Moment of Inertia Calculator — Free Ix and Iy for Any Shape

The moment of inertia (also called second moment of area) measures a cross-section's resistance to bending. It is the single most important property for beam design: it determines deflection, bending stress distribution, and whether a member is adequate for its span.

This page covers formulas, worked examples, and a reference table for the moment of inertia of common structural steel shapes. Use the section properties database to look up Ix and Iy for any AISC W-shape, HSS, UB, IPE, or HEA section instantly.

What Is Moment of Inertia?

The moment of inertia I of a cross-sectional area about a given axis is defined as:

I = ∫ y² dA

where y is the perpendicular distance from the axis to the infinitesimal area element dA. The units are length to the fourth power (in⁴, mm⁴, cm⁴, or m⁴).

In structural engineering, two values matter:

• Ix (strong axis): moment of inertia about the horizontal centroidal axis. Governs bending about the strong axis of a beam.
• Iy (weak axis): moment of inertia about the vertical centroidal axis. Governs weak-axis bending and lateral-torsional buckling resistance.

A higher moment of inertia means less deflection under the same load. This is why I-beams are shaped the way they are: material is concentrated at the flanges (far from the neutral axis) to maximize Ix while minimizing weight.

Formulas for Common Shapes

Rectangle (breadth b, depth d)

Circle (diameter D)

Hollow Circle / Pipe (outside D, inside d)

Triangle (base b, height h)

Semicircle (radius r)

I-Beam / Wide Flange Formulas

For a W-shape with flange width bf, flange thickness tf, web height h, and web thickness tw:

Ix = (bf × d³)/12 - (bf - tw)(d - 2tf)³/12

where d = h + 2tf is the total depth.

Iy = 2 × (tf × bf³)/12 + (h × tw³)/12

These formulas assume the section is symmetric about both axes, which is true for standard W-shapes.

T-Section Formulas

For a T-section cut from a W-shape (stem height hs, stem thickness ts, flange width bf, flange thickness tf):

1. Find the centroid: ȳ = (A_flange × tf/2 + A_stem × (tf + hs/2)) / (A_flange + A_stem)
2. Use the parallel axis theorem to compute Ix about the centroid

This is more involved because the centroid is not at the mid-depth. See the parallel axis theorem section below.

Parallel Axis Theorem

The parallel axis theorem transfers a moment of inertia from a centroidal axis to a parallel axis:

I = Ic + A × d²

where:

• Ic = moment of inertia about the centroidal axis
• A = area of the cross-section
• d = perpendicular distance between the two parallel axes

This is essential for built-up sections, composite members, and any shape that is not symmetric about the axis of interest.

Example: Two Plates Welded to Form a T

Consider a T-section made from a 6" × 3/8" flange plate and a 8" × 3/8" stem plate:

1. Flange area: 6 × 0.375 = 2.25 in²
2. Stem area: 8 × 0.375 = 3.0 in²
3. Total area: 5.25 in²
4. Centroid from bottom of stem: ȳ = (2.25 × 8.375 + 3.0 × 4.0) / 5.25 = 5.83 in
5. Ix_flange (own centroid) = 6 × 0.375³/12 = 0.0264 in⁴
6. Ix_stem (own centroid) = 0.375 × 8³/12 = 16.0 in⁴
7. Ix_total = 0.0264 + 2.25 × (8.375 - 5.83)² + 16.0 + 3.0 × (5.83 - 4.0)² = 35.3 in⁴

Worked Example: W12x26

A W12x26 is one of the most popular light beam sections. Let's verify its moment of inertia from the AISC database.

Given (from AISC Table 1-1):

• d = 12.22 in (total depth)
• bf = 6.490 in (flange width)
• tf = 0.570 in (flange thickness)
• tw = 0.260 in (web thickness)
• AISC listed Ix = 204 in⁴

Calculation:

Web height: hw = d - 2tf = 12.22 - 2(0.570) = 11.08 in

Ix = (6.490 × 12.22³)/12 - (6.490 - 0.260)(11.08³)/12 = (6.490 × 1824.7)/12 - (6.230 × 1360.4)/12 = 987.5 - 706.9 = 280.6 in⁴

Wait — this doesn't match the AISC value of 204 in⁴. The discrepancy arises because W-shapes have fillet radii at the web-flange junction that remove material. The simplified rectangular formula overestimates Ix by about 38%. This is why you always use the AISC published value for design, not a hand calculation.

Takeaway: For real W-shapes, always look up Ix from the section properties database. The formulas above are exact for rectangles and circles but approximate for rolled shapes with fillets.

