Download - Tolerance Analysis 09.04.03
-
7/29/2019 Tolerance Analysis 09.04.03
1/26
TOLERANCE ANLAYSIS
INTRODUCTION:
Any two parts produced, can never be exactly alike even under most closely
controlled conditions. Variation from work piece to work piece, will always be
found as long as an Inspector can measure their differences. Interchangeability of
parts is the basic requirement of modern production methods, which means that
individual parts can be produced at many locations and brought together for
assembly without bench fitting or rework.
The responsibility to select an economic tolerance for the product is
shouldered by the product designer. The process engineer selects the correctprocess, establishes the necessary controls, to meet the design specifications
economically. If relative tolerances are well specified on the drawing, many of the
common causes of excessive cost and waste manufacturing efforts can be curtailed.
A current economic picture of selecting proper tolerance can only be
understood, when full knowledge of the process and its capabilities are known.
ADDITION OF TOLERANCES
Fig.1
Consider A and B as two linear dimensions which are to be added, each
having bilateral tolerance of a1, a2 and b1, b2 respectively as shown in Fig 1.
The sum of the two dimensions A and B is equal to C and the effect on the
tolerance of C is to be analysed.
1
-
7/29/2019 Tolerance Analysis 09.04.03
2/26
The value of C is maximum and minimum when the dimensions A and B
are also maximum and minimum respectively.
Cmax = A max + B max
Cmin = A min + B min.
Cmax = ( C + c2 )= (A + a2) + (B + b2) ------- ( 1 )
Cmin = ( C - c1 )= (A a1) + (B b1) ------- ( 2 )
Subtracting C min ( 2 ) from C max ( 1 )
We get c2 + c1 = (a1+a2) + (b1 + b2)
If ( a1 + a2 ) = T a
( b1 + b2 ) = Tb
( c1 + c2 ) = T c
The new form of the tolerances will be T c = Ta + Tb
SUBTRACTION OF TOLERANCES
Fig 2
Considering the dimension B being subtracted from A where a1, a2 and b1,
b2 are bilateral tolerance of the respective dimensions as shown in fig 2.
The dimension C is the resultant of subtraction of (A-B) and the effect of
the tolerances on C is to be analysed.
Dimension C is maximum when A is maximum and B is minimum,
Dimension C is minimum when A is minimum and B is maximum.
Thus C max = A max B min
C min = A min B max;
2
-
7/29/2019 Tolerance Analysis 09.04.03
3/26
A max = A + a2 ; B min = B b1
A min = A a1 ; B max = B + b2Therefore
C max = ( C + c2 )= ( A + a2) ( B b1 ) ----------- ( 3 )
C min = ( C c1 ) = ( A a1) ( B + b2 ) ---------- ( 4 )
Subtracting C min ( 4 ) from C max ( 3 )
(c 2 + c 1 ) = ( a2 + a1 ) + ( b2 + b1)
If ( a2 + a1 ) = Ta
( b2 + b1 ) = Tb
( c2 + c1 ) = Tc
The Tc = Ta + TbHere again the tolerance on the dimension C is equal to the sum of the
tolerances on the dimensions A and B.
So, the sum of the tolerances on the both sides of the equation is equal in
addition and as well as in subtraction.
In the following cases, the value of X and tolerances x1, x2 are un-known.
Case-1x2 a2 z2x1 a1 z1
X A = Z
x2 a2 x2 - a1 z2x1 a1 x1 - a2 z1
X A = ( X A ) = Z
Therefore x2a1 = z2
x1 a2 = z1Example:
x2x1 0.1 0.2
X 5 = 10
x2 ( 0.1 ) = + 0.2
x2 + 0.1 = + 0.2
x2 = + 0.2 0.1 = + 0.1
x1 ( + 0.1 ) = 0.2x1 0.1 = 0.2
x1 = 0.1
x2x1 0.2 0.1 0.1
X = 10 + 5 = 15
Case-2 a2 x2 z2a1 x1 z1
A X = Z
a2 x1 z2a1x2 z1
( A - X ) = ZThe tolerance of the equation can be equated.
