Cambridge AS & A Level9702

Deformation of solids

Physics 9702 Chapter Notes

What this chapter covers

Deformation of solids - Stress and strainDeformation of solids - Elastic and plastic behaviour
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1. Forces and Deformation

When forces are applied to a solid object, they can change its shape. This change in shape is called deformation. There are two main types of forces to consider. Tensile forces are pairs of forces acting outwards from the centre of an object, tending to stretch it and increase its length. Compressive forces are pairs of forces acting inwards, tending to squash the object and decrease its length. The change in length is called extension (for tensile forces) or compression (for compressive forces). For these forces to cause deformation, they must act in balanced pairs; a single unopposed force would simply cause the object to accelerate.

Key term

Deformation: The change in the shape or size of an object due to an applied force.

Examiner insight

Examiners expect you to correctly identify whether a force is tensile or compressive in a given scenario and use the correct terminology (extension or compression).

Common pitfall

Forgetting that deformation forces act in pairs. For example, a wire being stretched is pulled by a weight at the bottom and an equal and opposite force from the support at the top.

Worked example 12 marks

A concrete column with an original height of 3.000 m supports a heavy load. Its height under the load is measured to be 2.997 m.(a) Identify the type of force acting on the column.(b) Calculate the compression of the column.

  1. 1

    Step 1: Identify the change in shape. The column's height has decreased, so it has been squashed.

  2. 2

    Step 2: Relate the change in shape to the force type. A squashing force is a compressive force. So, the force is compressive.

  3. 3

    Step 3: Calculate the compression. Compression is the difference between the original length and the new length.

  4. 4

    Compression = Original length - New length = 3.000 m - 2.997 m = 0.003 m (or 3 mm).

Recap

  • Tensile forces stretch an object, causing extension.
  • Compressive forces squash an object, causing compression.
  • Deformation refers to any change in shape caused by forces.
  • Forces causing deformation typically act in balanced pairs.

Quick check

  1. A rubber band is stretched between two fingers. Are the forces tensile or compressive?1 mark

2. Hooke's Law and Elasticity

When you apply a force to a spring, it stretches. For many materials, the extension is directly proportional to the applied force, as long as the force isn't too large. This relationship is known as Hooke's Law. We can write this as F = kx, where F is the applied force, x is the extension, and k is a constant called the spring constant. The spring constant (or force constant) is a measure of the stiffness of the spring; a stiffer spring has a larger k. If you plot a graph of force (y-axis) against extension (x-axis), the gradient of the straight-line section is the spring constant k. The point where the graph stops being a straight line is the limit of proportionality. Up to a certain force, called the elastic limit, the material will return to its original shape when the force is removed (elastic deformation). If you stretch it beyond the elastic limit, it will be permanently stretched (plastic deformation).

F = kx

k = F / x

Key term

Elastic Limit: The maximum force that can be applied to an object without causing permanent (plastic) deformation.

Examiner insight

Marks are often awarded for correctly distinguishing between the limit of proportionality (where the F-x graph is no longer linear) and the elastic limit (where permanent deformation begins).

Common pitfall

Confusing the limit of proportionality with the elastic limit. A material can be stretched past its limit of proportionality but still be within its elastic limit, meaning it will return to its original length even though F=kx no longer applies.

Worked example 14 marks

A spring extends by 8.0 cm when a load of 12 N is hung from it.(a) Calculate the spring constant.(b) What force would be needed to produce an extension of 15 cm, assuming the limit of proportionality is not exceeded?

  1. 1

    Step 1 (a): Convert the extension to SI units. x = 8.0 cm = 0.080 m.

  2. 2

    Step 2 (a): Use Hooke's Law to find k. k = F / x = 12 N / 0.080 m = 150 N m⁻¹.

  3. 3

    Step 3 (b): Convert the new extension to SI units. x = 15 cm = 0.15 m.

  4. 4

    Step 4 (b): Rearrange Hooke's Law to find the force. F = kx = 150 N m⁻¹ × 0.15 m = 22.5 N.

Recap

  • Hooke's Law states that force is directly proportional to extension (F ∝ x).
  • The spring constant, k, is the force per unit extension and represents stiffness.
  • Elastic deformation is temporary; the object returns to its original shape.
  • Plastic deformation is permanent; the object does not return to its original shape.
  • The limit of proportionality is where F is no longer proportional to x.
  • The elastic limit is the point beyond which plastic deformation occurs.

Quick check

  1. State Hooke's Law in words.1 mark
  2. A spring has a large spring constant. Is it stiff or flexible?1 mark

3. Stress, Strain, and Young Modulus

To compare the 'stretchiness' of materials without worrying about their shape and size, we use the concepts of stress and strain. Stress (σ) is a measure of how much force is applied per unit of area. It's defined as the force acting perpendicular to the cross-sectional area, A. So, stress σ = F/A. Its unit is the pascal (Pa) or N m⁻². Strain (ε) is a measure of the fractional change in length. It is the extension, x, divided by the original length, L. So, strain ε = x/L. Since it's a ratio of two lengths, strain has no units. For a material obeying Hooke's law, stress is proportional to strain. The constant of proportionality is called the Young modulus, E. It is a measure of the material's stiffness. The Young modulus E = stress / strain. A material with a high Young modulus, like steel, is very stiff and deforms very little under stress.

