Centroids and centres of gravity
Weight does not act at a point; it acts on every particle of a body at once. That whole cloud of forces collapses to one arrow, and the single point it passes through is the only geometry most of the rest of statics needs.
Free-body diagrams have been drawing a body’s weight as one arrow through one point since the beginning, and nobody has yet said why that is allowed or how the point was chosen. It is allowed for the same reason a distributed load on a beam may be replaced by a single force, and the point is chosen the same way: by demanding that the moments come out right. Everything in this topic is that one argument, carried from a line to an area and then to a solid.
Where the weight acts
Gravity is a body force. It does not reach in through a hook and pull; it pulls on every particle of the body separately, so the honest picture of a weighted body is a cloud of tiny parallel arrows, one per particle, each of magnitude \(\Delta W_i\). There is no single force there at all until you make one.
Making one is a reduction of exactly the kind carried out in force system resultants. The arrows are parallel and all point the same way, so their sum cannot vanish, and a parallel system whose sum is not zero reduces to a single force with no couple left over, provided that force is put on the right line. The magnitude is the easy half:
The line comes from the moment condition. Take moments about some convenient axis and insist that the single force reproduce them:
with \(\tilde{x}_i\) the coordinate of the \(i\)th particle and \(\bar{x}\) the coordinate of the resultant. Repeat for \(y\) and \(z\) and you have a point, \(G\), the centre of gravity. Its defining property is not that the weight is concentrated there (nothing is concentrated anywhere), but that about \(G\) the leftover moment of all those little weights is zero.
One subtlety hides in taking the \(y\) and \(z\) equations. A parallel force system fixes only a line of action, not a point on it, so one orientation of the body gives you a vertical line through \(G\) and no more. Turn the body on its side and repeat, and the two lines cross at the point. That the same point comes out whichever way you turn it is not obvious from the picture, but it falls out of the algebra: the sums above depend only on the body, never on which way gravity happened to be pointing.
Two footnotes. Replace \(\Delta W\) by \(\Delta m\) and the identical sum gives the centre of mass, which is the version that survives into dynamics. And the two coincide only while \(g\) is the same everywhere in the body, true to many decimal places for a bridge, not quite true for a structure a hundred kilometres tall, which is a fact of no practical use and some conceptual value: the centre of gravity is a statement about a force field, the centre of mass is a statement about the body alone.
The weight is not one force. It is one force per particle.
From weight to geometry
Sums over particles are a fiction; bodies are continuous. Write the weight of an element as \(dW = \gamma\,dV\), with \(\gamma\) the specific weight, and the sums become integrals:
Now watch what happens when the material is uniform. \(\gamma\) is a constant, so it comes out of both integrals and cancels between them:
There is no physics left in that expression. It refers to a shape and nothing else: not to what the shape is made of, not to how heavy it is, not to gravity. The point it locates is called the centroid of the volume, and it is a purely geometric object that happens to coincide with the centre of gravity whenever the density is uniform.
The same cancellation runs one dimension lower and then lower again. For a flat plate of constant thickness \(t\), \(dV = t\,dA\) and the thickness cancels too, leaving the centroid of the area. For a slender rod or wire of constant cross-section, \(dV = A\,dL\) and you are left with the centroid of a line. These are the three you will actually use, and the area version is much the most common, because structural sections are areas.
Keep the distinction sharp, because problems are set to test it. A centre of gravity is a physical point that depends on how the material is distributed; a centroid is a geometric point that depends only on shape. Bolt a steel plate to an identically shaped aluminium one and the composite has a centroid at the obvious place and a centre of gravity nowhere near it. Whenever the density varies, the \(\gamma\) stays inside the integral and there is nothing to cancel.
Nothing in any of this requires the answer to lie in the body. The centroid of a ring is the hole in the middle; the centroid of an angle section sits in the empty square between its legs; the centroid of a horseshoe is fresh air. That feels wrong the first few times and is entirely ordinary. The point is defined by a weighted average, and a weighted average of positions need not be a position anything occupies.
Load one end with denser material and G moves towards it.
The integral definition
For a plane area the definition in full is a pair of ratios, one per coordinate:
The denominator is just the area. The numerators have a name of their own, the first moment of area about an axis, and they carry units of length cubed, which is worth noticing because it is the first sign that these quantities are not areas and not lengths. The rearranged form is the one to remember:
The first moment of an area about an axis through its own centroid is zero. That single property is what the centroid is for, and it is why the same point turns up again in bending, in shear flow, and in the next topic.
Setting up the integral is a matter of choosing the element well. The rule is to pick a strip whose own centroid you already know, so that \(\tilde{x}\) and \(\tilde{y}\) are expressions rather than further integrals. A vertical strip of height \(y\) and width \(dx\) has \(dA = y\,dx\), its centroid sits at \(\tilde{x} = x\), and at \(\tilde{y} = y/2\), halfway up, not at \(y\). Forgetting that halving is the most reliable error in this whole subject.