3
-
7/29/2019 Tolerance Analysis 09.04.03
4/26
a2 - x1 = z2x1 = a2 - z2a1 - x2 = z1x2 = a1 - z1
Thereforex2 a1 - z1
x1 a2- z2X = ( A - Z )Example:
x2 0.1 x1 0.2
15 - X = 10
x2 - 0.1- (-0.2)x1 + 0.1- (+ 0.2)
X = ( 15 - 10 )
x2 x1 0.1
X = 5
Fig 3
What should be the dimension X of the component shown in Fig 3 if it is
to be made on a capstan lathe. Also analyze the effect of tolerance if the reference
is changed.
Considering MM as referencea2 x2 b2a1 x1 b1
A - X = B
+ 0.0 x2- 0.1 x1 0.1
30 - 8 = 22
+0.0- x1-0.1 - x2 0.1
( 30 8 ) = 22
0.0 x1 = + 0.1
x1 = 0.1
4
-
7/29/2019 Tolerance Analysis 09.04.03
5/26
0.1 - x2 = 0.1
x2 = 0.0
x2 + 0. 0
x1 -0.1The value of X = 8
Considering LL as referencea2 b2 x2a1 b1 x1A B = X
+ 0.0 x2- 0.1 0.1 x1
30 22 = 8
x2 = + 0.0 ( 0.1) = + 0.1
x1 = 0.1 ( +0.1) = 0.2
x2 + 0.1x1 - 0.2
X = 8
MULTIPLICATION BY A CONSTANT NUMBER
Any dimension with tolerances, if multiplied by a positive constant number, the
tolerance of the result also increases proportionately. Also when divided by a
constant number, the tolerance of the result, similarly decreases
proportionately
a2 a1When A is multiplied by a constant K , it becomes
a2 a2 * K a1 a1 * K
( A ) * K = ( A * K )
Example of multiplication: + 0.1 (+0.1) 3 + 0.3 (+0.3) 3
4 * (3) = (4 * 3) +0.3
+0.9 = 12
DIVISION BY A CONSTANT NUMBER a2 a1When A is divided by a constant K , it becomes
a2 a2 / K a1 a1 / K
( A ) K = (A K)
Example of division:
+0.3 0.3/3 +0.9 0.9/3
(12 ) 3 = ( 12 / 3 ) +0.1
+0.3
5
-
7/29/2019 Tolerance Analysis 09.04.03
6/26
= 4
Fig 4Example 1 : In the above Figure 4, shaft is placed between plates 1 & 2.For the
dimensions given in the Fig , taking LL as reference , find out the variation of
dimension x.
x2 +0.0 +0.1 +0.0 +0.0 x2x1 -0.2 -0.0 -0.2 -0.1 x1
X = 50 5 38 5 = 2
x 2 = + 0.0 ( 0.0 ) ( 0.2 ) ( 0.1 ) = + 0.3
x 1 = 0.2 ( + 0.1 ) 0.0 - 0.0 = 0.3
x2 +0.3
x1 -0.3 0.3X = 2 = 2
Verification: Ta + Tb + Tc + Td = T x
Ta + Tb + Tc + Td = 0.2 + 0.1 + 0.2 + 0.1= 0.6
T x = 0.3+0.3 = 0.60.6 = 0.6
Since LHS = RHS, the result is correct.
