Spar Pin

The wing fold mechanisms is designed so it rotates around an AN-6 bolt at the top and is held with a special spar pin at the bottom. These two bolts have to take all the shear and the torsion from the outer section. The cord-vise torsion of the wing, due to aileron deflections for instance, is taken by a spar pin at the aft spar. If the bolt or the spar pin breaks, the wing collapses. When looking at the spar pin, it looks very tiny. It is 9.7 mm in diameter, less than my little finger.

How is it possible the tiny little pin and the similar sized bolt can take all that load? It does, and Sonex has tested the Onex wing to more than 11g without collapse. I had to find out anyway, so I did a fast calculation.

The pins and bolt are held in double shear. The forces from the outer wing and the resulting forces acting on the main spar pin/bolt is shown below. "1" is the position of the AN6 bolt and "2" is the main spar pin.

The difficult part was to find the "T" and "Shear". This was difficult because finding the span-vise lift distribution is essential. Doing a Google search gives lots of results, but most of those only show the span-vise lift distribution as "some elliptical shape". This may very well be true, but it doesn't really help much, because what is needed is a simple mathematical description/approximation that can be used to actually calculate the torque and the shear. The figure below shows the "typical" distribution. The red line is for taper ratio of 1.0, as it is for the Onex.

But the internet is a great place, and as it happened rather detailed hand calculations as well as "first order" approximations is to be found in Civil Aeronautics Manual 04. This is an old document from 1941, and is the FAR-23 or CS-23 from the old days. This is exactly what is needed, because in the old days they didn't have computers and CAD and FEM and CFD, things were solved by graphs and tables. I only need the "first order" approximation. The detailed calculation found in that manual could easily be implemented in a spread sheet, but this detail is not needed here.

The manual specify that both situations shown above should be investigated. However, with a taper ratio of 1.0 only the the one marked with a red square is an approximation to the real situation. With that approximation finding the torque and shear was easy. What I did was to scale the C_N in the above figures so the lift on the entire wing surface was equal to 6 times MTOW (6g). I did not include any lifting effect of the fuselage.

Spar pinLong wingShort wingAN6 BoltLong wingShort wing
D_pin [mm]9.709.70D_AN6 (3/8'')0.370.37
A_pin [mm2]73.9073.90D_AN6 [mm]9.429.42
Shear [N/mm2]302.50262.37A_pin [mm2]69.7469.74
Shear [PSI]43874.1938053.01Shear [n/m2]299.57257.92
1.5xshear [N/mm2]453.75393.55Shear [PSI]43449.3737408.29
1.5xshear [PSI]65811.2857079.511.5xshear [N/mm2]449.36386.88
Pin shear [PSI] ?75000.0075000.001.5xshear [PSI]65174.0556112.43
Real Factor Pin ?1.711.97AN6_shear [PSI]75000.0075000.00
g Spar pin ?10.2611.83Real Factor AN61.732.00
g AN610.3612.03

The calculations show that there will be a difference between the long wing and the short wing option. Only the resulting shear forces on the bolt/pin is shown above. I don't know what the actual pin material is, but I would assume it is similar to an AN bolt, 125,000 PSI in tension and 75,000 PSI in shear.

With these assumption the long wing has a safety factor of 1.71 for the pin and 1.73 for the bolt. The short wing has 1.97 for the pin and 2.00 for the bolt. Hence the wing folding mechanism is capable of at least 10.26 g for the long wing and 11.83 g for the short wing. I am more than satisfied, those tiny pins are not that tiny after all, and these calculations are conservative. A typical loop is 4g.

No comments: