True strain

Materials Science · FE Reference Handbook section

Materials Science

18 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 strain 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 strain describes physically and when it applies.
State every one of the 18 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 strain 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 strain.
Wikimedia Commons, public domain

Materials Science — True strain: 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 18 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

fT	Quantity produced by "fT = L" — read its definition and unit from the handbook line directly above the equation.
εT	Quantity produced by "εT = true strain" — read its definition and unit from the handbook line directly above the equation.
dL	Quantity produced by "dL = differential change in length" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = initial length" — read its definition and unit from the handbook line directly above the equation.
Water	Quantity produced by "Water = 1000 at 0°C" — 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
Proper�es of Metals
Electrical
Density ρ Resistivity
Atomic Melting Melting Specific Heat (W/(m˙K))
Metal Symbol (kg/m3) (10−8 Ω˙m)
Weight Point (°C) Point (°F) (J/(kg˙K)) at 0°C
(273.2 K)
(273.2 K)
Aluminum Al 26.98 2,698 660 1,220 895.9 2.5 236
Antimony Sb 121.75 6,692 630 1,166 209.3 39 25.5
Arsenic As 74.92 5,776 subl. 613 subl. 1,135 347.5 26 −
Barium Ba 137.33 3,594 710 1,310 284.7 36 −
Beryllium Be 9.012 1,846 1,285 2,345 2,051.5 2.8 218
Bismuth Bi 208.98 9,803 271 519 125.6 107 8.2
Cadmium Cd 112.41 8,647 321 609 234.5 6.8 97
Caesium Cs 132.91 1,900 29 84 217.7 18.8 36
Calcium Ca 40.08 1,530 840 1,544 636.4 3.2 −
Cerium Ce 140.12 6,711 800 1,472 188.4 7.3 11
Chromium Cr 52 7,194 1,860 3,380 406.5 12.7 96.5
Cobalt Co 58.93 8,800 1,494 2,721 431.2 5.6 105
Copper Cu 63.54 8,933 1,084 1,983 389.4 1.55 403
Gallium Ga 69.72 5,905 30 86 330.7 13.6 41
Gold Au 196.97 19,281 1,064 1,947 129.8 2.05 319

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 strain

A 0.650 in diameter bar of original length 3 in carries 15,000 lb and stretches 0.34 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 15,000 lb
d₀ = 0.650 in
L₀ = 3 in
ΔL = 0.34 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
$A_{0} = (π /4) (0.650)^{2} = 0.33183 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 15000/0.33183 = 45, 204 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.34/3 = 0.11333$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 45, 204 (1 + 0.11333) = 50, 327 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.11333) = 0.10736$

Answer: σ_eng = 45,204 psi, ε_eng = 0.1133; σ_true = 50,327 psi, ε_true = 0.1074

Why the other options are there

σ_true = 40,602 psi (divided instead of multiplied)
ε_true = 0.1133 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 2

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

A 0.475 in diameter bar of original length 7 in carries 7,000 lb and stretches 0.18 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 7,000 lb
d₀ = 0.475 in
L₀ = 7 in
ΔL = 0.18 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
$A_{0} = (π /4) (0.475)^{2} = 0.17721 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 7000/0.17721 = 39, 502 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.18/7 = 0.02571$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 39, 502 (1 + 0.02571) = 40, 518 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.02571) = 0.02539$

Answer: σ_eng = 39,502 psi, ε_eng = 0.0257; σ_true = 40,518 psi, ε_true = 0.0254

Why the other options are there

σ_true = 38,512 psi (divided instead of multiplied)
ε_true = 0.0257 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 3

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

A 0.530 in diameter bar of original length 8 in carries 23,000 lb and stretches 0.05 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 23,000 lb
d₀ = 0.530 in
L₀ = 8 in
ΔL = 0.05 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
$A_{0} = (π /4) (0.530)^{2} = 0.22062 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 23000/0.22062 = 104, 252 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.05/8 = 0.00625$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 104, 252 (1 + 0.00625) = 104, 904 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.00625) = 0.00623$

Answer: σ_eng = 104,252 psi, ε_eng = 0.0063; σ_true = 104,904 psi, ε_true = 0.0062

Why the other options are there

σ_true = 103,605 psi (divided instead of multiplied)
ε_true = 0.0063 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 4

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

A 0.650 in diameter bar of original length 6 in carries 15,000 lb and stretches 0.31 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 15,000 lb
d₀ = 0.650 in
L₀ = 6 in
ΔL = 0.31 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
$A_{0} = (π /4) (0.650)^{2} = 0.33183 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 15000/0.33183 = 45, 204 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.31/6 = 0.05167$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 45, 204 (1 + 0.05167) = 47, 539 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.05167) = 0.05038$

Answer: σ_eng = 45,204 psi, ε_eng = 0.0517; σ_true = 47,539 psi, ε_true = 0.0504

Why the other options are there

σ_true = 42,983 psi (divided instead of multiplied)
ε_true = 0.0517 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 5