Example 2 :-0.040 +0.013 +0.012 +0.000 -0.006 x2
+0.040 -0.073 -0.009 +0.001 -0.019 -0.017 x1150 - 20 + 95 - 15 + 65 + 15 = 290
x2=(0.040)-(0.073)+(0.013)-(0.001)+(0.000)+(-0.006) = + 0.119
x1=(0.000)-(-0.040)+(-0.009)-(0.012)+(-0.019)+(-0.017) = 0.017
x1 +0.119x2 -0.017
X = 290
Verification: Ta+ Tb+ Tc + Td + Te + Tf = Tx
Ta+ Tb+ Tc + Td + Te + Tf
= 0.040+ [(0.040)(0.073)] + [0.013(0.009)] +
[(+0.0120(+0.001)] + (0.019)+ [(0.006)( 0.017)] = 0.136T x = [0.119(0.017)] = 0.136
6
-
7/29/2019 Tolerance Analysis 09.04.03
7/26
0.136 = 0.136
Since LHS = RHS , the result is correct.
Example 3 : +0.046 + 0.030 +0.021 +0.030 x2+ 0.011 +0.008 +0.011 x1
240 +75 20 60 = X
X = (240 + 75 20 60) = 235
x2 = + 0.046 + 0.030 ( + 0.008 ) ( + 0.011 ) = + 0.057
x1 = + 0.000 + 0.011 ( + 0.021 ) ( + 0.030 ) = 0.040
X2 +0.057x1 -0.040
235 = 235
Verification:
Ta + Tb + T c + Td = Tx.
Ta + Tb+ T c+ Td = 0.046 + ( 0.030 0.011 ) + ( 0.021 0.008 ) +( 0.030 0.011 )
= 0.097
Tx = 0.097
0.097 = 0.097
Since LHS = RHS, the result is correct.
Example 4 : +0.012 +0.009 +0.013 +0.008 x2-0.007 -0.004 +0.063 -0.009 -0.003 x1
65 -20 +450 -120 +12 = 387
x2 = + 0.012 ( 0.004 ) + 0.063(0.009) + 0.008 = + 0.096x1 = 0.007 ( + 0.009 ) + ( + 0.000 ) ( + 0.013) + (0.003)
= 0.032 x2 +0.0 96
x1 -0.032387 = 387
Verification: Ta+Tb+Tc+Td+Te = Tx
Ta+Tb+Tc+Td+Te =0.012(0.007)+ 0.009 (0.004)+0.063+0.013
(0.009)+0.008(0.003)
= 0.128
Tx = 0.096 (0.032) = 0.1280.128 = 0.128
Since LHS = RHS, the result is correct.
7
-
7/29/2019 Tolerance Analysis 09.04.03
8/26
Fig 5
The Fig 5 is of a gear box, it is required to have the axial clearance between
bearing and the shaft, as 1.25 0.25 mm (D), what is the variation of, value of X,
if the values of A, B, C and E are known.
Considering LL as reference, the value of X must lie as the difference
between the sum of the dimension of B, C, D, and E to A.
Thus the equation is :x2 b2 c2 d2 e2 a2x1 b1 c1 d1 e1 a1
X = B + C + D + E - A
x2 +0.0 +0.00 +0.25 +0.00 +0.25x1 -.05 -0.10 -0.25 -0.05 -0.25
X = 5 + 140 + 1.25 + 5 - 100
x2 x2 x2 x1 x1 x1
X = ( 151.25 100 ) = ( 51.25 )
x2 = + 0.0 + 0.00 + 0.25 + 0.00 ( 0.25 )
= + 0.5
x1 = 0.05 0.1 0.25 0.05 ( + 0.25 ) = 0.7 x1 +0.5
x2 -0.7
(51.25) = 51.25
8
-
7/29/2019 Tolerance Analysis 09.04.03
9/26
Verification: Tb+Tc+Td+Te+Ta = Tx
Tb+Tc+Td+Te+Ta = +0.05 + 0.1 + 0.25 +0.25 +
0.05 +0.25 + 0.25 =1.2
Tx = 1.21.2 =1.2
Since the LHS = RHS, the result is correct.
Fig 6
In Fig 6, the dimension P1, P3 advances one side and may be considered as
positive where as the dimension P2 and N advances in an opposite direction to the
former is considered as negative.
