Why Does the Polar Moment of Inertia Use r^2 in Its Formula?

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Discussion Overview

The discussion revolves around the formula for calculating the polar moment of inertia, specifically the use of ##r^2## in the integral ##I_z=\int \int r^2 dA##. Participants explore the implications of this formula in the context of bending stresses and the meaning of the double integral.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants clarify that the formula relates to the moment of inertia of a two-dimensional region being rotated around the z-axis.
  • There is a question about the appropriateness of using ##r^4## instead of ##r^2## in the integral.
  • One participant suggests that integrating ##r## would yield something related to the area, implying a distinction between the two integrals.
  • Another participant explains that the use of ##r^2## in the integral is related to the momentum of the tension force on the area element dA, which depends on the distance from dA to the center of the beam.
  • It is noted that the force on the surface dA is proportional to how much the material has been stressed or compressed, necessitating the multiplication by distance to obtain the momentum of this force.

Areas of Agreement / Disagreement

Participants express differing views on the use of ##r^2## versus ##r## in the integral, and there is no consensus on whether ##r^4## could be appropriate. The discussion remains unresolved regarding the implications of these choices.

Contextual Notes

Participants reference the integration process and the implications of different variables in the context of bending stresses, but there are no settled assumptions or definitions provided.

Andrea Vironda
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TL;DR
Derivation of the formula for the calculation, among other things, of bending stresses
Hi,
A well-known part of the formula for calculating the deflection stress is ##I_z=\int \int r^2 dA##
Usually a moment of inertia is something related to how difficult is to move an object. In this case is understandable but i don't understand the meaning of the double integral.
Using ##r^4## wouldn't be the same?
 
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Andrea Vironda said:
Summary:: Derivation of the formula for the calculation, among other things, of bending stresses

Hi,
A well-known part of the formula for calculating the deflection stress is ##I_z=\int \int r^2 dA##
Don't you mean moment of inertia rather than deflection stress?

I read the formula you wrote as the moment of inertia of some two-dimensional region in the x-y plane that is being rotated around the z-axis. dA represents the area of some infinitesimal region, with an implied mass of 1 unit of some kind. If the integral is replaced by an iterated Cartesian or rectangular integral, dA will become dxdy or dydx, depending on the order of integration. If the integral is replaced by an iterated polar integra, dA will be replaced by ##rdrd\theta##, so the iterated integral could look like ##\int_{\theta}\int_r r^2 r dr~d\theta##, or ##\int_{\theta}\int_r r^3 dr~d\theta##, assuming the integration is performed first on r. In both integrals the mass of the region dA would be 1 unit, by implication.
Andrea Vironda said:
Usually a moment of inertia is something related to how difficult is to move an object. In this case is understandable but i don't understand the meaning of the double integral.
Using ##r^4## wouldn't be the same?
 
Mark44 said:
Don't you mean moment of inertia rather than deflection stress?
Yeah, only part of deflection stress formula.

Ok i think i understood. but why ##r^2## into the integral and not simply ##r##?
If i integrate ##r## on ##dA## i will get something related to the area
 
Andrea Vironda said:
Ok i think i understood. but why r2 into the integral and not simply r?
It is about the momentum of the tension force on dA. You get r2 in the integral because the force on the surface dA depends on how much the material has been stressed or compressed, which is proportional to the distance from dA to the center of the beam.
To get the momentum of this force, you have to multiply with this distance again.
 
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