Free Moment of Inertia Calculator — Ix, Iy, and J for Any Cross-Section

The moment of inertia (second moment of area) is the single property that determines how much a beam deflects under load and how bending stress distributes through its cross-section. If you are selecting a beam, checking a deflection limit, or designing a built-up section, the number you reach for first is Ix.

This page is the definitive reference for calculating the moment of inertia of structural steel shapes. It covers formulas for every common cross-section, the parallel axis theorem with fully worked examples, the distinction between moment of inertia and section modulus, composite section analysis, and the polar moment of inertia J for torsion. Every formula is accompanied by a numeric example using real dimensions.

For instant lookups of published AISC values, use the section properties database. For custom geometry calculations, use the moment of inertia calculator.

What Is Moment of Inertia and Why Does It Matter

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

I = integral of y-squared dA

where y is the perpendicular distance from the axis to each infinitesimal area element dA. Because the distance is squared, material placed far from the axis contributes disproportionately to I. This is the entire reason I-beams exist: the flanges carry most of the area and are positioned at the maximum practical distance from the neutral axis, producing a high Ix for a given weight.

Units are length to the fourth power: in^4 in US customary, mm^4 or cm^4 in SI. Conversion: 1 in^4 = 416,231 mm^4 = 41.62 cm^4.

Three values matter in steel design:

• Ix (strong-axis moment of inertia): about the horizontal centroidal axis. Governs vertical bending deflection and strong-axis flexural stress. For a typical W-shape beam spanning horizontally, this is the controlling property.
• Iy (weak-axis moment of inertia): about the vertical centroidal axis. Governs weak-axis bending and, critically, influences lateral-torsional buckling resistance through the ratio Iy/Ix.
• J (torsional constant, sometimes called the polar moment of inertia): resists twisting about the longitudinal axis. Open sections (W, C, L) have very small J values. Closed sections (HSS, pipe) have large J values and are far superior in torsion.

The moment of inertia is a purely geometric property. It does not depend on material grade, yield strength, or modulus of elasticity. A W12x26 has Ix = 204 in^4 whether it is A992 steel, A36 steel, or 6061-T6 aluminum. The material affects capacity (through Fy and E), but I is shape alone.

Moment of Inertia Formulas for Common Shapes

Solid Rectangle (breadth b, depth d)

The most fundamental formula in structural engineering. Every other shape's I is built from rectangles.

The d^3 term means depth is dramatically more influential than breadth. Doubling the depth multiplies Ix by 8. Doubling the breadth only doubles it. This is why beams are deep, not wide.

Numeric example: A solid rectangular bar 4 in wide by 10 in deep has Ix = 4 x 10^3 / 12 = 4 x 1000 / 12 = 333.3 in^4. The same bar turned on its side (10 in wide by 4 in deep) has Ix = 10 x 64 / 12 = 53.3 in^4 -- a factor of 6.25 reduction.

Solid Circle (diameter D)

Numeric example: A 4-inch diameter solid round bar has I = pi x 4^4 / 64 = pi x 256 / 64 = 12.57 in^4. J = pi x 256 / 32 = 25.13 in^4.

Hollow Circle / Pipe (outside diameter D, inside diameter d)

Numeric example: A 6-inch Schedule 40 pipe has D = 6.625 in, d = 6.065 in (0.280 in wall). I = pi x (6.625^4 - 6.065^4) / 64 = pi x (1926 - 1353) / 64 = 28.1 in^4. This is only about 12% less than a solid 6.625-inch bar (I = 94.7 in^4), despite being 83% lighter.

Hollow Rectangular / Square Section (outside B x D, wall thickness t)

For an HSS rectangular section with uniform wall thickness t:

Ix = [B D^3 - (B - 2t)(D - 2t)^3] / 12

Iy = [D B^3 - (D - 2t)(B - 2t)^3] / 12

Numeric example: HSS 8x4x1/4 has B = 4 in, D = 8 in, t = 0.25 in. Inside dimensions: B_inner = 3.5 in, D_inner = 7.5 in. Ix = [4 x 8^3 - 3.5 x 7.5^3] / 12 = [4 x 512 - 3.5 x 421.9] / 12 = [2048 - 1476.6] / 12 = 47.6 in^4. The published AISC value is approximately 48.5 in^4, with the small difference due to the exterior corner radii.

