Young's modulus — stretching materials — Physics

Before this

Hooke's law
Elastic potential energy
Pressure — force spread over area · coming soon

This leads to

The mass-spring system · coming soon
Properties of metals, explained by bonding
Diamond and graphite

Two steel wires that disagree

A guitar string and a cable that holds up a bridge can both be made of steel. Hang the same load on each, and they behave completely differently.

Take a thin guitar string, 0.25 mm across and 0.65 m long. Its spring constant is about 15 000 N/m. Now take a steel cable 50 mm across and 100 m long. Its spring constant is about 3 900 000 N/m, roughly 260 times larger. (These are typical sizes, chosen for the example.)

So which number tells you how stiff steel is? Neither. The spring constant $k$ from the Hooke's law lecture describes one particular object. Change its thickness or its length, and $k$ changes, even though the material has not changed at all.

Engineers need one number that belongs to the material itself, whatever shape it is made into. This lecture builds that number. It is called the Young modulus, or Young's modulus. The idea is simple once you see where the thickness and the length go.

Where the thickness and the length go

You already know how springs combine. In the Hooke's law lecture, two identical springs side by side (in parallel) shared the load, so the pair was twice as stiff: $2 k$ . Two identical springs end to end (in series) each felt the full load and each stretched, so the pair was half as stiff: $\frac{1}{2} k$ .

A wire behaves in exactly the same way.

A wire with twice the cross-sectional area is like two thin wires side by side. Each carries half the load, so the thick wire stretches half as much. Its $k$ is twice as large. So $k$ is proportional to the area $A$ .

A wire twice as long is like two short wires end to end. Each part stretches by the same amount, so the long wire stretches twice as much. Its $k$ is half as large. So $k$ is inversely proportional to the length $L$ .

Put the two together, and $k$ is proportional to $\frac{A}{L}$ . The number left over, once the area and the length are taken out, depends only on the material. We call it $E$ :

k = \frac{E A}{L}

The Hooke's law lecture also promised that a solid's atoms behave like tiny springs. Here is that picture made exact. A wire is a bundle of chains of atoms, joined by bonds that act like springs. A thicker wire has more chains side by side. A longer wire has more bonds in each chain. Neither changes the springs themselves. $E$ describes the springs: how stiff the bonds are, and how closely the atoms are packed.

Stress and strain: sharing out the force and the stretch

The formula $k = \frac{E A}{L}$ can be split into two ideas that are useful on their own. Start from Hooke's law, $F = k x$ , and replace $k$ :

F = \frac{E A}{L} x

Divide both sides by $A$ :

\frac{F}{A} = E \times \frac{x}{L}

Each side of the second equation has a name.

Stress is the force on each square metre of cross-section. Its symbol is the Greek letter $σ$ (sigma):

σ = \frac{F}{A}

Stress is measured in N/m², which is called the pascal (Pa). It has the same unit as pressure, because it is the same kind of quantity: a force spread over an area. Real stresses are large, so you will usually see MPa ( $1 0^{6}$ Pa) and GPa ( $1 0^{9}$ Pa).

Strain is the extension as a fraction of the original length. Its symbol is $ε$ (epsilon):

ε = \frac{x}{L}

Here $x$ is the extension, as in the Hooke's law lecture. Some books write it $Δ L$ or $e$ . Strain is a length divided by a length, so it has no unit. A strain of 0.001 means the wire has stretched by one thousandth of its length, or 0.1%.

Now the equation reads $σ = Eε$ . That is it. That is the whole idea. Stress shares the force out over the area. Strain shares the stretch out over the length. What is left is a property of the material alone.

The Young modulus

$E = \frac{stress}{strain} = \frac{σ}{ε} = \frac{F / A}{x / L} = \frac{F L}{A x}$

It is valid up to the limit of proportionality. Its unit is the pascal, because strain has no unit.

Stop and think

Wire A is steel, 1 mm across. Wire B is steel, 2 mm across. Both are 2 m long. Which has the larger spring constant, and which has the larger Young modulus?

Show the answer

Wire B has the larger spring constant. Doubling the diameter makes the area four times larger, because area depends on the radius squared. So $k_{B} = 4 k_{A}$ . But both wires are steel, so they have the same Young modulus. The spring constant belongs to the object. The Young modulus belongs to the material. This difference is the reason the Young modulus exists.

Test this on four wires below. Three are steel, of different sizes. One is copper.

