I / H Section Properties Calculator

Enter the depth, flange width, web thickness and flange thickness of a doubly-symmetric I or H section — a rolled IPE, HE, UB or W beam without its root radii, or a welded plate girder — and get its area, second moments, section moduli, radii of gyration and torsion constant, drawn to scale.

Section dimensions

Enter the dimensions in the selected unit — the drawing and the results update as you type. The drawing uses centroidal coordinates: the bold gridlines are the x and y axes through the centroid C.

Length unit

Section

-150-100-50050100150-150-100-50050100yyxxb = 100b = 100h = 200h = 200tf = 10tf = 10tw = 6tw = 6CC
  • Section with its dimensions
  • Centroid C, centroidal axes x and y
Results
Results in
Area & centroid
Cross-sectional areaA3,080mm²
Centroid from the left edge50mm
Centroid from the bottom edgeȳ100mm
Second moments of area (centroidal axes)
About the x axisIx20,983,000mm⁴
About the y axisIy1,669,900mm⁴
Product of inertiaIxy0mm⁴
Polar moment of areaIx + IyIp22,653,000mm⁴
Principal axes
Maximum principal momentI120,983,000mm⁴
Minimum principal momentI21,669,900mm⁴
Angle from x to axis 1 (CCW)axis 1 = x, axis 2 = yθp0.00°
Section moduli
Elastic section modulus about xSx209,830mm³
Elastic section modulus about ySy33,398mm³
Plastic section modulus about xZx238,600mm³
Plastic section modulus about yZy51,620mm³
Radii of gyration
About the x axisrx82.538mm
About the y axisry23.285mm
Distances to the extreme fibres
Centroid to top / bottom fibrecy100mm
Centroid to left / right fibrecx50mm
Torsion & shear centre
Torsion constantThin-walled open-section approximation, J ≈ Σ b·t³/3 (no fillets).J79,627mm⁴

The I section is the most efficient open shape for bending about its strong axis: the two flanges carry the bending stress at the maximum lever arm and the web only has to carry shear. With the parallel-axis theorem the section is simply the outer rectangle b × h minus the two rectangles beside the web, Ix = [b·h³ − (b − tw)·(h − 2tf)³]/12.

The price is a weak axis: Iy is typically 5 to 20 times smaller than Ix, and as an open section the torsion constant J ≈ Σb·t³/3 is tiny — which is why I-beams twist easily and need lateral-torsional buckling checks. The calculator reports all of this, including the plastic modulus used in plastic design (Mpl = Z·fy).

How the i / h section properties are calculated

Area
A = 2·b·tf + (h − 2·tf)·tw
Second moment about x (strong axis)
Ix = [b·h³ − (b − tw)·(h − 2·tf)³] / 12
Second moment about y (weak axis)
Iy = [2·tf·b³ + (h − 2·tf)·tw³] / 12
Elastic section moduli
Sx = 2·Ix / h, Sy = 2·Iy / b
Plastic section modulus about x
Zx = b·tf·(h − tf) + tw·(h − 2·tf)² / 4
Torsion constant (thin-walled)
J ≈ [2·b·tf³ + (h − 2·tf)·tw³] / 3

Principal moments and axes follow from Ix, Iy and Ixy with I1,2 = (Ix + Iy)/2 ± √[((Ix − Iy)/2)² + Ixy²] and tan 2θp = −2·Ixy/(Ix − Iy); the radii of gyration are r = √(I/A) and the elastic section moduli S = I/c for each extreme fibre. Definitions of every property are in the glossary on the section properties overview.

Assumptions. Equal flanges, sharp corners, no root radius and no flange taper: rolled sections have slightly larger A, I_x and J than this idealisation (typically 1–4% for I_x, more for J). The torsion constant is the thin-walled open-section approximation.

Other sections

Frequently asked questions

How close is this to the catalogue values of an IPE or HEA?

Close for A, I_x and S_x — within a few percent, the difference being the root radius between web and flange (an IPE 200 entered as 200 × 100 × 5.6 × 8.5 gives I_x = 1 866 cm⁴ against 1 943 cm⁴ in the tables, about 4% low). The torsion constant J is more sensitive: the fillets add material exactly where the thin-walled formula is weakest, and catalogue J values can be 20–40% higher.

What is the plastic section modulus used for?

For the plastic moment resistance M_pl = Z_x·f_y of a ductile section, the basis of plastic design in Eurocode 3 and AISC. It assumes the whole section has yielded, half in tension and half in compression about the equal-area axis. For an I beam the ratio Z_x/S_x (the shape factor) is around 1.10–1.15, because the flanges already carry most of the elastic moment.

Why is the torsion constant so small compared with the polar moment?

Because the I section is open. Its polar moment I_p = I_x + I_y describes the distribution of area around the centroid, but a twisted open section develops shear stress only across the thickness of each plate, not around the profile, so J ≈ Σb·t³/3 — often less than 1% of I_p. Warping torsion carries the rest, which is why open sections are so much stiffer when their flanges are restrained.

Can I enter an unequal-flange beam?

Not on this page, which assumes equal flanges so that the centroid stays at mid-depth. A monosymmetric section with one wider flange can be built from a T section plus a rectangle with the parallel-axis theorem; the channel and T pages show how the calculator handles a centroid that is not at mid-depth.

References & further reading

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