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Increased tension in wire rope

Increased tension in wire rope

Increased tension in wire rope

Hi Guys,

I'm a little stuck on some calculations and hopefully some clarity can be found.

I'm looking to calculate the additional tension increase of a prestressed wire rope under deflection.

The wire rope is 6.5m in length and prestressed to 100Kgs a load of 8.5kgs is applied to the center of the rope and will deflect by 150mm.

Through physical testing i know that the wire tension increase is 5-10kgs (as accurate as i can measure with my tensiometer), however i am struggling to put this into a calculation.

(this is the fist time i have been involved with wire ropes).

I have (I think) exhausted all of the references relating to catenarys and cannot seem to make the calculation work for me (the catenary tells me that the new load acting on the cable is 200kgs+ this is not correct).

I fear that i am over complicating this? but if anyone can point me in the correct direction to the answer i would be forever in your debt.



(p.s. I am a new user and registered specifically to ask this question!)

RE: Increased tension in wire rope

Catenaries are fairly horrible to work with as they are very sensitive to the end conditions. So the first question I'd ask is the sag in the unloaded cable what you'd expect from the standard equations?


Greg Locock

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

RE: Increased tension in wire rope

The sag (i think) can be ignored? (unless this needs to be considered? in which case the below statement applies)

the actual sag (from testing) is 2mm.

RE: Increased tension in wire rope

try with this :
Google >>> "USS wire rope engineering handbook" : there are many load conditions for steel cables

RE: Increased tension in wire rope

it may well be?

100Kg "(tension preapplied to cable)" x ((sin(angle LH))+(sin(angle RH))) - finds the load required to deflect the cable by X (X=150mm).

100Kgs x (sin(2.47°)+sin(2.47°)= 8.61Kgs

Load required to deflect cable by 150mm = 8.61Kgs

However this calculation does not tell me the increase/new cable tension due to the deflection.

robyengIT - I have looked there already, but unfortunately not had any success finding what i need

RE: Increased tension in wire rope

ok, i'm not sure I understand.

I think you have a steel cable (weight not stated), 6.5m long ... so that's your first condition (self weight).

Then you tension this cable to a load of 100 kgs ... be careful with your units.

Then you apply a point load, 8.5 kgs, at the mid-span. And this load causes an increase in cable tension of 5-10 kgs.

Where did 2.47deg come from ? I think you need to determine the angle of the support reactions.

If you know the answer (5-10 kgs) for 4.25 kgs reacted at each end ... that's a surprising result atan(4.25/7.5) = 15deg.

Can you repeat the test (lab?) and measure this ?

another day in paradise, or is paradise one day closer ?

RE: Increased tension in wire rope

Good grief. This is like the first week of statics. He stated that the cable's sag under its own weight was only 2mm. Therefore it can probably be ignored. Draw a FBD of the center of the cable. The cable is a 2-force member. Geometry is all given, angle of the cables can be calculated. Sum forces in vertical to get vertical component of new cable tension. A little trig to get the actual new cable tension. The 100kgf is spurious data when calculating the new cable tension. If you want the increase, subtract the 100kgf from the newly calculated tension.

-handleman, CSWP (The new, easy test)

RE: Increased tension in wire rope

Handleman... if I understand the problem, it's not so trivial...it's deceptive. The cable has an initial load of 100 Kg and an initial sag; there is no reference to the cable weight or material (assumed to be steel and fy and Es not known) and a length of 6.5 m. The 8.5 Kg load applied to the center causes it to deflect 150mm. I'm not sure how that is arrived at, by measurement? In a pinch, you can establish the dia of the cable from the above info, but, it's very complicated. With a horizontal (nearly) cable, any load applied at mid span increases the tension substantially and as it deflects this tension is relieved a bit. If at the end of the day, the total deflection is 150 mm it is a difficult solution. It could be that the weight of the cable approaches, or exceeds the weight in the middle and the weight of the cable cannot be eliminated. Definitely not a simple question problem.


RE: Increased tension in wire rope

Thank you for all of your help guys! much appreciated.

RE: Increased tension in wire rope

A.smith... glad it's not J.smith... obvious alias... Are you fully confused now?


