# Comparison of 4130 and 2024 Tubing, Composite Tubes too.

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#### wsimpso1

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There was interest in comparing the strengths of a variety of tubes. While the goal is to look at composites in comparison to steel tube structures, the initial trip through some calculations is metal tubes. Let’s try to keep the discussion on topic – what the calculated results mean to our designs and how to do the calculations.

I picked round tubing for this first go around, and included the math for it. I am going to do some square tubing and include the math in a later post. No whining that I did not do your stuff – I have no desire to do your work for you, I have enough of my own. Feel free to contribute by doing analytical work and posting results...

I decided upon 1x035 4130 tubing as my benchmark. It has a yield strength of 63000 psi and a shear strength of 36,400 psi. I am doing the analysis for strength in column loading, tensile loading, torsion, and bending, then selecting stock 2024 T3 tubing that is commonly available…

Methodology

Just to make sure that this is understood, all of the math in here came directly out of theoretical analysis done prior to the 20th Century. While the math is considered theoretical, it has ALL been empirically verified and found to give excellent agreement in real world structures. You really can design structures with it using appropriate factors of safety. In metal airplane parts, FOS of 1.5 is customary, while in composites 2.0 is used. This will mean that steel tubes will be stressed at 2/3 of their yield strength for the stated loads.

Column loading came right out of Chapter 3 of Shigley, Third Edition. I used Euler and Johnson equations with the crossover point between Euler and Johnson at Sy/2. I did use them exactly as published and assumed free ends.

Free ends is generally assumed in welded tube fuselage design, as fixed ends really does require them to be VERY FIRMLY fixed, and that is just not approachable in most structures. Free ends is slightly conservative, and is commonly used for welded or fastener connected trusses.

Tensile yield is F/A where F is the tensile load and A is the cross sectional area of the tube. Stress has to be less than yield strength.

Torsional loading is Tr/J where T is torsion, r is max radius of the tube, and J is summarized below.

Bending moment is My/I where M is bending moment, y is max distance from neutral axis, and I is covered below.

Sy is yield strength of the material

For circular tubing with uniform wall thickness:

A = PI()/4*(Do^2-Di^2)

I= PI()/64*(Do^4-Di^4)

J = PI()/32*(Do^4-Di^4)

For Column Loading we have long Columns and short columns. First step is to determine which one we have…

L/k is the slenderness ratio, where L is the free length of the column, usually the intersection of the members. k is radius of gyration, which represents how far from centerline the area of the section is.

k = sqrt(I/A)

L/ksub1 is the break point between short and long column solutions.

L/ksub1 = sqrt(2*PI()^2*E/Sy)

Where PI() is the Excel function for Pi;

E is Young’s modulus for the material in question.

If L/k is less the L/ksub1, use the Johnson equation, otherwise use the Euler equation.

Johnson Equation

Pcr = A* (Sy-(Sy/(2*PI()))^2*(L/k)^2/E)

Euler Equation

Pcr = PI()^2*E*I/L^2

To check your work, at L/ksub1, both equations should give the same Pcr.

Now to Results - the table did not copy. I will attach it separately.

I did not bother to calculate exactly what load each could carry, just confirmed if each tube was stronger or weaker than the baseline. This is plenty of work as it is. Want to know the margins? You can contribute to the thread, do the math, extract the data, and update us on it.

Notice that the same 1 x 035 tube will carry 5998 pounds when it is 15” long, but make it 60” long and it can only carry 678 pounds. Length changes things a bunch.

So, if you start with a 1 x 035 piece of 4130 tube 15” long in column loading or in tensile or torsion, it can be replaced with a 1 x 065 tube. You can save weight as long as your method of attaching the ends is not heavier than the difference in tube weights.

Want to replace that 1 x 035 tube that is 60” long in column loading and bending as well as tension and torsion, and the lightest 2024 aluminum tube is 1-3/8 x 049. It is lighter than the baseline steel tube, but you have a lot less margin to come up with attachments and still be lighter than just welding steel to steel.

