# Moment of Inertia - Parallel Axis Help (Statics)

#### Sidelines

Hey all
I don't really have a specific problem to ask about, but I'm having trouble with the concept of the parallel axis theorem. Mainly when using the equation to solve for Ix/Iy, what Ix' and Iy' are equal to.

Just for an example, when calculating Iy of this problem: And splitting the object into 4 rectangles (two on the left, one center, one right)

The two on the left the Iy' is (1/3)base*height, but the other two it's (1/12)*base*height.

The difference I've been told is that the two on the left touch the Y axis, and then you can use the formula involving 1/3.
I understand that, but I seem to still be getting lost.

So I just have a few questions,
If there is no x and y axis, but there's a x' and y', can I treat x' and y' as normal x and y?
If objects/rectangles are touching the x' and y' axis' can I use the Ix' of (1/3)b*h (for rectangles) or do I still have to use 1/12*b*h?

Does anyone have a good explanation of how to use the parallel axis theorem?

This is all very confusing for me. An explanation would be amazing!
All help is greatly appreciated!

#### ChipB

PHF Helper
The moment of inertia for a rectangle (or any shape) depends on where the axis of measurement is. If the axis is down the center of the rectangle and goes through its center then I = (1/12) bh^3. If instead the axis is along one edge then the moment of inertia is larger, equal to (1/3)bh^3. Intuitively you may be able to visualize that if the axis is along the center line then all the rectangle is within b/2 of the axis, whereas if the axis is along the edge then half of the rectangle is further than b/2 away from the axis. This effect is captured with its larger value of I.

The parallel axis theorem lets you easily calculate I for cases where the axis is not through the object's center. The formula is:

where I_c is the moment of inertia through the center, 'A' is the area of the object, and 'd' is the distance from the center of A to the new axis. Example: you know that I_c for a rectangle is (1/12)bh^3; Since the edge is b/2 from the center then I for the edge is (1/12)bh^3 + (bh)(b/2)^2 = (1/3)bh^3, which agrees with what you already knew.

For your problem to calculate Iy you would add the I's for the 4 parts as you decribes:

1. Two rectangles with b=1 and h = 4, so Iy= 2 x bh^3/3 = 2(1)(4^3)/3
2. One rectangle with center at d= 4, A = 6, b= 8 and h = 1, so Iy = 6(1)63/12 + 6 x 4^2
3. One rectangle with center at d = 7.5, A = 4, b = 1 and h = 4, so Iy = 1(4^3)/12 + 4(7.5^2)

Add these to get the total value for Iy.

I don't know what you mean by iX' and Iy' - are these different axes than shown in the drawing?

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#### Sidelines

Thanks so much ChipB

#### sheon

what if we are calculating Ix of this problem? we have to use 1/12bh^3 or 1/3bh^3?

#### ChipB

PHF Helper
You can use (1/3)bh^3 for each of the four rectangles, since each has an edge on the x axis.