Take the area under \(y = kx^{2}\) from \(0\) to \(a\), with height \(h = ka^{2}\) at the right-hand end. The area is \(\int_0^a kx^{2}\,dx = ah/3\), and
Three quarters of the way along, not half: the area is bunched towards the tall end, and the weighted average follows it. Sketching the shape and asking which side looks heavier will catch a wrong answer faster than checking the algebra will.
Horizontal strips are sometimes far easier: when the region is described as \(x\) in terms of \(y\), or when a vertical strip would change its definition partway across. The two routes give the same answer, so choose on convenience, and change your mind early rather than late.
An area bounded by a curve. Slice it into strips.
Composite bodies
Almost nothing you meet in practice needs an integral. Real sections are assembled out of rectangles, triangles, circles and semicircles, and the centroid of each of those is already known. Integration is additive, so an integral over the whole is the sum of the integrals over the parts, and each of those is a part’s area times its known centroid:
The method is bookkeeping, and it repays being done as bookkeeping. Fix one origin and one set of axes before starting, and never move them. Then for each part record its area and the two coordinates of its centroid measured from that origin, form the products \(\tilde{x}_iA_i\) and \(\tilde{y}_iA_i\), and total the three columns. Most mistakes in composite problems are not conceptual; they are a coordinate measured from the edge of one part instead of from the origin.
Holes are the interesting case, and they are handled by a trick that sounds like cheating and is not. A plate with a circular hole is a full plate plus a circle of negative area. Integration does not care that no such object exists; subtracting the integral over the hole from the integral over the full plate is exactly what removing the material does, and the negative area implements that subtraction in both the numerator and the denominator at once.
Both at once is the part that gets dropped. The hole must be subtracted from the total area and its moment \(\tilde{x}A\) must be subtracted from the total moment. Take one and not the other and the answer is not merely a little off: it is the centroid of a body that could not exist. Signing the area once, at the moment you write it into the table, and then letting the arithmetic carry the sign through is the safest habit.
There is a good check available. When every part has positive area, \(\bar{x}\) is a weighted average of the \(\tilde{x}_i\) and must therefore land somewhere between the smallest and the largest of them. If it does not, the arithmetic is wrong, and no argument about the shape will save it. Once a hole is in the table that guarantee lapses. A negative weight can push the average outside the range, and often should, since removing material from one side drives the centroid towards the other.
A shape made of pieces whose centroids you already know.
Symmetry
Before computing anything, look at the shape. If an area has an axis of symmetry, its centroid lies on that axis, and one of the two coordinates is free.
The argument is a pairing. Measure \(\tilde{x}\) from the axis itself. Every element of area at \(+a\) has an identical partner at \(-a\), so their two contributions \(+a\,dA\) and \(-a\,dA\) cancel; the whole area pairs off this way, the first moment about the axis is zero, and by the definition above that is precisely the statement that the axis passes through the centroid. A shape with two axes of symmetry therefore has its centroid where they cross (a rectangle, a circle, an ellipse, an I-section with equal flanges), and needs no calculation at all. A shape with point symmetry, such as a parallelogram or a Z-section, has its centroid at the centre of symmetry for the same reason.
A T-section is the honest case. It is symmetric about a vertical axis, so \(\bar{x}\) is handed to you: the centroid is on the centreline of the stem. There is no horizontal axis of symmetry, so \(\bar{y}\) has to be earned by the composite sum over the flange and the stem. And \(\bar{y}\) is the coordinate that matters, because it fixes the neutral axis in bending. Symmetry usually gives you the coordinate you did not need.
Two traps. Symmetry of the outline is not symmetry of the area if the material is arranged differently on the two sides: a channel with unequal flanges looks nearly symmetric and is not, and “nearly” buys nothing here. And symmetry of shape is not symmetry of weight: a uniform disc has its centre of gravity at its centre, but the same disc cast half in lead does not, even though its centroid has not moved a millimetre. Whenever density varies, symmetry arguments apply to the centroid only.
None of the machinery here is new, only more general. When a distributed load was collapsed to a single force, the ratio that placed it was this same ratio, applied to the one shape that mattered there: the load diagram was an area, and the resultant passed through its centroid. The only difference is that the areas here are cross-sections rather than graphs.
And the property established in passing above, that the first moment of an area about a centroidal axis is zero, is what makes the centroid the axis worth measuring from in the first place. Having found the point, the natural next question is how the area is spread around it. That quantity is the second moment of area, it is what actually decides how stiff a beam is, and it is quoted about centroidal axes and nowhere else.