Ultimate Bearing Capacity

Geotechnical · FE Reference Handbook section

Geotechnical

6 formulas

10 exam-style examples

~57 min

All Geotechnical lectures

Learning objectives

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

This chapter section covers Ultimate Bearing Capacity 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 ultimate bearing capacity describes physically and when it applies.
State every one of the 6 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. Ultimate Bearing Capacity 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: ultimate bearing capacity.
Capstone Studio instructional photograph

Geotechnical — Ultimate Bearing Capacity: 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 6 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

qULT	Quantity produced by "qULT = cNc + γ ' Df Nq + 2 γ ' BNγ" — read its definition and unit from the handbook line directly above the equation.
Nc	Quantity produced by "Nc = bearing capacity factor for cohesion" — read its definition and unit from the handbook line directly above the equation.
Nq	Quantity produced by "Nq = bearing capacity factor for depth" — read its definition and unit from the handbook line directly above the equation.
Nγ	Quantity produced by "Nγ = bearing capacity factor for unit weight" — read its definition and unit from the handbook line directly above the equation.
Df	Quantity produced by "Df = depth of footing below ground surface" — read its definition and unit from the handbook line directly above the equation.
B	Quantity produced by "B = width of strip footing" — 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.

where

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

Allowable bearing pressure of a strip footing

A 1.5 m wide strip footing sits 1.2 m deep in sand with γ = 18 kN/m³, Nq = 18.4 and Nγ = 22.4 (c = 0). Use FS = 3 to find the allowable bearing pressure.

Given

B = 1.5 m, Df = 1.2 m
γ = 18 kN/m³
Nq = 18.4, Nγ = 22.4
FS = 3

Find

q_all

Start with the thinking

Cohesionless soil drops the cNc term.
Net versus gross matters — this is a gross allowable pressure.

Step-by-step solution

Surcharge — q = γDf = 18(1.2) = 21.6 kPa
Surcharge term
$q N q = 21.6 (18.4) = 397.4 k P a$
Width term — 0.5γBNγ = 0.5(18)(1.5)(22.4) = 302.4 kPa
Ultimate
$q_{u} l t = 397.4 + 302.4 = 699.8 k P a$
Allowable
$q_{a} l l = 699.8/3$
Result
$q_{a} l l = 233 k P a$

Answer: q_all ≈ 233 kPa

Why the other options are there

700 kPa (FS omitted)
133 kPa (width term dropped)

Reference: FE Reference Handbook — Geotechnical — Bearing capacity

Example 2

Traffic density from flow and speed

A freeway lane carries 1,800 veh/h at an average speed of 90 km/h. Find the density and the average spacing.

Given

q = 1,800 veh/h
u = 90 km/h

Find

Density k and spacing

Start with the thinking

q = ku is the fundamental relation.
Spacing is the reciprocal of density.

Step-by-step solution

Density
$k = q / u = 1, 800/90$
Evaluate
$k = 20.0 v e h / k m / l an e$
Spacing
$s = 1, 000/ k = 1, 000/20.0$
Result
$k = 20 v e h / k m, s p a c in g = 50.0 m$

Answer: k = 20 veh/km/lane, spacing = 50 m

Why the other options are there

162,000 (q multiplied by u)
k = 0.05 (units not converted)

Reference: FE Reference Handbook — Transportation — Traffic flow relationships

Example 3

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity

A strip footing of width B = 6.5 ft is founded at Df = 2.0 ft in soil with γ = 116.0 pcf and c = 401 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 6.5 ft
Df = 2.0 ft
γ = 116.0 pcf
c = 401 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$401 (25) = 10, 025 p s f$
Surcharge term
$116.0 (2.0) (12) = 2, 784 p s f$
Width term
$0.5 (116.0) (6.5) (10) = 3, 770 p s f$
Ultimate
$q u l t = 16, 579 p s f$
Allowable
$q a l l = q u l t / F S = 16, 579/3 = 5, 526 p s f$

Answer: qult ≈ 16,579 psf, qall ≈ 5,526 psf

Why the other options are there

qult = 20,349 psf (½ omitted on the γB term)
qall = 16,579 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

Example 4

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (2)

A strip footing of width B = 5.5 ft is founded at Df = 4.5 ft in soil with γ = 124.0 pcf and c = 470 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 5.5 ft
Df = 4.5 ft
γ = 124.0 pcf
c = 470 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (2)

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$470 (25) = 11, 750 p s f$
Surcharge term
$124.0 (4.5) (12) = 6, 696 p s f$
Width term
$0.5 (124.0) (5.5) (10) = 3, 410 p s f$
Ultimate
$q u l t = 21, 856 p s f$
Allowable
$q a l l = q u l t / F S = 21, 856/3 = 7, 285 p s f$

Answer: qult ≈ 21,856 psf, qall ≈ 7,285 psf

Why the other options are there

qult = 25,266 psf (½ omitted on the γB term)
qall = 21,856 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

Example 5

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (3)

A strip footing of width B = 6.5 ft is founded at Df = 3.0 ft in soil with γ = 112.0 pcf and c = 780 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 6.5 ft
Df = 3.0 ft
γ = 112.0 pcf
c = 780 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (3)

