pic: Beam Bending Example

[kramarczyk]

loading photo ...

[=Martin=Taylor=]

So basically what I am seeing here is that most of the stress results along the sides, which are forced to stretch when a force is applied. Right? So there really isn't that much stress on the top? How would it look if there were a bunch of holes everywhere (on the top and sides)?

[bigbeezy]

what software was this done with? Inventor maybe???

[Lowfategg]

Originally Posted by bigbeezy

what software was this done with? Inventor maybe???

Yup its inventor. I tried to do this with my beams but I could not get the results right. No matter how little force I applied it looked like a train hit the beam. I guess I did something wrong.

[MrForbes]

For the sake of making the discussion even more interesting, how about making several different shapes and sizes of beam, and applying the same (relatively small) load to them and showing them all? For example, you might have rectangular tubing with the load applied to the narrow side, and to the wide side, and round and square tubing, and an I beam, and a channel.

Also it would be nice if you'd give a quick explanation of how to set up the constraints in Inventor so others can do this, as it's not obvious to the novice Inventor user.

[Madison]

Originally Posted by Lowfategg

Yup its inventor. I tried to do this with my beams but I could not get the results right. No matter how little force I applied it looked like a train hit the beam. I guess I did something wrong.

Inventor exaggerates the deformation considerably, as you're meant to use the scale displayed on screen to determine the appropriate values rather than a visual representation.

[DonRotolo]

Originally Posted by Hachiban VIII

So basically what I am seeing here is that most of the stress results along the sides, which are forced to stretch when a force is applied. Right? So there really isn't that much stress on the top? How would it look if there were a bunch of holes everywhere (on the top and sides)?

The stresses are highest on the top and bottom. Along the sides (cool blue) there isn't much stress at all.

Holes will have a different effect depending on where they are placed. If in the blue areas, hardly any effect (so long as structrual rigidity is maintained) - while along the red areas (top and bottom) these holes will affect the strength of the beam, depending on their size and shape.

[sanddrag]

The question on holes got me curious. Here is a summary of my test and results

1"x1"x1/16" wall 6061 Aluminum Box Tube, 18" long. 0.625" holes down opposite sides, spaced at 1" centers
End faces fixed constraint
500 psi pressure applied to a 2" long patch, on the top side center of the tube.

For the given loading and fixed constraints, the part exhibited a 29% higher maximum stress with the holes than without. The part with holes had approximately 15% less mass than the part with holes.

Anyhow, if your holes are on the faces 90 degrees to the one the force is being applied, you don't lose all that much strength. This is the principal behind I-beams. The more cross sectional area you put further from the center, the stronger the beam will be, because it will have a higher moment of inertia. This is why beams with a taller cross section are much stronger in bending, and why I beams have so much material so far away from the center.

[kramarczyk]

Originally Posted by squirrel

For the sake of making the discussion even more interesting, how about making several different shapes and sizes of beam, and applying the same (relatively small) load to them and showing them all? For example, you might have rectangular tubing with the load applied to the narrow side, and to the wide side, and round and square tubing, and an I beam, and a channel.

Also it would be nice if you'd give a quick explanation of how to set up the constraints in Inventor so others can do this, as it's not obvious to the novice Inventor user.

I'll see what time I can get to run a series of these examples, first as the base sections, then we can send them to Jenny Craig. I'll also throw the kitbot rail in as a reference point. I used the left vertical edges as my constraints and fixed them only in two dimensions... on the axis parallel to the load, and on the axis that is both perpendicular to both the load and the axis of the beam. They were left free on the axis that would allow them to get closer to each other. A picture would help, I'll add that to the list. <edit> The part used was a 38" long piece of 2x2 inch aluminum with at 1/8" wall </edit>

Originally Posted by M. Krass

Inventor exaggerates the deformation considerably, as you're meant to use the scale displayed on screen to determine the appropriate values rather than a visual representation.

I have yet to see a system that can't scale the displayed deformation; deflections of a few thousands of a inch aren't terribly visible otherwise. You're right though, go with the numbers.

Originally Posted by sanddrag

The question on holes got me curious. Here is a summary of my test and results

1"x1"x1/16" wall 6061 Aluminum Box Tube, 18" long. 0.625" holes down opposite sides, spaced at 1" centers
End faces fixed constraint
500 psi pressure applied to a 2" long patch, on the top side center of the tube.

For the given loading and fixed constraints, the part exhibited a 29% higher maximum stress with the holes than without. The part with holes had approximately 15% less mass than the part with holes.

Got any screen shots? With the end faces held fixed I would expect to see some hot spots out near the end.

