Beam Theory Explained - How Spars Work

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wsimpso1

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Note that, when I said 6000 lbs of shear, I got that from total w @ a cantilever beam support end= shear
1000 lbs (half the max gross weight) x 4Gs = 4000 lbs x 1.5 factor = 6000 lbs.
I agree with the 6000 pounds of shear.

My point is the big load on the bolted joints is not the shear. It is the bending moment at someplace around 350 - 400,000 in-lb. If the two lines of bolts are 6" apart, the linear force on each line is about 59 -67,000 pounds. The shear at 6000 is at right angles to the bending reaction, so they add by sum of squares (sqrt(a^2 + b^2) = c) or still about 59-67,000 pounds...

Billski
 

geosnooker2000

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So, here's something to think about. If the inboard end of the spar transitioned to a hollow tube (not pipe {round}, but rectangular tube), and the 8 bolts were moved 90 degrees to bolt the top and bottom flanges in a vertical fashion, would the shear forces be any different? The top joint would be in compression, the bottom joint in tension, but the result would still be two pieces of metal trying to slide past each other putting a shearing force on the bolts.
 

wsimpso1

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So, here's something to think about. If the inboard end of the spar transitioned to a hollow tube (not pipe {round}, but rectangular tube), and the 8 bolts were moved 90 degrees to bolt the top and bottom flanges in a vertical fashion, would the shear forces be any different? The top joint would be in compression, the bottom joint in tension, but the result would still be two pieces of metal trying to slide past each other putting a shearing force on the bolts.
You are playing "what if" on fitting and joint design on a thread about how beams work. This is thread drift.

If you want to discuss fitting and joint design, please start a thread on the topic and ask there. In the meanwhile, arm yourself with some knowledge - see if you can find a book on Strength of Materials and/or Shigley's book on Mechanical Engineering Design. Libraries can get them on interlibrary loan or you can buy (used) earlier editions pretty inexpensively. With a little study of the chapters on beam theory and joint design, you will be able to propose and answer many "what if" scenarios yourself. Understand that beams and joints and fasteners were figured out long before the Wrights invented airplanes. The possibility of figuring out something new in these topics is remote. You will be well served by being able to figure out beams and joints and fasteners as we understand them.

Billski
 

geosnooker2000

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I will shut up. But understand, I have had three semesters of structures in Architecture school, I know how to calculate and size beams, I have done it for 34 years. I have designed steel buildings, wood frame buildings, concrete buildings (including parking garages, footings, bridges, etc.). But this (wing design/spar design) is far more complicated than earthbound buildings of any sort. Because not only does the wing have a vertical load on it, which is the same as any cantilevered beam I've designed in the past, it also has a horizontal load from drag forces on the plane which is not the same as a building. On a building, those forces (such as seismic and wind loads) are always negated by bracing. The only bracing against drag forces is the skin of the wing (and possibly internal "x" braces which are not used in all wings). Is that enough? Is that what is totally relied upon to negate side load on the spar? I don't know. This is what makes me think this is not as simple as a cantilevered balcony with 2 steel beams holding up a floor joist system between them.

Personally I like conversation. I am self-employed and have very little human contact other than my wife and my clients, so I prefer to converse about things with people whom share my interests (mostly on message boards) instead of study them in books. Not to say that I won't. I just prefer the former.

If you think discussing the end connections of a beam is a "thread drift" in a thread about BEAMS, then I don't know what to tell you...
 

mcrae0104

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Geo, grab a copy of Stress Without Tears. It will pick up where your structural education left off. You are right that wings are not as simple as cantilevered balconies, but the good news is that once you really understand the underlying principles, you are equipped to solve either problem.
 

wsimpso1

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Please pardon my direction to go to the books. We can not hope to teach big chunks of undergraduate semesters on a forum. Topics are just too big to do without text, examples, and then practice.

I do feel that folks who follow this forum should be able to find the topics of discussion by title. The genesis of this thread was a number of folks talking about beams while it was obvious that they did not understand the pieces of beam interact... I am happy to have another thread talking about how to connect things together, and another on how wings carry loads, but they are different topics. Trying to meld beam design and hardpoint design in one thread, well, it complicates learning for those at the beginning of the process. Let's just do one thing per thread...

I am starting a thread on wing and spar interrelationships...

Billski
 

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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