


Ortrographic Views
Should have mentioned, previous views were perspective.







Quote:
The Rhino pictures look great! Regards, Andrés 






Version 0.1 available
Hi,
I used the morning to calculate the load distributions. I included a graph into the project of the
I assumed a load factor of n=20, which is quiet a lot: full pullup at 40 m/s. Use it, therefore, only as a reference and decide yourself what you want it to withstand. The load distributions were determined along the main spar shown in yellow in the DXF file. The distribution is shown against the spanwise coordinate. Do not get confused, the loads are still along the spar! Best ragrds, Andrés 





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A free viewer of Rhino 3D files is available here by simply downloading the demo version of Rhino 3D. That demo is a full working version. As a techinical note, I have created the existing NURBS model from your profiles with no modification whatsoever. That model uses the power of NURBS to blend the airfoils you have provided per the planform. From this point, I will extend and modify that surface for the tip, which will be largely hand shaped. Also, the structural elements may be arranged differently to suit other requirements. Regards, Kent 






Quote:
Surprised to see the torsional moment is 1/4 of the bending moment. I was assuming less. I'm familiar with the graphs and the engineering principles. At the risk of starting a full blown structural design class, I do have a few questions, if you please. Does load factor of n=20 mean 20 times the dead weight of the plane? (20 G's) 4.5 kg plane x 20 = 90 kg? Is the cross spar is ignored in these graphs? The bending moment graph appears to ignore the cross spar but the shear graph appears to consider it. Maybe this is just a function of the load distribution. Thanks, Kent 


Last edited by Knoll53; Jan 06, 2012 at 10:20 AM.
Reason: removed Newton questions





Nm  newton metres ie a torque of 1Nm is 1 Newton at a distance of 1 metre or 0.5N at 2m etc etc
9.81N = 1kg (under conditions of standard gravity). 1metre = 39.37inches Stand by for imperial  metric fun!!! ;) 





Quote:
For me: 1 Newton/meter = 0.7376 lb./ft Thanks, Kent 


Last edited by Knoll53; Jan 06, 2012 at 10:37 AM.





Hi Kent,
SourceForge is a hoster for open source projects. Just a sort of dowload host service. If you create an account, I could include you as a developer and you could upload your things directly using git. Yes, a load factor n=20 means 20 G and, hence, 90 kg! That is what you get when you fly 40 m/s und make a full pull. You could fly faster, but you should not make a full pull then One kg is subjected to a force of 1 kg * G = 9.81 Newton (N) or 1 N ≈ 0,225 lbf. For the torque/moment we have 1 Nm = 0,7375621 ft lb. I hope that helps I attached graphs in US customary units. The high torsional moment is the price one pays for stability: outer sections have to pull up the inner lift producing ones. In a tailed airplane this is done by the long tail. A good property of an allwing is that the bending moment is smaller than in a tailed airplane, because mass is better distributed and is not virtually a pointwise load as in tailed airplanes. The high torsional moment compared to the bending moment comes from this two reasons: price of stability and good mass/lift balance. The cross spar is included in the calculation. You see it best in the bending moment Mz around zaxis (last image). The values of the cutting loads are determined in a local coordinate system along the spar. The xaxis points along the spar, the yaxis aftwards and the zaxis to the top. The convention used is as follows. The forces and moments are shown, which arising inside the structure and counteract the external forces. Take for example the bending moment My. It is positive telling us that the external forces produce a negative moment. Hence, the mass in the center forces the tips of the wing to point upwards, as one expects if a full pull at 40 m/s is made. Regards, Andrés 

Last edited by andrecillo76; Jan 06, 2012 at 11:49 AM.




Thanks for the new graphs and explanation. It all is clear. Any units are fine with me..... even Newtons.
The upward bending moment graph for the spar "My" is a wild one. Lucky thing the wing is so thick at the center. It's like this plane is designing itself. Take a look at my last image. Did I show the max. moment, "My" direction, at the right location? I was thinking that the cross spar might cause a spike at it's connection to the main spar. Thanks, Kent 





Kent,
your pictures look fine. The direction correspond to the definitions I used. The discontinuities appear at the other moment and forces. I use a discretized beam approach: small beams between the ribs have an own coordinate system. The beams touch at a joining surface and exchange stresses. At the surface the moments and forces, which are vectors and do not care about coordinate systems, are transformed into the new cantilever local coordinate system and are used as a boundary condition. I calculate for this small cantilever the forces and moment and plot them against the spanwise coordinate at which the cantilever center is located. A finite element method would actually be needed to calculate the stresses correctly at a joint like the one between the spars... Greetings, Andrés 





Thanks Andres,
I agree that FEA would be needed to fully analyze this framing. With the variety of ever changing loads coupled with the angled framing layout, it does not lend itself to a simplified analysis. I can imagine the cross spar floating up with load and not contributing to bending resistance of the main spar. This would not consider the connection between the main and cross spar as provided by the ribs and plywood skins. Of course the entire structure works as one structural element. I am happy to take my best guess and test fly. If it does not flutter at 90 mph, I'll call that success. Although I will attempt to make moment connections at the cross spar to the main spar, I'm sure that it will act as a pinned connection. I will use your maximum bending moment and carry it out past the cross beam connection. The spar cap will be carbon tow, the spar will be deep, so it will be cheap and easy and stiff. My wing joiner is past the cross spar connection and the bending moment looks quite light out there. Thanks again for your analysis, Kent 





Kent,
Norman's comment helped me to understand why you are confused (thanks Norm!). The analysis considers only one spar, which can be kinked. So either I decide for the "/\" as the main spar, or for "/\" but the analysis does not cover the full "A". The analysis was made for the "/\" case, considering that "" acts as the main spar at the center. I made the same calculation for the "/\" as the main spar. See attached figures. The torsional moment is substantially smaller now, because the complete spar is located very near to the local AC. For the "/\" case, the torsional moment at the cross spar increased due to the lifting force acting far away from it and creating an additional torque. Regards, Andrés 

Last edited by andrecillo76; Jan 07, 2012 at 05:43 AM.




Quote:
After sketching out some details and looking at the construction sequence, I've decided to build the center section ( 40" wing span ) as one piece which will allow the carbon tow to be install as continuous curved spar caps. This removes the kink at the center. The main spar is of course tied to the plywood Dtube. I'm happy with this. The main spar appears to more closely match the center of lift line at the root area better with the curved spar. Thanks, Kent 






Kent,
good idea! We've made very good experiences with main curved spars. The small and large Schapel have this type of spars and both have been subjected to very large stresses . We used also the cross spar, though. I'm sure that the structure will be stiff. The thickness of the foil will do the trick and 3/8" would also be my choice for this wing. Regards, Andrés 


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