I'm a chemist functioning as a Process/Materials engineer and have a problem that is well beyond my mechanical engineering skill set. We have a continuous production process where stranded aluminum or copper conductor is insulated. The conductor naturally twists over sheaves and a twisting motion is imposed during one section. I need to design a method to connect two ends of the conductor with a device that can handle the torsional load imposed in the process and be equal to or less than the diameter of the conductor. I don't know how to calculate the load imposed by the process or determine what load the device can handle. I know the conductor will rotate 8 times over 600 ft. worst case assuming it is fixed at each end. It isn't totally fixed at each end but if I can design as if it is I have a safety factor built in. Can the problem be simplified for example if the device is fixed relative to the longitudinal axis of each end of the conductor, does the device simply have to be the same or stronger than any section of the conductor? Sean
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designing for torsional load 2
- Thread starter Shylanel
- Start date
In five years you haven't been able to adequately define the problem you say you are trying to solve, one that's outside your field of expertise and have managed to gather no data about the problem, but expect people to help you while wasting their time playing a guessing game?
Reasons to believe you don't know what you are talking about:
I have been looking for off the shelf solutions to this of and on for about 5 years now.
I obviously didn't explain it well.
Why don't you ask me for the information needed for the calculation?
This is where you led me to think you are angry and stupid:
I did not start out asking for a design solution. If I had, the post would have been much different defining the various requirements of the design.
This is where you made sure:
spew insults sit on a pedestal keep your pompous arrogance to yourself
Reasons to believe you don't know what you are talking about:
I have been looking for off the shelf solutions to this of and on for about 5 years now.
I obviously didn't explain it well.
Why don't you ask me for the information needed for the calculation?
This is where you led me to think you are angry and stupid:
I did not start out asking for a design solution. If I had, the post would have been much different defining the various requirements of the design.
This is where you made sure:
spew insults sit on a pedestal keep your pompous arrogance to yourself
- Thread starter
- #22
Again, my problem isn't the design yet. I'm looking for how much load is applied to the cable between the capstans and I tried to simplify it to a point that it can be calculated. A 600 ft piece of cable (copper or aluminum conductor) between two points under tension is rotated once every 75 feet. How much torque load is now on that cable? Assume solid copper if it helps. If my ultimate design will work for solid copper, it will work for strand. Once I understand the load, I will move to the design. I'm assuming I will move forward with steel. What cross sectional area of steel is required to be equal to or stronger than a given copper (of Al) conductor? Is the relationship linear from small to large OD's?
I've already started figuring out design options. I'm trying to put some engineering data behind it instead of engineering judgment with experimentation.
The overall problem and process is a very lengthy explanation. Since everyone wants to know, I'll try to give a cliff notes version. We make medium and high voltage power cable from 2AWG to 2000 MCM, 5KV to 69KV using a Continuous Catenary Vulcanization (CCV) process. The overall line length is over 1800 ft spanning over 4 stories. The smallest cable is less than 0.5" and the largest > 3". The process is continuous but the conductor comes on reels (not continuous) so there has to be a method to join each end of the conductor without stopping the line. This is done with welds or crimps depending on the conductor type. When in continuous mode, the crimps or welds do not have to be disconnected. They simply have to be smaller than the conductor so they will pass through the extrusion tooling. When changing sizes of conductor, the extrusion tooling has to be changed requiring a stop in the process but you don't want to waste the strand between the extruder and payoff so the different conductors have to be joined at the payoff. When the new size gets to the extruder, the line is stopped and the two conductors are disconnected. The old conductor is pulled through the tooling. The tooling is changed. The new conductor is then put through the new tooling and rejoined with the old conductor for the line to be restarted. Knots can be used at the extruder but they are notoriously unreliable, take time and are difficult to make small enough as they have to go through rubber seals that can only open so much. The entire connection also has to be capable of sealing water at 150 psi from the strand. This is can be handled with what is called a water block weld when possible. When the insulation thickness is large relative to the conductor OD we apply a twister after the seals to rotate the cable as it exits the extruder to prevent the hot insulation from drooping. This keeps the dimensions in spec. There are multiple capstans in the process to pull the cable through the line. The primary capstan is a belt capstan which is basically a very large sheave with a rubber belt on it. A second belt is held in tension against the sheave. The cable passes between the belts and is pulled with friction. The other capstans are all caterpillar capstans which pull the cable through between two belts propelled with a caterpillar mechanism like a tank but these are horizontal. This is just the mechanical path of the process leaving out other auxiliary steps not germane to this problem. I also haven't touched the extrusion or curing process' which are the real challenge to making the cable. This cable is made by the mile by various companies all over the world every day. I can continue making the cable this way and everything will be fine. I'm trying to find a better and more efficient method that will reduce changeover time, increase reliability and reduce scrap.