One important thing to remember about reverse engineering with alternative materials and stock sizes:

An original designer working with known and/or calculated loads can only rarely select from stock and be exactly “on” for max stresses – typically the designer finds one size to be too small, the other options to be too big, and must select the too big size to preclude failure;

An original designer may have substantial uncertainty about the exact levels of working loads and will bump to the next size of stock materials to accommodate this uncertainty;

Then you come along, you can not know what the original designer knew, all that you can know is the strength of the part. Then you look at alternative material stocked by your suppliers, and select one that is at least as strong as what was originally used. The result is you now have something that was beefed up once or twice and is now beefed up second or third time. This is why it can be tough to do much better on weight when reverse engineering parts and assemblies.

In future posts I will look into some round composite tubes made from braided tubes over mandrels, then square tubes and how to calculate them.

I do need a source of aluminum square tubing to select from – Please recommend a web site with a set of stock sizes.

I will include a couple standard disposable mandrels from my shop, and then custom build to exact sizes too. Have patience, I am volunteering at this, I have to shop for health care insurance and clinics, renew my gun club membership, assist in assembling and operating the leaf gadget for the mower, and make progress on my homebuilt airplane.

Updated Table to correct a typo and include the strengths of all mentioned tubes under the loadings.

Billski

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#### Vigilant1

##### Well-Known Member
Thanks, Billski!

I did not bother to calculate exactly what load each could carry, just confirmed if each tube was stronger or weaker than the baseline. This is plenty of work as it is. Want to know the margins? You can contribute to the thread, do the math, extract the data, and update us on it.
I think I can do this for the other cases, but it will take a few days due to other stuff on my plate right now. I'm not a spreadsheet pro, but I can do the basics (e.g I don't know how to make it do an if>then comparison of Euler/Johnson, so I'll probably just do both and pick the most conservative). If anyone else would like to jump to do it sooner that's fine with me.

Column loading came right out of Chapter 3 of Shigley, Third Edition. I used Euler and Johnson equations with the crossover point between Euler and Johnson at Sy/2. I did use them exactly as published and assumed free ends.

Free ends is generally assumed in welded tube fuselage design, as fixed ends really does require them to be VERY FIRMLY fixed, and that is just not approachable in most structures. Free ends is slightly conservative, and is commonly used for welded or fastener connected trusses.
"Free ends": What value did you use for "k"--I'd like to be consistent with what you've done already. When I ran your cases through a Willford spreadsheet I have, the numbers came out different, and his notes say his values are for "pinned" ends, but I didn't tear his sheet apart to see what value he used for "k". I assume that may be the reason for the different values.

Thanks again.

#### Vigilant1

##### Well-Known Member
Trouble right out of the gate:
Torsional loading is Tr/J where T is torsion, r is max radius of the tube, and J is summarized below.

Bending moment is My/I where M is bending moment, y is max distance from neutral axis, and I is covered below.
These aren't making sense to me, but I'll reverse-engineer the results in your spreadsheet or consult another source until I understand it (which will be good for me). Specifically:
Equation 1) If we are solving for torsional loading, it's not clear where the value for T would come from.
Equation 2) If we are solving for bending moment, then bending moment can't be one of the terms on the other side of the equation, so I'm not understanding something that is probably very fundamental. Also, "y" would be 1/2" for this symmetrical round tube of 1" Do?

#### wsimpso1

##### Super Moderator
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Log Member
Thanks, Billski!

I think I can do this for the other cases, but it will take a few days due to other stuff on my plate right now. I'm not a spreadsheet pro, but I can do the basics (e.g I don't know how to make it do an if>then comparison of Euler/Johnson, so I'll probably just do both and pick the most conservative). If anyone else would like to jump to do it sooner that's fine with me.

"Free ends": What value did you use for "k"--I'd like to be consistent with what you've done already. When I ran your cases through a Willford spreadsheet I have, the numbers came out different, and his notes say his values are for "pinned" ends, but I didn't tear his sheet apart to see what value he used for "k". I assume that may be the reason for the different values.

Thanks again.
I think I used the easy way:
Calculate L/ksub1 in Q4 and L/k in R4;
Next column is =IF(R4>=Q4,(equation for Pce), "") so that if l/k is bigger than l/ksub1 we do Euler calc, if not, the conditional puts in a null answer;
Next column is =IF(R4<Q4,(equation for Pcj), "")so that if l/k is smaller than l/ksub1 we do Johnson calc, if not, the conditional puts in a null answer;
Next column is = Max(S4,T4) so we get the only one that has an answer.

