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Engineering Graphics and CAD · Lesson 14 of 35

Limits and fits: holes and shafts (ISO 286)

Specify how a shaft and hole go together (clearance, transition, interference) using the ISO code system.

Readiness check

Learning objectives

By the end of this lesson you can:

Define clearance, transition, and interference fits.
Read an ISO fit designation such as 25H7/g6.
Explain hole-basis and shaft-basis systems.
Compute the clearance of a fit from its limit values.
Select a fit class from a feature's function.

Check your starting point

Five to ten minutes.

If a shaft is always slightly smaller than its hole, will it slide or stick?
In "25H7/g6," which part do you think refers to the hole and which to the shaft?
Why might you want a shaft slightly larger than its hole?

Interpretation.

Q1: Slide (a clearance fit). If the shaft were larger, it would need pressing in.
Q2: The capital letter (H) is the hole, the lowercase (g) is the shaft. This lesson confirms the convention.
Q3: To hold it permanently by friction (an interference/press fit), for example a bushing that must not rotate.

You need L13 (tolerances and limits).

0 or 1 weak itemsContinue with this lesson.

2 weak itemsReview Lesson 13, then return.

3 or more weak itemsWork through the prerequisite examples before continuing.

The core idea

What it is. A fit is the relationship between the toleranced sizes of a mating hole and shaft. ISO 286 is a code system that specifies both tolerances compactly so the fit is standard and repeatable.

Why an engineer needs it. Countless mechanical joints are a shaft in a hole: bearings, dowels, bushings, pins. Whether the joint slides freely, locates precisely, or grips permanently depends entirely on how the two tolerances relate. ISO 286 lets a designer specify that relationship with a short code instead of writing four limit numbers.

What problem it solves. It standardizes mating tolerances so "25H7/g6" means the same clearance fit to every engineer and shop worldwide, and so hole and shaft tolerances are chosen consistently.

What goes wrong when it is ignored. Ad hoc tolerances on mating parts give unpredictable fits: a joint meant to slide may seize, or a press fit may fall out. Using the wrong basis or misreading the code produces the wrong assembly relationship.

A simple mechanical example. A 25 mm shaft in a 25 mm bushing. If the hole is 25.000 to 25.021 and the shaft is 24.980 to 24.993, the shaft is always smaller, so it rotates freely: a clearance fit. If instead the shaft were 25.022 to 25.035, it is always larger, so it must be pressed in and grips: an interference fit.

The three fit types:

Clearance fit: the shaft is always smaller than the hole; there is always a gap (it slides).
Interference fit: the shaft is always larger than the hole; they grip (press fit).
Transition fit: depending on where each lands in its tolerance, there may be a small clearance or a small interference (precise location, easy to assemble).

Reading the ISO 286 code (for example 25H7/g6):

25 is the basic (nominal) size.
H7 is the hole: the letter sets the position of the tolerance zone, the number sets its width (the IT grade). Capital letters are holes.
g6 is the shaft: lowercase letter for the shaft's zone position, number for its IT grade.
Hole basis (letter H, whose lower deviation is zero) is the usual default: the hole is fixed and different shaft letters give different fits. Shaft basis (letter h) fixes the shaft instead and is used less often.

Part 3: Dimensioning and technical definition.

Check: explain the decision in your own words before using a CAD command.

The concept. Tolerance zones show allowable variation directly. Their relative positions determine whether mating parts always clear, may transition, or interfere.

The skills, taught in order

Skill 14.1 - Classify the three fits

Concept. Clearance, transition, and interference describe whether the shaft is smaller, either, or larger than the hole. Terminology. Clearance, transition, interference. Procedure. Compare the tolerance zones: shaft entirely below the hole means clearance; entirely above means interference; overlapping means transition. Reasoning. The relative position of the two zones fully determines the fit type. Failure mode. Judging by nominal size alone, ignoring the tolerance zones. Check. Given two zones, state the fit type.

