Chapter B, Problem B.56P

Determine the mass moment ofinertia of the steel fixture of Probs. B.35 and B.39 with respect to the axis through the origin that forms equal angles with the x, y, and z axes.

To determine

The mass moment of the inertia of the steel fixture passes through the origin.

Answer to Problem B.56P

The mass moment of the inertia of the steel fixture passes through the origin is $3.46\times {10}^{-6}\text{kg}\cdot {\text{m}}^2$.

Explanation of Solution

Given information:

Given information:

The density of the steel is $7850\text{kg}/{\text{m}}^{\text{3}}$.

The following figure represents the given system.

Figure-(1)

Write the expression for the mass of component 1.

$m_1=\rho V_1$   ....(I)

Here, the mass of component 1 is $m_1$, the density of component $\rho$ and the volume of component 1 is $V_1$.

Write the expression for the volume of the component 1.

$V_1=abc$   ....(II)

Here, the sides of the component is $a\text{or}b\text{and}c$.

Write the expression for the mass moment of inertia for component 1.

${\left(I_x\right)}_1=\left({\overline{I}}_x\right)+m_1{d}^2$   ....(III)

Here, the mass moment of inertia for component 1 is ${\left(I_x\right)}_1$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_x\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\overline{I}}_x=\frac{m_1\left(a^2+c^2\right)}{12}$   ....(IV)

Write the expression for the centroidal axis from the reference axis.

$d=\left(\frac{a}{2}+\frac{c}{2}\right)$   ....(V)

Write the expression for the mass moment of inertia for component 2.

${\left(I_x\right)}_2=\left({\overline{I}}_{x2}\right)+m_2{d}^2$   ....(VI)

Here, the mass moment of inertia for component 2 is ${\left(I_x\right)}_2$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_{x2}\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\overline{I}}_{x2}=\frac{m_2\left(e^2+d^2\right)}{12}$   ....(VII)

Write the expression for the centroidal axis from the reference axis.

$d^2={\left(c-\frac{e}{2}\right)}^2+{\left(c+\frac{d}{2}\right)}^2$   ....(VIII)

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\left({\overline{I}}_x\right)}_3=m_3\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right)f^2+\frac{g}{12}$   ....(IX)

Here, the mass moment of inertia for component 3 is ${\left(I_x\right)}_3$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_x\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia for component 3.

${\left(I_x\right)}_3={\left({\overline{I}}_x\right)}_3+m_3{d}^2$   ....(X)

Write the expression for the mass of component 2.

$m_2=\rho V_2$   ....(XI)

Here, the mass of component 2 is $m_2$, the density of component $\rho$ and the volume of component 2 is $V_2$.

Write the expression for volume of the component 2.

$V_2=bde$ …….. (XII)

Here, the sides of the component is $d\text{or}b\text{and}e$.

Write the expression for the mass of component 3.

$m_3=\rho V_3$   ....(XIII)

Here, the mass of component 3 is $m_3$, the density of component $\rho$ and the volume of component 3 is $V_3$.

Write the expression for volume of the component 3.

$V_3=\frac{\pi f^{2}g}{4}$ …….. (XIV)

Here, the diameter of the circle is $f$ and the width the component 3 is $g$.

Write the expression for the distance of the component 3.

$d^2={\left(c-\left(\frac{4f}{3\pi }\right)\right)}^2+{\left(a-\frac{g}{2}\right)}^2$ …….. (XV)

Write the expression for the mass moment of inertia with respect to $x$ axis.

$\left(I_x\right)={\left(I_x\right)}_1-{\left(I_x\right)}_2-{\left(I_x\right)}_3$   ....(XVI)

Write the expression for the mass moment of inertia for component 1.

${\left(I_y\right)}_1=\left({\overline{I}}_y\right)+m_1{d}^2$   ....(XVII)

Here, the mass moment of inertia for component 1 is ${\left(I_y\right)}_1$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_y\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\overline{I}}_y=\frac{m_1\left(a^2+b^2\right)}{12}$   ....(XVIII)

Write the expression for the centroidal axis from the reference axis.

$d=\left(\frac{a}{2}+\frac{c}{2}\right)$   ....(XIX)

Write the expression for the mass moment of inertia for component 2.

${\left(I_y\right)}_2=\left({\overline{I}}_{y2}\right)+m_2{d}^2$   ....(XX)

Here, the mass moment of inertia for component 2 is ${\left(I_y\right)}_2$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_{y2}\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\overline{I}}_{y2}=\frac{m_2\left(b^2+d^2\right)}{12}$   ....(XXI)

Write the expression for the centroidal axis from the reference axis.

$d^2={\left(\frac{b}{2}+c+\frac{d}{2}\right)}^2$   ....(XXII)

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\left({\overline{I}}_y\right)}_3=\frac{m_3\left(3f^2+g^2\right)}{12}$   ....(XXIII)

Here, the mass moment of inertia for component 3 is ${\left(I_y\right)}_3$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_y\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia for component 3.

