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Beam curvature from deformed condition

Beam curvature from deformed condition

Beam curvature from deformed condition

(OP)
Hi, I apologize form my english, which is not the best.
I need to find geometrically the curvature of a beam of which I know exaclty the geometry in the deformed condition (I know the coordinates of every point of the beam). I've tried few methods but the results differ one from another, so I would like to know of you would find the curvature. I've attached a file showing the deformed beam and in yellow you can see all the points of which i know the coordinates.
thanks,

Andrea

RE: Beam curvature from deformed condition

If you know the coordinates of every point on the beam, doesn't that give you the "curvature"? Maybe I am a little rusty on a mathematical definition of curvature, but to me it simply implies a curved member. You can calculate the slope between adjacent points and define the curvature in terms of linear segments. What exactly are you trying to do?

RE: Beam curvature from deformed condition

(OP)
The curvature is defined by 1/R, where R is the radius of the circle that approximate the curved line in one point of a curved line. I guess I could find a curvature value for every 3 cosecutive points of the beam (there is only one circle that can contain 3 points),but in theory the results should all converge to a similar value. So, I want to know the lenght of the radius R that defines the arc that resemble my curved beam.

RE: Beam curvature from deformed condition

There are standard formulas for the 'end slope' of beams (Google it). It looks as though your example is simply supported with a UDL, therefore the end slope at each support will be equal and opposite. If you know the end slope and beam length, you can work out what the radius of the relevant circle will be.

I'm not sure how to do this mathematically - I would sketch the beam span in AutoCAD, draw in the end slopes (at the correct angle), then draw tangents to the end slopes to where they meet 'above' the mid point of the beam. That would give you the radius of the circle that approximates the curvature in the beam.

RE: Beam curvature from deformed condition

As RandomTaskkk noted, AutoCAD would be one way to do it. Just draw a line through the center points of the dots in the radial direction and do it a couple of times. Then extend the lines until they all meet together; that should be the center of it. Then draw a circle from the center point to the outermost dots. That should give you any property you want.

If you don't have access to a CAD package then get out a mathematical formula book; it should be in there somewhere.

RE: Beam curvature from deformed condition

isn't the curvature related to the deflected shape by differentiating the equation for deflection twice with respect to x along the member. You of course need the deflection equation to do this.

If v is the deflection, then curvature is d^2v/dx^2.

RE: Beam curvature from deformed condition

Ah, yes, of course. I just quickly assumed we were talking about an arc of a circle.

RE: Beam curvature from deformed condition

Alternatively, if you take the fundamental beam equation 1/R = M/EI = -d^2v/dx^2 and you have the moment diagram you can work it out directly?

It's pretty simple to derive the moment at a distance 'x' along a beam using statics.

RE: Beam curvature from deformed condition

If you have the deflection at designated points but do not know the loading and do not have an equation for the deflection curve then the slope, curvature, moment, shear and load can be found approximately by reversing Newmark's Numerical Procedure.

BA

RE: Beam curvature from deformed condition

(OP)
I don't have any moment diagram or equation because the bending is induced by a non uniform temperature gradient. Thats why I need to do it geometrically. I think I've manged to solve the problem with the geometry and i've set up the excel sheet to automate the process for more cases, but i guess i can use autocad to check if my method is right. Thank you all for your help.
Andrea

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