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Discussion in 'General Talk' started by rlev, Sep 27, 2015.
Anyone know what the max load is for the C-Beam Linear Actuator?
How about the other types?
Oh, boy. I say that as it is very close to my question about their 20x20 though mine was deflection based. I never did get an answer and Solidworks I don't trust that I did something right because 100 pounds in the middle of a 500mm 20x20 span attached at each end should deflect way more than 0.1mm.
Max load until failure would use some of the same sets of equations as I attempted and I never could get an answer so I am not sure anyone has tried to make these max load fail or measure the deflection based on weight in any sort of laboratory environment as they would on glass. If you get an answer let me know as I would love to know about the deflection as well.
Edit 11-5-15: just noticed I lost all the superscript formating ... just replaced with "^" ...
I did do a little math but I'm not 1000% sure I have it right ...
Don't remember where I found every thing and some of the text is copied or paraphrased but here's my work
Columns fail by buckling when their critical load is reached. Long columns can be analyzed with the Euler column formula
F = n π^2 E I / L^2
F = allowable load (lb, N)
n = factor accounting for the end conditions
E = modulus of elastisity (lb/in^2, Pa (N/m^2))
L = length of column (in, m)
I = Moment of inertia (in^4, m^4)
We need to solve for F the critical load where bucking will start to occur
for n -- will assume that both ends of the lead screw are hinged (n=1) ... in reality they might be considered closer to fixed (n=4) if there was no play on the bushings on the ends
for E -- the modulus of elasticity for stainless Steel (AISI 302) is 180x10^9 N/m^2 OR 180 GPa
find I --Area Moment of Inertia
Ix = ∫ y^2 dA
Ix = area moment of inertia (m^4, mm^4, inches^4)
y = the perpendicular distance from axis x to the element dA (m, mm, inches)
dA = an elemental area (m^2, mm^2, inches^2)
The cross section of the lead screw if a circle so
Ix = π r^4 / 4
= π d^4 / 64
For our 8mm stainless steel Acme lead screw
The threads are 8mm but what we need to consider is the rod part which is really only 6mm
So, d= .006m giving
Ix = π (.006)^4 / 64
going back to our bucking force equations F = n π^2 E I / L^2
F = (1)(π^2)(180x10^9)(π (.006)^4 / 64) / L^2
F = 355055841.35 / L^2
In theory a perfect screw with no other forces on it will start to buckle at
length max newtons max pounds
250mm 5680 1277
500mm 1420 319
1000mm 355 79.8
if the ends are fixed you can multiple these numbers by 4
Since I'm fairly conservative I want a factor of 5x to 10x of safety
Especially since I'm unsure of if I actually have perfectly fixed ends.
So, I would multiple the above numbers by a factor of 4/10ths and limit myself to no more then 32lbs on a 1000mm screw.
Maybe less if I don't trust n = 4
This does not take into account if the threads or block can hold that weight. It also does not take into account what happens when you drive the screw and what material wear may happen ... so the right number maybe much lower.
I'm sure If screwed up (pun intended) my math or thinking ... someone will post a correction ...
You need to understand the difference between strength and serviceability. Strength of these materials is irrelevant as the system will be totally unusable long before it fails. Take for instance an extrusion spanning 1500mm with a rolling load which is not heavy enough to fail the beam but is heavy enough to cause it to deflect 100mm at the center. Is the system serviceable with a 100mm deflection? Do you have enough extra motor power to be able to move it what is now uphill to get back to the end? This is serviceability. With most of the systems shown in this site, 1mm of deflection may be all that is necessary to make them unusable and this level of deflection may occur at a very small fraction of what they are capable of strengthwise.
So to answer your original question, nobody knows as nobody goes there. There is no reason to. You need to first define the parameters of serviceability then work towards a machine that meets those parameters. You'll find the maximum load will ultimately be the max load which works within the parameters not the load that fails the members.
Rick is spot on and why I am looking at the deflection values because if my print head suddenly sank .05mm and went back up to normal the layer would be destroyed and the print job (3d printer) would be a catastrophic failure (especially when using .1mm layer height). This is why I was trying to get Solidworks to tell me what to expect in my previous answer's scenario but something is wrong since 100 pounds of dead weight in the middle Solidworks says (using 6063-T5 20x20 v-rail) will be only .1mm of deflection. I can't believe that is all it would be with that much weight. It might be for all I know but thinking about having my nephew stand on a 500mm beam in the middle I envision it buckling and him hurting himself not just a measly 1/10 of 1mm of difference. I am sure it is something I am just not doing right.
Right ... I agree ... I'm not designing a CNC or 3d printer. I'm really looking into designing lifts for TVs or desks ... or for the 8mm screw ... moving a picture ...
I don't think this is good for lifts for tvs and desks. For that sort of stuff I normally use Unistrut and while it is not as versatile as extrusion it is steel and not aluminum.
Probably should have started with that information. The TR8*8 screws won't likely work for such projects as the load can actually turn the screw and move along the extrusion once the stepper de-energizes. M8x1.25 should be workable though. You'll want to hang the load however rather than putting the screw in compression.
If you'll give us a better idea where you are going with this, we might be able to give you a better idea if it is workable.