For a square HSS (B = D), Ix = Iy.

I-Beam / W-Shape (flange width bf, flange thickness tf, total depth d, web thickness tw)

The W-shape is approximated as three rectangles: two flanges and one web. Subtract the rectangles that are not present using the void method:

Ix = [bf x d^3 - (bf - tw) x (d - 2tf)^3] / 12

Iy = [2 x tf x bf^3 + (d - 2tf) x tw^3] / 12

These are approximate. Rolled W-shapes have fillet radii at the web-flange junction and tapered flange tips that remove material. Published AISC values include these effects and are 2-15% lower than the three-rectangle approximation, with the larger discrepancies occurring on smaller, lighter sections where the fillets represent a larger fraction of the total area.

Channel (C-Shape)

A channel has one axis of symmetry (y-axis through the web). The centroid in the x-direction is offset from the web face.

Ix = [bf x d^3 - (bf - tw) x (d - 2tf)^3] / 12 (same form as W-shape)

Iy: Requires finding the centroidal y-axis first. The centroid is located at distance x-bar from the back of the web:

x-bar = [tw x d x (tw/2) + 2 x tf x (bf - tw) x (tw + (bf - tw)/2)] / A_total

Iy is then computed about this centroidal axis using the parallel axis theorem for all three rectangles.

Angle (L-Shape)

An angle has no axis of symmetry. Both principal axes are rotated relative to the leg axes. The product moment of inertia Ixy is nonzero, and principal moments I_max and I_min must be found through Mohr's circle transformation:

I_max, I_min = (Ix + Iy)/2 +/- sqrt[((Ix - Iy)/2)^2 + Ixy^2]

where Ix and Iy are computed about the centroidal axes parallel to the legs. The principal axis angle theta_p = 0.5 x atan2(-2 x Ixy, Ix - Iy).

For single-angle struts and ties, the principal axis moments of inertia govern buckling and bending. Published AISC values for angles provide Ix, Iy, Ixy, Iz (minor principal), and Iw (major principal) in Table 1-7.

T-Section (WT, ST)

A T-section cut from a W-shape requires finding the centroid first. The centroid is always below the flange-web junction because the flange has more area per unit depth. The calculation proceeds in three steps:

1. Determine the area of each element and the total area.
2. Locate the centroid: y-bar = sum of (Ai x yi) / A_total, measured from the bottom of the stem.
3. Compute Ix about the centroid using the parallel axis theorem for each element.

Full worked example below.

Triangle (base b, height h)

Parallel Axis Theorem

The parallel axis theorem transfers a moment of inertia from a centroidal axis of an individual element to any parallel axis. It is the workhorse of composite section analysis.

I = I_c + A x d^2

where I_c is the moment of inertia of the element about its own centroidal axis, A is the area of that element, and d is the perpendicular distance between the element's centroidal axis and the target axis.

The d^2 term means that even a small area placed far from the target axis can contribute substantially to I. This explains why adding cover plates to the flanges of a beam is far more effective than thickening the web: the cover plates are at the maximum distance d from the neutral axis.

Worked Example 1: Built-Up T-Section from Two Plates

A T-section is fabricated by welding a 6-inch by 3/8-inch flange plate to an 8-inch by 3/8-inch stem plate. Compute Ix about the centroidal axis of the built-up section.

Step 1: Areas and element centroids

Flange: A_f = 6 x 0.375 = 2.25 in^2. Its own centroid is at y_f = 8 + 0.375/2 = 8.1875 in from the bottom of the stem (measured upward).

Stem: A_s = 8 x 0.375 = 3.00 in^2. Its own centroid is at y_s = 8/2 = 4.0 in from the bottom.

Total area: A = 2.25 + 3.00 = 5.25 in^2.

Step 2: Overall centroid from bottom

y-bar = (2.25 x 8.1875 + 3.00 x 4.0) / 5.25 = (18.422 + 12.000) / 5.25 = 5.795 in.

Step 3: Own moments of inertia (about each element's own centroid)

I_c,flange = 6 x 0.375^3 / 12 = 6 x 0.05273 / 12 = 0.0264 in^4.

I_c,stem = 0.375 x 8^3 / 12 = 0.375 x 512 / 12 = 16.00 in^4.

