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Concentrated Load on a Prestressed Cable

Concentrated Load on a Prestressed Cable

Concentrated Load on a Prestressed Cable

(OP)
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.

http://imgur.com/fz8Y1" border="0" alt="" />

RE: Concentrated Load on a Prestressed Cable

(OP)
http://i.imgur.com/fz8Y1.jpg" border="0" alt="" />

I apologize, I can not figure out how to edit my initial post as I see that the picture did not work correctly.

RE: Concentrated Load on a Prestressed Cable

After posting, you cannot edit further. What you can do is post again with the correct picture.
With a span of 262' and a sag of only 2', you would be close enough to assume that the maximum moment, M is WL/8 + PL/4 where W is the total uniform load, P is 20# and L is 262'.
The tension in the cable is M/s where s is the sag or in this case, 2'.

BA

RE: Concentrated Load on a Prestressed Cable

Could you tell us the weight per ft of the cable and the elasticity of it?

I'm a bit surprised BAretired has brought moments into it, but that may be a cunning shortcut I haven't thought of.

The few references I have seen for plug and chug on cables have been approximate (particularly the shape of the catenary), and I would be very concerned in the case of such a taut system that errors could rapidly get out of hand. I'm a bit surprised you can ignore the horizontal stiffness of the anchors, for example.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: Concentrated Load on a Prestressed Cable

Greg,
Yes, BA is clever. The moment is not the moment in the cable, which is taken to be zero, but the couple formed by the cable tension and the anchor force.

RE: Concentrated Load on a Prestressed Cable

Thanks hokie. More precisely, M/s is the horizontal component of the cable tension. The actual cable tension at the supports is slightly larger to account for the cable slope.

BA

RE: Concentrated Load on a Prestressed Cable

Yes, I realized after posting that I had left out a word.

Further to the OP's problem, it turns out that 20 pounds is not a "relatively small concentrated load", as that length of 5 mm cable only weighs about the same, or a little less. So the sag will be approximately tripled by the concentrated load. The problem is complicated by the inherent properties of wire rope.

RE: Concentrated Load on a Prestressed Cable

I should have said the cable force will be approximately tripled for the same sag. Knowing the actual mass of your cable, you can get a more accurate answer. With that information and the cable properties, you could figure the change in length of the cable and a new catenary length. The actual additional sag would be somewhere between the catenary ordinate and the ordinate defined by two straight lines totalling the catenary length.

RE: Concentrated Load on a Prestressed Cable

(OP)
Thanks for the responses,

hokie66, what I meant by "relatively small load" is that it is less than the weight of the wire required, as opposed to a lifeline system where the concentrated load is two orders higher.

The method BA mentioned is in the transmission line design manual, however it does not say what assumptions are made when using this formula. What I'm concerned about is that it might not properly take into account the inital prestress required to obtain a 2ft sag for the 262' span.

Wwire = 0.065lb/ft
E = 4.18e9 psf (200 MPa)
Area = 1.33e-4 sqft (There are two cables though, so double this area in axial deflection calcs)

I cannot seem to get the picture to work, however you can access it from this link
http://imgur.com/fz8Y1

RE: Concentrated Load on a Prestressed Cable

The shape of the loaded cable will be similar to the shape of the bending moment diagram of a 262' long beam. Finding the sag and hence the tension with the added 20# load could be an iterative process.
Another issue: wind load on the cable might influence cable tension more than the 20# hanging load.

BA

RE: Concentrated Load on a Prestressed Cable

i think BA's expression is for a beam in bending (with a distributed load "w" and a point load "P"). but i'd question if the small cable can carry much load in bending.

the cable's catenary shape is how it reacts the cable weight (with cable tension) ... the same way a cyclinder reacts pressure with membrane stress (pR/t).

is it reasonable to assume that the cable reacts the point load with tension ? (so there'd be a kink at the load point) ... you could you geometry and strain relationships to relate the tension to the change in length to the angle created.

then superimpose the two tensions.

RE: Concentrated Load on a Prestressed Cable

rb1957:
The cable carries no load in bending. It follows the funicular shape which happens to be geometrically similar to the bending moment diagram of a beam.

