Porosity

Geotechnical · FE Reference Handbook section

Geotechnical

1 formulas

10 exam-style examples

~47 min

All Geotechnical lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Porosity within Geotechnical. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what porosity describes physically and when it applies.
State every one of the 1 relation the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: unit weight in pcf with depth in ft gives stress in psf, not psi.

Lecture

Why this section exists. Porosity is the part of Geotechnical that lets you connect a layered soil profile beneath a footing or wall to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a phase-diagram quantity, an effective stress, or a bearing capacity. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. unit weight in pcf with depth in ft gives stress in psf, not psi. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Truck-mounted drill rig taking a soil boring beside a bridge, with sample jars in the foreground. — Photo 1. Where this shows up in practice: porosity.
Capstone Studio instructional photograph

Geotechnical — Porosity: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a layered soil profile beneath a footing or wall. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 1 relation on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

n	Quantity produced by "n = VV = 1 + e" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Soil phase relationships from field data

A sample has moist unit weight 19.2 kN/m³, water content 16% and Gs = 2.68. Find the dry unit weight, void ratio and degree of saturation.

Given

γ = 19.2 kN/m³
w = 0.16
Gs = 2.68
γw = 9.81 kN/m³

Find

γ_d, e, S

Start with the thinking

Dry unit weight first — everything else follows.
Se = wGs closes the saturation calculation.

Figure for Soil phase relationships from field data

Step-by-step solution

Dry unit weight
$γ_{d} = γ / (1 + w) = 19.2/1.16 = 16.55 k N / m^{3}$
Void ratio — e = Gsγw/γ_d − 1 = 2.68(9.81)/16.55 − 1
Evaluate
$e = 26.29/16.55 - 1 = 1.589 - 1 = 0.589$
Saturation
$S = w G s / e = 0.16 (2.68) /0.589$
Result
$γ_{d} = 16.6 k N / m^{3}, e = 0.589, S = 0.728 (72.8$

Answer: γ_d = 16.6 kN/m³, e = 0.589, S = 72.8%

Why the other options are there

γ_d = 22.3 kN/m³ (multiplied by 1 + w)
S = 100% (assumed saturated)

Reference: FE Reference Handbook — Geotechnical — Phase relationships

Example 2

Effective stress below a water table

A profile has 3 m of moist sand (γ = 18.0 kN/m³) over saturated sand (γ_sat = 20.5 kN/m³) with the water table at 3 m. Find the effective stress at 8 m depth.

Given

0–3 m: γ = 18.0 kN/m³
3–8 m: γ_sat = 20.5 kN/m³
Water table at 3 m

Find

σ′ at 8 m

Start with the thinking

Total stress accumulates layer by layer.
Pore pressure is measured from the water table, not the ground surface.

Step-by-step solution

Total stress
$σ = 18.0 (3) + 20.5 (5) = 54.0 + 102.5 = 156.5 k P a$
Pore pressure — u = γw(8 − 3) = 9.81(5) = 49.1 kPa
Effective stress
$σ^{'} = σ - u = 156.5 - 49.1$
Result
$σ^{'} = 107.4 k P a$

Answer: σ′ = 107 kPa

Why the other options are there

156.5 kPa (pore pressure ignored)
78.0 kPa (u measured from the surface)

Reference: FE Reference Handbook — Geotechnical — Effective stress

Example 3

Phase relations for a compacted fill — Porosity

A soil has Gs = 2.62, void ratio e = 0.78 and water content w = 15.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.62
e = 0.78
w = 15.5%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.62(62.4)/(1+0.78) = 91.85 lb/ft³
Moist unit weight — γ = γd(1+w) = 91.85(1+0.155) = 106.1 lb/ft³
Porosity
$n = e / (1 + e) = 0.78/1.78 = 0.438 = 43.8$
Saturation
$S = w G s / e = 0.155 (2.62) /0.78 = 0.521 = 52.1$

Answer: γd ≈ 91.8 lb/ft³, γ ≈ 106.1 lb/ft³, n ≈ 43.8%, S ≈ 52.1%

Why the other options are there

γd = 163.5 lb/ft³ (voids ignored)
S = 192.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Example 4

Phase relations for a compacted fill — Porosity (2)

A soil has Gs = 2.65, void ratio e = 0.54 and water content w = 9.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.65
e = 0.54
w = 9.5%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity (2)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.54) = 107.4 lb/ft³
Moist unit weight — γ = γd(1+w) = 107.4(1+0.095) = 117.6 lb/ft³
Porosity
$n = e / (1 + e) = 0.54/1.54 = 0.351 = 35.1$
Saturation
$S = w G s / e = 0.095 (2.65) /0.54 = 0.466 = 46.6$

Answer: γd ≈ 107.4 lb/ft³, γ ≈ 117.6 lb/ft³, n ≈ 35.1%, S ≈ 46.6%

Why the other options are there

γd = 165.4 lb/ft³ (voids ignored)
S = 214.5% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Example 5

Phase relations for a compacted fill — Porosity (3)

