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

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#### orion

##### Well-Known Member
Two quick comments here: One needs to be very careful in designing with the Graphlite rods since it is tempting to use the design/strength values as published. The problem with this though is that those properties are as tested in a laboratory setting. Now don't get me wrong, Graphlite rods are great but if you're really expecting over 300ksi in strength you might be surprised.

First off, many folks tend to use the round rods rather than the rectangular ones, but then calculate their spar as a solid rather than a matrix with a bunch of voids. This of course reduces the effective working area and so is not a proper representation of the structure. Furthermore, there is a big difference often times between the lab specimen and the behavior of the high strength elements in actual application, especially in compression. And that of course might be further exaserbatd when discussing structures built in a typical "homebuilt" environment. For that reason it is very advisable to be substantially more conservative in the design of these type of multi-element spars in order to account for all the less than ideal variables.

Regarding the use of Nomex honeycomb in homebuilts, the lack of application is generally due tot he material's characteristics. Structurally it's excellent however incorporating it in the structure in a sound and efficient way pretty much requires the use of prepregs and oven cures. Folks that have tried to use it within the homebuilt field using wet layup techniques were often disappointed with the result. The most common problems are an insufficient bond and an overweight structure (the latter from having the resin fill up the honeycomb voids).

#### bmcj

##### Well-Known Member
HBA Supporter
Orion, what fudge (derate) factor would you suggest using for the Graphlite rods? Would a 2:1 ratio (similar to that used for composite structures) be good?

#### orion

##### Well-Known Member
I was afraid someone would ask that: I really don't have a firm, documented number but in the past I've used 180ksi for the compression side and 200ksi for the tensile (both would then represent the ultimate strength value for the material - limit value would of course be half that). Yes, it's conservative but the Gaphlite products are so light and efficient the conservativeness doesn't really cost all that much in the final application.

#### Autodidact

##### Well-Known Member
Aircraft Spruce & Specialty also sells this for $100 to$150 per 100' roll and claim that the typical small aircraft requires approx. 1,000' of it (seems about right, actually). I've never tried to figure up material costs for an aircraft, but \$1500 for spar caps seems a lot. That's a buck-buck fifty a foot; maybe it can be had for less. At first, I thought this material was sort of a fad but it looks like a good solution strengthwise - if used correctly. I like the rectangular section rods but most availability seems to be for the round rods. I suppose you could put much smaller rods into the voids between the bigger ones, maybe?

#### Tom Nalevanko

##### Well-Known Member
The rectangular cross-section rods have their problems. They tend to twist for long sections. I have been told by people in the know that Rutan evaluated them a long time ago and rejected them. A circular cross-section gives the greatest linear surface area per volume. This is advantageous as this surface area is where they are loaded via shear and adhesive. So you need less length with the circulars to have the same pullout force.

The best work done with Graphlite rods was done by Bell Helicopter. You can find some good papers on its use if you look around...

#### mz-

##### Well-Known Member
Thanks for the tips, though a cylinder has less surface area per volume than a box.

#### Michealvalentinsmith

##### Well-Known Member
If I recall you want a little resin as possible in any composite and the function of the resin is simply to tie the fibers together to prevent slide of one fiber on another. (If you bend a bunch of twigs you see what I mean)

Seems having as many fibers run in the same direction as possible is the best - so unidirectional CF has better bending and compression strength - but shear and torsional strength suffers as you only have the resin itself to resist forces.

So if you're using graphlite rod in a resin matrix - all you've done really is beef up the size of your fibers - so your still kind of doing a hand lay up with resin and graphlite instead of matt. So you still want as little resin as possible, and to me that would be the rectangular section over the round.

As for adhesion to the the rods apparently the graphlite is made without any oils in pultrusion so it needs no preparation at all for a good bond - so I don't see that being a major issue with the box section.

If you've ever done wet hand layups you'll see the advantages of graphlite right away. Pre preg is OK but I doubt many home builder want to shell out for an autoclave. I know some guys get away with heat lamps etc but I wouldn't want my wing coming off on a hot day.

#### Tom Nalevanko

##### Well-Known Member
My initial statement was about rectangular rods, and my comment about area above was mis-stated. But it is not a matter of surface area per volume for pullout strength. It is a matter of cross-section area for rod strength and surface area for a unit length shape for the shear strength of an adhesive. This will determine how many inches of rod you will have to fasten with adhesive to assure that the pullout strength at least matches the tensile strength. If you look at a circular vs a square cross section where the Diameter D is the same as a side of a square D, the ratio of a one unit length piece is D/4 for both cases. For a rectangle with one side D and the other 2 X D. The ratio is 2/3 D. So Tension/Shear Area is 1/4 for a Square & Circle and 2/3 for a Rectangle with aspect ratio 2:1. This means that you will need 166% more length for the Rectangle than for a Square or Circular cross section rod.

So if tension is on the order of 200 ksi and the adhesive strength is on the order of 1 ksi, you will need 50X the Diameter of a Circular or Square Rod. For a 0.125 in. rod; 6.25 in. For a 0.125 x 0.250 rectangular rod you will need 133X the smaller dimension of 0.125 or 16.66 in. Perhaps this is a good reason to avoid the rectangular rods?

In the cases I know of at Bell Helicopter, they always used circular rods. This may also be due to the consideration of how the rods are packed together.

