Solving Second Moment of Area for Beam w/ Missing Circle

[ganondorf29]

Homework Statement

I'm having a problem calculating the I value for this beam. I'm not sure on how to account for the missing circle in the front square and how to account for the beam changing from a 1" x 1" square to a 5" x 0.5" beam

Homework Equations

Ixx = bh^12/12

The centroid of the beam is around x = 2.89711 in

The thickness of the beam is the same and so is it's density

The Attempt at a Solution

Here is a picture of the beam


**If there was no missing circle, could I just add I_square + I_rect = I_total? **

 

[SteamKing]

ganondorf29:
The reason why you need to calculate I (2nd moment of area) is not clear. Please include a complete statement of the problem you are trying to solve.

 

[rock.freak667]

The Attempt at a Solution

Here is a picture of the beam


**If there was no missing circle, could I just add I_square + I_rect = I_total? **

First thing, just get I for the square (with no hole in it) and then subtract the I for a cylinder. That will give you the I for the shape.

Second thing: The I that you get will be about its own centroid, not the centroid of the beam. So you need apply parallel axis theorem for the square (with the hole) and the rectangular section before you add them up.

 

[pongo38]

Do you want the I about the x-axis, or the y-axis? If about the x-axis, the parallel axis theorem as suggested by rockfreak does not apply. You made a mistake in your formula for Ixx in post #1. Can you see what the error is?

 

[rock.freak667]

First thing, just get I for the square (with no hole in it) and then subtract the I for a cylinder. That will give you the I for the shape.

Second thing: The I that you get will be about its own centroid, not the centroid of the beam. So you need apply parallel axis theorem for the square (with the hole) and the rectangular section before you add them up.

EDIT: Pongo is right, forget my parallel axis thing, I keep substituting I with J.