|Dec 14, 2007, 07:35 AM|
Whenever I design a wing, I always think about how the wing might fail if I overload in the air. I have the feeling that generally the upper spar may be prone to buckling before the lower spar will fail in tension. I think that the use of vertical grain spar webs will avoid this. If its an electric glider, maybe it is not so critical, but if you want to use a hi-start launch then you probably want every bit of strength you can get. Now I can buy carbon fibre strips at my LHS I am going to try laminating the top spar with a strip of carbon on the top. I don't know if it is necessary or not, but I can't stand having a wing failure after the amount of effort I put into a build.
|Dec 14, 2007, 05:27 PM|
Joined Jun 2005
Good info, John. I always have pictured the tension failure and gave no thought to buckling on top and the webs were added in my mind to keep the top and bottom spars from moving in opposite directions. I use the carbon fiber tape on the bottom where the dihedral changes and on top for inverted flying stress. I was admiring the wing from my Campbell 54" Quaker ( available from Penn Valley Hobbies) and just may use some of his design ideas. It has spruce center spars, my change, with no webbing and weighs just 111 grams or 3.9 oz. There are 4 spars, lots of ribs and washed out tips. The model weighs 1.25 pounds I imagine, and has been through lots of dives, loops and rolls. The wing is like a spring flexing with transparent covering. For the record, we are talking wing design for a canard sailplane. I hope to have pictures of the Quaker wing soon. Charles
|Dec 14, 2007, 06:42 PM|
Excuse Me! I'm to keep from laughing at your shear web siuation!
I know that the props guys have different structure than sailplanes, and they are not pulling the 25 G's that sailplanes are, but if you have a desire to increase your strength to weight ratio by20X, I strongly sugest to take the slow, tedious road When I decomisioned the original Counselor wing, it took 87 lbs to crack it, supported at the tip rib and 1/2" out board of root rib.
Kids and students having bridge building contests with 4 oz. and 36" of span have a gas!
I showed my middle step son a sailplane structure, and by adding camber, he could even increase his chance of winning! He got 2nd place at 54 lbs!@ 3.8 oz! I believe that is a strength to weight of 227/1, if my math is correct!I was shocked when I saw a tv guy remove root shear webs for an airplane repair
Also, what is actually happening when you scatter shear webs 'wherever' is you will actually create stress risers on both sides of the web! go ahead, try an experiment , spend the balsa and 1/16 web stock the weight is next to zero and the strength gai is how much?????
|Dec 14, 2007, 09:11 PM|
Charles, if you haven't had wing failures you must be doing something right, at least as far as the power models are concerned.
Even after a failure occurs, its hard to know which spar has failed first. My theory is that if the top spar is broken in two places, it is likely due to buckling. If the top spar has only failed in one place then maybe the bottom spar failed in tension first. I have only had one failure so far. That was with a hand-launch glider on a bungee. From memory it took about 20 or 30 launches before failing.
|Dec 14, 2007, 09:36 PM|
United States, IL, Chicago
Joined Dec 1996
Have to agree with Johnny on sheer webs - my usual OD model wing structure starts at the centre and a bay either side with sheer webs the same thickness as the spars - usually 1/4" x 1/8" spruce. The next couple of bays are 1/8", the rest to the tips are 1/16". Grain vertical, webs placed centrally on the spar's widths for an I beam.
These have long taken all I could sling at them in numerous models, and few accuse me of flying sedately for long I haven't had a broken wing as far back as I can remember (says he, touching laminated imitation plastic wood [Made in China], surreptiously)
|Dec 15, 2007, 06:39 AM|
Joined Jun 2005
Thanks Guys, I got a large charge from your three replies. It just made my day!! I GIVE UP !!! You must have it right and the shear web situation will be a on my mind from now on. I will be absent for a day and will talk more then. Thank you for the great conversation. Charles
|Dec 15, 2007, 01:19 PM|
United States, IL, Chicago
Joined Dec 1996
Hi there, reversed donkey fans
While Charles is off doing whatever he'd up to, anyone remember this?
I was scratting through old mags - have too many of them physically, so am reducing them to scans on old FRED (my computer, don't ask what FRED stands for, it would blow the PCP's minds )
I have vague memories of looking over this kit at some point, and have a feeling that it would make a better electric now we had stuff that would fit in its skinny fuselage. The mag it came out of was early 1990s - I wasn't taking this too seriously, so didn't note the exact issue down.
Guess something this wild would be bound to fade away in these times ... Must be the odd kit lurking in a garage collection someplace though.
|Dec 15, 2007, 04:58 PM|
Sounds like time for me to chime in here again.
