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Chapter 9 · 8 hours

Limits, Fits and Tolerance

Practice questions

Practice questions and answers

6 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 8 marks

Define the following terms with a neat sketch: basic size, actual size, limits of size, deviation (upper and lower), fundamental deviation, tolerance, allowance, hole and shaft, zero line, and clearance and interference.

Answer

   +  ------------------ upper limit
      |  tolerance zone |   (upper deviation)
      ------------------ lower limit
 0 ============================ zero line
      (basic size)                (lower deviation
   -                               if below the line)

Deviations are measured from the zero line: positive above, negative below.

TermMeaning
Basic (nominal) sizeThe size from which limits are derived; the theoretical size on the drawing, e.g. 40 mm
Actual sizeThe size found by measurement of the finished part
Limits of sizeThe two extreme permissible sizes: maximum limit and minimum limit
DeviationAlgebraic difference between a size and the basic size: upper deviation = max. limit - basic size (ES for hole, es for shaft), lower deviation = min. limit - basic size (EI, ei)
Fundamental deviationThe deviation (upper or lower) nearest to the zero line; it fixes the position of the tolerance zone, given by a letter (A to ZC for holes, a to zc for shafts)
ToleranceDifference between the maximum and minimum limits = upper deviation - lower deviation; the permissible variation in size. Size of zone is given by IT grade (IT01 to IT16)
AllowanceThe intentional difference between the dimensions of the mating parts; the minimum clearance (positive) or the maximum interference (negative). It is the difference between the maximum material limits of hole and shaft
Hole / shaftThe internal and external features of a part; a hole is not necessarily round (e.g. a keyway width), a shaft includes any external feature
Zero (reference) lineThe line representing the basic size; deviations are positive above and negative below
ClearanceThe positive difference between the hole size and the shaft size (hole larger than shaft)
InterferenceThe negative difference (shaft larger than hole before assembly)

Tolerance zone: the zone between the upper and lower limits, defined by the grade (size) and the fundamental deviation (position), as in 40 H740\ \text{H7}.

  • Practice · 8 marks

Explain clearance, transition and interference fits with sketches and give one application of each. Write the hole basis and shaft basis fits in the ISO system with an example for each type.

Answer

A fit is the relationship between a hole and a shaft that are assembled, resulting from the difference of their sizes before assembly.

Types

  Clearance    Transition    Interference
   |hole|       |hole|         |shaft|
   |    |       |    |         |     |
                |shaft|        |hole|
   |shaft|      |     |
   (hole zone   (zones         (shaft zone
    above shaft  overlap)       above hole)
  1. Clearance fit: the shaft is always smaller than the hole, so there is always a gap (minimum clearance ≥0\ge 0). The parts can slide or rotate. Example: shaft in a plain bearing (H7/g6, H8/f7), sliding fit of a piston in a cylinder (running fits).
  2. Interference (force/press) fit: the shaft is always larger than the hole, so the assembly needs force (pressing, or heating or cooling the parts) and they are fixed together by friction. Example: bush in a housing, gear on a shaft, wheel on an axle (H7/p6, H7/s6).
  3. Transition fit: the sizes overlap, so, depending on the actual dimensions, there may be a small clearance or a small interference. It is used for accurate location without much load: pulley on shaft with a key, spigot location, coupling hub (H7/k6, H7/n6).

Hole basis and shaft basis

  • Hole basis system: the hole has zero lower deviation (H); the fits are obtained by changing the shaft's fundamental deviation. Examples: H7/g6 (clearance), H7/k6 (transition), H7/p6 (interference).
  • Shaft basis system: the shaft has zero upper deviation (h); the fits are obtained by changing the hole's deviation. Examples: G7/h6 (clearance), K7/h6 (transition), P7/h6 (interference).

The minimum clearance is the difference between the lower limit of hole and the upper limit of shaft; the maximum clearance is the difference between the upper limit of hole and the lower limit of shaft.

  • Practice · 5 marks

Differentiate between the hole basis system and the shaft basis system of fits. Why is the hole basis system more commonly used?

