a5 8 tolerance stackup analysis jvr v1.1

© 2009 Infotech Enterprises Limited. All Rights Reserved <30-09-09>

Tolerance Stack up

Design Life Cycle and fundamental design calculations

Confidential © 2009 Infotech Enterprises Limited 2

Tolerance stackup analysis is used to deal with dimensional

tolerances in one-dimension, the resultant tolerance is always the

sum of the component tolerances. Analysis and control of

dimensional tolerances are relatively well developed compared to

those for geometric tolerances.

The stackup of geometrical tolerances was usually ignored or

replaced by the stackup of component tolerances.

Tolerance Stack up

Confidential © 2009 Infotech Enterprises Limited 3

A and B two linear dimensions to be added

a1 , a2 tolerance on A

b1 , b2 tolerance on B

A+B = C

The tolerance on C to be analysed.

Addition of Tolerances

Confidential © 2009 Infotech Enterprises Limited

C is max. when A and B are max.

C is min. when A and B are min.

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 )

c1, c2 tolerance on C

Subtracting C min ( 2 ) from C max ( 1 )

c1 + c2 = (a1+a2) + (b1 + b2)

If ,

( a1 + a2 ) = T a

( b1 + b2 ) = T b

( c1 + c2 ) = T c

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

Addition of Tolerances


A and B two linear dimensions to be subtracted

a1 , a2 tolerance on A

b1 , b2 tolerance on B

A - B = C

The tolerance on C to be analysed.

Subtraction of Tolerances


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 max - B min

C min = A min - B max ;

A max = A + a2

B min = B - b1

A min = A - a1

B max = B + b2

C max = ( C + c2 ) = ( A + a2) - ( B - b1 ) ----------- ( 3)

C min = ( C - c1 ) = ( A - a1) - ( B + b2 ) ---------- ( 4 )

Subtracting C min ( 4 ) from Cmax ( 3 )

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

If ( a2 + a1 ) = Ta

( b2 + b1 ) = Tb

( c2 + c1 ) = Tc

The Tolerance on the remaining length T c = Ta + Tb

Subtraction of Tolerances


The tolerances are getting added both in addition and subtraction.

Further slides deals with how to calculate unknown dimensions.

Addition & Subtraction of Tolerances


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

x2 a2 z2 x1 a1 z1 X - A = Z

x2 a2 x2 - a1 z2

x1 a1 x1 - a2 z1

(X) - (A) = ( X - A ) = Z

Therefore,

x2 - a1 = z2

x2 = a1 + z2

x1 - a2 = z1

x1 = a2 +z1

Case - 1

Therefore, x2 a1 + z2

x1 a2 + z1

X = ( A + Z )



x2 + 0.1 + 0.2 x1 - 0.1 - 0.2

X - 5 = 10

x2 - ( - 0.1 ) = + 0.2x2 + 0.1 = + 0.2x2 = + 0.2 - 0.1 = + 0.1

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

x2 + 0.1 + 0.2 x1 - 0.1 - 0.2

(X - 5 ) = 10

Example - 1

+0.1 + 0.1 + 0.2 - 0.1 - 0.1 - 0.2 Therefore , (15 - 5 ) = 10



a2 x 2 z2 a1 x1 z1 A - X = Z a2 - x1 z2 a1 - x2 z1

( A - X ) = Z The tolerances of the equation can be equated.

a2 - x1 = z2

x1 = a2 - z2

a1 - x2 = z1

x2 = a1 - z1

Therefore,

x2 a1 - z1 x1 a2 - z2 (X ) = ( A - Z )

Case – 2



+ 0.1 x2 + 0.2 - 0.1 x1 - 0.2 15 - X = 10

The tolerances of the equation can be equated.