Step 4: Parallel axis transfer to the overall centroid

d_f = 8.1875 - 5.795 = 2.393 in. d_s = 5.795 - 4.000 = 1.795 in.

Flange contribution: I_c,flange + A_f x d_f^2 = 0.0264 + 2.25 x 2.393^2 = 0.0264 + 12.88 = 12.91 in^4.

Stem contribution: I_c,stem + A_s x d_s^2 = 16.00 + 3.00 x 1.795^2 = 16.00 + 9.67 = 25.67 in^4.

Step 5: Total Ix

Ix = 12.91 + 25.67 = 38.6 in^4.

Notice that the flange's own I_c (0.0264 in^4) is negligible compared to its Ad^2 term (12.88 in^4). The flange's contribution comes almost entirely from its distance from the neutral axis. The stem contributes through both its own I_c (16.00 in^4) and the transfer term (9.67 in^4) because a significant portion of the stem is close to the neutral axis.

Worked Example 2: W12x26 -- Hand Calculation vs Published Value

A W12x26 has the following nominal dimensions from AISC Table 1-1:

• Total depth d = 12.22 in
• Flange width bf = 6.490 in
• Flange thickness tf = 0.570 in (average)
• Web thickness tw = 0.260 in
• Published Ix = 204 in^4

Three-rectangle approximation for Ix:

Web depth (clear between flanges): h = d - 2 x tf = 12.22 - 1.14 = 11.08 in.

Flange rectangles: area = 2 x 6.490 x 0.570 = 7.40 in^2. Web rectangle: area = 11.08 x 0.260 = 2.88 in^2. Total area: 10.28 in^2.

Ix by the void method: Ix = [6.490 x 12.22^3 - (6.490 - 0.260) x 11.08^3] / 12.

= [6.490 x 1825.0 - 6.230 x 1360.4] / 12 = [11844 - 8476] / 12 = 3368 / 12 = 280.7 in^4.

Why the discrepancy? The three-rectangle approximation returns 280.7 in^4, while AISC lists 204 in^4 -- a 38% overestimate. Three factors cause this:

1. Fillet radii: W-shapes have a radius at the web-flange junction. For a W12x26, the fillet radius k_det is approximately 0.625 in. The fillets remove material that was counted in the rectangular model.
2. Tapered flange tips: W-shape flanges are not perfectly rectangular. The outer tips are thinner than the web-flange junction. The average flange thickness tf already accounts for this somewhat, but the geometric simplification loses accuracy on lighter sections.
3. Actual web-flange intersection geometry: The k1 distance (toe of fillet) and k distance (center of fillet) define a curved transition zone that the rectangular model cannot capture.

The practical rule: for preliminary sizing, the three-rectangle formula gives a reasonable upper bound. For final design, always use the published AISC value. Use the section properties database to look up verified Ix and Iy for any shape.

Worked Example 3: Cover-Plated W12x26

Sometimes a beam must be strengthened after erection by welding cover plates to the flanges. Add a 6 in x 3/8 in cover plate to each flange of a W12x26. What is the new Ix?

Given: W12x26 Ix = 204 in^4, A = 7.65 in^2, d = 12.22 in. Cover plates: 6 x 0.375, A_cp = 2.25 in^2 each.

Step 1: Centroid check. The section remains doubly symmetric. The centroid is unchanged at mid-depth. Therefore only Ad^2 terms are needed for the cover plates.

Step 2: Cover plate own I. I_c,cp = 6 x 0.375^3 / 12 = 0.0264 in^4 (negligible, as expected for a thin flat plate).

Step 3: Distance from centroid to cover plate centroid. The centroid of each cover plate is at d/2 + t_cp/2 = 12.22/2 + 0.375/2 = 6.11 + 0.1875 = 6.298 in from the neutral axis. Each cover plate contributes: I_c,cp + A_cp x d^2 = 0.0264 + 2.25 x 6.298^2 = 0.0264 + 89.1 = 89.2 in^4.

Step 4: Total Ix. Ix_total = 204 + 2 x 89.2 = 382.4 in^4.

