Retaining Walls

Geotechnical · FE Reference Handbook section

Geotechnical

39 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 Retaining Walls 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 retaining walls describes physically and when it applies.
State every one of the 39 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. Retaining Walls 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: retaining walls.
Capstone Studio instructional photograph

Geotechnical — Retaining Walls: 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 39 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

FS overturning	Quantity produced by "FS overturning = M R" — read its definition and unit from the handbook line directly above the equation.
FSsliding	Quantity produced by "FSsliding = /" — read its definition and unit from the handbook line directly above the equation.
FS sliding	Quantity produced by "FS sliding = Pa cos a" — read its definition and unit from the handbook line directly above the equation.
q toe	Quantity produced by "q toe = B c1 + B m" — read its definition and unit from the handbook line directly above the equation.
e	Quantity produced by "e = 2 −e o" — read its definition and unit from the handbook line directly above the equation.
B	Quantity produced by "B = width of base" — read its definition and unit from the handbook line directly above the equation.
MR	Quantity produced by "MR = resisting moment" — read its definition and unit from the handbook line directly above the equation.
MO	Quantity produced by "MO = overturning moment" — read its definition and unit from the handbook line directly above the equation.
FR	Quantity produced by "FR = resisting forces" — read its definition and unit from the handbook line directly above the equation.
FD	Quantity produced by "FD = driving forces" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = vertical forces" — read its definition and unit from the handbook line directly above the equation.
δ	Quantity produced by "δ = k1φ2" — 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.

/ FR
/V 6 e
/ MR − MO
where
k1 and k2 are given, ranging from 1/2 to 2/3
GROUND
SURFACE WM
TMOB ASSUMED
PLANAR SLIP
T FF SURFACE
SLOPE FAILURE
ALONG PLANAR SURFACE
where
AASHTO Soil Classification
GENERAL CLASSIFICATION GRANULAR MATERIALS ( 35% OR LESS PASSING 0.075 SIEVE ) SILT-CLAY MATERIALS
( MORE THAN 35% PASSING 0.075 SIEVE )
A-1 A-2 A-7-5
GROUP CLASSIFICATION A-3 A-4 A-5 A-6 A-7-6
A-1-a A-1-b A -2-4 A -2-5 A -2-6 A -2-7
SIEVE ANALYSIS, PERCENT PASSING:
CHARACTERISTICS OF FRACTION PASSING
0.425 SIEVE (No. 40):
USUAL TYPES OF CONSTITUENT MATERIALS STONE FRAGM'TS, FINE SILTY OR CLAYEY GRAVEL AND SAND SILTY SOILS CLAYEY SOILS
GRAVEL, SAND SAND

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

Rankine active thrust on a retaining wall

A 5.0 m wall retains dry granular backfill with γ = 18.5 kN/m³ and φ = 32°. Find the active thrust per metre and its point of application.

Given

H = 5.0 m
γ = 18.5 kN/m³
φ = 32°
Dry backfill

Find

P_a and its location

Start with the thinking

Ka = tan²(45 − φ/2) for a smooth vertical wall with level backfill.
A triangular pressure diagram acts at H/3.

Step-by-step solution

Coefficient
$K a = tan^{2} (4 5^{\circ} - 1 6^{\circ}) = tan^{2} (2 9^{\circ}) = 0.307$
Thrust — P_a = ½KaγH²
Substitute
$P_{a} = 0.5 (0.307) (18.5) (25) = 0.5 (0.307) (462.5)$
Result
$P_{a} = 71.0 k N p er m e t r eo f w a l l$
Location
$H /3 = 1.67 mab o v e t h e ba se$

Answer: P_a = 71.0 kN/m acting 1.67 m above the base

Why the other options are there

231 kN/m (Kp used instead of Ka)
Acting at 2.5 m (uniform pressure assumed)

Reference: FE Reference Handbook — Geotechnical — Lateral earth pressure

Example 2

Rankine active thrust on a retaining wall — Retaining Walls

A 19.0 ft high frictionless wall retains cohesionless backfill with φ = 38° and γ = 123.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 19.0 ft
φ = 38°
γ = 123.0 pcf
(check: tan²(45−φ/2) = 0.238)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.616) / (1 + 0.616) = 0.238$
Base pressure — σa = KaγH = 0.238(123.0)(19.0) = 555.9 psf
Thrust — Pa = ½KaγH² = 0.5(0.238)(123.0)(19.0)² = 5,281 lb/ft
Location
$y ˉ = H /3 = 6.33 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.238, Pa ≈ 5,281 lb/ft acting 6.33 ft above the base

