Chapter B, Problem B.71P

For the component described in the problem indicated,determine (a) the principal mass moments of inertia at the origin (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

Probs. B.35 and B.39

To determine

(a)

The principal mass moment of inertias at the origin.

Answer to Problem B.71P

The principal mass moment of inertias at the origin are $4.1443\times {10}^{-3}{\text{kg-m}}^{\text{2}}$, $29.784\times {10}^{-3}{\text{kg-m}}^{\text{2}}$ and $32.2541\times {10}^{-3}{\text{kg-m}}^{\text{2}}$.

Explanation of Solution

Given information:

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

The following figure represents the given system.

Figure-(1)

Concept used:

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 are $d\text{or}b\text{and}e$.

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 of component 1.

$m_1=\rho V_1$ ....... (1)

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 mass moment of inertia for component 1.

${\left(I_{xy}\right)}_1={\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_1+m_1\left({\overline{x}}_1\right)\left({\overline{y}}_1\right)$ ....... (2)

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.

${\left(I_{yz}\right)}_1={\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_1+m_1\left({\overline{z}}_1\right)\left({\overline{y}}_1\right)$ ....... (3)

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.

${\left(I_{zx}\right)}_1={\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_1+m_1\left({\overline{z}}_1\right)\left({\overline{x}}_1\right)$ ....... (4)

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 of component 2.

$m_2=\rho V_2$ ....... (5)

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 the mass moment of inertia for component 2.

${\left(I_{xy}\right)}_2={\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_2+m_2\left({\overline{x}}_2\right)\left({\overline{y}}_2\right)$ ....... (6)

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.

${\left(I_{yz}\right)}_2={\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_2+m_2\left({\overline{z}}_2\right)\left({\overline{y}}_2\right)$ ....... (7)

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.

${\left(I_{zx}\right)}_2={\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_2+m_2\left({\overline{z}}_2\right)\left({\overline{x}}_2\right)$ ....... (8)

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 of component 3.

$m_3=\rho V_3$ ....... (9)

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 the mass moment of inertia for component 3.

${\left(I_{xy}\right)}_3={\left({\overline{I}}_{x^{\prime }{y}^{\prime }}\right)}_3+m_3\left({\overline{x}}_3\right)\left({\overline{y}}_3\right)$ ....... (10)

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.

${\left(I_{yz}\right)}_3={\left({\overline{I}}_{y^{\prime }{z}^{\prime }}\right)}_3+m_3\left({\overline{z}}_3\right)\left({\overline{y}}_3\right)$ ....... (11)

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.

${\left(I_{zx}\right)}_3={\left({\overline{I}}_{z^{\prime }{x}^{\prime }}\right)}_3+m_3\left({\overline{z}}_3\right)\left({\overline{x}}_3\right)$ ....... (12)

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.

$I_{xy}={\left(I_{xy}\right)}_1-{\left(I_{xy}\right)}_2-{\left(I_{xy}\right)}_3$ ....... (13)

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

Write the expression for the mass product of inertia.

$I_{yz}={\left(I_{yz}\right)}_1-{\left(I_{yz}\right)}_2-{\left(I_{yz}\right)}_3$ ....... (14)

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

Write the expression for the mass product of inertia.

$I_{zx}={\left(I_{zx}\right)}_1-{\left(I_{zx}\right)}_2-{\left(I_{zx}\right)}_3$ ....... (15)

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

Write the expression for the volume of the component 1.

$V_1=abc$ ...... (16)

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

Write the expression for volume of the component 2.

$V_2=bde$ ........ (17)

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

Write the expression for volume of the component 3.

$V_3=\frac{\pi f^{2}g}{4}$ ........ (18)

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

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).

$\begin{array}{c}{d}^2=\left({\left(\frac{0.16}{2}\right)}^2+{\left(\frac{0.16}{2}\right)}^2\right)\\ =0.007025\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $0.007025$ for $d^2$ and $0.011764\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_x$ in Equation (III).

$\begin{array}{c}{\left(I_x\right)}_1=0.011764\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right)0.007025{\text{m}}^2\\ =0.011764\text{kg}\cdot {\text{m}}^2+0.0352936\text{kg}\cdot {\text{m}}^2\\ =0.0470576\text{kg}\cdot {\text{m}}^2\end{array}$

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).

