Area A
-Ix (about horizontal axis)
-Iy (about vertical axis)
-Sx = Ix / cx
-Sy = Iy / cy
-Radius of gyration rx / ry
-Torsion constant and equivalents
Formulas used
All sections are taken about their centroidal axes; x is the horizontal axis (bending about it is bending in the height direction). Corners are sharp. The formulas are textbook results and can be checked by hand.
| Section | A | Ix | Iy |
|---|---|---|---|
| Rectangle b × h | b h | b h³ / 12 | h b³ / 12 |
| Solid circle d | π d² / 4 | π d⁴ / 64 (J = π d⁴ / 32) | |
| Round tube D, d | π (D² − d²) / 4 | π (D⁴ − d⁴) / 64 (J = π (D⁴ − d⁴) / 32) | |
| Rectangular tube B × H, wall t | B H − b h | (B H³ − b h³) / 12 | (H B³ − h b³) / 12 |
| I-section, flange bf × tf, web tw, depth H | 2 bf tf + tw hw | [bf H³ − (bf − tw) hw³] / 12 | [2 tf bf³ + hw tw³] / 12 |
Inner dimensions: b = B − 2t, h = H − 2t for the rectangular tube; hw = H − 2 tf for the I-section. Section modulus S = I / c with c the distance from the centroid to the outer fibre (H/2 or B/2 or D/2). Radius of gyration r = √(I / A). Bending stress is σ = M / S, and a simply supported beam deflects according to the second moment of area in the beam deflection calculator.
Worked example: 50 mm × 100 mm rectangle
b = 50 mm, h = 100 mm. A = 5000 mm². Ix = 50 × 100³ / 12 = 4.1667 × 10⁶ mm⁴. Iy = 100 × 50³ / 12 = 1.0417 × 10⁶ mm⁴. Sx = Ix / 50 = 83 333 mm³. With a bending moment of 2 kN·m about x, σ = 2 × 10⁶ / 83 333 = 24 MPa. Enter the values to check the calculator against this.
Questions
What is the moment of inertia of a rectangle?
For a rectangle of width b and height h, the second moment of area about the horizontal centroidal axis is Ix = b h³ / 12. About the vertical axis it is Iy = h b³ / 12. The calculator gives both.
What is the difference between moment of inertia and second moment of area?
In structural and machine design, "moment of inertia" usually means the second moment of area, in mm⁴ or in⁴. It describes how a cross-section resists bending. The mass moment of inertia, in kg·m², is a different quantity used for rotating bodies.
How do I calculate section modulus?
Section modulus is the second moment of area divided by the distance from the neutral axis to the outer fibre: S = I / c. Bending stress is then σ = M / S. For a rectangle S = b h² / 6; for a solid circle S = π d³ / 32.
What is the torsion constant J of a tube?
For a round bar or tube, J is the polar second moment of area, J = π (D⁴ − d⁴) / 32. Shear stress in torsion is τ = T r / J. The calculator reports J for circles and round tubes only; rectangular and I-sections need a different torsion constant.
Why does a tube beat a solid bar for bending?
Material near the neutral axis carries little bending stress. A tube whose inner diameter is 0.8 times its outer diameter has only 36 percent of the area of a solid bar of the same outside diameter, yet 59 percent of its second moment of area and 59 percent of its section modulus.
Does it include corner radii and fillets?
No. Rectangular tubes and I-sections use sharp corners, which slightly underestimates the area and inertia of rolled or cold-formed sections. For a catalogue section, use the catalogue values.
Engineering aid for guidance. Sharp-corner idealised sections; rolled and cold-formed catalogue sections have corner radii and tolerances, so use the supplier values for final design. Verify critical designs; you remain responsible for the design. Nothing you enter leaves your browser except through a share link you copy yourself.