Factor of safety against seepage liquefaction

Geotechnical · FE Reference Handbook section

Geotechnical

15 formulas

10 exam-style examples

~60 min

All Geotechnical lectures

Learning objectives

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

This chapter section covers Factor of safety against seepage liquefaction 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 factor of safety against seepage liquefaction describes physically and when it applies.
State every one of the 15 relations 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. Factor of safety against seepage liquefaction 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: factor of safety against seepage liquefaction.
Capstone Studio instructional photograph

Geotechnical — Factor of safety against seepage liquefaction: 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 15 relations 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

FSs	Quantity produced by "FSs = ic/ie" — read its definition and unit from the handbook line directly above the equation.
ic	Quantity produced by "ic = (γsat – γW)/γW" — read its definition and unit from the handbook line directly above the equation.
ie	Quantity produced by "ie = seepage exit gradient" — read its definition and unit from the handbook line directly above the equation.
e0	Quantity produced by "e0 = initial void ratio (prior to consolidation)" — read its definition and unit from the handbook line directly above the equation.
∆e	Quantity produced by "∆e = change in void ratio" — read its definition and unit from the handbook line directly above the equation.
p0	Quantity produced by "p0 = initial effective consolidation stress σ'0" — read its definition and unit from the handbook line directly above the equation.
pc	Quantity produced by "pc = past maximum consolidation stress σ'c" — read its definition and unit from the handbook line directly above the equation.
∆p	Quantity produced by "∆p = induced change in consolidation stress at center of consolidating stratum" — read its definition and unit from the handbook line directly above the equation.
I	Quantity produced by "I = Stress influence value at center of consolidating stratum" — read its definition and unit from the handbook line directly above the equation.
qs	Quantity produced by "qs = applied surface stress causing consolidation" — read its definition and unit from the handbook line directly above the equation.
DH	Quantity produced by "DH = change in thickness of soil layer" — 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.

RANGE OF
RECOMPRESSION RANGE OF
VIRGIN
COMPRESSION
e0 CR
VOID RATIO, e
p0 pC p0 + ∆p
PRESSURE (LOG10 SCALE)
SOIL CONSOLIDATION CURVE OVER CONSOLIDATED CLAY
where
where
H p + Dp
o o
H p + Dp
o o
Ho pc p + Dp
o o c

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

Seepage discharge through a sand layer

Water flows through a 2.4 m long sand sample of area 0.030 m² under a 1.5 m head difference, with k = 4.0 × 10⁻⁴ m/s. Find the discharge.

Given

L = 2.4 m
Δh = 1.5 m
A = 0.030 m²
k = 4.0 × 10⁻⁴ m/s

Find

Start with the thinking

Gradient is dimensionless: head loss over flow path length.
Darcy velocity is not the pore velocity.

Step-by-step solution

Gradient — i = Δh/L = 1.5/2.4 = 0.625
Darcy velocity
$v = k i = 4.0 \times 1 0^{- 4} (0.625) = 2.50 \times 1 0^{- 4} m / s$
Discharge
$Q = v A = 2.50 \times 1 0^{- 4} (0.030)$
Result
$Q = 7.50 \times 1 0^{- 6} m^{3} / s = 0.0075 L / s$

Answer: Q = 7.5 × 10⁻⁶ m³/s

Why the other options are there

1.2 × 10⁻⁵ m³/s (gradient inverted)
2.5 × 10⁻⁴ m³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical — Darcy's law

Example 2

Darcy seepage through a soil sample — Factor of safety against seepage liquefaction

A permeameter has k = 0.0035 ft/s, cross-sectional area A = 5.0 ft², sample length L = 6.0 ft and head difference Δh = 3.5 ft. Find the seepage discharge.

Given

k = 0.0035 ft/s
Δh = 3.5 ft
L = 6.0 ft
A = 5.0 ft²

Find

Discharge Q and discharge velocity v

Start with the thinking

Gradient is dimensionless: head lost divided by the flow path length.
Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

Gradient — i = Δh/L = 3.5/6.0 = 0.583
Darcy velocity
$v = k i = 0.0035 (0.583) = 2.042 e - 3 f t / s$
Discharge
$Q = v A = 2.042 e - 3 (5.0) = 1.021 e - 2 f t^{3} / s$
Per day
$Q = 882.0 f t^{3} / d a y$

Answer: Q ≈ 1.02e-2 ft³/s (882.0 ft³/day)

Why the other options are there

Q = 6.12e-2 ft³/s (length omitted from the gradient)
Q = 2.04e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 3

Infinite slope factor of safety (dry cohesionless soil) — Factor of safety against seepage liquefaction

A long, dry cohesionless slope stands at β = 18° in soil with φ = 34°. Find the factor of safety against sliding and the maximum stable slope angle.