Stress, σ = F / A

Strain, ε = x / L

Young modulus, E = σ / ε

Key term

Young Modulus (E): A measure of a material's stiffness, defined as the ratio of stress to strain within the limit of proportionality.

Common pitfall

Forgetting to convert units to standard SI units before calculation. Diameter is often given in mm and must be converted to metres, halved to get the radius, and then used in A = πr².

Fun fact

Graphene has an incredibly high Young modulus of about 1000 GPa (1 TPa), making it the strongest material ever tested, yet it is also extremely lightweight.

Worked example 15 marks

A steel wire of length 2.5 m and diameter 0.80 mm is stretched by a force of 100 N. It extends by 2.5 mm. Calculate(a) the stress,(b) the strain, and(c) the Young modulus of the steel.

  1. 1

    Step 1: List all quantities in SI units. F = 100 N. L = 2.5 m. Extension x = 2.5 mm = 2.5 × 10⁻³ m. Diameter d = 0.80 mm = 0.80 × 10⁻³ m.

  2. 2

    Step 2: Calculate the cross-sectional area, A. Radius r = d/2 = 0.40 × 10⁻³ m. A = πr² = π × (0.40 × 10⁻³)² = 5.027 × 10⁻⁷ m².

  3. 3

    Step 3 (a): Calculate the stress. σ = F / A = 100 N / (5.027 × 10⁻⁷ m²) = 1.989 × 10⁸ Pa ≈ 1.99 × 10⁸ Pa.

  4. 4

    Step 4 (b): Calculate the strain. ε = x / L = (2.5 × 10⁻³m) / 2.5 m = 1.0 × 10⁻³.

  5. 5

    Step 5 (c): Calculate the Young modulus. E = σ / ε = (1.989 × 10⁸ Pa) / (1.0 × 10⁻³) = 1.989 × 10¹¹ Pa ≈ 2.0 × 10¹¹ Pa.

Recap

  • Stress is the force applied per unit cross-sectional area (σ = F/A).
  • Strain is the fractional change in length (ε = x/L).
  • The Young modulus is the ratio of stress to strain (E = σ/ε).
  • Stress has units of Pascals (Pa), while strain is dimensionless.
  • The Young modulus is a property of a material, indicating its stiffness.

Quick check

  1. What are the SI units for stress?1 mark
  2. If two wires of the same material but different diameters are subjected to the same tensile force, which will experience greater stress?1 mark

4. Measuring the Young Modulus

To measure the Young modulus of a material, typically in the form of a wire, a specific experiment is performed. A long sample wire (e.g., 2 metres) is hung from a rigid support with a measurement scale (e.g., a vernier scale) attached. A control wire is often hung nearby to compensate for temperature changes. First, the original length (L) of the wire is measured with a tape measure. The diameter(d) is measured at several points along the wire using a micrometer screw gauge and an average is taken; this is used to calculate the cross-sectional area (A = π(d/2)²). Then, known weights are added incrementally to apply a force (F = mg). For each weight, the new position on the vernier scale is read to find the extension (x). A graph of stress (σ = F/A) on the y-axis against strain (ε = x/L) on the x-axis is plotted. The gradient of the straight-line portion of this graph gives the Young modulus, E.

E = gradient of stress-strain graph

E = (F/A) / (x/L) = FL / Ax

Key term

Micrometer Screw Gauge: A precision instrument used to measure small distances, such as the diameter of a wire, with high accuracy.

Examiner insight

In describing the experiment, examiners reward specific details about the instruments used for each measurement (e.g., micrometer for diameter) and a clear explanation of how the data is processed, preferably using a graphical method.

Worked example 13 marks

In an experiment to find the Young modulus of a metal, a student plots a graph of force F (y-axis) against extension x (x-axis). The gradient of the best-fit line is 4.5 × 10³ N m⁻¹. The wire has an original length of 2.2 m and a cross-sectional area of 6.0 × 10⁻⁷ m². Calculate the Young modulus of the metal.

  1. 1

    Step 1: Understand the relationship between the graph and the Young Modulus. The Young Modulus E = FL / Ax.

  2. 2

    Step 2: Rearrange the formula to relate it to the gradient. E = (F/x) × (L/A). The term (F/x) is the gradient of the force-extension graph.

  3. 3

    Step 3: Substitute the given values into the rearranged formula.

  4. 4

    Gradient = F/x = 4.5 × 10³ N m⁻¹.

  5. 5

    L = 2.2 m.

  6. 6

    A = 6.0 × 10⁻⁷ m².

  7. 7

    Step 4: Calculate E. E = (4.5 × 10³) × (2.2 / (6.0 × 10⁻⁷)) = 1.65 × 10¹⁰ Pa.

