Truss design - are these ones OK?

[RohanBeckett]

well done!

how much strain is there, on the tracks going all the way around?

Might need more than 1 motor to drive it?  

[Blakbird]

19 hours ago, Captainowie said:

That's a 32x32 baseplate, for scale. The structure measures 170cm across the bottom and 136cm in height, with a gap of 109cm between the base supports. At the top, the horizontal truss spans 130cm, and has 124cm clearance from the ground. It sags approximately 8mm in the middle, which increases to about 10mm in the absence of the diagonal braces. It contains over 500 15- and 11-long beams (plus about 50 other beams of various lengths and shapes), and more than 2100 pins. It weighs a bit less than 4.4kg.

It's huge, but it's not a truss.  A truss is made of all triangles, and this looks like all rectangles.  If you added some diagonal members, especially in the horizontal part, it would be much stiffer and would sag less.

[Captainowie]

13 hours ago, Doug72 said:

To prevent sag add a King post or Queen posts above the horizontal beam.

See: https://en.wikipedia.org/wiki/King_post

Well I thought about that, but decided in the end that 8mm over 1.3m was an acceptable level of sag, compared to the engineering effort required to eliminate it.

40 minutes ago, RohanBeckett said:

well done!

how much strain is there, on the tracks going all the way around?

Might need more than 1 motor to drive it?  

I've not tested it yet, but I'm confident one motor will suffice. I can turn the driving axle with my fingers without too much effort. In any case, the clutch gear doesn't slip, and the gear would slip long before the motor runs out of torque.

6 minutes ago, Blakbird said:

It's huge, but it's not a truss.  A truss is made of all triangles, and this looks like all rectangles.  If you added some diagonal members, especially in the horizontal part, it would be much stiffer and would sag less.

I'm not convinced. I found this reference to a truss not made of triangles (https://en.wikipedia.org/wiki/Truss#Vierendeel_truss), although I'll concede the technicality - even though it's called a truss, that article says "it does not fit the strict definition of a truss (since it contains non-two-force members)".

Regardless, I don't think that adding diagonals would help. The reason you see triangles all the time (in trusses and elsewhere) is that they're a stable shape when all you have are joints and rods. But I'm using 5x7 beams, which are a stable square all by themselves.

Having said that, I'm willing to give it a try. I'll report back.

Owen.

[Blakbird]

1 hour ago, Captainowie said:

I'm not convinced. I found this reference to a truss not made of triangles (https://en.wikipedia.org/wiki/Truss#Vierendeel_truss), although I'll concede the technicality - even though it's called a truss, that article says "it does not fit the strict definition of a truss (since it contains non-two-force members)".

Regardless, I don't think that adding diagonals would help. The reason you see triangles all the time (in trusses and elsewhere) is that they're a stable shape when all you have are joints and rods. But I'm using 5x7 beams, which are a stable square all by themselves.

"Truss" and "triangle" are a matched set.  Building from triangles is the thing that makes a structure a truss.  You can't call something a truss which is not made from triangles, which Wikipedia concedes in the example you referenced.  More technically, a truss is made from all two-force members.  This means that a free-body diagram of any element of a truss has only two forces, and they are along the direction of the member.  The result is that all truss stresses are only tension or compression.  This results in light, efficient structure which cannot deform unless something buckles.  The way you built your structure is also stable, but it relies on bending stresses in the corners of those 5x7 frames to become so.  Bending produces vastly more concentrated stress risers than tension or compression, and therefore much more material is needed to build a structure of equivalent stiffness.  By using triangles instead of frames, the structure is not only stiffer but can also be much lighter which, cycling back to the beginning, means less deflection.

Don't get me wrong here, I think you've achieved something remarkable structurally with the rigidity of the frame at this size.  I'm merely pointing out that "truss" has a very specific engineering meaning and this isn't a truss.  Even adding some diagonal members won't make it a truss because the presence of those frames means that you have bending stresses in your structure.  I do think that some well placed diagonals will help though.  You might even choose to pick a length that is slightly too short which will force the horizontal member to be slightly bowed up without gravity, and therefore closer to level when supporting weight.

[dr_spock]

Nice.  Look forward to seeing it in action.  Would it be considered a conveyor bridge then?  