Skill 14.2 - Read the code

Concept. The letter sets zone position, the number sets zone width; capitals are holes, lowercase shafts. Terminology. Fundamental deviation (the letter) and IT grade (the number). Procedure. Split the code into basic size, hole (capital letter and number), and shaft (lowercase letter and number). Read letter as position, number as width. Reasoning. The compact code fully specifies both tolerances once you know the convention. Failure mode. Reading the number as position or confusing hole and shaft. Check. In 25H7/g6, identify basic size, hole, and shaft, and say what each letter and number controls.

Skill 14.3 - Apply hole basis

Concept. Hole basis fixes the hole (H, lower deviation zero) and varies the shaft to set the fit. Terminology. Hole basis (H), shaft basis (h). Procedure. Default to hole basis: keep the hole H, choose the shaft letter for the desired fit (g for a small clearance, k for transition, p for interference, among others). Reasoning. Holes are often made with fixed tooling (drills, reamers), so fixing the hole and varying the shaft is economical. Failure mode. Mixing bases or assuming shaft basis by default. Check. State which member is fixed in hole basis.

Skill 14.4 - Compute the clearance

Concept. Clearance is hole size minus shaft size, evaluated at the limits for the extremes. Terminology. Maximum clearance equals largest hole minus smallest shaft; minimum clearance equals smallest hole minus largest shaft. Procedure. From the limit values, subtract to get maximum and minimum clearance; positive means gap, negative means interference. Reasoning. The extreme clearances bound how the assembled joint behaves. Failure mode. Subtracting the wrong pair of limits. Check. Given hole and shaft limits, compute maximum and minimum clearance.

Worked example 1: clearance of 25H7/g6

Problem. For 25H7/g6, using 25H7 equal to 25.000 to 25.021 and 25g6 equal to 24.980 to 24.993, classify the fit and compute its minimum and maximum clearance. (These limit values are illustrative; verify exact deviations against ISO 286-2.)

Planning. Compare zones for the fit type, then subtract limits for the clearances.

Solution.

Hole limits. 25H7: lower 25.000, upper 25.021 (H means the lower deviation is zero; IT7 gives the 0.021 width for this size range, per ISO 286, flagged for verification).
Shaft limits. 25g6: upper 24.993, lower 24.980 (g gives a small negative deviation, so the shaft sits just below nominal; IT6 gives the 0.013 width, flagged for verification).
Fit type. The largest shaft (24.993) is smaller than the smallest hole (25.000), so the shaft is always smaller than the hole: a clearance fit.
Minimum clearance. Smallest hole minus largest shaft: 25.000 minus 24.993 equals 0.007.
Maximum clearance. Largest hole minus smallest shaft: 25.021 minus 24.980 equals 0.041.
Interpretation. The joint always has between 0.007 and 0.041 of gap, a light running clearance suitable for a shaft that must rotate.

Result. 25H7/g6 is a clearance fit with clearance from 0.007 to 0.041 mm, a light running fit.

Why the method works. Comparing the extreme limits both classifies the fit (shaft always smaller) and bounds its behavior (min and max gap).

How to verify independently. The clearance range width equals the sum of the two tolerances: hole tolerance 0.021 plus shaft tolerance 0.013 equals 0.034, and indeed 0.041 minus 0.007 equals 0.034. The arithmetic is consistent.

Worked example 2: choosing a fit for two functions

Problem. On the same 25H7 hole, compare a g6 shaft (clearance) with a p6 shaft (interference) and select a fit for (a) a bushing that must rotate on the shaft and (b) a dowel pin that must stay pressed in permanently. Treat the p6 values qualitatively (verify against ISO 286-2). The complication is matching the fit class to two opposite functions.

Planning. Recall that g gives a shaft below the hole (clearance) and p gives a shaft above the hole (interference), then match to function.

