# Young's Modulus

## Young's Modulus Revision

**Young’s Modulus**

This section looks at some other terms that can be used to describe and quantify the effects of a force upon a material. We will look at **stress**, **strain** and **Young’s modulus** and how they are calculated.

**Tensile Stress and Strain**

For a material put under load, the **tensile stress** is equal to the** force applied per unit of cross-sectional area**. Tensile forces are applied to an object when a **force** is applied in opposite directions to each end of the object, causing a stretching effect.

**Tensile stress** can be calculated using the equation:

\sigma = \dfrac{F}{A}

- \sigma is the
**tensile stress**in pascals \left(\text{Pa}\right) - F is the
**force**in Newtons \left(N\right) - A is the
**cross-sectional area**in metres squared \left(m^2\right)

The **tensile strain** of an object under load is defined as the **extension**** per unit of original length**. As an equation, this is written as:

\epsilon = \dfrac{\Delta L}{L}

- \epsilon is the
**tensile strain**of the material. It is unitless. - \Delta L is the
**change in length**of the material in metres \left(m\right). - L is the
**original length**of the material in metres \left(m\right).

**Example: **A wire of radius 0.01 \: \text{m}, original length of 2 \: \text{m} is put under a tensile force of 75 \: \text{N}. The wire extends by 1.5 \: \text{m}. Calculate the tensile stress and strain of the wire.

**[2 marks]**

Finding the stress:

\begin{aligned}\sigma &= \dfrac{F}{A} \\ \\ \sigma &= \dfrac{75}{\pi \times 0.01^2} = \boldsymbol{238853.5} \: \textbf{Pa} \: \text{or} \: \boldsymbol{0.24} \: \textbf{MPa} \end{aligned}

Finding the strain:

\begin{aligned} \epsilon &= \dfrac{\Delta L}{L} \\ \\ \epsilon &= \dfrac{1.5}{2} = \boldsymbol{0.75} \end{aligned}So the **tensile stress** =\boldsymbol{0.24} \: \textbf{MPa} and the **tensile strain** = \boldsymbol{0.75}

**Stress-Strain Graphs**

A lot can be interpreted when **stress **is plotted against **strain **in a **stress-strain graph**. Here is an example:

As labelled, there are several distinguishable features of a** stress-strain graph**:

**Limit of proportionality** – the point at which the graph first starts to curve. This shows the point where the material no longer obeys **Hooke’s law**.

**Yield strength**– this is the first turning point of the graph and represents where the material **yields**. At this point an increase in** strain **can be seen without an increase in stress showing the material** yielding**.

**Maximum tensile stress** – the highest point on the graph. This shows the maximum **force per unit cross sectional area**.

**Breaking point** – the endpoint of the curve. At this point the material has snapped and therefore has no **stress** being applied.

On a **stress-strain graph** a **brittle** material could be identified as a material that has no or very little curved area of the graph. It would also tend to be displayed as a steep line with very low maximum **strain** as** brittle **materials do not stretch or bend very well without snapping. A **ductile** material would be represented by a long curved graph as** ductile** materials can be stretched well without.

**Elastic Strain Energy**

The **elastic strain energy** is the energy stored in the material due to the **work** being done on the material in stretching it. The **elastic strain energy** can be calculated in two ways:

**First method:**

- If the material obeys
**Hooke’s law**(anytime the graph is linear) we can use an equation:

\text{Elastic Strain Energy} = \dfrac{1}{2} F \Delta L or \text{Elastic Strain Energy} = \dfrac{1}{2} k \Delta L^2

- F is the
**force applied**in Newtons \left(\text{N}\right). - \Delta L is the
**change in the length**in metres \left(\text{m}\right) - k is the
**spring constant**in Newtons per metre \left(\text{Nm}^{-1}\right)

**Example:** Using the graph, calculate the **elastic strain energy** in the part of the graph that obeys **Hooke’s Law **

**[1 mark]**

The linear section is approximately from 0 \: \text{mm} to 2 \: \text{mm} of extension. So this part of the graph obeys **Hooke’s Law**.

\text{Elastic strain energy} = \dfrac{1}{2} F \times \Delta L

\text{Elastic strain energy} = \dfrac{1}{2} \times 14 \: \text{N} \times 0.002 \: \text{m} = \boldsymbol{0.0014} \: \textbf{J}

**Second method:**

The second method below can be applied to both instances where a material obeys **Hooke’s law** or if it has passed its** limit of proportionality **and is behaving **plastically**.

- The
**elastic strain energy**can be determined by calculating the area under a**Force-extension graph**.

**Example:** Estimate the total amount of elastic strain energy stored in the material over the whole graph.

**[4 marks]**

Firstly, split the graph into easy to calculate sections.

