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.

To find the moment of inertia of the T-section about its centroidal X-X axis, we will divide the section into two rectangles: the flange and the web. Flange (Rectangle 1): Width (b_1) = 120 mm Depth (d_1) = 30 mm Web (Rectangle 2): Width (b_2) = 30 mm Depth (d_2) = 120 mm Step 1: Calculate the area and centroidal position of each rectangle from the bottom reference axis. For Rectangle 1 (Flange): A_1 = b_1 × d_1 = 120 mm × 30 mm = 3600 mm^2 The centroid y_1 from the bottom of the web is 120 mm + 30 mm2 = 135 mm. For Rectangle 2 (Web): A_2 = b_2 × d_2 = 30 mm × 120 mm = 3600 mm^2 The centroid y_2 from the bottom of the web is 120 mm2 = 60 mm. Step 2: Calculate the centroid (y) of the entire T-section from the bottom reference axis. y = (A_1 y_1 + A_2 y_2)/(A_1 + A_2) y = (3600 mm^2 × 135 mm) + (3600 mm^2 × 60 mm)3600 mm^2 + 3600 mm^2 y = 486000 mm^3 + 216000 mm^37200 mm^2 = 702000 mm^37200 mm^2 = 97.5 mm The centroidal X-X axis is 97.5 mm from the bottom of the web. Step 3: Calculate the moment of inertia of each rectangle about its own centroidal axis (I_xx,c) and the distance from its centroid to the section's centroid (h). The moment of inertia of a rectangle about its own centroidal X-axis is I = (bd^3)/(12). For Rectangle 1 (Flange): I_xx1,c = (b_1 d_1^3)/(12) = 120 mm × (30 mm)^312 = (120 × 27000)/(12) mm^4 = 270000 mm^4 h_1 = |y_1 - y| = |135 mm - 97.5 mm| = 37.5 mm For Rectangle 2 (Web): I_xx2,c = (b_2 d_2^3)/(12) = 30 mm × (120 mm)^312 = (30 × 1728000)/(12) mm^4 = 4320000 mm^4 h_2 = |y_2 - y| = |60 mm - 97.5 mm| = 37.5 mm Step 4: Use the parallel axis theorem to find the moment of inertia of each rectangle about the T-section's centroidal X-X axis. 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 mm^4 + (3600 mm^2 × (37.5 mm)^2) I_XX1 = 270000 mm^4 + (3600 × 1406.25) mm^4 = 270000 mm^4 + 5062500 mm^4 = 5332500 mm^4 For Rectangle 2 (Web): I_XX2 = I_xx2,c + A_2 h_2^2 = 4320000 mm^4 + (3600 mm^2 × (37.5 mm)^2) I_XX2 = 4320000 mm^4 + (3600 × 1406.25) mm^4 = 4320000 mm^4 + 5062500 mm^4 = 9382500 mm^4 Step 5: Sum the moments of inertia of the individual parts to get the total moment of inertia of the T-section about its centroidal X-X axis. I_XX = I_XX1 + I_XX2 = 5332500 mm^4 + 9382500 mm^4 = 14715000 mm^4 Expressing this in scientific notation: I_XX = 14.715 × 10^6 mm^4 = 14715 × 10^3 mm^4 The moment of inertia of the T-section about its centroidal X-X axis is 14715 × 10^3 mm^4. Send me the next one 📸