Stress, Strain and Young's Modulus – Moments and Equilibrium of Bodies (Senior 4 Physics) | Ganzaa

Moments and Equilibrium of Bodies

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Senior 4 · Physics

Moments and Equilibrium of Bodies

4 lessons

Lesson 4 of 4

Stress, Strain and Young's Modulus

25 min

How much does a steel bar stretch?

Every concrete column and beam in a modern building is reinforced with steel bars, because concrete is strong when squeezed but cracks easily when stretched. The lift cables in a tall building and the wire ropes of a crane carry tonnes. Engineers must know how much each of these stretches under load, and at what load it will be permanently bent or snap. To compare a thin wire with a thick bar, they use quantities that do not depend on the size of the sample: stress, strain and Young's modulus.

Hooke's law, stress and strain

Hang loads on a wire or spring and the extension grows in proportion to the load, up to a point. This is Hooke's law:

F = k\,\Delta L \qquad (\text{up to the limit of proportionality})

where $k$ is the stiffness (spring constant) in $\mathrm{N\,m^{-1}}$ . But $k$ depends on the sample: a wire twice as long stretches twice as much under the same load, and one with twice the cross-sectional area stretches half as much. Dividing out the size gives quantities that describe the material:

\text{tensile stress}\;\; \sigma = \frac{F}{A} \;(\mathrm{Pa}), \qquad \text{tensile strain}\;\; \varepsilon = \frac{\Delta L}{L} \;(\text{no unit}), \qquad \text{Young's modulus}\;\; E = \frac{\sigma}{\varepsilon} = \frac{F L}{A\,\Delta L}

Comparing with Hooke's law, $k = \frac{EA}{L}$ : a stiffer sample is one made of a material with a large $E$ , thick and short.

Definition

Tensile stress $\sigma = \frac{F}{A}$ is the force per unit cross-sectional area (unit: pascal, $1\,\mathrm{Pa} = 1\,\mathrm{N\,m^{-2}}$ ). Tensile strain $\varepsilon = \frac{\Delta L}{L}$ is the extension per unit original length; it has no unit. The Young modulus $E = \frac{\sigma}{\varepsilon}$ is the ratio of stress to strain within the limit of proportionality (unit: Pa). Example: steel has $E = 2.1 \times 10^{11}\,\mathrm{Pa}$ (210 GPa).

Material	Young modulus $E$ ( $10^{9}\,\mathrm{Pa}$ )
steel	210
cast iron	100
brass	90
aluminium	70
glass	70
concrete	20
bone (in tension)	16
hardwood	15
nylon	5

Table 4.1: Young moduli of some materials (values from OpenStax College Physics, Table 5.3).

Worked example 1: stretching a steel wire

A steel wire 2.00 m long and 0.50 mm in diameter carries a load of 50 N. Find the stress, strain and extension ( $E = 2.1 \times 10^{11}\,\mathrm{Pa}$ ).

A = \pi r^2 = \pi (0.25 \times 10^{-3})^2 = 1.96 \times 10^{-7}\,\mathrm{m^2}, \qquad \sigma = \frac{50}{1.96 \times 10^{-7}} = 2.5 \times 10^{8}\,\mathrm{Pa}

\varepsilon = \frac{\sigma}{E} = \frac{2.55 \times 10^{8}}{2.1 \times 10^{11}} = 1.2 \times 10^{-3}, \qquad \Delta L = \varepsilon L = 1.21 \times 10^{-3} \times 2.00 = 2.4\,\mathrm{mm}

Worked example 2: a reinforcing bar

A steel reinforcing bar 3.0 m long and 12 mm in diameter carries a tension of 20 kN. By how much does it stretch?

A = \pi (6.0 \times 10^{-3})^2 = 1.13 \times 10^{-4}\,\mathrm{m^2}, \qquad \Delta L = \frac{FL}{AE} = \frac{2.0 \times 10^{4} \times 3.0}{1.13 \times 10^{-4} \times 2.1 \times 10^{11}} = 2.5 \times 10^{-3}\,\mathrm{m}

Only 2.5 mm over 3 m, a strain of less than 0.1%. Steel's large Young modulus is why a thin bar can carry the tension that would crack plain concrete.

