Centroid of a Composite Area (ȳ = ΣAȳ / ΣA)

Before you can compute anything about a built-up section — its second moment of area, its section modulus, the stress at the top fibre — you have to know where its neutral axis is. For a symmetric shape you know by inspection. For a T, an angle, a channel, a flitch beam or a plate welded to a rolled section, you have to find it, and the way you find it is an area-weighted average: $\bar{y} = \frac{\sum A_i y_i}{\sum A_i}$.

Work a T-section: a 200 × 20 mm flange sitting on a 20 × 180 mm stem, measured from the bottom of the stem. The flange has $A_1 = 4000$ mm² with its own centroid at $y_1 = 190$ mm; the stem has $A_2 = 3600$ mm² at $y_2 = 90$ mm. So $\bar{y} = (4000 \times 190 + 3600 \times 90)/7600 = 1\,084\,000/7600 = 142.6$ mm from the bottom. The neutral axis sits well up in the stem, nowhere near mid-depth, which is exactly why a T-section is so much stronger one way up than the other.

Three things go wrong with this calculation, and none of them announces itself. The datum has to be the same for every part — mixing "from the bottom" for one piece and "from the top" for another produces a perfectly plausible wrong answer. The $y_i$ are the parts' own centroids, not their edges. And more than two parts is handled by applying this twice: combine any two, treat the result as a single part of area $A_1 + A_2$ at $\bar{y}$, then bring in the third. Cut-outs enter as negative areas in the general method; here, subtract them by hand and keep the datum straight. Once $\bar{y}$ is right, the parallel axis theorem measures every transfer distance from it, and an error here propagates squared into everything downstream.