Chain dimensioning is a group of dimensions, which are determined by the
links of the chain and are interconnected in definite succession also have a close
loop characteristic.
+0.05 - 0.13- 0.03 - 0.20
In Fig 7, the dimension A = 100 . 0 , B = 60 calculate the nominal size
and the variation for the dimension C.
Fig 7
Nominal size C = Nominal size A Nominal size B
C = 100 60 = 40c2 a2 b2c1 a1 b1
C = A B
9
-
7/29/2019 Tolerance Analysis 09.04.03
10/26
c2 + 0.05 0.13c1 0.03 0.20
40 = 100 60
c2 = + 0.05 ( 0.20 ) = + 0.25
c1 = 0.03 ( 0.13 ) = + 0.1 c1 +0.25
c2 + 0.1
C = 40
Tc = Ta +Tb
Ta + Tb = + 0.05 ( 0.03 ) + ( 0.13 ) ( 0.2 )
= 0.15
Tc = 0.15
0.15 = 0.15
Since, LHS = RHS , the result is correct.
However it must be noted that the tolerance should not be given on a
concluding dimension like C as the cumulative tolerances of B and C may exceedthe tolerance given on the dimension A.
Fig.8
A hole 16H6 has to be drilled and reamed as shown in Fig 8. The
dimension is 79 0.1 from one side of the component. In a box jig plate the tolerance
for the dimension from one end, is to be calculated.
Here the component is butting against a rest pin which can be maintained
within close tolerances as it needs surface grinding. The resultant dimension here is
79 0.1. From the Fig, the nominal dimension of the hole from A A reference
is, 20 + 12 + 79 = 111. On the jig plate, the same dimension can be maintainedwithin a close tolerance of 111 0.05. The jig plate thickness can be very easily
maintained to a dimension of 20 - 0.05. Now the tolerance of 12 is to be calculated in
such a way that the resultant dimension should be 79 0.1.
10
-
7/29/2019 Tolerance Analysis 09.04.03
11/26
+0.00 X2 0.1 0.05 - 0.05 X1
79 = 111 20 12
Max. limit : + 0.1 = + 0.05 ( 0.05 ) x1x1 = 0
Minimum limit : 0.1 = - 0.05 ( + 0 )x2
x2 = 0.05 + 0.05 +0.00+0.00 - 0.05
The tolerance for the dimension 12 is 12 or 12.05
Verification: Td = Ta + Tb + Tc + 0.0 + 0.05
0.01 0.05 0.05 - 0.00
79 = 111 - 20 12
Td = 0.2
Ta + Tb + Tc = (0.05)-(-0.05)+0.0-(-0.05)+(0.05)-(-0.0)
= 0.2
0.2 = 0.2
Since LHS = RHS, the result is correct.
Fig 9
A hole of 16H7 [Fig 9 (a)] is to be made with reference to another
predrilled and reamed hole 10H7 which lies on a perpendicular plane as shown in
the Fig 9,
A box jig is a most suitable solution and the dimension of the hole 10H7
with reference to one side of the box jig is given as 245 0.05 as shown in Fig 9.b.
The tolerance for the dimension 97 must be found to position the drill jig
bush to drill and ream dia 16H7 hole. x2x1
Dimension should be 148 0.1 = 245 0.0597x2x1
148 0.1 + 97 = 245 0.05
+0.1 ( x1 ) = + 0.05
x1 = + 0.05
0.1( x2 ) =0.05
x2 =0.05The tolerances for the dimension 97 should be 0.05
11
-
7/29/2019 Tolerance Analysis 09.04.03
12/26
Verification: 245 0.05 - 97 0.05 = 148 0.1
Ta +Tb = 0.1+ 0.1 = 0.2
Tc = 0.2
0.2 = 0.2
Since , LHS = RHS the result is correct.
