Micrometer dispute on a ±0.010 mm dowel pin lot
Your company receives 500 hardened ground dowel pins, drawing size Ø12.000 ± 0.010 mm, from a new supplier whose inspection certificate reports every dimension in tolerance. Receiving inspection is done on the shop floor next to a heat-treatment furnace; ambient temperature at the time of checking is 33 °C. The inspector uses a 0–25 mm outside micrometer with 0.01 mm graduations, zeroed on its setting rod just before use, and records readings between 12.004 and 12.014 mm. He rejects the lot. The supplier's quality engineer counters that his digital micrometer, reading to 0.001 mm in an air-conditioned room at 20 °C, gave 11.996 to 12.006 mm. Steel's coefficient of thermal expansion is about 11.5 × 10⁻⁶ per °C. Production needs the pins within four days and the only alternative stock is a costlier grade in another material.
1. Work out whether the shop-floor temperature can explain the 0.008–0.010 mm disagreement, and judge whether either instrument is fit for a ±0.010 mm tolerance.
Read the worked answer
Thermal effect, calculated. Unrestrained expansion over the 13 °C difference: ΔL = 12 mm × 11.5 × 10⁻⁶ × 13 ≈ 0.0018 mm, under 0.002 mm. In a comparative measurement it is even smaller, because the micrometer frame, the setting rod and the pins are all steel and warm together, so most of the expansion cancels. Temperature can account for at most about 0.002 mm of the 0.008–0.010 mm gap - it is not the explanation.
What actually explains it. The shop instrument has 0.01 mm graduations across a 0.020 mm total tolerance band. The usual rule is that gauge discrimination should be about one-tenth of tolerance, i.e. 0.002 mm; this micrometer sits at one-half. Reading uncertainty of ±0.005 mm, zeroing on a setting rod at 33 °C, anvil flatness and parallax, and inconsistent measuring force together easily scatter by 0.008 mm. At the other end, the supplier's 0.001 mm micrometer at 20 °C is far better matched to the job, but its calibration status still needs to be seen.
Conclusion. The shop-floor micrometer cannot discriminate conforming from non-conforming pins on this tolerance, so the rejection is not defensible on that evidence - and equally, the supplier's certificate cannot be accepted on trust. The argument is about measurement systems, not about parts.
2. Set out how you would settle the dispute, and give the acceptance decision you would take to production.
Read the worked answer
Settlement plan.
- Move the check to the inspection room; soak the 500 pins on a steel surface plate for 24 h at 20 ± 2 °C, well away from the furnace.
- Measure 30 pins with a calibrated 0.001 mm micrometer zeroed on a Ø12 setting ring: two operators, two repetitions, both ends. Run a quick gauge R&R - if it exceeds 10% of the 0.020 mm tolerance, the measurement system is the problem, not the parts.
- Plot the distribution against the limits and apply a guard band, accepting at 12.000 ± 0.007 mm (about 70% of tolerance) so supplier risk is covered, and treating anything in the outer 0.003 mm as concession material.
Decision. The supplier's own data centre near 12.001 mm with a spread of 11.996–12.006 mm - inside limits with margin at 20 °C. The shop-floor readings of 12.004–12.014 are consistent with the same pins measured by a coarse gauge in a hot room next to a furnace, so the lot should not be rejected. Release it on the 20 °C data plus a sorted sample and a deviation note for the guard zone; no need to buy the costlier grade, and production gets its pins in four days.
Afterwards: recalibrate or retire the shop micrometer, keep a 0.001 mm instrument dedicated to close-tolerance work, and stop doing receiving inspection beside a heat-treatment furnace.