# Tension in a wire Vs. force applied by a tensioned wire

Discussion in 'Physics' started by strantor, Sep 20, 2014.

1. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Ok, I make 2 statements:
1. This is not an academic exercise, but a real life engineering problem which I'm not at leisure to disclose all the finer details of.
2. This problem will get more complicated as the discussion evolves.

Background:
1. There are two hydraulic cylinders with sheaves mounted on either end, and the cylinder/sheave assemblies are mounted on trollies which can be moved in and out, either in unison or independently. For the beginning of the discussion we will assume they move in unison.
2. There is a wire wrapped around the 4 sheaves
3. There is a load cell between one of the sheaves and one of the cylinders
4. There are two possible modes of operation of the cylinders: 1. they can be controlled by a hydraulic pressure regulator, or 2. by a load holding hydraulic circuit that once desired tension is reached, closes and locks tension value in. For the beginning of the discussion we will assume that the cylinders are controlled by regulator.
5. I need to calculate 1. Tension in the wire and 2. force applied to an object by the wire, as the trollies move forward and press the wire against the object. For the beginning of the discussion, we will assume that I've already got #1 figured out and we are only concerned with #2.

Here is a basic diagram, before the trollies move forward:

The trollies can move (in unison) up toward the top of your page and down toward the bottom. They cannot move side-to-side to any appreciable degree other than that which is caused by minimal flexing of the structure they are mounted on (not shown).

So, we assume that the cylinders are controlled by a hydraulic pressure regulator. We set the regulator to 100PSI and for example this results in 100lbs of tension in the wire. If we were to grab the wire from any side and drag it in toward the center, the cylinders would give way as the fluid inside is compressed >100PSI and is relieved by the pressure regulator. So if you were strong enough, you could pull the wire all the way until the cylinders bottom out, and then you would not be able to pull any further.

Here is a basic diagram of the cylinders moving in (driving up toward the top of your page in unison) and pressing the wire against an object:

Note that the cylinders have compressed a little bit, and there is now a 30 degree angle of deflection on both sides of the object. While the cylinders did compress, the travel of the cylinders was not as great as the travel of the trollies.

Now, if there were no preexisting tension in the wire, the tension in the wire should be exactly equal to the force applied to the object by the wire, because of the 30 degree angles. The two forces will/should be equal at (and ONLY at) this specific angle of 30 degrees, as demonstrated by this load rigging diagram:

BUT we cannot assume that the same holds true if there is preexisting tension in the wire (as in, hydraulic cylinders are pressurized by the regulator to 100PSI/100LBS tension before moving in).

Here is the formula that I'm currently using to calculate the force applied by the wire:
Code (Text):
1. Force_Applied_to_Object:= (Tension_Lbs-Preexisting_Tension)*(SIN(Left_Deflection_Angle)+SIN(Right_Deflection_Angle));
I have tested this formula with an array of load cells and proven it accurate in calculating the force given the tension and the deflection angles, but only when preexisting tension is zero. I do not currently have the hydraulic pressure regulator described above. I only have the hydraulic locking circuit, so the cylinders cannot give way. Without the regulator, I can only achieve about 10 degrees of deflection, and this is only possible by the structure flexing and the wire stretching. If I use any preexisting tension with this setup, my formula is wildly inaccurate.

I am looking to upgrade to the pressure regulator setup, and my first question is: If I install the regulator so that tension remains constant as the cylinder compress, will this formula [specifically the part (Tension_Lbs-Preexisting_Tension)] accurately describe the force applied to the object?

2. ### BR-549 Distinguished Member

Sep 22, 2013
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What is a shear? Shear assembly? Are there 4 pulleys on the cylinders? Is there any friction on the wire? Do you have any pictures of apparatus? Is the wire at constant length? Does wire run in a groove at any time?

3. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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I think you misread; I said sheave (as in, like a pulley).
Yes. two per cylinder, one mounted at either end.
For the sake of discussion, no. The sheaves/pulley are free-rotating.
I'll PM you.
Yes. it is a continuous seamless loop of wire rope.
Not sure what you're getting at; the only groove it runs in is the groove of the sheave/pulleys. I have taken the math out of it by showing the wire wrapped around the inner diameter of sheave/pulley and not showing the outer diameter.

4. ### studiot AAC Fanatic!

Nov 9, 2007
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Force_Applied_tbject:= (Tension_Lbs-Preexisting_Tension)*(SIN(Left_Deflection_Angle)+SIN(Right_Deflection_Angle));

I don't understand why you are using this formula when operating pre-tension?

Surely the bit I have underlined makes the first term always zero, without pre-tension?

