If you are a regular user of GD&T, you probably know that the ASME standard was recently revised (for details, see the blog entry below dated March 28, 2009). In today’s column, I’d like to introduce you to one of the changes.

The term “maximum material condition” or MMC has been around for a long time. This concept is invoked when the circled M symbol is placed in a feature control frame. Well, a new item for the 2009 standard is something very similar, called “maximum material boundary,” or MMB. Yet it is invoked using the same circled M symbol.

The reason this was introduced was to eliminate confusion when the M symbol is modifying a datum that has its own geometric tolerance. Consider the following example:

The position tolerance references datum feature B with the M symbol. But here’s the key: think about the size of a gage pin that would be inserted into the center hole (it should be attached to a flat plate that simulates datum A). It would not really be simulating the MMC of 22.0; instead, it would be 21.8 in order to accommodate the perpendicularity tolerance of 0.2. Thus the confusion — people would say “MMC” when discussing the datum, but had to understand that it wasn’t really the true MMC that would be simulated.

So the new standard uses the term “maximum material boundary” to differentiate it from the true “maximum material condition.” The picture above would not change, but in talking to someone you would say “the position tolerance of the OD is 0.4 at its MMC, relative to datum A and datum B at MMB.”

I suppose if there were no perpendicularity tolerance on the center hole, then you could interchangeably say “MMB” or “MMC,” since they would be the same (22.0). And by the way, the other choices of LMC or RFS on a datum reference also have their counterparts of “LMB” and “RMB” in the new standard.

I found this to be very informative. As a product instructor for CMM’s and Vision products it is imperative that we have a good working grasp of these conceps.

Jacob Grant

Great explanation. Thank you. Even after reading the standard 3 times I did not understand this topic quite as clearly as I do now.

Thanks.

This is same basic concept as virtual condition rolled up into the MMC Concept when it applies to a datum feature of size modified with th M.

That’s right, John. In fact, a certain automaker that I do a lot of GD&T work with used to have their own modifier which was a circled V. This could be used after a feature-of-size datum to indicate virtual condition. But it got to be too confusing because the national standard had M as meaning virtual condition.

So it’s really just semantics…

I was reading your post and thank you for that, I am still confused, in your example the drawing is showing in the FCF for the TOP the datum B as a secondary datum so the MMB will be 21.8 and it would be 22.0 only if the drawing changes the datum B to be the primary datum. is that correct?

Yes, that is true. Because both callouts have datum A as the primary datum, we include the perpendicularity number of 0.2 when calculating the MMB. But if the position tolerance of the outside diameter only referenced B(M), then 22.0 would be the MMB.

For the same part can you please provide me the minimum and maximum edge distance calculation

For the picture shown (a washer) the maximum wall thickness between the inner diameter (ID) and outer diameter (OD) is 16.7 mm.

This can be calculated a couple of ways, but here’s one way of showing the factors — first, the maximum:

+27.4 (biggest OD radius)

-11.0 (smallest ID radius)

+0.2 (OD’s stated pos tol in one direction)

+0.1 (datum shift due to ID at smallest size)

That adds up to 16.7. Notice that I include the datum shift because of the MMB modifier on the position tolerance. However, I do not include any bonus tolerance coming from the MMC modifier on the position tol. This is because bonus only occurs when the size dims are different than maximum material.

Conversely, the minimum answer is 14.7 mm:

+27.0 (smallest OD radius)

-11.3 (biggest ID radius)

-0.2 (OD’s stated pos tol in one direction)

-0.4 (OD’s bonus tol when made at smallest size)

-0.4 (datum shift due to ID at largest size)

The total is 14.7. Notice that this includes both bonus and shift. (If you’re familiar with stack-ups using a spreadsheet like the one we have in the Free Downloads, then I could also show it to you in that format.)

Thank you,

considering the diameter 22.0-22.6 is an external feature of size

can you please provide me the minimum and maximum edge distance

im confused on how the datum shift values of 0.1 and 0.4 were calculated?

