-------------------- | ----------------------------------------------------------------- | ------------------------------------------------ | ------------------------------------------------------------------------------------- | | Rectangle | bh^3/12 | hb^3/12 | b = width, h = height. The h^3 term means depth dominates stiffness. | | Solid Circle | pi*d^4/64 | pi*d^4/64 (same — symmetric) | d = diameter. Ix = Iy for circular sections. | | Hollow Circle | pi*(D^4 - d^4)/64 | pi*(D^4 - d^4)/64 | D = outer diameter, d = inner diameter. Subtract the hole. | | I-Beam (W-Shape) | Ix = sum of flange + web contributions (see worked example below) | Iy = 2*(tf*bf^3/12) + (d-2tf)*tw^3/12 | Flanges provide ~80% of Ix; web contributes ~80% of shear area. | | T-Section | Ix computed via centroid + parallel axis theorem | Iy = tf*bf^3/12 + (d-tf)*tw^3/12 | Centroid is closer to flange; Ix is asymmetric. | | Channel (C) | Ix computed via centroid + parallel axis theorem | Iy computed via centroid + parallel axis theorem | Centroid is outside the web plane; Iy calculation uses the shifted web axis. | | Equal-Leg Angle (L) | Ix = Iy = Iu = Iv (about centroidal axes) | Same as Ix (equal legs) | Principal axes are at 45 degrees to the legs. Imin is about the minor principal axis. |

Key insight: Because I is proportional to h^3 for rectangles, doubling the depth of a rectangular beam increases its stiffness 8 times. This is why deeper beams are dramatically stiffer and why I-beam flanges are placed as far apart as practical — the parallel axis theorem contribution (A × d^2) dominates the section's total moment of inertia.

Common Steel Section Ix Values — W-Shapes (US Imperial)

Actual Ix values from AISC Shapes Database v16.0 for common W-shapes in A992 steel. Values include the contribution of fillet radii at the flange-web junction, which adds approximately 2-5% to Ix compared to the three-rectangle approximation.

Observations from the table:

• Ix ranges from 118 in^4 (W10x22) to 20,300 in^4 (W36x300) — a factor of 172x across the W-shape range.
• For a given nominal depth, heavier sections have proportionally larger Ix due to thicker flanges (the A × d^2 contribution).
• rx (radius of gyration about the strong axis) is approximately 0.4d for most W-shapes.
• Iy is typically 10-20% of Ix for W-shapes, reflecting the section's strong directional preference.

Parallel Axis Theorem — Theory and Worked Example

The parallel axis theorem transfers a section's moment of inertia from its own centroidal axis to any parallel axis. For built-up and composite sections, the total moment of inertia is the sum of each component's transferred value.

Theorem statement:

Ix = Icx + A × d^2

Where:

• Ix = moment of inertia about the desired axis
• Icx = moment of inertia of the component about its own centroidal axis (parallel to the desired axis)
• A = cross-sectional area of the component
• d = perpendicular distance from the component's centroid to the desired axis

Worked example — Built-up T-section:

A T-section is formed by welding a 10 in × 1 in flange plate to a 8 in × 0.5 in web plate:

Step 1 — Component properties:

• Flange: A1 = 10 × 1 = 10 in^2, Icx1 = 10 × 1^3/12 = 0.833 in^4 (about its own centroid)
• Web: A2 = 8 × 0.5 = 4 in^2, Icx2 = 0.5 × 8^3/12 = 21.33 in^4 (about its own centroid)

Step 2 — Find the composite centroid (measured from top of flange): y_bar = [10 × 0.5 + 4 × (1 + 4)] / (10 + 4) = [5 + 20] / 14 = 1.786 in from top

Step 3 — Compute distances from composite centroid to each component centroid:

• Flange: d1 = 1.786 - 0.5 = 1.286 in
• Web: d2 = (1 + 4) - 1.786 = 3.214 in

Step 4 — Apply parallel axis theorem: Ix = (Icx1 + A1 × d1^2) + (Icx2 + A2 × d2^2) Ix = (0.833 + 10 × 1.654) + (21.33 + 4 × 10.33) Ix = (0.833 + 16.54) + (21.33 + 41.32) Ix = 17.37 + 62.65 = 80.0 in^4

Verification: The total Ix is dominated by the web's own Icx (21.33) and the web's parallel axis contribution (41.32). The flange contributes only 17.37 in^4 despite having more area, because its distance from the neutral axis is small. This illustrates why I-beam flanges are placed at the extreme fibers — the A × d^2 term grows with d^2.

How Moment of Inertia Affects Beam Deflection

Moment of inertia (I) appears in the denominator of every beam deflection formula. A larger I directly reduces deflection for a given load, span, and material. The relationship is inversely proportional: doubling I halves the deflection.