Recap

  • To measure E, you need to measure original length (L), diameter (d), force (F), and extension (x).
  • Use a tape measure for L, a micrometer for d, weights for F, and a vernier scale for x.
  • A long, thin wire is used to produce a measurable extension for a reasonable force.
  • Plot a graph of stress vs. strain; the gradient is the Young modulus.
  • A control wire is used to account for thermal expansion.

Quick check

  1. Why is a micrometer, rather than a ruler, used to measure the wire's diameter?1 mark
  2. Why should the wire be long and thin?1 mark

5. Elastic Potential Energy

When you stretch or compress a material that obeys Hooke's law, you do work on it. This work is stored as energy within the material, called elastic potential energy or strain energy. As long as the deformation is elastic, this energy can be fully recovered when the force is removed. The work done in stretching a spring is not simply force × extension, because the force is not constant; it increases from zero to its final value. The work done is equal to the area under the force-extension graph. For a material that obeys Hooke's Law, this graph is a triangle. The area of a triangle is ½ × base × height, so the energy stored is E_p = ½Fx. Since F = kx, we can also write the stored energy as E_p = ½k(x)x = ½kx². If a material is stretched beyond its elastic limit, some of the work done goes into permanently rearranging the atoms (plastic deformation), and this energy is not recovered as elastic potential energy; it is mostly dissipated as heat.

E_p = ½ Fx

E_p = ½ kx²

Key term

Strain Energy: The potential energy stored in an object when it is elastically deformed, equal to the work done to deform it.

Common pitfall

Using the formula E_p = ½Fx for a region where the force is not proportional to the extension. In such cases, you must find the area under the graph by counting squares or using calculus.

Fun fact

The Achilles tendon is the human body's most powerful spring. It can store and release elastic energy with about 90% efficiency, which significantly reduces the metabolic cost of running.

Worked example 14 marks

A spring has a spring constant of 250 N m⁻¹. It is stretched from its natural length by 10 cm.(a) Calculate the force required to hold it at this extension.(b) Calculate the elastic potential energy stored in the spring.

  1. 1

    Step 1: List all quantities in SI units. k = 250 N m⁻¹. x = 10 cm = 0.10 m.

  2. 2

    Step 2 (a): Calculate the force using Hooke's Law. F = kx = 250 N m⁻¹ × 0.10 m = 25 N.

  3. 3

    Step 3 (b): Calculate the stored energy using E_p = ½Fx. E_p = 0.5 × 25 N × 0.10 m = 1.25 J.

  4. 4

    Step 4(b) (Alternative): Calculate the stored energy using E_p = ½kx². E_p = 0.5 × 250 N m⁻¹ × (0.10 m)² = 0.5 × 250 × 0.01 = 1.25 J.

Recap

  • Elastic potential energy is the energy stored in a deformed object.
  • The energy stored is equal to the work done to deform the object.
  • Work done is the area under the force-extension graph.
  • For a material obeying Hooke's Law, stored energy E_p = ½Fx or E_p = ½kx².
  • If stretched beyond the elastic limit, not all work done is stored as recoverable potential energy.

Quick check

  1. If you double the extension of a spring, by what factor does the stored energy increase?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define stress and strain, stating the SI unit for each.2 marks
  2. A spring of original length 20.0 cm extends to 28.0 cm when a 5.0 N load is attached. Calculate the spring constant of the spring.3 marks
  3. With the aid of a sketch of a force-extension graph, distinguish between elastic deformation and plastic deformation. Label the elastic limit on your graph.4 marks
  4. A rock climber with a mass of 80 kg is supported by a climbing rope. The rope has a spring constant of 1.2 kN m⁻¹. Calculate the extension of the rope when the climber is hanging stationary. (Use g = 9.81 N kg⁻¹)4 marks
  5. An aluminium wire has a length of 3.0 m and a diameter of 1.0 mm. A load of 80 N is attached to the wire. Given the Young Modulus for aluminium is 7.0 × 10¹⁰ Pa, calculate (a) the stress in the wire, and (b) the extension of the wire.5 marks
  6. For the stretched wire in the previous question, calculate the elastic potential energy stored within it.2 marks
  7. Describe an experiment to determine the Young modulus of a metal in the form of a wire. Your description should include the measurements you would take, the instruments used, and how you would use the data to obtain a value for the Young modulus.6 marks
  8. A lift is supported by a steel cable of length 40 m and cross-sectional area 4.5 × 10⁻⁴ m². The total mass of the lift and its occupants is 1500 kg. Calculate the total extension of the cable. (Young Modulus for steel = 2.0 × 10¹¹ Pa, g = 9.81 N kg⁻¹)4 marks
  9. The force-extension graph for a particular material is a straight line from the origin to the point F=200N, x=2.0mm. The line then curves to the point F=300N, x=5.0mm. Estimate the total work done in stretching the material to an extension of 5.0 mm.4 marks
  10. Two springs, P and Q, are joined in parallel and support a single load. Spring P has a spring constant of 150 N m⁻¹ and spring Q has a spring constant of 250 N m⁻¹. If a load causes the parallel combination to extend by 5.0 cm, what is the total mass of the load? (Use g = 9.81 N kg⁻¹)5 marks

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