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Cylindrical pipe holding up a screen (real world project) 
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#1
Jan3113, 10:25 AM

P: 8

This is a real world project. I'm building a giant rollup window shade to be used for a special effects green screen.
The first real model attempt (without math) failed. See the two pictures of the tube structure being held up by chairs. The only axle was about 3ft of 1.5" pipe at each end. The rest is 8" diameter concrete tubes. We double walled the tubes by slicing the length of one and using it as a coupler inside to hold two tubes end to end. We made about twelve 3/4" wooden disks to help support the structure and secure to the axle. Deflection in the middle was too great. Second attempt in next post. 


#2
Jan3113, 10:36 AM

P: 8

Second attempt will be similar but with an axle pipe running the length of the tube.
I'm nervous this will deflect also. Here are my drawings of the stages to completed. Basic drawing: This might be the easiest place to start. The screen only weights 23lb total and will be even across the entire beam. Coupler drawing: Because I can't order large lengths of pipe, sections will need to be welded. This adds weight and weak points. I have some samples of pipe here. Two ends can be welded, then a coupler pipe slid over and welded also. Outer shell drawing: Because the screen needs to be rolled up and down, we need an 8" diameter concrete tube which will take 910 revolutions to wrap it up. An option here would be to use even larger diameter tubes (note: wall thickness is the same for all diameter tubes). Math to follow on next post. 


#3
Jan3113, 11:24 AM

P: 8

Moment is force times distance.
My thinking is because it's supported at both ends, I can run math for half of the beam. But what is my force? Weight: Support to middle of center pipe: 10.75' x 1.343lb = 13.43lb One coupler: 1' x 2.73lb = 2.73lb Outer shell: 10' x 1.2lb = 12.0lb 1/2 of green screen weight: 11.5lb total half weight of beam: 39.66lb Moment (M) = Force x Distance M = 39.66lb x 129in, == 5116.14inlb Let me consider the stress of the basic single pipe/beam model: Stress (S) = M x C/I C = outside radius of pipe (1") I = moment of Inertia of pipe I = pi x OD^4/64 = 3.14159 x 2^4/64 == 0.7854 in^4 and S = 6514.06psi What will be the deflection? D = (F x L^3) / (3 x E x I) Where: D = Deflection at the center of beam F = Force at center of beam (39.66lb) L = Length to center (129in) E = Young's modulus for steel material (30,000,000 psi) I = moment of Inertia (0.7854 in^4) D = (39.66 x 129^3) / (3 x 30,000,000 x 0.7854) == 1.205 inches This basic model, if I'm correct, is bending quite a bit. There is probably some other weight I didn't factor in so this will be a minimum deflection. Other things to consider: Strength of welded coupler. Could this increase deflection or decrease it? Strength of the concrete tubes (with wooden disks, note: there will be concrete tube couplers here as well, just not the entire length like in the first attempt). My thinking is the concrete tubes will add to the deflection making this attempt a failure on paper. Any ideas are very welcome. There is another model for rolling up the this green screen. Remember, the screen is very light. The problem is that its very wide: 20ft and needs to be rolled up somehow. Thanks, ~Shawn 


#4
Feb113, 12:22 AM

P: 54

Cylindrical pipe holding up a screen (real world project)
You could use some thick rigid foam panels (the kind used for insulation) to stiffen the tube. 1 sheet of it should do the job nicely, and they don't cost that much. A Stanley Sureform will be useful for shaping the foam so it fits snugly inside the sonotube.
You can make gussets for the joints out of salvaged door skin mahogany (try craigslist for a door, or go to a Habitat For Humanity ReStore if you have one nearby) and glue them on with construction glue. 


#5
Feb113, 08:53 AM

P: 8

I just found this company and it's 4" aluminum pipe with 0.5" max deflection:
http://www.phototechinc.com/102612_p..._pwrroller.htm costs $400 plus another $175 shipping, ouch Wish I thought of the rigid foam idea, just don't want to waste any money building another tube and have it fail when I could of just bought one. Thanks for all the help. ~Shawn 


#6
Feb213, 01:51 AM

P: 128

You want to get the structure radially outwards as far as possible to increase the I. List the constraints of your project, because steel is certainly cheaper than Alum. 


#7
Feb213, 02:35 AM

PF Gold
P: 330

Shouldn't you be using this equation, but with full beam weight? That'd give you a computed deflection less than 0.5 inches. What's your design limit? 


#8
Feb213, 02:46 AM

PF Gold
P: 330




#9
Feb213, 02:50 AM

PF Gold
P: 330

Besides that's the worst case scenario: A concentrated center load. A uniformly loaded beam will deflect a lot less I think. Probably ~60% of the concentrated center load case?
Also, wouldn't supporting it on chairs exaggarate the deflection than a real mount? How are you going to mount the ends? A fixed support end condition should give perhaps ~30% of the beam center deflection as a simply supported end point. I'm not sure which one is a realistic assumption for your mounts but I suspect the chairmount misleads you. PS. Apologies if any of this is wrong; I'm no structural expert. I'm glad to be corrected. 


