---------------------------- | -------------------- | ----------------------------- | -------------------------- | --------------- | | Elastic section modulus | Table 1-1 (Sx) | Tabulated (Zx in AS notation) | Tabulated (Wel) | Tabulated (Sx) | | Plastic section modulus | Table 1-1 (Zx) | Tabulated (Sx in AS notation) | Tabulated (Wpl) | Tabulated (Zx) | | Compact section limit | Table B4.1b | Cl 5.2 (Ze/Z ratio) | Table 5.2 (Class 1-4) | Table 2 | | Effective section (slender) | E7 (Qa, Qs) | Cl 6.2.4 (kf) | Cl 5.5.2 (effective width) | Cl 13.3.5 | | Radius of gyration | Used in E3 (KL/r) | Used in Cl 6.3 | Used in Cl 6.3.1 | Used in Cl 13.3 | | Transformed section (composite) | I3 (lower bound Itr) | AS 2327 Cl 6.3 | EN 1994-1-1 Cl 6.2 | CSA S16 Cl 17 |

Notation warning: AISC uses Sx for elastic section modulus and Zx for plastic section modulus. AS 4100 uses the reverse convention: Zx for elastic and Sx for plastic. This is a common source of confusion when comparing calculations across codes. Always verify which notation the standard is using.

Step-by-Step Example

Problem: Calculate the strong-axis moment of inertia (Ix) of a built-up section consisting of a W12x26 beam with a 10" x 3/4" cover plate welded to the bottom flange.

Step 1 -- Component properties: W12x26: Abeam = 7.65 in^2, Ix_beam = 204 in^4, d = 12.2 in. Cover plate: A_plate = 10.0 * 0.75 = 7.50 in^2, Iplate_own = 10.0 * 0.75^3 / 12 = 0.352 in^4.

Step 2 -- Locate composite centroid (measuring from bottom of cover plate): y*beam = 0.75 + 12.2/2 = 0.75 + 6.10 = 6.85 in (centroid of W12x26 above plate bottom). y_plate = 0.75/2 = 0.375 in (centroid of cover plate above its bottom). y_bar = (7.65 * 6.85 + 7.50 _ 0.375) / (7.65 + 7.50) = (52.40 + 2.81) / 15.15 = 55.21 / 15.15 = 3.645 in from bottom of plate.

Step 3 -- Apply parallel axis theorem: d_beam = 6.85 - 3.645 = 3.205 in (beam centroid above composite centroid). d_plate = 3.645 - 0.375 = 3.270 in (composite centroid above plate centroid).

Ix*total = (204 + 7.65 * 3.205^2) + (0.352 + 7.50 _ 3.270^2) = (204 + 7.65 _ 10.27) + (0.352 + 7.50 _ 10.69) = (204 + 78.6) + (0.352 + 80.2) = 282.6 + 80.6 = 363.2 in^4.

Step 4 -- Section modulus: c_top = (0.75 + 12.2) - 3.645 = 9.305 in. S_top = 363.2 / 9.305 = 39.0 in^3. c_bot = 3.645 in. S_bot = 363.2 / 3.645 = 99.6 in^3.

Result: Ix = 363 in^4 (78% increase over bare W12x26 at 204 in^4). The cover plate shifts the neutral axis downward, giving a much larger section modulus at the bottom fiber (99.6 in^3) than the top (39.0 in^3). For positive bending, the bottom fiber is in tension and the large S_bot is beneficial.

