Question 11.11: Calculate the principal moments of inertia about the center ...
Calculate the principal moments of inertia about the center of mass and the corresponding principal directions for the bent rod in Fig. 11.16. Its mass is uniformly distributed at 2 kg/m.
The mass of each of the four slender rod segments is
m_1=2 \cdot 0.4=0.8 kg \quad m_2=2 \cdot 0.5=1 kg \quad m_3=2 \cdot 0.3=0.6 kg \quad m_4=2 \cdot 0.2=0.4 kg (a)
The total mass of the system is
m=\sum_{i=1}^4 m_i=2.8 kg (b)
The coordinates of each segment’s center of mass are
x_{G_1}=0 \quad y_{G_1}=0 \quad z_{G_1}=0.2 m
x_{G_2}=0 \quad y_{G_2}=0.25 m \quad z_{G_2}=0.2 m (c)
x_{G_3}=0.15 m \quad y_{G_3}=0.5 m \quad z_{G_3}=0
x_{G_4}=0.3 m \quad y_{G_4}=0.4 m \quad z_{G_4}=0
If the slender rod in Fig. 11.15 is aligned with, say, the x axis, then a = \ell and b = c = 0, so that according to Eq. (11.63),
\left[ I _G\right]=\left[\begin{array}{ccc}0 & 0 & 0 \\0 & \frac{1}{12} m \ell^2 & 0 \\0 & 0 & \frac{1}{12} m \ell^2\end{array}\right]
That is, the moment of inertia of a slender rod about the axes normal to the rod at its center of mass is m\ell²/12, where m and \ell are the mass and length of the rod, respectively. Since the mass of a slender bar is assumed to be concentrated along the axis of the bar (its cross-sectional dimensions are infinitesimal), the moment of inertia about the centerline is zero. By symmetry, the products of inertia about the axes through the center of mass are all zero. Using this information and the parallel axis theorem, we find the moments and products of inertia of each rod segment about the origin O of the xyz system as follows:
Rod 1:
\left.I_x^{(1)}=I_{G_1}^{(1)}\right)_x+m_1\left(y_{G_1}^2+z_{G_1}^2\right)=\frac{1}{12} \cdot 0.8 \cdot 0.4^2+0.8\left(0+0.2^2\right)=0.04267 kg \cdot m ^2
\left.I_y^{(1)}=I_{G_1}^{(1)}\right)_y+m_1\left(x_{G_1}^2+z_{G_1}^2\right)=\frac{1}{12} \cdot 0.8 \cdot 0.4^2+0.8\left(0+0.2^2\right)=0.04267 kg \cdot m ^2
\left.I_z^{(1)}=I_{G_1}^{(1)}\right)_z+m_1\left(x_{G_1}^2+y_{G_1}^2\right)=0+0.8(0+0)=0
\left.I_{x y}^{(1)}=I_{G_1}^{(1)}\right)_{x y}-m_1 x_{G_1} y_{G_1}=0-0.8(0)(0)=0
\left.I_{x z}^{(1)}=I_{G_1}^{(1)}\right)_{x z}-m_1 x_{G_1} z_{G_1}=0-0.8(0)(0.2)=0
\left.I_{x z}^{(1)}=I_{G_1}^{(1)}\right)_{y z}-m_1 y_{G_1} z_{G_1}=0-0.8(0)(0)=0
Rod 2:
\left.I_x^{(2)}=I_{G_2}^{(2)}\right)_x+m_2\left(y_{G_2}^2+z_{G_2}^2\right)=\frac{1}{12} \cdot 1.0 \cdot 0.5^2+1.0\left(0+0.25^2\right)=0.08333 kg \cdot m ^2
\left.I_y^{(2)}=I_{G_2}^{(2)}\right)_y+m_2\left(x_{G_2}^2+z_{G_2}^2\right)=0+1.0(0+0)=0
\left.I_z^{(2)}=I_{G_2}^{(2)}\right)_z+m_2\left(x_{G_2}^2+y_{G_2}^2\right)=\frac{1}{12} \cdot 1.0 \cdot 0.5^2+1.0\left(0+0.5^2\right)=0.08333 kg \cdot m ^2
\left.I_{x y}^{(2)}=I_{G_2}^{(2)}\right)_{x y}-m_2 x_{G_2} y_{G_2}=0-1.0(0)(0.5)=0
\left.I_{x z}^{(2)}=I_{G_2}^{(2)}\right)_{x z}-m_2 x_{G_2} z_{G_2}=0-1.0(0)(0)=0
\left.I_{y z}^{(2)}=I_{G_2}^{(2)}\right)_{y z}-m_2 y_{G_2} z_{G_2}=0-1.0(0.5)(0)=0
Rod 3:
\left.I_x^{(3)}=I_{G_3}^{(3)}\right)_x+m_3\left(y_{G_3}^2+z_{G_3}^2\right)=0+0.6\left(0.5^2+0\right)=0.15 kg \cdot m ^2
\left.I_y^{(3)}=I_{G_3}^{(2)}\right)_y+m_3\left(x_{G_3}^2+z_{G_3}^2\right)=\frac{1}{12} \cdot 0.6 \cdot 0.3^2+0.6\left(0.15^2+0\right)=0.018 kg \cdot m ^2
\left.I_z^{(3)}=I_{G_3}^{(3)}\right)_z+m_3\left(x_{G_3}^2+y_{G_3}^2\right)=\frac{1}{2} \cdot 0.6 \cdot 0.3^2+0.6\left(0.15^2+0.5^2\right)=0.1680 kg \cdot m ^2