 

[Captainowie]

22 minutes ago, Blakbird said:

"Truss" and "triangle" are a matched set.  Building from triangles is the thing that makes a structure a truss.  You can't call something a truss which is not made from triangles, which Wikipedia concedes in the example you referenced.  More technically, a truss is made from all two-force members.  This means that a free-body diagram of any element of a truss has only two forces, and they are along the direction of the member.  The result is that all truss stresses are only tension or compression.  This results in light, efficient structure which cannot deform unless something buckles.  The way you built your structure is also stable, but it relies on bending stresses in the corners of those 5x7 frames to become so.  Bending produces vastly more concentrated stress risers than tension or compression, and therefore much more material is needed to build a structure of equivalent stiffness.  By using triangles instead of frames, the structure is not only stiffer but can also be much lighter which, cycling back to the beginning, means less deflection.

Don't get me wrong here, I think you've achieved something remarkable structurally with the rigidity of the frame at this size.  I'm merely pointing out that "truss" has a very specific engineering meaning and this isn't a truss.  Even adding some diagonal members won't make it a truss because the presence of those frames means that you have bending stresses in your structure.  I do think that some well placed diagonals will help though.  You might even choose to pick a length that is slightly too short which will force the horizontal member to be slightly bowed up without gravity, and therefore closer to level when supporting weight.

Ok, so I may have been a bit guilty of sloppy language. Point taken. I still contend, however, that a 5x7 frame is lighter and stiffer than anything of comparable size that you could make with LEGO triangles, if only for the little bit of give that's present in the pin joints.

Nonetheless, I added some diagonal members to the horizontal ... section, which has reduced the sag by about 2mm. I think I'll keep it like this:

 

It is hard, though, to accurately measure the sag. I'm trying to measure a difference of a couple of mm, using a tape measure that's tricky to keep vertical from a long way off the table.

[DrJB]

Looks good. Though, looking back through history, most bridges have an 'arch' geometry for maximum strentgth/stiffness (minimum sag). A design is as strong as it's weakest components. In this case, I think the pins might be the weakest. The only way to 'reinforce' the design further, is to add as many pins as there are holes aligned.

[Blakbird]

I love that Futurama quote!  I use it myself from time to time.  You may very well be right that the flexibility of the pins is more important than what you get from the diagonal members.  Glad to hear it made some small improvement.  If the deflection under the weight of the horizontal beam alone is noticeable, then it is likely that the diagonals will help even more when you are suspending a load.

Interestingly, as long as we are talking about "technically correct", almost no real world trusses are actually trusses.  Even a steel truss bridge does not actually have pinned joints.  The joints are usually welded plates or have many rivets.  In either case, they introduce bending into the members and therefore they are no true trusses. A real truss is remarkably hard to build because you need very strong pins which connect a stack of many members.  And LEGO sets which have apparent trusses, such as the boom in 8288, are not true trusses either.  Many of the pin joints in that boom are in the middle of members, and true trusses only have pins at the end. 

[Erik Leppen]

Actually, why did you choose to go for a rectangular gate? The forces concentrate in the corners. Why not go for a semicircular gate? There is a reason many bridges have round arches. (I know bridge arches are not circular, but a circle is easier to build than a parabola.)

 

Also, I think 16L links are perfect tension members, as they are very light.

[MIchaelF]

2 hours ago, Blakbird said:

  And LEGO sets which have apparent trusses, such as the boom in 8288, are not true trusses either.  Many of the pin joints in that boom are in the middle of members, and true trusses only have pins at the end. 

For someone with zero subject matter knowledge (i.e me!) would the boom in 42042 be considered to have true trusses?

[Captainowie]

9 hours ago, Blakbird said:

I love that Futurama quote!  I use it myself from time to time.  You may very well be right that the flexibility of the pins is more important than what you get from the diagonal members.  Glad to hear it made some small improvement.  If the deflection under the weight of the horizontal beam alone is noticeable, then it is likely that the diagonals will help even more when you are suspending a load.

Interestingly, as long as we are talking about "technically correct", almost no real world trusses are actually trusses.  Even a steel truss bridge does not actually have pinned joints.  The joints are usually welded plates or have many rivets.  In either case, they introduce bending into the members and therefore they are no true trusses. A real truss is remarkably hard to build because you need very strong pins which connect a stack of many members.  And LEGO sets which have apparent trusses, such as the boom in 8288, are not true trusses either.  Many of the pin joints in that boom are in the middle of members, and true trusses only have pins at the end. 

