Tolerance interpretation
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Tolerance interpretation. Dr. Richard A. Wysk IE550 Fall 2008. Agenda. Introduction to tolerance interpretation Tolerance stacks Interpretation. Tolerance interpretation. Frequently a drawing has more than one datum How do you interpret features in secondary or tertiary drawing planes?

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Tolerance interpretation

Tolerance interpretation

Dr. Richard A. Wysk

IE550

Fall 2008


Agenda

Agenda

  • Introduction to tolerance interpretation

  • Tolerance stacks

  • Interpretation


Tolerance interpretation1

Tolerance interpretation

  • Frequently a drawing has more than one datum

    • How do you interpret features in secondary or tertiary drawing planes?

    • How do you produce these?

    • Can a single set-up be used?


Tolerance stacking

Case #1

1

4

2

3

1.0±.05

1.5±.05

1.0±.05

?

TOLERANCE STACKING

What is the expected dimension and tolerances?

D1-4= D1-2 + D2-3 + D3-4

=1.0 + 1.5 + 1.0

t1-4 = ± (.05+.05+.05) = ± 0.15


Tolerance interpretation

Case #2

1

4

2

3

1.0±.05

1.5±.05

3.5±.05

TOLERANCE STACKING

What is the expected dimension and tolerances?

D3-4= D1-4 - (D1-2 + D2-3 ) = 1.0

t3-4 =  (t1-4 + t1-2 + t2-3 )

t3-4 = ± (.05+.05+.05) = ± 0.15


Tolerance interpretation

Case #3

1

4

2

3

?

1.0’±.05

1.00’±0.05

3.50±0.05

TOLERANCE STACKING

What is the expected dimension and tolerances?

D2-3= D1-4 - (D1-2 + D3-4 ) = 1.5

t2-3 = t1-4 + t1-2 + t3-4

t2-3 = ± (.05+.05+.05) = ± 0.15


Tolerance interpretation

Case #1

1

4

2

3

1.0±.05

1.0±.05

1.0±.05

?

From a Manufacturing Point-of-View

Let’s suppose we have a wooden part and we need to saw.

Let’s further assume that we can achieve  .05 accuracy per cut.

How will the part be produced?


Tolerance interpretation

Let’s try the following (in the same setup)

-cut plane 2

-cut plane 3

3

2

Mfg. Process

Will they be of appropriate quality?


So far we ve used min max planning

So far we’ve used Min/Max Planning

  • We have taken the worse or best case

  • Planning for the worse case can produce some bad results – cost


Expectation

PROCESS

DIMENSIONAL ACCURACY

POSITIONAL ACCURACY

DRILLING

+ 0.008

- 0.001

0.010

REAMING

+ 0.003

(AS PREVIOUS)

SEMI-FINISH BORING

+ 0.005

0.005

FINISH BORING

+ 0.001

0.0005

COUNTER-BORING (SPOT-FACING)

+ 0.005

0.005

END MILLING

+ 0.005

0.007

Expectation

  • What do we expect when we manufacture something?


Size location and orientation are random variables

2.45

2.5

2.55

Size, location and orientation are random variables

  • For symmetric distributions, the most likely size, location, etc. is the mean


What does the process tolerance chart represent

What does the Process tolerance chart represent?

  • Normally capabilities represent + 3 s

  • Is this a good planning metric?


An example

An Example

Let’s suggest that the cutting process produces  (, 2) dimension where (this simplifies things)

=mean value, set by a location

2=process variance

Let’s further assume that we set = D1-2 and that =.05/3 or 3=.05

For plane 2, we would surmise the 3of our parts would be good 99.73% of our dimensions are good.


Tolerance interpretation

.95

1.0

1.05

2.45

2.5

2.55

D1-2

We know that (as specified)

D2-3 = 1.5  .05

If one uses a single set up, then

(as produced)

and

D1-2

D1-3

D2-3 = D1-3 - D1-2


Tolerance interpretation

1.4

1.5

1.6

What is the probability that D2-3 is bad?

P{X1-3- X1-2>1.55} + P{X1-3- X1-2<1.45}

Sums of i.i.d. N(,) are normal

N(2.5, (.05/3)2) +[(-)N(1.0, (.05/3)2)]= N (1.5, (.10/3)2)

So D2-3


Tolerance interpretation

The likelihood of a bad part is

P {X2-3 > 1.55}-1 P {X2-3 < 1.45}

(1-.933) + (1-.933) = .137

As a homework, calculate the likelihood that

D1-4 will be “out of tolerance” given the same logic.


What about multiple features

What about multiple features?

  • Mechanical components seldom have 1 feature -- ~ 10 – 100

  • Electronic components may have 10,000,000 devices


Suppose we have a part with 5 holes

Suppose we have a part with 5 holes

  • Let’s assume that we plan for + 3 s for each hole

  • If we assume that each hole is i.i.d., the

    P{bad part} = [1.0 – P{bad feature}]5

    = .99735

    = .9865


Success versus number of features

Success versus number of features

1 feature = 0.9973

5 features = 0.986

50 features = 0.8735

100 features = 0.7631

1000 features = 0.0669


Should this strategy change

Should this strategy change?


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