Chapter 9, Problem 9.87RP

To determine

(a)

The other principal centroidal moment of inertia I₁.

Answer to Problem 9.87RP

  ${\overline{I}}_1=6.459\times {10}^{6}m{m}^4$

Explanation of Solution

Given information:

The inertial properties of $L150\times 100\times 10-mm$ :

  $\begin{array}{l}\overline{x}=23.8mm,\overline{y}=48.8mm,A=2400m{m}^2\\ {\overline{I}}_x=5.58\times {10}^{6}m{m}^4\\ {\overline{I}}_y=2.03\times {10}^{6}m{m}^4\end{array}$

The angle a locating the axis of minimum centroidal moment of inertia is 24.0°.

And the corresponding radius of gyration is ${\overline{k}}_2=21.9mm$.

Calculations:

  $\begin{array}{l}\text{From the relation}:\left(I=A{k}^2\right)\\ {\overline{I}}_2=A{\overline{k}}_2^2=\left(2400\right){\left(21.9\right)}^2=1.151\times {10}^{6}m{m}^4\\ \text{using the relation: }\left({\overline{I}}_1+{\overline{I}}_2={\overline{I}}_x+{\overline{I}}_y\right)\\ \therefore {\overline{I}}_1={\overline{I}}_x+{\overline{I}}_y-{\overline{I}}_2\\ {\overline{I}}_1=\left(5.58+2.03-1.151\right)\times {10}^6\\ \Rightarrow {\overline{I}}_1=6.459\times {10}^{6}m{m}^4\end{array}$

Conclusion:

The other principal centroidal moment of inertia is ${\overline{I}}_1=6.459\times {10}^{6}m{m}^4$.

To determine

(b)

  ${\overline{I}}_{xy}$.

Answer to Problem 9.87RP

  ${\overline{I}}_{xy}=-1.972\times {10}^{6}m{m}^4$

Explanation of Solution

Given information:

The inertial properties of $L150\times 100\times 10-mm$ :

  $\begin{array}{l}\overline{x}=23.8mm,\overline{y}=48.8mm,A=2400m{m}^2\\ {\overline{I}}_x=5.58\times {10}^{6}m{m}^4\\ {\overline{I}}_y=2.03\times {10}^{6}m{m}^4\end{array}$

The angle a locating the axis of minimum centroidal moment of inertia is 24.0°.

And the corresponding radius of gyration is ${\overline{k}}_2=21.9mm$.

Calculations:

  $\begin{array}{l}\text{From equation 9}\text{.23:}\\ {\overline{I}}_1=\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)+R\\ R={\overline{I}}_1-\frac{{\overline{I}}_x+{\overline{I}}_y}{2}=\left(6.459-\frac{5.58+2.03}{2}\right)\times {10}^6=2.654\times {10}^{6}m{m}^4\\ \text{Now, using equation 9}\text{.21:}\\ \sin 2{\theta }_2={\overline{I}}_{xy}/R\\ \therefore {\overline{I}}_{xy}=R\sin 2{\theta }_2\\ But\\ {\theta }_2={\theta }_1+{90}^o={24.0}^o+{90}^o={114.0}^o\\ \therefore 2{\theta }_2={228.0}^o\\ \therefore {\overline{I}}_{xy}=\left(2.654\times {10}^6\right)\sin {228.0}^o\\ \Rightarrow {\overline{I}}_{xy}=-1.972\times {10}^{6}m{m}^4\end{array}$

Conclusion:

For the given structural shape, ${\overline{I}}_{xy}=-1.972\times {10}^{6}m{m}^4$.