How to find moment of inertia of a beam?

Three paths. (1) Look it up: for a steel W-, S-, HP-, or HSS-shape, AISC Manual Part 1 lists I_x and I_y in in⁴ — W18×35 is I_x = 510 in⁴. For dressed sawn lumber, NDS Supplement Table 1B lists I per nominal size — a 2 × 12 is I_x = 178 in⁴. (2) Use a closed-form formula for primitive shapes: rectangle b × h is I_x = bh³/12; I-beam (B·H³ − b·h³)/12. (3) For a built-up section, split into rectangles, compute each piece's centroidal I = bh³/12, add A·d² parallel-axis correction, sum.

What is the moment of inertia formula for a rectangular beam?

For a rectangle of width b and depth h about the centroidal x-axis (perpendicular to the depth direction): I_x = b·h³/12. About the y-axis: I_y = h·b³/12. The factor of h³ in the strong-axis formula is why deeper beams are dramatically stiffer than wider ones — doubling depth gives 8× the stiffness, while doubling width gives only 2×.

What is the moment of inertia of a 2x12?

A nominal 2 × 12 sawn-lumber joist has dressed dimensions of 1.5 × 11.25 in (NDS rule). I_x about the strong (bending) axis = 1.5 × 11.25³ / 12 = 178.0 in⁴ (74.1 × 10⁶ mm⁴). I_y about the weak axis = 11.25 × 1.5³ / 12 = 3.16 in⁴. The strong-axis value is 56× the weak-axis — orient joists with the long dimension vertical, always.

How does the moment of inertia affect beam deflection?

Deflection is inversely proportional to I — double the moment of inertia and the deflection halves at the same load. For a simply-supported beam under uniform load w: δ_max = 5·w·L⁴/(384·E·I). The L⁴ in the numerator says span dominates; the I in the denominator says cross-section depth is the second lever. For a 20-ft floor joist on a 40 psf live load, jumping from a 2 × 10 (I = 98.9 in⁴) to a 2 × 12 (I = 178 in⁴) cuts deflection by 45 %.

How to find moment of inertia of a beam?

Three paths. (1) Look it up: for a steel W-, S-, HP-, or HSS-shape, AISC Manual Part 1 lists I_x and I_y in in⁴ — W18×35 is I_x = 510 in⁴. For dressed sawn lumber, NDS Supplement Table 1B lists I per nominal size — a 2 × 12 is I_x = 178 in⁴. (2) Use a closed-form formula for primitive shapes: rectangle b × h is I_x = bh³/12; I-beam (B·H³ − b·h³)/12. (3) For a built-up section, split into rectangles, compute each piece's centroidal I = bh³/12, add A·d² parallel-axis correction, sum.

What is the moment of inertia formula for a rectangular beam?

For a rectangle of width b and depth h about the centroidal x-axis (perpendicular to the depth direction): I_x = b·h³/12. About the y-axis: I_y = h·b³/12. The factor of h³ in the strong-axis formula is why deeper beams are dramatically stiffer than wider ones — doubling depth gives 8× the stiffness, while doubling width gives only 2×.

What is the moment of inertia of a 2x12?

A nominal 2 × 12 sawn-lumber joist has dressed dimensions of 1.5 × 11.25 in (NDS rule). I_x about the strong (bending) axis = 1.5 × 11.25³ / 12 = 178.0 in⁴ (74.1 × 10⁶ mm⁴). I_y about the weak axis = 11.25 × 1.5³ / 12 = 3.16 in⁴. The strong-axis value is 56× the weak-axis — orient joists with the long dimension vertical, always.

How does the moment of inertia affect beam deflection?

Deflection is inversely proportional to I — double the moment of inertia and the deflection halves at the same load. For a simply-supported beam under uniform load w: δ_max = 5·w·L⁴/(384·E·I). The L⁴ in the numerator says span dominates; the I in the denominator says cross-section depth is the second lever. For a 20-ft floor joist on a 40 psf live load, jumping from a 2 × 10 (I = 98.9 in⁴) to a 2 × 12 (I = 178 in⁴) cuts deflection by 45 %.

