Copper

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

56 formulas

10 exam-style examples

~60 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Copper 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 copper describes physically and when it applies.
State every one of the 56 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: psi vs ksi and kip vs lb decide the answer choice.

Lecture

Why this section exists. Copper 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: copper.
Wikimedia Commons, public domain

Mechanics of Materials — Copper: 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 56 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

θ1	Quantity produced by "θ1 = v = v= (L – b – x )" — read its definition and unit from the handbook line directly above the equation.
θ2	Quantity produced by "θ2 = 0≤x≤a" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v= (x 2 – 3 L x + 2L2)" — read its definition and unit from the handbook line directly above the equation.
vmax	Quantity produced by "vmax = – wx 3 wL2" — read its definition and unit from the handbook line directly above the equation.
θ max v max	Quantity produced by "θ max v max" — read its definition and unit from the handbook line directly above the equation.
at x	Quantity produced by "at x = 0.5193L" — read its definition and unit from the handbook line directly above the equation.
θmax	Quantity produced by "θmax" — read its definition and unit from the handbook line directly above the equation.
x θmax	Quantity produced by "x θmax = Mmax ^at all x h = M 0" — read its definition and unit from the handbook line directly above the equation.
θ max	Quantity produced by "θ max" — 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.

Alloys Bronze C86100 0.319 15.0 5.6 50 50 — 95 95 — 20 0.34 9.60
Magnesium 0.066 6.48 2.5 22 22 — 40 40 22 1 0.30 14.3
Alloy [ Am 1004-T611]
Structural A36 0.284 29.0 11.0 36 36 — 58 58 — 30 0.32 6.60
Steel
Alloys Stainless 304 0.284 28.0 11.0 30 30 — 75 75 — 40 0.27 9.60
Tool L2 0.295 29.0 11.0 102 102 116
— 116 — 22 0.32 6.50
Titanium [ Ti-6Al-4V] 0.160 17.4 6.4 134 134 — 145 145 — 16 0.36 5.20
Alloy
Nonmetallic
Low Strength 0.086 3.20 — — — 1.8 — — — — 0.15 6.0
Concrete
High Strength 0.086 4.20 — — — 5.5 — — — — 0.15 6.0
Plastic Kevlar 49 0.0524 19.0 — — — — 104 70 10.2 2.8 0.34 —
Reinforced 30% Glass 0.0524 10.5 — — — — 13 19 — — 0.34 —
Wood Douglas Fir 0.017 1.90 — — — — 0.30c 3.78d 0.90d — 0.29c —
Select Structural
Grade White Spruce 0.130 1.40 — — — — 0.36c 5.18d 0.97d — 0.31c —
a SPECIFIC VALUES MAY VARY FOR A PARTICULAR MATERIAL DUE TO ALLOY OR MINERAL COMPOSITION, MECHANICAL WORKING OF THE SPECIMEN, OR HEAT TREATMENT. FOR A MORE EXACT VALUE REFERENCE
BOOKS FOR THE MATERIAL SHOULD BE CONSULTED.
THE YIELD AND ULTIMATE STRENGTHS FOR DUCTILE MATERIALS CAN BE ASSUMED EQUAL FOR BOTH TENSION AND COMPRESSION.
c MEASURED PERPENDICULAR TO THE GRAIN.
d MEASURED PARALLEL TO THE GRAIN.

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

Bulk modulus and volumetric strain under hydrostatic pressure — Copper

A metal specimen has E = 30.0×10⁶ psi and Poisson's ratio ν = 0.28. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 34,000 psi.

Given

E = 30.0×10⁶ psi
ν = 0.28
p = 34,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 30.0 \times 1 0^{6} / [3 (1 - 2 \times 0.28)] = 22.727 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 30.0 \times 1 0^{6} / [2 (1 + 0.28)] = 11.719 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −34000/22,727,273 = -1.4960 × 10⁻³

Answer: K = 22.73×10⁶ psi, G = 11.72×10⁶ psi, ΔV/V = -0.1496%

Why the other options are there

K = 6.41×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 2

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (2)

A metal specimen has E = 29.0×10⁶ psi and Poisson's ratio ν = 0.33. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 32,000 psi.

