Limits, fits and tolerances
Limits, deviations, tolerances and allowances; clearance, transition and interference fits; hole- and shaft-basis systems; ISO grades and fundamental deviations; and tolerance stack-up with worked ISO-fit calculations.
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Why it matters
A gudgeon pin must slide in its piston boss but be firm in the small end; a bearing must be pressed onto a shaft yet slide into its housing; a gearbox must accept a replacement gear from any supplier. All of this is specified with limits and fits. A correct tolerance gives function and interchangeability at the lowest cost; a needlessly tight one multiplies machining cost, and a loose one causes noise, wear or seizure.
Key ideas
Basic vocabulary (for a hole and a shaft).
- Basic (nominal) size – the size from which limits are derived (e.g. 40 mm).
- Limits of size – the maximum and minimum permissible sizes.
- Deviation – limit minus basic size. Upper deviation: ES for holes, es for shafts. Lower deviation: EI for holes, ei for shafts.
- Tolerance – maximum limit − minimum limit = upper deviation − lower deviation (always positive).
- Fundamental deviation – the deviation closest to the zero line; it fixes the tolerance zone's position and is denoted by a letter.
- Allowance – the intentional difference between the maximum material limits of mating parts: minimum clearance (positive) or maximum interference (negative).
- Unilateral tolerance lies on one side of the basic size (e.g. 40 +0.025/0); bilateral on both (40 ± 0.02).
Fits.
- Clearance fit – the smallest hole is larger than or equal to the largest shaft; there is always clearance (or zero in the limiting case). Bearings, sliding pins.
- Interference fit – the largest hole is smaller than the smallest shaft; assembly needs pressing or shrinking (heating the hub or cooling the shaft). Valve seats, gear hubs, bearing races on rotating shafts.
- Transition fit – depending on actual sizes there may be a small clearance or a small interference. Location of gears, couplings and spigots.
Hole-basis vs shaft-basis. In the hole-basis system the hole has a fixed position (H: lower deviation zero) and different shafts give different fits; it is preferred because holes are made with fixed-size tools (drills, reamers) and checked with fixed plug gauges, whereas shafts are easily turned or ground to any size. Shaft-basis (h shaft: upper deviation zero) is used when one standard shaft (e.g. bright drawn bar, a bearing outer race) must mate with several holes.
ISO system (ISO 286, adopted in India as IS 919). A fit is written as basic size + hole letter and grade / shaft letter and grade, e.g. 40 H7/g6.
- Letters give the fundamental deviation: capital for holes (A … ZC), lower-case for shafts (a … zc). H and h have zero fundamental deviation. Shafts a–g lie below the zero line (clearance with H holes), js/j/k/m/n are transition, and p–zc lie above it (interference). The pattern is the mirror image for holes.
- Grades IT01, IT0, IT1 … IT18 give the tolerance magnitude: IT5–IT7 for precision grinding and reaming, IT7–IT9 for fine turning and boring, IT11–IT14 for rough work. In a fit the hole is usually one grade coarser than the shaft (holes are harder to make accurately).
- Tolerance values come from the standard tolerance unit i for the diameter step, multiplied by a grade factor (IT6 = 10i, IT7 = 16i, IT8 = 25i, IT9 = 40i, IT10 = 64i, IT11 = 100i for IT5–IT16 in the R5 progression).
- Diameter steps (e.g. 18–30, 30–50, 50–80 mm) are used with their geometric mean.
- In practice, read deviations directly from the standard tables.
Common hole-basis fits. H7/g6 (close running/sliding), H7/h6 (locational clearance), H7/k6 (transition, accurate location), H7/n6 (tighter transition), H7/p6 (light press), H7/s6 (medium drive).
Tolerance stack-up. In an assembly chain the worst-case tolerance of the resultant dimension is the sum of the individual tolerances. The statistical (root-sum-square) method assumes random, centred, independent variation and gives a smaller, more realistic band. Selective assembly (grading parts into groups and mating like with like) achieves tight fits from wider manufacturing tolerances.
Formulas
T_h = Hmax − Hmin = ES − EI T_s = Smax − Smin = es − ei
Hole and shaft tolerance (mm or µm).
