[Question] Calculating the force generated by a tarp hung between two walls

In summary, calculating the force generated by a tarp hung between two walls involves considering the tarp's weight, the angle of the tarp, and the tension created due to external factors like wind. The force can be determined using principles of static equilibrium, where the sum of forces in both horizontal and vertical directions must equal zero. This requires applying trigonometric functions to resolve the forces acting on the tarp and ensuring that all forces are balanced.
  • #1
DaGude
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TL;DR Summary
I'm trying to calculate the horizontal and vertical force that a tarp with downward pressure applies to the two walls it is attached to
Hi.
I have a problem I'm trying to solve.

I have a tarp that is stretched horizontally across two walls. The walls are 5.5m apart and 50m long.
There is a pressure of 10mBar above the tarp pushing it down, making it sag about 0.15m in the center.
I'm trying to calculate the force with which the tarp pulls the walls inward.
(and also the force at which the tarp pulls the walls down. So both the vertical and horizontal components of the force on the walls).

I thought the tarp would behave the same as a cable with a uniform downward force applied to it.
Similar to the cables on suspension bridges. I'm not sure if I can use the same formulas to calculate this though.

the formula I found for the cable is (w*L^2)/(8h) where L is the distance between the anker points. h is the amount the cable sags in the center and w is the force per meter on the cable.
(This assumes the force is acting vertically to the ground which in my example isn't the case since the pressure acts vertically on the tarp which is curved but the curve is minimal so I would assume the error to be negligible)

I also thought about calculating the angle at which the tarp meets the walls (assuming a parabolic arc for the tarp) and applying the force of the pressure to the walls at that angle as if the force was acting only in the center of the tarp

Would either of these methods work for what I'm trying to do or is there something I'm missing?

Thank you for any help in advance
 
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  • #2
Welcome to PF.

DaGude said:
I also thought about calculating the angle at which the tarp meets the walls (assuming a parabolic arc for the tarp) and applying the force of the pressure to the walls at that angle as if the force was acting only in the center of the tarp
The contact angle at the wall is a simple way to solve for the horizontal and vertical forces. You need to find the tension in the tarp, which is the hoop stress due to the differential pressure on a tarp. The tarp can be modelled as a circular arc, with a radius of curvature, that you must measure or estimate.

What is the application?

Flapping in the wind can cause sudden impulse forces to the supports, and damage the tarp fabric. A doubly curved surface, like a saddle, will not flap or flutter in the wind, but each panel needs four corners that are not co-planar, and the fabric should be slightly stretchy.
 
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  • #3
Baluncore said:
Welcome to PF.The contact angle at the wall is a simple way to solve for the horizontal and vertical forces. You need to find the tension in the tarp, which is the hoop stress due to the differential pressure on a tarp. The tarp can be modelled as a circular arc, with a radius of curvature, that you must measure or estimate.

What is the application?

Flapping in the wind can cause sudden impulse forces to the supports, and damage the tarp fabric. A doubly curved surface, like a saddle, will not flap or flutter in the wind, but each panel needs four corners that are not co-planar, and the fabric should be slightly stretchy.
Thank you for the answer. I had falsely assumed the tarp would have a parabolic shape which caused my calculations to be off. the tarp having a circular ark makes a lot more sense.

I need to know the force since I need to know how strong the walls and the tarp have to be. the whole construction will be protected from wind so flapping/fluttering won't be an issue.
 
  • #4
DaGude said:
I have a tarp that is stretched horizontally across two walls.
Is the tarp between, and attached to the two walls, or does it rest on the two walls, being pulled down to something beyond?
 