That was with a load (the track links). Unloaded, the diagonals within the horizontal section made about the same difference to the sag as the diagonal braces at either end, but there was no measurable difference using both as compared to using one or the other. But with the load I was able to measure a small difference - and then it was easier to decide to keep it as it was than to take the members out of the structure!

Ok then, here's a thought experiment for you. Suppose you did construct a structure that used very strong pins connecting a stack of many members. Suppose you then welded those pins in place such that they could no longer pivot. Is the result still a truss? If it is (because simply welding the pins in place doesn't change the forces acting on the members), does it then suddenly become not a truss once a load is placed on the structure? Can you design a bridge whose truss-ness depends on whether there's any traffic on it? Conversely, if it isn't (because you're requiring freely-rotating joints - even though their configuration is stable and don't actually rotate, but whatever), then what happens as the original truss rusts? The joints move over time from freely-rotating to rusted-stiff - does the structure similarly change continuously from truss to non-truss?

9 hours ago, Erik Leppen said:

Actually, why did you choose to go for a rectangular gate? The forces concentrate in the corners. Why not go for a semicircular gate? There is a reason many bridges have round arches. (I know bridge arches are not circular, but a circle is easier to build than a parabola.)

For flexibility of configuration. If I want to make it higher (shorter) or wider (narrower) I can just add (remove) length to (from) each section.

[Blakbird]

21 hours ago, MIchaelF said:

For someone with zero subject matter knowledge (i.e me!) would the boom in 42042 be considered to have true trusses?

Technically, no.  There are lots of joints in that boom that are not at the end of beams.  However, the diagonal members will always add a lot of stiffness and are helpful even if the structure cannot be categorized as a truss.

13 hours ago, Captainowie said:

Ok then, here's a thought experiment for you. Suppose you did construct a structure that used very strong pins connecting a stack of many members. Suppose you then welded those pins in place such that they could no longer pivot. Is the result still a truss? If it is (because simply welding the pins in place doesn't change the forces acting on the members), does it then suddenly become not a truss once a load is placed on the structure?

Whether or not a load is applied is not relevant to determining whether something is a truss.  The geometry and constraints will determine how forces are distributed, and as long as every member is a two-force member then it is a truss.  Your example with welded pins would not be a truss.  As soon as you have welded pins then the beams are not two force members because there is a moment constraint and therefore bending can be supported.  Releasing the moment constraint at the ends of beams by pinning them is a very important technique for reducing the stress in structures.  So welding the pins in place DOES change the forces acting on the members.  In fact, it has a very major effect on the overall stresses.

13 hours ago, Captainowie said:

Can you design a bridge whose truss-ness depends on whether there's any traffic on it? Conversely, if it isn't (because you're requiring freely-rotating joints - even though their configuration is stable and don't actually rotate, but whatever), then what happens as the original truss rusts? The joints move over time from freely-rotating to rusted-stiff - does the structure similarly change continuously from truss to non-truss?

As above, whether or not there is a load it is not relevant to whether or not the structure is a truss.  Unless the bridge is in zero gravity, it will always have to support at least its own weight anyway.

For your thought experiment, if a bridge was designed as a true truss with pinned joints and those pins and joints rusted and jammed, then that would indeed change the forces in the members and vastly decrease the strength of the bridge.  Of course, as I mentioned above, bridges don't actually use pinned joints anyway so there is no such degradation in the real world.  There are real trusses in other applications though like cranes, spacecraft, and aircraft.  The joints have to be inspected regularly.

Interesting engineering discussion folks!  I didn't expect this one on a LEGO forum, but it makes sense when designing such an ambitious model.

[Captainowie]

The thing that I really can't get past is this one:

13 hours ago, Blakbird said:

So welding the pins in place DOES change the forces acting on the members.

...

For your thought experiment, if a bridge was designed as a true truss with pinned joints and those pins and joints rusted and jammed, then that would indeed change the forces in the members and vastly decrease the strength of the bridge.

