Discussion in 'Aircraft Design / Aerodynamics / New Technology' started by wsimpso1, Dec 29, 2017.
As soon as I learn to read sideways, I'll throw away my 1950 Peery.
Sorry about that, BJC!
Here's a good place to get started. I posted about 60 videos for my students. Enjoy!
Hodges intent is good here, but I think we have TWO confusing things being mingled and we gotta figure them both out. I like the deck of cards analogy for leaf springs, then gluing the cards together to make stout beams.
When Hodges' leafs slip, that represents shear failure of the stack in the horizontal direction, but I have to resist the urge to talk about horizontal shear and vertical shear - as the vertical and horizontal components of shear always appear together in pairs of each and are always of equal magnitude... talking about them separately gets us into confusion. Got that? Vertical and horizontal components of shear show up in pairs and are simultaneous and identical in magnitude. more below.
Let's talk about BBerson's question. He was looking at an edition of Peery's Aircraft Structures that talks about beam loading. Peery blows through it in a couple pages. Timoshenko does a much more thorough job and might be more understandable. BBerson use "vertical shear" and "horizontal shear" in his question while the edition that talks about beam stresses in chapter 6.2 talks about vertical and horizontal forces, but not "vertical shear" and "horizontal shear" so things can get confusing.
One of the things that is confusing is the engineering use of the term shear. In beam theory, the beam's reason for being is to carry loads perpendicular to the beam's long axis. The internal load perpendicular to the beam that is carrying the loads is called shear. The folks studying this stuff never ended up needing a new word because the shear developed in the beam is largely carried by the web which is loaded in shear and sometimes called the shear web... If separate terms were invented, it might have helped us learn.
In material stresses, we imagine a little square in the material, with axial forces perpendicular to each face of the square and shear forces parallel to the faces. We can also rotate the square to places where the shear is zero and there are only axial forces. Look up Mohr's circle, stay in two dimensional stresses until you get it and then you can graduate to three dimensions.
Shear load is just the vertical force carried by the spar at that point. In a cantilever wing, it is equal to the total lift outboard of this same spot. Look at the web of the spar with zero load on it, then zoom in on a little square element of that web - let's imagine the element is square with horizontal and vertical sides. As we load up the wing with lift, the web carries most of the shear, the wing deflects under load and the little square element changes shape. If the lift is upward and the tip of the wing is to the right and root is to the left, the right edge of the square slides up a little ways, deforming the square into a parallelogram. And there are now forces along the four sides. There is a force from lower right and from upper left corners toward the upper right corner, and there are forces from lower right and upper left corners toward the lower left corner. These are the shear forces on this little element and all four have the same magnitude. And balancing these forces results in distribution of forces inside our beams.
One of the interesting things about our little square element of the web is it can be equally well presented as a square with the vertices at top and bottom and the flat sides at 45 degree angles. When the wing makes lift and the tip rises, the forces arising from shear on the faces of the element are pure tension on the upper right and lower left faces and pure compression on the other two, and of exactly the same magnitude. The deformation and stresses in the material of the web IS IDENTICAL no matter which way you think of it. Yep, 0-90 element in pure shear is deformed and stressed exactly the same as the +/-45 element under pure tension and compression... Look up Mohr's circle and put in pure shear on 0-90 axes, then put in the alternating compression and tension in the +/- 45 axes.
Going back to the original discussion, using shear as simultaneous forces horizontally and vertically makes resolving the forces inside a beam easier to handle.
Now to Hodges' comment. A beam under pure bending moves as a unit. Our wing beam with lift has a neutral axis with zero bending stresses. Apply a moment at each end, and the deflection increases, starting from zero at the neutral axis. The material is under compression above the neutral axis and in tension below the neutral axis. The deformation (and stresses) increase linearly with distance from the neutral axis. Can this be envisioned as that ream of paper as Hodges pointed out? Nope - if the sheets slipped on each other, the beam would have bending capability only equal to the bending capability of an individual sheet times the number of sheets. Not much. Now imagine the same stack of sheets with a little glue on each sheet. Now the top sheet (in engineering we call this a lamina) gets the most compression and the bottom sheet gets the most tension. And all of the bending stiffness we get turns out to be from the E and I of that shape with the plane cross sections of the beam staying plane. This math describes beam behaviour VERY well, and it all comes directly from first principles.
Now just to confuse things, when BBerson talked about horizontal shear, I know that shear is comprised of four loads on an element. You can not really talk about horizontal shear and vertical shear like they are different things. They are not separable. But in a 3D structure like an airplane wing, you can have loads and shear applied to the wing beam in a vertical axis (from lift) and in a horizontal axes - but which direction is "horizontal" in a 3D airplane? Hmmm. Out toward the wing tips will occur if you are rolling fast (zero g rolls) or yawing fast (spins) or both (snap rolls), and that will make tension in the spars. Maybe he meant drag effects, which are horizontal to the rear. This is adding load in another direction, the spars (or whole wing if built with a structural skin) will also handle this using standard beam theory with neutral axis and I and the rest calculated in that direction. Total stresses and strains at any one place can be calculated by superimposing these things at each critical location. Blah, blah, blah, but now I think that I understand that is not what BBerson had in mind...