The Ix of the built-up section is 87% higher than the bare W12x26. Adding 4.5 in^2 of steel (a 59% area increase) produces an 87% stiffness increase because the added area is placed at the maximum possible distance from the neutral axis. This is the parallel axis theorem in action: Ad^2 dominates.

Section Modulus vs Moment of Inertia

The moment of inertia I governs deflection. The section modulus S governs bending stress. They are related by the distance c from the neutral axis to the extreme fiber:

S = I / c

For a doubly symmetric section (W-shape, HSS), c = d/2 and Sx = 2 Ix / d.

Elastic section modulus Sx is used in allowable stress design (ASD) and service-level checks. The elastic bending stress is fb = M / Sx.

Plastic section modulus Zx is used in LRFD and limit states design. It represents the full plastic moment capacity Mp = Fy Zx. Zx is always larger than Sx because it accounts for stress redistribution after first yield.

For a rectangle: Sx = b d^2 / 6, and Zx = b d^2 / 4. The ratio Zx/Sx = 1.5, meaning the plastic moment is 50% higher than the yield moment.

For W-shapes: Zx/Sx typically ranges from 1.10 to 1.16. The flanges carry most of the plastic moment, so the shape factor is lower than for a rectangle.

Numeric example: W12x26: Sx = 33.4 in^3, Zx = 37.2 in^3, ratio = 1.114. Published values per AISC Table 1-1.

Numeric example: W24x68: Sx = 154 in^3, Zx = 177 in^3, ratio = 1.149. The slightly higher ratio reflects the deeper section's relatively thinner web.

The moment of inertia, section modulus, and plastic modulus are available for all 500+ shapes in the section properties database.

Moment of Inertia for Composite Steel-Concrete Sections

When a steel beam acts compositely with a concrete slab (through shear studs), the concrete is transformed to an equivalent steel area using the modular ratio:

n = Es / Ec

where Es = 29,000 ksi (structural steel) and Ec = 33 w_c^1.5 sqrt(f'c) per ACI 318 (or 57,000 sqrt(f'c) for normal-weight concrete in psi units).

For f'c = 4 ksi normal-weight concrete: Ec = 57,000 x sqrt(4000) = 3,605 ksi. The modular ratio n = 29,000 / 3,605 = 8.04, typically rounded to 8.

Transformed section method:

1. Divide the effective concrete flange width by n to get the equivalent steel width: b_eq = b_eff / n.
2. Treat the transformed slab as a steel rectangle of width b_eq and thickness t_slab.
3. Compute the overall centroid of the composite section (steel beam plus transformed slab).
4. Compute Ix of the composite section using the parallel axis theorem for each element.

Numeric example: A W18x55 (Ix = 890 in^4, A = 16.2 in^2, d = 18.11 in) is made composite with a 4.5-inch concrete slab of effective width 90 inches. f'c = 4 ksi, n = 8.

Transformed slab width: b_eq = 90 / 8 = 11.25 in. Transformed slab area: A_slab = 11.25 x 4.5 = 50.6 in^2. Total transformed area: A_total = 16.2 + 50.6 = 66.8 in^2.

Centroid of transformed slab above bottom of steel: y_slab = 18.11 + 4.5/2 = 20.36 in. Centroid of steel section: y_steel = 18.11/2 = 9.055 in. Overall centroid: y-bar = (16.2 x 9.055 + 50.6 x 20.36) / 66.8 = (146.7 + 1030.4) / 66.8 = 17.62 in from bottom.

I_slab (own): b_eq x t_slab^3 / 12 = 11.25 x 4.5^3 / 12 = 11.25 x 91.13 / 12 = 85.4 in^4. d_slab from composite centroid: 20.36 - 17.62 = 2.74 in. Slab contribution to Ix: 85.4 + 50.6 x 2.74^2 = 85.4 + 380.0 = 465.4 in^4.

d_steel from composite centroid: 17.62 - 9.055 = 8.57 in. Steel contribution to Ix: 890 + 16.2 x 8.57^2 = 890 + 1188.9 = 2079 in^4.

Composite Ix = 465.4 + 2079 = 2544 in^4.

Compare: bare steel Ix = 890 in^4. Composite Ix = 2544 in^4. The concrete slab nearly triples the flexural stiffness, which is why composite design dramatically reduces floor depth and steel tonnage in building construction.

