Centroid of Composite Areas: Method and Examples

What the centroid is

The centroid is the geometric centre of an area: the point where it would balance if it were cut from a thin, uniform plate. For a plate of uniform thickness and density it is also the centre of gravity. For a beam cross-section it is where the neutral axis passes in bending, which is why every bending stress and deflection calculation starts by finding it.

Any axis of symmetry contains the centroid. A shape with two axes of symmetry has its centroid where they cross, so you only need to compute the coordinate that symmetry does not give you.

The composite-area table

Set up one row per part with columns for A, x̄ᵢ, ȳᵢ, Aᵢx̄ᵢ and Aᵢȳᵢ. Measure every x̄ᵢ and ȳᵢ from the same reference axes, chosen at a corner or edge so most coordinates are positive.

Rectangle b × h: centroid at b/2, h/2 from a corner.
Right triangle: centroid one third of the base from the right-angle leg and one third of the height from the base.
Circle: at its centre. Semicircle of radius r: 4r/(3π) from the diameter.
Holes and cut-outs: enter the area with a minus sign, keep its centroid coordinate positive.

From the centroid to the moment of inertia

The second moment of area about the centroidal axis uses the same table: add each part’s own I to Aᵢdᵢ², where dᵢ is the distance from the part’s centroid to the composite centroid. That is the parallel axis theorem. The moment of inertia calculator shows both steps for standard sections, so you can check a hand table against it.

Key formulas

Formulas used on this page, with the textbook each is cited from
Formula	Expression	Where	Cited from
Centroid of a composite area	`x̄ = ΣAᵢx̄ᵢ / ΣAᵢ, ȳ = ΣAᵢȳᵢ / ΣAᵢ`	Aᵢ = area of part i (mm²), negative for holes; x̄ᵢ, ȳᵢ = coordinates of the centroid of part i from the reference axes (mm)	Hibbeler, Engineering Mechanics: Statics
Centroid of a semicircle	`ȳ = 4r / (3π)`	r = radius (mm); ȳ measured from the diameter	Beer & Johnston, Vector Mechanics for Engineers: Statics
Parallel axis theorem (areas)	`I = Σ(Īᵢ + Aᵢdᵢ²)`	Īᵢ = second moment of part i about its own centroidal axis (mm⁴), dᵢ = distance from that axis to the composite centroidal axis (mm)	Meriam & Kraige, Engineering Mechanics: Statics
Rectangle about its centroidal axis	`Ī = b·h³ / 12`	b = width parallel to the axis (mm), h = depth perpendicular to it (mm)	Hibbeler, Engineering Mechanics: Statics

Worked examples

Example: Centroid and moment of inertia of a T-section

Given

A T-section made of a 120 mm wide × 20 mm thick flange on top of a 20 mm wide × 100 mm deep web, centred on the web. Total depth 120 mm.
Reference axis at the bottom of the web.

Find

The height of the centroid ȳ above the bottom, and the moment of inertia about the horizontal centroidal axis.

Solution

The vertical axis of symmetry contains the centroid, so only ȳ is needed. Part areas and centroid heights:Web: A₁ = 20 × 100 = 2000 mm², ȳ₁ = 50 mm; Flange: A₂ = 120 × 20 = 2400 mm², ȳ₂ = 100 + 10 = 110 mm
Sum the table:ΣA = 4400 mm², ΣAȳ = 2000(50) + 2400(110) = 364 000 mm³
Centroid height:ȳ = 364 000 / 4400 = 82.73 mm above the bottom (37.27 mm below the top)
Distances from each part’s centroid to the composite centroid:d₁ = 82.73 − 50 = 32.73 mm, d₂ = 110 − 82.73 = 27.27 mm
Parallel axis theorem, part by part:Web: 20(100)³/12 + 2000(32.73)² = 1.667 × 10⁶ + 2.142 × 10⁶; Flange: 120(20)³/12 + 2400(27.27)² = 0.080 × 10⁶ + 1.785 × 10⁶
Add the four terms:I = 5.674 × 10⁶ mm⁴

Answer: ȳ = 82.73 mm above the bottom of the web, on the axis of symmetry, and I = 5.67 × 10⁶ mm⁴ about the horizontal centroidal axis. The centroid sits in the web, close to the heavier flange. The bending stress lesson uses this section.