Several issues involved on this one.
There are two kinds of washout, geometric and aerodynamic.
Geometric is what you're all probably familiar with, having the physical angle of incidence of the tip at a lower angle than at the root, so that at a given angle of the aircraft to the "relative wind" (the direction of the freestream airflow coming at the plane), the angle of the root puts it closer to its stall angle than the angle at the tip.
The problem with geometric washout is that there are other effects involved that can skew the results (more on that in a moment).
Aerodynamic washout means having any or all of a variety of factors adjusted so that the stall margin at the tip is greater than at the root. Typically this involves having a different airfoil at the tip, with a greater stall angle, than the airfoil at the root.
Complicating all of this is the fact that there are at least three different common references from which to measure the angle of attack (the angle relative to the incoming airflow) or angle of incidence (the angle relative to the rest of the airframe) of a given airfoil section along the wing:
In the case of a "flat bottomed" airfoil, there is the angle relative to the bottom surface.
Then there is the angle measured from the chord line, a line running from the furthest-forward point on the airfoil (typically a line that runs through the center of the leading edge radius) to the furthest-aft point (if the trailing edge is not sharp, then a line through the middle of its thickness). On a typicall flat-bottomed airfoil, this line will be maybe two or three degrees above the angle of the bottom surface.
Finally, there is the "zero-lift" line. This is the only one of the three that directly considers the aerodynamic properties of that airfoil shape. The zero-lift line is the line where the lift coefficient is exactly zero if the angle between the relative wind and this line is zero.
Use of the chord line is probably the most common in the model airplane world and in manufacturing. In wing design, the zero-lift line is the best starting point, then switching to use of the chord line as a reference once the airfoils along the span are determined.
Getting back to washout (now that we have defined the various refeences to measure it from), it could be for a variety of reasons. The one commonly citedis to try to make the root stall before the tip, but geometric washout could be used to fine-tune the aerodynamic washout resulting from use of different airfoils along the span, or (as in the case of a canard) to account for distorted inflow due to the downwash from a flying surface ahead of the wing.
In that last case, if you don't include some washout in the wing, then the downwash of the canard reduces the local angles of attack on the inboard portions of the wing, resulting in more load carried by the outer portions. This distorted lft distribution increases the induced drag, and also makes tip stalls more likely if the wing does stall. Of course with a canard, any kind of wing stall is bad.
With a glider, things get more complicated, particularly at lower wing loadings. As the wing loading goes down, so does the turning radius in a thermal turn. The smaller the turn radius, the more airspeed difference between the inside wingtip and the outside wingtip, as well as between the inside wingtip and the root. This means more lift coefficient needed at the inside wingtip to keep everything in balance about the roll axis.
Since an airfoil stalls because you asked it to make more lift coefficient than it was capable of doing, this difference in airspeed along the span can force a tip stall. Geometric washout is a poor solution, since the only way it can fix things is by forcing both tips to unload so much that the root does all the work, which makes your wing act like it has only 2/3 the effective span. It also means that the tips will be lifting downwards in high-speed flight, increasing induced drag at a portion of the flight envelope where the induced drag should be negligible. In an extreme case, it's even possible to get a negative stall from one tip in high-speed flight!
Geometric washout forces a reduction in the demand for lift-making by the tip. Conversely, aerodynamic washout (properly used) increases the lift-making capability of the tip, a much better solution to this problem, allowing the lift distribution at high angles of attack to remain elliptical, minimizing induced drag at the place in the flight envelope where induced drag is the most important.
Speaking of elliptical lift distribution along the span, yes, that does result in the lowest induced drag FOR A GIVEN SPAN. However, if the criteria is the lowest bending moment at the wing root (which corresponds to the lowest structural weight in the wing) then a bell-shaped lift distribution, as described by Prandtl and applied by the Horten brothers in their flying wing designs.
Of course, having an elliptical lift distribution does NOT mean that the planform has to be elliptical. IN fact, if you consider the effects of Reynolds number on those narrow tips, you can be pretty certain that an elliptical planform will not have an eeliptical lift distribution. Lift distribution is a function of planform, but also of the geometric and aerodynamic washout. By properly considering all of those factors, as well as things like drooping the ailerons by a certain ratio relative to the flap deflection, you can achieve an elliptical lift distribution over essentially the entire flight envelope. However, it isn't easy, it requires the right design tools and a lot of hard work on the part of the designer.
|Dec 15, 2007, 08:08 PM|
Now, regarding spar design:
First of all, the key parameter is the bending moment at each location along the span. If you know the bending moment at a given point, and the depth of the wing spar at that point, you can calculate the tensile and compressive stresses in the lower and upper spar caps at that point. Then the sum of the moment forces made by the spar caps has to equal the bending moment imposed on the spar.
