Original Posted By: "taildrags"
Terry, Brent-I am an engineer, so I can get carried away with numbers and formulas, but usuallywhen that happens people just quietly walk away and leave me muttering tomyself, so it's OK if you do too ;o) When we talk about a spar being "strong",we have to define what kind of strength we're talking about. I'll be over-simplifyinghere, but when it comes to beams (like our wing spars), there is bendingstrength and there is shear strength. We are interested in bending strengthso let's look at our Air Camper wing to see what's what.Structurally, there are three elements to our wing. Simplifying, of course. Thecentersection is one element, the section of the wing outboard from the cabanesto the lift strut attach points is another, and the part outboard of thelift strut attach points is a third. Of those three, the centersection is themost lightly loaded and the easiest to analyze structurally because it is supportedand restrained at both ends by the four cabane uprights. The next section,from the cabanes out to the lift strut attach points, is more heavily loadedthan the centersection but it, too, is restrained at both ends (by the cabanesat one end and by the lift struts at the other). The outboard-most sectionof wing is the part that has to work the hardest because it is only supportedat the lift strut end and the other end is cantilevered out into thin air.That part of the wing is what needs the most strength in bending, because that'swhat it has to work hardest at.The heavies that Terry flies have wings that are completely cantilevered out fromthe fuselage, so when you are at maximum takeoff weight with full fuel, pax,and baggage and you pull that baby into climb attitude on takeoff, the wingsbend gracefully in an upward sweep out at the tips as they do their hardest work.In a spectacular example of this taken to the extreme, check out the 787Dreamliner's wingtips bending upward about 25 feet at the tips (during testing,of course), here:
https://www.wired.com/2010/03/boeing-78 ... x-test/The bending stress is at its maximum at the wing roots of a cantilevered wing,diminishing as you go out spanwise to the tips, where it is zero. So, in theory,we can save weight by tapering the wing spars to follow the profile of thediminishing stress as we go out to the tips, but that adds complexity to thespar construction, and if we also taper the wing chordwise like the heavies are,it means every rib is different from its neighbor and everything is harderto make. So, we build our wing Hershey bar-style with a constant chord and constantspar size for simplicity and ease of construction. Consequently, the wingcentersection is very much overbuilt, the section from the cabanes to thelift struts is still plenty overbuilt, and the outboard section is probably justabout right but progressively more and more overbuilt as you go from the liftstrut attach points out to the wing tips where the bending stress is (theoretically)zero.Bottom line, we analyze the forces in an Air Camper wing where the bending stressis the highest: on the outboard section, where the lift struts attach. Inboardof that, the spars are restrained from deflecting in bending and outboardof that, they are not.Now for the math. Feel free to leave the room quietly ;o)The bending stress in a beam (which our wing spars are) is calculated by knowingtwo things: the moment (which is the product of the lift force on the wing timesthe distance where it acts on the wing), divided by the section modulus.Forget about two of those things for a minute because we are looking for themaximum "strength" of our wing and we know the properties of the wood (or metal)that the spar is made of so we know its maximum allowable stress, and we alsoknow the lift forces on the wing because we are designing it for a certainmaximum gross weight, G-force, and safety factor. That just leaves the sectionmodulus, which is a property of the beam section. It's different for a tube,a rod, an I-beam, a C-section, rectangle, square, bar, whatever. In the caseof a typical rectangular-section wood beam, the section modulus is equal tothe width (thickness) times the height (spar depth) squared, divided by six.As H. Ross Perot used to say during his talks, "stay with me now..." ;o)The amount of stress that the wood in our spar is able to take without failingis essentially constant and is a property of the wood. So, the larger we canget the section modulus of our spar to be, the "stronger" it is (the more bendingstress it can take, which means higher Gs or a higher gross weight, or both).So, lets look at the section modulus. We have two things we can work withto determine it: the thickness and the height. If we double the thickness,we also double the section modulus but if we double the HEIGHT, we increase thesection modulus by a factor of FOUR because the height gets squared in the equation(2x2 = 4). Since we're trying to build our spars as light as we can,why make the spars twice as thick to double the section modulus (but double theweight), when we can do the same thing by simply increasing the spar depth byabout 2 and the weight only goes up by about 40% instead of 100%? In resistingbending force on the spars, you get more bang for the buck by going tallerrather than fatter.So (and this has nothing to do with where you slept last night), your example ofa spar that is 1" thick and 5" tall means that its section modulus is about4.17 but if we make it 3 times that tall, 15", its section modulus skyrocketsto 37.5 and the spar isnt 3 times stronger, its 9 times stronger in bending thanthe 5" tall one. There's that squaring effect at play.But wait, because there are many reasons why it would be impractical to actuallybuild something like that due to buckling from "slenderness" and fiber shearin the web, twisting, and other things. However, the point is that in a wingspar that is subjected to bending loads, depth is everything. I'm not certainof the Riblett airfoil nomenclature, but if the 612 has 12% thickness and thewing chord is 60", the airfoil would be about 7.2" tall at its maximum thicknessso maybe 7" at the main spar. (Terry, how tall are the metal J-3 spars?)For a 1" thick rectangular Riblett 612 spar, the section modulus might be 8.17.The stock unrouted 1 thick Piet spar is 4-3/4" tall so it might have a sectionmodulus of 3.76 and the deeper 612 spar, if it's a rectangular shape likethe Piet's, is then theoretically competent to take more than twice the bendingstress than the stock Piet spar, so maybe (to temper what Wese said aboutthe Air Camper wing spar), maybe the 2-G Piet becomes a 4-G "Super Piet" ;o)All of this is theoretical and none of the above should be taken as a rigorousanalysis of either spar, or of the wing, or anything other than a general discussionto point out that calculated design can sometimes beat eyeball engineering.On the other hand, many things that are exquisitely designed, cannot readilybe constructed on the floor of a wooden barn using hand tools. Thus we havethe timeless and understated technical elegance of the Pietenpol Air Camper,which does what it does very well and without a lot of analysis or re-engineering.If you choose to depart from Mr. Pietenpols design, take your time analyzingwhat you do.--------Oscar ZunigaMedford, ORAir Camper NX41CC "Scout"A75 power, 72x36 Culver propRead this topic online here:
http://forums.matronics.com/viewtopic.p ... ___Subject: Pietenpol-List: Re: Ragwing Ultra-Piet