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Beam Theory Explained - How Spars Work

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BBerson

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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.
 

wsimpso1

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My book shows horizontal shear is maximum at the neutral axis of the I beam and zero at outer cap lamina.
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.

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. :)
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.

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.
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.

Billski
 

BBerson

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If you want a real forum at Oshkosh, contact the Forum director Mark Forss at mforss@eaa.org
Might be late but can't hurt to ask.
Or set a time at the HBA meeting shelter area.
 

wsimpso1

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If you want a real forum at Oshkosh, contact the Forum director Mark Forss at mforss@eaa.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.

Billski
 

mcrae0104

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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.
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.)

01.jpg
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.

02.jpg
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).

03.jpg
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.)

04.jpg
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.
 

BBerson

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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.
 

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wsimpso1

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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...

Billski
 

BBerson

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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.
 
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wsimpso1

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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 opposeing 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...

Billski
 

BBerson

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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 :D)
 

mcrae0104

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I think semi-tension fields are also covered at an introductory level in Stress Without Tears, if you have a copy of that handy.
 

BBerson

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Semi-tension fields is in Peery also, but haven't got to it yet. Don't have Stress Without Tears.
 

Speedboat100

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So, for the dummies in class, I think it would be instructive for someone who has done one before, to run through an example calculation on an actual spar in an actual wing, showing how the loading diagram was arrived at. For my part, I will jump out and show my ignorance by guessing for y'all's amusement how a spar is loaded, just to demonstrate what some of us need help in understanding.

Keep in mind, I have a masters in Architecture, which means I had 3 semesters of structures, so I am coming at this with calculating beams mostly intended to withstand only forces in the Z axis (with the exception of steel columns - something about radius of gyration that I forgot as soon as I passed the class). We DID get into truss design, so I "had" a good grasp of vector addition, but boy do I need a refresher...

Example. a low-wing aluminum airplane with a span of 32'. Fuselage is 4' wide. Leaves us with a 14' cantilevered beam (Wing Spar). Assuming a tapered wing design, the loads in the positive G category would be a gradual increase as they reach the attached end. There would not be a point load on the outboard end of the beam diagram, even though I imagine picking up a model airplane by the tips of the wings to see if it is within acceptable CG - That is not how the air (wind) will pick up the plane. it will pick it up at every instance along the underside of the spar.

So my first thought was, yeah, it's just a unequal uniform load consisting of half the weight of the plane times however many G-forces you are designing for (times 1.5 safety margin). Then I thought, no... the wing ribs will intersect and attach at, for example, every 16" o.c. Maybe it should be a series of point loads, each point load conveying the area each rib caries. In a building, I would not make the distinction, as wood and steel are far to stiff to worry about buckling @ 16" o.c. braced, but we are talking about aluminum here.

Well, that should be enough to start a beat-down-fest of my ignorance. Have I got ANY of that right?

Many modern aeroplane and wind turbine wings are shell structures...with a beam or double beam side as a spar...and few wing formers. Usually for homebuilder wings are tested with sandbags for whatever Gs they are designed for. Flutter analysis is another ball game altogether.

Martin Hollmann was a wizard in this field: https://www.aircraftdesigns.com/product/modern-aerodynamic-flutter-analysis/
 

wsimpso1

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Speedboat100,

The topic of this thread is how beams work, which is plenty for one thread. Your post is serious thread drift - please stay on topic. If you want to talk internal structure design and flutter, there are threads about those topics on HBA.com where you can talk about them to your heart's content.

Billski
 

Hot Wings

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Stickies in the old software were locked - no posting except what was vetted by the moderators and the sticky manager. I'd like to see it go back to this if possible and start a second related thread for open discussion.

Yup, this is off topic as well and is being self reported to the moderators.
 
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