Pretension in Wire Cable Wall

[jgraham114]

I'm currently working on a project that involves a stainless steel mesh wall. Basically, it is a woven wire wall suspended with cables. I have the USS Wire Rope Engineering Handbook. I can calculate the tension on a cable with the wind load pressure applied and a deflection limit.

The issue I'm having is calculating the required pretension in the cable to meet the deflection limit at max wind load. The USS Wire Rope Engineering Handbook states getting the pretension of horizontally suspended wire but not vertically.

Any ideas on where to look to get more information on this. Any other help or tips is greatly appreciated.

Let me know if you need more information.

 

[rb1957]

vertical wires ? won't pre-tension the mesh, so won't help stiffen. you can see how if the wires were in-plane with the mesh, that they'd stretch it and so stiffen it.

i imagine that you've got these wires on both sides of the mesh, which i guess looks quite odd. but if they're only above it (sounds natural) then they won't support any up load. so the wires will support the mesh where they attach, and the mesh will deflect as much as it wants to.

Quando Omni Flunkus Moritati

 

[jgraham114]

I don't think I described it very well. The cables run vertically at .688" on center, supported at each end and an intermediate support. The wire is woven between these cables.

The cables should have some pretension as to not exceed the desired maximum deflection from wind load. How do I calculate the pretension required in the cable?

 

[rb1957]

yeah, i was wondering about that as i reread your post ... "wall" ...

you're looking at the lateral deflection of these veritcal wires under a lateral wind load ?
your wire design book shows you a suggested pre-tension to react the weight of the cable, yes?

how does the wind load compare to the weight of the cable ? cable weight something like 0.3*pi*D^2/4 lbs/in

Quando Omni Flunkus Moritati

 

[jgraham114]

Lateral deflection with lateral wind load.

The design book tells how to calculate tension from distributed load and deflection desired (I've calculated the tension with the WL applied).

It also shows how to calculate the tension of an empty span (horizontal) based on the tension calculated before and self weight. The problem I'm running into is that an empty vertical cable has no deflection or self weight in the lateral direction.

There has got to be something that I'm over looking or over thinking.

 

[rb1957]

i imagine FWIW that they're saying first calc the tension required to meet your deflection criteria for wind loading, then add in the extra deflection due to weight, the weight adds to the wind load so you need a higher tension to meet the same deflection critieria.

in your case weight doesn't add, so tension due to wind loads is sufficient.

actually weight will have a small affect ... the lateral defliection due to lateral wing loads will move the CG off the axis of the cable, creating a small moment in the cable (increasing the tension in the upper 1/2, reducing it in the lower 1/2).

Quando Omni Flunkus Moritati

 

[jgraham114]

What the handbook demonstrates is how to calculate the tension in a cable with a given deflection criteria and loading. From that you can calculate what the erection tension and deflection would be based on the change in length and total tension under full load.

Start on pg. 14:

http://www.tramway.net/Tiger Rope.pdf

I guess where I'm getting confused is that the more pretension that is added to a cable the less deflection there will be from the same distributed load as opposed to installing the cable with no pretension.

How do I determine that pretension?

 

[dhengr]

Jgraham114
For your vert. application of wire rope, I would use the std. catenary equations in the handbook for a horiz. cable. But, substitute wind load per cable per inch for the cable weight in the handbook formulas. I wonder how you’ve figured the wind load per cable, how close together the perpendicular wires are woven and how tightly (tensioned). Remember, no matter how high the wire rope tension, it takes a very small load to cause a fairly large deflection of the cable. So you need a substantial perimeter support system in the plane of the cable net if you can’t tolerate much deflection. The wire rope or cable net shines once you allow some deflection so it really starts acting like a tension net or a tension field.

If you know your load per inch of cable and you know the deflection limit, you should be able to calc. the tension or horiz. reaction needed to meet that criteria. If you pretension the cable to something greater than that calc’d. T you should stay within your deflection limit. If the load increases beyond your load, it will cause a higher T and deflection. Also note, that different wire ropes stretch slightly differently. There is the elastic strain of the wires which make up the rope and there is also some mechanical stretching as the entire cable tightens on itself. You may want prestretched wire rope.

 

[jgraham114]

The deflection limit is 6" due to not wanting it to hit the support structure behind.

The wind load used is the full wind load assuming the woven fabric is impermeable.

Your second paragraph doesn't make sense. The tension calculated in the handbook (link in previous post) from the wind load for the 6" deflection would be the max tension when loaded. IMO, it is not a calculation for the amount of pretension required as it doesn't take into consideration the rope properties.

Equation for Tension in cable (according to USS Wire Engineering Handbook)

t=ws^2/8y

where:
t = Horizontal component of cable tension (in my case vertical)
w = distributed load
y = deflection at center of span

Now there is an equation that I tried but got a negative answer for the tension

L-L1=P(t-te)

where:
L = Length of cable when loaded
L1 = Length of cable unloaded (I used the span length)
te = Erection Tension
P = L/AE (A = net cross sectional area, E = Modulus of elasticity in tension)

 

[paddingtongreen]

I need more information as the end connections.

