Mechanical Properties of Solids • Topic 2 of 3

Moduli of Elasticity

Hooke's law tells us that stress divided by strain is a constant — the modulus of elasticity. Because a solid can be deformed in three distinct ways (changing length, volume or shape), there are three corresponding moduli. Each measures how stiff a material is against that particular kind of deformation: a large modulus means the material is hard to deform.

Young's modulus ($Y$) measures resistance to a change in length. It is the ratio of longitudinal (tensile or compressive) stress to longitudinal strain:

  • $Y=\frac{\text{Longitudinal stress}}{\text{Longitudinal strain}}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\,\Delta L}$.
  • SI unit $\text{N/m}^2$ (Pa); for steel $Y\approx2\times10^{11}\ \text{N/m}^2$, larger than for copper or aluminium, so steel is stiffer.
  • The extension of a loaded wire follows $\Delta L=\frac{FL}{AY}$ — a longer or thinner wire stretches more for the same load.

Bulk modulus ($B$) measures resistance to a change in volume when a body is squeezed equally from all sides by a pressure $\Delta P$:

  • $B=\frac{\text{Volume (hydraulic) stress}}{\text{Volume strain}}=-\frac{\Delta P}{\Delta V/V}$.
  • The negative sign shows the volume decreases as pressure increases, keeping $B$ positive.
  • Solids have a large bulk modulus, liquids smaller, gases the smallest — which is why gases are so easily compressed.

Compressibility ($k$) is simply the reciprocal of the bulk modulus, $k=\frac{1}{B}$. A highly compressible substance (a gas) has a small bulk modulus and a large compressibility.

Shear (rigidity) modulus ($\eta$ or $G$) measures resistance to a change in shape. It is the ratio of shear stress to shear strain:

  • $\eta=\frac{\text{Shear stress}}{\text{Shear strain}}=\frac{F/A}{\theta}=\frac{F}{A\theta}$, where $\theta$ is the angle of shear in radians.
  • Only solids possess rigidity (a non-zero shear modulus); fluids cannot sustain a shear stress at rest, so they flow.

Poisson's ratio ($\sigma$) captures a familiar fact: stretch a rubber band lengthwise and it gets thinner. The longitudinal extension is accompanied by a lateral (sideways) contraction. Poisson's ratio is defined as:

  • $\sigma=-\frac{\text{Lateral strain}}{\text{Longitudinal strain}}=\frac{\Delta d/d}{\Delta L/L}$ (taken as a positive number).
  • It is a pure number with no unit; for most metals $\sigma$ lies between about $0.2$ and $0.4$. Theoretically $\sigma$ can range from $-1$ to $+0.5$.
Wire of length L under a tensile load showing extension delta L for Young's modulusFLdelta LY = F L / (A delta L)
Solved Examples
1
Worked Example
A steel wire of length 2 m and cross-sectional area $1\times10^{-6}\ \text{m}^2$ stretches by 0.4 mm under a load of 40 N. Find Young's modulus of the wire.
Solution
  1. Step 1: Use $Y=\frac{FL}{A\,\Delta L}$ with $\Delta L=0.4\ \text{mm}=4\times10^{-4}\ \text{m}$.
  2. Step 2: Substitute: $Y=\frac{40\times2}{1\times10^{-6}\times4\times10^{-4}}=\frac{80}{4\times10^{-10}}$.
  3. Step 3: Compute: $Y=2\times10^{11}\ \text{N/m}^2$.

Answer: $Y=2\times10^{11}\ \text{N/m}^2$.

2
Worked Example
A load of 5 kg is attached to a wire of length 1 m and area $2\times10^{-7}\ \text{m}^2$ with $Y=2\times10^{11}\ \text{N/m}^2$. Find the extension produced. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: Use $\Delta L=\frac{FL}{AY}$ with $F=mg=5\times10=50\ \text{N}$.
  2. Step 2: Substitute: $\Delta L=\frac{50\times1}{2\times10^{-7}\times2\times10^{11}}=\frac{50}{4\times10^{4}}$.
  3. Step 3: Compute: $\Delta L=1.25\times10^{-3}\ \text{m}=1.25\ \text{mm}$.

