When you were a young child you may have played with very soft balls. You could squash them, stretch them and even twist them. However, when you released the ball it would return to its original shape. Well, this ball is an example of an elastic material; while objects such as springs, rubber bands, and trampolines can also fall into this category. A force can be applied to deform the shape of an object, but once that force is removed the object returns to its original shape. This process is called elastic deformation and is what we are going to investigate in this article.
- We are first going to define elasticity.
- We will then explore elastic deformation, as well as its counterpart, inelastic deformation.
- After learning the foundational knowledge we will perform the stretch test, and then investigate the results throughout the article.
- Then we will discuss the extension of materials and the equation to determine extension.
- Following this, we will explore how we can use Hooke's law in relation to extension.
- Finally, we will explore elastic potential energy and its corresponding equation.
Elastic deformation definition
Elasticity can be seen in materials that can deform. One material might be stretched by an applied force and then returned to its original shape after it is released. While a different material could be stretched, but then remain deformed after the applied force is released. But only of these is an elastic material.
Deformation is the change of shape of an object due to an applied force acting on it.
The first scenario that we mentioned is elastic deformation. These are objects that can change their shape due to stretching, twisting, compression and bending. But once released from the force applied to them, the object returns to its original shape. An example of this is a rubber band. If you stretch it and then release it, it goes back to its original shape. Therefore, rubber bands deform elastically.
In the diagram above you can compare the length of spring at three different stages. The first stage is before any force has been applied, the second is when a force is being applied and the third is when the applied force has been removed. You can observe that the spring has gone back to its original length, showing that elastic deformation took place.
Inelastic Deformation
When an object is deformed but cannot return back to its original shape, then we call it inelastic deformation. An aluminium drink can should be an example of this that most of you are familiar with. If you squeeze the can with enough force it will compress and crumple, when you release your grip on the can it will remain in its new shape.
Look at the diagram above showcasing inelastic deformation, you will notice it is very similar to the diagram in the previous section illustrating elastic deformation. However, this time the spring did not return to its original length after being deformed. But why is this if an identical spring was used both times? You might have mistakenly thought that a material can only exclusively be elastic or inelastic.
The answer lies in the force applied to the spring in both scenarios. The load is \(5\,\mathrm{g}\) more massive than the first scenario, so the weight (or applied force) in the second scenario will be doubled. You can see in the diagram above that when the heavier applied force was removed from the spring that it did not return to its original length. This shows that inelastic deformation has taken place as the length without any force applied has gone from \(10\,\mathrm{cm}\) to \(12\,\mathrm{cm}\).
The elastic limit is the maximum deformation a material or object can be exposed to before it begins to inelastically deform.
The reason why the spring deformed inelastically in the second scenario was that the applied force was so large that it breached the elastic limit.
If the applied force on a material is too great then the stress the material is under reaches its breaking point, at which point it will... break. For example, if you stretch a rubber band too far it will snap. Releasing all the elastic potential energy stored in it as mostly kinetic energy, with some sound energy and heat energy too.
Stretch test
A simple experiment can be carried out to determine how far an object or material can be deformed elastically before it begins to deform inelastically or breaks. It is called the stretch test and is what we are going to explore in this section. The results will help inform our understanding of deformation throughout the rest of the article.
To carry out this test, we need a few things:
- A spring
- A clamp
- A weight hanger
- A ruler
- Multiple small weights of the same mass
In the diagram above you can compare the length of spring at three different stages. The first stage is before any force has been applied, the second is when a force is being applied and the third is when the applied force has been removed. You can observe that the spring has gone back to its original length, showing that elastic deformation took place.
Inelastic Deformation
When an object is deformed but cannot return back to its original shape, then we call it inelastic deformation. An aluminium drink can should be an example of this that most of you are familiar with. If you squeeze the can with enough force it will compress and crumple, when you release your grip on the can it will remain in its new shape.
Look at the diagram above showcasing inelastic deformation, you will notice it is very similar to the diagram in the previous section illustrating elastic deformation. However, this time the spring did not return to its original length after being deformed. But why is this if an identical spring was used both times? You might have mistakenly thought that a material can only exclusively be elastic or inelastic.
