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statics explained

how to find the centroid

Centroid = balance point, the area average, on axes of symmetry, x̄ = ΣAx/ΣA weighted average, composite split, standard-shape centroids

Every area has exactly one balance point. It is the centroid, the average position of all the pieces of area, found from symmetry when you can and otherwise from the area-weighted average x̄ = Σ(A x)Σ A. That one point is where the shape is supported.

Drag the support and a flat shape balances only when the support sits directly under the centroid, and it tips anywhere else. That balance point is the centroid. It is the same place as holding a plate on one finger.

Change the shape and the worth of symmetry shows. An axis of symmetry must pass through the centroid, so a symmetric shape gives its center for free, and no integration is needed. Draw the axes and the point is already fixed.

Drag the area on the right and you see which way the centroid moves. The centroid is the area-weighted average x̄ = Σ(A x)ΣA, so it always leans toward the larger piece. The heavier side pulls the average toward itself.

A complicated shape is split into simple rectangles, each centroid is found, and they are combined with the same weighted average. Drag the dimensions and the centroid of the whole moves to match. The sum of the pieces is the place of the whole.

A few are worth memorizing. A rectangle balances at its center, a triangle at 1/3 of the height from the base, and a semicircle a little above the flat edge, and the toggle shows each one. These three are the basic pieces of a composite.

In PracticeThe centroid is the balance point of a shape, the average position of its area. With symmetry it is fixed at once, because the centroid always lies on every axis of symmetry. Otherwise it is the area-weighted average x̄ = Σ(A x)ΣA, ȳ = Σ(A y)ΣA, and the value leans toward the larger piece. A real shape is split into simple parts, a rectangle at its center, a triangle at 1/3 of the height, a semicircle just above the edge, and then combined. The resultant of a distributed load acts at this very place, and weighting the same area by the square of distance leads to the next step, the 2nd moment of area.
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