How to Calculate Moment of Inertia
Learn how to calculate the moment of inertia for common shapes, composite sections, and rotational systems. Includes the parallel axis theorem and worked examples.
What Is Moment of Inertia?
Moment of inertia (I) is a measure of a body's resistance to angular acceleration about a given axis. For rotating masses, I = Σ(m_i · r_i²), where each mass element m_i is multiplied by the square of its distance r_i from the rotation axis. In structural engineering, the area moment of inertia (second moment of area) I = ∫y² dA quantifies a beam cross-section's resistance to bending. The same symbol I is used for both; context determines which is meant. Units are kg·m² for mass moment and m⁴ for area moment.
Area Moment of Inertia for Common Shapes
For a solid rectangle (width b, height h) about its centroidal horizontal axis: I = b·h³/12. For a solid circle of diameter d about its centroid: I = π·d⁴/64. For a hollow rectangle (outer b×h, inner b_i×h_i): I = (b·h³ − b_i·h_i³)/12. For a hollow circle (outer diameter D, inner d): I = π·(D⁴ − d⁴)/64. These are the centroidal values — the minimum I for an axis parallel to the chosen direction. Structural tables and CAD software provide I directly for standard sections.
Parallel Axis Theorem
To find the moment of inertia about any axis parallel to the centroidal axis, use the parallel axis theorem: I = I_c + A·d², where I_c is the centroidal moment of inertia, A is the cross-sectional area, and d is the perpendicular distance between the two parallel axes. For example, a rectangle with I_c = b·h³/12 and area A = b·h placed with its centroid at distance d = h/2 from the base gives I_base = b·h³/12 + b·h·(h/2)² = b·h³/3. This theorem is essential for composite section analysis.
Composite Sections
Real structural cross-sections are often built-up from simpler shapes (flanges, webs, gussets). To find I for a composite section about a common axis: (1) find the centroid of the composite section using ȳ = Σ(A_i · ȳ_i) / ΣA_i; (2) apply the parallel axis theorem to each sub-shape about that centroid: I_total = Σ(I_c,i + A_i · d_i²), where d_i is the distance from each sub-shape's centroid to the composite centroid. Subtracting hollow regions works the same way with negative areas.
Mass Moment of Inertia for Rigid Bodies
For a solid cylinder (mass m, radius R) rotating about its own axis: I = ½·m·R². For a solid sphere: I = 2/5·m·R². For a slender rod of length L about its center: I = m·L²/12; about one end: I = m·L²/3. For a rectangular plate (sides a and b) about its centroid perpendicular to the plate: I = m·(a² + b²)/12. These formulas are derived by integrating ∫r² dm over the body volume and are used in dynamics problems involving angular acceleration, kinetic energy (KE = ½·I·ω²), and angular momentum.
Radius of Gyration
The radius of gyration k (or r_g) is defined as k = √(I/A) for area moments or k = √(I/m) for mass moments. It represents the equivalent distance from the axis at which the entire area (or mass) could be concentrated to give the same I. Structural steel tables report the radius of gyration because column buckling capacity depends on (k·L)/r, where k·L is the effective length. A larger radius of gyration relative to length means higher buckling resistance and is achieved by placing material far from the centroid, as in I-beams and hollow tubes.
Practical Applications
Area moment of inertia drives beam deflection (δ = F·L³/48·E·I for midspan load) and bending stress (σ = M·c/I). Maximizing I for a given area is why I-beams, H-sections, and box sections are preferred over solid rectangles for structural members. Mass moment of inertia governs motor sizing for rotary systems: the required torque to achieve angular acceleration α is τ = I·α. For a robotic arm or flywheel, minimizing I (lightweight design near the axis) reduces drive torque while maximizing I (flywheel rim mass) stores more rotational energy.
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