Moment Of Inertia I Beam⚠ unverified
Mechanical / Beams · Compute the second moment of area for a symmetric I-beam about the strong axis
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| bf | bf | m | 1.0 | Flange width |
| tf | tf | m | 1.0 | Flange thickness |
| hw | hw | m | 1.0 | Web height (clear distance between flanges) |
| tw | tw | m | 1.0 | Web thickness |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | I | m**4 | Second moment of area about the strong (centroidal) axis, in metres^4 (m**4) |
The science & history
Understanding the Parameters
-
Flange width $b_f$ and thickness $t_f$ — set the flange area $b_f t_f$. Because the flanges sit at distance $d = h_w/2 + t_f/2$ from the neutral axis, their contribution is dominated by the parallel-axis term $b_f t_f d^2$, which scales with the square of that distance. Wide, thick flanges placed far apart are what make an I-beam stiff.
-
Web height $h_w$ — the distance the web spans between flanges. It has a double role: it provides the web's own $t_w h_w^3/12$ inertia and it sets how far the flanges are pushed from the neutral axis (through $d$). Increasing $h_w$ is the most powerful lever on $I$ — a taller beam of the same weight is far stiffer.
-
Web thickness $t_w$ — the web mainly resists shear, not bending, so it is kept thin to save weight; its bending contribution $t_w h_w^3/12$ is usually a minority share of $I$. Too thin, though, and the web buckles.
The three terms map exactly onto the formula: web-own-inertia, flange-own-inertia (small), and flange-transfer (large).
Derivation (Approaching a Proof)
The section is decomposed into three rectangles — one web, two flanges — and the total second moment of area about the common centroidal axis is the sum of each part's inertia about that axis. For any part, the parallel-axis theorem (Parallel Axis Theorem) gives
$$I_{\text{part}} = I_{\text{part,centroid}} + A_{\text{part}}\, d_{\text{part}}^2,$$
where $d_{\text{part}}$ is the distance from the part's own centroid to the section neutral axis.
-
Web. Its centroid coincides with the section centroid ($d=0$), so it contributes only its own inertia: $\dfrac{t_w h_w^3}{12}$.
-
Each flange. A rectangle $b_f \times t_f$ has its own inertia $\dfrac{b_f t_f^3}{12}$ (small, because $t_f$ is thin) and sits at $d = \dfrac{h_w}{2} + \dfrac{t_f}{2}$ from the neutral axis, adding $b_f t_f\left(\dfrac{h_w}{2}+\dfrac{t_f}{2}\right)^2$. The factor of 2 accounts for the top and bottom flanges, which are symmetric.
Summing gives the stated formula. The symmetry (equal flanges) keeps the centroid at mid-height, so no net shift term is needed.
Dimensional check. Each term is (length)(length³) or (area)(length²) $=\text{m}^4$. ✓
History and Development
Rolled wrought-iron I-sections appeared in the mid-19th century (Zores in France, 1849; American mills shortly after) and rolled-steel wide-flange beams followed with the Grey mill process (early 1900s), becoming the workhorse of steel-framed buildings and bridges. The additive parallel-axis computation of their section properties is standard in every structural handbook; the AISC Steel Construction Manual tabulates $I$, $S$, and $Z$ for hundreds of standard rolled shapes computed exactly this way.
Related Concepts: Rectangular Moment of Inertia, Parallel Axis Theorem, Section Modulus, Beam Bending Stress, Plastic Section Modulus, Beam Buckling Load
Notes: Assumes a symmetric I-section (equal flanges) and strong-axis bending. Uses $h_w$ as the clear web height between flanges (flange centroids at $h_w/2 + t_f/2$); confirm this matches your dimension convention, as some references define the overall depth instead.