Question

Product of Inertia The product of inertia for an area about inclined axes is given by the formula

$I_{uv}=I_x\sin \theta \cos \theta -I_y\sin \theta \cos \theta +I_{xy}\left({\cos }^2\theta -{\sin }^2\theta \right)$

Show that this is equivalent to

$I_{uv}=\frac{I_x-I_y}{2}\sin \left(2\theta \right)+I_{xy}\cos \left(2\theta \right)$

Source: Adapted from Hibbeler, Engineering Mechanics: Statics, 10th ed., Prentice Hall @ 2004 .

Short answer

The product of inertia is$I_{uv}=\frac{I_x-I_y}{2}\sin \left(2\theta \right)+I_{xy}\cos \left(2\theta \right)$

Step-by-step solution

Given information

Given the product of inertia is$I_{uv}=I_x\sin \theta \cos \theta -I_y\sin \theta \cos \theta +I_{xy}\left({\cos }^2\theta -{\sin }^2\theta \right)$

Using trigonometry identity and converting

Using trigonometric identity, we get

$$\begin{gathered} I_{uv}=I_x\sin \theta \cos \theta -I_y\sin \theta \cos \theta +I_{xy}\left({\cos }^2\theta -{\sin }^2\theta \right) \\ {I}_{uv}=(I_x-I_y)\sin \theta \cos \theta I_{xy}\left({\cos }^2\theta -{\sin }^2\theta \right) \\ {I}_{uv}=\frac{I_x-I_y}{2}\left(2\sin \theta \cos \theta \right)I_{xy}\left(\cos 2\theta \right) \\ {I}_{uv}=\frac{I_x-I_y}{2}\left(\sin 2\theta \right)I_{xy}\left(\cos 2\theta \right) \end{gathered}$$