Lear Tolerance Stackup Course

Welcome to a Course On Tolerance Stack-up Analysis using Co-ordinate System of Dimensioning and GD&T For

Lear Corporatio Corporation n – Philippin Philippine e Engineering Engineering and Technology Center, Cebu

How is Course Organized?

 

    

Total 12 Sessions; 3days Pre-defined objectives at the beginning of each session Classroom exercises at the end of each session Homework Extended hours as necessary Assumption : Understanding of GD&T controls Feel free to interrupt and ask Questions

How is Course Organized?

 

    

Total 12 Sessions; 3days Pre-defined objectives at the beginning of each session Classroom exercises at the end of each session Homework Extended hours as necessary Assumption : Understanding of GD&T controls Feel free to interrupt and ask Questions

Classical Approach to Tolerance Stack-up Analysis …

What is Tolerance Stack-up Analysis?

Tolerance Stack-up Analysis (also called as Gap Analysis / Loop Diagrams / Circuit Analysis or COD (Chain of Dimensions)) is the process of calculating minimum and maximum airspaces or wall thickness or material interferences in a single part or assemblies. It ’s a Decision making tool and helps designer to answer one or more questions shown in next slides. It is a logical process divided in few steps …

Typically, Tolerance Stack-up Provides answers to …



Will these two surfaces touch in their worst case? If so, how much they will interfere?



What is maximum thickness of the two parts that must fit in the slot? Will the pin fit within the hole? How do I know if the worst case assembly will satisfy its dimensional objectives.

 

 

 

If we reduce the size of clearance hole, will the parts still assemble? Will the dimensioning and tolerancing scheme used on the parts, allow too much variation at assembly? Should the drawing be re-dimensioned and re-toleranced to reduce the accumulation of tolerances? …. ….

Why Perform Tolerance Stack-up?

A Tolerance Stackup allows the designer to: – – – – –

– – –

Optimize the tolerances of parts and assemblies in a new design. Balance accuracy, precision and cost with manufacturing process capability Determine part tolerances required to satisfy a final assembly condition. Determine the allowable part tolerances if the assembly tolerance is known. Determine if parts will work at their worst-case or with the maximum statistical variation. Troubleshoot malfunctioning existing parts or assemblies. Determine effect of changing a tolerance will have on assembly function Explore design alternatives using different or modified parts or tooling/fixturing methods.

Factors affecting Tolerance Stack-up Analysis There are four major factors that determine which dimensions and tolerances are included in a Tolerance Stack-up: 



 

The geometry of parts and assemblies that contribute to the distance (objective) being studied in the Tolerance Stack-up. The Dimensioning and Tolerancing schemes on the drawing of the parts and assemblies in the Tolerance Stack-up. The assembly process: how and and which order the parts are assembled? The direction of tolerance stack-up and direction of the dimensions and tolerances.

Basic Assumptions in Tolerance Stack-up Analysis : Problem Idealization Tolerance Stack-ups are preformed with following assumptions: –

All parts are considered in a static state. The tolerance stack-up allows parts to adjust (translate/rotate) relative to one another during assembly process, but the analysis is performed in a static condition. 

–

If more than one position or configuration of part/assembly to be studied (such as linkage or mechanism), then, tolerance stack-up should be done for the considered parts at each required position or orientation/configuration.

Tolerance Stack-ups are performed at a specified temperature. Unless specified otherwise, Tolerance stack-ups are performed at ambient temperature – the temperature at which the parts are assembled or inspected. 

If parts are assembled at one temperature and operate at different temperatures, it is important to study both conditions, as the parts must be assembled before they can operate.

Steps in Tolerance Stack-up Analysis



Step #1: –



Identify objectives: what are your end requirements? Such as flushness between features or gaps around a feature or alignment of features

Step #2: –

Identify all dimensions that contribute to your objectives as defined in step #1 and convert them to equal bilateral toleranced dimensions; as necessary




…

Step #3: –

Assign each dimension a +ve or –ve value. For Radial stacks (going up and down); start at the bottom of gap and end up at the top of gap – –

–

Down direction is –ve (top of gap to bottom) Up direction is +ve (bottom of gap to top OR towards end)

Stacks that go left and right in the assembly, start at the left side of gap and end up at the right side of the gap. – –

Left direction is –ve (right of gap to left) Right direction is +ve (left of gap to right OR towards end)

Remember to work on one part at a time; so deal with that part ’s pertinent features before moving to next part. This approach is best to work with assemblies having many parts




…

Step #4 (tips): –

Remember that one set of mating features between parts creates the variable or objective you are working for. Variables are either minimum gap or maximum gap or maximum overall assembly dimension. One set mating features creates it. So, though multiple routes may have to be evluated to find this most significant set of features, only one set creates worst case, from one part to next.