Why the other options are there

Pa = 10,563 lb/ft (½ omitted)
ȳ = 9.50 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 3

Rankine active thrust on a retaining wall — Retaining Walls (2)

A 13.5 ft high frictionless wall retains cohesionless backfill with φ = 28° and γ = 122.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 13.5 ft
φ = 28°
γ = 122.0 pcf
(check: tan²(45−φ/2) = 0.361)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.469) / (1 + 0.469) = 0.361$
Base pressure — σa = KaγH = 0.361(122.0)(13.5) = 594.6 psf
Thrust — Pa = ½KaγH² = 0.5(0.361)(122.0)(13.5)² = 4,014 lb/ft
Location
$y ˉ = H /3 = 4.50 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.361, Pa ≈ 4,014 lb/ft acting 4.50 ft above the base

Why the other options are there

Pa = 8,027 lb/ft (½ omitted)
ȳ = 6.75 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 4

Rankine active thrust on a retaining wall — Retaining Walls (3)

A 16.5 ft high frictionless wall retains cohesionless backfill with φ = 33° and γ = 122.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 16.5 ft
φ = 33°
γ = 122.0 pcf
(check: tan²(45−φ/2) = 0.295)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.545) / (1 + 0.545) = 0.295$
Base pressure — σa = KaγH = 0.295(122.0)(16.5) = 593.4 psf
Thrust — Pa = ½KaγH² = 0.5(0.295)(122.0)(16.5)² = 4,896 lb/ft
Location
$y ˉ = H /3 = 5.50 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.295, Pa ≈ 4,896 lb/ft acting 5.50 ft above the base

Why the other options are there

Pa = 9,792 lb/ft (½ omitted)
ȳ = 8.25 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 5

Rankine active thrust on a retaining wall — Retaining Walls (4)

A 18.5 ft high frictionless wall retains cohesionless backfill with φ = 28° and γ = 108.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 18.5 ft
φ = 28°
γ = 108.0 pcf
(check: tan²(45−φ/2) = 0.361)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.469) / (1 + 0.469) = 0.361$
Base pressure — σa = KaγH = 0.361(108.0)(18.5) = 721.3 psf
Thrust — Pa = ½KaγH² = 0.5(0.361)(108.0)(18.5)² = 6,672 lb/ft
Location
$y ˉ = H /3 = 6.17 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.361, Pa ≈ 6,672 lb/ft acting 6.17 ft above the base

Why the other options are there

Pa = 13,345 lb/ft (½ omitted)
ȳ = 9.25 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 6

Rankine active thrust on a retaining wall — Retaining Walls (5)

A 19.0 ft high frictionless wall retains cohesionless backfill with φ = 37° and γ = 118.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 19.0 ft
φ = 37°
γ = 118.0 pcf
(check: tan²(45−φ/2) = 0.249)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.602) / (1 + 0.602) = 0.249$
Base pressure — σa = KaγH = 0.249(118.0)(19.0) = 557.3 psf
Thrust — Pa = ½KaγH² = 0.5(0.249)(118.0)(19.0)² = 5,295 lb/ft
Location
$y ˉ = H /3 = 6.33 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.249, Pa ≈ 5,295 lb/ft acting 6.33 ft above the base

Why the other options are there

Pa = 10,589 lb/ft (½ omitted)
ȳ = 9.50 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 7

Rankine active thrust on a retaining wall — Retaining Walls (6)

A 8.0 ft high frictionless wall retains cohesionless backfill with φ = 30° and γ = 125.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 8.0 ft
φ = 30°
γ = 125.0 pcf
(check: tan²(45−φ/2) = 0.333)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.500) / (1 + 0.500) = 0.333$
Base pressure — σa = KaγH = 0.333(125.0)(8.0) = 333.3 psf
Thrust — Pa = ½KaγH² = 0.5(0.333)(125.0)(8.0)² = 1,333 lb/ft
Location
$y ˉ = H /3 = 2.67 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.333, Pa ≈ 1,333 lb/ft acting 2.67 ft above the base

Why the other options are there

Pa = 2,667 lb/ft (½ omitted)
ȳ = 4.00 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 8

Rankine active thrust on a retaining wall — Retaining Walls (7)

A 12.0 ft high frictionless wall retains cohesionless backfill with φ = 31° and γ = 115.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 12.0 ft
φ = 31°
γ = 115.0 pcf
(check: tan²(45−φ/2) = 0.320)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.515) / (1 + 0.515) = 0.320$
Base pressure — σa = KaγH = 0.320(115.0)(12.0) = 441.7 psf
Thrust — Pa = ½KaγH² = 0.5(0.320)(115.0)(12.0)² = 2,650 lb/ft
Location
$y ˉ = H /3 = 4.00 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.320, Pa ≈ 2,650 lb/ft acting 4.00 ft above the base

