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Using a carbonfiber rectengular tube as a wing spar

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Lendo

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I'm researching Pultrusion Carbon Tow and found a reliable Carbon Supplier here in Australia (Gurit), They supply Hyosung H2550 Standard Modulus Tow that comes from Sth. Korea This Company makes their own Precurser through to Carbon Tow they have a massive facility (with massive investment) and have an International QA- their Carbon is on Par with the Japanese high quality Tow, from my perspective as I'm looking to use it in Spar Caps.
The US Carbon Rods are good but too expensive down here.

Stanislavz, There is an European manufacturer (can't remember the name) equally as good and as expensive, Exchange and transport costs are a killer for us here in Australia.

Geraldc, I researched those products and although very good their also High Modulus= High Stiffness= and more brittle. Their strength is complete Tension IMHO and we need some flex in Wings for obvious reasons.

Standard Modulus is the key to lower cost, not too stiff (lets face it Carbon is stiff enough) not too brittle, and importantly always at a better price.
George
 

Merlin

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You can buy two cf spars, build one wing, test it to failure, and if you're happy with the result, build the second wing exactly the same, and hope the second spar is similar enough to the first one :)
Thats a big "hope"
 

tr7v8

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Here is an example of wing shear and bending diagram for a strut braced wing. Example airplane is 1320 pounds, carrying 6 g, 360 Inch span, elliptical lift distribution, root mount is at BL24, while strut is placed at BL120.6" to minimize bending moment in spar set. Note the moment at the strut and at the worst spot between strut and root have about the same bending moment, but of opposite sense - 22261 in-lb. Strut is under 5888 pounds tension, and the spar set must carry 5358 pounds compression over its roughly 96" length.

While a cantilever wing will have about 10 times as much moment at its root than the max bending moment in this strut braced wing, it is not zero and must be designed for while also carrying the compression resulting from the strut loading.

Billski
Interesting Bill, inrested in the formula to calculate this, to knock up a spreadsheet with. Any chance of the basics please?
 

tr7v8

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Fantastic thanks, units I can understand as well! NO work getting done tomorrow as I play with this!
 

stanislavz

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Stanislavz, There is an European manufacturer (can't remember the name) equally as good and as expensive, Exchange and transport costs are a killer for us here in Australia.
I am buying it from aircraft building company, so do really know quality is ok, and they sell me a small amount + only 100 km of drive.

Did buy some rolls of ud cf to with job lot - but it is just not worth it.
 

stanislavz

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And PS. If you plan to use carbon fiber pultrusion - strutted wing will be only marginally lighter - ~ 130m for strut braced and 230m for cantilever of 3.1x3.1mm rods.. Or ~4 kg of spar for strut braced and 8kg for cantilevered. Stressed low at 150 000 spi / 10000 kg/cm2 for wood spar like deflection . 4.8g * 1.5 reserve. If stressed closer to ultimate , reserve should be higher.

High wing cantilvere is a bit complicated, but - you will get a bonus roll-cage around cockpit..
 

wsimpso1

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Interesting Bill, inrested in the formula to calculate this, to knock up a spreadsheet with. Any chance of the basics please?
Several analytical tools are needed to do it.

Lift distribution - I just used a spanwise elliptical for the diagram I showed. That is usually pretty darned good, but you have to do something to add in aileron effect before you are ready for the next step. Folks love the extension of it called Schrenck's Approximation. There is a whole set of arguments over which one to use...

Then we build up the shear diagram using numerical integration to sum loads outboard of any spot along the entire semi-span. This is rigorously correct for cantilever wings as is. We usually stop accumulating lift from the wing at the fuselage. Externally braced wings will have loads from the bracing "drop in" at their connection points, and so are handled by any of several approaches. I like the two pass method that I will go through below.

Then we build up the bending moment diagram again using numerical integration to sum moments starting at the tip and working inboard. This is rigorously correct for cantilever wings as is. Externally braced wings require another pass through the math, which follows.

