Question 9.SP.12: A steel forging consists of a 6 × 2 × 2-in. rectangular pris......
A steel forging consists of a 6 × 2 × 2-in. rectangular prism and two cylinders with a diameter of 2 in. and length of 3 in. as shown. Determine the moments of inertia of the forging with respect to the coordinate axes. The specific weight of steel is 490 lb/ft³.
STRATEGY: Compute the moments of inertia of each component from Fig. 9.28 using the parallel-axis theorem when necessary. Note that all lengths should be expressed in feet to be consistent with the units for the given specific weight.
MODELING and ANALYSIS:
Computation of Masses.
Prism
V=(2 \text { in. })(2 \text { in. })(6 \text { in. })=24 ~\mathrm{in}^3
W=\frac{\left(24 ~\mathrm{in}^3\right)\left(490 ~\mathrm{lb} / \mathrm{ft}^3\right)}{1728 ~\mathrm{in}^3 / \mathrm{ft}^3}=6.81 ~\mathrm{lb}
m=\frac{6.81 ~\mathrm{lb}}{32.2 ~\mathrm{ft} / \mathrm{s}^2}=0.211 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}
Each Cylinder
V=\pi(1 \text { in. })^2(3 \text { in. })=9.42 ~\mathrm{in}^3
W=\frac{\left(9.42 ~\mathrm{in}^3\right)\left(490 ~\mathrm{lb} / \mathrm{ft}^3\right)}{1728 ~\mathrm{in}^3 / \mathrm{ft}^3}=2.67 ~\mathrm{lb}
m=\frac{2.67 ~\mathrm{lb}}{32.2 ~\mathrm{ft} / \mathrm{s}^2}=0.0829 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}
Moments of Inertia (Fig. 1).
Prism
I_x=I_z=\frac{1}{12}\left(0.211 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left[\left(\frac{6}{12} \mathrm{ft}\right)^2+\left(\frac{2}{12} \mathrm{ft}\right)^2\right]=4.88 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2
I_y=\frac{1}{12}\left(0.211 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left[\left(\frac{2}{12} \mathrm{ft}\right)^2+\left(\frac{2}{12} \mathrm{ft}\right)^2\right]=0.977 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2
Each Cylinder
\begin{aligned}I_x=\frac{1}{2} m a^2+m \bar{y}^2= & \frac{1}{2}\left(0.0829 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left(\frac{1}{12} \mathrm{ft}\right)^2 \\& +\left(0.0829 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left(\frac{2}{12} \mathrm{ft}\right)^2=2.59 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2\end{aligned}
\begin{aligned}I_y=\frac{1}{12} m\left(3 a^2+L^2\right) & =m \bar{x}^2=\frac{1}{12}\left(0.0829 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left[3\left(\frac{1}{12} \mathrm{ft}\right)^2+\left(\frac{3}{12} \mathrm{ft}\right)^2\right] \\& +\left(0.0829 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left(\frac{2.5}{12} \mathrm{ft}\right)^2=4.17 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2\end{aligned}
\begin{array}{c}I_z=\frac{1}{12} m\left(3 a^2+L^2\right)+m\left(\bar{x}^2+\bar{y}^2\right)=\frac{1}{12}\left(0.0829 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left[3\left(\frac{1}{12} \mathrm{ft}\right)^2+\left(\frac{3}{12} \mathrm{ft}\right)^2\right] \\+\left(0.0829 ~\mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)\left[\left(\frac{2.5}{12} \mathrm{ft}\right)^2+\left(\frac{2}{12} \mathrm{ft}\right)^2\right]=6.48 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2\end{array}
Entire Body. Adding the values obtained for the prism and two cylinders, you have
I_x=4.88 \times 10^{-3}+2\left(2.59 \times 10^{-3}\right) \quad I_x=10.06 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2
I_y=0.977 \times 10^{-3}+2\left(4.17 \times 10^{-3}\right) \quad I_y=9.32 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2
I_z=4.88 \times 10^{-3}+2\left(6.48 \times 10^{-3}\right) \quad I_z=17.84 \times 10^{-3} ~\mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2
REFLECT and THINK: The results indicate this forging has more resistance to rotation about the z axis (largest moment of inertia) than about the x or y axes. This makes intuitive sense, because more of the mass is farther from the z axis than from the x or y axes.
