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bending torsion instability of steel pipe 4

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robyengIT

Mechanical
Dec 20, 2013
894
Eurocode says that a steel pipe item (part of a structure) is not affected by a bending+torsion instability (i.e. when the torsion stiffness is bigger than the bending stiffness ??). I am looking of literature discussing and explaining such a statement. Thanks
 
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I'm not familiar with the Eurocode, but I believe the question you're asking is can a pipe be subjected to lateral-torsional or flexural-torsional buckling, right? The answer to those is no. Consider that the radius of gyration is the same in all directions.
 
Bmart is correct. A cruciform (X-Shape) is most susceptible to torsional instability under pure compression because it will twist under compression only. A pipe will not twist under compression, but it may buckle laterally. This concept also applies to lateral-torsional buckling.
 
I don't understand. Can someone explain me why a section with the same radius of gyration in each direction is not subjected to lateral buckling?
 
Good question.

A pipe is geometrically stable. For instance, a I-beam under load will deflect downward until reaching a certain point (buckling limit), the tension flange will then rotate sideway and introduce twist, thus instability results. The pipe does not possess this behavior, when stretched to strength limit, it continues deflect downward and simply fail in flexural bending.
 
Yes, I know, I agree. But, is it possible to explain it with math? I studied the buckling behavior of thin walled beams with open cross section, but not of beams with closed cross section. Is it possible to prove analitically that lateral buckling doesn't occur in pipe?
 
I'm not aware there is formulation for this phenomenon. I think it is more or less by observation.
 
I'm sure you could do it analytically. Add some infestimal rotational displacement, and see if that makes the section any less stable. It wouldn't because the beam is a perfect circle. But you see with a wide flange section, there will be a tendency for the compression flange to want to keep rotating.
 
Hollow sections can undergo lateral torsional buckling.


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Thank you retrograde, but that page report information about rectangular hollow section, not about circolar hollow sections. I'd like to find the formulation for circolar hollow sections
 
get the link of Retrograde and read the doc
... and attached file too
 
Billie93 said:
I don't understand. Can someone explain me why a section with the same radius of gyration in each direction is not subjected to lateral buckling?
We are talking about TORSION lateral Buckling and not lateral buckling
of course a pipe can do lateral buckling...but due to high torsional stiffness no torsional lateral buckling

 
I'm sure there is some good literature out there, and I recommend you continue to seek it out, but the easiest way to sort through this is to consider the behavior of beams in a general sense. And yes, klaus is correct that it's important to distinquish between LTB and what you learn in undergraduate as Euler (column) buckling. This conversation relates to LTB, not column buckling. Column buckling applies to all unbraced shapes of significant length.

Consider a rectangular beam bent about its MAJOR (strong) axis. As you load the beam, it will reach a point where the beam becomes unstable and wants to return to a lower energy state. In this case, the lower state is bending about the MINOR axis. That is why the instability results in a lateral bend about the minor axis. Now consider the same beam bent about its MINOR axis. In this case, no matter how much you stress the beam, it will never buckle by bending about the MAJOR axis. The reason for this is that the beam is already in its lower energy state (minor axis bending) and will not suddenly jump to its higher energy state (major axis bending). Note that rx and ry (radii of gyration for major and minor axes) are not equal... this is a numerical representative of the higher and lower energy states alluded to.

Now consider a circular cross section beam (hollow or solid). There is no strong axis or weak axis because the radius of gyration is the same no matter which way you take it. As you begin to bend the beam, it will not try to tip or buckle about another axis because the moment of inertia is the same in every direction. It is always in the lowest energy state.

Hopefully this helps illustrate why one situation applies while the other doesn't.
 
robiengIT
Oh, I missed the attached file, I'll read it as soon as I can

Klaus
Yes, of course I know the difference, I've studied the problem of lateral torsional buckling during university (in the case of open cross section). But we called "lateral torsional buckling" as "laterale buckling"

BMart006
Good explanation, thank you! But it is possible to find it out by math?
 
Billie93 said:
But it is possible to find it out by math?

We all know that a sphere ball on a dome is at a state of "limited stable", we can prove it becomes unstable mathematically, when it moves a little to the right or left, but is there a mathematics proof when it sits perfectly in the middle? The only expressions I can think of are ΣH = 0 and ΣV = 0, thus there is not possible for the ball to move, and the proof stands.

I think the explanation below is quite simple and straight forward.

Lateral Torsional Buckling occurs in unrestrained beams. A beam is unrestrained when its compression flange is free to displace laterally and rotate. When I sections are used as beams or beam columns the compression flange is under compressive stress and has a tendency to buckle but it is attached to the tension flange which resists the buckling giving rise to torsion within the beam section. This torsion twists and warps the unrestrained part of beam leading to lateral torsional buckling.

Will a circular shape behave the same - deflected out of shape that given raise to torsion?



 
Retired13
The ball on a dune is in a point of unstable equilibrium. The equilibrium is stable only in the points where the total potential energy is minimum. Why? If a body (in a minimum point of total potential energy) change his position, the kinetic energy must decrease (because the total energy, in the case of conservative forces, doesn't change): the kinetic energy decrease until the velocity becomes 0, then the velocity becomes negative and the body turn back in his original position. This is stability.
There is always a mathematical explanation.
 
Billie93,

You explained well. I wonder why you couldn't get over the LTB phenomenon, as BMart006 explained well too, mathematically speaking. You can follow suite to develop theory for circular pipe too, I believe.
 
This is my point. In literature, I find pretty easily the analytical description of lateral torsional buckling for open cross section subjected to constant flexural moment: the buckling moment (in a section without warping, as a pipe) is proportional to the square root of the moment of inertia in the weak direction mulyiplied by the polar moment of inertia; is this formulation valid even for the buckling moment of a pipe (which is, of course, a closed section and not an open cross section)?

Imagine a pipe subjected to constant moment Mx. The beam will deflect in y direction.
I understand (I hope) that, if you have the same radius of inertia in each direction, when something perturbs equilibrium and makes the section rotate along his longitudinal axis, nothing change, because you don't give any second order moment along a weak direction.
BUT, if something perturbs the equilibrium and makes the section deflect in x direction, it will generate a second order torsional moment. If it is big enough, the beam could buckle. Am I wrong? I know that the torsional rigidity of a pipe is very big, but I would like to understand if in strange structures (with effective length of 60 meters, for example) this phenomena could happen or not.

Thank you for your attention :)
 
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