For each spar cap's moment force, take the distance from the middle of that spar cap to the "neutral axis" times the tensile (lower cap) or compressive (upper cap) force in the spar cap. Divide the force by the cross-sectional area of the cap and you have the stress in the spar cap, which you can compare to the compressive or tensile strength of the spar cap material (minus an appropriate factor of safety).
When you bend a beam (such as a wing spar) some portions get stretched (the tensile side, for positive "G" the lower side) and some get squished (the compressive side, the upper cap for positive G). That means that there is some point in the thickness in between the two where the tensile and compressive stresses are exactly zero. That location is the neutral axis. The distance between the neutral axis and any given location in the spar cross-section tells us how much leverage the material in that location has, and how much deflection it sees (and therefore the amount of strain in the material) when the spar is under a bending moment.
If the upper and lower caps are identical (same shape, cross-sectional areas and material), then the neutral axis is halfway between them.
If you stiffen only one spar cap (beef it up, or add a doubler or some carbon to it), then the increased stiffness in that cap pulls the neutral axis closer to it. For a given applied bending moment in the spar at that location, this reduces the leverage of that spar cap, and increases the leverage of the other spar cap. The reinforced spar cap sees a benefit because of the extra strength, but not as much as you might expect, because of the loss of some leverage. The corresponding increase in leverage on the other side helps make life easier for the opposite spar cap. This is what happens when you add carbon to only one spar cap. It's generally more effective to add carbon to both spar caps, keeping the leverage betweeen them roughly where it was.
That said, on planes that see max loads mainly from the positive-G direction (such as winch or Hi-start launched sailplanes), you generally need more beef in the upper spar caps. This is because the compressive strength of most materials is less than the tensile strength, even without considering buckling failures. The difference is fairly pronounced in wood (as much as about 2:1 in many cases), but (contrary to popular belief), significant but not that great in materials like carbon or steel.
The big exception is Kevlar. The tensile strength of Kevlar is fantastic, actually higher than typical varieties of carbon, but its compressive strength is truly pathetic, less than half that of garden-variety hardware store E-glass. Rumor has it that the aramid molecule itself has a kink in it, and the low compressive strength is the result of buckling at the molecular level. In any case, the compressive strength is truly awful, to the point that unless the part involved does not see any significant compressive load (such as blades for most full-scale propellers, which see mostly tension from centrifugal forces), you're usually better off choosing some other material.
Oddly enough (and supporting the molecular buckling theory), the tensile strength of Kevlar after it's been failed in compression is nearly as great as it was before the failure. What typically happens in a compressive failure is the fibers lose their load-carrying ability, dumping the load onto the matrix material (epoxy/polyester/whatever), which then crumbles away, leaving a loose "rope" of nearly undamaged Kevlar fibers. Successive loading and unloading of the loose fibers rubs them against each other to the point that they eventually saw through themselves, but at least initially the broken pieces at least stay connected to each other.
So far we've been talking about compressive failure, pure crushing failure like what happens when you squash a ball of clay between your palms. Buckling is a different type of compressive failure entirely. Buckling typically happens at far less than the pure compressive strength of the material, unless the structure includes enough support along the length of the piece to prevent it. When you crush a soft-drink can, it fails in buckling, suddenly, catastrophically, and at far less than the pure compressive strength of the material.
If your spar failed in buckling, it means that your design did not take advantage of the entire compressive strength available from the material.
Buckling is a function of the stiffness of the material, not the material's strength. That's right, if you made a part out of the strongest heat-treated tool steel, and that same part out of the softest, cheapest, low-carbon coat-hanger-wire steel, both would fail in buckling at the same load, because they have almost exactly the same stiffness, despite their huge difference in tensile and compressive strength.
Buckling prevention is one of the two functions of the shear web in a typical spar. This is also why we typically orient the grain of the shear web material vertically, to maximize the support for the compressively-loaded spar cap. If the part is supported along its length enough to prevent it from flexing off to one side, it will prevent it from buckling.
The other function of a shear web is to connect the upper and lower spar caps together. One of those is in tension, the other is in compression, so something has to connect them together so one has something to pull on, and the other has something to push on. Without a means of communication between the two of them, their entire structural relationship would quite literally fall apart. In addition, the wing is making lift, an upward force. The loads in the spar caps are spanwise. They're handling the tension and compression caused by bending moments resulting from those upward lift forces times the moment arm created by the span, not the lift itself. The vertical lift force has to be collected along the span and transferred to the fuselage. This is done by the vertical components of the shear loads acting in the shear web.