Explore — four wires, one material number

Choose a wire and add a load. First look at the force–extension graph: the four wires give four different lines. Then press the button to plot the same data as stress against strain. What happens to the three steel lines?

Wire:

Load: 30 N

Four wires with a load of 50 N, all within the limit of proportionality. Steel has E = 200 GPa and copper has E = 117 GPa.

Wire	Size	Extension	k in N/m	Stress	Strain
A, steel	1.0 mm, 2.0 m	≈ 0.64 mm	≈ 78 500	≈ 63.7 MPa	≈ 3.18 × 10⁻⁴
B, steel	2.0 mm, 2.0 m	≈ 0.16 mm	≈ 314 000	≈ 15.9 MPa	≈ 7.96 × 10⁻⁵
C, steel	1.0 mm, 1.0 m	≈ 0.32 mm	≈ 157 000	≈ 63.7 MPa	≈ 3.18 × 10⁻⁴
D, copper	1.0 mm, 2.0 m	≈ 1.09 mm	≈ 45 900	≈ 63.7 MPa	≈ 5.44 × 10⁻⁴

The spring constants differ by a factor of about 7. But for every steel wire, stress ÷ strain = 200 GPa. On a stress–strain graph, the three steel wires lie on one line, and copper lies on a less steep line.

What should you have seen? On the force–extension graph, the four wires gave four lines with four different gradients, because each wire has its own $k$ . On the stress–strain graph, the three steel lines fell exactly on top of each other. Their sizes disappeared, and only the material was left. The gradient of that line is the Young modulus. Copper's line is less steep, because copper is less stiff than steel.

Some real values

Here are typical values of the Young modulus for some materials, from one engineering table. Real samples vary, especially for glass, wood and rubber.

Material	E in GPa
Rubber (small strains)	0.01 to 0.1
Nylon	2 to 4
Oak, along the grain	11
Glass	50 to 90
Aluminium	69
Copper	117
Structural steel	200
Diamond	1220

The range is enormous. Diamond is stiffer than rubber by a factor of more than ten thousand. The reason is in the bonds. In diamond, every carbon atom is held by four strong covalent bonds in a rigid network, so stretching the crystal means stretching those bonds. Metals are held by metallic bonding, which is strong but not quite so rigid. Rubber is different in kind. It is made of long, tangled chains. When you pull it, the chains mostly uncoil and straighten. Hardly any bonds are stretched, so a small stress gives a large strain.

The whole journey: the stress–strain graph

A testing machine can pull a sample until it breaks and record the stress and strain all the way. For a ductile metal such as copper or mild steel, the graph has the shape below. The points on it are easy to confuse, so take them one at a time.

The stress–strain graph of a ductile metal (shape only, not to scale). The straight part is really much steeper and shorter than it can be drawn here. Its gradient is the Young modulus.

P, the limit of proportionality. Up to P, the graph is a straight line through the origin. Stress is proportional to strain, and $E = σ / ε$ applies.
L, the elastic limit. Up to L, the wire returns to its original length when the load is removed. L is at P or a little beyond it. These are two different tests: one asks "is the graph straight?", the other asks "does it come back?".
Y, the yield point. Here the metal starts to flow. Layers of atoms slide over each other, and the strain grows a lot for little extra stress. The stretch from here on is permanent: plastic deformation.
U, the ultimate tensile stress (UTS). The highest stress the material can take. Some books call it the tensile strength. Beyond U, the wire narrows at one place, called a neck, and the force it can carry falls.
B, the breaking point. The wire snaps.

Different materials have different journeys.

Two kinds of material on the same axes (shapes only). A brittle material, such as glass or cast iron, breaks at the end of its straight line, with almost no plastic region. A ductile metal, such as copper, has a long plastic region, which is why it can be pulled into wire.

So the graph shows three different properties. Stiff means a steep straight line (a large Young modulus). Strong means a high top (a large ultimate tensile stress). Ductile means a long plastic region, and brittle means almost none.

The energy stored in each cubic metre

The last lecture promised that the area under a stress–strain graph gives the energy stored in each cubic metre of a stretched wire. Here is why.

