Question

Active strain gauges are connected in a half-bridge configuration to measure strain in a cantilever element. When powered by a voltage source of \(10\) V, the voltage at the output of the bridge is \(1\) mV. If the gauge factor is \(2.5\), the strain in the cantilever element is ________ microstrain.

• A. \(40\)
• B. \(80\)
• C. \(20\)
• D. \(160\)

Hint:

For a half-bridge with two active gauges, \(V_o/V_{ex} = (GF \cdot \varepsilon)/2\); solve for \(\varepsilon\) and convert to microstrain.

Answer 1

The Correct Option is B

Step 1: Recall the half-bridge output formula for strain gauges.
A cantilever in bending stretches on one face and compresses on the opposite face by nearly equal amounts. Mounting one active gauge on each face and placing them in adjacent arms of the Wheatstone bridge gives a half-bridge configuration, where the two gauges' resistance changes add rather than cancel. For this arrangement, the bridge output voltage \(V_o\) relates to the excitation voltage \(V_{ex}\), the gauge factor \(GF\), and the strain \(\varepsilon\) as
\[ \frac{V_o}{V_{ex}} = \frac{GF \cdot \varepsilon}{2} \] The factor of \(2\) in the denominator, rather than \(4\) for a single active gauge or \(1\) for a full bridge with four active gauges, reflects that exactly two of the four bridge arms respond to the strain here.

Step 2: Compute the normalized bridge output.
\[ \frac{V_o}{V_{ex}} = \frac{1 \text{ mV}}{10 \text{ V}} = \frac{1 \times 10^{-3}}{10} = 1 \times 10^{-4} \]
Step 3: Solve for the strain.
\[ 1 \times 10^{-4} = \frac{2.5 \cdot \varepsilon}{2} \] \[ \varepsilon = \frac{2 \times 1 \times 10^{-4}}{2.5} = \frac{2 \times 10^{-4}}{2.5} = 0.8 \times 10^{-4} = 8 \times 10^{-5} \]
Step 4: Convert to microstrain.
Since \(1\) microstrain \(= 1 \times 10^{-6}\),
\[ \varepsilon = 8 \times 10^{-5} = 80 \times 10^{-6} = 80 \text{ microstrain} \]
Step 5: Why the other options are wrong.
Option (A), \(40\) microstrain, would follow if the full-bridge formula \(V_o/V_{ex} = GF \cdot \varepsilon\) (no factor of \(2\)) were used by mistake, treating all four arms as active. Option (D), \(160\) microstrain, would follow from the quarter-bridge formula \(V_o/V_{ex} = GF \cdot \varepsilon /4\), which assumes only one gauge is active and is not what a half-bridge does. Option (C), \(20\) microstrain, does not correspond to any of the standard bridge formulas and undershoots the correct half-bridge sensitivity.

Final Answer:
The strain in the cantilever element is 80 microstrain. \[ \boxed{\varepsilon = 80 \text{ microstrain}} \]