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: $$I = \frac.$$ I have found that this comes from: $$ I_x = \iint_R z^2dA = \int_>^>\int_>^> z^2dxdz = b\int_>^> z^2dz = \frac. $$ Since we're deflecting the beam down (along the z axis), wouldn't it make sense to use the z inertia instead?
$\begingroup$ 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). $\endgroup$
Commented Nov 5, 2017 at 22:41$\begingroup$ 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
$\begingroup$ 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: $$\begin
Bending is due to internal moments (torques) that cause movement along a plane. On this case bending is along the yz plane, and therefore the moment responsible has to be along the x axis. You can view bending as kind of rotation, whose axis is perpendicular to the plane of movement.
Consider a diagram of the situation
The point of the area moment $I_x$ is to related the applied moment $M_x = \ell \, F_z$ to the slope $K$ of stress field on the cross section. This slope is used to find stress value at each height $z$ along the cross section $$ \sigma_y(z) = K\,z $$
This slope depends on two things. One is the applied moment $M_x$ and the other is the shape of the cross section. The shape contribution is called the area moment and it is used as follows
The last part is the calculation of $I_x$ . The sum of the torques due to the stress distribution about the x-axis at the origin is evaluated by slicing the cross section at each height $z$ .
$$ M_x = \int z\, \sigma_y A = \int K z^2 A = K \, \int z^2 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.
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 $A$ and sum it up $\rightarrow \int z^2 A$ .
