Posted by duncan webster on 09/09/2020 22:18:13:
well you asked for it!
just taking some numbers out of fresh air: diameter 10mm, overhang 100mm, tip load 1 N, use property values for steel
Twisting
Shear modulus = 79.6 GN/m^2
tool tip to centre of bar = 6 mm
Torque = 1 N * 6 mm = 6 N.mm
angular deflection = 32 * 100 * 6 / (79.6 * 1000 * pi * 10^4) = 7.7e-6 rads **LINK**
This gives rise to a tangential deflection of 0.0064 * 5 = 3.8e-5 mm
Bending Modulus = 207 N/mm^2
Second moment of area = pi * 10^4 / 64 = 490 mm^4
end deflection = 1 * 100^3 / (3 * 207 * 490 ) = 0.0033 mm
With a much shorter bar the difference would be smaller, if the length was about 10mm they would be the same, but I don't think we'd care. If anyone wants this sum in SMath studio send me a pm, I'll not hold my breath
Edited By duncan webster on 09/09/2020 22:19:46
No one has spotted it yet, (well Michael sort of did) I got my gigas and megas mixed up, which means that the calculated values are 1000 times too high. I've corrected the above. The deflections are very small, but so is the applied load.
Edited By duncan webster on 10/09/2020 05:21:53
Edited By duncan webster on 10/09/2020 05:52:33