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Civil
Horizontal tension in a parabolic suspension cable
Constant horizontal tension of a flexible cable between supports at the same level, with midspan sag f and a uniform vertical load w per metre of horizontal span L.
- Ref
- CIV-40
- Rev
- v1
- Field
- Civil
- Result
- H [N]
For a flexible cable under a uniform vertical load w per unit horizontal length, moment equilibrium of half the span about the lowest point gives H·f = w(L/2)(L/4) = wL²/8, so H = wL²/(8f). The cable takes a parabolic shape. Doubling the sag halves the horizontal tension, but the cable must then span a deeper gap.
Constant horizontal tension of a flexible cable between supports at the same level, with midspan sag f and a uniform vertical load w per metre of horizontal span L. Moment equilibrium of half the span about the low point gives wL²/(8f). The cable has no bending stiffness, so H rises sharply as the sag shrinks.
Calculator
- Total vertical load, wL60,000 N
- Sag to span ratio, f/L0.1
- Tension if the sag were doubled37,500 N
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The horizontal tension is 75 kN, with a sag-to-span ratio of 1/10.
- Cables are normally designed with sag-to-span ratios of about 1/8 to 1/12 as a compromise; this one is within or deeper than that range.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| L Span | 5 to 500 m | 2,083 | 2.083e7 | 2.083e7 | +10% | +21 % |
| w Load per unit span | 10 to 100,000 N/m | 375 | 3,750,000 | 3,750,000 | +10% | +10 % |
| f Midspan sag | 0.5 to 30 m | 7,500 | 450,000 | 442,500 | +10% | −9.09 % |
Sweep of Midspan sag
| Midspan sag (m) | Horizontal tension (N) |
|---|---|
| 0.5 | 450,000 |
| 4.188 | 53,730 |
| 7.875 | 28,570 |
| 11.56 | 19,460 |
| 15.25 | 14,750 |
| 18.94 | 11,880 |
| 22.63 | 9,945 |
| 26.31 | 8,551 |
| 30 | 7,500 |
Sweep one variable
Variables and how each one contributes
Horizontal tension
What the equation gives you. Everything else feeds it.
Load per unit span
The vertical load per unit horizontal length of the span, in N/m. Raises the tension in direct proportion. Raising it by 10 % (to 2,200) moves the result by +10 %, from 75,000 to 82,500 N.
Span
The horizontal span, in m. Raises the tension with the square of the span. Raising it by 10 % (to 33) moves the result by +21 %, from 75,000 to 90,750 N.
Midspan sag
The midspan sag below the supports, in m. A smaller sag raises the tension in inverse proportion. Raising it by 10 % (to 3.3) moves the result by −9.09 %, from 75,000 to 68,180 N.
Worked example with the values above
- L² = 900 m², and 8f = 24 m.
- Numerator: wL² = 2,000 × 900 = 1,800,000.
- H = 1,800,000 / 24 = 75,000 N.
Assumptions and limits
- Supports at the same level and vertical loads only, for arches and for cables.
- A flexible cable with negligible bending stiffness and a load uniform along the horizontal span (parabola).
Common mistakes
- Using the cable formulas for a cable under its own weight, which hangs as a catenary.
- Using sag measured from the wrong reference line.
- Forgetting the vertical component of tension at the supports.
Statistics and uncertainty
Cable tension is inversely proportional to sag, so the formulas are very sensitive to the measured sag: a 5 % error in sag gives a 5 % error in tension. Real cables also stretch and have some bending stiffness, which shifts results by a few per cent.
Where it is used
- Sizing cables, anchors, abutments and ties.
- Finding thrust and moments in three-hinged arches.
Maths you need
- Funicular shape: the shape that carries load by axial force only
- Three-hinge condition: the moment at the crown hinge is zero
- Horizontal tension: the constant horizontal component along a loaded cable