So that term is incorrect when there is pretension present.

I am still trying to ger my head around the operation of the assembly, but are you sure you have the sign correct in your formula?

5. ### #12 Expert

Nov 30, 2010
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I'm having difficulty with this statement: The forces should be equal only at 30 degrees.
Your riggers drawing shows a 4% increase in the total tension compared to the weight of the barrel.
This reminds me of The Law of Tangents which is in charge of how much force is required to bend a guitar string.
A slight deflection causes a lot of increase in the tension of the cable.
Here is a drawing that might help people picture this differently.
I have not come to any conclusions yet.

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6. ### atferrari AAC Fanatic!

Jan 6, 2004
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Strantor, I am having difficulty to understand how it works:

I am not sure what part/s is/are fixed wrt the rest.

Wire length, is it fixed?

I find hard to understand how the load cell works maybe because of the above. Sorry.

7. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Actually I think you just made me realize that my assumptions were wrong.
I have been thinking like this (for example there is 100lbs pretension): "Since there is tension in the wire before the wire ever contacts the object, that amount of tension (preexisting tension) will never contribute to the force applied to the object. Only that amount of tension which is added to the preexisting tension as a result of running the wire into the object will result in force applied to the object" - hence the "tension lbs - preexisting tension"
But then above, I stated that any additional pressure in the cylinders caused by increasing tension would be bled off by the regulator, hence the tension in the wire would remain a constant 100lbs until the cylinders bottom out.
So these two ideas don't mix.
I think my mind has been stuck in a rut since as I mentioned I don't have the regulator in place, and until now I've been dealing with a taut wire that has no give to it.
So, if I fit a regulator, the formula should be modified to:
Code (Text):
1. Force_Applied_to_Object:= (Tension_Lbs)*(SIN(Left_Deflection_Angle)+SIN(Right_Deflection_Angle));
Would you agree?

8. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Here, I've modified your drawing to illustrate the 30 degrees I am referring to.

Sorry I wasn't clear; I said 30 degrees, referring to the angle of deflection from 90 degrees (a taut wire) and then showed an image that illustrated my point using 120 degrees - same triangle, just specifying different corners of it.

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9. ### inwo Well-Known Member

Nov 7, 2013
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The force applied to the object is independent of wire angles. No?

The cylinders as shown are always perpendicular to the load.
Restrained from other motion, they will always share the load. (not equally)

I may be confused, as usual.
It seems there is no mechanical advantage in a closed loop. As soon as the weight in the 2nd picture, exceeds the cylinder (constant) pressure it will move all the way down.

10. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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I say "No." But if you're seeing something I'm not seeing, please explain. The way I see it, and the way that the math I've done so far indicates, the more the wire deflects, the more pressure will be applied to the object for a given (constant) tension in the wire. - this assumes the pressure in the cylinders is controlled by a pressure regulator, and the tension can actually be set to a constant tension.

The way you explain, fits what I observe now, with this current system where tension canNOT actually be set to a fixed value. The fixed value in my current setup is the dimensions between all the sheaves. So, once the cylinders extend and take up tension (for example 100lbs) then the dimensions get all locked into place, and any pressure applied by the wire results in EXTREME tension increase in the wire, for a very little amount of pressure applied to the object.
You're referring to #12's diagram? Again, I say "No." I say, if the 100lb barrel is hung from the tensioned wire, it will sag the wire to the point of 30 degrees deflection and that is where it will stop, because that is the point at which the weight of the barrel=the tension in the wire=the weight of the weight. But, again, this is a discussion and not a lecture, and if I'm off the mark, let me know

11. ### #12 Expert

Nov 30, 2010
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I think you're on the right track, but somebodies signature is, "One experiment is worth a hundred theories".
You are smart enough to be able to set up an experiment for this in your garage. Some 1/8th inch steel cable, a couple of pulleys, a couple of exercise weights, and a scale like you use to weigh a fish should do the trick.

I have figured out how to calibrate my Harbor Freight torque wrenches in the shed. You should be able to figure out how to test deflection angles and pressure.

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12. ### inwo Well-Known Member

Nov 7, 2013
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Picturing weight rather than forces.
The weight of object (apparatus is weightless and frictionless),
Weight of object equals the sum of scales under each cylinder support.

Last edited: Sep 21, 2014
13. ### inwo Well-Known Member

Nov 7, 2013
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Obviously you have spent much more time thinking about this than I have in the last 30M.