–

Errors could creep in if you follow one route from one set of mating features (hole/pin pairs) then continue the same route through another set. Only one of these sets shall create the smallest or largest gap or maximum/minimum overall dimension, Once you spot it, others become non-factors in analysis.

–

While reaching end objectives or goals, using more than one set of features within same two parts, will most likely produce incorrect results – and tolerances from other features may contribute to the critical set you are searching for. For example: when datum features are referenced at MMC or when more than one set of datum features are assembly features.




…

Step #5 (Basic Rules): –

When a single feature or a pattern of features are controlled by multiple Geometric Tolerances (such as orientation refined with position), the analyst must determine which, if either is contributing factor to variable. It is likely that none of geometric tolerance is a factor and instead size dimensions are factors.

–

The Designer must evaluate which factors are relevant through diagrams and logical reasoning.

–

The judgment of designer is critical in these determinations.

Beginning Tolerance Stack-up Analysis



Its important to arrange all the features and parts in the directions that will create the max or min gap / or variable you are searching for. This is to allow your loop always pass through material and you do not jump over an air space unnecessarily in analysis



You should position the features of the parts against each other so that you will get extreme configurations and make clear to you the correct path with +ve v/s –ve designations for each dimension.

Session #1 : The Basics



Objectives: 

 

Calculating mean dimensions with equal Bilateral Tolerances Calculating Inner and Outer Boundaries Virtual and Resultant Condition of features

Finding Mean Dimensions

Few Important Concepts of Tolerance Stack-up Analysis: –

–

–

There is NO difference between equal, unequal or unilaterally toleranced dimension. There is NO difference between a limit dimension and a plus or minus toleranced dimension. They all have extremes and they all have means. So, first thing is to change any dimension to an equal bilateral toleranced dimension.


Limit dimensions:

n 22 - n 20

Upper limit =

n

22, Lower limit =

Now, sum the limits :

n

n

20

22 + n 20 = n 42. Take the mean of sum =

Take the difference of limits:

n

22 - n 20 =

Therefore, limit dimension of as n 21` 1

n 22- n 20

n

n

21

2. Take the mean of difference =

n

1

is expressed as equal bilateral toleranced dimension


Unequal bilateral toleranced dimensions: n 50

+1 -3

So, Upper limit = Lower

limit =

n 5 0+ 1 =

n 50- 3=

Now, sum the limits :

n51

n 47

n 51

+

n 47

Then, take the difference of limits :

=

n 51

n 9 8. -

Mean of sum is

n 47

Therefore, unequal bilateral toleranced dimension of converted to equal bilateral toleranced dimension is

=

n 4.

n 50

n 9 8/ 2

=

n49

Mean of difference is +1 -3

n 49` 2

n 4/ 2

=

n2

Finding Mean Dimensions :

Exercise

Convert following Dimensions to an equal bilateral toleranced dimensions

1.

 100 13

2.

 150   155 2

3.

200 0

4.

 30  0.47

5.

0

 0.26

500 0.37

Boundaries

Boundaries are generated by collective effects of size and Geometric tolerances applied to feature(s) and often referred to as simply inner and outer boundaries There are two types of boundaries:  

Virtual Condition Boundary (VCB) Resultant Condition Boundary (RCB)

Virtual Condition Boundaries

(Refer ASME Y14.5M-1994

section 2.11)



FCFs that use m (MMC symbol), generate constant boundaries (VCB) for features under consideration and are calculated as: –

VCB for internal FOS such as hole = MMC Size Boundary – Geometric Tolerance value

–

VCB for external FOS such as pin = MMC Size boundary + Geometric Tolerance

VC Boundaries are Constant and do not vary based upon actual mating size of the feature

Virtual Condition Boundaries

(Refer ASME Y14.5M-

1994 section 2.11)



FCFs that use l (LMC symbol), generate constant boundaries (VCB) for features under consideration and are calculated as: –

VCB for internal FOS such as hole = LMC Size Boundary + Geometric Tolerance value

–

VCB for external FOS such as pin = LMC Size boundary Geometric Tolerance.

