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Geotechnical Engineering

10% of exam

Index properties and soil classification, phase relations, effective stress, lateral earth pressure, shear strength, bearing capacity, foundations, consolidation and settlement, and slope stability.

7 concepts

A. Index properties and soil classifications

Index Properties and Soil Classification

Grain-size curves, Atterberg limits, and the USCS/AASHTO systems that turn lab numbers into a soil name and an engineering expectation.

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B. Phase relations

Phase Relations and Effective Stress

The three-phase diagram, unit weights, and the principle of effective stress (σ' = σ − u) that governs every strength and settlement calculation in soil.

Materials

Fluid Mechanics

Surveying

Water Resources and Environmental

Structural Engineering

Geotechnical Engineering

Transportation Engineering

Construction Engineering

Soil is a three-phase material — solid grains, water, and air — and almost every geotechnical quantity is some ratio of the volumes or weights of those phases. The FE Reference Handbook's Phase Relationships page is one you will open repeatedly, because phase relations feed directly into compaction, seepage, settlement, and shear strength. The crowning idea is Terzaghi's principle of effective stress: soil strength and volume change are controlled not by the total stress but by the stress carried grain-to-grain,

\sigma' = \sigma - u

. Master the phase diagram and effective stress and a large block of the geotechnical exam becomes arithmetic.

The three-phase diagram

Draw a block split into solids (bottom), water (middle), and air (top). On the volume side,

V = V_s + V_w + V_a

and the voids are

V_v = V_w + V_a

. On the weight side, the air weighs essentially nothing, so

W = W_s + W_w

. Every index property below is just a ratio of two of these six quantities, and almost every phase problem is solved by assuming a convenient volume (often

V_s = 1

) and filling in the block.

Void ratio, porosity, water content, saturation

Four ratios describe how the phases are packed. Void ratio

e = V_v/V_s

and porosity

n = V_v/V

both measure void space (note

n = e/(1+e)

). Water content

w = W_w/W_s

is measured by oven-drying and is reported as a percent of the dry solid weight. Degree of saturation

S = V_w/V_v

tells what fraction of the voids holds water; the master identity

S\,e = w\,G_s

ties all four together with the specific gravity of solids.

e = \dfrac{V_v}{V_s}, \quad n = \dfrac{V_v}{V} = \dfrac{e}{1+e}, \quad S\,e = w\,G_s

Unit weights: dry, moist, saturated, submerged

The same soil has several unit weights depending on what water it carries. Moist (total) unit weight

\gamma = W/V

includes whatever water is present; dry unit weight

\gamma_d = W_s/V

excludes the water and is what compaction specs are written against; saturated

\gamma_{sat}

assumes

S = 100\%

. Below the water table, buoyancy reduces the effective weight to the submerged (effective) unit weight

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

. A relation you will use constantly is

\gamma_d = \gamma/(1+w)

.

\gamma_d = \dfrac{\gamma}{1+w} = \dfrac{G_s\,\gamma_w}{1+e}, \qquad \gamma' = \gamma_{sat} - \gamma_w

Specific gravity of solids

The specific gravity

G_s

compares the unit weight of the mineral grains to that of water,

G_s = \gamma_s/\gamma_w

. Most soils cluster near

G_s = 2.65

to

2.72

(quartz sand ≈ 2.65, clays slightly higher), so a value far outside this range should make you re-check your work.

G_s

is what links the volume side of the phase diagram to the weight side, and it appears in nearly every unit-weight formula.

G_s = \dfrac{\gamma_s}{\gamma_w} = \dfrac{W_s/V_s}{\gamma_w}

Total stress, pore pressure, and effective stress

At a depth in the ground, the total vertical stress

\sigma

is the weight of everything above per unit area — sum each layer's unit weight times its thickness. The pore water pressure is hydrostatic below the water table,

u = \gamma_w h_w

, where

h_w

is the height of water above the point. The effective stress that the grain skeleton actually feels is the difference. Raising the water table (or pumping it down) changes

u

and therefore

\sigma'

even when the total stress is unchanged — the engine behind settlement from dewatering and the loss of strength in liquefaction.

\sigma' = \sigma - u, \qquad u = \gamma_w\,h_w

Exam strategy

For a phase problem, assume a basis (

V_s = 1

or

V = 1\,\text{m}^3

), build the six-box diagram, and fill it using

G_s

,

w

, and

e

. Keep

\gamma_w = 9.81\,\text{kN/m}^3

(or

62.4\,\text{lb/ft}^3

) at hand. For a stress profile, work top-down accumulating

\gamma H

for total stress, then subtract

u = \gamma_w h_w

only below the water table — above it,

u = 0

(ignore capillary suction unless told otherwise). When buoyant weights are in play, you can instead add

\gamma' H

directly to get effective stress and skip computing

u

separately.

Key equations

Void ratio and porosity

e = \dfrac{V_v}{V_s}, \qquad n = \dfrac{V_v}{V} = \dfrac{e}{1+e}

Two measures of void space; $e$ referenced to solids, $n$ to total volume.

Water content

w = \dfrac{W_w}{W_s}\times 100\%

Master saturation identity

S\,e = w\,G_s

Links saturation $S$ , void ratio $e$ , water content $w$

Total (moist) unit weight

\gamma = \dfrac{W}{V} = \dfrac{(G_s + S e)\,\gamma_w}{1+e}

Dry unit weight

\gamma_d = \dfrac{\gamma}{1+w} = \dfrac{G_s\,\gamma_w}{1+e}

Saturated unit weight

\gamma_{sat} = \dfrac{(G_s + e)\,\gamma_w}{1+e}

Submerged (effective) unit weight

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

Effective stress

\sigma' = \sigma - u

Terzaghi's principle; grain-skeleton stress governs strength and settlement.