I want to use a crimping tool normally used for hydraulic hose that gives a stronger, smaller and faster crimp than what's normally used in this industry. This style is used for EHV cable. This crimper can most easily be installed and applied at the payoff for various reasons. To take full advantage of this crimper I want to develop a device that can connect two different size cables at the payoff, be disconnected at the extruder and then reconnected after the extruder during changeovers. There exist devices that allow for rotation. That is fine except when I want to apply the twister. If I can develop a device that is rigid and as strong as the conductor on either side then I can put the twister on immediately at startup and save several hundred feet of scrap. The working thought is to have crimps for each size conductor but with a common interface to each other or a third tool. I can then crimp the end pieces on each conductor, connect them, run the cable to the extruder, stop the line, disconnect the two ends, pull the old conductor through, change the tooling, pull the new conductor through, reconnect and start the line. Again whatever device is designed to connect the two crimps has to be smaller than the biggest conductor to make it through the seals.
In general when changing sizes it will be from a smaller conductor to a larger conductor. We obviously won't be going from 2 AWG to 2000 MCM either. We will jump several sizes at once though. I don't need one device that will work from 2AWG (~0.25" OD) and 2000 MCM (~1.6"OD). I expect to have 4-6 sizes covering various ranges of conductor to span the entire range. To know what size range of conductor each size device can handle I will also need to know what load they can handle. The crimps will be throw away. The connector between the two will be reusable though not indefinitely.
I hope this is enough detail for everyone to digest and understand what I'm asking. If you care to offer a design ideas in addition to the load calculation I will be grateful.
I've already started figuring out design options. I'm trying to put some engineering data behind it instead of engineering judgment with experimentation.
The overall problem and process is a very lengthy explanation. Since everyone wants to know, I'll try to give a cliff notes version. We make medium and high voltage power cable from 2AWG to 2000 MCM, 5KV to 69KV using a Continuous Catenary Vulcanization (CCV) process. The overall line length is over 1800 ft spanning over 4 stories. The smallest cable is less than 0.5" and the largest > 3". The process is continuous but the conductor comes on reels (not continuous) so there has to be a method to join each end of the conductor without stopping the line. This is done with welds or crimps depending on the conductor type. When in continuous mode, the crimps or welds do not have to be disconnected. They simply have to be smaller than the conductor so they will pass through the extrusion tooling. When changing sizes of conductor, the extrusion tooling has to be changed requiring a stop in the process but you don't want to waste the strand between the extruder and payoff so the different conductors have to be joined at the payoff. When the new size gets to the extruder, the line is stopped and the two conductors are disconnected. The old conductor is pulled through the tooling. The tooling is changed. The new conductor is then put through the new tooling and rejoined with the old conductor for the line to be restarted. Knots can be used at the extruder but they are notoriously unreliable, take time and are difficult to make small enough as they have to go through rubber seals that can only open so much. The entire connection also has to be capable of sealing water at 150 psi from the strand. This is can be handled with what is called a water block weld when possible. When the insulation thickness is large relative to the conductor OD we apply a twister after the seals to rotate the cable as it exits the extruder to prevent the hot insulation from drooping. This keeps the dimensions in spec. There are multiple capstans in the process to pull the cable through the line. The primary capstan is a belt capstan which is basically a very large sheave with a rubber belt on it. A second belt is held in tension against the sheave. The cable passes between the belts and is pulled with friction. The other capstans are all caterpillar capstans which pull the cable through between two belts propelled with a caterpillar mechanism like a tank but these are horizontal. This is just the mechanical path of the process leaving out other auxiliary steps not germane to this problem. I also haven't touched the extrusion or curing process' which are the real challenge to making the cable. This cable is made by the mile by various companies all over the world every day. I can continue making the cable this way and everything will be fine. I'm trying to find a better and more efficient method that will reduce changeover time, increase reliability and reduce scrap.
