Weight bearing strength of 5 ft. of black iron pipe

y velvet drapes spanning a window greater than 12 feet long without the nee d for a center support? I?ll anchor ends using flanges (nipples and elbows) screwed into solid wooden header above window. Seems like it shoul d be more than strong enough for such a girly thing.
Strong enough, perhaps, but for sag-resistance you want diameter, but not w all thickness (like a water/air pipe intended for machine-threading the ends). Fence rail ite ms

have less pounds per foot of metal, but are cold-worked steel and should pe rform acceptably; they join end-to-end (and with a bit of Liquid Nails applied, t he joints will be rigid enough for most purposes).
Designs of curtain rails of formed sheet metal will allow a 12 foot span, w ith a brace at 6 feet, that doesn't interfere with the drape support elements ( sliding plastic in a steel channel). A central brace in these designs is usually not visi ble with the curtain closed.
Reply to
whit3rd
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That's Euler-Bernoulli beam - very clear and exact. You can have the load it can take and the deflection at that load to great accuracy. If you set-up the load on the finished article, you'd find the deflection matched to within a millimetre or something like that. The end supports cannot be anything like rigid enough against rotation to benefit the load capacity and stiffness against deflection. You've got a "simple supported beam" ("double-supported beam"). A fair and reasonable conservative assumption would be to put the entire weight of the curtain in the middle of the "curtain rail" when doing the Euler-Bernoulli calculation for what tube to specify. So that truly is the "simply supported centrally-loaded beam" case.
I do a lot of these calculations. eg.
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Rich Smith
Reply to
Richard Smith
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That's Euler-Bernoulli beam - very clear and exact. You can have the load it can take and the deflection at that load to great accuracy. If you set-up the load on the finished article, you'd find the deflection matched to within a millimetre or something like that. The end supports cannot be anything like rigid enough against rotation to benefit the load capacity and stiffness against deflection. You've got a "simple supported beam" ("double-supported beam"). A fair and reasonable conservative assumption would be to put the entire weight of the curtain in the middle of the "curtain rail" when doing the Euler-Bernoulli calculation for what tube to specify. So that truly is the "simply supported centrally-loaded beam" case.
I do a lot of these calculations. eg.
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Rich Smith --------------------------
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Water pipe attached to a vertical wall with elbows, close nipples and floor flanges is Simply Supported because the threads can rotate within the flange.
Pipe and fence tubing longer than 10' is hard to find, and transport. A center support makes this problem MUCH simpler.
Reply to
Jim Wilkins
Interesting liittle project - O like your reasoning and explanation - particularly your "tap stick" I've run into a lot of your "welders" who would make the thing so heavy it could hardly support itself!!! (or they would fasten it to the rig instead of the barge and it would be so heavy it would tip the rig before it would bend - - -
Reply to
Clare Snyder
Speaking of welders, how about making a truss? Those can be quite lightweight, have a decorative design, and still be very rigid.
Reply to
Robert Nichols
I'll take that brief comment as warm praise coming from your background of experience. I wanted "inherent safety" - that there would be "clue sticks" along any way the equipment could be abused.
Reply to
Richard Smith
[digression from OP's purpose]
If the strength had been any less / if it hadn't felt right when assembled, I'd have probably done that. Form a "tent-shape" triangular truss part, rising diagonally from the deck to the top of the first "stanchion" then sloping down diagonally to about the middle of the deck? Ultimately - the strength was enough, while the secondary steelwork with the "clue-sticks" incorporated drew-out the safety margin desired.
Reply to
Richard Smith
I was thinking more along the lines of a double rail, about 3 inches apart, with diagonal members welded in. I made something like that out of square tubing a while back, and it turned out about 10X stronger than it needed to be.
Reply to
Robert Nichols
If the strength had been any less / if it hadn't felt right when assembled, I'd have probably done that. Form a "tent-shape" triangular truss part, rising diagonally from the deck to the top of the first "stanchion" then sloping down diagonally to about the middle of the deck? Ultimately - the strength was enough, while the secondary steelwork with the "clue-sticks" incorporated drew-out the safety margin desired.
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The most effective place to add support to a cantilever structure is about 1/4 of the way in from the end, as long as its beams are equally strong in tension and compression.
Reply to
Jim Wilkins
Good point. As a cantilever beam, highest moment is where structural meets the barge. So add a "doubler" in that region would "do the trick".
Reply to
Richard Smith
Answer for the Original Poster...
I took the case from this - you will see what I've taken as your situation fromteh numbers I settle on.
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I note Ray Hayes' explanation, where he looks to have used the same methodology Ray Hayes posted on February 23, 2018, 2:31 pm # replying to mjacobsen925, Ray Hayes wrote: Your reply is # foolish. This appears to be a valid question about using a pipe # section to support a hoist - an overhead lifting design that # requires calculation. 1.5" black iron pipe is the support beam # suggested for available electric garage hoists.
# For any who care, stress = M*c/I = M/s. Bending moment = M = WL/2, s # = 0.326 in^3 for 1.5" sch 40 pipe. Assume yield strength0,000 # psi. L`". stress = M/s = W*L/(2*s). If Safety Factor = 2.0 (low for # a lifting operation - F.S> should be 6 for overhead lifting): # Yield/F.S = 30000 psi/2.0 = 15000 psi = W*60/(2*0.326); W = 163 # lbs. Yield Strength (30,000 psi) is the load where the pipe will # bend without springing back to straight. Original request was for a # 1" deflection at the center. This is way overloaded - typical beam # limit might be L/360 = 0.167". But just for yucks: max deflection = # y = W*L^3/(48*E*I); E)e6, I = 0.310 in^4 (1.5" sch 40 pipe). y/W = # 60^3/(48*29e6&*0.310) = 0.0005, or W/y = 2000, or W = 2000*y. So to # get a 1.0" deflection on a 60" pipe (if it did not yield) would be W # 00 * 1.0 = 2000 lbs. *BUT the stress with that 2000# load would be # M/s = W*L/(2*s) 2000*60/(2*.326) = 184,000 psi - 6 times the yield # strength (meaning the pipe would bend to failure). *For reference, # limiting deflection to L/360 would allow a load of: W = 2000*y = # 2000*(60/360) = 333 lbs. This corresponds to a F.S. = about 1.0.
Dimensions for 1~1/2" Sched 40 tube Nom OD ID w-thk 1.5 1.900 1.61 0.15
(* 1.9 25.4) ;; 48.26 (* 0.15 25.4) ;; 3.8099999999999996
If I'm not mistaken, that's the dimensions of scaffold tube as we know it here in the UK.
I can tell you - a 6m length (- (/ 6e3 25.4) (* 12 19)) ;; 8.220472440944889 19ft 8in length !!! of scaffold tube supported on its ends will just take my weight in the middle - about 87kg (/ 87 0.4536) ;; 191.7989417989418 192lb
I'm going to use a yield strength of 235MPa There is every likelihood the yield is higher than that. So the value I'll calculate is the minimum load-bearing possible.
The Young's modulus at 2.1e11Pa is almost independent of steel strength, so the deflection prediction is invariant of steel grade.
Using my functions ;; moment cap (* 235e6 (beam-sect-mod-z-d ;; args I, d (ma2nd-annulus-dia-wthk-cx 48.3e-3 3.8e-3) ; 1.3245821667837055e-07 ; m^4 48.3e-3) ; 5.484812284818656e-06 ; m^3 ) ; 1288.9308869323843 ; N.m
So moment capacity is 1289 Newton-metres
(* 60 25.4 1e-3) ;; 1.524 ;; m length
(/ (simple-support-dblbeam-loadcap ;; M_cap, l 1288.9308869323843 1.524) ;; 3383.0207006099326 ;; N (Newtons) 9.81) ; 344.8543017951001 ; kg-f
(/ 345 0.4536) ;; 760.5820105820105 So your curtain rail will bear 760lb at the middle That's more than enough for a kid swinging on it.
Deflection - which is your question...
(dblsupport-centralload-beam-deflect ;; F(central), l, E, I (* 344.8543017951001 9.8) ;; 3379.5721575919815 ;; N 1.524 ;; m length of curtain rail 2e11 ;; Elastic modulus of steel (Pa) (ma2nd-annulus-dia-wthk-cx 48.3e-3 3.8e-3) ;; 1.3245821667837055e-07 ;; m^4 ma-2nd ) ;; 0.00940733235828569 ;; m of deflection at onset of deformation
(/ (* 0.00940733235828569 1e3) ;; 9.40733235828569 ;; mm deflection 25.4) ;; 0.370367415680539
(*
0.370367415680539 8) ;; 2.962939325444312 ;; about 3/8th-inch
Even at maximum loading of 345kg / 760lb the middle of the curtain rail will be only 9.4mm / 3/8th"
I hope that's the answer to the question.
Reply to
Richard Smith
With a 12ft length of "scaffold pole" (as I'd perceive it) curtain-rail...
Maximum central load before permanent bend = 143.7kg-f = 317lb-f
At that load, at centre 54 mm deflection = 2~1/8th inch deflection
If two children were climbing the curtains at the middle, you'd have higher concerns than whether the bottom of the curtains is still exactly level with the floor for the duration of their adventure...
Rich Smith
Reply to
Richard Smith

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