Welcome to a Course On Tolerance Stack-up Analysis using C o -ordinate Dimensioning and GD&T For
Satyam Venture Engineering Services Pvt.Ltd., Secunderabad, INDIA 1
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About iSquare
iSquare (InterOperability & InterChangeability Solutions) Pune, INDIA
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Focus Areas:
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CAD Data InterOperability : Consistent representation of 3D CAD data in variety of CAD/CAM/CAE applications and platforms.
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InterChangeability: Predicting Dimensional Variations, its impact and causes at the product and assembly level at early design stage.
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Relationships: l
InterOperability: –
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With International TechneGroup Incorporated, USA having more than 20 years of Experience in CAD Data InterOperability technology, solutions and services.
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Relationships: • InterChangeability: • With Dimensional Control Systems, USA having more than 15 years of experience in Dimensional Control techniques, solutions and Services.
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Our Offerings: •CAD Data InterOperability: •Focused & Customized Training Programs on: •CAD/CAM/CAE Data Exchange : Problems and Solutions from CAD, CAE , CAM Perspective. •CAD Model Quality Assessment : CAD Model Quality evaluation from downstream application perspective
•Software Solutions For: •Effective Data exchange between heterogeneous CAD/CAM systems: R egardless of source, target application, standard and formats !! Solutions Include CA Dfix, IGES/Works,CAD/IQ. •Model Quality Assessment from Downstream application perspective
•Quality Services for: •Data Exchange, Data Migration, Lower version to higher or vice -a-versa •‘Vendor – Supplier’ data integration : ensuring effective data exchange with minimal / NO rework at either ends.
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Our Offerings: •InterChangeability: •Focused & Customized Training Programs on: •Dimensional Management : Understanding and appreciation of computer aided tools for. Takes participants thru evolution, various approaches and real l ife problems from their application areas.
•Software Solutions For: •Dimensional Management / Stack Analysis: Solutions embedded in C ATIA V4/V5 as Gold Partner and also Stand Alone solutions for data coming from othe r CAD platforms !! Solutions Include 1-DCS, DCS-DFC, 3DCS-SA, 3DCS-CAA V5 Designer, 3DCS-CAA V5 Analyst, GDM3D
•Quality Services for: •Dimensional Engineering / Management : Base Line tolerance mode l creation, reporting with suggestions and recommendations. Follow -on consulting •Per requirement, includes 1D, 1D with GD&T, Full 3D simulations, Piece – part variations, assembly variation prediction against desired objectives.
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Training Programs in Dimension Management / Engineering l
Training Programs Launched: –
Fundamentals of GD&T based upon ASME Y14.5M : ~36hrs
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Tolerance Sta ck up Analysis: A logical approa ch to solve assembl y build problems : ~30hrs
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Advanced GD&T: Concepts and Applications as per ASME Y14.5M : ~30hrs Tolerance Stack up Analysis using DCS (Dimensional Control Syste ms, USA) Software Solutions (1DCS, DCS-DFC, 3DCS-SA): ~36hrs
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’05) Training Programs Under Development (Tentative release by Oct –
GD&T Workshop and Practice (15% theory, 85% working on various problems): ~24hrs
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The Role of Probability and Statistics in Mechanical Tolerance A nalysis: ~20hrs
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Measurement of GD&T and Functional Gauging Techniques : ~24hrs
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Metrology: Me asurement Unce rtainty and Analysis : ??
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Customers l l l l l l l l l l
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TATA Motors TATA Technologies TATA Auto Plastics TATA Auto Components Ashok Leyland Mahindra & Mahindra – Auto Godrej & Boyce Mfg Ltd. GE Infotech Enterprises TATA Johnson Control Automotive Kinetic Engineering Research & Development Establishment (Engrs) Armament Research & Development Establishment Bhabha Atomic Research Center
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Bajaj Auto Bajaj Tempo Brakes India Emerson Climate Technologies Grupo Antolin Mahindra Engg Design Develop Center Kirloskar Copeland Mahindra Engineering Services Onward Technologies Space Applications Center TATA Consultancy Lear Seatings Pvt. Ltd. Atlas Copco Jayahind Industries L&T Satyam Venture Engg Services
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Th at’ s about iSquare
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How is Course Organized?
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Total 11 Sessions; 3days (June 23,24 and 25 , 2005)
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Pre-defined objectives at the beginning of each session
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Classroom exercises at the end of each session Homework
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Extended hours as necessary
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Assumption : Understanding of GD&T controls
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Feel free to interrupt and ask Questions
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Session #1 : The Basics l
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Objectives: Ø
How to calculate mean dimensions with equal
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Bilateral Tolerances Calculating Inner and Outer Boundaries
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Virtual and Resultant Conditions
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What is Tolerance Stack-up Analysis?
Tolerance Stack-up Analysis (also called as Gap Analysis, Loop Diagrams or Circuit Analysis) is the process calculating minimum and maximum airspacesofor wall thickness or material interferences in a single part or assemblies It’s a logical process broken in few steps …
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Steps in Tolerance St ack-up Analysis l
Step #1: –
Identify objectives: for example, you want to test if no interference is possible at a certain place in an assembly, then you set your requirement as “Gap must be equal to or greater than zero”
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Step #2: –
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Identify all dimensions that contribute to your objectives as defined in step #1 (gap) and convert them to equal bilateral toleranced dimensions; if they are not already
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Steps in Tolerance Stack-up Analysis l
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 – –
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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. –
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Left direction is –ve (right of gap to left) Right direction is +ve (left of gap to right OR towards end)
Remember that you are working one part at a time; sodeal with one part’s significant features before jumping to next part. This is the best way to work with assemblies having many parts
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Steps in Tolerance Stack-up Analysis l
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Step #4 (Basic Rules): –
Remember that one set of mating features between parts creates t he variable you are looking for. Variable in this case is eithe r minimum gap or maximum gap or maximum overall assembly dimension. One set mating featur es creates it. So, though multiple routes may have to be investigated to fi nd this most significant set of features, only one set creates worst case , fr om one part to next.
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Its often mistake to follow one route from one set of mating fea tures (holes/shaft, hole/pin) then continue the same route through ano ther set. One of these sets creates the smallest or biggest gap or maximum overall dimension, Once you find, which it is, others become non -factors in analysis.
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Using more than one set of features within same two parts, will most likely produce wrong results. Still tolerances from other features may contribute to the critical se t you are using. For exa mple: when datum f eatures are referenced at MMC or when more than one set of datum features co me into effect.
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Steps in Tolerance Stack-up Analysis l
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Step #5 (Basic Rules): –
When a single feature or a pattern of features are controlled by more than one Geometric Tolerance (such as orientation combined with position), the designer must determine which, if either is contributing factor to variable. It is also possible t hat none of geometric tolerance is a factor and instead size dimensions are factors.
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The Designer must deduce what factors are pertinent through sketches and reasoning.
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The judgment of designer is critical in these determinations. i2
Beginning Tolerance Stack-up Analysis
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Add all +ve and – ve dimensions which will calculate your mean gap. If mean gap is – ve number, your requirement of ‘ no material interference’ (or ‘clearance’ in other words) is already violated!
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Then we must add sum of equal bilateral tolerances (1/2 of total tolerance) to the mean dimension (or gap) to determine maximum gap.
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Then we must subtract the sum of equal bilateral tolerances (1/2 of total tolerance) from the mean dimension (or gap) to determine minimum gap.
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Again any – ve value for minimum or maximum gaps indicate interference situation
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Maximum gaps are maximum clearance (or in case of interference fits, minimum interference)
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Minimum gaps are minimum clearance (or in case of interference fits, maximum interference)
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Beginning Tolerance Stack-up Analysis
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Its important to mentally shove all the features and parts in the directions that will create the max or min gap (variable).This is to allow your routes always pass through material and you don ’t want to jump over an air space unnecessarily in analysis
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You should position the features of the parts against each other so that you will get extremes and make clear to you the correct path and +ve v/s –ve designations for each number.
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Finding Mean Dimensions l
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Few Important Concepts of Tolerance Stack-up Analysis: –
There is NO difference between equal, unequal or unilaterally toleranced dimension.
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There is NO difference between a limit dimension and a plus or minus toleranced dimension
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They all have extremes and they all have means. So, first thing is to change any dimension to anequal bilateral toleranced dimension
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Finding Mean Dimensions
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Finding Mean Dimensions
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Finding Mean Dimensions :
Exercise
Convert following Dimensions to an equal bilateral toleranced dimensions
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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
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Virtual Condition boundary (VCB)
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Resultant Condition Boundary (RCB)
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Virtual Condition Boundaries
(Refer ASME Y14.5M sec tion
2.11)
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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
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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
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Virtual Condition Boundaries
(Refer ASM E Y14.5M
secti on 2.11)
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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
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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 26
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Resultant Condition Boundaries
(Refer ASME
Y14.5M secti on 2.11)
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RC Boundaries are non constant in nature and are generated on opposite side of the virtual conditions.
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When RFS (Regardless of Feature Size) concept applies to FOS, they generate only non-constant or RC boundaries.
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Case#1: Internal FOS controlled at MMC
Hole – MMC Concept
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Case#1: Calculating VC & RC boundaries
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
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Case#1: Creating equal Bilateral Toleranced Dimension from VCB and RCB
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Case#2: Internal FOS controlled at LMC
Hole – LMC Concept
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Case#2: Calculating VC & RC boundaries
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
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Case#2: Creating equal Bilateral Toleranced Dimension from VCB and RCB
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Case#3: Internal FOS controlled at RFS
Hole – RFS Concept
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Case#3: Calculating RC boundaries Since it’s a RFS Callout, no virtual condition boundaries exist and all boundaries are non-constant
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Case#3: Assumption about feature form in case of RFS callout Only if the hole has a significant depth, might this median line curvature (out of straightness) be a consideration. For thin parts, such as sheet metal, it is probably not of concern in these analyses. In fact many designers would agree that a banana shaped hole is not likely to occur on most products.
Therefore we are ignoring “axially out of straightness consideration from the analyses.
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Case#3: Creating equal Bilateral Toleranced Dimension from Inner and Outer Boundaries
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Case#4: External FOS Controlled at MMC
Shaft – MMC Concept
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Case#4: Calculating VC & RC boundaries
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
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Case#4: Creating equal Bilateral Toleranced Dimension from VCB and RCB
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Case#5: External FOS controlled at LMC
Shaft – LMC Concept
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Case#5: Calculating VC & RC boundaries
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
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Case#5: Creating equal Bilateral Toleranced Dimension from VCB and RCB
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Case#6: External FOS controlled at RFS
Shaft – RFS Concept
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Case#6: Calculating RC boundaries
Inner Boundary
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Case#6: Assumption about feature form in case of RFS callout In case of RFS Callout, one may want to consider additional deviation arising ‘out of form ’ to determine absolute worst case inner boundary.
This applies only if shaft length is significant. For very shirt shafts / pins, it is probably not of concern in the analysis.
