Today's Message Index:
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1. 10:07 AM - Re: Wing Strut Attachment (taildrags)
2. 10:41 AM - Re: Wing Strut Attachment (taildrags)
3. 10:50 AM - Re: Wing Strut Attachment (taildrags)
4. 02:04 PM - Re: Re: Wing Strut Attachment (helspersew@aol.com)
5. 03:45 PM - Re: Wing Strut Attachment (taildrags)
6. 04:05 PM - Re: Re: Wing Strut Attachment (John C Black)
7. 07:33 PM - Re: Re: Wing Strut Attachment (Clif Dawson)
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| Subject: | Re: Wing Strut Attachment |
John; I'm running some calculations to determine where the weakest link is in the
string of things between the wing attachment fitting for the lift strut and
the attachment to the fuselage at the bottom end. I'll post the results here
when I'm done.
In the meantime, I'm intrigued by your mention of building a test stand to load-test
a sample to failure in tension. This is easily done if you want to "backyard"
it. Simple hardware and lumber, but be prepared to be surprised at what
it takes to fail the assembly. Let's just pick one of the numbers that you
threw out there, the Heim joint at 14,000 lbs. If you have some nice heavy-wall
steel pipe out there, you can develop that kind of load by using a 5 ft chunk
of that pipe, mounting one end on something very sturdy using a bolt or other
attachment that will allow the pipe to pivot, and then attaching your test
piece 1 ft away from the pivot end of the pipe. You now have a 4:1 lever and
you can get 14,000 lbs on the test piece by loading the long end of your pipe
with 2800 lbs of "something". One such "something" is a fairly common 2-ton "come-along"
(Maasdam Pow'r-Pull). I have one in my truck. Using a come-along,
you can gradually load the test piece but since you don't know what's going
to fail or when it's going to fail, you should use an extension on the come-along
handle to get you some distance away and put up a plywood shield or something
you can work behind when you conduct the test.
If you have something stronger than pipe, you can make the lever arm longer to
get more advantage. For example, if you have an 8 ft section of steel I-beam,
you could develop a 7:1 advantage and only need to put 1,750 lbs on the long
end, which you can now do with just a 1-ton come-along. If you wanted to know
how much load you were putting on it as you loaded it, you could attach a 200
gallon container to the long end and start filling it with water (8.34 lbs/gal),
with marks on the container to let you know how much load was accumulating.
I know, this is getting ludicrous, but you get the point... you don't need
a special load testing apparatus to run your test.
I prefer to just run the calculations instead of conducting exciting load tests
like this, especially since there are so many Air Campers flying around safely
out there with all sorts of lift strut arrangements.
--------
Oscar Zuniga
Medford, OR
Air Camper NX41CC "Scout"
A75 power, 72x36 Culver prop
Read this topic online here:
http://forums.matronics.com/viewtopic.php?p=477122#477122
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| Subject: | Re: Wing Strut Attachment |
John: also, in regard to thread engagement, there are some general rules of thumb
as to how many threads need to be engaged in order for the full strength of
the connection to be developed. The question you're asking has to do with a
steel fastener threading into an aluminum bar, so the question has two answers...
one for the steel fork end fitting and another for the bar it's threaded into.
The *general* rule for a steel part is that it must be engaged at least
1x its diameter in order to develop full strength. The fork end in question has
a 3/8-24 thread, so at least 3/8" has to be engaged in the threads of the mating
part for the fork end to develop its full strength in tension. For a 24
threads-per-inch threading, that would be at least 9 threads engaged.
However, for *aluminum*, the rule of thumb is 1.5x the diameter due to the lower
shear strength of the material. Again for the same thread pitch and diameter,
that would be about 11 threads engaged, or roughly 1/2" of full engagement
for the aluminum bar to receive the full load being transferred to it by the fork
end. More threads are better though ;o)
Speaking of rules for threads in general, the first thread always takes the most
load and it drops off with successive threads as you come out. Here's the theoretical
breakdown as you add threads: 1st thread takes 34%; 1st and second
combined take 57%; 1st, 2nd, and 3rd combined take 73%; 1st through 4th take 84%,
1st through 5th take 91%; 1st through 6th threads take 98%. It would seem,
then, that all you would need is 6 or 7 threads engaged even in aluminum, but
the problem is the shear strength because when a threaded fitting fails by stripping,
once the first thread strips the following ones strip in rapid succession
so it's important to prevent a pull-out failure by thread stripping. If
you've worked in the shop any length of time, you know that sickening feeling
when you're putting some muscle onto a ratchet on a nut and it strips. It all
goes mushy and that fastener is done for.
