I am trying to calculate the torsion in a rectangular bar.
I have found closed form calculations in Shigley, Roark, and a NZ structural code which all agree (within 10% or less).
However, the solid works simulation of the same geometry is higher:
b = 4"
t = 1"
L = 3"
T = 10,000 in*lbf
Tau (Shigley) = 8.63 ksi
Tau (Roark) = 8.63 ksi
Tau (NZS 4404) ~9 ksi
Tau (FEA) = 11.65 ksi
The FEA results are 35% higher.
Has anyone every experienced this
Results are attached below:
Have you tried this as a beam analysis in Solidworks?
I have not tried to model as a beam element.
Wouldn't the beam element be limited to to the
cross sectional properties of the 1D element?
This torsional stress in rectangular bars appears to be a complex distribution.
Your thoughts?
Jody is correct, I didn't see the short length.
However, Roark's, et al., are exact formulas based on geometry. You can take them to the bank. FEA on the other hand is a little more iffy but handles odd geometry better. The tetras in SW can't rotate very well (i.e. are quite stiff) so you'll want to use curvature based mesh. Can't really tell the size mesh you are using but you'll probably want to make it fairly fine.
I will post graphics of the boundary conditions and the mesh plot (density).
I will also try to mesh using circular based mesh (that is an interesting comment).
Thanks,
Steve
Check your boundary conditions and loads...
I'd expect the highest shear stress to be on the outer fibers, not in the center of the bar.
Also, roarke and friends are equations for long bars (where the local effects of attachment at the ends can be ignored), but your bar has L shorter than B, which will affect the results.
Initially, I applied a torque on one end and a fixed restraint on the other end.
However, I then applied opposing torque on both ends with a
3-2-1 restraint on one end at 3 nodes (corner vertex points).
I will post the results later (on the road).
Analytical calculations are always an approximation, so the result may be different. The FEA result should be the most accurate.
Also important are stress singularities at the fixture. If the fixation is too rigid the stress values will not converge to a constant value when refining the mesh. The right interpretation is crucial when you compare the result with analytical calculation.
Janko Stellaard
Cadmes BV
The Netherlands
Janko, I initially tried a bar with a b/t ratio of 10 to match the thin plate assumption.
The results were still ~ 30% higher than theory. I must have an improper restraint.
Janko, have to disagree with you, it's the other way around. FEA can approach the exact answer that formulas can provide. The difficulty is that real world isn't perfect.
As James said, it is the exact opposite.
In many cases it is, because not all the effects can be included in a analytical calculation.
A good example for that is the calculation of Beam stress. The shear is neglected in the analytical calculation.
I can't tell how simplified this calculation is, but there is a big chance it is. But I have to correct my last post and replace "always" with "in many situations".
Mesh (global = 0.23"/ control (ends )= 0.025"):
Error Plot:
Restraints (3-2-1):
Loads - opposing torsion (10,000 in*lbf) on each end of bar:
Stress plot (using longitudinal axis as reference):
Note: circular based mesh gave the same results.
Stress variation across the half width of the bar (tauxymax ~ 13.35 ksi):
Do the restraints and loads seem reasonable?
A couple of things I see...
All the view of your model are from the same angle, so you are applying both loads AND restraints to the near face. I would restrain one end, and load the other. The simulation will calculate the reaction forces at the restraints.
How is the load actually applied to the bar..is it completely welded to something solid at each end, bolted, or is it captured in a slot like a flat blade screwdriver? Each of these would produce different stresses.
Lastly, for long bars, the maximum stress should be at the center of the long sides of the bar, on the outside surface. The stress at the center and on the corners should be zero. If your stress distribution is not doing that, something is wrong.
I think I figured out the problem.
I was using the incorrect shear component and the length is a factor
I will post my results over the weekend.
Apparently, the torsional shear stress is influenced by the stresses in bars with a short length.
I increased the length of the bar to 12" long and checked the torsion at a plane located at the mid length and the stresses
agree closely (with ~ 2%) with Roark and Shigley.
Initially, I made 2 mistakes.
1. I was checking stresses at the ends instead of the mid length.
2. I was checking the shear stresses in the wrong directions.
Here are my new results:
TauYZ is the shear in the Z-direction acting along the X-Z plane.
The Y axis is orthogonal to the X-Z plane, thus TauYZ notation.
Apparently, the torsional shear stress is influenced by the stresses in bars with a short length.
I increased the length of the bar to 12" long and checked the torsion at a plane located at the mid length and the stresses
agree closely (with ~ 2%) with Roark and Shigley.
Initially, I made 2 mistakes.
1. I was checking stresses at the ends instead of the mid length.
2. I was checking the shear stresses in the wrong directions.
Here are my new results:
TauYZ is the shear in the Z-direction acting along the X-Z plane.
The Y axis is orthogonal to the X-Z plane, thus TauYZ notation.