Calculating moment of inertia about z axis

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SUMMARY

The moment of inertia about the z-axis (Iz) is calculated using the formula Iz = 1/3 * 90 * 60^3 + 1/3 * 60 * 20^3 + 1/3 * 30 * 100^3. This differs from the moment of inertia about the y-axis (Iy), which is calculated as Iy = 1/12 * 80 * 90^3 + 1/12 * 80 * 30^3. The factor of 1/3 in the Iz equation arises from the standard equation for the second moment of inertia when the axis lies along the edge of a rectangle, I = 1/3 bh^3. Understanding the parallel-axis theorem is crucial for applying these formulas correctly.

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Homework Statement
The couple M is applied to a beam of the cross section shown in a plane forming an angle with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.
Relevant Equations
I_rectangle = (1/12)bh^3
Screenshot 2025-02-11 at 12.51.50 AM.png

Hi, I'm a little confused on calculating the moment of inertia about the z-axis.
For calculating the moment of inertia about the y-axis, I did the following (verified to be correct):
Iy = 1/12 * 80 * 90^3 + 1/12 * 80 * 30^3

I did the same for the z-axis but it turned out to be wrong.
The correct equation for Iz is 1/3 * 90 * 60^3 + 1/3 * 60 * 20^3 + 1/3 * 30 * 100^3

This is my first time seeing a second moment of inertia equation like that with a denominator of 3.
If someone could post an explanation regarding the equation for Iz, I'd greatly appreciate it.

Thanks!
 

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cherry said:
Homework Statement: The couple M is applied to a beam of the cross section shown in a plane forming an angle with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.
Relevant Equations: I_rectangle = (1/12)bh^3

View attachment 357146
Hi, I'm a little confused on calculating the moment of inertia about the z-axis.
For calculating the moment of inertia about the y-axis, I did the following (verified to be correct):
Iy = 1/12 * 80 * 90^3 + 1/12 * 80 * 30^3

I did the same for the z-axis but it turned out to be wrong.
The correct equation for Iz is 1/3 * 90 * 60^3 + 1/3 * 60 * 20^3 + 1/3 * 30 * 100^3

This is my first time seeing a second moment of inertia equation like that with a denominator of 3.
If someone could post an explanation regarding the equation for Iz, I'd greatly appreciate it.

Thanks!
Please go ahead and show your work for ##I_z##. Also see LaTeX Guide to format the equations.
 
cherry said:
Relevant Equations: I_rectangle = (1/12)bh^3
The above equation gives the (area or second) moment of inertia of a rectangle (measuring ##b \times h##) about an axis when:

a) the axis passes through the rectangle's centroid and lies in the plane of the rectangle;

b) the sides of length ##b## are parallel to the axis (so the sides of length ##h## are perpendicular to the axis).

If the above conditions do not apply, you need a different formula.
 
cherry said:
If someone could post an explanation regarding the equation for Iz, I'd greatly appreciate it.
I suggest reviewing the parallel-axis theorem to understand the factor of 1/3.
 
We've not heard back from the OP. However, note that the 'official' answer given in Post #2 as:
cherry said:
The correct equation for Iz is 1/3 * 90 * 60^3 + 1/3 * 60 * 20^3 + 1/3 * 30 * 100^3
can be written down immediately.

We consider the shape broken into several pieces such that the z-axis lies along an edge of each piece:

1739538277043.png

The standard equation for the second moment of inertia when the axis lies along the edge of a rectangle is ##I = \frac 13 bh^3## where ##b## is the length of the side along the axis. (The formula, if not supplied, can be derived using the parallel axis theorem.)

To get the 'official' answer (above) the two yellow rectangles are treated as a single one measuring 60 mm x 20 mm.
 
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