Think about a mating fixture or gage which this part will drop onto.

When the actual datum feature (the center hole) on this part is made at its smallest size, there is still the opportunity for the part to “shift” on the fixture because of the perpendicularity tolerance. That perpendicularity tolerance of diameter 0.2 becomes a radial shift of 0.1 in the stack-up that is shown above.

On the other hand, when the actual datum feature is made at its largest size, there is a lot more opportunity for the part to “shift,” because there is the perpendicularity amount of diameter 0.2 along with the growth in size, which is diameter 0.6. So the total shift is diameter 0.8, which becomes 0.4 for the radial stack shown above.

The datum shift explanation makes sense; thank you.

My next question is for bonus tolerances.

For max wall thickness

OD= LMC……….Bonus tolerance= LMC – MMC= 54.8 – 54.0 = 0.8 dia…….0.4 radial

datum B= MMC…….Bonus tolerance= 0 (size is at MMC)

For min wall thickness

OD = MMC…….Bonus tolerance= 0 (size is at MMC)

datum B = LMC…….Bonus tolerance= LMC- MMC= 22.6- 22.0 = 0.6 dia………0.3 radial

Why was bonus tolerance of 0.4 for the OD added into the stackup for the max wall thickness?

Why wasnt bonus tolerance of 0.3 for datum B added into the stackup for min wall thickness?

Here is how i calculated my tolerance stack up John.

Let me know if there are any errors in my analysis.

Radius-OD@LMC= 54.8/2=R27.4

Radius-OD@MMC= 54.0/2=R27.0

Radius-ID@LMC= 22.6/2=R11.3

Radius-ID@MMC= 22.0/2=R11.0

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Position tolerance of OD-two sides= 0.4

Position tolerance of OD- one side = 0.4/2 = R0.2

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bonus tolerance-OD @ MMC = 54.0 – 54.0 = 0 (R0.0)

bonus tolerance-OD @ LMC = 54.8 – 54.0 = 0.8 (R0.4)

bonus tolerance-ID @ MMC= 22.0 – 22.0 = 0 (R0.0)

bonus tolerance-ID@ LMC= 22.6 – 22.0 = 0.6 (R0.3)

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datum shift(ID @ MMC)= bonus tolerance (ID @ MMC) + perpendicularity =0+0.2 = 0.2 (R0.1)

datum shift(ID @ LMC)= bonus tolerance (ID @ LMC) + perpendicularity =0.6+0.2 = 0.8 (R0.4)

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Min wall thickness = (Radius-OD@LMC) -(Radius-ID@MMC) + (Position tolerance of OD- one side) + (bonus tolerance-OD @ LMC) +(bonus tolerance-ID @ MMC) +(datum shift(ID @ MMC)= 27.4 -11.0 +0.2 + 0.4 + 0.3 + 0.1 = 17.4

Max wall thickness = (Radius-OD@MMC) -(Radius-ID@LMC) + (Position tolerance of OD- one side) + (bonus tolerance-OD @ MMC) +(bonus tolerance-ID @ LMC) +(datum shift(ID @ LMC)= 27.0 -11.3 -0.2 – 0.0 -0.0 – 0.4 = 15.1

Assuming: OD= external feature of size ID= internal feature of size

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For the max wall thickness: OD= largest diameter=external fos=MMC… bonus not allowed

ID= smallest diameter=internal fos=MMC… bonus not allowed

Max wall thickness = (Radius-OD@MMC) -(Radius-ID@MMC) + (Position tolerance of OD- one side) + (bonus tolerance-OD @MMC) +(bonus tolerance-ID @ MMC) +(datum shift(ID @ MMC)= R27.4 -R11.0 +R0.2 + R0.0 + R0.0 + R0.1 = R16.7

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For the min wall thickness: OD= smallest diameter=external fos=LMC… bonus allowed