I hope this helps more than it hurts...
If your beam is a solid rectangular section, Shear Stress in the beam is a function where you are vertically on the beam. Shear stress is a parabola with zero shear stress at the top and bottom surfaces and a maximum of 1.5*V/A where V is the shear load at that point and A is the beam cross section area. As I pointed out above, the shear stress is four forces around any tiny element you are looking at...
If the beam is wide flange beam, it gets more complicated. Shear load derivations get complicated involving the first area moments of the whole beam and the portions above and below any point of interest... See Timoshenko for details on this. The big thing to know in flanged beams (oriented normally to the load) is the flanges carry little shear stress but since they are big can carry a fair amount of the total. The web carries larger shear stress and a big chunk of the total shear. Stress is still bigger at mid web than near the caps, but the difference is generally small. In the end, the web can almost always be treated conservatively as having tau = V/Aweb everywhere, then superimpose the axial strains from the caps to finish the stress picture.
Now this whole thing of "horizontal shear"... Within Chapter 6.2 of Peery, he talked about horizontal forces from shear, but let's keep remembering that shear is two horizontal and two vertical forces across 0-90 faces of our material elements. Or two tension and two compression forces across +/-45 faces... Horizontal shear and vertical shear are related and the same within any one little element. Since it is so easily handled by either hand calcs or FEA, we do not normally get wrapped up in it. If you want details, Timoshenko's chapters on beams is where to go.
My book shows horizontal shear is maximum at the neutral axis of the I beam and zero at outer cap lamina.
I don't think that book has quite enough detail to understand, so is confusing like most structures books.
I did order a Timoshenko book last year and found it was just some sort of course workbook that didn't work for me, so I threw it out.
I see the deck of cards beam (loosely bonded) failing in what I would call horizontal shear at the neutral axis as that card bond fails and slips horizontally. It's easy to see the slip in a bending deck of cards.
I guess a different Timoshenko book is needed, if there is one that explains this with common words from the beginning. I looked into Mohr's circle and it made no sense yet. I do get the little web square analogy that rotates a bit and gets more parallelogram. (from Peery)
Need all this explained at an Oshkosh forum with a deck of cards and other props.
The bottom line is I measured the web on a 1-23 glider wing. It was .020” last 9 feet at wing tip and much thicker at the root.
In any beam, shear is zero at top and bottom surfaces and the shear follows a parabolic curve elsewhere. Shear is low through the caps or flanges, and much higher in the web, still curved, but dropping only a little as you get near the caps. The webpage I got of Peery chapter 6.2, figure 6.6 showed the shear as I describe. Usually the V/Aw is a little conservative on max shear stress, so let's check it. We have shear load on the beam of 40,000 lbs and the spar is 6" deep with 1" caps, so the web is 4" and its width is 1", so the web area is 4 in^2. V/Aw = 40,000/4 = 10,000 psi. For figure 6.6, max shear is 8780 psi. Sounds like using V/Aw represents the shear in the web OK.
You are definitely on for that!
Mohr's circle allows us to see what the total stress picture is for a load case. Mohr's Circle plot has shear on the y axis and axial stress on the x axis. So if you have a 0-90 element in pure shear, it plots at (0,tau) and (0, -tau). Making circle of it, it is a circle radius tau based at the origin (0,0). Pure shear for the +/- 45 element plots as points at (tau,0) and (-tau,0), which is the same circle... We can do this sitting at a table with a pencil and paper and you will get it.
Maybe I need to run an Excel file on this and share it.
Makes sense to me. Probably found they just can not go thinner than 0.020" and then shear goes up bigtime as you go toward the root fitting.
If you want a real forum at Oshkosh, contact the Forum director Mark Forss at firstname.lastname@example.org
Might be late but can't hurt to ask.
Or set a time at the HBA meeting shelter area.
Oshkosh forum on mechanics? Thank you no... Introduction to Mechanics of Materials is a 13 week class meeting three hours a week and demanding another 4 to 6 hours on reading and doing homework and studying every week. Beam theory and total stress state is easily 9 to 12 class hours out of that class, and that assumes Calculus and Statics and Dynamics are already in the rearview mirror. And for an audience of 3 or 4 at OSH? Nope. We will just chat in twos and threes at HBHQ and get the concepts right.
Bill will explain this more ably in a few weeks I'm sure.
I find the deck of cards is unhelpful for explaining "horizontal shear" except that the cards are shearing against each other in the horizontal plane. It shows the way a bunch of beams stacked loosely on top of each other act but it does not at all show the way the stresses work inside a single beam.
A stack of cards looks like the second diagram below. If that were our model, we would expect there is more shear near the ends, and that shear stress is the same from top to bottom at any particular location on the beam. But beams don't act this way.
