Percent Reduction in Area (RA)

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

1 formulas

10 exam-style examples

~47 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Percent Reduction in Area (RA) within Mechanics of Materials. 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 percent reduction in area (ra) describes physically and when it applies.
State every one of the 1 relation 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: psi vs ksi and kip vs lb decide the answer choice.

Lecture

Why this section exists. Percent Reduction in Area (RA) is the part of Mechanics of Materials that lets you connect an axially loaded, bent or twisted member 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 stress or deformation at one point of one member. 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. psi vs ksi and kip vs lb decide the answer choice. 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: percent reduction in area (ra).
Wikimedia Commons, public domain

Mechanics of Materials — Percent Reduction in Area (RA): 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 an axially loaded, bent or twisted member. 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 1 relation 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The % reduction in area from initial area, Ai, to final area, Af , is:
A - Af

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

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA)

A steel tensile coupon starts with a 0.435 in diameter and necks to 0.270 in at fracture. The 2 in gauge length becomes 2.280 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.435 in
d_f = 0.270 in
L₀ = 2 in, L_f = 2.280 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.435)^{2} = 0.14862 i n^{2}, A_{f} = (π /4) (0.270)^{2} = 0.05713 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 61.6%, %elongation = 14.0%

Why the other options are there

38.0% (diameters instead of areas)
38.4% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 2

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (2)

A steel tensile coupon starts with a 0.445 in diameter and necks to 0.303 in at fracture. The 2 in gauge length becomes 2.500 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.445 in
d_f = 0.303 in
L₀ = 2 in, L_f = 2.500 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.445)^{2} = 0.15553 i n^{2}, A_{f} = (π /4) (0.303)^{2} = 0.07192 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 53.8%, %elongation = 25.0%

Why the other options are there

32.0% (diameters instead of areas)
46.2% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 3

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (3)

A steel tensile coupon starts with a 0.475 in diameter and necks to 0.371 in at fracture. The 2 in gauge length becomes 2.460 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.475 in
d_f = 0.371 in
L₀ = 2 in, L_f = 2.460 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.475)^{2} = 0.17721 i n^{2}, A_{f} = (π /4) (0.371)^{2} = 0.10781 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 39.2%, %elongation = 23.0%

Why the other options are there

22.0% (diameters instead of areas)
60.8% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 4

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (4)

A steel tensile coupon starts with a 0.525 in diameter and necks to 0.315 in at fracture. The 2 in gauge length becomes 2.300 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.525 in
d_f = 0.315 in
L₀ = 2 in, L_f = 2.300 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.525)^{2} = 0.21648 i n^{2}, A_{f} = (π /4) (0.315)^{2} = 0.07793 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 64.0%, %elongation = 15.0%

Why the other options are there

40.0% (diameters instead of areas)
36.0% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 5

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (5)

A steel tensile coupon starts with a 0.405 in diameter and necks to 0.227 in at fracture. The 2 in gauge length becomes 2.600 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.405 in
d_f = 0.227 in
L₀ = 2 in, L_f = 2.600 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.405)^{2} = 0.12882 i n^{2}, A_{f} = (π /4) (0.227)^{2} = 0.04040 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 68.6%, %elongation = 30.0%

Why the other options are there

44.0% (diameters instead of areas)
31.4% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 6

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (6)

A steel tensile coupon starts with a 0.590 in diameter and necks to 0.413 in at fracture. The 2 in gauge length becomes 2.340 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.590 in
d_f = 0.413 in
L₀ = 2 in, L_f = 2.340 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.590)^{2} = 0.27340 i n^{2}, A_{f} = (π /4) (0.413)^{2} = 0.13396 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 51.0%, %elongation = 17.0%

Why the other options are there

30.0% (diameters instead of areas)
49.0% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 7

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (7)

A steel tensile coupon starts with a 0.530 in diameter and necks to 0.419 in at fracture. The 2 in gauge length becomes 2.380 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.530 in
d_f = 0.419 in
L₀ = 2 in, L_f = 2.380 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.530)^{2} = 0.22062 i n^{2}, A_{f} = (π /4) (0.419)^{2} = 0.13769 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 37.6%, %elongation = 19.0%

Why the other options are there

21.0% (diameters instead of areas)
62.4% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 8

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (8)

A steel tensile coupon starts with a 0.490 in diameter and necks to 0.402 in at fracture. The 2 in gauge length becomes 2.380 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.490 in
d_f = 0.402 in
L₀ = 2 in, L_f = 2.380 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.490)^{2} = 0.18857 i n^{2}, A_{f} = (π /4) (0.402)^{2} = 0.12680 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 32.8%, %elongation = 19.0%

Why the other options are there

18.0% (diameters instead of areas)
67.2% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 9

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (9)

A steel tensile coupon starts with a 0.595 in diameter and necks to 0.405 in at fracture. The 2 in gauge length becomes 2.680 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.595 in
d_f = 0.405 in
L₀ = 2 in, L_f = 2.680 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.595)^{2} = 0.27805 i n^{2}, A_{f} = (π /4) (0.405)^{2} = 0.12857 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 53.8%, %elongation = 34.0%

Why the other options are there

32.0% (diameters instead of areas)
46.2% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 10

Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (10)

A steel tensile coupon starts with a 0.505 in diameter and necks to 0.389 in at fracture. The 2 in gauge length becomes 2.640 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.505 in
d_f = 0.389 in
L₀ = 2 in, L_f = 2.640 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.505)^{2} = 0.20030 i n^{2}, A_{f} = (π /4) (0.389)^{2} = 0.11876 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 40.7%, %elongation = 32.0%

Why the other options are there

23.0% (diameters instead of areas)
59.3% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

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 an axially loaded, bent or twisted member, 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

Percent Reduction in Area (RA) contains 1 relation; you must be able to find this page in under 15 seconds.
Exam style: a stress or deformation at one point of one member.
Unit rule: psi vs ksi and kip vs lb decide the answer choice.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

psi vs ksi and kip vs lb decide the answer choice
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.