# Composite wing spar design: Shear web/spar cap interface

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#### Autodidact

##### Well-Known Member
I think what Tom is trying to say is that the cross section of the rectangle increases faster than the surface area, i.e., you can double the cross-sectional area (and so the tensile strength) while only increasing the surface area by 1.5 times so the rod becomes stronger but its resistance to "pull out" reduces comparatively. I think you can also flatten out a rectangle to the point where it becomes a flat plane with no cross sectional area so that it has surface area but no tensile strength. I think what Tom is saying is that you want to "balance" tensile strength with pull-out strength as much as possible.

#### mz-

##### Well-Known Member
Yes, that the area increases faster than circumference is trivial for all shapes.

#### Birdmanzak

##### Well-Known Member
Your tensile strength will not dictate any cross-sectional area of X. The rods are of fixed dimension. You use what you can buy. For more tensile strength you use more rods. But the tensile strength and shear strength to hold a rod are directly related by geometry (given material strengths). For both a square and round cross-section of diameter/side length D, the tensile/shear area ratios are D/4 for a one unit length object. See the math below and you will find this to be true. For a rectangle, the numerator grows faster than the denominator; a disadvantage for fixing the rod because you will need more length in adhesive.

Rod: (Pi*(D/2)**2)/Pi*D = D/4 Square: D*D/4D=D/4

That square root stuff is higher math. LOL

Blue skies,

Tom

The thing is, if you've got a circular rod of diameter D which is strong enough to take the load, and you change it for a square rod of side length D, the square rod will also be strong enough (27% percent stronger than the circular rod), AND it will have 27% MORE surface area available to react the SAME shear load.

#### Tom Nalevanko

##### Well-Known Member
Yup; but they don't make any Graphlite square rod and you would be only adding weight if the circular rod did the job. And I don't like all the stress concentrations on the corners. So even if they made it, I would not use it. I just used the example of the square and round cross section to show that their proportion of tensile to shear areas always remained constant at D/4 and that the situation was less advantageous for a rectangular shape. By introducing the square I thought that explaining it would be simpler.

#### Tom Nalevanko

##### Well-Known Member
"Yes, that the area increases faster than circumference is trivial for all shapes."

You are somewhat missing the point. The ratio of the areas stay proportionate; irrespective of dimension. Talking about something increasing faster has nothing to do with the argument. For a circle and a square, the ratio stays constant at D/4.

#### Tom Nalevanko

##### Well-Known Member
Perhaps we should all agree to disagree or whatever and move on to building great aircraft? Thanks for contributing to the discussion.

#### mz-

##### Well-Known Member
Okay, I said that wrong, you can't compare an area and a length anyway, but it still doesn't make sense to me. I'll read all this stuff again later.

#### Autodidact

##### Well-Known Member
This is a little bit difficult to visualize, so I drew a picture, but I'm still not quite sure what's going on here. If you have a rectangle with one variable side and the other side equal to 1, then as the variable side increases from 0 to infinity, the Area/Perimeter ratio converges to 1/2, from an initial value of zero. The pic below shows what happens with a square and a rectangle of the same cross-sectional area, i.e., if the cross section stays the same, the Area/Perimeter ratio appears to decrease. If the area is allowed to grow, the ratio increases:

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#### ultralajt

##### Well-Known Member
Uff, you guys went deep into details...

My five cents here:

As the compression and tensile strength of the carbon pultruded rods used for the spar caps are amazingly high, the cross section of those spar caps became insane small. So, if I design a such spar, I use in concern also the fact that thin spar cap should be keept perfectly in line with help of the web, ribs and wing skin. If spar cap can "drift" from straight line it will fail instantly. So using a bit larger caps cross section and support against buckling by properly spaced ribs, properly dimensioned web and possible stiff skin, will resulted in a strong spar, without headache about delamination of cap from web etc... The second concern is then just a proper transfer of the loads into a wing hardware (wing, strut, fuselage, landing gear..connections).

When in doubt, I would d rather use already proven design approach, than try to invent something that were not yet used on spar design.

Thats why I really like to study different aeroplane designs, and already bought couple homebuild plans, and download some of them from net. Also reading of design books (I recomend Alex Strojnik books) helps a lot to step out from the darkness.. :nervous:

Mitja

#### Tom Nalevanko

##### Well-Known Member
Strojnik's books are good but a bit dated. And Alex is no longer around to update them.

Practically speaking with the rods, strength will be usually be adequate while deflection may be excessive for many structures. So you will add rods to meet your deflection goals. But they are so light, that this is usually not a big deal, compared to the alternatives.