I completely accept that welding (or rusting) the pins in place change the forces that CAN act on the members, but I can't see how it necessarily DOES. If I have a true truss whose members are experiencing only compression or tension, and then I fix the joints in place, the members are still experiencing only compression or tension, aren't they? If there's no change to the external forces involved (what I meant above by "adding a load"), then since there's no acceleration there must also be no change to the internal forces involved, right?

The only explanation I can think of is something that would require the joints to pivot - i.e. a (small) change in geometry. The only way I can see that happening is if we allow the members to deform with applied forces: members under tension get slightly longer, and members under compression get slightly shorter. But given that we're talking strict definitions and theories, I would have assumed ideal members (uniform density, completely inflexible, etc).

[allanp]

Members will stretch and compress slightly which, due to trigonometry, will cause a change in the triangles corners. If a joint cannot be allowed to pivot to compensate this will cause a bending force in the members which would weaken it. But I would think the difference is so minimal as to be of very little practical use. Many real life structures have their truss members welded into one solid piece.

Edited March 3, 2017 by allanp

[Blakbird]

16 hours ago, Captainowie said:

I completely accept that welding (or rusting) the pins in place change the forces that CAN act on the members, but I can't see how it necessarily DOES. If I have a true truss whose members are experiencing only compression or tension, and then I fix the joints in place, the members are still experiencing only compression or tension, aren't they? If there's no change to the external forces involved (what I meant above by "adding a load"), then since there's no acceleration there must also be no change to the internal forces involved, right?

Nope, the forces (actually stresses) are completely different once the pins are locked, even with the same load.  I can't really think of a good way to describe it in text, but it becomes really obvious when you draw a free body diagram of the forces and moments.

Consider this as a simple example.  It is not a truss, but it shows the principle.  Imagine a straight horizontal beam made from two members, pinned at the ends and in the middle.  Now hang a weight from the middle joint.  The beams can't support any moment, so they start to rotate.  At that point, they start acting like a rope with a weight in the middle and go into tension.  Now apply the same weight but this time lock the middle pin.  Now the beams CAN support a moment and they do NOT go into tension.  Instead they go into bending.  Depending on the shape of the cross section, the bending stress may be vastly greater than the tension stress.

The same thing applies to a truss.  The bottom surface of a truss is made from many individual links, generally in tension.  If you weld or lock them together, then they are instead one continuous beam and not individual links.  Every place those vertical or diagonal truss members tie in to that horizontal beam, they will be introducing an out-of-plane load which gets reacted as bending.  When every joint is pinned, bending cannot occur so that load gets transferred out to adjacent members in tension or compression instead.

That's the best I can do without diagrams.  Check out any Static or Mechanics of Materials textbook and you'll find plenty of equations and examples that demonstrate the principle.  One of the big analytical benefits of a truss is that it is statically determinate.  What that means is that if I write out the equations of equilibrium, there will always be a greater number of equations than unknowns, and therefore I can solve for the exact forces in every member no matter how many members it has.  On the other hand, if the members are fixed then the structure is statically indeterminate.  This means that I can't solve it with Statics because there are not enough equations.  I need to use Deformable Body Mechanics to take into account the stiffness and deflection of each member to solve.  It is vastly more complicated to do by hand, and also represents a less efficient structure.  I had to solve an 8 member system like that by hand in college and it took half a notebook.

It is theoretically possible to design a very simple structure in which the boundary conditions don't matter.  For example, imagine a vertical link of 5 members with a weight hanging from them (a chain).  In this case, it doesn't matter if the links are pinned or fixed, you'd still get exactly the same forces and all the stresses would be tension.  But as soon as you have a load that is applied in a direction that is not parallel to every link, then the answer does change with the boundary conditions.

I always marvel at how far we can go down into the weeds in analyzing even the simplest structure.  The very first thing you learn about in Statics is analyzing beams, and the simplest combination of beams is the truss.  However, in the real world it is impossible to ever have a real truss.  A real truss has load applied only at the ends, but in the presence of gravity every member needs to support its own weight which is a distributed load and NOT applied at the end, so there is always bending present to some degree.  In most engineering disciplines you can ignore this and say "close enough for all practical purposes" because you have a nice big safety factor on top of the biggest load you think you'll ever see.  But in other disciplines like aerospace the safety factor is small and you can't ignore much of anything or you end up sending a billion dollar mission to Mars which crashes because the legs didn't deploy.

If I've gone too far into the theory here, feel free to tell me to go away.  I find it interesting, but others may find it annoying.

