1. Jul 25, 2009

### arigger

I have been asking this question of everyone I meet and can't find a straight answer.

Imagine a wire rope anchored at both ends, spanning 50ft [15m].
That line is 'pre-tensioned' to... say 500 lbs. [2kN].
The idea is that it is pulled fairly 'flat'.
What happens to the line tension when I apply a load vertically at the mid-point? How about three loads at quarter points?
Do I need to consider the catenary effect or can I work this out free-body style?
Is there one end-all be-all formula for this? Will I be able to read it/ perform the functions required?

I know several ways to get the answer if I know how much sag is in the line or if there is a known vertical component, what entertainment riggers call a 'bridle'. But starting with a 180 degree triangle seems to get me nowhere.

Thanks for looking
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2. Jul 27, 2009

### jmb

If the force (load) applied is much larger than the weight of the rope itself, then to a good approximation you ought to be able to ignore catenary like effects and treat the geometry of the problem as triangular. (If the load is not much more than the weight of the rope itself, and the rope is strongly pretensioned already to keep it very flat, then I wouldn't imagine the load would increase the tension by a significant amount anyway).

Applying any load will cause the rope to deviate from flat slightly, otherwise there would be no upward component of force in the rope to balance the load. The problem is you don't know the value (angle) of this deviation.

If in addition to this assumption (load much greater than rope weight) you can also make the following assumptions:

• The anchors are rigid (otherwise you would have to model their elastic response too).
• The rope is perfectly elastic (otherwise the problem is non-linear, you will need a good material model and will probably need to model the whole thing numerically).

then I believe you can solve the problem without having to physically measure the deviation (angle) from horizontal caused by the load...

Using the terminology defined in the attached diagram one can proceed as follows:

Equating the applied load with the rope's upward force gives:

$$F=2T\sin{\theta}$$ [Equation 1]

In addition, if we write $$l$$ for the actual length of the rope $$l_0$$ for the untensioned length of the rope and $$k$$ is the 'spring constant' of the rope (which can be determined from its dimensions and elastic modulus), we can also write:

$$T=k\left(l-l_0\right)=k\left(\Delta l+l_{\rm p}-l_0\right)$$ [Equation 2]

where we have further split $$l$$ into the change in length due to the application of the load, $$\Delta l$$, and the length of the rope before the load was applied but after pretensioning it, $$l_{\rm p}$$.

From the attached diagram we can see that:

$$\Delta l=l\left(1-\cos{\theta}\right) \Rightarrow \Delta l = \left(\Delta l+l_{\rm p}\right)\left(1-\cos{\theta}\right)$$

Solving the above for $$\Delta l$$ gives:

$$\Delta l = l_{\rm p}\left(\sec{\theta}-1\right)$$

which we can substitute into equation 2 to get:

$$T=k\left(l_{\rm p}\sec{\theta}-l_0\right)$$ [Equation 3]

We then note that we can rearrange equation 1 to get:

$$\sin{\theta}=\frac{F}{2T} \Rightarrow \cos{\theta}=\sqrt{1-\frac{F^2}{4T^2}}$$

substituting this into equation 3 and rearranging, we eventually get a quartic equation for the tension:

$$4T^4+8kl_0T^3+\left[4k^2\left(l_0^2-l_{\rm p}^2\right)-F^2\right]T^2-2kl_0F^2T-k^2l_0^2F^2=0$$

There is a general solution for quartic equations (see http://en.wikipedia.org/wiki/Quartic_function" [Broken]) although it is not pretty! So in theory one should be able to get a closed form answer to the value of $$T$$ from this if one knows the values of $$F,k,l_0\;\text{and}\;l_{\rm p}$$.

I would imagine the same approach is possible when there are several different loads, although I don't know whether this would still yield a closed form solution.

CAVEAT: I did this all with pen and paper so I would strongly suggest you verify it yourself rather than taking my word for it. Equally I can't claim to be certain that I haven't made a mistake in the physics either!

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3. Jul 27, 2009

### arigger

Thanks,
It'll take me longer to vet these calcs than it took you to conceive them... Will check back in if they seem right.
BTW does anyone have a reliable source for the K value of wire rope? The best I could find was http://www.latchandbatchelor.co.uk/rope-info/rope-data/" [Broken].

Last edited by a moderator: May 4, 2017
4. Jul 27, 2009

### arigger

The first thing I'm running up against is the l0 vs. lp. How are those two values substantially different? If I pull a line from point to point that should be the pretension length regardless of the amount of tension I apply. If l0 is the slack length then how does that affect the parts between anchors at pretension? My model has fixed anchors 50' apart and slack line is drawn out of the system. If my questions seem dense... Well, I am a little dense.

