Cables
Every other member in statics arrives with its shape already decided. A cable does not. It can carry a pull and nothing else, so it has no choice but to hang in whatever curve makes a pure pull sufficient — which means the shape is not given to you, it is part of the answer.
A beam is a beam whatever you do to it. Load it, and it bends, but it stays the straight thing you drew. A cable is different in a way that is easy to state and hard to absorb: it has no bending stiffness at all, so it cannot resist a load by deforming a little and pushing back. It resists by moving until its own geometry does the job. Ask what shape a loaded cable takes and you are not asking about a detail. You are asking the central question of the topic.
A member that can only pull
The idealisation is the one from equilibrium of a particle, taken seriously and pushed further. A cable is perfectly flexible and inextensible: it carries no bending moment, no shear, and no compression. Cut it anywhere and the internal force exposed is a single tension directed along the cable at that point — that is, along the tangent to whatever curve the cable happens to be lying in.
Everything else in the subject follows from that sentence. A beam can be loaded transversely and stay straight because it develops shear and bending to carry the load across. A cable has neither available, so the only way it can equilibrate a transverse load is to have a slope that changes at the load — to kink, or to curve. The load bends the cable into a shape from which a pure axial pull, resolved into components, happens to balance it.
The immediate practical consequence is that the shape is an unknown. In a beam problem you are handed the geometry and asked for the internal forces. In a cable problem the geometry and the internal forces come out together, which is why cable problems feel harder than their equation count suggests. You are usually given one piece of geometry — a sag, a support height, the position of one point on the cable — and that single fact is what closes the system.
It also means a cable cannot be relied on in the way a strut can. A negative tension is not a compression; it is a signal that the cable has gone slack and that the configuration you assumed does not exist. If a cable in your solution comes out with \(T < 0\), the arithmetic is not wrong and the sign does not simply flip. The model has failed, and the real structure has found some other shape.
One more thing worth fixing early. Because the tension acts along the cable, its direction is decided by the slope, and the slope changes from point to point. The tension is therefore not a single number for the whole cable, the way it is for a straight cord in a particle problem. It varies along the length. What does not vary is one particular component of it, and that turns out to be the most useful fact in the topic.
Stretched between two anchors, an unloaded cable is a line.
Cables under concentrated loads
Start with the case where the cable’s own weight is negligible beside the loads it carries, and those loads hang from a few discrete points. Between the load points there is nothing acting on the cable at all, so its slope cannot change there: each stretch is a straight line. The cable becomes a chain of straight segments, kinked only where the loads are. The resulting shape is called a funicular polygon, from the Latin for rope.
Each kink is then exactly the particle problem you already know. Three forces act at the joint — the tension in the segment on each side and the load hung there — and they are concurrent, so
at every joint in turn. Since the horizontal equation at each joint says the horizontal components of the two tensions are equal and opposite, the horizontal pull is carried unchanged from one segment to the next, all the way from one support to the other. Call it \(H\); it is the same number everywhere in the cable, and it is usually the unknown worth solving for first.
Count the unknowns before you start, because this is where cable problems catch people. With \(n\) loads there are \(n+1\) segments, each with its own tension and its own slope, plus the vertical positions of the \(n\) joints. Joint equilibrium supplies \(2n\) equations, which is one short. The missing equation is always a geometric fact you are given: the total length of the cable, or the sag at one specified point. Without it the problem is not determinate, and hunting for another force equation will not help.
The shape depends on the loads, and only on the loads. Double the middle weight and every joint drops, the segment slopes all change, and a different polygon appears between the same two anchors. This is the sense in which a cable has no shape of its own. Reverse all the loads in your mind and the same polygon, turned upside down, is a perfectly efficient arch — which is the observation Robert Hooke made about hanging chains, and it is still how funicular arch forms are found.
Three loads, hung at known places along the span.
Load spread along the horizontal
Now let the loads become continuous. There are two distinct ways a distributed load can be spread over a cable, they give two different curves, and confusing them is the classic error in this topic. Take first the case where the load is uniform per unit of horizontal distance: \(w\) newtons for every metre of span, regardless of how steeply the cable happens to be running there.
