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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 teh purpose of doing so , i dont understand ...

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- Thread starter foo9008
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- #1

- 678

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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 teh purpose of doing so , i dont understand ...

- #2

SteamKing

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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-bar## 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 teh purpose of doing so , i dont understand ...

## Homework Equations

## The Attempt at a Solution

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

- #3

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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)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-bar^{2}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 theentire figureabout 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).

- #4

SteamKing

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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 Adnormally , 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

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.

- #5

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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?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 Ad^{2}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 Ad^{2}, 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.

- #6

SteamKing

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No. Each of the sections A, B, and C is a different area and centroidal location from the reference baseline.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?

You must calculate and sum the I

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