## Axis Theorem

In the figure below, what is the moment of inertia about the x’ axis in m^4?

##
__
__**Hint**

**Hint**

Parallel Axis Theorem:

$$$I_x=I_{x_{c}}+d_{y}^{2}A$$$

where
$$d_y$$
is the distance between the new axis and the object’s centroid,
$$I_{x_{c}}$$
is the moment of inertia about the centroid axis,
$$A$$
is the total cross sectional area, and
$$I_x$$
is the moment of inertia about the new axis.

##
__
__**Hint 2**

**Hint 2**

The moment of inertia about the centroid axis of a triangle:

$$$I_{x_{c}}=\frac{bh^3}{36}$$$

where
$$b$$
and
$$h$$
are defined as:

Parallel Axis Theorem:

$$$I_x=I_{x_{c}}+d_{y}^{2}A$$$

where
$$d_y$$
is the distance between the new axis and the object’s centroid,
$$I_{x_{c}}$$
is the moment of inertia about the centroid axis,
$$A$$
is the total cross sectional area, and
$$I_x$$
is the moment of inertia about the new axis.

Most online tables (including the FE handbook) for area moment of inertia and centroids will have a figure similar to this:

where the centroid and moment of inertia about the centroid axis are respectively defined as:

$$$y_c=\frac{h}{3}$$$

$$$I_{x_{c}}=\frac{bh^3}{36}$$$

Keep in mind that the problem is not identical to the reference figure (it’s flipped vertical). Therefore, the triangle’s centroid relative to the x-axis is actually:

$$$y_c=\frac{2h}{3}=\frac{2(1m)}{3}=\frac{2m}{3}=0.667\:m$$$

Since
$$d_y$$
is the distance between the new axis and the object’s centroid:

$$$d_y=x'-y_c=1.25m-0.667m=0.5833\:m$$$

Fortunately, the moment of inertia about the centroid axis equation remains the same:

$$$I_{x_{c}}=\frac{(0.9m)(1m)^3}{36}=\frac{0.9m^4}{36}=0.025\:m^4$$$

Thus, the moment of inertia about the x’ axis is:

$$$I_x=0.025m^4+(0.5833m)^2[\frac{1}{2}(0.9m)(1m)]=0.025m^4+0.34m^2(\frac{0.9m^2}{2})$$$

$$$=0.025m^4+0.153m^4=0.18\:m^4$$$

$$$0.18\:m^4$$$