# Second moment of area - Need walkthrough

1. Mar 12, 2009

### peleus

Hi all,

I'm at uni starting off engineering, and I'm looking for a walk through in this algebra involved in the second moment of area. While I know it's probably pretty basic I'm undertaking a bridging course to try and keep up with my maths.

http://home.exetel.com.au/peleus/smoa.jpg

Here's a picture of the most relevant lecture slide discussing the problem.

I'll type out the steps they undertook to get the final formula for the second moment of area for a rectangle.

On the next page we take the integral of this, which I can do fine.

This gives

1. $$I = \frac{b}{3}[y^3]$$ with limits +h/2 and -h/2

Taking it further we end up with

2. $$I = \frac{b}{3}[\frac{h^3}{8}-(-\frac{h^3}{8})]$$

and finally we take it to the step

$$I = \frac{bh^3}{12}$$

- Why are the limits h/2 and -h/2, isn't this simply the middle of the rectangle?
- Why is $$dI = y^2 dA$$, where does the $$y^2$$ come from?

Any help is greatly appreciated.

Last edited by a moderator: Apr 24, 2017
2. Mar 12, 2009

### minger

Because its the second moment of area. The first moment of area is defined as:
$$M_x = \int_A y\,dA$$
and is commonly used to find the centroid of an object.

It's y² simply by definition.

edit: I don't quite understand your first question. I'll take a stab though. The limits are h/2 because you're finding the moment around the centroid of the object. If you were finding it about an axis on the base of the part, the limits would be 0->h. For practice, try finding the second moment of area about an axis other than the the centroidial axis.

hint: the area moment of inertia is 4 times greater about the base rather than the centroid (having y squared helps!).

3. Mar 12, 2009