View Full Version : Torsion in hollow rectangular steel tube
franznad
Nov4-09, 12:49 PM
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?
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?
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.
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.
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