${\left(I_y\right)}_3={\left({\overline{I}}_y\right)}_3+m_3{d}^2$   ....(XXIV)

Write the expression for the distance of the component 3.

$d^2=\left(\frac{b}{2}\right)+\left(a-\frac{g}{2}\right)$ …….. (XXV)

Write the expression for the mass moment of inertia with respect to $y$ axis.

$\left(I_y\right)={\left(I_y\right)}_1-{\left(I_y\right)}_2-{\left(I_y\right)}_3$   ....(XXVI)

Write the expression for the mass moment of inertia for component 1.

${\left(I_z\right)}_1=\left({\overline{I}}_z\right)+m_1{d}^2$   ....(XXVII)

Here, the mass moment of inertia for component 1 is ${\left(I_z\right)}_1$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_z\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\overline{I}}_z=\frac{m_1\left(c^2+b^2\right)}{12}$   ....(XXVIII)

Write the expression for the centroidal axis from the reference axis.

$d=\left(\frac{b}{2}+\frac{c}{2}\right)$   ....(XXIX)

Write the expression for the mass moment of inertia for component 2.

${\left(I_z\right)}_2=\left({\overline{I}}_{z2}\right)+m_2{d}^2$   ....(XXX)

Here, the mass moment of inertia for component 2 is ${\left(I_z\right)}_2$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_{z2}\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\overline{I}}_{z2}=\frac{m_2\left(b^2+e^2\right)}{12}$   ....(XXXI)

Write the expression for the centroidal axis from the reference axis.

$d^2={\left(\frac{b}{2}+c+\frac{e}{2}\right)}^2$   ....(XXXII)

Write the expression for the mass moment of inertia with respect to centroidal axis.

${\left({\overline{I}}_z\right)}_3=m_3\left(\frac{1}{2}-\frac{16}{9{\pi }^2}\right)f^2$   ....(XXXIII)

Here, the mass moment of inertia for component 3 is ${\left(I_z\right)}_3$, the mass moment of inertia with respect to centroidal axis is $\left({\overline{I}}_z\right)$ and the distance between the centroidal axis and parallel axis is $d$.

Write the expression for the mass moment of inertia for component 3.

${\left(I_z\right)}_3={\left({\overline{I}}_z\right)}_3+m_3{d}^2$   ....(XXXIV)

Write the expression for the distance of the component 3.

$d^2={\left(\frac{b}{2}\right)}^2+{\left(c-\frac{4f}{3\pi }\right)}^2$ …….. (XXXV)

Write the expression for the mass moment of inertia with respect to $y$ axis.

$\left(I_z\right)={\left(I_z\right)}_1-{\left(I_z\right)}_2-{\left(I_z\right)}_3$   ....(XXXVI)

Write the expression for the mass moment of inertia for component 1.

${\left(I_{xy}\right)}_1={\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_1+m_1\left({\overline{x}}_1\right)\left({\overline{y}}_1\right)$   ....(XXXVII)

Here, the products of the inertia of the body with respect to centroidal axis for component 1 is ${\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_1$, the coordinates for the centre of gravity of the component 1 is $\left({\overline{x}}_1\right)\text{and}\left({\overline{y}}_1\right)$.

Write the expression for the mass moment of inertia for component 1.

${\left(I_{yz}\right)}_1={\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_1+m_1\left({\overline{z}}_1\right)\left({\overline{y}}_1\right)$   ....(XXXVIII)

Here, the products of the inertia of the body with respect to centroidal axis for component 1 is ${\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_1$, the coordinates for the centre of gravity of the component 1 is $\left({\overline{z}}_1\right)\text{and}\left({\overline{y}}_1\right)$.

Write the expression for the mass moment of inertia for component 1.

${\left(I_{zx}\right)}_1={\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_1+m_1\left({\overline{z}}_1\right)\left({\overline{x}}_1\right)$   ....(XXXIX)

Here, the products of the inertia of the body with respect to centroidal axis for component 1 is ${\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_1$, the coordinates for the centre of gravity of the component 1 is $\left({\overline{z}}_1\right)\text{and}\left({\overline{x}}_1\right)$.

Write the expression for the mass moment of inertia for component 2.

${\left(I_{xy}\right)}_2={\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_2+m_2\left({\overline{x}}_2\right)\left({\overline{y}}_2\right)$   ....(XL)

Here, the products of the inertia of the body with respect to centroidal axis for component 2 is ${\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_2$, the coordinates for the centre of gravity of the component 2 is $\left({\overline{x}}_2\right)\text{and}\left({\overline{y}}_2\right)$.

Write the expression for the mass moment of inertia for component 2.

