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.)

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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)

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