tolerance analysis 09.04.03

7/29/2019 Tolerance Analysis 09.04.03

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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.

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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;

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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.

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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

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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

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= 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

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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


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


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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

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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

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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.

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+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


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

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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.

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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.

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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`

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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.

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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

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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

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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

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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

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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

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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.

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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

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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.

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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.

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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

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tolerance analysis 09.04.03

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