${\left(I_{yz}\right)}_2={\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_2+m_2\left({\overline{z}}_2\right)\left({\overline{y}}_2\right)$   ....(XLI)

Here, the products of the inertia of the body with respect to centroidal axis for component 2 is ${\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_2$, the coordinates for the centre of gravity of the component 2 is $\left({\overline{z}}_2\right)\text{and}\left({\overline{y}}_2\right)$.

Write the expression for the mass moment of inertia for component 2.

${\left(I_{zx}\right)}_2={\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_2+m_2\left({\overline{z}}_2\right)\left({\overline{x}}_2\right)$   ....(XLII)

Here, the products of the inertia of the body with respect to centroidal axis for component 2 is ${\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_2$, the coordinates for the centre of gravity of the component 2 is $\left({\overline{z}}_2\right)\text{and}\left({\overline{x}}_2\right)$.

Write the expression for the mass moment of inertia for component 3.

${\left(I_{xy}\right)}_3={\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_3+m_3\left({\overline{x}}_3\right)\left({\overline{y}}_3\right)$   ....(XLIII)

Here, the products of the inertia of the body with respect to centroidal axis for component 3 is ${\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_3$, the coordinates for the centre of gravity of the component 3 is $\left({\overline{x}}_3\right)\text{and}\left({\overline{y}}_3\right)$.

Write the expression for the mass moment of inertia for component 3.

${\left(I_{yz}\right)}_3={\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_3+m_3\left({\overline{z}}_3\right)\left({\overline{y}}_3\right)$   ....(XLIV)

Here, the products of the inertia of the body with respect to centroidal axis for component 3 is ${\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_3$, the coordinates for the centre of gravity of the component 3 is $\left({\overline{z}}_3\right)\text{and}\left({\overline{y}}_3\right)$.

Write the expression for the mass moment of inertia for component 3.

${\left(I_{zx}\right)}_3={\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_3+m_3\left({\overline{z}}_3\right)\left({\overline{x}}_3\right)$   ....(XLV)

Here, the products of the inertia of the body with respect to centroidal axis for component 3 is ${\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_3$, the coordinates for the centre of gravity of the component 3 is $\left({\overline{z}}_3\right)\text{and}\left({\overline{x}}_3\right)$.

Write the expression for the mass product of inertia.

$I_{xy}={\left(I_{xy}\right)}_1-{\left(I_{xy}\right)}_2-{\left(I_{xy}\right)}_3$   ....(XLVI)

Here, the mass product of inertia is $I_{xy}$.

Write the expression for the mass product of inertia.

$I_{yz}={\left(I_{yz}\right)}_1-{\left(I_{yz}\right)}_2-{\left(I_{yz}\right)}_3$   ....(XLVII)

Here, the mass product of inertia is $I_{yz}$.

Write the expression for the mass product of inertia.

$I_{zx}={\left(I_{zx}\right)}_1-{\left(I_{zx}\right)}_2-{\left(I_{zx}\right)}_3$   ....(XLVIII)

Here, the mass product of inertia is $I_{zx}$.

Write the expression for the position vector of $OA$.

$\overrightarrow{OA}=40\text{i}-35\text{j}$

Write the expression for the magnitude of the position vector $\overrightarrow{OA}$.

$d_{OA}=\sqrt{2825}$

Write the expression for the unit vector.

${\lambda }_{OA}=\frac{\overrightarrow{OA}}{d_{OA}}$   ....(XLIX)

Write the expression for the mass moment of the inertia of the steel fixture passes through the origin.

$I_{OA}=I_x{\lambda }_x^2+I_y{\lambda }_y^2-2I_{xy}{\lambda }_x{\lambda }_y$   ....(L)

Calculation:

Substitute $160\text{mm}$ for $a$, $80\text{mm}$ for $b$ and $50\text{mm}$ for $c$ in Equation (II).

$\begin{array}{c}{V}_1=\left(160\text{mm}\right)\left(80\text{mm}\right)\left(50\text{mm}\right)\\ =640000{\text{mm}}^3\left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\\ =6.4\times {10}^{-4}{\text{m}}^3\end{array}$

Substitute $7850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $6.4\times {10}^{-4}{\text{m}}^3$ for $V_1$ in Equation (I).

$\begin{array}{c}{m}_1=\left(6.4\times {10}^{-4}{\text{m}}^3\right)\left(7850\text{kg}/{\text{m}}^{\text{3}}\right)\\ =5.024\text{kg}\end{array}$

Substitute $38\text{mm}$ for $e$, $80\text{mm}$ for $b$ and $70\text{mm}$ for $d$ in Equation (XII).

$\begin{array}{c}{V}_2=\left(38\text{mm}\right)\left(80\text{mm}\right)\left(70\text{mm}\right)\\ =212800{\text{mm}}^3\left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\\ =2.28\times {10}^{-4}{\text{m}}^3\end{array}$

Substitute $7850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $2.28\times {10}^{-4}{\text{m}}^3$ for $V_2$ in Equation (XI).

