Moment of inertia of a cantilever beam

classical-mechanicsmomentmoment of inertiarotational-dynamicsstructural-beam

As part of a project, I'm calculating deflection of a cantilever beam, something like the image below where the beam bends due to a load.

enter image description here

To do so requires its moment of inertia, and I'm thinking of using the beam's rectangular cross section. However, all the sources I've looked at only describe moment of inertia about the x or y axes and rotating about the center of the object–not one end of the beam and the z axis as is shown with the cantilever beam.

So, would the moment of inertia in this case still be $I=\frac{1}{12}bh^3$ as is the case for all the other rectangular cross section moment formulas I've seen, or would it be different and why?

Best Answer

Note that the $I(x)$ term in the beam deflection formula is the area moment of inertia of a cross-section of the beam about an axis perpendicular to the plan of the cross-section. $I(x)$ may be a function of the distance $x$ along the beam - although in your example its is not, as the cross-section of the beam is the same at all points along it.

The perpendicular axis theorem tells us that the area moment of inertia of a two-dimensional shape about an axis perpendicular to its plane is the sum of its moments of inertia about two perpendicular axes within its plane. For a rectangle with height $h$ and breadth $b$ this gives

$\displaystyle I_z = I_x + I_y = \frac 1 {12} bh^3 + \frac 1 {12} b^3h = \frac 1 {12} bh(h^2+b^2)$