For the L-shaped area in Fig. 1 of Example C-2, (a) determine the orientation of the centroidal principal axes and show the orientation on a sketch. (b) Determine the principal moments of inertia.

I_y = \frac{5894}{147}t^4 = 40.10t^4, I_z = \frac{33,103}{294}t^4 = 112.60t^4I_{yz} = \frac{-270}{7}t^4 = -38.57t^4

Step-by-Step

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(a) From Eq. (C-23),

\tan 2θ_p = \frac{- I_{yz}}{\left(\frac{I_y – I_z}{2}\right)} (C-23)

\tan 2θ_p = \frac{- I_{yz}}{\left(\frac{I_y – I_z}{2}\right)} = \frac{-\left(\frac{-270}{7}\right)}{\frac{11,788 – 33,103}{2(294)}} = -1.0642θ_{p1} = 133.22°, 2θ_{p2} = -46.78°

Then, as illustrated in Fig. 1,

θ_{p1} = 66.6°, θ_{p2} = -23.4°

(b) From Eq. (C-26a),

\begin{aligned} & I_{\max } \equiv I_{p_1}=\frac{I_y+I_z} {2}+\sqrt{\left(\frac{I_y-I_z}{2}\right)^2+I_{y z}^2} \\ & I_{\min } \equiv I_{p_2}=\frac{I_y+I_z}{2}-\sqrt{\left(\frac{I_y-I_z}{2}\right)^2+I_{y z}^2} \end{aligned} (C-26)

I_{p_1}=\frac{I_y+I_z} {2}+\sqrt{\left(\frac{I_y-I_z}{2}\right)^2+I_{y z}^2}= \frac{40.10t^4 + 112.60t^4}{2} + \sqrt{\left(\frac{40.10t^4 – 112.60t^4}{2} \right)^2+ (-38.57t^4)^2}

= 129.28t^4

or

I_{p1} = 129.3t^4Similarly, from Eq. (C-26b),

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