$\begin{array}{c}{m}_2=\left(2.28\times {10}^{-4}{\text{m}}^3\right)\left(7850\text{kg}/{\text{m}}^{\text{3}}\right)\\ =1.67048\text{kg}\end{array}$

Substitute $24\text{mm}$ for $f$ and $40\text{mm}$ for $g$ in Equation (XIV).

$\begin{array}{c}{V}_3=\frac{\pi {\left(24\text{mm}\right)}^2\left(40\text{mm}\right)}{2}\\ =36191.147{\text{mm}}^3\left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\\ =3.6191\times {10}^{-5}{\text{m}}^3\end{array}$

Substitute $7850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $3.6191\times {10}^{-5}{\text{m}}^3$ for $V_3$ in Equation (XIII).

$\begin{array}{c}{m}_3=\left(3.6191\times {10}^{-5}{\text{m}}^3\right)\left(7850\text{kg}/{\text{m}}^{\text{3}}\right)\\ =0.2841\text{kg}\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $50\text{mm}$ for $c$ and $160\text{mm}$ for $c$ in Equation (IV).

$\begin{array}{c}{\overline{I}}_x=\frac{5.024\text{kg}\times \left({\left(160\text{mm}\right)}^2+{\left(50\text{mm}\right)}^2\right)}{12}\\ =\frac{5.024\text{kg}\times \left({\left(160\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(50\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =\frac{5.024\text{kg}\times \left(0.0281{\text{m}}^2\right)}{12}\\ =0.011764\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $50\text{mm}$ for $c$ and $160\text{mm}$ for $a$ in Equation (V).

$\begin{array}{c}d=\left(\frac{160\text{mm}}{2}+\frac{50\text{mm}}{2}\right)\\ =\left(80\text{mm}+25\text{mm}\right)\\ =105\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =0.105\text{m}\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $0.105\text{m}$ for $d$ and $0.011764\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_x$ in Equation (III).

$\begin{array}{c}{\left(I_x\right)}_1=0.011764\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right){\left(0.105\text{m}\right)}^2\\ =0.011764\text{kg}\cdot {\text{m}}^2+0.05538\text{kg}\cdot {\text{m}}^2\\ =0.06715\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $50\text{mm}$ for $c$, $38\text{mm}$ for $e$ and $70\text{mm}$ for $d$ in Equation (VIII).

$\begin{array}{c}{d}^2={\left(50\text{mm}-\frac{38\text{mm}}{2}\right)}^2+{\left(50\text{mm}+\frac{70\text{mm}}{2}\right)}^2\\ ={\left(69\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(85\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ =0.008186{\text{m}}^2\end{array}$

Substitute $1.67048\text{kg}$ for $m_2$, $38\text{mm}$ for $e$ and $70\text{mm}$ for $d$ in Equation (VII).

$\begin{array}{c}{\overline{I}}_{x2}=\frac{1.67048\text{kg}\times \left({\left(38\text{mm}\right)}^2+{\left(70\text{mm}\right)}^2\right)}{12}\\ =\frac{1.67048\text{kg}\times \left({\left(38\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =0.0008831\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.008186{\text{m}}^2$ for $d^2$, $0.0008831\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{x2}$ and $1.67048\text{kg}$ for $m_2$ in Equation (VI).

$\begin{array}{c}{\left(I_x\right)}_2=0.0008831\text{kg}\cdot {\text{m}}^2+\left(1.67048\text{kg}\right)\left(0.008186{\text{m}}^2\right)\\ =0.0008831\text{kg}\cdot {\text{m}}^2+0.0136745\text{kg}\cdot {\text{m}}^2\\ =0.0145576\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $24\text{mm}$ for $g$, $40\text{mm}$ for $f$ and $0.2841\text{kg}$ for $m_3$ in Equation (IX).

$\begin{array}{c}{\left({\overline{I}}_x\right)}_3=0.2841\text{kg}\times \left[\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right){\left(40\text{mm}\right)}^2+\frac{24\text{mm}}{12}\right]\\ =0.2841\text{kg}\times \left[\begin{array}{l}\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right){\left(40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ +\frac{{\left(24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}{12}\end{array}\right]\\ =0.2841\text{kg}\times \left[0.0001117{\text{m}}^2+0.000048{\text{m}}^2\right]\\ =0.00059035\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $24\text{mm}$ for $g$, $40\text{mm}$ for $f$, $50\text{mm}$ for $c$ and $160\text{mm}$ for $a$ in Equation (XV).

