Limits, fits and tolerances; GD&T basics

Limits, deviations, tolerances and fits (clearance, transition, interference), the ISO/IS 919 IT-grade system, and GD&T basics including datums, MMC/LMC, bonus tolerance and virtual condition.

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Why it matters

No machine makes two parts exactly the same size, so every drawing must say how much variation is acceptable and how mating parts should behave together: free running, locating, or pressed. Limits and fits make mass-produced parts interchangeable, and GD&T controls form, orientation and position, which size limits alone cannot. Getting this right decides whether a bearing seat, gear bore or dowel pin works on the first assembly or ends up as scrap.

Key ideas

Size terms

  • Basic (nominal) size: the size from which limits are set, e.g. Ø50 mm.
  • Limits: the maximum and minimum permissible sizes.
  • Tolerance: the difference between the upper and lower limits. It is always positive and has no sign.
  • Deviation: limit minus basic size. Upper deviation (ES for holes, es for shafts) and lower deviation (EI for holes, ei for shafts). Deviations carry a sign.
  • Fundamental deviation: the deviation closest to the zero line. It fixes the position of the tolerance zone and is shown by a letter (capital for holes, small for shafts).
  • Allowance: the intentional difference between the maximum material limits of mating parts (minimum hole minus maximum shaft). It is the tightest condition of the fit: positive for clearance, negative for interference.

Fits

  • Clearance fit: the smallest hole is larger than or equal to the largest shaft, so there is always clearance (or zero). Example: H7/g6 (location with small running clearance), H8/f7 (running fit).
  • Interference fit: the largest hole is smaller than the smallest shaft, so there is always interference. Example: H7/p6, H7/s6 (press fits for bushes and gears).
  • Transition fit: the tolerance zones overlap, so an assembled pair may have small clearance or small interference. Example: H7/k6, H7/n6 (location of gears, pulleys and couplings on keyed shafts).

Hole-basis and shaft-basis systems

  • Hole-basis: the hole is always an H hole (lower deviation EI = 0, so the minimum hole size equals the basic size). Different fits are obtained by changing the shaft letter. This is the preferred system because holes are made with fixed-size tools (drills, reamers) and checked with fixed plug gauges.
  • Shaft-basis: the shaft is always an h shaft (upper deviation es = 0, so the maximum shaft size equals the basic size), and the hole letter changes. Used where one bright-drawn shaft carries several parts with different fits.

ISO / IS 919 system

  • A tolerance is written as a letter plus a grade number: in Ø50 H7/g6, H7 is the hole and g6 the shaft.
  • There are 20 international tolerance grades IT01, IT0, IT1 … IT18. A smaller number means a tighter tolerance. Roughly: IT01–IT4 for gauges, IT5–IT7 for precision fits, IT8–IT11 for general machining, IT12–IT18 for rough work.
  • Tolerance grades IT5 to IT16 are multiples of the standard tolerance unit i, which grows with size because larger parts are harder to hold to the same absolute accuracy.
  • The diameter steps in the tables (e.g. 18–30, 30–50, 50–80 mm) are used with the geometric mean of the step as D.
  • Fundamental deviations are given by empirical formulas in the standard. Take exact values from the IS 919 / ISO 286 tables in your data book; the formulas reproduce them to within rounding.