VC Boundaries are Constant and do not vary based upon actual mating size of the feature

Resultant Condition Boundaries

(Refer ASME

Y14.5M-1994 section 2.11)



RC Boundaries are non constant in nature and are generated on opposite side of the virtual conditions.



When RFS (Regardless of Feature Size) concept applies to FOS, they generate only non-constant or RC boundaries.

Case#1: Internal FOS controlled at MMC

Case#1: Calculating VC & RC boundaries Hole Size  GTol

 VirtualCondition( FixedBound ary )

 49  1   48  50  2   48  51  3   48  Worst case inner boundary Hole Size  GTol  Re sul tan tConditon(VariableBoundary)

 49  1   50  50  2   52  51  3   54  Worst case outer boundary VCB for internal FOS (such as hole) controlled at MMC = MMC Size Boundary – Geometric Tolerance value VCB for external FOS (such as pin) controlled at MMC = MMC Size boundary + Geometric

Case#1: Creating equal Bilateral Toleranced Dimension from VCB and RCB

Resultant condition of hole

 54

+ Virtual condition of hole

 48  102

SUM


 54

- Virtual condition of hole

 48

DIFFERENCE

 6

So,  51  3

Then, 102 2



51

&

6 2

3

Is an equal bilateral expression of the dimension and its tolerance

Case#2: Internal FOS controlled at LMC

Case#2: Calculating VC & RC boundaries Hole Size  GTol  VirtualCondition( FixedBound ary )

 51  1   52  50  2   52  49  3   52  Worst case outer boundary Hole Size  GTol



Re sul tan tConditon(VariableBoundary )

 51  1   50  50  2   48  49  3   46  Worst case inner boundary VCB for internal FOS (such as hole) controlled at LMC = LMC Size Boundary +Geometric Tolerance value VCB for external FOS (such as pin) controlled at LMC = LMC Size boundary -



 46

+ Virtual condition of hole

 52

SUM

 98


 46

- Virtual condition of hole

 52

DIFFERENCE

 6

So,  49  3

Then, 98 2



49

&

6 2

3


Case#3: Internal FOS controlled at RFS

Case#3: Calculating RC boundaries Since it’s a RFS Callout, no virtual condition boundaries exist and all boundaries are non-constant

Hole Size  GTol



InnerBoundry

 49  1   48  Worst case Inner boundary  50  1   49  51  1   50 Hole Size  GTol



OuterBoundary

 49  1   50  50  1   51  51  1   52  Worst case Outer boundary

Case#3: Creating equal Bilateral Toleranced Dimension from Inner and Outer Boundaries

Outer Boundary of hole

 52

+ Inner Boundary of hole

 48  100

SUM

Outer Boundary of hole

 52

- Inner Boundary of hole

 48

DIFFERENCE

 4

So,  50  2

Then, 100 2

 50

&

4 2



2


Case#4: External FOS Controlled at MMC

Case#4: Calculating VC & RC boundaries Shaft Size  GTol  VirtualCondition( FixedBound ary )

 47  1   48  46  2   48  45  3   48  Worst case outer boundary Shaft Size  GTol  Re sul tan tConditon(VariableBoundary )

 47  1   46  46  2   44  45  3   42  Worst case inner boundary VCB for internal FOS (such as hole) controlled at MMC = MMC Size Boundary – Geometric Tolerance value VCB for external FOS (such as pin) controlled at MMC = MMC Size boundary + Geometric Tolerance value


Resultant Condition of Shaft

 42

+ Virtual Condition of Shaft

 48

SUM

 90


 42

- Virtual Condition of Shaft

 48  6

DIFFERENCE

So,  45  3

Then, 90 2



45

&

6 2



3


Case#5: External FOS controlled at LMC

Case#5: Calculating VC & RC boundaries Shaft Size  GTol  VirtualCondition( FixedBound ary )

 45  1   44  46  2   44  47  3   44  Worst case inner boundary Shaft Size  GTol  Re sul tan tConditon(VariableBoundary )

