Young's Modulus

Young's modulus, or elastic modulus, describes the ability of materials to resist changes in length when they undergo tension or compression. This modulus is very useful in engineering as it provides details about the elastic properties of materials, such as their tensile strength and stiffness.

Young's Modulus Young's Modulus

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Table of contents

    Defining Young's modulus

    The Young's modulus is equal to the longitudinal stress being applied divided by the strain. The stress and strain of an object undergoing tension can be also be expressed as follows: when a metal object is pulled by a force F at each end, the object will be stretched from the original length L0 to a new stretched length Ln.

    Since the object is stretched the cross-sectional area will be decreased. The stress can be expressed as the tensile force applied (F) measured in newton (N), divided by the cross-sectional area (A) measured in m2, as seen in the equation below. The resulting unit of stress is N/m2.

    \[\sigma[N/m^2] = \frac{F}{A}\]

    The strain, also known as relative deformation, is the change in length caused by tension or compression divided by the original length L0. Strain is dimensionless, as both terms of the fraction are measured in metres and can be calculated from the following equation.

    \[\varepsilon = \frac{L_n[m]-L_0}{L_0[m]} = \frac{\Delta L}{L_0}\]

    Young's modulus formula

    The elastic modulus E can be expressed as the stress divided by the strain as shown in the formula below.

    \[\text{Young's modulus }[N/m^2] = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon} = \frac{\frac{F}{A}}{\frac{\Delta L}{L_0}} = \frac{FL_0}{A \Delta L}\]

    The units of Young's modulus units are the same as the stress, N/m2 which is equivalent to Pa (pascal). Since the elastic modulus is usually a very large number of the magnitude of 109 it is often expressed in Giga Pascals, shown as GPa.

    \[\text{Young's modulus} = \frac{FL_0}{A \Delta L} = \frac{F[N] L_0[m]}{A[m^2]\Delta L[m]} = \frac{N}{m^2} = Pa\]

    A metal bar experiences a load of 60 N applied at the end. The metal bar has a cylindrical shape and cross-sectional area of 0.02 mm2. The length of the bar increases by 0.30 percent. Find the elastic modulus of the bar.

    Solution:

    Since Young's modulus is required, we need to find the stress and the strain first (remember, Young's modulus is the ratio of stress over strain). We apply the stress formula to find stress.

    \[\sigma = \frac{F}{A} = \frac{60 N}{0.02 \cdot 10^{-6} m^2} = 3 \cdot 10^9 Pa\]

    Then we apply the strain formula to find the strain:

    \[\varepsilon = \frac{L_n - L_0}{L_0} = \frac{\Delta L}{L_0} = 0.3 \% = 0.003\]

    Then finally we divide stress over strain to find Young's modulus.

    \[\text{Young's modulus} = \frac{\text{stress}}{\text{strain}} = \frac{3 \cdot 10^9}{0.003} = 1 \cdot 10^{12} Pa\]

    How is Young's modulus related to Hooke's Law?

    Hooke's law states that the force acting on a body or spring creating a displacement Δx, is linear to the displacement created as shown by the equation below, where k is a constant relating to the spring's stiffness. Hooke's law can be applied to situations where a body deforms elastically.

    \[F = k \Delta x\]

    Similar to Hooke's law, where the extension or compression of a spring is linearly proportional to the force applied; the stress that is applied on a body is linearly proportional to Young's modulus E, as shown in the rearranged equation below.

    \[\sigma = E \varepsilon\]

    Graphically calculate Young's modulus

    Since the elastic modulus of a material is linearly proportional to the stress applied divided by the strain, Young's modulus can also be calculated from a stress-strain graph, which describes a linear deformation as laid out by Hooke's law. Young's modulus is equal to the slope of the linear region of the stress vs strain curve, as shown in the figure below.

    Young's modulus, Stress vs strain graph, StudySmarter

    Young's modulus calculation from stress-strain graph

    Experiments to show Young's modulus

    In order to measure Young's modulus of a metal, several experiments can be constructed. Varying loads will be applied to a copper wire - the resulting extension will be measured to construct a stress-strain graph. According to the stress and strain equations, the required parameters will be measured with the following equipment.

    • Wire

    • Micrometre

    • Pulley

    • Metre ruler

    • Caliper

    • Weights

    • Clamp

    • Wooden block

    • Bench

    Methodology

    1. Using the ruler, we measure the initial length of the wire. We use the micrometre to measure the diameter of the wire at three points along its length. This is for the average diameter to be used in the area calculation.

    2. We connect one end of the wire to the pulley that is clamped to a bench, and the other end to a clamped wooden block.

    3. The extension of the wire due to the force of the weights is measured and recorded. The difference between the new length and the original length prior to extension is used in the strain calculation.

    4. Repeat the process to get 5 to 10 more readings. Various measurements with various weights are recommended to be taken to reduce errors.

    5. Then plot stress vs strain and find the elastic modulus by taking the gradient of the line.

    Results analysis

    The aim of the experiment is to estimate Young's modulus. This can be found by estimating stress and strain. The following steps are required to find stress and strain.

    1. Find the area of the wire using the measured diameter and the equation \(A= \pi r^2\) where \(r=\frac{R}{2}\).

    2. Rearrange Young's modulus formula and solve it for F. This will give us \(F=\frac{(E\cdot A)}{L} \cdot \Delta L\).

    3. The rearranged equation is similar to the equation of a line of the form y = ax, where y is F and the slope is the coefficient of ΔL.

    4. A graph of Force vs ΔL is constructed from the recorded force and extension points so that the slope can be found. The slope \(\frac{\Delta F}{\Delta L}\) is found and multiplied by the original length L0 and divided to area A, to estimate the value of Young's modulus.

    Stress-strain graph and material characteristics

    Some important characteristics of materials are shown in the figure below.

    • The red region indicates the elastic region where it deforms according to Hooke's law, and stress and strain are proportional to each other. The red point indicates the elastic limit or yield point. This is the point to which a material can still hold its original length after the load is applied.

    • The green region indicates the plastic region where the material is unable to return to its initial state and has undergone permanent deformation. The green point indicates the tensile strength point, until which the material can withstand the maximum load per unit without breaking.

    • The blue point indicates the breaking point or breaking stress, where the material will break.

    Young's modulus, Stress vs strain graph material characteristics, StudySmarter

    A stress-strain graph of a material's elastic limit, plastic region, and breaking points

    Young's Modulus - Key takeaways

    • Young's modulus is a material's ability to resist change in length under tension or compression.

    • Young's modulus can be calculated graphically using a stress-strain graph.

    • Experimental calculation of Young's modulus is possible by constructing a load-length difference.

    • Using the stress and strain graph, a material's tensile strength and breaking point can be found.

    Frequently Asked Questions about Young's Modulus

    How do you determine Young's modulus experimentally?

    You can determine Young's modulus experimentally by applying a load to a material, then creating a load vs change in length graph, and calculating the slope of the graph under study.

    What is Young's modulus?

    Young’s modulus is the ability of materials to retain their original length/shape under load.

    Why does aluminium have a lower Young's modulus than steel?

    Because steel is stiffer than aluminium, hence it will be more likely to retain its shape under load.

    What does a higher Young's modulus indicate? 

    A higher Young's modulus indicates that the material is stiffer and requires higher force to be deformed. 

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