Blue skies,

Tom

#### Lucrum

##### Well-Known Member
Log Member
So what (I think) I've learned from the most recent exchanges on this thread.
Is that when using the Graphlite round rods I need to add 27.37% more rods than if the rods were rectangular, to make up for the void between the circular shape of the rods. For example a 10 x 10 matrix of 0.063" rods would measure 0.63" x 0.63". If the rods were flat sided then the cross section area of the graphlite rod material would be (0.63in * 0.63 in) 0.3969 in^2. But if using round rods the combined cross sectional area of the rods in a 10 x 10 rod matrix would only be 0.3116 in^2. (R^2*pi * 100 rods) Even though outer dimension of the block of rods itself measures the same as the 10 x 10 matrix of flat sided rods. (for simplicity I assumed square rods)

And I'm not going to achieve the published strength numbers. Although I was thinking, apparently erroneously, that one of graphlite rods strengths was much less scatter of finished structures strengths because of the pultrusion process and the resulting alignment of carbon fibers. As opposed to normal wet layups of fabric in someones garage.

#### Autodidact

##### Well-Known Member
OK, back to the original topic, i.e., how to deal with an understrength shear web. I did another experiment, this time using maths instead of grabbing some stuff out of the garage and stretching it this way and that.
I drew a beam section, cantilevered it out from a wall 12", applied a concentrated vertical up load, and then increased it until the outer surfaces of the caps were at yield for the material, in this case 1144 steel @ 100,000 psi, and 58% of that for shear strength (fig."A"sub1). Then I used the vertical force (Fsubv) to calculate normal (comp/tensn) and shear at the middle (point "a") and at the top (point "b") of the shear web, see fig."B"sub1. The maximum shear at the middle ("a") is the calculated shear at that point because there is no normal stress there. So I concentrated on the top of the web (point "b") where there is significant normal stress as well as shear stress and these combine to create a greater amount of normal and shear stress at various angles to the beam axis. These stresses can be found with a graphical device called Mohr's circle (fig."C"sub1). In Mohr's circle, the calculated normal stress is laid along the horizontal axis (not the x-axis of the beam, though) and the shear stress is measured, to the same scale, from the normal stress endpoint vertically up for positive or down for negative. The center point of the normal stress line is found and then a circle drawn with that point as its center and the endpoint of the shear stress on the circumference. Then, a diameter is drawn through the shear stress endpoint and the midpoint of the normal stress. The maximum normal stress is on the horizontal axis from the origin ("0") to point "u" on the circumference; here there is no shear stress. The maximum shear stress is equal to the radius of Mohr's circle.
The maximum normal and shear stress at point "b" in the shear web, (fig."C"sub1) is seen to be less than the yield of the 1144 steel. But now I replace the web with a fictional material called "staluminite" (could have called it RCS, for really crappy steel) which has the weight by volume of steel, but the strength charactaristics of 6061-t6 aluminum, i.e., 40,000 psi tensile and 58% (or so) of that for shear. Now my web is severely understrength. I tried three ways to bring it back up to strength: increase the size of the caps, increase the size of the web, or, as per Billski's admonition, increase both the web and the caps in some balanced manner.

1) In fig."D"sub1, I tried increasing the size of the caps but the shear gradually went up and I could not strengthen the web this way (duh!).

2) In fig."E"sub1, I enlarged the web and was able to bring it back up to strength, but not without making the web actually wider than the caps, possibly because I chose a material with 40% of the strength of the caps. Makes for an odd looking "I" beam.

3) I increased both the web and the caps and was able to bring the beam back up to strength and still have a reasonable looking beam section. Also, the cross-sectional area is about half that of the one in fig."E"sub1, and so half the weight. I also found the angle at which the maximum stresses would act; not so important for steel and aluminum, but would be of interest when using FRP.

So, I have tried to replicate the "trend" of having to increase web and caps when the web is understrength as when using orthotropic materials such as FRP composites, but by using isotropic material so as to avoid doing all of that matrix algebra with pencil and paper. Did I succeed? I think I may have, but I thought I had it after the last experiment, too! Hopefully, Billski will be kind enough to evaluate this one too .

