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!)

 

[itsmoked]

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

 

[chicopee]

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.

 

[handleman]

dik,
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.

chicopee,
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)

 

[Denial]

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.

 

[moon161]

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.

 

[rb1957]

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 ?

 

[Denial]

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.

 

[gruntguru]

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

 

[rb1957]

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 ?

 

[moon161]

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.

 

[gruntguru]

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

 

[rb1957]

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 ?

 

[racookpe1978]

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.

 

[Nescius]

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.