Young modulus and elastic strain energy

Stress, Strain and the Young Modulus

Key Equations

Stress:

$$\sigma =\frac{F}{A}$$

Variables:

• F = applied tensile force (N)
• A = cross-sectional area of the sample (m²)
• σ = tensile stress (Pa)

SI unit: pascal (Pa), equivalent to N m⁻²

Rearrangements: $F=\sigma A$ and $A=\frac{F}{\sigma }$

ProportionalityStress is directly proportional to the applied force when the cross-sectional area is constant.

Strain:

$$\epsilon =\frac{\Delta x}{L}$$

Variables:

• Δx = extension of the sample (m)
• L = original length of the sample (m)
• ε = tensile strain (dimensionless)

SI unit: none (a ratio)

Rearrangements: $\Delta x=\epsilon L$ and $L=\frac{\Delta x}{\epsilon }$

ProportionalityStrain is directly proportional to the extension when the original length is constant.

Young modulus:

$$E=\frac{\sigma }{\epsilon }=\frac{FL}{A \Delta x}$$

Variables:

• E = Young modulus (Pa)
• σ = stress (Pa)
• ε = strain
• F, L, A, Δx as defined above

SI unit: pascal (Pa)

Rearrangements:

$$\sigma =E\epsilon $$

$$F=\frac{EA \Delta x}{L}$$

$$\Delta x=\frac{FL}{EA}$$

ProportionalityWithin the linear region, stress is directly proportional to strain for a given material.

The Young modulus is a property of a material that measures its stiffness when stretched or compressed within its elastic limit. It is defined as the ratio of stress to strain in the linear region of the stress–strain graph. A stiffer material has a larger Young modulus, so it resists deformation more strongly for a given stress.

Stress describes how the load is shared across the cross-section of the sample. Strain describes how the extension compares to the original length. Because both are normalised quantities, they depend only on the material, not on the dimensions of the sample. This makes the Young modulus a true material property, allowing different samples to be compared directly.

The Young modulus is measured experimentally by stretching a thin wire and recording the extension produced by known loads. Plotting stress against strain yields a straight line in the elastic region. The gradient of this line equals the Young modulus.

MisconceptionA material with a larger Young modulus is not necessarily stronger, only stiffer. Stiffness describes resistance to deformation. Strength describes the maximum stress before breaking.

Exam TipState “stiffer,” not “stronger,” when describing a larger Young modulus.

CORE PRACTICAL 3: Determining the Young Modulus of a Material

Aim

Determine the Young modulus of a metal in the form of a wire by measuring how its extension varies with applied tensile force.

Method

1. Measure the original length L of the wire from the clamp to the marker using a metre ruler.
2. Measure the diameter d of the wire at five points along its length using a micrometer screw gauge. Calculate the mean diameter and then the cross-sectional area using $A=\pi {\left(\frac{d}{2}\right)}^{2}$.
3. Apply a small initial load to remove kinks. Record the position of the marker as the zero reading.
4. Add a known mass to the hanger. Record the new marker position to find the extension Δx.
5. Repeat step 4, increasing the load in equal steps, until at least eight readings are taken.
6. Calculate stress and strain for each load. Plot stress on the y-axis against strain on the x-axis.
7. Determine the gradient of the linear region of the graph. This gradient equals the Young modulus E.

Variables

• Independent variable (IV): Applied tensile force F (N), increased in equal steps using slotted masses of known weight.
• Dependent variable (DV): Extension Δx (m), measured using a metre ruler against a fixed marker on the wire.
• Control variables (CV): Original length L (kept constant by using the same wire), cross-sectional area (same wire throughout), material of the wire, room temperature.

Expected results

A straight line through the origin on the stress–strain graph, with gradient equal to the Young modulus of the wire. The line curves at high stresses once the limit of proportionality is exceeded.

Precaution

Safety goggles must be worn because the wire may snap suddenly under high load. A soft mat or padded box should be placed below the masses to prevent injury. Systematic error arises if the wire is not initially straight, so a small preload is applied first. Random error in diameter is reduced by averaging readings at several points.

Measurement technique

Use a micrometer screw gauge (resolution 0.01 mm) for diameter and a metre ruler (resolution 1 mm) for length and extension. Check the micrometer for zero error before each reading. Record the marker position from the same viewing angle each time to avoid parallax error. Plotting stress against strain rather than F against Δx allows the gradient itself to be the Young modulus.

SafetyThe wire stores significant elastic strain energy and may recoil if it snaps. Wear goggles and place padding beneath the masses.

Worked Example: Calculating the Young Modulus from CP3 Data

Scenario

A student stretches a copper wire of original length 2.50 m and diameter 0.45 mm. Under a load of 65 N, the wire extends by 8.5 mm. Calculate the Young modulus of the copper.

