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Old 10-05-2017, 09:31 AM   #1

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Question Newbie Question: Composite I-Beam

I'm trying to wrap my head around the simplest "real" composite structure I can think of, the bane of physic and engineering students the world over - The I-Beam. And am I completely befuddled. I can't tell if I made a math error, a formula error, or if I am trying to apply the rules of football to a baseball game.

Base material: A&P's QISO Light, .28mm thick/ply, 272gm/m^2, Long Tensile 800MPa, Modulus 47GPa, Transverse 800 MPa, 44 GPa. Resin ignorable.

100 mm wide, 100 mm tall, 1 meter long I-Beam. 2 layers of fabric (web and flange)
a=.56mm, b=100mm, H=100mm, h=98.88mm, L=1000mm.
Total fabric area: 2 plies of .1m*1m, 2 plies of .2m*1m = .6m^2, or 163.2 grams

Load conditions: Cantilevered beam, unsupported end load, (A) 1000N, (B)1N

Moment of Inertia: Ix = (ah^3/12) + (b/12)(H^3-h^3) = (.0056*.001)/12+(.00416*(.001-.00000097) = .001457 = Call it .0015m^4 for simplicity.

Load(A) = 1000N*1 meter = 1000Nm
Load(B) = 1N * 1 meter = 1Nm

Bending Moment (A) = My/I (using Ix as I)= (1000Nm*.05m)/.0015m^4 = 33333.3 Pa, or .033MPa

Bending Moment (B) = My/I (using Ix as I) = (1Nm*.05m)/.0015m^4=666.7Pa or .000667Mpa

And here is where I get stumped:

Deflection of a end-loaded cantilevered beam: FL^3/3EI
(Case A): (1000N * 1m )^3/3(47GPa*.0015m^4)= 4.2728 meters
(Case B): (1Nm)/3(47GPa*.0015m^4)= 4.72E-9 (uhh.. nano-meters?)

If my math is right, this beam fails catastrophically long before 1000N and completely ignores a load of 1N. Am I completely lost?

Thank you,

Edit: Or should I be using MPa instead of GPa for deflection?

Last edited by Clueless; 10-05-2017 at 09:40 AM. Reason: Clarification...
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Old 10-08-2017, 09:27 AM   #2

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I would forget about all those formulas. There are too many variables in the real world for them to be useful without a practical test.

Once you have completed a practical test, you will know for sure and there will be no more need for any confusion.

As an FYI, there are a few videos on YouTube with some guy testing a carbon fiber I-beam and showing it's failiure point.

Just remember that cf is different to metals. If five guys make a cf part of a similar shape with the same weight of cf, there will be a large variance in the strength and stiffness of each part.

I can't pretend to understand all those formulas but all of the charts and tables I have seen on the Internet stating the relative strength of cf to steel and aluminum have been inaccurate, meaningless, without context and lacking crucial details.

What is the actual aim on this exercise? Are you building something?
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Old 10-10-2017, 07:57 PM   #3

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Practical tests are great for giving "yes, this I-Beam is strong enough, and broke here at N Newtons - but I don't know why" answers.

A practical test that confirms the idea of "This I-Beam will break between 100mm and 115mm from the supported end when a point load of N Newtons is applied to the unsupported end" is a completely different result.

I want to get to the point where I understand the why of latter (predicting the behavior) - and not the how of the former, which is reacting to the behavior.

Thank you,

Last edited by Clueless; 10-10-2017 at 07:59 PM. Reason: Formatting
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