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.If you have a stack of notebook paper and lift up on the ends, the sheets of paper will slide across each other. That's the horizontal shear. In this same scenario, vertical shear would be displayed if all of the sheets were ripped like these guys with super strong grips rip phone books.
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...The vertical shear force in your example is 50 pounds. What is the horizontal shear force in your example bridge I beam (not a truss)? Let's say it is 40 feet total span.
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.My book shows horizontal shear is maximum at the neutral axis of the I beam and zero at outer cap lamina.
You are definitely on for that!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.
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.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.
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 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.
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.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.
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?