Tolerance Accumulation

7/27/2019 Tolerance Accumulation

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THE INSTITUTE FOR ENHANCEMENT OF TECHNOLOGY

1

TOLERANCE ACCUMULATION

AND

ANALYSIS


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ADDITION OF TOLERANCES

A and B two linear dimensions to be added

a1, a2 tolerance on Ab1, b2 tolerance on B

A+B = C

The tolerance on C to be analysed.


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C is max. when A and B are max.

C is min. when A and B are min.

Cmax = A max + B max

Cmin

= Amin

+ Bmin

.

Cmax

= ( C + c2

)= (A + a2) + (B + b

2) ------- ( 1 )

Cmin

= ( C - c1

)= (Aa1) + ( Bb

1) ------- ( 2 )

c1, c2 tolerance on C

Subtracting Cmin

( 2 ) from Cmax

( 1 )

c2

+ c1

= (a1+a

2) + (b

1+ b

2)

If ( a1

+ a2

) = Ta

( b1

+ b2

) = Tb

( c1

+ c2

) = Tc

The Tolerance on the total length will be Tc = Ta + Tb


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SUBTRACTION OF TOLERANCES

A and B two linear dimensions to be subtracted

a1, a2 tolerance on Ab1, b2 tolerance on B

A-B = C

The tolerance on C to be analysed.


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C is max. when A is max. and B is min.

C is min. when A is min. and B is max.

C max = A maxB minC

min= A

minB

max;

A max = A + a2B

min= Bb

1

Amin

= Aa1

Bmax

= B + b2

Cmax

= ( C + c2

) = ( A + a2)( Bb

1) ----------- ( 3 )

C min = ( Cc1 ) = ( Aa1) ( B + b2 ) ---------- ( 4 )

Subtracting Cmin

( 4 ) from Cmax

( 3 )

(c 2 + c 1 ) = ( a2 + a1 ) + ( b2 + b1)If ( a2+ a

1) = T

a

( b2+ b

1) = T

b

( c2 + c1 ) = Tc

The Tolerance on the remaining length Tc = Ta + Tb


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The tolerances are getting addedboth in addition and subtraction.


7/80THE INSTITUTE FOR ENHANCEMENT OF TECHNOLOGY

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TO CALCULATE UNKNOWN DIMENSIONS

EXAMPLES



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Case-1

x2 a2 z2x1 a1 z1

X A = Z

x2 a2 x2 - a1 z2x1 a1 x1 - a2 z1

X A = ( X A ) = Z

Therefore x2a1 = z2x2 = a1 + z2

x1a2 = z1

x1 = a2 +z1

Therefore

x2 a1 + z2

x1 a2 +z1X = ( A + Z )

The value of X and tolerances x1, x2 are un-known.



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Example:

x2

+ 0.1 + 0.2

x1

- 0.1 - 0.2

X 5 = 10

x2

(0.1 ) = + 0.2

x2 + 0.1 = + 0.2

x2

= + 0.20.1 = + 0.1

x1 ( + 0 . 1 ) =0.2x1 0.1 =0.2

x1

= 0.1

x2 + 0.1 + 0.2

x1 - 0.1 - 0.2

X 5 = 10

+0.1 + 0.1 + 0.2

-0.1 - 0.1 - 0.2

15 5 = 10




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Case-2 a2 x2 z2a1 x1 z1

A X = Z

a2x1 z2a

1x

2z

1

( A - X ) = Z

The tolerances of the equation can be equated.a2 - x1 = z2x1 = a2 - z2

a1 - x2 = z1x2 = a1 - z1

Therefore

x2 a1 - z1x

1a

2 -z

2

X = ( A - Z )




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Example:

+ 0.1 x2 + 0.2

- 0.1 x1 - 0.2

15 - X = 10

+0.1x1 +0.2

-0.1 x2

-0.2

( 15 - X ) = Z

The tolerances of the equation can be equated.

