Cable tension and elongation under a load

A cable that is straight and has no weight carries one thing, a tension T along its length, the same at every point. It cannot push, so T is never negative; where equilibrium would ask a cable to push, the page says so and gives no number. The three problems are solved separately and share one cable, described by its area A, its modulus E and its coefficient of thermal expansion α.

In a sling a weight W hangs from the point where two legs meet, each leg a cable running up to its anchor at an angle from the horizontal. The point is in equilibrium when the two tensions, the weight and any horizontal force H sum to nothing along x and along y: two equations for two tensions. They come from statics alone, and the size and the material of the cable never enter them. For two legs at the same angle θ from the horizontal, with no H, each carries T = W/(2 sin θ). That grows without limit as the legs flatten: at 30° each leg carries the whole of W, and the nearer the legs come to vertical the nearer each tension comes to half of W. Two legs exactly vertical lie on one line and the load cannot be split between them, so they are refused. With unequal angles, or with an H, the same two equations are solved for the two tensions. A single leg has no second equation to meet: it takes the direction of the combined pull of the weight and the horizontal force, so its angle comes out of the calculation and T = √(W² + H²).

With its tension known, the rest is elastic arithmetic. The stress is σ = T/A and the strain ε = σ/E. Over a length L the cable lengthens by δ = TL/(AE), which is εL. For a rope, A is the metallic area, a fraction of the circle its diameter draws: the page takes A = kd², with a factor k for each construction, and πd²/4 for a solid round.

The materials, ropes and strands offered carry typical values of modulus, coefficient of expansion and metallic-area factor. They are for a first estimate and no more; the value to use is the manufacturer’s for the cable you have, typed in place of the typical one. A wire rope is the case to watch: its modulus in service is about half that of the steel of its wires, because the strands straighten and bed against each other as the load comes on. With the modulus of the steel, the stretch of a wire rope comes out about half of what it really is.

A change of temperature ΔT lengthens a cable that is unrestrained by δ = αLΔT, and the stretch the page gives is that plus the elastic one. If both ends cannot move at all, the cable cannot change its length and the change of temperature shows up as a force, EAαΔT: compression when it warms, which a cable cannot carry, so it goes slack, and tension when it cools. The page gives this force for each cable as if the ends of that cable were held, and adds it to the tension from the load to show what the cable would carry.

Each leg of a sling lengthens by its own δ, and the load point moves. To first order the move d obeys −u · d = δ for each leg, where u is the unit vector from the load point to the anchor: the distance from an anchor grows by the stretch of its leg. The page also gives the exact position, the point where two circles meet, centred on the anchors with the stretched lengths of the legs as radii, the one near the starting point. The two agree when the stretch is small beside the legs, and their difference is the error of the first-order method. Because the legs have turned, the tensions are no longer those of the first geometry; the page works them out again in the displaced position, as a second-order result, and gives the change. A single leg moves along itself by its stretch, exactly.

To pull a block against friction, the block weighs W and rests on a surface that rises at θ along the direction of the pull; the cable pulls it at an angle φ above the surface, which is not the angle above the horizontal (that is θ + φ). The cable lifts the block by T sin φ, so the force pressing it on the surface is N = W cos θ − T sin φ, and friction is μN. Along the surface, with the block about to move or moving at constant speed, T cos φ = W sin θ + μN. Eliminating N gives T = (sin θ + μ cos θ)W/(cos φ + μ sin φ). Two coefficients are used: the static one for the pull that starts the motion and the kinetic one for the pull that keeps it going. With an acceleration a along the path, the inertia a/g joins the friction term in the numerator: T = (sin θ + μ cos θ + a/g)W/(cos φ + μ sin φ). Where the cable holds the block back as it moves the other way, down a slope for instance, friction acts the other way too, the terms in μ and in a/g change sign, and a is taken in the direction of travel: T = (sin θ − μ cos θ − a/g)W/(cos φ − μ sin φ).

Only the denominator depends on φ, and it is greatest where tan φ = μ. The pull is least, then, with the cable at the friction angle above the surface, φ = arctan μ, whatever the slope and whatever the acceleration. There the denominator is √(1 + μ²) and T = (sin θ + μ cos θ)W/√(1 + μ²). The curve of T against φ that the page draws is flat near that minimum, so a few degrees either way change little.

Three things can make the formula mean nothing, and the page says which. If the T it asks for makes N negative, the block would come off the surface and there is no friction to speak of: the case is refused, with the angle past which the cable lifts the block, which at rest or at constant speed is the angle at which the cable is vertical. If cos φ + μ sin φ is zero or less, as when the cable presses the block into the surface at a steep angle, the harder it pulls the harder friction grips, no tension moves the block, and the page says the cable jams. And where gravity along the surface already beats friction in the direction of travel, no pull is needed and the block moves by itself: the page then gives the tension of a cable that holds it back.

The page solves up to three pulls: the one that starts the motion (static coefficient), the one that keeps it moving at constant speed (kinetic coefficient) and, when an acceleration is entered, the one that gives it that acceleration. The largest is the one that governs, and the stress and the stretch of the cable are worked out at that tension, with the section, the length and the temperature change given.

A cable wrapped over a drum that does not turn, through a wrap angle β in radians, is on the point of slipping when the tensions at its ends obey T_big = e^(μβ) T_small. It comes from one small element: over an arc dβ the tension changes by dT = μT dβ, and adding up the arcs gives the exponential. The radius of the drum does not appear, only the angle wrapped and the friction coefficient, static where the cable is about to slip and kinetic where it slides. Any ratio up to that one is held by friction; it is the limit. Every full turn multiplies the ratio by the same factor, e^(2πμ). Exactly one of the four quantities, the two tensions, μ and the wrap, is left blank and the page solves for it. The force of the cable on the drum is the vector sum of the two tensions.

Three cases that have a closed form, for checking the answers by hand.

A cable of length L, area A and modulus E hangs vertically with a weight W: T = W, σ = W/A and δ = WL/(AE). Held at both ends and cooled by ΔT, the same cable carries a tension EAαΔT. In a sling of two legs at θ above the horizontal, one each side, T = W/(2 sin θ) in each leg, and at 30° that is W.

A block pulled along level ground at the best angle, φ = arctan μ, takes T = μW/√(1 + μ²), and the force pressing it on the ground is N = W/(1 + μ²).

On a drum the ratio for n turns is e^(2πnμ): the ratio for two turns is the square of the ratio for one.