--------
Oscar Zuniga
Medford, OR
Air Camper NX41CC "Scout"
A75 power, 72x36 Culver prop
Read this topic online here:
http://forums.matronics.com/viewtopic.php?p=477124#477124
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| Subject: | Re: Wing Strut Attachment |
John: found just the thing for your test rig: Northern Tool has a 3-ton hydraulic
jack for only $60! No problem developing a couple of thousand pounds of load
on that test rig ;o)
--------
Oscar Zuniga
Medford, OR
Air Camper NX41CC "Scout"
A75 power, 72x36 Culver prop
Read this topic online here:
http://forums.matronics.com/viewtopic.php?p=477125#477125
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| Subject: | Re: Wing Strut Attachment |
Here is an example that has been flying for 7 years. Upper strut attachment. Carlson
"small" strut, and Mcmaster-Carr 7071 alloy insert, with two AN4 bolts.
Dan Helsper
Loensloe Airfield
Puryear, TN
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| Subject: | Re: Wing Strut Attachment |
Ive taken a look at the forces in the wing lift struts of the Air Camper. To run
the analysis, Ive used the dimensions given in the Orrin Hoopman (1933-34)
plans to create the geometry, but theres not a lot of difference in any of the
others that are commonly used. The main lift strut length is given as 89-3/4
between the bolt holes, the front cabane length is given as 21-1/4, and the distance
from the cabane mounting point to the lift strut mounting point at the
bottom of the fuselage side is 22-3/4, so that side of the triangle is 44. With
those two dimensions, we find that the angle that the lift strut makes with
the bottom of the wing is 29.4 degrees.
Assuming an aircraft gross weight of 1,088 lb and a design loading of +3.8G (normal
category aircraft), we get a total load to be supported by the wing of 4,134
lb, and each half of the wing thus has to support 2,067 lb. According to
the source quoted in my spar strength article in the BPA Newsletter (Noel Becars
analysis in a 1963 Sport Aviation), the main spar in a strut-braced monoplane
at a high positive angle of attack carries about 84% of the total load, so
that results in 1,737 lbs that each front lift strut carries.
If the load on the spar where the lift strut attaches is 1,737 lbs and the strut
is at an angle of 29.4 degrees to the wing, the tensile force in the strut will
be 1737/(sin 29.4) = 3,538 lbs. This is the force that every element in the
lift strut assembly must meet or exceed in order for the wing to sustain +3.8G
loading.
Lets start with the item that everybody is asking about, which is the small Carlson
extruded aluminum strut itself. I have graphically checked the value that
Carlson gives for the cross-sectional area of aluminum material in the strut,
0.431 sq.in., and my number comes very close to theirs so Ill use theirs. Assuming
that there are no obvious stress concentrating aspects of the cross-section
and that it all gets evenly loaded by transfer from the attachments at the
ends, and that the tensile strength of 6061-T6 aluminum is 45,000 psi, the
strut should be able to carry a tensile load of 19,395 lbs. The Carlson literature
gives it as 18,012 lbs so well use their number. This is about 5 times
more than what it will need to carry at +3.8 G of wing loading. The shape is
important in the other regime of *negative* G loading (compression) and in bending,
but thats a separate topic from just the loading in tension under wing lifting
condition. I agree with the person who stated that the information that
Carlson gives about round aluminum tubing is confusing and of no use in strength
analysis. They compare their strut to a 12.75 diameter, 0.065 wall tube...
absurd. The closest match to their strut and a round tube with that wall thickness
is a 2-1/4 OD 6061-T6 tube, which has a metal area of 0.446 sq.in., a
weight of 0.533 lb/lin.ft., and a tensile strength of 20,079 lb. It is obvious
by inspection that the Carlson strut will have significantly lower aerodynamic
drag than a round 2-1/4 tube, which along with cost is just about the only
useful comparison between the two.