+0.1 - x1 = +0.2x1 = +0.1 - 0.2 = -0.1

-0.1 - x2 = -0.2x2 = -0.1 +0.2 = +0.1

Therefore

+ 0.1 x2 + 0.2 - 0.1 x1 - 0.2 15 - X = 10

+ 0.1 +0.1 + 0.2 - 0.1 -0.1 - 0.2 15 - 5 = 10

+0.1 - x1 +0.2 - 0.1 - x2 - 0.2

( 15 - X ) = 10

Example - 1


+ 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.

Example - 2


Nominal size ‘ C ’ = Nominal size ‘ A ’ - Nominal size ‘ B ’ C = 100 - 60 = 40

c2 a2 b2 c1 a1 b1

C = A - B

c2 + 0.05 - 0.13 c1 - 0.03 - 0.2040 = 100 - 60

c2 = + 0.05 - ( - 0.20 ) = + 0.25c1 = - 0.03 - ( - 0.13 ) = + 0.1

c2 +0.25 c1 +0.1C = 40

Verification: Tc = Ta +Tb Ta + Tb = + 0.05 - ( - 0.03 ) + ( - 0.13 ) - ( - 0.20)

= 0.15 Tc = 0.15 0.15 = 0.15

Example - 2


Calculate the dimension ‘x’ and tolerance on ‘x’ with ‘M-M’ and ‘L-L’ as references.

Example - 3


Considering MM as reference

a2 x2 b2

a1 x1 b1 A - X = B

+ 0.0 x2 - 0.1 x1 ±0.130 - 8 = 22

+0.0- x1 -0.1 - x2 ±0.1 ( 30 - 8 ) = 22

0.0 - x1 = 0.1 x1 = - 0.1 -0.1 - x2 = 0.1 x2 = 0.0

x2 + 0.0x1 - 0.1

The value of X = 8

Example - 3


Considering LL as reference

a2 b2 x2 a1 b1 x1 A - B = X

+ 0.0 x2 - 0.1 ± 0.1 x1 30 - 22 = 8

x2 = + 0.0 - ( - 0.1) = + 0.1x1 = - 0.1 - (+ 0.1) = - 0.2

x2 + 0.1 x1 - 0.2X =

8

Example - 3


“Any dimension with tolerances, if multiplied by a positive constant

number, the tolerance of the result also increases proportionately.

When divided by a constant number, the tolerance of the result,

similarly decreases proportionately”

Multiplication & Division by a constant number


a2 a1

When A is multiplied by a constant K , it becomes

a2 a2 * K a1 a1 * K( A ) * K = ( A * K )

Example :

+ 0.1 (+0.1) 3 + 0.3 (+0.3) 34 * (3) = (4 * 3)

+0.3 +0.9 = 12

Multiplication by a constant number


a2

a1

When 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

Division by a constant number


Find out the variation of dimension ‘x’ with reference to ‘LL’

Example : 1


x2 +0.0 +0.1 +0.0 +0.0 x2 x1 -0.2 -0.0 -0.2 -0.1 x1

X = 50 - 5 - 38 - 5 = 2

x2 = + 0.0 - ( - 0.0 ) - ( - 0.2 ) - ( - 0.1 ) = + 0.3x1 = - 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 = Tx

Ta + Tb + Tc + Td = 0.2 + 0.1 + 0.2 + 0.1= 0.6

Tx = 0.3+0.3 = 0.6 0.6 = 0.6

Example : 1


-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 = TxTa+ Tb+ Tc + Td + Te + Tf = 0.040 + [(-0.040)-(-0.073)] + [(0.013)-(-0.009)] + [(+0.012-(+0.001)] + [(-0.000)-(-0.019)] + [(-0.006)-( -0.017)] = 0.136 Tx = [0.119-(-0.017)] = 0.136 0.136 = 0.136

Example : 2


+ 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 +0.057 x1 - 0.040 235 = 235 Verification :

Ta + Tb + Tc + Td = Tx

Ta + Tb + Tc + 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 : 2


+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 +0.096 x1 - 0.032 387 = 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) = 0.128

0.128 = 0.128

Example : 3


Calculate the value of ‘x’ to have the axial clearance between bearing

and the shaft, as 1.25 ± 0.25 mm

Example : 4


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.