Common Design Mistakes

• Measuring parallel axis distance from the wrong reference: The distance d in the parallel axis theorem must be measured from each component's own centroid to the COMPOSITE centroid, not from the base or top of the section. Using the wrong reference point produces errors that scale with d^2.
• Forgetting to recompute the centroid when adding components: Adding a cover plate or stiffener shifts the neutral axis. The new composite centroid must be computed first, then the parallel axis distances updated. Using the original W-shape centroid ignores this shift and overestimates Ix.
• Confusing AISC Zx with AS 4100 Zx: AISC uses Zx for the plastic section modulus. AS 4100 uses Zx (or Ze) for the elastic section modulus and Sx for the plastic section modulus. Mixing these up can cause a 10-15% error in capacity calculations.
• Using Ix when Iy governs: For columns and lateral bracing checks, Iy (weak-axis moment of inertia) controls. For a W12x26, Ix/Iy = 204/17.3 = 11.8. Using Ix for a weak-axis bending or buckling check overestimates stiffness by an order of magnitude.
• Not accounting for bolt holes in the net moment of inertia: For tension-controlled members with bolt holes, the net section Ix should account for removed material. While this rarely changes Ix significantly for flanges (holes are near the neutral axis for web-bolted connections), it matters for splice plates and gussets.
• Ignoring the shape factor when using elastic vs. plastic modulus: The shape factor Zx/Sx ranges from about 1.10 for W-shapes to 1.50 for solid rectangles. Using the elastic section modulus with the full Fy (without the shape factor) underestimates the plastic moment by 10-50% depending on the cross-section.

Frequently Asked Questions

How does moment of inertia affect beam deflection? Beam deflection under a given load is inversely proportional to the product EI, where E is the elastic modulus and I is the moment of inertia about the bending axis. For a simply supported beam with a midspan point load, the deflection formula is δ = PL³/(48EI); doubling I halves the deflection. This is why selecting a deeper W-shape — which increases I dramatically — is the primary tool for controlling deflection in long-span beams. For steel (E = 200 GPa or 29,000 ksi), I is the dominant design variable since E is fixed.

How does the parallel axis theorem work for composite sections? The parallel axis theorem states that the moment of inertia of a sub-element about any axis equals its own centroidal moment of inertia plus the product of its area and the square of the distance from its centroid to the reference axis: I = I_own + A·d². To apply it to a composite section (such as a cover-plated beam or a welded built-up section), first locate the composite centroid, then sum I_own + A·d² for each part. The most common mistake is forgetting to first find the composite centroid and instead measuring d from the bottom fibre directly.

What is the difference between Ix and Iy for a W-shape? For a wide-flange section, Ix is the second moment of area about the major (strong) axis — the horizontal axis through the centroid — and governs bending in the typical gravity-load orientation. Iy is the second moment of area about the minor (weak) axis and is much smaller because the flanges, which carry most of the area, are positioned close to that axis. A W18×35, for example, has Ix = 510 in⁴ and Iy = 15.3 in⁴; the ratio of roughly 33:1 explains why weak-axis bending and lateral-torsional buckling are critical when a beam is unbraced.

What is radius of gyration and why does it matter for column buckling? Radius of gyration r = √(I/A) is the distance from the neutral axis at which the entire cross-sectional area could be concentrated to give the same moment of inertia. It appears in the slenderness ratio KL/r that governs column buckling: a higher r means a less slender column for the same unbraced length, yielding higher buckling strength. For W-shapes under axial load, the minor-axis radius of gyration ry typically controls because it is smaller than rx, and the effective length KL about the weak axis usually governs unless bracing is provided at different intervals for each axis.

What is section modulus S and how does it relate to bending stress? Section modulus S = I/c, where c is the distance from the neutral axis to the extreme fibre. The elastic bending stress at the extreme fibre is fb = M/S (or equivalently fb = Mc/I). For a W-shape that is symmetric about the neutral axis, the tension and compression fibres are at equal distances from the centroid, so St = Sc = S. Section modulus is the direct link between the applied bending moment M and the resulting extreme-fibre stress; selecting a section with adequate S = M/Fb (where Fb is the allowable bending stress) is the fundamental step in flexural design.

What are the moment of inertia values for a W12×26, and how do they affect beam selection? The W12×26 has tabulated AISC values of Ix = 204 in⁴ (strong axis) and Iy = 17.3 in⁴ (weak axis) — a ratio of nearly 12:1. The elastic section modulus Sx = 33.4 in³ and plastic section modulus Zx = 37.2 in³, giving a shape factor Zx/Sx = 1.11 (typical for W-shapes). At Fy = 50 ksi, the plastic moment Mp = 50 × 37.2 = 1,860 kip-in = 155 kip-ft. For deflection, a 20-ft span under 2 kip/ft uniform load gives δ_max = 5wL⁴/(384EI) = 5×(2/12)×(240)⁴/(384×29,000×204) = 0.86 in, which at L/360 = 240/360 = 0.67 in would fail serviceability — a deeper section such as a W14×30 (Ix = 291 in⁴) should be considered.