\left.I_{x y}^{(3)}=I_{G_3}^{(3)}\right)_{x y}-m_3 x_{G_3} y_{G 3}=0-0.6(0.15)(0.5)=-0.045 kg \cdot m ^2
\left.I_{x z}^{(3)}=I_{G_3}^{(3)}\right)_{x z}-m_3 x_{G_3} z_{G_3}=0-0.6(0.15)(0)=0
\left.I_{y z}^{(3)}=I_{G_3}^{(3)}\right)_{y z}-m_3 y_{G_3} z_{G_3}=0-0.6(0.5)(0)=0
Rod 4:
\left.I_x^{(4)}=I_{G_4}^{(4)}\right)_x+m_4\left(y_{G_4}^2+z_{G_4}^2\right)=\frac{1}{12} \cdot 0.4 \cdot 0.2^2+0.4\left(0.4^2+0\right)=0.06533 kg \cdot m ^2
\left.I_y^{(4)}=I_{G_4}^{(4)}\right)_y+m_4\left(x_{G_4}^2+z_{G 4}^2\right)=0+0.4\left(0.3^2+0\right)=0.0360 kg \cdot m ^2
\left.I_z^{(4)}=I_{G_4}^{(4)}\right)_z+m_4\left(x_{G_4}^2+y_{G_4}^2\right)=\frac{1}{12} \cdot 0.4 \cdot 0.2^2+0.4\left(0.3^2+0.4^2\right)=0.1013 kg \cdot m ^2
\left.I_{x y}^{(4)}=I_{G_4}^{(4)}\right)_{x y}-m_4 x_{G 4} y_{G 4}=0-0.4(0.3)(0.4)=-0.0480 kg \cdot m ^2
\left.I_{x z}^{(4)}=I_{G_4}^{(4)}\right)_{x z}-m_4 x_{G_4} z_{G_4}=0-0.4(0.3)(0)=0
\left.I_{y z}^{(4)}=I_{G_4}^{(4)}\right)_{y z}-m_4 y_{G_4} z_{G_4}=0-0.4(0.4)(0)=0
The total moments of inertia for all the four rods about O are
I_x=\sum_{i=1}^4 I_x^{(i)}=0.3413 kg \cdot m ^2 \quad I_y=\sum_{i=1}^4 I_y^{(i)}=0.09667 kg \cdot m ^2 \quad I_z=\sum_{i=1}^4 I_z^{(i)}=0.3527 kg \cdot m ^2 (d)
I_{x y}=\sum_{i=1}^4 I_{x y}^{(i)}=0.0930 kg \cdot m ^2 \quad I_{x z}=\sum_{i=1}^4 I_{x z}^{(i)}=0 \quad I_{y z}=\sum_{i=1}^4 I_{y z}^{(i)}=0
The coordinates of the center of mass of the system of four rods are, from Eqs. (a) through (c),
x_G=\frac{1}{m} \sum_{i=1}^4 m_i x_{G_i}=\frac{1}{2.8} \cdot 0.21=0.075 m
y_G=\frac{1}{m} \sum_{i=1}^4 m_i y_{G_i}=\frac{1}{2.8} \cdot 0.71=0.2536 m (e)
z_G=\frac{1}{m} \sum_{i=1}^4 m_i z_{G_i}=\frac{1}{2.8} \cdot 0.16=0.05714 m
We use the parallel axis theorems to shift the moments of inertia in Eq. (d) to the center of mass G of the system
I_{G_x}=I_x-m\left(y_G^2+z_G^2\right)=0.3413-0.1892=0.1522 kg \cdot m ^2
I_{G_y}=I_y-m\left(x_G^2+z_G^2\right)=0.09667-0.02489=0.17177 kg \cdot m ^2
I_{G_z}=I_z-m\left(x_G^2+y_G^2\right)=-0.3527-0.1958=0.1569 kg \cdot m ^2
I_{G_{x y}}=I_{x y}+m x_G y_G=-0.093+0.05325=-0.03975 kg \cdot m ^2
I_{G x z}=I_{x z}+m x_G z_G=0+0.012=0.012 kg \cdot m ^2
I_{G_{y z}}=I_{y z}+m y_G z_G=0+0.04057=0.04057 kg \cdot m ^2
Therefore, the inertia tensor, relative to the center of mass, is
[ I ]=\left[\begin{array}{ccc}I_{G_x} & I_{G_{x y}} & I_{G_{x z}} \\I_{G_{x y}} & I_{G_y} & I_{G_{y z}} \\I_{G_{x z}} & I_{G_{y z}} & I_{G_z}\end{array}\right]=\left[\begin{array}{crl}0.1522 & -0.03975 & 0.012 \\-0.03975 & 0.07177 & 0.04057 \\0.012 & 0.04057 & 0.1569\end{array}\right] \left( kg \cdot m ^2\right) (f)
To find the three principal moments of inertia, we may proceed as in Example 11.9, or simply enter the following lines in the MATLAB Command Window:
IG = [ 0.1522 –0.03975 0.012
–0.03975 0.07177 0.04057
0.012 0.04057 0.1569];
[eigenVectors, eigenValues] = eig(IG)
to obtain
eigenVectors =
0.3469 –0.8482 –0.4003
0.8742 0.1378 0.4656
–0.3397 –0.5115 0.7893
eigenValues =
0.0402 0 0
0 0.1658 0
0 0 0.1747
Hence, the three principal moments of inertia and their principal directions are
\lambda_1=0.04023 kg \cdot m ^2 e ^{(1)}=0.3469 \hat{ i }+0.8742 \hat{ j }-0.3397 \hat{ k }
\lambda_2=0.1658 kg \cdot m ^2 \quad e ^{(2)}=-0.8482 \hat{ i }+0.1378 \hat{ j }-0.5115 \hat{ k }
\lambda_3=0.1747 kg \cdot m ^2 \quad e ^{(3)}=-0.4003 \hat{ i }+0.4656 \hat{ j }+0.7893 \hat{ k }