second moment of area - triangle

[jonnburton]

I am trying to derive the equation for the second moment of area of a trangle about its central axis (Y-Y).

I was wondering if anyone could give me some advice on how to approach this. This is what I have done so far:



The base of the triangle is of width b, the height h.

The expression for the slope on the RHS of the image is $\displaystyle y=h(1-\frac{x}{2b})$ and the expression for the opposite slope on the LHS is $\displaystyle y=h(\frac{x}{2b}+1)$

I know the first thing is to take an elementary strip of width dx.

the next step which I've seen people do (havign looked it up online) is to say the second moment of area of this strip about its own centroid is $\displaystyle \frac{BH^3}{12}$, using capitals to denote the dimensions refer to the strip, and not the triangle.

(I can't say I see where that previous step comes from. When we take the second moment of area of a rectangular plate, we don't start off by giving an expression for the second moment of an elementary strip about its own centroid, so why is it done with the triangle?) Or is it because in a plate the centroid of the plate is the same height as the centroid of all the strips..?

Anyway, be that as it may, the expression for the second moment of area of the ith strip is given by:

$\displaystyle I_n = \frac{BH^3}{12}+Ad^2$. where d is the distance from the elementary strip to the centreline of the triangle.

Plugging in the values of the slope for H:

$\displaystyle I_n = \frac{1}{12}\left(h-\frac{h}{2b}x\right)^3+dx\left(h-\frac{h}{2b}x\right)d^2$


However, this is where I have more doubts. There aretwo slopes; above I plugged in the slope to the right of 0. So now I am not sure how to take this further, or even if what I've done so far is the right way of going about this.