Moving along the strut to the ends, the fork ends wont be a problem... with nominal
AN6 ends (3/8) and 125,000 psi steel strength, theyre good for at least 13-14,000
pounds in tension... about 4 times the needed strength. The catalog
touts them as high strength, but no strength value is given. The Heim ends appear
to have the same strength in tension as the forks, so they shouldnt be a
problem either.
The trickiest part is the attachment between the fork (or Heim) ends and the lift
strut. Looking at the 1 x 3/4 6061-T6 bar that Carlson uses, when undrilled
it has a cross-sectional area of 0.75 sq.in. so its good for about 33,750 pounds
in tension but at the spot where the first mounting hole is drilled through
it to attach it to the strut, the cross-sectional area is reduced to about
0.5625 sq.in. and the capacity drops to 25,312 pounds. Still about 7 times whats
needed. The aluminum bar is just fine; steel is not necessary.
Where the fork ends are threaded into the bar stock, the edge distance of the hole
in the end of the bar is about 3/16 on the short side, so that shouldnt be
a limiting factor. If a sufficient number of threads of the fork end are engaged
into the threaded hole in the bar, that wont be the weak point either. The
problem then comes down to the end of the bar that slides up into the hollow
end of the strut and how the two are connected. Since the bar is much thicker
than the wall of the lift strut, the weakest point of the connection is at
the mounting bolt holes through the sides of the strut. Although there are rigorous
methods for determining the load distribution between the bolts in a multi-bolt
connection, a fairly safe assumption in this case is that if the bolts
are not placed too close to one another or too close to the edges of the material,
at ultimate loading all of the bolts will be carrying about the same amount
of load. Lets look at the load at the bolt holes through the strut wall
nearest to the end of the strut. There is no need to look at shear strength of
the AN4 connecting bolts.... they will be far stronger in shear than the thin
walls of the strut.
There are at least two possible modes of failure at the connecting bolts but Ill
look only at shear out of the bolt, where the steel bolt pulls or tends to elongate
the hole in the softer aluminum as it yields. In that mode, the force
that the bolt can carry before the hole in the aluminum shears out is equal to
the shear strength of the material (extruded 6061-T6 is about 24,000 psi) times
the area that the bolt bears on. For AN4 bolts and a strut wall thickness
of 0.119 at its thinnest point, with the first bolt in the row being about 3/4
from the end of the strut (such as in the image Piet_construction_449.jpg that
Dan Helsper shared earlier), that works out to about 8,568 lbs for the first
bolt in the row. That one alone should be able to carry twice the required
load, so two bolts through each end of the strut should be more than adequate
but if they give you three holes, use three bolts. The distance of the first
bolt hole from the end of the strut is very important though... if the edge distance
is reduced from 3/4 to 1/2 (for example), the pull-out load drops to about
5,712 lbs... only about 1.6 times the required capacity. With at least two
attachment bolts, the connection should be adequate, but by all means keep the
first bolt hole comfortably away from the end of the strut.
--------
Oscar Zuniga
Medford, OR
Air Camper NX41CC "Scout"
A75 power, 72x36 Culver prop
Read this topic online here:
http://forums.matronics.com/viewtopic.php?p=477134#477134
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| Subject: | Re: Wing Strut Attachment |
Oscar,
WELL DONE !!
Thank God you are interested in Pietenpols !!
This really helps.
John
PS Regarding doing a destruction test to confirm I think I have borrowed 6 Jersey
barriers weighing about 2 tons each to anchor my test piece to the ground.
Several guys around here have cranes and other construction equip that can lift
10 tons. Creating a test piece is no problem. The problem is how to measure
how much force is being applied as we lift. Some cranes estimate force.
All this is crude. Im trying to find a universtity that has a machine set up
to do this with some accuracy.