Another way to look at it, is to see if adding weight will reach equilibrium, with fixed hyd. pressure..
As the cable deflects, more of the force vector is directed down. By the changing wire angles.
I see it continuing to bottom.
I'd be glad to see different. Then I'd learn something.

ps.
In my thought experiments. The apparatus must be supported someplace.
I choose the lower sheave axles.
I have no idea how side loads are supported.

14. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Thanks Inwo, I've got to go to bed now, so I'll give you a thoughtful reply in the morning. In the meantime, if you're a night owl, I've sent you some material in a PM.

15. ### inwo Well-Known Member

Nov 7, 2013
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Couple things.
1. Measurement (and control) of hyd. pressure could replace load cell.

2. Cable tension could be measured rather than calculated.
Use load cell, on a fixed angle deflection sheave,to measure dynamic cable tension.
There may be a place with fixed dimensions and angles, existing.

Or drive torque measurement if cable is in motion.

I know it's not your job to redesign. Only to program what you have.
Don't see how it's possible without more data points. ??????

edit:
Known hyd. pressure and known deflection angle should give cable tension.
IOW adjust hyd. pressure to achieve desired deflection angle.

No matter what you decide. You will need a way to test cable tension for program calibration. Math only goes so far.

Last edited: Sep 21, 2014

Nov 9, 2007
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17. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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if your pressure regulator is maintaining a constant tension in your wire then you will be maintaining a constant force on your target object. Only by allowing the pressure to increase can you affect an increase of force against your target. Regulating pressure is the wrong course of action I believe.
The function of force applied and pressure increase, if kept reasonably small vs the tensile strength of your cable, will be determined by sq in area of your pistons and length of displacement in same. having two pistons and a pulley arrangment will complicate the math but not too much. pretension would subtracted out from the final pressure reading to find applied force

18. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Why is this? It seems to me that pressure in the cylinder should be directly proportional to tension in the wire, assuming a pressure regulator is used, and the cylinders are not bottomed out (somewhere between fully extended and fully retracted). I see it as such: The piston inside the cylinder has a fixed surface area, and if a fixed Pressure per Square Inch is applied to it, the cylinder will extend with a fixed amount of force, which will result in a fixed amount of tension in the wire. For example; piston has a 1in^2 surface area; 100PSI is applied to it, so 100lbs of force is generated at the end of the rod. This results in 200lbs of tension in the wire as there are two cylinders.

That's what I'm trying to do; calculate the force applied by the wire using known tension and known deflection angle.

19. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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if cable tension released, as in, say cable stretches, cylinders will move to account for this and pressure will resume its selected level. the force applied will remain constant for a specific pressure and piston area. the only situation where angles would enter the picture would be in regard to side forces not in same plane and direction as piston movement

20. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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This is true assuming that the cylinders never bottom out, and the tension is always a function of hydraulic pressure. BUT that is not always true; the trolleys will be driven in to the point that the cylinders bottom out; at that point there will be spike in tension with no reflected spike in pressure.

Cable tension is measured by load cell. I drew the load cell in the above drawings. The physical location of the load cell is between the end of the left side cylinder and the left side bottom sheave (I drew it in the middle of the cylinder to illustrate the following point: )
The raw value from the load cell can be scaled directly into cable tension if, and only if, the trolleys are perfectly aligned. If all the sheaves are at perfect 90 degree angles to each other, then the compressive force measured by the load cell has a 1:1 relationship to that which it would have if a single wire were hung from it with an equal weight. BUT when the trolleys are misaligned (as they regularly will be in normal mode of operation) that 1:1 relationship goes away.

I have calculated and proven through testing, that an angular trolley offset of 11 degrees (left side trolley up closer to the top of your screen than right side) will cause a tension readout error of 25% in the more dangerous direction (ex: 1000lbs actual tension would read out as 750lbs).

The reason is that due to the placement of the load cell, it only measures that amount of force which is in-line with the cylinder. The tension force which exists in the x-axis (bottom edge and top edge of the wire going from left to right) is not seen by the load cell.

I have explained this with 2 force vectors; one in the Y-axis, which always has a value of 1, and one in the X-axis which is variable to offset angle, and these two vectors are added to generate a tension correction factor. My formula is this:
Code (Text):
1. (*triangle side A*)
2. Position_Offset_In:= Left_Position-Right_Position;
3. (*pythagorean theorem ( (A^2)+(B^2)=(C^2) ), calculate hypotenuse: *)
4. (*C = sqrt( (A^2)+(B^2) *)
5. (*Side B = 242" (distance from sheave center to sheave center across structure B^2 = 58,564*)
6. Hypotenuse:= SQRT(58564.0+(Position_Offset_In*Position_Offset_In));
7. (*calculate trolley offset angle*)