Answer

In the hole basis system the basic size is the minimum limit (lower limit) of the hole, so the lower deviation of the hole is zero (letter H). Different fits are obtained by varying the size of the shaft. In the shaft basis system the basic size is the maximum limit of the shaft, so the upper deviation of the shaft is zero (letter h). Different fits are obtained by varying the size of the hole.

PointHole basisShaft basis
Fixed memberHole (H, lower deviation 0)Shaft (h, upper deviation 0)
Varying memberShaftHole
Example40 H7/g6, H7/k6, H7/p640 G7/h6, K7/h6, P7/h6
Tools for the fixed partStandard reamers, broaches and plug gauges are limited in numberStandard shafts, ground bars
CostLower for general useHigher: many hole sizes need many reamers
UsedMachine tools, general engineeringWhere long drawn or ground bars of a standard size carry many parts, e.g. line shafting, textile and agricultural machines

Why hole basis is preferred

  • Holes are made by drills, reamers, boring and broaches whose sizes are fixed; one reamer size of H7 covers all fits.
  • Shafts are easily turned and ground to any size, so the changing member is the cheaper one to change.
  • Fewer tools and gauges are needed, so the cost of production and inspection is lower.

The shaft basis is used when a single shaft carries several parts with different fits (for example a bar on which several bearings, collars, and pulleys are fitted) because cutting steps on a shaft would weaken it and cost more.

  • Practice · 5 marks

Differentiate between unilateral and bilateral tolerance with examples and state when each is used. Convert the bilateral dimension 40.00±0.0340.00 \pm 0.03 mm into a unilateral form with the same limits, and state the type of tolerance in 60+0.02+0.0560^{+0.05}_{+0.02} mm.

Answer

PointUnilateralBilateral
DefinitionTolerance on one side of the basic size onlyTolerance on both sides of the basic size
Example250+0.0325^{+0.03}_{0} or 25−0.02025^{0}_{-0.02}25±0.0225 \pm 0.02 or 25−0.01+0.0325^{+0.03}_{-0.01}
Basic sizeOne of the limitsLies between the limits
UseInterchangeable parts, hole and shaft systems (the basic size is the datum of fits); allows fixed tools to be used and the tolerance to be changed without changing the datumWhere variation on either side is acceptable, e.g. positions of holes, centre distances, and in machining and assembly dimensions
GaugingEasier to adjust setting and to change toleranceSymmetrical; the setting at the mean size

Conversion: the limits of 40.00±0.0340.00 \pm 0.03 are 39.9739.97 and 40.0340.03 mm; the total tolerance is 0.060.06 mm. As a unilateral dimension with the same limits:

39.97 0+0.06 mm39.97^{+0.06}_{\ 0}\ \text{mm}

Type of 60+0.02+0.0560^{+0.05}_{+0.02}: both deviations are positive, so the limits are 60.0260.02 and 60.0560.05 mm and the basic size 60 mm lies outside the tolerance zone. It is a unilateral tolerance (both limits are on the same side of the basic size), with a tolerance of 0.05−0.02=0.030.05 - 0.02 = 0.03 mm.

  • Practice · 8 marks

For a 40 mm H7/g6 fit calculate the limits of the hole and the shaft, the tolerances, the maximum and minimum clearance, and state the type of fit. Use the standard tolerance unit i=0.45D3+0.001Di = 0.45\sqrt[3]{D} + 0.001D (μ\mum, DD in mm), IT6 =10i= 10i, IT7 =16i= 16i and the fundamental deviation for shaft 'g' =−2.5D0.34= -2.5D^{0.34} (μ\mum); DD is the geometric mean of the diameter step 30 to 50 mm. Round tolerances and deviation to the nearest standard value.

