Lecture 2 Fits & Tolerances

Lecture 2

Fits & Tolerances.

By the end of this lecture, you'll learn: • The meaning of

tolerances, • Range of Tolerances for

different production processes ,

• Types of Fits, • Differences between

clearance & interference cases,

• Differences between basic hole & basic shaft systems.

Machine design – 2nd year Mechanical - Lecture 2 1

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Tolerances

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• Grades of Tolerances IT 1 IT 16

Where IT= international tolerances

IT 1-2-3-4 IT 5 IT 6 ……….. IT 16 ]<ìˆãq_Œ^{éÏÖ < ê×e<á^¹æ…

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• <{Ö]<‹{ËßÖ ITÖ]<ì^{èê<‡æ^{r{jÖ]<Ü{éÎ<]î<<{size E<<†{ŞÎ–<Ùç{<DJJJJ

• Where i is tolerance unit

Mean Dia. (mm)

e.g.: 10 – 18 mean = 14 18 – 30 mean = 24

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IT 6 IT 10

IT 6 IT 7 IT 8 No of units

10 i 16 i 25 i

10 -- 18 11μ 18μ 27μ

size

18 – 30 13μ 21μ 33μ

3i 0.45 D 0.001* D= +

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Grades of tolerances IT1 to IT16


Fits

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Fit

Clearance fit interference fit çÛÃÖ]<àÚ<†Ç‘_<gÏm< << << << << << << <çÛÃÖ]<àÚ<Ò_<gÏmد

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A Z except Q O L I <“é’¡<Üjèæ<< <

Capital letter hole Small letter shaft

<Ä•æ<°fè<Í†u<ØÒblock of tolerance <{Ö]<àÚ<(nominal size) Zero line ¼{v{<æ_<”ç×}<kÞ^Ò<ð]ç‰<íÖ^£]<‚è‚ <Üjè<äßÚ<æ< <

H/a Clearance z/Z Interference


CLEARANCE case

Shaft tolerance

Hole tolerance

Zero deviation line

Shaft center line


INTERFERENCE case `

Shaft tolerance

Hole tolerance

Shaft center line

Zero deviation line




We have two systems: Basic hole system Basic shaft system Capital / Small Small / Capital H / g h / P

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Example 1

Φ50 H7 / g6

• Nominal size Φ50 mm

• Hole accuracy is according to IT7

• Shaft accuracy is according to IT6

• Basic hole system

• Clearance fit

o Summary: H / a h Clearance fit H / j,k,m transition fit H / n z interference fit

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Example 2 Φ60 h6 / P7

• Nominal size Φ60 mm • Shaft IT6 • Hole IT7 • Basic shaft system

• interference fit o Summary:

h / A H clearance fit h / J,K,M transition fit h / N Z interference fit

Note

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Example 3 Find min., max. Interference or clearance:-

[I] Φ50 H7 / g7

Solution

Φ50 H7 min hole = 50 + zero = 50 mm max hole = 50 + 0.025 = 50.025 mm

Φ50 g6

From graph: Φ50 g7 Min shaft = 50 – 0.034 = 49.966 mm Max shaft = 50 – 0.009 = 49.991 mm Max clearance = max hole – min shaft Min interference = min hole - max shaft

[II] Φ50 h6 / P6

Solution

Φ50 h6 min shaft = 50 -0.016 = 49.984 mm max shaft = 50 - 0 = 50 mm

Φ50 P7 =

From fig.: Φ50 P6 = Min hole = 50 – 0.017 = 49.983 mm Max hole = 50 – 0.033= 49.967 mm Max int. = min hole – max shaft Min int. = max hole - min shaft

∵0.0250Φ50 mm

∴

∵0.0090.025Φ50 mm

−−

0.0090.034Φ50 mm

−−

∴

∴

∵0

0.016Φ50 mm−

∴

∵0.0170.042Φ50 mm

−−

0.0170.033Φ50 mm

−−

∴

∴

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Lecture 2 Fits & Tolerances