I want to use a crimping tool normally used for hydraulic hose that gives a stronger, smaller and faster crimp than what's normally used in this industry. This style is used for EHV cable. This crimper can most easily be installed and applied at the payoff for various reasons. To take full advantage of this crimper I want to develop a device that can connect two different size cables at the payoff, be disconnected at the extruder and then reconnected after the extruder during changeovers. There exist devices that allow for rotation. That is fine except when I want to apply the twister. If I can develop a device that is rigid and as strong as the conductor on either side then I can put the twister on immediately at startup and save several hundred feet of scrap. The working thought is to have crimps for each size conductor but with a common interface to each other or a third tool. I can then crimp the end pieces on each conductor, connect them, run the cable to the extruder, stop the line, disconnect the two ends, pull the old conductor through, change the tooling, pull the new conductor through, reconnect and start the line. Again whatever device is designed to connect the two crimps has to be smaller than the biggest conductor to make it through the seals.
In general when changing sizes it will be from a smaller conductor to a larger conductor. We obviously won't be going from 2 AWG to 2000 MCM either. We will jump several sizes at once though. I don't need one device that will work from 2AWG (~0.25" OD) and 2000 MCM (~1.6"OD). I expect to have 4-6 sizes covering various ranges of conductor to span the entire range. To know what size range of conductor each size device can handle I will also need to know what load they can handle. The crimps will be throw away. The connector between the two will be reusable though not indefinitely.
I hope this is enough detail for everyone to digest and understand what I'm asking. If you care to offer a design ideas in addition to the load calculation I will be grateful.
- Thread starter
- #23
Well Dave I hope you've sufficiently boosted your ego to get you through the weekend. I'm done for today but if you need some more boosting via vain attempts to make me feel less of a human than you, feel free to stop back on Monday.
I have to admit I didn't expect to find trolls of your nature on an engineering forum.
I have to admit I didn't expect to find trolls of your nature on an engineering forum.
Why?
You've stated that you can crimp.
You've stated that you can crimp steel tube.
You've stated that crimped steel tube connections work.
You've stated that you can throw crimps away.
So why won't a steel tube that's sized for the smaller cable on one end and expanded to the size of the larger cable on the other end be crimped on and be done with it.
I'm imagining the device that you seem to be imagining as obscenely expensive and unreliable - if it's even possible.
You've stated that you can crimp.
You've stated that you can crimp steel tube.
You've stated that crimped steel tube connections work.
You've stated that you can throw crimps away.
So why won't a steel tube that's sized for the smaller cable on one end and expanded to the size of the larger cable on the other end be crimped on and be done with it.
I'm imagining the device that you seem to be imagining as obscenely expensive and unreliable - if it's even possible.
- Thread starter
- #25
Those would work fantastic if the crimp didn't have to come apart at the extrusion head when going from one size to another. I can even use a steel braid between the two to accommodate the size difference. The tooling cannot be changed with the conductor in place. The crimp then has to be reconnected after the tooling is changed. The crimper chosen cannot be easily used or installed near the extruder plus it doubles the cost. Crimpers that can, don't have the tonnage or design needed to make the connections "fool proof".
The following picture is something that is used to pull these cables overhead in the field. They aren't very expensive and I would use them in a heartbeat if I wanted something capable of rotation. I'm trying to get past that. Even if they didn't rotate I would be concerned about uneven load on the pins during rotation.
These don't have to be dirt cheap either. They will pay for themselves at upwards of $500 apiece. I'm looking for around $100 or less though.
The following picture is something that is used to pull these cables overhead in the field. They aren't very expensive and I would use them in a heartbeat if I wanted something capable of rotation. I'm trying to get past that. Even if they didn't rotate I would be concerned about uneven load on the pins during rotation.
