MethodWhat is calculated?
The amplifier displays the bridge voltage ratio. Gauge factor and bridge factor turn it into strain; for the quarter bridge the exact eq. 4.7 yields a slightly larger strain than the linear approximation at large signals. Converting to stress is valid only for a uniaxial stress state – biaxial states need two strains and Hooke's law per eqs. 9.27/9.28.
Equations
ε = 4·(U_M/U_B)/(k·B) (aus Keil Gl. 4.8)
Viertelbrücke exakt: U_M/U_B = ΔR/(2(2R₀+ΔR)) ⇒ ΔR/R₀ = 4s/(1−2s) (Keil Gl. 4.7)
σ = E·ε (einachsig)
ProcedureStep by step
- Enter the signal in mV/V, gauge factor and bridge factor of the circuit used.
- For quarter bridges above 1 mV/V use the exact value.
- Adopt σ = E·ε only for uniaxial loading along the gauge.
Typical mistake
Using the amplifier's fixed gauge factor 2.00 although the gauge has 2.08, or entering the signal in mV instead of mV/V.
PracticeApplication and limits
Evaluating raw signals from data loggers without a strain-gauge mode, converting transducer sensitivities and checking amplifier settings (gauge factor, bridge factor).
Lead resistance, transverse sensitivity and temperature response are excluded; the non-linearity correction applies to one active arm only (B = 1).
SourceTechnical basis
Keil, Dehnungsmessstreifen, 2nd ed. 2017, sec. 4.2, eqs. (4.7), (4.8), fig. 4.2; sec. 9.4, eqs. (9.27), (9.28).
The source supports the equation structure and worked examples; this calculator does not replace calibration of the measuring chain.
FAQFrequently asked questions
How large is the non-linearity at 0.5 mV/V?
ΔR/R₀ = 4·0.0005/(1−0.001) = 2.002 ‰ instead of 2.000 ‰, i.e. 0.1 %.
What if the amplifier reads µm/m?
Then it has already applied its gauge and bridge factors; only deviations need correcting (ε = ε_reading·k_instrument/k_gauge).
Can I compute σ this way for bending?
Yes, the outer-fibre stress is uniaxial; ε must be the bending strain of one gauge, not B times it.
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