Estimate the area of each section:

**From** \boldsymbol{0-20} \: \textbf{N}: 0.5 \times 20 \times 0.003 = \boldsymbol{0.03} \: \textbf{J}

**At** \boldsymbol{20} \: \textbf{N}: 20 \times 0.001 = \boldsymbol{0.02} \: \textbf{J}

**From** \boldsymbol{20 – 35} \: \textbf{N}: \left(21 \times 0.002\right) + \dfrac{1}{2} \left(14 \times 0.002\right) = \boldsymbol{0.056} \: \textbf{J}

**Whole area** = 0.03 + 0.02 + 0.056 = \boldsymbol{0.106} \: \textbf{J}

**Young’s Modulus**

Another quantity that can be calculated using **stress** and **strain** is the Y**oung’s modulus**. **Young’s modulus** is the ratio of **tensile stres**s divided by tensile** strain**. This can be written in the form:

\text{Young's Modulus} = \dfrac{\text{Stress}}{\text{Strain}}

We can take our equations from the previous sections and write this as:

\text{Young's Modulus} = \dfrac{F/A}{\Delta L / L} = \dfrac{FL}{A \Delta L}

**Example: **The Young’s modulus of aluminium is 68 \: \text{GPa}. If the tensile stress on aluminium is 15 \: \text{GPa}, what is the strain?

**[2 marks]**

\text{Young's Modulus} = \dfrac{\text{Tensile Stress}}{\text{Tensile Strain}}

\textbf{Tensile Strain} = \dfrac{\textbf{Tensile Stress}}{\textbf{Young's Modulus}}

\text{Tensile Strain} = \dfrac{68 \times 10^9}{15 \times 10^9} = \boldsymbol{4.5}

**Required Practical 4**

**Determination of the Young’s modulus of different pieces of wire.**

This experiment can be carried out using any type of wire. The experiment may also be done using a vertical piece of wire suspended from the ceiling.

**Doing the experiment:**

- Firstly, before any equipment is set up, the diameter of the wire needs to be measured. A
**micrometer**can be used for this. Measure the diameter of the wire in three different places and orientations along the wire before taking an average. Be careful not to squash the wire by over tightening the micrometer. - Set up the equipment as shown in the diagram above.
- Make the wire tight but with no mass on the mass hanger and measure the
**original length**of the wire. Mark a point on the wire using a marker. This could be a pointed piece of tape that points downwards to the ruler. - Begin by
**adding a 100 \: \text{g} mass**onto the mass hanger. Record this mass and also record the**extension**by observing how far the marker has moved from the original position. Record the extension in metres. - Repeat the above step 10 times until 1 \: \text{kg} has been added.

**Analysing the results:**

To analyse the results, all masses must be converted to newtons by multiplying by g \left(9.81 \: \text{N/kg}\right) and all lengths/diameters must be converted to metres.

**Young’s modulus** is given by the equation:

\text{Young's Modulus} = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{FL}{A \Delta L}

where F is the **force** applied to the wire, L is the** original length** of the wire, A is the** cross-sectional area** of the wire and \Delta L is the** extension** of the wire.

In our experiment we have recorded the **force** on the wire \left(F\right) and the corresponding **extension** of this wire \left(\Delta L\right). Therefore, we can plot a graph of **force** on the y-axis and **extension **on the x-axis:

The **gradient** of this graph will give \dfrac{F}{\Delta L}.

Our** Young’s Modulus** equation is :

\text{Young's Modulus} = \dfrac{FL}{A \Delta L}

So we can rewrite this as:

\text{Young's Modulus} = \text{gradient} \times \dfrac{L}{A}

So to work out our **Young’s Modulus**, we just multiply the **gradient **of the graph by the **original length** and divide by the **cross-sectional area**.

You should have measured the diameter of your wire in the experiment. We can use this and the area of a circle equation A = \pi r^2 to calculate the **cross-sectional area** of the wire, A.

## Young's Modulus Example Questions

**Question 1: **Describe the stress-strain graph you would expect to see for a brittle material.

**[2 marks]**

A **brittle material is a material that cannot stretch or bend very much without breaking.**

Therefore, the stress strain graph will mostly be** linear with a low breaking stress and strain**.

**Question 2: **Young’s modulus for aluminium is 70 \: \text{GPa} and 170 \: \text{GPa} for Iron. What does this tell you about these two materials?

**[2 marks]**

Young’s modulus is a measure of how good a material is at coping with a high load, without deforming.

Therefore, as iron has a much higher Young’s modulus, we would expect **iron to be better at withstanding high stresses whilst deforming very little compared to aluminium. **

**Question 3: **Young’s modulus for steel is 110 \: \text{GPa} and can withstand a tensile strength of 1.8 \: \text{GPa}. Calculate the strain when the steel undergoes this tensile strength.

**[2 marks]**

## Young's Modulus Worksheet and Example Questions

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