GD&T basics

  • Size limits control only size. GD&T adds control of geometry through a feature control frame: symbol, tolerance value (with Ø if the zone is cylindrical), material modifier, then datum references in order of precedence (primary, secondary, tertiary).
  • Characteristic groups: form (straightness, flatness, circularity, cylindricity, no datum needed); orientation (parallelism, perpendicularity, angularity); location (position, and in older standards concentricity and symmetry); runout (circular and total runout); profile (of a line, of a surface).
  • Datums: theoretically exact planes, axes or points simulated by inspection equipment (surface plate, mandrel). The 3-2-1 principle: the primary datum is contacted at three points, the secondary at two, the tertiary at one, removing all six degrees of freedom.
  • Basic dimensions (boxed) give the exact true position; the tolerance comes from the frame, not from ± values.
  • Material conditions: MMC is the condition with most material (smallest hole, largest shaft); LMC has least material (largest hole, smallest shaft); RFS means the geometric tolerance applies whatever the size.
  • Bonus tolerance: when a position tolerance is specified at MMC, the tolerance grows by the amount the feature departs from MMC. This lets functional parts pass and is the basis of fixed functional gauges.
  • Virtual condition: the worst-case boundary of size plus geometry. For a hole at MMC it is MMC size minus the geometric tolerance; for a shaft it is MMC size plus the geometric tolerance. Mating parts assemble if the hole's virtual condition is at least the shaft's.
  • Envelope principle (Rule #1, ASME): a feature of size at MMC must have perfect form, so form errors must fit inside the size tolerance unless stated otherwise.

Formulas

Tolerance = Upper limit − Lower limit = ES − EI (hole) = es − ei (shaft)

  • Limits and deviations in mm or μm. Tolerance is always positive.

Maximum clearance = Hole_max − Shaft_min Minimum clearance (allowance) = Hole_min − Shaft_max

  • A negative result means interference. For an interference fit, maximum interference = Shaft_max − Hole_min.
  • Clearance fit if Hole_min ≥ Shaft_max; interference fit if Hole_max ≤ Shaft_min; otherwise transition.

i = 0.45·∛D + 0.001·D

  • i = standard tolerance unit (μm); D = geometric mean of the diameter step (mm), D = √(D₁·D₂). Applies to sizes up to 500 mm.

IT5 = 7i, IT6 = 10i, IT7 = 16i, IT8 = 25i, IT9 = 40i, IT10 = 64i, IT11 = 100i

  • From IT6 onward each grade is about 1.6 times the previous one (the R5 preferred-number series).

Shaft fundamental deviations (upper deviation es, μm, D in mm), as given in IS 919 tables: d: es = −16·D^0.44 e: es = −11·D^0.41 f: es = −5.5·D^0.41 g: es = −2.5·D^0.34 h: es = 0

  • For holes in the hole-basis system, H has EI = 0. Other shaft letters (k, n, p, s …) are read from the data book.

Bonus tolerance = |Actual size − MMC size| (for a modifier at MMC) Virtual condition (hole) = MMC size − geometric tolerance at MMC Virtual condition (shaft) = MMC size + geometric tolerance at MMC

Worked examples

Example 1 (standard): classifying a fit. Given: hole Ø50 mm with limits +0.05 / 0.00 mm, shaft Ø50 mm with limits 0.00 / −0.05 mm.

  1. Hole limits: Hole_max = 50.05 mm, Hole_min = 50.00 mm. Hole tolerance = 0.05 mm.
  2. Shaft limits: Shaft_max = 50.00 mm, Shaft_min = 49.95 mm. Shaft tolerance = 0.05 mm.
  3. Maximum clearance = Hole_max − Shaft_min = 50.05 − 49.95 = 0.10 mm.
  4. Minimum clearance = Hole_min − Shaft_max = 50.00 − 50.00 = 0.00 mm.
  5. Since Hole_min ≥ Shaft_max, there is never interference. This is a clearance fit with maximum clearance 0.10 mm and zero allowance (a sliding fit, not a transition fit).

Example 2 (GATE level): limits of Ø50 H7/g6 from the formulas. Given: Ø50 mm lies in the 30–50 mm step. Use i = 0.45·∛D + 0.001·D, IT7 = 16i, IT6 = 10i, and for g: es = −2.5·D^0.34.