These don't have to be dirt cheap either. They will pay for themselves at upwards of $500 apiece. I'm looking for around $100 or less though.
What is the value of the scrapped cable length?
If it is enough $$ to warrant, just use two crimpers. Crimp on the above picture (tailpipe adaptor tube I presume, great visual) at the unroller, cut it off at the extruder, and crimp on a new one after the extruder. Only throw away 10 ft of product.
Forces, I think you have to test. You are dealing with torsional but it appears catenary tension forces also. Getting x degrees of torsion in L length depends on the cable tension as well as the torsional stiffness of the material. certanly solid acts different than stranded
If it is enough $$ to warrant, just use two crimpers. Crimp on the above picture (tailpipe adaptor tube I presume, great visual) at the unroller, cut it off at the extruder, and crimp on a new one after the extruder. Only throw away 10 ft of product.
Forces, I think you have to test. You are dealing with torsional but it appears catenary tension forces also. Getting x degrees of torsion in L length depends on the cable tension as well as the torsional stiffness of the material. certanly solid acts different than stranded
Forgot to add, does this device have to bend? Can you live with a fairly long stiff solid section? If so, make a male and female version of the picture in various sizes to fit the cables. Have a universal spline, or stub shaft separately splined, or something to connect that is torsiaonlly rigid. . Crimp one half to each cable end join and unjoin them.
With the forces you are describing, I would not look at this as a reusable tool connected to the end, but as a crimped/swaged/welded device that is cut out later and disposable.
With the forces you are describing, I would not look at this as a reusable tool connected to the end, but as a crimped/swaged/welded device that is cut out later and disposable.
What you are asking for is elastic properties and torsion of a beam.
TL/jG = torsion amount. T is the torque, L is the length, j is the polar moment of inertia, and G is the shear modulus of the material. Torsion is measure in radians as long as the units cancel out. For copper G is listed as 48GPa. j -> and ; pi * R^4/2 for a circular section.
I do wonder how the control system handles it when a larger diameter transitions to a smaller one. The much stiffer larger diameter will try to backdrive the built-up torsion into the short span of smaller diameter or the change-over from copper to aluminum. I wonder why they don't coordinate the turn from one end to the other like a rotissorie. Not for five years, so I'm not expecting an answer, but I'm guessing it's because the torsion load is miniscule and can be ignored.
Don't feel less human, you are simply undereducated. You should find a licensed structural engineer to review whatever you come up with and some anger management support as well.
The strength requirements are primarily driven by how tight the cable is in the catenary. Those will exceed the torsion load by one or two orders of magnitude.
The question you should have asked - how much tension load is on 1800 feet of cable with x feet of droop that weighs y pounds per foot, assuming both ends of the catenary are at the same elevation? explains how calculus is used to figure this out.
Only caring about torsion in this application is a good way to get people killed or do a lot of damage to property.
TL/jG = torsion amount. T is the torque, L is the length, j is the polar moment of inertia, and G is the shear modulus of the material. Torsion is measure in radians as long as the units cancel out. For copper G is listed as 48GPa. j -> and ; pi * R^4/2 for a circular section.
I do wonder how the control system handles it when a larger diameter transitions to a smaller one. The much stiffer larger diameter will try to backdrive the built-up torsion into the short span of smaller diameter or the change-over from copper to aluminum. I wonder why they don't coordinate the turn from one end to the other like a rotissorie. Not for five years, so I'm not expecting an answer, but I'm guessing it's because the torsion load is miniscule and can be ignored.
Don't feel less human, you are simply undereducated. You should find a licensed structural engineer to review whatever you come up with and some anger management support as well.
The strength requirements are primarily driven by how tight the cable is in the catenary. Those will exceed the torsion load by one or two orders of magnitude.
The question you should have asked - how much tension load is on 1800 feet of cable with x feet of droop that weighs y pounds per foot, assuming both ends of the catenary are at the same elevation? explains how calculus is used to figure this out.
Only caring about torsion in this application is a good way to get people killed or do a lot of damage to property.
GregLocock
Automotive
"piece of cable (copper or aluminum conductor) between two points under tension is rotated once every 75 feet. How much torque load is now on that cable?"