That sounds artificial until you look at a suspension bridge. The weight being carried is the deck, which is horizontal and uniform, and it is delivered to the cable through closely spaced hangers. Each metre of deck weighs the same and hands its load up to the cable, so the cable sees a load uniform in \(x\). The cable itself is far lighter than the deck, so ignoring its own weight is reasonable.
Take as a free body the piece of cable from the lowest point out to a general station \(x\). Three forces hold it: the horizontal pull \(H\) at the low point, where the cable is flat and the tension has no vertical part; the tangential tension \(T\) at the cut; and the load \(wx\), acting down through the midpoint of that span. Taking moments about the cut kills \(T\) and leaves
measuring \(y\) upward from the low point. It is a parabola, and it is exact for this loading — not an approximation to anything. Putting in the support, where \(x = L/2\) and \(y = h\), gives the horizontal pull directly:
which is worth staring at for a moment. The pull goes as the inverse of the sag. Halve the sag to make the bridge look sleeker and you double the force in the cable and in everything anchoring it. A cable pulled nearly straight develops enormous tension for a modest load, and the limit as \(h \to 0\) is infinite — which is why no cable, however tight, is ever truly straight under load.
A deck hung from a cable loads it evenly per metre of span.
Load spread along the cable
The other case is a cable carrying its own weight and nothing else: a power line between pylons, a mooring chain, a rope slung between two trees. Here the load is uniform per unit length of the cable, not per unit of span. Every metre of chain weighs the same, but near the supports, where the chain runs steeply, a metre of chain covers much less ground than a metre near the bottom.
So the load per horizontal metre is not constant: it is heaviest where the cable is steepest. The two loadings differ by the factor relating arc length to horizontal distance,
and carrying that factor through the same free-body argument gives a differential equation rather than an algebraic one. Its solution is the catenary, from catena, a chain:
with \(x\) measured from the low point and \(y\) from a horizontal line one distance \(a\) below it. That reference line is not decorative. Measured from it, the tension at any point is simply \(T = w y\), which is the tidiest result in the subject and the fastest way to find the tension at the supports.
The two curves are genuinely different functions, and the difference is not a rounding error in principle. But expand the hyperbolic cosine and the leading terms are \(\cosh u = 1 + u^{2}/2 + u^{4}/24 + \dots\): the first two terms are a parabola, and the correction is fourth order in the slope. So for a shallow cable the two coincide almost exactly. Below a sag-to-span ratio of about one in ten the catenary and the parabola of the same span and sag differ by well under one per cent of the sag, and the parabola — being algebraic rather than transcendental — is used without apology.
The rule to carry away is about which load dominates, not about which formula looks harder. If the cable is holding something much heavier than itself, spread along the horizontal, use the parabola. If it is holding only itself, use the catenary and treat the parabola as an approximation to be checked. And if the sag is large, do not approximate at all: a slack mooring chain is nowhere near parabolic, and its supports feel the difference.
A chain carries its own weight, spread along its own length.
Where the tension is worst
One fact runs through every case above and deserves stating on its own. Take any piece of a cable as a free body. If the loads on it are vertical — weights, deck loads, its own mass — then horizontal equilibrium of that piece involves only the two tensions at its ends, and it forces their horizontal components to be equal:
The horizontal component of tension is the same at every point of a cable under vertical loading, whether the cable is a funicular polygon, a parabola, or a catenary. It is not an approximation and it does not depend on the loading being uniform. It is the reason one symbol, \(H\), can carry an entire cable analysis.
The vertical component does vary. It is zero at the lowest point, where the cable is horizontal and \(T = H\) exactly, and it grows as you move outward, because more and more accumulated load has to be carried up to the supports. Combining the two,
so the tension is smallest where the cable is flattest and largest where it is steepest. Since the cable is steepest at its supports, the maximum tension is always at a support — at the higher support, if the two are at different levels. That is where the cable is sized, where the anchorage is designed, and where a cable fails.
This is also the practical answer to why sag is a design variable rather than an aesthetic one. A flat cable has a small \(\theta\) everywhere, which sounds harmless, but \(H = wL^{2}/8h\) has already gone up in proportion as the sag came down, and the maximum tension goes up with it. Sag costs headroom and buys tension. Every real cable structure is a settlement between those two.