Reference · Structural · AISC · NDS · Euler–Bernoulli

Moment of Inertia in Beams — Formulas, Steel & Wood Tables

The moment of inertia of a beam is the cross-section property that controls both bending stress (σ = M·c/I) and deflection (δ ∝ 1/I). This page covers the closed-form formulas for rectangular wood beams, AISC W-shape steel beams, and HSS structural tubes; lists I-values for the common sections from sawn lumber to W24×62; and walks through a worked floor-joist deflection example. Reviewed by a licensed PE.

Moment of inertia in beams calculator

For built-up beam cross-sections (flitch, glulam, hybrid steel-wood), the composite-centroid calculator computes I, S, and the centroid coordinates from a list of rectangles. For a single primitive (rectangular wood beam, steel I, hollow tube), the dedicated moment-of-inertia tool uses the closed-form formulas. Both calculators include parallel-axis correction.

→ Composite centroid & inertia calculator · → Single-shape inertia calculator

Beam moment of inertia formulas

Eq. 01 — Rectangular beam (centroidal axis) SI

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

·: b = width (perpendicular to load)
·: h = depth (parallel to load)
·: Sawn-lumber dressed dims per NDS (2 × 12 → 1.5 × 11.25 in)

Eq. 02 — Standard steel I-beam (W-shape) 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
·: Or look up directly in AISC Manual Part 1

Eq. 03 — Hollow rectangular tube (HSS) SI

I_x = \frac{B \, H^3 - (B - 2t)(H - 2t)^3}{12}

·: B, H = outer width and depth
·: t = wall thickness (use 0.93 × nominal for ASTM A500, full for A1085)
·: AISC HSS tables list pre-computed I for every standard section

Eq. 04 — Bending stress (Euler–Bernoulli) SI

\sigma = \frac{M \, c}{I}

·: M = bending moment at the section
·: c = distance from neutral axis to extreme fibre
·: I = beam moment of inertia about the bending axis

Eq. 05 — Maximum deflection (simply supported, uniform load) SI

\delta_{max} = \frac{5 \, w \, L^4}{384 \, E \, I}

·: w = uniform load per unit length
·: L = span
·: E = modulus of elasticity (29 000 ksi steel, 1 600 ksi typical SYP wood)

Standards governing beam design

Document	Scope
AISC 360-22	Steel building specification — Chapter F (flexural members) consumes I via M_n = Fy·Z and elastic checks
AISC Steel Construction Manual (16th ed.)	Section properties (I_x, I_y, S, Z, r) for every W, S, HP, HSS, C, MC, L
ANSI/AWC NDS-2018	National Design Specification for Wood Construction — sawn lumber, glulam, LVL section properties
NDS Supplement	Table 1A (rough) and Table 1B (dressed) sawn-lumber dimensions, including I and S
IBC 1604.3 / Table 1604.3	Deflection limits: floor live load L/360, roof live load L/240, total load L/240 floor / L/180 roof
EN 1993-1-1 (Eurocode 3)	European steel design — uses I_y / I_z (different axis convention from AISC)

Reference: I_x for common beam sections

Beam	Material	Depth (in)	I_x (in⁴)	S_x (in³)
2 × 6 sawn (1.5 × 5.5)	Wood (NDS)	5.5	20.80	7.56
2 × 8 sawn (1.5 × 7.25)	Wood	7.25	47.63	13.14
2 × 10 sawn (1.5 × 9.25)	Wood	9.25	98.93	21.39
2 × 12 sawn (1.5 × 11.25)	Wood	11.25	177.98	31.64
11-7/8" LVL 1.75"	Engineered wood	11.875	244	41.1
W8×10	Steel	7.89	30.8	7.81
W12×26	Steel	12.22	204	33.4
W18×35	Steel	17.7	510	57.6
W21×50	Steel	20.8	984	94.5
W24×62	Steel	23.7	1 550	131
HSS 8×4×1/4	Steel	8.0	43.5	10.9
HSS 12×4×3/8	Steel	12.0	137	22.8

