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Formula for deflection of 6061 T6 hollow tube, please help.

  1. Feb 6, 2012 #1
    Hello all, new member and amateur engineer from Tucson, AZ. I am working on a homebuilt, part 103 legal, ultralight project. I am trying to find the formula for deflection of an aluminum tube. I need to know how much weight will cause permanent deformation of a tube suspended between two points. I have the material properties of 6061 T6 aluminum, but none of the formulas seem to take into account the hollow section of the tube and wall thickness. I know a simple formula exists, I just cant find it.
    This is what I know about 6061 T6;

    Ultimate Tensile Strength= 310 MPa, 45000 psi
    Tensile Yield Strength= 276 MPa, 40000 psi
    Modulus of Elasticity= 68.9 GPa, 10000 ksi
    Ultimate Bearing Strength= 607 MPa 88000 psi
    Bearing Yield Strength= 386 MPa 56000 psi
    Poisson's Ratio= 0.33
    Shear Modulus= 26 GPa 3770 ksi
    Shear Strength= 207 MPa 30000 psi

    source;http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA6061t6
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2012 #2
    There are various equations for deflection you can use. All simple, but they vary depending on the loading of the member.

    You need the Ixx, Youngs Modulus, Load and Effective Length for all of them.

    You can compare the deflection from the equations to the maximum allowable and that tells you if you are within specification.

    I'd recommend for rough estimates if you aren't sure what you're doing, to use one of the many available calculators online: http://easycalculation.com/mechanical/deflection-round-tube-beams.php

    A general equation that may be suitable is shown at the bottom of that page.
     
  4. Feb 6, 2012 #3

    AlephZero

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    I think the answer depends very much on how the tube is fixed at the ends. You may get permanent deformation that makes the tube go out of round, before you permanently "bend" it along its length.

    The simplistic way to analyse this (ignoring the above warning!) is to find the bending moment corresponding to an the maximum allowable stress (i.e. use [itex]\sigma = My/I[/itex]) and then look at the shear-force and bending-moment diagrams to see what load will produce that max bending moment. But I certainly wouldn't want to fly in anything designed like that :eek:
     
  5. Feb 7, 2012 #4
    Greatly appreciate the info, I think I'm getting this. I'm trying to determine size and diameter of the wing tubing. It has to withstand +6 and -4 g's as a safety factor, with the operating limits of the airplane being +4 and -2 g's. The aircraft fully loaded will weigh 480lbs, so I have two aluminum tubes running the entire 28' span, supported by struts(top and bottom) at the mid point and stainless steel tension wires at the ends, which leaves eight 7' lengths that must support a total weight of 2880 lbs(480lbs at 6 g's). That is 360lbs per 7', unsupported length. The beam deflection calculator doesn't specify type of aluminum or temper. It also doesn't specify if it is uniform load or not.
     
  6. Feb 7, 2012 #5

    AlephZero

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    Many beam structures (iincluding yours if I understood it correctly) are statically determinate, so the forces and stresses are indepedent of the material.

    Obviously the amount of bending depends on the flexibility of the material, but your OP was about permanent deformation, i.e. exceeding the elastic stress limit.

    If you use that criterion to size the structure, the second step in the design is to check if the spar is stiff enough not to bend too much.

    But I don't know how you would set a criterion for an acceptable amount of bending - presumably an over-flexible spar could mess up the aerodynamics and/or reduce the effectiveness of control input forces, but you need to convert that idea into a number somehow.
     
  7. Feb 7, 2012 #6
    I believe that the "bearing yield strength" is the pressure at which permanent deformation occurs(the beam does not return to it's original shape)and the "ultimate bearing strength" is the pressure at which failure will occur. The 7' segment of tube which is suspended on both ends by strut and tension cable, and strut and aluminum billet block in the middle of the wing. The bearing yield strength is 56000psi for 6061 T6, so I think this means that any force under 56000psi will deflect the tube, but it will return to its shape, or does this mean the tube will not bend until it reaches 56000psi? The scary thing is that most people that build ultralights don't run these numbers!
     
  8. Feb 7, 2012 #7
    I don't need exact numbers, just numbers that are well under the limits of what the material can withstand, efficiency is not a factor in an ultralight aircraft. I just have never worked with aluminum, and I don't have a working knowledge of its strength. This aircraft will have a ballistic parachute.
     
  9. Nov 1, 2013 #8
    Copy someone who has already built it.

    The number one thing you guys are all over looking is the fatigue strength.
    Google S-N curve.
    You will not be building your craft to last forever. Aluminum does not have an infinite fatigue life....even a minute repetitive stress will cause it to fail in the long run 10^8 deflections. Aluminum is not like steel, it does not have an endurance limit at 10^6 cycles.

    You have to look at the fatigue life you want.

    1 cycle....100 cycles....10,000 cycles etc.
     
  10. Jul 27, 2014 #9
    Dear metal249, i am disturbed by your incompatible units of kips and psi. Dimensional analysis will convert and normalize these units!!! FORMER A.S.C.E. MEMBER
     
    Last edited: Jul 27, 2014
  11. Jul 27, 2014 #10
    1 ksi=1,000 psi
     
    Last edited: Jul 27, 2014
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