L

L

Example : 4


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 x1

X = (151.25 - 100) = (51.25) x2 = + 0.0 + 0.00 + 0.25 + 0.00 - ( - 0.25 ) = + 0.50

x1 = - 0.05 - 0.10 - 0.25 - 0.05 - ( + 0.25 ) = - 0.70

x2 +0.50 x1 -0.70(51.25) = 51.25

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 =1.20

Tx = 1.20 1.20 = 1.20

L

L

Example : 4


A hole 16H6 has to be drilled

and reamed

The dimension is 79 0.1 from

one side of the component.

Calculate the tolerance for the

dimension 12 from one end in a

box jig plate to obtain the

resultant dimension 79 0.1

Example : 5


+0.00 x2

± 0.10 ± 0.05 - 0.05 x1 79 = 111 - 20 - 12

Max. limit : + 0.10 = + 0.05 - ( - 0.05 ) - x1

x1 = 0.00

Min. limit : - 0.1 = - 0.05 - ( + 0.00 ) - x2

x2 = 0.05

The tolerance for the dimension 12 is + 0.05 +0.00

+ 0.00 - 0.05 12 or 12.05

Verification:

Td = Ta + Tb + Tc +0.00 +0.05 ± 0.10 ± 0.05 - 0.05 - 0.00

79 = 111 - 20 - 12 Td = 0.20 Ta + Tb + Tc = (0.05)-(-0.05)+0.00-(-0.05)+(0.05)-(-0.00) =

0.20 0.20 = 0.20

Example : 5


A hole of 16H7 is to be made with reference to another predrilled and reamed hole 10H7 which lies on a perpendicular plane

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

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

Example : 6


x1 0.1 0.05 x2

Dimension should be 148 = 245 - 97

x2 0.1 x1 0.05 148 + 97 = 245+0.1 - ( x1 ) = +0.05 x1 = +0.05

-0.1 - ( x2 ) = - 0.05 x2 = - 0.05

The tolerances for the dimension 97 is 0.05

Verification :

0.05 0.05 0.1

245 - 97 = 148

Ta +Tb = 0.1 + 0.1 = 0.2

Tc = 0.2

0.2 = 0.2

Example : 6


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 the overall dimensions 180 0.05 are toleranced.

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

Case 1 : considering ‘ LL’ as reference

x2 x1 0.05 0.1 0.2X = 180 - 50 - 100

x2 = + 0.05 - ( - 0.1 ) - ( - 0.2 ) = + 0.35x1 = - 0.05 - ( + 0.1 ) - ( + 0.2 ) = - 0.35 x2 x1 0.35 X = 30

Case 2 : considering ‘AA’ as reference

x1

x2 0.05 0. 2 X = 130 - 100

x2 = + 0.05 - ( - 0.2 ) = + 0.25x1 = - 0.05 - ( + 0.2 ) = - 0.25

x1 x2 0. 25X = 30

Example : 7


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 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.

Example : 8


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 0.0 5 -0.1430 - 10 = 20

a2 - ( - 0.05 ) = + 0.08a2 = 0.08 - 0.05 = + 0.03a1 - ( + 0.05 ) = - 0.14a1 = - 0.09

a2 +0.03 a1 -0.09 A = 30

+0.03 -0.09

The depth of hole is = 30

Example : 8


A set up for the worst case of interchangeability.

Case 1 : Sizes of the holes and shafts are identical.

‘d’ is the diameter of the shafts. ‘D’ is the diameter of holes in two plates.

Tolerance on dimensions between centers of two holes


1-1 = Axis of holes in the plate12-2 = Axis of the holes in the plate 2CC = Axis of the shafts mating the holesM = 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 - 2d2m = 2 ( D-d )2m = 2Lm = 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



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.



Case 2 : The size of the holes and shafts vary

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.



1-1 = Axis of holes in the plate 12-2 = Axis of holes in the plate 2c1 -c1 = Axis of the shaft of diameter d1c2 -c2 = Axis of the shaft of diameter d2M = 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) = L2

m = ( L1 + L2 ) ÷ 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.



