES PE Civil Reference Handbook — §3.1 Lateral Earth Pressures
NCEES PE Civil Reference Handbook — §3.3 Effective and Total Stresses
FHWA-NHI-06-088/089 Soils and Foundations Reference Manual — Rankine/Coulomb development and retaining-wall design

B. Soil consolidation and compaction

Compaction: Proctor, Percent Compaction and Relative Density

Standard vs modified Proctor, maximum dry unit weight and optimum moisture, relative compaction, relative density of granular soils, and the zero-air-voids limit.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

C. Bearing capacity

Bearing Capacity of Shallow Foundations (General Equation)

The general bearing-capacity equation, the Nc/Nq/Ngamma factors, shape/depth/water-table corrections, and gross vs net, ultimate vs allowable with a factor of safety.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

D. Settlement

Effective Stress, Consolidation and Settlement

The effective-stress principle, the e-log p' curve, Cc and Cr, normally vs over-consolidated soils (OCR), primary consolidation settlement, and the time rate via Tv and cv.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

E. Slope stability

Slope Stability: Factor of Safety, Charts and Method of Slices

Infinite-slope and circular-arc failure, the factor-of-safety definition, drained vs undrained (phi=0) analysis, Taylor charts, the ordinary method of slices, and pore-pressure effects.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

Smooth vertical wall, horizontal cohesionless backfill. Dimensionless; $K_a<1$ .

Maximum resistance state; $K_p = 1/K_a$ . Mobilizing full $K_p$ requires large wall movement.

Lateral pressure at depth $z$ (psf). For cohesionless soil drop the $2c'\sqrt{K}$ term. Use $\gamma'$ below the water table.

Resultant per unit wall length (lb/ft) for a triangular distribution; acts at $H/3$ above the base.

Depth over which active pressure is tensile (ignored in design) for a cohesive backfill.

at mid-height.

.

$\delta$ = wall friction, $\theta$ = wall batter from vertical, $\beta$ = backfill slope. Reduces to Rankine when $\delta=\theta=\beta=0$ .

(check:

\tfrac{1-\sin32^\circ}{1+\sin32^\circ} = \tfrac{0.470}{1.530} = 0.307

). Pressure at base (

z = H = 18\,\text{ft}

):

p_a = K_a\gamma H = 0.307(120)(18) = 664\,\text{psf}

. Resultant:

P_a = \tfrac12 K_a\gamma H^2 = \tfrac12(0.307)(120)(18)^2 = 5{,}970\,\text{lb/ft}

, acting at

H/3 = 6.00\,\text{ft}

above the base. Sanity check:

P_a

is also the triangle area

\tfrac12(664)(18) = 5{,}980\,\text{lb/ft}

(rounding) — consistent. Final:

K_a = 0.307

,

P_a \approx 5{,}970\,\text{lb/ft}

at

6.00\,\text{ft}

.

). A uniform surcharge

q = 300\,\text{psf}

acts on the surface. Find the total horizontal thrust per foot of wall.

Solution.

K_a = \tan^2(45^\circ-15^\circ) = \tfrac{1}{3} = 0.333

. Use effective stress for soil; add water separately. At the surface

p_a = K_a q = 0.333(300) = 100\,\text{psf}

. At the water table (8 ft):

\sigma'_v = 120(8)+300 = 1{,}260\,\text{psf}

, so

p_a = 0.333(1{,}260) = 420\,\text{psf}

. At the base (20 ft):

\sigma'_v = 120(8) + (125-62.4)(12) + 300 = 2{,}011\,\text{psf}

, so

p_a = 0.333(2{,}011) = 670\,\text{psf}

; water

u = 62.4(12) = 749\,\text{psf}

. Areas: upper strip

\tfrac12(100+420)(8) = 2{,}080

; lower soil strip

\tfrac12(420+670)(12) = 6{,}540

; water triangle

\tfrac12(749)(12) = 4{,}490\,\text{lb/ft}

. Total

= 2{,}080+6{,}540+4{,}490 = 13{,}100\,\text{lb/ft}

. Sanity check: the water alone (4,490 lb/ft) is over a third of the total — exactly why neglecting it is dangerous. Final:

P \approx 13{,}100\,\text{lb/ft}

.

,

K_0

, and

K_p

, and find the passive resultant.

Solution.

K_a = \tan^2(45^\circ-18^\circ) = 0.260

;

K_0 = 1-\sin36^\circ = 1-0.588 = 0.412

;

K_p = \tan^2(45^\circ+18^\circ) = 3.85

. Note

K_a < K_0 < K_p

and

K_p/K_a = 3.85/0.260 = 14.8

— the full active-to-passive range. Passive resultant:

P_p = \tfrac12 K_p\gamma H^2 = \tfrac12(3.85)(125)(10)^2 = 24{,}100\,\text{lb/ft}

, at

H/3 = 3.33\,\text{ft}

. Sanity check:

K_p = 1/K_a = 1/0.260 = 3.85

confirms the reciprocal relationship. Final:

K_a = 0.260

,

K_0 = 0.412

,

K_p = 3.85

,

P_p \approx 24{,}100\,\text{lb/ft}

.

. Wall friction is

\delta = \tfrac23\phi' = 22.7^\circ

. Compare the Coulomb active coefficient with the Rankine value.

Solution. Rankine:

K_a = \tan^2(45^\circ-17^\circ) = 0.283

. Coulomb (with

\theta=\beta=0

): numerator

\cos^2 34^\circ = 0.687

. Denominator:

\cos(\delta)= \cos22.7^\circ = 0.923

; the bracket

\big[1+\sqrt{\tfrac{\sin56.7^\circ\,\sin34^\circ}{\cos22.7^\circ\,\cos0^\circ}}\big]^2 = \big[1+\sqrt{\tfrac{0.836\cdot0.559}{0.923}}\big]^2 = [1+0.711]^2 = 2.93

. So

K_a = 0.687/(0.923\cdot2.93) = 0.254

. Sanity check: wall friction lowers the active coefficient (0.254 vs 0.283, about 10% less) because friction helps support the failure wedge — the expected direction. Final: Coulomb

K_a = 0.254

vs Rankine

0.283

.

double-counts the water.

under-predicts the load.

.

and

K_p = 1/K_a

before proceeding.

; that tension cannot be relied on and is usually discarded.