The answer lies in the force applied to the spring in both scenarios. The load is \(5\,\mathrm{g}\) more massive than the first scenario, so the weight (or applied force) in the second scenario will be doubled. You can see in the diagram above that when the heavier applied force was removed from the spring that it did not return to its original length. This shows that inelastic deformation has taken place as the length without any force applied has gone from \(10\,\mathrm{cm}\) to \(12\,\mathrm{cm}\).
The elastic limit is the maximum deformation a material or object can be exposed to before it begins to inelastically deform.
The reason why the spring deformed inelastically in the second scenario was that the applied force was so large that it breached the elastic limit.
If the applied force on a material is too great then the stress the material is under reaches its breaking point, at which point it will... break. For example, if you stretch a rubber band too far it will snap. Releasing all the elastic potential energy stored in it as mostly kinetic energy, with some sound energy and heat energy too.
Stretch test
A simple experiment can be carried out to determine how far an object or material can be deformed elastically before it begins to deform inelastically or breaks. It is called the stretch test and is what we are going to explore in this section. The results will help inform our understanding of deformation throughout the rest of the article.
To carry out this test, we need a few things:
- A spring
- A clamp
- A weight hanger
- A ruler
- Multiple Small weights of the same mass
Now onto how we carry out the test:
- We first use the clamp to secure the spring by clamping it to the top.
- Then attach the empty weight hanger to the bottom of the spring.
- Record the length of the spring using the ruler, we can note this down as our original length.
- Add a weight to the hanger then measure the length of the spring and record it. Repeat this step for multiple weights.
- The weight added per measurement is the dependent variable.
- The length of the spring after each subsequent force is applied is the independent variable.
- Using our measured lengths with different applied forces (weights) alongside the original length we can calculate the extension.
Remember to clamp the stand to a secure table, this will prevent the weights from falling off and avoid any potential injuries.
The results from this experiment will allow us to explore the elasticity of the spring. We should be able to determine the extension of the spring under different magnitudes of applied force (weight in this case).
Extension
We mentioned extension in our short practical above, but what is this exactly?
The extension is the difference between the increased length from the original length.
To calculate the extension of a material we can use the following equation and the results that we noted down during the stretch practical.
\[\text {extension} = \text{length} - \text{original length}.\]
Here we have an example of the results of a stretch test we carried out on a spring:
Weight (N) | Length (mm) |
0 | 120 |
1 | 140 |
2 | 160 |
3 | 180 |
4 | 200 |
5 | 218 |
6 | 233 |
7 | 244 |
Table 1. - Results of a stretch test where it is shown the weight vs. the extension (length).
Question
Calculate the extension when \(2\,\mathrm{N}\) of applied force acts on the spring.
First, let us see what figures we need to calculate the extension from the equation:
\[\text {extension} = \text{length} - \text{original length}.\]
Solution
To solve this problem, we need the original length and the length when \(2\,\mathrm{N}\) was added.
Our length at \(2\,\mathrm{N}\) is \(160\,\mathrm{mm}\) and our original length at \(0\,\mathrm{N}\) is \(120\,\mathrm{mm}\).
We can substitute these values into our equation: \[\text {extension} = 160\,\mathrm{mm} - 120\,\mathrm{mm} = 40\,\mathrm{mm}\]
Therefore, our extension at an applied force of \(2\,\mathrm{N}\) is \(40\,\mathrm{mm}\) for this spring.
Elastic deformation graph
To better visualise our results we can plot a graph. The y-axis shows the weight added (of force applied) in newtons \(\mathrm{N}\) and on the x-axis, we have the extension in \(\mathrm{mm}\). Looking at the graph below, you can see that for the extension of the spring, the relationship between the weight and extension is directly proportional. There is a linear relationship in dark blue between extension and applied force between \(0\,\mathrm{N}\) and \(4\,\mathrm{N}\).
Fig. 4 - Graph of experimental results from the stretch test of weight applied against the spring's extension.
The pink dot represents the approximate weight that the linear relationship between the force applied and extension. The green line is a curve of best fit showing that the spring's extension increases less after each subsequent weight is added past \(5\,\mathrm{N}\). If we kept adding more weight to our spring system it would likely break entirely!