The stress–strain curve

Figure 4.1: Stress against strain for a ductile metal (solid) and a brittle material (dashed).

O to P: stress is proportional to strain (Hooke's law). The gradient is the Young modulus.
P to E: no longer proportional, but still elastic: remove the load and the wire returns to its original length. E is the elastic limit.
Beyond E: plastic deformation: the wire is permanently stretched. At the yield point Y it suddenly "gives" and stretches a lot for little extra stress.
U: the ultimate tensile stress (tensile strength), the greatest stress the material can take. After U the wire narrows at one place (a "neck") and breaks at B.

A ductile material (copper, mild steel) has a large plastic region and can be drawn into wires; a brittle material (glass, cast iron, concrete in tension) breaks with almost no plastic deformation.

Energy stored in a stretched wire

The work done in stretching is the area under the force–extension graph. Within Hooke's law the graph is a straight line, so

W = \tfrac{1}{2}F\,\Delta L = \tfrac{1}{2}k(\Delta L)^2, \qquad \text{energy per unit volume} = \tfrac{1}{2}\sigma\varepsilon

Experiment 4.1: measuring the Young modulus of a wire

Apparatus: two long, identical wires hung from a rigid support, a vernier scale joining them (Figure 4.2), a micrometer screw gauge, a metre rule, slotted 1.0 kg masses, safety goggles.

Figure 4.2: A test wire and a reference wire. The reference wire cancels the sag of the support and changes of temperature.

Hang enough load on both wires to straighten them. Measure the length $L$ of the test wire from the support to the vernier.
Measure the diameter with the micrometer at several places and in two directions at right angles; average, and find $A = \frac{\pi d^2}{4}$ .
Add loads in equal steps, reading the vernier each time; then remove them in steps and read again. Equal readings on loading and unloading show the wire stayed elastic.
Plot load $F$ against extension $\Delta L$ . The gradient is $k$ , and $E = \frac{kL}{A}$ .

$F$ (N)	10	20	30	40	50
$\Delta L$ (mm)	0.79	1.60	2.39	3.21	3.98

Table 4.2: Illustrative results for a wire with $L = 2.000\,\mathrm{m}$ and $d = 0.40\,\mathrm{mm}$ .

Figure 4.3: Load against extension for Table 4.2. The shaded area is the energy stored at 50 N.

Warning

Safety. Wear goggles: a stretched wire that snaps whips back and can blind. Put a box of sand or a soft mat under the load so that falling masses cannot land on feet, and stand to one side. Do not load the wire beyond the planned maximum.

Worked example 3: analysing the experiment

From Figure 4.3 the gradient is $k = \frac{50\,\mathrm{N}}{3.98 \times 10^{-3}\,\mathrm{m}} = 1.26 \times 10^{4}\,\mathrm{N\,m^{-1}}$ . With $A = \pi (0.20 \times 10^{-3})^2 = 1.26 \times 10^{-7}\,\mathrm{m^2}$ :

E = \frac{kL}{A} = \frac{1.26 \times 10^{4} \times 2.000}{1.26 \times 10^{-7}} = 2.0 \times 10^{11}\,\mathrm{Pa}

The energy stored at 50 N is $\tfrac{1}{2} \times 50 \times 3.98 \times 10^{-3} = 0.10\,\mathrm{J}$ . The largest uncertainty comes from the diameter: it is squared in $A$ , so a 2% error in $d$ gives a 4% error in $E$ . That is why the diameter is measured many times with a micrometer.

Warning

Common mistakes. (1) Using the diameter instead of the radius in $A = \pi r^2$ , which makes $A$ four times too big. (2) Forgetting to convert mm to m and $\mathrm{mm^2}$ to $\mathrm{m^2}$ ( $1\,\mathrm{mm^2} = 10^{-6}\,\mathrm{m^2}$ ). (3) Calling the elastic limit and the limit of proportionality the same point. (4) Saying a material with a high Young modulus is necessarily strong: $E$ measures stiffness; strength is the ultimate tensile stress. Glass is stiff but brittle.

