Question C.3: For the L-shaped area in Example C-2, use the composite-area......
For the L-shaped area in Example C-2, use the composite-area procedure to determine the centroidal product of inertia, I_{yz}. (Note: Here the (y, z) reference frame is a centroidal reference frame for the whole area; (y_1, z_1) and (y_2, z_2) are centroidal reference frames for the constituent areas A_1 and A_2, respectively.)
The centroidal product of inertia relative to the (y, z) axes is given by
I_{yz} = (I_{y_1z_1} + A_1\bar{y}_1\bar{z}_1) + (I_{y_2z_2} + A_2\bar{y}_2\bar{z}_2)It is very important to note that \bar{y}_1, \bar{z}_1 etc., are signed values, that is, some of them could be negative. By Eq. (C-17), I_{y_1z_1} = I_{y_2z_2} = 0. Therefore,
I_{yz} = 0 (C-17)
I_{yz} = A_1\bar{y}_1\bar{z}_1 + A_2\bar{y}_2\bar{z}_2= (6t^2) \left(-\frac{43}{14} t +\frac{1}{2} t\right) \left(3t – \frac{22}{14} t\right)
+ (8t^2) \left(5t – \frac{43}{14} t\right) \left(-\frac{22}{14} t + \frac{1}{2} t\right)
or
I_{yz} = – \frac{270}{7} t^4Note that, since the centroid C_1 lies in the second quadrant and C_2 lies in the fourth quadrant, both A_1 and A_2 make negative contributions to I_{yz}.