Since the wing in question is externally braced, the root connection is usually a single pin or bolt connecting each spar to the fuselage. No moment can be reacted from the wing to the fuselage at the root. Instead, the entire wing moment is carried by the external brace.

Take the moment about the root connection point and divide the moment by the lateral distance from root connection to strut mounting. This is the vertical force dropped in at the strut connection.

Make a new shear diagram - the shear from tip to strut mount is taken from before. The shear diagram from the strut mount on the wing to the root connections is the original shear minus the vertical component of the strut load.

To get the new moment diagram, you repeat the numerical integration like you used above, but now numerically integrate the adjusted shear diagram. The moment curve then has a cusp at the strut mount and should be zero at the root mount. You can numerically integrate starting at the root end towards the strut mount, with moment at the root equal to zero - Moment at the strut mount should be same working from top to the mount and working root to the mount.

The external brace can not carry much moment at all, and is most appropriately analyzed as a pin connected joint. As such, it can only carry load along its length. We know the vertical component (calculated from wing moment) and its angle to the fuselage end mount is determined, then simple trig gives us the horizontal component of the load and the load along its length. The horizontal component of the brace load shows up as compression in a high wing under positive g. Add that to the spar's load case. Under negative g or bracing from above, the sense of these loads is reversed.

Where loads are tensile, a simple calculation will show many struts and even some spars appear to be over built. I urge thoroughness here - we are supposed to design airplanes for negative g flight as well. I have had loose navigational materials fly about the cabin on a couple occasions and am glad the airplane stayed together through them. Anyway, any time the flight loads put compression loads on structures, the designer must also consider buckling in its many forms (Euler, Johnson, crippling, and wrinkling being the major ones).

While I specified numerical integration to proceed from lift distribution to shear and them moment diagrams, this can also be done using integral calculus. If one enjoys integral calculus, knock yourself out. It does get kind of interesting when the lift is elliptically distributed, using Schrenk's approximation, and when adding in effects of a deflected aileron.

My reference for shear and bending analysis is Mechanics of Materials by Timoshenko and Gere, chapters 4-6, there are other texts that cover this just fine. Buckling using Euler and Johnson methods is in all versions of Shigley, while crippling/wrinkling/etc you go to Bruhn or other aircraft structures texts.

Billski
 
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wsimpso1

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Could you explain please?
The biggest issue is that unless you can maintain a high degree of fiber straightness from it being laid down through debulk and cure, carbon fiber structures will not achieve "book" strengths. Some researchers indicate that wet laminated carbon fiber will generally be on the order of half or less of "book" strengths - only modestly stronger than S-glass. To achieve the promise of "book" values, you need to work in prepreg materials, machine application, vacuum bagging, and autoclave cure cycles.

We homebuilders thus resort to other means. While we can build spar caps using unidirectional tape of E-glass and S-glass and open layup methods that make book strengths, the best we can do for weight savings seems to be pultruded carbon rod, with Marske Graphlite being most well known. The fiber volume fraction in Graphlite is as good as many prepregs, and the strength is close too. Fiber straightness is built in and rod alignment within the assembly is pretty well assured too.

Billski
 

wsimpso1

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I think i'm going to be using wood spar reinforced with unidirectional carbon fiber covered with fabric.
I will do one destructive test to make sure i'm within my safety limit.
It has been done. It will either be a wooden spar with some expensive graphite fiber decoration or a graphite spar with a heavy and expensive wooden core. We have talked about it quite a bit in the past... I suspect it will appear when using the Advanced Search tool.

Billski
 

ragflyer

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Billski has the details right to the general solution..... you can simplify though in many cases.
1. On lift distribution: For the kind of airplane 99% of folks are designing we can ignore Schreck or even elliptical distribution. The current UK CAA section s regs allows you to just assume the load distributions as proportional to the chord.In other words if the wing is not tapered then all you have is a constant load distributed evenly across the wing. Mind you schreck or elliptical is not super complicated, but you will find the above quick and conservative without significant weight penalty.

2. Assuming 1 and constant chord wing you can greatly simplify things by using the following to calculate the bending moment and reaction at the strut and root. Note, x is the fraction of the semispan that overhangs the strut attachment.

