Stress and strain

1. Overview

Deformation in solid materials is the result of external forces acting to change the object's physical dimensions. In A-Level Physics, we focus on one-dimensional deformation, where forces act along the longitudinal axis of a material. When a material is subjected to a load, it responds by either stretching (extension) or squashing (compression).

The fundamental relationship governing this behavior is Hooke’s Law, which describes how the extension of a material is directly proportional to the applied force, provided the material remains within its limit of proportionality. While Hooke's Law describes the behavior of a specific sample (like a particular spring or wire), the concepts of stress, strain, and the Young modulus allow us to define the intrinsic mechanical properties of the material itself, independent of its geometry. The Young modulus is a critical measure of a material's stiffness and its ability to resist deformation.

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Key Definitions

• Tensile Force: A pair of equal and opposite forces acting on an object that tend to stretch it. These forces act away from each other along the same line of action.
• Compressive Force: A pair of equal and opposite forces acting on an object that tend to squash or shorten it. These forces act toward each other along the same line of action.
• Extension ($x$ or $\Delta L$): The increase in length of an object when a tensile force is applied. It is calculated as $x = L - L_0$, where $L$ is the new length and $L_0$ is the original length.
• Compression: The decrease in length of an object when a compressive force is applied. It is calculated as $x = L_0 - L$.
• Load ($F$): The external tensile or compressive force applied to a material, measured in newtons (N).
• Limit of Proportionality: The point beyond which the extension of a material is no longer directly proportional to the applied load.
• Hooke’s Law: The principle that the extension of a body is directly proportional to the applied force, provided the limit of proportionality is not exceeded.
• Tensile Stress ($\sigma$): The force per unit cross-sectional area acting normally to the area.
• Tensile Strain ($\epsilon$): The extension per unit original length. It is a ratio and therefore has no units.
• Young Modulus ($E$): The ratio of tensile stress to tensile strain for a material, provided the limit of proportionality is not exceeded. It represents the stiffness of the material.

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Content

3.1 Forces and Deformation in One Dimension

Deformation occurs when the atoms within a solid are displaced from their equilibrium positions by external forces.

• Tensile deformation: Caused by tensile forces. The atoms are pulled further apart.
• Compressive deformation: Caused by compressive forces. The atoms are pushed closer together.
• One-dimensional assumption: In this syllabus, we assume forces act along a single axis (e.g., a wire being pulled vertically). We do not consider shear forces or three-dimensional volume changes.

3.2 Hooke’s Law and the Spring Constant

For many materials, particularly metals and certain polymers, the relationship between load and extension is linear at low forces. This is expressed by Hooke's Law:

$F \propto x$ $F = kx$

Where:

• $F$ is the applied load (N).
• $x$ is the extension (m).
• $k$ is the spring constant (or stiffness constant).

Properties of the Spring Constant ($k$):

• Unit: $\text{N m}^{-1}$.
• Physical Meaning: $k$ represents the force required per unit extension. A higher $k$ value indicates a stiffer object that is harder to deform.
• Graphical Interpretation: On a graph of Force ($F$) vs. Extension ($x$), the gradient of the linear section passing through the origin is equal to $k$.

The Limit of Proportionality: Every material has a limit to its linear behavior. On an $F-x$ graph, this is the point where the line begins to curve. Beyond this point, Hooke's Law no longer applies, and the extension is no longer proportional to the force.

3.3 Tensile Stress and Strain

To compare materials of different sizes, we must "normalize" the force and extension.

Tensile Stress ($\sigma$): Stress accounts for the thickness of the material. A thick wire requires more force than a thin wire of the same material to achieve the same extension. $\sigma = \frac{F}{A}$

• $A$ is the cross-sectional area ($\text{m}^2$). For a circular wire, $A = \frac{\pi d^2}{4}$ or $A = \pi r^2$.
• Unit: pascal (Pa) or $\text{N m}^{-2}$.

Tensile Strain ($\epsilon$): Strain accounts for the original length of the material. A long wire will extend more than a short wire under the same force. $\epsilon = \frac{x}{L}$

• $L$ is the original length (m).
• $x$ is the extension (m).
• Unit: Dimensionless (often expressed as a decimal or a percentage).

3.4 The Young Modulus ($E$)

The Young Modulus is a fundamental property of a material. Unlike the spring constant $k$, which depends on the shape and size of the sample, the Young Modulus $E$ is the same for any sample of a given material (e.g., all steel samples have roughly the same $E$).

