Tension force on banners under wind load. 1

[Mina Bekhit]

Hi

I am new to this forum.

I am assigned to design a structural support for a mesh vinyl banner with dimensions of 40'x120'. The banner will be one way span in the 40 ft direction. This banner will be located in New Jersey where the wind speed can reach 99 mph. Based on that the banner will be subject to 45 psf wind load.

I am trying to calculate the tension force in the banner under wind load and transfer this load to my structure. I am using this formula
T = WL2/8xS
W = 45 psf wind load. L = 40 feet. S = banner sag (missing information)
Based on banner sag of 3", the Tension force will be 36kip/ft
Based on banner sag of 24", the Tension force will be 4.5kip/ft
I am also missing the banner modules of elasticity and strain. I tried to obtain this information from the banner company with no luck of getting it.

Can any one help on the above calculation and what is a reasonable modules of elasticity and strain to be used? Is there any other way of calculating the tension in the banner? What is a reasonable sag range for a 40 ft banner span?

Thanks in advance.

 

[JLNJ]

Are you looking for dead load sag or the horizontal deflection in the banner under the 99 mph wind load?

I imagine that the dead load effect will be minimal compared to the force on the banner acting as a sail. You will need something crazy stout to anchor it to.

(Is that 40 inches by 120 feet or is it actually 40 feet tall?)

 

[Mina Bekhit]

I am looking for the deflection of the banner under 99 mph wind speed. I will use this deflection to obtain the tension in the banner that will transfer to my structure.

The banner is 40 ft x 120 ft. One way span in the 40 ft direction.

 

[dhengr]

Mina Bekhit:
You come to E-Tips hoping to find some magician who can answer your question, maybe, maybe not. Isn’t your “T” the horiz. (one of the, vert. in your case?) reaction components at the support, not actually the banner tension? Don’t give up on the banner company yet, they can’t be making these things without this kind of info., their banners have to be supported by someone, something. And, someone there has this info. or can get it, or can get you in touch with their material supplier who knows their own product. That assumes they want to sell you a banner. Don’t do this kind of info. gathering by e-mail, and don’t take the word of the first person who picks up the phone, ask to speak with an engineer, or someone (the co. pres., if needs be) who has this knowledge, or will work a bit to get it. The e-mail or phone answering monkeys, these days, are meant to get rid of 98% of the pests who have a question, without spending much on customer service. They should be able to give you some general guidance on how these should be supported, without assuming your EOR responsibility. What are reasonable sags on a 40’ span, what is initial tension on the banner material for proper support, support grommet spacing and unit strength, etc. etc? Don’t these banners usually have some sort of pressure relief patches sewn into them, which break loose at some point to reduce the total loading?

 

[rb1957]

1) I think dh has an excellent practical suggestion of "blow-out" panels to prevent overloading the structure … yes, not what you asked for but really good idea !
Also good practical suggestion about talking to the banner designer … they should know what their banner is capable of, and how much load goes into the reaction points.

2) Are you supporting the 40' side, or the 120' side ?

3) It's easy to calc the normal reaction (due to the force of the wind). The difficult reaction is the "in-plane" component, reacting the tension induced by the deflection of the banner. I don't recognise your equation … is it L^2 or 2*L ? I would look up catenary loads. Yes, I know this is not a catenary, but it is a close approximation with your distributed load and large deflection.

another day in paradise, or is paradise one day closer ?

 

[Mina Bekhit]

Yes, I can try to go to the manufacture. However, it seems there no engineers that can answer questions regarding the material they are using in this type of industry.

I am providing supports along the 120 ft sides. The banner will act as a one way slab in the 40 ft direction.

I meant L^2 not Lx2.

 

[BAretired]

For a banner that size, it is not practical to span with banner material alone. Provide a structure to support the banner. This is a small banner with structure behind:

This is a larger banner; still nowhere nearly as big as 40'x120'.

For a 40' high banner, a structure is required to resist the wind load.

BA

 

[RHTPE]

You stated "mesh vinyl banner" - what is its porosity? Does it behave as a solid surface or does some wind go through the material? A 40' wide X 120' H banner, even when printed, would be expected to have porosity greater than expected. Even printing on a mesh with 1" square openings is going to appear (visually) as a solid surface.

ASCE 7 Wind Load Section has factors for fencing or perforated material. From that you can calculate wind load forces. Then you can go on to cable analysis to determine tension in the material and reactions at the supported edge.

