Mechanical Properties of Solids • Topic 1 of 3

Stress, Strain & Hooke's Law

Push or pull on a rigid body and, strictly, no body is perfectly rigid — it changes shape or size, however slightly. When an external force acts on a body and tends to change its length, volume or shape, we call it a deforming force. The body resists this change because the molecules, pulled away from their equilibrium positions, set up internal restoring forces. The property by which a body regains its original shape and size once the deforming force is removed is called elasticity.

Stress is the internal restoring force set up per unit area of cross-section. In magnitude it equals the applied force per unit area:

  • $\text{Stress}=\frac{F}{A}$, SI unit $\text{N/m}^2$ or pascal $(\text{Pa})$; dimensional formula $[\text{ML}^{-1}\text{T}^{-2}]$.
  • Tensile stress — force stretches the body (longitudinal, length increases).
  • Compressive stress — force squeezes the body (longitudinal, length decreases).
  • Shear (tangential) stress — force acts parallel to a face, sliding one layer over another.
  • Hydraulic (volume) stress — a fluid presses equally from all sides, changing the volume.

Strain is the measure of deformation — the fractional change produced. Being a ratio of two like quantities, strain is a dimensionless number with no unit:

  • Longitudinal strain $=\frac{\Delta L}{L}$ (change in length / original length).
  • Volumetric strain $=\frac{\Delta V}{V}$ (change in volume / original volume).
  • Shear strain $=\frac{\Delta x}{L}=\tan\theta\approx\theta$ (relative sideways shift / height) for small angles.

Hooke's law is the foundation of this whole chapter. It states that, within the elastic limit, the stress developed in a body is directly proportional to the strain produced:

  • $\text{Stress}\propto\text{Strain}$, so $\frac{\text{Stress}}{\text{Strain}}=\text{constant}=E$, the modulus of elasticity.
  • The law only holds for small deformations — beyond the elastic limit, stress and strain are no longer proportional.

The full story is told by the stress–strain curve obtained by stretching a wire to breaking point. The straight portion from O obeys Hooke's law up to the proportional limit (A). Just beyond lies the elastic limit (B), the maximum stress from which the body still returns to its original shape. Past the yield point (C) the material deforms plastically — strain increases with little extra stress, and the deformation becomes permanent. The curve rises to the maximum ultimate tensile strength and finally reaches the fracture point (D) where the wire snaps. A large gap between the yield and fracture points means the material is ductile; a small gap means it is brittle.

Stress-strain curve with proportional limit, elastic limit, yield point and fractureStrainStressA prop. limitB elastic limitC yieldultimateD fracture
Solved Examples
1
Worked Example
A steel wire of cross-sectional area $2\times10^{-6}\ \text{m}^2$ is stretched by a force of 100 N. Calculate the tensile stress in the wire.
Solution
  1. Step 1: Use $\text{Stress}=\frac{F}{A}$.
  2. Step 2: Substitute: $\text{Stress}=\frac{100}{2\times10^{-6}}$.
  3. Step 3: Compute: $\text{Stress}=5\times10^{7}\ \text{N/m}^2$.

Answer: $\text{Stress}=5\times10^{7}\ \text{N/m}^2=5\times10^{7}\ \text{Pa}$.

2
Worked Example
A wire of original length 2 m is stretched so that its length increases by 1 mm. Find the longitudinal strain produced.
Solution
  1. Step 1: Use $\text{Strain}=\frac{\Delta L}{L}$ with $\Delta L=1\ \text{mm}=1\times10^{-3}\ \text{m}$ and $L=2\ \text{m}$.
  2. Step 2: Substitute: $\text{Strain}=\frac{1\times10^{-3}}{2}$.
  3. Step 3: Compute: $\text{Strain}=5\times10^{-4}$ (no unit).

Answer: $\text{Strain}=5\times10^{-4}$ (dimensionless).

