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Max load bending on steel tube.

  1. Jan 5, 2010 #1
    I'm trying to calculate the max load of a rack made of 3X3X1/4 steel tube. It is 8ft long and 4ft wide. I just want to calculate the max load (distributed load) in bending cross the 8ft and also in compression for the legs. i can't find the right formulas. can anyone help?

    so far i what i think to be correct

    MOI (I)=3.495 in^4
    Section modulus=2.330 in^3
    MOE (E)=29,000,000 psi ( i think this is correct but i'm not sure. just want to use low grade steel)

    I've looked though many sources but end up with

    stress=6M / (bh^2)

    σ - the bending stress
    M - the moment about the neutral axis
    b - the width of the section being analyzed
    h - the depth of the section being analyzed

    Not sure what M is here?


    σ = (yqL^2) / (8I)


    σ = maximum stress (Pa (N/m2), N/mm2, psi)
    y = Perpendicular distance from to neutral axis (m, mm, in)
    q = uniform load (N/m, N/mm, lb/in)
    L = length of beam (m, mm, in)
    I = moment of Inertia (m4, mm4, in4)

    so what i don't have is "q" I want to know how much "q" i can have. right?
  2. jcsd
  3. Jan 5, 2010 #2


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    It really helps to see a print or diagram, but if you want to estimate this, you can use basic beam formulas. The one you would probably be looking for is the very first one:

    http://www.structsource.com/analysis/types/beam.htm [Broken]

    M is the bending moment at a particular location along the beam.
    Last edited by a moderator: May 4, 2017
  4. Jan 5, 2010 #3
    i'm not sure what to use for x in your equations. I tried x=48" (first equations = 0 in this case) and tried x=0" but that still left me with 4 unknowns and 3 equations. i didn't have V, M, w, or d(deflection)

    Honestly i've been looking at this for a long time. giving my more equations to work with isn't going to help much. i've given the information that i have to work with. can someone help me apply it in the correct formula and get an answer?

    Length is: 8ft (96in)
    made of: 3"x3"x1/4" square tube made of cheap structural steel
    E=29,000,000 psi
    I=3.495 in^4
    supported on both ends fixed (welded)
    I need to know the max allowable load.

    how do i use the numbers given to find the max allowable load of this tube? (please help me solve this problem not just give me an equation)
  5. Jan 5, 2010 #4


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    Are you familiar with beam bending? That may be a source of the problem here. X is the distance along the beam. You also have to understand what your boundary conditions are. At x=0 and X=48 you will not have any deflection. W is the load per unit length. M is the moment.

    You'll need to give us a diagram of some kind. I think that is the only way to avoid confusion. Also, please state the purpose of this. I will not do engineering calculations for something like this if there is even a remote chance this could injure someone or damage property.
  6. Jan 5, 2010 #5


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    drewjohn: Your second formula in post 1 is correct. Use it to solve for q. Your listed numbers are correct. Let sigma in your second formula equal the steel tube material tensile yield strength (Sty), divided by factor of safety (FS), such as sigma = Sty/FS = (275 MPa)/FS. Now solve for q, which will be the maximum allowable uniformly distributed load on each square tube in the rack top.
  7. Jan 6, 2010 #6
    (emphasis mine)

    No. It's a tube, not a beam. That was a great link you provided, Fred, but wrong geometry.

    That checks, as well as the rest of your comment.
  8. Jan 6, 2010 #7


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    I was not aware that:
    \sigma = \frac{My}{I}
    didn't work for sections with holes in them. Has something changed since I passed Machine Design I?
  9. Jan 6, 2010 #8

    where did the 275 mpa come from? searched for a chart but didn't find one. Is this where the beam goes from elastic to plastic bending? I don't think i need to incorporate the safety factor in this calculation because i'm not planning on getting any were near the max "q" I just need to know what it is.
  10. Jan 6, 2010 #9


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    It's correct, but I doubt anything has changed since you machine design class; the formula is derived from the neutral axis and the moment-curvature relationship for a beam. According to my mechanics of materials text book (Gere, James M. (2004). Mechanics of Materials, 6th Edition. Toronto, ON: Thompson Learning, Inc.) the relationship is called the flexure formula, which calculates the bending stresses or flexural stresses.

    [tex]M[/tex] is the bending moment at a position 'x' in the beam, [tex]y[/tex] is the distance from the neutral axis, and [tex]I[/tex] is the second moment of area (a.k.a. the MOI).

    As to the fact that it doesn't work for beams with holes in them, it's because a hole is a discontinuity in the beam's cross-section. You would have to either use a stress concentration factor for the hole, or try to derive an equivalent Second Moment of Area for the beam if there are equally spaced holes along its length.
  11. Jan 6, 2010 #10

    1) I have not done calculations like these in years, but i'm still vaguely familiar.
    2) it's a rack that has been used for 20yrs and someone decided to do some calculations on it. It is actually much beefier than i have stated. it has angle welded inside the tubes. I'm also doing the calculations for one tube and not both. I'm just trying to get a rough estimate for now not final numbers for that we will get someone certified.

    Attached Files:

    • rack.pdf
      File size:
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  12. Jan 6, 2010 #11


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    drewjohn: That is just my estimate of the tensile yield strength. Yes, tensile yield strength is where the beam goes from elastic to plastic bending.
    Completely eliminate this concept from your thinking. The only strength value you should have in your mind is allowable bending stress, Sb, where Sb = Sty/FS. Your bending stress cannot exceed the allowable bending stress, Sb. Racks generally see some dynamic activity, so FS should probably be at least 2.0. Therefore, e.g., Sb = 275/2.0 = 137 MPa. Let sigma in your second formula in post 1 equal Sb, then solve for q, which will be the maximum allowable uniformly-distributed load on each square tube in the rack top (unless the stress level on the tube weldment is higher than the above stress level at the tube midspan).
  13. Jan 6, 2010 #12
    sorry. i didn't mean it like that. I will be using a safety factor. it's just i believe it to be already there. built in by never even coming close to the max load. i don't mean that i don't need one.

    here is what i have

    33000(psi)=((1.5in)(q)(96in)^2))/((8)(3.495)) = 66.74lb/in

    66.74*96in = 6407lbs - total distributed weight on one tube

    times 2 tubes = 12800lbs - total distributed weight on two tubes

    do these numbers look right?
  14. Jan 6, 2010 #13


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    I was referring to the "hole" as the inner part of the tube. It was a silly, semi-sarcastic post in response to mugaliens proposing that tubes are not beams.
  15. Jan 6, 2010 #14


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    We totally need a sarcasm smiley or something like that. If it helps minger, I always add a bit of sillyness or sarcasm to your posts.

    As for the OP, I think you are fine with the bending stresses in the beam. The only thing to also check on in the beam is the shear flow in the section of the beam. I'll take a look at the beam tonight .
  16. Jan 7, 2010 #15


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    drewjohn: Do use the factor of safety. For this type of structure and analysis, it is generally not allowed in any design code to omit the factor of safety. In other words, it is required that you use Sb in your calculation, not Sty, to compute the allowable load.
    Last edited: Jan 7, 2010
  17. Jan 7, 2010 #16
    I just wonder how you can make calculations, drawings with all those foots, pounds.
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