Finds the shape, length and forces of a cable between two supports at any heights, under its own weight, uniform loads in plan and point forces, fixed by a sag, the lowest point, the horizontal force, the length or the cut length, with the stretch of the cable, temperature and the worst arrangement of the variable loads. Every step shows its formula.

What it checks

The chord A–B: l_c = √(l² + Δh²) and its slope.
The shape: the funicular of the loads as entered, fixed by the one condition you choose. Once that is known the cable is statically determinate. The axial stiffness is E·A.
The loads: the vertical load as entered without the own weight, ΣF, and ΣF_var for the worst-case loads all present.
The worst arrangement: a cable is non-linear, its shape changes with the loads, so every arrangement is analysed on its own, N_arr in all, with the temperature changes one at a time. The module names the arrangement that gives the largest cable force and the one that gives the largest sag.
Horizontal force and shape: H from T·cos θ = H everywhere; the lowest point x_low; y_low, where the cable is horizontal and T = H; the length S = ∫ √(1 + y′²) dx.
Cut length: S₀ = ∫ ds/(1 + T/(E·A)) at the reference temperature, and the stretch S − S₀.
Support reactions: the horizontal pull H, R_y, T = √(H² + R_y²) and the angle θ of the cable, as entered and as largest and smallest, with T_max. Equilibrium: ΣR_y equals the loads as entered plus g_c·S.
Along the cable: the segments between the forces and a table of equally spaced points.
Cable force, optional: T_max ≤ T_R.

What you enter and what you get

You enter the span l and the height Δh of the right support above the left one, negative when B is lower. Then what fixes the shape: a sag f below the chord at a position x_f, the depth of the lowest point below the lower support, the horizontal force, the length of the cable, or its unstressed (cut) length. The stretch is neglected, or included from E and A.

For the loads: the own weight g_c per metre of cable, up to four uniform loads per metre in plan over the whole span or part of it, up to eight vertical forces, and up to two temperature changes with α_T, each as entered or in the worst case. Then the resistance T_R and the number of points for the table.

You get the shape, the lengths, the forces at the supports and along the cable, and a pass or fail for the cable force.

Assumptions and limits

A cable has no bending stiffness. It takes the shape of its loads, straight between forces, a parabola under a load spread evenly in plan, a catenary under its own weight, and carries them in tension only.
The loads as entered fix the cable. The worst-case loads and the temperature changes then act on that same cable: the same length, or the same cut length if it stretches.
Warming lengthens the cable: it sags more and its force drops. Two temperature changes in the worst case are alternatives, a warm and a cold day, never taken together.
T_R is the breaking force divided by the safety factor of the code you design to, or a design resistance, with the loads entered to match.
The usual design sag is about l/8 to l/12.

Worked example

A cable of 120 m span with the right support 8 m lower, a sag of 12 m at mid-span, stretch included with E = 195,000 N/mm² and A = 6,000 mm², own weight 0.55 kN/m and 30 kN/m in plan over the whole span as entered; in the worst arrangement 20 kN/m over the first 60 m, 250 kN at x = 90 m, and +30 °C or −30 °C with α_T = 12·10⁻⁶/°C; a resistance of 8,000 kN. The cable force check passes. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Span between the supports l	120 m
Height of the right support above the left one Δh	-8 m
The shape is fixed by	A sag you choose
Position of the chosen sag x_f	60 m
Sag at that position, below the chord f	12 m
Stretch of the cable under its force	Included — from E and A
Modulus of elasticity of the cable E	195000 N/mm²
Metallic cross-section area A	6000 mm²
Own weight per metre of cable g_c	0.55 kN/m
Number of uniform loads per metre in plan	2
Over	Whole span
Intensity per metre in plan q₁	30 kN/m
Arrangement	As entered
Over	Part of the span
Start x_a,2	0 m
End x_b,2	60 m
Intensity per metre in plan q₂	20 kN/m
Arrangement	Worst case
Number of vertical forces	1
Position x₁	90 m
Force (downward positive) P₁	250 kN
Arrangement	Worst case
Number of temperature changes	2
Coefficient of thermal expansion α_T	12 10⁻⁶/°C
Change ΔT₁	30 °C
Arrangement	Worst case
Change ΔT₂	-30 °C
Arrangement	Worst case
Check the cable force against a resistance	Yes
Resistance of the cable T_R	8000 kN
Points of the table along the cable	11

