Weight bearing strength of 5 ft. of black iron pipe

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.

Rich 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.

Rich Smith

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Pipe and fence tubing longer than 10' is hard to find, and transport. A center support makes this problem MUCH simpler.

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 - - -

Speaking of welders, how about making a truss? Those can be quite lightweight, have a decorative design, and still be very rigid.

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.

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.

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.

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

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".

Not quite getting that. Anywhere you can put up a sketch?

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.

# 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

(* 1.9 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

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

(* 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.

Maybe you could counter weight the ends, but your supports better be up to the job then.

And trim the bottom so it looks level.

With a 12ft length of "scaffold pole" (as I'd perceive it) curtain-rail...

Maximum central load before permanent bend =

At that load, at centre

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