$E = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\epsilon}$

Derivation of the Combined Formula: Substituting the definitions of stress and strain into the Young Modulus equation:

1. $E = \frac{F/A}{x/L}$
2. $E = \frac{F}{A} \times \frac{L}{x}$
3. $E = \frac{FL}{Ax}$

Graphical Interpretation: On a Stress-Strain graph, the gradient of the linear region (from the origin to the limit of proportionality) is equal to the Young Modulus.

3.5 Experiment to Determine the Young Modulus

The syllabus requires you to describe a practical method to find $E$ for a metal wire.

Apparatus:

• Wire: A long (at least $2.0 \text{ m}$), thin wire.
• Support: A rigid fixed beam or a second reference wire (Searle’s apparatus) to eliminate errors from support sagging.
• Micrometer Screw Gauge: To measure the diameter $d$.
• Metre Rule: To measure the original length $L$.
• Weights and Hanger: To apply a known load $F = mg$.
• Vernier Scale or Traveling Microscope: To measure the small extension $x$ accurately.

Procedure:

1. Measure Diameter: Use the micrometer to measure the diameter $d$ at several points along the wire and at different orientations (rotating the wire $90^\circ$). Calculate the average $d$ and then the area $A = \frac{\pi d^2}{4}$.
2. Measure Original Length: Use the metre rule to measure the length $L$ from the fixed support to the point where the extension is being measured (the marker).
3. Apply Initial Load: Apply a small "pre-load" to straighten the wire and remove any kinks. Record this as the zero position.
4. Incremental Loading: Add weights (e.g., $0.5 \text{ kg}$ increments). For each mass, record the new position of the marker.
5. Calculate Extension: For each load, $x = \text{Current Position} - \text{Initial Position}$.
6. Unloading: Gradually remove weights to ensure the wire returns to its original length, confirming the limit of proportionality was not exceeded.

Data Analysis:

1. Plot a graph of Load $F$ (y-axis) against Extension $x$ (x-axis).
2. Draw a line of best fit through the linear portion.
3. Calculate the gradient ($m = \frac{\Delta F}{\Delta x}$).
4. Calculate $E$ using the relationship: $E = \text{gradient} \times \frac{L}{A}$.

Minimizing Uncertainties:

• Why a long wire? To make the extension $x$ as large as possible, reducing the percentage uncertainty in the measurement of $x$.
• Why a thin wire? To produce a larger extension for a given force (since $x = \frac{FL}{AE}$).
• Why multiple diameter readings? To account for the fact that the wire may not be perfectly cylindrical (non-uniform cross-section).

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Worked Example 1 — Calculating Young Modulus from Experimental Data

A copper wire has an original length of $2.40 \text{ m}$ and a diameter of $0.80 \text{ mm}$. When a mass of $5.0 \text{ kg}$ is hung from the wire, it extends by $1.2 \text{ mm}$. Calculate the Young modulus of copper.

Step 1: Convert all values to SI units.

• $L = 2.40 \text{ m}$
• $d = 0.80 \times 10^{-3} \text{ m}$
• $m = 5.0 \text{ kg} \implies F = mg = 5.0 \times 9.81 = 49.05 \text{ N}$
• $x = 1.2 \times 10^{-3} \text{ m}$

Step 2: Calculate the cross-sectional area $A$. $$A = \frac{\pi d^2}{4} = \frac{\pi \times (0.80 \times 10^{-3})^2}{4}$$ $$A = 5.027 \times 10^{-7} \text{ m}^2$$

Step 3: Use the combined Young Modulus formula. $$E = \frac{FL}{Ax}$$ $$E = \frac{49.05 \times 2.40}{(5.027 \times 10^{-7}) \times (1.2 \times 10^{-3})}$$ $$E = \frac{117.72}{6.032 \times 10^{-10}}$$ $$E = 1.95 \times 10^{11} \text{ Pa}$$

Answer: $1.95 \times 10^{11} \text{ Pa}$ (or $195 \text{ GPa}$).

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Worked Example 2 — Comparing Two Wires

Wire A and Wire B are made of the same material. Wire B has twice the length and half the diameter of Wire A. If both wires are subjected to the same tensile force $F$, calculate the ratio: $\frac{\text{extension of Wire B}}{\text{extension of Wire A}}$.

Step 1: Express extension $x$ in terms of other variables. From $E = \frac{FL}{Ax}$, we get $x = \frac{FL}{AE}$.

