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In many references, mostly civil engineering, the second moment of area is referred as Moment of Inertia. Is that "really" correct? From my understanding, moment of inertia is analogous to mass in rotational motion, not the second moment of area.

Qmechanic
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ukg
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  • possible duplicate of System of Particles and Moment of Mass – DumpsterDoofus Mar 10 '14 at 13:22
  • Technically, moment of inertia is the second moment of density. Therefore, it's correct whenever density $\rho$ is uniform. – DumpsterDoofus Mar 10 '14 at 13:24
  • @DumpsterDoofus i did not get the right answer from the post you referred. My doubt is regarding the reason why the second moment of area is commonly called moment of inertia and is it a technically correct usage. – ukg Mar 10 '14 at 13:45
  • The linked post explains the interpretation of moment of inertia as a second moment of $\rho$, from which a reinterpretation in the context of the second moment of area is pretty clear. 2nd moment of area equaling moment of inertia is technically incorrect in general, but it is correct (up to a constant factor) whenever $\rho$ is constant, and exact when $\rho=1$. Write the formulas for second moment of area and moment of inertia side by side, substitute $\rho(\mathbf{r})=1$ and see for yourself. I can post it as an answer if you'd like. – DumpsterDoofus Mar 10 '14 at 13:54
  • Related : http://physics.stackexchange.com/q/48068/392 – John Alexiou May 09 '14 at 20:42

2 Answers2

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I agree, the terminology is a total mess.

  1. Statical Moment of Area or First Moment of Area $\int x {\rm d}A$
  2. Second Area of Moment $\int x^2 {\rm d}A$
  3. First Moment of Mass $\int x {\rm d}m$
  4. Mass Moment of Inertia or Second Moment of Mass $\int x^2 {\rm d}m$

So what is correct? You can distinguish between area moments and mass moments, and then annotate if it is the first, second or polar version.

John Alexiou
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For a 2D object with density $\rho(\mathbf{r})$, the second moment of area is $$A=\iint\mathbf{r}^2d\mathbf{r}$$ and the moment of inertia is $$I=\iint\rho(\mathbf{r})\mathbf{r}^2d\mathbf{r}.$$ When $\rho(\mathbf{r})=\rho_0$ you get $$I=\rho_0A$$ and when $\rho_0=1$ you get equality. Otherwise, they are not necessarily the same.

DumpsterDoofus
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  • When we want to determine the tension on a beam while bending we use the second moment of area instead of the moment of inertia, but why? If you know the answer leave a response to my question that I asked here – bruno Oct 31 '14 at 10:33
  • Hello @DumpsterDoofus, can you tell me which book contains the information of your answer please? I'm very interested in. – Gennaro Arguzzi Feb 06 '19 at 10:34