Solution.

g6 (clearance). As computed, the shaft sits below the hole, giving a 0.007 to 0.041 gap. The parts slide; good where relative motion is needed.
p6 (interference). The p shaft zone sits above the hole zone, so the shaft is larger than the hole and must be pressed in; the joint grips and resists motion. (Exact p6 limits from ISO 286-2; qualitatively, minimum interference is small and positive, maximum interference larger.)
Rotating bushing. Needs to turn freely, so it needs clearance: choose 25H7/g6 (or a similar clearance fit). An interference fit would seize it.
Pressed dowel. Must not move, so it needs interference: choose 25H7/p6 (or a similar interference fit). A clearance fit would let it fall out.
Same hole, different shaft. Hole basis shines here: the 25H7 hole is unchanged; only the shaft letter changes (g for slide, p for grip), so one reamer makes both holes.

Comparison. g6 and p6 use the identical hole but produce opposite behaviors: g6 slides, p6 grips. Function dictates the shaft letter.

Result. Use 25H7/g6 for the rotating bushing (clearance) and 25H7/p6 for the pressed dowel (interference); hole basis keeps the hole common.

Independent check. For the rotating bushing, confirm minimum clearance is positive (0.007, a gap); for the dowel, confirm the shaft's lower limit exceeds the hole's upper limit (interference). Opposite signs confirm opposite fits.

Misconceptions and diagnostics

Misconception	Why it seems reasonable	Why it is wrong	Evidence that reveals it	Correction	Diagnostic question
"The letter sets the size."	Letters look like the main code.	The letter sets the zone position; the number (IT grade) sets its width.	Two fits with the same letter but different numbers have different widths.	Read letter as position, number as width.	"Does the letter give position or width?"
"Shaft basis is the normal system."	Shafts are the moving part.	Hole basis (H) is the usual default, because holes use fixed tooling.	Most catalog fits are quoted hole-basis.	Default to hole basis unless told otherwise.	"Is the hole fixed at H here?"
"Nominal size tells you the fit."	Both are 25.	The fit depends on the tolerance zones, not the shared nominal.	Same 25 nominal gives clearance or interference by zone.	Compare the zones, not the nominal.	"Where do the two tolerance zones sit relative to each other?"

Practice ladder

Level A - Recognition

Task. Classify eight given fits (from their limit values) as clearance, transition, or interference. Deliverable. Eight classifications. Success criteria. At least six correct, including a transition case. Answer guidance. Compare the shaft zone to the hole zone. Common errors. Calling an overlapping (transition) fit a clearance fit. Difficulty. Low.

Level B - Guided application

Task. Compute minimum and maximum clearance for three fits from supplied limit values, with prompts. Deliverable. A table of min/max clearance. Success criteria. Correct subtractions; signs interpreted (gap vs interference). Answer guidance. Min clearance equals smallest hole minus largest shaft; max equals largest hole minus smallest shaft. Common errors. Subtracting the wrong limits. Difficulty. Medium.

Level C - Independent application

Task. Select a fit class for several stated functions (running bushing, located gear, pressed bearing race) and justify each. Deliverable. A fit choice per function with justification. Success criteria. Clearance for motion, transition for location, interference for permanence, each justified. Answer guidance. Motion needs clearance; permanence needs interference; precise location needs transition. Common errors. Choosing interference for a rotating part. Difficulty. Medium to high.

Level D - Transfer and design

Task. For two interfaces of the bearing block (Project P4), specify fits and compute their clearances from provided ISO 286-2 table extracts, then justify against function. Deliverable. Two specified fits with computed clearances and rationale. Success criteria. Correct code reading; correct clearance arithmetic; fits match function; table values sourced. Answer guidance. Use the provided table extracts; do not invent deviations. Common errors. Guessing deviation values instead of using the table. Difficulty. High.

Working with AI, and proving it yourself

Use AI as a tutor

Useful AI support:

Ask it to explain the difference between g6 and p6 conceptually, then confirm with the zone positions.
Ask it to check your clearance arithmetic from limit values you provide.
Ask it to suggest a fit class for a function, then verify against the clearance you compute.

