Example: A welded T-section
Given
- A flange plate 100 mm wide × 20 mm thick sits on top of a web plate 20 mm thick × 80 mm tall, centred on it.
- Origin at the bottom-left corner of the section’s bounding box: web corner at (40, 0) mm, flange corner at (0, 80) mm.
Find
- The centroid.
- Ix and Iy about the centroidal axes.
- The section moduli to the top and bottom fibres.
Solution
- Split the section into two rectangles and find each area and centroid height.
Flange A₁ = 2000 mm², y₁ = 90 mm; web A₂ = 1600 mm², y₂ = 40 mm - The section is symmetric about x = 50 mm, so x̄ = 50 mm. Find ȳ.
ȳ = (2000 × 90 + 1600 × 40) / 3600 = 244 000 / 3600 = 67.78 mm - Apply the parallel-axis theorem to each part about the centroidal x axis.
Ix = (100 × 20³/12 + 2000 × 22.22²) + (20 × 80³/12 + 1600 × 27.78²) = 3.142 × 10⁶ mm⁴ - Both centroids are on x = 50 mm, so there are no transfer terms for Iy.
Iy = 20 × 100³/12 + 80 × 20³/12 = 1.720 × 10⁶ mm⁴ - The top fibre is 100 − 67.78 = 32.22 mm above the centroid and the bottom fibre 67.78 mm below.
Sx,top = 3.142 × 10⁶ / 32.22 = 97 520 mm³, Sx,bottom = 3.142 × 10⁶ / 67.78 = 46 360 mm³
Answer: Centroid at (50, 67.78) mm, Ix = 3.142 × 10⁶ mm⁴, Iy = 1.720 × 10⁶ mm⁴, Sx,top = 97 520 mm³ and Sx,bottom = 46 360 mm³.