Therefore we are ignoring ‘out-of-straightness ’ consideration from our analysis. If your product runs a risk of banana shaped shafts, you may wish to consider illustration on the left in your calculations.
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Case#6: Creat ing equa l Bilate ral Tole rance d Dimension from Inner and Outer Boundaries
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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
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Finding Inner & Outer Boundaries :
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Exercise
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Session #2: Analyzing a Box Assembly l
Objectives:
To determine min and max gap for a simple eleven parts assembly l
Perform the calculations
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Create a Loop Analysis Diagram Create a Number Chart
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Box Assembly
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Box Assembly: Part #1 l
Cavity in part#1 has limit dimensions (392.43 – 384.81) ; so we need to convert these to mean with equal bilateral toleranced dimensions … •
Add limit dimensions : 392.43 + 384.81=777.24
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Find Mean dimension: 777.24/2=388.62 Find total tolerance by subtracting limit dimensions: 392.43 – 384.81=7.62 Find equal bilateral tolerance=7.62/2=3.81
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Finally express limit dimensions as equal bilaterally toleranced dimension as: 388.62` 3.81 i2
Box Assem bly: Part #2- #11
parts MMC … (1)
parts LMC … (2)
Add (1) and (2) …(3)
Subtract (1) and (2)…(4) Half of (3) …(5) Half of (4) …(6) (5) (6)
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Box Ass embly: Par t #1 & Part #2- #11 put together
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Box Assembly : Loop Analysis Diagram Loop Diagram begins by showing Gap to be calculated at the top l
So, loop diagram begins at Part#11 (plate) and progresses downwardconstantly through material until it reaches at the last plate at the bottom of an assembly (ie. Part#2 or plate#2) l
The sum of all these– ve mean dimensions, which run from top to bottom is381and has total tolerance of` 3.81 l
The loop then reverses and progresses up through cavity (ie. Part #1). l
This portion of the loop is +ve since it progresses frombottom to top l
The logic of +ve and –ve is simple… material removes airspace (therefore –ve) and cavity which lacks material adds to airspace (therefore +ve) l
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Box Assembly : Loop Analysis Diagram The mean dimension of cavity is 388.62 and has a total tolerance of` 3.81 l
So, in numbers chart, we add means: (+)388.62 + ( )381.00 = 7.62…(1) l
If this number is –ve, it would have proven that even mean sizes of parts, when produced result in l
interference. Now that sum is +ve, we can proceed…
Next step is to add, charted plus or minus tolerances : 3.81+3.8 1 = 7.62 …(2)
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Next step is to calculate min and max gaps (airspace or interfer ence):
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Mean dimensions difference + sum of tolerances = (1) + (2)= (+)7 .62+(+)7.62=+15.24 (max gap)
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Mean dimensions difference - sum of tolerances = (1) - (2)= (+)7.62-(+)7.62=0 (min gap)
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Box Assembly : Alternate Method to calculate min / max gap From situations suc h as this, it is easier to simply calculate th e MMC of the cavity and the collective MMCs of the plates and subtract them to get minimum gap.
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Box Assembly : Loop Analysis Diagram
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Session #2:
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Exercises
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Session #3: Loop Analysis for Features of Size (FOS) l
Objectives: –
Using Loop Analysis Technique; determine Max and Min gap in Horizontal and Vertical Directions
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Determine proper start and End points for stack-ups Graph the numbers calculated into Loop Diagram
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Analyzing FOS: Problem Description
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Analyzing FOS: Charts to be used
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Analyzing FOS: Steps Involved (Horizontal Direction) l
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Convert horizontal limit dimensions of part #1 to equal bilateral toleranced dimension. (26.615` 0.405) Convert horizontal limit dimensions of part #2 to equal bilateral toleranced dimension. (25.705` 0.105) Graph the loop from left-to-right through material, using appropriate signs (+ve / -ve) (- 25.705 and + 26.615) Add these mean dimensions (don’t forget signs) to get difference between mean dimensions ((-) 25.705 + (+) 26.615)=+0.910 Add plus and minus tolerances for part#1 and part#2 to get total plus and minus tolerance (0.105+0.405)=0.510 Max gap in horizontal direction is = sum of difference between mean dimensions and total of plus and minus tolerances(0.910+0.510=1.42) Min gap in horizontal direction is = difference (subtraction) of difference between mean dimensions and total of plus and minus tolerances(0.9100.510=0.4)
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Analyzing FOS: Steps Involved (Vertical Direction) l
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Convert vertical limit dimensions of part #1 to equalbilateral toleranced dimension. (26.615` 0.405) Convert vertical limit dimensions of part #2 to equalbilateral toleranced dimension. (24.390` 0.610) Graph the loop from top-to-bottom through material, using appropriate signs (+ve / -ve) (- 24.390 and + 26.615) Add these mean dimensions (don’t forget signs) to get difference between mean dimensions ((-) 24.390 + (+) 26.615)=+2.225 Add plus and minus tolerances for part#1 and part#2 to get total plus and minus tolerance (0.610+0.405)=1.015 Max gap in vertical direction is = sum of difference between mean dimensions and total of plus and minus tolerances(2.225+ 1.015)=3.24 Min gap in vertical direction is = difference (subtraction) of difference between mean dimensions and total of plus and minus tolerances(2.2251.015)=1.21
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Analyzing FOS: Easier Method using MMC and LMC Calculating min and max gaps ma y be easier as discussed before ( slide #48); by subtracting the MMCs for minimum gaps and LMC s for maximum gaps as shown below:
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Analyzing FOS: Charts and Loops with dimensions …
Part #1, vertical direction Part #1, horizontal direction
Part #2, horizontal direction
Part #2, vertical direction
Material side (-ve) Part #2 thickness
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Session #3:
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Exercise
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Session #4: Analysis of an assembly with Plus and Minus tolerancing l
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Objectives: l
Calculate the airspaces and interferences for a plus and minus toleranced assembly
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Performing multiple loop analyses on an assembly
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Assembly with plus and minus tolerances : Problem Description
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Assembly with plus and minus tolerances : Charts to be used
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Steps Involved in calculating stack in Horizontal Direction l
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Convert horizontal limit dimension 1 male (31.75 – 34.90) to equal bilateral toleranced dimension. (33.325 ` 1.575) Convert horizontal limit dimension 2 air (21.41 – 19.84) to equal bilateral toleranced dimension. (20.625 ` 0.785) Convert horizontal limit dimension 3 male (16.67 – 15.09) to equal bilateral toleranced dimension. (15.88 ` 0.79) Graph the loop from left-to-right through material, using appropriate signs (+ve / ve) (- 33.325, +20.625 and + 15.880) Add these mean dimensions (don’ t forget signs) to get difference between mean dimensions ((-) 33.325 + (+) 20.625 + (+)15.880)=+3.180 Add plus and minus tolerances for dimensions 1, 2, 3 to get tota l plus and minus tolerance (0.785+0.790+ 1.575 = 3.150) Max gap in horizontal direction is = sum of difference between mean dimensions and total of plus and minus tolerances (3.180+ 3.150=6.330) Min gap in horizontal direction is = difference (subtracti on) of difference between mean dimensions and total of plus and minus tolerances (3.180-3.150=0.030)
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Steps Involved in calculating stack in Vertical Direction
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Convert vertical limit dimension 1 air (26.21-25.40) to equal bilateral toleranced dimension. (25.805 ` 0.405)
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Convert vertical limit dimension 2 material (25.40 – 24.59) to equal bilateral toleranced dimension. (24.995 ` 0.405)
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Graph the loop from top-to-bottom through material, using appropriate signs (+ve /
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-ve) (- 24.995, +25.8 05) Add these mean dimensions (don’ t forget signs) to get difference between mean dimensions ((-) 24.995 + (+) 25.805)=+0.8 10
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Add plus and minus tolerances for dimensions 1, 2, 3 to get tota l plus and minus tolerance (0.405+0.405 = 0.810)
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Max gap in horizontal direction is = sum of difference between mean dimensions and total of plus and minus tolerances (0.810+0.810=1.620)
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Min gap in horizontal direction is = difference (subtracti on) of difference between mean dimensions and total of plus and minus tolerances (0.810-0.810=0.0)
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Easier Method for calculating stacks using MMC and LMC Calculating min and max gaps ma y be easier as discussed before ( slide #56); by subtracting the MMCs for minimum gaps and LMC s for maximum gaps in horizontal direction as shown below:
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Assembly Analysis: Charts and Loops with dimensions …
Vertical Loop
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Horizontal Loop
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Session #4:
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Exercise
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Session #5: Analyzing a Floating Fastener Assembly with Geometric Controls l
Objectives: l
Calculate Virtual and Resultant conditions (Inner / Outer Boundaries) for GD&T callouts
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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
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Graph the numbers in tolerance stack-up diagram
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Determine all unknown gaps in the assembly
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Floating fastener assembly sketch with GD&T
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Steps involved in analyzing floating fastener assembly
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Calculate Virtual and Resultant Condition for each holes (holes #1 thru #4)
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For each hole, calculate difference between resultant condition and virtual condition boundaries. This difference represents total size tolerance for each hole. Take half of the difference which is represents equal bilateral tolerance value.
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For each hole, add resultant condition and virtual condition boundaries; and take mean of the sum. This mean represents the mean diameter of that hole (for analysis purpose)
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Again, for each hole, take mean of values in step #2 and #3. This new mean represents mean radius` mean radial tolerance
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Defining Virtual and Resultant Condition Boundaries As per ASME Y14.5M -1994, l
Virtual Condition is defined as a constant value outer locus (for external FOS specified at MMC or internal FOS specified at LMC ) and a constant value inner locus (for internal FOS specified at MMC and external FOS specified at LMC ).
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Resultant conditions are in opposite direction to virtual conditions and are non-constant in nature. They are the worst case inner locus and worst case outer locus of FOS
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Charting values calculated per steps #1 through #4
Step #1:Calculate Virtual and Resultant Condition for each holes (holes #1 thru #4)
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Charting values calculated per steps #1 through #4 Step #2:For each hole, calculatedifference between resultant condition and virtual condition boundaries. This difference representstotal size tolerance for each hole. Take half of the difference which is represents equal bilateral tolerance value.
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Charting values calculated per steps #1 through #4 Step #3, #4:For each hole, add resultant condition and virtual condition bou ndaries; and take mean of the sum. This mean represents themean diameter of that hole(for analysis purpose) Again, for each hole, take mean of values in step #2 and #3. Thi s new mean represents mean radius` mean radial tolerance
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Step #1: values printed in the chart Step #1:Calculate Virtual and Resultant Condition for each holes (holes #1 thru #4)
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Step #2: Values printed in the Chart Step #2:For each hole, calculatedifference between resultant condition and virtual condition boundaries. This difference representstotal size tolerance for each hole. Take half of the difference which is represents equal bilateral tolerance value.
