Step 1 We are asked to calculate the moment of inertia of shaded area about the x-axis.
Step 2 The centroid for each section is shown as:
We have the height of first section which is ${h_1} = \left( {150\;{\rm{mm}} + 150\;{\rm{mm}}} \right) = 300\;{\rm{mm}}$.
We have the base of first section which is ${b_1} = \left( {100\;{\rm{mm}} + 100\;{\rm{mm}}} \right) = 200\;{\rm{mm}}$.
We have the height of second section which is ${h_2} = \left( {150\;{\rm{mm}} + 150\;{\rm{mm}}} \right) = 300\;{\rm{mm}}$.
We have the base of second section which is ${b_2} = 150\;{\rm{mm}}$.
We have the radius of third section which is $r = 75\;{\rm{mm}}$.
Step 3 To calculate the area of first section we use the formula:
\[{A_1} = {h_1}{b_1}\]
Step 4 Substitute the known values in the formula:
\begin{array}{c} {A_1} = \left( {300\;{\rm{mm}}} \right)\left( {200\;{\rm{mm}}} \right)\\ = 60000\;{\rm{m}}{{\rm{m}}^2} \end{array}
Step 5 To calculate the area of second section we use the formula:
\[{A_2} = \frac{1}{2}{h_2}{b_2}\]
Step 6 Substitute the known values in the formula:
\begin{array}{c} {A_2} = \frac{1}{2}\left( {300\;{\rm{mm}}} \right)\left( {150\;{\rm{mm}}} \right)\\ = 22500\;{\rm{m}}{{\rm{m}}^2} \end{array}
Step 7 To calculate the area of third section we use the formula:
\[{A_3} = \pi {r^2}\]
Step 8 Substitute the known values in the formula:
\begin{array}{c} {A_3} = \left( {3.1416} \right){\left( {75\;{\rm{mm}}} \right)^2}\\ = {\rm{17671}}{\rm{.5}}\;{\rm{m}}{{\rm{m}}^2} \end{array}
Step 9 To calculate the distance of centroid of the first segment from x-axis we use the formula:
\[{y_1} = \frac{{{h_1}}}{2}\]
Step 10 Substitute the known values in the formula:
\begin{array}{c} {y_1} = \frac{{\left( {300\;{\rm{mm}}} \right)}}{2}\\ = 150\;{\rm{mm}} \end{array}
Step 11 To calculate the distance of centroid of the second segment from x-axis we use the formula:
\[{y_2} = \frac{{{h_2}}}{3}\]
Step 12 Substitute the known values in the formula:
\begin{array}{c} {y_2} = \frac{{\left( {300\;{\rm{mm}}} \right)}}{3}\\ = 100\;{\rm{mm}} \end{array}
Step 13 From the given diagram, the centroid of third section is given as:
\[{y_3} = 150\;{\rm{mm}}\]
Step 14 To calculate the moment of inertia of first section about an axis parallel to x-axis we use the formula:
\[{\bar I_1} = \frac{1}{{12}}{b_1}h_1^3\]
Step 15 Substitute the known values in the formula:
\begin{array}{c} {{\bar I}_1} = \frac{1}{{12}}\left( {200\;{\rm{mm}}} \right){\left( {300\;{\rm{mm}}} \right)^3}\\ = {\rm{450000000}}\;{\rm{m}}{{\rm{m}}^4} \end{array}
Step 16 To calculate the moment of inertia of second section about an axis parallel to x-axis we use the formula:
\[{\bar I_2} = \frac{1}{{36}}{b_2}h_2^3\]
Step 17 Substitute the known values in the formula:
\begin{array}{c} {{\bar I}_2} = \frac{1}{{36}}\left( {150\;{\rm{mm}}} \right){\left( {300\;{\rm{mm}}} \right)^3}\\ = {\rm{112500000}}\;{\rm{m}}{{\rm{m}}^4} \end{array}
Step 18 To calculate the moment of inertia of third section about an axis parallel to x-axis we use the formula:
\[{\bar I_3} = \frac{\pi }{4}{r^4}\]
Step 19 Substitute the known values in the formula:
\begin{array}{c} {{\bar I}_3} = \frac{{\left( {3.1416} \right)}}{4}{\left( {75\;{\rm{mm}}} \right)^4}\\ = {\rm{24850546}}{\rm{.9}}\;{\rm{m}}{{\rm{m}}^4} \end{array}
Step 20 To calculate the total moment of inertia of the shaded area we use the formula:
\[I = \left( {{{\bar I}_1} + {{\bar I}_2} - {{\bar I}_3}} \right) + \left( {{A_1}y_1^2 + {A_2}y_2^2 - {A_3}y_3^2} \right)\]
Step 21 Substitute the known values in the formula:
\[\begin{array}{c} I = \left\{ \begin{array}{l} \left( {450000000\;{\rm{m}}{{\rm{m}}^4} + {\rm{112500000}}\;{\rm{m}}{{\rm{m}}^4} - 24850546.9\;{\rm{m}}{{\rm{m}}^4}} \right) + \\ \left( \begin{array}{l} \left( {60000\;{\rm{m}}{{\rm{m}}^2}} \right){\left( {150\;{\rm{mm}}} \right)^2} + \left( {225000\;{\rm{m}}{{\rm{m}}^2}} \right){\left( {100\;{\rm{mm}}} \right)^2} - \\ \left( {17671.5\;{\rm{m}}{{\rm{m}}^2}} \right){\left( {150\;{\rm{mm}}} \right)^2} \end{array} \right) \end{array} \right\}\\ = {\rm{537649453}}{\rm{.1}}\;{\rm{m}}{{\rm{m}}^4} + {\rm{1350000000}}\;{\rm{m}}{{\rm{m}}^4} + {\rm{225000000}}\;{\rm{m}}{{\rm{m}}^4} - {\rm{397608750}}\;{\rm{m}}{{\rm{m}}^{\rm{4}}}\\ {\rm{ = 1715040703}}{\rm{.1}}\;{\rm{m}}{{\rm{m}}^{\rm{4}}} \end{array}\]