[Captainowie]

3 hours ago, Blakbird said:

If I've gone too far into the theory here, feel free to tell me to go away.  I find it interesting, but others may find it annoying.

I very much doubt you'd find anyone on this forum willing to tell Blakbird to go away! Besides, this is a topic about truss design and we're talking about trusses.

 

3 hours ago, Blakbird said:

Imagine a straight horizontal beam made from two members, pinned at the ends and in the middle.  Now hang a weight from the middle joint.  The beams can't support any moment, so they start to rotate.  At that point, they start acting like a rope with a weight in the middle and go into tension.  Now apply the same weight but this time lock the middle pin.  Now the beams CAN support a moment and they do NOT go into tension.  Instead they go into bending.  Depending on the shape of the cross section, the bending stress may be vastly greater than the tension stress.

Ok, so assume we have two ideal (massless, rigid) members of unit length, pinned together (with massless pins). Their ends are securely fixed at the same height to immovable walls two units apart. The geometry of the situation allows that they will form a straight horizontal beam. Then we add the weight. In order to remain in equilibrium, there needs to be an upwards force to counter the weight of the weight. The members rotate slightly, so that the tension in the rods has a slight upwards component, and everything balances. But in order for that to happen, the distance between the ends of the rods must be less than the sum of the length of the rods. So either the rods deform (violating one assumption) or the walls move slightly inwards (violating another assumption). Or, I guess, the mass of the weight distorts spacetime enough that you can have a non-degenerate triangle with side lengths of 1, 1, 2. :-) This is doing my head in!

In any case, I get that with the locked pin there's a bending force on what is now acting like one single member. Ok, so we unlock the pin again, and let the members rotate (the paragraph above notwithstanding!). Once they've reached equilibrium, there is nothing making the joint want to rotate. So we lock the pin again. How does that induce a bending moment? If you have a joint that could move but doesn't because its forces are balanced, how does preventing it from moving from that position all of a sudden make the forces unbalanced? (I recognise that my use of "balanced" and "unbalanced" here are probably not correct, but I hope you can still see what I'm getting at).

Putting it another way, consider the two rods with a fixed joint in a v-shape, connected at the ends of the rods to the wall at the same height. I can imagine that if I pull down on the vertex that there will be some bending stress that will want to make the angle of the V become smaller. If I now move the walls a smidge closer together, I create a bending stress that wants to make the V angle wider. What happens when I pull on it now? The joint has a force that wants to make it wider, and a force that wants to make it narrower. Surely if I pull exactly hard enough, I can make those forces balance each other out, and there's no bending stress in the joint. Can you see why this is a reasonable (if ultimately futile) objection to your statements?

I have made an attempt at drawing the freebody diagrams that should make it "obvious", but I don't know how to do that for an angled shape. I can see how they could be different, but what I can't get my head around is how a system can transition from one state to the other. Maybe there's something in the fact that I'm thinking about forces rather than stresses (I studied physics, not engineering!), but Wikipedia tells me that stress is just force per unit area, so I can't see how that would make a difference.

[Blakbird]

8 hours ago, Captainowie said:

Ok, so assume we have two ideal (massless, rigid) members of unit length, pinned together (with massless pins). Their ends are securely fixed at the same height to immovable walls two units apart. The geometry of the situation allows that they will form a straight horizontal beam.

In this particular ideal example, there is no stable solution.  Two massless, rigid links oriented horizontally cannot support a vertical load, as you correctly surmised.  It's is impossible because there is no restraint in that direction.  Of course, there is no such thing as massless or rigid which is why it works in real life.

Quote

In any case, I get that with the locked pin there's a bending force on what is now acting like one single member. Ok, so we unlock the pin again, and let the members rotate (the paragraph above notwithstanding!). Once they've reached equilibrium, there is nothing making the joint want to rotate. So we lock the pin again. How does that induce a bending moment? If you have a joint that could move but doesn't because its forces are balanced, how does preventing it from moving from that position all of a sudden make the forces unbalanced? (I recognise that my use of "balanced" and "unbalanced" here are probably not correct, but I hope you can still see what I'm getting at).

