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How to deal with non uniform cross section bending?

by Vr6Fidelity
Tags: bending, cross, deal, uniform
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Vr6Fidelity
#1
Apr19-09, 06:24 PM
P: 11
How to deal with non uniform cross section bending?

How does one approach a solution for bending of a shaft having multiple diameters.

I have one particular shaft in mind that has 11 diameters to get to the midpoint.

I know the deflection at the midpoint, and I would like to calculate the force required to cause that deflection.

Since the load is a point load, The moment diagram is a nice simple triangle. I have calculated I for all sections, I just need to know how to approach the solution.

Some of the 11 sections are tapered.

Any words from the wise? Ill my equations in my bag of tricks are for uniform cross section.
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ericgrau
#2
Apr20-09, 12:38 PM
P: 20
I'm too lazy to derive it at the moment. But if you learn the generic beam equation you can derive it for any cross section, even one that was continuously changing:
http://en.wikipedia.org/wiki/Beam_equation

It'll help you conceptually at the very least, even if you don't need to do the calculus to solve it.
Vr6Fidelity
#3
Apr20-09, 01:00 PM
P: 11
Quote Quote by ericgrau View Post
words....
Ok I worked on this till 5 am last night. here is what I came up with, let me know if i have gone terribly wrong somewhere.

The shaft has variable sections but only one point load in the center (imbalance force).

Therefore the moment diagram is a rather simple triangle.

Now here is where it gets interesting:

If i break every section of shaft down to individual little shafts, I can apply the fraction of the total moment present at that section. Basically I non dimensionalized the moment from zero to one.

So then I use this moment to find the bending deflection in terms of THETA in radians.

Where Theta = (ML)/(2EI)

That is the deflection of one end. Total angular deflection is 2 Theta per section.

Since there is no axial load, the length of the sections remains constant so the deflection if you were to keep one end from moving, would be:

2 Theta * (PI/180) to go to degrees

then deflection = Sin(degrees)*L where L is the section length (hypotenuse)

So if you are following me I now have the deflection for each section, and it seems valid.

I then Sum all the deflections to get the total deflection.

I can now run this program in excel and guess values of moment to achieve the known deflection. It seems to work extremely well.

I can then take that moment I just figured out, and find the bending stress anywhere.

Seem valid to you?

ericgrau
#4
Apr29-09, 11:13 AM
P: 20
How to deal with non uniform cross section bending?

Seems like it should be right, though I haven't double checked it thoroughly. Just remember that the beam equation is only valid for small deflections. For bending stress I assume you're using S = Mc/I.
Vr6Fidelity
#5
Apr29-09, 04:11 PM
P: 11
Yeah, It is right. I already submitted the paper and presented it.


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