2
$\begingroup$

For a isosceles triangle with base b and height h the surface moment of inertia around tbe z axis is $\frac{bh^3}{36}$ (considering that our coordinate system has z in the horizontal and y in the vertical axis and got it's 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 in 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 can for the love of god not figure out how to derive the formular from that. How is it done?

Any help would be highly appreciated!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

enter image description here

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

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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
    Nov 10, 2016 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.