How to calculate area moment of inertia?

For a single closed-form shape, use the standard formula: circle I = πD⁴/64, rectangle I = bh³/12 about the centroidal axis, hollow circle I = π(D⁴ − d⁴)/64, I-beam I = (B·H³ − b·h³)/12. For a composite cross-section, split it into rectangles, compute each rectangle's centroidal I = bh³/12, then add the parallel-axis correction A·d² for each piece (where d is the distance from that piece's centroid to the overall neutral axis), and sum. The same procedure works whether you call it area moment of inertia or second moment of area — they are identical.

How to calculate second moment of area?

Same calculation as area moment of inertia — the two terms describe the identical geometric property and use the same formulas. UK and Eurocode practice prefers "second moment of area"; US practice prefers "area moment of inertia". Both have units of length⁴ (in⁴, mm⁴, m⁴) and both feed the same bending-stress equation σ = M·c/I.

How to find area moment of inertia?

Three ways. (1) Look it up: AISC Steel Construction Manual Part 1 lists I_x and I_y for every standard W-, S-, HP-, HSS-, channel-, and angle-shape. Eurocode product tables do the same in mm⁴. (2) Use a closed-form formula for simple shapes — see the formula card above. (3) For composite or built-up sections, decompose into rectangles, apply the parallel-axis theorem, sum.

How to find second moment of area?

Same workflow as area moment of inertia. Either look up I in a section table (AISC, Eurocode, or manufacturer), use a closed-form formula for primitive shapes, or split a composite into rectangles and apply Huygens' parallel-axis theorem. The composite-centroid calculator at /inertia/ does the latter automatically once you enter the rectangles.

What are the units of area moment of inertia?

Length to the fourth power. In US customary: in⁴ (inches to the fourth). In SI: mm⁴ for steel and concrete sections (most product tables), m⁴ for whole-structure dynamics, cm⁴ in some European product literature. Conversions: 1 in⁴ = 416 231 mm⁴ = 41.62 cm⁴ = 4.162 × 10⁻⁷ m⁴.

What is the area moment of inertia of a circle?

For a solid circle of diameter D: I = πD⁴/64 about any centroidal axis (the value is the same for every diameter through the centre because the circle is rotationally symmetric). For a hollow tube/pipe with outer diameter D and inner diameter d: I = π(D⁴ − d⁴)/64. A 4-inch (100 mm) solid circle has I = π·100⁴/64 = 4 909 000 mm⁴ = 11.79 in⁴.

Reference · Structural · AISC · Eurocode · Huygens

Area Moment of Inertia — Formulas, Units & Cross-Section Reference

Area moment of inertia (also called the second moment of area) is the geometric quantity that determines a beam\'s resistance to bending. This page lists the closed-form formulas for circle, rectangle, I-beam, T-beam, and pipe sections, gives the parallel-axis theorem for composite shapes, summarises the units in use (in⁴, mm⁴, m⁴), and walks through a worked beam example using AISC W-shape data. Reviewed by a licensed PE.

Figure 1 — Definition: each differential area element dA contributes y²·dA to the moment of inertia; integrate over the section.

Area moment inertia calculator

For built-up cross-sections (assemblies of rectangles, I-shapes, plates, and round bars) the composite-centroid / inertia calculator computes I_x, I_y, J, and the centroid coordinates for any combination. For single primitive shapes (circle, rectangle, I-beam, pipe) the dedicated moment-of-inertia tool covers the closed-form formulas. Both calculators include the parallel-axis correction automatically.

→ Open the composite centroid & inertia calculator · → Open the single-shape inertia calculator

Area moment of inertia formulas

Eq. 01 — Definition (general integral) SI

I_x = \int_A y^2 \, dA, \qquad I_y = \int_A x^2 \, dA

·: I_x, I_y = second moments about the x- and y- axes (length⁴)
·: dA = differential area element
·: y, x = perpendicular distance from the corresponding axis to dA

Eq. 02 — Solid circle (diameter D) SI

I = \frac{\pi D^4}{64}

·: I = area moment of inertia about any centroidal axis
·: D = diameter
·: Same value for every centroidal axis (rotational symmetry)

Eq. 03 — Rectangle (b × h, centroidal axis) SI

I_x = \frac{b h^3}{12}, \qquad I_y = \frac{h b^3}{12}

·: b = width, h = height (h is measured along the bending axis)
·: For a square b = h, both axes give I = b⁴/12
·: About the base, not centroid: I = bh³/3

Eq. 04 — I-beam (B × H outer, b × h inner cavity) SI

I_x = \frac{B H^3 - b h^3}{12}

·: B = full flange width; H = full depth
·: b = (B − web thickness); h = clear web height between flanges
·: Subtraction works because both rectangles share the centroidal axis

