Catenary Cable Tension
Catenary Cable Tension
(OP)
All,
I've been searching for a way to calculate the tension in a cable assuming catenary configuration due to two vertical point loads, each at the third point of the span.
I thought I analyze each point separately and treat that point as a tried analyzing it similar to method of joints. But my analysis says with 1 k vertical load, the tension is tremendously larger, and it doesn't seem to make sense.
Some information that may be useful:
Sag- 21 ft
Span- 1100 ft
Beta- I calculated this to be 4.68 degrees, may be wrong.
Any insight is greatly appreciated! Thanks
I've been searching for a way to calculate the tension in a cable assuming catenary configuration due to two vertical point loads, each at the third point of the span.
I thought I analyze each point separately and treat that point as a tried analyzing it similar to method of joints. But my analysis says with 1 k vertical load, the tension is tremendously larger, and it doesn't seem to make sense.
Some information that may be useful:
Sag- 21 ft
Span- 1100 ft
Beta- I calculated this to be 4.68 degrees, may be wrong.
Any insight is greatly appreciated! Thanks






RE: Catenary Cable Tension
RE: Catenary Cable Tension
Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Catenary Cable Tension
thread507-246123: Catenaries with point loads
thread507-323861: Concentrated Load on a Prestressed Cable
RE: Catenary Cable Tension
RE: Catenary Cable Tension
Good luck
RE: Catenary Cable Tension
http://www.maths.lth.se/na/courses/Documents/papin...
Interesting as well to see how structural design continues driving some mathematical efforts as in the early XIX century.
RE: Catenary Cable Tension
With L = 1100', h = 21', T = P*1100/63 = 17.5P
If the weight of cable is known per lineal foot, the additional moment is approximately wL2/8 and the added tension is roughly wL2/(8*21).
That method should be sufficiently accurate for all practical purposes.
BA
RE: Catenary Cable Tension
tookattempted to take, the ratio of the slope length and the sag as the multiplier. While technically correct it makes no practical difference to BA's answer.Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Catenary Cable Tension
RE: Catenary Cable Tension
BA
RE: Catenary Cable Tension
RE: Catenary Cable Tension
Assumed:
Cable: 7/16 inch EHS 7 Strands Steel, Section 0.1156 in^2, Weight 0.399 lb/ft
Temperature initial and final 60F
Initial condition:
Span 1100ft, sag 21 ft
Tensions: Horizontal 2872.2 lbs, Span Max 2880.5 lbs
Final condition (loads added):
Load 1k each 1/3 of span
Tensions: Horizontal 10500.9 lbs, Span Max 10571.4 lbs
Max sag: 40.57 ft
Resultats account real behaviour of the 7/16 inch EHS 7 Strands Steel Cable
RE: Catenary Cable Tension
BA
RE: Catenary Cable Tension
L = 44’ Sag = 4’-0’’
Tension = wL^2 / 8d d=Sag
WATCH UNITS
T = (115 plf)(44’)(44’) / 8(4’)
T =6960 lbs.
RE: Catenary Cable Tension
OK, now we have:
Assumed:
Cable: 7/16 inch EHS 7 Strands Steel, Section 0.1156 in^2, Weight 0.399 lb/ft
Initial condition (loads added):
Load 1k each 1/3 of the span
Span 1100ft, sag 21 ft
Tensions: Horizontal 20294.5 lbs, Span Max 20331.0 lbs (98% of the ultimate tension capacity, NG for chosen type of the cable)
For cable no large load plastic deformation or creep accounted.
Merci!
RE: Catenary Cable Tension
M = PL/3 + wL2/8
= 1000*1100/3 + 0.399(1100)2/8 = 427,000'#
H = 427,000/21 = 20,334#
which agrees closely with 20,294.5 calculated by Baires92 above.
Actually, I would have expected the precise answer to be a little larger than my approximation because the length of cable must be longer than 1100'. Instead, it is a little smaller. Why would that be? Any comments?
BA
RE: Catenary Cable Tension
Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Catenary Cable Tension
Wire: ASTM A475 7/8" (0.581 lb/ft according to ASTM).
BA, is there a guidance/reference for your equation? I wanted to do a little more reading.
Thanks!
RE: Catenary Cable Tension
32huynh, in response to your question, the shape of a cable under various types of load follows the shape of the bending moment curve of a beam with the same span. That is an exact relationship. To draw the bending moment diagram of a loaded beam is to draw the configuration of a cable under the same load to some scale. If you stipulate the maximum sag, in your case 21', the geometry of the cable can be found at every point along the cable.
1. A weightless cable under the action of a number of concentrated loads assumes the shape of the bending moment diagram of a beam of the same length. It consists of a series of straight line segments, sometimes called the funicular polygon.
2. A cable having a uniform weight per lineal foot of horizontal span follows a parabolic curve and has a moment of wL2/8 where w is the uniform load per foot horizontally.
3. A cable having a uniform weight per lineal foot of curve follows a catenary which you can find on a number of Internet sites.
4. When the sag of a cable is small (21' in 1100' is small), the load per curved foot is not too far from the same load per horizontal foot, so a reasonable approximation may be found by assuming a parabolic curve but it is an approximation.
5. Unless there are horizontal forces applied in the length of cable, the horizontal force, H is constant over its full length but the cable tension varies according to the slope, getting larger toward the supports.
BA