Measuring tension in cables of Wind barrier structure supports fabric 1

[engABZ]

Hello,
I'm facing a problem in how to measure the tension developed in cables of Wind-barrier structure, that supports fabric cloth, I need it to control the tension that the cantilevered columns supporting the fence should handle without failing.

The fence to be 15 meter high and covered laterally with fabric cloth of 2m wide and spanning 25m between the two columns. Each fabric to be supported by two cables (top and bottom end.

Photos from google

http://www.nettingus.com/images/wind_barrier.JPG

http://www.nettingus.com/images/windscreen_9.JPG

I have thought of using (Catenary (sag-tension) approach) but couldn't really know how to apply it in my case. Also there will almost be no sag in the cables supporting the fabric, the real sag due to wind blowing, will be developed in the fabric (Almost 0 in top and bottom of each fabric and max. at the middle).

Any suggestions?
Regards

 

[STH003]

I would look at it like tributary loading. Figure out what the pressure profile due to wind on your barrier will be for the given return period you need. Multiply this pressure by the area of fabric each cable supports to get the tension force in each cable.

 

[SlideRuleEra]

I'm not certain from the wording of your post is you want to "measure" tension in existing cables (at ambient conditions) or "calculate" tension under an assumed (high wind?) loading.

To measure tension, perhaps an instrument like this would do it:

http://www.checkline.com/cable_tension_meters/CTM

To calculate tension (for high wind speed), one way is to make some simplifying assumptions:

1. The fabric & cables are "weightless" (compared) the force of the wind. Therefore all loading on the cables is horizontal.

2. As STH003 suggested, each cable is loaded by the tributary area of half the fabric panel above it and half the fabric panel below it.

3. Load (PSF) of the wind on any one panel row is uniform. However, each panel row MAY have a different loading to simulate increased wind speed as elevation increases.

Then you have have anchored, uniformly loaded, level span conditions. See page 39 of "USS Wire Rope Engineering Hand Book - 1968" on this page of my website:

http://www.slideruleera.net/miscellaneous.html

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[cancmm]

Realistically the cable has to have some measure of sag or the tensile force will calculated at infinity. I agree that the catenary is the way to go. Can you get a realistic idea as to the initial tension in the lines when they are installed? Maybe even make a conservative assumption? Then, knowing the self weight, you could iteratively calculate the sag under self-weight alone and relate this to the deflection realized under wind loads. Then the tension could be calculated. Haven't thought this through entirely so I may be mistaken.

 

[STH003]

What is the ultimate goal to finding the tension in these cables? Looks like the structure is already standing. Is it to check the design so that it doesn't fall down or do you need the loads because you are modifying what is already there?

 

[engABZ]

Thanks a lot everybody for the suggestions

I should have written ("Calculate" instead of "Measure")
The objective of calculating the Tension in the cable, is to design the columns such that it will not fail due to unbalanced tension developed when there is non-uniform wind load blowing into the two sides supported by the column.

So it looks like the easiest way is to consider the tributary loading, but would it be accurate enough (is it conservative assumption)?

 

[STH003]

In terms of conservative/unconservative, well it depends.

If you neglect the weight of the fabric and self weight of the cables, then it is unconservative.

Who knows how "permeable" your fabric is. From the picture it looks like some air can slip through, but since this would be hard to calculate, the full psf value from the wind would be appropriate (I wouldn't call it conservative)

 

[SlideRuleEra]

Whether you use a "precise" (software) simulation of the wind loading or simplifying assumptions the most important factor will be the pressure that you assign for the design wind speed. Select too low a value and the calculated result will NOT be accurate or conservative. Otherwise, the "accuracy" of the two methods should be comparable (perhaps within a few percent of each other).

To (slightly) increase the accuracy of either approach you could consider the cable tension caused by ONLY the dead load of the fabric and cables under no-wind conditions. Then vectorially add this to the horizontal tension from the wind.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[rb1957]

i guess the difficulty is that the airload on the barrier can be reacted either by tension in the cables or by fixed end moment in the cantilever posts.

how much bending can the posts absorb ? ie assume there ae no cables. this'll deflect the end of the cantilever (a predictable amount, using Roark, a cantilever with a UDL). This'll strain the cables, note each post is supported by two cables at an angle. this'll put tension in the cables which will relieve the bending in the posts, and add compression to the posts. this'll cahnage the tip deflection of the posts ... a few cycles of iteration should get you there. "there" will be a balanced load solution, next, are the elements (the posts and the cables)strong enough ??