Calculation: The cable
Step	Value	Ref.
Chord A–B: length and slopel_c = √(l² + Δh²); slope -3.8°	l_c = 120.266 m
What fixes the shapewith the loads as entered; the cable is statically determinate once this is known (the funicular shape of its loads)	f = 12.000 m at x = 60.00 m
Axial stiffness	E·A = 1,170.0 MN

Calculation: The loads
Step	Value	Ref.
Vertical load as entered, without the own weight	ΣF = 3,600.00 kN
Worst-case loads, all present	ΣF_var = 1,450.00 kN

Calculation: The worst arrangement
Step	Value	Ref.
Loads that may each be present or absentuniform load 2, force 1, temperature change 1, temperature change 2	k = 4
Arrangements analysed one by oneevery combination; the temperature changes one at a timea cable is non-linear (its shape changes with the loads): every arrangement is analysed on its own	N_arr = 12
Largest cable force with	uniform load 2, force 1, temperature change 2
Largest sag with	uniform load 2, temperature change 1

Calculation: Shape and horizontal force
Step	Value	Ref.
Horizontal component of the cable force, loads as enteredfrom the condition: T·cos θ = H everywhere	H = 4,583.76 kN
Lowest point, loads as enteredwhere the cable is horizontal and T = H	x_low; y_low = 70.00; -16.333 m
Length of the cable, loads as enteredS = ∫ √(1 + y′²) dx; 2.585 % longer than the chord	S = 123.375 m
Unstressed (cut) lengthS₀ = ∫ ds/(1 + T/(E·A)), at the reference temperaturecut the cable to this length (before end fittings and tolerances)	S₀ = 122.880 m
Stretch under the loads as entered	S − S₀ = 495.3 mm

Calculation: Support reactions
Step	Value	Ref.
Equilibrium: sum of the vertical reactionsequals the loads as entered and the own weight g_c·S = 3,667.86 kN	ΣR_y = 3,667.86 kN

Step	1Ax = 0	2Bx = 120.00 m
Horizontal pull on the support, loads as entered H kNtowards the span	4,583.76	4,583.76
Vertical reaction, loads as entered R_y kNupward positive	2,139.79	1,528.06
Cable force at the support, loads as entered T kNT = √(H² + R_y²)	5,058.61	4,831.75
Angle of the cable to the horizontal, loads as entered θ °downward into the span positive	25.02	18.44
Horizontal pull, largest / smallest H kN	6,302.34 / 4,552.99	6,302.34 / 4,552.99
Vertical reaction, largest / smallest R_y kN	3,216.98 / 2,137.75	1,906.23 / 1,525.97
Cable force, largest T_max kN	7,075.91	6,582.79

Calculation: Segments between the forces, loads as entered
Step	Lengthm	Slope θ°, rising +	T at startkN	T at endkN	Ref.
Segment 1x = 0.00 to 90.00 m	92.523	-25.02	5,058.61	4,624.31
Segment 2x = 90.00 to 120.00 m	30.852	7.59	4,624.31	4,831.75

Calculation: Along the cable
Step	Sag fm, as entered	Sag fm, largest	Sag fm, smallest	TkN, largest
x = 0.00 m	0.000	0.000	0.000	7,075.91
x = 12.00 m	4.321	4.918	4.151	6,821.30
x = 24.00 m	7.681	8.608	7.412	6,612.88
x = 36.00 m	10.081	11.070	9.785	6,455.07
x = 48.00 m	11.520	12.304	11.269	6,351.56
x = 60.00 m	12.000	12.333	11.864	6,304.98
x = 72.00 m	11.520	11.726	11.188	6,305.03
x = 84.00 m	10.080	10.529	9.490	6,326.36
x = 90.00 mat a force	9.000	9.593	8.366	6,378.70
x = 96.00 m	7.680	8.125	7.059	6,409.54
x = 108.00 m	4.320	4.513	3.896	6,486.37
x = 120.00 m	0.000	0.000	0.000	6,582.79

Calculation: Check — cable force
Step	Value	Ref.
Largest cable forceT_max ≤ T_R	η = 0.884 · Pass88 % of the capacity used

How this module is verified

Verified against independent calculations used on real projects, before release.

Questions

Which length do I cut the cable to?

The unstressed length S₀, before end fittings and tolerances. In the worked example it is 122.88 m, against 123.37 m under the loads as entered.

Why is each arrangement analysed on its own?

A cable is non-linear: its shape changes with the loads, so the arrangements cannot simply be added. The worked example has 12 of them.

Can I print the calculation?

The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.

Page updated: 14 November 2026

Cable between two supports

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