cantilevered tube failre 1

[sreek333]

I have a cantilevered ANSI 4130 steel tube 1.0” dia X .095” thk, with 10” overhang. I applied load and it failed at 1350 lbs. The failure is localized buckling not pure bending. I did the same test on .75” dia X .250” thk 4130 steel tube with same 10" long. It sustained up to 1800 lbs and the failure is pure bending. If you compare the suction modulus (Z), 1.0” dia tube has more than the .75” tube, then what causes the failure. Is there any specific formula to determine the minimum thickness for pure bending failure than buckling? I don’t want unpredictable localized buckling failure in my structure. Thanks for the help in advance.

 

[JAE]

Are your diameters that of the inside dia. or the outside dia?

 

[sreek333]

Both are Outside Diameters.

 

[JAE]

How was the tube cantilevered? Did it sit on a ledge such that there would be non-uniform bearing across the bottom of the tube?

 

[JAE]

...also - what is the yield of the material?

 

[sreek333]

Actuallt it was in a linear bearing.
You can asuume a 15" long tube left end locked with rigid clamps for 2" and a linear ball bearing in a ridid housing for 1.5". The distance between clamps and bearing is 1.5" so that the remaing overhang length is 10". Load applied at the right side extreem end.

 

[sreek333]

yield is 180 to 200 KSI

 

[JAE]

For 1" tube
Z = 0.0781 in^3
d/t = 1 / .095 = 10.53
S = 0.0559 in^3

For 3/4" tube
Z = 0.0677 in^3
d/t = .75 / .25 = 3
S = 0.0409 in^3

Per AISC HSS specification

[λ]p = 0.0714 x E / Fy
= 0.0714 x 29000 / 200 = 10.35

[λ]r = 0.309 x E / Fy
= 0.309 x 29000 / 200 = 44.81

The 1" tube is non-compact. The 3/4" tube is compact

So for the 1" tube:
[φ]Mn = 0.9 x ((.0207)(E)/(d/t) / Fy) + 1)FyS
= 12.93 in-kips

For the 3/4" tube:
[φ]Mn = 0.9 x Fy x Z
= 12.18 in-kips

This shows that the 1" should be a bit greater...but I would think that the concentrated force from the bearings on the wall of the pipe is what initiated a local buckling, and a comparison of the moment capacities doesn't reflect the failure mode.

AISC also has section 8.1 of the HSS Specification that deals with this - and one of the variables is b1, which is the thickness of the "plate" which places a concentrated load on the tube in a line perpendicular to the axis of the tube. The equations are a bit more involved...I'd suggest reviewing that section to see how they compare.

 

[sreek333]

Thanks JAE. I will go through AISC section 8.1.

 

[JAE]

I forgot to mention that the b1, being the plate thickness applying the load (at the ball bearing support in your case) - would be small if the bearings connect to the tube at a single point (i.e. a round bearing touching a straight tube wall at a single point). So using the bearings may be the problem.