What they do is in the third diagram, where the neutral axis stays the same length and it's shortened on top and stretched on the bottom. (I know you know this, BB, but it sets up the next part for others who might be reading this thread.)
Next, let's take some sample elements out of the third beam. One near the top, one at the neutral axis, one near the bottom. Now we see where these tension and compression forces are working on our little blocks: perpendicular to the surface.
But we also know there is shear (vertical shear, if you like) acting on each element. Let's look at that top piece that's in compression, and this time draw it with some shear (parallel to the sides).
But now we've created a problem: this little bit of mass we've drawn isn't in static equilibrium. It wants to spin like a pinwheel because of the shear forces, and we know that beams don't spin, so we must be missing something. To complete the picture of forces acting on this little element, we add "horizontal" shear and the complete picture looks like the one below. (Note that there could be forces acting up and down, normal to the surface, but I've left them out of the example.)
Now here is the whole point of this post. The "horizontal" shear is the same magnitude as the "vertical shear." That's why the distinction isn't necessary.
The next step in examining this little chunk of material would be to take a diamond-shaped sample out of the beam, and through the magic of vector addition, resolve all of these same forces (shear and normal) into forces that are normal to the faces. Then they're known as principal forces and we can see what kind of stress the material is really seeing at its most basic level. That's the Mohr's circle thing Billski is talking about, but you don't need Mohr's circle to see that "horizontal shear" is equal to "vertical shear" and really is just plain old shear.
It looks like V (vertical shear) at root is the same as P (load) as shown here for an even loaded cantilever beam (number 6). Typical wing spar in second image.
I am not sure if the little square is just 1" or if it is a full square section of the web from cap to cap.
Because the horizontal shear is greatest at the neutral axis. So it must be more complex.
Yeah, our wings usually look like number 6 except that our loading drops off towards the tips, like your second page shows.
Note how the order of magnitude of wing loading is lowest, shear is much higher, and bending moment is much larger again.
Size of the stress element? Well, we usually go for infinitesimally small. If you make it bigger, it misleads you because the shear at top and bottom of the element are different, so is the tension or compression, and the math to do that gets more involved. We are trying to simplify things a little so we can understand what is going on. Once you accept that the stresses resisting bending and shear vary along the height of the beam, you enter the origin of how to calculate the shear everywhere.
Oh, and the deck of cards only represents beams when the cards are glued together...
As I recall, when the beam is loaded with sandbags the web gets a big temporary 45° wrinkle across each panel between stiffeners. Almost like a tension member in a truss. It makes me think it's similar to a truss with tension across one side and compression across the opposite of the panel causing the buckle. I visualize it more as combined tension and compression across the web. Couldn't really visualize web shearing so I just now grasped a paper envelope on each edge and pushed with one hand and pulled with the other hand to simulate the cap motion under load. It sort of makes the 45° wrinkles appear.
I guess they call that opposing cap force reaction in the web the shear.
You now have your proof and may have already had your epiphany here! That wrinkled sheet showed you the tension at one 45 and compression (to the point of buckling) on the other 45. When the loads are small enough to not buckle the sheet, both directions carry some of the beam shear and the little square elements work as we described before. The 0-90 square element under shear IS THE SAME AS the +45/-45 element under tension across one pair of faces and compression on the other pair.
Once you buckle the side in compression, the load being carried by the compression direction drops to a lower value, maybe even approaching zero. IFF (and that is a BIG If and only If) the spar is stable enough to not collapse, the tension direction will still carry load until you approach yield strength on the tensile diagonal, so you can get to higher shear loads this way. This is Tension Field Design, invented before WWII and implemented most famously in the Mitsubishi A6M Zero. Look at the fuselage skins of most B-52's and you can see diagonal wrinkles.
The whole trick in Elastic Stability is making the structure stable (not allowing it to collapse) when parts wrinkle due to local elastic instability.
Usually in beam shear webs we try to fully stabilize the web and keep out of wrinkles and crippling. Plywood webs are put on with grain at +/- 45 and is usually pretty thick and thus stable by itself. Composite webs usually have it split in two on both sides of a foam core. Metal webs get stiffeners, big flanges around lightening holes, and other features. If you prevent wrinkling or other crippling, you can approach full shear strength in the web.
The rest of the stress picture in beams is when we take the web deforming in shear and then add in the top cap's shortening and the bottom cap's lengthening. mcrae0104 covered it nicely...
I have the outer wing spar made with .016" web at tip and .020" at root. Seems impossibly light but might work if loaded in pure tension since tension is optimal when possible.
I thought it was just called a Wagner beam. Never thought about the tension.
Will study chapter 15.2 Pure Tension Field Beams (Peery) today.
I have noticed the wrinkles on the B-52. (funny how age does that )
I think semi-tension fields are also covered at an introductory level in Stress Without Tears, if you have a copy of that handy.
Semi-tension fields is in Peery also, but haven't got to it yet. Don't have Stress Without Tears.
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