Up to the limit of proportionality, the energy stored in a wire is $\frac{1}{2} F x$ , the area of the triangle under the force–extension graph. Now write the force and the extension in terms of stress and strain: $F = σ A$ and $x = ε L$ . Then

E_{e} = \frac{1}{2} F x = \frac{1}{2} (σ A) (ε L)

E_{e} = \frac{1}{2} σ ε \times A L

The product $A L$ is the volume of the wire. So the energy stored in each cubic metre is $\frac{1}{2} σ ε$ , which is the area of the triangle under the stress–strain line. Check the unit: pascals times no unit gives N/m², and N/m² is the same as N m/m³, which is J/m³. The same rule as before, area under the graph, works on both graphs. On the force–extension graph it gives joules. On the stress–strain graph it gives joules per cubic metre.

Before the worked examples, make a prediction.

Predict — a longer, thicker wire

Wire P and wire Q are made of the same metal. Q is twice as long as P and has twice the diameter. The same load hangs on each, within the limit of proportionality. How does the extension of Q compare with the extension of P?

The prediction: wire Q is the same metal as wire P, twice as long, with twice the diameter, and carries the same load.

The answer is that Q stretches half as much as P. The extension is $x = \frac{F L}{A E}$ . Doubling the length doubles the extension. Doubling the diameter makes the area four times larger, which divides the extension by 4. Together: $2 \div 4 = \frac{1}{2}$ . The common wrong answer is "the same", which forgets that area depends on the diameter squared.

Worked examples

Worked Example 1 — The Young modulus of a steel wire

A steel wire is 2.0 m long and has a cross-sectional area of $3.0 \times 1 0^{- 6}$ m². A force of 600 N stretches it by 2.0 mm. Find the stress, the strain and the Young modulus.

First convert the extension to metres: $x = 2.0 mm = 2.0 \times 1 0^{- 3}$ m.

Stress: $σ = \frac{F}{A} = \frac{600}{3.0 \times 1 0 ^{- 6}} = 2.0 \times 1 0^{8}$ Pa, which is 200 MPa.

Strain: $ε = \frac{x}{L} = \frac{2.0 \times 1 0 ^{- 3}}{2.0} = 1.0 \times 1 0^{- 3}$ , with no unit.

Young modulus: $E = \frac{σ}{ε} = \frac{2.0 \times 1 0 ^{8}}{1.0 \times 1 0 ^{- 3}} = 2.0 \times 1 0^{11}$ Pa, which is 200 GPa.

Check: this matches the table value for structural steel. Also $\frac{F L}{A x} = \frac{600 \times 2.0}{3.0 \times 1 0 ^{- 6} \times 2.0 \times 1 0 ^{- 3}} = 2.0 \times 1 0^{11}$ Pa. ✓

Worked Example 2 — How far will it stretch?

A steel wire is 1.5 m long and 0.80 mm in diameter. A 4.0 kg mass hangs from it. The Young modulus of steel is 200 GPa. Take $g = 9.8$ N/kg. Find the extension.

Force: $F = m g = 4.0 \times 9.8 = 39.2$ N.

Area: the radius is half the diameter, 0.40 mm $= 0.40 \times 1 0^{- 3}$ m. So $A = π r^{2} = π (0.40 \times 1 0^{- 3})^{2} \approx 5.03 \times 1 0^{- 7}$ m².

Stress: $σ = \frac{39.2}{5.03 \times 1 0 ^{- 7}} \approx 7.80 \times 1 0^{7}$ Pa.

Strain: $ε = \frac{σ}{E} = \frac{7.80 \times 1 0 ^{7}}{200 \times 1 0 ^{9}} \approx 3.90 \times 1 0^{- 4}$ .

Extension: $x = ε L \approx 3.90 \times 1 0^{- 4} \times 1.5 \approx 5.85 \times 1 0^{- 4}$ m, which is about 0.6 mm.

Check: a stress of 78 MPa is far below the stress at which structural steel starts to yield, about 250 MPa, so the straight-line rule applies. A copper wire of the same size would stretch further, about 1.0 mm, because copper is less stiff. But copper starts to yield at only about 70 MPa, so this load would already be too much for it.

Worked Example 3 — Reading a stress–strain graph

A metal's stress–strain graph is a straight line from the origin to a strain of 0.002 at a stress of 400 MPa. It then curves, reaches a highest stress of 500 MPa, and breaks at a strain of 0.25. Find the Young modulus and the UTS, describe the material, and find the energy stored per cubic metre at the limit of proportionality.

Young modulus: the gradient of the straight part, $E = \frac{400 \times 1 0 ^{6}}{0.002} = 2.0 \times 1 0^{11}$ Pa $= 200$ GPa.

UTS: the highest stress on the graph, 500 MPa.

