"James Waldby" wrote in message
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On Sun, 09 Apr 2017 18:08:14 -0400, Jim Wilkins wrote:
"James Waldby" wrote in message
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On Sun, 09 Apr 2017 10:11:22 -0400, Jim Wilkins wrote:
"James Waldby" ... wrote in message
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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=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