Moment of Inertia of a T-Section

Step 1: area and centroid of each part, measured up from the base. Flange: A₁ = b × h = 8 × 2 = 16 in². It sits on top of the 6-in web, so its centroid is y₁ = 6 + 2/2 = 7 in. Web: A₂ = 2 × 6 = 12 in². Its centroid is at half its height, y₂ = 6/2 = 3 in. Step 2: centroid of the whole section. ȳ = ΣAy / ΣA = (A₁y₁ + A₂y₂) / (A₁ + A₂) = (16 × 7 + 12 × 3) / (16 + 12) = 148/28 = 5.29 in. Step 3: parallel-axis theorem. I = Σ(I_c + A·d²), where I_c = bh³/12 for each rectangle (h is the side perpendicular to the axis) and d = distance from the part's centroid to ȳ. Flange: I_c1 = 8(2)³/12 = 5.33 in⁴, d₁ = 7 − 5.29 = 1.71 in, A₁d₁² = 16(1.71)² = 47.0 in⁴, so 52.3 in⁴. Web: I_c2 = 2(6)³/12 = 36.0 in⁴, d₂ = 5.29 − 3 = 2.29 in, A₂d₂² = 12(2.29)² = 62.7 in⁴, so 98.7 in⁴. Step 4: I_x = 52.3 + 98.7 = 151 in⁴. Adding only the I_c terms gives 41.3 in⁴, which forgets the A·d² terms.