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Aileron test
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Topic: Aileron test (Read 14480 times)
docmatt
doc matt
2015 Donor
Posts: 56
Total likes: 4
DE E-04 under construction
OS:
Windows Vista/Server 2008
Browser:
MS Internet Explorer 9.0
Eagle Type: DE
Re: Aileron test
«
Reply #15 on:
May 14, 2015, 09:22:59 PM »
I'm not sure it is really a difference in glue strength I was seeing. Because of the way the diagonals fit, there is no good way to use a gusset or corner block, so the glue area is very small and at least partially on end grain. The T88 fills gaps better and because of its viscosity, can create a bit of its own "gusset". I found the Titebond III convenient when the joints are nice and tight with lots of area. Also Titebond prevents one from trying to stretch the last bit of a batch of epoxy. Nothing worse than a joint with too little epoxy that is already starting to set up.
The more I think about it, the more I think the fabric will do the work of the diagonals.
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doc matt
DE E-04 under construction
Mineral Point, WI
Vince Carucci
Sr. Member
Posts: 220
Total likes: 123
EAA 324
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Eagle Type: LE: T-53
Re: Aileron test
«
Reply #16 on:
May 15, 2015, 06:58:23 PM »
Tightbond when applied correctly creates a bond stronger than the wood itself. If you glued two pieces of together with the appropriate clamping force and then tried to break the wood apart, you would find the wood split and not the glue joint. But this works best when the grains are glued in parallel. If you were to glue the end grains together, it would barely hold. This might be the case with the diagonals in Doc's ailerons. T-88 has no such requirement; it's just a powerful epoxy.
To Murray:
Material Properties of Sitka Spruce:
The compressive yield strength of properly dried Spruce is about 5,500 psi parallel to the grain, and a little over 300 psi perpendicular to the grain. I found similar numbers from numerous sources (US/UK/Canadian).
To put this into perspective, the maximum bending stress experienced by Lenard's wing when he loaded it with 1,000 lbs of bricks was, 3,100 psi compression and 2,325 psi tension on the front spar top and bottom respectively. The rear spar top/bottom was 1,950 psi compression/ 2,040 psi tension. Normal flying loads are much less. I ran these numbers several times and always get nearly the same results.
P.S. If there are any engineers reading this, I wouldn't mind a second look at these numbers. I can explain the formulas and assumptions.
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docmatt
doc matt
2015 Donor
Posts: 56
Total likes: 4
DE E-04 under construction
OS:
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Browser:
MS Internet Explorer 9.0
Eagle Type: DE
Re: Aileron test
«
Reply #17 on:
May 15, 2015, 08:45:21 PM »
Vince, I think you are exactly right about my diagonals. It was a case of poor joint configuration. I looked again at how I did it today. When adding the nose plywood, I extended it about an inch over the cap strips to create a gusset at the caps and diagonals. Now it is confidence inspiring. Even more so, now that it looks like the diagonals may be unnecessary. ; )
Now if I could just get back to construction after my usual spring stall/spin...
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doc matt
DE E-04 under construction
Mineral Point, WI
Murray Randall
2015 Donor
Posts: 91
Total likes: 1
Murray
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Eagle Type: Cabin mod XL
Re: Aileron test
«
Reply #18 on:
May 16, 2015, 10:43:02 AM »
.
You're quoting some uncommonly low strength values from various sources, Vince.
Thats me trying to be polite which is most unnatural for me.
The easy way to determine the bending strength of a a piece of wood is to clamp lightly in bench vise measure up a distance, pull it to failure with a fish scale.
I did that and got 11,780 psi one direction and 8,330 psi the other.
There should be pic attached.
Then weigh a little piece on the digital scale.
Anyone can do this with a material they see in lumber yard.
I'd say the accuracy of my crude set up is 10%. If my explanation below is not clear to anyone, please pipe up. My point, if there is one, is: test YOUR wood, test YOUR glued joints .
The applicable equations for the size sample you use are:
I = b times h cubed divided by 12
I = moment of inertia in lbs/sq in
b = base dim in inches
h = height dim in inches
S = M times c divided by I
S = stress in lbs/sq inch
M = bending moment = dist from vice up to spring scale times the spring scale reading in lbs
c = half the distance from the compression side to the tension side of your piece
My sample of Stika Spruce from Aircraft Spruce was .5 by .262 failed with 1.125 lb in the weaker axis and 1.5 lb in the stronger axis, applied 6 in up. The fish scale doesn't show as such in the pic
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Cabin Mod to XL
Murray Randall
2015 Donor
Posts: 91
Total likes: 1
Murray
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Eagle Type: Cabin mod XL
Re: Aileron test
«
Reply #19 on:
May 16, 2015, 10:48:08 AM »
Damn Sorry I said my test piece failed at 1.125 lb and 1.5 lb That should 11.25 lb and 15.0 lb Murray
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Cabin Mod to XL
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