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Queensferry

Module 5 · Lesson 5.1

Cables carrying point loads

Why the horizontal pull is constant, and how to analyse a loaded cable.

Why this matters

A cable is the most efficient structural element there is: every fibre is working at the same stress, in tension, with nothing wasted. Understanding it also explains the arch, which is the same structure turned upside down.

By the end of this lesson you should be able to

  • Say why a cable can only carry axial tension
  • Prove that the horizontal component of tension is constant
  • Find the tension in a cable carrying concentrated loads

What you should already know

  • Resolving forces and equilibrium (Module 2)
  • Taking moments about a chosen point (Module 2)

A cable is perfectly flexible. It has essentially no bending stiffness, so it cannot resist a moment anywhere along its length. That single fact does all the work in this module.

If a cable cannot carry moment, then at every point the internal force must act along the cable. It can only pull. And because it can only pull, the cable has no choice about its shape: it hangs in whatever geometry makes the load path pure tension. Change the loading and the cable physically moves to a new shape.

From first principles

The horizontal component of tension is constant

We want to show: that H is the same at every point along a cable carrying only vertical loads.

Take any short piece of the cable and look at it on its own. The only things acting on it are the cable tensions pulling at each end, and whatever vertical load it carries. Now resolve horizontally. The vertical loads contribute nothing, because they are vertical. So the horizontal pull coming in at one end must exactly equal the horizontal pull going out at the other. That has to be true for every piece you choose, so the horizontal component never changes anywhere along the cable.

A cable spanning 20 m with a 30 kN load at midspan, sagging 2 m, with the horizontal and vertical force components annotated30 kNdip 2 mL = 20 mH = 75 kN · V = 15 kN · Tmax = 76 kN
The cable of the worked example below. The shape is not drawn by eye — it is computed as the bending moment diagram of the equivalent beam divided by H, which is what a perfectly flexible cable must adopt.

Worked example

Cable carrying a single central load

Given

  • A cable is anchored at two points 20.0 m apart at the same level
  • A load of 30 kN hangs at mid-span
  • The cable sags 2.0 m below the anchorage line at the load point

Find

The horizontal component of tension and the tension in the cable.

Assumptions

  • The cable is perfectly flexible
  • Cable self-weight is negligible beside the 30 kN load

    Predict first

    If the same 30 kN load were hung from a cable with only 1.0 m of sag instead of 2.0 m, what would happen to H?

    Practice

    A cable spans 20 m between level supports and carries a single 60 kN load at mid-span. The sag at mid-span is 2.0 m. What is the horizontal component of the cable tension, in kN?

    Practice

    For the same cable, what is the maximum tension, in kN? (It occurs in the segment next to a support.)

    Practice

    If the same cable is slackened so the mid-span sag doubles to 4.0 m, what is the new horizontal thrust, in kN?

    Summary

    • A flexible cable carries no moment, so its internal force acts along the cable
    • With vertical loads only, H = T cos θ is constant everywhere
    • Tension is greatest where the cable is steepest, at the supports
    • Reducing the sag increases H in inverse proportion
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    This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint