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 several 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 will act as a shelf for the sink. I keep thinking the the strength of the pipe would be greater for a smaller size considering that an arc is quite good 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.

On Saturday, April 8, 2017 at 11:02:21 AM UTC-7, christine wrote:
I am installing a 100 lb farm sink. The specs suggest supporting it for 300 lbs. I have...lengths of 12 gauge strut channel or .5 or .75 galvanized pipe.

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.

On Sat, 08 Apr 2017 11:02:20 -0700, christine wrote:

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 several 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 will act as a shelf for the sink. I keep thinking the the strength of the pipe would be greater for a smaller size considering that an arc is quite good 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.

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.

--
jiw

"James Waldby" wrote in message
news
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=29,000,000. 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

On 04/09/2017 12:42 AM, James Waldby wrote:
On Sat, 08 Apr 2017 11:02:20 -0700, christine wrote:

....

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
ofhttp://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.
....


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.

--

On 04/09/2017 9:11 AM, Jim Wilkins wrote:
"James wrote in message
news
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 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.

--

On Saturday, April 8, 2017 at 11:02:21 AM UTC-7, christine wrote:
I am installing a 100 lb farm sink. The specs suggest supporting
it for 300 lbs. I have...lengths of 12 gauge strut channel or .5 or .75 galvanized pipe.

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 ...

To be on the safe side, you could get plumbing estimates and then switch to one or two steel bay replacements for behind the wall. Steel studs behind walls support far more weight than wooden 2x4's.

On 04/09/2017 10:22 AM, dpb wrote:
....

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.

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.

--


--

"dpb" wrote in message
news
On 04/09/2017 9:11 AM, Jim Wilkins wrote:
"James wrote in message
news
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 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.

--

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

On Sun, 09 Apr 2017 10:11:22 -0400, Jim Wilkins wrote:

"James Waldby" ... wrote in message news
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=29,000,000. 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.
....

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=96; Pcr=pi*pi*E*I/(K*L)**2; print 'Pcr={:.1f} at od={:.3f}, id={:.3f}, I={:.5f}'.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=96; Pcr=pi*pi*E*I/(K*L)**2; print 'Pcr={:.1f} at od={:.3f}, id={:.3f}, I={:.5f}'.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=96; Pcr=pi*pi*E*I/(K*L)**2; print 'Pcr={:.1f} at od={:.3f}, id={:.3f}, I={:.5f}'.format(Pcr, od, id, I)

Here are the results:
Pcr=9808.7 at od=1.900, id=1.770, I=0.31583
Pcr=8433.0 at od=1.900, id=1.790, I=0.27154
Pcr=16761.6 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.

--
jiw

On 04/09/2017 10:45 AM, dpb wrote:
On 04/09/2017 9:11 AM, Jim Wilkins wrote:
"James wrote in message
news
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 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.

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.

--

On 04/09/2017 3:16 PM, James Waldby wrote:
....

....

... 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=96; ...

Here are the results:
Pcr=9808.7 at od=1.900, id=1.770, I=0.31583
Pcr=8433.0 at od=1.900, id=1.790, I=0.27154
Pcr=16761.6 at od=2.375, id=2.265, I=0.53971
....

I just typed in command window w/ Matlab...

OD=1.9;ID=1.77;
E=29E6;
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.

--

"James Waldby" wrote in message
news
On Sun, 09 Apr 2017 10:11:22 -0400, Jim Wilkins wrote:

"James Waldby" ... wrote in message
news
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=29,000,000. One end is pinned and loaded more or less
equally on both sides, the other is effectively a ball and socket
on
the centerline.

...

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:

Here are the results:
Pcr=9808.7 at od=1.900, id=1.770, I=0.31583

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.

That's the one I used, with E=29000000, I=0.158 and L=96.

"I" came from he
http://www.engineersedge.com/calcula...re_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

On Sun, 09 Apr 2017 18:08:14 -0400, Jim Wilkins wrote:

"James Waldby" wrote in message
news
On Sun, 09 Apr 2017 10:11:22 -0400, Jim Wilkins wrote:

"James Waldby" ... wrote in message
news
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=29,000,000. One end is pinned and loaded more or less
equally on both sides, the other is effectively a ball and socket
on the centerline.
...
....
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:

Here are the results:
Pcr=9808.7 at od=1.900, id=1.770, I=0.31583
....

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.

That's the one I used, with E=29000000, I=0.158 and L=96.

"I" came from he
http://www.engineersedge.com/calcula...re_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.

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_a rea
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.

--
jiw

On Sun, 09 Apr 2017 15:51:39 -0500, dpb wrote:
On 04/09/2017 3:16 PM, James Waldby wrote:
... 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=96; ...

Here are the results:
Pcr=9808.7 at od=1.900, id=1.770, I=0.31583
...

I just typed in command window w/ Matlab...

OD=1.9;ID=1.77;
E=29E6;
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.

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_a rea

With that fixed we get the same results. Thanks for the help!

--
jiw

"James Waldby" wrote in message
news
On Sun, 09 Apr 2017 18:08:14 -0400, Jim Wilkins wrote:

"James Waldby" wrote in message
news
On Sun, 09 Apr 2017 10:11:22 -0400, Jim Wilkins wrote:

"James Waldby" ... wrote in message
news
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=29,000,000. One end is pinned and loaded more or less
equally on both sides, the other is effectively a ball and socket
on the centerline.
...
...
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:

Here are the results:
Pcr=9808.7 at od=1.900, id=1.770, I=0.31583
...

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.

That's the one I used, with E=29000000, I=0.158 and L=96.

"I" came from he
http://www.engineersedge.com/calcula...re_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.

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_a rea
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.

--
jiw

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