Brodhead Pietenpol Association

Original Posted By: "Greg Cardinal"
There three stresses that exist in a spar. A spar in a Pietenpol, and theentire wing structure, for that matter, can be considered a continuous beam(talking about the original one piece wing here) with a cantileveredoverhang at each end. With a normal positive loading on the wing (ie innormal flight) the upper half of the spar is undergoing compressivestresses. The top of the tip is trying to move toward the centerline of theaircraft. Wood is quite strong in compression. The compression stress isgreatest along the upper surface of the spar and decreases to zero at aboutthe point halfway down the spar. The point where it is zero is called theneutral line.The lower half of the spar is under a tensile load. It is trying tostretch. The bottom of the spar above your head in the cockpit is trying tomove out to the tip. The greatest tensile stresses are on the lower surfaceof the spar. Wood is nowhere near as strong in tension as it is incompression. As you move up the spar from its bottom surface, the tensilestress becomes less and less until you get to zero which is at the neutralline, the same line as you found when you came down from the top of thespar.The wood above the line is trying to compress and the wood below the line istrying to stretch. This sets up a horizontal shear along the entire lengthof the spar right at the neutral line. In normal positive G flight, theupper half of the spar is trying to move toward the centerline of theaircraft while the lower half is trying to move toward the wing tips.Another type of stress in a spar is vertical shear. This is very obvious toall of us as we visualize the wing trying to pull up and the wing strutstrying to pull down. We can visualize the outer half of an understrengthwing spar tearing straight up at the wing struts due to more weight than thespar can handle.We have introduced the three types of stresses in a spar and have definedtwo subtypes:1) Compression stress2) Tensile stress3) Shearing stresses 3a) Vertical shear 3b) Horizontal shearWe normally orient our solid spars so that we can take advantage of thegrain orientation to maximize the resistance to vertical shear. Wetypically size the spar so that the lower half of the spar which is normallyin tension, does not fail due to over stretching. We rarely think of thehorizontal shear load since the other three stresses (1, 2, and 3a) usuallycause the spar to be robust enough that horizontal shear is really no longera major consideration.That is until we are designing for minimum weight with minimum requiredmaterial. In this case, grain orientation of a plywood web can have adistinct advantage. A solid piece of wood (not plywood) has a drawback inthat it cannot handle horizontal shear very well because the wood can splitalong the longitudinally oriented grain. If you were to use a router on asolid spar to give it an I beam section, you would aggravate the situationby providing very little cross section to resit the horizontal shear.You can immediately imagine the advantage of a thin solid piece of wood withanother thin solid piece of wood glued to it at a 90 degree angle. Doingthis would greatly increase the ability of the spar to resist the horizontalshear stress. This is plywood and this is the main reason why it wasdeveloped. A piece of plywood made of an even number plies of the samethickness of the same wood would theoretically have the same strength fromboth top to bottom and left to right although not as strong in one directionas the nonply piece.But plywood is made with an ODD number of plies so it has more strength oneway than another. If we were to use a three ply piece of plywood with thetwo outer faces pointing left and right relative to the span of the spar, wewould have two plies resisting vertical shear but only one ply (the middlevertical one) resisting horizontal shear. This may not be enough withoutgoing up a few sizes. If we oriented the plies so that the two face plieswere up and down, we would have the two face plies resisting the horizontalshear but that would leave only the center ply to resist the vertical shear.Surely not enough strength without going up a few sizes.To minimize the thickness of the plywood web, we have an option.If we orient the odd number of ply plywood's grain 45 degrees to the span ofthe spar, then we maximize the benefit of the plywood strength whileminimizing the amount of material used. Horizontal shear and vertical shearstresses placed 45 degrees to the grain of the plies of the plywood web areable to be handled much better. WIthout going into more detail, it is thesuperior way to do it.ANC-19 (help me Greg or Wayne) indicates that it is more desireable to havethe face grain of the plywood start at the bottom of the spar toward thecenterline of the aircraft and point toward the top of the spar and the tip.This would mean that at the centerline of the aircraft, the grain on the webwould be scarfed together as to form a V in the face grains.Comments, Cy?----- Original Message -----

Original Posted By:> "Christian Bobka"
Let me humbly add a little to the mix here. What you stated about theforces on the wing is correct for a one piece wing attached only at thecenter, a cantilevered wing, or cantilevered wingtips. What was not takeninto consideration is the wing struts and what they do the the compressiveand tensile loads on the spar.Most wingstruts are attaced at the center of lift of a wing (This usuallyhappens slightly inboard of the center due to the fact that the wingtip isless efficient at producing lift). If they are placed in the exact centerof lift, the wingstruts will support the entire lifting load. The wingattachments at the centerline of the aircraft will have no vertical load.If these struts are oriented at an angle of 45 degrees exactly, for everypound of lift supported they would impose an equal pound of compression inthe section of spar between the strut attachment point and the centerattachment point. Usually, they are at less of an angle horizontally thanthat so they actually produce more than a pound for a pound of lift.Fortunately, there is a wing on the other side of the aircraft pushingequally in the opposite direction.What I mean to say is that between the wingstrut and the root there is anadditional compression force to be added. As I write this, I wonder if it'slarge enough to be concidered, so forgive the mind numbing calculations thatare to follow...Assume a 14' wing with 750lbs lifting load (half of the total 1500lbs). Theoutboard cantlevered section of the wing would support half of that, 375lbs,in 7' of wing. Assume that the center of lift for this section is at it'smidpoint, this would produce 1313ft.lbs. of moment (1/2 of 7' x 375) at thestrut attachment point. Assume that this strut is 6" x 1" (0.5' x 0.083')at the cross section and half of that moment would be taken by compressionin the top half and tension in the bottom half. So the compression force,which I'll assume is at the center point of the top half, is 2625lbs(1/2 of1313ft.lbs. divided by 1/2 of 0.5'). So there is 2625lbs of compression and2625 of tension in the upper and lower halfs of the spar, respectively. Theinboard section of wing would have the same resultant forces witht theexception that the compression of the wing strut is added. Assume that thehorizontal load imposed by this strut is 1000lbs, the inboard section ofspar would see 3125lbs of compression and 2125lbs of tension.So it's a force to be considered but its only a fraction (20% in this case).So just outboard of the wing strut, the spar would see 126000psi ofcompression in the top, 126000psi of tension in the bottom, and 18000psi ofvertical shear. The horizontal shear would simplistically calculated to4500psi (2625lbs in 1" x 7' of spar horizontal cross-section) but wouldrequire a calculus equasion due to the elastic deformation along that largelength. It would surely be much larger than that at that point.Regardless, this leads me to believe that a failure in the wing would mostlikely happen at the wingstrut where the combination of the largest forcesoccur.Robert HainesMurphysboro, IllinoisP.S. - I don't want to sound like a know-it-all, just trying to keep theconversation going.