Moment of Inertia: Why Did Author Change Formula?

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Discussion Overview

The discussion revolves around the moment of inertia, specifically addressing the differences in formulas presented by an author in two images. Participants are trying to understand the reasoning behind the changes in the formulas and the application of the Parallel Axis Theorem in calculating the moment of inertia for composite shapes.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants explain that the moment of inertia is initially calculated about the base of the figure, which requires adding A ⋅ y-bar² values for each section to account for their individual centroids.
  • Others argue that for bending stress calculations, the moment of inertia should be about the centroid of the entire figure, necessitating the use of the Parallel Axis Theorem to adjust the calculations.
  • A participant questions why the second formula uses subtraction instead of addition for the A(d²) term, suggesting confusion over the application of the theorem.
  • Another participant clarifies that the total moment of inertia must be corrected to the centroid of the whole figure by subtracting A(d²), where d is the distance from the centroid to the base.
  • Some participants express uncertainty about whether it is necessary to calculate the individual contributions of each section before making the correction for the entire figure.
  • One participant asserts that each section has different areas and centroidal locations, indicating that individual calculations are necessary before applying the correction.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether it is necessary to calculate the individual moments of inertia for each section before applying the correction for the entire figure. There are competing views on the necessity and methodology of these calculations.

Contextual Notes

Participants express uncertainty regarding the assumptions behind the formulas and the specific definitions of terms used, particularly in relation to the Parallel Axis Theorem and the treatment of different sections of the figure.

foo9008
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Homework Statement


we know that the moment of inertia is given by the formula of I = (ab^3) / 12 + A(y^2) ... why in the second photo , the author make it as I = 1443333- A(y^2) ? what is the purpose of doing so , i don't understand ...

Homework Equations

The Attempt at a Solution

 

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foo9008 said:

Homework Statement


we know that the moment of inertia is given by the formula of I = (ab^3) / 12 + A(y^2) ... why in the second photo , the author make it as I = 1443333- A(y^2) ? what is the purpose of doing so , i don't understand ...

Homework Equations

The Attempt at a Solution

In the first picture, the moment of inertia is being calculated about the base of the figure, which is why sections A, B, and C all have an A ⋅ y-bar2 value added to the MOI of each section about its own centroid.

However, the MOI of the entire figure about the base is not what you want to use when calculating bending stress, for example. You want to calculate the MOI of the entire figure about its own centroid. This is why the Parallel Axis Theorem is applied on page 2, to correct the MOI of the figure to its composite centroid location, as calculated on the first page. (y-bar for the figure is located 28.07 mm from the base).
 
SteamKing said:
In the first picture, the moment of inertia is being calculated about the base of the figure, which is why sections A, B, and C all have an A ⋅ y-bar2 value added to the MOI of each section about its own centroid.

However, the MOI of the entire figure about the base is not what you want to use when calculating bending stress, for example. You want to calculate the MOI of the entire figure about its own centroid. This is why the Parallel Axis Theorem is applied on page 2, to correct the MOI of the figure to its composite centroid location, as calculated on the first page. (y-bar for the figure is located 28.07 mm from the base).
normally , the total moment of inertia is I = I(center of mass) + A(d^2) , where d = distance from the centroid of entire mass to the axis,right? why n the 2nd photo , it's I = I(center of mass) - A(d^2)https://en.wikipedia.org/wiki/Parallel_axis_theorem
 
chetzread said:
normally , the total moment of inertia is I = I(center of mass) + A(d^2) , where d = distance from the centroid of entire mass to the axis,right? why n the 2nd photo , it's I = I(center of mass) - A(d^2)https://en.wikipedia.org/wiki/Parallel_axis_theorem
In the first page, the moments of inertia of the individual sections A, B, and C are calculated about the individual centroids of A, B, and C, respectively, and then transferred to the base of the by adding Ad2 terms. Once the total inertia is calculated about the base, then the moment of inertia must be corrected to the centroid of the whole figure by subtracting Ad2, where A is the total cross sectional area and d is the location of the centroid w.r.t. the base, in this instance d = 28.07 mm.

This is done so that the correct moment of inertia of the entire figure about its centroid is calculated and can then be used to determine bending stress.
 
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SteamKing said:
In the first page, the moments of inertia of the individual sections A, B, and C are calculated about the individual centroids of A, B, and C, respectively, and then transferred to the base of the by adding Ad2 terms. Once the total inertia is calculated about the base, then the moment of inertia must be corrected to the centroid of the whole figure by subtracting Ad2, where A is the total cross sectional area and d is the location of the centroid w.r.t. the base, in this instance d = 28.07 mm.

This is done so that the correct moment of inertia of the entire figure about its centroid is calculated and can then be used to determine bending stress.
cann we just minus the A(d^2) , where a = total area of figure when we are calculating the moment about the centorid of the entire figure ? there's no need to calculate the d one by one just like the author did in the first figure?
 
chetzread said:
cann we just minus the A(d^2) , where a = total area of figure when we are calculating the moment about the centorid of the entire figure ? there's no need to calculate the d one by one just like the author did in the first figure?
No. Each of the sections A, B, and C is a different area and centroidal location from the reference baseline.

You must calculate and sum the Io + Aidi2 for each piece before making the correction for the entire section.
 

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