The tolerance 0.05 can be maintained in a tool room.
Fig 10
The component shown in Fig.10 is dimensioned by two methods, out of
which the more practical one is to be analyzed. On the component drawing in case
1 the dimensions, 50 0.1; 100 0.2 , and the overall dimensions 180 0.05 are
toleranced and in case 2 , dimensions 100 0.2 , 130 0.05and 30 0.05are toleranced .
In case 1 considering LL as the reference X2
x1
X = 180 0.05 50 0.1 100 0.2
x2 = + 0.05 - ( - 0.1 ) - ( - 0.2 ) = + 0.35
x1 = - 0.05 - ( + 0.1 ) - ( + 0.2 ) = - 0.35
x2X1
X = 30 0.35
In case 2 considering AA as reference X2
Xx1 = 130 0.05 - 100 0. 2
x2 = + 0.05 - ( - 0.2 ) = + 0.25
x1 = - 0.05 - ( + 0.2 ) = - 0.25X2
Xx1 = 30 0. 25
In neither of the two cases the tolerances of 30 0.05 is achievable. Thus in
practice, taking the datum as AA, the given tolerances is easily achievable in jig
boring.
12
-
7/29/2019 Tolerance Analysis 09.04.03
13/26
Fig 11
In the above Fig 11, the designer reflects the dimension C from the
functional reference LL . +0.08
It is impracticable to measure the depth ( 20 - 0.14 ) of the hole having
diameter D1 from the functional reference LL , hence an auxiliary referencemust be chosen to facilitate the measurement of the depth of the hole D1. The best
auxiliary reference selected in this case is MM , which is also reference for
manufacturing. In view of this new reference the dimensions of A, B, C and their
tolerances are evaluated as follows.
Nominal dimension of A = Nominal dimension B+ Nominal dimension C
A = B + C
A = 20 +10 =30
Total Tolerance of C = Total tolerance of A-Total tolerance of B
The equation will be A - B = C a2 b2 c2
a1 b1 c1
A - B = C a2 + 0.08
a1 - 0.14
30 10 0.0 5 = 20a2 ( 0.05 ) = + 0.08
a2 = 0.08 - 0.05 = + 0.03
a1 ( + 0.05 ) = 0.14
a1 = 0.09 a2 +0.03
a1 -0.09
A = 30+0.03
The depth of hole is = 30 -0.09.
It is however difficult to calculate the resulting dimensions if the
component or the assembly has number of dimensions. In such cases compensating
elements or spacers or adjusting shims are to be employed.
13
-
7/29/2019 Tolerance Analysis 09.04.03
14/26
Tolerancing Dimensions between centers
There are three methods of dimensioning between centers of holes, which are as
follows:1. Series or chain dimensioning
2. Parallel dimensioning of each hole with respect to the datum.
3. Series and parallel a combination of the above two methods.
The methods mentioned are shown in Fig12.
Fig 12
Tolerance on Dimensions between centers of two holes`
14
-
7/29/2019 Tolerance Analysis 09.04.03
15/26
Fig 13.1
Fig 13.2
On two plates two holes are made at a distance of M apart, with a tolerance
of m and if the plates are to be assembled with the aid of screws they should
assemble freely , or the tolerance should be of critical magnitude critical value to
achieve 100% interchangeability.
Case1: When sizes of the holes and shafts are identical. Ifd is the diameter of
the shafts and D is the diameter of holes in two plates, a set up for the worst case
of interchangeability is shown in Fig no. 13.1.
1-1 = Axis of holes in the plate1
2-2 = Axis of the holes in the plate 2
CC = Axis of the shafts mating the holes.
M = The distance between the centers of holes
m = The tolerance on M on the respective holes
( D-d ) = Clearance between hole and shaft = L
Considering a datum line of no clearance from the figure, the following
chain of dimension can be placed keeping in view the signs, an equation for the
expression of tolerance is as 0 = D/2 - ( M+ m ) + D/2- d+ D/2 + ( M-m ) + D/2 -d
0 = 2D - 2m - 2d
2m = 2 ( D-d )
2m = 2L
or m = L, as this is a critical value, m can be less than L but not more, as 100%
interchangeability is to be achieved.