Moment of Inertia for Top 10 Popular W-Shapes

For the full database of 500+ shapes, use the free section properties tool.

Section Modulus vs Moment of Inertia

The section modulus S is derived from the moment of inertia:

S = I / c

where c is the distance from the neutral axis to the extreme fiber. For a symmetric section, c = d/2.

• Sx (elastic section modulus): used for allowable stress design and elastic analysis
• Zx (plastic section modulus): used for LRFD plastic design. Always larger than Sx.

For a rectangle: Sx = bd²/6 and Zx = bd²/4.

The relationship I → S → bending capacity is the core of beam design. Higher I means less deflection. Higher S means higher allowable moment.

Composite Section Moment of Inertia

When calculating I for a composite section (steel beam + concrete slab, or built-up sections), use the modular ratio n = Es / Ec where Es is the steel elastic modulus (29,000 ksi) and Ec is the concrete elastic modulus.

Transform the concrete flange to an equivalent steel width: beq = bc / n

Then calculate I using the transformed section with standard formulas.

Typical values: for f'c = 4 ksi concrete, Ec ≈ 3,605 ksi, so n ≈ 8. For f'c = 5 ksi, n ≈ 7.

Radius of Gyration

The radius of gyration r relates I to the cross-sectional area:

r = √(I/A)

This is used in column design to determine the slenderness ratio KL/r. A higher r means the section is more resistant to buckling.

For W-shapes, rx (strong axis) is typically much larger than ry (weak axis). This is why columns buckle about the weak axis unless braced.

Frequently Asked Questions

What is the moment of inertia of a W12x26? Ix = 204 in⁴ and Iy = 17.3 in⁴ per AISC Table 1-1. These values include the effect of fillet radii.

How do I calculate moment of inertia for an irregular shape? Break the shape into simple rectangles and circles. Calculate I for each piece about its own centroid, then use the parallel axis theorem (I = Ic + Ad²) to transfer to the overall centroid.

What is the difference between Ix and Iy? Ix is about the horizontal centroidal axis (strong axis for a typical beam). Iy is about the vertical centroidal axis (weak axis). Ix is always larger for W-shapes because flanges are far from the strong axis.

What units is moment of inertia measured in? In US customary units: in⁴. In SI: mm⁴, cm⁴, or m⁴. Conversion: 1 in⁴ = 416,231 mm⁴.

Does moment of inertia depend on material? No. Moment of inertia is a purely geometric property. It depends only on the shape and dimensions of the cross-section, not on whether it is steel, aluminum, concrete, or wood.

What is the parallel axis theorem used for? It transfers a moment of inertia from a centroidal axis to any parallel axis. Essential for built-up sections, composite beams, and finding I about non-centroidal axes.

Why is Ix so much larger than Iy for an I-beam? Because the flanges (where most of the area is) are located far from the strong (x) axis but close to the weak (y) axis. The I = ∫y²dA integral grows with the square of distance.

How does moment of inertia affect deflection? Deflection is inversely proportional to EI. Double the moment of inertia, halve the deflection. This is why deeper beams deflect less.

Can I use the section properties tool instead of hand calculations? Yes. The section properties database has Ix, Iy, Sx, Sy, Zx, Zy, rx, ry, and J for all standard shapes. Always use published values for design.

Related Pages

• Section Properties Database — Ix, Iy, Sx, Sy, Zx, Zy for 500+ shapes
• Moment of Inertia Calculator Tool — Interactive calculator for custom shapes
• Steel Beam Sizes Chart — Dimensions and properties for W, UB, IPE, HEA
• Beam Deflection Calculator — Uses Ix in deflection checks
• Beam Capacity Calculator — Flexure checks using Sx and Zx

Moment of Inertia in Deflection Calculations

The moment of inertia directly controls beam deflection through the flexural stiffness EI. For a simply supported beam with uniformly distributed load:

Δ = 5wL⁴ / (384EI)

If you need to limit deflection to L/360:

I_required = 5wL³ × 360 / (384E) = 5wL³ / (1.067E)

For E = 29,000 ksi:

I_required (in⁴) = 5wL³ / (1.067 × 29,000) = wL³ / 6,189

where w is in kips/in and L is in inches. Or with w in kips/ft and L in feet:

I_required = 125wL³ / 48EI × conversion factors

This is the most common use of moment of inertia in practice: selecting a beam deep enough (large enough I) to meet deflection limits.

Worked Deflection Example

A simply supported beam spans 25 feet and carries a service live load of 0.8 kips/ft. Deflection limit is L/360.