Hooke's law and elastic deformation
We have discussed in the previous section that the relationship between the weight added and extension is directly proportional, up to a certain limit. This point is called the limit of proportionality.
The limit of proportionality is when the amount of force used to deform an elastic material is too large, and the relationship between weight and extension is no longer directly proportional.
The idea that the extension is directly proportional unless the limit of proportionality is reached is known as Hooke’s law and can be represented with the following equation
\[F=ke,\]
where \(F\) is the force applied in Newtons \(\mathrm{N}\), \(k\) is the spring constant in Newtons per meter \(\frac{\mathrm{N}}{\mathrm{m}}\), and \(e\) is the extension in \(\mathrm{m}\).
We can determine the spring constant of the spring used in the stress test using only the graph. If you scroll up back to the graph you will find that force is on the y-axis and the extension is on the x-axis. A straight line should be produced where Hooke's law holds true (before the limit of proportionality). The gradient of this line is the spring constant!
Limit of proportionality against the elastic limit
It is crucial to note that the limit of proportionality and the elastic limit is not the same thing. Our spring definitely breached the limit of proportionality in our experiment, as the relationship between the applied force and extension became non-linear somewhere between \(4\,\mathrm{N}\) and \(5\,\mathrm{N}\) of weight.
Using only the results we recorded, it is impossible to know if the spring deformed elastically or inelastically past this point. Interestingly, an object can deform elastically past the limit of proportionality, and still return to its original shape after the applied force is removed.
To improve our experiment we should have followed a better experimental procedure. After we applied a new weight and recorded the extension, we should then have removed all masses from the hanger and recorded the length of the spring each time. This way we could check to see if the spring returned to its original shape or not, and therefore see if it deformed elastically or inelastically.
Elastic potential energy
Imagine that we elastically deform an object. We actually do work on the object as we deform it. In simple cases such as stretching or compression, the work done \(W\) in Joules \(\mathrm{J}\) is equal to the product of the applied force \(F\) in Newtons \(\mathrm{N}\) over a distance \(x\) in \(\mathrm{m}\).
\[W=Fx.\]
When an object is elastically deformed in this way, elastic potential energy is stored in the object. The amount of energy stored in the spring \(E_e\) in Joules \(\mathrm{J}\) can be determined by the following equation
\[E_e=\frac 1 2 \Delta kx^2,\]
where \(k\) is the spring constant in Newtons per meter \(\frac{\mathrm{N}}{\mathrm{m}}\), and \(\Delta x^2\) is the change in the position (or distance) squared in \(\mathrm{m^2}\).
Interestingly, the elastic potential energy stored in an object that is elastically deformed is equal to the work done to deform the object. Therefore the following equation is true:
\[W=E_e.\]
Elastic deformation example
To consolidate our knowledge of elastic deformation, let us go through a couple of example questions relevant to the topic with written solutions.
Question
A spring has a constant of \(22\,\frac{\mathrm{N}}{\mathrm{m}}\). Calculate the force required to make it extend by \(0.1\,\mathrm{m}\).
Solution
We can apply the equation from Hooke’s law to solve this question.
\begin{align*}F&=kx,\\[6pt]F&=22\,\frac{\mathrm{N}}{\mathrm{m}} \times 0.1\,\mathrm{m},\\[6pt]F&=2.2\,\mathrm{N}.\\\end{align*}
Question
Jerry does \(4\,\mathrm{J}\) of work on an old mattress spring, attempting to compress it. The spring constant is \(50\,\frac{\mathrm{N}}{\mathrm{m}}\). By what length does Jerry compress the spring?
Solution
We know that the work done on an elastically deformed object is equal to the elastic potential energy,
\[W=E_e=\frac 1 2 kx^2.\]
This equation can be rearranged so that the change of position (length) becomes the subject of the equation.
\begin{align*}2W&=kx^2,\\[6pt]\frac{2W}{k}&=x^2,\\[6pt]x&=\sqrt{\frac{2W}{k}}.\end{align*}Now, we simply have to substitute our known values of work done and the spring constant to calculate our answer.\[x=\sqrt{\frac{2 \times 4\,\mathrm{J}}{50\,\frac{\mathrm{N}}{\mathrm{m}}}}=0.4\,\mathrm{m}.\]
We can determine the spring constant of the spring used in the stress test using only the graph. If you scroll up back to the graph you will find that force is on the y-axis and the extension is on the x-axis. A straight line should be produced where Hooke's law holds true (before the limit of proportionality). The gradient of this line is the spring constant!
