Predicting elastic and plastic wheel deformation

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SUMMARY

This discussion focuses on predicting elastic and plastic deformation of a steel wheel with a diameter of 6 inches and thickness of 2 inches under a contact force of 1000 lbs. The material properties include an elastic modulus (E) of 30,000,000 psi and a Poisson ratio of 0.27. The conversation explores methods for calculating stress distribution using finite layers and shear modulus, as well as deriving contact pressure equations. Key formulas discussed include the maximum pressure calculation and stress in the z-direction, emphasizing the complexity of accurately predicting conditions for plastic deformation.

PREREQUISITES
  • Understanding of material mechanics, specifically elastic and plastic deformation.
  • Familiarity with stress distribution calculations in cylindrical geometries.
  • Knowledge of finite element analysis principles.
  • Proficiency in applying material properties such as Young's modulus and Poisson's ratio.
NEXT STEPS
  • Research finite element analysis (FEA) techniques for modeling wheel deformation.
  • Learn about contact mechanics and stress distribution in cylindrical bodies.
  • Study the effects of shear modulus on deformation predictions.
  • Explore advanced material properties and their influence on plastic deformation thresholds.
USEFUL FOR

Mechanical engineers, materials scientists, and researchers focused on wheel design and deformation analysis will benefit from this discussion.

RayRoc
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I am trying to predict elastic wheel deformation. I am not quite sure how to do it. How might one approach the problem?

Take a steel wheel: 6 in. diameter, 2 in. thick, with a contact force of 1000lbs

E = 30,000,000 psi
Poisson ratio: .27

Would one calculate this in finite layers? Considering sections at a time and applying shear modulus through the layers to find the finite stress of the steel?

I have seen the stress distribution of a steel wheel against a track, and it looked like a strait coulomb from the contact area to the axial. Can it be as simple as:

Contact pressure = (volume of wheel displaced/ volume of .5 the wheel cross section) * Poisson ratio * E

There are of course many effects not realized by such a simplistic view especially when one wants to find the conditions that will cause plastic deformation at the contact area; which is ultimately what I want to know.

The gears in my head have failed, the more I think about this problem the louder the grinding noises.
 
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Contact stresses for a cylinder on a cylinder are as follows. The area of contact is a narrow rectangle of width 2b, and length l, where l is the length of the cylinders, and
b = \sqrt{ \frac{2F}{\pi l} \frac{ (1-\nu^2_1)/E_1 + (1- \nu^2_2)/E_2}{1/d_1 + 1/d_2} }
With the maximum pressure being
p_{max} = \frac{2F}{\pi b l}
You can apply this to a cylinder in contact with a plane by making d_2 = \infty

edit: Just seen that you're looking for deformation, not stress. Let's see...stress in the z-direction, perpendicular to the contact line is:
\sigma_z = \frac{-p_{max}}{ \sqrt{1+ z^2/b^2}}
Perhaps you could integrate that from z = 0, to z = r to calculate the summation of strain?
 

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