Vertical reaction at strut R1 = TotalLoad/2*(1+x/(1-x))

Vertical reaction at root R2 = TotalLoad/2*(1-x/(1-x))

Hence struct load is R1/SIN(strut angle with spar)

Spar Axial load is R1/TAN(strut angle with spar)

Bending moment at strut attach M1 = TotalLoad*semiSpan/2*x^2

Max Bending moment between root and strut attach M2 = TotalLoad*semiSpan/8*(1-x/(1-x))^2

The optimal location for strut attach is very close to 70% from root (x =0.293) where M1 = M2 = 0.0429*TotalLoad*Semispan^2

An equivalent cantilever spar would have max bending moment = 0.5 *TotalLoad*Semispan^2
Hence an optimal strut (assuming uniform load distribution as 1) will have a max bending moment that is 8.6% less than a cantilever spar.

3. for highly tapered wings the above may not be accurate but you could very easily build a spreadsheet. Start at the wing tip shear = BM = running load. Then you incrementally get close to root and the shear = shear of previous station + running load and BM is previous BM + current shear *station interval. For strutted wings at the appropriate station you subtract the strut reaction. Of course you would have to calculate R1 and R2. To get R1 assume cantilever and find max BM (at root) using the spreadsheet. Then R1 = BM/ l*(1-x) and R2 = TotalLoad - R1. Sounds complicated but takes a few mins to build and do in a spreadsheet. For constant chord wings you can just use the formulas above.
 
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stanislavz

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Or just buy Jim Marske book. But again - with pultrusion an normal wing aspect ratio cantilever is not big mass penalty. Especially for single piece wing with removable wingtips if necessery..
 

Lendo

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Ragflyer, The load from root to Strut is uniform but from Strut to Tip it's Elliptical.
George
 

ragflyer

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Ragflyer, The load from root to Strut is uniform but from Strut to Tip it's Elliptical.
George
Not sure I understand- comment or question?? The reality of the loading is of course not uniform across the whole span, not just after the attachment. My point was you could assume uniform load across the entire span. This would be conservative and simple and is endorsed by the current CAA section s regs for the class of aircraft most of us design and fly. The formula I included are based on this assumption.
 

wsimpso1

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The current UK CAA section s regs allows you to just assume the load distributions as proportional to the chord.In other words if the wing is not tapered then all you have is a constant load distributed evenly across the wing. Mind you schreck or elliptical is not super complicated, but you will find the above quick and conservative without significant weight penalty.
That all depends on what you think is a "significant weight penalty". I ran a quick estimate of spar weight needed using uniform and elliptical lift distributions with strut braced and cantilever wings. Strut braced wings using uniform lift distribution instead of elliptical drives overdesign of the main spar by about 12%. In cantilever wings, the same simplification drives an overdesign penalty of about 30%.

Is 12% or 30% "significant"? On a 30 pound set of spars for Buttercup, you might be able to save 3.5 pounds, and you might agree this is no big deal. My bird with a 2150 gross weight and a cantilever wing with a 100 pound spar set, and adding 30 pounds does seem significant. Maybe you are hard up against a weight target and saving every pound is important. And you get to decide of you need to do a downtown job or if the rough but conservative method is OK.

If you are writing an Excel spreadsheet to perform the numerical integration for the shear, bending moment, and deflection anyway, making the lift distribution pretty accurate seems to be a pretty small thing.

Billski
 
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mcrae0104

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As a point of reference for another material (wood), I'm coming up with a bit over 50 lb for 26' span, 1150 lb aerobatic weight using Schrenk and ANC-18. This is for single-piece design and does not include fittings and addl material when I convert it to a 3-pc spar.
 

ragflyer

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That all depends on what you think is a "significant weight penalty". I ran a quick estimate of spar weight needed using uniform and elliptical lift distributions with strut braced and cantilever wings. Strut braced wings using uniform lift distribution instead of elliptical drives overdesign of the main spar by about 12%. In cantilever wings, the same simplification drives and overdesign penalty of about 30 %.