 45  1   46  46  2   48  47  3   50  Worst case outer boundary VCB for internal FOS (such as hole) controlled at LMC = LMC Size Boundary +Geometric Tolerance value VCB for external FOS (such as pin) controlled at LMC = LMC Size boundary Geometric Tolerance value



 50

+ Virtual Condition of Shaft

 44

SUM

 94


 50

- Virtual Condition of Shaft

 44

DIFFERENCE

 6

So,  47  3

Then, 94 2



47

&

6 2



3


Case#6: External FOS controlled at RFS

Case#6: Calculating RC boundaries Since it’s a RFS Callout, no virtual condition boundaries exist and all boundaries are non-constant

Shaft Size  GTol  OuterBound ry

 45  1   46  46  1   47  47  1   48  Worst case Outer boundary Shaft Size  GTol  InnerBound ary

 45  1   44   46  1   45  47  1   46

Worst case Inner boundary

Case#6: Creating equal Bilateral Toleranced Dimension from Inner and Outer Boundaries

Outer Boundary of Shaft

 48

+ Inner Boundary of Shaft

 44

SUM

 92

Outer Boundary of Shaft

 48

- Inner Boundary of Shaft

 44

DIFFERENCE

 4

So,  46  2

Then, 92 2



46

&

4 2



2


Formulae to Remember…

For Internal FOS controlled at MMC / LMC: VCB at MMC (IB) = MMC Size Boundary – Geometric Tolerance value at MMC VCB at LMC (OB) = LMC Size Boundary + Geometric Tolerance value at LMC

For External FOS controlled at MMC / LMC: VCB at MMC (OB) = MMC Size boundary + Geometric Tolerance value at MMC VCB at LMC (IB) = LMC Size boundary - Geometric Tolerance value at LMC

Finding Inner & Outer Boundaries :

Exercise

Calculate Inner and Outer boundary for features having following specifications

Session #2: Analyzing a “C” Channel Assembly

Objectives: To determine min and max gap for a simple eleven parts assembly.   

Perform the calculations Create a Loop Analysis Diagram Create a Number Chart

“C”

Channel Assembly

“C”

Channel Assembly : Loop Analysis Diagram Up Direction

Down Direction

(+ve)

(-ve)

`

Tolerance

Remarks

GAP 188.4 188.4+/-1.5

255.67 255.67

188.4

255.67 – 188.4 255.67+/-0.1

1.5

All 10 blocks

0.1

Channel inner

1.6

Totals

67.27 – 1.6 = 65.67 67.27 + 1.6 = 68.87

Min GAP Max GAP

Session #2: Exercises


6

5

4

3 2

1

Loop #

Up Direction / Right Direction

Down Direction / Left Direction

(+ve)

(-ve)

`

Tolerance

1

56.62

0.1

2

34.74

0.1

3

10

0.1

4

90

0.15

5

30

0.15

6

230.58

240.58

Remarks

0.2

211.36

0.8

Totals

(29.22 - 0.8) = 28.42 (29.22 + 0.8) = 30.02

Min GAP / Max GAP



5 2

1

9 8 3

7

6

4

Loop #



(+ve)

(-ve)

`

Tolerance

1

26

0.1

2

23

0.15

3

235

0.2

4

23

04.35.15

5

51

0.5

6

22

8

0.1 60.2

7 28

9

285

Remarks

0.1 0.1

39

0.15

222.2

1.55

Totals

(62.8 – 1.55) = 61.25 (62.8 + 1.55) = 64.35

Min GAP / Max GAP

Session #3: Loop Analysis for Box and Cavity



Objectives: –

– –

Using Loop Analysis Technique; determine Max and Min gap in Horizontal and Vertical Directions Determine proper start and End points for stack-ups Graph the numbers calculated into Loop Diagram

Problem Description

Calculate: MIN / MAX Horizontal Gap MIN / MAX Vertical Gap

Loop Diagram

1 1

2

2

Horizontal Direction

Vertical Direction

Number Chart Horizontal Direction

Loop #



(+ve)

(-ve)

1 2

`

Tolerance

25.9

0.1

26.75 26.75

Remarks

0.5 25.9

0.6

Totals

(0.85 – 0.6) = 0.25 (0.85 + 0.6) = 1.45

Min GAP / Max GAP

Vertical Direction

Loop #



(+ve)

