Moment of Inertia - Tube with Slot

[PSteven]

Hi,

I'm trying to determine the failure point for a tube with a slot. The tube has a piston sliding back and forth, and is loaded laterally by an external weight attached to the piston through the slot.

I can't seem to find an equation representing the moment of inertia for a tube with a slot. In my application the slot does not run the length of the tube, but to simplify the math this would be an acceptable approximation.

The moment of inertia for a tube is:

I =pi * (do^4 - di^4)/64.

Does anyone know the case for a tube with a slot?

Thanks

 

[electricpete]

Where is the slot in relation to the axis of bending?

For example, if the axis of bending is 12:00 - 6:00, is the slot constrained to be at 6:00 or 9:00 or can it be anywhere?

Is the arc of the slot relatively small?

One quick simple approximation (for small slot) would be to treat the slot as a rectangle. Use moment of inertia for rectange, transfer it to the correct location by parallel axis theorem, and subtract it from moment of inertia of the tube.

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[GBor]

I think I would be inclined to approximate it much like a channel compared to box tubing. Your bending axis will be off-center due to the loss of section at the loading point.

Sounds like you may have another issue, though, and that is the bending (widening) of the slot trying to pry it open. You also have a dynamic loading situation, so you can't really use any plane stress or plane strain assumptions. May be time for a simulation or, at a minimum, a dynamic analysis.

Not a bad little problem...

Garland E. Borowski, PE
Star Aviation

 

[electricpete]

Here is a brute force solution of the general geometry using the definition of area MOI:

http://home.comcast.net/~electricpete/eng-tips/AreaMoi_TubeWithSlot.pdf

It looks correct to me and checks as expected for the specific case of hollow cylinder, but you might want to double check it for yourself.


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[electricpete]

My pdf file states:
"# R1 is tube ID"
"# R2 is tube OD"

Obviously R stands for radius, so should be ID/2, OD/2, respectively.

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[desertfox]

Hi pschranz

Well I was wondering whether you can use the moment of inertia for the tube and use a stress concentration factor
of say 3 for the area of the tube with the slot in it.
You give no sizes or loadings so its difficult to say how you might actually analyse it.

desertfox

 

[PSteven]

Thanks everyone.

These are my dimensions:

Tube OD = 1"
Tube ID = 0.875"
Tube Length = 6"
Slot Width = 0.5"
Slot Length ~ 4"

electricpete:

The piston is normally fixed at either end of the slot, and travels between these locations. The piston is loaded when in a fixed position, and is not substantially loaded when it travels between the two endpoints.

The tube is supported relatively speaking at both ends of the slot.

I appreciate your equation, it's a great help. It does seem to reduce to the simple case of a hollow cylinder (tube) when theta1 = 0 and theta2=2*pi. I need to read up on first principles of moments of inertia to verify if it is correct in the general sense. I think I need to do this exercise.

I calculate as follows:

Tube with Slot:

theta1 = 0
theta2 = 296deg = 360 - 2 invsin (0.5/(2*r)) where I used the mean radius of the tube

I = 0.287613

Normal Tube:

I = 0.267232


This appears to make sense.

 

[PSteven]

Ooops.. I treated ID and OD in the equation as diameters, but as electricpete says should be the inner and outer radius respectively.

Tube with Slot:
I=0.017975

Tube:
I=0.02031

My ultimate goal is to determine the weight at the yield point for the tube. Using the equation for the maximum stress for a circular beam:

sigma = M * r / I

M - bending moment
r - radius of beam
I - moment of inertia

It appears that the maximum stress (sigma) increases when the tube has a slot in it. Does this make any sense?

 

[PSteven]

My mistake, the maximum stress is set to the yield tensile strength (or ultimate tensile strength) for the metal being used, which in this case is aluminum 6061-T6.


Using the following:

sigma = M * r / I

M = WL/4 (I'm not sure why this equation)

and solving for W

W = (4I*sigma) / (L * r)

sigma = 40ksi (yield tensile strength)

http://www.onlinemetals.com/merchant.cfm?pid=1212&step=4&showunits=inches&id=71&top_cat=60

Tube With 0.5" Slot:

W = (4 * 0.017975*40k)/(6*0.5) = 959lbs

Tube:

W = (4 * 0.02031*40k)/(6*0.5) = 1083lbs

It makes sense that the maximum load should decrease.

 

[khardy]

The easy way to find I of an iregular shape is with AutoCAD. Look into the "region" comand and the "mass properties" comand.

 

[GBor]

From AutoCAD, I=0.0180. Looks like the formula is spot on.

 

[PSteven]

Thanks khardy & GBor:

That's good to know! Especially that it matches the results from the equation electricpete provided.

Do you know if SolidWorks or Alibre has the same capability?

 

[GBor]

In SolidWorks, you can model your part, but you will need a face somewhere in the area of the slot. Click on the face. Select "Tools: Section Properties" and it will give you a variety of references for the area moment of inertia. The problem is making sure you are looking at the correct one...it will give you "Ix" and "Iy" relative to the global axes, local axes, the section centroid...all useful for various applications. Work a simple one (like a rectangular section) with a geometric hanging out in space (not centered at the origin) and look at the results. You can hand calc it for the rectangle with (1/3) b h^3 for the inertia relative to the centroid of the shape. Campare that answer, but make sure you are selecting the proper "b" and "h"...may want to make it a square so that it doesn't matter.

I haven't used Alibre in a long time, but I'm sure it has a similar capability.

 

[JStephen]

That should be something that can be integrated directly. It may be easier to find the moment of inertia about the center of the tube, and then use the parallel axis theorem to transfer that to the neutral axis, which will be shifted from the centerline.

One issue is that the allowable bending stress of the open area will be much reduced from that of a closed tube, and I'm not aware of any references that give an allowable stress for that case.

 

[desertfox]

Hi pschranz

If the tube is in bending as you suggest it is then the bending stress will be much higher than that calculated by the formula even with the correct moment of inertia value.
The reduced material at tube surface will mean the stress lines in that area will crowd together creating a stress concentration. If it were me and the thing isn't life threatening I would use a stress concentration factor of 3 and multiply the stress obtained by your formula by this factor and then compare yield stress for the material with the calculated figure.

regards

desertfox

 

[GBor]

If this slot is sufficiently long, you also need to take a look at whether the edges of the slot bend or buckle, but I've been under the impression that this is a working sliding cylinder, so the concept is already proven. If not, there are dynamic loads and other considerations. The safety factor of 3 should cover it, but there are plenty of interesting things happening structurally.

 

[PSteven]

desertfox, GBor:

The tube should not be significantly in bending. We are using a steel tube right now with 1/8" wall thickness and there is no apparent bending.

We are moving towards using aluminum, and a smaller wall thickness, so I want to crunch the math to see if it makes sense theoretically. The mild steel we are using I believe has comparable yield tensile strength values to 6061-T6 aluminum.


My formula calculated the weight at which point the tube yields. I can then apply the FOS of 3 to this value. Right?

Also, the load is not exactly in the middle of the tube. For the 6" tube the load is centered at 1.8" in one position, and 5" in a second position.


Regards

 

[GBor]

I would think so. Divide the load you calculated by 3 and that is how much you can place on the tube with the slot.

 

[chicopee]

It is relatively easy to calculate moment of inertias of tubes w/ or w/o slot. In your case, if you know how to use ACAD, you can graphic the slotted tube and do a LI command.