Concentrated Load on a Prestressed Cable 1

[DickHallorann]

Hi all,

I've got a situation where there is a 262ft span of 5mm steel cable with a relatively small concentrated load in the center. In the unloaded condition, the cable forms a catenary shape with a maximum sag of 2ft. A load of 20lb is then applied to the center of the span, pulling it out of catenary shape. I've been using a couple different Transmission Line Design Manuals to provide insight into analysing this problem (this system is not a transmission line, however the manuals have semi-useful catenary formulas). It is unclear in these manuals how to include the tension induced by the concecntrated load, with the pre-stress required to maintain an unloaded sag of 2ft.

I have also been unable to find any useful HLL (Horizontal Lifeline) design manuals to help with this problem. In the end I'd like to be able to find the reactions at the supports as well as the maximum tension in the cable. Any help into the matter would be greatly appreciated. I've attached a photograph of the system, the concentrated load is a flow monitoring boat.

 

[BAretired]

A tension of 41# is not sufficient to cause a strain of 1.1" so the final sag is less than 4' and the final tension is larger than previously calculated.
Iteration could be used to find a more precise solution. I don't know of a closed form solution.

BA

 

[slickdeals]

Have you looked at the US Wire Rope Handbook, particularly pages 44 through 48. You can download it from

http://www.slideruleera.net/USSWireRopeEngrHandBook.zip

It’s no trick to get the answers when you have all the data. The trick is to get the answers when you only have half the data and half that is wrong and you don’t know which half - LORD KELVIN

 

[slickdeals]

Another reference would be from PTI on prestressed barrier cables.

http://www.post-tensioning.org/Uploads/Technote14.pdf

It’s no trick to get the answers when you have all the data. The trick is to get the answers when you only have half the data and half that is wrong and you don’t know which half - LORD KELVIN

 

[prex]

As I see it, this problem must be solved in two steps:
1)Self weight only, a suitable tension must be applied to the cable to obtain a sag of 2 ft
2)Now, by maintaining the same distance of support points, a load is applied in the middle.
Step 1 can be calculated by a sheet in the first site below, for beams with tension, here.
Simply set the moment of inertia to a low value (not too low though, as the iterative calculation will fail and you get all zeroes) and set the tension to tentative values until you get the desired sag.
This sheet gives a tension of 265 lb for the 2 ft sag.
Step 2 can be solved by a sheet for beams with held ends, here.
Again the moment of inertia is set to a low value, the uniform load is here set to zero, and the point load to 10 lb (assuming double cable means that each cable will take 10 lb).
The tension is now 185 lb and the sag 3.3 ft.
Now the step 3, the most important one. As the sag in step 1 is very small compared to length, I think we can assume that the geometry of the cable in step 1 will not influence step 2: my conclusion is that the two steps can be simply added, giving a total tension (per cable) of 450 lb and a total sag of 5.3 ft.

prex
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[rb1957]

wonde why prex's tension for weight only is less than 1/2 BA's ?
looking at prex's link, it shows a typical beam problem with a support on a roller ... not really this problem, yes?

 

[prex]

By formula (weight only):
tension H=wL[sup]2[/sup]/8f=0.065*262[sup]2[/sup]/8/2=279 N (against 265 from the sheet above)

prex
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[DickHallorann]

BA, I believe you are correct where you recieve 615# of tension. That is what I have in my calcs as well, however I'm still uncertain as to whether that properly accounts for the initial prestress required for the unloaded condition. Hypothetical, say the initial sag is desired to be 0.1ft. The horizontal tension in this case is much higher, and one would expect that the application of a point load would result in an axial tension significantly higher than 615#. However, using the iterative method to calculate point load sag (3.44ft), and employing the above method, a tension of 710# is calculated. This does not seem correct to me but I am not too familiar with cables.

If we modified the formula slightly so that 's' is changed to 'sINITIAL' and 'sFINAL', would this be allowed? This comes back to my initial concern of can the tensions simply be added together.

H = (1/sINITIAL)*(wL^2/8) + (1/sFINAL)*(pL/4)

Your insight is appreciated

 

[BAretired]

rb1957, I think prex is treating the problem as two cables. I have been treating it as a single cable having the weight of two. Also, the unit weight he is using is not quite the same as a total of 35# so that may be the reason for some differences.
DickHalloran, When the double cable hangs with a two foot sag and no concentrated load, the tension is 35*262/(8*2)= 573# (or 286.5# per single cable).
I do not believe the addition of a 20# concentrated load causes the double cable to sag to 4'. If we assume the final sag is 2.5', then the horizontal component H = 2456/2.5 = 982# per double cable (491# per single).
On this assumption, the added tension is 982-573 = 409# which causes a strain in the cable proportionate to 1/E which you have not yet given us.

If we modified the formula slightly so that 's' is changed to 'sINITIAL' and 'sFINAL', would this be allowed? This comes back to my initial concern of can the tensions simply be added together.

s(initial) is known to be 2' whereas s(final) is unknown and has to be solved by an iterative process.
No, tensions cannot be simply added together. Superposition does not apply to this problem.


BA

 

[BAretired]

If the cable load is treated as a series of point loads, a good approximation can be obtained as shown in the attachment below. In (A), the cable is considered without the point load. In (B), the point load is considered by itself and in (C), the results are combined. Trial sags of 4.0, 3.0 and 3.5 are considered and the length of straight line segments are computed.

In the final trial, the change in unit strain is found to be (131.0493-131.0191)/131 = 0.000230. The change in horizontal component of cable tension is 129#, giving a tensile stress of 129/0.038 or 3395psi. This would be consistent with E if approximately 14.7e6 psi which seems in the ballpark for a cable.

So it appears that the final sag is in the order of 3.5' and the cable tension is about 702#.

BA

 

[GregLocock]

I took another approach, using a spreadsheet model of 2 ft sections, and the standard trick with catenaries, which is that the horizontal component of the tension is constant. From this it is easy to calculate the angle of each section of the cable, given symmetry.

I ignored compliance, since the OP didn't nominate a useful number.

The unladen case needs a tension of 566 lbf to get a 2.00 ft sag, and the total length is 262.0404 ft, at 0.13 lbf/ft cable weight.

The same length of inelastic cable has a sag of 2.21 ft and a tension of 1100 lbf when loaded with 20 lb at midspan.

Obviously ignoring compliance is going to lead to an overestimate of tension and an underestimate of sag, but these numbers are worryingly different to the various options in the thread above.

Cheers

Greg Locock


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[BAretired]

If the cable is inextensible and the sag at mid-span is taken as 2.21' then the sag at the quarter points would be 1.36'. The length of the first two diagonals would be 65.5141 and 65.5055 respectively for a total of 131.0196 in half the length of cable. This compares closely with the value of 131.0191' obtained for Case (A) Cable Weight Alone.

The corresponding H value would then be 2456/2.21 = 1111# which agrees closely with the value found by GregLocock above.

The total strain in the 262' length of cable, taking an E value of, say 15,000,000 psi would be 262*1111/(0.038E) = 0.51' or about 6" which cannot be ignored.

BA

 

[GregLocock]

Thanks, I'll noodle away at adding compliance.

Cheers

Greg Locock


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[GregLocock]

OK, using E*A of pi*5^2/4*2*57.3 kN mm-2 (ie 2 off 7x7 laid cables) I get 3.54 ft of deflection and 686 lbf tension for the 20lbf load. For the unladen case 565 lbf and 2.00ft deflection. The curve of deflection for the latter agrees well with the parabola in the links given above.





Cheers

Greg Locock


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[BAretired]

Thanks Greg.

BA