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.

 

[DickHallorann]

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

 

[BAretired]

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

 

[GregLocock]

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


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[hokie66]

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.

 

[BAretired]

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

 

[hokie66]

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.

 

[hokie66]

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.

 

[DickHallorann]

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

 

[BAretired]

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

 

[rb1957]

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.

 

[BAretired]

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

 

[rb1957]

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 ??

 

[DickHallorann]

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.

 

[rb1957]

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 ?

 

[BAretired]

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

 

[hokie66]

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.

 

[BAretired]

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

 

[GregLocock]

E is not what I need, the elasticity of a cable is much greater than that of the steel from which it is made.

Cheers

Greg Locock


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[paddingtongreen]

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.