11-08-2010 #11these are very much rough and ready, others (Ross, routercnc, et al) can do the more detailed analysis... and tell me I am wrong!
Im still struggling with the spindle design math, I think I would just model the structure and put it in FEA as I'm getting lazy...:heehee:
Irving have you got a link to that catalogue? I cant find it,
I Looked at the individual section specs and the dimms are in mm's but the Ixx and Iyy is cm^4 and vol cm3, could be the problem with the decimal places.......
Firstly just looking at the section sizes, given that it will be 1200mm long, would seem a reasonable start. It will depend on how far down the cutting is taking place away from this section, since this will lead to a twisting torque, along with the deflections Irving mentioned above. Any ideas on the height down to the workpiece?
I need to dash off now, so will get back to you with some proper analysis. Really you need to use what is called the 'parallel axis theorem' which works out how to combine the two shapes (or any number of shapes), and find the neutral axis. This will be slightly below 'half way' in your L case. Then you can calculate the total section 'I' value, second moment of area, in mm4. From here it is then possible to get the deflection, and given a bit more data from you, the twisting.
The Hepco 40x80 shows an Ixx value of 61.4cm4, which if my brain is working is 61.4x1e4 mm^4. This is a start, hope to be back with some better answers tomorrow unless someone jumps in . . .
Ah, Ross, you jumped in while I was typing. Here is the Hepco catalogue link:
Yes, watch those units. Should be consistant really.
Last edited by routercnc; 12-08-2010 at 06:14 AM.
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Hoping for about a cutting height of 150mm ideally. I never realised there was so much to consider. Are these forces transferred through the z-axis or exerted on the z-axis as well?
Basically I need to design a couple of different x-axis then work out the deflection to minimise it? Ooo the planning is going to take a while!!
The answer is both, the z axis also needs to be rigid enough to transmit the forces, which will then be transfered to the y axis and will be max at full z travel. This will cause a moment on the y axis, (both directions) so like routercnc said your main concern is the torsional stiffness, beam deflection will mainly be from the weight of the x axis and carriage (static and dynamic) but some uplift could arise if you are plunge routing....
Maybe wait for Irving or routercnc, they are better at explaining than me.
Think about where the tip of the cutter will be when cutting...it'll be about 200mm below the y-axis (depending on how much y-axis clearance over the workpiece you want. So any sideways force on the cutter will be trying to bend the cutter.. but if that was solid it would be trying to bend the z-axis... but if that was solid it would be trying to twist the y-axis - the worst scenario being the twist in the X direction perpendicular to Y. Each of those components will deflect to some extent and transmit the forces back up to the component above them in the chain... and then the y-axis transmits them to the gantry sides and thence to the x-rails...
The easiest way to get a handle on the forces is to draw stick diagrams... some forces can be ignored generally, e.g. the twisting motion of the Y-axis on the Y-axis due to a sideways force on the cutter parallel to Y, although that force is also trying to push Y sideways and that is trying to rotate the bearings on the x-rails and that can't be ignored... at least initially... good bracing and gussetting helps, but you need also to keep the moving weight down...
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Irving where did you find the force allowances for wood? I could do with investigating aluminium as well, as I can almost guarantee I'm going to end up wanting to make something in aluminium!
ummm... off the top of my head i dont recall... google for 'specific cutting power'... its 17 for aluminium... that allows you to work out for a given material removal rate how much power is needed and therefore also, given revs the torque, which given a cutter size, gives the force applied and thats a guesstimate for the cutting force at the cutter... its rough and ready so we've generally doubled it... some hardwoods are nearly as tough as ali!
See what I mean Irving's much better than me. I can see it but I'm buggered if I can explain it :whistling:
With regard to cutting forces look here (ignore the spindle stuff)
Scroll down to the Niagara site as well and I'm sure there is a section on wood too, only just found it so haven't had a Reilly good look but there is plenty on metals
Going back to your machine have you looked at the 80 x 80 box stuff. over the long distance you need torsional stiffness, and as Irving pointed out lightness. (after all you have to move it) the box offers the best resistance to torsion, has equal strength in both axis and is light as there is no/little material in the center.
With the composite beam you have a lot of material (weight) close to the neutral axis (no stress) which again going back to parallel axis theory you want all the material at the extremes, like a tube or next best thing a box.
hope this helps
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