Ch02tol
-
Upload
vadi-velan -
Category
Documents
-
view
212 -
download
0
description
Transcript of Ch02tol
![Page 1: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/1.jpg)
Tolerance interpretation
Dr. Richard A. WyskISE316
Fall 2010
![Page 2: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/2.jpg)
Agenda
• Introduction to tolerance interpretation• Tolerance stacks• Interpretation
![Page 3: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/3.jpg)
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?
![Page 4: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/4.jpg)
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
1.0±.05 1.0±.05?
1.5±.05
1
2 3
4
Case #1
![Page 5: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/5.jpg)
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
1.0±.05 1.5±.05
1
2 3
4
3.5±.05
Case #2
![Page 6: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/6.jpg)
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
1.0’±.05 ?
1
2 3
4
3.50±0.05
Case #3
1.00’±0.05
![Page 7: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/7.jpg)
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?
1.0±.05 1.0±.05?
1.0±.05
1
2 3
4
Case #1
![Page 8: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/8.jpg)
Mfg. ProcessLet’s try the following (in the same setup)
-cut plane 2
-cut plane 3
Will they be of appropriate quality?
3
2
![Page 9: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/9.jpg)
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
![Page 10: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/10.jpg)
Expectation• What do we expect when we manufacture
something?PROCESS DIMENSIONAL
ACCURACYPOSITIONAL 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
![Page 11: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/11.jpg)
Size, location and orientation are random variables
• For symmetric distributions, the most likely size, location, etc. is the mean
2.45 2.5 2.55
![Page 12: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/12.jpg)
What does the Process tolerance chart represent?
• Normally capabilities represent + 3 s• Is this a good planning metric?
![Page 13: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/13.jpg)
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 3of our parts would be good 99.73% of our dimensions are good.
An Example
![Page 14: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/14.jpg)
We know that (as specified)
D2-3 = 1.5 .05
If one uses a single set up, then
(as produced)
D1-2
and
D1-3
.95 1.0 1.05D1-2
2.45 2.5 2.55
D2-3 = D1-3 - D1-2
![Page 15: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/15.jpg)
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
1.4 1.5 1.6
![Page 16: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/16.jpg)
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.
![Page 17: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/17.jpg)
What about multiple features?
• Mechanical components seldom have 1 feature -- ~ 10 – 100
• Electronic components may have 10,000,000 devices
![Page 18: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/18.jpg)
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
![Page 19: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/19.jpg)
Success versus number of features
1 feature = 0.99735 features = 0.98650 features = 0.8735100 features = 0.76311000 features = 0.0669
![Page 20: Ch02tol](https://reader036.fdocuments.net/reader036/viewer/2022083013/5695d3e41a28ab9b029f8c58/html5/thumbnails/20.jpg)
Should this strategy change?