$\begin{array}{c}{\left({\overline{I}}_x\right)}_3=0.2841\text{kg}\times \left[\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right){\left(24\text{mm}\right)}^2+\left(\frac{{\left(40\text{mm}\right)}^2}{12}\right)\right]\\ =4.93047\times {10}^{-5}\text{kg}\cdot {\text{m}}^2\end{array}$

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 $d^2$, $4.93047\times {10}^{-5}\text{kg}\cdot {\text{m}}^2$ for ${\left({\overline{I}}_x\right)}_3$ and $0.2841\text{kg}$ for $m_3$ in Equation (X).

$\begin{array}{c}{\left(I_x\right)}_3=4.93047\times {10}^{-5}\text{kg}\cdot {\text{m}}^2+\left(0.2841\text{kg}\right)\left(0.02118{\text{m}}^2\right)\\ =0.00059035\text{kg}\cdot {\text{m}}^2+0.006018\text{kg}\cdot {\text{m}}^2\\ =0.0060673\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.0060673\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_3$, $0.0145576\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_2$ and $0.0470576\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_1$ in Equation (XVI).

$\begin{array}{c}\left(I_x\right)=0.0470576\text{kg}\cdot {\text{m}}^2-0.0145576\text{kg}\cdot {\text{m}}^2-0.0060673\text{kg}\cdot {\text{m}}^2\\ =0.04598\text{kg}\cdot {\text{m}}^2\\ =26.43\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

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

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

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

$\begin{array}{c}{d}^2=\left({\left(\frac{0.16}{2}\right)}^2+{\left(\frac{0.08}{2}\right)}^2\right)\\ =0.008{\text{m}}^2\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $0.008{\text{m}}^2$ for $d^2$ and $0.013877\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_y$ in Equation (XVI).

$\begin{array}{c}{\left(I_y\right)}_1=0.013877\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right)0.008{\text{m}}^2\\ =0.013877\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right){\left(0.0144\text{m}\right)}^2\\ =0.054069\text{kg}\cdot {\text{m}}^2\end{array}$

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).

$\begin{array}{c}{d}^2=\left({\left(\frac{80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2+{\left(50\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)+\frac{70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2\right)\\ =0.008825{\text{m}}^2\end{array}$

Substitute $0.008825{\text{m}}^2$ 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).

$\begin{array}{c}{\left(I_y\right)}_2=0.00157296\text{kg}\cdot {\text{m}}^2+\left(1.6704\text{kg}\right)\left(0.008825{\text{m}}^2\right)\\ =0.00157296\text{kg}\cdot {\text{m}}^2+0.0261\text{kg}\cdot {\text{m}}^2\\ =0.01631424\text{kg}\cdot {\text{m}}^2\end{array}$

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).

$\begin{array}{c}{d}^2={\left(\frac{80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2+{\left(160\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)-\frac{40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2\\ =0.0212{\text{m}}^2\end{array}$

Substitute $0.0212{\text{m}}^2$ 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).

$\begin{array}{c}{\left(I_y\right)}_3=0.00007879\text{kg}\cdot {\text{m}}^2+\left(0.2841\text{kg}\right)\left(0.0212{\text{m}}^2\right)\\ =0.00610171\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.00610171\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_3$, $0.01631424\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_2$ and $0.054069\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_1$ in Equation (XXVI).

$\begin{array}{c}\left(I_y\right)=0.054069\text{kg}\cdot {\text{m}}^2-0.01631424\text{kg}\cdot {\text{m}}^2-0.00610171\text{kg}\cdot {\text{m}}^2\\ =0.03165305\text{kg}\cdot {\text{m}}^2\\ \approx 31.2\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

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).

$\begin{array}{c}{d}^2=\left({\left(\frac{80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2+{\left(\frac{50\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2\right)\\ =0.002225{\text{m}}^2\end{array}$

Substitute $5.024\text{kg}$ for $m_1$, $0.002225{\text{m}}^2$ for $d^2$ and $0.003726\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_z$ in Equation (XXVII).

