Moment of inertia
The second moment of area, usually called the moment of inertia I in structural work, measures how the area of a cross-section is spread about an axis. It is not about mass or motion here; it is purely geometry. Area far from the axis counts much more than area close to it, because each small piece contributes its area times the square of its distance. That is why an I-section, with most of its steel in the flanges, is so much stiffer in bending than a solid square bar of the same weight.
I appears in almost every beam and column formula. Bending stress is M y / I, beam deflection is inversely proportional to E I, and the buckling load of a column is proportional to E I. The section modulus Z = I / y_max combines the shape and the extreme fibre distance into one number for stress checks. The radius of gyration r = √(I/A) tells how slender a column is.
For a rectangle of width b and depth d about its centroidal axis, I = b d³ / 12, so doubling the depth gives eight times the stiffness while doubling the width only doubles it. For a circle, I = π D⁴ / 64. For built-up sections, the parallel axis theorem moves each part's own I to the common centroidal axis by adding its area times the square of the shift: I = I_c + A h². Add the parts, subtract any holes.
The polar moment of inertia J, the sum of the two perpendicular second moments, governs torsion of circular shafts. Products of inertia and principal axes become important for angles and unsymmetrical sections, which bend sideways as well as downward under vertical load. Mass moment of inertia, a different but related quantity, enters rotational dynamics of machine foundations.
Rectangle: I = b d³ / 12Circle: I = π D⁴ / 64; polar J = π D⁴ / 32Parallel axis: I = I_c + A h²Section modulus: Z = I / y_max; radius of gyration r = √(I / A)
Applying the parallel axis theorem from an axis that is not the part's own centroidal axis.
Problem. Find I about the horizontal centroidal axis of a symmetric I-section: flanges 200 × 20 mm, web 10 × 260 mm, overall depth 300 mm.
- Centroid is at mid-depth (150 mm) by symmetry.
- Each flange: own I = 200 × 20³ / 12 = 133 333 mm⁴; area 4 000 mm² at h = 140 mm, so A h² = 4 000 × 140² = 78 400 000 mm⁴. One flange total = 78.53 × 10⁶ mm⁴.
- Web: 10 × 260³ / 12 = 14.65 × 10⁶ mm⁴.
- Total I = 2 × 78.53 + 14.65 = 171.7 × 10⁶ mm⁴.
Answer. I ≈ 1.72 × 10⁸ mm⁴
Quick quiz
1By what factor does I of a rectangle rise if depth is doubled?
Eight times, because I depends on d³.
2What does the parallel axis theorem add?
The area of the part times the square of the distance between the two axes.
3Which quantity governs torsion of a solid shaft?
The polar moment of inertia J.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- F. P. Beer, E. R. Johnston, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, latest edition.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, latest edition.
- S. Timoshenko, D. H. Young, Engineering Mechanics, McGraw Hill, 4th edition, 1956.
Free and official sources
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