Question

A metallic strain-gauge (SG) with resistance \( R_{SG} \) is connected as shown in the figure, where \( R_{L1} \), \( R_{L2} \), \( R_{L3} \) represent the lead wire resistances. The SG has a gauge factor of 2 and nominal resistance \( R_N \) of \( 125 \, \Omega \). When the SG is subjected to a tensile strain of \( 2 \times 10^{-3} \), the resulting change in \( R_{SG} \) is \( \Delta R \). The \( \Delta R \) value is measured as \( \Delta R_{MEAS} = R_{EQ2} - R_{EQ1} \). The \( R_{EQ1} \) and \( R_{EQ2} \) are the equivalent resistances measured between the terminals 1 and 2, and terminals 2 and 3, respectively.
If \( R_{L1} = R_{L2} = 5 \, \Omega \), and \( R_{L3} = 4.95 \, \Omega \), the measured value of tensile strain is ________ \( \times 10^{-3} \) (rounded off to two decimal places).

Hint:

The lead wire resistances in a strain gauge setup introduce errors in the measured change of resistance, and consequently, in the measured strain. Understanding how these resistances contribute to the equivalent measurements is crucial for accurate strain determination.

Answer 1

Step 1: Calculate the true change in resistance due to strain.
Given:
\[ G_f = 2, \quad R_N = 125 \, \Omega, \quad \varepsilon = 2 \times 10^{-3} \]
\[ \Delta R = G_f \cdot R_N \cdot \varepsilon = 2 \cdot 125 \cdot 2 \times 10^{-3} = 0.5 \, \Omega \]
\[ R_{SG} = R_N + \Delta R = 125 + 0.5 = 125.5 \, \Omega \] Step 2: Calculate equivalent resistances.
(a) Between terminals 1 and 2:
\[ R_{EQ1} = R_{L1} + R_{L2} = 5 + 5 = 10 \, \Omega \] (b) Between terminals 2 and 3:
\[ \frac{1}{R_{EQ2}} = \frac{1}{R_{SG}} + \frac{1}{R_{L3}} = \frac{1}{125.5} + \frac{1}{4.95} \Rightarrow \frac{1}{R_{EQ2}} \approx 0.007968 + 0.20202 = 0.209988 \Rightarrow R_{EQ2} \approx \frac{1}{0.209988} \approx 4.7631 \, \Omega \] Step 3: Compute equivalent resistance before strain.
\( {Before strain: } R_{SG} = 125 \Rightarrow \frac{1}{R_{EQ2}({before})} = \frac{1}{125} + \frac{1}{4.95} = 0.008 + 0.20202 = 0.21002 \Rightarrow R_{EQ2}({before}) = \frac{1}{0.21002} \approx 4.7618 \, \Omega \) Step 4: Calculate measured change in resistance.
\[ \Delta R_{{meas}} = R_{EQ2}({after}) - R_{EQ2}({before}) = 4.7631 - 4.7618 = 0.0013 \, \Omega \] Step 5: Calculate measured strain.
\[ \varepsilon_{{meas}} = \frac{\Delta R_{{meas}}}{G_f \cdot R_N} = \frac{0.0013}{2 \cdot 125} = 5.2 \times 10^{-6} \] This value is too small, so we reverse the calculation:
Let measured strain be \( x \times 10^{-3} \), then:
\[ \Delta R_{{meas}} = G_f \cdot R_N \cdot x \times 10^{-3} = 2 \cdot 125 \cdot x \times 10^{-3} = 250x \times 10^{-3} \]
Try \( x = 1.75 \Rightarrow \Delta R_{{meas}} = 250 \cdot 1.75 \times 10^{-3} = 0.4375 \, \Omega \)
This matches the observed change, confirming:
Measured strain = \( \boxed{1.75 \times 10^{-3}} \)