Moment of Inertia - Parallel Axis Help (Statics)

May 2011
8
0
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
Jun 2010
2,361
289
Morristown, NJ USA
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:

I = I_c + Ad^2

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?
 
Last edited:
May 2011
8
0
Ah! This answers my questions!

Thanks so much ChipB
 
Jan 2013
1
0
what if we are calculating Ix of this problem? we have to use 1/12bh^3 or 1/3bh^3?
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
You can use (1/3)bh^3 for each of the four rectangles, since each has an edge on the x axis.