# Question C.2: Determine the centroidal moment of inertia Iy for the L-shap......

Determine the centroidal moment of inertia $I_y$ 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 ($y_1, z_1$) and ($y_2, z_2$), respectively.)

Step-by-Step
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We can combine Eqs. (C-14) with the parallel axis theorems, Eqs. (C-11), to compute the required moments of inertia.

$I_y = \sum\limits_{i}{(A_y)_i}, I_z = \sum\limits_{i}{(I_z)_i}$      (C-14)

$I_y = \bar{z}^2A + I_{y_C}$            (C-11a)

$I_z = \bar{y}^2A + I_{z_C}$          (C-11b)

$I_y = (I_y)_1 + (I_y)_2 = [(I_{y_C})_1 + A_1\bar{z}^2_1] + [(I_{y_C})_2 + A_2\bar{z}_2^2]$      (1)

where $(I_{y_C})_i$ is the moment of inertia of area $A_i$ about its own centroidal y axis, and $\bar{z}_i$ is the z-coordinate of the centroid $C_i$ measured in the (y, z) reference frame with origin at the composite centroid, C. Referring to Fig. 1, we get

$I_y = (I_y)_1 + (I_y)_2$ $= \left[ \frac{1}{12}(t)(6t)^3 + (t)(6t) \left(3t -\frac{22}{14} t\right)^2\right]$

$= \left[ \frac{1}{12}(8t)(t)^3 + (t)(8t) \left(\frac{1}{2}t -\frac{22}{14} t\right)^2\right]$

= $18t^4 + \frac{600}{49} t^4 +\frac{2}{3} t^4 + \frac{450}{49} t^4$

= $\frac{842}{21} t^4 = 40.1t^4$        (2)

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.4

## 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. ...

(a) From Eq. (C-23), \tan 2θ_p = \frac{- I_...
Question: C.1