L/360 = 25 × 12 / 360 = 0.833 in

I_required = 5 × (0.8/12) × (25×12)⁴ / (384 × 29000) = 5 × 0.0667 × 324,000,000 / 11,059,200 = 108,000,000 / 11,059,200 = 97.6 in⁴

A W12x26 with Ix = 204 in⁴ works with comfortable margin (204 > 97.6). The actual deflection would be:

Δ = 5 × 0.0667 × (300)⁴ / (384 × 29000 × 204) = 97.6 × (97.6/204) ≈ 0.388 in

Well within L/360 = 0.833 in.

Moment of Inertia for Hollow Structural Sections (HSS)

HSS sections are commonly used as columns and beams where torsional stiffness matters.

Square HSS (outside B × B, wall thickness t)

Ix = (B⁴ - (B-2t)⁴) / 12

Rectangular HSS (outside B × D, wall thickness t)

Ix = (BD³ - (B-2t)(D-2t)³) / 12 Iy = (DB³ - (D-2t)(B-2t)³) / 12

Round HSS (outside D, wall thickness t)

Ix = π(D⁴ - (D-2t)⁴) / 64

For example, HSS6x6x3/8:

• B = 6 in, t = 0.375 in, inside = 5.25 in
• Ix = (6⁴ - 5.25⁴)/12 = (1296 - 759.3)/12 = 44.7 in⁴
• AISC lists Ix = 46.1 in⁴ (slightly higher due to corner radii)

Moment of Inertia and Torsion

The torsional moment of inertia J (also called the polar moment of inertia for circular sections) resists twisting:

• Solid circle: J = πD⁴/32
• Hollow circle: J = π(D⁴ - d⁴)/32
• Thin-walled tube: J = 2πr³t
• Rectangle: J ≈ βbt³ (where β depends on b/t ratio)
• W-shapes: J is very small and published in AISC Table 1-1

For W-shapes, J is typically 0.1 to 2.0 in⁴, much smaller than Ix. This is why open sections (W, C, L) are poor in torsion compared to closed sections (HSS, pipe).

Using the Section Properties Database

The free section properties tool provides instant lookup for:

• All 148 AISC W-shapes (W4 through W44)
• 73 HSS shapes (square and rectangular)
• 26 UB shapes (AS 4100)
• 18 IPE shapes (EN 1993)
• 38 HEA/HEB shapes (EN 1993)

For each shape, the database lists Ix, Iy, Sx, Sy, Zx, Zy, rx, ry, J, and the torsional constant Cw. No manual calculation needed.

Frequently Asked Questions (Extended)

Why does the hand-calculated Ix not match the AISC value for W-shapes? W-shapes have fillet radii at the web-flange junction. These rounded corners remove a small amount of material compared to the idealized rectangular model. The difference is typically 2-5% for large sections and up to 15-20% for smaller sections.

What is the product moment of inertia Ixy? Ixy = ∫xy dA. For symmetric sections (W-shapes about both axes), Ixy = 0. For asymmetric sections (angles, channels), Ixy is nonzero and must be considered in biaxial bending calculations.

How does I change when a beam is oriented at an angle? Use the rotation formula: Ix' = (Ix + Iy)/2 + (Ix - Iy)/2 × cos(2θ) - Ixy × sin(2θ). For W-shapes with Ixy = 0, this simplifies.

What is the difference between second moment of area and mass moment of inertia? Second moment of area (I, in⁴) resists bending. Mass moment of inertia (I_mass, lb·in·s²) resists angular acceleration. They share the same mathematical structure but apply to different physical problems.

How do I compute I for a plate girder with varying web depth? Break the girder into segments with constant depth. Compute I for each segment. At transition points, check both the deeper and shallower section.

What is the warping constant Cw? Cw measures a thin-walled open section's resistance to warping torsion. For W-shapes, Cw = (tf × bf³ × (d-tf)²) / 24 approximately. Important for lateral-torsional buckling checks.

Can I calculate I for a composite steel-concrete beam? Yes. Transform the concrete slab to equivalent steel using n = Es/Ec (typically 7-9). Then compute I of the transformed section using standard formulas and the parallel axis theorem.

Related Pages

• Section Properties Database — Ix, Iy, Sx, Sy, Zx, Zy for 500+ shapes
• Moment of Inertia Calculator Tool — Interactive calculator for custom shapes
• Steel Beam Sizes Chart — Dimensions and properties for W, UB, IPE, HEA
• Beam Deflection Calculator — Uses Ix in deflection checks
• Beam Capacity Calculator — Flexure checks using Sx and Zx

Disclaimer

This is a calculation tool, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in construction, fabrication, or permit documents. The user is responsible for the accuracy of all inputs and the verification of all outputs.