Related pages

• Section properties database
• Steel Beam Deflection Check
• Beam calculator
• Steel weight calculator
• Unit converter
• Tools directory
• Reference tables directory
• Guides and checklists
• How to verify calculator results
• Disclaimer (educational use only)
• steel beam capacity per AISC and AS 4100
• Mohr's Circle Calculator

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Moment of Inertia Formulas by Cross-Section Shape

I-Section (Wide Flange) Second Moment of Area

For a doubly-symmetric I-section with flanges and web:

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

Parameters: d = total depth, bf = flange width, tf = flange thickness, tw = web thickness

Example — W12x26: d=12.2”, bf=6.49”, tf=0.380”, tw=0.230”

• Gross Ix ≈ 204 in⁴ (tabulated AISC value; use calculator for derived sections)

The moment of inertia about the weak axis (Iy):

Iy = 2 × (tf × bf³)/12 + (d - 2tf) × tw³/12

→ Calculate I-Section Properties →

T-Section Second Moment of Area

A T-section consists of a flange on top of a web. The centroid must be located first using the composite area method.

Step 1: Find centroid (ȳ from bottom)

ȳ = (A_web × ȳ_web + A_flange × ȳ_flange) / (A_web + A_flange)

Step 2: Apply parallel axis theorem

Ix = (I_web_own + A_web × d_web²) + (I_flange_own + A_flange × d_flange²)

Where d_web and d_flange are distances from each part's centroid to the composite centroid.

Example — 150×150×10 T-section (flange 150×10, web 140×10):

• A_flange = 1500 mm², ȳ_flange = 145 mm from bottom
• A_web = 1400 mm², ȳ_web = 70 mm from bottom
• Composite ȳ = (1400×70 + 1500×145) / 2900 = 109 mm from bottom
• Ix ≈ 14.7 × 10⁶ mm⁴

Angle Section Second Moment of Area

Equal-leg and unequal-leg angles have their centroid off the corner. The principal axes are rotated from the geometric axes.

For an equal-leg angle (L b×b×t):

Ix = Iy = (t × b³)/3 - (b - t) × (b - t)³/3 × (1/4)

The centroid distance from the back of the leg:

ȳ = (b² × t - (b-t)² × t/2) / (2×b×t - t²)

The principal axis moment of inertia (Imax, Imin):

Imax,min = (Ix + Iy)/2 ± √[((Ix - Iy)/2)² + Ixy²]

For equal-leg angles, Ixy is non-zero which means the section has a product of inertia term — important for combined bending.

Rectangular Section Second Moment of Area

The most fundamental formula:

Ix = b × h³ / 12     (about centroidal axis parallel to b)
Iy = h × b³ / 12     (about centroidal axis parallel to h)

For a rectangle at a distance d from the reference axis (parallel axis theorem):

I = b × h³/12 + A × d²     where A = b × h

Example — 200×50 plate (b=200mm, h=50mm):

• Ix = 200 × 50³ / 12 = 2.08 × 10⁶ mm⁴
• Iy = 50 × 200³ / 12 = 33.3 × 10⁶ mm⁴

Circular and Hollow Circular Section

Solid circle (diameter D):

Ix = Iy = π × D⁴ / 64

Hollow circle (outer D, inner d):

Ix = Iy = π × (D⁴ - d⁴) / 64

Example — 100mm solid bar:

• Ix = π × 100⁴ / 64 = 4.91 × 10⁶ mm⁴

Example — 100mm OD × 80mm ID hollow section:

• Ix = π × (100⁴ - 80⁴) / 64 = 2.90 × 10⁶ mm⁴

Parallel Axis Theorem

All composite section calculations rely on the parallel axis theorem:

I_total = Σ(I_own + A × d²)

Where:

• I_own = moment of inertia of each part about ITS OWN centroidal axis
• A = area of each part
• d = distance from each part's centroid to the COMPOSITE centroidal axis

The most common mistake: using d as the distance from the reference axis rather than from the composite centroid. Always find the composite centroid first.

→ Open the Moment of Inertia Calculator to compute Ix, Iy for any composite section.