$\begin{array}{c}{\left(I_z\right)}_1=0.003726\text{kg}\cdot {\text{m}}^2+\left(5.024\text{kg}\right)\left(0.002225{\text{m}}^2\right)\\ =0.0149044\text{kg}\cdot {\text{m}}^2\end{array}$

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

$\begin{array}{c}{\overline{I}}_{z2}=\frac{1.6704\text{kg}\times \left({\left(80\text{mm}\right)}^2+{\left(38\text{mm}\right)}^2\right)}{12}\\ =\frac{1.6704\text{kg}\times \left({\left(80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(38\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)}{12}\\ =0.00109188\text{kg}\cdot {\text{m}}^2\end{array}$

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

$\begin{array}{c}{d}^2=\left({\left(\frac{80\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2+{\left(50\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)-\frac{38\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}^2\right)\\ =0.002561{\text{m}}^2\end{array}$

Substitute $0.002561{\text{m}}^2$ for $d^2$, $1.6704\text{kg}$ for $m_2$ and $0.00109188\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{z2}$ in Equation (XXX).

$\begin{array}{c}{\left(I_z\right)}_2=0.00109188\text{kg}\cdot {\text{m}}^2+\left(1.6704\text{kg}\right)\left(0.002561{\text{m}}^2\right)\\ =0.0053697\text{kg}\cdot {\text{m}}^2\end{array}$

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).

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

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 ${\left(I_z\right)}_3$, $0.0053697\text{kg}\cdot {\text{m}}^2$ for ${\left(I_z\right)}_2$ and $0.0149044\text{kg}\cdot {\text{m}}^2$ for ${\left(I_z\right)}_1$ in Equation (XXXVI).

$\begin{array}{c}\left(I_z\right)=0.0149044\text{kg}\cdot {\text{m}}^2-0.0053697\text{kg}\cdot {\text{m}}^2-0.000957\text{kg}\cdot {\text{m}}^2\\ =0.013675\text{kg}\cdot {\text{m}}^2\\ =8.5773\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

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 (2).

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

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 (3).

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

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 (4).

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

Substitute $38\text{mm}$ for $e$, $80\text{mm}$ for $b$ and $70\text{mm}$ for $d$ in Equation (17).

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

Substitute $7850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $2.28\times {10}^{-4}{\text{m}}^3$ for $V_2$ in Equation (5).

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

Substitute $1.67048\text{kg}$ 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 (6).

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

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 (7).

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

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 (8).

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

Substitute $24\text{mm}$ for $f$ and $40\text{mm}$ for $g$ in Equation (18).

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

Substitute $7850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $3.6191\times {10}^{-5}{\text{m}}^3$ for $V_3$ in Equation (9).

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

Substitute $0.2841\text{kg}$ 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 (10).

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

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 (11).

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

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 (12).

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

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 (13).

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

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 (14).

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

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 (15).

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

The principal mass moment of inertia at the origin is calculated as follows:

$\begin{array}{l}{K}^3-\left(I_x+I_y+I_z\right)K^2+\left(I_x{I}_y+I_y{I}_z+I_z{I}_x-{\left(I_{xy}\right)}^2-{\left(I_{yz}\right)}^2-{\left(I_{zx}\right)}^2\right)K\\ -\left(I_x{I}_y{I}_z-I_x{\left(I_{yz}\right)}^2-I_y{\left(I_{zx}\right)}^2-I_z{\left(I_{xy}\right)}^2-2\left(I_{xy}\right)\left(I_{yz}\right)\left(I_{zx}\right)\right)=0\\ K^3-\left(26.43+31.2+8.5773\right)\times {10}^{-3}{K}^2+\\ \left(26.43\times 31.2+31.2\times 8.5773+8.5773\times 26.43-{\left(2.504\right)}^2-{\left(3.996\right)}^2-{\left(8.801\right)}^2\right){10}^{-6}K\\ -\left(\begin{array}{l}26.43\times 31.2\times 8.5773-26.43{\left(3.996\right)}^2-31.2{\left(8.801\right)}^2-8.5773{\left(2.504\right)}^2\\ -2\left(2.504\right)\left(3.996\right)\left(8.801\right)\end{array}\right)\times {10}^{-9}=0\\ K^3-\left(66.1824\times {10}^{-3}\right)K^2+\left(1217.76\times {10}^{-6}\right)K-\left(3981.23\times {10}^{-9}\right)=0\end{array}$

After solving above equation,

$\begin{array}{l}{K}_1=4.1443\times {10}^{-3}{\text{kg-m}}^{\text{2}}\\ K_2=29.784\times {10}^{-3}{\text{kg-m}}^{\text{2}}\\ \text{and}\\ K_3=32.2541\times {10}^{-3}{\text{kg-m}}^{\text{2}}\end{array}$

Conclusion:

The principal mass moment of inertias at the origin are $4.1443\times {10}^{-3}{\text{kg-m}}^{\text{2}}$, $29.784\times {10}^{-3}{\text{kg-m}}^{\text{2}}$ and $32.2541\times {10}^{-3}{\text{kg-m}}^{\text{2}}$.

