Chapter 9, Problem 9.51P

To determine

(a)

The principal moments of inertia and the principal directions at the centroid C for the semicircular region.

Answer to Problem 9.51P

The principal moments of inertia:

  $I_1=81.43\times {10}^{6}m{m}^4$

  $I_2=22.77\times {10}^{6}m{m}^4$

Principal directions are along x and y axes

Explanation of Solution

Given information:

The semicircular region:

Calculations:

Because to symmetry, the x- and y- axes are the principal axes at C.

Hence,

  $\begin{array}{l}{I}_1=I_y=\frac{\pi }{8}{R}^4=\frac{\pi }{8}{\left(120\right)}^4\\ \Rightarrow I_1=81.43\times {10}^{6}m{m}^4\\ I_2=I_x=0.1098R^4=0.1098{\left(120\right)}^4\\ \Rightarrow I_2=22.77\times {10}^{6}m{m}^4\end{array}$

Conclusion:

The principal moments of inertia at the centroid C for the semicircular region shown are $I_1=81.43\times {10}^{6}m{m}^4$ and $I_2=22.77\times {10}^{6}m{m}^4$. And the principal directions are along x and y axes.

To determine

(b)

The moments and the products of inertia about the u-v-axes for the semicircular region shown.

Answer to Problem 9.51P

Moments of inertia:

  $I_u=33.2\times {10}^{6}m{m}^4$

  $I_v=71.0\times {10}^{6}m{m}^4$

Products of inertia:

  $I_{uv}=22.5\times {10}^{6}m{m}^4$

Explanation of Solution

Given information:

For the semicircular region shown:

  $I_1=I_y=81.43\times {10}^{6}m{m}^4$

  $I_2=I_x=22.77\times {10}^{6}m{m}^4$

Calculations:

  $\begin{array}{l}\frac{1}{2}\left(I_x+I_y\right)=\frac{1}{2}\left(22.77+81.43\right)\times {10}^6=52.10\times {10}^{6}m{m}^4\\ \frac{1}{2}\left(I_x-I_y\right)=\frac{1}{2}\left(22.77-81.43\right)\times {10}^6=-29.33\times {10}^{6}m{m}^4\\ \text{Moments of inertia about the }u-v-\text{axes, using the relations:}\\ I_u=\frac{1}{2}\left(I_x+I_y\right)+\frac{1}{2}\left(I_x-I_y\right)\cos 2\theta -I_{xy}\sin 2\theta \\ I_u=\left[52.10-29.33\cos \left(-{50}^o\right)-0\right]\times {10}^6\\ \Rightarrow I_u=33.2\times {10}^{6}m{m}^4\\ I_v=\frac{1}{2}\left(I_x+I_y\right)-\frac{1}{2}\left(I_x-I_y\right)\cos 2\theta +I_{xy}\sin 2\theta \\ I_v=\left[52.10+29.33\cos \left(-{50}^o\right)+0\right]\times {10}^6\\ \Rightarrow I_v=71.0\times {10}^{6}m{m}^4\\ Hence,\text{ Products of inertia about the }u-v-\text{axes}:\\ I_{uv}=\frac{1}{2}\left(I_x-I_y\right)\sin 2\theta +I_{xy}\cos 2\theta \\ I_{uv}=\left[-29.33\text{ sin}\left(-{50}^o\right)+0\right]\times {10}^6\\ \Rightarrow I_{uv}=22.5\times {10}^{6}m{m}^4\end{array}$

Conclusion:

For the semicircular region, the moments of inertia about the u-v axes are $I_u=33.2\times {10}^{6}m{m}^4$ and $I_v=71.0\times {10}^{6}m{m}^4$. And the products of inertia about the u-v axes is $I_{uv}=22.5\times {10}^{6}m{m}^4$.