2. exact size is impossible to achieve
every time in manufacturing
establish boundaries within which deviation
from exact size of a component is allowed
enable interchangeability of components
during assembly
3. Interchangeability
interchangeability occurs when one part in an assembly can be
substituted for a similar part which has been made to the same
drawing
by interchangeability, we mean that identical components
manufactured by different personnel under different
environment can be assembled and replaced without any
rectification during assembly stage without affecting the
functioning of the component
universal interchangeability
it means all the parts to be assembled are made from different
manufacturing unit
local interchangeability
it means all the parts to be assembled are made in the same
manufacturing unit
4. Y
X
X = BASIC SIZE
Y = UPPER RANGE
Z = LOWER RANGE
X - Z = ?
X +Y = ?
UL
LL
Z
5. FOUR
TWENTY
F I V E
BASIC SIZE + UPPER RANGE = UPPER LIMIT
BASIC SIZE - LOWER RANGE = LOWER
LIMIT
ONE
25 4 29
25 1 24
+ =
- =
6. BASIC SIZE
it is the size of a part to which
all limits of variation are determined
ACTUAL SIZE
it is the actual measurement
dimension of a part manufactured
BASIC SIZE
ACTUAL
SIZE
ACTUAL
SIZE
AFTER
MANUFACTURING
25 25.2 24.8
7. LIMITS ?
The maximum and minimum permissible sizes
within which the exact size of a component lies
are called limits. They are called UL & LL.
IT is a difference b/n MAXIMUM LIMIT and
MINIMUM LIMIT of shaft or hole.
TOLERANCE ?
8. ZERO LINE
it is the straight line
representing the basic size of component
OR
LINE OF ZERO DEVIATION
12. UNI-LATERAL
ONLY ONE RANGE
UPPER RANGE
OR
LOWER RANGE
BI-LATERAL
TWO RANGES
UPPER RANGE
AND
LOWER RANGE
UPEER
RANGE
BASIC SIZE
LOWER
RANGE
BASIC SIZE
UPEER
RANGE
BASIC SIZE LOWER
RANGE
-
TOLERANCE TYPES
OR
+
+
-
-
+
-
-
+
16. FITS
it is defined as
a degree of alignment
[looseness / tightness / interference]
between two mating parts
to perform predetermined motion
when they are assembled together
17. CLEARANCE FIT
DHOLE > DSHAFT
INTERFERENCE FIT
DSHAFT > DHOLE
TRANSITION FIT
DHOLE = DSHAFT
DH=DS
19. H O L E B A S I S S Y S T E M
if the system of assembly of shaft and hole is
considering on the basis of HOLE, then that type
of system is known as hole basis system
BASIC SIZE = LOWER LIMIT OF HOLE
ZERO LINE
BASIC
SIZE
LL
H S
LINE
ZERO
BASIC
SIZE
H S
20. hole dia. constant
and shaft dia. varies
[> = <]
When dH > dS Clearance Fit Loose Fitting
When dH = dS Transition Fit Tight Fitting
When dH < dS Interference Fit No Fitting
R E L AT I O N O F F I T S I N H O L E B A S I S S Y S T E M
22. if the system of assembly of shaft and hole is
considering on the basis of SHAFT, then that type
of system is known as shaft basis system
BASIC SIZE = UPPER LIMIT OF SHAFT
ZERO LINE
BASIC
SIZE
UL
H S
LINE
ZERO
BASIC
SIZE
H S
S H A F T B A S I S S Y S T E M
23. shaft dia. constant
and hole dia. varies
[< = >]
When dS < dH Clearance Fit Loose Fitting
When dS = dH Transition Fit Tight Fitting
When dS > dH Interference Fit No Fitting
R E L AT I O N O F F I T S I N S H A F T B A S I S S Y S T E M
25. A 75 mm journal rotates in a bearing. The tolerance for both
journal and bearing is 0.075 mm and the required allowance is
0.10 mm. Determine the dimensions of the journal and the bearing
bore with the basis hole standard. Also represent graphically all
the findings.
basic size = 75.00 mm
as per basis hole standard
basic size = lower limit of hole
Lower limit of hole = 75.00 mm
Upper limit of hole = 75.07 mm
Upper limit of shaft = 74.90 mm
Lower limit of shaft = 74.82 mm
Journal or shaft diameter = 75 mm
Shaft tolerance = 0.075 mm
Bearing or hole tolerance = 0.075 mm
Allowance = 0.10 mm
Shaft and hole dimensions i.e. LL, UL = ?
GRAPHICAL PRESENTATION IN NEXT SLIDE
A N S W E R S
G I V E N D ATA
27. Find the value of allowance, hole tolerance, shaft tolerance and type of
tolerance for the following dimensions of metal parts according to base hole
system. Hole = 37.50 mm, 37.52 mm and Shaft = 37.47 mm, 37.45 mm.
Also represent graphically all the findings OWN.
Hole = 37.50 and 37.52
HLL = 37.50 and HUL = 37.52
Shaft = 37.47 and 37.45
SUL = 37.47 and SLL = 37.45
allowance, tolerances [S & H]
and tolerance type = ?
Allowance = 0.03 mm
Hole tolerance = 0.02 mm
Shaft tolerance = 0.02 mm
Tolerance type = unilateral
A N S W E R S
G I V E N D ATA
28. A 100 mm dia. Journal and bearing assembly has a clearance fit with the
following specifications: Tolerance on bearing = 0.005 mm
Tolerance on journal = 0.004 mm
Allowance = 0.002 mm
Determine the size of the bearing and journal under [a] hole basis system
[b] shaft basis system. Con. unilateral system of tolerance.
A N S W E R S
Basic size = 100 mm
Hole tolerance = 0.005 mm
Shaft tolerance = 0.004 mm
Allowance = 0.002 mm
Size of journal and bearing = ?
HBS
BEARING
LL = 100 mm
UL = 100.005 mm
JOURNAL
UL = 99.998 mm
LL = 99.994 mm
JOURNAL
UL = 100 mm
LL = 99.996 mm
BEARING
LL = 100.002 mm
UL = 100.007 mm
SBS
G I V E N D ATA
29. In a limit system the following limits are specified to given a
clearance fit between a shaft and a hole.
30
-0.005
-0.018
mm
SHAFT 30
+0.020
-0.000
mm
HOLE
Find the following elements:[A] basic size [B] shaft tolerance [C]
hole tolerance [D] shafts limits [E] hole limits [F] maximum
clearance [G] minimum clearance.
30. 0.013 mm
0.020 mm
29.995 mm and 29.982 mm
30.020 mm and 30.00 mm
0.038 mm
0.005 mm
30 mm
?
?
?
? ? ?
?
?
A
B
C
D
E
F
G
31. [b] shaft tolerance = 0.013 mm
[c] hole tolerance = 0.020 mm
[d] shafts limits = 29.995 mm and 29.982 mm
[e] hole limits = 30.020 mm and 30.00 mm
[f] maximum clearance = 0.038 mm
[g] minimum clearance = 0.005 mm
[a] basic size = 30 mm
SOLUTION
STEPS
32. In a limit system the following limits are specified to given a
clearance fit between a shaft and a hole.
30
-0.005
-0.018
mm
shaft 30
+0.020
-0.000
mm
hole
Find the following elements:[A] basic size [B] shaft tolerance [C]
hole tolerance [D] shafts limits [E] hole limits [F] maximum
clearance [G] minimum clearance.
Answer the following questions on basis of solution
of the above question
1. A + B = ?
2. F - C = ?
3. G2 = ?
4. E - D = ? & ?
5. A + B + C + F + G = ?