3
Worked Example
A force of 50 N is applied tangentially to the top face of a cube of side 0.1 m fixed at its base. Find the shear stress on the top face.
Solution
  1. Step 1: Shear stress $=\frac{F}{A}$ where $A$ is the area of the face on which the tangential force acts.
  2. Step 2: Area $A=(0.1)^2=0.01\ \text{m}^2$; substitute: $\text{Shear stress}=\frac{50}{0.01}$.
  3. Step 3: Compute: $\text{Shear stress}=5000\ \text{N/m}^2$.

Answer: $\text{Shear stress}=5\times10^{3}\ \text{N/m}^2$.

4
Worked Example
The top face of a 0.5 m tall block is displaced sideways by 2 mm relative to the fixed bottom face. Find the shear strain.
Solution
  1. Step 1: Shear strain $=\frac{\Delta x}{L}$ (sideways shift / height), with $\Delta x=2\times10^{-3}\ \text{m}$ and $L=0.5\ \text{m}$.
  2. Step 2: Substitute: $\text{Shear strain}=\frac{2\times10^{-3}}{0.5}$.
  3. Step 3: Compute: $\text{Shear strain}=4\times10^{-3}\ \text{rad}$.

Answer: $\text{Shear strain}=4\times10^{-3}$ (the angle $\theta$ in radians).

5
Worked Example
Within the elastic limit a wire shows a stress of $2\times10^{8}\ \text{N/m}^2$ when the strain is $1\times10^{-3}$. Find the modulus of elasticity of the wire material.
Solution
  1. Step 1: By Hooke's law the modulus $E=\frac{\text{Stress}}{\text{Strain}}$.
  2. Step 2: Substitute: $E=\frac{2\times10^{8}}{1\times10^{-3}}$.
  3. Step 3: Compute: $E=2\times10^{11}\ \text{N/m}^2$.

Answer: $E=2\times10^{11}\ \text{N/m}^2$ (this is the order of $Y$ for steel).

6
Worked Example
Two wires P and Q of the same material have the same length but the radius of P is twice that of Q. If equal forces are applied to each, compare the tensile stresses in P and Q.
Solution
  1. Step 1: Stress $=\frac{F}{A}=\frac{F}{\pi r^2}$, so for equal $F$ the stress $\propto\frac{1}{r^2}$.
  2. Step 2: With $r_P=2r_Q$, $\frac{\text{Stress}_P}{\text{Stress}_Q}=\frac{r_Q^2}{r_P^2}=\frac{r_Q^2}{(2r_Q)^2}=\frac{1}{4}$.
  3. Step 3: Therefore the stress in P is one-fourth of that in Q.

Answer: $\text{Stress}_P:\text{Stress}_Q=1:4$.

Key Points

  • Stress is the internal restoring force per unit area, $\text{Stress}=\frac{F}{A}$, measured in $\text{N/m}^2$ (Pa); it may be tensile, compressive, shear or hydraulic.
  • Strain is the fractional deformation and is dimensionless: longitudinal $\frac{\Delta L}{L}$, volumetric $\frac{\Delta V}{V}$, shear $\frac{\Delta x}{L}$.
  • Hooke's law: within the elastic limit, $\text{Stress}\propto\text{Strain}$, so $\frac{\text{Stress}}{\text{Strain}}=E$ (modulus of elasticity).
  • On the stress–strain curve: proportional limit (Hooke's law holds), elastic limit (returns to shape), yield point (permanent set), then fracture.
  • A wide yield-to-fracture region means a ductile material; a narrow one means a brittle material.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The SI unit of stress is the same as that of:
Explanation: Stress is force per unit area, $\text{N/m}^2$, the same unit as pressure (pascal).
Q2.Strain is:
Explanation: Strain is a ratio of two like quantities (e.g. $\frac{\Delta L}{L}$), so it has no unit and no dimensions.
Q3.Hooke's law is valid only:
Explanation: Stress is proportional to strain only up to the elastic (proportional) limit; beyond it the law fails.
Q4.When a force is applied tangentially (parallel) to a face of a body, the stress produced is:
Explanation: A tangential force sliding one layer over another sets up shear (tangential) stress.
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