Answer

Step 1: diameter step

D=30×50=38.73 mmD = \sqrt{30 \times 50} = 38.73\ \text{mm}

Step 2: tolerance unit

i=0.45 (38.73)1/3+0.001(38.73)=0.45(3.383)+0.039=1.561 μmi = 0.45\,(38.73)^{1/3} + 0.001(38.73) = 0.45(3.383) + 0.039 = 1.561\ \mu\text{m}
  • IT7 =16i=24.98 μ= 16i = 24.98\ \mum, standard value 25 μ\mum
  • IT6 =10i=15.6 μ= 10i = 15.6\ \mum, standard value 16 μ\mum

Step 3: fundamental deviation of shaft g

es=−2.5 D0.34=−2.5(38.73)0.34=−2.5(3.468)=−8.67 μm≈−9 μmes = -2.5\,D^{0.34} = -2.5(38.73)^{0.34} = -2.5(3.468) = -8.67\ \mu\text{m} \approx -9\ \mu\text{m}

Step 4: limits

Hole 40 H7 (lower deviation zero):

  • Lower limit =40.000= 40.000 mm
  • Upper limit =40.000+0.025=40.025= 40.000 + 0.025 = 40.025 mm
  • Tolerance =0.025= 0.025 mm

Shaft 40 g6 (upper deviation −0.009-0.009 mm):

  • Upper limit =40−0.009=39.991= 40 - 0.009 = 39.991 mm
  • Lower limit =39.991−0.016=39.975= 39.991 - 0.016 = 39.975 mm
  • Tolerance =0.016= 0.016 mm

Step 5: clearances

Cmax=hole max−shaft min=40.025−39.975=0.050 mmC_{max} = \text{hole max} - \text{shaft min} = 40.025 - 39.975 = 0.050\ \text{mm} Cmin=hole min−shaft max=40.000−39.991=0.009 mmC_{min} = \text{hole min} - \text{shaft max} = 40.000 - 39.991 = 0.009\ \text{mm}

Check: Cmax−Cmin=0.041=0.025+0.016C_{max} - C_{min} = 0.041 = 0.025 + 0.016.

Both clearances are positive, so this is a clearance fit (close running or sliding fit).

Answer: hole 40.000 to 40.025 mm; shaft 39.975 to 39.991 mm; CmaxC_{max} = 0.050 mm, CminC_{min} = 0.009 mm; clearance fit.

  • Practice · 6 marks

The limits of size of three hole-shaft pairs are given below. For each, find the tolerances, the maximum and minimum clearance (or interference), and state the type of fit. (a) Hole 30.000 to 30.021 mm; shaft 29.967 to 29.980 mm. (b) Hole 50.000 to 50.025 mm; shaft 50.002 to 50.018 mm. (c) Hole 25.000 to 25.021 mm; shaft 25.035 to 25.048 mm.

Answer

Use: Cmax=C_{max} = hole max −- shaft min; Cmin=C_{min} = hole min −- shaft max. A negative value is interference.

(a)(b)(c)
Hole tolerance0.0210.0210.0250.0250.0210.021
Shaft tolerance0.0130.0130.0160.0160.0130.013
CmaxC_{max} = hole max - shaft min30.021−29.967=+0.05430.021 - 29.967 = +0.05450.025−50.002=+0.02350.025 - 50.002 = +0.02325.021−25.035=−0.01425.021 - 25.035 = -0.014
CminC_{min} = hole min - shaft max30.000−29.980=+0.02030.000 - 29.980 = +0.02050.000−50.018=−0.01850.000 - 50.018 = -0.01825.000−25.048=−0.04825.000 - 25.048 = -0.048
Fit typeClearanceTransitionInterference

Results

  • (a) Clearance fit: the clearance is always between 0.020 mm and 0.054 mm. The shaft is always smaller than the hole, so it is a running fit (like H7/f7).
  • (b) Transition fit: the maximum clearance is 0.023 mm and the maximum interference is 0.018 mm. Depending on actual sizes it may be a clearance or an interference. Used for accurate location (like H7/k6).
  • (c) Interference fit: the minimum interference is 0.014 mm and the maximum is 0.048 mm. Always a press fit (like H7/s6).

The three pairs are on the hole basis, as the hole lower limit equals the basic size in all of them (lower deviation zero).

Answer: (a) clearance, 0.020 to 0.054 mm; (b) transition, +0.023 to -0.018 mm; (c) interference, 0.014 to 0.048 mm.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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