Objective :

To calculate the tolerance stackup on the clearance between the Reel

Assembly and Bedbar-Bedknife assembly by Worst Case analysis and

Statistical analysis.Need for Tolerance stack up :

The clearance between the Reel Assembly and Bedbar-Bedknife

assembly determines the quality of grass being cut.

Tolerance Stack up Case Study


Reel Mower Cutting Unit :

Bedbar-Bedknife Assembly

Reel Assembly

Plate-Side Support

Plate-Side Support



Reel Mower Cutting Unit-side View :

Clearance



Assumptions :

Reel sub assembly center axis is aligned to the mounting hole axis of

Plate-Side Supports(107-3256).

Bedbar-bedknife assembly(107-3274) center axis is aligned to the

mounting hole axis of Plate-Side supports(107-3256).

Title block tolerances are considered wherever applicable

Clearance between the Reel sub assembly and Bedbar-Bedknife sub

assembly is assumed to be .0008



A1A2

G2

D2D1

Bearing DimensionsShaft & Plate-support Dimensions



Clearance

G1

E1

F1

F2



C1

B1



Tolerance Stack up Calculations :

Tolerance accumulated at the clearance = ( Tolerance accumulated at Bed knife cutting edge ) –

( Tolerance accumulated

at Reel cutting edge )

= (G1+E1+F1) – F2

(G1 + E1 + F1) = Height between shaft axis to top of the Bed Knife.

F2 = Height between shaft axis to bottom of Reel Assembly.

Clearance between the Bed Knife and Bed Reel Assembly is :

(G1 + E1 + F1) – F2



As per the Tolerance – Stack - Calculation sheet :

G1 = 1.5940E1 = 1.8317F1 = 0.1025F2 = 3.5274

G1 + E1 + F1 = 1.5940 + 1.8317 + 0.1025 = 3.5282

Therefore, Clearance = 3.5282 - 3.5274

= 0.0008

As per tolerance analysis :

±0.0018G1 = 1.5940

±0.0100E1 = 1.8317

±0.0150

F1 = 0.1025

+0.0160 -0.0190

F2 = 1.5940

Tolerance Stack up calculations



+ 0.0160 x2 ±0.0018 ±0.0100 ±0.0150 - 0.0190 x1 1.5940 + 1.8317 + 0.1025 - 3.5274 = X

0.0018 + 0.0100 + 0.0150 – ( - 0.0190 ) x2 -0.0018 - 0.0100 - 0.0150 – ( 0.0160 ) x1

0.0008 = X

Therefore :

Nominal clearance X = 0.0008Upper value x2 = 0.0458Lower value x1 = -0.0428

Max clearance : X + x2 = 0.0008 + 0.0458 = 0.0466Min clearance : X + x1 = 0.0008 - 0.0428 = -0.0420



As per statistical analysis :

Probable tolerance on G1 = 0.00180 E1 = 0.01000 F1 = 0.01500 F2 = 0.01745(G1 + E1 + F1) = 0.00180 + 0.01000 +

0.01500= 0.02680

Therefore, clearance

0.02680 0.01745 x2 +0 +0 x1 3.5282 - 3.5274 = X

0.02680 x2 -0.01745 x1

0.0008 = X

Max clearance = 0.0008+0.02680 = 0.02760Min clearance = 0.0008-0.01745 = -0.01665

Probable clearance = 0.02760 – (-0.01665)= 0.04425



Conclusions :

As per Worst Case Analysis :

Nominal Clearance between the Reel Assembly(107-4044) and the

Bedknife Assembly(107-3274) is 0.0008

Maximum Clearance is 0.0466

Minimum Clearance is -0.0420

As per Statistical Analysis :

Probable Clearance between the Reel Assembly and the Bedknife Assembly

is 0.04425



Reference Books


Thank you

a5 8 tolerance stackup analysis jvr v1.1

Documents