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. Deter-mine the moments of inertia of the forging with respect to the coordinate axes. The specific weight of steel is 490 lb/ft³.

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Accepted Answer

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

[latex]\begin{aligned}V &=(2 ext { in.})(2 ext { in.})(6 ext { in.})=24 ext { in }^{3} \W &=\frac{\left(24 ext { in }^{3} ight)\left(490 \mathrm{lb} / \mathrm{ft}^{3} ight)}{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}\end{aligned}[/latex]

Each Cylinder

[latex]\begin{aligned}V &=\pi(1 \mathrm{in} .)^{2}(3 \mathrm{in} .)=9.42 \mathrm{in}^{3} \W &=\frac{\left(9.42 \mathrm{in}^{3} ight)\left(490 \mathrm{lb} / \mathrm{ft}^{3} ight)}{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}\end{aligned}[/latex]

Moments of Inertia (Fig. 1).
Prism

[latex]\begin{aligned}&I_{x}=I_{z}=\frac{1}{12}\left(0.211 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft} ight)\left[\left(\frac{6}{12} \mathrm{ft} ight)^{2}+\left(\frac{2}{12} \mathrm{ft} ight)^{2} ight]=4.88 imes 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} ight)\left[\left(\frac{2}{12} \mathrm{ft} ight)^{2}+\left(\frac{2}{12} \mathrm{ft} ight)^{2} ight]=0.977 imes 10^{-3} \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^{2}\end{aligned}[/latex]

Each Cylinder

[latex]\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} ight)\left(\frac{1}{12} \mathrm{ft} ight)^{2} \+\left(0.0829 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft} ight)\left(\frac{2}{12} \mathrm{ft} ight)^{2}=2.59 imes 10^{-3} \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^{2} \I_{y}=\frac{1}{12} m\left(3 a^{2}+L^{2} ight)=m \bar{x}^{2}=\frac{1}{12}\left(0.0829 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft} ight)\left[3\left(\frac{1}{12} \mathrm{ft} ight)^{2}+\left(\frac{3}{12} \mathrm{ft} ight)^{2} ight] \+\left(0.0829 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft} ight)\left(\frac{2.5}{12} \mathrm{ft} ight)^{2}=4.17 imes 10^{-3} \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^{2} \I_{z}=\frac{1}{12} m\left(3 a^{2}+L^{2} ight)+m\left(\bar{x}^{2}+\bar{y}^{2} ight)=\frac{1}{12}\left(0.0829 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft} ight)\left[3\left(\frac{1}{12} \mathrm{ft} ight)^{2}+\left(\frac{3}{12} \mathrm{ft} ight)^{2} ight] \+\left(0.0829 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft} ight)\left[\left(\frac{2.5}{12} \mathrm{ft} ight)^{2}+\left(\frac{2}{12}\mathrm{ft} ight)^{2} ight]=6.48 imes 10^{-3} \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^{2}\end{aligned}[/latex]

Entire Body. Adding the values obtained for the prism and two cylinders, you have

[latex]\begin{array}{lrl}I_{x} & =4.88 imes 10^{-3}+2\left(2.59 imes 10^{-3} ight) & I_{x}=10.06 imes 10^{-3} \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^{2} \I_{y} & =0.977 imes 10^{-3}+2\left(4.17 imes 10^{-3} ight) & I_{y}=9.32 imes 10^{-3} \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^{2} \I_{z} & =4.88 imes 10^{-3}+2\left(6.48 imes 10^{-3} ight) & I_{z}=17.84 imes 10^{-3} \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^{2}\end{array}[/latex]

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.

Question 9.SP.12: A steel forging consists of a 6 × 2 × 2-in. rectangular pris...

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