How can I draw a free body diagram of a rotating wheel?

AI Thread Summary
To analyze the A356 Aluminum alloy wheel under rotational loading, a free body diagram is essential to identify stress concentrations. The wheel experiences a maximum load of 650 kgf and a torque of 4600 Nm, leading to cracks after 1.3 million rotations. The applied torque must balance with frictional forces for consistent rotation. While initial hand calculations can indicate stress locations, the complexity of the wheel suggests that Finite Element Analysis (FEA) would provide more accurate results. Understanding the stress distribution is crucial for suggesting design modifications to prevent future failures.
Gevorg
Messages
7
Reaction score
0
I am currently performing failure analysis on an A356 Aluminum alloy wheel subjected to rotational loading, with a maximum load of 650 kgf and a torque of 4600 Nm. Cracks formed in the wheel after 1.3*10^6 rotations.

My job is to perform a stress analysis to determine the regions of maximum stress concentration under such loading conditions. I figured it is first necessary to draw a free body diagram of the entire wheel and then apply the method of sections, in which I would make cuts along specific regions of the wheel and determine the forces acting on them.

Below I've drawn a free body diagram of the wheel. The sum of the forces acting on the 5 holes should equal 650 kgf. The up pointing arrow is equal in magnitude to their sum. The applied torque, shown in blue, should be equal in magnitude to the torque generated by the left pointing force of friction in order for the wheel to rotate at a constant angular velocity.

I still feel that something is missing and can't quite put my finger on it. Could someone provide some insight?
upload_2017-11-12_18-42-27.png
 

Attachments

  • upload_2017-11-12_18-42-27.png
    upload_2017-11-12_18-42-27.png
    30.7 KB · Views: 2,187
Physics news on Phys.org
Your diagram looks good for a metal wheel running on a hard (steel or concrete) surface. The moment could be applied as five forces on the bolt holes, each force in a circumferential direction. If that wheel has a pneumatic tire on it, the vertical and tangential forces are applied around the rim.
 
After further thought, the cracks tell you the location(s) of the maximum fatigue stresses. And the total number of load cycles until cracks appear tells you the magnitude of the stresses.

If this wheel is on a motor vehicle, then the maximum stresses may come from cornering forces. That would require a completely different free body diagram.

The complexity of the wheel make hand calculations difficult, and of questionable accuracy. This problem is best solved using FEA. It does not need a high end FEA package, the SolidWorks Premium package will easily handle this job.
 
  • Like
Likes Gevorg
JRMichler said:
After further thought, the cracks tell you the location(s) of the maximum fatigue stresses. And the total number of load cycles until cracks appear tells you the magnitude of the stresses.

If this wheel is on a motor vehicle, then the maximum stresses may come from cornering forces. That would require a completely different free body diagram.

The complexity of the wheel make hand calculations difficult, and of questionable accuracy. This problem is best solved using FEA. It does not need a high end FEA package, the SolidWorks Premium package will easily handle this job.

You are right, JRMichler; cracks form at the locations of maximum stress. To the best of my knowledge, the wheel was not subject to cornering forces. I unfortunately don't have any background in FEA; however, simple calculations should be enough to prove that stress concentrations occur where the cracks formed and allow me to suggest design modifications.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top