BA

RE: Concentrated Load on a Prestressed Cable

BA,
that's what i meant by "the cable's catenary shape is how it reacts the cable weight (with cable tension)" ... ie as opposed to a beam reacting transverse loads by bending

and that's why i was a little surprised by your original post (M = WL/8+PL/4)

practically will the cable (weighing some 150 lbs) notice an extra 20 lbs ??

RE: Concentrated Load on a Prestressed Cable

(OP)
rb,

The cable is only about 35lb (0.065lb/ft, 2 times cable length as there are two of them). By my calculations, (using trigonometry and (TL/AE)=deflection, then iterating until a result is found) the additional point load increases the sag from 2ft to 4ft, thereby having a significant effect on tension in the cable. My problem is that I do not know if this tension is now ADDED to the initial tension required to hang in the unloaded condition or if this tension is the sole answer.

RE: Concentrated Load on a Prestressed Cable

woops ... one too many decimal places !

i think superposition should work ... so you get 2' deflection for the 20 lbs point load and 2' from the catenary ?

RE: Concentrated Load on a Prestressed Cable

W = 35#
P = 20#
1. Cable only, s = 2'
M = 35*262/8 = 1146'#
H = M/s = 573#
R = 35/2 = 17.5#
Tmax = (573^2 + 17.5^2)^0.5= 573#

2. Cable plus point load
M = 1146 + 20*262/4 = 2456#
Now, we must calculate the revised sag. I have not done that, but assuming it changes from 2' to 4' per DickHalloran then:
H = 2456/4 = 614#
Tmax = (614^2 + 27.5^2)^0.5 = 615#
So the added point load makes a considerable change to the sag but adds only 41# to the tension.
To check the revised sag, it would be necessary to calculate the length of cable without the point load, then adding the stretch and finally, computing the revised sag consistent with the shape of the BM diagram.

BA

RE: Concentrated Load on a Prestressed Cable

I think BA has it about right. If the same sag of 2ft is to be maintained, then the 20 lb load approximately doubles the tension force required in the cables (I said before that the force would be tripled, but I thought there was only one cable).

rb1957, the relation of the shape of a hanging cable to the moment diagram is the basis for the method of load balancing in post-tensioned concrete structures. Neat trick which you might not use in your field.

RE: Concentrated Load on a Prestressed Cable

The cable length is about 262.04' for a sag of 2' (no point load).
The cable length is approximately 262.13' for cable plus point load assuming a sag of 4' at midspan.
This means that the cable stretched 0.09' or about 1.1" over 262 feet under an additional tension of 41#. Does that sound about right?

BA

RE: Concentrated Load on a Prestressed Cable

I doubt that number for the additional sag. I think there is not sufficient wire. If you imagine the curve as circular (not too far out)the radius must be large and the curved cable (length of arc) not much longer than the straight line from the support to the low point.

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

RE: Concentrated Load on a Prestressed Cable

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

RE: Concentrated Load on a Prestressed Cable

Have you looked at the US Wire Rope Handbook, particularly pages 44 through 48. You can download it from
http://www.slideruleera.net/USSWireRopeEngrHandBoo...

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

RE: Concentrated Load on a Prestressed Cable

Another reference would be from PTI on prestressed barrier cables.
http://www.post-tensioning.org/Uploads/Technote14....

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

RE: Concentrated Load on a Prestressed Cable

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Concentrated Load on a Prestressed Cable

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?

RE: Concentrated Load on a Prestressed Cable

By formula (weight only):
tension H=wL2/8f=0.065*2622/8/2=279 N (against 265 from the sheet above)

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Concentrated Load on a Prestressed Cable

(OP)
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

RE: Concentrated Load on a Prestressed Cable

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.

Quote (DickHalloran)

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

RE: Concentrated Load on a Prestressed Cable

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

RE: Concentrated Load on a Prestressed Cable

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


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: Concentrated Load on a Prestressed Cable

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

RE: Concentrated Load on a Prestressed Cable

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


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: Concentrated Load on a Prestressed Cable

Thanks Greg.

BA

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