(-ve)

1 2

24.425 26.75 26.75

`

Tolerance

Remarks

0.575 0.5

24.425

1.075

Totals

(2.325 – 1.075) = 1.25 (2.325 + 1.075) = 3.4

Min GAP / Max GAP

Session #4: Analysis of an assembly with Limit tolerancing



Objectives: 

Calculate the airspaces and interferences for a plus and minus toleranced assembly



Performing multiple loop analyses on an assembly

Assembly with limit tolerancing : Problem Description

Assembly with limit tolerancing : Loop Diagrams

2

3

1

2

1

Assembly with limit tolerancing : Number Chart Horizontal Direction Up Direction Right Direction

Down Direction Left Direction

Loop #

(+ve)

(-ve)

1

20.84

0.66

2

15.8

0.75

3 36.64

`

Tolerance

Remarks

32.7

1.2

32.7

2.61

Totals

(3.94 – 2.61) = 1.33 (3.94 + 2.61) = 5.55

Min GAP / Max GAP Max / Min Overall Dim

Vertical Direction

Loop #

Up Direction Right Direction


(+ve)

(-ve)

1 2

`

25.125

Remarks

0.375

25.975

25.975

Tolerance

0.575

25.125

0.95 (0.85

0.95) = -0.1

Totals Min GAP / Max GAP

Session #5: Analyzing a Floating Fastener Assembly



Objectives: 

 

  

Calculate Virtual and Resultant conditions (Inner / Outer Boundaries) for GD&T callouts Determine mean of all these boundaries Convert all FOS (diameters and widths) to mean radii with equal bilateral tolerance Mixing FOSs (widths and diameters) in number chart Graph the numbers in tolerance stack-up diagram Determine all unknown gaps in the assembly

Floating fastener assembly sketch with GD&T

Floating fastener Part sketches with GD&T

140 140

 

6-7

6-7

3.5+/-0.5

 

5.5+/-0.5  

300

Floating fastener Assembly with parts shoved towards center

VCB of holes in top plates = (MMC – Gtol) = (6-0.5) = 5.5 RCB of holes in top plates = (LMC + Gtol + Btol) = (7+0.5+1) = 8.5 Mean Dia with equal bilateral representation of these holes is: 7+/-1.5

VCB of holes in base plate = (MMC – Gtol) = (5.5-0.5-0) = 5 RCB of holes in base plate = (LMC + Gtol + Btol) = (5.5+0.5+0+1) = 7 Mean Dia with equal bilateral representation of these holes is: 6+/-1

Loop Diagram with values printed… 2

1

9 8

3

7

4 5

6



(+ve)

(-ve)

Loop #

`

Remarks

Tolerance

1

140

0

Basic dimension

2

3.5

0.75

Over radius of top plate hole

0.5

Over pin dia

0.5

Over radius of base plate hole

0

Basic dimension

0.5

Over radius of base plate hole

0.5

Over pin dia

0.75

Over radius of top plate hole

3

3.5

4 5

3 300

6 7

3 3.5

8

3.5

9

140

307

293

Basic dimension

3.5

Totals

Can you imagine a configuration for MAX Gap? And then calculate MAX Gap

Loop #



(+ve)

(-ve)

`

Tolerance

Remarks

Totals Min GAP / Max GAP

Session#6: Analyzing an Assembly for Gaps and Overall Dimensions (Fixed Fastener Case)





Objectives:   

Calculate assembly overall MAX and MIN dimensions Calculate MAX and MIN gaps within assembly as shown Calculate boundaries using various GD&T controls

Min Gap and Min Overall Dimensions Configuration VCB of hole = (MMC – Gtol) = (13-0.03-0.05) = 12.92 RCB of hole = (LMC + Gtol + Btol) = (13+0.03+0.05+0.06) = 13.14 Mean Dia with equal bilateral representation of this hole is: 13.03+/-0.11

Start Point of Loop End Point of Loop

VCB of pin = (MMC – Gtol) = (12.5-0.03-0.05) = 12.22 RCB of pin= (LMC + Gtol + Btol) = (12.5+0.03+0.05+0.06) = 12.64 Mean Dia with equal bilateral representation of this pin is: 12.43+/-0.21

