Question A.6: Determine the moment of inertia Ix of the area shown with re......

Location of Centroid. The centroid C of the area has been located in Concept Application A.2 for the given area. From this, C is located 46 mm above the lower edge of area A.

Computation of Moment of Inertia. Area A is divided into two rectangular areas A_1 \text { and } A_2 (Fig. A.15b), and the moment of inertia of each area is found with respect to the x axis.

Rectangular Area \mathbf{A _1}. To obtain the moment of inertia \left(I_x\right)_1 of A _1 with respect to the x axis, first compute the moment of inertia of A _1 with respect to its own centroidal axis x′. Recalling the equation in part a of Concept Application A.4 for the centroidal moment of inertia of a rectangular area,

\left(\bar{I}_{x^{\prime}}\right)_1=\frac{1}{12} b h^3=\frac{1}{12}(80 \ mm )(20 \ mm )^3=53.3 \times 10^3 \ mm ^4

Using the parallel-axis theorem, transfer the moment of inertia of A _1 from its centroidal axis x′ to the parallel axis x:

\begin{aligned}\left(I_x\right)_1=\left(\bar{I}_{x^{\prime}}\right)_1+A_1 d_1^2 & =53.3 \times 10^3+(80 \times 20)(24)^2 \\& =975 \times 10^3 \ mm ^4\end{aligned}

Rectangular Area \mathbf{A _2}. Calculate the moment of inertia of A _2 with respect to its centroidal axis x” and use the parallel-axis theorem to transfer it to the x axis to obtain

\begin{aligned}& \left(\bar{I}_{x^{\prime \prime}}\right)_2=\frac{1}{12} b h^3=\frac{1}{12}(40)(60)^3=720 \times 10^3 \ mm ^4 \\\left(I_x\right)_2= & \left(\bar{I}_{x^{\prime \prime}}\right)_2+A_2 d_2^2=720 \times 10^3+(40 \times 60)(16)^2 \\= & 1334 \times 10^3 \ mm ^4\end{aligned}

Entire Area A. Add the values for the moments of inertia of A_1 \text { and } A_2 with respect to the x axis to obtain the moment of inertia \bar{I}_x of the entire area:

\begin{aligned}& \bar{I}_x=\left(I_x\right)_1+\left(I_x\right)_2=975 \times 10^3+1334 \times 10^3 \\& \bar{I}_x=2.31 \times 10^6 \ mm ^4\end{aligned}