Example: A bracket plate with a triangular extension and a hole

Given

A plate made of a 160 mm wide × 120 mm tall rectangle with its lower-left corner at the origin.
A right triangle attached to its right edge: vertical leg 120 mm along x = 160 mm, base 60 mm along the bottom edge from x = 160 to 220 mm.
A 50 mm diameter hole centred at (60 mm, 70 mm).

Find

The coordinates (x̄, ȳ) of the centroid of the plate.

Solution

Rectangle:A₁ = 160 × 120 = 19 200 mm², x̄₁ = 80 mm, ȳ₁ = 60 mm
Triangle (centroid one third of the base from the vertical leg, one third of the height from the base):A₂ = ½ × 60 × 120 = 3600 mm², x̄₂ = 160 + 60/3 = 180 mm, ȳ₂ = 120/3 = 40 mm
Hole, as a negative area:A₃ = −π(25)² = −1963.5 mm², x̄₃ = 60 mm, ȳ₃ = 70 mm
Total area:ΣA = 19 200 + 3600 − 1963.5 = 20 836.5 mm²
First moments:ΣAx̄ = 1 536 000 + 648 000 − 117 810 = 2 066 190 mm³; ΣAȳ = 1 152 000 + 144 000 − 137 445 = 1 158 555 mm³
Divide by the total area:x̄ = 2 066 190 / 20 836.5 = 99.16 mm, ȳ = 1 158 555 / 20 836.5 = 55.60 mm

Answer: The centroid is at (99.16 mm, 55.60 mm). The triangle pulls it right of the rectangle’s centre (80 mm), and the hole, above and to the left of that centre, pulls it further right and down.

Common mistakes

Measuring part centroids from different reference axes. Pick one origin for the whole shape and measure every x̄ᵢ and ȳᵢ from it.
Adding the hole instead of subtracting it. A hole is a negative area. Its first moment Aᵢx̄ᵢ is negative too, because Aᵢ is.
Putting the triangle’s centroid at one third from the wrong side. It is one third of each leg measured from the right angle (equivalently, two thirds from the opposite vertex).
Overlapping parts. Split the shape so the parts do not overlap, or subtract the overlap once. For a T, the web stops where the flange begins.
Leaving out Aᵢdᵢ² when moving on to I. Each part’s own Ī is about its own centroid. You must add Aᵢdᵢ² to move it to the composite centroid.

In industry

Frequently asked questions

Is the centroid the same as the centre of mass?

For a plate of uniform thickness and density, yes. If density or thickness varies, weight each part by its mass instead of its area.

Can the centroid be outside the shape?

Yes. The centroid of an L-shape, a ring or a channel can fall in empty space. It is a point of balance, not necessarily a point on the material.

Why use negative areas instead of splitting around the hole?

It is much simpler. A plate with a circular hole would need curved pieces otherwise; subtracting a full circle gives the same answer.

How many decimal places should I keep?

Keep at least four significant figures in the table and round at the end. The parallel axis terms use d², so rounding the centroid early magnifies the error in I.

Keep learning

All Statics lessons · All engineering subjects

Sources

R. C. Hibbeler, Engineering Mechanics: Statics, Pearson
F. P. Beer, E. R. Johnston, D. F. Mazurek, Vector Mechanics for Engineers: Statics, McGraw Hill
J. L. Meriam, L. G. Kraige, Engineering Mechanics: Statics, Wiley

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Centroids of Composite Areas

What the centroid is

The composite-area table

From the centroid to the moment of inertia

Key formulas

Worked examples

Example: Centroid and moment of inertia of a T-section

Given

Find

Solution

Example: A bracket plate with a triangular extension and a hole

Given

Find

Solution

Common mistakes

In industry

Frequently asked questions

Keep learning

See also

Sources

Generate yours automatically from your existing docs