Given

β = 18°
φ = 34°
c = 0, dry (no seepage)

Find

FS and the limiting slope angle

Start with the thinking

For a dry cohesionless infinite slope the unit weight and depth cancel.
FS = 1 corresponds to β = φ, the angle of repose.

Step-by-step solution

Infinite slope (c = 0, dry)
$F S = tan ϕ / tan β$
Numerator
$tan 3 4^{\circ} = 0.6745$
Denominator
$tan 1 8^{\circ} = 0.3249$
Factor of safety
$F S = 2.08$
Limiting angle
$β_{m} a x = ϕ = 3 4^{\circ} (F S = 1.0)$

Answer: FS ≈ 2.08; the slope is stable up to β = 34°

Why the other options are there

FS = 1.89 (angles divided instead of their tangents)
FS = 0.48 (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 4

Darcy seepage through a soil sample — Factor of safety against seepage liquefaction (2)

A permeameter has k = 0.0019 ft/s, cross-sectional area A = 17.0 ft², sample length L = 18.5 ft and head difference Δh = 2.5 ft. Find the seepage discharge.

Given

k = 0.0019 ft/s
Δh = 2.5 ft
L = 18.5 ft
A = 17.0 ft²

Find

Discharge Q and discharge velocity v

Start with the thinking

Gradient is dimensionless: head lost divided by the flow path length.
Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

Gradient — i = Δh/L = 2.5/18.5 = 0.135
Darcy velocity
$v = k i = 0.0019 (0.135) = 2.568 e - 4 f t / s$
Discharge
$Q = v A = 2.568 e - 4 (17.0) = 4.365 e - 3 f t^{3} / s$
Per day
$Q = 377.1 f t^{3} / d a y$

Answer: Q ≈ 4.36e-3 ft³/s (377.1 ft³/day)

Why the other options are there

Q = 8.08e-2 ft³/s (length omitted from the gradient)
Q = 2.57e-4 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 5

Infinite slope factor of safety (dry cohesionless soil) — Factor of safety against seepage liquefaction (2)

A long, dry cohesionless slope stands at β = 26° in soil with φ = 37°. Find the factor of safety against sliding and the maximum stable slope angle.

Given

β = 26°
φ = 37°
c = 0, dry (no seepage)

Find

FS and the limiting slope angle

Start with the thinking

For a dry cohesionless infinite slope the unit weight and depth cancel.
FS = 1 corresponds to β = φ, the angle of repose.

Step-by-step solution

Infinite slope (c = 0, dry)
$F S = tan ϕ / tan β$
Numerator
$tan 3 7^{\circ} = 0.7536$
Denominator
$tan 2 6^{\circ} = 0.4877$
Factor of safety
$F S = 1.55$
Limiting angle
$β_{m} a x = ϕ = 3 7^{\circ} (F S = 1.0)$

Answer: FS ≈ 1.55; the slope is stable up to β = 37°

Why the other options are there

FS = 1.42 (angles divided instead of their tangents)
FS = 0.65 (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 6

Darcy seepage through a soil sample — Factor of safety against seepage liquefaction (3)

A permeameter has k = 0.0032 ft/s, cross-sectional area A = 2.0 ft², sample length L = 18.5 ft and head difference Δh = 11.5 ft. Find the seepage discharge.

Given

k = 0.0032 ft/s
Δh = 11.5 ft
L = 18.5 ft
A = 2.0 ft²

Find

Discharge Q and discharge velocity v

Start with the thinking

Gradient is dimensionless: head lost divided by the flow path length.
Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

Gradient — i = Δh/L = 11.5/18.5 = 0.622
Darcy velocity
$v = k i = 0.0032 (0.622) = 1.989 e - 3 f t / s$
Discharge
$Q = v A = 1.989 e - 3 (2.0) = 3.978 e - 3 f t^{3} / s$
Per day
$Q = 343.7 f t^{3} / d a y$

Answer: Q ≈ 3.98e-3 ft³/s (343.7 ft³/day)

Why the other options are there

Q = 7.36e-2 ft³/s (length omitted from the gradient)
Q = 1.99e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 7

Infinite slope factor of safety (dry cohesionless soil) — Factor of safety against seepage liquefaction (3)

A long, dry cohesionless slope stands at β = 17° in soil with φ = 36°. Find the factor of safety against sliding and the maximum stable slope angle.

