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^2Di^2)
I= PI()/64*(Do^4Di^4)
J = PI()/32*(Do^4Di^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 13/8 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
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^2Di^2)
I= PI()/64*(Do^4Di^4)
J = PI()/32*(Do^4Di^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 13/8 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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