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
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
bending torsion instability of steel pipe 4
- Thread starter robyengIT
- Start date
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.
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.
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?
canwesteng
Structural
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.
Retrograde
Structural
Hollow sections can undergo lateral torsional buckling.
- Thread starter
- #12
- Thread starter
- #13
-
4
- #15
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.
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?
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:
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.
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.
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.
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
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
- Status
Similar threads
- Question
- Locked
- Question