Identify the beam cross-section Steel: pull the AISC shape designation (W18×35, HSS 6×4×3/8, C9×15). Wood: read the dressed dimensions (a "2 × 12" is actually 1.5 × 11.25 in per NDS). Composite (e.g. flitch beam, glulam, LVL): list each lamination as its own rectangle.
Look up or compute I_x For standard steel shapes, use AISC Manual Part 1 — W18×35 has I_x = 510 in⁴. For sawn lumber, use NDS Supplement Table 1B — a 2 × 12 has I_x = 178 in⁴. For a built-up section, decompose into rectangles and apply Huygens parallel-axis theorem.
Apply Euler–Bernoulli beam theory For a simply-supported beam under point load P at midspan: maximum moment M = PL/4; maximum bending stress σ = M·c/I; maximum deflection δ = PL³/(48·E·I). For uniform load w: M = wL²/8; δ = 5wL⁴/(384·E·I).
Check stress against allowable Steel LRFD: Mu ≤ φMn = φ·Fy·Z (Z is plastic section modulus, AISC Eq F2-1). Steel ASD: M ≤ Mn/Ω. Wood: σ_b = M/S ≤ F'b adjusted with NDS adjustment factors (CD, CM, Ct, CL, etc.).
Check deflection against serviceability Live-load deflection limit: L/360 for floors, L/240 for roof (IBC 1604.3 / Table 1604.3). Solve δ ≤ L/360 for the required I; then choose the smallest section whose I exceeds that.

Worked example — 16 ft floor joist deflection

A 16 ft (4.88 m) simply-supported residential floor joist carries 40 psf live load + 10 psf dead load on 16 in (406 mm) o.c. spacing. Pick the smallest 2× sawn-lumber size that meets L/360 live-load deflection.

Loads per joist: w = (40 + 10) psf × 16/12 ft = 66.7 plf total; w_LL = 53.3 plf.
Live-load deflection limit: L/360 = 16 × 12 / 360 = 0.533 in.
Required I (E = 1 600 000 psi for SYP No. 2): δ = 5·w·L⁴/(384·E·I) → solve for I → I_required = 5 × (53.3/12) × (192)⁴ / (384 × 1.6e6 × 0.533) = 92.8 in⁴.
Pick a section: 2 × 10 (I = 98.9 in⁴) works; 2 × 12 (I = 178) gives 45 % more margin.
Verify bending stress: M = wL²/8 = 66.7 × 16² / 8 = 2 134 ft·lb = 25 600 in·lb; σ = M·c/I = 25 600 × 4.625 / 98.9 = 1 197 psi → less than F\'b ≈ 1 200 psi adjusted for SYP No. 2. Marginal — go with 2 × 12 for safety.

Comparison — wood beam vs. steel beam vs. engineered LVL

Aspect	Sawn lumber 2 × 12	LVL 1.75 × 11-7/8	Steel W12×14
I_x (in⁴)	178	244	88.6
E (ksi)	1 600	2 000	29 000
E·I (in²·kip)	285 × 10³	488 × 10³	2 569 × 10³
Self-weight (plf)	4.3	5.6	14.0
Best for	Residential up to 16 ft	Long single-family spans, headers	Commercial / industrial > 20 ft

Variants and related queries

Inertia of a beam — same property, different name

"Inertia of a beam" is engineering shorthand for the area moment of inertia of the beam\'s cross-section. The same property also goes by "second moment of area" (UK / Eurocode practice) and "area moment of inertia" (US engineering). All three describe the same I = ∫y² dA integral and use the same formulas.

Beam — what counts as one

In structural engineering, a beam is any horizontal or near-horizontal member that carries transverse loads through bending. Joists (closely spaced beams supporting floor or roof sheathing), girders (large beams supporting joists), purlins (roof beams supporting decking), lintels (beams over openings), and rafters (sloped roof beams) are all beams. Each gets the same M·c/I bending check.

