2nd Moment of Area of a ring

Question

This is my first question here. I am having trouble finding the 2nd moment of area of a ring with a thickness 't' and radius 'r'. I seem to be a factor of 2 out.

What I do is I consider an area like a piece of pie, and say that this is dA. I believe

$$dA=r*t*dθ$$

where r is the radius, t is the thickness and dθ is the small angle over which dA is encompassed. Then I integrate with respect to area:

$$\int r^2 dA$$ $$\int_0^{2\pi} r^2 * r*t*d\theta$$ $$\int_0^{2\pi} r^3*t*d\theta$$ $$2\pi*r^3*t$$

which is a factor of 2 out. Any help would be appreciated Thanks Matt

Answer 1

You have done everything right! You calculated the polar moment of inertia $J$, which is the sum of second moment of areas $I_i$. But here you have a symmetric object so:

$$J = I_x+I_y$$ and $$I_x=I_y=\frac{J}{2} = \pi r^3t$$.

Answer 2

You can't use the thickness as a constant. That is a circular section and has to be integrated. look at the shaded T on your diagram, the geometry is like a trapezoid.

There is an easy way to do this which is subtracting the I of the inner disk from that of the outer disk.

But following your way, we have.

$$J_z= \int_{}^{} \int_{}^{} r^2 da= \int_{0}^{2\pi}\int_{r1}^{r2} r^2(rdrd\theta) = \int_{0}^{2\pi}\int_{r1}^{r2}r^3drd\theta \\ \int_{0}^{2\pi} \left [ \frac{r_2^4 -r_1^4}{4} \right]d\theta = \pi/2(r_2^4-r_1^4)$$