Ok I'll look at Unistrut ... I was just looking for a ballpark number or way to tell how much weight the screw or belt could hold ... I knew in most cases it was under sized to use directly for lift but I was thinking I might be able to use the v-rail counter balanced for some projects. I'll probably just start experimenting ...
My applications are
1) to lift 200b cabinet about 1 to 2 feet at about 1 to 2 inches a second
2) to move pictures vertically or horizontally
3) 60" tv lift and
4) motorized curtains
BTW equations for figuring out deflection of a beam are easy to find on the web ... and I think open builds gives all the numbers you need to calculate it. (if you didn't see them already).
I saw them and we (on my thread) did them but neither of our findings were accurate I think and besides running a simulation in Solidworks with the exact material, and model, should give accurate results. *shrug*
For your case my deal isn't so much a lead screw but the extrusion itself and while I don't know how yours would be built I can say in my case 200 lbs would crush my machine. At 60 inches of lift meaning the beam will be at least 70 inches long I think it would simply bow out and collapse but that is just what I expect from such a soft piece of extrusion. The unistrut is what they use for holding up solar panel arrays outside that handle 90 mph winds with a couple of hundred pounds on them and if you know solar panels that is basically like a kite so almost full force of the weight and push from the wind at 90mph and the twisting torque and they easily handle it but put 6063-t5 aluminum extrusion to that test and bye bye to the structure and your solar array.
2, 3, and 4 are possible but don't get your hopes up on #1. I wouldn't touch that with anything less than 80/20 15-series (or equivalent) extrusion.
You think even that would handle that?
Depends on the configuration but at 50# per corner, it should be fine for a 2' straight lift. Could also easily move up to 1530 section and that would be more than enough. And while both would be more expensive than Unistrut, it would also be easier to remove any play from the system as these are better designed for such movements.
I agree Unistrut is for loading and staying put but what kind of TV? Tube (CRT) type? Plasma? LCD? 60" I suspect would be an LCD type but I have seen 60" Plasma as well and those are nut busters (not like the old CRT widescreen Sony Trinitron which I have one and it takes two or three to lift it) but 60 inch LCD averages about 50 pounds in total these days, or so I read.
I guess if the extrusion was anchored into a stud say every 12 inches then I would have no fear of 50 pounds on one but a free standing extrusion I would be a tad afraid.
The problem here is that I didn't state my initial question very well ... actually rereading its obvious that I'm asking wrong question ... although with more information it would be correctly stated. The second problem was not telling people how I'm using the v-slot rail.
For the vertical rise there is very little load on the beam. It's on whatever drive mechanism I decide to use. The beam is only a guide. In these applications I'm moving 20 to 200lbs, so if I use a screw it will probably need to be way bigger then 8mm. For horizontal movement the beam will be bearing the load but I'm only moving 1 to 10 pounds. I will probably use a belt or cable to drive it in this case. For both cases if beam deflection becomes an issue I have the option of adding extra support a long it's length. As you can see I'm not too worried about beam deflection in either case. In the vertical screw drive case I am worried about sizing the screw (or screws) to support the load.
At two hundred pounds that is asking a lot out of the bearings and even the extreme wheels. I think even the extreme wheels would break down or go flat if left in the same spot for too long.
Now for your C-Beam situation I have to ask could the C-Beam actuator mechanism handle a load in the horizontal position that weighs 200 pounds without digging into the beam? I don't know but by looking at it I never thought it was made for any sort of heavy loads in the vertical position and in the horizontal an 8mm acme rod wouldn't be strong enough to keep that slider off of the beam with such a load for sure.
The lifts I have seen have been scissor type lifts that lock a set positions with linear supports, tbh the c beam isn't designed for heavy lift the likely fail point(s) will be the nut block then the bearings as they are not designed for that sort of loading (on the face) where a tapper roller would be more suitable
tr8x8 screw seems to have a bit of whip at speed @1000mm, a thicker screw may be an idea but then you get into the loading on the stepper and/or driver
Yes, A scissor lift is the other way I am thinking of doing the 200lb project ... mainly because I'm having trouble finding a closed belt long enough, strong enough, and cheap enough to drive two screws from a centralized point. For the TV lift I'm thinking a single 1/2" threaded rod about 3' long.
BTW, I was not planning on using a stepper for either project.
1/2 inch all threaded rod should work in that case but the 8mm in the C-Beam wouldn't for sure.
Your original question was about the C-Beam Linear Actuator which uses a stepper motor and the 8mm Acme threaded rod but now you say you aren't going to use a stepper so how were you going to drive the C-Beam linear actuator?
Yes, I was interested if the C-Beam or the belt driven methods could be used in any of my projects ... I was thinking of the lighter load ones ... i.e. moving a picture (just getting a data point). I think everyones eye jumped to the 200lb project (even though I never considered to use the 8mm c-beam for that case ... didn't think it needed to be stated as it was obvious).
I was not planning on using a stepper because if I could get away with a reversible ac motor (driving two screws on either side of the lift) and a few switches I can eliminate a power supply and all that extra logic. If I have the processor and power supply there for other things then I might think about steppers. I have not calculated the torque and if I can get the speed I want yet in either case. I also don't think I'd drive the screw directly.
Speed vs torque is always a big issue to find the sweet spot. Can always gear it to help the torque which lowers the speed.