$\begin{array}{c}{d}^2={\left(50\text{mm}-\left(\frac{4\left(24\text{mm}\right)}{3\pi }\right)\right)}^2+{\left(160\text{mm}-\frac{40\text{mm}}{2}\right)}^2\\ ={\left(39.81\text{mm}\right)}^2+{\left(140\text{mm}\right)}^2\\ ={\left(39.81\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(140\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ =0.02118{\text{m}}^2\end{array}$

Substitute $0.02118{\text{m}}^2$ for $d^2$, $0.00059035\text{kg}\cdot {\text{m}}^2$ for ${\left({\overline{I}}_x\right)}_3$ and $0.2841\text{kg}$ for $m_3$ in Equation (X).

$\begin{array}{c}{\left(I_x\right)}_3=0.00059035\text{kg}\cdot {\text{m}}^2+\left(0.2841\text{kg}\right)\left(0.02118{\text{m}}^2\right)\\ =0.00059035\text{kg}\cdot {\text{m}}^2+0.006018\text{kg}\cdot {\text{m}}^2\\ =0.006608\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.006608\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_3$, $0.0145576\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_2$ and $0.06715\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_1$ in Equation (XVI).

$\begin{array}{c}\left(I_x\right)=0.06715\text{kg}\cdot {\text{m}}^2-0.0145576\text{kg}\cdot {\text{m}}^2-0.006608\text{kg}\cdot {\text{m}}^2\\ =0.04598\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $160\text{mm}$ for $a$ and $50\text{mm}$ for $c$ in Equation (XVII).

$\begin{array}{c}{\overline{I}}_y=\frac{\left(5.024\text{kg}\right)\left({\left(160\text{mm}\right)}^2+{\left(50\text{mm}\right)}^2\right)}{12}\\ =\frac{\left(5.024\text{kg}\right)\left({\left(160\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =0.013877\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $160\text{mm}$ for $a$ and $80\text{mm}$ for $b$ in Equation (XVIII).

$\begin{array}{c}d=\left(\frac{160\text{mm}}{2}+\frac{80\text{mm}}{2}\right)\\ =80\text{mm}+\text{40}\text{mm}\\ =1\text{20}\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =0.120\text{m}\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $0.120\text{m}$ for $d$ and $0.013877\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_y$ in Equation (XVI).

$\begin{array}{c}{\left(I_y\right)}_1=0.013877\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right){\left(0.120\text{m}\right)}^2\\ =0.013877\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right){\left(0.0144\text{m}\right)}^2\\ =0.013877\text{kg}\cdot {\text{m}}^2+0.0749376\text{kg}\cdot {\text{m}}^2\\ =0.08632\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $1.6704\text{kg}$ for $m_2$, $80\text{mm}$ for $b$ and $70\text{mm}$ for $d$ in Equation (XX).

$\begin{array}{c}{\overline{I}}_{y2}=\frac{1.6704\text{kg}\left({\left(80\text{mm}\right)}^2+{\left(70\text{mm}\right)}^2\right)}{12}\\ =\frac{1.6704\text{kg}\left({\left(80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =0.00157296\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $50\text{mm}$ for $c$, $80\text{mm}$ for $b$ and $70\text{mm}$ for $d$ in Equation (XXI).

$\begin{array}{c}{d}^2={\left(\frac{80\text{mm}}{2}+50\text{mm}+\frac{70\text{mm}}{2}\right)}^2\\ ={\left(40\text{mm}+50\text{mm}+35\text{mm}\right)}^2\\ ={\left(125\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ =0.015625{\text{m}}^2\end{array}$

Substitute $0.015625{\text{m}}^2$ for $d^2$, $0.00157296\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{y2}$ and $1.6704\text{kg}$ for $m_2$ in Equation (XIX).

$\begin{array}{c}{\left(I_y\right)}_2=0.00157296\text{kg}\cdot {\text{m}}^2+\left(1.6704\text{kg}\right)\left(0.015625{\text{m}}^2\right)\\ =0.00157296\text{kg}\cdot {\text{m}}^2+0.0261\text{kg}\cdot {\text{m}}^2\\ =0.02767296\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $24\text{mm}$ for $f$, $40\text{mm}$ for $g$ and $0.2841\text{kg}$ for $m_3$ in Equation (XXII).

$\begin{array}{c}{\left({\overline{I}}_y\right)}_3=\frac{0.2841\text{kg}\times \left(3{\left(24\text{mm}\right)}^2+{\left(40\text{mm}\right)}^2\right)}{12}\\ =\frac{0.2841\text{kg}\times \left(3{\left(24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =0.00007879\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $160\text{mm}$ for $a$, $80\text{mm}$ for $b$ and $40\text{mm}$ for $g$ in Equation (XXIV).

$\begin{array}{c}{d}^2=\left(\frac{80\text{mm}}{2}\right)+\left(160\text{mm}-\frac{40\text{mm}}{2}\right)\\ =40\text{mm}+\text{120}\text{mm}\\ =160\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =0.160\text{m}\end{array}$

Substitute $0.160\text{m}$ for $d^2$, $0.00007879\text{kg}\cdot {\text{m}}^2$ for ${\left({\overline{I}}_y\right)}_3$ and $0.2841\text{kg}$ for $m_3$ in Equation (XXIII).