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
Shylanel:
Are you intentionally putting the twist in the conductor cable every 75', and is that a full revolution or something less? Why are you doing this and what torque are you applying? Or, does the pulling equipment (capstans right?) cause the twisting because of the lay and construction properties of the cable? This same twist is induced in the conductor under load (self wt., ice, wind), when it is hung and it just tends to tighten the lay or cable construction a bit, it is not a major stress issue on cable under normal conditions. I think I’m getting a vague idea what some of your issues are, contact me rwhaiatcomcastdotnet. You must have to stop things a foot before the extrusion tooling/die, make the disconnect, pull the lead cable through the die, change the die, pull/push the trailing cable through the die (how, with some intermediate device?), make up the cable end connection again, and then continue the process on the new reel of cable, right? And, you do waste those few feet of cable in the end. What are your pulling forces? You should be able to determine them or have some feel for them. Crimping or swagging should work as long as the crimp sleeve has an OD slightly smaller than the insulation dia. I would take MintJ’s crimping sleeve, cut it at the dia. transition, and make it a crimped end cap for each cable dia., with a cap pl. on the end. On one end cap I would have a single pin pl. and on the next end cap I would have a mating double pin pl. and you pull the pin/bolt and nut to make the disconnect, and this would provide some torsional strength and pulling strength to be determined. Gotta know more about what you have there to comment further.
Are you intentionally putting the twist in the conductor cable every 75', and is that a full revolution or something less? Why are you doing this and what torque are you applying? Or, does the pulling equipment (capstans right?) cause the twisting because of the lay and construction properties of the cable? This same twist is induced in the conductor under load (self wt., ice, wind), when it is hung and it just tends to tighten the lay or cable construction a bit, it is not a major stress issue on cable under normal conditions. I think I’m getting a vague idea what some of your issues are, contact me rwhaiatcomcastdotnet. You must have to stop things a foot before the extrusion tooling/die, make the disconnect, pull the lead cable through the die, change the die, pull/push the trailing cable through the die (how, with some intermediate device?), make up the cable end connection again, and then continue the process on the new reel of cable, right? And, you do waste those few feet of cable in the end. What are your pulling forces? You should be able to determine them or have some feel for them. Crimping or swagging should work as long as the crimp sleeve has an OD slightly smaller than the insulation dia. I would take MintJ’s crimping sleeve, cut it at the dia. transition, and make it a crimped end cap for each cable dia., with a cap pl. on the end. On one end cap I would have a single pin pl. and on the next end cap I would have a mating double pin pl. and you pull the pin/bolt and nut to make the disconnect, and this would provide some torsional strength and pulling strength to be determined. Gotta know more about what you have there to comment further.
- Thread starter
- #32
Thank you all for the productive responses.
kcj - The scrap cost ranges from $1 to $30 per foot of cable. I can easily justify a second crimper however the only crimpers with enough tonnage are 'in process' crimpers in that the conductor must run through the crimper after being crimped. There is no place to put this after the extrusion head as there is a vulcanization tube pressed to the crosshead that is much larger in OD than the maximum ID of the crimper. Also there is no such thing as 10 ft of startup scrap with this process. After starting it takes on a great day 100 ft to get the vulcanization conditions and product dimensions in spec before it can be called good. On large KV products or if we can't get a good water seal, there is a minimum of 600 ft of scrap after start. On bad days it takes 2000 ft to get a good start. The purpose of this is to keep the startup closer to 100 ft.
I'm expecting the solution to be rigid but has to go over 10 6' OD sheaves prior to the extruder so it would have to be strong enough to handle that. There would be limits on the length. The goal is reusable and I've seen it done with rotation allowed but disposable is doable if necessary.
Mint - something like that scaled up could work. How would the two ends be connected in between?