I called them free ends, but the Shigley's text calls them "rounded". n=1.

K? K is sqrt(I/A)for the section. For the 1x035 that is 0.341

I would be really careful about the Wilford spreadsheets. They have been handed about a lot, and you do not know what is in them. I have only a checked a few places in the Wilford spreadsheets, but I found mistakes.

Billski

#### wsimpso1

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Equation 1) If we are solving for torsional loading, it's not clear where the value for T would come from.
Torsion for the steel tube is just the one that makes shear stress equal to shear yield on the steel tube by iteration on the equation shown. Then for the aluminum tubes, I used that torsion and checked if the stress exceed that material's yield.

Equation 2) If we are solving for bending moment, then bending moment can't be one of the terms on the other side of the equation, so I'm not understanding something that is probably very fundamental. Also, "y" would be 1/2" for this symmetrical round tube of 1" Do?
Bending moment for the steel tube is just the one that makes tensile stress equal to tensile yield on the steel tube by iteration of the equation shown. Then for the aluminum tubes, I used that bending moment and checked if the stress exceed that material's yield.

• Second row pertains to a 1x035 tube in various loading modes:
• Force where the subject tube 15" long in compression will elastically collapse or yield;
• Force where the subject tube 60" long in compression will elastically collapse or yield;
• Force where the subject tube will yield in tension;
• Torque where the subject tube will yield in torsion;
• Bending moment where the subject tube will yield in bending;
• And last is the estimate of the weight of the subject tube per foot of length.
• Third and subsequent rows are stock size tubes in 2024 T3 and whether they can stand in for the 1x035 steel tube.
For column loading, I calculated Pcr for the steel tube, then picked standard aluminum tubes and checked if Pcr was bigger or smaller than yield for that tube. If Pcr is bigger than for steel, it got a Y for yes.

I only included a tube size if it was adequate in at least one of the five loading situations. There are a lot of tubes that are inadequate or really oversize - I tried to stick with those that are usable.

For tension, torsion, and bending, I used the stress equations I listed above, put in the load limits I found for the steel tube, and checked if the stresses for the subject tube exceeded material yield. In tension and bending we are calculating tensile stress and checking against the tensile yield. I torsion we are calculating shear stress and checking it against the shear yield.

My shear yield strength used a standard 0.577 times tensile yield which comes out of strain energy failure theory (von Mises stress, etc).

Clear?

Billski

#### Pops

##### Well-Known Member
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I use free ends (c-1) to be on the conservative side. The welded tubes will have a certain amount of movement under load and can not be considered fixed ends.
I'll just stand in the background and learn.

#### Vigilant1

##### Well-Known Member
Bending moment for the steel tube is just the one that makes tensile stress equal to tensile yield on the steel tube by iteration of the equation shown. Then for the aluminum tubes, I used that bending moment and checked if the stress exceed that material's yield.
Thanks, I'm glad I asked. I didn't know that it was safe to assume that the bending failure would be due to tensile overload (on the tension side of the beam).
My shear yield strength used a standard 0.577 times tensile yield which comes out of strain energy failure theory (von Mises stress, etc).
That's not something I forgot . . . because I never knew it.
Getting clearer, thanks. Sorry to require the handholding. When I can sit down and work on a proper spreadsheet it will come together. Or I'll ask additional basic questions. Before I'm done I'm hoping to use that Avogadro's number that has been wasting RAM space in my cranium for the last 3 decades.

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#### wsimpso1

##### Super Moderator
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I didn't know that it was safe to assume that the bending failure would be due to tensile overload (on the tension side of the beam).
In metals, tensile and compressive yield are essentially the same. With a beam that is symmetric about its neutral axis, you only have one distance from neutral axis to worry over. Rounds, squares, rectangles, all are kind of convenient that way. Get into non-symmetric beams and you have more places you have to check. Get into non-symmetric loading (differing positive and negative g limits qualify) and composite materials and you have even more things to check on every iteration.

