Steam King and nvk. Many thanks. It really is appreciated. Third time lucky! Here we go….
1. I calculate torsional rigidity = GJ.
2. J = 4tA^2/S. 4 x 0.003 (wall thickness) x (0.147x0.107)^2 (enclosed area) / 0.508 (the perimeter .147+.147+.107+.107). So J = 5.84E-6 m^4 (correct...
Steam King, thanks.
1. So I should not have used second moment of inertia or indeed parallel axis theorem because they do not apply to torques only to bending moments.
2. I should have used torsional rigidity = GJ. I have G so all I need to do is to calculate J by using your equation. In...
Phrak and Steam King - thanks. Whilst it would be interesting to see if Eular Buckling occurred sooner than material failure I am confident that I am expected to find the material failure figures.
This brings me to Steam King's equation "tau = T / (2At), where T = torque, t = wall...
The approach to the question is supposed to be very basic in order to help us with revision (it needs to be for my level of understanding) so I doubt it will get into 3-d .
I think that to get the twist angle I need torque (as above) but to get torque I need Torque = 2 x area x shear...
Phrak, Thanks. What equation would I use to find the twist angle that uses the information that I already have?
I think it may be twist angle = (torque x length) / (torsional rigidity). I have torsional rigidity (above - if it is correct?) and I have length (3m) but I am not sure about torque...
Revision that is turning into a learning from scratch session. Some help would be welcome! (I have laso put this on the materials section of engineering).
- I have a hollow rectangular beam. 0.150m by 0.110m with uniform thickness walls at 0.003m. It is 3m long.
- It has a modulus of...
Revision that is turning into a learning from scratch session. Some help would be welcome!
- I have a hollow rectangular beam. 0.150m by 0.110m with uniform thickness walls at 0.003m. It is 3m long.
- It has a modulus of rigidity (G) of 27 GigaPascals.
- It has a max permitted shear...