vortex00

My Feedback: (22)

How do I figure what the max bending load is for a 1.5" 6061-T6, .058 wall tube? Formula?

Thanks,
Jack

Ollie

Calculate the bending load for a solid bar of the same OD and subtract the bending load of a solid bar of the same ID. With such a thin wall compared to the diameter, the tube will buckle before it fails in tension or compression. Therefore, the maximum calculated load needs to be derated for the buckling mode of failure unless the tube is filled with a material that prevents buckling. I would suggest that it be derated by a factor of two ot three to be safe.

vortex00

My Feedback: (22)

Thanks Ollie, I'll look at the buckling loads.

Jack

Johng

My Feedback: (4)

Buckling loads are key. That can be solved by filling the tube with the right size (tight fit) dowel rod. It's alot easier than trying to calculate critical buckling loads.

Then the bending stress will be the critical factor. Strength of material are given as stress levels in pounds per square inch (PSI). Yield stress is the stress at which the metal begins to deform. The yield stress in tension is usually the critical (lower) value. SO, step one is to look up the yield stress in tension of the type of aluminum you are using.

stress = M * y/ I

where M is the bending moment applied, I is the cross section moment of inertia(natural resistance to bending of the particular shape), and y is the distance from the neutral axis to the point being calculated. In this case y is the outside radius of the tube.
You would need to look up I in a structures book - or how to calculate I, anyway.

To get the max bending load, just rearrange the formula;

M = stress * I/y

and use the yield stress value for 6061-T6, I for a hollow circular section, and y is the radius. Getting the Moment - you can then back figure the amount of lift the wing can produce before bending the tube, neglecting the bending strength of the wood that you added for buckling. Sound like fun??

vortex00

My Feedback: (22)

Hi John,
I'm sure what I have will work, I'm trying to run the numbers to come up with the same answers I got from the experts. What I don't have is the Moment of Inertia for 6061 Aluminum tubes.
Right now I have 250 ft-lb of Moment at the root.

Thanks for the reply

Jack

Rotaryphile

A 1-1/2" O.D. tube with wall thickness 0.058" has a section modulus of 0.09121 inches squared. Applying a moment of 250 pound feet, which equals 3000 pound inches will result in a bending stress of 3000 divided by 0.09121=.32,891 psi. This should be considered close the the maximum; with the tube failing, probably by local buckling, at around 35 to 37,000 PSI, unless the shear load is uncommonly high.

Mike James

A bit of related trivia...

I recently read an article on a model web site (Sorry I didn't record the URL) by some folks who tested various tubes, including carbon fiber. These were static tests, with weight applied to both wings equally.

What was interesting to me was that all the tubes failed in the center. So, apparently the fuselage sides don't create the kind of shear load I would've expected. Maybe someone could develop wing tubes that were thicker (inside) in the center and thinner at the ends, for more efficiency.

Live and learn.

Ollie

I beams and box beams have the best strength to weight ratios and the most efficient use of material to resist bending loads. The tops and bottoms are sized to carry the tension and compression loads. The sides or web are sized to carry the shear loads and to resist buckling on the compression side.

The reasons round shapes are popular are ready availability and ease of construction.

For cantilevered wing spars, the tops and bottoms can be tapered as can the vertical element for additionad weight saving. A spar that is tapered so that the material crossections match the local bending load will weigh aproximately 1/3 the weight of a spar of uniform material crossections.

Tapered unidirectional carbon spar elements are commercially available from at least two sources. These commercially available shapes are linearly tapered and save about 1/2 the weight of a uniform element.

Rotaryphile

6061-T6 is a good general purpose aluminum alloy, particularly if welding is required, but I would try to find a piece of 2024 or 7075. 6061 has a yield strength of about 40,000 PSI, versus about 70,000 for 2024-T3, and 7075 is stronger still, but hard to find.

I agree with Ollie 100% regarding tapered box spars - I always use these, and I get about 1/3 the weight of a constant section spar, for the same strength. Tubes are frequently used to join two piece wings, but the convenience of two piece wings comes with quite a significant weight penalty; in my calculations - somewhere around 10 percent at best. Making the joiner tube as long as possible, and machining it to taper the wall thinner at the outboard ends could help significantly.

vortex00

My Feedback: (22)

OK guys, Thanks for the detailed replys.
Rotaryphile- How did you figure the section modulus of the 1.5" tube?
I got my tubes from Wicks Aircraft, various sizes available but the 2024 and 7075 are not available in that size. I'm OK with the 6061-T6.
I'm building a 1/6 scale U-2, 13.5' w/s. The 10" dia. fuse. will have 16" stub wings and 5.5' plug-in wings using the tubes. Stubs will have a box spar 42" of 1/4" 5-ply plywood with 1/4" spruce caps. plug-in tube will insert the approx. 18" into the stub and fuse at the 50% point. I just need to work on securing the rotation of the outer wing yet.
Initial calculations are based on full elev. deflection at 100mph generating 25lb/sq.ft. 25lb gross weight.
Any other recommendations/considerations?

Jack

Rotaryphile

Hi Jack: I just plucked the figure from a table I had on hand of tube section properties. The section modulus is easy to calculate, and once you have it, just divide the moment in pound inches by the section modulus to find the bending stress. I would try to find a piece of 2024 in preference to 6061, since it is quite a common alloy, and is about 75% stronger. I use 6061 only where extensive forming is needed, or welding. 2024 is hard to weld, while 6061 is a breeze. You might try this website for section modulus - (Sx): http://www.onlineengineer.ca/circle3.htm
It asks for the dimensions in millimeters, but if you plug in inches instead, it should work just fine, and the section modulus it calculates will be in inches cubed, not mm cubed. I recently helped a local modeler make the same calculation for a big AT-6, and the recommended wing joiner tube was a bit shy of strength, when I calculated the moment that the wing would develop at maximum expected speed for snap roll entry. Incidentally, I get about 38 pounds lift per square foot at 100 mph, using safe lift coefficient of 1.5, which should be about right for the scale wing section. When calculating the moment developed by the wing, you can usually subtract around 20% to allow for the negative moment developed by the weight of the wing, so using 25 pounds per square foot to figure the actual moment on the tube might not be too far off.

vortex00

My Feedback: (22)

Rotaryphile-Good info thanks, I got the site for tubes. Just what I wanted. Also, further research on Wicks aircraft site revealed that they do carry the 2024-T3 tubes. I will use these instead. Similar weights and stronger.

Jack

Rotaryphile

Glad to hear, Jack. 2024-T3 should cost very little, if any more, than 6061-T6, and it is a helluva lot stronger. A good all-purpose aircraft alloy of WW2 vintage, and still hard to beat for all-round usefulness. I routinely form 2024, but you have to be careful to avoid sharp bends, such as when forming dural landing gear. A little heating, up to around 300 F. or so helps, but avoid going much hotter or you will lose strength, although the material will slowly age harden, and you can reharden it quickly if you really want to get into aluminum heat treating, which doesn't need high temperature anyway - nothing like those involved in hardening steel.