4
$\begingroup$

I have a question that is annoying me for a long time. I know that I can calculate the moment of inertia of a rectangular cross section around a given axis located on its centroid by the following formulas:

enter image description here

I also know that more generically, the moment of inertia is given by the integer of an area times the square of the distance from its centroid to the axis.

So lets say I have a rectangular section with a height of 200 mm and a width of 20 mm.

If I use the formulas of the first method, in relation to an x axis parallel to the width:

$$I_x=\frac{bh^3}{12}=\frac{20\cdot200^3}{12}=1333.33\text{ cm}^4$$

Using the second method, why do I get different results when calculating twice the area of half a section, multiplied by the square of the distance from its centroid to the x axis.

$$I_x= 2A_{half\ section}d^2 = 2\cdot(200/2\cdot20)*(200/4)^2= 1000\text{ cm}^4$$

$\endgroup$
7
$\begingroup$

You have misunderstood the parallel axis theorem.

The moment of inertia of an object around an axis is equal to

$$I = \iint\limits_R\rho^2\text{d}A$$

where $\rho$ is the distance from any given point to the axis. In the case of a rectangular section around its horizontal axis, this can be transformed into

$$\begin{align} I_x &= \int\limits_{-b/2}^{b/2}\int\limits_{-h/2}^{h/2}y^2\text{d}y\text{d}x \\ I_x &= \int\limits_{-b/2}^{b/2}\left.\dfrac{1}{3}y^3\right\rvert_{-h/2}^{h/2}\text{d}y\text{d}x \\ I_x &= \int\limits_{-b/2}^{b/2}\dfrac{1}{3}\dfrac{h^3}{4}\text{d}x \\ I_x &= \left.\dfrac{1}{3}\dfrac{h^3}{4}x\right\rvert_{-b/2}^{b/2} \\ I_x &= \dfrac{bh^3}{12} \end{align}$$

Now, what if we wanted to get the inertia around some other axis at a distance $r$ from our centroid? In this case, all we have to do is:

$$I = \iint\limits_R(\rho+r)^2\text{d}A$$ $$I = \iint\limits_R\left(\rho^2 + 2\rho r + r^2\right)\text{d}A$$ $$I = \iint\limits_R\rho^2\text{d}A + 2r\iint\limits_R\rho\text{d}A + r^2\iint\limits_R\text{d}A$$

The first component $\iint\limits_R\rho^2\text{d}A$ is simply equal to the original moment of inertia. The second component $2r\iint\limits_R\rho\text{d}A$ is equal to zero since we're integrating around the centroid (it'll become a function of $y^2$, which when integrated from $-h/2$ to $h/2$ gives zero). The third component is equal to $Ar^2$. So, in the end, we get:

$$I' = I + Ar^2$$


So, if you want to calculate the moment of inertia of a rectangular section by considering each of its halves (half above the centroid, half below), you need to do:

$$\begin{align} I_{half} &= \dfrac{b\left(\dfrac{h}{2}\right)^3}{12} \\ I'_{half} &= I_{half} + b\left(\dfrac{h}{2}\right)\left(\dfrac{h}{4}\right)^2 \\ &= \dfrac{bh^3}{96} + \dfrac{bh^3}{32} = \dfrac{bh^3}{24} \\ I_{full} &= 2I'_{half} = \dfrac{bh^3}{12} \end{align}$$

Which is the original value for the full section. QED.

$\endgroup$
2
$\begingroup$

The following sentence is not correct:

the moment of inertia is given by the integer of an area times the square of the distance from its centroid to the axis

You have to add to that, the moment of inertia of the area around its own centroid. That is what the parrallel axis theorem is all about: $$ I = I_o + A\cdot d^2 $$

where: - Io the moment of inertia around centroid - I is the moment of inertia around any parallel axis and - d the distance between the two axes

So applying the above to your example, each half area (below and above centroidal axis) should have a moment of inertia equal to:

$$ I_{half} = \frac{b (h/2)^3}{12} + \frac{bh}{2}\cdot\left(\frac{h}{4}\right)^2 $$ $$ I_{half} = \frac{b h^3}{96} + \frac{b h^3}{32} $$ $$ I_{half} = \frac{b h^3}{24} $$

Therefore, for the whole section, due to symmetry: $$ I = 2 I_{half} = \frac{b h^3}{12} $$

Demonstrating the example with your numbers: $$ I = 2\left(\frac{20 (100)^3}{12} + 20\cdot 100\cdot\left(50\right)^2 \right)\,mm^4$$ $$ I = 2\left(1666666.7 + 5000000 \right) \,mm^4 $$ $$ I = 13333333.3 \,mm^4 = 1333.33 cm^4 $$

Usually in enginnereing cross sections the parallel axis term $Ad^2$ is much bigger than the centroidal term $I_o$. It is rather acceptable to ignore the centroidal term for the flange of an I/H section for example, because d is big and flange thickness (the h in the above formulas) is quite small. In other circumstances however this is not accepteble.

Calculation of moment of inertia for parallel axes

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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