RE: Increased tension in wire rope

OP already stated that the measured deflection due to sag is 2mm. Compared with the 150mm measured deflection under 8.5KGF load, the sag is negligible.

Again, draw the FBD of the center point of the cable. Is it moving? No? Then statics has to work. You know the geometry. 150mm rise, 3.25m run. Each half of the cable pulls "upward" with half the applied load, or 4.25kgf. T times 0.15/3.25 = 4.25. T=92kgf.

Again, this is statics. It is true. Therefore, something is incorrect with the OP's setup. Let's examine the likely culprits. 6.5mm span? Not so likely. Easy to measure. Same with deflection amount and actual mass of applied weight. That leaves tension. OP has already stated that tensometer accuracy is questionable, because the measured increase was "5-10kgf". Therefore, the obvious conclusion is that the initial 100kgf measurement is inaccurate.

Next question: What is the real initial tension? That is unknown. You can either measure it (which we have already established that you can't) or, IF you know the elastic modulus of the cable, you can calculate the length difference between straight and the hypotenii and back-calculate from the tension found by statics to the real pre-weight tension. Of course, this assumes perfectly rigid supports and perfect clamping. If the cable has, for example, looped ends that stretch differently than the bulk cable then this goes out the window too.

Conclusion: The problem was presented somewhat backward. Statics gives us the real, true final tension. Elastic modulus and measured stretch give us the true initial tension.

-handleman, CSWP (The new, easy test)

RE: Increased tension in wire rope

No alias here... i am related to a J.smith and i know of another J.smith too...

i'm much less confused than i was when i stated this post, you guys have been super! thank you!

RE: Increased tension in wire rope

handleman's 92kgs is the tension required to support the weight, not the additional tension.
to fine tune this, the tension is the hypotenuse, so T = 4.25/sin(atan(0.15/3.25)) = 92.2kg

maybe the first mistake was thinking this is a catenary ... 6.5m span, 0.15m deflection.

one thing to look at maybe is the strain energy in the cable (about .1% strain) ?

I get an angle of 1.32deg (about 1/2 the 2.47 mentioned earlier) ?

another day in paradise, or is paradise one day closer ?

RE: Increased tension in wire rope

6.5 meter rope with an initial sag of 2mm? How was this measured? Relative to the floor?

RE: Increased tension in wire rope

The OP specifically stated point load at the center. This is most certainly NOT a catenary. Catenary is the shape created by the distributed weight load of a hanging rope/chain due to its own weight. OP specifically stated that the measured sag (which is a catenary) before weight was added is 2mm. A 2mm catenary over 6.5m. This is pretty negligible for our purposes, as compared with the 150mm deflection induced by the weight.

But, just for fun, we can find the approximate weight of the 6.5m rope if it makes us feel any better. We're already doubting the initial tension statement, but, giving a window of 90 to 110kgf actual tension, our window for total mass of the 6.5m rope is somewhere between 0.23kg and 0.28kg. (verification left as an exercise) Again, insignificant compared to the 8.5kg weight hanging from the center.

As rb1957 said, the 92kgf is the total tension after the weight is added. It includes the initial tension, plus the added tension due to the weight. This is statics! What does that tell us? It tells us that something that we thought we knew about the initial conditions before weight was added is WRONG. It doesn't matter what the material is, how stretchy it is, what the initial tension was, nothing. The only thing that matters when calculating the final tension is the geometry of the system and the weight of the point load in the center. Again, this is ignoring the insignificant cable weight.

So, given that we know the final tension, how can we find the initial tension? Again, as rb mentioned, it has everything to do with the strain of the rope.

