Area Moment of Inertia & Section Modulus
Compute the area moment of inertia I and the section modulus W of any cross-section. Pick a base shape (rectangle, circle, circular tube, hollow rectangle) and read A, I_x, I_y and W_x live – or stack arbitrary rectangles into a composite profile in the parallel-axis assembler. The calculator finds the common centroid, the total moment of inertia and the governing section modulus from the outer-fibre distance.
Section property calculator (I and W)
Pure property calculator for second moments of area of homogeneous plane cross-sections. It computes area, area moment of inertia I and section modulus W about the bending axis. No stress or deflection check – use the beam calculator for that. The composite mode assumes axis-parallel rectangles sharing a common y-axis.
Results
Calculating …
Formulas and fundamentals
The area moment of inertia I (second moment of area, unit mm⁴) describes the bending stiffness of a cross-section about an axis through the centroid: I = ∫ y² dA. Closed-form expressions apply to the base shapes: rectangle I_x = b·h³/12, circle I = π·d⁴/64, circular tube I = π·(D⁴−d⁴)/64 and hollow rectangle I_x = (B·H³−b·h³)/12 as the difference of outer and inner rectangle. Deflection is governed by the axis with the smaller I; for the rectangle I_x grows with the cube of the depth (h³), which makes the orientation of the section decisive.
The section modulus W = I/e (unit mm³) links I with the outer-fibre distance e and governs the bending stress σ = M/W. For doubly symmetric base shapes the centroid lies at the centre, so e = h/2 or e = d/2: rectangle W_x = b·h²/6, circle W = π·d³/32, circular tube W = π·(D⁴−d⁴)/(32·D). A common mistake is σ = M/I – it is a division by W, not by I.
Composite sections are assembled from part rectangles using the parallel-axis (Steiner) theorem. First the common centroid follows from y_s = Σ(A_i·y_i)/ΣA_i, then the total moment of inertia I = Σ(I_own,i + A_i·d_i²) with the distance d_i = y_i − y_s of each part from the overall centroid. The Steiner term A·d² dominates for areas far from the axis and explains why I-beams and box sections are so stiff in bending. The section modulus follows from the larger of the two outer-fibre distances e_o and e_u, where the highest edge stress occurs.
Worked example
A rectangular section with b = 100 mm and h = 200 mm has an area A = 20,000 mm². The moment of inertia about the strong axis is I_x = b·h³/12 = 100·200³/12 = 66,666,667 mm⁴, the section modulus W_x = b·h²/6 = 100·200²/6 = 666,667 mm³.
A circular section with d = 100 mm yields I = π·d⁴/64 = 4,908,739 mm⁴ and W = π·d³/32 = 98,175 mm³. For the same area the rectangle is far stiffer in bending when its depth lies in the load direction.
Stacking two 100 × 20 mm rectangles with their centroids at y = +60 mm and y = −60 mm (a shallow twin flange), the Steiner theorem gives I = 2·(100·20³/12 + 2000·60²) = 14,533,333 mm⁴. The own inertia of the slender rectangles is tiny; almost all of I comes from the Steiner term A·d² – the principle behind every I-beam.
Frequently asked questions
What is the difference between area moment of inertia I and section modulus W?
I (in mm⁴) describes bending stiffness and governs deflection (f ~ 1/I). W (in mm³) describes the utilisation of the outer fibre and governs bending stress σ = M/W. They are linked through the outer-fibre distance e: W = I/e. Do not confuse them – σ = M/I is wrong.
About which axis is the calculation made?
The bending axis x runs horizontally through the centroid; the depth h is perpendicular to it. I_x is therefore the moment governing bending about that axis. For the base shapes I_y about the perpendicular axis is also reported. For bending the axis with the smaller I is always critical.
How does the parallel-axis (Steiner) theorem work?
It shifts the moment of inertia from a part's own centroidal axis to a parallel reference axis: I = I_own + A·d². Here I_own is the part's own inertia about its own centroid, A its area and d the distance to the overall centroid. The calculator first determines y_s and then sums all part contributions.
Why are I-beams so stiff in bending?
Because their flanges sit far from the centroid and the Steiner term A·d² grows with the square of the distance. Material at the edge contributes far more to stiffness than material near the centre. That is why area is concentrated in the flanges and joined by a thin web.
Which section modulus applies to unsymmetric sections?
The smaller section modulus governs, i.e. W = I/e using the larger of the two outer-fibre distances e_o (top) and e_u (bottom). The highest bending stress occurs at that fibre. The calculator reports both outer-fibre distances and forms W from the larger one.