Therefore m = L ------------------------- ( 1 )
CONCLUSION:
If 100% interchangeability is to be achieved in the above case, the
tolerance on dimension between the two holes cannot be more than twice the
diametrical clearance between the holes and shafts. It could be less or equal to the
clearance.
15
-
7/29/2019 Tolerance Analysis 09.04.03
16/26
Case II: The size of the holes and shafts vary: Fig 14
Fig 14
Let D1 be the diameter of the holes in plates 1 and 2, for the mating shaft d1.
Let D2 be the diameter of the holes in plates 1 and 2 for the mating shaft d2 .
A worst case set-up for 100% interchangeability is shown in the Fig.
1-1 = Axis of holes in the plate 1
2-2 = Axis of holes in the plate 2
c1 -c1 = Axis of the shaft of diameter d1
c2 -c2 = Axis of the shaft of diameter d2
M = The distance between the centers of holes
m = The tolerance on M.
( D1- d1 ) , ( D2 - d2 ). =The clearance between the holes and
Shafts.
Starting from a point of no clearance, a set of chain dimensions can be
placed as follows, keeping in view the directional sign and equating it to zero.
0 = D2/2 - ( M+ m )+ D1/2 - d1 + D1/2 + ( M - m ) + D2/2 - d20 = - 2m + ( D1 + D2 ) - ( d1 + d2 )
2m = ( D1 - d1 ) + ( D2 - d2 )
2m = ( L1 + L2 ). Since ( D1 - d1 ) = L1 and ( D2 - d2) = L2m = ( L1 + L2 ) 2 --------------------------- ( 2 )
CONCLUSION:
The critical tolerance m on the centers of two holes in the above case
should be equal to half the sum of clearances in both holes and could be less for
100% interchangeability between the two plates.
The principle involved in framing the equation is that the relativedisplacement of a point is zero, when it travels between two fixed points and
returns by the same path. Starting from a point of no clearance, a set of chain
16
-
7/29/2019 Tolerance Analysis 09.04.03
17/26
dimensions can be framed as above, keeping in view the directional sign and
equating it to zero.
Tolerance on Dimensions between centers of three HolesSeries Dimensioning.
Fig 15
If D is the diameter of holes in the plates 1 and 2 and d the diameter of mating
shafts, Fig 15 shows the holes displaced according to the tolerance for the worstcase.
1-1 = Axis of the holes in plate 1
2-2 = Axis of the holes in plate 2
c-c = Axis of the shafts
M = Dimension between holes
m = The tolerance of the dimensions.
0 = D/2 + ( M- m ) + ( M m ) + D/2 d + D/2 - ( M+ m )- ( M + m ) +D/2 - d
0 = 2D - 2d - 4m
4 m = 2 ( D d ) = 2 L. Since ( D - d ) = L
m = L/2
Similarly
For 4 holes dimensioned in series m=L/3
For 5 holes dimensioned in series m=L/4
If n holes are in series m= L/n-1------------------------- ( 3 )
For n holes dimensioned in series tolerance between the holes should be
equal to or less than L/n-1 for 100% interchangeability. Fig 16
17
-
7/29/2019 Tolerance Analysis 09.04.03
18/26
Fig 16
CONCLUSION:
As the number of holes increases in case of series dimensioning, thetolerance, between the holes decreases. Hence the cost of production increases and
it is difficult to make. If same tolerances are given to the large dimensions as that
of small dimensions, it will be in many instances impracticable.
Parallel Dimensioning
Fig 17
18
-
7/29/2019 Tolerance Analysis 09.04.03
19/26
If D is the diameter of holes in the plates 1 and 2 and d is the diameter of the
shafts mating with, Fig 17 shows the holes displaced according to the tolerances
for the worst case.
