Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

CIRCULAR / TOTAL RUNOUT EFFECT ON DIAMETER-SIZE

Status
Not open for further replies.

JJAV1983

Mechanical
Nov 7, 2012
35
Hope you can help me with this very simple question I think.

I have a cylinder which OD.= 7.998+/- 0.001, if I want to stablish a total runout of .002", will this mean that the cylinder OD could be 8.001" if we take into account that the part was produced at 7.999" + .002" (runout)?

Thank you.

JJAV




















 
Replies continue below

Recommended for you

Shortly - runout does not control size. It tells you how much size "wobbles" in relation to certain axis.
 
I am just confirming CH's statement - in case you need double confirmation.
 
No. Total runout controls roundness, straightness, taper, coaxiality, but not size. Size is controlled by, hmm, size. Your part is contained by ø7.999 envelope and actual local size is between 7.997 to 7.999.
 
Thank you all for your answers, I got it now.
 
Hi guys,

I missed the forum discussion but couldn't help to ask you guys on one of design practice in my company.
Currently when we are calculating radial clearance between a rotating shaft and its housing, we are adding the run out by half to the shaft size to ensure when housing bore is made with enough [highlight #FCE94F]radial clearance[/highlight] to the shaft.

hence, max shaft sizes (radial) is calculated as follow (shaft OD : 10 +/- 0.5)/2 +(run out : 0.1)/2 = 5.25 + 0.05 = 5.30 mm. (radial)

So... is the calculation is off as you guys mention - runout did not control sizes?

Thanks,
James. Kim
Power Transmission Engineer
 
Popootv,
What you are calculating is not a “size” but rather what is called “tolerance stack-up” – combined effect of size variation and runout.
Naturally, without some sort of a sketch we cannot say if you are doing it right. :)
 
Thanks CH, it is a tolerance stack calculation. Without sketch, I hope the explanation below would enlightened everyone to my question:-

Stack calculation : Minimum radial clearance between rotating shaft (Path A) and static housing (Path B)

Path A
Shaft Nominal diameter : 10 +/- 0.5 mm
Shaft Run out : 0.05 mm
Radial max path A stack : (10.5/2)+(0.05/2)=5.50 mm.

Path B
Housing Bore Nominal diameter : 12 +/- 0.5 mm
Radial min path B stack :(12.5/2)=6.25 mm

Min radial clearance between path B and path A : 1.25mm

My next question is should the run out calculated divided by two or just remain as whole in the stack calculation?

Thanks,
James Kim
 
Sorry correction to the value below :-

Stack calculation : Minimum radial clearance between rotating shaft (Path A) and static housing (Path B)

Path A
Shaft Nominal diameter : 10 +/- 0.5 mm
Shaft Run out : [highlight #FCE94F]0.10 mm[/highlight]
Radial max path A stack : [highlight #FCE94F](10.5/2)+(0.10/2)=5.30 mm.[/highlight]

Path B
Housing Bore Nominal diameter : 12 +/- 0.5 mm
Radial min path B stack :(12.5/2)=6.25 mm

Min radial clearance between path B and path A : [highlight #FCE94F]0.95mm[/highlight]

My next question is should the run out calculated divided by two or just remain as whole in the stack calculation?

Thanks,
James Kim
 
popootv,
I think I know what you are trying to say, but let me ask the question:
When you say “shaft has runout of 0.10 mm” is it runout of what in relation to what?
Also, what is relationship between the “housing bore” and the rest of the housing – is there any tolerance involved?
 
popootv,
1. Your calculation is incorrect. If you search for min clearance, the bore diameter should 11.5 not 12.5. This would result in 0.45 min clearance. However...

2. The clearance is probably even smaller, because your stack-up does not take any locational relationship between shaft and bore. A relationship between axis of the bore and axis of the shaft must exist and must be taken into account in the stack-up. This is what probably CH is asking for.
 
Yes,
The way I understood the question, we are dealing with assembly tolerance stack-up, BUT we ignore most of the assembly, or consider it made to theoretically exact dimensions.
Also, given that runout is compound control, the shaft could be perfectly round and coaxial, but have conicity of 0.10.
So I assume shaft being perfect size and shape and the entire amount of runout contribute to positional error A.K.A. eccentricity.
I even made a picture of said assembly, and if the picture is correct, then yes, in RADIAL calculation you should add ½ of MMC diameter to ½ of runout value.
And yes, pmarc, bore should be at MMC as well
 
 http://files.engineering.com/getfile.aspx?folder=79f61208-d811-4bfd-a519-125270d01997&file=Draw1.JPG
The parts you drawn would not only have to be perfect, but there would have to be no gap either between ID of bearings and the shaft, or between OD of bearings and the housing.
 
Yes, exactly.
That’s why I also called them “theoretically exact” A.C.A. “BASIC”.
So, both shape and size are perfect.
There nothing wrong per se about simplifying your model, especially when you make calculations “for yourself”, on the back of the old envelope; but if you ask total strangers for advise, you better provide more detailed information, agreed?
 
Size and runout (any GTOL) are independent of each other.



 
Status
Not open for further replies.

Part and Inventory Search

Sponsor