Dave - As was described the line is 1800 ft over several floors. The catenary is only 600 ft of this length and is about a 100 ft drop. I'd have to look up the exact drop for this line. It travels through a 8" tube under temperature and pressure shaped in a 26° catenary drop. The control is handled between the capstans described in this case based on the position of the cable in the tube which is measured inductively. There are limitations when going between extreme sizes. In those cases the cable has to be dragged down the tube with extreme care to avoid breakage. The twister is not applied in these situations. The tension between the cables is essentially the weight of the cable which can range from 1 to maybe 8 lbs/ft. That is a measured value-not theoretical. The torsional load is significant as I've seen steel bolts get sheared in two with only the normal torsional load from traveling over sheaves and the strand. They sheared when the rotational motion designed failed. Adding torsional load is significant.
dheng - yes you pretty much have the idea except it is the tip of the extruder that is the problem. It is 10 to 30 mils larger in OD than the conductor depending on the size. There also isn't a crimper with enough tonnage that I can use after the extruder due to the vulcanization tube. The pulling force is essentially the weight of the cable between the capstans.
There is a small natural twist between the two capstans. When the insulation thickness is the same or greater than the OD of the conductor, the insulation will droop out of the extruder. To counteract that, the cable is rotated in the direction of the lay to take advantage of gravity. This is why vertical CV lines are built for EHV cable. I can wait until the new cable is out of the pull out capstan before turning the twister on but that almost guarantees an extra 300 feet of scrap. That's why I want to design something that can handle the extra load. Generally the twister is set at 1/4 turn every 20 feet or 1 full turn every 75-80 feet. This is from experience and not something I've ever experimentally determined as the optimal twist. The twister is essentially a caterpillar capstan where one belt can be adjusted at an angle.
On the strand - I understand strand will behave differently than a solid rod. I could be totally off but I think in this case it will behave closer to a solid than strand most folks think of. This strand is alternate lay in all but the smallest. It is not bunch strand where all the strands are twisted at once. It is a layered strand where each successive layer is the opposite lay so in a 3 layer strand the inner layer will be left hand lay, the middle layer right hand and the outer layer left hand again. The larger cable can have upwards of 6 passes of alternating lays. The largest conductors are actually segmented cable. Each layer is also compressed significantly so friction should eliminate movement between the layers relative to each other. The end result in my mind is that each layer will counteract the natural tendency to shorten the lay experienced in a unilay or bunched strand. While more flexible than solid I would think it would behave more similar to a solid in terms of torsion. I could be wrong. I'm thinking that the torsional load of a solid will be greater than the strand so if the design can handle the load on a solid rod it can handle the load on the strand. If the values for solid are known I don't have to figure out how to measure the load on the strand.
General CCV line
The first picture is a vertical line for EHV cable. The second picture further down is a CCV line.
kcj - The scrap cost ranges from $1 to $30 per foot of cable. I can easily justify a second crimper however the only crimpers with enough tonnage are 'in process' crimpers in that the conductor must run through the crimper after being crimped. There is no place to put this after the extrusion head as there is a vulcanization tube pressed to the crosshead that is much larger in OD than the maximum ID of the crimper. Also there is no such thing as 10 ft of startup scrap with this process. After starting it takes on a great day 100 ft to get the vulcanization conditions and product dimensions in spec before it can be called good. On large KV products or if we can't get a good water seal, there is a minimum of 600 ft of scrap after start. On bad days it takes 2000 ft to get a good start. The purpose of this is to keep the startup closer to 100 ft.
I'm expecting the solution to be rigid but has to go over 10 6' OD sheaves prior to the extruder so it would have to be strong enough to handle that. There would be limits on the length. The goal is reusable and I've seen it done with rotation allowed but disposable is doable if necessary.
Mint - something like that scaled up could work. How would the two ends be connected in between?
Dave - As was described the line is 1800 ft over several floors. The catenary is only 600 ft of this length and is about a 100 ft drop. I'd have to look up the exact drop for this line. It travels through a 8" tube under temperature and pressure shaped in a 26° catenary drop. The control is handled between the capstans described in this case based on the position of the cable in the tube which is measured inductively. There are limitations when going between extreme sizes. In those cases the cable has to be dragged down the tube with extreme care to avoid breakage. The twister is not applied in these situations. The tension between the cables is essentially the weight of the cable which can range from 1 to maybe 8 lbs/ft. That is a measured value-not theoretical. The torsional load is significant as I've seen steel bolts get sheared in two with only the normal torsional load from traveling over sheaves and the strand. They sheared when the rotational motion designed failed. Adding torsional load is significant.
dheng - yes you pretty much have the idea except it is the tip of the extruder that is the problem. It is 10 to 30 mils larger in OD than the conductor depending on the size. There also isn't a crimper with enough tonnage that I can use after the extruder due to the vulcanization tube. The pulling force is essentially the weight of the cable between the capstans.
