I have a question / Question from Don day GD&THierarchy Workbook 1

[huni]

Hello Gents

I am preparing for GDTP cert now.
so,I have been studying ASME standard and Don day's Hierachy workbook.
I have a question.

This question came from Hierachy workbook.
The GD&T Hierarchy Y14.5-2009 workbook.
The question number is C.5.6
(I have a solution note also..but there is no explain how to solve it)

Q : C.5.6
answer: a) 0.15

I only understand that hole is produced at 20mm (so, perpendicularity is increased to 0.4 mm / but actual out of perpendicular is 0.1mm-->??
Did I understand correctly?)
and Question : The pattern of features( 4 X ∅10 +/-0.2)may be displaced radially from the datum axis by

a : 0.15

Is anybody knows about this question?
Would you explain how to solve it?

Please help me.

I attached the captured photo of question.

Thanks

 

[Woosang]

huni,

this is about "datum feature displacement", or called "datum shift"

have a look at 5.3.2.2 in 1994 or 4.11.9 in 2009

in your example, you should consider virtual condition of datum feature B and its related actual mating envelope because it relates to datum A and is referenced as secondary datum.

virtual condition is ø19.6 and related actual mating envelope is ø19.9 so datum feature axis can be displaced in cylinder of ø0.3 (0.15 radially)
thus, your pattern of holes has additional position error allowed in relation to datum feature B axis.

 

[greenimi]

"The maximum amount of datum feature B shift that you are allowed to use is a half of difference between size of actual related mating envelope (RAME) of datum feature B (this envelope is perfectly oriented and located, if applicable, relative to primary datum A) and the size of virtual condition (also called maximum material boundary, MMB) of datum feature B."

In order to know how much datum shift B is allowed for position control of pattern of 4 holes, it is not enough to know the size of the hole B. You must know the size of Related Actual Mating Envelope (RAME) of hole B - this is the envelope that is perfectly oriented and/or located with respect to preceding datum. So in this case you must know size of RAME that is perfectly perpendicular to datum plane A, because this is the only datum that precedes B in position FCF applied to 4 holes.

If the hole Ø 20 has no perpendicularity error the answer would be different. Because if it has any perpendicularity, the size of RAME of hole B will be smaller than Ø 20 and the amount of datum shift allowed will also become smaller.
20 - 0.1 = 19.9 (RAME)
19.8 (MMC) - 0.2 (perpendicularity) = 19.6 (VC at MMB)
19.9 - 19.6 = 0.3; 0.3 / 2 = 0.15
Same answer as Woosang.

 

[CheckerHater]

The question was "how much the pattern can be displaced", not "how much datum can shift"

So, we take diameter 0.30 from FCF, divide it by 2 and get 0.15 The same result. Datum shift is NOT a bonus (according to Don)

"For every expert there is an equal and opposite expert"
Arthur C. Clarke Profiles of the future

 

[greenimi]

CH,
So you are not using in your calculation datum feature B “ ACTUAL” size (20mm)?
Also you are not using in your calculation the ACTUAL perpendicularity error (0.1mm)?
If not, why not? Does not matter? I am asking, please be gentle!

 

[axym]

huni,

Keep in mind that the current GDTP certification exams are based on ASME Y14.5M-1994. Also keep in mind that third-party textbooks paraphrase the content of the ASME standard, with different emphasis and wording.

Regarding question C.5.6, I would say that Woosang and greenimi's calculations are correct and capture the intent of the question.

But I also share CH's concern about the wording of the question in terms of how much "the pattern of features may be displaced radially from the datum axis". I would say that this wording is very misleading, and possibly incorrect. It should say "the datum feature may be displaced radially from the datum axis". This would follow what I believe the intent to be, which was to quantify the datum feature shift/displacement that greenimi described.

I happened to look at question C.5.5, and I believe that there is a problem there as well. The question uses the term "axis of datum feature B", which is defined as the axis of the unrelated actual mating envelope. In order for the axis of datum feature B to have zero deviation with respect to the four holes' basic location, it would have to be perfectly perpendicular to datum A. The thing that would have zero deviation with respect to the four holes' basic location is the axis of the related actual mating envelope of B (because B is the secondary datum feature, not the primary). Again, misleading and possibly incorrect.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[CheckerHater]

Axym, thanks for supporting my concern.

Greenimi, there are several aspects to datum shift forming sort of a "murky" area that is not always considered or fully understood.

First, yes, datum shift is real. If you imagine the most simple situation, say bushing with OD serving as datum MMD and ID controlled MMC wrt said datum, the shift will even act as bonus allowing grater eccentricity.

BUT as soon as part becomes more complicated, it is easy to create the situation when part cannot even "shift" although the datum feature is actually "loose". Here comes the:

Second, no, datum shift is not bonus, you cannot simply add it to other tolerances.

Third: for practical reason it is safer to ignore datum shift, especially if you are using CMM. This is why authors of several books on GD&T recommend to be very careful around datum shift.

I doubt if simple straightforward solution will be found anytime soon.

"For every expert there is an equal and opposite expert"
Arthur C. Clarke Profiles of the future

 

[Woosang]

Guess this example is in the context of the wording from 1994-5.3.2.2: "since the axis of the datum feature's actual mating envelope must serve as the origin of measurements for the pattern of features, the features are therefore viewed as if they, as a group, had been displaced relative to the axis of the datum feature's actual mating envelope"

But can't fine such description in 2009-4.11.9 Datum Feature Shift/Displacement.