Min overall Dimension Loop diagram (4) Min Left bottom gap Loop diagram (6) Min Right top gap Loop diagram (6)

Chart the values

(Min overall Dim)



Loop #

(+ve)

(-ve)

1

105

2

6.515

`

Tolerance

Remarks

0

Basic dimension

0.055

Over radius of hole

3

6.215

0.105

Over radius of pin

4

65

0

Basic dimension

0.16

Totals

169.54

Min Overall Dim

176.215

6.515

Chart the values

Loop #



(+ve)

(-ve)

1 2

(Min left bottom gap)

15 105

3

6.515

`

Tolerance

Remarks

0.1 0

Basic dimension

0.055

Over radius of hole

4

6.215

0.105

Over radius of pin

5

65

0

Basic dimension

6 176.215

143.5

0.7

165.015

0.96

Totals

10.24

Min left bottom gap

Chart the values

Loop #



(+ve)

(-ve)

1 2

(Min Right top gap)

140 105

3

6.515

`

Tolerance

Remarks

0.7 0

Basic dimension

0.055

Over radius of hole

4

6.215

0.105

Over radius of pin

5

65

0

Basic dimension

6 176.215

15

0.1

161.515

0.96

Totals

15.66

Min right top gap

Max Gap and Max Overall Dimensions Configuration

Start Point of Loop End Point of Loop

Max overall Dimension Loop diagram (4) Max Left bottom gap Loop diagram (6) Max Right top gap Loop diagram (6)

Chart the values

Loop #

…



(+ve)

(-ve)

`

Tolerance

Remarks

Totals Min GAP / Max GAP Max / Min Overall Dim

Session #7: Calculating MAX overall Diameter for a Revolving Assembly

MAX?

Detailed Part Drawing with GD&T Controls

Part 1

Part2

Determine factors and non-factors affecting objectives with logical reasoning

Step#3: Create a Loop Diagram OB = MMC + Gtol = 250+0.2+0.15 = 250.35 IB = LMC – Gtol = 250-0.2-0.15 = 249.65 Mean dia with equal bilateral tolerance = 250+/-0.35

OB = MMC + Gtol = 250+0.2+0.15 = 250.35 IB = LMC – Gtol = 250-0.2-0.15 = 249.65 Mean dia with equal bilateral tolerance = 250+/-0.35

Step#4: Chart the values … Up Direction Right Direction


Loop #

(+ve)

(-ve)

1

125

0.175

2

25.11

-

LMC of hole / 2

-

LMC of spigot / 2

3 4

25.035 125

275.11

`

Tolerance

Remarks

0.175

25.035

0.35

Totals

250.425

Max Assembly Dia

Session #8: Analyzing a Guide Assembly with Fixed fasteners

Assembly

Part #1 & #2: Detailed Drawing

Session #7: Analyzing a Guide Assembly with Fixed fasteners



Objectives:     

Calculate Boundaries for Threaded features Work with multiple Geometric Controls on a single feature GD&T Controls affecting and non-affecting stack-up Calculate desired gaps Use product knowledge / experience and Assembly conditions in stack-up analysis

Locating parts to create MIN Gap Configuration

One line contact

CL of clearance hole in block

CL of threaded hole in slot

Chart the values Up Direction Right Direction


Loop #

(+ve)

(-ve)

1

13.5

0.15

Over 50% width of slot

2

3.95

0.12

Over radius of screw

`

Tolerance

Remarks

3

4.1325

0.055

Over radius of clearance hole

4

12.25

0.15

Over 50% width of block

16.3825

0.475

Totals

0.5925

Min GAP / Max GAP Max / Min Overall Dim

17.45

OB of Slot = LMC + Gtol =27.2+0.1 = 27.3 IB of Slot = MMC – Gtol = 26.8-0.1 = 26.7 Mean width of slot with equal bilateral tolerance = 27+/-0.3

OB of threaded hole/screw = MMC + Gtol =8+0.14 = 8.14 IB of threaded hole/screw = LMC – Gtol = 7.8-0.14 = 7.66 Mean dia with equal bilateral tolerance = 7.9+/-0.24

OB of Block = MMC + Gtol =24.7+0.1 = 24.8 IB of Slot = LMC – Gtol = 24.3-0.1 = 24.2 Mean width of block with equal bilateral tolerance = 24.5+/-0.3