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Step #3,4: Values printed in the Chart Step #3, #4:For each hole, add resultant condition and virtual condition bou ndaries; and take mean of the sum. This mean represents themean diameter of that hole(for analysis purpose) Again, for each hole, take mean of values in step #2 and #3. Thi s new mean represents mean radius` mean radial tolerance
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Locating / Orienting parts in an Assembly to create MIN gap …
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The loop diagram begins by pushing all the parts is such a manner that the parts configuration (position) in an assembly would create a minimum gap. As shown in figure left this has had the effect of trapping the fasteners (or pins) between hole #1 and #3 and Hole #2 and #4.
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The pin on the left passing thru hole 1,3 is trapped by pushing part having hole 1 to the right and the pin on right passing thru holes 2 and 4 is trapped by pushing part having hole 2 to the left (both these operations closes gap between the parts having holes 1, 2)
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Since this is a floating fastener situation, pin on the left (passing thru holes 1,3) and pin on the right (passing thru holes 2,4) continues to slide unless the pins are held against the left and right side of respective holes as shown in figure
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Construct a Loop Diagram
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The Loop begins at the the face on the left side of the gap, it proceeds towards left (thru material), designated as– ve numbers thru the basic dimension of 125mm to the center of hole #1.
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Go 3mm left (-ve) thru the radius of hole #1 ( as calculated with its VC and RC boundaries). We are done with hole #1 and exhausted part #1
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Now go over the left pin trapped between holes 1, 3 in right direction (+ve) 3mm
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Now we come across hole#3 and we now go from right side of hole#3 towards it center in left direction (-ve) 3mm.
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Step 4 takes us to center of hole 3. From there we go towards left (+ve) to the center of hole 4; 260mm. We are done with hole #3. This step reverses the route going left (-ve) thru 3mm radius of hole #4. We are done with hole #4.
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Here we again reverse the loop and go right direction (+ve) thru the right pin diameter 3mm
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This step begins on last hole #2, the route goes from right side edge of this hole towards center (-ve) 3mm
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Go in the same direction from center of hole #2 to the end of the gap (inner right side face of part) (-ve) 125mm
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Logic Behind Loop Diagram The logic behind this loop route was to proceed from left edge/face of gap through all features having an effect on the Minimum Gap, to the right of gap. To begin, the parts were shoved to create a min gap configuration and in this case this is the only logical route to take. The route went left and right and involved all related features until loop was complete. The related feature list includes four holes #1 thru #4, pins on left and right . Hole radii were used because the pertinent dimensions binding the gap to holes and holes to each other, went to the hole centers.
Key in this table the – ve and +ve route values
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The full pin diameters were used as the pins got trapped between hole edge faces. The basic dimensions were used to allow a route from left side of gap to the center of hole 1 and then center of hole 3 to to the center of hole 4 and at the last from center of hole 2 to the right side face of the gap
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Loop Diagram with values printed …
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Can you Locate / Orient parts in an Assembly to create MAX gap …
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Loop Diagram for MAX gap with values printed …
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Session #5:
Exercise
Calculate MIN and MAX Gap for the assembly shown in figure at left
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Session#6: Analyzing an Assembly with Tab and Slot (Fixed Fastener)
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Session#6: Analyzing an Assembly with Tab and Slot l
Objectives:
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Calculate assembly overall MAX and MIN dimensions Calculate MAX and MIN gaps within assembly as shown
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Calculate boundaries using variousGD&T controls
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Steps involved in analyz ing Tab -Slot assembly
1.
2.
Calculate Virtual and Resultant Condition for Tab and Slot. They work on similar principals as hole and pin and are controlled at MMC. For Slot and Tab, calculatedifference between resultant condition and virtual condition boundaries. This difference represents total size tolerance for Slot or Tab. Take half of the difference which is represents equal bilateral tolerance value.
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3.
For Slot and Tab, add resultant condition and virtual condition boundaries; and take mean of the sum. This mean represents the mean width for either Slot or Tab(for analysis purpose)
4.
Again, for Slot and Tab, take mean of values in step #2 and #3. This new mean represents mean radius` mean radial tolerance i2
Charting values calculated per steps #1 through #4 Step #1: CalculateVirtual and Resultant Condition for Tab and Slot. They work on similar principals as hole and pin and are controlled at MMC
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Charting values calculated per steps #1 through #4
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2.
For Slot and Tab, calculate difference between resultant condition and virtual condition boundaries. This difference represents total size tolerance for Slot or Tab. Take half of the difference which is representsequal bilateral tolerance value.
3.
For Slot and Tab, add resultant condition and virtual condition boundaries; and take mean of the sum. This mean represents the mean width for either Slot or Tab(for analysis purpose)
4.
Again, for Slot and Tab, take mean of values in step #2 and #3. This new mean representsmean radius` mean radial tolerance
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Step #1: values printed in the chart Step #1: CalculateVirtual and Resultant Condition for Tab and Slot. They work on similar principals as hole and pin and are controlled at MMC
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Step #2,3,4: Values printed in the Chart
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2.
For Slot and Tab, calculate difference between resultant condition and virtual condition boundaries. This difference represents total size tolerance for Slot or Tab. Take half of the difference which is representsequal bilateral tolerance value.
3.
For Slot and Tab, add resultant condition and virtual condition boundaries; and take mean of the sum. This mean represents the mean width for either Slot or Tab(for analysis purpose)
4.
Again, for Slot and Tab, take mean of values in step #2 and #3. This new mean representsmean radius` mean radial tolerance
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Locating parts in an Assembly to create MIN overall Dimension and creating a Loop Diagram …
1
Before we begin constructing loop diagram for minimum overall dimension, we must imagine the parts being shoved together such that configuration creates minimum overall assembly dimension
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This means left side of tab is pushed against left side of slot
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Loop Diagram follows this route …
2
3 4
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1.
We know now to start with left side face/edge of part#1 to the center of slot.
2.
Then back to the left side of slot and tab
3.
Then back to the center of tab and then
4.
To the right side face of part#2
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Loop Diagram with values printed …
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Can you Locate parts in an Assembly to create MAX overall dimension and create a Loop Diagram? …
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Loop Diagram for MAX Overall Dimension with values printed …
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Calculating MAX, MIN values for Lower-Left and Upper-Right gaps
Case #1: Min Lower-Left Gap
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Case #2: Max Lower-Left Gap
Case #3: Min Upper-Right Gap
Case #4: Max Upper-Right Gap
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Case #1: Min Lower-Left Gap
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Case #2: Max Lower-Left Gap
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Case #3: Min Upper-Right Gap
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Case #4: Max Upper -Right Gap
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Calculating MAX overall Diameter for a coupling
Assembly
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Detailed Part Drawing with GD&T Controls Notice the controls used and study the drawing
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Factors and Non-Factors in calculating overall Diameter
Will the controls and dimensions circled in color will participate in overall diameter calculations l
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Will the controls and dimensions circled in color will participate in overall diameter calculations OR
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Will dimensions circled in both colors participate or None? l
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Step #1: Calculate Virtual condition and Resultant Condition boundaries for Threaded holes
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Threaded Hole on Crank Shaft (consider a stud here now): §
VC= n 8.0+0.44=n 8.44
§ §
RC=n 8.0-0.44=n 7.56 Sum of RC+VC=n 16; half of this=n 8
§
Difference of VC and RC=n 8.44-n 7.56=0.88; half of this is 0.44
§
So, threaded hole expressed in equal bilateral toleranced dimens ion is: n 8 ` 0.44
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Step #1: Calculate Virtual condition and Resultant Condition boundaries for Clearance holes
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Clearance Hole on Coupling: § § § § §
VC= n 8.66-0.22=n 8.44 RC=n 8.90+0.22=n 9.12 Sum of RC+VC=n 17.56; half of this=n 8.78 Difference of VC and RC=n 9.12-n 8.44=0.68; half of this is 0.34 So, clearance hole expressed in equal bilateral toleranced dimension is: n 8.78` 0.34
Now, Calculate differencebetween biggest clearancehole diamete r and biggest threaded hole diameter =n 9.12- n 8.44= n 0.68= clearance
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Step #2: Calculate Clearances between Datum Feature Diameters ‘D ’ and ‘B ’ l
In this case the perpendicularity tolerance callout on crankshaf t’s center bore and couplings center shoulder are ignored since the maximum clearanc e (and thus “play”) between these two features would occur at when both feature s a re at their LMC sizes and perfectly perpendicular to their datum planes.
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Center Bore on Crankshaft “Datum feature D ” :
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Center Shoulder on Coupling “Datum feature B ” :
§
§
LMC = 50.10 LMC = 49.97
Subtracting these two values, we get clearance of 50.10 – 49.97 = 0.13, which is less than 0.68 clearance calculated on threaded and clearance hole in previous slide. This means in this case, the threaded/clearance holes are not the factors in stack -up and we would consider only offset between datum features B and D due to their respective LMC sizes.
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Step#3: Create a Loop Diagram All Dimensions printed in the Loop are Nominal (bolt holes are ignored now onward)
+115
+25.05
-24.985
+115
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Step#4: Chart the values
Bottom to Top
Top to Bottom
(+ve)
(-ve)
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-
` Tolerance
` .15 (Size tol of ` 0.2/2 and Gtol of`
Remarks From start to center of
25.05
-
0.1/2) radial calculations -
-
24.985
-
From edge of center bore/shoulder to center of coupling
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-
` .15 (Size tol of ` 0.2/2 and Gtol of`
From center of coupling to end
0.1/2) radial calculations 255.05
24.985
255.05 - 24.985 = 230.065
116
…
0.30 230.065 + 0.30 =230.365
crankshaft From center of crankshaft to edge of center bore/shoulder
Totals
Max Dimension
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Session #6:
117
Exercis e #1
Calculate MAX/MIN Overall dimensions, Calculate MIN/MAX Gaps
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Session #6:
Exercis e #2
Calculate MAX overall diameter of assembly
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Session #7: Analyzing a Rail Assembly having Fixed fasteners
Assembly
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Part #1: Detailed Drawing
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Part #2: Detailed Drawing
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Session #7: Analyzing a Rail Assembly having Fixed fasteners l
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Objectives: l
Calculate Boundaries for Threaded features
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Work with multiple Geometric Controls on a single feature Determine effect of Projected Tolerance Zone on Stack -up
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GD&T Controls affecting and non-affecting stack-up
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Calculate Clearance and Interference
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Use product knowledge / experience and Assembly conditions in stack-up analysis
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Observations from Assembly and Part drawings …
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A classic example of fixed fastener assembly: a threaded hole in rail and a clearance hole in block Note that datum features B on both part are given a flatness tolerance since they are mating Note the refinement frame for Datum feature B on rail. Consider objectives: to calculate max/min gap between rail and block in assembled condition. This means we need to calculate pertinent boundaries for features that affect objectives such as boundaries for slot in rail, width of block, screws when mounted in rail, clearance holes in block.
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Steps involved in analyzing Rail assembly
1.
