Statics · Centroids
Statics · Centroids
Where does the weight act?
So far forces have been arrows with a clear point of application. The own weight of a component is not: it is distributed over the whole body. Nevertheless one calculates with one force at one point — the centroid. This module shows why this is permitted and how to find it, using the base plate of a crane column: an L-shaped plate with a hole for the cable feed-through.
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The base plate of the crane post
On the right lies an L-shaped plate of constant thickness: a horizontal leg, a vertical one on top of it, and in the horizontal one a circular cable feed-through. The question of this chapter is: at which point may the entire weight of this plate be thought of as concentrated without changing its static effect?
Adjust the hole and the leg lengths with the sliders. The crosshair in the view moves along — not arbitrarily, but according to a rule that the next chapters develop.
The centroid of a body is that point at which the spatially distributed weight may be thought of as concentrated, without the static effect changing. That is precisely the justification for drawing the own weight as a single force — as was already done for the tailgate in the previous module.
The hole sits to the right of the centre of the horizontal leg. How does it affect the centroid?
Centroid of a group of parallel forces
The centroid is not a new idea but an application of the previous chapter. A group of parallel forces can be replaced by a single resultant; its direction is that of the forces, and its position follows from the equivalence of the moments.
Consider a weightless rod loaded by parallel individual forces and held in equilibrium by a holding force . From the two conditions
the required position follows immediately:
In the textbook this point is called the centre of forces. In space a second coordinate is added, obtained in exactly the same way — from the moment condition about the other axis.
The formula is a weighted mean: each position counts with the weight of its force . Large forces pull the centroid towards themselves, small ones hardly at all. The same structure — sum of the products divided by the sum — recurs throughout the chapter, only with areas, volumes or lengths instead of forces.
Continuously distributed line and area loads
Up to here these were individual forces. A distributed load — with the dimension force per length — is conceptually replaced by infinitely many infinitesimal individual forces: over the length the individual load acts. The sums become integrals:
In practice one rarely has to integrate for this, because the statement is easy to remember: The magnitude of the resultant equals the area under the load curve, and it acts at the centroid of that area. For a uniformly distributed load this means rectangular area and midpoint; for a triangular load, half the rectangular area and one third or two thirds of the length.
The view on the right shows both — switch between uniform and triangular load and watch how the magnitude and point of application of the resultant jump.
A triangular load grows from zero to over the length . How large is the resultant and where does it act?
Centre of gravity and centre of mass of a body
For an extended body the weight of the volume elements takes the place of the individual forces. With the density arises
Analogous definitions hold for the centre of mass and the centre of volume. At homogeneous bodies the density is constant and cancels out — then all three points coincide and it suffices to integrate over the volume.
If the body consists of part bodies with known centroids, no integral is needed at all: the sum again takes the place of the integral. The base plate is calculated in exactly this way — only in the plane.
If the sub-bodies consist of different materials, one must not calculate with volumes alone — then each partial volume has to be weighted with its density ( the ). Only when the density is the same everywhere do the centre of gravity, the centre of mass and the centroid of volume coincide.
Centroid of an area and centroidal axes
The base plate is a thin disc of constant thickness. For it the thickness cancels out, and the volume integral becomes an area integral:
The integrals arising here are given their own name: they are called first moments of area or static moments.
Mind the indices: contains and vice versa — the name refers to the axis the moment relates to.
If the origin is placed at the centroid, and become zero, and with them the static moments vanish as well. Axes through the centroid are called centroidal axes, and the following theorem holds:
For every area element at a positive distance from an axis of symmetry there is an equally large one at a negative distance — in the sum they cancel. The static moment about an axis of symmetry is therefore zero, and hence every axis of symmetry is a centroidal axis. In practice this means: if a cross-section has an axis of symmetry, the centroid lies on it and only a coordinate has to be calculated.
A cross-section has two axes of symmetry. What follows for the centroid?
Composite cross-sections and negative areas
If an area consists of part areas whose centroids are known, no integration is required. Sums take the place of the integrals — every partial area counts with its area as weight:
And the trick of this chapter: for areas with cut-outs the cut-outs are regarded as "negative" areas . The hole in the base plate therefore enters the sum with its area and its centroid — only with a minus sign. Nothing is calculated differently; only a sign is set.
The calculation is best carried out as a Table: one row per partial area with , , and the products. The derivation below builds this table for the base plate row by row — the view on the right highlights the partial area in question.
| i | Aᵢ / mm² | xᵢ / mm | yᵢ / mm | xᵢAᵢ / 10³ mm³ | yᵢAᵢ / 10³ mm³ |
|---|
Start the tabular method — the rows appear step by step.
Check the base plate: the centroid lies at — and therefore outside the plate, namely in the notch between the two legs. This is not a calculation error but a property of non-convex areas: for a ring it lies in the hole, for a horseshoe in mid-air. The centroid is a calculation point, not a material point — an L-plate cannot be supported there, but it can be calculated as if its weight acted there.
Hands-on task: Move the hole so that the centroid of the plate stays at the height remains, but lies exactly at lies.
centroid of a line
The same structure a third time: replacing the area element by the line element and the area by the length , one obtains the centroid of a line:
For a straight line the centroid lies on it at the midpoint; for a curved line it generally lies outside — for a semicircular arc, for instance, in mid-air between its ends. Applications are bent wires and folded sheet-metal strips, but also the position of the resultant of forces distributed uniformly along a line.
If the line is composed of segments of known length, the sum form applies again — with the lengths as weights:
From now on the centroid is a tool rather than a topic: in the following modules the own weight of a component acts there as a matter of course, and the resultant of a distributed load stands at the centroid of its load area.
Where does the centroid of a homogeneous wire bent into a semicircle lie?
Control accounts
All results for the example values of the guiding system — for checking by hand.
Total area
The hole is subtracted — it is the “negative” area.
centroid at x
Without the bore it would lie at 90 mm — the bore pulls it to the left.
centroid at y
The vertical leg lifts the centroid above the horizontal one.
Sources: Gross/Hauger/Schröder/Wall, Technische Mechanik 1 (Springer), chapter 4 · Engineering Mechanics I course (HSO, summer term 2026).
On to the three exercises — from the concept check through the centroid of a section to the plate with a cut-out.
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