In a uniaxial compressive strength test, a \(120\ \Omega\) strain gauge of gauge factor 2.0 is pasted on the rock sample as shown. At the end of the test, the change in resistance of the strain gauge is \(0.5\ \Omega\). The longitudinal deformation of the sample, in \(mm\), is . (rounded off to two decimal places)
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Use the gauge factor definition to turn the measured change in resistance into a strain value, then apply that strain to the sample's original length.
Here is the same result worked through percentages instead of decimal fractions, which some students find easier to follow.
The strain gauge starts at $120\ \Omega$ and its resistance changes by $0.5\ \Omega$ during the test. As a percentage, that change is $\dfrac{0.5}{120} \times 100 = 0.4167\%$.
The gauge factor tells us how many times bigger the percentage resistance change is compared to the percentage strain: $GF = \dfrac{\%\Delta R}{\%\varepsilon}$. Rearranging, the percentage strain is $\%\varepsilon = \dfrac{\%\Delta R}{GF} = \dfrac{0.4167}{2.0} = 0.2083\%$.
This percentage strain means the sample shortens by $0.2083\%$ of its original length. The figure gives the original gauge length as $110\ \text{mm}$, so the deformation is $\Delta L = \dfrac{0.2083}{100} \times 110 = 0.2292\ \text{mm}$, which rounds to $0.23\ \text{mm}$.
Let's summarize:
Convert the resistance change to a percentage change first.
Divide by the gauge factor to get the percentage strain.
Apply that percentage directly to the sample's original length to find the deformation.
So the sample deforms by about 0.23 mm under the compressive load.