Calculate Inner and Outer Boundaries for Slot and Width of block. (Will there be VCB or just RCBs?)
2.
For Slot and Width of block, calculate difference between Inner and Outer boundaries. This difference represents total size tolerance for Slot or Width of block. Take half of the difference which is represents equal bilateral tolerance value.
124
3.
For Slot and Width of block, add Inner and Outer boundaries; and take mean of the sum. This mean represents the mean width for either Slot or Width of block (for analysis purpose)
4.
Again, for Slot and Width of block, take mean of values in step #2 and #3. This new mean representsmean radius` mean radial tolerance i2
Steps involved in analyzing Rail assembly 5. 6.
7.
8.
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Calculate Inner and Outer Boundaries for Threaded Hole and Clearance hole. (Will there be VCB or just RCBs?) For Threaded Hole and Clearance hole, calculate difference between Inner and Outer boundaries. This difference represents total size tolerance for Threaded Hole and Clearance hole . Take half of the value. difference which is represents equal bilateral tolerance For Threaded Hole and Clearance hole, add Inner and Outer boundaries; and take mean of the sum. This mean represents the mean width for either Threaded Hole and Clearance hole(for analysis purpose) Again, for Threaded Hole and Clearance hole, take mean of values in step #2 and #3. This new mean represents mean radius` mean radial tolerance i2
Step #1: Boundaries calculations for slot and width 1.
Calculate Inner and Outer Boundaries for Slot and Width of block.
Note that slot in the rail has refinement frame. Apositional to lerance is refined by a orientation (perpendicularity) tolerance. So, is positional tole rance a factor in stack -up or an orientation tolerance? Draw tolerance zone shapes / boundaries for each frame and discu … ss
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Step #2,3,4: Calculating Mean Radius / tolerance for slot and width of block 2. Outer Boundary of Slot =
Outer Boundary of Block =
- Inner Boundary of Slot =
- Inner Boundary of Block =
--------------------------------------
--------------------------------------
Difference =
Difference =
½ Difference of Slot =
½ Difference of Block=
Outer Boundary of Slot = +Inner Boundary of Slot =
+Inner Boundary of Block =
--------------------------------------
--------------------------------------
Sum
Sum
For Slot and Width of block, calculate difference between Inner and Outer boundaries. This difference representstotal size tolerance for Slot or Width of block. Take half of the difference which is represents equal bilateral
Outer Boundary of Block =
=
½ of Sum of OB & IB of Slot =
½ of Sum of OB & IB of Block =
½ Sum `
½ Sum `
½ Difference of Slot =
½ of ½ Sum `
½ of ½ Diff of Slot =
3.
=
½ Difference of Block =
½ of ½ Sum `
½ of ½ Diff of Block =
4.
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tolerance value. For Slot and Width of block, add Inner and Outer boundaries; and take mean of the sum. This mean represents themean width for either Slot or Width of block (for analysis purpose) Again, for Slot and Width of block, take mean of values in step #2 and #3. This new mean represents mean radius` mean radial tolerance
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Step #1: Boundaries calculations: Values printed in the chart 1.
Calculate Inner and Outer Boundaries for Slot and Width of block.
Note that we have ignored positional tolerance on the slot in rai l.Only orientation (perpendicularity) is accounted for in the analysis.
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Step #2,3,4: Calculating Mean Radius / tolerance : values printed in the chart Outer Boundary of Slot = 1.510
Outer Boundary of Block = 1.444
- Inner Boundary of Slot = 1.502
- Inner Boundary of Blo ck = 1.436
- -------------------------------------
--------------------------------------
Difference = 0.008
Difference = 0.008
½ Difference of Slot = 0.004
½ Difference of Block = 0.004
Outer Boundary of Slot = 1.510 +Inner Boundary of Slot = 1.502
+Inner Boundary of Block = 1.436
- -------------------------------------
--------------------------------------
Sum
Sum
2.
Outer Boundary of Block = 1.444
= 3.012
½ of Sum of OB & IB of Block = 1.440
½ Sum `
½ Sum `
½ Difference of Slot = 1.506
` 0. 00 4
½ of ½ Sum ` ` 0.002
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½ of ½ Diff of Slot = 0.753
tolerance value. 3.
For Slot and Width of block, add Inner and Outer boundaries; and take mean of the sum. This mean represents themean width for either Slot or Width of block (for analysis purpose)
4.
Again, for Slot and Width of block, take mean of values in step #2 and #3. This new mean represents mean radius` mean radial tolerance
= 2.880
½ of Sum of OB & IB of Slot = 1.506
½ Difference of Block = 1.440
` 0. 004
½ of ½ Sum ` 0.720 ` 0. 002
For Slot and Width of block, calculate difference between Inner and Outer boundaries. This difference representstotal size tolerance for Slot or Width of block. Take half of the difference which is represents equal bilateral
½ of ½ Diff of Block =
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Step #5: Boundaries calculations for Threaded and Clearance Hole 5. Calculate Inner and Outer Boundaries for Threaded Hole and Clearance hole. (Why not VC and RC Boundaries?)
What is projected tolerance, why it is important? Explain …
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Step #6,7,8: Calculating Mean Radius / tolerance for Threaded & Clearance Hole Outer Boundary of Screw =
Outer Boundary of Hole =
- Inner Bound ary of Screw =
- Inner Boundary of Hole =
--------------------------------------
--------------------------------------
Difference =
Difference =
½ Difference of Screw =
½ Difference of Hole =
Outer Boundary of Screw = +Inner Boundary of Screw =
+Inner Boundary of Hole =
--------------------------------------
--------------------------------------
Sum
Sum
6.
For Threaded Hole and Clearance hole, calculate difference between Inner and Outer boundaries. This difference representstotal size tolerance for Threaded Hole and Clearance hole. Take half of the difference which is represents equal bilateral tolerance value.
7.
For Threaded Hole and Clearance hole, add Inner and Outer boundaries; and takemean of the sum. This mean represents the mean width for either Threaded Hole and Clearance hole(for analysis purpose)
8.
Again, for Threaded Hole and Clearance hole, takemean of values in step #2 and #3. This new mean representsmean radius` mean radial tolerance
Outer Boundary of Hole =
=
=
½ of Sum of OB & IB of Screw =
½ of Sum of OB & IB of Hole =
½ Sum `
½ Sum `
½ Diffe rence of Screw =
½ of ½ Sum `
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½ of ½ Diff of Sc rew =
½ Difference of Hole =
½ of ½ Sum `
½ of ½ Diff o f Hole =
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Step #1: Boundaries calculations: Threaded & Clearance Hole: Values printed in the chart 5. Calculate Inner and Outer Boundaries for Threaded Hole and Clearance hole.
Inner Boundary of Screw Mounted in Rail =n 0.2408 (LMC Major Dia) – 0.0140 = n 0.2268 Outer Boundary of Screw Mounted in Rail = n 0.250 + 0.0140 = n 0.264 Outer Boundary of Hole in Block = n 0.286+ 0.015 = n 0.301 Inner Boundary of Hole in Block = n 0.276- 0.005 = n 0.271
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Step #6,7,8: Calculating Mean Radius / tolerance for Threaded & Clearance Hole: values printed in the chart Outer Boundary of Screw = 0.2640
Outer Boundary of Hole = 0.301
- Inner Bound ary of Screw = 0.2268
- Inner Boundary of Hole = 0.271
- -------------------------------------
--------------------------------------
Difference = 0.0372
Difference = 0.030
½ Difference of Screw = 0.0186
½ Difference of Hole = 0.015
Outer Boundary of Screw = 0.2640 +Inner Boundary of Screw = 0.2268
+Inner Boundary of Hole = 0.271
- -------------------------------------
--------------------------------------
Sum
Sum
6.
For Threaded Hole and Clearance hole, calculate difference between Inner and Outer boundaries. This difference representstotal size tolerance for Threaded Hole and Clearance hole. Take half of the difference which is represents equal bilateral tolerance value.
7.
For Threaded Hole and Clearance hole, add Inner and Outer boundaries; and takemean of the sum. This mean represents the mean width for either Threaded Hole and Clearance hole(for analysis purpose)
8.
Again, for Threaded Hole and Clearance hole, takemean of values in step #2 and #3. This new mean representsmean radius` mean radial tolerance
Outer Boundary of Hole = 0.301
= 0.4908
= 0.572
½ of Sum of OB & IB of Screw = 0.2454
½ of Sum of OB & IB of Hole = 0.286
½ Sum `
½ Sum `
½ Diffe rence of Screw = 0.2454
½ Difference of Hole = 0.286
` 0.0186
` 0.015
½ of ½ Sum ` ½ of ½ Diff of Sc rew = 0.1227 ` 0.0093
½ of ½ Sum ` ½ of ½ Diff o f Hole = 0.1430 ` 0. 007 5
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Locating parts to create MIN Gap Configuration
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Steps in Creating a Loop Diagram 1.
The Loop begins on the left edge/face of slot and proceeds in the +ve (right ) direction to the center of threaded hole (or screw).
2.
Then it continues in same(+ve) direction until right side of outer boundary of the
1 3
threaded Notewe thathave we must staytoon the same hole. part until to jump next mating part 3.
Now the loop reverses. Jumping to mating part features in– ve direction (left) to the center of clearance hole.
4.
Finally, from center of clearance hole we continue in same – ve direction to the end of loop ie. Left edge/face of the block.
2 4
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Loop Diagram : Values Printed Step 1: +0.753` 0.002 Slot Step 2: +0.1227` 0.0093 Mounted Screw
1 2
Step 3: -0.1430` 0.0075 Clearance Hole Step 4: -0.720` 0.002 Block
3 4
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Calculate MIN gap
…
MIN GAP -
+
MIN GAP ` Tolerance
-
+
0.0020
0.1227
0.0093
0.1430 0.7200 0.8630
` Tolerance
0.7530
0.0075 0.0020 0.8757
0.0208 (Totals)
0.8757- 0.8630 = 0.0127 = Mean Gap 0.0127 – 0.0208 = -0.0081 (Interference Max)
Since the min gap value is a – ve number (-0.0081), we know that there is interference possible. Discuss on possibility of occurring such interference in practic e. Not the configuration under which the interference occurred. Can such configuration exist? A nd is avoidable?
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Locating parts to create MAX Gap Configuration
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Steps in Creating a Loop Diagram
1.
The Loop begins on the left edge/face of slot and proceeds in the +ve (right ) direction to the center of threaded hole (or screw).
2.
Then it continues in opposite (-ve)
4 3
direction until lefthole. side of outer of the threaded Note thatboundary we must stay on the same part until we have to jump to next mating part 3.
Now the loop reverses. Jumping to mating part features in +ve direction (left) to the center of clearance hole.
4.
Finally, from center of clearance hole we continue in same – ve direction to the end of loop ie. Left edge/face of the block.
