Increased tension in wire rope 1

[A.smith]

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.

Thanks

Ash

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

 

[GregLocock]

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?



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[A.smith]

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.

 

[robyengIT]

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

 

[GregLocock]

If you can ignore the deflection due to self weight then why isn't it just triangle of forces?


Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[A.smith]

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

 

[jlnsol]

Catenary sag with respect to the other variables can easily be calculated here:

http://www.wolframalpha.com/input/?i=catenary+sag

Or go to the wolframalpha site and type in "catenary sag"

 

[rb1957]

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 ?

 

[handleman]

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)

 

[dik]

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.

Dik

 

[A.smith]

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

 

[dik]

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

Dik

 

[handleman]

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)

 

[A.smith]

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!

 

[rb1957]

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 ?

 

[Nescius]

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

 

[handleman]

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)

 

[dik]

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

Dik

 

[handleman]

https://www.google.com/search?q=define:+insignificant

https://www.google.com/search?q=define:+negligible

-handleman, CSWP (The new, easy test)

 

[dik]

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

Dik