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If someone is interested, I did a calculation of the main spar pin and the AN6 bolt for the wing fold mechanism. The only problem was to find an approximation for the wing lift distribution. I finally found the "old" (FAR-23 or CS-23) which had a good and simple approximation, although I would think it is very conservative for high AoA. I am an engineer, master and PhD, but I don't work in aviation, I work in the energy sector, turbines mostly.

What I found is the wing spar pin and bolt is more than adequate. They will not break before at least 10g for the long wing and 11g for the short. Because the calculations are conservative regarding wing load distribution, they most probably will hold considerably more than this as well. This should be good to know for people wondering about the strength of these rather tiny bolts and pins (as I did).

Sonex has of course tested the wing to 11g without it breaking, but still, for me it was nice to see that basic calculations show the exact same thing.

You can find it here: http://onex-svingenb.blogspot.no/p/spar-pin.html.

Bjørnar,
I did a similar thing for the hummelbird years ago. They use 1/2" bolts to pin the wing. They are way stronger than they need to be. The other thing in your analysis to consider is the strength of the aluminum. The steel is good for 120ksi or so but the compressive allowables for aluminum or less. I have faith in the design too, but the exercise is fun. Nice work!

Edit: I would analyze the aluminum for all potential failure modes- not just compressive. There are edge tear out modes (shear), and break across the hole modes (tension).

Bryan Cotton wrote:I would analyze the aluminum for all potential failure modes- not just compressive. There are edge tear out modes (shear), and break across the hole modes (tension).

That is true - in general. But on the Onex it is all steel. The steel parts are bolted and riveted onto the aluminum spars. So the spar pin and the bolt is sliding through steel. It really is very strong.

Steel on steel is good!

I was not able to see previous actual calculations so I off my calculations and opinions for comment

My Analysis of Connection
First off I’m building a ONEX (and having a lot of fun at it) and plan to fly it in normal and utility category, not acrobatic category.

For general information my background is as a licensed Professional Engineer , but my practice was not in anything regarding aircraft design and I was in management my last 25 years so I cannot say I’m currently practicing (although I still do my continuing education and maintain my multi state licensing). So I’m not an expert at all in aircraft design, but also not a shade tree mechanic.

The other analysis in my opinion contained a few inappropriate assumptions. The couple distance between the Bolt and Pin is not the distance between them. My assumption is the load is perpendicular to the spar (as in the load test) and as such the couple distance is the “vertical” distance between them. The two simple failure modes of a bolted connection are the bolt itself and the structure they are attached to.

My Analysis:
Per Sonex maximum moment 120,223 in# at joint ( I got this info from Jeremy months ago)
Per Sonex maximum Shear 3124 # at joint
(Note: my calculation of the moment and shear from photos of 10g load test achieve very similar results)

Couple force result of moment on bolt and pin: 120,223 in# / 6.3" (vertical distance between bolt & pin) = 19084 #
Vector resultant force on pin or bolt= Sq. root of (19084 ^2 +
3124^2) = 19147#
Just so everybody has an understanding of this from practical perspective this is three (3) Chevy Suburbans hanging from one bolted connection – would you stand under the three SUVs?

Bearing load outboard Plates (folding wing section over 3/8” thick so not governing for this test):
19147# / (.375" pin dia. X (2 X .187" thickness of plates)) = 136,000 psi
or 136,000 X (6/10) = 81,600 psi @ 6 g’s
or 136,00 X (4.4/100) = 60,000 psi @ 4.4 g’s(utility category)

Shear Load on Pin (double shear):
19174#/ (2 [double shear]((3/8 [dia of pin]/2 )^ 2)X PI X .1875) = 95000 psi
or 95000 X (6/10) = 57,000 psi @ 6 g’s
or 95,000 X (4.4/100) = 41,800 psi @ 4.4 g’s (utility category)

4140 Steel
Annealed Yield Strength 60,000 to 70,000 psi
Tensile Strength 110,000 psi

Notes
1. Heat-treat of plate and pin unknown – Was able to drill plates with standard high-speed bit – My Pin tested in the very low 20’s on the Rc scale, both appear to be annealed.
2. The modules of elasticity of 6061 T6 is approx. 1/3 of steel so it will not carry any real loads at these interfaces
3. Pin will see Stress Reversal but was considered in this analysis (will reduce allowable stress levels)
4. Impact forces not considered in this analysis (will reduce allowable stress / safety factor)
5. Tear out of the joint not considered in this analysis (I have noted some people have been putting in larger dia. Bolts - I would not do this without a redesign of the joint as this could result in actually a weaker structure.

My Conclusions / Comments:
The joint should be capable of withstanding the forces induced in Utility Category operations (as long as not to many max stress reversals are applied). The joint, in my opinion, would be very questionable if it could withstand many stress reversal cycles at the Acrobatic Category levels.
If somebody with a better understanding of the joint, materials, and forces can enlighten me to a better conclusion I would be very interested in understand were my analysis and conclusion is in error.
My comments on stress reversal are only based on my limited knowledge on the subject, but as a practical matter we typically only used 50% of yield as a maximum allowable stress level when the material was subject to stress reversal although we also anticipated more cycles then one would expect in an aircraft such as this.
If the joint became loose / sloppy then impact forces come into play and then the joint would be questionable even for normal category, again in my opinion.

paulonex wrote:Shear Load on Pin (double shear):
19174#/ (2 [double shear]((3/8 [dia of pin]/2 )^ 2)X PI X .1875) = 95000 psi

Are you sure about "X .1875" here? I'm unclear why the thickness matters here.

Thanks for catching my typing error and I also found I transposed two digits from a calculation above (the .1875 was the 3/8" pin dia / 2 to get the radius - just typing and not thinking and was not used in the actual calculation)

Corrected values below:
19147#/ (2 [double shear]((3/8 [dia of pin]/2 )^ 2)X PI) = 87000 psi
or 87000 X (6/10) = 52,000 psi @ 6 g’s
or 87,000 X (4.4/100) = 38,000 psi @ 4.4 g’s (utility category)

Values went down a little but my conclusion in my opinion is still the same

Paul,
Fun stuff. One question on your assumptions- you say the vertical distance is relevant, and not the space between the pins. If we take this to the extreme, and have them side by side, that would translate to no strength. Not true however because if you consider that we are trying to rotate about a pin when load is applied, each pin has to provide a load to counteract this torque.

I haven't dug into the rest of your analysis yet.

I think one other error is the analysis seems to run all the load due to torque through one bolt. For the forces and moments to add to zero they pretty much will see the same load due to bending. I see division by 2 for double shear but it should happen again for two bolts.

please review my previous post
The forces are of equal value but in opposite direction
In a positive g load the top joint (the bolt) is seeing a compressive type force and the pin a pulling force.