Do I have to integrate to find a centroid?

Almost never for engineering cross-sections. Decompose the shape into rectangles and circles whose centroids you know, then use the composite-body sum. Reserve integration for genuinely curved shapes (a parabolic spandrel, a semi-circle) where no decomposition helps.

What if my pieces overlap?

Either trim the overlapping piece so the union is covered exactly once (as the horizontal leg of the L was started at x = 20), or add the overlap back as a negative-area hole. Both give the same answer; pick whichever is less bookkeeping.

Centroid of an Area: Statics Tutorial

Where does a shape's area effectively act? Learn the first-moment method, the composite table, and the trick of holes as negative areas.

StaticsMechanical Engineering Year 1Free preview

⏱️ About 16 min

Balance a ruler on one finger and the balance point is its centroid — the single point where the ruler's whole area seems to act. Find that point for any flat shape and you have done something quietly powerful: you know where a distributed force (the ruler's weight, a beam's self-weight, the pressure on a submerged plate) effectively pushes. The good news is that almost no real engineering shape needs calculus. Instead you break it into simple rectangles, look up each one's centroid, and combine them with a tidy table. This lesson builds that table method from scratch — and ends with the counter-intuitive truth that the centroid is not always inside the material at all.

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The big idea: The centroid of an area is the point at which the whole area could be concentrated without changing its first moment about any axis. It is found from the first moments of area Qx = ∫ y dA and Qy = ∫ x dA as x̄ = Qy/A and ŷ = Qx/A. For a shape built from simple parts (the composite method), the integrals become sums: x̄ = Σ Ai x̄i / Σ Ai, and likewise for ŷ. Holes and cut-outs enter with negative area.

🎯 By the end, you'll be able to

Define the centroid via the first moment of area, and locate standard shapes by symmetry
Apply the composite-body table method: list each part's area and centroid, then sum
Treat holes and cut-outs as negative areas
Work a full L-section example and recognise when the centroid falls outside the material

📎 Helpful to know first

2D Equilibrium: ΣF=0, ΣM=0
Equivalent Force Systems & Resultants

The centroid: where area balances

For a flat area, the centroid is the geometric centre — the point about which the area is evenly distributed. A rectangle's centroid is at its centre, a circle's at its centre, a triangle's a third of the way up from its base. For these standard shapes you never compute; you use symmetry.

The reason the centroid matters so much in statics is that a distributed force spread over an area — the self-weight of a beam, the pressure on a submerged wall, the load from a column of soil — has the same external effect as a single resultant force acting through the centroid of the loaded area. Locate the centroid and you have replaced a spread-out load with one tidy arrow. That is the whole game.

🔑 First moment of area and the centroid

Define the first moments of area about the x and y axes:

Q_x = ∫ y dA, Q_y = ∫ x dA.

The centroid (x̄, ŷ) is the point that reproduces those first moments when the whole area A is concentrated there:

x̄ = Q_y/A = ∫ x dA / A, ŷ = Q_x/A = ∫ y dA / A.

Notice the first power of distance — this is what distinguishes centroids (first moment) from moments of inertia (second moment, covered in Lesson 3). A first moment can be positive, negative, or zero (it is zero about any axis through the centroid).

The composite-body table method

Few engineering shapes are a single rectangle. But almost all of them can be decomposed into rectangles — an I-beam into three, a channel into three, an angle into two. The centroid of the assembly follows from the centroids of its parts by a weighted average, with area as the weight:

x̄ = Σ A_i x̄_i / Σ A_i, ŷ = Σ A_i ŷ_i / Σ A_i.

The discipline is to organise the work as a table with one row per part and four columns: A_i, x̄_i, ŷ_i, and the products A_i x̄_i, A_i ŷ_i. Sum the columns; divide the moment sums by the area sum. The table keeps the bookkeeping honest and makes errors easy to spot.

✨ Holes are just negative areas

A cut-out is a piece of area that is missing, so it enters the table with a negative A_i (and the corresponding negative A x̄ and A ŷ products). Nothing else changes — the same two formulas still give the centroid of the perforated shape. This is by far the cleanest way to handle slots, bolt-holes, and hollow boxes.

A symmetric L-angle from two rectangles. Each part's centroid is dotted; the composite centroid (red crosshair) lands at (27.1, 27.1) mm, in the open notch and therefore outside the material.

Worked L-section: build the table

Take a symmetric L-angle with legs 20 mm thick and 80 mm long, oriented in the corner of an x–y system at the bottom left. Decompose it into two rectangles that do not overlap (the horizontal leg starts at x = 20, so the shared 20×20 corner is counted once):

Vertical leg 20×80: A₁ = 1600 mm², centroid (x̄₁, ŷ₁) = (10, 40).
Horizontal leg 60×20: A₂ = 1200 mm², centroid (50, 10).

Now sum. Because the two legs have the same thickness and the same reach, the section is symmetric about the 45° line — so we expect x̄ = ŷ. The full computation is in the example below.

📝 Worked example: For the symmetric L-angle above (vertical leg 20×80 with centroid (10, 40), horizontal leg 60×20 with centroid (50, 10)), find the composite centroid (x̄, ŷ).

