ASTM Grain Size

Materials Science · FE Reference Handbook section

Materials Science

8 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 ASTM Grain Size 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 astm grain size describes physically and when it applies.
State every one of the 8 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. ASTM Grain Size 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: astm grain size.
Wikimedia Commons, public domain

Materials Science — ASTM Grain Size: 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 8 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

SV	Quantity produced by "SV = 2PL" — read its definition and unit from the handbook line directly above the equation.
^ − h	Quantity produced by "^ − h" — read its definition and unit from the handbook line directly above the equation.
N`0.0645 mm2 j	Quantity produced by "N`0.0645 mm2 j = 2 n 1" — read its definition and unit from the handbook line directly above the equation.
PL	Quantity produced by "PL = number of points of intersection per unit length between the line and the boundaries" — read its definition and unit from the handbook line directly above the equation.
N	Quantity produced by "N = number of grains observed in an area of 0.0645 mm2" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = grain size (nearest integer > 1)" — 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.

Nactual N
Actual Area _0.0645 mm 2 i
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

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size

A steel component measures 163 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 5.

Given

BHN = 163
ASTM grain size n = 5

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 163 = 81, 500 p s i = 81.5 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 5 - 1) = 16 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 82 ksi; N = 16 grains/in²

Why the other options are there

0.33 ksi (ratio inverted)
32 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 2

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (2)

A steel component measures 351 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 7.

Given

BHN = 351
ASTM grain size n = 7

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 351 = 175, 500 p s i = 175.5 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 7 - 1) = 64 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 175.5 ksi; N = 64 grains/in²

Why the other options are there

0.70 ksi (ratio inverted)
128 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 3

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (3)

A steel component measures 352 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 5.

Given

BHN = 352
ASTM grain size n = 5

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 352 = 176, 000 p s i = 176.0 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 5 - 1) = 16 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 176.0 ksi; N = 16 grains/in²

Why the other options are there

0.70 ksi (ratio inverted)
32 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 4

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (4)

A steel component measures 388 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 6.

Given

BHN = 388
ASTM grain size n = 6

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 388 = 194, 000 p s i = 194.0 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 6 - 1) = 32 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 194.0 ksi; N = 32 grains/in²

Why the other options are there

0.78 ksi (ratio inverted)
64 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 5

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (5)

A steel component measures 322 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 7.

Given

BHN = 322
ASTM grain size n = 7

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 322 = 161, 000 p s i = 161.0 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 7 - 1) = 64 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 161.0 ksi; N = 64 grains/in²

Why the other options are there

0.64 ksi (ratio inverted)
128 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 6

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (6)

A steel component measures 394 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 7.

Given

BHN = 394
ASTM grain size n = 7

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 394 = 197, 000 p s i = 197.0 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 7 - 1) = 64 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 197.0 ksi; N = 64 grains/in²

Why the other options are there

0.79 ksi (ratio inverted)
128 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 7

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (7)

A steel component measures 245 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 4.

Given

BHN = 245
ASTM grain size n = 4

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 245 = 122, 500 p s i = 122.5 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 4 - 1) = 8 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 122.5 ksi; N = 8 grains/in²

Why the other options are there

0.49 ksi (ratio inverted)
16 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 8

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (8)

A steel component measures 246 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 4.

Given

BHN = 246
ASTM grain size n = 4

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 246 = 123, 000 p s i = 123.0 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 4 - 1) = 8 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 123.0 ksi; N = 8 grains/in²

Why the other options are there

0.49 ksi (ratio inverted)
16 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 9

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (9)

A steel component measures 209 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 7.

Given

BHN = 209
ASTM grain size n = 7

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 209 = 104, 500 p s i = 104.5 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 7 - 1) = 64 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 104.5 ksi; N = 64 grains/in²

Why the other options are there

0.42 ksi (ratio inverted)
128 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

Example 10

Hardness-to-strength conversion and ASTM grain size — ASTM Grain Size (10)

A steel component measures 184 BHN. Estimate its tensile strength from the hardness relationship, then determine how many grains per square inch at 100× correspond to an ASTM grain size number of 5.

Given

BHN = 184
ASTM grain size n = 5

Find

Approximate tensile strength and grains per in² at 100×

Start with the thinking

For steels the empirical relation between hardness and tensile strength is TS ≈ 500 × BHN in psi.
The ASTM grain size number doubles the grain count for every increment of one.

Step-by-step solution

Formula — TS (psi) ≈ 500 × BHN
Substituting
$T S = 500 \times 184 = 92, 000 p s i = 92.0 k s i$
Formula
$N = 2^{(} n - 1) g r ain s p er i n^{2} a t 100 \times$
Substituting
$N = 2^{(} 5 - 1) = 16 g r ain s / i n^{2}$
Interpretation — finer grains (higher n) raise yield strength through the Hall-Petch effect

Answer: TS ≈ 92 ksi; N = 16 grains/in²

Why the other options are there

0.37 ksi (ratio inverted)
32 grains/in² (exponent off by one)

Reference: FE Reference Handbook — Materials Science → ASTM Grain Size

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

ASTM Grain Size contains 8 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.

ASTM Grain Size

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

SV=2PL

−h

N‘0.0645mm2​j=2n1

=Needs technical review

SV=grain−boundarysurfaceperunitvolume

PL=numberofpointsofintersectionperunitlengthbetweenthelineandtheboundaries

N=numberofgrainsobservedinanareaof0.0645mm2​

n=grainsize(nearestinteger>1)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section