# Cylindrical pipe holding up a screen (real world project)

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P: 2,907
 Quote by rollingstein @Q_Goest What's the difference between your two ideas? I like them, but they somehow seemed very similar to me. Is it just the difference between elastic and permanent strain?
Thanks. It's kinda hard to explain without pictures. But with the first one, you don't allow the bar to rotate and instead you have the screen rotate around the bar. The bar would be bent prior to installation and then when under load it would straighten out, kinda like a flatbed truck.

For the other idea, you have a straight bar that you allow to rotate but you tweak the ends slightly to help eliminate the sag. So you actually put a moment on the ends of the bar which helps to cancel the moment created in the bar by the linearly distributed load. The stresses would never exceed yield so the bar would be straight if you removed it from the installation and laid it on the ground for example.
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P: 330
 Quote by Q_Goest Thanks. It's kinda hard to explain without pictures. But with the first one, you don't allow the bar to rotate and instead you have the screen rotate around the bar. The bar would be bent prior to installation and then when under load it would straighten out, kinda like a flatbed truck.
If the inner bar won't rotate it may be more cost-efficient to go for an I beam / box or some similar non symmetric section?

You only need a high MI to resist deflection along the direction gravity acts. A cylinder might be wastefully exuberant.
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P: 7,177
 Quote by rollingstein A cylinder might be wastefully exuberant.
If you need a cylinder to roll the screen around, you might as well use it to provide stiffness as well.

The bottom line is that I for a thin cylinder is approximately ##\pi r^3 t## but the mass is proportional to ##rt##. Also the deflection is proportional to w/I.

So ignoring the weight of the screen compared with the cylinder, changing the thickness will not reduce the deflection (w/I stays the same) but the deflection is proportional to ##1/r^2##.

So what you need is a large radius cylinder.

If you try to make an internal non-rotating support, it will need to have a significant depth to get the stiffness, so you will still need a big cylinder radius to fit around it.

If you could split the screen into two parts with the two rollers at a shallow angle, so the two screens overlap when they are unrolled, you could have a central support, and halve the length of each roller, which would make a big difference.
P: 8
 Quote by Q_Goest For the other idea, you have a straight bar that you allow to rotate but you tweak the ends slightly to help eliminate the sag. So you actually put a moment on the ends of the bar which helps to cancel the moment created in the bar by the linearly distributed load. The stresses would never exceed yield so the bar would be straight if you removed it from the installation and laid it on the ground for example.
This idea is great. By deflecting the ends with a special bracket, it will put an opposite deflection in the beam. When the beam has a load it will straighten out. Only problem is the beam will bow when the screen is unrolled, so better to make it as stiff as possible.

Going through the numbers again for a simple supported beam:

Pipe/beam:
OD: 2in
ID: 1.87in
Length between simple supports: 258in
Total weight/load: approx 80lbs (includes all materials) or 0.2899lb/in
L = Length (258in between supports)
E = Young's modulus for steel material (30,000,000 psi)
I = pi/64 * (OD^4 - ID^4); I = 0.18514in^4

Deflection: d = w * OD^4 / (384 * E * I)
d = 0.2899lb/in * (2^4) / (384 * 30,000,000 * 0.18514in^4)
d = an incredibly small number!

making a mistake somewhere....ugh
so deflection has nothing to do with Stress or Moment?
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 Quote by massta This idea is great. By deflecting the ends with a special bracket, it will put an opposite deflection in the beam. When the beam has a load it will straighten out. Only problem is the beam will bow when the screen is unrolled, so better to make it as stiff as possible. Going through the numbers again for a simple supported beam: Pipe/beam: OD: 2in ID: 1.87in Length between simple supports: 258in Total weight/load: approx 80lbs (includes all materials) or 0.2899lb/in L = Length (258in between supports) E = Young's modulus for steel material (30,000,000 psi) I = pi/64 * (OD^4 - ID^4); I = 0.18514in^4 Deflection: d = w * OD^4 / (384 * E * I) d = 0.2899lb/in * (2^4) / (384 * 30,000,000 * 0.18514in^4) d = an incredibly small number! making a mistake somewhere....ugh so deflection has nothing to do with Stress or Moment?
Sorry, my bad. I had Do instead of L in the deflection equation. I've edited the post. You should get a deflection of 0.8967"
 P: 8 Pipe/beam: OD: 2in ID: 1.87in Length between simple supports: 258in Total weight/load: approx 80lbs (includes all materials) or 0.2899lb/in L = Length (258in between supports) E = Young's modulus for steel material (30,000,000 psi) I = pi/64 * (OD^4 - ID^4); I = 0.18514in^4 Deflection: d = w * L^4 / (384 * E * I) d = 0.2899lb/in * (258in^4) / (384 * 30,000,000psi * 0.18514in^4) d = 94.9 in How did you get 0.8967? hmmm... I was using this eq. in my original math: D = (F x L^3) / (3 x E x I)
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P: 2,907
 Quote by massta Deflection: d = w * L^4 / (384 * E * I) d = 0.2899lb/in * (258in^4) / (384 * 30,000,000psi * 0.18514in^4)
Check again. You should get 0.6022. I accidentally input 285" in length and got .8967".

The Superbowl is being distracting! lol

Oh... and when deflecting the ends of the beam you would maintain a moment on the ends of the beam so that as it rotates it remains relatively straight. You might imagine a couple of rolling element bearings on the end of the beam so that they tend to point the beam upwards toward the center. Those bearings would simply maintain a moment on the end of the beam tending to flatten the curvature. The beam wouldn't change shape as it rotates so the sag in the beam wouldn't change as it turns. Think about it...