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Torsion in hollow rectangular steel tube

  1. Nov 4, 2009 #1
    A horizontal hollow rectangular mild steel tube 30mm x 50mm with 4mm wall thickness and 350mm long, which is part of a frame has 2 lengths of studding 300mm long , 180mm apart and equidistant from the ends, welded to the 50mm face of the tube.
    A total perpendicular load of 400Kgs is applied to the ends of the studding.
    By what amount the tube will twist?
    I am only interested in the twisting or torsion of the horizontal tube, as this is the weakest part of the frame. I also feel that it might simplify calculation if the load is considered at the midpoint of the tube.
    I have looked at various equations in Wikipedea but I am even more confused now than when I started to look at this problem. Is there a simple equation /s which I could use to calculate this?
     

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  3. Nov 4, 2009 #2

    nvn

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    franznad: Are you sure your studs are 300 mm long? They do not look that long. The location and direction of your applied loads is unclear. You need to draw the applied load vectors correctly, showing the location and direction that each applied load acts. Is your stud welded to only one face of the tube, without passing through the other face? Or is your stud passing through both 50 mm faces and bolted? And what is the diameter and thread size of your stud?
     
  4. Nov 5, 2009 #3
    This is a fairly straightforward problem in structural statics. You can construct a free body diagram of the tube and determine the torque applied to the rectangular section. You can then calculate the shear flow from T = 2Aq, where T is the torque, A is the cross sectional area, and q is the shear flow. Once you know the shear flow, you can calculate the shear stresses in the member and then use Bredt's formula to calculate the rate of twist.
     
  5. Nov 12, 2009 #4

    nvn

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    franznad: I understand a nut is welded to each 50 mm face of your tube, such that each stud passes through two nuts.

    Could you label the dimensions on the top rectangular tube of your frame? We do not know exactly where the 350 mm is measured from and to. I am currently assuming 350 mm is the distance from inner face to inner face of the two vertical tubes shown in your diagram. We actually need the exact distance from the stud to the vertical tube centre line, which I am currently assuming is L = [(350 + 2*25) - 180]/2 = 110 mm. Please correct me if wrong.

    It is also somewhat unclear whether the applied load on each stud is P = 1962 N, or P = 3924 N. The total load for two studs is 2*P. Strictly speaking, the text in post 1 says the applied load on each stud is P = 1962 N. Please correct me if wrong.

    Nonetheless, assuming no local distortion of the tube face occurs, the torsional rotation of the rectangular tube at your studs, in units of degrees, would be theta = [1.4059e-6 deg/(N*mm)]*P*L, where P = applied load per stud (N), and L = distance from stud to vertical tube centre line (mm). Therefore, if P = 1962 N and L = 110 mm, then theta = [1.4059e-6 deg/(N*mm)](1962 N)(110 mm) = 0.3034 deg. Therefore, because the tip of your stud is 300 mm from the rectangular tube longitudinal centre line, the vertical deflection at the tip of your studs due to a torsional rotation of theta = 0.3034 deg (not including bending of your studs themselves) would be y = -1.59 mm.
     
  6. Nov 24, 2009 #5
    Nvn-Thank you for your explanation and apologies for the delay in replying, but my modem has been down. I have put in another drawing which, I think you will find, is clearer. The total length of the tube is 350mm, so the distance from each stud to the centre of the verticals will be
    {[350—2*25]—180}/2 = 60mm. Each stud has a force of 1962 N. However, I am not certain of the equation you give for theta. Could you please explain what is 1.4059e-6 ?
     

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  7. Nov 27, 2009 #6

    nvn

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    franznad: Nice diagram. The 1.4059e-6 means 1.4059*10^-6, which means 0.000 001 405 9. If P = -1962 N and L = 60 mm, then theta = [0.000 001 405 9 deg/(N*mm)](-1962 N)(60 mm) = -0.1655 deg. Therefore, because the tip of your stud is 300 mm from the rectangular tube longitudinal centre line, the vertical deflection at the tip of your studs (not including bending of your studs themselves) would be y = -0.867 mm.
     
    Last edited: Nov 28, 2009
  8. Nov 27, 2009 #7
    Just consider a straight beam with a rectangular cross section with two torsion points. Eg. T=(1962N)*(0.3m) and fixed ends. Assuming the ends will not deflect as they are stiffened from the vertical beams and bolted (bracketed) to the wall.
    Now, you will have to check the method of your bolt attachment. Holes drilled straight through the beam with a nut on both sides?
    A single bolt and nut welded to one side is a buckling problem, not necessarily torsion. You will not have even shear flow on the full cross section. It will bend one side of your beam as your wall is thin (eg. 4 mm). Also, always good to check the bearing stress (eg. bearing stress = Load/bearing area = Load/(hole diameter*thickness)
     
  9. Dec 4, 2009 #8
    Nvn—Thanks very much for the result. I was apprehensive when designing this frame that the amount of distortion would have been greater than would have been acceptable when the frame was in use, but am relieved with the minimal result
    I follow the last part of the calculation of y but I am still puzzled by the first part and how you arrived at 0.1655 deg. I gathered that your notation, 1.4059*e-6 was 0.0000014059. I have come back to this problem many times but I can’t see as to how you got it.
     
  10. Dec 4, 2009 #9
    darkside 00—Can you please elaborate on T =[1962N]*[0.3m]
    There are nuts welded on both 50mm faces.
    I am interested in the bearing stress [I presume you mean on the nuts]. You say
    Bearing stress=load/bearing area=load/[hole dia*thickness]. Do you mean thickness of the nut or thickness of the tube?
     
  11. Dec 4, 2009 #10

    Mech_Engineer

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    To calculate the moment (torque) applied to the tube, you multiply the force (1962 N) by its lever arm (300mm -> 0.3m). It works out of 588.6 N*m each.
     
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