# need strength

I am installing a 100 lb farm sink. The specs suggest supporting it for 300 lbs. I have a span of 30 inches in a cabinet. I would like to use either s
everal lengths of 12 gauge strut channel or .5 or .75 galvanized pipe. I will have both the ends of either of these sunk into 1 inch thick wood. The load will be over the entire length of the steel and not be moving. They w ill act as a shelf for the sink. I keep thinking the the strength of the p ipe would be greater for a smaller size considering that an arc is quite go od at supporting a load. Can any one suggest what would be the best choice and how many lengths that they think i should use. Oh, the sink is 29.75 x 17.875 and 10 inches tall.
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On Saturday, April 8, 2017 at 11:02:21 AM UTC-7, christine wrote:

Pipe is subject to buckling if used in compression, so a 2x4 would be better. The prospect of rotation at joints means that pipe is rarely useful, and only with crossbracing (like the turnbuckles on an old screen door). Nails and construction glue will make a good ell in wood, but strut or pipe will require a gusset plate (or welding) to do the same. I'd use wood.
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On Sat, 08 Apr 2017 11:02:20 -0700, christine wrote:

Presumably you mean that several pieces of channel or pipe would run across under the sink, from one cabinet side wall to the other. Some beam deflection calculators are available online.
For example, putting length 30, diameter 1.05, wall thickness 0.113, and force 75# (assuming 4 beams each carrying 1/4 of 300#) into the "Calculate Deflection for Round Tube Beams" calculator at the bottom of <http://www.engineering.com/calculators/beams.htm shows deflection of about 0.61" which might be unsafe as unequal loading could lead to a cascade of failures.
That calculator probably assumes a point load at the center of the beam. But supporting a sink, it seems like much of the load would be out near the ends of the beam, putting much of the load into shear instead of bending.
For the atc-mechanical.com calculator (also available as a phone app) at <http://www.atc-mechanical.com/calculators/tube-size-using-structural-properties/ you can specify a length of 30, "Select Tube based on Simple Beam LOAD, but limit deflection to Length/360", etc., and get a list of the loads that common sizes of tubes and rectangles can carry. This calculator may also assume a point load at the center of the beam.
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jiw

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Could you double-check my use of on-line calculators to determine the column buckling load of a pinned-end chain link fence post 1.90" OD, 1.77" ID (16 gauge) and 96" long as 4900 lbs?
I entered E),000,000. One end is pinned and loaded more or less equally on both sides, the other is effectively a ball and socket on the centerline.
I've been using 2-3/8" posts for a lifting tripod with a 3/4 ton chain hoist, after proof-testing the tripod to (much more) by pulling a stump with a larger hoist. The calculator I (mis?)used gave their buckling load at around 8000 lbs.
The 2-3/8" post tripod requires one trip through the woods for it and another with the rest of the gear. Someone gave me three leftover 1.90" posts which I can carry with one arm, but I'm not yet comfortable with their safety factor, assuming one leg supports most of the load as when moving boulders sideways.
I left out the details of the end fittings because I don't want anyone who can't proof test it copying the design. One 2-3/8" post already had to be replaced.
-jsw
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On 04/09/2017 9:11 AM, Jim Wilkins wrote:

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I get ~1600 lb and that's assuming fully axial loading with no moment...will be significantly less with side loading. Seems on first blush at a reasonable number, 8'L/2"D is a pretty long and slender wand...and 16 ga ain't that stout.
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I've put about that much load on a tripod of 1-5/8" fence posts, to proof test it. Normally it carries a 1/4 ton HF hoist.
Originally I used wood posts and happened to hit the buckling limit exactly -- the post would remain at whatever deflection I pushed it to.
Today's hoist project is to pull up some bushes and saplings. -jsw
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On 04/09/2017 10:45 AM, dpb wrote:

Oh, crud...I had a typo and missed a factor of pi....the Euler formula does say 4900. I'm surprised it's that high; the other number surely doesn't seem out of line.
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On Sun, 09 Apr 2017 10:11:22 -0400, Jim Wilkins wrote:

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As I'm not a trained engineer, except for a few brief classes long ago that outlined some mechanical and civil engineering stuff, I don't know the ins and outs of column design.
Anyhow, when I enter your numbers into Euler's formula for pinned-end critical load, I get values about twice as high as yours. Maybe a safety factor of 2 was used where you did the calculation? Or possibly a different K (column effective length factor) ? I calculated via the formula shown at <https://en.wikipedia.org/wiki/Euler%27s_critical_load with K=1 for the case "rotation free and translation fixed" at both ends of the column. I think that's correct for the bottom ends of your columns, but wasn't sure that the top is translation fixed. Here is the Python code for several cases:
from math import pi, sqrt od=1.9; id=1.77; r1=id/2; r2=od/2; I=(r2**4 - r1**4)*pi/2; K=1; L�; Pcr=pi*pi*E*I/(K*L)**2; print 'Pcr= at od=, id=, I='.format(Pcr, od, id, I) od=1.9; id=od-2*0.055; r1=id/2; r2=od/2; I=(r2**4 - r1**4)*pi/2; K=1; L�; Pcr=pi*pi*E*I/(K*L)**2; print 'Pcr= at od=, id=, I='.format(Pcr, od, id, I) od=2.375; id=od-2*0.055; r1=id/2; r2=od/2; I=(r2**4 - r1**4)*pi/2; K=1; L�; Pcr=pi*pi*E*I/(K*L)**2; print 'Pcr= at od=, id=, I='.format(Pcr, od, id, I)
Here are the results: Pcr�08.7 at od=1.900, id=1.770, I=0.31583 Pcr�33.0 at od=1.900, id=1.790, I=0.27154 Pcr761.6 at od=2.375, id=2.265, I=0.53971
Some calculators below require I as an input, which is the reason for printing it via the Python code. The 1.9" OD case is calculated two ways because the following link shows wall thickness of 0.055" for "Standard Residential" chain-link-fence posts (vs. 0.065 for Heavy Residential): <https://fence-material.com/fence-posts-galvanized-16ga/
Anyhow, those numbers above agree with results from calculator <http://www.engineersedge.com/calculators/ideal-pinned-column-buckling-calculator-1.htm which is a fairly compact page with a few small images on it. <http://www.engineersedge.com/column_buckling/column_ideal.htm also is mostly text, with images for some formulas. It explains the inputs, distinguishes slender vs short columns, and has an Effective Length Constant table for various fixed/guided/pinned cases.
<http://www.efunda.com/formulae/solid_mechanics/columns/calc_column_critical_load.cfm also gets the same results.
<http://www.tech.plym.ac.uk/sme/desnotes/buckling.htm is all text except for two formula images and its associated calculator at <http://www.tech.plym.ac.uk/sme/desnotes/buccalc.htm looks like all text too. It wants metric data, eg 200 GPa for Young's modulus.
<https://mechanicalc.com/reference/column-buckling discusses pinned columns and the Euler Formula vs Johnson Formula. It has a link to <https://mechanicalc.com/calculators/column-buckling/ which computes Pcr and Scr and draws two plots, a "Critical force curve" and a "Critical Stress Plot" to make the safety-factor size easy to see at different loads and lengths. For this calculator, ASTM A572 appears to be the most suitable material to specify. However, I don't see 1.9" OD and 2.375" OD tube on the cross section list presented to non-subscribers; it appears that subscribers can create custom tube sizes and can select some standard sizes from a cross-section database.
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jiw

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On 04/09/2017 3:16 PM, James Waldby wrote: ...
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I just typed in command window w/ Matlab...
>> OD=1.9;ID=1.77; >> E)E6; >> L=8*12; >> I=pi*(ID^4-OD^4)/64; >> Pcr=pi^2*E*I/L^2; >> num2str(Pcr,'%.0f') ans 4904 >>
Your I is off by a factor of
>> Ipy=((OD/2)^4 - (ID/2)^4)*pi/2 Ipy 0.3158 >> I/Ipy ans 0.5000 >>
which accounts for the difference. When you used radii instead of diameters missed one factor of 2 in denominator it seems.
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On Sun, 09 Apr 2017 15:51:39 -0500, dpb wrote:

Instead of the problem being due to radii vs. diameters, it was due to using wrong-axis I. As shown in following link, I_z (which I used by mistake) is twice as big as either of I_x or I_y. <https://en.wikipedia.org/wiki/List_of_second_moments_of_area#Second_moments_of_area
With that fixed we get the same results. Thanks for the help!
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jiw