Pore water pressure

u = \gamma_w\,h_w

Hydrostatic; $h_w$

Specific gravity of solids

G_s = \dfrac{W_s/V_s}{\gamma_w}

Worked examples

Filling in the phase diagram

Problem. A moist soil sample has total unit weight $\gamma = 19.2\,\text{kN/m}^3$ , water content $w = 12\%$ , and $G_s = 2.68$ . Find the dry unit weight, void ratio, porosity, and degree of saturation.

Solution. Dry unit weight first:

\gamma_d = \gamma/(1+w) = 19.2/1.12 = 17.1\,\text{kN/m}^3

S = \dfrac{w\,G_s}{e} = \dfrac{(0.12)(2.68)}{0.534} = 0.602

Effective stress below the water table

Problem. A profile has 2 m of moist sand ( $\gamma = 17.5\,\text{kN/m}^3$ ) above the water table, then 4 m of saturated sand ( $\gamma_{sat} = 19.5\,\text{kN/m}^3$

Common pitfalls

•Mixing up porosity and void ratio: $n = V_v/V$ (referenced to total volume), $e = V_v/V_s$ (referenced to solids). They are related by $n = e/(1+e)$ , never equal.
•Computing pore pressure above the water table. Hydrostatic $u = \gamma_w h_w$ applies only below the water table; above it (ignoring capillarity) $u = 0$ , and total = effective stress.
•Using moist or saturated unit weight for the depth below the water table when you want effective stress directly — use the submerged weight $\gamma' = \gamma_{sat}-\gamma_w$
•Applying $\gamma_d = \gamma/(1+w)$ with $w$ as a percent instead of a decimal. $w = 12\%$
•Forgetting that water content is per dry solid weight, not per total weight: $w = W_w/W_s$ . It can legitimately exceed 100% for soft clays.
•Carrying $\gamma_w$ in the wrong system: $9.81\,\text{kN/m}^3$ in SI but $62.4\,\text{lb/ft}^3$

References

NCEES FE Reference Handbook — Civil Engineering (Geotechnical, Phase Relationships)
ASTM D854 — Specific Gravity of Soil Solids — Source of measured $G_s$
Terzaghi, Peck & Mesri, Soil Mechanics in Engineering Practice — Effective stress principle

E. Stability of retaining structures

Lateral Earth Pressure and Retaining Walls

At-rest, active, and passive states, the Rankine and Coulomb coefficients, the pressure distribution, and checks for sliding, overturning, and bearing.

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F. Shear strength

Shear Strength and Bearing Capacity

The Mohr-Coulomb failure law, drained vs undrained strength, and the ultimate bearing-capacity equation with the Nc, Nq, Nγ factors and a factor of safety.

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H. Foundation types

Foundation Types

Shallow foundations (spread, combined, mat), deep foundations (piles, drilled shafts), pile capacity from skin friction and end bearing, group action, and selection criteria.

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I. Consolidation and differential settlement

Consolidation and Settlement

Primary consolidation of clays via the e-log p curve, Cc/Cr indices, normally vs over-consolidated behavior, settlement magnitude, and the Tv–cv time rate.

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J. Slope stability

Slope Stability and Soil Stabilization

Factor of safety against sliding, the infinite-slope and method-of-slices models, the critical surface, compaction control, and chemical/geosynthetic stabilization.

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Weight of water per weight of dry solids; from oven-drying.

, and specific gravity

G_s

. At full saturation

e = wG_s

.

Includes pore water present. $\gamma_w = 9.81\,\text{kN/m}^3 = 62.4\,\text{lb/ft}^3$ .

Solids weight per total volume; the target of compaction control.

All voids full of water ( $S=100\%$ ).

Buoyant unit weight below the water table; use to get

\sigma'

directly.

= height of water above the point (zero above the water table).

Grain density relative to water; typically 2.65–2.72.

. Void ratio from

\gamma_d = G_s\gamma_w/(1+e)

:

1+e = G_s\gamma_w/\gamma_d = (2.68)(9.81)/17.14 = 26.29/17.14 = 1.534 \Rightarrow e = 0.534

. Porosity:

n = e/(1+e) = 0.534/1.534 = 0.348

, i.e.

34.8\%

. Saturation from

Se = wG_s

:

S = wG_s/e = (0.12)(2.68)/0.534 = 0.602 = \textbf{60.2\%}

. Results:

\gamma_d = 17.1\,\text{kN/m}^3

,

e = 0.534

,

n = 34.8\%

,

S = 60.2\%

. Sanity check:

S < 100\%

is consistent with a 'moist' (not saturated) sample, and

e \approx 0.5

is reasonable for a medium-dense soil.

) below it. Find the total stress, pore pressure, and effective stress at 6 m depth.

Solution. Total vertical stress (sum

\gamma H

for each layer):

\sigma = (17.5)(2) + (19.5)(4) = 35 + 78 = 113\,\text{kPa}

. Pore pressure (only the 4 m below the water table contributes):

u = \gamma_w h_w = (9.81)(4) = 39.2\,\text{kPa}

. Effective stress:

\sigma' = \sigma - u = 113 - 39.2 = \textbf{73.8 kPa}

. Cross-check with submerged weight:

\sigma' = (17.5)(2) + (19.5-9.81)(4) = 35 + 38.76 = 73.8\,\text{kPa}

— same answer, confirming the buoyant-weight shortcut.

\sigma' = \sigma - u = 113 - 39.2 = 73.8\ \text{kPa}

, not

\gamma_{sat}

.

means divide by

1.12

, not

1+12

.

in USCS. A units swap here scales every stress.