That's not something I forgot . . . because I never knew it.
Shigley covers distortion energy theory that has come to be definitive of failure criteria in metals. In composites, it is also distortion energy related (Tsai and Hahn, Jones)

Getting clearer, thanks. Sorry to require the handholding. When I can sit down and work on a proper spreadsheet it will come together. Or I'll ask additional basic questions. Before I'm done I'm hoping to use that Avogadro's number that has been wasting RAM space in my cranium for the last 3 decades.
No worries, I am doing this to help folks understand why things are built the way they are. As for Avagadro's number (IIRC, 6.23E23) the book What If - Serious Scientific Answers to Absurd Hypothetical Questions answers the question "What if you had a mole of moles?". Definitely worth a dash to the public library..

Billski

##### Well-Known Member
Avocado's number : the count of atoms in a guacamole.

When I get to work I'll have a play with the spreadsheet : see what's available in Aus (metric).
It would be interesting to tweak it to produce a substitution list : ie given a wall thickness, what's the min. dia ally tube that exceeds the given 4130.
I'll have a think about that: some limits (ie torsion) may be less important in some circs.

Log Member
Tube charts

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##### Well-Known Member
Notice that the same 1 x 035 tube will carry 5998 pounds when it is 15” long, but make it 60” long and it can only carry 678 pounds. Length changes things a bunch.
I'm implementing those equations, and I get 678 with a 1.5 FOS for a 60" tube (Euler), and 5998 without the FOS for a 15" tube (Johnson). I get 3999 with a 1.5 FOS.
Am I missing something?

#### wsimpso1

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I'm implementing those equations, and I get 678 with a 1.5 FOS for a 60" tube (Euler), and 5998 without the FOS for a 15" tube (Johnson). I get 3999 with a 1.5 FOS.
Am I missing something?
Might be me. I will check my spreadsheet programming.

Billski

#### wsimpso1

##### Super Moderator
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Typo, read off the wrong column for that one. Thanks for checking it AdrianS!

I will post the fixed number for the steel and the actual max working loads of the aluminum tubes sometime today. Maybe get to some composite tubes later this week.

Anyone have a source of square aluminum tubes I can use?

Billski

#### BBerson

##### Well-Known Member
HBA Supporter
onlinemetals.com has square aluminum tubes. Very limited size and wall thickness options.

##### Well-Known Member
Typo, read off the wrong column for that one. Thanks for checking it AdrianS!

I will post the fixed number for the steel and the actual max working loads of the aluminum tubes sometime today. Maybe get to some composite tubes later this week.

Anyone have a source of square aluminum tubes I can use?

Billski
As a long-time professional programmer, having an independant implementation can be handy as a back-check. I also compared results to other published formulae, but the other formulas are just rearrangements of the ones you posted.

What version of Excel are people using?
I'm playing with dropdown material and section selections : I think I can keep it Excel 97 compatible (and no macros).

#### Vigilant1

##### Well-Known Member
What version of Excel are people using?
I'm playing with dropdown material and section selections : I think I can keep it Excel 97 compatible (and no macros).
I'm building a spreadsheet now, about 50% complete, I might get it done tomorrow, maybe later (I burned up a bearing on a blade of my riding mower today, so now another thing to do tomorrow. Trying to finish the last mow of the year.) I use LibreCalc, but save it in Excel 97.

Yes, I'm keeping things very simple, but using "comments" when needed to explain what is going on or options. First columns are inputs (Material, OD, wall thickness, column length), the next columns are the results (Design compression load, Design tensile load, design torsion load, design bending moment, weight of part), and there are a lot of columns to the right (available for reference/checking if desired) with the intermediate calculations and reference values for that case. So far I've needed : inside dia, area, density, weight per inch, Sy, I, J, E, k, L/k, L/ksub1, Euler buckling load, Johnson buckling load--there will be more.

The square tubes and (esp) the composites will require other math, I'm not sure it is worthwhile to try to make a master sheet that will work with anything. I'd probably keep it simple and use separate sheets for composites and maybe for the squares.

I'll keep chugging along, but don't wait on me. I make a lot of mistakes, but at least I'm slow . . .

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##### Well-Known Member
I've got dropdowns working for material selection and section : I put in all the AN standards from pops' post.
When someone posts equivalent aluminum sizes I can add them quite quickly, also different alloys etc.

I intend to add square tubes eventually, but I will need an engineer to define the equations

##### Well-Known Member
Progress so far : it's a bit short on comments, and needs more tube sizes
Excel 97 format

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#### wsimpso1

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Update initial post - the table has had a typo corrected and actual tube strengths were added.

Composite tubes in process..