Limit of proportionality against the elastic limit
It is crucial to note that the limit of proportionality and the elastic limit is not the same thing. Our spring definitely breached the limit of proportionality in our experiment, as the relationship between the applied force and extension became non-linear somewhere between \(4\,\mathrm{N}\) and \(5\,\mathrm{N}\) of weight.
Using only the results we recorded, it is impossible to know if the spring deformed elastically or inelastically past this point. Interestingly, an object can deform elastically past the limit of proportionality, and still return to its original shape after the applied force is removed.
To improve our experiment we should have followed a better experimental procedure. After we applied a new weight and recorded the extension, we should then have removed all masses from the hanger and recorded the length of the spring each time. This way we could check to see if the spring returned to its original shape or not, and therefore see if it deformed elastically or inelastically.
Elastic potential energy
Imagine that we elastically deform an object. We actually do work on the object as we deform it. In simple cases such as stretching or compression, the work done \(W\) in Joules \(\mathrm{J}\) is equal to the product of the applied force \(F\) in Newtons \(\mathrm{N}\) over a distance \(x\) in \(\mathrm{m}\).
\[W=Fx.\]
When an object is elastically deformed in this way, elastic potential energy is stored in the object. The amount of energy stored in the spring \(E_e\) in Joules \(\mathrm{J}\) can be determined by the following equation
\[E_e=\frac 1 2 \Delta kx^2,\]
where \(k\) is the spring constant in Newtons per meter \(\frac{\mathrm{N}}{\mathrm{m}}\), and \(\Delta x^2\) is the change in the position (or distance) squared in \(\mathrm{m^2}\).
Interestingly, the elastic potential energy stored in an object that is elastically deformed is equal to the work done to deform the object. Therefore the following equation is true, \[W=E_e.\]
Elastic deformation example
To consolidate our knowledge of elastic deformation, let us go through a couple of example questions relevant to the topic with written solutions.
Question
A spring has a constant of \(22\,\frac{\mathrm{N}}{\mathrm{m}}\). Calculate the force required to make it extend by \(0.1\,\mathrm{m}\).
Solution
We can apply the equation from Hooke’s law to solve this question.
\begin{align*}F&=kx,\\[6pt]F&=22\,\frac{\mathrm{N}}{\mathrm{m}} \times 0.1\,\mathrm{m},\\[6pt]F&=2.2\,\mathrm{N}.\\\end{align*}
Question
Jerry does \(4\,\mathrm{J}\) of work on an old mattress spring, attempting to compress it. The spring constant is \(50\,\frac{\mathrm{N}}{\mathrm{m}}\). By what length does Jerry compress the spring?
Solution
We know that the work done on an elastically deformed object is equal to the elastic potential energy,
\[W=E_e=\frac 1 2 kx^2.\]
This equation can be rearranged so that the change of position (length) becomes the subject of the equation.
\begin{align*}2W&=kx^2,\\[6pt]\frac{2W}{k}&=x^2,\\[6pt]x&=\sqrt{\frac{2W}{k}}.\\\end{align*}Now, we simply have to substitute our known values of work done and the spring constant to calculate our answer.\[x=\sqrt{\frac{2 \times 4\,\mathrm{J}}{50\,\frac{\mathrm{N}}{\mathrm{m}}}}=0.4\,\mathrm{m}.\]
Elastic Deformation - Key takeaways
- Elastically deformable objects can change their shape due to stretching, twisting, compression and bending. But once released from the force applied to them, the object returns to its original shape.
- Inelastic deformation is when an object is deformed, but cannot return back to its original shape.
- To determine how far an object or material can stretch we can use the stretch test.
- The extension is the difference between the length of an object and its original length.
- The extension is directly proportional to the applied force unless the limit of proportionality is reached.
- The energy stored in a stretched spring can be calculated using this equation: \(E_e=\frac 1 2 \Delta kx^2\).
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