Step 2: Set up the ratio. Since the material is the same, $E_A = E_B$. Since the force is the same, $F_A = F_B$. $$\frac{x_B}{x_A} = \frac{F L_B / (A_B E)}{F L_A / (A_A E)} = \frac{L_B}{L_A} \times \frac{A_A}{A_B}$$

Step 3: Substitute the relationships between dimensions.

• $L_B = 2 L_A$
• $d_B = 0.5 d_A \implies A_B = \frac{\pi (0.5 d_A)^2}{4} = 0.25 A_A$

Step 4: Calculate the final ratio. $$\frac{x_B}{x_A} = \frac{2 L_A}{L_A} \times \frac{A_A}{0.25 A_A} = 2 \times \frac{1}{0.25} = 2 \times 4 = 8$$

Answer: The extension of Wire B is 8 times the extension of Wire A.

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Key Equations

Equation	Symbols	SI Units	Status
$F = kx$	$F$: Force, $k$: Spring constant, $x$: Extension	$F$ (N), $k$ ($\text{N m}^{-1}$), $x$ (m)	Memorise
$\sigma = \frac{F}{A}$	$\sigma$: Stress, $F$: Force, $A$: Area	$\sigma$ (Pa), $F$ (N), $A$ ($\text{m}^2$)	Memorise
$\epsilon = \frac{x}{L}$	$\epsilon$: Strain, $x$: Extension, $L$: Original length	$\epsilon$ (None), $x$ (m), $L$ (m)	Memorise
$E = \frac{\sigma}{\epsilon}$	$E$: Young Modulus, $\sigma$: Stress, $\epsilon$: Strain	$E$ (Pa)	Data Sheet
$E = \frac{FL}{Ax}$	Combined Young Modulus formula	$E$ (Pa)	Memorise
$A = \frac{\pi d^2}{4}$	Cross-sectional area of a wire	$A$ ($\text{m}^2$), $d$ (m)	Memorise

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Common Mistakes to Avoid

• ❌ Wrong: Using the diameter directly in the area formula as $A = \pi d^2$.
• ✅ Right: Use $A = \frac{\pi d^2}{4}$ or calculate the radius first ($r = d/2$) and use $A = \pi r^2$.
• ❌ Wrong: Forgetting to convert area units correctly. For example, thinking $1 \text{ mm}^2 = 10^{-3} \text{ m}^2$.
• ✅ Right: $1 \text{ mm}^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$. Similarly, $1 \text{ cm}^2 = 10^{-4} \text{ m}^2$.
• ❌ Wrong: Using the total length of the wire instead of the extension in Hooke's Law or the Young Modulus formula.
• ✅ Right: Always subtract the original length from the new length to find $x$ ($x = L_{new} - L_{original}$).
• ❌ Wrong: Giving the Young Modulus the unit $\text{N m}^{-1}$.
• ✅ Right: The Young Modulus is a measure of stress/strain. Since strain has no units, $E$ has the same units as stress: Pascals (Pa) or $\text{N m}^{-2}$.
• ❌ Wrong: Assuming the gradient of any force-extension graph is the Young Modulus.
• ✅ Right: The gradient of a Force-Extension graph is the spring constant $k$. The gradient of a Stress-Strain graph is the Young Modulus $E$.

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Exam Tips

1. Precision in Definitions: When defining Hooke's Law, you must mention the "limit of proportionality." Simply saying "force is proportional to extension" will often lose the mark.
2. The "Why" of the Experiment: Exam questions often ask why a long, thin wire is used. Your answer should focus on reducing percentage uncertainty by making the extension $x$ larger and more measurable.
3. Significant Figures: In 9702 papers, pay close attention to the precision of the data provided. If the diameter is given as $0.50 \text{ mm}$ (2 s.f.), your final Young Modulus should typically be rounded to 2 or 3 s.f.
4. Graph Analysis: If a graph does not pass through the origin, there may be a "systematic error" (like a pre-load not being accounted for). However, the gradient will still give you the correct value for $k$ or $E$.
5. Unit Consistency: Before starting any calculation, list your variables and convert them all to SI base units (m, kg, s, N). This prevents the most common source of error in Stress/Strain questions.
6. Stress-Strain vs. Force-Extension: Always check the axes of the graph provided.
   • $F-x$ graph: Area under the graph = Work done (Elastic Potential Energy).
   • $\sigma-\epsilon$ graph: Area under the graph = Energy stored per unit volume. (Note: This is covered in topic 6.2, but it's useful to distinguish them now).