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Could any of you Maths or Physics wizards tell me how much deflection I could expect on two sections of 25mm precision rail supported at both ends by 50mm with 500mm between supports with a carriage of (L)200mm X (W)300mm (4 bearings) fitted. The mass on the carriage is approx. 25Kg? Oh and please keep it simple if possible.:confused:
Thanks in advance
yes I can tell you...
oh you mean like now?
when I get home.... sometime this weekend
Many thanks ... couldn't ask for more :beer:Quote:
Originally Posted by irving2008
Since I've just had my structural analysis exam today I had vowed not to do any more calcs for at least a week........:nope:
So use this site http://www.engineeringtoolbox.com/be...on-d_1312.html
use the metric option Beam Supported at Both Ends, Load at Center
q= load = mass x gravity (25kg x 9.81= 245.25N) add the weight of the carrige as well
l = length in mm
I= moment of inertia = pi times r^4/4 = 19174.76 mm^4 (for a circle)
E= modulus of elasticity (find out from supplieror or use 200000 N/mm^2 for steel )
y= radius (25mm)12.5mm (thanks Irving)
This is only for a simply supported beam so will be worst case, If the supports are nice and solid then you could go for a fixed beam. I'll have to digg out the formulas and do an excel sheet for all the options.:smile:
I think most of this is correct, my brain is pretty fried at the mo tho:heehee:
Good luck
If you would Ross that would be great, i would like to add that to the site as a feature. Its also a question i get asked quite often on ebay.Quote:
Originally Posted by Ross77
huh, beat me to it...
deflection of a fixed-fixed beam with a central point load (worst case) is:
d = P (l^3)/(192 E I) = 250 (500^3)/(192 * 2e5 * pi/4 * 12.5^4) = .042mm
Ross - sanity check - radius is 12.5 not 25 - they were 25mm beams.
duh me stupid :rolleyes:Quote:
Ross - sanity check - radius is 12.5 not 25 - they were 25mm beams.
I just spent 3 weeks revising, slope defection by equation and integration, stress and strain, and dynamics:- centripetal force, Dynamic amplification factor, Euler rule etc etc. and the question was how to calculate the Natural freq.........F**kin typical
told you my brain was fried.......
ok so r= 12.5 :redface: didn't you read the caveat :smile: ( see first paragraph)
Lee ,Sure I will sort something out for you. But better get Irving to check it eh......
Yea sure make it collaboration if you like, maybe even put it into a .exe and keep it going as a MYCNCUK work in progress program, then you could add things to it as you come up with them, like Irving’s motor calc sheet. Would be cool, I think we may need a "MYCNCUK Browser Toolbar as well what you think?, I was looking at making a taskbar app as well for the site, you could install it and it would inform you of new PM’s and reply’s and so on.Quote:
Originally Posted by Ross77
The cals will be pretty easy to do as there are only the 3 beam types, cantilever, simply suporrted and fixed. the trick will be explaining how to use each one and the conditions required to meet them as i'm sure some of the bar mounts sold couldnt be classed as fixed.......
I'll do a draft and Pm it to you and Irving(if he's interested that is)
What happened to the motor calc sheet? i thought you hade a tutorial section but I cant find it anywhere. Browser tool bar sound good if it makes finding things easier.
P.S what with all this 'serving the masses'??? have you got a summer job at Burger King?:heehee:
You need to decide whether your end supports are rigid enough to justify treating the rail as 'encastre'. By this I mean are they strong enough to resist the rotation of the end of the rail when it is loaded. I suspect that it would be fairer to treat the ends of the rail as being simply supported. In which case, the deflection is given by P(L^3)/48EI This will give a deflection four times greater than Irving's.
I would treat Irving's solution as the best possible case and mine as the worst possible case. The true answer would lie somewhere inbetween.
Thats kind of what I said as assuming simply supported will be to lenient, and in some cases fully fixed will never be meet. the other issue is whether you assume a central point load or two point loads at the edge of the carriage and use the reduced deflection at these points as the actual deflection (ignoring max deflection in the middle). Its all a case of how far do you go?:rolleyes:
I'm pretty sure that there wont be much difference in the deflection answers and the overall performance has more to do with how they are used. As such I don't believe they should be used as the structural part of the axis as any flex in the side supports will be transfered to the rails and exaggerate the defection, add to this the vibration at resonance and the bar will soon resemble a guitar string:naughty:
There are plenty of deflection calculators already out there but unless you know what all the parts mean and can rearrange the units to be all the same the answers can vary. So I'm going to model some of the supports and typical axis lengths and put them in a FEA program to test my answers.