ID= largest diameter=internal fos= LMC… bonus allowed

Min wall thickness = (Radius-OD@LMC) -(Radius-ID@LMC) – (Position tolerance of OD- one side) – (bonus tolerance-OD @ LMC) -(bonus tolerance-ID @ LMC) -(datum shift(ID @ LMC)= R27.0 -R11.3 -R0.2 – R0.4 -R0.3 – R0.4 = R14.4

(Q) Why was the bonus tolerance of ID @ LMC= R0.3 not used in your min wall thickness stackup?

is it because the datum shift of ID@ LMC already has bonus tolerance of ID @ LMC accounted for thus making it redudant to have both the (datum shift & bonus tolerance) for ID in the stackup

Max wall thickness = (Radius-OD@LMC) -(Radius-ID@MMC) + (Position tolerance of OD- one side) + (bonus tolerance-OD @ LMC) +(bonus tolerance-ID @ MMC) +(datum shift(ID @ MMC)= 27.4 -11.0 +0.2 + 0.4 + 0.3 + 0.1 = 17.4

Min wall thickness = (Radius-OD@MMC) -(Radius-ID@LMC) – (Position tolerance of OD- one side) – (bonus tolerance-OD @ MMC) -(bonus tolerance-ID @ LMC) -(datum shift(ID @ LMC)= 27.0 -11.3 -0.2 – 0.0 -0.0 – 0.4 = 15.1

No– the max wall thickness occurs when the OD and ID are both at MMC. (Recall that the definition of MMC is “maximum material condition,” which equates to maximum wall thickness.)

See my comment of March 13, 2016 for the calculation.

No bonus tolerance is included when calculating the max wall. But the bonus, and the shift, are included in the min wall calculation.

You are right MMC’s do not need Bonus tolerances. That was a typo on my part.

My real question is

(Q)Is Bonus tolerance of Datum B @ LMC (R0.3) considered redundant and not included separately in the min wall thickness stack up because Datum Shift of Datum B @ LMC (R0.4) already takes into account accounted Bonus tolerance of Datum B @ LMC (R0.3)?

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Including Bonus tolerance of Datum B @ LMC (R0.3) on top of Datum Shift of Datum B @ LMC (R0.4):

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Min wall thickness =R27.0 Radius of OD @ LMC -R11.3 Radius of Datum B @ LMC

-R0.2 Position tolerance of OD-One-sided

-R0.4 Bonus tolerance of OD @ LMC

-R0.3 Bonus tolerance of Datum B @ LMC

-R0.4 Datum Shift of Datum B @ LMC

=R14.4

Not Including Bonus tolerance of Datum B @ LMC (R0.3) separately and only including Datum Shift of Datum B @ LMC (R0.4) because it already takes bonus tolerance of Datum B @ LMC (R0.3) into account:

Min wall thickness =R27.0 Radius of OD @ LMC -R11.3 Radius of Datum B @ LMC

-R0.2 Position tolerance of OD-One-sided

-R0.4 Bonus tolerance of OD @ LMC

-R0.4 Datum Shift of Datum B @ LMC

=R14.7

For your first question:

“Is Bonus tolerance of Datum B @ LMC (R0.3) considered redundant and not included separately in the min wall thickness stack up because Datum Shift of Datum B @ LMC (R0.4) already takes into account accounted Bonus tolerance of Datum B @ LMC (R0.3)?”

The answer is is yes. What I called shift tolerance is just datum B’s total variation, consisting of the size tolerance (R0.3) and the geometric tolerance (R0.1). So it was already rolled into the calculation.

For your two calculations:

The R0.4 datum shift of Datum B @ LMC in the first calculation is incorrect; that 0.4 seems to come from the the R0.3 size of datum B and the R0.1 perpendicularity of datum B. But you already included the R0.3 in the previous line.

The second calculation is the correct one — Answer = 14.7 min.