On Jan 6, 2018, at 3:45 PM, taildrags <taildrags@hotmail.com> wrote:
Ive taken a look at the forces in the wing lift struts of the Air Camper. To run
the analysis, Ive used the dimensions given in the Orrin Hoopman (1933-34)
plans to create the geometry, but theres not a lot of difference in any of the
others that are commonly used. The main lift strut length is given as 89-3/4
between the bolt holes, the front cabane length is given as 21-1/4, and the distance
from the cabane mounting point to the lift strut mounting point at the
bottom of the fuselage side is 22-3/4, so that side of the triangle is 44. With
those two dimensions, we find that the angle that the lift strut makes with
the bottom of the wing is 29.4 degrees.
Assuming an aircraft gross weight of 1,088 lb and a design loading of +3.8G (normal
category aircraft), we get a total load to be supported by the wing of 4,134
lb, and each half of the wing thus has to support 2,067 lb. According to
the source quoted in my spar strength article in the BPA Newsletter (Noel Becars
analysis in a 1963 Sport Aviation), the main spar in a strut-braced monoplane
at a high positive angle of attack carries about 84% of the total load, so
that results in 1,737 lbs that each front lift strut carries.
If the load on the spar where the lift strut attaches is 1,737 lbs and the strut
is at an angle of 29.4 degrees to the wing, the tensile force in the strut will
be 1737/(sin 29.4) = 3,538 lbs. This is the force that every element in the
lift strut assembly must meet or exceed in order for the wing to sustain +3.8G
loading.
Lets start with the item that everybody is asking about, which is the small Carlson
extruded aluminum strut itself. I have graphically checked the value that
Carlson gives for the cross-sectional area of aluminum material in the strut,
0.431 sq.in., and my number comes very close to theirs so Ill use theirs. Assuming
that there are no obvious stress concentrating aspects of the cross-section
and that it all gets evenly loaded by transfer from the attachments at the
ends, and that the tensile strength of 6061-T6 aluminum is 45,000 psi, the
strut should be able to carry a tensile load of 19,395 lbs. The Carlson literature
gives it as 18,012 lbs so well use their number. This is about 5 times
more than what it will need to carry at +3.8 G of wing loading. The shape is
important in the other regime of *negative* G loading (compression) and in bending,
but thats a separate topic from just the loading in tension under wing lifting
condition. I agree with th!
e person who stated that the information that Carlson gives about round aluminum
tubing is confusing and of no use in strength analysis. They compare their
strut to a 12.75 diameter, 0.065 wall tube... absurd. The closest match to their
strut and a round tube with that wall thickness is a 2-1/4 OD 6061-T6 tube,
which has a metal area of 0.446 sq.in., a weight of 0.533 lb/lin.ft., and a
tensile strength of 20,079 lb. It is obvious by inspection that the Carlson strut
will have significantly lower aerodynamic drag than a round 2-1/4 tube, which
along with cost is just about the only useful comparison between the two.
Moving along the strut to the ends, the fork ends wont be a problem... with nominal
AN6 ends (3/8) and 125,000 psi steel strength, theyre good for at least 13-14,000
pounds in tension... about 4 times the needed strength. The catalog
touts them as high strength, but no strength value is given. The Heim ends appear
to have the same strength in tension as the forks, so they shouldnt be a
problem either.
The trickiest part is the attachment between the fork (or Heim) ends and the lift
strut. Looking at the 1 x 3/4 6061-T6 bar that Carlson uses, when undrilled
it has a cross-sectional area of 0.75 sq.in. so its good for about 33,750 pounds
in tension but at the spot where the first mounting hole is drilled through
it to attach it to the strut, the cross-sectional area is reduced to about
0.5625 sq.in. and the capacity drops to 25,312 pounds. Still about 7 times whats
needed. The aluminum bar is just fine; steel is not necessary.