Tip

$E = \frac{FL}{A\,\Delta L}$ : "Full Load over Area times stretch". Check the order of magnitude: metals $10^{10}$ to $10^{11}\,\mathrm{Pa}$ ; plastics about $10^{9}\,\mathrm{Pa}$ .

Young's modulus of copper, a practical (Physics Online). Watch how the wire's diameter is measured with a micrometer, how the extension is read as loads are added, and how the Young modulus is found from the gradient of the graph.

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Practice

Define stress, strain and Young modulus, giving the SI unit of each.
An aluminium wire 1.5 m long and 1.0 mm in diameter carries a load of 100 N. Find its extension ( $E = 7.0 \times 10^{10}\,\mathrm{Pa}$ ).
A nylon fishing line 2.0 m long stretches 30 mm under a stress of $7.5 \times 10^{7}\,\mathrm{Pa}$ . Find the Young modulus of nylon.
A steel wire of diameter 2.0 mm has an ultimate tensile stress of $4.0 \times 10^{8}\,\mathrm{Pa}$ . What is the largest load it can carry?
Two wires of the same material carry the same load. Wire A is twice as long and has twice the diameter of wire B. Find the ratio of their extensions.
A spring of stiffness 400 N m⁻¹ is stretched by 5.0 cm. Find the energy stored.
A student finds $E = 3.2 \times 10^{11}\,\mathrm{Pa}$ for a steel wire. Suggest two likely errors and how to detect them.

Show answersHide answers

Stress $\frac{F}{A}$ , Pa; strain $\frac{\Delta L}{L}$ , no unit; Young modulus $\frac{\text{stress}}{\text{strain}}$ , Pa.
$A = \pi(0.50 \times 10^{-3})^2 = 7.85 \times 10^{-7}\,\mathrm{m^2}$ ; $\Delta L = \frac{100 \times 1.5}{7.85 \times 10^{-7} \times 7.0 \times 10^{10}} = 2.7 \times 10^{-3}\,\mathrm{m} = 2.7\,\mathrm{mm}$ .
Strain $= \frac{0.030}{2.0} = 0.015$ ; $E = \frac{7.5 \times 10^{7}}{0.015} = 5.0 \times 10^{9}\,\mathrm{Pa}$ .
$F = \sigma A = 4.0 \times 10^{8} \times \pi(1.0 \times 10^{-3})^2 = 1.3 \times 10^{3}\,\mathrm{N}$ .
$\Delta L \propto \frac{L}{d^2}$ : A has $\frac{2}{4} = \frac{1}{2}$ the extension of B, so $\Delta L_A : \Delta L_B = 1 : 2$ .
$W = \tfrac{1}{2} \times 400 \times 0.050^2 = 0.50\,\mathrm{J}$ .
The value is about 50% above the accepted $2.1 \times 10^{11}\,\mathrm{Pa}$ , too much for random scatter, so either $A$ or $\Delta L$ was measured too small. (1) The diameter may have been underestimated, for example by a micrometer with a zero error or by measuring at one thin spot; re-measure $d$ at several points with a zeroed micrometer. (2) The extensions may have been under-read on the vernier; repeat each reading while loading and again while unloading. A graph of $F$ against $\Delta L$ that is straight and passes through the origin helps to show which readings are faulty.

Quick check

Key points

$\sigma = \frac{F}{A}$ , $\varepsilon = \frac{\Delta L}{L}$ , $E = \frac{\sigma}{\varepsilon} = \frac{FL}{A\Delta L}$ ; Hooke's law $F = k\Delta L$ with $k = \frac{EA}{L}$ .
Stress–strain curve: limit of proportionality, elastic limit, yield point, ultimate tensile stress, breaking point; ductile versus brittle.
Energy stored $= \tfrac{1}{2}F\Delta L$ = area under the force–extension graph.
Measure $E$ with a long wire and a reference wire; the diameter dominates the uncertainty.

1How much does a steel bar stretch?

2Hooke's law, stress and strain

3The stress–strain curve

Energy stored in a stretched wire

4Experiment 4.1: measuring the Young modulus of a wire

How much does a steel bar stretch?

Hooke's law, stress and strain

The stress–strain curve

Experiment 4.1: measuring the Young modulus of a wire