+0.1 - x1

= +0.2

x1 = +0.1 - 0.2 = -0.1

-0.1 - x2

= -0.2

x2

= -0.1 +0.2 = +0.1

Therefore

+ 0.1 x2 + 0.2

- 0.1 x1 - 0.215 - X = 10

+ 0.1 +0.1 + 0.2

- 0.1 -0.1 - 0.2

15 - 5 = 10



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+ 0.05- 0.03

Dimension A = 100 . 0

- 0.13

- 0.20

Dimension B = 60.0

Calculate the nominal size and the variation for the dimension C.



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Nominal size C = Nominal size A Nominal size B

C = 10060 = 40

c2 a2 b2c1 a1 b1

C = A B

c2 + 0.05 0.13

c1 0.03 0.20

40 = 100 60

c2 = + 0.05(0.20 ) = + 0.25

c1 =0.03(0.13 ) = + 0.1

c2 +0.25

c1 + 0.1

C = 40

Verification: Tc = Ta +TbTa + Tb = + 0.05(0.03 ) + (0.13 )(0.2 )

= 0.15

Tc = 0.15

0.15 = 0.15



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Calculate the dimension x and tolerance on x with M-M

and L-L as references



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Considering MM as reference

a2 x2 b2a

1x

1b

1

A - X = B

+ 0.0 x2- 0.1 x1 0.1

30 - 8 = 22

+0.0 - x1-0.1 - x2 0.1

( 308 ) = 22

0.0x1 = + 0.1

x1

= 0.1

0.1 - x2 = 0.1

x2 = 0.0

x2 + 0. 0x

1 - 0.1

The value of X = 8



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Considering LL as reference

a2 b2 x2a1 b1 x1

A B = X

+ 0.0 x2- 0.1 0.1 x

1

30 22 = 8

x2 = + 0.0

(

0.1) = + 0.1x1

= 0.1( +0.1) =0.2

x2 + 0.1

x1 - 0.2

X = 8



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When divided by a constant number, thetolerance of the result, similarly decreasesproportionately

MULTIPLICATION & DIVISION BY A CONSTANTNUMBER

Any dimension with tolerances, if multipliedby a positive constant number, the tolerance of the result also increases

proportionately.



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a2

a1When A is multiplied by a constant K , it becomes

a2 a2 * Ka1 a1 * K

( A ) * K = ( A * K )

MULTIPLICATION BY A CONSTANT NUMBER

Example :

+ 0.1 (+0.1) 3

+ 0.3 (+0.3) 3

4 * (3) = (4 * 3)

+0.3

+0.9

= 12



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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 :

+0.3 0.3/3

+0.9 0.9/3

( 12 ) 3 = ( 12 / 3 )

+0.1

+0.3

= 4



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Find out the variation of dimension x with reference to LL

EXAMPLE 1



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x2 +0.0 +0.1 +0.0 +0.0 x2

x1 -0.2 -0.0 -0.2 -0.1 x1X = 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 0.3

X = 2 = 2

Verification: Ta + Tb + Tc + Td = T xTa + Tb + Tc + Td = 0.2 + 0.1 + 0.2 + 0.1

= 0.6

T x = 0.3+0.3 = 0.6

0.6 = 0.6



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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 x1

150 - 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

x2 + 0.119

x1 - 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)] = 0.136

T x = [0.119(0.017)] = 0.136

0.136 = 0.136



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Example 3 : + 0.030 +0.021 +0.030 x2+0.046 + 0.011 +0.008 +0.011 x1

240 +75 20 60 = X

X = (240 + 752060) = 235x2 = + 0.046 + 0.030( + 0.008 )( + 0.011 ) = + 0.057

x1 = + 0.000 + 0.011( + 0.021 ) ( + 0.030 ) =0.040

x2 +0.057

x1 -0.040

235 = 235

Verification:

Ta + Tb + T c + Td = Tx.

Ta

+ Tb

+ Tc

+ Td

= 0.046 + ( 0.0300.011 ) + ( 0.0210.008 ) + ( 0.0300.011 )

= 0.097

Tx = 0.097

0.097 = 0.097


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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 = + 0.096

x1 =0.007( + 0.009 ) + ( + 0.000 )( + 0.013) + (0.003)

= 0.032

x2 +0.0 96

x1 -0.032

387 = 387

Verification: Ta+Tb+Tc+Td+Te = TxTa+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) = 0.128

0.128 = 0.128


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Calculate the value of x to have the axial clearancebetween bearing and the shaft, as 1.25 0.25 mm


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Considering LL as reference

The value of X must lie as the difference between the sum of thedimension of B, C, D, and E to A.

