Question

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)

Hint:

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.

Answer 1

Correct Answer: 0.23

Step 1: Recall what the gauge factor means.
The gauge factor (GF) of a strain gauge is the ratio of the fractional change in its resistance to the strain it experiences: \[ GF=\frac{\Delta R/R}{\varepsilon} \] where \(\varepsilon=\Delta L/L\) is the longitudinal strain of the sample it is bonded to.

Step 2: Find the fractional change in resistance.
\(\Delta R/R=0.5/120=0.004167\).

Step 3: Find the strain from the gauge factor.
\(\varepsilon=(\Delta R/R)/GF=0.004167/2.0=0.002083\).

Step 4: Find the deformation using the sample's original length.
From the figure, the gauge length of the rock sample is \(L=110\ \text{mm}\).
\(\Delta L=\varepsilon \times L=0.002083 \times 110=0.2292\ \text{mm}\), which rounds to \(0.23\ \text{mm}\).

Final Answer:
The sample shortens by about \(0.23\ \text{mm}\) under the load. \[ \boxed{0.23} \]