Description: the plastic region runs from a strain of about 0.002 to 0.25, more than a hundred times the elastic strain. So the material is ductile.

Energy per cubic metre: $\frac{1}{2} σ ε = \frac{1}{2} \times 400 \times 1 0^{6} \times 0.002 = 4.0 \times 1 0^{5}$ J/m³.

Check: a wire of this metal with a volume of 1 cm³ ( $1 0^{- 6}$ m³) would store $4.0 \times 1 0^{5} \times 1 0^{- 6} = 0.40$ J at that point. That is about the energy needed to lift a 100 g apple by 40 cm.

Worked Example 4 — Why rubber needs care

A rubber cord is 0.20 m long, with a square cross-section 2.0 mm by 2.0 mm. A force of 2.0 N stretches it by 10 mm. Estimate its Young modulus.

Area: $A = (2.0 \times 1 0^{- 3})^{2} = 4.0 \times 1 0^{- 6}$ m². Stress: $σ = \frac{2.0}{4.0 \times 1 0 ^{- 6}} = 5.0 \times 1 0^{5}$ Pa. Strain: $ε = \frac{0.010}{0.20} = 0.050$ . So $E \approx \frac{5.0 \times 1 0 ^{5}}{0.050} = 1.0 \times 1 0^{7}$ Pa, which is 10 MPa or 0.01 GPa.

Now the care. A rubber cord can be stretched to several times its length, but its stress–strain graph curves almost from the start. So $σ \div ε$ is only a fair estimate of $E$ for small strains, like the 5% here. If you stretched this cord to 80% strain and divided, you would get a number, but it would not be the Young modulus. Rubber also gets noticeably thinner as it stretches, so the area does not stay fixed.

Check: 0.01 GPa is at the bottom of the table's range for rubber, and about 20 000 times smaller than steel.

Three classic mistakes

Trap 1 — Diameter for radius, and millimetres for metres

For a round wire, $A = π r^{2}$ with $r$ half the diameter, in metres. A wire 0.80 mm across has $r = 0.40 \times 1 0^{- 3}$ m. Using the diameter makes the area 4 times too big. Forgetting to convert millimetres makes it a million times too big, because the conversion is squared.

Trap 2 — Using E beyond the straight line

$E = σ / ε$ holds only up to the limit of proportionality. Past it, stress ÷ strain is just a number, not the Young modulus. Take the gradient from the straight part of the graph, and never from a point in the plastic region.

Trap 3 — Mixing up stiff, strong and ductile

Stiff means a large Young modulus. Strong means a large ultimate tensile stress. Ductile means a long plastic region. Glass is stiff but brittle. A material can be any mix of the three, so answer the question that was asked.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

An aluminium wire is 3.0 m long and 1.2 mm in diameter. It carries a load of 30 N. The Young modulus of aluminium is 69 GPa. Find the extension in millimetres.

Hint 1
Find the area first. The radius is 0.60 mm, which is $0.60 \times 1 0^{- 3}$ m.

Hint 2
$A \approx 1.13 \times 1 0^{- 6}$ m². Then $x = \frac{F L}{A E}$ .

Full answer
$A = π (0.60 \times 1 0^{- 3})^{2} \approx 1.13 \times 1 0^{- 6}$ m². Stress $= \frac{30}{1.13 \times 1 0 ^{- 6}} \approx 2.65 \times 1 0^{7}$ Pa. Strain $= \frac{2.65 \times 1 0 ^{7}}{69 \times 1 0 ^{9}} \approx 3.84 \times 1 0^{- 4}$ . Extension $= 3.84 \times 1 0^{- 4} \times 3.0 \approx 1.15 \times 1 0^{- 3}$ m, about 1.2 mm. Check with one formula: $\frac{30 \times 3.0}{1.13 \times 1 0 ^{- 6} \times 69 \times 1 0 ^{9}} \approx 1.15 \times 1 0^{- 3}$ m. ✓
Wire B is made of the same metal as wire A. It is twice as long and has half the diameter. The same load hangs on each. How many times larger is the extension of B?

Hint 1
The extension is proportional to $\frac{L}{A}$ .

Hint 2
Half the diameter gives $(\frac{1}{2})^{2} = \frac{1}{4}$ of the area.