C_max = Hmax − Smin C_min = Hmin − Smax
Maximum and minimum clearance (mm); a negative value is an interference.
I_max = Smax − Hmin
Maximum interference (mm), for interference and transition fits.
T_fit = C_max − C_min = T_h + T_s
Fit tolerance (variation of clearance).
i = 0.45·D^(1/3) + 0.001·D
Standard tolerance unit (µm) for sizes up to 500 mm; D = geometric mean of the diameter step, D = √(D₁·D₂) (mm).
es (g) = −2.5·D^0.34 es (f) = −5.5·D^0.41 es (e) = −11·D^0.41 es (d) = −16·D^0.44
Empirical formulas for the upper deviation of common clearance shafts (µm), as given in IS 919/ISO-based textbooks; use the standard tables for exact rounded values.
T_worst = Σ T_k T_RSS = √(Σ T_k²)
Worst-case and root-sum-square stack-up of a chain of dimensions (T_k = ± half-tolerances, or full tolerances used consistently).
Worked examples
Example 1 (standard) – limits of a 40 H7/g6 fit.
- Diameter step 30–50 mm: D = √(30 × 50) = 38.73 mm.
i = 0.45·D^(1/3) + 0.001·D= 0.45 × 3.383 + 0.039 = 1.561 µm.- IT7 = 16i = 25.0 → 25 µm; IT6 = 10i = 15.6 → 16 µm.
- H hole: EI = 0, ES = +25 µm ⇒ hole 40.000 – 40.025 mm.
- g shaft: es = −2.5 × 38.73^0.34 = −8.7 → −9 µm; ei = −9 − 16 = −25 µm ⇒ shaft 39.975 – 39.991 mm.
- C_max = 40.025 − 39.975 = 0.050 mm; C_min = 40.000 − 39.991 = 0.009 mm – a clearance fit. These match the tabulated ISO values (+25/0 and −9/−25 µm).
Example 2 (GATE level) – stack-up in a slot. Three blocks of 20 ± 0.02, 30 ± 0.03 and 50 ± 0.05 mm are placed in a slot of 100.20 ± 0.04 mm. Find the gap.
- Block stack: nominal 100.00, worst-case ± (0.02 + 0.03 + 0.05) = ± 0.10 ⇒ 99.90 to 100.10 mm.
- Maximum gap = 100.24 − 99.90 = 0.34 mm; minimum gap = 100.16 − 100.10 = 0.06 mm (worst case: nominal 0.20 ± 0.14).
- Statistical estimate: ±√(0.02² + 0.03² + 0.05² + 0.04²) = ±0.073 mm ⇒ gap ≈ 0.20 ± 0.073 mm, i.e. about 0.13 to 0.27 mm.
Example 3 – clearance from given deviations. Hole 40 (+0.04/−0.02), shaft 40 (+0.01/−0.03): C_max = 40.04 − 39.97 = 0.07 mm; C_min = 39.98 − 40.01 = −0.03 mm, i.e. up to 0.03 mm interference – a transition fit.
Common mistakes
- Calling the hole basis "the hole is at basic size": it means the hole's lower limit is the basic size (EI = 0).
- Confusing allowance (difference at maximum material conditions) with tolerance (variation of one part).
- Mixing µm deviations with mm sizes without converting.
- Using the actual size instead of the geometric mean of the diameter step in the i formula.
- Reading interference as positive clearance – keep the sign: C_min < 0 means interference.
- Assuming a transition fit is "half-way between" – it is defined by overlapping zones, and the result can be either clearance or interference.
For GATE ME
Expect numericals on limits, maximum/minimum clearance or interference, the i formula and IT grades, fundamental deviations for ISO fits, identification of fit type from given deviations, and worst-case stack-up. Conceptual questions cover hole-basis vs shaft-basis, letters and grades, and selective assembly. Practise one complete ISO fit calculation by hand.
Quick check
- In the hole-basis system, which deviation of the hole is zero?
- Hole 25.000–25.021, shaft 25.022–25.035 mm: what type of fit?
- What is IT7 in terms of i?
- Maximum clearance for hole 50.00–50.03 and shaft 49.95–49.98 mm?
