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Support tension of a uniformly loaded parabolic cable
Total cable tension at either support: the horizontal tension H and the vertical component wL/2 (half the total load) combine as a vector sum.
- Ref
- CIV-41
- Rev
- v1
- Field
- Civil
- Result
- Tend [N]
The cable tension has two components at a support: the constant horizontal tension H and a vertical component equal to half the total load, wL/2, since the supports share the load equally. They combine as a vector sum, Tend = √(H² + (wL/2)²). This is the maximum tension in the cable, because the horizontal component is constant and the vertical component is largest at the supports.
Total cable tension at either support: the horizontal tension H and the vertical component wL/2 (half the total load) combine as a vector sum. This is the largest tension in the cable. The supports are at the same level.
Calculator
- Vertical component at each support, wL/230,000 N
- Cable slope at the support, atan(wL/(2H))21.8 °
- Tension relative to H, Tend/H1.077
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 support tension is 80.78 kN, 7.703 % more than the horizontal component, with the cable meeting the support at 21.8° to the horizontal.
- Anchor fittings and saddles must be designed for this larger value, and the anchor must resist both components.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| H Horizontal tension | 100 to 1e7 N | 30,000 | 1e7 | 9,970,000 | +10% | +8.68 % |
| w Load per unit span | 10 to 100,000 N/m | 75,000 | 1,502,000 | 1,427,000 | +10% | +1.44 % |
| L Span | 5 to 500 m | 75,170 | 505,600 | 430,400 | +10% | +1.44 % |
Sweep of Horizontal tension
| Horizontal tension (N) | Tension at the support (N) |
|---|---|
| 100 | 30,000 |
| 421.7 | 30,000 |
| 1,778 | 30,050 |
| 7,499 | 30,920 |
| 31,620 | 43,590 |
| 133,400 | 136,700 |
| 562,300 | 563,100 |
| 2,371,000 | 2,372,000 |
| 1e7 | 1e7 |
Sweep one variable
Variables and how each one contributes
Tension at the support
What the equation gives you. Everything else feeds it.
Horizontal tension
The horizontal tension component, in N. Raises the support tension, most strongly when H dominates. Raising it by 10 % (to 82,500) moves the result by +8.68 %, from 80,780 to 87,790 N.
Load per unit span
The vertical load per unit horizontal span, in N/m. Raises the vertical component and so the tension. Raising it by 10 % (to 2,200) moves the result by +1.44 %, from 80,780 to 81,940 N.
Span
The horizontal span, in m. Raises the vertical component wL/2 in direct proportion. Raising it by 10 % (to 33) moves the result by +1.44 %, from 80,780 to 81,940 N.
Worked example with the values above
- Vertical component: wL/2 = 2,000 × 30 / 2 = 30,000 N.
- Squares: H² = 5.625×109, (wL/2)² = 9×108.
- Tend = √(6.525×109) = 80,777.5 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