2 1
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Loop Diagram : Values Printed Step 1: +0.753` 0.002 Slot
4
Step 2: -0.1227` 0.0093 Mounted Screw
3
Step 3: +0.1430` 0.0075 Clearance Hole Step 4: -0.720` 0.002 Block
2 1
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Calculate MAX gap
…
MIN GAP -
+
MIN GAP ` Tolerance
-
0.1227
0.0020 0.0093
0.1430
0.0075 0.0020
0.8960
0.0208 (Totals)
0.7200 0.8427
` Tolerance
+ 0.7530
0.8960 - 0.8427 = 0.0533 = Mean Gap 0.0533 – 0.0208 = -0.0741 (MAX GAP)
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Conclusions l
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142
…
The traditional methodology we used for MAX and MIN gap calculations may be misleading! It may arrive at a wrong decision if oneis t rying to determine from MIN gap value whether or not parts will actually fit together; if the route chosen assumes the screw WCOB touches WCIB of clearance hole, but it does not actually touch … If WCIB of holes (internal FOS) and WCOB of the shafts (external FOS) of all the mating features are compatible, we can assume that the parts are able to fit. For example, the WCIB of the slot in rail is 1.502 and WCOB of block is 1.444 (smaller); therefore they don’t interfere and there is still a play (clearance) within these boundaries. Similarly, we know that, WCOB of mounted scr ews is 0.264 and WCIB of clearance hole is 0.271; therefore we conclu de that they these two boundaries don’t interfere and there is play (clearance) within these boundaries Therefore, if the parts are allowed to assembly naturally (not pushed to extremes at the assembly stage), they will fit w/o interference.
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Looking at the case from Different Angle
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…
Once the WCB are proven compatible, another check is to determine minimum airspace between WCOB of mounted screw and WCIB of slot and the maximum wall thickness between WCIB of clearance hole and WCOB of block.
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143
If there is more airspace than wall thickness , no interference should occur when assembled naturally.
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Min Airspace V/s Max Wall Thickness The Loop analysis approach assumes the airspace between screw and clearance hole will be used to push the parts in mostundesirable assembly conditions,
However, if we assume that this airspace is used to push the parts into most desirable and practical condition; the analysis proves the parts will assemble
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Steps involved in calculating MIN Air space On Slot Part 1.
MMC of the Screw = + Geo Tolerance for Threaded Holes =
---------------------------------------------------------------------
Watch signs in this column
Sum = Virtual Condition of the Screw = ½ of Virtual Condition of Screw =
2.
MMC of the Slot = - Geo Tol for the Slot =
-----------------------------------------------------Difference = Inner Boundary Of Slot = ½ of Inner Boundary of Slot =
3.
½ of Inner Boundary of Slot =
- ½ of Virtual Condition of Screw = -------------------------------------------------------------Difference = Min Airspace between the screw surface & Slot wall=
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Steps involved in calculating MAX wall thickness on Block Part 1.
MMC of the Clearance Holes = - Geo Tolerance for Holes =
---------------------------------------------------------------------
Watch signs in this column
Difference = Virtual Condition of the Hole (IB) = ½ of Virtual Condition of Hole =
2.
MMC of the Block = + Geo Tol for the Block =
-----------------------------------------------------Sum = Outer Boundary of Block = ½ of Outer Boundary of Block =
3.
½ of Outer Boundary of Block =
- ½ of Virtual Condition of Hole = -------------------------------------------------------------Difference = MAXWall thickness between the H ole surface & Block wall =
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Steps involved in calculating MIN Air space and MAX wall thickness
Minimum Airspace = - Maximum Wall Thickness = ----------------------------------------------------Difference = Clearance between Rail and Block per side =
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Calculated MIN Air space 1.
MMC of the Screw =0.250 + Geo Tolerance for Threaded Holes =0.014
--------------------------------------------------------------------Sum = Virtual Condition of the Screw =0.264 ½ of Virtual Condition of Screw = 0.132
2.
MMC of the Slot =1.504 - Geo Tol for the Slot = 0.002
-----------------------------------------------------Difference = Inner Boundary Of Slot =1.502 ½ of Inner Boundary of Slot =0.751
3.
½ of Inner Boundary of Slot =0.751
- ½ of Virtual Condition of Screw =0.132 -------------------------------------------------------------Difference = Min Airspace between the screw surface & Slot wall =0.619
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Calculated MAX Wall thickness 1.
MMC of the Clearance Holes =0.276 - Geo Tolerance for Holes = 0.005
--------------------------------------------------------------------Difference = Virtual Condition of the Hole (IB) 0 =.271 ½ of Virtual Condition of Hole = 0.1355
2.
MMC of the Block =1.442 + Geo Tol for the Block =0.002
-----------------------------------------------------Sum = Outer Boundary of Block =1.444 ½ of Outer Boundary of Block =0.722
3.
½ of Outer Boundary of Block =0.7220
- ½ of Virtual Condition of Hole =0.1355 -------------------------------------------------------------Difference = MAXWall thickness between theHole surface & Block wall = 0.5865
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Now, calculate min clearance between rail and block per side
Minimum Airspace =0.6190 - Maximum Wall Thickness = 0.5865 ----------------------------------------------------Difference = Clearance between Rail and Block per side 0.0325 = This value also represents Minimum clearance with parts adjusted optimally for Assembly
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MAX Gap calculation: Discussion on ‘Perpendicularity ’ as factor l
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151
MAX gap can be viewed in two ways: –
First considers perpendicularity tolerance on both rail slot and block
–
Second, does not consider perpendicularity.
Second only approach that more uniform maximum gapLMC is created whenassumes both features (slot and block) are at their and are perfectly perpendicular to bottom faces. It seems logical that both are worth knowing.
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MAX Gap calculation: With & without ‘Perpendicularity ’ as factor
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MAX Gap calculation : without ‘Perpendicularity ’ as factor
-
+ 0.7540
0.1134
Mounted Screw 0.1505
0.7190
0.8324
Part/Feature Slot
Clearance Hole Block
0.9045
Totals
(0.9045 - 0.8324) = Min Ga p = 0.0721
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Session #7:
154
Exercises
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Session #8: Tolerance Stack-up Analysis for Single Part
Please read the part drawing and write down your findings
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Session #8: Tolerance Stack-up Analysis for Single Part l
Objectives: –
Perform variety of single-part tolerance stack-up calculations using… l
Two single stacked composite position controls
l
Datum features referenced at MMC Profile of surface
l
156
–
Analyze envelopes of perfect form and geometric tolerances of perfect orientation at MMC
–
Calculate minimum and maximum axialseparations
–
Understanding ‘SEP REQ’ notes effect on gaging requirements and effect of multiple DRFs in accumulating tolerances
–
Determine MAX and MIN wall thickness usingdifferent approaches
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What are we trying to find out in single part …? We are going to calculate minimum wall thickness between surface of small hole in bottommost row and bottom edge of part as shown in figure left. l
What factors control this wall thickness?
l
From where the bottom most hole controlled from? Is it edge of part? It is controlled from Datum feature B (a four way locator) l
Note that datum feature B also has positional tolerance of n 0.050 at MMC. This tolerance grow up to n 0.060 if datum feature B is produced atLMC l
Min wall
This means datum feature B axis can lie anywherein n 0.060 max when produced at LMC, lets assume it lies on circumferenceof n 0.060 position tolerance circle towards bottom edge of part (worst case). l
Note that wherever datum feature B hole ends up; small hole is measured from its (datum B ’s) axis. l
Small hole has no positional tolerance if produced at MMC, but i ts positional tolerance can grow up ton 0.010 if hole is produced at LMC. We will assume that it is worst case. l
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What are we trying to find out in single part …? The FCF for bottommost hole references B and C datums at MMC l
Though FCF references C; the datum features C ’s role is to prevent part rotation and since bottommost hole is positioned from datum feature B, C is not a factor in calculation.Datum C is two way control l
Since datum feature B is referenced at MMC, gage pin size for datum B will be of size = virtual condition for B = n 0.245. If datum B is produced at n 0.245, it will hug gage pin. But if datum B is produced at larger size (within size limits), the par t may shift aside until Datum B feature surface hits the gage pin. l
Min wall
Remember gage pin axis is datum feature B and not axis of AMS/AME. l
If datum feature B is produced at n 0.255 (LMC), the part can shift radially by 0.005 or diametrically 0.010 l
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What are we trying to find out in single part …? This means the bottommost hole can further shift towards bottom most edge by 0.005; thus thinning wall further! l
Such shift due to clearance between gage pin and datum feature of size (FOS) is called as Datum ShiftOR DRF displacement. l
Therefore, the bottommost hole shifts downward towards part
l
edge by an amount equal to = datum shift/2+positional tolerance of datum feature B at LMC/2= 0.010/2 + 0.060/2 = 0.035 Min wall
Find the resultant condition (outer boundary) size of bottom most hole = (0.130+.01) = 0.140. Mean of this = 0.070. l
Now, add 0.070+0.035 = 0.105 . This value indicateshow far the bottommost hole’s surface can go from its true position in worst case . Remember this number. Double of this = 0.210 which is outer boundary of hole in worst case. l
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Begin the loop now
…
Start at the bottom surface of bottommost smallest hole, go up ( away from start point) by 0.105 to the center of the hole (-0.105) l
From center of this hole go up by 1.375 (basic) to the center of datum feature B (-1.375)
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From center of datum feature B, go further up to the part edge (basic 0.375), ( -0.375)
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From top upper edge of part, loop reverses in downward direction (+ve) and goes until inner boundary of width datum feature G. Inner boundary for datum feature G is calculated as 1.950 (LMC Width)-0.100 (Geo tol at LMC) = +1.850 l
So,
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We used fundamental principals to calculate min wall thickness which accounted for datum shifts and bonus tolerances
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Same example, solved using traditional approach of chart and loop
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We calculated worst case outer boundary of bottommost hole as n 0.210.
l
Now, calculate worstin case inner boundary the same manner.
Expressed as equal bilateral toleranced dimension as n 0.130` 0.080 or in terms of radius as R0.065` 0.040
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Now construct Loop Diagram and print values in chart …
Using traditional loop and chart method, you can find both: minimum and maximum wall thickness!
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Using the same traditional approach to calculate right-side wall thickness
We know by now that for all 28 hole pattern, the mean radius ` mean tolerance is R0.065` 0.040 and this number will be used everywhere to calculate wall thickness or separation between holes Can you calculate min and max distance between center of hole and right side edge of part?
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Using the same traditional approach, calculate max/min wall thickness between datum features B,C
Steps:
164
1.
Calculate inner and outer WCB for datum features B and C (consider only lower segment in composite position callout. Why not upper?)
2.
How to calculate inner and outer boundaries for Basic Dimension 2.000 between datum features B and C??
3.
Create a loop and make a chart
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Using the same traditional approach, calculate max/min wall thickness between datum features B,C
Now, calculate min and max axial separation …
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Using the same traditional approach, calculate max/min wall thickness between datum features B and E
Steps:
166
1.
Calculate inner and outer WCB for datum features B and C (consider only upper segment in composite position callout. Why not lower?)