$\begin{array}{c}{\left(I_y\right)}_3=0.00007879\text{kg}\cdot {\text{m}}^2+\left(0.2841\text{kg}\right)\left(0.160\text{m}\right)\\ =0.00007879\text{kg}\cdot {\text{m}}^2+0.045456\text{kg}\cdot {\text{m}}^2\\ =0.04553479\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.04553479\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_3$, $0.02767296\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_2$ and $0.08632\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_1$ in Equation (XXVI).

$\begin{array}{c}\left(I_y\right)=0.08632\text{kg}\cdot {\text{m}}^2-0.02767296\text{kg}\cdot {\text{m}}^2-0.04553479\text{kg}\cdot {\text{m}}^2\\ =0.013116\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $80\text{mm}$ for $b$ and $50\text{mm}$ for $c$ in Equation (XXVIII).

$\begin{array}{c}{\overline{I}}_z=\frac{5.024\text{kg}\times \left({\left(50\text{mm}\right)}^2+{\left(80\text{mm}\right)}^2\right)}{12}\\ =\frac{5.024\text{kg}\times \left({\left(50\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =\frac{5.024\text{kg}\times \left(0.0089{\text{m}}^2\right)}{12}\\ =0.003726\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $80\text{mm}$ for $b$ and $50\text{mm}$ for $c$ in Equation (XXIX).

$\begin{array}{c}d=\left(\frac{80\text{mm}}{2}+\frac{50\text{mm}}{2}\right)\\ =40\text{mm}+\text{25}\text{mm}\\ =65\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =0.065\text{m}\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $0.065\text{m}$ for $d$ and $0.003726\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_z$ in Equation (XXVII).

$\begin{array}{c}{\left(I_z\right)}_1=0.003726\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right){\left(0.065\text{m}\right)}^2\\ =0.003726\text{kg}\cdot {\text{m}}^2+0.02122\text{kg}\cdot {\text{m}}^2\\ =0.0249524\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $80\text{mm}$ for $b$, $1.6704\text{kg}$ for $m_2$ and $38\text{mm}$ for $e$ in Equation (XXXI).

$\begin{array}{c}{\overline{I}}_{z2}=\frac{1.6704\text{kg}\times \left({\left(80\text{mm}\right)}^2+{\left(38\text{mm}\right)}^2\right)}{12}\\ =\frac{1.6704\text{kg}\times \left({\left(80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(38\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =0.0019188\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $80\text{mm}$ for $b$, $50\text{mm}$ for $c$ and $38\text{mm}$ for $e$ in Equation (XXXII).

$\begin{array}{c}{d}^2={\left(\frac{80\text{mm}}{2}+50\text{mm}-\frac{38\text{mm}}{2}\right)}^2\\ ={\left(40\text{mm}+50\text{mm}-19\text{mm}\right)}^2\\ ={\left(\text{71}\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ =0.005041{\text{m}}^2\end{array}$

Substitute $0.005041{\text{m}}^2$ for $d^2$, $1.6704\text{kg}$ for $m_2$ and $0.0019188\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{z2}$ in Equation (XXX).

$\begin{array}{c}{\left(I_z\right)}_2=0.0019188\text{kg}\cdot {\text{m}}^2+\left(1.6704\text{kg}\right)\left(0.005041{\text{m}}^2\right)\\ =0.0019188\text{kg}\cdot {\text{m}}^2+0.008420\text{kg}\cdot {\text{m}}^2\\ =0.01032\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.2841\text{kg}$ for $m_3$ and $24\text{mm}$ for $f$ in Equation (XXXIII).

$\begin{array}{c}{\left({\overline{I}}_z\right)}_3=0.2841\text{kg}\times \left(\frac{1}{2}-\frac{16}{9{\pi }^2}\right){\left(24\text{mm}\right)}^2\\ =0.2841\text{kg}\times \left(0.31987\right){\left(24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ =0.00005234\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $80\text{mm}$ for $b$, $50\text{mm}$ for $c$ and $24\text{mm}$ for $f$ in Equation (XXXV).

$\begin{array}{c}{d}^2={\left(\frac{80\text{mm}}{2}\right)}^2+\left(50\text{mm}-\frac{4\times 24\text{mm}}{3\pi }\right)\\ ={\left(40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(39.8140\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ =0.0031848{\text{m}}^2\end{array}$

Substitute $0.0031848{\text{m}}^2$ for $d^2$, $0.00005234\text{kg}\cdot {\text{m}}^2$ for ${\left({\overline{I}}_z\right)}_3$ and $0.2841\text{kg}$ for $m_3$ in Equation (XXXIV).

