For a isosceles triangle with base $b$ and height $h$, the surface moment of inertia around the $z$-axis is $\frac{bh^3}{36}$ (considering that our coordinate system has $z$ in the horizontal and $y$ in the vertical axis, and has its origin on the triangle's center of mass (which is at $\left\{\frac{b}{2},-\frac{h}{3}\right\} $ if you put your coordinate system's origin at the bottom left corner if the triangle).

I know that the formula for the moment of inertia around the $z$-axis is $I_z = \int_A{y^2 dA}$, but I cannot figure out how to derive the formula from that. How is it done?


1 Answer 1


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The infinitesimal area $\text{dA}$ is $2 z \text{dy}$.

The relationship between $z$ and $y$ can be got from the slope. $$\frac{z}{y-\frac{2 h}{3}}=\frac{0-\frac{b}{2}}{\frac{2 h}{3}+\frac{h}{3}}$$

Solving, we get $$z=\frac{b (2 h-3 y)}{6 h}$$

Thus $$I_{zz}=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 dA =\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 z \, dy =\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 \frac{b (2 h-3 y)}{6 h} \, dy=\frac{b h^3}{36}$$

  • $\begingroup$ Thanks for answering my query. I've since come to the same conclusion aa you and managed to solve the problem myself, but it's good to see that I was right with my assumptions. Thanks for helping me out! $\endgroup$
    – Skydiver
    Commented Nov 10, 2016 at 17:52

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