2.
How to calculate inner and outer boundaries for Basic Dimension 0.375 between datum features B and E??
3.
Create a loop and make a chart
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Using the same traditional approach, calculate max/min wall thickness between datum features B and E
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Using the same traditional approach, calculate max/min wall thickness between datum features B and E
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Same Example, but now with different GD&T Scheme
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Calculating MIN and MAX axial separation between datum features B and C These two holes are controlled toeach otherand datum plane A l
Their relationship with datum A is simply orientation, ie. The hole axis is perpendicular to datum A l
The analysis can be thought in terms of tolerance zonesor virtual conditions. l
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Calculating MIN and MAX axial separation between datum features B and C using tolerance zones A cylindrical tolerance zone exists at each end of basic dimension 2.000 l
The tolerance zone diameter is zero at holes’ MMC (0.245) l
The tolerance zone diameter can grow up to 0.01 at holes LMC (0.255) l
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Calculating MIN and MAX axial separation between datum features B and C using tolerance zones
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Calculating MIN and MAX axial separation between datum features B and C using Virtual Condition Boundaries
The Virtual Condition for each hole is 0.245 l
Since functional gage would be made at virtual condition size, we would consider gage pins of 0.245 size and these gage pins (virtual boundaries) would be 2.000 apart from center to center l
If the holes are produced at 0.245, they will hug the gage pins as shown. l
However, when holes size grow , they could be out of proportion i n any radial direction by an amount equa l to half of amount of departure from MMC (half of (AMS-MMC size)) l
If the holes are produced at LMC size (0.255), it could result i n following configurations … l
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Calculating MIN and MAX axial separation between datum features B and C using Virtual Condition Boundaries
OR
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Calculating MIN and MAX edge-axis separation between datum feature B and left edge of part
MIN-MAX
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Calculating MIN and MAX edge-axis separation between datum feature B and left edge of part The profile control relates the outside surface of part to two holes (datum feature B and C) l
Consider upper left corner of part and datum feature hole B and also profile callout l
We can think of profile (edge of part al around)
l
can grow or shrink by +/ -0.025 per surface. This could be thought of affecting two basic dimensions by ` 0.025 So, just considering profile tolerance of 0.050, we could see that the distance between the actually produced edge of part (profile) and axis of the hole is 0.500 ` 0.25 (0.475 - 0.525) and 0.375 ` 0.25 (0.350 – 0.400) l
Similarly, overall dimensions of 2.000 (1.625+0.375) and 3.125 a re affected by ` 0.025 per surface. Therefore overall dimensions of the part are 2.000 ` 0.050 x 3.125 ` 0.050 l
The datum features referenced on profile callout has no effect o n size of the part, but has effect on angle and location of part surfaces l
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Calculating MIN and MAX edge-axis separation between datum feature B and left edge of part But, now that the datum feature B is referenced at MMC, does have an effect on 0.500 and 0.375 dimensions beyond 0.050 profile tolerance due to pattern shift ( or datum shift effect). l
Following are the illustrations of extreme configurations:
l
OR
MAX DIST=0.025+0.500+0.005=0.530
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MIN DIST=0.500 -0 .025-0.005=0.470
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Calculating MIN and MAX wall thickness between datum feature B and left edge of part
MAX WALL THK=0.530-0.1275=0.4025
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MIN WALL THK=0.470-0.1275=0.3725
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Calculating MIN and MAX wall thickness between datum feature B and left edge of part using Loop Diagram
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Calculating MIN and MAX wall thickness between bottommost small hole and right edge of part Calculations for bottommost small hole The Loop analysis can be done with these numbers, but we must consider that the profile tolerance on right edge is` 0.025 with an additional shift of 0.005 ; which is a separate requirement from the 28 holes on another ` 0.005 (since the edges could shift one way and 28 hole pattern shifts opposite way)
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Calculating MIN and MAX wall thickness between bottommost small hole and right edge of part : Loop Diagram
OR
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Calculating MIN wall thickness between bottommost small hole and right edge of part : Alternate method
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Calculating MAX wall thickness between bottommost small hole and right edge of part : Alternate method
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Session #8:
184
Exercises
1.
Calculate MIN and MAX wall thickness and axis separation from left edge of the part the n 0.125 ` 0.005 hole closest to it.
2.
Calculate MIN and MAX wall thickness from the bottom edge of the part to then 0.125` 0.005 hole closest to it.
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Session #8:
Exercises
1.
Calculate MIN and MAX wall thickness between B and C holes when they are produced at LMC
2.
Calculate MIN and MAX wall thickness from the B hole to the top edge of the part
3.
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Calculate MIN and MAX wall thickness from the lower rightn 04.0-4.3 hole to the right edge and bottom edge of the part
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Session #8:
Exercises
1.
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Calculate MIN and MAX wall thickness between surface of one hole n 0.570-0.590 and the Outside Diameter
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Session #8:
Exercises
1.
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Calculate MIN wall thickness between hole pattern to Datum feature C
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Session #9: Tolerance Stack-up Analysis for a Five part assembly
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Session #9: Tolerance Stack-up Analysis for a Five part assembly
Objectives: l
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
l
Learn Simplifying a complex situation Calculate radial clearance and interference
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Part #1: Detailed Drawing
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Part #2: Detailed Drawing
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Part #3: Detailed Drawing
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Part #4: Detailed Drawing
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Part #5: Detailed Drawing
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Step#1: Check if housing inner width is sufficient to house part #2,3,4
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1.
Shortest pertinent length on part#1 is 250-0.5 = 249.5
2.
Seating length on part#5 forpart#1 = 324.5-51.0 = 273.5, therefore provides sufficient stability for part #1 when fastened with part#5.
3.
’s inner width of 249.5 So, the only important factor is housing
4.
The parallelism tolerance of 0.2 on datum E of part#1 is a factor
5.
Profile tolerance of 0.2 on datum A of part#1 relative to DatumD is factor
6.
since it controls attitude of datum A surfaces relative to datum D Attitude variations from #4,#5 tends walls to lean 0.2 each, but the size tolerance (249.5-250.5) on overall inner width cannot be violated
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Step#1: Check if housing inner width is sufficient to house part #2,3,4 Part #2 has MMC size = 12.8
l
Part #3 has MMC size = 100.0. Parallelism tolerance of 0.05 on datum plane B gets accommodated in MMC size limit automatically l
Part #4 : Start from datum plane A, travel towards right a basic dimension of 48.75 up to center plane of last width feature of 15` 0.1, and add ½ of is outer boundary = (15.1+0.0 (geo tol at MMC)) = 15.1; take ½ of it = 7.55 l
Now add all above: 12.8+100.0+48.75+7.55=169.1. This is the minimum space needed to house parts 2,3,4 l
Coming back to inner width of housing; which is 249.5 0.2 (attitude tolerance due to leaning of walls) = 249.3 l
So, the clearance = 249.3 – 169.1 = 80.2, so we have plenty of clearance to house all three parts.
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Step#2: Check if housing inner height is sufficient to house part #2,3,4: Radial gap study Parts 2,3,4 are pushed upwardto create a minimum gap configuration of assembly l
Note “one line” places in assembly where mating parts respective features coincide. l
l
#4: Outer boundary = (nPart 251.0mmc+ n 0.2 geo tol) = n 251.2, ½ of it = R125.5 Clearance hole: n 8.8LMC = R4.4
l
Part #3: Threaded hole treated as mounted screw: inner boundary of screw=n 7.76 (LMC of screw)-n 0.30 (geo tol)=n 7.46; n 7.46-0.05 (pattern shift due to Dm reference on threaded hole)=n 7.41 = inner boundary of screw with pattern shift; ½ of it = R3.705 l
LMC of D = n 99.95; ½ of it = R49.975 l
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Step#2: Check if housing inner height is sufficient to house part #2,3,4: Radial gap study Part #2: n 100.05 (LMC of D)+ n 0.10 (positional tolerance on D at LMC) = n 100.15. Take ½ of it = R50.075 l
ln
115 (LMC of E), take 1/5 of it = R57.5
Part #1: Center hole. n 115.52 (LMC of hole) + n 0.20 (Geo tol at LMC) = n 115.72 (outer l
boundary of hole). Take½ of it = R57.86
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Create Loop and Print values in the chart.
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Challenges in this example were: Factors and Non -factors in Gap analysis. Factors / Non-factors in Part#1:
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–
Is Datum C a factor? Yes, the inner boundary of datum C needs to be accounted since this will decide amount of gap. R137.25
–
The outer boundary of center clearance hole is also a factor, since movement of this hole affects the other parts related to it. The outer boundary of this hole is n 115.72 or R 57.86. The outer boundary allows the equal movement of shaft (Part#2)
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Challenges in this example were: Factors and Non -factors in Gap analysis. Factors / Non-factors in Part#2: –
Is Datum E a factor? Yes, The most airspace between datum feature E and clearance hole on part#1 would occur when datum E is at LMC and perfect perpendicular to C. So, perpendicularity tolerance is not a factor in this analysis. LMC of datum feature E isn 115 or R57.5
–
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Since datum feature D is alignment feature between part #2 and #3, the threaded holes are not factor in this analysis. Datum D has two geometric tolerances , but since location of datum feature D will determine location of part#3, the position callout is a factor and perpendicularity is not. Since the l argest movement and largest size of D allows largest movement of part#3, the outer boundary of D is n 100.05+n 0.1 = n 100.15 or R50.075
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Challenges in this example were: Factors and Non -factors in Gap analysis. Factors / Non-factors in Part#3:
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–
Is Datum D a factor? Yes, The most airspace between datum feature D and bore on part#2 would occur when datum D is at LMC and perfect perpendicular to C. So, perpendicularity tolerance is not a factor in this analysis. LMC of datum feature D isn 99.95 or R49.975
–
Since datum feature D is alignment feature between part #2 and #3, the clearance holes are not factor in this analysis. The threaded holes move the screws around and create alignment between part#3 and part#4, therefore pattern of four holes is a factor. If LMC screws are moved by the threaded holes as a group off of the datum axis D, it would eventually move part#4. Also since pattern of threaded holes reference datum feature D at MMC, the pattern may shift additional amount if D is produced at LMC. So the inner boundary of screws mounted in threaded holes and shifting as a group would be n 7.76(LMC Screw)-n 0.3 (geo tol)-0.05(pattern shift)=n 7.41 or R3.705
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Challenges in this example were: Factors and Non -factors in Gap analysis. Factors / Non-factors in Part#4: –
The clearance holes connect part#4 to threaded holes in part#3, but since these clearance holes are positioned only to each other and held perpendicular to datum A, and other pertinent features are related to axis of these four holes, the position tolerance on these four holes is not a factor. The movement between part #3 and 4 is governed by clearance between screws and clearance holes. The LMC of clearance holes is n 8.8; R4.4
–
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The OD of part#4 is also a major factor. The movement of this OD off the axis of four hole pattern (datum B), effectively increases the overall size, therefore runout tolerance is a factor. Thus the outer boundary of part#4 is n 251+0.2=n 251.2 or R125.6
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Challenges in this example were: Factors and Non -factors in Gap analysis. l
These are all factors in this analysis. We started w ith with factors on Part#4 until we exhausted those and then move d on to part#3 and so on…
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This way, many parts can be analyzed in a complex assembly without ’s why getting lost. Remember, its done ONE PART AT A TIME. That doing single part analysis helps engineer to do assembly analysis.