Originally Posted by sanddrag

Moment of inertia includes mass to the 1st power, and distance to the 2nd power. So, roughly speaking, when the "meat" of your beam is twice as far away from the central axis, then it gets 4 times stronger.

The applicable moment of inertia(actually the second moment of area, but commonly referred to as the Moment of Inertia, I. Don't ask me why, I have know idea why we confuse people like that.) formula for rectangular tubes is ((W*H^3)-(w*h^3))/12 Where W & H are the outer rectangular width and height respectively and w & h are the inner rectangular width and height respectively. Point being that mass is not generally involved in beam bending calculations unless it contributes a significant amount to the actual load. Additionally, the strength of the beam increases at the cube of the height of the beam, so doubling the beam depth increases the bending resistance by 8x.

http://en.wikipedia.org/wiki/List_of...nts_of_inertia
http://web.mst.edu/~mecmovie/index.html

[Daniel_LaFleur]

Originally Posted by Hachiban VIII

So basically what I am seeing here is that most of the stress results along the sides, which are forced to stretch when a force is applied. Right? So there really isn't that much stress on the top? How would it look if there were a bunch of holes everywhere (on the top and sides)?

Correct. But the stress on the sides is in compression or tension which is usually far stronger than in sheer which is what the top (and bottom) are in.

Holes would create 'stress concentrators' where the lines of flux (stress) would be concentrated between the edges of the holes and the sidewall. The stress loads here would be the same as in the full part but that section would only have the strength of the, now smaller, width.

[Rick TYler]

Originally Posted by kramarczyk

Point being that mass is not generally involved in beam bending calculations unless it contributes a significant amount to the actual load. Additionally, the strength of the beam increases at the cube of the height of the beam, so doubling the beam depth increases the bending resistance by 8x. [/url]

I've confessed in these pages before that I am not an ME, so treat this as a question rather than an argument, please.

Is the "strength" of the beam the same as "bending resistance?" I read in Dave Gerr's book on boat design that the stiffness (which I would think is the same as bending resistance) of a beam, all else being equal, goes up with the fourth power of thickness, not the third power. Of course, the failure strength might go up as the third power -- about this I know nothing.

[kramarczyk]

Originally Posted by Rick TYler

I've confessed in these pages before that I am not an ME, so treat this as a question rather than an argument, please.

Is the "strength" of the beam the same as "bending resistance?" I read in Dave Gerr's book on boat design that the stiffness (which I would think is the same as bending resistance) of a beam, all else being equal, goes up with the fourth power of thickness, not the third power. Of course, the failure strength might go up as the third power -- about this I know nothing.

Good question... strength and bending resistance are not the same. Strength is basically the material's ability to resist being pulled apart (breaking due to stress). Stiffness is the part's resistance to deflection. Stiffness is contributed to by both the material choice (Young's modulus, E) and the shape of the part (Moment of Inertia, I). In many cases devices fail due to being insufficient stiffness rather than not being strong enough... think of a long, small diameter PVC arm.

I'm not familiar with Gerr's book, so I can't speak to the context. <speculation> Since you reference 'thickness' he may be experiencing both an increase in height and width of the section, but I can't be certain.</speculation>

Not sure if I am clarifying or confusing things... pics to follow.

[kramarczyk]

Originally Posted by kramarczyk

I used the left vertical edges as my constraints and fixed them only in two dimensions... on the axis parallel to the load, and on the axis that is both perpendicular to both the load and the axis of the beam. They were left free on the axis that would allow them to get closer to each other.

Clarification... "One was left free on the axis that would allow them to get closer to each other, the other was fixed on this axis." Leaving both free causes the software to attach weak springs to contain that degree of freedom which is a bad habit to get into. I also failed to mention that I'm using 1000 lbf uniformly distributed loads as I look at these beams. It's a completely ficticious load, but since we're only looking at how the beams carry the loads and not the absolute magnitudes I don't see an issue.

Attached is the deflection diagram that goes with the original post...
3.475 lbm
8643 psi max stress
.133 in max deflection

[kramarczyk]

OK, I'm posting each of the beam sections seperately because I think it organizes the data better... holler if I should stop.

4" x 2" aluminum tube (wide)
1/8" wall
1000 lbf uniform load
5.3283 lbm
5158 psi max stress
.078 in max deflection

[kramarczyk]

Last one of this batch...

2" x 4" aluminum tube (tall)
1/8" wall
1000 lbf uniform load
5.3283 lbm
5096 psi max stress - Note this is a singularity out at the support where as the others are in the middle. The middle of this is somewhat lower than 3400psi
.026 in max deflection

Are we getting anything out of this? (i.e. Keep going?)