Fundamental deflection formulas showing the role of I:

Simply supported beam, uniform load w over span L:

delta_max = 5 w L^4 / (384 E I)

Simply supported beam, point load P at midspan:

delta_max = P L^3 / (48 E I)

Cantilever beam, point load P at tip:

delta_max = P L^3 / (3 E I)

Cantilever beam, uniform load w:

delta_max = w L^4 / (8 E I)

All four formulas share the same structure: deflection is proportional to load × span^n / (E × I). The span exponent (n = 3 or 4) makes span the most powerful lever on deflection, but I is the primary section property the designer controls.

Practical design implications:

• To reduce deflection by 50% without changing span or material, you must double I.
• Increasing beam depth by 26% roughly doubles I (since I relates to d^3 for rectangular sections).
• A W18x55 (Ix = 890 in^4) is 45% stiffer than a W16x40 (Ix = 518 in^4) — deflection reduces from, say, 1.20 in to 0.70 in for the same 28 ft span under uniform load.
• The section modulus Sx = Ix/c determines bending stress; Ix determines deflection. A section with adequate Sx may still have excessive deflection if Ix is insufficient.

Serviceability limits: IBC Table 1604.3 and ASCE 7 specify deflection limits as span ratios — L/360 for live load on floors, L/240 for total load, L/180 for roof members without brittle finishes. Matching the required Ix to the deflection limit is often the controlling design criterion for longer-span beams.

AISC 360 vs EN 1993 — Section Classification and Moment of Inertia

Both AISC 360 and EN 1993-1-1 use moment of inertia as a fundamental section property, but they differ in how section classification affects capacity and how I is used in stability calculations.

Section Classification Systems

Key Difference: Effective Section Properties

Under AISC 360, a slender-element section uses reduced effective widths but the gross Ix (using full section) is still used for deflection calculations at service load levels. Under EN 1993, a Class 4 section uses effective section properties (Ieff) for ALL calculations including deflection — which means a Class 4 section is penalized in deflection as well as strength.

Key Difference: Buckling Curves

EN 1993-1-1 uses five buckling curves (a0, a, b, c, d) to determine the LTB reduction factor chi_LT, selected based on the section type and fabrication method. AISC 360 uses a single unified LTB curve but modifies it with Cb for moment gradient. The EN 1993 approach is more granular but also more conservative for some section types — a hot-rolled I-section on curve 'a' or 'b' under EN 1993 will typically show slightly lower LTB resistance than the equivalent AISC 360 calculation for the same unbraced length.

Practical Example — W12x26 LTB Comparison

For a W12x26 (A992, Fy = 50 ksi) with Lb = 12 ft, Cb = 1.0:

• AISC 360: Lp = 6.6 ft, Lr = 21.0 ft. Inelastic LTB governs. phi*Mn ≈ 98 kip-ft.
• EN 1993 (equivalent IPE 300, S355): lambda_LT ≈ 1.15, chi_LT ≈ 0.51 (curve 'a'). Mb,Rd ≈ 0.51 × Wpl,y × 355 / 1.0 ≈ 87 kN-m (≈ 64 kip-ft).

The difference arises from the different buckling curve shape and partial safety factor (gamma_M1 = 1.0 for EN 1993 stability vs phi_b = 0.90 for AISC flexure). The gross Ix is the same in both codes, but the treatment of stability differs.

Practical Tips for Using the Moment of Inertia Calculator

When selecting a beam for deflection control: Start with the required Ix from the deflection formula rearranged: I_req = 5wL^4/(384 × E × delta_allow). For a 30 ft simply supported beam with w = 1.0 kip/ft and L/360 deflection limit: I_req = 5 × 1.0 × 360^4 / (384 × 29,000,000 × (360/360)) = 5 × 1.68e10 / (384 × 29,000,000 × 1.0) = 755 in^4. A W21x50 (Ix = 984 in^4) would satisfy with 30% reserve.

When checking an existing beam: Compare the beam's tabulated Ix to the computed I_req. If I_actual > I_req for deflection and S_actual > S_req for strength, the beam passes both checks. If deflection controls, consider a deeper section (larger d gives larger I for similar weight).

When analyzing a built-up section: Use the parallel axis theorem method shown above. The Ix of a built-up I-girder is dominated by the flange area times the square of the distance to the neutral axis. A 1-inch increase in web depth adds approximately 2 × Af × (d/2)^2/(d) to Ix through the parallel axis effect — much more than the web's own bh^3/12 contribution.