Given

β = 17°
φ = 36°
c = 0, dry (no seepage)

Find

FS and the limiting slope angle

Start with the thinking

For a dry cohesionless infinite slope the unit weight and depth cancel.
FS = 1 corresponds to β = φ, the angle of repose.

Step-by-step solution

Infinite slope (c = 0, dry)
$F S = tan ϕ / tan β$
Numerator
$tan 3 6^{\circ} = 0.7265$
Denominator
$tan 1 7^{\circ} = 0.3057$
Factor of safety
$F S = 2.38$
Limiting angle
$β_{m} a x = ϕ = 3 6^{\circ} (F S = 1.0)$

Answer: FS ≈ 2.38; the slope is stable up to β = 36°

Why the other options are there

FS = 2.12 (angles divided instead of their tangents)
FS = 0.42 (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 8

Darcy seepage through a soil sample — Factor of safety against seepage liquefaction (4)

A permeameter has k = 0.0023 ft/s, cross-sectional area A = 14.5 ft², sample length L = 16.5 ft and head difference Δh = 3.5 ft. Find the seepage discharge.

Given

k = 0.0023 ft/s
Δh = 3.5 ft
L = 16.5 ft
A = 14.5 ft²

Find

Discharge Q and discharge velocity v

Start with the thinking

Gradient is dimensionless: head lost divided by the flow path length.
Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

Gradient — i = Δh/L = 3.5/16.5 = 0.212
Darcy velocity
$v = k i = 0.0023 (0.212) = 4.879 e - 4 f t / s$
Discharge
$Q = v A = 4.879 e - 4 (14.5) = 7.074 e - 3 f t^{3} / s$
Per day
$Q = 611.2 f t^{3} / d a y$

Answer: Q ≈ 7.07e-3 ft³/s (611.2 ft³/day)

Why the other options are there

Q = 1.17e-1 ft³/s (length omitted from the gradient)
Q = 4.88e-4 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 9

Infinite slope factor of safety (dry cohesionless soil) — Factor of safety against seepage liquefaction (4)

A long, dry cohesionless slope stands at β = 15° in soil with φ = 28°. Find the factor of safety against sliding and the maximum stable slope angle.

Given

β = 15°
φ = 28°
c = 0, dry (no seepage)

Find

FS and the limiting slope angle

Start with the thinking

For a dry cohesionless infinite slope the unit weight and depth cancel.
FS = 1 corresponds to β = φ, the angle of repose.

Step-by-step solution

Infinite slope (c = 0, dry)
$F S = tan ϕ / tan β$
Numerator
$tan 2 8^{\circ} = 0.5317$
Denominator
$tan 1 5^{\circ} = 0.2679$
Factor of safety
$F S = 1.98$
Limiting angle
$β_{m} a x = ϕ = 2 8^{\circ} (F S = 1.0)$

Answer: FS ≈ 1.98; the slope is stable up to β = 28°

Why the other options are there

FS = 1.87 (angles divided instead of their tangents)
FS = 0.50 (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

Example 10

Darcy seepage through a soil sample — Factor of safety against seepage liquefaction (5)

A permeameter has k = 0.0016 ft/s, cross-sectional area A = 8.5 ft², sample length L = 8.5 ft and head difference Δh = 11.5 ft. Find the seepage discharge.

Given

k = 0.0016 ft/s
Δh = 11.5 ft
L = 8.5 ft
A = 8.5 ft²

Find

Discharge Q and discharge velocity v

Start with the thinking

Gradient is dimensionless: head lost divided by the flow path length.
Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

Gradient — i = Δh/L = 11.5/8.5 = 1.353
Darcy velocity
$v = k i = 0.0016 (1.353) = 2.165 e - 3 f t / s$
Discharge
$Q = v A = 2.165 e - 3 (8.5) = 1.840 e - 2 f t^{3} / s$
Per day
$Q = 1, 590 f t^{3} / d a y$

Answer: Q ≈ 1.84e-2 ft³/s (1,590 ft³/day)

Why the other options are there

Q = 1.56e-1 ft³/s (length omitted from the gradient)
Q = 2.16e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Factor of safety against seepage liquefaction

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

Factor of safety against seepage liquefaction contains 15 relations; 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.