Mechanical Properties of Solids • Topic 3 of 3

Elastic Behaviour & Applications

When you stretch a wire you do work against the internal restoring forces, and that work is stored inside the deformed body as elastic potential energy. Release the load and the energy is given back as the body springs to its original shape — exactly what makes a catapult, a bow or a trampoline work.

For a wire stretched by an amount $\Delta L$, the restoring force grows from $0$ to its final value $F$, so the average force is $\frac{F}{2}$. The work done — and hence the energy stored — is:

  • $U=\frac{1}{2}\times F\times\Delta L=\frac{1}{2}\times\text{Stress}\times\text{Strain}\times\text{Volume}$.
  • The elastic energy per unit volume (energy density) is $u=\frac{1}{2}\times\text{Stress}\times\text{Strain}=\frac{1}{2}\times\frac{(\text{Stress})^2}{Y}$.
  • This corresponds to the area under the stress–strain graph within the elastic region.

Real materials are not perfectly elastic. Two effects matter in engineering:

  • Elastic after-effect: some materials do not return to their original shape instantly when the load is removed — there is a small time delay before full recovery. Quartz and phosphor-bronze show almost no after-effect, which is why they are used in sensitive instruments.
  • Elastic fatigue: a material subjected to repeated cycles of stress gradually becomes weaker and may break at a stress far below its normal breaking value. This is why bridges, aircraft parts and railway tracks are inspected and replaced before failure, and why a wire bent back and forth many times eventually snaps.

Applications of elasticity turn these ideas into safe, economical design:

  • Cables and ropes: the maximum safe load on a rope is fixed by the breaking stress of its material. To lift heavy loads, engineers use a rope made of many thin strands twisted together (a stranded rope) rather than one thick wire, since this gives flexibility with the required cross-sectional area, and they apply a generous factor of safety.
  • Beams in bridges and buildings: a horizontal beam supported at its ends sags by a depression $\delta=\frac{W L^3}{4 b d^3 Y}$ under a central load $W$. The depression falls sharply (as $\frac{1}{d^3}$) when the depth $d$ is increased, so beams are made deep rather than wide, and an I-shaped (girder) cross-section puts material where it resists bending best while saving weight.
  • Cantilevers: a beam fixed at one end and free at the other (a balcony, a diving board) bends most at its free end; its load capacity is again governed by Young's modulus and the depth of the section.
  • Pillars and columns: the maximum height of a column is limited by the compressive strength of its material, which is why the legs of furniture and supporting columns have a large cross-section.

The recurring lesson is that a material with a high Young's modulus and a high elastic limit, used in a well-chosen shape, lets engineers build structures that are both strong and light.

Elastic potential energy as area under the stress-strain line and a sagging beamU = areaStrainStressWbeam sags by delta
Solved Examples
1
Worked Example
A wire is stretched by a force of 100 N producing an extension of 2 mm. Calculate the elastic potential energy stored in the wire.
Solution
  1. Step 1: Use $U=\frac{1}{2}\times F\times\Delta L$ with $\Delta L=2\times10^{-3}\ \text{m}$.
  2. Step 2: Substitute: $U=\frac{1}{2}\times100\times2\times10^{-3}$.
  3. Step 3: Compute: $U=0.1\ \text{J}$.

Answer: $U=0.1\ \text{J}$.

2
Worked Example
A wire of volume $1\times10^{-5}\ \text{m}^3$ is subjected to a stress of $4\times10^{7}\ \text{N/m}^2$ producing a strain of $2\times10^{-4}$. Find the elastic potential energy stored.
Solution
  1. Step 1: Use $U=\frac{1}{2}\times\text{Stress}\times\text{Strain}\times\text{Volume}$.
  2. Step 2: Substitute: $U=\frac{1}{2}\times4\times10^{7}\times2\times10^{-4}\times1\times10^{-5}$.
  3. Step 3: Compute: $U=\frac{1}{2}\times80=\frac{1}{2}\times(4\times10^{7}\times2\times10^{-4}\times10^{-5})=\frac{1}{2}\times8\times10^{-2}=4\times10^{-2}\ \text{J}$.

Answer: $U=4\times10^{-2}\ \text{J}=0.04\ \text{J}$.