Is 12% or 30% "significant"? On a 30 pound set of spars for Buttercup, you might be able to save 3.5 pounds, and you might agree this is no big deal. My bird with a 2150 gross weight and a cantilever wing has a 100 pound spar set, and adding 30 poundsi does seem significant. Maybe you are hard up against a weight target and saving every pound is important. And you get to decide of you need to do a downtown job or if the rough but conservative method is OK.

If you are writing an Excel spreadsheet to perform the numerical integration for the shear, bending moment, and deflection anyway, making the lift distribution pretty accurate seems to be a pretty small thing.

Billski
Sure Bill it is a judgement call but worth knowing so the designer can choose! As I said for most designs contemplated on the forum they are going to be closer in spec to a buttercup than your design and the designer is going to be have a more limited analytical skill set.

No doubt in some cases where weight is super critical it may be worth doing the more deeper analysis. But even for your design we are talking ~1.5% increase in weight (and no this will not snowball across all components) and will make no perceptible difference to performance.

Also keep in mind a 12% increase in bending moment does not automatically have to be a 12% increase in weight.
 

ragflyer

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As a point of reference for another material (wood), I'm coming up with a bit over 50 lb for 26' span, 1150 lb aerobatic weight using Schrenk and ANC-18. This is for single-piece design and does not include fittings and addl material when I convert it to a 3-pc spar.
I suspect using the simplified calculation you may add 5 pounds and get a slightly higher safety factor (~10%). Only the wood flanges would get slightly bigger, obviously would not change plywood shear webs and spacers.
 

wsimpso1

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I assumed that the cap area scaled with the bending moment and the web area scaled with the shear. I simply integrated the moment and shear along the span to get a figure proportional to spar weight, so yes, I got cap and web growth - the growth increases were close in both cap and web.

I do not get how this can not snowball. When we widen the flanges, we also get more adhesive to connect the flanges to the skin. If we hold cap width, the caps get thicker, driving cap centroids closer to the neutral axis, driving still more cap volume. Either way you get more mass outboard, increasing moment of inertia in roll and slowing roll response. Roll inertia will likely increase faster than wing stiffness does, lowering natural frequency in first flapping mode. Then if we are willing to add weight here that could be avoided by the simple task of substituting the equation of an ellipse for a constant, I would expect that analytical laziness would creep into other places, all of which does lengthen takeoff runs, reduce climb rates, drives bigger other structures, blah, blah, blah... In short, weight begets more weight.

Now if that is ok with the designer, hey, knock yourself out. I am just pointing out it is there.

Billski
 

ragflyer

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I do not get how this can not snowball. When we widen the flanges, we also get more adhesive to connect the flanges to the skin. If we hold cap width, the caps get thicker, driving cap centroids closer to the neutral axis, driving still more cap volume. Either way you get more mass outboard, increasing moment of inertia in roll and slowing roll response.
Billski
Ok with due respect this is getting ridicules.... you are telling me a bit of adhesive and flange width increase will critically affect roll rate for your average homebuilt design ...seriously?? It is like building to 6.6G rather than 6 G. Tons of airplanes (particularly composites) are (over) built to that degree.

Many many designs have been built using this assumption in the UK/World. Practical recreational homebuilt designs are not balanced/optimized on a knife edge. Even at the highly optimized airplane world you can see amazing increases in gross weights throughout its life span with relatively modest changes. Snowballing in more than a few cases is a good anecdote and it is a good way to shut a newbie from modifying a design.

Finally Schrenk's assumption is an approximation of reality. You could use VLM and get a still better approximation. Should you be doing that lest roll rate is affected?

Sometime what happens (in section s airplanes) is designers begin with this simplifications and during further refinement/load tests find they do not have enough margin (the designers are amateurs after all). All is not lost then as the regs allow you to use schreck to give you more margin.

It is all about engineering judgement. In some cases it makes sense to get a better approximation in others it does not. Experience shows in a typical homebuilt (CAA section S) the simplification may well be worth it. But then again choose what makes sense for your project.
 
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