# 2nd Moment of Area

• Apr 12th 2009, 08:06 AM
Mush
2nd Moment of Area
Hello. I'm having a bit of a problem calculating the 2nd moment of area of 2D shapes etc.

The Question:

Calculate $I_x$ for the section sketched in Figure 1 to within about 2%.

(I have attached the figure as a JPEG - all measurements are in millimetres).

Useful equations

In my notes, there are 3 moments of area defined:

$I_z = \int_A y^2 \,dA$

$I_y = \int_A z^2 \,dA$

$J = I_x = \int_A r^2 \,dA = \int_A (y^2+z^2)dA = I_z+I_y$

Where J is the POLAR 2nd moment of area.

My Instructor's Solution

$I_{zz} = \bigg[ \frac{1}{12} \times 300 \times 25^3 + 300 \times 25 \times \bigg(\frac{500-25}{2}\bigg)^2\bigg] \times 2 + \frac{1}{12} \times 12 \times 450^3$

$= 9.38 \times 10^8 \, mm^4$

My Confusion

1) In my instructors solution he calculated $I_{zz}$ which is a notation he has never used in his notes? Does this mean the same as $I_x$, which is what the question asked for? Or has he calculated what he calls $I_z$ in his notes?

2) In his solution he seems to be pulling figures straight out of his arse, and there is no integration involved at all, despite the fact that the definitions of 2nd moments of intertia are all integrals.

Can anyone offer an explanation?