4130 to Aluminum

Discussion in 'General Experimental Aviation Questions' started by TJay, Jan 12, 2017.

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  1. Jan 13, 2017 #21

    gtae07

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    I designed welded structures for test rigs for a while, and we used aluminum for a couple of reasons (primarily internal company politics...). We assumed O (annealed) material properties for areas anywhere close to a weld.

    You'd want to do a lot of testing to get a good statistical base. But you'd probably just be better using the steel.
     
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  2. Jan 13, 2017 #22

    wsimpso1

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    Actually, it gets to T3 or T4 just sitting around, but not to T5 or T6. The profile for T6 requires some pretty specific times and high temps. If you do not heat treat it, you will have substantially lower strengths, and if you do heat treat, you can expect distortion.

    Billski
     
  3. Jan 13, 2017 #23

    wsimpso1

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    You do not so much swap between these two materials in a fuselage frame, as you design from the beginning. Buckling dominates in compression, and buckling is a function of L**2/EI. You could easily end up with much different tube diameters and different truss bay lengths in the search for min weight in each... Then, if you start with a known steel tube fuselage, you do not know which tubes are barely big enough and which are almost adequate one size smaller, which have the potential for lighter tubes when you do the design in a different material...
     
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  4. Jan 13, 2017 #24

    TJay

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    Well after all that, Think Ill just go with 4130, If I would have to step up to 1 inch aluminum my weight savings is so little its not worth the 30 hours of tig welding, Thanks till next time,
     
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  5. Jan 13, 2017 #25

    BBerson

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    It is incorrect to use T3, T4, T6, etc. to describe a weld made with 4043 rod.
    The T numbers are shop processes involving quenching and are specific to each heat treatable alloy. 4043 welding rod is not a heat treat treatable alloy, it can be work hardened, but it can only be made permanently soft with heat. It will not age harden.

    A large tube could be welded with gussets to minimize some of the weld affected zone.
     
  6. Jan 13, 2017 #26

    12notes

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    Whoops. Sorry, lazy typing, I meant 6061-T6 every time, I just didn't type it. It would be a hell of a welding process that changed the alloy composition.
     
  7. Jan 13, 2017 #27

    lr27

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    Not if you don't use the same alloy in the filler rod. From what I've read, in the case of 6061, it's a bad idea, which may explain why there don't seem to be any such rods on the market.
     
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  8. Jan 13, 2017 #28

    lr27

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    Probably the right decision, but if you could heat treat the aluminum after welding....
     
  9. Jan 15, 2017 #29

    proppastie

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    Bruhn C4
     
  10. Jan 15, 2017 #30

    lr27

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    4th power for the OD -4th power for the ID
     
  11. Jan 15, 2017 #31

    wsimpso1

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    LR27, Please be careful about what you think you know. Rigorously, Dana is right:

    Beam theory has us believing that plane sections of beams remain plane while the beam is bent, and that gives rise to I, which is the integral of dA*y^2 where dA is the area and y is the vertical distance from the neutral axis for the dA. Sum them up to get I, do it with integral calculus and your get the familiar formulae. This term works beautifully for our estimates of strength and stiffness in a HUGE range of structures.

    Yes, I for a solid circle works out to PI()/64*OD^4, and for a tube that is PI()/64*(OD^4-ID^4), but remember that the ID is correlated to the OD, and that changes things. In tubes, holding wall thickness constant (we do this in airplane design as well as in suspension bridge deck trusses and other things where strength to weight ratios matter) and just changing the OD, y's scale with OD, and area scales with OD, so in dA*y^2, OD shows up three times, not four. Your I then goes only with the cube. As a check, you can run calcs for I for diameters from something small, say 3/8", through something large, say 3", all with a constant wall, and then run linest() for fit of OD^3 and OD^4 against I. I did this myself. Took about five minutes. Fit is 100.00% on OD^3, not so good on OD^4, and we use the one that fits perfect if it is not too much trouble.

    I can send my little spreadsheet for those interested in seeing how to do this...

    This little analysis follows in other shapes too. Generally, similar beams (same shape and section thickness) become stiffer with cube of depth and stronger with square of depth, and both stiffness and strength growing with first power of width.

    So much high theory. Now if you remembered the formulae for tubes and used them in your design calcs, you would select tubes correctly anyway. But let's just remember that beam stiffness scales with the cube of depth, even in round tubes.

    Billski
     
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  12. Jan 15, 2017 #32

    proppastie

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    charts Bruhn
     

    Attached Files:

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  13. Jan 15, 2017 #33

    lr27

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    If you can get away with the same wall thickness and a larger OD, it just means the wall was too thick in the first place. ;-p

    Consider a 1 inch, .1 wall tube and a 3 inch, .1 wall tube. According to you, the latter should be 27 times as stiff. (pi/64)*(1^4-.8^4) =.0290 (pi/64)*(3^4-2.8^4)=.9589 .9589/.0290=33.07 not 27 Let's try it with 2.0 OD and 1.8 ID. It should be 8 times as stiff as the 1 inch OD with same wall, if it really goes with the cube. (pi/64)*(2^4-1.8^4) =.2701 .2701/.0290=9.3

    I think, as the diameter gets bigger and the wall stays the same, it approaches proportionality to the cube in the limit. Say, 1 AU* with .1" wall vs. 2 AU with .1" wall. But by then, if you put a speck of dust on "top", it will buckle immediately**, though it might take a long time before it buckles enough to notice. ;-) Or is there a constant relating to wall thickness somewhere in the curve fit? I tried to figure one out, but it didn't work.

    Or, of course, I could be COMPLETELY wrong. I'm a bit sleepy today.

    --------
    *Astronomical Unit, distance between Earth and Sun
    **in a Newtonian universe
     
  14. Jan 15, 2017 #34

    lr27

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    Thanks for the chart. It could be very useful sometime.
     
  15. Jan 16, 2017 #35

    proppastie

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    Last edited: Jan 16, 2017
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  16. Jan 20, 2017 #36

    Armilite

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    Or you can just make it Simple and just look at some of the 750+ Kitplanes and Ultralights made, and see what similar Types to what you want to build and see what they used. Most Planes only use maybe 4-5 different sizes of tubing on Average:

    3/4"-.058" (Dash Area)
    1.0" or 1.125"(80% of the Airframe) in .058" or .065"
    1.25" in .065"(Landing Gear)
    2.0" in .065"(Support Tubes/Trailing Edge Wing Spars)
    3.0" in .065"(Leading Edge Wing Spars).
     

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