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6. Deformation of solids

Written by: Adhulan Rajkamal
Formatted by: Adhulan Rajkamal

Index

6.1 Stress and strain

  • Deformation is caused by tensile or compressive forces:
    • Tensile → Causes extension
    • Compressive → Causes compression
  • Load → The external force applied to a material, causing it to deform; measured in newtons (N)
    • Weight of the load on a spring → tensile force acting on the spring
  • Extension → The increase in length of a material when subjected to a tensile load
    • \( \text{Extension} = \text{Stretched Length} – \text{Original Length} \)
  • Compression → The decrease in length of a material when subjected to a compressive load (pushing force)
  • Hooke’s Law → Provided that the limit of proportionality is not exceeded, the extension of an object is proportional to the applied load
    • The law can be expressed as: \( F \propto e \)
    • Introducing a constant: \( F = k e \)
    • \( k \) is the spring constant, defined as force per unit extension: \( k = \frac{F}{e} \)
  • Limit of proportionality → The maximum point up to which the material obeys Hooke’s Law; beyond this point, deformation becomes non-linear

The Young Modulus

  • Stress (\( \sigma \)) → The force applied per unit cross-sectional area; measured in pascals (Pa): \[ \sigma = \frac{F}{A} \]
    • F → Force applied (N)
    • A → Cross-sectional area (m²)
  • Strain (\( \varepsilon \)) → The ratio of extension (or compression) to the original length of a material: \[ \varepsilon = \frac{e}{L_0} \]
    • e → Extension or compression (m)
    • L₀ → Original length (m)
    • Strain has no units
  • Young Modulus (\( E \)) → A measure of the stiffness of a material, defined as the ratio of stress to strain within the proportionality limit
  • In the region where the changes are proportional, it can be seen that:
    • \(\text{stress} \propto \text{strain}\)
    • Adding a constant to remove the proportionality sign: \[ \text{stress} = E \times \text{strain}\]
    • The constant \(E\) is known as the Young modulus of the material.
    • Rearranging the equation above gives the equation for calculating Young modulus: \[ E = \frac{\sigma}{\varepsilon} = \frac{F/A}{e/L_0} = \frac{F L_0}{e A} \]
      • Measured in pascals (Pa) → Same unit as stress (\(\sigma\))
      • Higher \(E\) indicates a stiffer material

Experiment – Young modulus

  • The following setup can be used to perform a simple experiment to measure the Young modulus of a wire.
    • Set-up: A wire is clamped at one end with a load carrier attached to the other; a paper flag marks the reference point.
    • Measurements:
      • Original length (\( L_0 \)) is measured from the clamp to the reference point.
      • Diameter (\( d \)) is measured using a micrometer, and the cross-sectional area is calculated as: \[ A = \frac{1}{4} \pi d^2 \]
      • Extension (\( e \)) is measured as masses (\( m \)) are added to the carrier.
    • Calculation:
      • Load (\( F \)) is found using: \[ F = mg \]
      • A graph of \( F \) (y-axis) against \( e \) (x-axis) is plotted (Force-extension graph).
      • The Young modulus (\( E \)) is determined from the gradient (\( \frac{F}{e} \)) as: \[ E = \frac{F L_0}{e A} = \frac{F}{e} \times \frac{L_0}{A} = \textit{gradient} \times \frac{L_0}{A} \]
  • Note: Ensure the wire does not exceed the limit of proportionality for valid results.

6.2 Elastic and plastic behaviour

  • Elastic limit → maximum stress or force that a material can withstand and still return to its original shape and size when the load is removed
  • Elastic deformation → Object returns to its original shape and size when the force on it is removed; occurs within elastic limit
  • Plastic deformation → Object does not return to its original shape and size when force is removed; occurs when the material is stressed beyond the elastic limit
  • Area under force-extension graph represents the work done
    • Elastic potential energy of a deformed material = work done to deform the material (within its limit of proportionality)
  • Equations to calculate the elastic potential energy of a deformed material:
    • \( E_p = \frac{1}{2} F e \) → Related to the area under the force-extension graph.
      • \( F \) → Force
      • \( e \) → Extension
    • \( E_p = \frac{1}{2} k e^2 \)
      Where:
      • \( k \) → Spring constant
      • \( e \) → Extension
  • Derivation of the second equation:
    • From the first equation: \( E_p = \frac{1}{2} F e \) and Hooke’s Law: \( F = k e \)
    • Substituting \( F = k e \) into the first equation:
      \( E_p = \frac{1}{2} (k e) e = \frac{1}{2} k e^2 \)
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