Chain catenary vertical reaction help 11

[Carcajou]

For dead load chain catenary with different y-axis end support positions, it's clear that as more chain is hanging from upper support A) both vertical attachment end reactions cannot be identical, and B) upper support vertical end reaction will be higher than lower support vertical end reaction. How are vertical end support reactions determined - losing my sanity a bit. I think chain horizontal component will be equal across chain length, and a cable would react identical to a chain. S.O.S.

Carcajou

 

[r13]

In my opinion, a chain in any position (except vertical), loaded through its weight center, each end will have a reaction equal to one half of the load. I may have misunderstood your question though.

 

[JStephen]

If you have a solution with both ends level, it seems you could cut a free-body diagram cutting the chain at any two points and have solutions for supports at those points.

 

[WARose]

Would you or someone in your office happen to have a Hibbler's book 'Structural Analysis' (3rd ed)? A solution is presented there.....but it's a little long for me to reproduce here.

 

[BridgeSmith]

If one end is significantly lower than the other, yes, the vertical reactions will be a little different. What will be the same is the horizontal reactions at the supports. You'll have to work out the slope of the chain at each end to get the vertical reactions. I hope that helps get you started.

I've done numerous catenary cable spans, usually with points loads (traffic signals) along the span, but I never was too concerned about the vertical loads - just the horizontal force and cable tension, which are much larger due to the small sag, and the primary source of stress in the poles supporting the cable span. That said, the incremental analysis that I use to approximate the catenary does calculate the vertical reactions. I'll take a look at my design spreadsheet and see if I can work out how the analysis would change for supports at different elevations.



Rod Smith, P.E., The artist formerly known as HotRod10

 

[IDS]

Consider two free-body diagrams with the cable cut at the point where it is horizontal.

Clearly there will be no vertical force at the cut end, so the vertical reaction at the support ends will be the weight of the length of the segment that it is supporting.

I have a catenary spreadsheet here:

https://newtonexcelbach.com/2009/06/25/a-catenary-function/

It generates the catenary profile from horizontal to support, so for supports at different levels you could set up two catenaries and use Goalseek to find the two spans that give the required total span and sags.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BridgeSmith]

I think I have worked out how to find the vertical reactions using my spreadsheet, but since I've set it up only for my own personal use, it wouldn't be very user-friendly for someone else. However, if you would be willing to post the particulars of your situation (span length, difference in elevation, sag, chain weight per foot, and the magnitude and location on the span for any point loads), I would be happy to run it and let you know what I come up with, so you can compare to your analysis.

Rod Smith, P.E., The artist formerly known as HotRod10

 

[BridgeSmith]

If you have a solution with both ends level, it seems you could cut a free-body diagram cutting the chain at any two points and have solutions for supports at those points.

That's the way to the solution, and essentially how I approached it. As I see it, it may take some iteration to get the sag for the section you cut to correspond to that of the actual span. Unless someone can suggest a direct solution?


I just realized that for the vertical reactions, the sag is irrelevant. JStephen's approach is the direct solution. You'd need the sag to calculate the horizontal reactions or tension in the chain, though.

Rod Smith, P.E., The artist formerly known as HotRod10

 

[IDS]

I just realized that for the vertical reactions, the sag is irrelevant. JStephen's approach is the direct solution. You'd need the sag to calculate the horizontal reactions or tension in the chain, though

The sag effects the length, so it does effect the vertical reactions.

I have updated my spreadsheet (attached) to generate a full catenary with different sags either end, and to adjust the position of the maximum sag and the canary constant using Solver, to generate the curve for any specified sags and cable weight.

Note that:
The left hand length must be less than the total length!
If the solver (right hand button) doesn't find a solution try just adjusting the "A" value only with the left hand button, then the right button to adjust the sags as well.
The solver needs to be enabled.

Let me know if any problems.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BridgeSmith]

The sag effects the length, so it does effect the vertical reactions.

You're right, of course. The sag for the section used in the analysis would need to be fairly close to the actual to get the length and total weight close.