Let's consider what we know to be true: 8.5kg weight hanging at 150mm deflection from a 6.5m span. Now, imagine 2 different ropes. First, consider a theoretically perfectly zero stretch rope. What happens when you remove the weight? What is the tension? Almost nothing. It will fall into a loose catenary with the same length as the 2 hypotenii. Interestingly enough, since we already calculated the weight of the actual rope is apx 0.25kgf, we can calculate the catenary sag amount as apx 130mm, and the cable tension somewhere around 1.6 kgf for this completely non-stretch rope (another exercise for the reader). Now let's consider a rubber band. It's a really long rubber band. and it has a "spring constant", if you will, of 1kgf/mm. When you remove the center weight, the band contracts and goes horizontal (ignoring its mass). We can easily calculate the length difference between the hypotenii and the horizontal as apx. 7mm. Since the spring constant is 1kgf/mm, the pre-weight tension is calculated to be 92kgf-7kgf=85kgf. What if the rubber band is a bit stretchier? Like 0.1kgf/mm? Then the initial tension would be only 92-70=22kgf.

Of course, as I mentioned earlier this all assumes perfect rigidity from the entire system, such that all of the geometry change between the 2mm catenary and the 150mm deflection is due to stretch of the rope. If your stand or anchors are deflecting, or the tensometer somehow has a not-insignificant spring rate, all bets are off.

-handleman, CSWP (The new, easy test)

RE: Increased tension in wire rope

I would suggest a catenary with a cusp at mid span... this would reflect the weight of the cable...


RE: Increased tension in wire rope

You could almost consider the weight of the suspended object as negligible also since the cable weight could be of the same magnitude...


RE: Increased tension in wire rope

The tensiometer probably has a lot of inherent error in it that can't give a reliable reading with the initial tension and added tension, it provides some solid confusion. Much better would be a spring scale on one end.

Keith Cress
kcress - http://www.flaminsystems.com

RE: Increased tension in wire rope

I think that the OP was badly stated "The wire rope is 6.5m in length and prestressed to 100Kgs a load of 8.5kgs is applied to the center...." There should have been a period after the 100Kgs then a different picture presents itself as "The wire rope is 6.5m in length and prestressed to 100Kgs. A load of 8.5kgs is applied to the center..." . The 100 kgs prestress is to stretch the rope only. Afterward the 8 kgs is applied in a cable that is inextensible. The cable, first is under gravity acting as a catenary and secondly under the point load of 8Kgs which can be solved by vectors as pointed above. The catenary evalustion may not be necessary depending of the unit weight of the cable. For a reference into solving this type of problem, here is the title and author as I don't have the link "; I don't have the link but here is the thesis and its author "ON SHAPE CONTROL OF CABLES UNDER VERTICAL STATIC LOADS-DANIEL PAPINI". Page 61 is the reference needed.

RE: Increased tension in wire rope

No. The cable weight has been solved for. It is apx 0.25kg, which is, what 3% of the added 8.5kg weight? Read my post.

The OP stated 2mm catenary with approximately 100kgf prestress before the load was applied. Read his second post. Tension, un-loaded catenary dip, and span distance are all you need to solve for the weight of the cable, which I showed above to be apx 1/4kg. It's most certainly NOT inextensible. You can't deflect 150mm from a 2mm catenary without stretching.

-handleman, CSWP (The new, easy test)

RE: Increased tension in wire rope

Ash.  On my web site (http://rmniall.com) you will find a free downloadable spreadsheet that gives a rigorous analysis for the problem of a (single) point load applied to a cable.  The cable can be inclined from horizontal, the load can be inclined from vertical, and the cable can be extensible.

RE: Increased tension in wire rope

I wish the spreadsheet worked for me, but no dice.

While simple statics reasonably gives 92 kg tension in the rope, a basic assumption of this is that the angle of the rope at the point load is defined by the rise and the run, and neglects any curvature of the rope, which we know exists.

Taking the angle as asin (W/2T) and T as 105kgf, I get 2.3*, which is a hair less than the 2.5* stated above. I'm happy with that.

RE: Increased tension in wire rope

T = 105kg is a small change in angle, but a reasonably measurable change in deflection (132mm vs 150mm).

the whole thing, like most tests (labs?), doesn't add up ...

the stated pretension of 100 kg should not have allowed a displacement of 150mm for the stated cable length and load applied.