Given

E = 29.0×10⁶ psi
ν = 0.33
p = 32,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 29.0 \times 1 0^{6} / [3 (1 - 2 \times 0.33)] = 28.431 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 29.0 \times 1 0^{6} / [2 (1 + 0.33)] = 10.902 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −32000/28,431,373 = -1.1255 × 10⁻³

Answer: K = 28.43×10⁶ psi, G = 10.90×10⁶ psi, ΔV/V = -0.1126%

Why the other options are there

K = 5.82×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 3

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (3)

A metal specimen has E = 30.0×10⁶ psi and Poisson's ratio ν = 0.30. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 26,000 psi.

Given

E = 30.0×10⁶ psi
ν = 0.30
p = 26,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 30.0 \times 1 0^{6} / [3 (1 - 2 \times 0.30)] = 25.000 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 30.0 \times 1 0^{6} / [2 (1 + 0.30)] = 11.538 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −26000/25,000,000 = -1.0400 × 10⁻³

Answer: K = 25.00×10⁶ psi, G = 11.54×10⁶ psi, ΔV/V = -0.1040%

Why the other options are there

K = 6.25×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 4

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (4)

A metal specimen has E = 16.0×10⁶ psi and Poisson's ratio ν = 0.34. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 5,000 psi.

Given

E = 16.0×10⁶ psi
ν = 0.34
p = 5,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 16.0 \times 1 0^{6} / [3 (1 - 2 \times 0.34)] = 16.667 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 16.0 \times 1 0^{6} / [2 (1 + 0.34)] = 5.970 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −5000/16,666,667 = -0.3000 × 10⁻³

Answer: K = 16.67×10⁶ psi, G = 5.97×10⁶ psi, ΔV/V = -0.0300%

Why the other options are there

K = 3.17×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 5

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (5)

A metal specimen has E = 29.0×10⁶ psi and Poisson's ratio ν = 0.34. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 38,000 psi.

Given

E = 29.0×10⁶ psi
ν = 0.34
p = 38,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 29.0 \times 1 0^{6} / [3 (1 - 2 \times 0.34)] = 30.208 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 29.0 \times 1 0^{6} / [2 (1 + 0.34)] = 10.821 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −38000/30,208,333 = -1.2579 × 10⁻³

Answer: K = 30.21×10⁶ psi, G = 10.82×10⁶ psi, ΔV/V = -0.1258%

Why the other options are there

K = 5.75×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 6

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (6)

A metal specimen has E = 29.0×10⁶ psi and Poisson's ratio ν = 0.27. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 7,000 psi.

Given

E = 29.0×10⁶ psi
ν = 0.27
p = 7,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 29.0 \times 1 0^{6} / [3 (1 - 2 \times 0.27)] = 21.014 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 29.0 \times 1 0^{6} / [2 (1 + 0.27)] = 11.417 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −7000/21,014,493 = -0.3331 × 10⁻³

Answer: K = 21.01×10⁶ psi, G = 11.42×10⁶ psi, ΔV/V = -0.0333%

Why the other options are there

K = 6.28×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 7

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (7)

A metal specimen has E = 29.0×10⁶ psi and Poisson's ratio ν = 0.26. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 24,000 psi.

Given

E = 29.0×10⁶ psi
ν = 0.26
p = 24,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 29.0 \times 1 0^{6} / [3 (1 - 2 \times 0.26)] = 20.139 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 29.0 \times 1 0^{6} / [2 (1 + 0.26)] = 11.508 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −24000/20,138,889 = -1.1917 × 10⁻³

Answer: K = 20.14×10⁶ psi, G = 11.51×10⁶ psi, ΔV/V = -0.1192%

Why the other options are there

K = 6.36×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 8

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (8)

A metal specimen has E = 10.4×10⁶ psi and Poisson's ratio ν = 0.31. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 7,000 psi.

