|Sep 08, 2014, 01:11 PM|
Actually this is one question I know the answer for...
Sorry to tell you but your 6.4 G figure is far too low. Because of the speeds and tight turn radius our models can achieve the actual limit you want to use for structural stress should be 30-40 G's. We have had this discussion regarding G forces several times on another message board, and engineers have repeatedly shown us figures that reach or exceed these ranges, even using trainers as examples.
I have folded several flying wings where I used 30 G's as the design limit, twice when pulling out of shallow dives I would not have thought I was even close to this limit.
|Sep 09, 2014, 11:18 AM|
My God, y're kidding !
things to consider ..
Let me do my same calculation on a classic spar setup, this G figure should be a good basis to compare to the carbon former G value ..
So then the G values are to be considered relative to eachother and not absolute .
|Sep 09, 2014, 01:36 PM|
I know a lot of DS guys have pegged the eagle tree limits of +/- 38 G's on a regular basis, they may have actually pulled way beyond that.
It's pretty dependent on speed and turn radius, I know many years ago they did a test on pattern ships that were pulling over +/- 25 G's back in the late 80's.
|Sep 09, 2014, 02:14 PM|
Ok, let's do a simplified math.
I just want to compare a usual classic spar design I always use, with the carbon former i plan to use.
1) The classic spar as a reference:
- Suppose you have only one spar system, with one spar top and one spar below, and webbing in between.
- the spars have 5x3mm cross section (15mm2 area)
- center spar to center spar distance is 25mm
2) Max compressive strength data of spar material: 40N/mm2
--> max compressive strength of spar = 40N/mm2 x 15mm2 = 600N
--> max moment on spar system = 600N x 25 mm = 15000Nmm
3) Take wing span of 1100mm and suppose that lift takes place a wing tip (=worst case approach), then the max allowable force at the wing tip would be = 15000Nmm /550mm = 27N.
Thus the max. lifting force would be 2 x 27N = 54N
Suppose the plane is 1.7kg heavy, the max G = 54N/(1.7kgx10) = 3.2G's
So what does this mean ?
That my classic spar setup, existing of 2 spars 5x3mm can pull about 3.2G's,
meaning the carbon former setup is stronger with it's max 6.4G's.
I cannot refer to the G's you mention, but only compare to the classic spar design I used several times before, which did fine.
|Sep 09, 2014, 09:54 PM|
The G levels I was talking about is the forces that can be generated during maneuvers at speed. During a discussion regarding the German Speed Cup rules in order for an aircraft to enter the course you have to make a turn with a radius of less than 50 meters at the 300+ mph these aircraft pull something like 50 G's. Now keep in mind these are extreme high performance airplanes but if you have a "quick airframe" and make a tight turn you can really pile on the G's quickly, granted this is more of an issue the faster you go. Your design has potential to be a fairly quick and nimble airframe that will generate G forces the like of a jet fighter rather than a Cessna. One thing I noticed you didn't add the strength the skin added when calculating your strength and it really adds a lot of strength we sometimes forget about.
I would say your delta probably has gone over 30 G's in some of the flying in your video. Either way what you have been doing has worked thus far. Heck that delta took a serious hit and came away with pretty minor damage.
|Sep 10, 2014, 12:13 PM|
What we all use frequently is a carbon rod, or 2 of them, for joining wings.
I'm surprised their strength is not so high:
SIGMA (y) = M* y / I
- SIGMA(y)= stress at location y (N/mm2)
- M = bending moment (Nmm)
- y = distance to neutral bending line of cross section (mm)
- I = moment of area (mm4)
--> max bending moment is then:
M_max = SIGMA_max / y_max * I
I of cirkel cross section = 1/4 * pi() * r^4
SIGMA_max for carbon = 720 N/mm2
Resulting max. bending moment:
Rod diameter = 8mm --> I = 201mm4--> M_max= 720 / 4 * I = 18.096Nmm
The same for the classic spar setup of before gives = 15.000Nmm
The same for the carbon former results in = 39.690Nmm
- A carbon rod of 8mm is as strong as a classic spar setup
- A carbon rod is 2 times weaker than the carbon former.
|Sep 10, 2014, 03:19 PM|
I think as a joiner we are using it as much for it shear strength as much as it bending strength, and a good portion of the compressive force on the wing is transferred to the body through the surface of the wing root itself. On my new plane I will probably be using dual joiners (top and bottom) and (forward and aft) to join the wing if I scale it up enough that they need to be removable because of the operational envelope and stresses it may encounter. But I really hope I can keep it down to a size where it is built all in one piece it will be much lighter. The final version is going to end up having retracts along with gear doors so it's probably going to have to be a bit larger than the first or second "test" airframes.
Of late I am leaning toward the classic spar setup using carbon as spar elements supported by a very strong shear web. Carbon is good under tension and using it as the top and bottom spar keeps the wings fairly rigid. I also like using two or three spar pairs in each wing as you saw me do with my delta. I can use much lighter elements for each spar since each is only carrying only a portion of the load a single spar setup has to endure. This method also makes it resist any torsional movement as well.
While the concept of a carbon tube or former in the center of the wing makes it fairly easy to keep everything aligned while building it isn't as strong as using smaller carbon elements within the classic spar/web configuration, at least that is my opinion. I used the tube/former in the center of the wing, and the classic spar with small carbon elements, and have found that the classic setup (using small carbon elements as far from the center of the airfoil as I can get them) is the strongest method by far.
If I have to use joiners for the wing I always have them as close to the spar as I can, if it is a swept wing I always make sure they at least tie in to the spar where they end.
|Sep 11, 2014, 11:22 AM|
I can only confirm what you said.
Smaller components as far from the center as possible, gives the most strength for the smallest weight.
You saw my design of the plane, I chose an open design: the bottom is completely removable for battery and gear hatch. Meaning the cross section of the main carbon spar is nothing than about 10x3mm in the middle of the fuselage.
With my calculations i posted here, i strongly believe that the 10x3mm section is large enough, thanks to the massive carbon strength, to withstand the forces. And as compared to a classic spar setup, it should be ok.
Concerning the sheeting, tensile strength is low, as the sheet is devided at the wing root.
Compressible strength could be counted for, ok, but as a moment can never exist without the two forces, you have to count with the weakest element, thus skin sheeting can not realy be counted for coping with bending moments (but surely for torsion).
When do you plan to start building you ship ?
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