- Which is usually the coarser grade in a fit, hole or shaft?
Answers: 1. The lower deviation EI; 2. Interference fit (smallest shaft exceeds largest hole); 3. 16i; 4. 0.08 mm; 5. The hole.
Interview questions
All Engineering Materials and Manufacturing Processes interview questionsTry answering each one aloud before you open it.
1.What are limits, fits, and tolerances in engineering?Concept
Limits, fits, and tolerances are terms used to describe the allowable variations in dimensions of manufactured parts. Limits define the maximum and minimum sizes of a part. Fits describe the relationship between two mating parts, such as a shaft and a hole. Tolerances specify the permissible variation in a dimension, ensuring parts fit together properly and function as intended.
2.Explain the difference between clearance fit, interference fit and transition fit.Concept
In a clearance fit the smallest permissible hole is at least as large as the largest permissible shaft, so there is always clearance – used for running and sliding parts. In an interference fit the largest hole is smaller than the smallest shaft, so the parts must be pressed or shrink-fitted together – used where torque or axial load must be carried without fasteners. In a transition fit the tolerance zones overlap, so a particular pair may have a small clearance or a small interference – used for accurate location, as in H7/k6.
3.Why are tolerances important in manufacturing processes?Application
Tolerances are crucial in manufacturing because they ensure that parts will fit together properly and function as intended. They allow for some variation in manufacturing processes while maintaining the quality and performance of the final product. Tolerances help in reducing waste, improving efficiency, and ensuring interchangeability of parts.
4.What could happen if a part is manufactured outside its specified tolerance?Application
If a part is manufactured outside its specified tolerance, it may not fit properly with other components, leading to assembly issues or malfunction. This can result in increased wear, reduced performance, or even failure of the product. It may also lead to increased costs due to rework or scrap.
5.How do engineers decide on the appropriate fit for a particular application?Application
Engineers consider factors such as the function of the assembly, the materials used, the operating environment, and the required precision. They also refer to standards and guidelines, such as ISO or ANSI, to determine the appropriate fit. The decision is based on ensuring the desired performance, reliability, and manufacturability of the product.
6.What is the role of a tolerance stack-up analysis in design?Application
Tolerance stack-up analysis is used to evaluate the cumulative effect of individual part tolerances in an assembly. It helps in identifying potential issues with fit and function before manufacturing. By analyzing the stack-up, engineers can ensure that the assembly will meet design requirements and function properly, reducing the risk of costly redesigns or rework.
7.Explain how a go/no-go gauge is used in quality control.Application
A go/no-go gauge is a tool used to check whether a part's dimensions fall within specified tolerances. The 'go' side of the gauge should fit the part if it is within tolerance, while the 'no-go' side should not fit if the part is out of tolerance. This simple pass/fail test helps ensure that parts meet quality standards without requiring precise measurements.
8.Calculate the maximum and minimum clearance for a shaft with a diameter tolerance of 50 ± 0.1 mm and a hole with a diameter tolerance of 50.2 ± 0.1 mm.Numerical
Maximum clearance = Maximum hole size - Minimum shaft size = (50.2 + 0.1) mm - (50 - 0.1) mm = 50.3 mm - 49.9 mm = 0.4 mm. Minimum clearance = Minimum hole size - Maximum shaft size = (50.2 - 0.1) mm - (50 + 0.1) mm = 50.1 mm - 50.1 mm = 0 mm.
9.A hole is specified with a basic size of 30 mm and a tolerance of +0.05/-0.02 mm. What are the maximum and minimum sizes of the hole?Numerical
Maximum size of the hole = Basic size + Upper tolerance = 30 mm + 0.05 mm = 30.05 mm. Minimum size of the hole = Basic size - Lower tolerance = 30 mm - 0.02 mm = 29.98 mm.
10.What is the significance of using ISO system of limits and fits in global manufacturing?Application
The ISO system of limits and fits provides a standardized approach to defining tolerances and fits, facilitating international trade and collaboration. It ensures compatibility and interchangeability of parts produced in different countries, reducing the need for custom adjustments. This standardization helps in maintaining quality, reducing costs, and improving efficiency in global manufacturing.
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