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Wing loading of carbon fiber per...?


Arugela

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Is there a simple way to figure out how much wing loading of carbon fiber per sqaure foot you can get per inch/ foot of thickness. I was looking at 1ft+ of carbon fiber then trying to figure out weight and cost of that much carbon fiber.

How much wing loading would a cubic foot of carbon be able to handle? Could it potentially handle a literal ton of wing loading per sqaure foot? How much overhead to you need past the expected maximum wing loading as a safety margin?

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I don't think you can go by measurements alone as the layup/orientation of the plies is integral to the application. Unlike aluminum or other metals that are monolithic (properties like strength, ductability, brittleness, etc are uniform throughout), CFRP can be created with varying properties throughout by changing how you orient the carbon fibre.

Carbon fibre is only strong along one axis, so for example if you cut two squares of it, rotated one 90* and laid it on the first and bonded them, it would be strong in two axes sort of similar to metal. Where it gets complex is you can place the plies at any angle, add in some CFD to find exactly where you need strength, and it quickly becomes much more akin to 3D printing than metallurgy.

Take all that with a grain of salt since I have not worked on CFRP in a while. I have definitely forgotten some details, and the technology must have come a long way since I used it.

Edited by Meecrob
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Looking at wikipedia, a 787 has 377 m^2 of wing area and a MTOM of 254,692 kg, making the wing loading 675 kg/m^2 in level flight. Add in whatever the margin is (usually something like 1.3 g for maneuvers plus some safety factor that I don't know). Anyway, at least 1000 kg/m^2.

Probably there are military airplanes with CFRP wings and much higher maximum wing loading during high-g maneuvers.

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You're not going to have just a solid block of carbon fiber. You're going to have structure, and that makes it complicated.

There are simple models you can use to get into the ballpark. For a tube with a given diameter and wall thickness, it's relatively straight forward to estimate how much bending load that tube can take. There are standard approaches used in construction that you can look up. If you have a simple, hollow wing with no significant reinforcement ridges and a fairly uniform cross-section, you can extend that approach to a non-circular cross-section get pretty close to the real value. For a model plane, I'd call it close enough, and simply expand the safety margin a bit to absorb the error. This might also be good enough if you're just trying to get a rough feasibility/cost estimate.

If you want to build an actual airliner, you're just going to have to load your design into a finite element model and run it and see what sort of stress you get. There are no shortcuts.

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1 hour ago, mikegarrison said:

Looking at wikipedia, a 787 has 377 m^2 of wing area and a MTOM of 254,692 kg, making the wing loading 675 kg/m^2 in level flight. Add in whatever the margin is (usually something like 1.3 g for maneuvers plus some safety factor that I don't know). Anyway, at least 1000 kg/m^2.

Probably there are military airplanes with CFRP wings and much higher maximum wing loading during high-g maneuvers.

Pretty sure the 40% margin is from an g load who will ground the plane for an strip down level inspection, not from operational limit, 1.3 g + 40% is 1.8 g. 
Fighter jets are rated for around 8 g  I think, but they probably don't have the 40% safely limit but they also have ejection seats. 
Its very high stung vehicles like race cars. Pretty sure most pilots from major powers ejected during training and exercise not combat after the Vietnam war. 

Edited by magnemoe
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