Answer: $\Delta L=1.25\ \text{mm}$.

3
Worked Example
A pressure of $2\times10^{6}\ \text{N/m}^2$ is applied to a liquid, reducing its volume by $0.1\%$. Find the bulk modulus of the liquid.
Solution
  1. Step 1: Use $B=\frac{\Delta P}{\Delta V/V}$ (magnitude). Here $\frac{\Delta V}{V}=0.1\%=1\times10^{-3}$.
  2. Step 2: Substitute: $B=\frac{2\times10^{6}}{1\times10^{-3}}$.
  3. Step 3: Compute: $B=2\times10^{9}\ \text{N/m}^2$.

Answer: $B=2\times10^{9}\ \text{N/m}^2$.

4
Worked Example
The bulk modulus of water is $2.2\times10^{9}\ \text{N/m}^2$. Find its compressibility.
Solution
  1. Step 1: Compressibility $k=\frac{1}{B}$.
  2. Step 2: Substitute: $k=\frac{1}{2.2\times10^{9}}$.
  3. Step 3: Compute: $k\approx4.55\times10^{-10}\ \text{m}^2/\text{N}$.

Answer: $k\approx4.5\times10^{-10}\ \text{m}^2/\text{N}$ (or $\text{Pa}^{-1}$).

5
Worked Example
A tangential force of 200 N is applied to the top face (area $0.02\ \text{m}^2$) of a block, producing a shear strain of $0.001\ \text{rad}$. Find the shear modulus of the material.
Solution
  1. Step 1: Use $\eta=\frac{F/A}{\theta}$.
  2. Step 2: Shear stress $=\frac{200}{0.02}=1\times10^{4}\ \text{N/m}^2$; then $\eta=\frac{1\times10^{4}}{0.001}$.
  3. Step 3: Compute: $\eta=1\times10^{7}\ \text{N/m}^2$.

Answer: $\eta=1\times10^{7}\ \text{N/m}^2$.

6
Worked Example
A wire of length 1 m and diameter 1 mm is stretched by 1 mm. Its diameter decreases by $0.0003$ mm. Find Poisson's ratio for the material.
Solution
  1. Step 1: Longitudinal strain $=\frac{\Delta L}{L}=\frac{1\times10^{-3}}{1}=1\times10^{-3}$.
  2. Step 2: Lateral strain $=\frac{\Delta d}{d}=\frac{0.0003}{1}=3\times10^{-4}$.
  3. Step 3: $\sigma=\frac{\text{lateral strain}}{\text{longitudinal strain}}=\frac{3\times10^{-4}}{1\times10^{-3}}=0.3$.

Answer: $\sigma=0.3$ (no unit).

Key Points

  • Young's modulus: $Y=\frac{FL}{A\,\Delta L}$ measures resistance to change in length; steel $Y\approx2\times10^{11}\ \text{N/m}^2$.
  • Bulk modulus: $B=-\frac{\Delta P}{\Delta V/V}$ measures resistance to change in volume; the negative sign keeps $B$ positive.
  • Compressibility is the reciprocal of bulk modulus, $k=\frac{1}{B}$.
  • Shear (rigidity) modulus: $\eta=\frac{F/A}{\theta}$ measures resistance to change in shape; only solids have rigidity.
  • Poisson's ratio $\sigma=-\frac{\text{lateral strain}}{\text{longitudinal strain}}$ is a dimensionless number, about $0.2$–$0.4$ for most metals.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Young's modulus is the ratio of:
Explanation: $Y=\frac{\text{longitudinal stress}}{\text{longitudinal strain}}=\frac{FL}{A\Delta L}$.
Q2.The bulk modulus relates to a change in:
Explanation: $B=-\frac{\Delta P}{\Delta V/V}$ measures resistance to a change in volume under uniform pressure.
Q3.Which type of modulus is possessed only by solids and not by fluids at rest?
Explanation: Fluids cannot sustain a shear stress at rest, so only solids have a non-zero shear modulus.
Q4.Poisson's ratio is:
Explanation: It is the ratio of two strains, hence dimensionless; for metals it is roughly $0.2$–$0.4$.
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