#10
Feb213, 03:33 AM

PF Gold
P: 330

Worst case, could you add a set of small support rollers at the center for support? The screen can squeeze between them and the main roller. Not sure if you know what I mean?



#11
Feb213, 01:59 PM

P: 1,418

I think that limits the deflection of the 2 half lengths to 1/16 of that of the longer beam if this technique is viable. 


#12
Feb213, 05:10 PM

P: 8

@rollingstein, you are correct. I made the mistake of using a solid beam.
For hollow tube: I = pi * (OD4  ID4) / 64 Where: pi = 3.14159 OD = the outside diameter (2 in) ID = the inside diameter (1.87 in) so I = 0.185142 in^4 plug in new I for Deflection: D = (F * L^3) / (3 * E * I) D = 5.11 in !!! answer: steel is too flexible @strawcat, I like your idea of using foam to support it. You've got me thinking. I wondering how much flex 2"x8"x20' flex. This could work if I cut and placed two 2"x3"x20" strips along the edge to make it a double Ibeam inside an 8 inch tube (like a cross). But foam doesn't come in super long pieces and would need to be connected some how. @jupiter6, some people have suggested PVC but it's very heavy and I suspect flex, but will pick them up again to check at the hardware store tomorrow. This adds to the force. @rollingstein and 256bits, a center support has been thought out. It would only work if it was spring loaded and moved up and down at the diameter changed when the screen unwound and wound. Thanks for all your help! The best answer I think is using foam. If I could get larger concrete tubes, say 12" diamter, then I could fit foam triangles inside to support, much like a bridge. 45 degree triangular support is the best strength to weight ratio. hmmm Seems like a ton of work when I can purchase a beam out of aluminum and not have to make anything, haha. Feel free to keep this going. 


#13
Feb213, 10:27 PM

PF Gold
P: 330

5 inches sounds like a lot. 


#14
Feb313, 01:38 AM

PF Gold
P: 330

Did some number crunching:
Uniform Load of 0.3 lb/in and fixed supports gives max deflection 0.6 inch Even if you consider all weight (80 lb) concentrated at beam centre I get a deflection of 1.2 inches with fixed supports. Main problem is roller supports; then deflections jump to the 35 inch range. 


#15
Feb313, 08:21 AM

P: 8

I'd like to see your formula.
The only support is at then end, I understand the force is even throughout the beam but it only has one support so the entire force (or 1/2 because there are two supports) are enacted here. That said, all of the weight isn't at the middle, it's spread out. So this is where I could use some help. ~Shawn 


#16
Feb313, 10:15 AM

Sci Advisor
HW Helper
PF Gold
P: 2,903

Hi Shawn. Welcome to the board,
For a tubular structure, moment of inertia I = [itex]\pi[/itex]/64 (D_{o}^{4}D_{i}^{4}) For a beam, uniformly loaded with SIMPLY SUPPORTED ends, the equations for stress and maximum deflection are: The maximum moment is m = w L^{2} / 2 where w = linearly distributed load in units of force per unit length. Make sure to add ALL contributions to weight including the pipe/tube and the stuff it's supporting. Stress = m D_{o}/(2 I) Deflection = 5 w L^{4} / (384 E I) where E = modulus of elasticity of the pipe/tube material. For a beam, uniformly loaded with FIXED ends, the equations for stress and maximum deflection are: The maximum moment is m = w L^{2} / 12 Stress = m D_{o}/(2 I) Deflection = w L^{4} / (384 E I) Obviously, fixing the ends so they can’t deflect will reduce the maximum deflection at the center of the span by 80%, so that's a lot better than simply supported ends. I'd suggest making up a spreadsheet to see how changing various inputs changes the output. If you need to reduce the deflection of the span further and make it essentially flat, you could use one of the equations above and make a beam that’s bent to that curve, fix the ends as per the equation and then when it’s loaded, it will flatten out.* Of course, if this bent bar were to rotate, it would be horrible, so you would want to keep the bar steady and have some way of having the screen rotate on the bar such that the bar doesn’t need to rotate. One last option would be to take a pipe/tube and deflect the ends slightly so that you put a moment on the ends of the bar so as to help reduce the sag in the middle. The bar could then rotate if you wish, unlike the other option above. In other words, imagine holding a thin plastic bar horizontally out in front of you and watching it sag in the middle, then twist your hands so the sag comes out of it. You could do the same here and allow the pipe/tube to rotate on bearings but the bearings would be canted slightly so the sag is reduced. Not sure if that would completely eliminate the deflection, I'd have to think about the equations, but that's another possibility. Additional information here: http://www.engineersedge.com/beam_calc_menu.shtml *For example, you may have seen 18 wheelers driving down the road pulling a flatbed with no load and noticed they often have a bow to them. The flatbed trailer will bend under load so that it ends up nearly flat. 


#17
Feb313, 11:07 AM

PF Gold
P: 330

Under load shouldn't the bent bar be essentially flat be design? i.e. it is only bent unloaded? 


#18
Feb313, 11:09 AM

PF Gold
P: 330

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? 


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