OB of clearance hole = LMC + Gtol =8.25+0.05+0.06 = 8.36 IB of clearance hole = MMC – Gtol = 8.19-0.05 = 8.14 Mean dia with equal bilateral tolerance = 8.265+/-0.11

Locating parts to create MAX Gap Configuration One line contact

CL of clearance hole in block CL of threaded hole in block

Chart the values

Loop #



(+ve)

(-ve)

`

Tolerance

Remarks

Totals Min GAP / Max GAP Max / Min Overall Dim

Form Tolerances in Tolerance Stack-up

Min

MIN / MAX?

Max

Orientation Tolerances in Tolerance Stack-up

Min

MIN / MAX?

Max

Part Stacks using Position (RFS)

Min

Find MAX and MIN Distance (1) between edges of two small holes.

Max


Min

Find MAX and MIN Distance “X”.

Max


Min

Find MAX and MIN Distance between Centerlines of Hole and Slot.

Max

Part Stacks using Position (Bonus)

Min

Find MAX and MIN Distance between Edges of two small holes.

Max


Min

Max

Find MAX and MIN Distance (2) between Centerlines of the two small holes.


Min

Find MAX and MIN wall thickness.

Max

Part Stacks using Position (Bonus & Shift)

Min

Max

Find MAX and MIN horizontal distance between edges of datum G and n 8.6-8.2 hole.

Part Stacks using Position (Bonus & Shift)

Min

Max

Find MAX and MIN distance between edge of the groove and side of the part.

Part Stacks using Profile

Min

Find the MAX and MIN distance.

Max

Part Stacks using Profile

Min

Find the MAX and MIN distance

Max

Part Stacks using Form/Orientation/Profile Form/Orientation/Profile

Min

Max

Part Stacks using Form/Orientation/Profile

Min

Max

Part Stacks using Form/Orientation/Profile

Min

Max

Part Stacks

–

Composite Position Control

Part Stacks

–

Composite Position Control

MIN / MAX? MIN / MAX?

Min

Max

MIN?

MIN?

Session #10: Tolerance Stack-up Analysis of an Assembly with Revolving Parts

GAP? Part 1 Part 4

Part 2 Part 5

Tolerance Stack-up Analysis of an Assembly with Revolving Parts

Objectives: 

 

Calculating tolerance stack-ups on a five part rotating assembly with a variety of geometric controls such as: position, perpendicularity, parallelism, profile, flatness, projected tolerance zones, runout, total runout, concentricity, positional coaxiality Learn Simplifying a complex situation Calculate radial clearance and interference

Part #1: Detailed Drawing

Part #2,3 : Detailed Drawing





Session #11: Trigonometry and Proportions in Tolerance Stack-up Analysis

Trigonometry and Proportions in Tolerance Stack-up Analysis



Objectives: 

 





Understanding the role of trigonometry and proportions in tolerance stack-up and geometric tolerancing Understanding the effect of “Unstable Datums features ” Know how vertical stacks affect horizontal envelope requirements. Mixing trigonometry and algebra determining stack-up results Consider the rules in Y14.5.1 (Math Standard) for constructing a valid Datum

Example of Rocking Datum and proportions …

Out of flatness is shown on datum A on one side of part center; since this is worst case than flatness tolerance being evenly spread on entire surface 

Y14.5.1 states that in order to be a valid primary datum feature, the points used to construct a datum plane (3 high points of contact minimum) must not lie solely in one of the outer thirds of the surface. So its possible to conceive of slightly worse situation than this, but we are restricting to rocking at center point of part 

The illustration shows that flatness tolerance allows datum A to lean by an amount equal to flatness tolerance = 0.002. If the part is inspected on surface that does not lean; but assembled on surface that leans, the pin will be forced to lean with with it, by an amount = 0.006 


Normally this is ignored while calculating worst mating conditions of features like 80 length pin. We normally calculate worst mating condition diameter = MMC size + geo tol at MMC = 20.2+0.4 = 20.6. 