Eq. 05 — Hollow tube / pipe (outer D, inner d) SI

I = \frac{\pi (D^4 - d^4)}{64}

·: D = outside diameter, d = inside diameter
·: Subtract the inner solid I from the outer solid I
·: Closed-section torsion uses the polar form J = π(D⁴ − d⁴)/32

Eq. 06 — Parallel-axis theorem (Huygens) SI

I = I_{centroid} + A \cdot d^2

·: I_centroid = inertia of sub-area about its own centroid
·: A = sub-area
·: d = perpendicular distance from sub-area centroid to the composite neutral axis

Standards and authoritative sources

Standard / source	Scope
AISC Steel Construction Manual (16th ed.)	Part 1 lists I_x, I_y, S, Z, r for every US W-, S-, HP-, HSS-, C-, MC-, L-shape (in⁴ units)
AISC 360-22	Specification for structural steel buildings — uses I in flexural strength (Chapter F) and deflection checks (Chapter L)
EN 10365 / Eurocode product tables	European IPE, HEA, HEB, UPN section properties in mm⁴
ANSI/AWC NDS-2018	National Design Specification for Wood Construction — moment of inertia tables for sawn lumber and engineered wood
ASCE 7-22	Loading standard whose deflection limits (L/360, L/240) consume the I value
ISO 80000-3	SI units and quantities — defines moment of inertia symbols and units

Reference: I_x for common cross-sections

Shape	Dimensions	I_x (in⁴)	I_x (mm⁴ ×10⁶)
Solid circle	D = 4 in (100 mm)	12.57	4.91
Hollow pipe (Sched 40)	4 in NPS (D = 4.5, t = 0.237)	7.23	3.01
Solid square	4 × 4 in (100 × 100 mm)	21.33	8.33
HSS rectangular	HSS 6 × 4 × 1/4	20.9	8.70
W-shape (steel I)	W12×26	204	84.9
W-shape (steel I)	W18×35	510	212
W-shape (steel I)	W24×62	1 550	645
IPE (Eurocode)	IPE 300	198	83.6
Wood joist (NDS)	2 × 12 (1.5 × 11.25 in)	177.98	74.1

Identify the cross-section shape and axis Pick the shape (circle, rectangle, I-beam, T-beam, pipe) and the bending axis (usually the strong axis x–x for beams). For composite sections, list each rectangular sub-shape and its centroid offset from the overall neutral axis.
Apply the closed-form formula For a solid circle of diameter D: I = πD⁴ / 64. For a rectangle b × h: I = bh³/12 about the centroidal axis. For an I-beam: I = (B·H³ − b·h³) / 12 where B,H are outer flange/depth and b,h are the open web cavity. Pull from the section table for standard W-, HSS-, IPE-shapes.
Add parallel-axis correction for off-centroid pieces Huygens' parallel-axis theorem: I = I_centroid + A · d², where d is the distance from each sub-area's own centroid to the composite section's neutral axis. Apply once per sub-piece, then sum. Without this step composite sections are systematically under-strength.
Convert units if needed I has units of length⁴ — in⁴ (US), mm⁴ (metric SI), m⁴ (large structures). Conversions: 1 in⁴ = 416 231 mm⁴ = 4.162 × 10⁻⁷ m⁴. The AISC manual lists I in in⁴; Eurocode tables in mm⁴.
Check against bending stress Compute σ = M·c / I, where M is the bending moment, c is the distance from neutral axis to extreme fiber. The whole point of computing I is so this stress check works. If σ exceeds Fy/Ω (ASD) or φFy (LRFD), the section is undersized.

Worked example — built-up T-beam

Compute I_x for a built-up T-beam: top flange 6 in × 1 in plate, web 1 in × 8 in plate. The T is symmetric left-right but not top-bottom; you must first locate the centroid before applying the inertia formulas.

Sub-areas: A₁ (flange) = 6 × 1 = 6 in²; A₂ (web) = 1 × 8 = 8 in². Total A = 14 in².
Centroid from the top fibre: ȳ = (6 × 0.5 + 8 × 5.0) / 14 = 3.07 in.
Centroidal I of each rectangle: I₁ = 6·1³/12 = 0.5 in⁴; I₂ = 1·8³/12 = 42.7 in⁴.
Distances to overall neutral axis: d₁ = 3.07 − 0.5 = 2.57 in; d₂ = 5.0 − 3.07 = 1.93 in.
Parallel-axis additions: 6·2.57² = 39.6 in⁴; 8·1.93² = 29.8 in⁴.
Total I_x: 0.5 + 42.7 + 39.6 + 29.8 = 112.6 in⁴.

The same answer drops out of the composite-centroid calculator in two clicks — but doing it by hand once is the only way to internalise why the parallel-axis correction matters.

Area moment of inertia vs. mass moment of inertia vs. polar moment

Property	Symbol	Units	Used for
Area moment of inertia	I, I_x, I_y	length⁴ (in⁴, mm⁴)	Bending stress / deflection of beams
Polar (are