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Beam Theory Explained - How Spars Work

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proppastie

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Well I just did an interesting "thought experiment" ......building up my rear spar the stiffeners need filler to bring them up to the level of the caps......it seems like a lot of wasted metal/weight. So anyway I decided to do a little FEA to see how much stronger the long stiffeners were vs the short stiffeners only on the web of the spar. What an eye-opener....

With the same load/boundary conditions, the stress on the web with the short stiffeners is 54k psi...way past yield....where the stress on the web with the long stiffeners is 900 psi.
 

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Dave Hodges

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If V is "horizontal shear" ( Maximum at neutral axis)
What is "vertical shear"? ( mentioned in Peery, chapter 6.2)
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.
 

BBerson

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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.
Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
 

Dave Hodges

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Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
The cap that is in compression will tend to buckle. The opposite cap will be in tension. Let's say we have a weightless beam that is supported on the ends (it's a bridge), and a hundred pounds is hanging from the middle of the beam. Each support is pushing up 50 pounds. So the vertical shear force between the left bridge support and the middle of the beam is 50 pounds. The vertical shear stress would be 50 pounds divided by the cross-sectional area of the beam. The bending stress (compression and tension), as Bilski said, gets greater as you get further from the centroid. So at the top of the beam you have the bending stress (compression) along one axis. Bilski called it the X axis. You also have vertical shear stress along the Z axis. When you have two perpendicular stresses, each allowable stress is determined by the other pependicular stress. These two stresses (vertical shear and compression, at the top half of the beam, or vertical shear and tension at the bottom half) fit inside of a piece of pie, so to speak. If one stress is maximim, you can't have any of the other. If the other is maximum, you can't have any of the one. But if one is about 70% of max, the other can also be about 70% of max. Think of their magnitudes as two lines that form a rectangle that has to stay inside the fourth of a pie.
 

Dave Hodges

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Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
1561261916483-1406192674.jpg The best way to understand horizontal shear is to look at a truss with a top flange, a bottom flange and zigzags connecting to flanges. Here's the first page of the structural design chapter of my book How To Build a Hodgepodge Lodge. If you can follow the first three pages of this chapter, you can see why the horizontal shear is what it is.
 

Dave Hodges

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Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
1561262405135-314641855.jpg
 

Dave Hodges

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Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
15612625092851044105422.jpg
 

Dave Hodges

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Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
1561262595608-174674305.jpg
 

Dave Hodges

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Ok, that makes sense. I would think vertical shear is much less, maybe ten times less or something? So vertical shear could be mostly ignored if the sheet metal web is designed for the horizontal shear and attached with enough rivets to the caps and vertical stiffeners at 6" spacing.
I suppose the vertical shear comes only from the caps attempting to buckle?
1561262689349-2009480059.jpg
 

BBerson

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Each support is pushing up 50 pounds. So the vertical shear force between the left bridge support and the middle of the beam is 50 pounds. The vertical shear stress would be 50 pounds divided by the cross-sectional area of the beam.
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.
 
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wsimpso1

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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.
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...

Billski
 
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wsimpso1

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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.
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.

Billski
 
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