Rod Smith, P.E., The artist formerly known as HotRod10

 

[BridgeSmith]

I'm thinking the vertical reaction at each end would be the tributary weight from the low point of the sag to each respective end, correct?

Rod Smith, P.E., The artist formerly known as HotRod10

 

[IDS]

I'm thinking the vertical reaction at each end would be the tributary weight from the low point of the sag to each respective end, correct?

Yes

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IRstuff]

This might be of interest:

https://www.researchgate.net/publication/239754721_Linking_Hanging_Chain_Models_to_Fabrication

http://home.earthlink.net/~w6rmk/math/catenary.htm

http://www.dredgingengineering.com/moorings/catenarya/asymetrische afleiding goed.htm

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[Denial]

You will find on my website (

http://rmniall.com)

a spreadsheet that will do exactly what you want: a catenary whose ends are at different elevations.[ ] (It also allows for further "complications", such as a point load somewhere along the span.)

 

[HS_PA_EIT]

Try your arrangement with graphical statics/funicular polygons. Search "funicular curve through three points" for a good powerpoint presentation. This can be used easily in Autocad to find catenary forms and reactions.

 

[Carcajou]

Thank you all for aligning my cranium on this. I was given this problem during a job interview just out of school. I had like an hour to answer 9 other similar problems and I'm sure it's why I didn't get the position. Always meant to get back at it when I had the time.

Carcajou

 

[IDS]

See:

https://newtonexcelbach.com/2020/01/31/asymmetric-catenary-function/

for download file, more details, and links to solving solver problems.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Celt83]

I have a great book on Cable-Suspended Roofs by Prem Krishna that has the centenary derivation, I've attached a PDF of the relevant pages.
Link

Amazon Link to current iteration of the book: Link

My Personal Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

Open Source Structural GitHub Group:

https://github.com/open-struct-engineer

 

[Bill West]

I took a historical interest in a 21 mile 11,000ft rise cableway in Chilecito, Argentina. It was built to reach the Mina La Mejicana in the Famatina region. In 1905 it set a world record for length and it still is for rise. I worked out the catenary for all 275 towers on a spreadsheet and put it into a KML file so I could take a ride down the line in Google Earth.

In the course of this activity I felt that some boiled down observations could bring catenaries down from higher math to levels that could be easily entered in Excel.

The horizontal force along a catenary cable is equal everywhere, pulling to left on the left of the low point and to the right on the right of it. The vertical force at a support tower is equal to the weight of the cable from it down to the low point. The vertical force at a point on the way down to the low point is equal to the weight of the remaining cable to the low point. The horizontal and vertical vectors at a given point give the angle of the cable at that point and the tension in it. The point tension increases from the low point to a maximum at the tower.

If a tower mount is not clamped then the horizontal tension in any adjacent catenary will be the same. This leads to using a spreadsheet to connect and adjust the sags in a series of catenaries over uneven terrain.

The span and the tower height are not the only givens, the ratio of the tension to the unit cable weight is also a user choice. The heavier the cable the deeper the low point will be pulled, varying the sharpness of the sag. If the cable had zero weight it would be a horizontal line with zero sag. If the total weight per half length equaled the tension it would drop straight down and rise straight back up to the next tower with a span of zero feet. The key point for all the curved shapes in between is that there is only one shape of hyperbolic curve, we are just using the tension/weight ratio and span dimensions to scale where our span fits across that curve.

To calculate a catenary by hand or in Excel, choose:
-a trial horizontal tension N
-a sample cable weight per unit length Q
-the horizontal span distance L

Working in feet & lb or in meters & kg, we find:
A = N/Q the tension to unit weight ratio.

If there is a difference in the elevation of the tower tops, delta He, we find:
M = A*asinh(((delta He/A)/2)/sinh((L/2)/A)) the low point's offset from middle of the span.
Otherwise we use the mid point of the span as the low point.

Either way, using Lp as the distance of a point of interest from the low point we find:
Hp = A(cosh(Lp/A)-1) the height of the point above the low point.
This is also described as the sag if it is figured for an end point.
It is valid for symmetrical spans, for unequal tower top elevations with the low point between them (the OP's question, also called an inclined catenary), for unequal tower top elevations with the low point to one side of them and for any intermediate point.