There is a small natural twist between the two capstans. When the insulation thickness is the same or greater than the OD of the conductor, the insulation will droop out of the extruder. To counteract that, the cable is rotated in the direction of the lay to take advantage of gravity. This is why vertical CV lines are built for EHV cable. I can wait until the new cable is out of the pull out capstan before turning the twister on but that almost guarantees an extra 300 feet of scrap. That's why I want to design something that can handle the extra load. Generally the twister is set at 1/4 turn every 20 feet or 1 full turn every 75-80 feet. This is from experience and not something I've ever experimentally determined as the optimal twist. The twister is essentially a caterpillar capstan where one belt can be adjusted at an angle.
On the strand - I understand strand will behave differently than a solid rod. I could be totally off but I think in this case it will behave closer to a solid than strand most folks think of. This strand is alternate lay in all but the smallest. It is not bunch strand where all the strands are twisted at once. It is a layered strand where each successive layer is the opposite lay so in a 3 layer strand the inner layer will be left hand lay, the middle layer right hand and the outer layer left hand again. The larger cable can have upwards of 6 passes of alternating lays. The largest conductors are actually segmented cable. Each layer is also compressed significantly so friction should eliminate movement between the layers relative to each other. The end result in my mind is that each layer will counteract the natural tendency to shorten the lay experienced in a unilay or bunched strand. While more flexible than solid I would think it would behave more similar to a solid in terms of torsion. I could be wrong. I'm thinking that the torsional load of a solid will be greater than the strand so if the design can handle the load on a solid rod it can handle the load on the strand. If the values for solid are known I don't have to figure out how to measure the load on the strand.
General CCV line
The first picture is a vertical line for EHV cable. The second picture further down is a CCV line.
As long as the twist is within the yield region of the material, the typical equation : a= Tl/JG applies; a= the angular deformation in radian, T = the torque, l = the length between ends, J = the polar moment of inertia which for a solid wire would be (pie)x(r^4)/2, r = the radius of the wire, G = the modulus of elasticity in shear for the materials of interest. This formula is found in any textbook of strength of material and machine design, ME handbook, or undubitably many other textbook and handbooks .
Now for the ends to be held stationary under the calculated torque and I am assuming (which can make an A** out you and me) that both ends are riding over grooved sheaves you may want to consider grooved clamps over the top of the exposed wires within their sheaves. Friction would be the holding force and a sketch of this idea is in the attachment. Obviously the clamp is nearly a point load on the wire, however, you may be able to lengthen a little the point of contact without deforming the ends over the circumference of the groove. A small contact area from the clamp is probably all that is needed to maintain the calculated torque on the wire.
Now for the ends to be held stationary under the calculated torque and I am assuming (which can make an A** out you and me) that both ends are riding over grooved sheaves you may want to consider grooved clamps over the top of the exposed wires within their sheaves. Friction would be the holding force and a sketch of this idea is in the attachment. Obviously the clamp is nearly a point load on the wire, however, you may be able to lengthen a little the point of contact without deforming the ends over the circumference of the groove. A small contact area from the clamp is probably all that is needed to maintain the calculated torque on the wire.
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- #36
In looking at the information you piece-meal leak out, it seems to me that you are fishing for an answer that is worth a very large amount of money. Anyone who comes up with a solution to this problem should keep it to secret and contact a patent attorney, one that can handle multi-national patent applications.
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- #38
Dave you really do like your electronic graffiti. I shouldn't even respond to this. Not sure why I am.