To determine

The principal axis about the origin.

Answer to Problem B.71P

The principal axis about the origin are ${\left({\theta }_x\right)}_1=67.8°$, ${\left({\theta }_y\right)}_1=80.1°$, ${\left({\theta }_z\right)}_1=24.6°$, ${\left({\theta }_x\right)}_2=31.6°$, ${\left({\theta }_y\right)}_2=71.4°$, ${\left({\theta }_z\right)}_2=114.5°$, ${\left({\theta }_x\right)}_3=68.8°$, ${\left({\theta }_y\right)}_3=158.8°$ and ${\left({\theta }_z\right)}_3=88.5°$.

Explanation of Solution

Given information:

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

The following figure represents the given system.

Figure-(1)

Calculation:

The direction cosine is calculated as follows:

$\begin{array}{l}\left(I_x-K_1\right){\left({\lambda }_x\right)}_1-I_{xy}{\left({\lambda }_y\right)}_1-I_{zx}{\left({\lambda }_z\right)}_1=0\\ \text{and}\\ -I_{xy}{\left({\lambda }_x\right)}_1+\left(I_y-K_1\right){\left({\lambda }_y\right)}_1-I_{yx}{\left({\lambda }_z\right)}_1=0\end{array}$

Substitute the values from the sub-part (a) in above equations as follows:

$\begin{array}{l}\left(26.4325-4.1443\right)\times {10}^{-3}{\left({\lambda }_x\right)}_1-\left(2.5002\times {10}^{-3}\right){\left({\lambda }_y\right)}_1-\left(8.8062\times {10}^{-3}\right){\left({\lambda }_z\right)}_1=0\\ \text{and}\\ -\left(2.5002\times {10}^{-3}\right){\left({\lambda }_x\right)}_1+\left(31.1726-4.1443\right)\times {10}^{-3}{\left({\lambda }_y\right)}_1-\left(4.0627\times {10}^{-3}\right){\left({\lambda }_z\right)}_1=0\end{array}$

After solving above equations,

$\begin{array}{l}{\left({\lambda }_z\right)}_1=2.4022{\left({\lambda }_x\right)}_1\\ \text{and}\\ {\left({\lambda }_y\right)}_1=0.45357{\left({\lambda }_x\right)}_1\end{array}$

Direction cosine in x direction is calculated as follows:

$\begin{array}{c}{\left(0.45357{\left({\lambda }_x\right)}_1\right)}^2+{\left({\lambda }_x\right)}^2{}_1+{\left(2.4022{\left({\lambda }_x\right)}_1\right)}^2=1\\ {\left({\lambda }_x\right)}_1=0.37861\end{array}$

Direction cosine in y and z direction are,

$\begin{array}{l}{\left({\lambda }_y\right)}_1=0.17173\\ \text{and}\\ {\left({\lambda }_z\right)}_1=0.9095\end{array}$

So, the direction is calculated as follows:

$\begin{array}{l}{\left({\theta }_x\right)}_1=67.8°\\ {\left({\theta }_y\right)}_1=80.1°\\ \text{and}\\ {\left({\theta }_z\right)}_1=24.6°\end{array}$

Again,

The direction cosine is calculated as follows:

$\begin{array}{l}\left(I_x-K_2\right){\left({\lambda }_x\right)}_2+\left(I_{xy}\right){\left({\lambda }_y\right)}_2-I_{zx}{\left({\lambda }_z\right)}_2=0\\ \text{and}\\ -I_{xy}{\left({\lambda }_x\right)}_2+\left(I_y-K_2\right){\left({\lambda }_y\right)}_2-\left(I_{yz}\right){\left({\lambda }_z\right)}_2=0\end{array}$

Substitute the values from the sub-part (a) in above equations as follows:

$\begin{array}{l}\left(26.4325-29.784\right)\times {10}^{-3}{\left({\lambda }_x\right)}_2-\left(2.5002\times {10}^{-3}\right){\left({\lambda }_y\right)}_2-\left(8.8062\times {10}^{-3}\right){\left({\lambda }_z\right)}_2=0\\ \text{and}\\ \left(-2.5002\right)\times {10}^{-3}{\left({\lambda }_x\right)}_2-\left(31.1726-29.784\right)\times {10}^{-3}{\left({\lambda }_y\right)}_2-\left(4.0627\times {10}^{-3}\right){\left({\lambda }_z\right)}_2=0\end{array}$

