Moment of Inertia

Area moment of inertia measures a section's resistance to bending; mass moment of inertia measures resistance to angular acceleration. Learn standard formulas, the parallel-axis theorem, polar moment, radius of gyration and section design.

Why is a plank stiffer on edge than flat? Why are beams I-shaped and shafts hollow? The answer is the second moment of area, I: material placed far from the bending axis contributes much more stiffness (it counts as distance squared).

The mass moment of inertia plays the same role for rotation: it tells you how hard it is to spin up a flywheel, wheel or robot arm.

An analogy

A figure skater spins faster by pulling in her arms: moving mass closer to the axis reduces her moment of inertia. A beam does the opposite — it puts material far away to resist bending.

Why distance squared matters

I = ∫y² dA: each bit of area counts by the square of its distance from the axis. A 100 × 200 mm rectangle has I = 66.7 × 10⁶ mm⁴ on edge but only 16.7 × 10⁶ mm⁴ flat — four times less for the same material.

Parallel-axis theorem

The inertia about any axis equals the inertia about the parallel centroidal axis plus A·d². Composite sections are built from parts using this, which is why I-beams (flanges far from the neutral axis) are so efficient.

Polar moment and torsion

For twisting, the polar moment J = I_x + I_y governs stiffness and stress. Hollow tubes put material at large radius and are far more efficient in torsion than solid bars — drive shafts are tubes for this reason.

Mass moment of inertia and dynamics

I_m = ∫r² dm determines rotational kinetic energy (½I_mω²) and the torque needed to accelerate (T = I_m α). Motor sizing, flywheel design and robot-arm dynamics all start with it; inertia ratios between load and motor shape servo performance.

Formulas

• Rectangle (about centroid): I = b·h³ / 12
• Solid circle: I = π d⁴ / 64, J = π d⁴ / 32
• Parallel-axis theorem: I = I_c + A·d²
• Radius of gyration: k = √(I / A)
• Mass MOI (solid disc): I_m = ½ m r²

Worked example: Rectangle on edge vs flat

Given: 100 × 200 mm section

Find: I for both orientations.

• On edge (h = 200): 100 × 200³/12 = 66.7 × 10⁶ mm⁴
• Flat (h = 100): 200 × 100³/12 = 16.7 × 10⁶ mm⁴
• Ratio = 4

Answer: On edge is 4× stiffer

Worked example: T-section by parallel-axis theorem

Given: Flange 100 × 20 (y = 90), web 20 × 80 (y = 40); ȳ = 67.8 mm

Find: I about the centroidal axis.

• Flange: 66 667 + 2000 × 22.2² = 1.054 × 10⁶
• Web: 853 333 + 1600 × 27.8² = 2.088 × 10⁶
• Total = 3.14 × 10⁶ mm⁴

Answer: I ≈ 3.14 × 10⁶ mm⁴

Common mistakes

• Wrong: Using h and b the wrong way round in bh³/12. — Right: h is the dimension perpendicular to the bending axis (it's cubed).
• Wrong: Adding centroidal I values of parts directly. — Right: Add I_c + A d² for each part about the common neutral axis.

Applications

• I-beams and box sections: Material concentrated in flanges gives high I per kilogram for floors, bridges and chassis rails.
• Drive shafts: Hollow tubes give high polar moment for torsion with low weight.
• Flywheels & motors: Mass MOI sets stored energy and the motor torque needed for acceleration.

Key takeaways

• I = ∫y² dA; rectangles bh³/12, circles πd⁴/64.
• Parallel axis: I = I_c + Ad².
• Depth and distance from the axis dominate stiffness.
• Mass MOI governs rotational dynamics: T = Iα.

Going further

• Product of inertia and principal axes: Unsymmetrical sections (angles) have a product of inertia I_xy; principal axes (rotated by θ with tan 2θ = −2I_xy/(I_x − I_y)) give max and min I — they bend about these.
• Inertia tensor: In 3D dynamics, mass properties form a 3 × 3 tensor; its principal axes explain why objects spin stably about their largest and smallest axes but not the middle one.
• Warping constant: Thin-walled open sections resist twisting through warping (C_w) as well as St Venant torsion — vital for lateral-torsional buckling of beams.