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-   -   Euler-Bernoulli Beam (http://www.physicsforums.com/showthread.php?t=669827)

aerowenn Feb6-13 11:33 AM

Euler-Bernoulli Beam
 
I'm trying to model a Euler-Bernoulli beam to gather the total angular torque it will provide on a hub on which it is anchored. The beam is a cantilever, and I'm using the standard deflection equations which represent behavior to an applied force on the tip.

You can solve for the force needed to get a specific deflection by using the delta max function with a known displacement. This, however, isn't my question.

The Euler-Bernoulli equation to show the shape of this beam (equation shown below) only takes into account the change along the y-axis. This is because the equation is made for rectangular coordinates. What I want is to also show the change in the tip position relative to the mounting location (it should stay roughly the same, and eventually get shorter with large deflections). I should add that I need to do this for the entire equation, not just the tip. The end goal is to model the torque provided by the beam moving under its own power (piezoelectric) so I need to model the relative positions of all infinitely small positions along the beam.

http://sphotos-a.xx.fbcdn.net/hphoto...12093515_n.jpg


The current equation only changes y, and thus makes the beam longer to account for the deflection. I've included a screenshot of a plot to illustrate this:

http://sphotos-a.xx.fbcdn.net/hphoto...83014146_n.jpg

Is anyone aware of a modified form of this equation that will account for this? I want to say it's as easy as changing from rectangular to polar coordinates, but I don't think so. Since the beam actually develops an arc shape, this would change depending on where the position being observed on the beam correct?

Any input would be greatly appreciated!

SteamKing Feb6-13 02:10 PM

Re: Euler-Bernoulli Beam
 
It's not clear what effect you are trying to account for. Remember, when a beam undergoes bending, part of the beam is in tension, and part is in compression. As a consequence of Hooke's Law, the fibers in tension will lengthen, while those in compression are shortened. The neutral plane of the beam has no bending stress, and thus has the same length as before the bending occurred.

aerowenn Feb6-13 02:29 PM

Re: Euler-Bernoulli Beam
 
Quote:

Quote by SteamKing (Post 4259719)
It's not clear what effect you are trying to account for. Remember, when a beam undergoes bending, part of the beam is in tension, and part is in compression. As a consequence of Hooke's Law, the fibers in tension will lengthen, while those in compression are shortened. The neutral plane of the beam has no bending stress, and thus has the same length as before the bending occurred.

Thanks! What I'm trying to do is develop an equation that will give me the torque the beam exerts on object to which it is mounted in the cantilever setup.

It's easy enough to make this a time dependent function by making P a sinusoid, and thus, the acceleration isn't hard to get after that. This function will then give me the acceleration at any position along the beam for a given time t. From there I need inertia, which could be modeled as a point mass in orbit around a central location (in this case the mounting point of the beam to the rigid structure).

My problem is that this equation gives the deflection in rectangular coordinates, I need to convert to polar so that I can eventually arrive at torque provided by each infinitely small portion of the beam. You're saying that Hooke's law keeps the beam the same length, I can see and agree with that, but will the individual infinitely small portions of the beam not change their distance from the mounting position if the beam deflects and remains the same length? If the beam curls, it seems like the tip would get closer to the mounting position.

If they don't, I suppose at an instance in time the relative motion of the individual particles of the beam could still be modeled in rectangular coordinates, since the subsequent torque provided by each is due to the tangential motion of that infinitely small portion (along the length axis since I keep mentioning it).

End game is to get the torque provided by the beam on it's mounting position, if you could imagine an axis coming out of the page where beam meets wall/rigid body.

aerowenn Feb6-13 02:41 PM

Re: Euler-Bernoulli Beam
 
To extend on my point. At neutral, if you take position x along the beam, the distance of x from the mounting location is simply x.

However, if you take x at full deflection, according to the equation I posted above, it now has a position (x, y) or (x, w) in the case of that equation. Since x is the same, the new distance is going to be the magnitude of those, sqrt(x^2 + w^2). Therefore it is now further from the mounting position.

AlephZero Feb6-13 04:05 PM

Re: Euler-Bernoulli Beam
 
Quote:

Quote by SteamKing (Post 4259719)
The neutral plane of the beam has no bending stress, and thus has the same length as before the bending occurred.

Yes, it has the same length as before, but it is now curved. Therefore the x coordinate of the tip is not zero (using the OP's defintion of x and y).

This effect is negligible unless the beam deforms with large rotations, but it is not neglibible if you take a piece of spring steel and bend it into a complete circle! (The strains remain small everywhere and the material is within its elastic limit, so the basic ideas of beam theory still apply).

The "classical" (latin) name for this problem was the "elastica", but Google willl probably find more about a rock band with the same name!

I don't think there are many closed form solutions to this, but this site http://www.ucl.ac.uk/~ucesgvd/ referenced in http://en.wikipedia.org/wiki/Elastica_Theory looks interesting.

aerowenn Feb6-13 08:40 PM

Re: Euler-Bernoulli Beam
 
Quote:

Quote by AlephZero (Post 4259879)
Yes, it has the same length as before, but it is now curved. Therefore the x coordinate of the tip is not zero (using the OP's defintion of x and y).

This effect is negligible unless the beam deforms with large rotations, but it is not neglibible if you take a piece of spring steel and bend it into a complete circle! (The strains remain small everywhere and the material is within its elastic limit, so the basic ideas of beam theory still apply).

The "classical" (latin) name for this problem was the "elastica", but Google willl probably find more about a rock band with the same name!

I don't think there are many closed form solutions to this, but this site http://www.ucl.ac.uk/~ucesgvd/ referenced in http://en.wikipedia.org/wiki/Elastica_Theory looks interesting.

Good deal, thanks! There doesn't seem to be a lot on this problem, I think because it may not have many practical uses. I'm looking into the feasibility of an attitude control system for small satellites with these piezoelectric beams.

But yes, the scenario you described with spring steel is what I was talking about, bend it enough, and it's distance will change significantly. I think for my purposes I can neglect as mentioned. I'm looking at a 40mm beam length and a tip deflection of 1mm in one direction, so 2mm during oscillation (in theory, this will be greater once a steady oscillation is reached, similar to pushing a swing).

It looks like I can just assume motion tangent to the mounting position and calculate torque using inertia and etc.

Thanks for the post!

AlephZero Feb6-13 09:02 PM

Re: Euler-Bernoulli Beam
 
Quote:

Quote by aerowenn (Post 4260199)
I'm looking at a 40mm beam length and a tip deflection of 1mm in one direction, so 2mm during oscillation (in theory, this will be greater once a steady oscillation is reached, similar to pushing a swing).

It looks like I can just assume motion tangent to the mounting position and calculate torque using inertia and etc.

With those numbers, ignoring it will probably be smaller than other unknowns, unless you are trying to do an insanely accurate experiment.

You can get a rough approximation to the size of the effect by assuming the beam is hinged at the "fixed" end, and stays straight. The axial movement inwards is then ##L(1 - \cos \theta)## where ##\sin \theta = y/L## at the tip. For small angles, that gives an axial movement of the order of ##y^2/(2L)##.

For your beam that is about 0.01mm compared with the length of 40mm. That's an "error" of about 0.025%. I bet you don't know the material properties of the beam to that accuracy!


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