Newbie Question: Composite I-Beam

General Composites Discussions Engineering Talk
#1

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 .1m1m, 2 plies of .2m1m = .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,
Clueless

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

#2

Yep. Almost all of the above.

  1. When calculating Area Moment of Inertia, use the correct values for b. Ixx=1.85105E-7
  2. I forgot to use the Section Modulus (Sx) equation ((bH^2)/6)-((ah^3/6H) =.00016575
  3. I actually did something right, but fed it the wrong numbers to do it.
    Case (A) deflection: 42.78 meters
    Case (B) deflection: 4.27E-08 meters

Thank you,
Clueless

#3

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?

#4

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,
Clueless

#5

I can understand why you would want that but the nature of composites makes it impossible to form the ability to make generalized predictions that would hold true for anything except a specific fabric and resin combo cured in specific controlled conditions. Even then, it would be rely on each batch of cf and resin being identical and I have not found this to be a certainty.

I bet you could get close within a margin of error if you fixed the part shape, the choice of fabric, the resin, the curing temp and curing time.

The problem can be illustrated fairly well by taking three pieces of identical cf and curing one of them as a flat sheet, one in a curved shape and the third one rolled into a tube. Even if you do everything else the same, there will be a massive difference in the properties you are interested in.

If you repeat the test using different resins, or different weaves and different curing cycles etc then you will also find meaningful differences.

My interest in this topic is a need for designing parts that are as light and stiff as possible. I experiment a lot to find ways of achieving this and improving my designs. I have seen first hand how much the properties can vary by altering the design of the layers between the laminates.

What I am saying is that it is easily possible to make two identical looking parts using the same materials and processes but give them drastically different properties using design features that are not visible in the finished part.

With something like steel or aluminum, it is easy to roughly predict the properties of a 1/4" x 6" x 24" plate of a specific alloy. This is not so with carbon fiber unless no effort is made to optimize the properties through design.

With all that being said, if you were aware of and understood all the variables, maybe it would be possible to predict the strength of a part using a complex 3D modelling program with a simulated environment. They exist for other materials.

If you do find a program that can accurately predict the strength of complex carbon fiber parts then let us know.