Why is the moment of inertia about the $x$-axis used in deflection?

I am following along this page for an experiment on the Young's Modulus of a beam by deflection, and I don't understand why they use the rotational inertia about the $$x$$ axis of the beam: $$I = \frac{bh^3}{12}.$$

I have found that this comes from:

$$I_x = \iint_R z^2dA = \int_{-\frac{h}{2}}^{\frac{h}{2}}\int_{-\frac{b}{2}}^{\frac{b}{2}} z^2dxdz = b\int_{-\frac{h}{2}}^{\frac{h}{2}} z^2dz = \frac{bh^3}{12}.$$

Since we're deflecting the beam down (along the z axis), wouldn't it make sense to use the z inertia instead?

• It's difficult to know exactly what you mean because neither you nor your link ever define a coordinate system. But it doesn't matter whether you call it the moment of inertia "about the $x$ axis" or the moment of inertia "for $z$-direction beam deflections". (Presumably then the beam points in the $y$ direction?) It's the same equation. The point is that deflection in a certain direction should be penalized much more if the beam cross section extends more in that direction. That's what the $h^3$ term achieves (corresponding to the cross section height, or the dimension in the $z$ direction). Nov 5 '17 at 22:41
• That is the area moment of inertia, which is different from the rotational moment of inertia. The rotational inertia of a plate of dimensions $h$ and $b$ with mass $m$ is $\frac{m}{12}(b^2 + h^2)$ about an axis through the center. May 21 '19 at 0:20
• $I_x$ is not rotational inertia. It is area moment of the section. Sep 19 '19 at 14:07
• Mr. Russ Elliott (the author of the article) is responsible for propagating a common confusion in this topic. His terminology of "moment of inertia" instead of "area moment" does not help to clarify the differences: $$\begin{array}{r|ccl} \text{Term} & \text{Definition} & \text{Units} & \text{Notes}\\ \hline \text{Moment of Inertia} & I_{xx}=\iiint\left(y^{2}+z^{2}\right){\rm d}m & {\rm kg\,m^{2}} & \text{used in dynamics with rotations}\\ \text{Area Moment} & I_{xx}=\iint y^{2}{\rm d}A & {\rm m^{4}} & \text{used in flexible beams } \end{array}$$ Sep 19 '19 at 14:58

Assuming that the length of your beam is along the $x$-axis and the deflection towards $z$-axis, what counts for the bending moment in pure bending of beams is the area moment of inertia of the cross-section of your beam, that is the area in the $yz$-plane according to my assumption above. If your cross-section is rectangular with width $b$ and height $h$ you find the expression you've indicated.

Bending is due to internal moments (torques) that cause movement along a plane. On this case bending is along the yz plane, and therefore the moment responsible has to be along the x axis. You can view bending as kind of rotation, whose axis is perpendicular to the plane of movement.

Consider a diagram of the situation The point of the area moment $$I_x$$ is to related the applied moment $$M_x = \ell \, F_z$$ to the slope $$K$$ of stress field on the cross section. This slope is used to find stress value at each height $$z$$ along the cross section $$\sigma_y(z) = K\,z$$

This slope depends on two things. One is the applied moment $$M_x$$ and the other is the shape of the cross section. The shape contribution is called the area moment and it is used as follows

\begin{aligned} K & = \frac{M_x}{I_x} & & \text{or} & \sigma_y(z) = \frac{M_x}{I_x}\,z \end{aligned}

The last part is the calculation of $$I_x$$. The sum of the torques due to the stress distribution about the x-axis at the origin is evaluated by slicing the cross section at each height $$z$$.

$$M_x = \int z\, \sigma_y {\rm d}A = \int K z^2 {\rm d}A = K \, \int z^2 {\rm d}A$$

therefore $$\boxed{ I_x = \int z^2 {\rm d}A }$$

Note also, the the bending happens in the yz plane and there it is considered to be along the x axis (think of it as a rotation). So the label $$I_x$$ is appropriate.

One last thought

The term "moment of inertia" in your post (and the linked paper) is incorrect. Inertia only refers to resistance to rotational motion in dynamics. The proper term is "second area moment", which describes exactly what it is. Take the second moment of an infinitesimal area $${\rm d}A$$ and sum it up $$\rightarrow \int z^2 {\rm d}A$$.

• Great answer, but $K$ should depend on the Young modulus of the material...The term "moment of inertia" is probably wrongly chosen, especially for static problems, but it is widely used! Sep 19 '19 at 19:16
• @user8736288 - The slope $K$ does not depend on modulus of elasticity. Young's modulus is used when calculating deflections and in this case we only consider the stress distribution. The slope depends solely on the shape and the applied moment. Sep 19 '19 at 19:36
• @user8736288 - You mean moment of inertia is widely misused. Sep 19 '19 at 19:57
• Sorry, you're completely right about $K$! Good luck on changing "moment of inertia"."moment of area" is hardly any better, I agree "second moment of area" is the correct one. Minor point - is there a particular convention as to why the represented stress seem to counteract $M_{x}$? I woud read $M_{x}= -z\sigma_{y}/I_{x}$ from your diagram. Sep 20 '19 at 6:58
• @user8736288 if stress was a vector we would use a cross product, bit because it is a tensor we are stuck with index notation. The moment is $${\rm d} M_x = (y \sigma_z - z \sigma_y) {\rm d}A$$ but we only have stress in one direction. The stress formula used in engineering is $$\sigma_y(z) = \frac{z\,M_x}{I_x}$$ or $$\boxed{ M_x = I_x \; \frac{ \sigma_y}{z}}$$ Sep 20 '19 at 12:37

Your beam defletcts in the $$z$$ direction, but the sections of your beam rotates around the $$x$$ axis, hence the name given.