  1. D = √(30 × 50) = 38.73 mm.
  2. i = 0.45 × 38.73^(1/3) + 0.001 × 38.73 = 0.45 × 3.383 + 0.0387 = 1.561 μm.
  3. IT7 = 16 × 1.561 = 24.98 μm ≈ 25 μm. IT6 = 10 × 1.561 = 15.6 μm ≈ 16 μm.
  4. Hole H7: EI = 0, ES = +25 μm. Hole limits: 50.000 to 50.025 mm.
  5. Shaft g: es = −2.5 × 38.73^0.34 = −8.67 μm ≈ −9 μm. ei = es − IT6 = −9 − 16 = −25 μm. Shaft limits: 49.975 to 49.991 mm.
  6. Maximum clearance = 50.025 − 49.975 = 0.050 mm. Minimum clearance = 50.000 − 49.991 = 0.009 mm.
  7. Result: hole 50.000/50.025 mm, shaft 49.975/49.991 mm; clearance fit with clearance 9 μm to 50 μm. These agree with the ISO 286 table values.

Example 3 (GD&T bonus tolerance). Given: hole Ø20 +0.10/0 mm with position tolerance Ø0.20 mm at MMC. A produced hole measures Ø20.08 mm.

  1. MMC of hole = 20.00 mm (smallest hole).
  2. Bonus = 20.08 − 20.00 = 0.08 mm.
  3. Allowed position tolerance = 0.20 + 0.08 = Ø0.28 mm.
  4. Virtual condition = 20.00 − 0.20 = Ø19.80 mm. Any mating pin whose virtual condition is at most Ø19.80 mm will always assemble, and a fixed Ø19.80 mm functional pin gauge checks size and position together.

Common mistakes

  • Calling a fit "transition" when the minimum clearance is exactly zero. Zero allowance with no possible interference is still a clearance fit.
  • Mixing the direction of subtraction: maximum clearance uses the largest hole and the smallest shaft; minimum clearance uses the smallest hole and the largest shaft.
  • Writing tolerance with a sign, or giving a "range" when the tolerance (a single number) is asked.
  • Using the nominal size instead of the geometric mean of the diameter step in the formula for i.
  • Forgetting that i is in micrometres while D is in millimetres.
  • Swapping MMC and LMC for holes: the MMC of a hole is its smallest size.
  • Assuming GD&T bonus tolerance applies when the frame says RFS (no modifier).

For GATE ME

Expect numericals on: limits and clearance or interference from given deviations; deciding the type of fit; computing IT grades and fundamental deviations from the standard tolerance unit formula (always use the geometric mean of the step); and, often in the same question, designing GO/NO-GO gauges for the computed limits (see the gauges topic). Conceptual MCQs test hole-basis versus shaft-basis, MMC/LMC, and which GD&T characteristics need a datum. Practise rounding to whole micrometres the way the tables do.

Quick check

  1. A hole is 25.000/25.021 mm and a shaft is 25.022/25.035 mm. What type of fit is this?
  2. What is the MMC size of a hole Ø12 +0.018/0 mm?
  3. In the hole-basis system, which deviation of the hole is zero?
  4. Using i = 0.45·∛D + 0.001·D, roughly how many times larger is IT8 than IT7?
  5. Does flatness need a datum reference?

Answers: 1. Interference fit (smallest shaft 25.022 mm exceeds largest hole 25.021 mm). 2. Ø12.000 mm. 3. The lower deviation EI = 0. 4. 25/16 ≈ 1.56 times. 5. No, it is a form tolerance and never references a datum.

Try answering each one aloud before you open it.

  1. 1.What are limits, fits, and tolerances in the context of mechanical engineering?Concept

    Limits are the maximum and minimum permissible sizes of a feature, and tolerance is their difference, the total permitted variation (always positive). A fit is the relationship between mating parts, usually a hole and a shaft, defined by the clearance or interference their limits produce: clearance, transition or interference. In the ISO system a fit is written like Ø50 H7/g6, where the letter fixes the position of the tolerance zone (fundamental deviation) and the number fixes its width (IT grade).

