# Deformation of materials

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## Introduction

In this chapter we will look at some mechanical (physical) properties of various materials that we use. The mechanical properties of a material are those properties that are affected by forces being applied to the material. These properties are important to consider when we are constructing buildings, structures or modes of transport like an aeroplane.

## Hooke's law

Deformation (change of shape) of a solid is caused by a force that can either be compressive or tensile when applied in one direction (plane). Compressive forces try to compress the object (make it smaller or more compact) while tensile forces try to tear it apart. We can study these effects by looking at what happens when you compress or expand a spring.

Hooke's Law relates the restoring force of a spring to its displacement from equilibrium length.

The equilibrium length of a spring is its length when no forces are applied to it. When a force is applied to a spring, e.g., by attaching a weight to one end, the spring will expand and become longer. The difference between the new length and the equilibrium length is the displacement.

## Hooke's law

Hooke's law is named after the seventeenth century physicist Robert Hooke who discovered it in 1660 (18 July 1635 - 3 March 1703).
Hooke's Law

In an elastic spring, the extension varies linearly with the force applied.

$F=-kx$

where $F$ is the restoring force in newtons (N), $k$ is the spring constant in $N·{m}^{-1}$ and $x$ is the displacement of the spring from its equilibrium length in metres (m).

## Experiment : hooke's law

Aim:

Verify Hooke's Law.

Apparatus:

• weights
• spring
• ruler

Method:

1. Set up a spring vertically in such a way that you are able to hang weights from it.
2. Measure the equilibrium length, ${x}_{0}$ , of the spring (i.e. the length of the spring when nothing is attached to it).
3. Measure the extension of the spring for a range of different weights. Note: the extension is the difference between the spring's equilibrium length and the new length when a weight is attached to it, $x-{x}_{0}$ .
4. Draw a table of force (weight) in newtons and corresponding extension.
5. Draw a graph of force versus extension for your experiment.

Conclusions:

1. What do you observe about the relationship between the applied force and the extension?
2. Determine the gradient (slope) of the graph.
3. Now calculate the spring constant for your spring.

A spring is extended by 7 cm by a force of 56 N.

Calculate the spring constant for this spring.

1. $\begin{array}{ccc}\hfill \mathrm{F}& =& -\mathrm{kx}\hfill \\ \hfill 56& =& -\mathrm{k}×0,07\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathrm{k}& =& \frac{-56}{0,07}\hfill \\ & =& -800\mathrm{N}·{\mathrm{m}}^{-1}\hfill \end{array}$

A spring of length 20cm stretches to 24cm when a load of 0,6N is applied to it.

1. Calculate the spring constant for the spring.
2. Determine the extension of the spring if a load of 0,5N is applied to it.
1. We know:

$F$ = 0,6 N

The equilibrium spring length is 20 cm

The expanded spring length is 24 cm

2. First we need to calculate the displacement of the spring from its equilibrium length:

$\begin{array}{ccc}\hfill x& =& 24\mathrm{cm}-20\mathrm{cm}\hfill \\ & =& 4\mathrm{cm}\hfill \\ & =& 0,04\mathrm{m}\hfill \end{array}$

Now use Hooke's Law to find the spring constant:

$\begin{array}{ccc}\hfill F& =& -kx\hfill \\ \hfill 0,6& =& -k·\phantom{\rule{0.222222em}{0ex}}0,04\hfill \\ \hfill k=-15\mathrm{N}.{\mathrm{m}}^{-1}\end{array}$
3. $F$ = 0,5 N

We know from the first part of the question that

$k$ = -15 $\mathrm{N}.{\mathrm{m}}^{-1}$

So, using Hooke's Law:

$\begin{array}{ccc}\hfill F& =& -kx\hfill \\ \hfill x& =& -\frac{F}{k}\hfill \\ & =& -\frac{0,5}{-15}\hfill \\ & =& 0,033\mathrm{m}\hfill \\ & =& 3,3\mathrm{cm}\hfill \end{array}$

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