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Cantelever Steel Tubing Beam

  1. Sep 13, 2011 #1
    Cantilever Steel Tubing Beam

    1. The problem statement, all variables and given/known data
    I am trying to design on paper a cantelever beam made of steel rectangle tubing. One 60" vertical section will be bolted to the concrete while a 2nd section 40" long will be welded to the top hanging over horizontally. At the end of the horizontal piece there is a cable holding 2000 lbs. I need to figure out what size steel rectangle tubing to use and the wall thickness for this load. I would also assume that using the longer side of the rectangle as the height would make it stronger, correct?

    2. Relevant equations
    I'm using E=2.9*10^7 lbf/in^2 for elasticity of steel
    I=[(b_o*h_o^3) - (b_i*h_i)]/12 where o is outside dimension and i is inside dimension for rectangle tubing
    To find the deflection of the horizontal member, I would use the equation y(x)=[-w/(6*E*I)]*(3*L*x^2-x^3)
    I don't know for sure where to go from here or where to even start with the vertical member.

    3. The attempt at a solution
    I assume you would break this problem into 2 parts, 1 for the horizontal and 1 for the vertical but without knowing the size of the tubing, you can't really find I and even if I substitute some values for I and come up with a horizontal beam deflection, what does that tell me or help me to find an adequate size tube. Is it true that an acceptable deflection is the beam's length divided by 250? So for my situation, 40"/250=0.16".
     

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    Last edited: Sep 13, 2011
  2. jcsd
  3. Sep 13, 2011 #2
    Determine an allowable stress for your horizontal beam. Start with 60% of the yield stress. Then use the formula for peak stress in a cantelever beam point loaded at the end. This will determine your wall thickness.

    You can take credit for your gusset: stress = M c / I. And I will include the square tube and gusset.
     
  4. Sep 13, 2011 #3
    Thank you 'edgepflow'. How do I determine an allowable stress for the horizontal beam? Something hasn't clicked yet for me. Also do you know the formula for peak stress in a cantilever beam point loaded at the end?
    For my gusset, Stress = M c / I, where c is the distance from the neutral axis to the extreme fiber, I is the moment of inertia of the beam cross section, and M is Moment? So I also need to calculate 'I' for the gusset whether its a piece of sheet metal or another section of rectangular tubing? I typed up a spreadsheet to calculate the deflection for different size tubes...
     
  5. Sep 13, 2011 #4
    Start with the yield stress of your material. For example, consider A36 structural steel. The yield strength is 36000 psi = 36 ksi. Take 60% of this for starters, so an allowable stress = 0.6 X 36 ksi = 21.6 ksi. Repeat this for your material.

    The max stress for a cantilever beam point loaded at the end is (see Machinery Handbook or many other sources):

    stress-MAX = W * L * FI / Z

    W = load at end of beam = 2000 lbf
    L = length of beam = 40 inches
    Z = Section Modulus = I / c
    FI = impact factor = 1.5

    I include the impact factor, FI, to account for the fact that when the load is first connected, it will accelerate / decelerate and momentarly weigh more than 2000 lb.

    You can calculate I for your gusset from the parallel axis theorem.

    BTW, your 1" deflection for every 250" beam length guideline is fine.
     
  6. Sep 14, 2011 #5
    Thanks again. How come sometimes I (moment of inertia) has mass and sometimes it doesn't?? Or am I doing something wrong? My units for I using I=[(b_o*h_o^3) - (b_i*h_i)]/12 where o is outside dimension and i is inside dimension for rectangular tubing is (in^4)...
    Also c is the distance from the centerline of the horizontal beam to the edge? So if I used a 3"x6" tube, longest side upright, my 'c' would equal 3"? and I would equal [(3" * 6"^3) - (2.5" * 5.5"^3)]/12 = 10.43 in^4 for 3" x 6" rectangular tubing with .25" wall thickness.
    My moment, M = P * L = 2000 lbf * 40 in = 80,000 lbf-in
    Using the Stress_max = W * L * FI / Z = 2000 lbf * 40 in * 1.5 / (10.43 in^4 / 3 in) = 34,506 lbf/in^2 = 34,506 psi
    In using the parallel axis theorem for the gusset: I_z = I_cm + m * r^2 or I_z = I_x + A * r^2
    I guess I'm pretty confused on if I need the mass moment of inertia or the area moment of inertia.
     
  7. Sep 14, 2011 #6
    The mass moment of inertia is used in rotation problems and is the same idea as mass in Newton's 2nd law: F = ma.

    For these strenght of materials calculations, your moment of inertia will have units of length^4.
     
  8. Sep 15, 2011 #7
    Remember to also compute your percentage of allowable stress:

    % Allowable Stress = Calculated Stress / Allowable Stress.

    Of course, this number must be less than 1.
     
  9. Sep 16, 2011 #8
    Ok thanks, so the 60% of the yield stress that we initially chose gives me my allowable stress of 2.6 ksi, why did you pick 60%? By doing the calculations the smallest rectangular tubing that would work is 6" x 4" by .375" thick...that seems plenty big enough. Since I'm realistically only needing to lift 1000 pounds, I said 2000 just to factor in some safety. Would you still use the 1.5 for impact factor in the Stress-Max formula? Since my allowable stress is 21.6 ksi and my calculated stress is 19.7 ksi. This does not include anything with the gusset. How do I factor in the vertical beam from here?
     
  10. Sep 18, 2011 #9
    The 60% of yield stress is a typical number used for members in tension. See AISC Steel Construction Manual for this value.

    If you already doubled your actual load, you do not need an additional impact factor.
     
  11. Sep 19, 2011 #10
    That allowable stress was supposed to be 21.6 ksi, typo...thanks. Do you have any help for me to factor in the vertical member or any more help on the gusset?
     
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