Displacement can be determined from energy, as the tension in the cable should increase (as it strains). The potential energy lost by the force should equal the strain energy gained by the cable, and check the final displacement statically. Else iterate ... the displacement of 150mm is balanced by 92kg tension (statically) but the tension in the cable is higher than this (due to strain); so guess a smaller displacement, 100mm?

or maybe the weight is wrong ... what weight would cause 150mm deflection in a cable with more than 100kg tension ?

or the pretension (most likely ?) ... if the final tension is 92kg, and the strain from the deflected cable increased the tension by ?

another day in paradise, or is paradise one day closer ?

RE: Increased tension in wire rope

Moon161.  "No dice"?  I'm always looking for ways to improve my spreadsheets, so if you can tell me what particular dice my cable spreadsheet lacks I might be able to accommodate whatever it is you are looking for.  Contact via the e-mail address in the spreadsheet and on my web site would be best.

RE: Increased tension in wire rope

rb1957. I think you are assuming the supports are rigid. Greg and handleman have it right.

Take half the cable and analyse the triangle of force vectors (side = 8.5/2 = 4.25, hypotenuse = 100+10 = 110).
This is similar (equi-anglular) to the displacement triangle so displacement = 4.25/110 x 3250 = 125.6 mm. This is slightly less than the measured displacement (150mm) because the rope has some bending stiffness and the centre section of the rope acts like a beam.

Therefore the projection of the tension forces will intersect slightly further out than the measured displacement at the centre of the rope.

je suis charlie

RE: Increased tension in wire rope

well charlie,
why "3750" ? I'm doing the same force/displacement triangle.

we know 1/2 the applied load is 4.25kg, and we think the tension with the load is something like 105kg ...
asin(4.25/105)=0.0405rad, then displacement should be tan(0.0405)*6500/2 = 132mm

or maybe the displacement of 150mm is correct, then atan(.15/3.25)=0.046rad;
and then the tension in the cable would be 4.25/sin(0.046)=92kg

another day in paradise, or is paradise one day closer ?

RE: Increased tension in wire rope

Most errors with the spreadsheet went away when I extracted the .xls from the zip to a folder instead of opening from the .zip file.

RE: Increased tension in wire rope

rb1957. Sorry typo. Post corrected. Doesn't change my analysis though. Beaming of the cable centre section will explain the difference.

Do you need to use trig tables and introduce rounding errors? Also not sure why you have used tan and asin for similar triangles. Again I guess you are assuming the supports are rigid and the cable has stretched? I am still surprised there is 5% difference in our deflection result (126 vs 132mm).

Oh wait - you used 5 kg tension increase and I used 10. (OP said "5 - 10")

je suis charlie

RE: Increased tension in wire rope

yes, I am assuming the supports are fixed and the cable stretches.

you could assume that the cable is inextensible ... that would change at atan(.15/3.25) to asin(.15/3.25) which should be a very small change in the result (as sin(theta) = tan(theta) = theta for small angles). But then if inextensible, then no change in tension, no?

another day in paradise, or is paradise one day closer ?

RE: Increased tension in wire rope

But the only way I can get the geometry to work (150 mm deflection at center after a 8.5 Kg weight is applied at the center) is if the tension is applied by a 200 Kg weight pulling the wire rope (cable) over a pulley. The minute extra expansion of the stretch of the cable cannot the whole length long enough to create that 2.64 degree angle from the straight case. The total wire length between two points 6500 mm long with a 150 mm vertical is 6506.9 mm. You'd be claiming the cable stretched 7 mm.
Now, the "cable" might stretch that much by the slipping and "unkinking" of the individual wire strands against each other, but the "metal" itself would not stretch that much if it were a solid cylinder of uniform cross-section.

And the orignal 200 Kg weight should remove much of that "mechanical" wire fiber unkinking and slip from the initial "unstressed" condition to its t=0 straight length before putting the 8.5 Kg weight on.

RE: Increased tension in wire rope

Quote (racookpe1978)

You'd be claiming the cable stretched 7 mm. Now, the "cable" might stretch that much by the slipping and "unkinking" of the individual wire strands against each other, but the "metal" itself would not stretch that much if it were a solid cylinder of uniform cross-section.

Why not? The cable is small. As handleman points out, the total mass of the 6.5m rope is only ~.25kg. My quick calculation shows that 7mm is plausible.

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