Step 1: Convert non-SI values to SI units

Diameter:

$$d=0.45 \text{mm}=4.5\times {10}^{-4} \text{m}$$

Extension:

$$\Delta x=8.5 \text{mm}=8.5\times {10}^{-3} \text{m}$$

Step 2: Find the cross-sectional area

Equation used

$$A=\pi {\left(\frac{d}{2}\right)}^{2}$$

Substitution:

$$A=\pi (2.25\times {10}^{-4}{)}^{2}$$

$$A=1.59\times {10}^{-7} {\text{m}}^{2}$$

Step 3: Find the stress

Equation used

$$\sigma =\frac{F}{A}$$

Substitution:

$$\sigma =\frac{65}{1.59\times {10}^{-7}}$$

$$\sigma =4.09\times {10}^{8} \text{Pa}$$

Step 4: Find the strain

Equation used

$$\epsilon =\frac{\Delta x}{L}$$

Substitution:

$$\epsilon =\frac{8.5\times {10}^{-3}}{2.50}$$

$$\epsilon =3.40\times {10}^{-3}$$

Step 5: Find the Young modulus

Equation used

$$E=\frac{\sigma }{\epsilon }$$

Substitution:

$$E=\frac{4.09\times {10}^{8}}{3.40\times {10}^{-3}}$$

$$E=1.20\times {10}^{11} \text{Pa}$$

Answer

$$E\approx 1.20\times {10}^{11} \text{Pa}$$

Interpretation

The result matches the accepted value for copper. This confirms that the wire was stretched within its linear elastic region during the experiment.

Elastic Strain Energy

Key Equations

Elastic strain energy (within Hooke’s law):

$${E}_{el}=1/2F\Delta x$$

Variables:

• Eel = elastic strain energy (J)
• F = force at extension Δx (N)
• Δx = extension (m)

SI unit: joule (J)

Rearrangements: $F=\frac{2{E}_{el}}{\Delta x}$ and $\Delta x=\frac{2{E}_{el}}{F}$

Equivalent form using stiffness:

$${E}_{el}=1/2k\Delta {x}^{2}$$

Variables:

• k = stiffness or spring constant (N m⁻¹)
• Δx = extension (m)
• Eel = elastic strain energy (J)

ProportionalityWhen Hooke’s law holds, doubling the extension doubles the force and quadruples the elastic strain energy.

The elastic strain energy is the work done on a material to deform it elastically, stored as potential energy until released. When a force F stretches a material by an extension Δx within its elastic limit, work is done on the material and energy is transferred to the elastic store. Releasing the material returns this energy to a kinetic or mechanical store.

The energy stored equals the area under the force–extension graph between zero and the final extension. For a linear graph (where Hooke’s law applies), this area is a triangle with base Δx and height F. The triangle area gives the relation ${E}_{el}=1/2F\Delta x$. Substituting F = kΔx produces the alternative form ${E}_{el}=1/2k\Delta {x}^{2}$.

Some materials, such as rubber, do not obey Hooke’s law and produce a curved force–extension graph. The triangle formula no longer applies. The energy stored is still the area under the curve, but this area must be estimated by counting squares on a gridded graph or by approximating the curve with several trapeziums and summing their areas.

Examiner Insight“Area under the force–extension graph” is the safe answer for any shape of graph. The half-formula only works for the linear region.

Exam TipFor non-linear graphs, state “estimate the area by counting squares” to earn the method mark.

Worked Example: Elastic Strain Energy Stored in a Stretched Wire

Scenario

A wire of length 2.0 m and cross-sectional area 5.0 × 10⁻⁷ m² is stretched by 5.0 mm. The Young modulus of the wire is 2.0 × 10¹¹ Pa. The wire remains in its linear elastic region. Calculate the elastic strain energy stored.

Step 1: Convert the extension to SI units

$$\Delta x=5.0 \text{mm}=5.0\times {10}^{-3} \text{m}$$

Step 2: Find the force producing this extension

Equation used

$$E=\frac{FL}{A \Delta x}$$

Rearranging for F:

$$F=\frac{EA \Delta x}{L}$$

Substitution:

$$F=\frac{(2.0\times {10}^{11})(5.0\times {10}^{-7})(5.0\times {10}^{-3})}{2.0}$$

$$F=250 \text{N}$$

Step 3: Find the elastic strain energy

Equation used

$${E}_{el}=1/2F\Delta x$$

Substitution:

$${E}_{el}=1/2(250)(5.0\times {10}^{-3})$$

$${E}_{el}=0.625 \text{J}$$

Answer

$${E}_{el}\approx 0.63 \text{J}$$

Interpretation

The wire stores 0.63 J of elastic potential energy. Releasing the load would return this energy as the wire snaps back to its original length.