Limits:

A text assistant may state ISO 286 deviation values from memory that are wrong; always confirm against the standard.
It does not know your assembly's function or loads.

Verify AI output against: ISO 286-2 tables for the exact deviations, hand clearance arithmetic, and the function-to-fit mapping (motion/clearance, location/transition, permanence/interference).

Prove it yourself

A plausible but incorrect AI answer, and how to catch it. You ask, "For a shaft that must rotate freely in a bushing, is 25H7/p6 a good fit?" and the assistant replies: "Yes, H7/p6 is a standard, precise fit, so it will work well for a rotating shaft."

This is wrong: p6 is an interference fit, which grips and would prevent free rotation. Detect it by checking the zone positions: the p shaft sits above the hole, so the shaft is larger and presses in. The evidence is the clearance sign: minimum clearance is negative (interference), so it cannot rotate freely. Correct conclusion: a rotating shaft needs a clearance fit such as H7/g6, not an interference fit like H7/p6.

Retrieval and spaced review

Define clearance, transition, and interference fits.
In an ISO fit code, what does the letter set, and what does the number set?
Which member is fixed in hole basis, and why is it the default?
How do you compute minimum and maximum clearance?
Which fit type suits a rotating shaft? A pressed dowel?
Why can two parts with the same nominal have different fits?
Cumulative (L13): How does the IT grade (the number) relate to the tolerance idea from L13?
Reconstruction task: From memory, compute the clearance range of 25H7/g6 from the limits given in Worked Example 1.

Answers. 1: clearance, shaft always smaller (gap); interference, shaft always larger (grip); transition, either small gap or small interference. 2: the letter sets the tolerance-zone position (fundamental deviation), the number sets its width (IT grade). 3: the hole is fixed at H (lower deviation zero); default because holes use fixed tooling. 4: min equals smallest hole minus largest shaft; max equals largest hole minus smallest shaft. 5: clearance for a rotating shaft; interference for a pressed dowel. 6: the fit depends on the tolerance zones, not the shared nominal. 7: the IT grade is the standardized tolerance (zone width) for that size and grade.

Suggested review intervals. 1 day, 3 days, 7 days.

Reference mapping and next step

Read further

Giesecke ch.11
ISO 286-1:2010, ISO 286-2:2010.

Standards details must be checked against the current official edition used by your institution or employer.

Finish the lesson

You can now: classify the three fit types; read an ISO 286 code; apply hole basis; compute clearance; and select a fit from function.

Self-assessment checklist.

I can classify a fit from its tolerance zones.
I can read 25H7/g6 into basic size, hole, and shaft.
I know hole basis fixes the hole at H.
I can compute minimum and maximum clearance.
I choose fits by function (motion, location, permanence).

Next lesson: L15 - Datums and introductory geometric tolerancing (ISO 1101). Why it follows: size tolerances and fits control how big features are, but not their shape, orientation, or exact location. Next you learn how datums and geometric tolerances control those, at an introductory level, completing the toleranced definition.

Required files or submissions: submit your Level D two-fit specification for Project P4 with computed clearances. Optional extension: for a real shaft-and-hole joint you can find, identify whether it is clearance, transition, or interference, and reason about why.

Part III continues in 13-part3-lessons-cont.md with L15 (Datums and introductory geometric tolerancing), L16 (Surface texture, threads, and standard features), and L17 (Reading and inspecting a complete part drawing).

# Engineering Graphics and CAD - Phase 4: Full Lesson Content, Part III (continued), L15-L17

Continues 12-part3-lessons.md. Holds L15 (datums and introductory geometric tolerancing), L16 (surface texture, threads, and standard features), and L17 (reading and inspecting a complete part drawing). Introductory GD&T only, per the scope boundary; advanced GD&T is named as later study. Surface-texture and thread standard numbers are flagged for verification against the current catalogue. No em dashes.