Jokingly a friend and I thought of a way to suspend a tarp-like material over my porch, very similar to they way a trampoline is sprung. However, this fabric is a rectangle and fits over arches.

My question is: I know the weight of the fabric and gravity therefore I kkq the load. I can account for snow accumulation in the winter and add than as part of the force. I don't know how to model the force.

I want a spring on each end with the fabric in the middle. I will then add the spring in parallel.

I have roughly 9 feet of fabric from the leftmost arch to the spring and 20 feet between the arches and 9 feet on the rightmost between the arch and springs. I have not calculated K (spring constant) because I'm unsure of the force.

I modeled the force in the X direction as that is the direction Gravity wants to pull the fabric but then, how do I account for the spring pulling the fabric tight?

To summarize, how do I account for the spring pulling the fabric tight...what is its force components?


It depends on how much you are willing to let the fabric sag.

The horizontal tension in the fabric will be approximately constant from one end to the other. When the fabric is horizontal it provides no lifting force as the tension can only pull in the direction the fabric is pointing. If the height of the fabric is modeled as $h=f(x)$ then the vertical tension, $T_v=\frac{dh}{dx}T_h$ , the horizontal tension.

By symmetry we know that in the middle of the two arches the fabric must be horizontal. We can call this the origin where $x=h=0$. We know by symmetry and force balancing that the vertical tension $x$ away from the center must lift the fabric and snow between the origin and $x$. Thus if we model the weight of the fabric and snow as a pressure $P$ we have $T_v=P\,x$

$$T_v=\frac{dh}{dx}T_h=P\,x$$ $$\frac{dh}{dx}=\frac{P}{T_h}x$$ $$h=\frac{P}{2\,T_h}x^2+C$$ plugging in for our origin gives $C=0$ so now we want to know the height at the arch (this will let us know how far the center has sagged)


And finally solving for horizontal tension


So plugging in some reasonable sounding numbers gives:

$$T_h=\frac{250 Pa\,(10ft)^2}{8\,(0.5ft)}=6250\,Pa\,ft=131 \frac{lbf}{ft}$$

So if you placed you springs every foot they would need to be able to handle 131 pounds of force to get 6 inches of sag at the center if you had about 1 foot of fluffy snow. If you double the number of springs you half the force. It would have to be some pretty strong fabric to handle this kind of tension. If you'd like to actually attempt this I would recommend getting fabric that has rated tensile strength and ensure that it exceeds your requirements by a factor of 5 or more (called a safety factor typically between 5-10). There are many effects not taken into account here (wind, oscillation, etc.) Also the attachment mechanism at the ends would likely be the failure point, so I would have an engineer take a look at your design before building anything.

Oh right springs. You'd want to make sure that when the snow all melts that your springs can pull the fabric taught again. So they must be able to pull up the slack when the fabric is straight across. This slack can be approximated with triangles and doubling it for a rough estimate:

$$D= 4(\sqrt{10^2+0.5^2}-10)2 = 0.1 ft = 1.2 in$$

The slack would of course be take up by both sides so you would need your springs to handle about half that, but if you want the fabric to be tight handling an inch on each side would seem reasonable. So if you were going with the spring every 6 inches and thus $66\,lbf$ when loaded with snow you would want a spring constant of no more than $66 \frac{lbf}{in}$ so that it would be stretched at least an inch when loaded.

A final note: the force of all of those springs would bee enough to tear down most walls, so the structure that you're attaching those springs to would have to be very sturdy. I would really recommend having an engineer/contractor take a serious look at this if you want to go down this route.


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