One other load in the shear web comes into play at dihedral joints. The spar caps coming into the joint from both sides are at an angle to each other. Because of this, the compression/tension forces in them do not exactly cancel, they each have a vertical component that shows up as a vertical compression or tension force in the shear web itself.
I heard the tragic story once of a new airliner design back around the WW II era, where the engineer designing the shear web for the wing spar forgot to take this load at the dihedral joint into account. Somehow this didn't get spotted in certification tests, and later resulted in the loss of an aircraft, along with a number of fatalities. According to the story, when the engineer learned of his error and its result, he couldn't live with himself. Utterly despondent with his feelings of guilt, he increased the number of fatalities related to the accident by one.
Generally the loads in the shear web are not as difficult to deal with as the loads in the spar caps. If the shear web is beefy and stiff enough to properly support the compressively-loaded spar cap, it's generally also strong enough to handle the shear loads imposed on it. However, it is possible (especially with the extensive use of carbon in spar caps these days) to have spar caps that are overstrong, and a shear web that isn't. Ideally, each of these components should be just the right size to do its job, plus the necessary safety factors. Anything less than that invites catastrophy, but anything more than that adds unnecessary weight, which also invites catastrophy.
|Dec 16, 2007, 11:50 AM|
More on spars:
OK, so we've covered the case where we have something like an "I" beam, with two concentrated spar caps at the top and bottom, and a vertical web between them to handle the shear. However, this approach doesn't work for a solid rectangular beam like the spar in a Gentle Lady or Sig Riser, or a tubular spar or solid round joiner rod. For those, we need something different.
Repeat after me:
"Sigma equals emm see over eye." Memorize that incantation, it's the magic incantation that makes sense out of all of this.
So much for phonetics. What that really stands for is the formula:
Sigma = M*c/I
It's the formula for bending stress in a beam.
Sigma is of course the bending stress. In my examples it will be in PSI (pounds per square inch), but it could also be in metric units, such as Newtons per square centimeter. However, I prefer my Newtons as cookies, so we'll stick with English units.
"M" is the bending moment. In our case, since we're expecting our answer to be in pounds per square inch, we need to keep all of our units in pounds and inches, so the moment is expressed in inch-pounds. We take the lift forces along the individual sections of the the wing, multiply each by the distance from the root, then add the moments up to find the total moment load on the wing root. The same technique works for any other location on the wing, just take the lift forces and moment arms for the portions of the wing outboard of that point.
"c" (in inches, in our examples) is the distance from the neutral axis to the "outermost fiber" of the spar, the point on the spar's cross-section that's furthest above or below the neutral axis. Of course if you wanted the stress for some other point in the spar cross section, plug in the distance for that. However, the highest stresses will be at the top and bottom edges.
"I" is confusingly called the "moment of inertia" because its formula is very similar to the formula used for figuring the "mass moment of inertia" used in calculating things like "flywheel effect". However, in this case it's a number that accounts for the structural effects of the size and shape of the spar's cross-section. The alternate name for it is "Section Modulus", which makes a lot more sense to me, so I prefer that term. In the English system it has units of inches to the fourth power ("in.^4").
For a rectangular cross section, the formula for section modulus is:
I = b*(h^3)/12, where b is the width of the spar cross section and h is the height.
For a circle, the formula for I is:
I = Pi*(D^4)/64
Let's try some examples:
Case 1. Assume we have a Sitka spruce spar (tensile strength = 10,200 PSI, compression parallel to the grain = 5610 PSI), 3/4" high and 1/4" wide, in a 2-meter wing that has to handle a Hi-start with 20 lbs tension at launch. Assume an elliptical lift distribution, so the total lift in the wing panel acts at an effective moment arm of about 16" out from the root. That gives us a moment at the root of 16*20/2 (remember, there are two wing panels dviding the 20 lbs lift between them), or 160 in-lbs.
M = 160 in-lbs
c = 3/4"/2, or .375"
I = .25*(.75^3)/12 = .008789 in^4
Sigma = (160)*(.375) / (.008789) = 6827 PSI
That's going to be fine in tension on the bottom, but that's more than the compressive strength on top.
Case 2. Same spar; how much do we have to beef it up to make it strong enough to not fail in compression?
We're coming up with 6827 PSI for a material that's only good for 5610 in compression. That means we need to reduce the stress to 5610/6827, or 82% of where we are now, which means increasing the strength of the spar by 1/.82, or 1.22 times.