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

A 0.500 in diameter bar of original length 8 in carries 24,000 lb and stretches 0.25 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 24,000 lb
d₀ = 0.500 in
L₀ = 8 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
$A_{0} = (π /4) (0.500)^{2} = 0.19635 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 24000/0.19635 = 122, 231 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.25/8 = 0.03125$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 122, 231 (1 + 0.03125) = 126, 051 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.03125) = 0.03077$

Answer: σ_eng = 122,231 psi, ε_eng = 0.0313; σ_true = 126,051 psi, ε_true = 0.0308

Why the other options are there

σ_true = 118,527 psi (divided instead of multiplied)
ε_true = 0.0313 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 6

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

A 0.490 in diameter bar of original length 7 in carries 20,000 lb and stretches 0.05 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 20,000 lb
d₀ = 0.490 in
L₀ = 7 in
ΔL = 0.05 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
$A_{0} = (π /4) (0.490)^{2} = 0.18857 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 20000/0.18857 = 106, 059 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.05/7 = 0.00714$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 106, 059 (1 + 0.00714) = 106, 817 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.00714) = 0.00712$

Answer: σ_eng = 106,059 psi, ε_eng = 0.0071; σ_true = 106,817 psi, ε_true = 0.0071

Why the other options are there

σ_true = 105,307 psi (divided instead of multiplied)
ε_true = 0.0071 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 7

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

A 0.615 in diameter bar of original length 5 in carries 8,000 lb and stretches 0.05 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 8,000 lb
d₀ = 0.615 in
L₀ = 5 in
ΔL = 0.05 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
$A_{0} = (π /4) (0.615)^{2} = 0.29706 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 8000/0.29706 = 26, 931 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.05/5 = 0.01000$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 26, 931 (1 + 0.01000) = 27, 200 p s i$
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 = 26,931 psi, ε_eng = 0.0100; σ_true = 27,200 psi, ε_true = 0.0100

Why the other options are there

σ_true = 26,664 psi (divided instead of multiplied)
ε_true = 0.0100 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 8

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

A 0.420 in diameter bar of original length 6 in carries 12,000 lb and stretches 0.22 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 12,000 lb
d₀ = 0.420 in
L₀ = 6 in
ΔL = 0.22 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
$A_{0} = (π /4) (0.420)^{2} = 0.13854 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 12000/0.13854 = 86, 615 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.22/6 = 0.03667$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 86, 615 (1 + 0.03667) = 89, 791 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.03667) = 0.03601$

Answer: σ_eng = 86,615 psi, ε_eng = 0.0367; σ_true = 89,791 psi, ε_true = 0.0360

Why the other options are there

σ_true = 83,551 psi (divided instead of multiplied)
ε_true = 0.0367 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 9

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

A 0.685 in diameter bar of original length 7 in carries 12,000 lb and stretches 0.04 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 12,000 lb
d₀ = 0.685 in
L₀ = 7 in
ΔL = 0.04 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
$A_{0} = (π /4) (0.685)^{2} = 0.36853 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 12000/0.36853 = 32, 562 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.04/7 = 0.00571$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 32, 562 (1 + 0.00571) = 32, 748 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.00571) = 0.00570$

Answer: σ_eng = 32,562 psi, ε_eng = 0.0057; σ_true = 32,748 psi, ε_true = 0.0057

Why the other options are there

σ_true = 32,377 psi (divided instead of multiplied)
ε_true = 0.0057 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

Example 10

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

A 0.435 in diameter bar of original length 7 in carries 8,000 lb and stretches 0.23 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Given

P = 8,000 lb
d₀ = 0.435 in
L₀ = 7 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
$A_{0} = (π /4) (0.435)^{2} = 0.14862 i n^{2}$
Formula
$σ_{e} n g = P / A_{0}$
Substituting
$σ_{e} n g = 8000/0.14862 = 53, 830 p s i$
Formula — ε_eng = ΔL/L₀
Substituting
$ε_{e} n g = 0.23/7 = 0.03286$
Formula
$σ_{t} r u e = σ_{e} n g (1 + ε_{e} n g)$
Substituting
$σ_{t} r u e = 53, 830 (1 + 0.03286) = 55, 598 p s i$
Formula
$ε_{t} r u e = ln (1 + ε_{e} n g)$
Substituting
$ε_{t} r u e = ln (1.03286) = 0.03233$

Answer: σ_eng = 53,830 psi, ε_eng = 0.0329; σ_true = 55,598 psi, ε_true = 0.0323

Why the other options are there

σ_true = 52,117 psi (divided instead of multiplied)
ε_true = 0.0329 (no logarithm)

Reference: FE Reference Handbook — Materials Science → True strain

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 strain contains 18 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 strain

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

fT=L

εT = true strainNeeds technical review

dL=differentialchangeinlength

L=initiallength

εT = ln (1 + ε)Needs technical review

HeatConductivityλ

Water=1000at0∘C

SiBIIIA600×1024

AIIIIA20×1024

GaIIIA40×1024

PVA1,000×1024

AsVA2,000×1024

SbVA70×1024

GeAlIIIA400×1024

GaIIIA500×1024

InIIIA4×1024

AsVA80×1024

SbVA10×1024

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section