That looks better Barry (not that there was anything wrong with the first one)
One point Ive just realized for both our sheets is that it only deals with one beam or bar when in reality two will be used, especially for a gantry!
The deflection is easy as you just halve it, but could be a bit more difficult for the torsion calcs. Rotation about the central point with torsion and horizontal deflection? nasty.
Prompted by a recent post asking about deflection of profile sections layed out in an 'L' shape, here is a spreadsheet to calculate I values for almost arbitrary shapes.
These I values can then be used to calculate deflections. I'm in the process of adding this to the earlier spreadsheet, plus a few more new features, and hope to have something out early next week. For now, here is this mini calculator on it's own:
Nice one Barry & Ross, I've just been through these sheets... This sort of stuff only usually gets done if you're spending someone else's money :)
I wish I'd run calcs like this on my Y beam before building. I don't really need to now - I already know the answer :-~
Those sheets are great!
I've just found that SolidWorks calculates moments of inertia, including second moment of area. I just drew out the SBR20 rail (20mm, supported) to try and compare it to unsupported rail. Here's the results for just the aluminium support on the rail:
Section properties of the selected face of SBR-20 supported round rail
Area = 400.80 millimeters^2
Centroid relative to assembly origin: ( millimeters )
X = 22.50
Y = 5.85
Z = 0.00
Moments of inertia of the area, at the centroid: ( millimeters ^ 4 )
Lxx = 8228.45 Lxy = 0.00 Lxz = 0.00
Lyx = 0.00 Lyy = 41748.33 Lyz = 0.00
Lzx = 0.00 Lzy = 0.00 Lzz = 49976.78
I for the 20mm steel bar is 7840mm^4 ... so how do I combine these two values? Solid edge will calculate them as one but of course doesn't take into account the different moduli(?) for aluminium and steel.
EDIT: I've just found out how to do it - 'Parallel axis theorem'. Why can't it be simple!
Last edited by Jonathan; 22-08-2010 at 11:34 PM.
Right I've worked out I for the 20mm supported rails in both directions.
In X direction I=53643mm^4
In Y direction I=22257mm^4
To account for the two materials used in the rails I've just scaled I for the bar by 67/200 so if you enter the above values and select steeel in the spreadsheets in this thread it should work.
For comparison the 20mm unsupported rail is 7854mm^4 - so it's roughly 2.8 to 6.8 times weaker than supported rail which is just 'floating'.
(I'm fairly confident that these values are correct, though I've literally only learnt about moments of inertia in the past 2 hours and it's like 1am so no guarantees!)
Last edited by Jonathan; 23-08-2010 at 01:06 AM.
Glad to see the spreadsheet being used. I think they could still do with making even more user friendly. They're ok if you are used to doing this stuff and just want to save doing it by hand, but if this is new territory then you could easily be put off, or worse still, try to use them in the wrong way and get meaningless answers.
Your method of mixing materials by scaling Young's modulus has got me thinking . . . think it's OK.
Your estimate of stiffness increase for supported rails over unsupported (2.8 to 6.8 times better) neglects the part that the supported rail is bolted to (often a thick plate for Z axis), which is the main reason for doing it. You'll find if you factor in the mounting plate that the differences are significantly greater still. Good work though at 1 in the morning!
Other CAD s/ware, such as AutoCAD, CATIA will give properties including second moment of area, so for complex shapes this is another way to do it.
Can you link me to somewhere that details how to calculate the twisting?
Other CAD s/ware, such as AutoCAD, CATIA will give properties including second moment of area, so for complex shapes this is another way to do it.[/QUOTE]
Ah sorry, missed what you meant by floating.
Ali vs steel parallel question. I've always thought that items like ali profile or aluminium extruded RHS etc. are pretty true due to the extrusion process used. Steel box can be resistance welded along one edge, which I would suspect is not so good. If going for steel sections on critical parts which held the rails then I'd be considering having the faces ground. But I'm not an expert to my suggestion is that you post this up for a better response!
Twisting calculations were from Roark's formulas for stress and strain, an old University text book I have about 1" thick full of tables and equations. Excellent book, but heavy going in places. If you search the internet for 'twisting constant K', which is a bit like I values for bending, you might get somewhere.
I started off down that path of calculating the I values of each bit in Autocad and the parallel axis etc...... but gave up on the torsion constant for irregular shapes, when I was looking on the net I kept seeing reference to to the 2 polar moment which CAD can also work out but this is only true for circular tube and bar. I only gave up because I can now use FEA to do it all for me. if you get any where then I sure would like to know.
I also found that the more you analise it the more time it takes and you soon realise that, as Barry pointed out, just doing a few basic cals on the main structure will be sufficient. adding the rails etc becomes the factor of safety that seems to be missing from a lot of calcs. :naughty:
Last edited by Ross77; 23-08-2010 at 10:44 PM.
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