Torsional Constant J

The torsional constant J (sometimes called St. Venant torsional constant) governs resistance to uniform torsion. It is not the same as the polar moment of inertia except for circular sections.

Solid circle: J = pi D^4 / 32

Hollow circle: J = pi (D^4 - d^4) / 32

Thin-walled open section (rectangle with b >> t): J = (b t^3) / 3

For a W-shape, J is the sum of J for each rectangular element (two flanges plus web):

J_w = (2 x bf x tf^3 + (d - 2tf) x tw^3) / 3

This produces very small values. A W12x26 has J = (2 x 6.490 x 0.570^3 + 11.08 x 0.260^3) / 3 = (2 x 6.490 x 0.185 + 11.08 x 0.0176) / 3 = (2.403 + 0.195) / 3 = 0.866 in^4. Compare to Ix = 204 in^4 -- J is 236 times smaller.

For closed sections (HSS), J is much larger. HSS 6x6x3/8 has J = 46.1 in^4 (same as Ix because the section is square). This is why HSS sections are used for members subjected to significant torsion, such as spandrel beams, canopy beams, and crane runway beams with eccentric loads.

The warping constant Cw is a related property that governs warping torsion resistance in open sections. For W-shapes, Cw = (Iy x (d - tf)^2) / 4 approximately. W12x26 Cw = 994 in^6 (per AISC). Warping torsion becomes significant in lateral-torsional buckling checks.

Radius of Gyration

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

r = sqrt(I / A)

In column design, the slenderness parameter is KL/r. A larger r means the section is more efficient in compression. Because Ix is larger than Iy for W-shapes, rx > ry and columns buckle about the weak axis unless braced.

Notice that ry is always much smaller than rx. For W12x26, rx/ry = 3.4. This ratio is why unbraced column length governs compression capacity for W-shapes unless weak-axis bracing is provided.

How to Use the Moment of Inertia Calculator

The moment of inertia calculator computes Ix, Iy, Sx, Sy, and the centroid location for custom geometries without manual integration:

1. Select a base shape (rectangle, circle, I-beam, T-section, channel, angle).
2. Enter dimensions in inches or millimeters.
3. The calculator computes Ix, Iy, Sx, Sy, and area. For asymmetric shapes, it also computes Ixy and the principal axis moments.
4. For built-up sections, add multiple elements and the tool applies the parallel axis theorem automatically.
5. Compare results against published section properties to verify your inputs.

For standard rolled shapes, skip the manual entry entirely and use the section properties database for instant lookup of verified AISC values.

Why Hand Calculation of Ix Does Not Match Published Values

This is the most common question from engineers new to steel design. The three-rectangle formula systematically overestimates Ix by 2-15% for W-shapes. The reasons are geometric, not errors in the formula:

1. Fillet radii (k_det and k1 in AISC nomenclature) remove material from the idealized rectangular cross-section. The fillet radius at the web-flange junction is typically 0.5 to 1.0 inches depending on the section size. For lighter sections, the fillets represent a larger fraction of the web depth.
2. Tapered flanges: W-shape flanges are thicker at the web junction than at the tips. The AISC publishes an average flange thickness tf that accounts for some of this taper, but the exact geometry cannot be captured by rectangular elements.
3. Toe radius: The flange tips have a small radius, not a sharp corner.

For cold-formed sections (C, Z, track), the corner radii are larger relative to section depth, increasing the discrepancy versus rectangular approximations.

Practical rule: Use the rectangular formulas for initial sizing and feasibility checks. Always use published values for final design. The section properties database provides AISC-verified Ix and Iy for all shapes.

Moment of Inertia Reference Table: Popular W-Shapes

All values from AISC Steel Construction Manual, 15th Edition, Table 1-1. For the full database of 500+ shapes including UB, IPE, HEA, and HEB, use the section properties database.

Frequently Asked Questions

What is the difference between Ix and Iy? Ix is the moment of inertia about the horizontal centroidal axis (strong axis for a beam). Iy is about the vertical centroidal axis (weak axis). For W-shapes, Ix is always larger -- typically 3 to 15 times Iy -- because the flanges are placed far from the strong axis.

What units is moment of inertia measured in? Length^4. In US customary: in^4. In SI: mm^4 (most common for steel sections), cm^4, or m^4. Conversion: 1 in^4 = 416,231 mm^4 = 41.62 cm^4 = 4.162 x 10^-7 m^4.

