Centroid and centre of gravity
The centre of gravity is the point through which the whole weight of a body can be taken to act. For a flat shape of uniform thickness and material, it coincides with the centroid, the geometric centre of the area. Civil engineers need centroids constantly: to find where the resultant of a distributed load acts, to locate the neutral axis of a beam section, to check that the resultant on a footing or dam falls within the safe zone, and to compute second moments of area.
For simple shapes the centroid is known: a rectangle at its middle, a triangle at one third of its height from the base, a semicircle at 4r/3π from the diameter. For a built-up shape, such as a T-beam or an angle, divide it into simple parts, take the area of each and the position of each part's centroid from a reference line, then take the area-weighted average. A hole or cut-out is handled as a negative area. Symmetry saves work: the centroid lies on every axis of symmetry.
The same averaging works for lines (useful for bent bars and cables), volumes (earthwork, concrete blocks) and weights (a wall made of masonry with a concrete coping). For non-uniform bodies, such as a dam with water in its pores or a composite steel and concrete section, weight each part by its density or by its modular ratio instead of by area alone.
At the advanced level the centroid is found by integration, x̄ = ∫x dA / ∫dA, and the idea extends to the plastic centroid of a reinforced concrete column and the shear centre of open thin-walled sections. Getting the centroid wrong shifts every moment calculation after it, so a quick sanity check, such as confirming the centroid sits nearer the heavier part, is worth the few seconds it takes.
Pick the reference line at the bottom edge so every distance is positive.
Composite area: ȳ = Σ(Aᵢ yᵢ) / Σ AᵢRectangle: centroid at h/2; triangle: h/3 from the baseSemicircle: 4r/(3π) from the diameterBy integration: x̄ = ∫x dA / ∫dA
Problem. A T-section has a flange 300 mm wide × 60 mm deep on top of a web 40 mm wide × 240 mm deep. Find the height of the centroid above the bottom of the web.
- Web: area = 40 × 240 = 9 600 mm², centroid at 120 mm.
- Flange: area = 300 × 60 = 18 000 mm², centroid at 240 + 30 = 270 mm.
- ȳ = (9 600 × 120 + 18 000 × 270) / (9 600 + 18 000) = (1 152 000 + 4 860 000) / 27 600.
- ȳ = 6 012 000 / 27 600 = 217.8 mm.
Answer. About 218 mm above the bottom of the web
Quick quiz
1Where is the centroid of a triangle?
On the median, one third of the height from the base.
2How is a hole handled in a composite centroid calculation?
As a negative area.
3When do centroid and centre of gravity differ?
When the body is not uniform in density or thickness.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- F. P. Beer, E. R. Johnston, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, latest edition.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, latest edition.
- S. Timoshenko, D. H. Young, Engineering Mechanics, McGraw Hill, 4th edition, 1956.
Free and official sources
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