When comparing sections across different steel grades: Ix and Sx are purely geometric properties — they do not depend on steel grade. A W16x40 has Ix = 518 in^4 whether it is A36 (Fy = 36 ksi), A992 (Fy = 50 ksi), or A913 (Fy = 65 ksi). The strength (phi*Mn) changes with Fy, but the stiffness (E × I) does not — all structural steels have E = 29,000 ksi.

Frequently Asked Questions

What is the difference between moment of inertia (I) and section modulus (S)?

Moment of inertia I (units: in^4 or mm^4) measures a section's resistance to bending deformation — it controls deflection and buckling. Section modulus S = I/c (units: in^3 or mm^3) measures a section's resistance to bending stress — it controls yielding. For a rectangular section: I = bh^3/12, S = bh^2/6, and the ratio S/I = 2/h. A beam can have adequate S (strength) but insufficient I (stiffness), resulting in excessive deflection. Long-span beams and beams with sensitive finishes are typically governed by I (deflection) rather than S (strength).

How do I calculate moment of inertia for a hollow rectangular section (HSS)?

For a rectangular HSS with outer dimensions B × H and wall thickness t: Ix = (B × H^3 - (B-2t) × (H-2t)^3) / 12. This is the outer rectangle's I minus the inner void's I. The parallel axis theorem is not needed because both the outer and inner rectangles share the same centroid. For a square HSS 8×8×1/2: Ix = (8 × 8^3 - 7 × 7^3) / 12 = (4096 - 2401) / 12 = 141.3 in^4. HSS sections are doubly symmetric so Ix = Iy. The torsional constant J for closed sections is much larger than for open sections, making HSS excellent for torsional resistance.

Why does a W-shape have such different Ix and Iy values?

A W-shape concentrates material in the flanges, which are located far from the strong-axis (x-x) centroid but close to the weak-axis (y-y) centroid. For Ix, the flanges contribute roughly A_flange × (d/2)^2 each via the parallel axis theorem — the distance term d/2 is large, making this contribution dominant. For Iy, the flanges contribute their own centroidal Iy (bf × tf^3/12 + small parallel axis term) — the distance from the y-y axis to the flange tip is only bf/2, which is much smaller than d/2. This asymmetry is intentional: beams are designed to bend about the strong axis, and the weak axis stiffness is only needed for lateral stability.

What is the polar moment of inertia (J) and how does it relate to Ix and Iy?

The polar moment of inertia J (also called the torsional constant) measures a section's resistance to twisting (torsion). For circular sections, J = Ix + Iy = pi*d^4/32. For non-circular sections, J is NOT equal to Ix + Iy. For open sections like W-shapes, J = sum(1/3 × b × t^3) for each plate element — this is typically very small because the t^3 term (web or flange thickness cubed) is tiny. A W16x40 has J = 0.794 in^4 compared to Ix = 518 in^4 — the torsional stiffness is only 0.15% of the bending stiffness. For closed sections (HSS tubes), J = 4A^2 / sum(b/t) where A is the enclosed area, giving much larger torsional stiffness.

Can I use the calculator for non-steel materials like aluminum or timber?

The moment of inertia I is a purely geometric property — it is the same regardless of material. You can use the Ix and Iy values from this calculator for aluminum, timber, concrete, or any other material. However, the deflection and stress calculations use the material's elastic modulus E. Steel has E = 29,000 ksi (200 GPa); aluminum has E = 10,100 ksi (70 GPa); timber has E ≈ 1,600 ksi (11 GPa) for Douglas fir. For the same I and load, an aluminum beam deflects 2.87 times more than steel, and a timber beam deflects 18 times more. The section property tables on this page are optimized for steel design codes (AISC 360, EN 1993) and use steel-specific notation.

How do I find the moment of inertia for an existing steel beam without calculation?

All standard rolled steel sections have tabulated Ix and Iy values published in their respective shape databases: AISC Shapes Database v16.0 (US W, S, HP, C, L, HSS shapes), ArcelorMittal Sections Handbook (European IPE, HEA, HEB), OneSteel/InfraBuild Hot Rolled Structural Steel Products (Australian UB, UC, PFC), and CISC Handbook of Steel Construction (Canadian W, WWF). The tabulated values account for the actual geometry including fillet radii at flange-web junctions, which add 2-5% to simplified three-rectangle calculations. Our section properties database provides pre-computed Ix and Iy for all common shapes across AISC, AS 4100, EN 1993, and CSA S16.

Related pages

• Section properties database
• Moment of inertia — composite section calculator
• Beam capacity calculator
• Beam deflection calculator
• Beam span table
• Custom section properties calculator
• Steel beam sizes reference
• Steel grades reference
• Column capacity calculator
• Portal frame design
• All tools directory
• All reference tables
• How to verify calculator results
• Home — Steel Calculator

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