maximum deflection - sanity check
Gents,
I'm trying to calculate the maximum deflection of an unsupported part of a shaft with the following data:
L total unsupported length = 0.18m (not mm;))
A cross section surface (round rod) = pi * (0.01m)^2
E Young's modulus = 180GPa
F force applied to the end of shaft = 1000N (random round number)
delta (max deflection) = F*L / E*A
Now popping the values in to the equation I get that max deflection is equal:
delta = F * 3.19*10^-9 [m/N]
which basically means that with F = 1000N the deflection is
delta = 3.19 * 10^-6 m = 3.19*10^-3 mm = 0.00319mm
That is rather small, considering that 1000N is pretty much the same as 100kg weight attached to one end.
Pack of biscuits to anyone who can point an error in the above and call be stupid:)
Regards,
dsc.
Re: maximum deflection - sanity check
0.18mm unsupported length?
Re: maximum deflection - sanity check
Ha, well spotted, although it's just a unit error.
After some reading I don't think the above equation for delta is correct as there's no moment of inertia involved.
Regards,
dsc.
Re: maximum deflection - sanity check
I just ran the figures and got 3183mm of deflection...
A = pi0.01^2 = 0.000314159
F*L = 1000 * 0.18 = 180
E*A = 180 * 0.0003blahblah = 0.05654866
180 / 0.0565blahblah = 3183.09886183
Re: maximum deflection - sanity check
Perhaps you are working out how far it will stretch if you pull on the end, not how far it will bend if you load it at 90 degrees?
Re: maximum deflection - sanity check
You could be right here Robin. Its been a very long time since I used Youngs Modulus but in this form I thought it was used for calculating a change in length due to compression or extension.
Re: maximum deflection - sanity check
Ah yes, it's all wrong!:)
Here's the right formula (taken from http://www.clag.org.uk/beam.html):
δ = FL3 ∕ 3EI
I = πr4 ∕ 4 = πd4 ∕ 64
(for round beam)
which means that deflection at a given F and L, for a 20mm dia beam is:
δ = FL3 * 0.0002358
Assuming L = 0.18m and F=100N the maximum deflection is:
δ = 100 * 0.005832 * 0.0002358 m = 0.0001375m = 0.1375mm
Which makes much more sense than the previously calculated value.
The funny thing is that changing the dia from 20mm to 22mm gives you a maximum deflection drop from 0.1375mm to 0.094mm. That's just by making the shaft 2mm thicker. Quite amazing me thinks.
Regards,
dsc.