A symmetrical channel section has width of its top and bottom flanges as 100 mm and thickness 10 mm. The web is 80 mm high between flanges and 10 mm thick. Its moment of inertia about a centroidal axis in its plane parallel to the flanges is 449.3 X 10⁴ mm⁴ . Calculate its moment of inertia about a parallel axis at the top face of the upper flange.

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RRB JE CBT 2 (Mechanical Engineering) ReExam Official Paper (Held On: 04 Jun, 2025)

Answer 1

Concept:

Flange width = 100 mm

Flange thickness = 10 mm

Web height = 80 mm, Web thickness = 10 mm

Total height of section = 10 + 80 + 10 = 100 mm

Centroidal MOI = 449.3 × 10⁴ mm⁴ = 4493000 mm⁴

Calculation:

Area of top flange, A₁ = 100 × 10 = 1000 mm²

Area of bottom flange, A₂ = 100 × 10 = 1000 mm²

Area of web, A₃ = 10 × 80 = 800 mm²

Total area, A = A₁ + A₂ + A₃ = 1000 + 1000 + 800 = 2800 mm²

Distance from centroid to top of section, d = 100 / 2 = 50 mm