The flange of an I-section is 100 mm wide and 10 mm thick, and has moment of inertia If about its own centroidal axis parallel to flange length, in the plane of the flange. Its centroidal axis is 50 mm from the centroidal axis X-X of the I-section normal to the web in the plane of the I-section. Area moment of inertia of the flange about axis X-X is:

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

Answer 1

Concept:

To calculate the moment of inertia of a flange about the X–X axis of the I-section, we use the Parallel Axis Theorem.

Given:

Flange width, b = 100 mm

Flange thickness, t = 10 mm

Distance between flange centroid and axis X–X, d = 50 mm

Calculation:

Area of flange, A = b × t = 100 × 10 = 1000 mm², d = 50 mm

Using Parallel Axis Theorem,

\( I_{X-X} = I_f + A \times d^2 \)

Substitute values:

\( I_{X-X} = I_f + 1000 \times 50^2 = I_f + 1000 \times 2500 \)

\( I_{X-X} = I_f + 2.5 \times 10^6~\text{mm}^4 \)