I'm not fishing for a solution. All I ever wanted was how to do determine the torsional load and if my assumptions were correct. All the information needed for that answer was in the first post. I have the calculation I was looking for. Thank you to all that contributed. Any brainstorming ideas for a solution are welcome and will be considered but I am not nor was I ever here to solicit a solution. If someone figures out how to do it and wants to keep it to yourself I wish you luck. Feel free to contact me for testing and I may be able to suggest a few companies that may be interested in buying it from you...Greenlee and Sherman & Reilly come to mind. There's also Davis Standard, Troester and Maillefer who are the primary suppliers of CCV lines internationally. This isn't an industry known for paying a premium for technology that can be done or duplicated in house without risk of detection for patent infringement. Most in this industry have no idea how patents work and are in awe of anyone with a patent. Without plant inspection how can a company determine if the technology is being used? Nothing in the final product could be tested. You would end up paying a lot of money with limited potential for profit on an unenforceable patent.
I'm not fishing for a solution. All I ever wanted was how to do determine the torsional load and if my assumptions were correct. All the information needed for that answer was in the first post. I have the calculation I was looking for. Thank you to all that contributed. Any brainstorming ideas for a solution are welcome and will be considered but I am not nor was I ever here to solicit a solution. If someone figures out how to do it and wants to keep it to yourself I wish you luck. Feel free to contact me for testing and I may be able to suggest a few companies that may be interested in buying it from you...Greenlee and Sherman & Reilly come to mind. There's also Davis Standard, Troester and Maillefer who are the primary suppliers of CCV lines internationally. This isn't an industry known for paying a premium for technology that can be done or duplicated in house without risk of detection for patent infringement. Most in this industry have no idea how patents work and are in awe of anyone with a patent. Without plant inspection how can a company determine if the technology is being used? Nothing in the final product could be tested. You would end up paying a lot of money with limited potential for profit on an unenforceable patent.
You can probably estimate the torsional strain created in the cable based on the relative diameters of the cable and capstan, and the number of turns the cable wraps around the capstan. The cable contact follows a helical path as it wraps around the capstan, and the constant axial shift of the contact between the cable OD and the capstan creates a torsional moment about the cable neutral axis. For example, if the cable OD is 1 inch and the OD of the capstan is 20 inches, then the twist in the cable would be approximately 115deg over a 65.9 inch length for each wrap around the capstan (ignoring slippage, etc).
That's my theory, but maybe someone can point out if my approach is flawed.
That's my theory, but maybe someone can point out if my approach is flawed.
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- #40
Thank you all for the help. I have something I think I can work with though I'm not fond of the some of the assumptions I'm making. The main one being the shear modulus of the conductor. Using the Shear Modulus for solid copper The strand should be twisting itself apart with our medium sized conductors up. Since I know this isn't the case and I can't find any values for strand or a relationship between strand and solid, I adjusted the Shear Modulus value down until the largest strand doesn't twist itself apart. This still should be a larger modulus than actual but I don't really have an easy way to get that number. If anyone knows how to get that number please share.
The tensile load in our process is essentially constant at 12.5 MPa and the torsional load (using the adjusted modulus) is always greater than the tensile ranging from 61% greater on small strand to nearly 10X greater on the large strand.
This can then be translated to determine minimum diameter steel is needed to work on a given conductor diameter. A better way to put it is given an OD of steel how much larger of an OD of conductor can it safely pull. The smaller sizes can handle up to an 80% increase in relative OD of the conductor. This decreases to about 40% for the larger conductors. This gives us plenty of range to work with.
This all passes the sniff test with experience. I'd feel better if all the data was known or measurable but as usual we aren't given those resources in this industry.
The tensile load in our process is essentially constant at 12.5 MPa and the torsional load (using the adjusted modulus) is always greater than the tensile ranging from 61% greater on small strand to nearly 10X greater on the large strand.
This can then be translated to determine minimum diameter steel is needed to work on a given conductor diameter. A better way to put it is given an OD of steel how much larger of an OD of conductor can it safely pull. The smaller sizes can handle up to an 80% increase in relative OD of the conductor. This decreases to about 40% for the larger conductors. This gives us plenty of range to work with.
This all passes the sniff test with experience. I'd feel better if all the data was known or measurable but as usual we aren't given those resources in this industry.
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