Are all of the verticals connected to a rigid body, top, bottom and in the middle? Are the ends of the horizontals connected to a rigid body?

If the ends of the horizontals are not connected to a rigid body, the end verticals will deflect in-plane towards the center as well as outward.

Think of a sail on a square rigged ship. Double curvature.

I think we can get to a solution if we have the information.

Surely, the verticals carry the weight of the horizontals via friction?

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[CANPRO]

As paddingtongreen noted, there is more information needed to really understand the problem. The following is my opinion on just a single span cable subjected to a uniform load and a pretension load. This is also assuming the cable is some initial length suspended between two supports, and then stretched to a new length and the supports held rigidly in their positions (also neglecting self-weight):

First t=ws^2/8y is not just according the USS Wire Rope Handbook, it is according to statics. Cut the cable in half and draw the free body diagram.

I think the best thing to do with this problem is break it down in steps instead of looking for the formula at the end of the road.

The cable will be suspended with some initial length and then will be pretensioned to some new length. Using these two lengths you can get an expression for the strain in the cable. Next, the wind will load the cable to your maximum allowable deflection. From this max allowable deflection you can calculate the new length of cable. Using the loaded length and the initial length you can find the new strain in the cable. Both expressions for the strain will have the initial length in them, you can rearrange and set the expressions equal to each other. This will leave you with an expression containing the pretensioned length and strain and the loaded length and strain.

I assume the pretensioned length has been specified, the loaded length is a function of the pretensioned length and the allowable deflection, and you specify the strain in the cable as the designer. You can now relate the loaded strain to the initial strain. You know the tension in the cable is a function of its span and deflection by t=ws^2/8y, and you also know the tension in the cable is a function of its strain. So you should be able to use the above logic to relate the initial preload to the deflection.

Its late, and I don't really feel like grinding this all out on paper right now...but I think you can get the idea. It might be more complicated than this because the strain in the cable will not be constant once its in its loaded position. Theres also the chance that could all be wrong, but I hope it helped.

 

[jgraham114]

That also makes sense. I do not know the relation between the initial length and the pretensioned length.

The vertical cables are 2mm 316SS 1x19 (each strand is 0.4mm), Max Tension=800N/mm^2(per Manufacture). Deflection limit is 6". Cable is a twin span with each span being 226" (I'm thinking I might need to add anchor to shorten span based on tensile strength). Uniform loading based on wind load is 0.5 lb/in.

I have calculated the tension at mid-span (226"/2) to be 540 lb and the reaction at the ends as 543 lb with the 0.5 lb/in loading for 6" deflection.

I just need to figure out how to determine what pretension would be required to achieve the 6" deflection.

I mean I can't be the first person to come across this situation... I must be missing something.

 

[CANPRO]

"I do not know the relation between the initial length and the pretensioned length" <---- you do know this relationship, it is the strain in the cable once it is pretensioned. You also have the same relationship between the initial length and the loaded length. That is the key to the problem. As I stated before, you need to look at this problem in steps.

You also stated "Now there is an equation that I tried but got a negative answer for the tension L-L1=P(t-te)"

If your cable naturally wants to deflect less than the 6" you're trying to limit it to, then this equation will not work...because you do not need a pretension load to maintain the deflection limit. This is why I suggested break the problem down into steps rather than try to get a cookbook formula at the end...if you develop your own solution based on basic principals you will get a real feel for what you're trying to accomplish.

Honestly, this problem drove me crazy the last 24 hours trying to figure it out...so I broke it down into steps and developed my own solution. See attached spreadsheet.

Note that this spreadsheet uses the average tension (which I have approximated) to calculate the average strain in the cable. This holds true when the deflection/span (or the horizontal reaction to vertical reaction) ratio is small because the tension in the cable only varies by a few pounds. If you use goal seek to find when the pretension is zero, by varying the deflection, you will get the natural sag of the cable. This is also neglecting self-weight of cable.

Please do not use my spreadsheet for your design. I am sharing this with you to get an idea of how you might approach this problem. Also, I wrote it while watching football and drinking beer, so it did not get my undivided attention and I make no guarantee that it is correct. As stated by others already, this problem depends largely on the support conditions of the cables...and an important note by dhengr about the cable and the mechanical stretching, if the wire rope is not pre-stretched it makes the problem more complex.

 

[CANPRO]

By the way...I just checked my spreadsheet with the equation your referenced "L-L1=P(t-te)" and it was within 1 lb.

So I'm fairly confident my spreadsheet works (for this geometry and loading anyway). I'm also more confident now that you had a negative answer before because you did not require a pretension to maintain your deflection limit.