You will have to get material properties from the manufacturer - while you're there, you should do as dhengr suggests.



Ralph
Structures Consulting
Northeast USA

 

[Mina Bekhit]

The manufacture vinyl mesh data sheet specify a 25% of air will pass through the mesh.

 

[RHTPE]

To use ASCE 7 you'll likely need fiber diameter and the clear opening through the fabric.


Ralph
Structures Consulting
Northeast USA

 

[BAretired]

I agree that porosity will affect the wind load on the banner, but a forty foot high wall is still a formidable structure to design, even for reduced wind pressure.

Is the banner to be attached to an existing structure capable of resisting wind reactions or is it a freestanding banner? If freestanding, the 120' long top support will need to be supported by columns which, at that height would likely need diagonal bracing to resist lateral pressure.

If the only concern is the 40' span, steel cables attached to the banner could be used to support it against wind pressure. The force in each steel cable would be F = wl[sup]2[/sup]/8s where w is the load per foot and s is the sag.

BA

 

[rb1957]

does a (vinyl) banner have to react 99mph winds ?

This is standing 120' high ?

Does your 45psf number include the porosity of the fabric (25%, ie 75% of the wind flows thru the banner) ?

What is "T" in your equation ? The units are psf*ft^2/ft = lb/ft … Tension per ft width ?

Where does your equation come from ? Have you looked into catenary equations ?

another day in paradise, or is paradise one day closer ?

 

[BAretired]

It's not a catenary.

BA

 

[bob33]

I am assuming the banner is 120' long and 40' high. One option is weights hung from the bottom of the banner. Two ties at each location could be used to go on both sides of a horizontal pipe section with the weight hanging below. The weights and pipe could take the expected wind load and the pipe alone the remainder. This would ensure a reasonable billowing out at ultimate load. I think you are using the correct equation for uniform load. The elastic modulus for PVC varies considerably and plastics creep under load. The manufacturers must have this information.

 

[BridgeSmith]

Maybe I'm not understanding why you say it's not a catenary span, BA. It's a very wide 'cable', but the analysis to get from the pressure applied to tension along the span, I think, should be the same.

 

[rb1957]

BA, I know it's not a catenary (as I noted in my previous post) … but I think catenary equations solve this uniformly loaded "cable" to reasonably determine the in-plane reactions.

another day in paradise, or is paradise one day closer ?

 

[BAretired]

@HotRod10,
A catenary exists when the dead weight of a cable acts in the same direction as the applied load, namely vertically down. In this case, the weight of vinyl acts down while the applied load, wind acts laterally. The OP has not indicated whether the 40' span is vertical or horizontal but in either case, it is not a catenary.

For small sags, the parabolic assumption is close enough for practical purposes so tension = wl[sup]2[/sup]/8s.


BA

 

[BAretired]

BA, I know it's not a catenary (as I noted in my previous post) … but I think catenary equations solve this uniformly loaded "cable" to reasonably determine the in-plane reactions.

I guess I missed where you noted it's not a catenary. So why use catenary equations?

BA

 

[rb1957]

I thought they solved a uniformly distributed load on a cable. I know a "true" catenary is loaded by self-weight, but I think this uniform airload is near enough. Probably no worse a solution to assuming a parabolic shape.

Anyways, you've given me the source of the OP's equation, for tension per ft width, depending on s. But the first question is how strong is the banner ? and second question (as asked for) E for the banner (to get stress for a known sag, or strain).

Surprising lack of support from banner manufacturer ?

another day in paradise, or is paradise one day closer ?

 

[Mina Bekhit]

Thank you everyone for your answers. Here is a clarification of all the asked questions:

The banner will be attached to an existing garage structure using outriggers from the garage columns. The top of the banner is at level 60’ above the ground and the bottom is at 20’ above the ground. The 40 ft span of the banner is running vertically.

The “T” in my calculation is the tension force in the banner when subjected to a wind load. Assuming the banner will act as a cable under the cable own weight. Yes, the “T” is tension per feet long of the 120 feet long banner. I investigated catenary action and believe that the equation I am using (WL^2/8xS) is close enough to determine the amount of tension in the banner.

Taking into consideration the prosperity action, the wind load will be 45 psf x 0.75 = 33.75 psf which is still a huge load to apply in a 40’ banner span and even excessive to be resisted by the structure.

The client wants it to be attached at the top and bottom only without intermediate support, but I am challenging that now because of the extremely high forces I am getting.

I really appreciate everyone's input.