When a solid bar, wire, cable, or rod is stretched or compressed, it deforms. In the linear elastic region, this
deformation is described by three main quantities: stress, strain, and Young’s modulus.
1. Stress
Normal stress is force per unit cross-sectional area:
\[
\sigma=\frac{F}{A}.
\]
Here, \(F\) is the axial force and \(A\) is the cross-sectional area. The SI unit is the pascal:
\[
1\ \mathrm{Pa}=1\ \mathrm{N/m^2}.
\]
Large engineering stresses are often written in megapascals:
\[
1\ \mathrm{MPa}=10^6\ \mathrm{Pa}.
\]
2. Strain
Longitudinal strain is the relative change in length:
\[
\varepsilon=\frac{\Delta L}{L_0}.
\]
Since strain is a ratio of two lengths, it is dimensionless. A strain of \(0.001\) means the object changes length
by \(0.1\%\) of its original length.
3. Young’s modulus
In the linear elastic region, stress is proportional to strain:
\[
\sigma=E\varepsilon.
\]
The proportionality constant \(E\) is Young’s modulus:
\[
E=\frac{\sigma}{\varepsilon}.
\]
Substituting \(\sigma=F/A\) and \(\varepsilon=\Delta L/L_0\), we get
\[
E=\frac{F/A}{\Delta L/L_0}.
\]
\[
\boxed{
E=\frac{F L_0}{A\Delta L}
}.
\]
4. Circular wire or cable area
If the object is a circular wire or cable with diameter \(d\), the cross-sectional area is
\[
A=\frac{\pi d^2}{4}.
\]
This is why a small change in diameter can strongly affect stress: area depends on \(d^2\).
5. Tension and compression sign convention
A common sign convention is:
- Tension: positive stress and positive strain.
- Compression: negative stress and negative strain.
Young’s modulus itself is reported as a positive material stiffness.
6. Solving for different variables
The same elastic relation can be rearranged depending on the unknown.
Force:
\[
F=EA\frac{\Delta L}{L_0}.
\]
Extension or shortening:
\[
\Delta L=\frac{F L_0}{AE}.
\]
Area:
\[
A=\frac{F L_0}{E\Delta L}.
\]
Strain from stress:
\[
\varepsilon=\frac{\sigma}{E}.
\]
Stress from strain:
\[
\sigma=E\varepsilon.
\]
7. Stress–strain graph
In the elastic region, a stress–strain graph is approximately a straight line through the origin:
\[
\sigma=E\varepsilon.
\]
The slope of this line is Young’s modulus:
\[
E=\frac{\Delta\sigma}{\Delta\varepsilon}.
\]
A steeper slope means a stiffer material. Steel, for example, is much stiffer than rubber because it has a much
larger Young’s modulus.
8. Elastic limit and safety factor
The linear model is valid only before the material reaches its elastic limit. A simple utilization estimate is
\[
\text{utilization}=\frac{|\sigma|}{\sigma_{\mathrm{limit}}}.
\]
The corresponding safety factor is
\[
\text{safety factor}=\frac{\sigma_{\mathrm{limit}}}{|\sigma|}.
\]
If the utilization is greater than \(1\), the entered stress is above the chosen elastic-limit estimate.
9. Typical Young’s modulus values
| Material |
Approximate Young’s modulus |
Meaning |
| Steel |
\(200\ \mathrm{GPa}\) |
Very stiff structural material |
| Aluminium |
\(69\ \mathrm{GPa}\) |
Less stiff than steel but lightweight |
| Copper |
\(117\ \mathrm{GPa}\) |
Moderately high stiffness |
| Nylon |
\(2{-}3\ \mathrm{GPa}\) |
Much more flexible than metals |
| Rubber-like materials |
\(\sim 0.01\ \mathrm{GPa}\) |
Very compliant; often nonlinear |
10. Worked example
A wire has
\[
L_0=2.5\ \mathrm{m},
\qquad
d=2.0\ \mathrm{mm},
\qquad
F=120\ \mathrm{N},
\qquad
\Delta L=1.8\ \mathrm{mm}.
\]
First compute the area:
\[
A=\frac{\pi d^2}{4}.
\]
\[
A=\frac{\pi(2.0\times10^{-3})^2}{4}
=
3.14\times10^{-6}\ \mathrm{m^2}.
\]
Stress:
\[
\sigma=\frac{F}{A}
=
\frac{120}{3.14\times10^{-6}}
=
3.82\times10^7\ \mathrm{Pa}.
\]
\[
\sigma=38.2\ \mathrm{MPa}.
\]
Strain:
\[
\varepsilon=\frac{\Delta L}{L_0}
=
\frac{1.8\times10^{-3}}{2.5}
=
7.20\times10^{-4}.
\]
Young’s modulus:
\[
E=\frac{\sigma}{\varepsilon}
=
\frac{3.82\times10^7}{7.20\times10^{-4}}
=
5.31\times10^{10}\ \mathrm{Pa}.
\]
\[
E=53.1\ \mathrm{GPa}.
\]
Key idea: stress measures load intensity, strain measures relative deformation, and Young’s modulus measures how
much stress is needed to produce a given strain.