 

[axym]

Woosang,

I went back and read the wording from 5.3.2.2 (and 5.3.4.2), and I need to temper my comments a bit. In Don's defense, the wording in the questions is more consistent with the wording in the 1994 standard than I had thought. The standard does describe displacement in terms of the pattern of features being displaced (personally I find this confusing). However, the standard says that they are displaced relative to the datum feature's actual mating envelope. Question C.5.6 says that they are displaced relative to the datum axis, which I do not think is correct.

The description of datum feature shift/displacement was clarified in Y14.5-2009 Section 4.11.9. But in Section 7.3.6.2.2 Departure of Datum Features From MMB, some of the original description remains:

"In Fig. 7-18, illustration (b), where datum feature B departs from MMB, relative movement can occur between datum axis B and the axis of the related actual mating envelope of datum feature B. See para. 4.11.9."

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[greenimi]

I know we might talk semantics here, but just I CANNOT agree with CH assessment such as “take diameter 0.30 from FCF, divide it by 2 and get 0.15 The same result.” I cannot take that as a correct answer for the OP question. I would say that is just flat out wrong. It is a just strange coincidence that for the position of the 4 holes the tolerance zone is Ø.030.
I do agree (fully agreement) that:
1.) datum shift is real.
2.) datum shift is not bonus, you cannot simply add it to other tolerances
3.) for practical reason it is safer to ignore datum shift, especially if you are using CMM

But all the above 3 reasons DOES not prove that just take Ø.030 tolerance zone and divide it by 2 (in order to get the “correct” answer for the OP question) is valid. Again, I just go as far and say it is just flat wrong.

 

[Belanger]

I agree with you, greenimi. The best explanation was woosang's. You simply take the difference between the virtual condition and the RAME of the actual hole (yes, I know RAME didn't exist in 1994 but it's a convenient way to say it). It's merely a coincidence that that difference of 0.3 happens to be the same as the holes' stated position tolerance.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

 

[gabimo]

Ch,
Would you mind to explain a bit why do you think your answer is the right one?
Just curious and confused about your replay from 2 days ago.
Thanks
Gabi

 

[3DDave]

Since C.5.6 gives no information about the condition of the holes, the only reference can be to the condition of datum B, which is at 20mm but tilted by .1mm vs a .2mm perpendicularity tolerance. Since the minimum diameter is 19.8mm, this gives 0.2+(20-19.8)-0.1 =0.3 diameter, or 0.15 radial deviation in the direction of the tilt.

Perpendicular to the tilt direction the pattern can move 0.2 + (20-19.9) = 0.4 diameter or 0.2 radially.

Since one typically wants to know the maximum displacement possible (minimum is always zero) the answer should be 0.2. If the question was rephrased as the displacement with the greatest limitation, that is 0.15.

I would also have worded it as the amount the A|B(M) datum reference frame can be shifted relative to the axis of the 20.0 hole. While more accurate describing the situation, I'm sure it doesn't parrot the words in the standard closely enough to be acceptable

Ignoring datum shift is a mistake. In many cases I have seen the greatest contribution to variation is due to bad datum selection leading to excessive shift. It might make inspection easier, but when the parts get installed the assemblers have to take time aligning unrelated features for workmanship.

 

[powerhound]

Dave,

I'm not following you when you imply that the direction of tilt affects the answer. A datum simulator will be 19.6 in diameter. The 20mm hole tilted at 0.1 will result in a 19.9 inner boundary relative to datum A. You will only get 0.15 of radial displacement no matter how you move the part. In one direction, the datum simulator will hit the top of the hole and in another it will hit the bottom.

John Acosta, GDTP Senior Level
Manufacturing Engineering Tech
SSG, U.S. Army
Taji, Iraq OIF II

 

[CheckerHater]

Gabi,

I don't think my answer is the right one.

Huni said 0.15 is the right answer.

I was just saying that pattern displacement may have nothing to do with datum shift, hence radial displacement is the half of allowable diametrical displacement.

I didn't take bonus tolerance into consideration, but then it wouldn't be 0.15

"For every expert there is an equal and opposite expert"
Arthur C. Clarke Profiles of the future

 

[3DDave]

PH - you'll just have to draw your own picture.

 

[huni]

Thank you for all of comments.

Thank you for everyone

I'm reading every user's opinion carefully.

Thank you!

 

[greenimi]

I do not get what 3DDave is trying to say, but maybe it is just me. I don’t even “imagine” what picture should I draw to comprehend. The level is way too high for me.

 

[axym]

Hi All,

Hats off to Dave on this one. He's absolutely right, and to see it you have to go a couple of levels deeper. Forget what the standard says for a minute, and look at the geometry itself.

The picture that needs to be drawn is three-dimensional, with the tilted hole interacting with the datum feature simulator. The tilt brings the surface of the hole closer to the simulator surface and decreases the radial dispacement to 0.15 - but only in the direction of the tilt. In the "transverse" direction, the surface of the hole is not brought closer to the simulator surface and the amount of shift is still 0.20.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[powerhound]

I was thinking the same thing. That's rocket surgery or brain science stuff.

John Acosta, GDTP Senior Level
Manufacturing Engineering Tech
SSG, U.S. Army
Taji, Iraq OIF II