4.1 Stress, strain and the Young modulus — Practise

Work through these on your own. Write the unit of every answer, and change millimetres and square millimetres to metres before you substitute. In the last two questions, explain your reasoning in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A wire of cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ carries a tensile force of $50\,\text{N}$ . Find the stress.
A wire of original length $1.60\,\text{m}$ extends by $0.80\,\text{mm}$ . Find the strain.
A material has a stress of $1.2 \times 10^{8}\,\text{Pa}$ and a strain of $6.0 \times 10^{-4}$ . Find its Young modulus.
A wire has a diameter of $0.60\,\text{mm}$ and carries a force of $30\,\text{N}$ . Find the stress in the wire.
A steel wire has $E = 2.0 \times 10^{11}\,\text{Pa}$ , cross-sectional area $5.0 \times 10^{-7}\,\text{m}^2$ and length $1.80\,\text{m}$ . A force of $40\,\text{N}$ stretches it. Find the extension.
A steel wire and a copper wire have the same length and area and carry the same force. Steel has $E = 2.0 \times 10^{11}\,\text{Pa}$ and copper has $E = 1.2 \times 10^{11}\,\text{Pa}$ . Find how many times larger the copper extension is than the steel extension.
A cable of cross-sectional area $1.2 \times 10^{-4}\,\text{m}^2$ supports a load of mass $900\,\text{kg}$ . Take $g = 9.81\,\text{N}\,\text{kg}^{-1}$ and find the stress.
A wire $2.50\,\text{m}$ long extends by $1.5\,\text{mm}$ . Find the strain, and write it as a percentage.
A wire of length $2.50\,\text{m}$ and area $4.0 \times 10^{-7}\,\text{m}^2$ extends by $1.0\,\text{mm}$ under a force of $32\,\text{N}$ . Find the Young modulus.
You know the force, the original length, the extension and the cross-sectional area of a stretched wire. Describe how to find the Young modulus, and state its unit.
Two steel wires have the same length. One has twice the diameter of the other. Both carry the same force. Explain what happens to the stress and to the extension of the thicker wire.
Show that $E = \dfrac{FL}{A\,\Delta L}$ gives the base units $\text{kg}\,\text{m}^{-1}\,\text{s}^{-2}$ .

Answers and concise solutions

1. The tensile stress

Answer: The stress is $\dfrac{50}{2.0 \times 10^{-6}} = 2.5 \times 10^{7}\,\text{Pa}$ .

2. The tensile strain

Answer: The extension is $0.80\,\text{mm} = 8.0 \times 10^{-4}\,\text{m}$ , so the strain is $\dfrac{8.0 \times 10^{-4}}{1.60} = 5.0 \times 10^{-4}$ . The strain has no unit.

3. The Young modulus

Answer: The Young modulus is the stress divided by the strain: $\dfrac{1.2 \times 10^{8}}{6.0 \times 10^{-4}} = 2.0 \times 10^{11}\,\text{Pa}$ .

4. The tensile stress

Answer: The radius is $0.30\,\text{mm} = 3.0 \times 10^{-4}\,\text{m}$ . The area is $\pi (3.0 \times 10^{-4})^2 = 2.8 \times 10^{-7}\,\text{m}^2$ . The stress is $\dfrac{30}{2.8 \times 10^{-7}} = 1.1 \times 10^{8}\,\text{Pa}$ .

5. The extension

Answer: The extension is $\dfrac{FL}{AE} = \dfrac{40 \times 1.80}{5.0 \times 10^{-7} \times 2.0 \times 10^{11}} = 7.2 \times 10^{-4}\,\text{m}$ , or $0.72\,\text{mm}$ .

6. The ratio of the extensions

Answer: The extension is inversely proportional to the Young modulus when the other quantities are fixed, so the ratio is $\dfrac{2.0 \times 10^{11}}{1.2 \times 10^{11}} = \dfrac{5}{3} = 1.7$ . The copper wire extends 1.7 times as far as the steel wire.

7. The tensile stress

Answer: The weight is $mg = 900 \times 9.81 = 8829\,\text{N}$ . The stress is $\dfrac{8829}{1.2 \times 10^{-4}} = 7.4 \times 10^{7}\,\text{Pa}$ to two significant figures.

8. The strain and the percentage strain

Answer: The strain is $\dfrac{1.5 \times 10^{-3}}{2.50} = 6.0 \times 10^{-4}$ . As a percentage this is $6.0 \times 10^{-4} \times 100\% = 0.060\%$ .

9. The Young modulus

Answer: The stress is $\dfrac{32}{4.0 \times 10^{-7}} = 8.0 \times 10^{7}\,\text{Pa}$ and the strain is $\dfrac{1.0 \times 10^{-3}}{2.50} = 4.0 \times 10^{-4}$ . The Young modulus is $\dfrac{8.0 \times 10^{7}}{4.0 \times 10^{-4}} = 2.0 \times 10^{11}\,\text{Pa}$ .

10. The method and the unit

Find the stress by dividing the force by the cross-sectional area.
Find the strain by dividing the extension by the original length.
Divide the stress by the strain to get the Young modulus.
The unit is the pascal, which is one newton per square metre.

Answer: Divide the force by the area to get the stress. Divide the extension by the original length to get the strain. Divide the stress by the strain to get the Young modulus, in pascals.

11. The stress and the extension of the thicker wire

The area of a circle depends on the square of the diameter, so doubling the diameter makes the area four times larger.
The force is the same, and stress is force divided by area, so the stress in the thicker wire is one quarter of the stress in the thinner wire.
The strain is the stress divided by the Young modulus, so it is also one quarter. The extension is the strain times the length, and the lengths are equal, so the thicker wire extends by one quarter as much.

Answer: Doubling the diameter makes the area four times larger. The same force then gives one quarter of the stress, so the strain is one quarter and the extension is one quarter as large.

12. A base-unit check

Answer: On the right, the newton is $\text{kg}\,\text{m}\,\text{s}^{-2}$ . The expression is $\dfrac{(\text{kg}\,\text{m}\,\text{s}^{-2}) \times \text{m}}{\text{m}^2 \times \text{m}} = \text{kg}\,\text{m}^{-1}\,\text{s}^{-2}$ , which is the pascal. Both sides match, so the equation is homogeneous.