Total area: A = A1 + A2 = 1600 + 1200 = 2800 mm^2.
First moment about the y-axis: Q_y = A1*x1 + A2*x2 = 1600*10 + 1200*50 = 16000 + 60000 = 76000 mm^3.
x_bar = Q_y / A = 76000 / 2800 = 27.14 mm.
First moment about the x-axis: Q_x = A1*y1 + A2*y2 = 1600*40 + 1200*10 = 64000 + 12000 = 76000 mm^3 (same value, as the 45 degree symmetry predicts).
y_bar = Q_x / A = 76000 / 2800 = 27.14 mm.
Interpretation: the centroid (27.14, 27.14) sits at x > 20 AND y > 20, i.e. in the open notch of the L. The centroid of an area need not lie inside the material.

✓ Centroid at (x_bar, y_bar) = (27.14, 27.14) mm, in the open notch and therefore outside the steel.

⚠️ The centroid is not always inside the material

Students often assume the centroid must lie within the shape. The L-section above shows it does not. A channel (C-section) can have its centroid outside the web; a ring with an off-centre hole can have its centroid in the hole. The centroid is a property of the area, not a point of the solid — it is perfectly allowed to fall in empty space. What you must not then do is assume a load through that point is reacted by material at that point; the load is distributed over the actual area.

🎮 Interactive: composite-section builder LIVE

Predict first: Load the L-section preset. Before reading the numbers, predict: is the centroid inside the steel, or out in the notch? Toggle a flange off and watch the crosshair move.

An interactive composite-section builder on a canvas. A dropdown selects a preset section (single rectangle, T, I, L, or C); checkboxes toggle each rectangle piece and an optional rectangular hole on or off. A red crosshair marks the live centroid, and a side panel reports total area, centroid coordinates, I_x, I_y and radius of gyration, updating instantly as pieces are toggled.

Assemble a cross-section from rectangle pieces (single bar, T, I, L, C) and toggle pieces on or off. The red crosshair is the live centroid; the readout shows area, centroid, I_x, I_y and more. Toggle a piece and the crosshair visibly shifts.

A T-section, by the same method

A T-section (a flange on top of a web) has a vertical line of symmetry, so x̄ is on that line for free; only ŷ needs computing. With a flange 120 mm wide × 20 mm thick (centroid 110 mm above the base) and a web 20 mm wide × 100 mm tall (centroid 50 mm above the base), the height of the centroid is a single weighted average — the practice problem below walks it. The same table method handles it with no new ideas.

✏️ Practice: For the symmetric L-section in the worked example, the first moment about the y-axis is Qy = A1 x̄1 + A2 x̄2 = 1600·10 + 1200·50 = 76000 mm³. The total area is A = 2800 mm². Compute x̄ = Qy/A, in mm (2 d.p.).

Solution

x_bar = Q_y / A = 76000 / 2800 = 27.1428... = 27.14 mm.
By the 45 degree symmetry of this section, y_bar has the same value, 27.14 mm.

✏️ Practice: A T-section has a flange 120 mm wide × 20 mm thick (area 2400 mm², centroid 110 mm above the base) and a web 20 mm wide × 100 mm tall (area 2000 mm², centroid 50 mm above the base). Find the centroid height ŷ = Σ Ai ŷi / Σ Ai, in mm (2 d.p.).

Solution

Total area: A = 2400 + 2000 = 4400 mm^2.
First moment about the base: Q = 2400*110 + 2000*50 = 264000 + 100000 = 364000 mm^3.
y_bar = Q / A = 364000 / 4400 = 82.727... = 82.73 mm.
The centroid sits high up — nearer the heavy flange — which is typical of a T.

Check your understanding

1. The centroid of an L-shaped angle is computed to be at (27.1, 27.1) mm. This point lies in the open notch of the L. What does that mean?

The centroid is a property of the area, not a point of the solid. For non-convex or open shapes it routinely falls in empty space; this is correct, not an error.

2. In the composite-body method, a rectangular hole is entered in the table as:

A hole is missing area, so it carries a negative A_i and the corresponding negative A*x_bar and A*y_bar products. The same two centroid formulas then give the perforated shape's centroid.

3. The first moment of area about an axis through the centroid is:

By definition the centroid is where the first moment vanishes: Q = A * (distance from centroid) = A * 0 = 0. That is exactly the condition used to locate it.

✅ Key takeaways

The centroid locates where an area's first moment vanishes: x_bar = (integral of x dA)/A, y_bar = (integral of y dA)/A.
For composite shapes, replace the integrals with sums over parts: x_bar = sum(A_i x_i)/sum(A_i).
Holes and cut-outs enter the table as negative areas.
Worked symmetric L-angle: centroid (27.14, 27.14) mm, which lies outside the material in the open notch.
The centroid is a property of the area, not the solid; it need not lie inside the shape.

➡️ The centroid tells you where an area acts. But ‘where’ is only half the story when the load is the body's own weight — there, the mass distribution, not the geometry, decides. That is the distinction between centroid, centre of mass, and centre of gravity, and it is the next lesson.

Want to test yourself on this? Try the Mechanical Aptitude test →

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