Find the moment of inertia of the lamina with a circular hole of 30 mm diameter about the axis AB as shown in Fig. 7.24.

For Rectangle 1 (Flange):

$$A_1 = b_1 \times d_1 = 120 \text{ mm} \times 30 \text{ mm} = 3600 \text{ mm}^2$$

The centroid $y_1$ from the bottom of the web is $120 \text{ mm} + \frac{30 \text{ mm}}{2} = 135 \text{ mm}$.

For Rectangle 2 (Web):

$$A_2 = b_2 \times d_2 = 30 \text{ mm} \times 120 \text{ mm} = 3600 \text{ mm}^2$$

The centroid $y_2$ from the bottom of the web is $\frac{120 \text{ mm}}{2} = 60 \text{ mm}$.

$$\bar{y} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2}$$ $$\bar{y} = \frac{(3600 \text{ mm}^2 \times 135 \text{ mm}) + (3600 \text{ mm}^2 \times 60 \text{ mm})}{3600 \text{ mm}^2 + 3600 \text{ mm}^2}$$ $$\bar{y} = \frac{486000 \text{ mm}^3 + 216000 \text{ mm}^3}{7200 \text{ mm}^2} = \frac{702000 \text{ mm}^3}{7200 \text{ mm}^2} = 97.5 \text{ mm}$$

The centroidal X-X axis is $97.5 \text{ mm}$ from the bottom of the web.

The moment of inertia of a rectangle about its own centroidal X-axis is $I = \frac{bd^3}{12}$.

For Rectangle 1 (Flange):

$$I_{xx1,c} = \frac{b_1 d_1^3}{12} = \frac{120 \text{ mm} \times (30 \text{ mm})^3}{12} = \frac{120 \times 27000}{12} \text{ mm}^4 = 270000 \text{ mm}^4$$ $$h_1 = |y_1 - \bar{y}| = |135 \text{ mm} - 97.5 \text{ mm}| = 37.5 \text{ mm}$$

For Rectangle 2 (Web):

$$I_{xx2,c} = \frac{b_2 d_2^3}{12} = \frac{30 \text{ mm} \times (120 \text{ mm})^3}{12} = \frac{30 \times 1728000}{12} \text{ mm}^4 = 4320000 \text{ mm}^4$$ $$h_2 = |y_2 - \bar{y}| = |60 \text{ mm} - 97.5 \text{ mm}| = 37.5 \text{ mm}$$

The parallel axis theorem states $I_{XX} = I_{xx,c} + A h^2$.

For Rectangle 1 (Flange):

$$I_{XX1} = I_{xx1,c} + A_1 h_1^2 = 270000 \text{ mm}^4 + (3600 \text{ mm}^2 \times (37.5 \text{ mm})^2)$$ $$I_{XX1} = 270000 \text{ mm}^4 + (3600 \times 1406.25) \text{ mm}^4 = 270000 \text{ mm}^4 + 5062500 \text{ mm}^4 = 5332500 \text{ mm}^4$$

For Rectangle 2 (Web):

$$I_{XX2} = I_{xx2,c} + A_2 h_2^2 = 4320000 \text{ mm}^4 + (3600 \text{ mm}^2 \times (37.5 \text{ mm})^2)$$ $$I_{XX2} = 4320000 \text{ mm}^4 + (3600 \times 1406.25) \text{ mm}^4 = 4320000 \text{ mm}^4 + 5062500 \text{ mm}^4 = 9382500 \text{ mm}^4$$

$$I_{XX} = I_{XX1} + I_{XX2} = 5332500 \text{ mm}^4 + 9382500 \text{ mm}^4 = 14715000 \text{ mm}^4$$

Expressing this in scientific notation:

$$I_{XX} = 14.715 \times 10^6 \text{ mm}^4 = 14715 \times 10^3 \text{ mm}^4$$

The moment of inertia of the T-section about its centroidal X-X axis is $\boxed{\text{14715} \times 10^3 \text{ mm}^4}$.