Edit: Normal stress is designated "sigma (lower case Greek symbol)", and shear stress is designated "tau (lower case Greek symbol)". "I" is moment of inertia of the beam cross-sectional area, "t" is thickness of the beam at the point in question, i.e., "t" = 1/2" in the shear web, or 4" in the caps, and "V", here, is equal to the vertical force "Fsubv" since it is the only vertical force to the left of the beams cross section at the root, and the "root" cross sectionis where we're calculating these forces to apply.
Also, in fig."A"sub1, in the equation for Fsubv = 56,075.8 (bottom right hand corner), the numerator of 100,000/1.7833 is supposed to have five zeros instead of four.

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#### Autodidact

##### Well-Known Member
Since the above post contains some of the basics of beam theory, I thought I would post some of the formulas - mainly the beam cross section moment of inertia:

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#### wsimpso1

##### Super Moderator
Staff member
Log Member
Autodidact,

Well, I did not check your numbers, but your method of analysis is right on. The big thing that gets missed so often is the total stress state in the web at the caps, and you captured it.

It all works very nicely with metal, and you can solve it in closed form.

In most airplane wings, the bending moments are much larger relative to the shear loads. For instance, in my bird, design shear load at the fuselage wall worked out to 6574 pounds, and bending moment is 401,678 in-lb. With the shear so small compared to the bending moment, you could end up with thin little shear webs and beefy caps. Hmmm.

This method could be applied using composites, using tensile and compressive strengths of unidirectional material for the caps and shear strength of composites at +/-45 fo the shear web, but when you get to Mohr's circle to figure out what is going on in the shear web out near the caps, well, it doesn't exactly work that way. The failure criteria is not so neat in composites. In metals, it is a nice symmetric oval centered on the origin. In composites, it is a pretty flat ellipse that is really big in the tension and compression in the directions of the fibers, really small across the fibers in tension, somewhat bigger in compression across the fibers, and the ellipse is not centered on the origin. To handle this, the failure criteria is written in strain space, so you have to calculate the beam strains and then the lamina strains and then do the transforms for any fibers not at zero degrees (matrix math, sorry) and then check the failure criteria.

Then, you would find (as Autodidact did) that the shear webs are not beefy enough, and go about the process of adding some shear web thickness and adding some spar cap thickness to satisfy that you won't fail anything, and then play with less or more of one, and how much of the other is needed to both satisfy the failure checks and result in minimum wieght.

The basic method is right, but you can get surprised on failures if you skip the full up analysis.

Billski

#### Autodidact

##### Well-Known Member
I very much appreciate your help, Billski. And no, I wasn't trying to find a way of not doing the matrix stuff, I know I will have to come to terms with it sooner or later. I just wanted to get my head around the basic idea. As can be seen from my original post, my intuitive approach was not enough to get "the picture" and it was only when I dragged myself over to the table and did some math that I was able to finally visualize more or less what was going on here. Even this simplified approach was very tedious and frustrating to do [the iteration process] on paper, thank goodness for spread sheets!:lick:

Edit: Geez, this is kind of embarrassing. I'm coming off like I'm writing some kind of tutorial which is absurd since I don't actually know anything. Billski is the tutor here.

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#### wsimpso1

##### Super Moderator
Staff member
Log Member
Autodidact,

Well, you captured the entire beam problem, and it does seem that you got all of the math, including Mohr's Circle. Good Job! If you can do this part, the extension to composites is doable. And you can do metal now. Don't worry about it sounding like a tutorial - This is all good because forums are read by lots of folks and now somebody else can learn from your writings.

Look up distortion energy failure criteria in Shigley, and think about the max stress in the x and y directions being the same. Von Mises stress expresses the failure envelope of metals. In unidirectional composites, the same type of failure diagram gets really flat and extends extra distance to the lower left (compression in fiber direction and across fiber direction). And then with fibers in other directions, well, matrix math helps you to keep track of all that.

One issue that you did not run across was that the wing also has pitching moment applied along the wing, so you get torsion in addition to bending and shear. The spreadsheets I built allow me to include the wing skins and do shear, bending and torsion together. This matters to the spar only in that everything sees the twist deformation, including the spar, even though torsion is primarily carried by the skins and then the forward and aft spars near the attachments. I taper the contribution of skin stiffness to zero as I approach the attachments - there is only the spars at the attachments. In composite structures with cored skins, the skin contributes quite a bit to bending. In metal, not so much, as the elastic stability comes into play.

Anyway, you have the basics, and can build on that!

Billski

#### Starman

##### Well-Known Member
Are those carbon rods any good for spars even though they are only four feet long?

#### bmcj

##### Well-Known Member
HBA Supporter
I think I read that they will custom cut longer lengths, but I could not find the details for it. I wonder if they are manufactured to the high standard / high fiber linearity that we expect for aircraft use.