Hookes law and deformation

Hooke’s Law and Stiffness

Key Equations

Hooke’s law:

$$\Delta F=k\Delta x$$

Variables:

• ΔF = applied force on the object (N)
• k = stiffness (also called spring constant or force constant) (N m⁻¹)
• Δx = extension or compression from the natural length (m)

SI unit: force in newtons (N)

Rearrangements:

$$k=\frac{\Delta F}{\Delta x}$$

$$\Delta x=\frac{\Delta F}{k}$$

ProportionalityForce is directly proportional to extension when stiffness is constant. Doubling the force doubles the extension.

Hooke’s law states that the force applied to an elastic object is directly proportional to its extension, provided the limit of proportionality is not exceeded.

The stiffness k measures how much force is required to produce a unit extension. A stiffer object has a larger value of k. A spring with k = 200 N m⁻¹ requires 200 N to extend it by one metre.

The law applies to both stretching (tensile) and squashing (compressive) deformations. Δx is positive for extension and the same equation describes compression by the same amount.

Hooke’s law is a linear relationship and only holds up to the limit of proportionality. Beyond this point, extension increases more rapidly than the applied force predicts.

Worked Example: Stiffness of a Loaded Spring

Scenario

A spring has a natural length of 12.0 cm. When a 250 g mass is hung from it, the new length is 18.5 cm. Calculate the stiffness of the spring. Take g = 9.81 N kg⁻¹.

Step 1: Convert the given values to SI units.

Mass conversion:

$$m=\frac{250}{1000}=0.250\text{ kg}$$

Extension conversion:

$$\Delta x=\frac{18.5-12.0}{100}=0.0650\text{ m}$$

Step 2: Find the weight of the mass, which is the force applied to the spring.

$$\Delta F=mg$$

$$\Delta F=0.250\times 9.81$$

$$\Delta F=2.4525\text{ N}$$

Step 3: Rearrange Hooke’s law for stiffness.

Rearranging for k:

$$k=\frac{\Delta F}{\Delta x}$$

$$k=\frac{2.4525}{0.0650}$$

$$k=37.7{\text{ N m}}^{-1}\text{ (3 s.f.)}$$

Interpretation

A stiffness of about 38 N m⁻¹ means roughly 38 N is needed to extend this spring by one metre.

Stress, Strain and the Young Modulus

Key Equations

Tensile or compressive stress:

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

Variables:

• σ = stress (Pa or N m⁻²)
• F = force applied perpendicular to the cross-section (N)
• A = cross-sectional area (m²)

SI unit: pascal (Pa)

Rearrangements:

$$F=\sigma A$$

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

ProportionalityFor a fixed force, stress is inversely proportional to cross-sectional area. Doubling the area halves the stress.

Tensile or compressive strain:

$$\epsilon =\frac{\Delta L}{{L}_{0}}$$

Variables:

• ε = strain (no unit, dimensionless)
• ΔL = change in length (m)
• L₀ = original length (m)

SI unit: dimensionless ratio (often given as a decimal or percentage)

Rearrangements:

$$\Delta L=\epsilon {L}_{0}$$

$${L}_{0}=\frac{\Delta L}{\epsilon }$$

ProportionalityFor a given extension, strain is inversely proportional to the original length.

Young modulus:

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

Variables:

• E = Young modulus (Pa)
• σ = stress (Pa)
• ε = strain (dimensionless)

SI unit: pascal (Pa)

Rearrangements:

$$E=\frac{F{L}_{0}}{A\Delta L}$$

$$\Delta L=\frac{F{L}_{0}}{AE}$$

ProportionalityFor a given material, stress is directly proportional to strain in the linear region.

Stress, strain and the Young modulus describe the deformation of materials in a way that does not depend on the size or shape of the sample.

Stress is the force applied per unit cross-sectional area. It tells us how concentrated the load is on the material. A thin wire experiences greater stress than a thick wire under the same force.

Strain is the fractional change in length of the material. Because it is a ratio of two lengths, strain has no units. A strain of 0.002 means the length has increased by 0.2% of its original value.

The Young modulus measures how stiff a material is, independent of the dimensions of the test sample. A material with a large Young modulus is hard to stretch. Steel has a Young modulus of about 2 × 10¹¹ Pa, while rubber has a value around 10⁷ Pa.

The Young modulus is found from the gradient of the linear region of a stress-strain graph. It is a property of the material itself, not the object.

Examiner InsightStiffness (k) describes an object, while the Young modulus (E) describes the material. Two wires made of the same metal share the same Young modulus but can have very different stiffnesses if their lengths and thicknesses differ.

