Structural Analysis Fundamentals / Cables
Distributed loads: parabola and catenary
Deriving the cable shape under a uniform load, and when the catenary matters.
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Why a flat cable pulls so hard
Sag and force trade against each other, and not gently. Flatten the cable and watch the tension climb — this is the whole of cable behaviour in one slider.
Model
- Sag
- 4.80 m
- Sag / span
- 1 : 12.5
- Horizontal pull H
- 750 kN
- Vertical at each support
- 240 kN
- Maximum tension (at supports)
- 787 kN
- Minimum tension (at mid-span)
- 750 kN
- T max / total load
- 1.64 ×
- Cable length
- 61.01 m
- Extra length over the span
- 1.01 m
- Catenary H (self-weight)
- 756 kN
- Parabola vs catenary
- 0.8 %
Things worth trying
- Start at 8% sag on a 60 m span. The horizontal pull is 750 kN to carry 480 kN of load — the cable already pulls sideways half again as hard as the load it is carrying.
- Now drag the sag down to 4%. H doubles to 1500 kN. The load has not changed at all: H = wL²/8h, so halving the sag doubles the force. This is why a nearly-level cable is a bad idea and why washing lines sag.
- Take the sag to 1%. H reaches 6000 kN — twelve and a half times the load being carried. Every anchorage, every foundation, has to take that horizontally.
- Go the other way, to 25%. H falls to 240 kN, half the load. Cheap in force, expensive in headroom: that is the trade a suspension bridge designer makes.
- Watch the T max / total load row as you move the sag. At 8% it is 1.64; at the slider's deepest 25% it has only fallen to 0.71. The floor is 0.5 — as the sag deepens H vanishes but each support still carries half the load vertically — but it is an asymptote, and a buildable cable never gets near it.
- Notice the maximum is at the SUPPORTS, not mid-span. The horizontal component is constant everywhere; what changes is the vertical, which is greatest where the cable is steepest.
- Now add the catenary. At small sag the two curves are indistinguishable and the difference in H is under 1% — the parabola is an excellent approximation for a deck-loaded cable.
- Push the sag to 25% and the gap opens up. The parabola assumes load uniform along the SPAN; the catenary assumes it uniform along the CABLE. When the cable is much longer than the span, that distinction starts to matter.
- Now the one that catches people. H = wL²/8h looks like a square law in span, and it is — at a FIXED SAG. Hold the sag RATIO instead and h grows with L too, so H = wL/8×(h/L) is merely LINEAR: 60 m gives 750 kN, 120 m gives 1500, 180 m gives 2250. Which law applies depends entirely on what you are holding constant.
Why this matters
Closely spaced hangers can transfer deck load approximately uniformly per horizontal metre. Under that loading an ideal flexible cable has a parabolic profile.
By the end of this lesson you should be able to
- Derive the parabolic cable profile from equilibrium
- Relate sag, span, load and horizontal thrust
- Say what changes when the load follows the cable rather than the span
From first principles
The parabolic cable
We want to show: that a cable under a load uniform along the span hangs as a parabola, with H = wL²/8h.
Think about what a cable has to do. It carries load purely by changing direction: the slope changes as you move along it, and that change in slope is what turns the load downwards into tension along the cable. Where the load is heaviest per metre, the slope has to change fastest. With a uniform load the slope changes at a constant rate, and a curve whose slope changes at a constant rate is a parabola. The formal derivation just puts numbers on that idea.
Uniform self-weight per unit cable length gives a catenary. A steep segment contains more cable, and therefore more weight, per horizontal metre than a shallow segment.
The parabola approximates the catenary at shallow sag. Check the error in the quantity of interest before using that approximation; combined deck loading and cable weight may require a more detailed model.
Worked example
Main cable of a footbridge
Given
- Span between towers 60.0 m, supports at the same level
- Mid-span sag 6.0 m
- Deck load carried by the cable 8.0 kN per horizontal metre
Find
The horizontal thrust, the maximum tension and where it occurs.
Assumptions
- Load uniform along the horizontal projection (closely spaced hangers)
- Cable self-weight neglected beside the deck load
- Perfectly flexible cable
Practice
A cable spans 40.0 m between level supports with a mid-span sag of 4.0 m, carrying 10 kN per horizontal metre. What is the horizontal component of tension?
Practice
For that same cable, what is the maximum tension? Give your answer in kN.
Summary
- A uniform load along the span gives a parabola: y = wx²/2H
- H = wL²/8h — thrust grows with the square of the span and falls with sag
- T is greatest at the supports and equals H at the lowest point
- Self-weight along the cable gives a catenary, which matters for deep sags