Catenary Cable Tension 1

[32huynh]

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

 

[ishvaaag]

One ready way to do this is with the cable element in SAP 2000; there maybe some formula somewhere, or just a derivation for the case at some book or closed form formula in a formulary. I may be looking for that tomorrow. At 1/52 sag to span it may show high stresses, anyway.

 

[paddingtongreen]

Not counting the weight of the wire, the the wire tension would be 27 times the point load.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[hokie66]

A couple of threads which may help:

thread507-246123
thread507-323861

 

[ishvaaag]

This worksheet (printout) solves the funicular shape that would have a particular sagitta and span (final shape) for some point loads and uniform loads in projection (not much different from cable weight distribution at so low separation for horizontal). The funicular is a scaled shape of the isostatic moment of the loads set upon a straight line joining supports.

 

[fcu45]

Use SAP2000 it is having very powerful analysis engine for non-linear cable analysis with lots of parameters that you can control.

Good luck

 

[ishvaaag]

Even if SAP2000 will be the simpler way for those that have worked enough the cable models, I have found a swedish thesis (in english) that seems to address the case. See section 6.2 at p. 48.

http://www.maths.lth.se/na/courses/Documents/papini.pdf

Interesting as well to see how structural design continues driving some mathematical efforts as in the early XIX century.

 

[BAretired]

paddington says 27 times P. Let's see, neglecting weight of cable, M = PL/3 and T = M/h or PL/3h.

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

 

[paddingtongreen]

I am not able to reproduce my number, it is obviously wrong. The tension in the cable is maximum between the support and the load so I took attempted 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.

 

[MiketheEngineer]

Google it - I just did a couple of days ago - a 1,000 sites!!

 

[BAretired]

Actually, the number I calculated is the horizontal component of the cable force. The cable force is maximum where the slope is maximum, namely at each support. But the slope is pretty small, so it doesn't make a big difference.

BA

 

[32huynh]

Thanks a lot everyone for your input and feedback. I'll try to wrap my head around these suggestions, will definitely be back if I have any confusion!

 

[Baires92]

The resultats of the FE analysis with PLS-CADD software:

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

 

[BAretired]

Baires92 has solved a different problem than the one given. In the problem given, the cable sag is 21' after application of third point loads, not before.

BA

 

[MiketheEngineer]

Close enoguhh answer----

L = 44’ Sag = 4’-0’’
Tension = wL^2 / 8d d=Sag

WATCH UNITS

T = (115 plf)(44’)(44’) / 8(4’)
T =6960 lbs.

 

[Baires92]

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!

 

[BAretired]

Using my earlier approximation with P = 1000# and w = 0.399#/':

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

 

[paddingtongreen]

@BA, I would guess that his 21" sag includes the strain and maybe an unwinding factor in the, shorter, starting cable.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[32huynh]

Thanks again for all the input. Here is the other parameter I forgot to include:

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!

 

[BAretired]

At 0.581#/' the horizontal component of the cable is approximately 21,645#.

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