After solving above equations,

$\begin{array}{l}{\left({\lambda }_z\right)}_2=-0.48713{\left({\lambda }_x\right)}_2\\ \text{and}\\ {\left({\lambda }_y\right)}_2=0.37525{\left({\lambda }_x\right)}_2\end{array}$

Direction cosine in x direction is calculated as follows:

$\begin{array}{c}{\left(0.37527{\left({\lambda }_x\right)}_2\right)}^2+{\left({\lambda }_x\right)}^2{}_2+{\left(-0.48713{\left({\lambda }_x\right)}_2\right)}^2=1\\ {\left({\lambda }_x\right)}_2=0.85184\end{array}$

Direction cosine in y and z direction are,

$\begin{array}{l}{\left({\lambda }_y\right)}_2=0.31967\\ \text{and}\\ {\left({\lambda }_z\right)}_2=-0.41496\end{array}$

So, the direction is calculated as follows:

$\begin{array}{l}{\left({\theta }_x\right)}_2=31.6°\\ {\left({\theta }_y\right)}_2=71.4°\\ \text{and}\\ {\left({\theta }_z\right)}_2=114.5°\end{array}$

Similarly,

The direction cosine is calculated as follows:

$\begin{array}{l}\left(I_x-K_3\right){\left({\lambda }_x\right)}_3-\left(I_{xy}\right){\left({\lambda }_y\right)}_3-I_{zx}{\left({\lambda }_z\right)}_3=0\\ \text{and}\\ -I_{xy}{\left({\lambda }_x\right)}_3+\left(I_y-K_3\right){\left({\lambda }_y\right)}_3-I_{yz}{\left({\lambda }_z\right)}_3=0\end{array}$

Substitute the values from the sub-part (a) in above equations as follows:

$\begin{array}{l}\left(26.4325-32.2541\right)\times {10}^{-3}{\left({\lambda }_x\right)}_3-\left(\left(2.5002\right)\times {10}^{-3}\right){\left({\lambda }_y\right)}_3-\left(8.8062\times {10}^{-3}\right){\left({\lambda }_z\right)}_3=0\\ \text{and}\\ \left(-2.5002\right)\times {10}^{-3}{\left({\lambda }_x\right)}_3+\left(\left(31.1726-32.2541\right)\times {10}^{-3}\right){\left({\lambda }_y\right)}_3-\left(4.0627\times {10}^{-3}\right){\left({\lambda }_z\right)}_3=0\end{array}$

After solving above equations,

$\begin{array}{l}{\left({\lambda }_z\right)}_3=0.071276{\left({\lambda }_x\right)}_3\\ \text{and}\\ {\left({\lambda }_y\right)}_3=-2.5795{\left({\lambda }_x\right)}_3\end{array}$

Direction cosine in x direction is calculated as follows:

$\begin{array}{c}{\left(-2.5795{\left({\lambda }_x\right)}_3\right)}^2+{\left({\lambda }_x\right)}^2{}_3+{\left(0.071276{\left({\lambda }_x\right)}_3\right)}^2=1\\ {\left({\lambda }_x\right)}_3=0.36134\end{array}$

Direction cosine in y and z direction are,

$\begin{array}{l}{\left({\lambda }_y\right)}_3=0.93208\\ \text{and}\\ {\left({\lambda }_z\right)}_3=0.025755\end{array}$

So, the direction is calculated as follows:

$\begin{array}{l}{\left({\theta }_x\right)}_3=68.8°\\ {\left({\theta }_y\right)}_3=158.8°\\ \text{and}\\ {\left({\theta }_z\right)}_3=88.5°\end{array}$

The sketch is shown below:

Conclusion:

So, the principal axis about the origin are ${\left({\theta }_x\right)}_1=67.8°$, ${\left({\theta }_y\right)}_1=80.1°$, ${\left({\theta }_z\right)}_1=24.6°$, ${\left({\theta }_x\right)}_2=31.6°$, ${\left({\theta }_y\right)}_2=71.4°$, ${\left({\theta }_z\right)}_2=114.5°$, ${\left({\theta }_x\right)}_3=68.8°$, ${\left({\theta }_y\right)}_3=158.8°$ and ${\left({\theta }_z\right)}_3=88.5°$.