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Engineering & Technology · Production Metrology, Tolerances & Assembly

Fits, tolerance stacks and assembly variation

A housing receives two spacers. Each spacer passes its own length inspection, yet the remaining gap varies from assembly to assembly. The gap belongs to the relationship among the housing and both spacers. It cannot be read from either spacer drawing by itself.

Use Measurement, Units & Engineering Evidence and Sketches, Models & Technical Communication for measurement and drawing ideas. Manufacturing Processes supplies context for process choices; the chapter explains its own elementary limit and stack calculations. These are original teaching cases, not drawing-standard compliance claims or equipment specifications.

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After reading, you can

Calculate an elementary worst-case clearance and explain why part limits alone do not describe assembly probabilities.

01

Begin with the dimension chain and its signs

For the original one-dimensional stack, let H be the available housing span, A and B the two spacer lengths, and G the remaining gap. Then G = H − A − B. Increasing H opens the gap; increasing either spacer closes it. This sign check is a useful way to catch a reversed equation before calculating.

Using the supplied nominal dimensions H = 30.0 mm, A = 10.0 mm and B = 19.5 mm gives G = 0.5 mm. The model ignores tilt, deformation, contact texture and temperature differences. It describes aligned rigid lengths under the same specified conditions.

The gap is a result of three dimensionsAn illustrative stack shows a housing span H30 mm, part A10 mm and part B19.5 mm. The nominal gap is 0.5 mm. A separate interval plot shows the worst-case gap range 0.3 to 0.7 mm.G = H − A − B · original one-dimensional stackH = 30.0 mmA = 10.0B = 19.5Gap 0.5 mm; width shown to scale0.20.30.50.70.8G (mm) · possible interval from the supplied limits
H = 30.0 ± 0.1 mm; A = 10.0 ± 0.05 mm; B = 19.5 ± 0.05 mm. G = H − A − B has nominal value 0.5 mm, minimum 29.9 − 10.05 − 19.55 = 0.3 mm, and maximum 30.1 − 9.95 − 19.45 = 0.7 mm. These are interval bounds, not a probability distribution.
02

Find the smallest and largest permitted combinations

Suppose H may differ by ±0.10 mm and each spacer by ±0.05 mm. The smallest gap combines the smallest housing with the largest spacers: 29.90 − 10.05 − 19.55 = 0.30 mm. The largest combines the largest housing with the smallest spacers: 30.10 − 9.95 − 19.45 = 0.70 mm.

This is a worst-case interval calculation. Every combination of the supplied independent bounds is allowed in the model, so it does not rely on a probability distribution. Statistical variation analysis asks a different question and needs justified distributions, dependence and process behaviour. Drawing limits alone do not supply those assumptions.

03

Connect the interval to the intended assembly

If the project explicitly requires 0.25 ≤ G ≤ 0.75 mm, the calculated interval lies inside that classroom requirement. If it instead requires 0.40 ≤ G ≤ 0.60 mm, the same dimension scheme does not guarantee compliance. Changing the assembly requirement changes the conclusion, even though every part has the same limits.

A positive scalar gap is only one aspect of a fit. Real surfaces also have shape, position, orientation and contact conditions. Thermal expansion can shift dimensions, and elastic deformation can change the relationship under load. Preserve the simple calculation as a model with stated scope rather than promoting it to a complete assembly approval.

Give a design change a quantitative purpose

Imagine reducing the housing allowance from ±0.10 to ±0.05 mm while keeping the spacer limits unchanged. The gap allowance becomes ±(0.05 + 0.05 + 0.05) = ±0.15 mm around 0.50 mm, giving 0.35 to 0.65 mm.

That is narrower than 0.30 to 0.70 mm but still does not guarantee the 0.40–0.60 mm requirement. A change should be evaluated against the actual target; being more precise is not the same as being sufficient. The process and cost implications also need evidence.

Try the reasoning

  1. Which combination gives the smallest G?

    Show the answer and reasoning

    The minimum H and maximum A and B, because both spacer lengths are subtracted.

  2. Does the 0.30–0.70 mm interval say that its midpoint is most likely?

    Show the answer and reasoning

    No. It is an interval-bounds calculation. No probability distribution has been specified.

  3. Can all parts satisfy their own limits while an assembly misses a gap requirement?

    Show the answer and reasoning

    Yes. The permitted combination of dimensions may not satisfy the assembly-level relationship. Tolerance design needs that relationship explicitly.

What to carry forward

Write the assembly relationship first. Worst-case bounds and statistical variation need different information.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. Massachusetts Institute of Technology2 sources
    1. Variation Buildup in Single Parts · 2.875, Fall 2004Daniel E. Whitney · MIT OpenCourseWareCourse notes · PDFocw.mit.eduRead source
    2. The Datum Flow Chain · 2.875, Fall 2004Daniel E. Whitney · MIT OpenCourseWareCourse notes · PDFocw.mit.eduRead source
  2. Uncertainties in Dimensional Measurements Made at Nonstandard TemperaturesDennis A. Swyt · NISTResearch paper · PDFnvlpubs.nist.govRead source