Physics A Level
Chapter 7: Matter and materials 7.5 Stretching materials
Physics A Level
Chapter 7: Matter and materials 7.5 Stretching materials
When we determine the force constant of a spring, we are only finding out about the stiffness of that particular spring. However, we can compare the stiffness of different materials. For example, steel is stiffer than copper, but copper is stiffer than lead.
Stress and strain
Figure 7.12 shows a simple way of assessing the stiffness of a wire in the laboratory. As the long wire is stretched, the position of the sticky tape pointer can be read from the scale on the bench.
Why do we use a long wire? Obviously, this is because a short wire would not stretch as much as a long one.
We need to take account of this in our calculations, and we do this by calculating the strain produced by the load. The strain is defined as the increase in length of a wire (its extension) divided by its the original length.
That is:
$\begin{array}{l}
strain = \frac{{extension}}{{original\,length}}\\
\varepsilon = \frac{x}{L}
\end{array}$
where $\varepsilon $ is the strain, x is the extension of the wire and L is its original length.
wire
For example, if a wire of length $1.500 m$ is stretched and the length becomes $1.518 m$, the extension is $0.018 m$ and the $strain = \frac{{0.018}}{{1.5000}} = 0.012$ .
Note that both extension and original length must be in the same units, and so strain is a ratio, without units. Sometimes, strain is given as a percentage. For example, a strain of 0.012 is equivalent to $1.2\% $.
Why do we use a thin wire? This is because a thick wire would not stretch as much for the same force.
Again, we need to take account of this in our calculations, and we do this by calculating the stress produced by the load.
The stress is defined as the force applied per unit cross-sectional area of the wire. That is:
$\begin{array}{l}
stress = \frac{{normal\,force}}{{cross - \sec tional\,area}}\\
\sigma = \frac{F}{A}
\end{array}$
where $\sigma $ is the stress, F is the applied force that acts normally (at right angles) on a wire of cross-sectional area A.
The units of stress are newtons per square metre (N m−2) or pascals (Pa), the same as the units of pressure:
$N\,{m^{ - 2}}$
The Young modulus
We can now find the stiffness of the material we are stretching. Rather than calculating the ratio of force to extension as we would for a spring or a wire, we calculate the ratio of stress to strain. This ratio is a constant for a particular material and does not depend on its shape or size. The ratio of stress to strain is called the Young modulus of the material. That is:
$\begin{array}{l}
Young\,\bmod ulus = \frac{{stress}}{{strain}}\\
E = \frac{\sigma }{\varepsilon }
\end{array}$
where E is the Young modulus of the material, σ is the stress and ε is the strain.
The unit of the Young modulus is the same as that for stress, $N\,{m^{ - 2}}$ or Pa. In practice, values may be quoted in MPa or GPa. These units are related as:
$\begin{array}{l}
1\,mPa = {10^6}\,Pa\\
1\,GPa = {10^{96}}\,Pa
\end{array}$
Usually, we plot a graph with stress on the vertical axis and strain on the horizontal axis (Figure 7.13).
It is drawn like this so that the gradient is the Young modulus of the material. It is important to consider only the first, linear section of the graph. In the linear section stress is proportional to strain and the wire under test obeys Hooke’s law.
Table 7.1 gives some values of the Young modulus for different materials.
first, straight-line section of the graph
| Young modulus / GPa | Material |
| 70 | aluminium |
| 90–110 | brass |
| 7–20 | brick |
| 40 | concrete |
| 130 | copper |
| 70–80 | glass |
| 200 | iron (wrought) |
| 18 | lead |
| 3 | Perspex® |
| 2.7–4.2 | polystyrene |
| 0.01 | rubber |
| 210 | steel |
| 50 | tin |
| 10 approx. | wood |
Questions
13) List the metals in Table 7.1 from stiffest to least stiff.
14) Which of the non-metals in Table 7.1 is the stiffest?
15) Figure 7.14 shows stress–strain graphs for two materials, A and B. Use the graphs to determine the Young modulus of each material.
16) A piece of steel wire, $200.0 cm$ long and having cross-sectional area of $0.50\,m{m^2}$, is stretched by a force of $50 N$. Its new length is found to be $200.1 cm$. Calculate the stress and strain, and the Young modulus of steel.
17) Calculate the extension of a copper wire of length $1.00 m$ and diameter $1.00 mm$ when a tensile force of $10 N$ is applied to the end of the wire. (Young modulus of copper $= 130 GPa$.)
18) In an experiment to measure the Young modulus of glass, a student draws out a glass rod to form a fibre $0.800 m$ in length. Using a travelling microscope, she estimates its diameter to be $0.40 mm$.
Unfortunately, it proves impossible to obtain a series of readings for load and extension. The fibre snaps when a load of $1.00 N$ is hung on the end. The student judges that the fibre extended by no more than 1 mm before it snapped.
Use these values to obtain an estimate for the Young modulus of the glass used. Explain how the actual or accepted value for the Young modulus might differ from this estimate.
19) For each of the materials whose stress–strain graphs are shown in Figure 7.15, deduce the values of the Young modulus.
PRACTICAL ACTIVITY 7.2
Determining the Young modulus
You must be able to describe this experiment in detail. Learn how to draw the diagram in Figure 7.11 and how to measure each of the quantities in:
$\begin{array}{l}
E = \frac{{FL}}{{Ax}}\\
= \left( {\frac{F}{x}} \right) \times \times \left( {\frac{{4L}}{{\pi {d^2}}}} \right)
\end{array}$
Metals are not very elastic. Normally, they can only be stretched by about 0.1% of their original length.
Beyond this, they become permanently or plastically deformed. As a result, some careful thought must be given to getting results that are good enough to give an accurate value of the Young modulus.
First, the wire used must be long. The increase in length is proportional to the original length, and so a longer wire gives larger and more measurable extensions. Typically, extensions up to $1 mm$ must be
measured for a wire of length $1 m$. To get suitable measurements of extension there are two possibilities: use a very long wire, or use a method that allows measurement of extensions that are a fraction of a millimetre.
The apparatus shown in Figure 7.12 can be used with a travelling microscope placed above the wire and focused on the sticky tape pointer. When the pointer moves, the microscope is adjusted to keep the pointer at the middle of the cross-hairs on the microscope. The distance that the pointer has moved can then be measured accurately from the scale on the microscope.
Second, the cross-sectional area of the wire must be known accurately. The diameter of the wire is measured using a micrometer screw gauge. This is reliable to within $ \pm 0.01\,mm$. Once the wire has been loaded in increasing steps, the load must be gradually decreased to ensure that there has been no permanent deformation of the wire.
A graph of F against x can be drawn and the gradient used to find an average value of , where F is the weight of the load and x is the extension shown by the distance moved by the pointer.
The area A is found from $A = \frac{{\pi {d^2}}}{4}$ , where d is the diameter of the wire.
The diameter should be measured at several points along the wire and the average value found. The length L is measured from the sticky pointer to the point where the wire is clamped.
Other materials, such as glass and many plastics, are also quite stiff and so it is difficult to measure their Young modulus. Rubber is not as stiff, and strains of several hundred per cent can be achieved.
However, the stress–strain graph for rubber is not a straight line. This means the value of the Young modulus found is not very precise, because it only has a very small linear region on a stress–strain graph.