I-Beam deflection to find E

[gigglin_horse]

I have a force vs displacement graph of an I-beam that is supported by 2 pin joints, and 2 equal forces on top causing it to bend.
I am required to find the Elastic modulus of the beam.
How do I do this?
Am I using the equation: E/R = σ/y = M/I, or turning the displacement into strain?
I also have no idea how to do this. Do I find the neutral axis and radius of curvature?
Can someone please help me?!
Thank you.

 

[WhoWee]

Let's start with what you DO understand...please take a closer look at your graph and post your findings.

 

[gigglin_horse]

Are you indicating a linear relationship?
Force/displacement for all data points doesn't give a constant relationship. Am I meant to average this?
Am I in the right direction??

 

[Studiot]

2 pin joints,

Are you sure it was 2?

 

[gigglin_horse]

There were 2 supporting the bottom, and 2 applying force at the top.

 

[Studiot]

If you have two pin joints you have indeterminate horizontal reactions.
Beams are normally supported with one end free to move ie on roller or sliding supports. Only one end is pinned.

 

[gigglin_horse]

Well, our experiment didn't.
How does this effect my calculations?

 

[Studiot]

How did you 'pin' an I beam?

 

[Studiot]

Can you draw a moment diagram for your beam using your measured loads?

You need to relate the area under the moment diagram to EIy where y is the measured deflection.

Have you heard about either the area-moment method or the double integration method?

 

[gigglin_horse]

This image is what our situation looked like:
http://www.physics.unc.edu/~aold/APPL150/MaterialsProject_files/image002.jpg

I have a moment diagram for the bending, which is also shown in the figure above.

The area-moment method, I do not know, but I'm assuming the double integration method is for find the BMD?

 

[Studiot]

Your beam is what is known as simply supported. It is not pinned.

You do not appear to have measured the deflection , at the centre of the beam. The central deflection is given by

$$EI\delta y = \frac{{Fa}}{{48}}\left( {3{L^2} - 4{a^2}} \right)$$

To calculate the deflection whre you are showing it, under one of the loads, requires one of the methods I have mentioned.
The double integration method calculates the equation of the 'elastic curve' which is basically a geometric equation describing the variation of the vertical position of the neutral axis along the beam , by doubly integrating the differential equation representing moment diagram.

You need to discuss these with your teacher if you need more than the central deflection.

 

[Studiot]

I make the expression for the deflection at the loads to be

$$EI\delta y = \frac{{F{a^2}{{\left( {L - a} \right)}^2}}}{{3L}}$$

 

[gigglin_horse]

Sorry of this seems stupid of me, but, we took several data points, recording the load and the deflection due to that load.
So, when the Force comes up in an equation, do I just take ANY load point and the corresponding deflection?

 

[gigglin_horse]

Ok, so now I have calculated M, I, y, and σ, but I still can't find E without R. Am I correct?
I'm not so good with integration to find R. Is there another way to find E?

 

[Studiot]

I cannot imagine any college would have its students carry out an experiment without studying the theoretical background.

Did you understand the formula I presented in posts 12 and 13?
If not you need to review your course material, perhaps you missed some part?

Where does my formula mention R?
Did you measure R?
Perhaps that was because you are looking in the wrong place for your relationship and do not actually need it.

Have you studied superposition in relation to beams?

 

[Studiot]

I don't know if this helps but if we take any beam and set up an x-y coordinate system, with y vertical and x starting at one end and running along the length of the beam.

Then at any section along the beam ie any particular x value

$$EI\frac{{{d^2}y}}{{d{x^2}}} = M$$

Integrate once

$$EI\frac{{dy}}{{dx}} = \int {Mdx + {C_1}}$$


Integrate a second time

$$EIy = \int {\int {Mdxdx} } + {C_1} + {C_2}$$

This is known as the double integration method.

The second equation yields the shear force (diagram) along the beam
The third equation yields the deflection along the beam

Now it is obvious that $$\int {Mdx}$$ is the area under the moment diagram so we can do this graphically by working on the moment diagram and measuring the area under it.

Engineers have also worked out standard loading conditions and superposition allows us to read these from tables and combine them.

 

[gigglin_horse]

Ahhh, ok. I think I have it now.
Thank you for all your help :)

 

[Studiot]

Hope your lab report goes well now and you can catch up fully on bending and beam theory.

 

[hotvette]

we took several data points, recording the load and the deflection due to that load. So, when the Force comes up in an equation, do I just take ANY load point and the corresponding deflection?

If the beam were always within it's linear elastic range, each pair of force / defection points should theoretically produce the same value for E. In reality it won't for several reasons: measurement error, imperfect linear elasticity, etc. You should calculate E based on each point and see how close the values are. You might then average the values.