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#### Birdmanzak

##### Well-Known Member
My initial statement was about rectangular rods, and my comment about area above was mis-stated. But it is not a matter of surface area per volume for pullout strength. It is a matter of cross-section area for rod strength and surface area for a unit length shape for the shear strength of an adhesive. This will determine how many inches of rod you will have to fasten with adhesive to assure that the pullout strength at least matches the tensile strength. If you look at a circular vs a square cross section where the Diameter D is the same as a side of a square D, the ratio of a one unit length piece is D/4 for both cases. For a rectangle with one side D and the other 2 X D. The ratio is 2/3 D. So Tension/Shear Area is 1/4 for a Square & Circle and 2/3 for a Rectangle with aspect ratio 2:1. This means that you will need 166% more length for the Rectangle than for a Square or Circular cross section rod.

So if tension is on the order of 200 ksi and the adhesive strength is on the order of 1 ksi, you will need 50X the Diameter of a Circular or Square Rod. For a 0.125 in. rod; 6.25 in. For a 0.125 x 0.250 rectangular rod you will need 133X the smaller dimension of 0.125 or 16.66 in. Perhaps this is a good reason to avoid the rectangular rods?

In the cases I know of at Bell Helicopter, they always used circular rods. This may also be due to the consideration of how the rods are packed together.

What? mz- is right. For a rod with a cross-section of 1 square Unit of Length, the perimeter of:

A circular rod is 3.54 UoL (ie. sqrt(4*pi))
A square rod is 4 UoL, and
A 2:1 rectangular rod is 4.24 UoL (ie. sqrt(18)).

The rectangular rod has the MOST surface area per unit length for any given tensile strength. Almost 20% more than the circular section. The diameter of the rod isn't important, the section area is.

#### Tom Nalevanko

##### Well-Known Member
What?? But no one makes or sells rods by equal cross-sectional areas? You have a fixed set of rods from which to choose.

I believe that my analysis above is correct. Your argument starts with a false premise. And I did consider tensile area vs shear area.

"The diameter of the rod isn't important, the section area is." The diameter and section area (for a circle) are directly related.

#### mz-

##### Well-Known Member
Huh? The cross sectional area (CSA) defines the strength (and weight) of the rod.
The circumference (P) defines the bonding area and thus the length you need to use glue on for a given strength.

If your circular rod has less cross sectional area (it is lighter per length) than the rectangular rod, then it has less strength. No wonder then you need less glue length (if the difference is more than tiny). That's no advantage. You could achieve something similar by using a smaller rectangular rod.

#### Tom Nalevanko

##### Well-Known Member
Yes; but you don't have a smaller rectangular rod. And you are talking generalities about the relationship between CSA and P. While it is the same for a square and round cross section, it is not the same for a rectangle. With the 2:1 aspect ratio rectangular rod, you have 2X the cross section of a square with its smallest dimension (and thus 2X the strength) but instead of a length of 1, you need a length of 2.66. If all were equal (according to your strength argument) you would need a length of 2. So there is a disadvantage re the increased length vs strength. Thus the round rod (or square if they existed) have an advantage.

#### mz-

##### Well-Known Member
You're saying rectangles are worse because they are stronger?
What about a bigger rod? It's also worse than a small rod? It's also stronger and has less bonding area per length per strength.

#### Tom Nalevanko

##### Well-Known Member
I am saying that a rectangular rod of the same tensile strength has a disadvantage over the circular (or square) rod in that it takes a longer rod length to fix it. This has practical effects at the end of spars, etc.

>>What about a bigger rod? It's also worse than a small rod? It's also stronger and has less bonding area per length per strength.

I did the analysis in terms of D (diameter or smaller side) initially and later for equal tensile strengths. Bigger or smaller makes no difference. However using many smaller rods gives you a certain flexibility that is desirable.

The Graphlite rods owe their high strength to the fact that the carbon fibers interior to the rod are aligned to within 0.02%.

One industry that understands this is the oil industry. They use pultruded fiberglass sucker rods in the grasshopper style pumps. Much lighter than steel and more reliable too. There are several patents about fastening the end of the rod to the pump. All use a pultruded round rod with various attachments schemes (some pretty ingenious). None use a rectangular rod although these do exist because the attachment and fixture would have to be longer, heavier, etc.

#### Tom Nalevanko

##### Well-Known Member
To make this as simple as possible -- take a 1 x 1 square. Its XS is 1 x 1 = 1. Its Perimeter is 1+1+1+1=4. The first number is proportional to tensile strength while the second number is proportional to shear strength for a unit length piece. Now say you go to a rectangle of 1 x 2. The XS is now 2 while the P is 1+1+2+2=6. So the tensile strength has doubled but the shear strength did not double to 8 but only went to 6. And because you can show that a circle and square has exactly the same relation, there is a disadvantage to the rectangle as I hope to have explained above.

It is well known that if you want to increase the surface area of a cube, your optimal approach is to cut it into many smaller cubes. This is the principle used in charcoal filters where a 1in. x 1in. x 1in cube would not absorb much because it has a relatively small surface area. But cut into cubes 0.001 on a side, the overall surface area increases by orders of magnitude.

Blue skies,

Tom

#### bmcj

##### Well-Known Member
HBA Supporter
Wait a minute Tom. If your tensile strength needs dictate a cross sectional area of X, then the radius of the round rod will need to be sqrt (X/pi), making the perimeter 2*sqrt (X*pi). A square section would have a side of sqrt X and a perimeter of 4*sqrt X. Divide the two and you find that the round rod has only 88% of the area that the square rod has.

#### Tom Nalevanko

##### 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

#### Rom

##### Well-Known Member
I really don't have a firm, documented number but in the past I've used 180ksi for the compression side and 200ksi for the tensile
Is this for the cross section of the Graflite/resin matrix or just the rods?
My calculations use the sum of of the rods.

#### orion

##### Well-Known Member
Just the net area of the rods

#### mz-

##### Well-Known Member
I give up, I don't understand what Tom is trying to say.