Exam TipState clearly whether a question is about an object or a material.

Worked Example: Young Modulus of a Metal Wire

Scenario

A metal wire has an original length of 2.50 m and a cross-sectional area of 0.40 mm². When a load of 85 N is applied, the wire extends by 3.2 mm. Calculate the Young modulus of the metal.

Step 1: Convert all given values to SI units.

Area conversion:

$$A=\frac{0.40}{{10}^{6}}=4.0\times {10}^{-7}{\text{ m}}^{2}$$

Extension conversion:

$$\Delta L=\frac{3.2}{1000}=3.2\times {10}^{-3}\text{ m}$$

Step 2: Calculate the stress.

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

$$\sigma =\frac{85}{4.0\times {10}^{-7}}$$

$$\sigma =2.125\times {10}^{8}\text{ Pa}$$

Step 3: Calculate the strain.

$$\epsilon =\frac{\Delta L}{{L}_{0}}$$

$$\epsilon =\frac{3.2\times {10}^{-3}}{2.50}$$

$$\epsilon =1.28\times {10}^{-3}$$

Step 4: Calculate the Young modulus.

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

$$E=\frac{2.125\times {10}^{8}}{1.28\times {10}^{-3}}$$

$$E=1.66\times {10}^{11}\text{ Pa (3 s.f.)}$$

Interpretation

The value is close to the Young modulus of steel, suggesting the wire is made of a steel-like alloy.

Force-Extension Graphs and Deformation Terms

Force-extension graphs and key deformation terms describe how an object responds to increasing load and reveal where its behaviour changes from elastic to plastic.

A force-extension graph plots the applied force on the y-axis against the extension on the x-axis. For a force-compression graph, the x-axis represents compression instead. The gradient of the linear region equals the stiffness k.

The graph begins as a straight line through the origin, showing that force is directly proportional to extension. This is the region where Hooke’s law applies. Beyond a certain point, the line curves and the relationship is no longer linear.

The key deformation terms are defined as follows:

The limit of proportionality usually occurs slightly before the elastic limit. Between these two points, the material is still elastic but no longer obeys Hooke’s law.

If the load is removed before the elastic limit, the object returns along the same line to the origin. If the load is removed after the elastic limit, the object follows a parallel line back, leaving a permanent extension at zero force.

MisconceptionThe limit of proportionality and the elastic limit are often confused. Hooke’s law fails at the limit of proportionality, but the material can still be elastic beyond this point until the elastic limit is reached.

Exam TipName the specific point being tested rather than vague phrases like “where Hooke’s law stops.”

Stress-Strain Graphs and Breaking Stress

Stress-strain graphs and the concept of breaking stress describe how a material — rather than a specific object — responds to load up to the point of failure.

A stress-strain graph has the same general shape as a force-extension graph, but the axes show stress (Pa) on the y-axis and strain (dimensionless) on the x-axis. Because stress and strain do not depend on the dimensions of the sample, the graph describes the material itself.

The gradient of the linear region equals the Young modulus of the material. The same key features appear: the limit of proportionality, the elastic limit, the yield point, and the regions of elastic and plastic deformation.

The breaking stress (also called the ultimate tensile stress) is the maximum stress that a material can withstand before it fractures. It corresponds to the highest point on the stress-strain graph. A material with a high breaking stress can support a large load per unit area before failing.

For brittle materials such as glass or cast iron, the graph is almost a straight line that ends abruptly at the breaking point, with no significant plastic region. For ductile materials such as copper or mild steel, there is a long plastic region after the yield point before the material breaks.

Worked Example: Predicting Wire Failure

Scenario

A steel wire has a cross-sectional area of 0.50 mm² and a breaking stress of 5.0 × 10⁸ Pa. A student wants to use it to support a load of 300 N. Determine whether the wire will fail.

Step 1: Convert the cross-sectional area to SI units.

$$A=\frac{0.50}{{10}^{6}}=5.0\times {10}^{-7}{\text{ m}}^{2}$$

Step 2: Calculate the stress in the wire under the load.

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

$$\sigma =\frac{300}{5.0\times {10}^{-7}}$$

$$\sigma =6.0\times {10}^{8}\text{ Pa}$$

Step 3: Compare with the breaking stress.

The applied stress is 6.0 × 10⁸ Pa. The breaking stress is 5.0 × 10⁸ Pa. The applied stress exceeds the breaking stress.

Interpretation

The wire will break under this load because the stress is greater than the material can withstand.