Why the other options are there

Pa = 5,301 lb/ft (½ omitted)
ȳ = 6.00 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 9

Rankine active thrust on a retaining wall — Retaining Walls (8)

A 10.0 ft high frictionless wall retains cohesionless backfill with φ = 29° and γ = 118.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 10.0 ft
φ = 29°
γ = 118.0 pcf
(check: tan²(45−φ/2) = 0.347)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.485) / (1 + 0.485) = 0.347$
Base pressure — σa = KaγH = 0.347(118.0)(10.0) = 409.4 psf
Thrust — Pa = ½KaγH² = 0.5(0.347)(118.0)(10.0)² = 2,047 lb/ft
Location
$y ˉ = H /3 = 3.33 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.347, Pa ≈ 2,047 lb/ft acting 3.33 ft above the base

Why the other options are there

Pa = 4,094 lb/ft (½ omitted)
ȳ = 5.00 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

Example 10

Rankine active thrust on a retaining wall — Retaining Walls (9)

A 13.5 ft high frictionless wall retains cohesionless backfill with φ = 30° and γ = 125.0 pcf. Find Ka, the total active thrust per foot of wall and its point of application.

Given

H = 13.5 ft
φ = 30°
γ = 125.0 pcf
(check: tan²(45−φ/2) = 0.333)

Find

Ka, Pa and the resultant location

Start with the thinking

Active pressure grows linearly with depth, so the thrust is a triangle.
The resultant acts at H/3 above the base, not at mid-height.

Step-by-step solution

Active coefficient
$K a = (1 - sin ϕ) / (1 + sin ϕ) = (1 - 0.500) / (1 + 0.500) = 0.333$
Base pressure — σa = KaγH = 0.333(125.0)(13.5) = 562.5 psf
Thrust — Pa = ½KaγH² = 0.5(0.333)(125.0)(13.5)² = 3,797 lb/ft
Location
$y ˉ = H /3 = 4.50 f t ab o v e t h e ba se$

Answer: Ka ≈ 0.333, Pa ≈ 3,797 lb/ft acting 4.50 ft above the base

Why the other options are there

Pa = 7,594 lb/ft (½ omitted)
ȳ = 6.75 ft (triangular distribution treated as uniform)

Reference: FE Reference Handbook — Geotechnical → Retaining Walls

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

Retaining Walls contains 39 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.

Retaining Walls

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

FSoverturning=MR

FSsliding=/

Vhtand+BCa+Pp

FSsliding=Pacosa

FSbearingcapacity=qULT

qtoe=Bc1​+Bm

e=2−eo

e=eccentricity

B=widthofbase

MR=resistingmoment

MO=overturningmoment

FR=resistingforces

FD=drivingforces

V=verticalforces

δ=k1​ϕ2​

Ca=k2​C2​

αS LSNeeds technical review

FS=factorofsafetyagainstslopeinstability

=TFF/TMOB

TFF=availableshearingresistancealongslipsurface

= cLS + WM cos αS tan φNeeds technical review

TMOB=mobilizedshearforcealongslipsurface

= WM sin αSNeeds technical review

LS=lengthofassumedplanarslipsurface

WM=weightofsoilaboveslipsurface

αS = angle of assumed slip surface with respect to horizontalNeeds technical review

2.00mm(No.10)≤50−−−−−−−−−−

0.425mm(No.40)≤30≤50≥51−−−−−−−−

0.075mm(No.200)≤15≤25≤10≤35≤35≤35≤35≥36≥36≥36≥36

LIQUIDLIMIT−−≤40≥41≤40≥41≤40≥41≤40≥41

PLASTICITYINDEX∗6maxNP≤10≤10≥11≥11≤10≤10≥11≥11

NP=Non−plastic(use"0").Symbol"−"meansthattheparticularsieveanalysisisnotconsideredforthatclassification.

soil.Example:A−6withGI=15islesssuitablethanA−6withGI=10.

GI=(F−35)[0.2+0.005(LL−40)]+0.01(F−15)(PI−10)

where:F=PercentpassingNo.200sieve,expressedasawholenumber.Thispercentage

LL=Liquidlimit

PI=Plasticityindex

IfthecomputedvalueofGI<0,thenuseGI=0.

COARSE−GRAINEDGravelsCleanGravelsCu≥4and1≤Cc≤3DGWWell−gradedgravelE

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section