  • #5
Baluncore said:
Is the tarp between, and attached to the two walls, or does it rest on the two walls, being pulled down to something beyond?
The tarp is attached to the walls and is sagging down between them due to the pressure.
Technically the tarp is stretched over the top of a "box". but the box is significantly longer than it is wide and I only want to know the force for a section in the middle so the shorter walls can be neglected and the system can be treated as if they weren't in place
 
  • #6
DaGude said:
The tarp is attached to the walls and is sagging down between them due to the pressure.
Technically the tarp is stretched over the top of a "box". but the box is significantly longer than it is wide and I only want to know the force for a section in the middle so the shorter walls can be neglected and the system can be treated as if they weren't in place
Do the side walls not contribute significantly to supporting the weight of the tarp?
 
  • #7
Baluncore said:
You need to find the tension in the tarp, which is the hoop stress due to the differential pressure on a tarp.
How would I calculate the hoop stress if I assume the tarp does not have any thickness?
 
  • #9
Chestermiller said:
Do the side walls not contribute significantly to supporting the weight of the tarp?
In the corners, they certainly would but the further you get from them the less they support the tarp and the more of the force has to be absorbed by the main(longer) walls. If the walls will have to sustain this force in the center anyway, it's not a problem to (wrongly)assume they will have to sustain it in the corners as well.
 
  • #10
Baluncore said:
Don't we divide by the wall thickness in this formula? If we assume a wall thickness of 0 then we can't use this formula. or am I misunderstanding something?

Edit: would I just not divide by the thickness and then multiply by the length to get the total Force f for the whole length?
 
  • #11
@DaGude
Do you have a picture or diagram of this? How much precision do you need out of these calculations?
 
  • #12
I THINK that your description can be summarized by this sketch:
Tarp.jpg

If so, the curve is a catenary (search the term). That assumes that the 10 mbar force (or do you mean pressure?) is evenly distributed over the entire tarp. If these assumptions are correct, then the total vertical force on the walls is equal to the total weight of the tarp plus the 10 mbar pressure. If you carefully measure the angle of the tarp relative to a true horizontal, then the horizontal force is easily calculated. The total force is the vector sum of the vertical and horizontal forces.

A diagram showing the angle would help considerably. Plus your calculation of the total vertical force component.
 
  • #13
jrmichler said:
If so, the curve is a catenary (search the term).
It is a catenary if the mass of the tarp only is being supported, but there is also the pressure difference, so the hyperbolic mathematics can get easier.

If the major force is a vertical pressure component, then the curve will be close to a parabola, more like the cables of a suspension bridge. The compromise approximation is part-way between the two, a circular arc, which makes it really simple, like a pipe with internal pressure.
 
  • #14
Juanda said:
Do you have a picture or diagram of this? How much precision do you need out of these calculations?
jrmichler said:
I THINK that your description can be summarized by this sketch:
Tarp.jpg
This is pretty much what it looks like.

The values don't need to be perfectly precise. They serve only as a "minimum requirement" for the strength of the walls/tarp. I did have a structural engineer type this problem into one of his fancy computer tools but I would really like a formula to calculate these forces for different parameters without having to run to the engineer every time I make a change. The vertical force I calculate is pretty close to the values he gave me but the horizontal force is still out a fair bit

I have also made a tool in desmos if you want to take a look at it: https://www.desmos.com/calculator/dqhe68sgrd
 
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  • #15
DaGude said:
The values don't need to be perfectly precise. They serve only as a "minimum requirement" for the strength of the walls/tarp.

I feel there are some extra details that are necessary to define. When ropes are hanged, they make a catenary shape. In those calculations, the deformation (change in length) of the catenary is not taken into account. Here I feel you have an initially flat tarp and then it deforms to the shape you described that sags 0.15 m. I say so because you didn't mention the original length or shape of the tarp.
On the other hand, if you hang a rope between two poles and the rope has the same length as the distance between poles, the solution gets closer and closer to a straight line with infinite tension if you reach the limit.

Since you provided some information about the sag, assuming the tarp deforms or that it sags because it's got some extra length will give the same result for the procedure shown ahead.