 

[STH003]

hmmm I guess there is a confusion as to which cables we are looking at.

I'd use the same process as above, and follow the load path of the tributary system to the next sequence, at the columns.

You wouldn't need to iterate between cables and column (unless you wanted to incorporate 2nd order effects... but no vertical load so ok here), the column will deflect, and the cables will stretch to accommodate this deflection. You can then set up your FBD and solve for the stresses in the cables

See attached schematic for what I mean

 

[rb1957]

if you simplify it so the post is pinned, then the component of the cable tension reacting the applied airload is 1/2 the airload (assuming a uniform distribution) and the max moment in th epost is 3/8*wL^2 (plus compression from the cable tension)

 

[paddingtongreen]

All of the above is fine, if you can establish the initial tension. I don't know how to do that because any test of one span will yield unrealistic answers if the cable is anchored at each column, because the columns only have to move slightly to put the whole thing off.

There is a chance since the end bay is longer than the rest, and the sag is greater. You could measure the sag at the long bay and the normal bay and compare them. If the cables are not anchored, if they just pass through holes or loops, they will have the same tension, if you ignore friction, and you can calculate that from the sags under vertical loading.

If the cables are connected to the columns. You might be able to test one cable with a vertical load and see how much it deflects but you would have to check that the columns did or didn't move to shorten the span.

I don't know if this all makes sense it does to me, but it is not easy to explain.



Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[engABZ]

Thanks a lot guys for the valuable information

Regarding the point mentioned by rb1957 about the Uniform load on the column, I have taken a UDL = wind load x 2 x span between columns/2.

And that load case will be used in addition with case of non-uniform load (developing lateral tension in the column).

My problem that I was uncertain of how to calculate the lateral tension. But now it is clear for me and makes sense.

Thanks again everybody

 

[paddingtongreen]

Either I can't find it here or something is being missed.

An important piece of information is missing: are the cables connected to the columns or just threaded through holes? Regardless, they must be connected at the end column. If the sag is small, the catenary approaches the parabola. The horizontal in-plane component (along the nominal centerline of the cable) equals the total wind force on the cable, divided by 2, multiplied by half the span, divided by the sag.

From this the cable pull on the end column will be several multiples of the wind force on that cable. If the sag was equal to 1/20th of the span, the connection to the column would be W/2 out of plane and 5W in plane. If the sag is only half as much, the in-plane component doubles to 10W. If the cable is simply threaded through the columns, the longer bays will sag more so that the tension in the cable is equal either side of an interior column.

Intermediate columns would see a more balanced in plane load either way.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[rb1957]

is the concern the cables supporting the columns (i thought it was) or cables in the plane of the material ? these will support the membrane some, and any tension would be a lateral load on the outer columns ... lateral bending.

random thought ... remember wind blows both ways ... it looks as though you've got bracing on one side only .. in one direction the wind will tension the cables supporting the posts (good) ... in the other direction it won't (bad)

 

[paddingtongreen]

rb1957, I assumed it was the in-plane cables, I don't know how catenary could be applicable to the bracing. The end column, in the first picture, appears to be unbraced in-plane.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[rb1957]

agreed, that's why there'd be (lateral) bending on the end columns

 

[engABZ]

hmmmm

The cable is supporting the fabric, and the cable is connected to the column through two plates (the cable is in between them) attached to the column (consider cable is continuous).

The sag in the cable is very small (Tensioned) I don't have the value of the tension, but I think half of the total weight of the cable along the span will be appropriate each side!

The cantilevered columns are not braced, and has no anchors.

Here are some other pictures:

http://files.engineering.com/getfil...bb-3c8f4f8d68e2&file=Crawford-wind-fence3.jpg

http://files.engineering.com/getfil...c6-639610be8522&file=2007_0314richply0003.jpg

 

[rb1957]

the last pic shows not bracing cables, the 1st pic showsa suspicion of cables at the end, supporting in the plane of the barrier.

so te columns, pretty impressive looking things, react the airload as cantilevers.

sort of begs the question ... "where're the cables ?"

 

[paddingtongreen]

Okay, one end shows in-plane bracing but one doesn't (first photo, above). I stand by what I posted in my post 15 Jul 10 9:58, above.

Michael.
Timing has a lot to do with the outcome of a rain dance.