5. Jul 28, 2009

### arigger

So It seems I am way over my head here, mathematically. I had never heard of a quartic function and can't make sense of the answers I get when I run my numbers on a quartic function calculator I found online. I guess I was expecting a single answer. I feel like there has to be a simpler answer. Any help?
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6. Jul 28, 2009

### jmb

I answered your other questions first (see below) then looked at this. I've left my answers below because I felt you might be interested in understanding how it all works. However looking at these numbers (although I'm not totally convinced about some of them and they've misprinted the equations) it seems that, unless you are considering very heavy loads (10,000 metric tons plus), these kind of ropes can't be treated as elastic, since there will be significant permanent deformation on loading.

In this case I think the best way to determine the tension (short of actually using a commercial tension meter) would be to determine the angle $$\theta$$ by measuring the deflection from horizontal caused by the load (e.g. rig a string tightly between the anchor points and then measure the vertical deviation from this of the loaded rope). If this deflection is only small then I don't think you're going to get an accurate answer though --- commercial tension meters may be the only solution.

The derivation I gave relied on being able to determine tension from rope extension which simply doesn't work if there is significant permanent deformation on loading.

One other possibility is prestressing the rope to eliminate future permanent deformations (something that the website you quote mentions), in this case you might be able to use the original method I suggested depending on how successful the prestressing was and on whether you could determine an accurate value of $$k$$ for the rope.

Are there any engineers (or more practical physicists than me!) who can offer any other advice?

$$\hline$$

Yes, but as you apply tension, you are pulling on the rope: thus the amount of rope between the two anchors decreases.

I'm afraid I mathematically simplified things a bit in my derivation. I was imagining a piece of rope that, when pretensioned, exactly fit between the two anchors (which were a distance $$l_{\rm p}$$ apart). Thus when the pretension is released (i.e. the rope is released from one of the anchors) the rope elastically contracts to some length $$l_0$$ that is actually less than the distance between the two anchor points.

Of course in reality this isn't possible. However it is still easy to calculate the correct values for $$l_{\rm p}$$ and $$l_0$$ for the more realistic case as follows:

• Attach the rope to the anchor points, pretension it, and make it fast.
• If the tension is enough to make the rope flat then, as you say, $$l_{\rm p}$$ will just be the length of rope between the anchor points (i.e. the length of that part of the rope under tension).
• Use some kind of marker to mark the position on the rope of the start and end anchors.
• Release the tension in the rope and measure between these two markers: this new, shorter, length is $$l_0$$.

To be accurate this requires that the rope be perfectly elastic (i.e. there is no permanent deformation from pretensioning it) and that you have a very clean way of attaching it to the anchor points (e.g. some kind of clamp or similar). If you are just knotting the rope it would be very difficult practically to determine precisely where the tension carrying part of the rope ends. Similarly if your anchors actually consist of pulleys and the rope is then made fast at some distance from these pulleys you would either need to rederive the solution taking this into account or instead formulate the answer in terms of the pretension (see below).

Lastly note that $$k$$ must be calculated for the untensioned length of rope $$l_0$$.

$$l_0$$ is the slack length of that part of the rope which when pretensioned exactly spans the gap between anchors (i.e. $$l_{\rm p}$$).

$$l_0$$ is being used solely to determine the value of the pretension (the more pretension you apply the more the rope extends and so the greater the difference between $$l_0$$ and $$l_{\rm p}$$). If you already know the value of the pretension then you can reformulate the problem without using $$l_0$$ as follows:

• Rewrite equation 2 as $$T=k\Delta l+T_0$$ where $$T_0$$ is the value of the pretension.
• Replace all the remaining $$l_{\rm p}$$ terms with $$D$$ (the distance between anchor points).
• Rework the derivation along the same lines as before.

This has the advantage that it will also work for the cases where the rope is not actually made fast at the anchor points (e.g. the anchor points are just pulleys) but at some fixed distance from them, provided $$k$$ is calculated for the entire tension carrying length of the rope not just that part of the rope lying between the pulleys.

Let me know if you need help with this. Also, if you want to apply this to a real situation, it would be good to know exactly how you are anchoring the rope...

The formulae from the general solution for the quartic will give four answers. However, all but one of these should be unphysical (i.e. negative or complex tensions) so you should be able to pull the correct one out.

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