Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work.

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
  #11   Report Post  
Old April 6th 21, 01:47 PM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Jun 2011
Posts: 5,844
Default Weight bearing strength of 5 ft. of black iron pipe

"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.


  #12   Report Post  
Old April 6th 21, 06:08 PM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Nov 2017
Posts: 4,369
Default Weight bearing strength of 5 ft. of black iron pipe

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? Ill 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 - - -
  #13   Report Post  
Old April 7th 21, 04:20 AM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Sep 2007
Posts: 101
Default Weight bearing strength of 5 ft. of black iron pipe

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"
  #14   Report Post  
Old April 7th 21, 09:57 AM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Apr 2016
Posts: 86
Default Weight bearing strength of 5 ft. of black iron pipe

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.
  #15   Report Post  
Old April 7th 21, 10:14 AM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Apr 2016
Posts: 86
Default Weight bearing strength of 5 ft. of black iron pipe

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.


  #16   Report Post  
Old April 7th 21, 02:05 PM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Sep 2007
Posts: 101
Default Weight bearing strength of 5 ft. of black iron pipe

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"
  #17   Report Post  
Old April 8th 21, 12:32 AM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Jun 2011
Posts: 5,844
Default Weight bearing strength of 5 ft. of black iron pipe

"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.

  #18   Report Post  
Old April 8th 21, 09:29 AM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Apr 2016
Posts: 86
Default Weight bearing strength of 5 ft. of black iron pipe

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".
  #19   Report Post  
Old April 8th 21, 09:30 AM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Apr 2016
Posts: 86
Default Weight bearing strength of 5 ft. of black iron pipe

"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?
  #20   Report Post  
Old April 8th 21, 10:25 AM posted to rec.crafts.metalworking
external usenet poster
 
First recorded activity by DIYBanter: Apr 2016
Posts: 86
Default Weight bearing strength of 5 ft. of black iron pipe

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.


Reply
Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules

Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Compressive strength 3/4" black steel (gas pipe) Simon Lang Metalworking 1 November 15th 20 01:43 PM
Strength to Weight Ratio -- Units Tim Wescott Metalworking 3 May 26th 15 07:24 PM
Sulphur Black 2BR 200%,Sulphur Black B 200%,Solubilised Sulphur Black,Liquid Sulphur Black renjiang Metalworking 2 November 12th 08 04:33 PM
Tenon Strength / Rail strength- Max? [email protected] Woodworking 2 August 21st 06 05:21 PM
How much weight can a 3/4" black iron pipe support hanging from its threads? Stu Metalworking 2 September 27th 03 09:12 PM


All times are GMT +1. The time now is 08:21 PM.

Powered by vBulletin® Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.
Copyright 2004-2021 DIYbanter.
The comments are property of their posters.
 

About Us

"It's about DIY & home improvement"

 

Copyright © 2017