Limits and Fits

Limits and Fits:

Leaving Cert. Notes

Limits and Fits:Introduction:In production engineering it is not possible to make components to an exact size every time. If a component cannot be made to an exact size then the amount by which it can be in error must be known and included with the dimension.

The maximum and minimum sizes of a component are known as Limits. The difference between the maximum and minimum sizes (limits) is known as the Tolerance.The ISO system of Limits and Fits

This system of limits and fits gives a range of sizes to which parts should be made if the type of fit is known. The following is an example of the types of fit in common use.

(a)Clearance Fit.

(b)Interference Fit.

(c)Transition Fit.

(a)Clearance Fit.

With this type of fit there is Space between the two parts. The shaft is always smaller than the part it fits into.

(b)Interference Fit.

In this assembly there is No Space between the parts. This means that force is required to assemble the parts.

(c)Transition Fit.

This is a range of fits which can be either clearance or interference depending on the tolerance. i.e. the shaft can be either smaller or larger than the part it fits into.

Tolerances:

The type of fit between two components depends on the size to which each component is made Since no size can be exact then each part must be made within two sizes. These two sizes are called the Limits.

e.g.20.02- Upper Limit.

19.98- Lower Limit.

The Tolerance is calculated by subtracting the lower limit from the upper limit.

e.g.20.02

19.98

Tolerance = 00.04Bilateral Tolerance.

This is when the tolerance is specified above and below the nominal diameter.

e.g.25 ( 0.02

Unilateral Tolerance.

This is when the specified tolerance lies to one side of the nominal diameter.

e.g.25 + 0.02

Systems of Fits.

There are two systems of fits in use.

(i)Hole Basis System.

(ii)Shaft Basis System.

The Hole Basis System.

This is the preferred system. It involves manufacturing the hole to a fixed size and the shaft size is varied.

The Shaft Basis System.

In this system the shaft size is fixed and the hole size is varied.

Limit Gauges.Limit gauges are use to ensure that components are within the specified limits. Two gauges are used.

One for the Upper Limit- GO Gauge

One for the Lower Limit- NOT GO Gauge.

Types of Gauges.

Gap Gauge

- used for checking Shafts

Ring Gauge- used for checking Shafts

Plug Gauge

- used for checking Holes.

Interchangeability.In this system the parts are manufactured within certain limits and when selected at random will fit together.

Selective Assembly.In this system the parts are measured before assembly. The parts that give the most satisfactory fit are assembled together

Example 1.Determine the limits on a shaft and a bearing of 20mm nominal diameter, which fit together with a clearance fit. The tolerance on the shaft is g6 and on the bearing H7

Calculate.

(a)The tolerance on the shaft.

(b)The tolerance on the bearing.

(c)The maximum clearance.

(d)The minimum clearance.

Solution:

Basic dia. = 20mm

From Tables.

Tolerance.

Bearing(H7)

= + 0.021, 0

Shaft (g6)

= - 0.007, - 0.020

Largest size

Minimum size

(Upper Limit)

(Lower Limit)

Bearing.20.021

20.000

Shaft19.993

19.980

Tolerance. (Difference between the limits)

Shaft.

Bearing.

19.993 (UL)

20.021 (UL)

19.980 (LL)

20.000 (LL)

0.013

0.021Maximum Clearance.

Largest hole size - Smallest shaft size.

20.021 - 19.980

= 0.041Minimum Clearance.

Smallest hole size - Largest shaft size

20.000 - 19.993

= 0.007Ans.

(a)0.013

(b)0.021

(c)0.041

(d)0.007

Example 2.

Determine the Limits on a shaft and a pulley of 40mm nominal diameter. The tolerance on the shaft is h6 and on the pulley is H7

Calculate.


(b)The tolerance on the pulley.



Solution:

Basic dia. = 40mm

From Tables.

Tolerance.

Pulley (H7)

= + 0.025, 0

Shaft (h6)

= - 0.016, 0

Largest size

Minimum size

(Upper Limit)

(Lower Limit)

Pulley.40.025

40.000

Shaft40.000

39.084


Shaft.

Pulley.

40.000 (UL)

40.025 (UL)

39.084 (LL)

40.000 (LL)

0.016



40.025 - 39.984



40.000 - 40.000

= 0.000Ans.

(a)0.016

(b)0.025

(c)0.041

(d)0.000

Example 3.

Determine the Limits on a shaft and a pulley of 45mm nominal diameter. The tolerance on the shaft is g6 and on the pulley is H7

Calculate.


(b)The tolerance on the pulley.



Solution:

Basic dia. = 40mm

From Tables.

Tolerance.

Pulley (H7)

= + 0.025, 0

Shaft (h6)

= - 0.025, - 0.009

Largest size

Minimum size

(Upper Limit)

(Lower Limit)

Pulley.45.025

45.000

Shaft44.991

44.975


Shaft.

Pulley.

44.991 (UL)

40.025 (UL)

44.975 (LL)

40.000 (LL)

0.016



45.025 - 44.975



45.000 - 44.991

= 0.009Ans.

(a)0.016

(b)0.025

(c)0.050

(d)0.009

Revision Questions - Limits and Fits:1.Explain the following terms.

(a)Limits.

(b)Tolerance.

(c)Upper Limit

(d)Lower Limit

(e)Nominal diameter.

2.State and explain the three common types of fit in use.

3.(a)Explain the difference between the Hole Basis System and the Shaft

Basis System.

(b)Give examples of where both might be used.

4.Explain the following.

(a)Bilateral Tolerance.

(b)Unilateral Tolerance.

5.The Limits of a Bearing and Shaft assembly are shown.

BearingShaft

40.028

40.0040.042 UL

40.030 LL

Determine:

(a)The Tolerance on each Part.

(b)The maximum clearance

(c)The minimum clearance

(d)The type of fit.

6.(a)Explain the difference between a GO and a NOT GO gauge.

(b)List three types of gauges and describe their use.

7.Explain the following.

(a)Interchangeability.

(b)Selective Assembly.

8.Discuss the reasons for using a system of limits and fits in precision

engineering.

Your answer should include the advantages and disadvantages.

Page - 6

Documents

Limits and Fits