 OK, I finally understand what you are trying to say.  If the structure were pinned and then allowed to strain into a stable static position, then you are correct that converting the pins to fixed would not change anything.  However, as soon as there was any variation in the load, including removal of the load, then the structure would respond in a different way.  So you are right, but in practice this situation cannot really exist.  Even the action of building the structure in the first place results in a variation of loading, and for something like a bridge, anything crossing it or even the wind would change the load.  For your LEGO structure, picking it up and moving it would change the load and it would respond to that differently once the pins had been fixed.  I get your point though.  If you could really change a joint from pinned to fixed once the structure was already at equilibrium and nothing else changed, then the stresses would not change either.

Quote

I have made an attempt at drawing the freebody diagrams that should make it "obvious", but I don't know how to do that for an angled shape. I can see how they could be different, but what I can't get my head around is how a system can transition from one state to the other. Maybe there's something in the fact that I'm thinking about forces rather than stresses (I studied physics, not engineering!), but Wikipedia tells me that stress is just force per unit area, so I can't see how that would make a difference.

Stress is indeed force per unit area, but the way it is distributed is different in different modes.  In tension, stress is uniform and a function of cross sectional area which is typically of the second power to geometry.  In bending, stress is non-uniform and a function of first moment of inertia which is fourth power to geometry.  This is why the bending stress can be locally much higher even for the same load.  Consider a 0.1x0.1m beam with a length of 1m and a load of 1N at the tip.  If this were hanging vertically, the stress would be pure tension.  The area is 0.01m^2 and the stress is 100Pa uniformly distributed across the section.  Now consider the same beam horizontally fixed at one end with the weight hanging from the other end.  The beam is now in bending and shear.  The moment of inertia is 8.33e^-6 m^4 and the moment is 1N-m.  The peak bending stress is 6000Pa at the extreme fiber, but varies to zero at the center and -6000Pa at the other side.  As you can see, in this example the stress is 60x higher when the beam is in bending.  This factor is not constant though, because it depends on the length of the beam and the shape of the section.  But it does illustrate why limiting a structure to only tension and compression is so efficient. 

I think we just describing two different problems.  If I draw a truss bridge and then put a load in the middle and solve for the forces and stresses in all the members, I'll get a totally different answer if the pins are fixed versus pinned.  However, that's not the problem you are talking about.  You are talking about changing the joints in the middle of the problem.  The pinned structure strains, reaches equilibrium and THEN you change to fixed.  In this case, no forces or stresses change.  I've never seen an engineering problem presented that way though.  The default solution methodology is to start from an unloaded state and then transition to a loaded state.

[Captainowie]

10 hours ago, Blakbird said:

I think we just describing two different problems.

Yep, in my experience that's what happens when two intelligent people seem to not be able to agree on something!

10 hours ago, Blakbird said:

I've never seen an engineering problem presented that way though.

Ah, but this wasn't supposed to be an engineering problem, it was supposed to be an edge case to try to nail down a definition. If fixing the joints in place after the structure reaches equilibrium doesn't change the forces involved, would you still call the resulting structure a truss? If yes, then you could, in theory, construct a structure whose truss-ness depends on its load. This is neither a good nor a bad thing, just a consequence of the definition.

[Blakbird]

44 minutes ago, Captainowie said:

Ah, but this wasn't supposed to be an engineering problem,

For me, everything is an engineering problem!

44 minutes ago, Captainowie said:

If fixing the joints in place after the structure reaches equilibrium doesn't change the forces involved, would you still call the resulting structure a truss? If yes, then you could, in theory, construct a structure whose truss-ness depends on its load. 

No, a truss is a truss based on the geometry and constraints.  It will always be a truss no matter how it is loaded.  I can apply a load in any direction at any joint and the whole thing is still two-force members.  A structure which happens to be composed of two-force members under a single load condition but would cease to be so if the load changed would not be called a truss.  I don't know what you would call it!  A structure which starts out as a truss but then rusts into a solid brick also has no name.  Maybe in Dutch....

[Captainowie]

22 hours ago, Blakbird said:

No ...  A structure which happens to be composed of two-force members under a single load condition but would cease to be so if the load changed would not be called a truss.

Very well. Thanks for an enlightening discussion.

Owen.

[Blakbird]

12 hours ago, Captainowie said:

Very well. Thanks for an enlightening discussion.

Now we can get back to talking about your awesome LEGO creation!  You know, your tr.....  frame.