Maximum Load on Tube Before Plastic Deformation

[PSteven]

Hello,

I have a seating application that involves a metal cylindrical tube, steel or aluminum most likely, which includes a metal cylinder sliding back and forth. The tube has a slot on top through which a rod extends from the cylinder to support a seat.

I have a prototype working, which was built adhoc in the shoppe. I would like to be more rigorous in the design and make a suitable selection for the tube.

Currently, I am using steel with an OD of 1.125" and an ID of 0.875". The tube is 6" long and the cylinder within is 1.5" long. This seems to work well with no noticeable plastic deformation, although the cylinder designed was a bit loose. For the next prototype I would like to try anodized aluminum, but I am unsure what size of tube and wall thickness to select.

How do I calculate the maximum load for a given length, OD and ID for an aluminum tube and a given length of cylinder for such an application?

Thanks,
Paul

 

[PSteven]

I found the following analysis related to a beam analysis. Would the final formula be correct?

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The ultimate loading of these tubes is of interest in view of possible misfortunes, such as
unauthorized visitors climbing on the tubes or tree limbs falling on them.
The maximum stress in a circular beam is

Maximum Stress (sigma) = Mr/I

where M is the bending moment, which for a simple beam is M=Wl/8, and r is the radius of the beam. Setting the maximum stress to the yield stress for aluminum tube, 28,000 psi,
and solving for the maximum load gives

W =8*sigma*I/(rl)


For a tube of outer diameter do and inner diameter di, the moment is

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

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Plugging my numbers in gives:

W = 8 * 28000 * 0.04977 * 2 / (1.125*6)
=3303lbs

 

[rb1957]

i'd be worried about the slot. it sounds like you have a seat attached to the cyclinder which slides inside the tube, so that the cyclinder loads the tube.

the slot weakens the tube; your equations are for a complete tube. also WL/8 is the maximum moment for a beam loaded by two forces (W/2) at the 1/4 points (L/4 and 3L/4). i think your geometry is alittle different ... that the moment arm is 3L/8, so the moment is 3WL/16. if you want to be conservative, you could use WL/4. this assumes that the cyclinder loads the tube only at the ends, if it is a tight fit it'll load the tube along it's length; this'll incease the moment in the tube and redcue the moment in the cyclinder.

worry about the shear load at the end of the cyclinder, particularly with the slot creating an open section.

 

[GregLocock]

Is this like a bicycle saddle? Have you seen how they fail?

What rb said, but more so.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[PSteven]

Thanks for your responses.

rb1957, currently the tube is loaded at either of the ends, so perhaps WL/4 is an appropriate ball park figure. Does this includes a compensation for the slot, which as you indicate weakens the entire tube?

Are you indicating that a tight fitting cylinder, still free to travel, will load the tube more than if it was loose fitting, causing the tube to fail with less stress?

This is a bicycle application. I've been using the steel prototype now for 2 months commuting to work 4 times a week, round trip 60km over rolling hills. I'm a 215lb rider, and the tube seems to be standing up to the abuse quite well.

Greg, can you point me to any examples of such applications failing? I'd be interested to see any previous examples of this.

Regards,
Paul