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That's the one I used, with E)000000, I=0.158 and L�.
"I" came from here: http://www.engineersedge.com/calculators/section_square_case_12.htm I entered D=1.9, d=1.77 (the wall is actually 0.070") and got 0.15792 in^4 for the area moment of inertia. The equation gives the same result.
The lower end is a cone in a hole in the foot plate, the upper end is a cross bolt with load-supporting chain hanging from both ends, not quite the way you might expect but free to rotate and equalize.
Maybe I need to dig out the Statics text for a refresher. Anyway, I proof test on a stump before lifting a load.
-jsw
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On Sun, 09 Apr 2017 18:08:14 -0400, Jim Wilkins wrote:

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As dpb pointed out, my values of I, second moment of area, were off by a factor of 2, which made the computed numbers twice too big. I was using I_z instead of I_x or I_y, and as shown in <https://en.wikipedia.org/wiki/List_of_second_moments_of_area#Second_moments_of_area I_z is twice as big as either of I_x or I_y.
Thus it looks like your calculations, dpb's corrected calculations, and my corrected calculations now match up, at about Pcr = 4900# for the 8' length of 1.9" tube. I imagine that leaves a good safety margin for most log lifting, but for stump pulling maybe not. It would be interesting to have a finite-elements analysis of what the load rating would increase to, using (for example) a re-bar triangle with sleeves at its corners to yoke leg midpoints together. Cutting unsupported length in half might multiply Pcr by 4, making it large enough that the tubes' compressive strength, instead of Euler buckling, is the issue.
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jiw

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I need these to fold compactly for transportation and storage so external bracing is out. Even chains joining the feet to keep them from spreading were more trouble than benefit. The plain round tubes themselves slowly get beat up and dented in use, I've replaced two already. Around here the remaining trees grow in places too steep and rough to build houses. Developers have to blast to stay in business.
Stump pulling doesn't need the safety margin of lifting because there's little stored energy, any give releases the tension immediately. That's why I use it to proof test. In this rocky soil a 2" tree I can carry away in one hand may take 2-3000 Lbs to break loose. -jsw
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On 04/09/2017 12:42 AM, James Waldby wrote:

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That seems awfully big; I couldn't get the link at that site to work with Firefox so I went to my old standby at <www.engineeringtoolbox.com>
I presume same 3/4" Sch 40 pipe which from the venerable Crane handbook shows I=0.037 in^4 and I'd assume there's only a place for the support on either edge so the load per each is 300/2-->150 lb --> 150/30 = 5 lb/in. uniform load.
The beam calculator with a 27E6 modulus gives
Total Load : 150 (lb) Length of Beam - L : 30 (in) Moment of Inertia - I : 0.037 (in4) Modulus of Elasticity - E : 27000000 (psi) Distance of extreme point off neutral axis - y : 0.53 (in) Support Force - R1 : 75 (lb) Support Force - R2 : 75 (lb) Maximum Stress - : 15963 (psi) Maximum Deflection - : 0.0845 (in)
which is much more like I'd expect, altho a tenth of an inch is more than I'd want to design for.
I agree with the other respondent who recommended just using wood; a tubafor of nominal SYP with an equivalent load is 0.002"/ft owing to the fact I==> 0.67 instead of 0.037 or 20X as stiff.
OTOH, if the OP were adamant about the pipe idea, for 1" I=0.087 and 1-1/4" = 0.195 or ~2X and 5X the 3/4", respectively. Either of the latter would be adequate although still a lot more of a hassle to deal with probably than just framing in the opening with construction lumber and appropriate supports for it.
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On 04/09/2017 10:22 AM, dpb wrote: ...

OBTW, I had cut E to 27E6; just looked up that from the Ductile Iron Pipe Research Association (who even knew there was such an organization? :) ) they say it's more like 24E6. That would raise above estimate to almost exactly the approximate 0.1".
This also is for free ends, not fixed; if used pipe flanges for the mounting that'd stiffen it up a fair amount...altho I'm not going back and recalculate; it's just so much more sensible to frame an appropriate support instead.
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