Strain Gauges and the Gauge Factor
Strain Gauges and the Gauge Factor
A bonded strain gauge converts mechanical deformation of a solid surface into a proportional change in electrical resistance. When a tiny grid of metal foil is stretched, its wire becomes slightly longer and thinner, and its resistance rises by a predictable amount. This effect, scaled up by clever bridge circuits, underlies nearly every modern load cell, pressure transducer, and torque sensor.
The book by Webster and Pallàs-Areny treats the strain gauge as the textbook example of a small-signal resistive transducer — the signal is only a few millivolts per volt of excitation, so amplification and temperature compensation are essential. In this lesson you will master the gauge-factor equation and the strain-to-stress link, preparing you for the Wheatstone bridge in the signal-conditioning module.
We will also look at how Poisson's ratio contributes to the gauge factor and why a typical metal foil gauge has $GF \approx 2.0$ while semiconductor gauges can reach $GF \approx 100$.
$$GF = \frac{\Delta R / R}{\epsilon}$$
where $\epsilon = \Delta L / L$ is the mechanical strain (dimensionless). For annealed constantan foil gauges, $GF \approx 2.0$; for silicon semiconductor gauges, $GF$ ranges from 50 to 200.
$$\sigma = E\,\epsilon, \qquad \epsilon = \Delta R / (R\cdot GF)$$
where $E$ is Young's modulus of the loaded material (e.g., $210\,\text{GPa}$ for steel, $70\,\text{GPa}$ for aluminium).
A foil gauge with $R = 350\,\Omega$ and $GF = 2.05$ is bonded to a steel beam experiencing a strain of $\epsilon = 500\,\mu\epsilon = 5\times10^{-4}$. Compute $\Delta R$.
$\Delta R = R\cdot GF\cdot \epsilon = 350 \cdot 2.05 \cdot 5\times10^{-4} = 0.359\,\Omega$.
A change of $\approx 0.36\,\Omega$ on 350 Ω — about 0.1 %. This tiny signal is why a Wheatstone bridge is mandatory.
The same beam is aluminium ($E = 70\,\text{GPa}$). For the 500 με strain above, compute the stress.
$\sigma = E\,\epsilon = 70\times10^9 \cdot 5\times10^{-4} = 3.5\times10^7\,\text{Pa} = 35\,\text{MPa}$.
Well below aluminium's yield stress of $\approx 250\,\text{MPa}$, so the beam is elastic.
A semiconductor gauge has $GF = 120$. For the same strain $\epsilon = 5\times10^{-4}$ and $R = 350\,\Omega$, compute $\Delta R$ and compare with the metal-foil result.
$\Delta R = 350 \cdot 120 \cdot 5\times10^{-4} = 21\,\Omega$.
This is 58× larger than the metal-foil response. Semiconductor gauges give much stronger signals — but they are also more temperature-sensitive and brittle, which is why metal foil remains the workhorse.
Practice Problems
Show Answer Key
1. $\Delta R = 120 \cdot 2.0 \cdot 10^{-3} = 0.24\,\Omega$.
2. $\sigma = 210\times10^9 \cdot 3\times10^{-4} = 6.3\times10^7\,\text{Pa} = 63\,\text{MPa}$.
3. To amplify small $\Delta R$ signals while cancelling lead resistance and temperature drift.
4. Piezoresistive effect in silicon changes resistivity itself, not just geometry, producing a far larger $\Delta R/R$ per unit strain.
5. $\epsilon = \Delta R / (R\cdot GF) = 0.5 / (350\cdot 2.1) = 6.80\times10^{-4} = 680\,\mu\epsilon$.
6. Roughly $\pm 5000\,\mu\epsilon$ (0.5 % strain); beyond this the bond and foil begin to fatigue.