L

L


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EQUATION :

x2 +0.0 +0.00 +0.25 +0.00 +0.25

x1 -0.05 -0.10 -0.25 -0.05 -0.25

X = 5 + 140 + 1.25 + 5 - 100

x2 x2 x2

x1 x1 x1X = ( 151.25100 ) = ( 51.25 )

x2 = + 0.0 + 0.00 + 0.25 + 0.00(0.25 ) = + 0.5

x1 = 0.050.10.250.05( + 0.25 ) = 0.7

x2 +0.5

x1 -0.7(51.25) = 51.25

Verification:Tb+Tc+Td+Te+Ta = TxTb+Tc+Td+Te+Ta = +0.05 + 0.1 + 0.25 +0.25+0.05 +0.25 + 0.25 =1.2

Tx = 1.2

1.2 =1.2


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A hole 16H6 has to be drilled and reamedThe dimension is 79 0.1 from one side of the component.Calculate the tolerance for the dimension 12 from one end in abox jig plate to obtain the resultant dimension 79 0.1


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+0.00 x2

0.1 0.05 - 0.05 x179 = 111 20 12

Max. limit : + 0.1 = + 0.05(0.05 )x1

x1 = 0Min. 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


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A hole of 16H7 is to be made with reference to anotherpredrilled and reamed hole 10H7 which lies on a perpendicularplane

The dimension of the hole 10H7 with reference to one side ofthe box jig is given as 245 0.05

Calculate the tolerance for the dimension 97 to position thedrill jig bush to drill and ream dia 16H7 hole.


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x2

Dimension should be 148 0.1 = 245 0.0597 x1x2

x1

148 0.1 + 97 = 45 0.05+0.1( x1 ) = + 0.05

x1 = + 0.05

0.1( x2 ) =0.05

x2 =0.05

The tolerances for the dimension 97 is 0.05Verification: 245 0.05 - 97 0.05 = 148 0.1

Ta +Tb = 0.1+ 0.1 = 0.2

Tc = 0.2

0.2 = 0.2


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The component shown in Fig. is dimensioned by

two methods. Analyses the more practical one.

In case 1 the dimensions, 50 0.1; 100 0.2 , and theoverall dimensions 180 0.05 are toleranced

In case 2 , dimensions 100 0.2 , 130 0.05and 30 0.05are toleranced .

In case 1 considering LL as the reference

X2

x1X = 180 0.05 50 0.1 100 0.2x2 = + 0.05 - ( - 0.1 ) - ( - 0.2 ) = + 0.35

x1 = - 0.05 - ( + 0.1 ) - ( + 0.2 ) = - 0.35

x2

x1

X = 30 0.35In case 2 considering AA as reference

x2

Xx1 = 130 0.05 - 100 0. 2x2 = + 0.05

- ( - 0.2 ) = + 0.25

x1 =

- 0.05 - ( + 0.2 ) = - 0.25

X2

Xx1 = 30 0. 25


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The dimension C from the functional reference LL .

+ 0.08

It is impracticable to measure the depth ( 20 - 0.14 ) of the holehaving diameter D1 from the functional reference LL

Hence the best auxiliary reference selected in this case is MM ,which is also reference for manufacturing. In view of this newreference the dimensions of A, B, C and their tolerances areevaluated as follows.


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Nominal dimension ofA = Nominal dimension B+ Nominal dimension C

A = B + C

A = 20 +10 = 30

Total Tolerance ofC = Total tolerance ofA-Total tolerance ofB

The equation will be A - B = C

a2 b2 c2a

1b

1c

1

A - B = C

a2 +0.08

30 a1 10 0.0 5 = 20 - 0.14

a2 (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.


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TOLERANCING DIMENSIONS BETWEEN CENTERS

Methods of dimensioning between centers of holes

Series or chain dimensioning

Parallel dimensioning of each hole with respect to the datum.

