Question C.4: For the L-shaped area in Fig. 1 of Example C-2, (a) determin......

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

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

$2θ_{p1}$ = 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)

= $\frac{40.10t^4 + 112.60t^4}{2} + \sqrt{\left(\frac{40.10t^4 – 112.60t^4}{2} \right)^2+ (-38.57t^4)^2}$

= 129.28$t^4$

or

Similarly, from Eq. (C-26b),