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The first and the foremost, one must decide o bjectives ( what gaps or overall dimensions or material thickness are to be calculated).Secondly assembly should be investigated to determine which parts, which part features, which sizes and which geometric tolerances are and are not factors in analysis.
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Session #9:
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Exercise
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Trigonometry and Proportions in Tolerance Stack-up Analysis
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Trigonometry and Proportions in Tolerance Stack-up Analysis l
Objectives: l
Understanding the role of trigonometry and proportions in tolerance stack-up and geometric tolerancing
l
Understanding the effect of “Rocking Datums” Know how skewed vertical stacks affect horizontal housing requirements. Mixing trigonometry and algebra determining stack-up results
l
l
l
207
Consider the rules in Y14.5.1 (Math Standard) for constructing a valid Datum
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Trigonometry and Proportions in Tolerance Stack-up Analysis l
The examples we worked up till included mainly addition and subtraction; however many situation calls for trigonometric aspe cts while calculating stack-ups.
l
Consider flatness, which controls rocking of datum features can affect stack-up analysis
208
l
Y14.5M states that the datum feature is to be rocked to an optimum assembly condition, in other words, if it rocks, rock it until the part checks good to simulate the manner in which assembler would rock the part until it is assembled.
l
So, its illogical to rock the part until it interfered in the as sembly, same as it would be illogical to rock it until the part checked bad in inspection. i2
Trigonometry and Proportions in Tolerance Stack-up Analysis
209
l
If the rock point is at the center of part, it is difficult to d etermine that one is to rock the part one way or other – which way the assembler will choose?
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To summarize, if rocking is the option chosen, (over say, shimming it up to equalize the work), that even if there were only two ways to rock the part, there is only 50% chance that the assembler and the inspector will choose the same way.
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l
In normal GD&T approaches, it is assumed that these chance will workout optimally. However, in tolerance stack-up analysis, the approach is exactly opposite. If datum feature part would is to beallow rocked interferes! So, has howrock, muchthe a rock partuntil to it lean (imperfect orientation) must be calculated to determine the amount it would contribute to the possibility of say, interference.
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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 l
Y14.5.1 states that in order to be a valid primary datum feature, the points used to l
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 l
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Example of Rocking Datum and proportions …
Normally this is ignored while calculating worst mating conditions of features like 6.000 length pin. We normally calculate worst mating condition diameter = MMC size + geo tol at MMC = 1.010+0.005 = 1.015. l
But with additional radial lean of 0.006, the
l
worst mating condition can be seen as 1.015 + 2x0.006=1.027 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, therefore½ of 1.027 = R0.5135 l
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Example of Rocking Datum and proportions …
Parallelism is also a factor that can be related to the problems that flatness creates. Parallelism when used on planer surfaces, controls flatness and angle to datums referenced. l
In the illustration on left, produced part has
l
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) l
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Example of Rocking Datum and proportions …
Initially, the three parts were aligned with center, left edge and right edge aligned, then the parts are either to left or right l
This would assume that interior part features such as holes (not shown here) have been positioned from one of these l
This much space would be needed if parts were stacked this way and allowed to rock this way
214
features as secondary datum feature. lEach 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. l
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Example of Rocking Datum and proportions …
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 eitherdirection. l
This much space would be needed if parts were stacked this way and allowed to rock in either direction
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Calculating overall housing dimension requirements To calculate overall housing space requirements, visualize parts stacked as shown. l
Part #1
Part #2
We need to calculate offset of bottom left most point on part #3 from the bottom left most point on part #1 l
l
Since part may rock on either side, the offset calculated must be doubled then added to length (400mm) of part#1.
Part #3
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Calculating overall housing dimension requirements
217
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Calculating overall housing dimension requirements
218
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Calculating overall housing dimension requirements
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Calculating overall housing dimension requirements
220
l
Therefore offset = a2+a6 = 0.04995+0.100=0.14995 (say 0.150)
l
Hence, housing size must increase by 0.150 on each side = 400 + (2 * 0.150) = 400.300
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Simplified Summary of previous Example
221
…
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Conclusions from the Exercise
222
…
l
Purpose of exercise was to show how complex the calculations can be when one assumes certain flatness, parallelism and perpendicularity problems may occur.
l
You may want to use computer software to simulate such situations, but you must study the software to make certain that they are sufficient to simulate wide range of possibilities that may occur and concerns you most
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Removing “out-of-flatness ” from our example … Assume we have three parts assembly and all parts are cylindrical. l
In a situation, where we had a shaft that passes / fits through the center hole of all these three parts, the parallelism of top and bottom of part sandwiched between other two would be a factor. l
The virtual condition of hole in part#3 would have to be increased beyond 0.240 virtual condition boundary shared by part#1, part#2 because of out-ofparallelism between top and bottom of part#2. l
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Removing “out-of-flatness ” from our example … This amount in increase in virtual condition diameter is directly proportional to parallelism tolerance and can be expressed as: l
Ptol/dia of part#2 = increase in VC for part#3/thickness of part#3 l
So, in this case, 0.002/7=x/1; ie. 0.0002857 increase in VC of part#3 ’s hole. l
So, the “? ” size in part#3’s position callout is = 0.2400+0.0003=0.2403 l
As is sometimes the case, the increase is so small, it may be that once calculated, may be ignored. l
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Session #10:
Exercise
Calculate the increase and the VC for the hole in part#3, given following changes to figure in previous slide:
225
–
The parallelism tolerance on part#2 to datum A = 0.020
–
The diameter of part#2 = 36.000
–
The thickness of part#3 = 16.000
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Session#11 : The Theory of Statistical Probability
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The Theory of Statistical Probability l
227
Objectives: l
Convert arithmetically calculated tolerances to statistically calculated tolerances.
l l
Use Root Sums Square (RSS) formula Comparing “Worst-case ” and “Statistical” tolerances
l
Reintegrating statistical tolerances into the assembly
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Background l
So far, in our examples, weassumed that all parts and tolerance s that participate in stack-up analysis are produced at their worst case tolerances and also assembled in worst case configurations!
l
The probability of producing features at their worst case assemb ly conditions is unlikely unless manufactures are targeting them and in most
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228
…
cases they are not. Although manufacturing practices differ place to place, even if they are aiming at smallest hole and largest shaft, they would unlikely be trying to use up all of the tolerances that affect the assembly.
l
We have seen, there are four things related to part geometry to come together to create worst caseassembly conditions. Theyare Size , Shape (form), angle (orientation) and position (location)
l
For example, in mating features that have size dimensions and also have position tolerances, all four would affect their worst case.(see next slide)
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Background
l
l
229
…
Under ASME Y14.5M-1994, size tolerance also controls feature’s form within size limits on all rigidparts. Position controls both: orientation (within position tolerance) and location.
A part feature size form,angle which i s also positioned hashaving to span alltolerance, of its size consequently tolerance, shape, a nd location to be made at the worst case assembly conditions.
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Background l
l
l
230
…
Manufacturers that use Statistical Process Control, are generall y believed to be in statistical control instead of Statistical cha os that it would take to produce features at their worst! Such manufacturers will produce parts which when measured; will found to follow a natural variation which forms a natural bell c urve distribution of part dimensions. Such distributions will depict that large % of produced parts measure close to average (nominal) dimension. The magnitude and spread of dimensions will vary from the nominal by an amount that can be represented in a graph known as Gaussian Frequency Curve; where in the area under the curve represents 100% of the parts produced. The height of the curves represents the times dimensions have been produced for variable individual component i2
Background
…
The dispersion of dimensions under the curve is described as “standard deviation ” and often represented by letter s (sigma), and calculated as:
l
s The arithmetic mean +or- one standard deviation (` 1 s ) is often described as containing 68.26% of the produced parts under this normal curve. By the same logic ` 2 s is 95.46% of the total production and ` 3 s is l
99.73%
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Root Sum Squares (RSS) Method l
So, the statistical probability can be applied to tolerancestac k-up analysis for assemblies both with and without geometric tolerances.
l
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:
l
Statistical probability has been practiced for many years and we ll documented. Statistical approaches are more reliable for volume production. For small production runs, the frequency curve tends to be skewed from its normal shape.
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Applying RSS: Steps Involved and Examples l
Steps: Once the worst case calculations (lets call it as “ 100% tolerance” stack-up analysis) are done; 1.
Using the RSS formula, calculate assembly tolerance (lets call it as statistical probability tolerance)
2.
Determine the percentage (%) ratio between statistical probability tolerance and 100% assembly tolerance Determine the increased statistical probability tolerances to be re-distributed to the assembly’s individual components.
3.
