4.1 Stress, strain and the Young modulus — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Write the unit at every step.

Warm-up: remember

Write $3.0\,\text{mm}^2$ in $\text{m}^2$ .
Write $2.4 \times 10^{-4}$ as an ordinary decimal.
Find the area of a circle of diameter $0.80\,\text{mm}$ , in $\text{m}^2$ , to two significant figures.

Answers

$3.0 \times 10^{-6}\,\text{m}^2$ . The millimetre is squared, so the factor is $10^{-6}$ .
$0.000\,24$ .
The radius is $0.40\,\text{mm} = 4.0 \times 10^{-4}\,\text{m}$ , so $A = \pi r^2 = 5.0 \times 10^{-7}\,\text{m}^2$ .

If question 1 goes wrong, look again at the prefix table in Section 1.1 and remember that the power applies to the prefix as well. If question 3 goes wrong, halve the diameter before you square it, and change the millimetre to a metre first.

Practise with help

1. A wire of cross-sectional area $4.0 \times 10^{-6}\,\text{m}^2$ carries a force of $80\,\text{N}$ . Find the stress.

Hint

Stress is force divided by cross-sectional area. The force is already in newtons and the area is already in square metres.

Next step

Write $\sigma = \dfrac{F}{A}$ with $F = 80$ and $A = 4.0 \times 10^{-6}$ , then work out the division.

Full working

\sigma = \frac{F}{A} = \frac{80}{4.0 \times 10^{-6}} = 2.0 \times 10^{7}\,\text{Pa}

The stress is $2.0 \times 10^{7}\,\text{Pa}$ , which is $20\,\text{MPa}$ .

2. A wire of original length $2.00\,\text{m}$ extends by $1.2\,\text{mm}$ . Find the strain.

Hint

The extension and the original length must have the same unit before you divide them. The answer is a ratio, so it has no unit.

Next step

Change $1.2\,\text{mm}$ to $1.2 \times 10^{-3}\,\text{m}$ , then divide by $2.00\,\text{m}$ .

Full working

\begin{aligned} \Delta L &= 1.2\,\text{mm} = 1.2 \times 10^{-3}\,\text{m} \\ \varepsilon &= \frac{\Delta L}{L} = \frac{1.2 \times 10^{-3}}{2.00} = 6.0 \times 10^{-4} \end{aligned}

The strain is $6.0 \times 10^{-4}$ , which is $0.060\%$ .

3. A material has a stress of $9.0 \times 10^{7}\,\text{Pa}$ and a strain of $4.5 \times 10^{-4}$ . Find its Young modulus.

Hint

The definition of the Young modulus divides the stress by the strain.

Next step

Write $E = \dfrac{\sigma}{\varepsilon}$ and substitute the two values.

Full working

E = \frac{\sigma}{\varepsilon} = \frac{9.0 \times 10^{7}}{4.5 \times 10^{-4}} = 2.0 \times 10^{11}\,\text{Pa}

The Young modulus is $2.0 \times 10^{11}\,\text{Pa}$ , which is the value for steel.

4. A wire has a diameter of $0.50\,\text{mm}$ and carries a force of $20\,\text{N}$ . Find the stress in the wire.

Hint

Find the cross-sectional area first. It is the area of a circle, so you need the radius, not the diameter.

Next step

Halve the diameter to get $0.25\,\text{mm}$ , change it to metres, then use $A = \pi r^2$ .

Full working

\begin{aligned} r &= 0.25\,\text{mm} = 2.5 \times 10^{-4}\,\text{m} \\ A &= \pi r^2 = \pi \times (2.5 \times 10^{-4})^2 = 1.96 \times 10^{-7}\,\text{m}^2 \\ \sigma &= \frac{F}{A} = \frac{20}{1.96 \times 10^{-7}} = 1.0 \times 10^{8}\,\text{Pa} \end{aligned}

The stress is $1.0 \times 10^{8}\,\text{Pa}$ , to two significant figures.

5. A copper wire has a Young modulus of $1.2 \times 10^{11}\,\text{Pa}$ , a length of $1.50\,\text{m}$ and a cross-sectional area of $3.0 \times 10^{-7}\,\text{m}^2$ . A force of $36\,\text{N}$ stretches it. Find the extension.

Hint

Use $\Delta L = \dfrac{FL}{AE}$ . The force and the length go on the top, and the area and the Young modulus go on the bottom.

Next step

Substitute $F = 36$ , $L = 1.50$ , $A = 3.0 \times 10^{-7}$ and $E = 1.2 \times 10^{11}$ into the rearranged formula.

Full working

\begin{aligned} \Delta L &= \frac{FL}{AE} \\ &= \frac{36 \times 1.50}{3.0 \times 10^{-7} \times 1.2 \times 10^{11}} \\ &= \frac{54}{3.6 \times 10^{4}} = 1.5 \times 10^{-3}\,\text{m} \end{aligned}

The wire extends by $1.5\,\text{mm}$ .

6. A steel wire and a copper wire are the same length and carry the same force. Their cross-sectional areas are equal. Explain which wire stretches more, and why. The Young modulus of steel is $2.0 \times 10^{11}\,\text{Pa}$ and of copper is $1.2 \times 10^{11}\,\text{Pa}$ .

Hint

When the length, the area and the force are fixed, the extension is inversely proportional to the Young modulus. The smaller value of $E$ gives the larger extension.

Next step

Write the ratio of the two extensions, then put the larger value of $E$ on the bottom of the fraction.

Full working

The two wires carry the same stress, because the force and the area are equal. The strain is the stress divided by the Young modulus, so the smaller Young modulus gives the larger strain. Copper has the smaller value, $1.2 \times 10^{11}\,\text{Pa}$ , so the copper wire stretches more. The ratio is

\frac{\Delta L_{\text{copper}}}{\Delta L_{\text{steel}}} = \frac{E_{\text{steel}}}{E_{\text{copper}}} = \frac{2.0 \times 10^{11}}{1.2 \times 10^{11}} = 1.7

The copper wire extends 1.7 times as far as the steel wire.