## Moment of Inertia

Calculate the moment of inertia of the trapezoid area about the x' axis. Assume the centroid axis line is 2.6 m away from the x-axis.

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__**Hint**

**Hint**

Parallel Axis Theorem: The moment of inertia of an area about any axis is defined as the moment of inertia of the area about a parallel centroidal axis plus a term equal to the area multiplied by the square of the perpendicular distance
$$d$$
from the centroidal axis to the axis in question.

$$$I_{x'}=I_{xc}+d_{y}^{2}A$$$

where
$$A$$
is the area of the trapezoid:

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__**Hint 2**

**Hint 2**

First, find the moment of inertia about the centroid axis:

$$$I_{xc}=\frac{h^3(a^2+4ab+b^2)}{36(a+b)}$$$

First, let's find the moment of inertia about the centroid axis:

$$$I_{xc}=\frac{h^3(a^2+4ab+b^2)}{36(a+b)}$$$

$$$I_{xc}=\frac{6^3(3^2+4(3)(6)+(6)^2)}{36(3+6)}=\frac{216(117)}{324}=78$$$

Parallel Axis Theorem: The moment of inertia of an area about any axis is defined as the moment of inertia of the area about a parallel centroidal axis plus a term equal to the area multiplied by the square of the perpendicular distance
$$d$$
from the centroidal axis to the axis in question.

$$$I_{x'}=I_{xc}+d_{y}^{2}A$$$

where
$$A$$
is the area of the trapezoid:

$$$A=\frac{h(a+b)}{2}=\frac{6(3+6)}{2}=27$$$

Finally:

$$$I_{x'}=78+(5-2.6)^{2}(27)=233.52\:m^4$$$

$$$233.52\:m^4$$$