But with additional radial lean of 0.376, the worst mating condition can be seen as 20.4 + 2x0.376=21.352 

Also, while calculating the minimum gap between this shaft and the housing into which it fits, as per procedure we used in previous sessions, we would probably be working with radii, therefore ½ of 21.352 = R10.676 

Simple Proportions: 0.2/42.5 = X/80 80*0.2 = 45.5*x 0.376 = x


Parallelism is also a factor that Parallelism t hat can be related to the problems that flatness creates. Parallelism when used on planer surfaces, controls flatness and angle to datums referenced. 

In the illustration on left, produced part has crest in middle (rock point) and surfaces sloping on either side of rock point. 

So, when two or more such parts are stacked on top of one another, and each having problem as shown, such assembly would exhibit a problem of not fitting other assemblies/housings or closing holes on parts into which pins ore screws had to fit. (see next slide) 


Initially, the three parts were aligned with center, left edge and right edge aligned, then the parts are either to left or right 

This would assume that interior part features such as holes (not shown here) have been positioned from one of these features as secondary datum feature. 

Each part during inspection has been adjusted 9shimmed up) to allow high point shown at the bottom center of part 1 and 2 to establish the datum plane, but during assembly parts have been rocked instead of equalized. 

This is just one speculation as what can happen due to out of flatness of bottom of parts 1,2. Many such scenarios are possible. 

This much space would be needed if parts were


Unlike previous configuration, this configuration calculates the space needed to house these parts if they were stacked with their edges aligned and then rocked in either direction. 

This much space would be needed if parts were stacked this way and allowed to rock in either direction

Session#12: The Theory of Statistical Probability

The Theory of Statistical Probability



Objectives: 

  

Convert arithmetically calculated tolerances to statistically calculated tolerances. Use Root Sums Square (RSS) formula Comparing “Worst-case” and “Statistical” tolerances Reintegrating statistical tolerances into the assembly

Background … The dispersion of dimensions under the curve is described as “standard deviation” and often represented by letter σ (sigma), and calculated as:



σ

The arithmetic mean +or- one standard deviation ( ` 1 σ ) is often described as containing 68.26% of the produced parts under this normal curve. By the same logic ` 2 σ is 95.46% of the total production and ` 3 σ is 

99.73%

Root Sum Squares (RSS) Method



The statistical probability can be applied to tolerance stack-up analysis for assemblies both with and without geometric tolerances.



Thus the tolerance of an assembly is expressed as “square root of the sum of squares of the individual component tolerances ” and is called as RSS formula:

T A 



T

2 1



2

T T 2



2 3

T

 ......

2

n

Statistical probability has been practiced for several years and well documented. Statistical approaches are more reliable for volume production. For small production runs, the frequency curve tends to be skewed from its normal shape.

Applying RSS: Steps Involved with Example



Method: once the worst case calculations are performed, 1. 2.

3.

Using the RSS formula, calculate assembly tolerance Determine the percentage (%) ratio between statistical probability tolerance and 100% assembly tolerance Determine the increased statistical probability tolerances to be re-assigned to the assembly ’s individual part features.

RSS Calculations: Example#1 Worst case table Up Direction

Down Direction

(+ve)

(-ve)

`

Tolerance

188.4

1.5

All 10 blocks

0.1

Slot inner

1.6

Totals

67.27 – 1.6 = 65.67 67.27 + 1.6 = 68.87

Min GAP Max GAP

255.67 255.67

188.4

255.67 – 188.4

Remarks

RSS case table Up Direction (+)

Down Direction (-)

`

188.4 255.67 255.67

188.4

`

Tolerance Squared

Remarks

1.5

2.25

All 10 blocks

0.1

0.01

Channel Inner

Tolerance

1.6

2.26 1.5033

Totals Sqrt of total tolerance



Previous slide shows that the worst case assembly tolerance is +/-1.6, while the assembly tolerance based on RSS calculations is +/-1.5033



It states that if the parts are produced under statistical control, the li kely tolerance on assembly is +/1.5033 and NOT +/-1.6.



If we calculate the ratio of worst case tolerance to RSS tolerance = 1.6/1.5033 = 1.064.



This ratio can be used to increase the individual part level to lerance, in short, you can multiply part tolerances by factor of 1.064.



Therefore the individual blocks will receive a new tolerance of 0.15 * 1.064 = 0.1596 and the channel will receive a new tolerance of 0.1 * 1.064 = 0.1064

Lear Tolerance Stackup Course

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