True stress

Materials Science · FE Reference Handbook section

Materials Science

24 formulas

10 exam-style examples

~60 min

All Materials Science lectures

Learning objectives

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

This chapter section covers True stress within Materials Science. 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 true stress describes physically and when it applies.
State every one of the 24 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: percent versus fraction in composition and strain.

Lecture

Why this section exists. True stress is the part of Materials Science that lets you connect a steel, concrete or polymer specimen under test 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 definition, a phase-diagram read, or a one-line property calculation. 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. percent versus fraction in composition and strain. 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.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: true stress.
Wikimedia Commons, public domain

Materials Science — True stress: 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 steel, concrete or polymer specimen under test. 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 24 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

vT	Quantity produced by "vT = A" — read its definition and unit from the handbook line directly above the equation.
σT	Quantity produced by "σT = true stress" — read its definition and unit from the handbook line directly above the equation.
F	Quantity produced by "F = applied force" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = actual cross-sectional area" — read its definition and unit from the handbook line directly above the equation.
σ	Quantity produced by "σ = Eε" — read its definition and unit from the handbook line directly above the equation.
where E	Quantity produced by "where E = elastic modulus" — read its definition and unit from the handbook line directly above the equation.
df	Quantity produced by "df = n − RT" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = stress sensitivity" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = activation energy for creep" — read its definition and unit from the handbook line directly above the equation.
R	Quantity produced by "R = ideal gas law constant" — read its definition and unit from the handbook line directly above the equation.
T	Quantity produced by "T = absolute temperature" — read its definition and unit from the handbook line directly above the equation.
where R	Quantity produced by "where R = stress ratio" — 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
The elastic modulus (also called modulus, modulus of elasticity, Young's modulus) describes the relationship between
engineering stress and engineering strain during elastic loading. Hooke's Law applies in such a case.
Key mechanical properties obtained from a tensile test curve:
• Elastic modulus
• Ductility (also called percent elongation): Permanent engineering strain after failure
• Ultimate tensile strength (also called tensile strength): Maximum engineering stress
• Yield strength: Engineering stress at which permanent deformation is first observed, calculated by 0.2% offset method.
Other mechanical properties:
• Creep: Time-dependent deformation under load. Usually measured by strain rate. For steady-state creep this is:
dt Av e
where
• Fatigue: Time-dependent failure under cyclic load. Fatigue life is the number of cycles to failure. The endurance limit is
the stress below which fatigue failure is unlikely.
vr B
where
• Fracture toughness: The combination of applied stress and the crack length in a brittle material. It is the stress intensity
when the material will fail.
where
The critical value of stress intensity at which catastrophic crack propagation occurs, KIc, is a material property.
Representative Values of Fracture Toughness
Material K Ic (MPa•m 1/2 ) K Ic (ksi-in 1/2 )
A1 2014-T651 24.2 22
A1 2024-T3 44 40

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

Engineering versus true stress and strain — True stress

A 0.430 in diameter bar of original length 4 in carries 17,000 lb and stretches 0.11 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 17,000 lb
d₀ = 0.430 in
L₀ = 4 in
ΔL = 0.11 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.430)^{2} = 0.14522 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 17000/0.14522 = 117, 064 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.11/4 = 0.02750$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 117, 064 (1 + 0.02750) = 120, 283 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.02750) = 0.02713$

Answer: σ_eng = 117,064 psi, ε_eng = 0.0275; σ_true = 120,283 psi, ε_true = 0.0271

Why the other options are there

σ_true = 113,931 psi (divided instead of multiplied)
ε_true = 0.0275 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 2

Engineering versus true stress and strain — True stress (2)

A 0.635 in diameter bar of original length 6 in carries 22,000 lb and stretches 0.20 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 22,000 lb
d₀ = 0.635 in
L₀ = 6 in
ΔL = 0.20 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.635)^{2} = 0.31669 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 22000/0.31669 = 69, 468 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.20/6 = 0.03333$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 69, 468 (1 + 0.03333) = 71, 784 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.03333) = 0.03279$

Answer: σ_eng = 69,468 psi, ε_eng = 0.0333; σ_true = 71,784 psi, ε_true = 0.0328

Why the other options are there

σ_true = 67,227 psi (divided instead of multiplied)
ε_true = 0.0333 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 3

Engineering versus true stress and strain — True stress (3)

A 0.710 in diameter bar of original length 4 in carries 25,000 lb and stretches 0.14 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 25,000 lb
d₀ = 0.710 in
L₀ = 4 in
ΔL = 0.14 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.710)^{2} = 0.39592 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 25000/0.39592 = 63, 144 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.14/4 = 0.03500$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 63, 144 (1 + 0.03500) = 65, 354 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.03500) = 0.03440$

Answer: σ_eng = 63,144 psi, ε_eng = 0.0350; σ_true = 65,354 psi, ε_true = 0.0344

Why the other options are there

σ_true = 61,009 psi (divided instead of multiplied)
ε_true = 0.0350 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 4

Engineering versus true stress and strain — True stress (4)

A 0.590 in diameter bar of original length 5 in carries 25,000 lb and stretches 0.25 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 25,000 lb
d₀ = 0.590 in
L₀ = 5 in
ΔL = 0.25 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.590)^{2} = 0.27340 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 25000/0.27340 = 91, 442 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.25/5 = 0.05000$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 91, 442 (1 + 0.05000) = 96, 014 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.05000) = 0.04879$

Answer: σ_eng = 91,442 psi, ε_eng = 0.0500; σ_true = 96,014 psi, ε_true = 0.0488

Why the other options are there

σ_true = 87,088 psi (divided instead of multiplied)
ε_true = 0.0500 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 5