Does material affect the moment of inertia? No. I is purely geometric. A W12x26 has Ix = 204 in^4 regardless of whether the material is A992 steel (Fy = 50 ksi), A36 steel (Fy = 36 ksi), aluminum, or even unreinforced plastic. The material affects allowable stress and deflection (through E), but not I.

How does moment of inertia affect beam deflection? Deflection is inversely proportional to I. For a simply supported beam under UDL: delta = 5wL^4 / (384EI). Double I, halve the deflection. This is the most direct engineering use of I: selecting a section deep enough to control deflection to L/360 or L/240.

What is the polar moment of inertia J? J is the torsional constant that resists twisting about the longitudinal axis. For circular sections, J = pi D^4 / 32 (exactly 2x the centroidal I for solid circles). For non-circular sections, J is not the same as the polar second moment of area; it is a torsional stiffness constant that depends on the cross-section shape.

How do I calculate I for a section with holes (bolt holes, openings)? Subtract the I of the hole area about the overall centroid. For a circular hole of diameter d_h at distance y_h from the neutral axis: I_loss = pi d_h^4 / 64 + (pi d_h^2 / 4) x y_h^2. The Ad^2 term is typically much larger than the own I_c of the hole. For multiple holes, subtract each one.

What is the product moment of inertia Ixy? Ixy = integral of xy dA. For doubly symmetric sections (W-shapes, HSS, pipe), Ixy = 0 and the centroidal axes are the principal axes. For singly symmetric sections (channels, T-sections) about the axis of symmetry, Ixy = 0. For asymmetric sections (angles, Z-shapes), Ixy is nonzero and the principal axes are rotated relative to the geometric axes. The principal moments are found through Mohr's circle: I_max,min = (Ix+Iy)/2 +/- sqrt[((Ix-Iy)/2)^2 + Ixy^2].

What is the warping constant Cw? Cw (units: in^6) governs a thin-walled open section's resistance to warping torsion. When a W-shape twists, the flanges bend laterally in opposite directions, creating a secondary warping stress pattern. Cw quantifies this resistance. For I-shapes, Cw is approximately Iy x (d - tf)^2 / 4. Values are tabulated in AISC Table 1-1 alongside J.

Can Iinterchange second moment of area and mass moment of inertia? No. Second moment of area I (in^4) resists bending and is a geometric property of a cross-section. Mass moment of inertia I_mass (kg x m^2 or lb x in x s^2) resists angular acceleration and is a dynamic property of a rigid body. They share the same mathematical structure (integral of r^2 dm vs integral of y^2 dA) but describe entirely different physical phenomena.

How do I compute I for a plate girder with varying flange thickness? Divide the girder into segments of constant cross-section. Compute I for each segment using the actual flange and web dimensions at that location. For deflection, use the average I weighted by segment length. For strength checks, use the minimum I in the region where the moment is highest.

Is the moment of inertia the same in all unit systems? The numerical value depends on the unit system, but the physical property is the same. A section with Ix = 100 in^4 in US units has Ix = 41,623,100 mm^4 in SI. The ratio Ix / Iy is dimensionless and independent of units.

Related Pages

• Section Properties Database -- Ix, Iy, Sx, Sy, Zx, Zy, rx, ry, J, and Cw for 500+ shapes
• Moment of Inertia Calculator Tool -- Interactive calculator for custom and built-up shapes
• Steel Beam Sizes Chart -- Dimensions and properties for W, UB, IPE, HEA sections
• Beam Deflection Calculator -- Deflection checks using Ix per AISC serviceability limits
• Beam Capacity Calculator -- Flexure and shear checks using Sx, Zx per AISC 360
• Radius of Gyration Reference -- Column buckling and slenderness using r = sqrt(I/A)
• Section Modulus Reference -- Elastic and plastic section modulus explained

Disclaimer

SteelCalculator.app is a calculation tool, not a substitute for professional engineering certification. All moment of inertia values and calculations presented on this page are for educational and preliminary design reference only. 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 bears full responsibility for verifying all outputs against applicable code provisions, confirming input accuracy, and obtaining professional review. See our full disclaimer and how to verify calculations.