Looking at our Sigma = M*c/I formula, we can see that stress is linear with the width of the spar. That means we would need to increase the 1/4" width to 1.22*.250, or .304". Spar stock 5/16" wide would be a little over that, so good enough. 3/4" x 3/8" would be even better, with some reasonable safety factor, but any more than that is starting to add noticeable weight.
The other approach, assuming our airfoil thickness allows, is to increase the 3/4" dimension. Looking again at our formula, we can see that the height of the spar shows up in the "c" in the numerator of the formula once, and cubed in the "I" in the denominator, so overall the stress is inversely propoertional to the square of the spar depth. This means that we only have to increase the spar depth by the square root of 1.22 times, or 1.10, resulting in a spar depth of .827". Going to 7/8" x 1/4" stock would increase the strength by (.875/.75)^2, or 1.36 times, more than enough to get the additional 1.22 times that we need. Adding material to the depth of the spar is obviously more effective than increasing the width!
Case 3. Now let's look at a joiner rod for the middle of the same wing. We look in the catalog for an aluminum joiner made from 7075-T6 alloy (yield strength 66,000 PSI), about as strong as aluminum gets. Will a 3/8" diameter joiner be enough, or do we need 1/2"?
For a 3/8" joiner:
I = 3.141592 * (.375^4) /64 = .000971
c = .375/2 = .1875"
Sigma = 160*.1875/.000971 = 30,900 PSI
That's well within what 7075-T6 can handle, and a 3/8" joiner will be enough for this airplane. Now, if we decide to try to winch launch it, the loads are much higher, so both the spar and the joiner will need to be considerably stronger.
Case 4. To save weight, we're considering a 3/8" o.d. x 1/4" i.d. tube of the same material. Will it be strong enough?
"c" is still equal to .375/2, or .1875".
"M" is still 160 in.lbs.
"I" is the .000971 for the 3/8" o.d., minus the "I" for the 1/4" hole in the middle.
The hole has an "I" of:
Pi*(.250^4)/64, or .000192 in^4.
This gives us a total "I" for the tube of:
I = .000971-.000192 = .000779
Plugging all of that back into our formula, we get:
Sigma = 160*.1875/.000779 = 38,498 PSI
which is still well within the capabilities of that alloy. When it comes to bending, it's obviously the material on the outside of the part that does most of the work.
Other cross sections can be analysed by the same method. The only catch is coming up with the location of the neutral axis (so you can measure the value for "c"), and determining the value of "I". That's where a good CAD program can be a big help. Many of those have an option for calculating the centroid of an arbitrary shape (or combination of shapes, so you can include the contributions of things like leading and trailing edge stock). The centroid of your cross section gives you the location of the neutral axis. The CAD program may also have an option for figuring the moment of inertia about an axis you specify (a horizontal axis through the centroid in this case), which gives you the value of "I".
Things get somewhat more complicated when you start mixing materials, such as a spruce spar with carbon reinforcements on top and bottom, with balsa leading and trailing edges. However, if there's a significant amount of carbon involved, that stuff is so stiff that in most cases the carbon will try to carry the entire load all by itself.
|Dec 17, 2007, 12:32 AM|
Don, Thanks for the lesson on beam theory. The calculations are good and I think I should probably learn to do the calculations as part of my design process. I am still a bit concerned about the buckling aspect. Depending on the accuracy of construction I can't help feeling there will be some stress concentrations that may influence the buckling behavoir when a spar is in compression. Maybe I would put in a factor of safetly of 200 - 300%? Anyway I should really do the numbers to correlate the theory to experience.
Here is what I have been working on so far. It is a small HLG wing and so far I have only made the inner panels that are each 36cm (14.1") long, with chord of 160mm (6.3"). These have small 1/8" spars - I think they may be basswood, but I don't know for sure. On the top spar I have laminated a 3mm x 1mm CF strip, glued using 5 minute epoxy. The shear webs are 3/32 balsa on the 3 bays each side closest to the root and 1/16 on the other bays. The shear webs are vertical grain, even though the gap between the spars is only 4.5mm. The tip panels will be 190mm (7.5") in length and have the 1/8" spars, but without the carbon lamination. These tip panels taper to 125mm (5") at the tip. The airfoil on the tip panels blends from AG36 (8.2%) at the polyhedral break to AG34 (9.3%) at the tips.
This is the first time I have used the CF to reinforce spars, the first time I have used the AG airfoils with the three flat facets on the back portion of the airfoil, and the first canard glider I have made. After flying this model, if I am still enthusiastic about canard gliders I may build a second canard glider around 2m or 2.2m wingspan. That design may build on my experience here, but would probably use flaps, ailerons, rudder and elevator, and that involves quite a lot more work.
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