1-1 = Axis of holes in plate 1
2-2 = Axis of holes in plate 2
c-c = Axis of the shafts
M1 = The distance of the second hole from the first
M2 = The distance of the third hole from the first
m = The tolerance of M1 dimension
m = The tolerance of M2 dimension
Considering a datum line of no clearance from the figure the followingchain of dimensions can be laid down taking care of the signs of the dimensions,
the expression for the tolerances is, as follows.
-d+D/2+(M1-m)+D/2-d+D/2-(M1+m)+(M2+m)-D/2+d-D/2-(M2-m)-D/2+d= 0
2 m 2 m = 0
m = m ----------------------------------- ( 4 )
Fig 18
19
-
7/29/2019 Tolerance Analysis 09.04.03
20/26
CONCLUSION:
As the number of holes increases the tolerance remains the same L/2
The tolerance between the first hole and second hole is equal to tolerance between
the first hole and third hole. We also conclude that, if the tolerance is equal, it has
no bearing on the number of holes .The same tolerance for small dimension
remains unaltered even for larger dimensions, but difficulties may occur to achieve
the tolerances on larger dimension
Comparison between series & parallel Dimensioning:By comparing the series and parallel dimensioning we understand that if the
number of holes exceeds 3, parallel dimensioning is only recommended. By series
dimensioning, tolerance decreases which is difficult to achieve and expensive.
Tolerance on Dimensions of holes in plates I and II from their
common edge for interchangeability:
Fig 19
Case1: Fastener having clearance in both the plates (Fig 19).
Two plates are to be assembled by a fastener of diad, passing through theholes of dia D. The locations of holes are with reference to the common edge
0.The distance of the holes from the common edge is M m , where m is the
tolerance.
The hatched square in the Fig 19 indicates the largest area in which the
hole centers can lie for complete interchangeability.
Let us consider a position in which the center of the hole in plate 1 is at a
distance of ( M + m) and the center of the hole in plate 2 is ( M m) on both axis.In the plan view a line joining the common edge and the centers makes an angle of
45 degrees and the figure shows the respective distance from the common edge.
Applying the principles as before,
0 = (D/2)-2 (M+m) +2(M-m) + (D/2)-d0 = (D-d) - 2 2m, Therefore 2*2m = (D d)
( D d ) = L
20
-
7/29/2019 Tolerance Analysis 09.04.03
21/26
2m = ( D d ) 2 = 0.7 * L
Case II: Fastener having clearance in plate 1 and press fit in plate 2.
Fig 20
For the position as shown in Fig 20.
0 = (D/2)-2(M+m)+2(M-m)-(d/2)
(D-d)/2 = 2*2m ; ( D d ) = L
0.35 L = 2m.
Tolerances of dimensions in two plates from common surface for
interchangeability
There are two plates and the position of the two holes in these plates is
determined by the distance between the datum surface. How much should be the
tolerance of these distances, if we want to have 100% interchangeability so that it
can assemble without any difficulty.
21
-
7/29/2019 Tolerance Analysis 09.04.03
22/26
Case I:
Fig 21
Fastener having clearance in both plates . Holes of dia D in plates 1and 2are at a distance of ( M + m ) and ( M m ) respectively and is assembled by a
fastener of dia d .
Applying the principle
0 = ( D/2 ) ( M m ) + ( M + m ) + ( D/2 ) d
D d = 2 m
D d = L,
L = 2m
0.5 L = m
for complete interchangeability
m 0.5 L
Case II: Fastener having clearance in one plate and press fit in other plate Fig 21
Fig 22
0 = D / 2 - ( M + m ) + ( M m ) ( d / 2 )
( D d ) / 2 = 2 m
( D d ) = L
L / 2 = 2 m
m = 0.25 L
For complete interchangeability m 0.25 L
22
-
7/29/2019 Tolerance Analysis 09.04.03
23/26
Calculation of tolerance for distance between holes on Drill Jigs
Fig 23
Two holes are to be drilled and reamed at a distance W 0 within the limits
T0 , by using a jig plate with slip bushes as shown in Fig 23
Let W1 T1 be the distance between the liner bushes in jig plate.