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RSS Calculations: Example#1
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RSS Calculations: Example#1
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RSS Calculations: Example#1 So, 259.08 – 254.00 = +5.08 = Mean Gap
l
Now, take square root of 7.096760 = √ 7.096760 =2.6639744 = Statistical (RSS) tolerance. l
Now, determine % ratio between statistical tolerance and worst case tolerance: l
l
2.6639744 (tolerance likely to be consumed) / 5.08 (worst case tolerance) = 0.5244044, so 2.6639744 is approx. 52% of 5.08 1 / 0.5244044 = 1.9069252 and 1.9069252 x 2.6639744 = 5.08 (srcinal worst case tolerance) l
Thus the worst case calculations allow a gap of 5.08` 5.08; ie. Max gap of 10.16 and min gap of 0. l
While the statistically calculated assembly tolerance allows a gap of 5.08` 2.66 and this is amount of tolerance likely to be consumed in a volume production scenario. l
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RSS Calculations: Example#1 So, if we want to consume` 5.08 tolerance, the piece part tolerance should be increased to 191% of its initial value. Thus the` .254 (the tolerance srcinally given to part#11 thru part#2) becomes 1.91x` 0.254 = ` 0.485. Similarly tolerance on part#1 becomes ` 4. 851 l
This is the answer to the problem of what the statistical tolerances for each part in the assembly would be if calculated by RSS method l
As per the statistical methods, the tolerance srcinally assigned would not be fully consumed, the worst case gap calculations given, 5.08 ` 5.08, becomes statistical probability within` 3 s of consuming only 5.08` 2.66. This results in max gap of 7.74 and min gap of 2.42 l
With the newly assigned statistically calculated tolerances, with each piece part given a tolerance of` 0.485 for parts part#11 thru part#2, for a total of` 4. 85 and part#1 with a tolerance of ` 4. 85 , we have probability of` 9.7. l
So, we would have a gap of 5.08 ` 9.7 for a max gap of 14.78 and min gap of –4.62. In other words, max interference of 4.62, but unlikely. l
Most likely is that out of ` 9.7 , only ` 5.08 tolerance will be consumed l
0.485² = 0.235225 … (part#2 thru part#11 tolerance) 0.235225 x 10 = 2.35225 4.85² = 23.5225 … (part#1’s tolerance) 2.35225 + 23.5225 = 25.87475 Square root of 25.87475 =√ 25.87475 = 5.08
So, by the same RSS method, we arrived` at 9.7 tolerance, we were able to calculate that the likely consumed amount of tolerance by the assembly will be ` 5.08, only so likely max gap is still 10.16 and likely min gap is still zero e ven though the individual component tolerance has been increased by 191%
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RSS Calculations: Example#2
Bottom Left MIN Gap Calculations: We converted all dimensions to equal bilateral toleranced dimensions. l
All basic dimensions had zero tolerances. l
We had slot and tab having both size and geometric tolerance of position which we converted to +/- toleranced dimension. l
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RSS Calculations: Example#2
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So, 158.385 – 154.595 = +3.790 = Mean Gap
Now, take square root of 0.50605 =√ 0.50605 =0.7113719 = Statistical (RSS) tolerance. l
Now, determine % ratio between statistical tolerance and worst case tolerance: l
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0.7113719(tolerance likely to= be consumed) 0.910 (worst case tolerance) 0.7817273, so / 0.7817273 is approx. 78% of 0.910 1 / 0.7817273 = 1.2792184 and 1.2792184 x 0.7113719 = 0.910 (srcinal worst case tolerance) l
Thus the worst case calculations allow a min gap of 3.790- 0.910 = 2.88. l
While the statistically calculated assembly tolerance allows a min gap of 3.790 - 0.71=3.08 and this is amount of gap likely to occur in a volume production scenario. l
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RSS Calculations: Example#2
So, if we want to consume` 0.910 tolerance, the piece part tolerance should be increased to 128% of its initial value. Thus the` .0.055 (the tolerance srcinally given to slot and tab) becomes 1.28x` 0.055 = ` 0.0704. Similarly tolerance on wall becomes 1.28x` 0.100= ` .128. And the tolerance on overall dimension becomes 1.28 x` 0.7= ` 0.896 l
This is the answer to the problem of what the statistical tolerances for each part in the assembly would be if calculated by RSS method l
As per the statistical methods, the tolerance srcinally assigned would not be fully consumed, the worst case MIN gap calculations given,3.790- 0.910 = 2.88, becomes statistical probability within` 3 s of consuming only 3.790- 0.71=3.08 MIN gap l
With the newly assigned statistically calculated tolerances, with each piece part given a tolerance as 0.070 (slot), + 0.070 (tab) + 0.128 (wall) + 0.896 (overall dim) = ` 1.164 l
So, we would have a min gap of 3.790-1.164 = 2.626, but unlikely. l
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Most likely is that out of ` 1.164 , only ` 0 to.910 tolerance will
be consumed 0.070² = 0.0049 … (slot’s new tolerance squared) 0.070² = 0.0049 … (tab’s new tolerance squared) 0.128² = 0.016384 … (wall’s new tolerance squared) 0.896² = 0.806404 … (overall dim’s new tolerance squared) Add all above = 0.832588 Square root of 0.832588 =√ 0.832588 = 0.910
So, by the same RSS method, we arrived` at 1.164 tolerance, we were able to calculate that the likely consumed amount of tolerance by the assembly will be ` 0.910, only so likely min gap is still 2.88 even though the individual component tolerance has been increased by 191%
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Assumptions Under RSS Calculations
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The statistical approach assumes a ‘ zero mean shift’ for all the dimensions being used. It is based upon manufacturing processes that are under st atistical control, not in statistical chaos!
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Those not employi ng statistical process control in manu facturing should not use the RSS tolerancing methodology described now.
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RSS method also assumes that par ts produced for assembl y have be en mixed and components are picked randomly for assembly.
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The logic of RSS model is interesting … it basically allows more tolerance for those manufacturers that need it least : those using SPC controls!
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It calculates that the chances of producing a part that spans its larger statistical tolerance (ST) are so small that if it does ha ppen, the random ly selected mating parts will make up for the potential problem by not spanning their tolerances.
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In fact, it presup poses that the mati ng part will be produc ed so much well than its tolerance extremes as to allow par ts to assemble well !!. If this is false assump tion, unacceptable functional conditions may arise such as interferenc e.
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Assumptions Under RSS Calculations
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The 100% tolerancing scares many engineers and they are uncomfortable when they see that a line fit possibility exist be tween mating features. Such possibility exists when inner boundary (of holes/slots) are same as outer boundary of mating parts such as shafts/tabs.
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If that makes them uncomfortable, then all owing more tolerances using RSS calculations, and consequently a greater possibility of interference should make them even more restless!
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Benderizing Tolerances l
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In 1968, a statistician: A. Bender wrote a paper for SAE entitled “Statistical Tolerancing as it relates to Quality Control and Designer” . In his paper, he suggested a safety factor added to RSS formula. Instead of taking mere square root of sum of squ ares of individual feature tolerances; he suggested a factor of 1.5 be multiplied by the answer of RSS calculations. In other words 1.5 times the square root of the sum of squares of individual feature tolerances
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Why? .. It was so that additional tolerance given to features of part in assembly was not so risky. It is known that, most cases, produci ng features at their worst case condition is unlikely, but it is also known that it happens!
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Some studies have shown that the RSS methodology doesn’t accurately reflect the production scenario,so having ‘cushion’would be wise.
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Reintegrating the Statistical Tolerance into the Assembly l
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Once the statistical tolerance is calculated, it has be integrated back (re-distribute to individual components and features) into assembly – which we have done in previous examples. In case of first example of 12 parts assembly, we had only +/tolerances and therefore reintegration was a simple process – the mean dimension remains the same and each tolerance gets multiplied by 1.91, ie. Tolerance of ` 0.254 got changed to ` 0.485. In second example, the process of reintegrating statistical tolerance is slightly difficult; since the inner and outer bound aries for slot and tab are calculated from collective effects of size and geometric tolerance (position); we need to apportion the statist ical tolerance to both: size and position. i2
Reintegrating the Statistical Tolerance into the Assembly l
Recollect that in second example, we increased worst case tolerances by approx 128%. Therefore the slot’s radial dimensions would be 6.095 ` 0.07 instead of previous 6.095 ` 0.055, similarly tab’s radial dimensions will be 5.985 ` 0.07 instead of previous 5.985 ` 0.055.
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Now, we need to apportion this increased tolerance to size and p osition. To do this, we simply reverse the process …
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Slot: –
` 0.07 x 2 = ` 0.14 = tolerance on slot width
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12.19 ` 0.14 are the slot dimensions.
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– 0.14 = 12.05 Therefore inner boundary of slot = 12.19
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And, outer boundary of slot = 12.19+0.14 = 12.33 Since srcinal TOP on slot was 0.05m , we will increase this also by 128%=0.064 m Similarly, TOP at slot LMC was 0.11 l , which when increased by 128% becomes 0.1408 l .
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6.095 x 2 = 12.19 = slot width
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Now, to get newMMC of slot, we add i(nner boundary of slot + TOPon slot at MMC)= 12.05+ 0.064m =12.114
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On the similar lines, to get LMC size of slot, we subtract (Outer boundary of slot - TOP on slot at LMC) = 12.33 – 0.1408 = 12.189
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Reintegrating the Statistical Tolerance into the Assembly l
Cross check the reintegration …. The inner boundary = 12.114 (MMC) – 0.064 (geo tol at MMC) = 12.05 The outer boundary = 12.189 (LMC) + 0.0.064 (geo tol at LMC) + 0 .075 (bonus tolerance at LMC = 12.189-12.114) = 12.33 So, the reintegration was successful. – –
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The key in this process was to use the % that all tol erances were increased to 128% in this case.
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There are other methods that can be used to reintegrate the tolerances that distributes them differently. Some try to help“difficult to manufacture” features by drawing tolerances from other features in assembly. This allows difficult to manufacture features to get more of the tole rances. But if that was a factor, it probably should have been thought of and handled when tolerances were being arithmetically calculated and before calculation of the StatisticalTolerances began.
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Using Statistical Tolerances on Drawings
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The statistical tolerances thus calculated are identified with symbol. When both the statistical tolerance and the smaller arithmetic tolerance are shown, only those facilities using SPC controls are to be allowed the larger tolerance.
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Methods for calculating Statistical Tolerances l
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Methods for calculating statistical tolerances vary from company to company and in case of complex assemblies, its done with the help of computer software.
One such known Carlo” method to simulating randommethod nature is of“Monte` manufacturing processand withrelates random number stream generation which will in turn simulate dimensions of the part of an assembly. Given a knowledge of manufacturing capability, random numbers are generated to simulate process results. Thereafter careful averaging and multiple simulations with varying samples, one arrives at the likely amount of tolerance t hat will be consumed.
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Summarizing Reintegration of Statistical Tolerances into Assembly …
Final Answer
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Session #11:
Exercises
Calculate Statistical Tolerances for the Wall, Slot, Tab and Overall dimension using RSS methodology
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Session #11:
Exercises
Using the drawing below from Session #5, Calculate Statistical T olerances to be reintegrated into assembly for all features used inminimum gap calculation. Instead of using the standard RSS formula, use the following RSS formula with “safety a 1.5 correction ” factor. TA = 1.5 x√ T12+ T22+ T3 2+ T42+ T52+ T62
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Miscellaneous
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GD&T Reference Chart
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Dimensioning Habits (?)
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Suggested Readings & References … l
ASME Y14.5M-1994 Geometric Dimensioning and Tolerancing
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ASME Y14.5.1M-1994 Mathematical Definition of Dimensioning and Tolerancing Pri ncipals
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Geometrics IIIm - Lowell W. Foster - The best book on GD&T from a technical point of view. Geometric Dimensioning and Tolerancing: Applications and Techniq ues for Use in Design, Manufacturing, and Inspection - James D. Meadows - Not so in depth but more practical
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Tolerance Design: A Handbook for Developing Specifications – Clyde M. Creveling More general approach, very academic but still Optimal a reference on the subject of tolerance analysis CAD/CAM Theory and Practice : Ibrahim Zeid (Dedicates a chapter on Mechanical Tolerancing) A good reference book.(< Rs.500/-) Interpretation of Geometric Dimensioning and Tolerancing : Danie l Puncochar. Good examples and explanation on various Geometric Tolerances. Dimensioning & Tolerancing Handbook : Paul Drake Jr.
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References from ETI Mailbag
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All books are priced in US$ 40 -US$125 range.
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Thank You! Rajendra Deshmukh Principal Consultant
i2 iSquare
(I nterOperabili ty & I nterChange ability So luti ons)
Pune, INDIA Telefax: 020-24250234 Cell: +91-98.900. 36625 Email:
[email protected]
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