Moment — distinguishing the moments

"Moment" can mean three different things in beam design: bending moment M (kip-ft, units of force × length), area moment of inertia I (in⁴, units of length⁴), and section modulus S = I/c (in³). They are linked by σ = M·c/I = M/S. When someone says "the moment in the beam is 50 kip-ft", they mean M; when they say "the moment of inertia is 510 in⁴", they mean I.

Laterally braced beam

A beam is "laterally braced" when its compression flange is restrained against lateral-torsional buckling. Concrete slabs, metal decking with positive shear connection, and discrete bracing at predetermined intervals all qualify. AISC F2 applies the full plastic capacity Mn = Fy·Z only when the unbraced length Lb ≤ Lp; longer Lb reduces capacity per AISC F2-2.

Frequently asked questions

How to find moment of inertia of a beam?: Three paths. (1) Look it up: for a steel W-, S-, HP-, or HSS-shape, AISC Manual Part 1 lists I_x and I_y in in⁴ — W18×35 is I_x = 510 in⁴. For dressed sawn lumber, NDS Supplement Table 1B lists I per nominal size — a 2 × 12 is I_x = 178 in⁴. (2) Use a closed-form formula for primitive shapes: rectangle b × h is I_x = bh³/12; I-beam (B·H³ − b·h³)/12. (3) For a built-up section, split into rectangles, compute each piece's centroidal I = bh³/12, add A·d² parallel-axis correction, sum.
What is the moment of inertia formula for a rectangular beam?: For a rectangle of width b and depth h about the centroidal x-axis (perpendicular to the depth direction): I_x = b·h³/12. About the y-axis: I_y = h·b³/12. The factor of h³ in the strong-axis formula is why deeper beams are dramatically stiffer than wider ones — doubling depth gives 8× the stiffness, while doubling width gives only 2×.
What is the moment of inertia of a 2x12?: A nominal 2 × 12 sawn-lumber joist has dressed dimensions of 1.5 × 11.25 in (NDS rule). I_x about the strong (bending) axis = 1.5 × 11.25³ / 12 = 178.0 in⁴ (74.1 × 10⁶ mm⁴). I_y about the weak axis = 11.25 × 1.5³ / 12 = 3.16 in⁴. The strong-axis value is 56× the weak-axis — orient joists with the long dimension vertical, always.
How does the moment of inertia affect beam deflection?: Deflection is inversely proportional to I — double the moment of inertia and the deflection halves at the same load. For a simply-supported beam under uniform load w: δ_max = 5·w·L⁴/(384·E·I). The L⁴ in the numerator says span dominates; the I in the denominator says cross-section depth is the second lever. For a 20-ft floor joist on a 40 psf live load, jumping from a 2 × 10 (I = 98.9 in⁴) to a 2 × 12 (I = 178 in⁴) cuts deflection by 45 %.

The Euler–Bernoulli beam equation — historic source

The curvature of an elastic line at any point is proportional to the moment of the bending forces at that point and inversely proportional to the product of the modulus of elasticity and the moment of inertia of the cross-section about the neutral axis.

Euler, L. — Methodus Inveniendi Lineas Curvas → Additamentum I, De Curvis Elasticis (Lausanne, 1744)

Related calculators and references

Sources and further reading

AISC 360-22 — Specification for Structural Steel Buildings, Chapter F (flexural members).
AISC Steel Construction Manual, 16th edition — Part 1 (section properties) and Part 3 (deflection design).
ANSI/AWC NDS-2018 — National Design Specification for Wood Construction; NDS Supplement Table 1B.
IBC 2021 §1604.3 and Table 1604.3 — beam deflection limits.
Hibbeler, R. C. Mechanics of Materials, 10th edition. Pearson, 2017 — Chapter 6 (bending) and Chapter 12 (deflection).
Euler, L. Methodus Inveniendi Lineas Curvas, Additamentum I. Lausanne, 1744 — original beam theory.