$\begin{array}{c}{\left(I_z\right)}_3=0.00005234\text{kg}\cdot {\text{m}}^2+\left(0.2841\text{kg}\right)\left(0.0031848{\text{m}}^2\right)\\ =0.00005234\text{kg}\cdot {\text{m}}^2+0.0009048\text{kg}\cdot {\text{m}}^2\\ =0.000957\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.000957\text{kg}\cdot {\text{m}}^2$ for ${\left(I_z\right)}_3$, $0.01032\text{kg}\cdot {\text{m}}^2$ for ${\left(I_z\right)}_2$ and $0.0249524\text{kg}\cdot {\text{m}}^2$ for ${\left(I_z\right)}_1$ in Equation (XXXVI).

$\begin{array}{c}\left(I_z\right)=0.0249524\text{kg}\cdot {\text{m}}^2-0.01032\text{kg}\cdot {\text{m}}^2-0.000957\text{kg}\cdot {\text{m}}^2\\ =0.013675\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $40\text{mm}$ for ${\overline{x}}_1$, $25\text{mm}$ for ${\overline{y}}_1$ and $0$ for ${\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_1$ in Equation (XXXVII).

$\begin{array}{c}{\left(I_{xy}\right)}_1=0+5.024\text{kg}\left(40\text{mm}\right)\left(25\text{mm}\right)\\ =5024\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =5.024\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $80\text{mm}$ for ${\overline{z}}_1$, $25\text{mm}$ for ${\overline{y}}_1$ and $0$ for ${\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_1$ in Equation (XXXVIII).

$\begin{array}{c}{\left(I_{yz}\right)}_1=0+5.024\text{kg}\left(80\text{mm}\right)\left(25\text{mm}\right)\\ =10048\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =0.010048\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $80\text{mm}$ for ${\overline{z}}_1$, $40\text{mm}$ for ${\overline{x}}_1$, and $0$ for ${\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_1$ in Equation (XXXIX).

$\begin{array}{c}{\left(I_{zx}\right)}_1=0+5.024\text{kg}\left(80\text{mm}\right)\left(40\text{mm}\right)\\ =16076.8\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =0.0160768\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $1.67048\text{kg}$ for $m_2$, $40\text{mm}$ for ${\overline{x}}_2$, $31\text{mm}$ for ${\overline{y}}_2$ and $0$ for ${\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_2$ in Equation (XL).

$\begin{array}{c}{\left(I_{xy}\right)}_2=0+1.67048\text{kg}\left(40\text{mm}\right)\left(31\text{mm}\right)\\ =2071.5648\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =2.071\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $1.67048\text{kg}$ for $m_2$, $85\text{mm}$ for ${\overline{z}}_2$, $19\text{mm}$ for ${\overline{y}}_2$ and $0$ for ${\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_2$ in Equation (XLI).

$\begin{array}{c}{\left(I_{yz}\right)}_2=0+1.67048\text{kg}\left(85\text{mm}\right)\left(31\text{mm}\right)\\ =4400.8252\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =4.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $1.67048\text{kg}$ for $m_2$, $85\text{mm}$ for ${\overline{z}}_2$, $40\text{mm}$ for ${\overline{x}}_2$, and $0$ for ${\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_1$ in Equation (XLII).

$\begin{array}{c}{\left(I_{zx}\right)}_2=0+1.67048\text{kg}\left(85\text{mm}\right)\left(40\text{mm}\right)\\ =5679.632\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =5.679\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.2841\text{kg}$ for $m_3$, $40\text{mm}$ for ${\overline{x}}_3$, $39.8\text{mm}$ for ${\overline{y}}_3$ and $0$ for ${\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_3$ in Equation (XLIII).

$\begin{array}{c}{\left(I_{xy}\right)}_3=0+0.2841\text{kg}\text{kg}\left(40\text{mm}\right)\left(39.8\text{mm}\right)\\ =452.288\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =0.452\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.2841\text{kg}$ for $m_3$, $140\text{mm}$ for ${\overline{z}}_3$, $39.8\text{mm}$ for ${\overline{y}}_3$ and $0$ for ${\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_3$ in Equation (XLIV).

$\begin{array}{c}{\left(I_{yz}\right)}_3=0+0.2841\text{kg}\left(140\text{mm}\right)\left(39.8\text{mm}\right)\\ =1607.02\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =1.607\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.2841\text{kg}$ for $m_3$, $140\text{mm}$ for ${\overline{z}}_3$, $40\text{mm}$ for ${\overline{x}}_3$, and $0$ for ${\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_3$ in Equation (XLV).

$\begin{array}{c}{\left(I_{zx}\right)}_3=0+0.2841\text{kg}\left(140\text{mm}\right)\left(40\text{mm}\right)\\ =1590.96\text{kg}\cdot {\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =1.591\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.452\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{xy}\right)}_3$, $2.071\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{xy}\right)}_2$ and $5.024\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{xy}\right)}_1$ in Equation (XLVI).