A soil has Gs = 2.74, void ratio e = 0.64 and water content w = 13.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.74
e = 0.64
w = 13.0%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity (3)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.64) = 104.3 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.3(1+0.130) = 117.8 lb/ft³
Porosity
$n = e / (1 + e) = 0.64/1.64 = 0.390 = 39.0$
Saturation
$S = w G s / e = 0.130 (2.74) /0.64 = 0.557 = 55.7$

Answer: γd ≈ 104.3 lb/ft³, γ ≈ 117.8 lb/ft³, n ≈ 39.0%, S ≈ 55.7%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 179.7% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Example 6

Phase relations for a compacted fill — Porosity (4)

A soil has Gs = 2.63, void ratio e = 0.83 and water content w = 8.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.63
e = 0.83
w = 8.5%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity (4)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.63(62.4)/(1+0.83) = 89.68 lb/ft³
Moist unit weight — γ = γd(1+w) = 89.68(1+0.085) = 97.30 lb/ft³
Porosity
$n = e / (1 + e) = 0.83/1.83 = 0.454 = 45.4$
Saturation
$S = w G s / e = 0.085 (2.63) /0.83 = 0.269 = 26.9$

Answer: γd ≈ 89.7 lb/ft³, γ ≈ 97.3 lb/ft³, n ≈ 45.4%, S ≈ 26.9%

Why the other options are there

γd = 164.1 lb/ft³ (voids ignored)
S = 371.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Example 7

Phase relations for a compacted fill — Porosity (5)

A soil has Gs = 2.65, void ratio e = 0.48 and water content w = 15.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.65
e = 0.48
w = 15.5%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity (5)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.48) = 111.7 lb/ft³
Moist unit weight — γ = γd(1+w) = 111.7(1+0.155) = 129.0 lb/ft³
Porosity
$n = e / (1 + e) = 0.48/1.48 = 0.324 = 32.4$
Saturation
$S = w G s / e = 0.155 (2.65) /0.48 = 0.856 = 85.6$

Answer: γd ≈ 111.7 lb/ft³, γ ≈ 129.0 lb/ft³, n ≈ 32.4%, S ≈ 85.6%

Why the other options are there

γd = 165.4 lb/ft³ (voids ignored)
S = 116.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Example 8

Phase relations for a compacted fill — Porosity (6)

A soil has Gs = 2.74, void ratio e = 0.74 and water content w = 14.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.74
e = 0.74
w = 14.0%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity (6)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.74) = 98.26 lb/ft³
Moist unit weight — γ = γd(1+w) = 98.26(1+0.140) = 112.0 lb/ft³
Porosity
$n = e / (1 + e) = 0.74/1.74 = 0.425 = 42.5$
Saturation
$S = w G s / e = 0.140 (2.74) /0.74 = 0.518 = 51.8$

Answer: γd ≈ 98.3 lb/ft³, γ ≈ 112.0 lb/ft³, n ≈ 42.5%, S ≈ 51.8%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 192.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Example 9

Phase relations for a compacted fill — Porosity (7)

A soil has Gs = 2.64, void ratio e = 0.57 and water content w = 18.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.64
e = 0.57
w = 18.0%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity (7)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.64(62.4)/(1+0.57) = 104.9 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.9(1+0.180) = 123.8 lb/ft³
Porosity
$n = e / (1 + e) = 0.57/1.57 = 0.363 = 36.3$
Saturation
$S = w G s / e = 0.180 (2.64) /0.57 = 0.834 = 83.4$

Answer: γd ≈ 104.9 lb/ft³, γ ≈ 123.8 lb/ft³, n ≈ 36.3%, S ≈ 83.4%

Why the other options are there

γd = 164.7 lb/ft³ (voids ignored)
S = 119.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Example 10

Phase relations for a compacted fill — Porosity (8)

A soil has Gs = 2.63, void ratio e = 0.72 and water content w = 17.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

Gs = 2.63
e = 0.72
w = 17.5%
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure for Phase relations for a compacted fill — Porosity (8)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.63(62.4)/(1+0.72) = 95.41 lb/ft³
Moist unit weight — γ = γd(1+w) = 95.41(1+0.175) = 112.1 lb/ft³
Porosity
$n = e / (1 + e) = 0.72/1.72 = 0.419 = 41.9$
Saturation
$S = w G s / e = 0.175 (2.63) /0.72 = 0.639 = 63.9$

Answer: γd ≈ 95.4 lb/ft³, γ ≈ 112.1 lb/ft³, n ≈ 41.9%, S ≈ 63.9%

Why the other options are there

γd = 164.1 lb/ft³ (voids ignored)
S = 156.4% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Porosity

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a layered soil profile beneath a footing or wall, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Porosity contains 1 relation; you must be able to find this page in under 15 seconds.
Exam style: a phase-diagram quantity, an effective stress, or a bearing capacity.
Unit rule: unit weight in pcf with depth in ft gives stress in psf, not psi.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

unit weight in pcf with depth in ft gives stress in psf, not psi
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Porosity

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

n=VV=1+e

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section