Area A
mm²
Angle section — Geometric properties
Calculate area, centroid, second moments of area, section moduli and radii of gyration with clearly defined dimensions.
Created v1.0
By Thibaut Grzelak, Mechanical Analysis Engineer
01
Choose a shape
03
Diagram
04
Results
Centroid x̄
mm
Centroid ȳ
mm
Second moment Iₓ
mm⁴
Second moment Iᵧ
mm⁴
05Show additional results
Minimum modulus Wₓ,min mm³
Minimum modulus Wᵧ,min mm³
Radius of gyration rₓ mm
Radius of gyration rᵧ mm
Product of inertia Iₓᵧ mm⁴
Principal moment I₁ (max.) mm⁴
Principal moment I₂ (min.) mm⁴
Principal-axis angle θₚ °
Torsion constant J mm⁴
Plastic modulus Wpl,x mm³
Plastic modulus Wpl,y mm³
Shear area Av,x mm²
Shear area Av,y mm²
Wpl is integrated exactly over the geometry. J is shown only where an exact implemented solution applies; code-dependent shear areas remain unavailable.
View calculation detailsFormulas, numerical substitution and resultsOpen ↓Close ↑
Calculation note
Full calculation details
Show additional results
- Wpl,x
- ∫A |y − yp| dA
- Wpl,y
- ∫A |x − xp| dA
- J
- Not available
- Av,x / Av,y
- Not available
Wpl is integrated exactly over the geometry. J is shown only where an exact implemented solution applies; code-dependent shear areas remain unavailable.
01Equations and symbols
Formulas used
A
tH + (B − t)tx̄, ȳ
ΣAᵢxᵢ/A ; ΣAᵢyᵢ/AIₓ, Iᵧ
Σ(Iᵢ + Aᵢdᵢ²)Iₓᵧ
Σ[Aᵢ(xᵢ−x̄)(yᵢ−ȳ)]I₁, I₂
(Iₓ+Iᵧ)/2 ± √[((Iₓ−Iᵧ)/2)²+Iₓᵧ²]θₚ
½ atan2(−2Iₓᵧ, Iₓ−Iᵧ)Wmin
I / cmax| Symbol | Meaning | Dimension |
|---|---|---|
| B | Width B | L |
| H | Height H | L |
| t | Common thickness t | L |
| A | Area without double counting | L² |
| x̄, ȳ | Centroid | L |
| Iₓ, Iᵧ | Centroidal second moments | L⁴ |
| Iₓᵧ | Signed centroidal product of inertia | L⁴ |
| I₁, I₂ | Maximum and minimum principal moments | L⁴ |
| θₚ | Counterclockwise angle from x to principal axis 1 | 1 |
| Wmin | Minimum section moduli | L³ |
02Assumptions and limits
Assumptions and limits
- Equal leg thickness.
- Sharp outer corner.
- Inner fillet ignored.
- x and y are not necessarily principal axes.
This calculation does not check strength, stability or code compliance.
03Calculator validation
Calculator validation
Two non-overlapping rectangles are combined using the parallel-axis theorem, then the centroidal tensor is diagonalized.
- B = 100, H = 150 and t = 10 mm give A = 2,400 mm².
- x̄ = 23.75 mm and ȳ = 48.75 mm.
- Iₓ = 5,576,250 mm⁴ and Iᵧ = 2,026,250 mm⁴.
- Iₓᵧ = −1,968,750 mm⁴ under the displayed convention.
- I₁ = 6,452,023.8 mm⁴, I₂ = 1,150,476.2 mm⁴ and θₚ = +23.9813°.
- For B = H, x̄ = ȳ and Iₓ = Iᵧ.
04References
References
- Roark’s Formulas for Stress and Strain, 9th edition, section-property tables.
- Young, Budynas & Sadegh, Roark’s Formulas for Stress and Strain, McGraw-Hill.
- BIPM, The International System of Units (SI), 9th edition.
Geometric properties
Second moment of area and section modulus: which value to use?
I describes material distribution and W connects geometry to bending stress. Values change with the x or y axis.
Second moment of area I
Also called area moment of inertia, I is used in deflection and buckling.
I [mm⁴, cm⁴, m⁴, in⁴]Elastic section modulus W
W uses distance c to the extreme fibre. For an unsymmetrical section, check both sides.
W = I / cChoose the correct shape
Rectangle, circle, tube, I, channel, T and angle sections require different geometries.