"Richard Smith" wrote in message ...

Clare Snyder writes:

......................

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.
http://www.weldsmith.co.uk/tech/stru...lat_calcs.html


Rich Smith
--------------------------
https://www.archtoolbox.com/material...imensions.html

https://amesweb.info/section/section...alculator.aspx

https://amesweb.info/Beam/beam-defle...alculator.aspx

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.

On Tue, 06 Apr 2021 13:06:00 +0100, Richard Smith
wrote:

Clare Snyder writes:

On Sun, 4 Apr 2021 21:19:14 -0500, Robert Nichols
wrote:

On 4/4/21 4:52 PM, Clare Snyder wrote:
On Sun, 4 Apr 2021 09:36:53 -0500, Robert Nichols
wrote:

On 4/3/21 11:47 PM, Clare Snyder wrote:
On Sat, 3 Apr 2021 19:38:26 -0700 (PDT), Linda Iverson
wrote:

On Friday, August 2, 2019 at 6:18:04 AM UTC-7, Goncalves wrote:
replying to mjacobsen925, Goncalves wrote:
Really mjacobsen925, a forum is where one asks questions not where one gets
censored.
--
for full context, visit https://www.polytechforum.com/metalw...pe-639655-.htm

What if I just want to use black steel pipe as a curtain rod to hang heavy velvet drapes spanning a window greater than 12 feet long without the need for a center support? I’ll anchor ends using flanges (nipples and elbows) screwed into solid wooden header above window. Seems like it should be more than strong enough for such a girly thing.
2 or 3 inch might work but not 3/4 or 1 inch unless you want them to
sag

The pipe, even 2 inch, will sag just from its own weight.
Sched 40 will sag - sched 80 MIGHT work in 2 inch but not with HEAVY
drapery

A 12 foot length of 2 inch Sched 80 steel pipe would weigh over 60 lbs.
That's quite some curtain rod!!
But it will support itself and some load over a 12 ft span. (about 1
1/2 times?? as much as sched 40.

Just did as bit of investigating and 1 1/4 inch sched 80 should be
adequate for a pretty heavy curtain - and weigh about 40 lb? - about
10 lb per meter) will sipport about 30 lb evenly didtributed with
about an inch of deflection.
Sched 40 will only handle around 20 lb with almost 1 1/2 inches of
deflection with a wight of ropughly 30 lb.

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.
http://www.weldsmith.co.uk/tech/stru...lat_calcs.html


Rich Smith
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 - - -

On 4/6/21 12:08 PM, Clare Snyder wrote:
On Tue, 06 Apr 2021 13:06:00 +0100, Richard Smith
wrote:

Clare Snyder writes:

On Sun, 4 Apr 2021 21:19:14 -0500, Robert Nichols
wrote:

On 4/4/21 4:52 PM, Clare Snyder wrote:
On Sun, 4 Apr 2021 09:36:53 -0500, Robert Nichols
wrote:

On 4/3/21 11:47 PM, Clare Snyder wrote:
On Sat, 3 Apr 2021 19:38:26 -0700 (PDT), Linda Iverson
wrote:

On Friday, August 2, 2019 at 6:18:04 AM UTC-7, Goncalves wrote:
replying to mjacobsen925, Goncalves wrote:
Really mjacobsen925, a forum is where one asks questions not where one gets
censored.
--
for full context, visit https://www.polytechforum.com/metalw...pe-639655-.htm

What if I just want to use black steel pipe as a curtain rod to hang heavy velvet drapes spanning a window greater than 12 feet long without the need for a center support? I’ll anchor ends using flanges (nipples and elbows) screwed into solid wooden header above window. Seems like it should be more than strong enough for such a girly thing.
2 or 3 inch might work but not 3/4 or 1 inch unless you want them to
sag

The pipe, even 2 inch, will sag just from its own weight.
Sched 40 will sag - sched 80 MIGHT work in 2 inch but not with HEAVY
drapery

A 12 foot length of 2 inch Sched 80 steel pipe would weigh over 60 lbs.
That's quite some curtain rod!!
But it will support itself and some load over a 12 ft span. (about 1
1/2 times?? as much as sched 40.

Just did as bit of investigating and 1 1/4 inch sched 80 should be
adequate for a pretty heavy curtain - and weigh about 40 lb? - about
10 lb per meter) will sipport about 30 lb evenly didtributed with
about an inch of deflection.
Sched 40 will only handle around 20 lb with almost 1 1/2 inches of
deflection with a wight of ropughly 30 lb.

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.
http://www.weldsmith.co.uk/tech/stru...lat_calcs.html


Rich Smith
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.

--
Bob Nichols AT comcast.net I am "RNichols42"

Clare Snyder writes:

On Tue, 06 Apr 2021 13:06:00 +0100, Richard Smith

.............



I do a lot of these calculations.
eg.
http://www.weldsmith.co.uk/tech/stru...lat_calcs.html


Rich Smith
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 - - -

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.

Robert Nichols writes:

...........

[digression from OP's purpose]

On 4/6/21 12:08 PM, Clare Snyder wrote:
I do a lot of these calculations.
eg.
http://www.weldsmith.co.uk/tech/stru...lat_calcs.html


Rich Smith
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.

--
Bob Nichols AT comcast.net I am "RNichols42"

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.

On 4/7/21 4:14 AM, Richard Smith wrote:
Robert Nichols writes:

...........

[digression from OP's purpose]

On 4/6/21 12:08 PM, Clare Snyder wrote:
I do a lot of these calculations.
eg.
http://www.weldsmith.co.uk/tech/stru...lat_calcs.html


Rich Smith
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.

--
Bob Nichols AT comcast.net I am "RNichols42"

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.

--
Bob Nichols AT comcast.net I am "RNichols42"

"Richard Smith" wrote in message ...

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.

------------------

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.

Robert Nichols writes:

.....

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.

--
Bob Nichols AT comcast.net I am "RNichols42"

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".

"Jim Wilkins" writes:

------------------

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.

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.

https://www.polytechforum.com/metalw...pe-639655-.htm

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.