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$780 (25) = 19, 500 p s f$
Surcharge term
$112.0 (3.0) (12) = 4, 032 p s f$
Width term
$0.5 (112.0) (6.5) (10) = 3, 640 p s f$
Ultimate
$q u l t = 27, 172 p s f$
Allowable
$q a l l = q u l t / F S = 27, 172/3 = 9, 057 p s f$

Answer: qult ≈ 27,172 psf, qall ≈ 9,057 psf

Why the other options are there

qult = 30,812 psf (½ omitted on the γB term)
qall = 27,172 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

Example 6

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (4)

A strip footing of width B = 4.5 ft is founded at Df = 5.5 ft in soil with γ = 122.0 pcf and c = 858 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 4.5 ft
Df = 5.5 ft
γ = 122.0 pcf
c = 858 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (4)

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$858 (25) = 21, 450 p s f$
Surcharge term
$122.0 (5.5) (12) = 8, 052 p s f$
Width term
$0.5 (122.0) (4.5) (10) = 2, 745 p s f$
Ultimate
$q u l t = 32, 247 p s f$
Allowable
$q a l l = q u l t / F S = 32, 247/3 = 10, 749 p s f$

Answer: qult ≈ 32,247 psf, qall ≈ 10,749 psf

Why the other options are there

qult = 34,992 psf (½ omitted on the γB term)
qall = 32,247 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

Example 7

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (5)

A strip footing of width B = 4.0 ft is founded at Df = 2.0 ft in soil with γ = 112.0 pcf and c = 734 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 4.0 ft
Df = 2.0 ft
γ = 112.0 pcf
c = 734 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (5)

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$734 (25) = 18, 350 p s f$
Surcharge term
$112.0 (2.0) (12) = 2, 688 p s f$
Width term
$0.5 (112.0) (4.0) (10) = 2, 240 p s f$
Ultimate
$q u l t = 23, 278 p s f$
Allowable
$q a l l = q u l t / F S = 23, 278/3 = 7, 759 p s f$

Answer: qult ≈ 23,278 psf, qall ≈ 7,759 psf

Why the other options are there

qult = 25,518 psf (½ omitted on the γB term)
qall = 23,278 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

Example 8

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (6)

A strip footing of width B = 7.5 ft is founded at Df = 4.0 ft in soil with γ = 124.0 pcf and c = 790 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 7.5 ft
Df = 4.0 ft
γ = 124.0 pcf
c = 790 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (6)

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$790 (25) = 19, 750 p s f$
Surcharge term
$124.0 (4.0) (12) = 5, 952 p s f$
Width term
$0.5 (124.0) (7.5) (10) = 4, 650 p s f$
Ultimate
$q u l t = 30, 352 p s f$
Allowable
$q a l l = q u l t / F S = 30, 352/3 = 10, 117 p s f$

Answer: qult ≈ 30,352 psf, qall ≈ 10,117 psf

Why the other options are there

qult = 35,002 psf (½ omitted on the γB term)
qall = 30,352 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

Example 9

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (7)

A strip footing of width B = 4.5 ft is founded at Df = 3.5 ft in soil with γ = 124.0 pcf and c = 677 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 4.5 ft
Df = 3.5 ft
γ = 124.0 pcf
c = 677 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (7)

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$677 (25) = 16, 925 p s f$
Surcharge term
$124.0 (3.5) (12) = 5, 208 p s f$
Width term
$0.5 (124.0) (4.5) (10) = 2, 790 p s f$
Ultimate
$q u l t = 24, 923 p s f$
Allowable
$q a l l = q u l t / F S = 24, 923/3 = 8, 308 p s f$

Answer: qult ≈ 24,923 psf, qall ≈ 8,308 psf

Why the other options are there

qult = 27,713 psf (½ omitted on the γB term)
qall = 24,923 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

Example 10

Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (8)

A strip footing of width B = 3.0 ft is founded at Df = 2.5 ft in soil with γ = 113.0 pcf and c = 221 psf. Using Nc = 25, Nq = 12, Nγ = 10, find qult and qall with FS = 3.

Given

B = 3.0 ft
Df = 2.5 ft
γ = 113.0 pcf
c = 221 psf
Nc=25, Nq=12, Nγ=10

Find

qult and qall

Start with the thinking

Terzaghi strip equation has three terms: cohesion, surcharge, and width.
The width term carries the ½ factor — a classic FE trap.

Figure for Ultimate and allowable bearing capacity of a strip footing — Ultimate Bearing Capacity (8)

Step-by-step solution

Terzaghi (strip) — qult = cNc + γDf Nq + ½γB Nγ
Cohesion term
$221 (25) = 5, 525 p s f$
Surcharge term
$113.0 (2.5) (12) = 3, 390 p s f$
Width term
$0.5 (113.0) (3.0) (10) = 1, 695 p s f$
Ultimate
$q u l t = 10, 610 p s f$
Allowable
$q a l l = q u l t / F S = 10, 610/3 = 3, 537 p s f$

Answer: qult ≈ 10,610 psf, qall ≈ 3,537 psf

Why the other options are there

qult = 12,305 psf (½ omitted on the γB term)
qall = 10,610 psf (factor of safety not applied)

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

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

Ultimate Bearing Capacity contains 6 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.