$\begin{array}{c}{I}_{xy}=\left[\begin{array}{c}5.024\times {10}^{-3}\text{kg}\cdot {\text{m}}^2-2.071\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ -0.452\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}\right]\\ =5.024\times {10}^{-3}\text{kg}\cdot {\text{m}}^2-2.543\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ =0.002504\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.160\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{yz}\right)}_3$, $4.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{yz}\right)}_2$ and $0.010048\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{yz}\right)}_1$ in Equation (XLVII).

$\begin{array}{c}{I}_{yz}=\left[\begin{array}{l}0.010048\text{kg}\cdot {\text{m}}^2-4.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ -1.607\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}\right]\\ =0.010048\text{kg}\cdot {\text{m}}^2-0.006\text{kg}\cdot {\text{m}}^2\\ =0.003996\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $1.591\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{zx}\right)}_3$, $5.679\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{zx}\right)}_2$ and $0.0160768\text{kg}\cdot {\text{m}}^2$ for ${\left(I_{zx}\right)}_1$ in Equation (XLVIII).

$\begin{array}{c}{I}_{zx}=\left[\begin{array}{l}0.0160768\text{kg}\cdot {\text{m}}^2-5.679\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ -1.591\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}\right]\\ =0.0160768\text{kg}\cdot {\text{m}}^2-0.00727\text{kg}\cdot {\text{m}}^2\\ =0.008801\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $\sqrt{2825}$ for $d_{OA}$ and $40\text{i}-35\text{j}$ for $\overrightarrow{OA}$ in Equation (XLIX).

$\begin{array}{c}{\lambda }_{OA}=\frac{\text{40i}-35\text{j}}{\sqrt{2825}}\\ =\frac{\text{40i}}{\sqrt{2825}}-\frac{35\text{j}}{\sqrt{2825}}\end{array}$   ....(LI)

From Equation (LI) the unit vector along the different axis.

$\begin{array}{c}{\lambda }_x=\frac{40\text{mm}}{\sqrt{2825\text{mm}}}\\ =\frac{40\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{\sqrt{2825\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}}\\ =\frac{0.04\text{m}}{\sqrt{2.825\text{m}}}\end{array}$

$\begin{array}{c}{\lambda }_y=\frac{30\text{mm}}{\sqrt{2825\text{mm}}}\\ =\frac{30\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{\sqrt{2825\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}}\\ =\frac{0.035\text{m}}{\sqrt{2.825\text{m}}}\end{array}$

Substitute $0.04598\text{kg}\cdot {\text{m}}^2$ for $I_x$, $\frac{0.04\text{m}}{\sqrt{2.825\text{m}}}$ for ${\lambda }_x$, $0.013116\text{kg}\cdot {\text{m}}^2$ for $I_y$, $\frac{0.035\text{m}}{\sqrt{2.825\text{m}}}$ for ${\lambda }_y$ and $0.002504\text{kg}\cdot {\text{m}}^2$ for $I_{xy}$ in Equation (L).

$\begin{array}{c}{I}_{OA}=\left\{\begin{array}{l}\left(0.04598\text{kg}\cdot {\text{m}}^2\right){\left(\frac{0.04\text{m}}{\sqrt{2.825\text{m}}}\right)}^2+\\ \left(0.013116\text{kg}\cdot {\text{m}}^2\right){\left(\frac{0.035\text{m}}{\sqrt{2.825\text{m}}}\right)}^2-\\ 2\left(0.002504\text{kg}\cdot {\text{m}}^2\right)\left(\frac{0.04\text{m}}{\sqrt{2.825\text{m}}}\right)\left(\frac{0.035\text{m}}{\sqrt{2.825\text{m}}}\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\left(0.04598\text{kg}\cdot {\text{m}}^2\right)\left(\frac{1.6\times {10}^{-3}{\text{m}}^2}{2.825\text{m}}\right)+\\ \left(0.013116\text{kg}\cdot {\text{m}}^2\right)\left(\frac{1.225\times {10}^{-3}{\text{m}}^2}{2.825\text{m}}\right)-\\ 2\left(0.002504\text{kg}\cdot {\text{m}}^2\right)\left(\frac{1.4\times {10}^{-3}{\text{m}}^2}{2.825\text{m}}\right)\end{array}\right\}\\ =0.260\times {10}^{-6}\text{kg}\cdot {\text{m}}^2+5.68\times {10}^{-6}\text{kg}\cdot {\text{m}}^2-2.48\times {10}^{-6}\text{kg}\cdot {\text{m}}^2\\ =3.46\times {10}^{-6}\text{kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

The mass moment of the inertia of the steel fixture passes through the origin is $3.46\times {10}^{-6}\text{kg}\cdot {\text{m}}^2$.