Question 3 of 7: Strain-Gauge Hysteresis and Dynamic Settling Time

Question 3: Strain-Gauge Hysteresis and Dynamic Settling Time (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) The gauge only reads strain that has actually been transmitted through the bond line to its grid, so the fidelity of that transfer depends entirely on the bond being rigid. If the polymer creeps — flows slowly under sustained shear — the transmitted strain lags behind the member's true, instantaneous strain: part of the deformation is absorbed by viscous flow in the glue rather than being carried through instantly. While the load is increasing, the gauge therefore under-reads (it hasn't "caught up" to the true strain yet); while the load is decreasing, the bond likewise does not relax back instantly, so residual creep strain the polymer has not yet given up makes the gauge over-read relative to the true unloading path. Plotted as gauge output vs. applied load, the loading and unloading branches trace two different paths through the same load range — a lagging loop that is, by definition, hysteresis — even though the strain gauge element itself is behaving ideally; the non-ideality lives entirely in the (viscoelastic) bond.

[Figure not reproduced: Fig. 1 — Underdamped 2nd-order step response of the beam+load-cell system (redrawn from the source graph). Peaks decay geometrically toward the steady 2.0 kg reading; the dashed red lines mark the ±2% settling band and the settling time $t_s$. See the official exam paper.]

b) The trace is a classic lightly-damped second-order step response: a fast initial rise overshoots the final value, then rings through several decaying oscillations before settling near 2.0 kg. Reading the mass immediately (or at any of the early peaks/troughs) would give a value that could be off by 50-70% of the final reading, so the beam must be allowed to settle to within some acceptable tolerance band of its final value before the reading is trusted — the standard engineering criterion is a ±2% (or, less strictly, ±5%) band around the final steady value, with $t_s$ defined as the last time the response leaves that band and never re-enters it.

Given. From the graph: steady-state reading $M_{ss}=2.0$ kg, first overshoot peak $M_1=3.4$ kg at $t_1\approx0.33$ s, second peak $M_2\approx2.65$ kg one damped period later at $t\approx0.95$ s (the following peaks fall at $t\approx1.58$ and $2.22$ s, reading ≈2.3 and ≈2.15 kg).

Find. The settling time $t_s$ (2% criterion) and the damping parameters that justify it.

Approach. Fit the underdamped second-order model $M(t)=M_{ss}\left[1-e^{-\zeta\omega_n t}\left(\cos\omega_d t+\dfrac{\zeta}{\sqrt{1-\zeta^2}}\sin\omega_d t\right)\right]$ to the two peaks read off the graph, then use the standard settling-time formula for that model.

1. Damping ratio from the first-peak overshoot. Fractional overshoot $OS=(M_1-M_{ss})/M_{ss}=(3.4-2.0)/2.0=0.70$ (70%). Since $\ln(OS)=-\zeta\pi/\sqrt{1-\zeta^2}$, $$\zeta=\dfrac{-\ln(OS)}{\sqrt{\pi^2+\ln^2(OS)}}=\dfrac{-\ln(0.70)}{\sqrt{\pi^2+\ln^2(0.70)}}\approx0.113.$$ A lightly damped system, consistent with the many visible rings before the trace flattens.
2. Natural frequency from the ringing period. The peaks are one damped period $T_d$ apart; the four visible peaks sit at $t\approx0.33,\,0.95,\,1.58$ and $2.22$ s, i.e. $T_d\approx(2.22-0.33)/3\approx0.63$ s, so $\omega_d=2\pi/T_d\approx9.97$ rad/s and $$\omega_n=\dfrac{\omega_d}{\sqrt{1-\zeta^2}}\approx10.0\ \text{rad/s}.$$ Substituting these back into the step-response formula reproduces the 3rd and 4th peaks (≈2.34 kg near $t\approx1.57$ s, ≈2.16 kg near $t\approx2.20$ s) to within the graph's own reading precision, confirming the fit in both amplitude and time.
3. Settling time. For the ±2% band, $t_s=4/(\zeta\omega_n)$; for the looser ±5% band, $t_s=3/(\zeta\omega_n)$: $$\boxed{t_{s,2\%}=\dfrac{4}{(0.113)(10.0)}\approx3.5\ \text{s}, \qquad t_{s,5\%}\approx2.6\ \text{s}.}$$ This matches the graph — the trace is flat to within the drawn line width (around 2.0 kg) from about 3.5 s onward, the last visible ring peaking near $t\approx3.5$ s.

c) A fixed "wait N seconds" rule from part (b) only holds for the one mass shown, because a real beam+sensor is a spring-mass-damper system whose natural frequency depends on the total oscillating mass, $\omega_n=\sqrt{k_{beam}/(m_{beam,eff}+m_{load})}$: adding a heavier load lowers $\omega_n$ (and generally shifts $\zeta$ too, since the damping mechanism is rarely mass-independent), so heavier loads will in general settle in a different time than the one derived above. A computer DAQ system can determine the correct settling time for each load automatically, rather than relying on one pre-recorded curve, using an algorithm such as:

1. Capture the full transient. Sample the sensor output at a rate well above the expected ringing frequency (Nyquist, part 2g) for several seconds after the load is applied, for that specific mass.
2. Estimate the steady-state value. Take a moving average (or median) of the last several samples once the trace has visibly flattened, as the working estimate of $M_{ss}$ for that load.
3. Scan backward for the last band exit. Define a tolerance band (e.g. ±2% of $M_{ss}$) and scan the recorded trace from the end backward to find the last time index at which the reading was still outside that band; that time is $t_s$ for this specific load.
4. Release the reading only after $t_s$. In real-time operation (rather than post-processing a recorded trace), the same idea runs forward: keep a short rolling buffer of the most recent samples and declare the reading "settled" the first moment all samples in that buffer fall within the tolerance band of their own running average — at that point, take (or release) the mass reading. Because this re-evaluates $\zeta$, $\omega_n$ and $M_{ss}$ from the live data every time, it automatically gives the correct settling time for whatever mass is actually on the beam, without needing a separate calibration curve for each possible load.