Sp = A*sinh(Lp/A) is the running length along the cable from the low point.
2x the Sp for the end point of a symmetrical catenary finds the whole span.

Hp is the y where our Lp fits the x of our hyperbolic curve and Sp is the arc length along the curve.

For graphs just generate an incremented table of Lp points as + or - offsets from the low point, copy down the Hp height calculation and graph Lp & Hp as x & y. Start and finish with the Lp for the actual top points to make the top data calcs exact. For the inclined catenary I plotted a second calc moving the lower tower out to where it would be the same height as the taller one. Made this a dotted line and then it was easier to see how the subject span was just part of a longer hyperbolic curve.

The calcs can be done directly in Excel. Goal seek would be needed if you want to go from a desired sag re ground clearance back to the allowable tension or weight. But the one goal seek can solve an entire span or string of spans. If your terrain is extreme you can get locations where the upper tower is actually trying to lift the lower tower (the mid point calc gives an answer outside your span). Then you have to reduce N until the sag droops away from the low tower. Or have a receiving tower design that can resist the uplift, this complicates the cable attachment if it is a cableway. Overall there must have been a lot of trial and error in manual calculation days. I recollect once seeing Plexiglas curve templates being used by transmission line designers to solve fitting to route profiles.

For a point load at the low point, figure out an amount of L about that point that has a cable weight equal to your load. Cut that length out of your graph and slide the two pieces back together. The vertical tension left at the cable ends will equal your load. Could probably just goal seek an Lp offset back up the sag until the point's vertical tension/Sp cable weight matches half your cargo. Oops, actually you have to get back to having the original span. Major distributed point loads such as suspension bridge suspenders would be more complicated. Loads distributed uniformly along the run of the cable are just a heavier cable.

Okay, that's enough recapping of what many of you already know better than I do. Let's go for a ride on the Chilecito cableway, it's in the attached KML below. Set Google Earth's Tools/Options/Touring line settings to Camera tilt 75deg, Camera range 15 meters and Speed 175. Then select the "Top to Bottom" line in Places and click the Play Tour button at the bottom of the Places box.

Sagen.AT has a paper the builder's (Adolf Bleichert & Co) Senior Engineer wrote on the project. Note that the plan views are flipped top to bottom. Google translate will fix the language.
[link "Sagen.At"]

http://www.sagen.at/doku/Eisenbahn/Drahtseilbahn_Kordilleren.html

[/url]
Wikipedia has articles on the cableway
[link "Mina La Mejicana"]

https://es.wikipedia.org/wiki/Mina_La_Mejicana

[/url]
[link "Materialseilbahn Chilecito-La Mejicana"]

https://de.wikipedia.org/wiki/Materialseilbahn_Chilecito-La_Mejicana

[/url]
The site is now a tourist attraction, search Google Image for Chilecito Cable Carril for site pictures.

Bill

 

[Denial]

Celt83

Thanks for that Prem Krishna reference.[ ] My first reaction was "that's a useful set of formulae", so I set about checking them against a problem for which I knew the answers.[ ] However in doing this I discovered possible/probable/definite errors.[ ] Have you managed to confirm the formulae in some way?

The formulae I would query are:

(2-6) for T".[ ] If I change alpha to beta I get agreement, but that might just be a coincidence.

(2-8) for S".[ ] Self evidently, for dimensional reasons alone, the final 1 (that's the number 1) cannot be correct.[ ] If I change it to l (the lower case 12th letter of the alphabet) then I get agreement with my test problem, which again might just be a coincidence.

(2.12) for elastic stretch.[ ] Self-evidently incorrect because of dimensional inconsistency.[ ] If I change it to
[ ][ ][ ][ ](Hl²/X)(1/EA)
I get an answer that is in reasonable agreement with my test problem, but since I have no theoretical justification for this change that might be a fluke as well.

Note that my investigations were entirely arithmetical, based on the one single example.[ ] I have done no algebraic / mathematical investigation.

Any comments?