Series and Parallel A combination of the above two methods

TOLERANCE ON DIMENSIONS


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TOLERANCE ON DIMENSIONS

BETWEEN CENTERS OF TWO HOLES

d is the diameter of the shaftsD is the diameter of holes in two platesA set up for the worst case of interchangeability

Case1: Sizes of the holes and shafts are identical.


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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 as0 = D/2 - ( M+ m ) + D/2- d+ D/2 + ( M-m ) + D/2 -d

0 = 2D - 2m - 2d

2m = 2 ( D-d )

2m = 2L

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


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If 100% interchangeability is to be achieved in theabove case, the tolerance on dimension between thetwo holes cannot be more than twice the diametricalclearance between the holes and shafts. It could be lessor equal to the clearance.

CONCLUSION

TOLERANCE ON DIMENSIONS BETWEEN CENTERS OF TWO HOLES


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Case II: The size of the holes and shafts vary

TOLERANCE ON DIMENSIONS BETWEEN CENTERS OF TWO HOLES

D1 be the diameter of the holes in plates 1 and 2, for the mating shaft d1.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.


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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 beplaced 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 - d2

0 = - 2m + ( D1 + D2 ) - ( d1 + d2 )2m = ( D1 - d1 ) + ( D2 - d2 )

2m = ( L1 + L2 ).

Since ( D1 - d1 ) = L1 and ( D2 - d2) = L2m = ( L1 + L2 ) 2


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The critical tolerance m on the centers of two holes inthe above case should be equal to half the sum of clearancesin both holes and could be less for 100% interchangeabilitybetween the two plates.

CONCLUSION

TOLERANCE ON DIMENSIONS BETWEEN CENTERS


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TOLERANCE ON DIMENSIONS BETWEEN CENTERSOF THREE HOLES

D - is the diameter of holes in the plates 1 and 2

d - the diameter of mating shafts

The holes displaced according to the tolerance for the worstcase, is shown.

Series Dimensioning


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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 holesm = The tolerance of the dimensions.

0 = D/2 + ( M- m ) + ( Mm ) + D / 2d + D/2 - ( M+ m )- ( M + m ) +D/2 - d

0 = 2D - 2d - 4m

4 m = 2 ( Dd ) = 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

For n holes dimensioned in series tolerance between the holes should be equal

to

or less than L/n-1 for 100% interchangeability.

Series or Chain dimensions between


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For n holes dimensioned in series tolerance between the holes

should be equal to or less than L/n-1 for 100% interchangeability.

Series or Chain dimensions betweencenters


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As the number of holes increases in case of series dimensioning,

the tolerance, between the holes decreases. Hence the cost of

production increases and it is difficult to make. If same tolerancesare given to the large dimensions as that of small dimensions, it

will be in many instances impracticable.

CONCLUSION

TOLERANCE ON DIMENSIONS BETWEEN CENTERS


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TOLERANCE ON DIMENSIONS BETWEEN CENTERSOF THREE HOLES

Parallel Dimensioning

D - is the diameter of holes in the plates 1 and 2

d - is the diameter of the shafts

Fig . shows the holes displaced according to the tolerances for the worst case.


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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 = 0m = m


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Parallel Dimensioning


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As the number of holes increases the tolerance remains the same L

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 .

CONCLUSION


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FOUR-HOLE GROUP WITH COORDINATE TOLERANCES


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Comparison between series & parallel Dimensioning:

If the number of holes exceeds 3, parallel dimensioning is only

recommended. By series dimensioning, tolerance decreases which isdifficult to achieve and expensive.

TOLERANCE ON DIMENSIONS OF HOLES IN PLATES I AND


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TOLERANCE ON DIMENSIONS OF HOLES IN PLATES I ANDII FROM THEIR COMMON EDGE FOR INTERCHANGEABILITY:

Case1: Fastener having clearance in both the plates

Two plates are to be assembled by a fastener ofdiad, passing through the holes

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.


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Center of the hole in plate 1 is ( M + m )

Center of the hole in plate 2 is ( Mm )

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 = (Dd)

( Dd ) = L

2m = ( Dd ) 2 = 0.7 * L

TOLERANCE ON DIMENSIONS OF HOLES IN PLATES I AND


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TOLERANCE ON DIMENSIONS OF HOLES IN PLATES I ANDII FROM THEIR COMMON EDGE FOR INTERCHANGEABILITY:

Case II: Fastener having clearance in plate 1 and press fit in plate 2.