Engineering versus true stress and strain — True stress (5)

A 0.505 in diameter bar of original length 6 in carries 11,000 lb and stretches 0.13 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 11,000 lb
d₀ = 0.505 in
L₀ = 6 in
ΔL = 0.13 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.505)^{2} = 0.20030 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 11000/0.20030 = 54, 919 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.13/6 = 0.02167$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 54, 919 (1 + 0.02167) = 56, 109 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.02167) = 0.02144$

Answer: σ_eng = 54,919 psi, ε_eng = 0.0217; σ_true = 56,109 psi, ε_true = 0.0214

Why the other options are there

σ_true = 53,754 psi (divided instead of multiplied)
ε_true = 0.0217 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 6

Engineering versus true stress and strain — True stress (6)

A 0.625 in diameter bar of original length 5 in carries 25,000 lb and stretches 0.16 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 25,000 lb
d₀ = 0.625 in
L₀ = 5 in
ΔL = 0.16 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.625)^{2} = 0.30680 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 25000/0.30680 = 81, 487 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.16/5 = 0.03200$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 81, 487 (1 + 0.03200) = 84, 095 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.03200) = 0.03150$

Answer: σ_eng = 81,487 psi, ε_eng = 0.0320; σ_true = 84,095 psi, ε_true = 0.0315

Why the other options are there

σ_true = 78,961 psi (divided instead of multiplied)
ε_true = 0.0320 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 7

Engineering versus true stress and strain — True stress (7)

A 0.695 in diameter bar of original length 8 in carries 16,000 lb and stretches 0.08 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 16,000 lb
d₀ = 0.695 in
L₀ = 8 in
ΔL = 0.08 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.695)^{2} = 0.37937 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 16000/0.37937 = 42, 176 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.08/8 = 0.01000$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 42, 176 (1 + 0.01000) = 42, 597 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.01000) = 0.00995$

Answer: σ_eng = 42,176 psi, ε_eng = 0.0100; σ_true = 42,597 psi, ε_true = 0.0100

Why the other options are there

σ_true = 41,758 psi (divided instead of multiplied)
ε_true = 0.0100 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 8

Engineering versus true stress and strain — True stress (8)

A 0.505 in diameter bar of original length 6 in carries 7,000 lb and stretches 0.23 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 7,000 lb
d₀ = 0.505 in
L₀ = 6 in
ΔL = 0.23 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.505)^{2} = 0.20030 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 7000/0.20030 = 34, 948 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.23/6 = 0.03833$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 34, 948 (1 + 0.03833) = 36, 288 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.03833) = 0.03762$

Answer: σ_eng = 34,948 psi, ε_eng = 0.0383; σ_true = 36,288 psi, ε_true = 0.0376

Why the other options are there

σ_true = 33,658 psi (divided instead of multiplied)
ε_true = 0.0383 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 9

Engineering versus true stress and strain — True stress (9)

A 0.535 in diameter bar of original length 3 in carries 9,000 lb and stretches 0.35 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 9,000 lb
d₀ = 0.535 in
L₀ = 3 in
ΔL = 0.35 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.535)^{2} = 0.22480 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 9000/0.22480 = 40, 035 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.35/3 = 0.11667$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 40, 035 (1 + 0.11667) = 44, 706 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.11667) = 0.11035$

Answer: σ_eng = 40,035 psi, ε_eng = 0.1167; σ_true = 44,706 psi, ε_true = 0.1103

Why the other options are there

σ_true = 35,853 psi (divided instead of multiplied)
ε_true = 0.1167 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

Example 10

Engineering versus true stress and strain — True stress (10)

A 0.605 in diameter bar of original length 4 in carries 8,000 lb and stretches 0.21 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 8,000 lb
d₀ = 0.605 in
L₀ = 4 in
ΔL = 0.21 in

Find

σ_eng, ε_eng, σ_true and ε_true

Start with the thinking

Engineering stress uses the original area; true stress uses the instantaneous area.
Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

Area
$A_{0} = (π /4) (0.605)^{2} = 0.28748 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 8000/0.28748 = 27, 828 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.21/4 = 0.05250$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 27, 828 (1 + 0.05250) = 29, 289 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.05250) = 0.05117$

Answer: σ_eng = 27,828 psi, ε_eng = 0.0525; σ_true = 29,289 psi, ε_true = 0.0512

Why the other options are there

σ_true = 26,440 psi (divided instead of multiplied)
ε_true = 0.0525 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True stress

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 steel, concrete or polymer specimen under test, 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

True stress contains 24 relations; you must be able to find this page in under 15 seconds.
Exam style: a definition, a phase-diagram read, or a one-line property calculation.
Unit rule: percent versus fraction in composition and strain.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

percent versus fraction in composition and strain
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.

True stress

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

vT=A

σT = true stressNeeds technical review

F=appliedforce

A=actualcross−sectionalarea

σ=Eε

whereE=elasticmodulus

df=n−RT

A=pre−exponentialconstant

n=stresssensitivity

Q=activationenergyforcreep

R=idealgaslawconstant

T=absolutetemperature

R=vmin

whereR=stressratio

ForR=−1andhighcyclefatigue,basedontheBasquinequation:

N=dn

N=cyclestofailure

σr = completely (fully) reversed stressNeeds technical review

AandB=materialconstants

KIC=Yvra

KIC=fracturetoughnessa

σ=appliedengineeringstress

a=cracklength

Y=geometricalfactorEXTERIORCRACK(Y=1.1)INTERIORCRACK(Y=1)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section