  2. 2.Explain the importance of Geometric Dimensioning and Tolerancing (GD&T) in manufacturing.Concept

    Size limits do not control form, orientation or location, so a part can be within size yet still not assemble or function. GD&T uses feature control frames and datums to state exactly which geometry matters and how much it may vary relative to a defined reference frame. This makes the drawing unambiguous for machining and inspection, ties tolerances to function, and with MMC modifiers allows bonus tolerance and functional gauging.

  3. 3.What is the difference between a clearance fit and an interference fit?Concept

    In a clearance fit the smallest hole is at least as large as the largest shaft, so there is always a gap (or zero) and the parts can slide or rotate, e.g. H7/g6 or H8/f7. In an interference fit the largest hole is smaller than the smallest shaft, so the shaft is always oversize relative to the hole and the parts must be pressed or shrink-fitted together, e.g. H7/p6 or H7/s6. The interference creates contact pressure that transmits torque or axial load without keys.

  4. 4.Why is it important to specify tolerances in engineering drawings?Application

    No process makes exact sizes, so a dimension without a tolerance cannot be accepted or rejected objectively. The tolerance tells the machinist which process is needed and tells the inspector the accept/reject limits. It guarantees interchangeability and function of mating parts. Tolerances should be as wide as function allows, because cost rises steeply as tolerance tightens.

  5. 5.How does GD&T contribute to reducing manufacturing costs?Application

    GD&T tolerances only what affects function, so non-critical features can have open tolerances. Cylindrical position zones give about 57% more area than the equivalent square ± zone, and MMC modifiers add bonus tolerance as a feature departs from MMC, so fewer good parts are rejected. Clear datums also mean the part is fixtured and inspected the same way it functions, which cuts disputes, rework and scrap.

  6. 6.What happens if a part is manufactured outside its specified tolerance?Application

    It is nonconforming and is rejected at inspection: oversize shafts or undersize holes can often be reworked, while undersize shafts and oversize holes are usually scrap. If it escapes inspection it may fail to assemble, run loose, overheat or wear early. A deviation request can accept it only if engineering confirms the function is unaffected. Repeated out-of-tolerance parts mean the process capability is inadequate and must be fixed at source.

  7. 7.Explain the concept of a transition fit and provide an example of its application.Concept

    In a transition fit the hole and shaft tolerance zones overlap, so a given pair may have small clearance or small interference. It gives accurate location while still allowing assembly with light pressure or a mallet. Typical ISO choices are H7/k6 and H7/n6, used for gears, pulleys, couplings and bearing inner rings located on shafts, usually with a key carrying the torque.

  8. 8.Calculate the maximum and minimum clearance for a hole-shaft assembly where the hole is 50 mm with tolerance +0.1/0 mm and the shaft is 49.8 mm with tolerance 0/−0.05 mm.Numerical

    Hole limits are 50.00 to 50.10 mm and shaft limits are 49.75 to 49.80 mm. Maximum clearance = largest hole − smallest shaft = 50.10 − 49.75 = 0.35 mm. Minimum clearance = smallest hole − largest shaft = 50.00 − 49.80 = 0.20 mm. Both are positive, so it is a clearance fit.

  9. 9.What is the role of a datum in GD&T?Concept

    A datum is a theoretically exact plane, axis or point, derived from a real datum feature, that establishes the reference frame for orientation and location tolerances. The datum order in the feature control frame (primary, secondary, tertiary) tells how the part is located: 3 points on the primary, 2 on the secondary and 1 on the tertiary, fixing all six degrees of freedom. Good datums are functional surfaces, so the part is measured the way it is assembled.

  10. 10.If a shaft is designed to have an interference fit with a hole, what considerations should be made during assembly?Application

    Check the maximum interference against the press force and the hub and shaft stresses (thick-cylinder Lamé analysis), so the hub does not yield or crack. For larger interference use shrink fitting (heat the hub) or expansion fitting (cool the shaft, e.g. liquid nitrogen) instead of pressing. Chamfer the leading edges, lubricate, keep the parts coaxial during pressing to avoid galling, and remember the hole bore and shaft surface finish reduce the effective interference.

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