Full answer
Twice the length multiplies the extension by 2. A quarter of the area multiplies it by 4. So B stretches $2 \times 4 = 8$ times as much as A. Check by numbers: if A is 1 m long with area 4 units, $L / A = 0.25$ ; B is 2 m long with area 1 unit, $L / A = 2$ , which is 8 times larger. ✓
A metal's stress–strain graph is straight up to a stress of 300 MPa at a strain of 0.0015. A wire of this metal is 2.0 m long with a cross-sectional area of $5.0 \times 1 0^{- 7}$ m². Find the energy stored per cubic metre at the limit of proportionality, and then the energy stored in the wire. Check your answer with $\frac{1}{2} F x$ .

Hint 1
Energy per cubic metre is the area of the triangle under the stress–strain line: $\frac{1}{2} σ ε$ .

Hint 2
The volume is $A L = 1.0 \times 1 0^{- 6}$ m³. For the check, $F = σ A$ and $x = ε L$ .

Full answer
$\frac{1}{2} \times 300 \times 1 0^{6} \times 0.0015 = 2.25 \times 1 0^{5}$ J/m³. Volume $= 5.0 \times 1 0^{- 7} \times 2.0 = 1.0 \times 1 0^{- 6}$ m³, so the energy is $2.25 \times 1 0^{5} \times 1.0 \times 1 0^{- 6} = 0.225$ J. Check: $F = 300 \times 1 0^{6} \times 5.0 \times 1 0^{- 7} = 150$ N and $x = 0.0015 \times 2.0 = 3.0 \times 1 0^{- 3}$ m, so $\frac{1}{2} F x = \frac{1}{2} \times 150 \times 3.0 \times 1 0^{- 3} = 0.225$ J. ✓
A lift car of mass 800 kg hangs from a steel cable 40 m long. To keep a large safety margin, the designer decides that the stress in the cable must not exceed 100 MPa, well below the steel's yield stress. Take $g = 9.8$ N/kg and $E = 200$ GPa. Find the smallest diameter the cable can have, and the extension of the cable at that stress. Why would a designer choose a stress so far below the yield point?

Hint 1
The smallest area is the one that gives exactly 100 MPa: $A = F \div σ$ .

Hint 2
$F = 7840$ N, so $A = 7.84 \times 1 0^{- 5}$ m². Then $d = 2 A / π$ . For the extension, use $ε = σ / E$ .

Full answer
$F = 800 \times 9.8 = 7840$ N. $A = \frac{7840}{100 \times 1 0 ^{6}} = 7.84 \times 1 0^{- 5}$ m². Then $r = A / π \approx 5.0 \times 1 0^{- 3}$ m, so the diameter must be at least about 10 mm. Strain $= \frac{100 \times 1 0 ^{6}}{200 \times 1 0 ^{9}} = 5.0 \times 1 0^{- 4}$ , so the extension is $5.0 \times 1 0^{- 4} \times 40 = 0.020$ m, or 2.0 cm. Check: $\frac{F L}{A E} = \frac{7840 \times 40}{7.84 \times 1 0 ^{- 5} \times 200 \times 1 0 ^{9}} = 0.020$ m. ✓ The margin covers what the simple calculation leaves out: passengers add mass, the lift speeds up and slows down (which raises the tension), the cable wears and rusts, and it is bent over pulleys thousands of times. A stress far below the yield point keeps the cable on the straight, elastic part of its graph for its whole working life.

The bigger idea

The Young modulus is named after the British scientist Thomas Young, but he was not the first to use the idea. Leonhard Euler described it in 1727, and Giordano Riccati made the first experiments with it in its modern form in 1782, about 25 years before Young's work. The idea is older than the name.

The powerful move is this: divide out the size of the object, and what is left belongs to the material. Physics does this again and again. Density is mass with the size divided out. Resistivity, later in the course, is electrical resistance with the length and area divided out, exactly as here. And a material number points down to the atoms. One number, measured with a wire and a ruler, tells you about forces between atoms you will never see.

Key concepts in this lesson

The spring constant depends on the size of an object, because $k = \frac{E A}{L}$ , while the Young modulus $E$ depends only on the material.
Stress is force per unit cross-sectional area, $σ = F / A$ , measured in pascals.
Strain is extension divided by original length, $ε = x / L$ , and it has no unit.
The Young modulus is stress divided by strain, which is the gradient of the straight part of a stress–strain graph.
A stress–strain graph shows the limit of proportionality, the elastic limit, the yield point, the ultimate tensile stress and the breaking point.
The area under a stress–strain graph is the energy stored per cubic metre, which is $\frac{1}{2} σ ε$ up to the limit of proportionality.