L1, L2 = Clearance (around) between the slip bush and liner bush for the first and
second holes respectively.
e1, e2 = eccentricity of inside diameter of slip bush with respect to its outsidediameter for the two bushes respectively.
So that W0 = f (w1, L1, L2, e1, e2.).
CONDITION 1:
When the distance between liner bushes in the jig plate is at maximum
distance (W1+T1) and slip bushes at far off position.
O1, O2 are centers of the slip bushes which are eccentric with respect to
their outside diameter of slip bushes by e1 and e2.
OO1 = L1 + e1OO2 = L2 + e2.
23
-
7/29/2019 Tolerance Analysis 09.04.03
24/26
They are in extreme positions as shown in the figure.
This distance between holes on the component
W 1 + T1 + OO1 + OO2
The sum of these distance should be less than the maximum distance given on the
component.
( W1 + T1 + L1 + e1 + e2 ) W0 + T0.
CONDITION 2:
When the distance between liner bushes in jig plate is at minimum distance(W1-T1) and when slip bushes are also at minimum distance:
O1, O2 are the centers of the slip bushes, which are eccentric with respect to outside
by e1 and e2 respectively.
OO1 = L1 + e1OO2 = L2 + e2
The distance between the holes obtained on the component
( W1 + T1 ) = ( W1 - T1 ) - OO1 - OO2
= ( W1 - T1 ) - ( L1 + e1 ) - ( L2 + e2 )= ( W1 - T1 - L1 - L2 - e1 - e2 )
The sum of these distances should be more than the minimum distance
( Wo - To ) given on the component
( W1 - T1 ) - L1 - L2 - e1 - e2 ( W o - To )
Note: The clearance between the drill and the slip bushes has its effect on
the distance between the holes on the component.
The drill size is minimum , bushes and the drill are at far off position.
O1, O2 are the positions of the centres of the slip bush when the clearance betweenthe drill and the slip bush are considered.
With the effect of clearance between the drill and the slip bushes, the maximum
extreme dimension obtained is as
= W1 + T1 + L1 + L2 + e1 + e2 + OO1 + OO2
= W1 + T1 + L1 + L2 + e1 + e2 + ( d b + Tb ) - ( d - Td )
The sum of these distances should be less than or equal to Wo + To.
24
-
7/29/2019 Tolerance Analysis 09.04.03
25/26
Fig 24
Fig 25
CASE II: When drill size is minimum the bushes position is also at minimum
distance , the drill is at minimum distance.
OO1 = OO2 = ( d b - Tb ) - ( d - Td ) 2
Actual distance between holes on component
= W1 - T1 - L1 - L2 e1 - e2 - OO1 - OO2
= W1 - T1 - L1 - L2 - e1 - e2 - ( d b + Tb ) + ( d - Td )
This should be more than the minimum distance ( W0 - T0 ) given on component.
W1 - T1 - L1 - L2 - e1 - e2 - ( d b + Tb ) - ( d - Td ) > W0 - T0Final conditions are
W1 + T1 + L1 + L2 + e1 + e2 + ( db +Tb ) - ( d - Td ) ( W0 + T0 )
W1 - T1 - L1 - L2 - e1 e2 - ( db + Tb ) + ( d - Td ) > ( W0 - T0 )
As per Indian standard the tolerance on drill is = h9 The tolerance of the hole in slip bush is = F 7
The clearance between the liner bush and the slip bush is = F 7 / h 6
25
-
7/29/2019 Tolerance Analysis 09.04.03
26/26