Moment of inertia

1. Jun 26, 2016

foo9008

1. The problem statement, all variables and given/known data
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. Relevant equations

3. The attempt at a solution

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2. Jun 26, 2016

SteamKing

Staff Emeritus
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).

3. Jun 26, 2016

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

4. Jun 26, 2016

SteamKing

Staff Emeritus
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.

5. Jun 26, 2016

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?

6. Jun 26, 2016

SteamKing

Staff Emeritus
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.