Given

E = 10.4×10⁶ psi
ν = 0.31
p = 7,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 10.4 \times 1 0^{6} / [3 (1 - 2 \times 0.31)] = 9.123 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 10.4 \times 1 0^{6} / [2 (1 + 0.31)] = 3.969 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −7000/9,122,807 = -0.7673 × 10⁻³

Answer: K = 9.12×10⁶ psi, G = 3.97×10⁶ psi, ΔV/V = -0.0767%

Why the other options are there

K = 2.14×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 9

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (9)

Given

E = 29.0×10⁶ psi
ν = 0.33
p = 18,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 29.0 \times 1 0^{6} / [3 (1 - 2 \times 0.33)] = 28.431 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 29.0 \times 1 0^{6} / [2 (1 + 0.33)] = 10.902 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −18000/28,431,373 = -0.6331 × 10⁻³

Answer: K = 28.43×10⁶ psi, G = 10.90×10⁶ psi, ΔV/V = -0.0633%

Why the other options are there

K = 5.82×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

Example 10

Bulk modulus and volumetric strain under hydrostatic pressure — Copper (10)

A metal specimen has E = 16.0×10⁶ psi and Poisson's ratio ν = 0.27. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 29,000 psi.

Given

E = 16.0×10⁶ psi
ν = 0.27
p = 29,000 psi

Find

K, G and the volumetric strain

Start with the thinking

The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

Formula
$K = E / [3 (1 - 2 ν)]$
Substituting
$K = 16.0 \times 1 0^{6} / [3 (1 - 2 \times 0.27)] = 11.594 \times 1 0^{6} p s i$
Formula
$G = E / [2 (1 + ν)]$
Substituting
$G = 16.0 \times 1 0^{6} / [2 (1 + 0.27)] = 6.299 \times 1 0^{6} p s i$
Formula — ΔV/V = −p/K
Substituting — ΔV/V = −29000/11,594,203 = -2.5013 × 10⁻³

Answer: K = 11.59×10⁶ psi, G = 6.30×10⁶ psi, ΔV/V = -0.2501%

Why the other options are there

K = 3.46×10⁶ psi (sign in the bracket flipped)
K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Copper

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

Copper contains 56 relations; 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.

Copper

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

22θmax=vmax=v=(3L2​−4x2​)Mmaxatcenterh=4

θmax vmax 0 ≤ x ≤ L/2Needs technical review

θ1​=v=v=(L−b−x)

6EILx=a6EIL(L−b−a)6EIL

θ1​θ2​

xPab(L+a)Mmaxa​tpointofloadi=L

θ2 = 0≤x≤aNeeds technical review

−M0​Lvmax=−M0​x

L θ1 =Needs technical review

v=(x2−3Lx+2L2​)

θ1​θ2​Mmaxatx=0h=M0​

θ2 =Needs technical review

vmax=−wx3wL2​

θ max =Needs technical review

24EIv=(x−2Lx2+L3)Mmaxatcenterh=8

θmaxvmax

θ1​=vx=L/2=768EIv=16x−24Lx2​+9L3​)

θ2​0≤x≤L/2

θ2​=7wL

−wLMmaxcatx=8lm=128wL2​

xvmax=−0.006563wL

v=384(8x3​−24Lx2​

θ1​

22atx=0.4598LL/2≤x<L

θ1 =Needs technical review

−7w0​L3​vmax=−0.00652w0​L

v=4224Mmaxdatx=n=0

θ2​=w0​L

atx=0.5193L

θ1​θ2​45EI

a b vmax v=Needs technical review

−Pa2​−Pa2​(3L−a)6EIMmaxatx=0h=Pa

x θmax = vmax =Needs technical review

2EI6EI−Px2​(−x+3a),forx≤a

v=Needs technical review

θmaxNeeds technical review

θmax =Needs technical review

− wL 3 vmax =Needs technical review

xv=(x−4Lx+6L2​)Mmaxatx=0h=2

L θmaxNeeds technical review

vθmax

vmaxM0​Lvmax=M0​L

x θmax = Mmax ^at all x h = M 0Needs technical review

M0 EI 2EI v=Needs technical review

v=24EIx2−2Lx+32L2​(

−−0≤x≤L/2

θmax = wL vmax = 7wLNeeds technical review

vmax48EI384EIMmaxatx=0h=8

LLv=(4x−L/2)

2 2 θmax L/2 ≤ x ≤ LNeeds technical review

θmax = vmax = Mmax ^at x = 0 h = 06Needs technical review

x24EI30EIv=120EIL(10L3​−10L2x+5Lx2​−x3)

L θmaxNeeds technical review

θmaxNeeds technical review

θmax

ν max ν Mmax (at x = 0) =Needs technical review

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section