# Weight bearing strength of 5 ft. of black iron pipe

If I fully support 60 in. of 1-1/2 inch diameter black iron pipe (bought at Lowes or Home Depot), approximately how much weight can it support in
the center before it begins to bend more than 3/4 in. - 1 in. ?
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On Thursday, July 9, 2015 at 7:18:04 PM UTC-7, Maddog wrote:

All this calculating and discussion over a piece of pipe bending? It's not a crane lifting a load over a construction site. Get a grip!
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Hello, i read the whole thread.... But what if i was trying to build a cran e over a job site and had a similar idea using the same materials? How coul d i improve it?... Big factors to keep in mind are portability (easy setup and take down), least weight vs. Highest strength of steel piping, safety, and lowest possible cost.
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Hello, i read the whole thread.... But what if i was trying to build a crane over a job site and had a similar idea using the same materials? How could i improve it?... Big factors to keep in mind are portability (easy setup and take down), least weight vs. Highest strength of steel piping, safety, and lowest possible cost. =============================================This column calculator takes pipe 2" and up: https://courses.cit.cornell.edu/arch264/calculators/example7.3/index.html
"Pinned at both ends" means the load is applied directly in line with the pipe, without sideways force that tends to bow it. That restricts your design considerably, all the commercial hoists I've studied apply a cantilevered load to the columns which requires more complex calculations. http://www.strucalc.com/cantilever-column/
W sections are better than pipe for the bending loads on horizontal beams. http://www.engineeringtoolbox.com/american-wide-flange-steel-beams-d_1319.html
It's possible to build a portable gantry crane with a simple geometry that uses only pinned- or ball-end columns but I can't prove the safety of my design and won't describe it. I test everything for deflection with a measured proof load, and sometimes find unexpected weakness or bad material. -xyz
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http://digitalcommons.calpoly.edu/cgi/viewcontent.cgi?article 02&context=braesp
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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.
Take away - pipe is not as strong as it looks. Don't stand under that hoist, be careful it does not snag on something that lets it go to its full cable tension unless the support beam can handle that weight.
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in message

I bolted the HF 1300# electric hoist directly to a gantry trolley I welded from 1/4" thick angle iron, with wheels, axles and bearings copied from the HF 2000# trolley hoist, except that their axles are pressed into the frame while mine pass through it with nuts on the outside.
The hoist starts up with a substantial jerk that may apply double or more the steady load to the support.
I've seen 25000 psi suggested as the yield for imported water pipe made from random scrap steel.
https://www.ebay.com/i/232626917798?chn=ps
-jsw
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On Saturday, February 24, 2018 at 1:05:51 PM UTC-5, Jim Wilkins wrote:

ASTM specs black iron at 30.000 psi yield but your point about imported Chinese pipe is valid. BTW - I snuck in an extra 2x F.S. by using "bending moment M=WL/2" - the calculated value id WL/4.
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=I use this to quickly determine what available material is or isn't practical before refining the design: http://www.amesweb.info/StructuralBeamDeflection/SimpleBeamConcentratedLoadAtAnyPoint.aspx
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On Sunday, February 25, 2018 at 9:17:18 AM UTC-5, Jim Wilkins wrote:

afety

163

(1.5"

r W

S.

f-5-ft-of-black-iron-pipe-639655-.htm

That web site is a good resource. The deflection of Length/360 is a good st arting point. Note the original question was looking at 1" deflection, or L /60 - much too high a bend. For steel, E)e6 psi, and a 1st guess at Yie ld Strength would be 30,000 psi. For Aluminum, E36 psi, Yield (untempe red) = 10,000 psi. For overhead lifting, a safety factor of 6 should be u sed.
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On Sunday, February 25, 2018 at 9:17:18 AM UTC-5, Jim Wilkins wrote:

afety

163

(1.5"

r W

S.

f-5-ft-of-black-iron-pipe-639655-.htm

That site is a good reference. General design would look at Length/360 as a deflection (original post asked about 1" deflection on a 60" pipe, or L/60 - way too high). For steel, E)e6 psi, Yield0,000 psi (mild steel l ike black iron pipe). For Aluminum, Ee6 psi, Yield (untempered),0 00 psi.
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replying to mjacobsen925, Goncalves wrote: Really mjacobsen925, a forum is where one asks questions not where one gets censored.
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replying to mjacobsen925, Goncalves wrote: Really mjacobsen925, a forum is where one asks questions not where one gets censored.
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On Friday, August 2, 2019 at 6:18:04 AM UTC-7, Goncalves wrote:

ts

-bearing-strength-of-5-ft-of-black-iron-pipe-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 e lbows) screwed into solid wooden header above window. Seems like it should be more than strong enough for such a girly thing.
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On Sat, 3 Apr 2021 19:38:26 -0700 (PDT), Linda Iverson

2 or 3 inch might work but not 3/4 or 1 inch unless you want them to sag
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On 4/3/21 11:47 PM, Clare Snyder wrote:

The pipe, even 2 inch, will sag just from its own weight.
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Bob Nichols AT comcast.net I am "RNichols42"

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On Sun, 4 Apr 2021 09:36:53 -0500, Robert Nichols

Sched 40 will sag - sched 80 MIGHT work in 2 inch but not with HEAVY drapery
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On 4/4/21 4:52 PM, Clare Snyder wrote:

A 12 foot length of 2 inch Sched 80 steel pipe would weigh over 60 lbs. That's quite some curtain rod!!
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Bob Nichols AT comcast.net I am "RNichols42"

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

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.
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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. http://www.weldsmith.co.uk/tech/struct/210314_ebbeam_drillplat/19_drillplat_calcs.html
Rich Smith