Where the fork ends are threaded into the bar stock, the edge distance of the hole
in the end of the bar is about 3/16 on the short side, so that shouldnt be
a limiting factor. If a sufficient number of threads of the fork end are engaged
into the threaded hole in the bar, that wont be the weak point either. The
problem then comes down to the end of the bar that slides up into the hollow
end of the strut and how the two are connected. Since the bar is much thicker
than the wall of the lift strut, the weakest point of the connection is at
the mounting bolt holes through the sides of the strut. Although there are rigorous
methods for determining the load distribution between the bolts in a multi-bolt
connection, a fairly safe assumption in this case is that if the bolts
are not placed too close to one another or too close to the edges of the material,
at ultimate loading all of the bolts will be carrying about the same amount
of load. Lets look at the load at !
the bolt holes through the strut wall nearest to the end of the strut. There is
no need to look at shear strength of the AN4 connecting bolts.... they will
be far stronger in shear than the thin walls of the strut.
There are at least two possible modes of failure at the connecting bolts but Ill
look only at shear out of the bolt, where the steel bolt pulls or tends to elongate
the hole in the softer aluminum as it yields. In that mode, the force
that the bolt can carry before the hole in the aluminum shears out is equal to
the shear strength of the material (extruded 6061-T6 is about 24,000 psi) times
the area that the bolt bears on. For AN4 bolts and a strut wall thickness
of 0.119 at its thinnest point, with the first bolt in the row being about 3/4
from the end of the strut (such as in the image Piet_construction_449.jpg that
Dan Helsper shared earlier), that works out to about 8,568 lbs for the first
bolt in the row. That one alone should be able to carry twice the required
load, so two bolts through each end of the strut should be more than adequate
but if they give you three holes, use three bolts. The distance of the first
bolt hole from the end of the strut i!
s very important though... if the edge distance is reduced from 3/4 to 1/2 (for
example), the pull-out load drops to about 5,712 lbs... only about 1.6 times
the required capacity. With at least two attachment bolts, the connection should
be adequate, but by all means keep the first bolt hole comfortably away from
the end of the strut.
--------
Oscar Zuniga
Medford, OR
Air Camper NX41CC "Scout"
A75 power, 72x36 Culver prop
Read this topic online here:
http://forums.matronics.com/viewtopic.php?p=477134#477134
Message 7
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| Subject: | Re: Wing Strut Attachment |
Now if the rest of our beloved beast matched these figures ----- goodby
Sukhoi. :-)
Attached are pics of my strut ends and the jury struts on my Hemlock struts.
Just for comparison, you understand.
Clif
"Hell, there are no rules here; we're trying to accomplish something."
(Thomas Edison)
Handle every Stressful situation like a dog.
If you can't eat it or play with it,
Piss on it and walk away.
I?Tve taken a look at the forces in the wing lift struts of the Air Camper.
To run the analysis, I?Tve used the dimensions given in the Orrin Hoopman
(1933-34) plans to create the geometry, but there?Ts not a lot of
difference in any of the others that are commonly used. The main lift strut
length is given as 89-3/4? between the bolt holes, the front cabane length
is given as 21-1/4?, and the distance from the cabane mounting point to
the lift strut mounting point at the bottom of the fuselage side is
22-3/4?, so that side of the triangle is 44?. With those two
dimensions, we find that the angle that the lift strut makes with the bottom
of the wing is 29.4 degrees.
Assuming an aircraft gross weight of 1,088 lb and a design loading of +3.8G
(normal category aircraft), we get a total load to be supported by the wing
of 4,134 lb, and each half of the wing thus has to support 2,067 lb.
According to the source quoted in my spar strength article in the BPA
Newsletter (Noel Becar?Ts analysis in a 1963 Sport Aviation), the main spar
in a strut-braced monoplane at a high positive angle of attack carries about
84% of the total load, so that results in 1,737 lbs that each front lift
strut carries.
If the load on the spar where the lift strut attaches is 1,737 lbs and the
strut is at an angle of 29.4 degrees to the wing, the tensile force in the
strut will be 1737/(sin 29.4) = 3,538 lbs. This is the force that every
element in the lift strut assembly must meet or exceed in order for the wing
to sustain +3.8G loading.
--------
Oscar Zuniga
Medford, OR
Air Camper NX41CC "Scout"
A75 power, 72x36 Culver prop
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