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

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.

$I_y = \frac{5894}{147}t^4 = 40.10t^4, I_z = \frac{33,103}{294}t^4 = 112.60t^4$

$I_{yz} = \frac{-270}{7}t^4 = -38.57t^4$
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)

$\tan 2θ_p = \frac{- I_{yz}}{\left(\frac{I_y – I_z}{2}\right)} = \frac{-\left(\frac{-270}{7}\right)}{\frac{11,788 – 33,103}{2(294)}} = -1.064$

$2θ_{p1}$ = 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)

$I_{p_1}=\frac{I_y+I_z} {2}+\sqrt{\left(\frac{I_y-I_z}{2}\right)^2+I_{y z}^2}$

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

= 129.28$t^4$

or

$I_{p1} = 129.3t^4$

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

$I_{p2} = 23.4t^4$

Question: C.5

## (a) Draw the Mohr’s circle for the centroidal moments and products of inertia for the L-shaped area in Fig. 1 of Example C-2, given that: ...

(a) Sketch Mohr’s circle and calculate the princip...
Question: C.2

## Determine the centroidal moment of inertia Iy for the L-shaped section in Example C-1. (Here, in Fig. 1, the origin of the (y, z) reference frame is at the centroid of the composite area. The centroidal reference axes for the rectangular “legs” of the L-shaped area are (y1, z1) and ...

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Question: C.1