For the position as shown

0 = (D/2)-2(M+m)+2(M-m)-(d/2)(D-d)/2 = 2*2m ; ( Dd ) = L0.35 L = 2m.

TOLERANCES OF DIMENSIONS IN TWO PLATES FROM


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COMMON SURFACE FOR INTERCHANGEABILITY

DDia. of hole in Plates 1 & 2

( M + m ) and ( Mm ) - distance of holes

ddia. of fastener

Case I : Fastener having clearance in both plates

Applying the principle

0 = ( D/2 ) ( M + m ) + ( M - m ) + ( D/2 )d

Dd = 2 m

Dd = L,

L = 2m0.5 L = m

for complete interchangeability

m 0.5 L

TOLERANCES OF DIMENSIONS IN TWO PLATES FROM


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Case II: Fastener having clearance in one plate and press fit in other plate

COMMON SURFACE FOR INTERCHANGEABILITY

0 = D / 2 - ( M + m ) + ( Mm ) ( d / 2 )

( Dd ) / 2 = 2 m

( Dd ) = LL / 2 = 2 m

m = 0.25 L

For complete interchangeability m 0.25

CALCULATION OF TOLERANCE FOR DISTANCE BETWEENHOLES ON DRILL JIGS


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HOLES ON DRILL JIGS

Two holes are to be drilled and reamed at a distance W0 within the limits T0 , by using a jig

plate with slip bushes as shown in Fig.

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 outside diameter for the

two bushes respectively.

So that W0 = f ( w1, L1, L2, e1, e2.).


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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.

This distance between holes on the component

W 1 + T1 + OO1 + OO2

The sum of these distance should be less than the maximum distance givenon the component.

( W1 + T1 + L1 + e1 + e2 ) W0 + T0.

CONDITION 2:


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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

+ e1

OO2 = L2 + e2

The distance between the holes obtained on the component

( W1 + T1 ) = ( W1 - T1 ) - O O1 - OO2= ( W

1- T

1) - ( L

1+ e

1) - ( L

2+ e

2)

= ( W1 - T1 - L1 - L2 - e1 - e2 )

The sum of these distances should be more than the minimum distance( W

o- T

o) given on the component

( W1- T

1) - L

1- L

2- e

1- e

2 ( W o - To )

EFFECT OF CLEARANCE BETWEEN THE DRILL AND THE SLIP BUSHES ONDISTANCE BETWEEN THE HOLES ON THE COMPONENT


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DISTANCE BETWEEN THE HOLES ON THE COMPONENT

Case I : 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 between the

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+ L

1+ L

2+ e

1+ e

2+ OO

1+ OO

2

= 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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CASE II: When drill size is minimum the bushes position is also at minimum

distance , the drill is at minimum distance.

OO1

= OO2

= ( db

- Tb

) - ( d - Td

) 2

Actual distance between holes on component

= W1- T

1- L

1- L

2e

1- e

2- OO

1- OO

2

= W1- T

1- L

1- L

2- e

1- e

2- ( d

b+ T

b) + ( d - T

d)

This should be more than the minimum distance ( W0- T

0) given on component.

W1- T

1- L

1- L

2- e

1- e

2- ( d

b+ T

b) - ( d - T

d) > W

0- T

0

Final conditions are

W1+ T

1+ L

1+ L

2+ e

1+ e

2+ ( d

b+T

b) - ( d - T

d) ( W

0+ T

0)

W1- T

1- L

1- L

2- e

1e

2- ( d

b+ T

b) + ( d - T

d) > ( W

0- T

0)

As per Indian standard the tolerance on drill is = h 9

The tolerance of the hole in slip bush is = F7

The clearance between the liner bush and the slip bush is = F7/ h

6

(If the liner bush is used for drilling also)


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1

2


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3


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4


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6


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DRAWING


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DRAWING


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70

10


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20


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72

30


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7340


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74

50


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60


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7670


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80


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78

90


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100


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Tolerance Accumulation

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Transcript of Tolerance Accumulation