Where this fits
in engineering
A shaft has to turn inside a bore, and the drawing states how much gap is allowed. But a figure written to one decimal place is a band, not a point.
True
Size
A shaft goes into a bore. The inspector writes down 24.6 mm and 24.8 mm, both to one decimal place. Those two numbers are not sizes. They are ranges.
What 24.6 to 1 d.p. really means
Do the same for the bore
Combine the bounds, not the readings
24.55 ≤ d < 24.65
Half a unit either side of the reading.
24.75 ≤ D < 24.85
Does it fit?
− 24.55
− 24.65
Both bounds sit inside it, so every possible shaft fits every possible bore.
Three things to carry into the exam
- Bounds are half a unit of the stated accuracy either side
- Adding keeps the bounds together; subtracting swaps them over
- Answer with both bounds, and say what they mean for the job
24.85 − 24.615
= 0.235 mm
The engineering
context and termsA shaft has to turn inside a bore. Too tight and it seizes, too loose and it rattles and wears. The gap between them is the clearance, and the drawing states the range it has to fall in. But no measurement is exact: a figure written to one decimal place is a band, not a point. An inspector who signs off on the written numbers alone is guessing. Working with bounds is how she can be certain the fit is safe whatever the true sizes turn out to be.
- Shaft and bore
- The round part, and the round hole it has to fit and turn inside.
- Clearance
- The gap between them, found by subtracting the shaft size from the bore size.
- Tolerance
- The range of sizes the drawing allows. A part inside tolerance is acceptable even though it is not exact.
- Vernier caliper
- A hand measuring instrument. The one in this video resolves to 0.02 mm.
- First-off inspection
- Checking the first part of a run before the rest are made, so an error is caught early rather than repeated.
Questions
4 questions · 13 marks-
A length is recorded as 18.4 mm, correct to 1 decimal place. Write down the error interval for the length, L. [2]18.35 ≤ L < 18.45
1 mark for both bounds, 1 for correct inequality signs. The upper bound is strict. -
A steel bar is 250 mm long, to the nearest millimetre. Eight bars are laid end to end. Calculate the upper bound of the total length. [3]Upper bound of one bar = 250.5 mm
8 × 250.5 = 2004 mm
A frequent error is 8 × 250.4999… or adding 0.5 once at the end. -
Using the figures in the video, find the least possible clearance between the shaft and the bore. Explain why you cannot simply work out 24.8 − 24.6. [3]24.75 − 24.65 = 0.10 mm
Both written values are rounded, so the true clearance is not fixed at 0.2 mm. The tightest case pairs the smallest possible bore with the largest possible shaft.
The explanation mark needs the idea that the readings are rounded, not just “because of bounds”. -
A rectangular plate measures 320 mm by 180 mm, each measured to the nearest 10 mm. Calculate the upper and lower bounds of its area. [5]Bounds: 315 ≤ length < 325 and 175 ≤ width < 185
Upper bound = 325 × 185 = 60,125 mm²
Lower bound = 315 × 175 = 55,125 mm²
2 marks for both sets of bounds, 1 each for the two products, 1 for units. Note that multiplication keeps the bounds together, unlike the subtraction in question 3.
Controls. Tap the stage or press space to play and pause. Drag the amber circle along the line to scrub, or use the arrow keys to jump five seconds. The chapter buttons drop you straight into each stage of the working, which is the bit worth pausing on with a class. Captions start off; CC turns them on. Everything is inside this one file, so it runs offline. Screen-record the stage at 1920×1080 if you need an MP4 with a voiceover.
Demo developed by Lee Guthrie for EAL