I'd then calculate it assuming a circular shape. From the information you provided, there's only 1 circle that fits so the angle it makes with the wall is constrained. It's possible to obtain the value algebraically but I just drew it.
(Number 2000 is meaningless. I added it just so it's not shown in blue)

1696178667036.png


I'd simplify it further to something like this:
1696178976771.png
Where
##P##: Difference in pressure
##S##: Surface area
##W##: Weight
Since the problem is symmetric, you can apply the equilibrium of forces and find the reactions at the supports/walls. If you parameterize the problem (find ##\alpha## in terms of the sag), you should be able to solve it for many input values using Excel without involving the structural engineer as you suggested in #14.

I don't know what you're really working with but I'd typically multiply the input loads with a big number to have a margin of safety.
I think the circular shape approach is OK in this case because:
  1. You don't need accurate numbers. FEM analysis is what I think really would give you a precise answer to what you're asking (assuming the tarp is initially flat and it deforms because of its weight and the pressure difference).
  2. You didn't mention the weight of the tarp and you assume the tarp has 0 thickness so I guess the weight is almost meaningless.
  3. With a uniform difference in pressure and near-zero weight, I'm pretty sure the shape will be circular.
  4. I think the circle will have a flatter angle than the comparable catenary so the horizontal forces will be greater. Since that's the thing worrying you the most because it'd damage the walls, using a model with bigger side loads seems convenient.

By the way, I also tried deriving the shape from applying equilibrium to an infinitesimal chunk of the tarp/rope similar to how it's done with catenaries but I couldn't get anything useful. This is the picture in case anyone knows how to solve such a thing.
1696179691610.png


My plan was to follow the same procedure as with catenaries but add the effect of the perpendicular force. I tried finding the center of curvature for that chunk so that I could work out the angle of ##Pds## but it's too hard. By the way, that assumes the tarp does not deform and the sag comes from the tarp being longer than the space between walls.

Still convinced I'd solve it, I tried discretizing the problem and solving it numerically in the time domain with springs and some dissipative force. I was somewhat successful when only gravity is acting but as soon as I input the effect from pressure, the software crashes on me. I've been trying for quite a while but I can't make it work. Here is a pic of the result with only gravity acting on the rope/tarp with unrelated input values.
1696180145559.png


To sum up, I hope the circular shape approach is enough because I'm out of bullets here.
 
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FAQ: [Question] Calculating the force generated by a tarp hung between two walls

How do I calculate the tension in a tarp hung between two walls?

To calculate the tension in a tarp hung between two walls, you need to consider the weight of the tarp, any additional load it carries, and the angle at which it is hung. Use the formula T = (W / (2 * sin(θ))), where T is the tension, W is the weight of the tarp and any additional load, and θ is the angle between the tarp and the horizontal.

What factors affect the force exerted by a tarp hung between two walls?

The force exerted by a tarp hung between two walls is affected by the weight of the tarp, any additional load it carries, the distance between the walls, and the angle at which the tarp is hung. Environmental factors like wind can also influence the force.

How do I account for wind load when calculating the force on a tarp?

To account for wind load, you need to consider the wind pressure and the surface area of the tarp. The wind load can be calculated using the formula F_wind = 0.5 * ρ * V^2 * A * Cd, where ρ is the air density, V is the wind velocity, A is the surface area of the tarp, and Cd is the drag coefficient. Add this force to the weight of the tarp and any additional load to get the total force.

What is the impact of the angle of the tarp on the tension force?

The angle of the tarp significantly impacts the tension force. As the angle decreases (the tarp becomes more horizontal), the tension increases. Conversely, as the angle increases (the tarp becomes more vertical), the tension decreases. This is because the vertical component of the force increases with a steeper angle, reducing the horizontal tension.

How do I ensure the tarp does not tear under the calculated force?

To ensure the tarp does not tear, you need to choose a tarp material with a tensile strength greater than the calculated tension force. Additionally, reinforce the edges and corners of the tarp, use strong attachment points, and distribute the load evenly to prevent localized stress points.

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