3
Worked Example
Find the elastic energy density (energy per unit volume) in a material where the stress is $2\times10^{8}\ \text{N/m}^2$ and $Y=2\times10^{11}\ \text{N/m}^2$.
Solution
  1. Step 1: Use $u=\frac{1}{2}\times\frac{(\text{Stress})^2}{Y}$.
  2. Step 2: Substitute: $u=\frac{1}{2}\times\frac{(2\times10^{8})^2}{2\times10^{11}}=\frac{1}{2}\times\frac{4\times10^{16}}{2\times10^{11}}$.
  3. Step 3: Compute: $u=\frac{1}{2}\times2\times10^{5}=1\times10^{5}\ \text{J/m}^3$.

Answer: $u=1\times10^{5}\ \text{J/m}^3$.

4
Worked Example
A steel cable can withstand a maximum (breaking) stress of $3\times10^{8}\ \text{N/m}^2$. What is the maximum load it can support if its cross-sectional area is $2\times10^{-4}\ \text{m}^2$?
Solution
  1. Step 1: Maximum load $F=\text{Breaking stress}\times A$.
  2. Step 2: Substitute: $F=3\times10^{8}\times2\times10^{-4}$.
  3. Step 3: Compute: $F=6\times10^{4}\ \text{N}$.

Answer: $F=6\times10^{4}\ \text{N}$ (engineers would apply a factor of safety and rate it lower).

5
Worked Example
A beam supported at its two ends sags by $\delta$ under a central load. If the depth $d$ of the beam is doubled (all else fixed), by what factor does the sag change? Use $\delta=\frac{W L^3}{4 b d^3 Y}$.
Solution
  1. Step 1: From the formula, $\delta\propto\frac{1}{d^3}$ when $W$, $L$, $b$ and $Y$ are fixed.
  2. Step 2: Doubling $d$: $\delta'\propto\frac{1}{(2d)^3}=\frac{1}{8d^3}$.
  3. Step 3: Therefore $\delta'=\frac{\delta}{8}$.

Answer: The sag reduces to one-eighth — which is why beams are made deep rather than wide.

6
Worked Example
Explain why thick ropes used in cranes are made of a number of thin wires twisted together rather than a single solid wire of the same cross-section.
Solution
  1. Step 1: The safe load depends on the total cross-sectional area (load $=\text{safe stress}\times A$), which a bundle of thin wires can match.
  2. Step 2: A stranded rope is far more flexible — it bends easily over pulleys and drums without developing dangerous bending stresses, unlike a stiff solid wire.
  3. Step 3: If one strand develops a flaw it can be detected and the rope replaced before all strands fail, improving safety against elastic fatigue.

Answer: Stranded ropes give the same load capacity with much greater flexibility and safety than a single thick wire.

Key Points

  • Elastic PE stored in a stretched wire: $U=\frac{1}{2}\times F\times\Delta L=\frac{1}{2}\times\text{stress}\times\text{strain}\times\text{volume}$.
  • Elastic energy density (per unit volume): $u=\frac{1}{2}\times\text{stress}\times\text{strain}=\frac{1}{2}\times\frac{(\text{stress})^2}{Y}$, the area under the stress–strain line.
  • Elastic after-effect is the delay before a body fully recovers; elastic fatigue is the weakening under repeated stress cycles.
  • Beams sag by $\delta=\frac{W L^3}{4 b d^3 Y}$; since $\delta\propto\frac{1}{d^3}$, beams are made deep and use an I-section.
  • Ropes use many thin twisted strands for flexibility and a factor of safety; column height is limited by compressive strength.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The elastic potential energy stored in a stretched wire is:
Explanation: The force grows from 0 to $F$, so the work stored is $\frac{1}{2}F\,\Delta L=\frac{1}{2}\times\text{stress}\times\text{strain}\times\text{volume}$.
Q2.The elastic energy per unit volume equals:
Explanation: Energy density $u=\frac{1}{2}\times\text{stress}\times\text{strain}$, the area under the stress–strain line.
Q3.The gradual weakening of a material under repeated cycles of stress is called:
Explanation: Elastic fatigue is the loss of strength under repeated loading; the after-effect is merely a delay in recovery.
Q4.Girders used in bridges are given an I-shaped cross-section mainly to:
Explanation: An I-section places material where bending stress is largest, giving high stiffness with minimum weight.
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