Structural Analysis Fundamentals / Cables
Cables carrying point loads
Why the horizontal pull is constant, and how to analyse a loaded cable.
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Read the lesson register and outstanding review notesWhy this matters
An ideal flexible cable carries load through axial tension. Its shape and anchorage forces follow from equilibrium, which also provides the basis for the inverted cable analogy used in arch analysis.
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)
The ideal cable model neglects bending stiffness, so the internal force acts along the cable and is tensile. The cable adopts a shape that equilibrates the applied loads. Its length, support geometry and extension determine which equilibrium shape is possible.
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.
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