Spar Failure

Homebuilt Aircraft & Kit Plane Forum

Help Support Homebuilt Aircraft & Kit Plane Forum:

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
5,256
Location
NJ
This is real...Background: Phil in Ireland is building a Carbon Dragon and testing his spar had a failure.

http://www.ihpa.ie/carbon-dragon/index.php/forum/carbon-dragon/809-progress-on-phil-lardner-s-all-carbon-cd?start=60

He has decided to splice the center with Aluminum. This splice is similar to what I am doing with my Aluminum Dragon.

Now comes the question.:

When doing the stress analysis of a wing all the references show the moments for 1/2 the wing.

Is it possible the moment at the exact center is twice that.

Shown below is my "lineal FEA" (Grape)

sparcarrythrough1.jpg

The load on the wing is represented by 1919# at centroid of 110" shown is the whole wing.

The reaction at node 8 is 3838 (2x 1919) ....so the moment is 422180 in-lb at the exact center?
 

mcrae0104

Well-Known Member
HBA Supporter
Log Member
Joined
Oct 27, 2009
Messages
4,074
Location
KBJC
Is it possible the moment at the exact center is twice that.
No, the moment does not double at the middle. You could sketch a load, shear, and moment diagram to satisfy yourself that that is not the case. I'm not familiar with your software so I can't comment on that.
 

BBerson

Light Plane Philosopher
HBA Supporter
Joined
Dec 16, 2007
Messages
15,268
Location
Port Townsend WA
The moment isn't twice.
But a failed cap is serious business.
Aluminum bandaid is not a standard fix.
I don't know of any proper fix of a failed cap.
 

plncraze

Well-Known Member
HBA Supporter
Joined
May 11, 2006
Messages
2,113
One of the most critical aspects of testing and measuring is set up and understanding what you are looking for in the process of testing. When you look at Strojnik's books you will find pictures of the inboard root rib and the need for over-dimensioning the part due to high loads. When the wing spar is in the fuselage it rides in the fuselage slot and the root rib. Jim Marske throws a whole lot of extra layers of glass there to make himself feel better. I believe he doubles the skin thickness around the spar slot compared to the surrounding area of the fuselage. In the pictures of the Dragon spar test shown above in the link it seems the spar is lacking the support the fuselage would give in the real world application. The moment diagram of in the fuselage wings shows that the moment drops dramatically once inside the fuselage. Since the above test does not have a fixture to simulate this I am not sure if this is a valid test. Also since the original Dragon was wood it had different buckling characteristics than one build from thinner materials. The spar was not held rigidly enough to accurately simulate what would happen in flight.
 

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
5,256
Location
NJ
No, the moment does not double at the middle. You could sketch a load, shear, and moment diagram to satisfy yourself that that is not the case. I'm not familiar with your software so I can't comment on that.
Doing the shear and moment diagram the moments shifts from high positive to high negative at the exact center. ( I think)

When looking at the individual elements using my program the moment is 422180/2 or as is traditionally shown for 1/2 the wing.

One could use the picture I show as just a representation and ignore the program. The math is pretty simple, 1919 x 110 or 3838 x 110.
 

BBerson

Light Plane Philosopher
HBA Supporter
Joined
Dec 16, 2007
Messages
15,268
Location
Port Townsend WA
The horizontal root shear might be high when testing a single wing with a weak support setup.
But those two wings together the shear cancels to zero at center.
 

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
8,327
Location
Saline Michigan
To the OP's question, nope, the moment coming in from one wing and the moment from the other wing are NOT additive. See comments below on four-point bending, then go look up the topic. Have confidence...

Now as to the carbon spar failure on test, where to start? Several comments:

His text indicates he only tested at the failure load, with no lower loading cases, so we do not know how much the wing was capable of. Nor did he tell us how long it stood the one load case before it failed;

If the spar were to go into elastic instability between the inner ribs, we should all expect it to buckle and break at or near the centerline - the element under compression deforms and then the section at centerline moves off center and has the largest additional moments due to the deformation there at centerline. If the buckling is unconstrained, it will buckle past the failure strains of the materials and stuff will be broken;

The top cap is in compression in this scenario. When buckling is not considered and loading is symmetric, the entire beam between the supports sees the same bending moment and zero shear load. How is this possible? The simple case is called four point bending, you can look this up. Put a beam across two supports. Place equal loads at two spots an equal distance out from the fixed supports on each side. The beam sees shear between the load location and the support equal to the load, the shear is picked off at the support. Bending moment grows from where the load is applied until you reach the supports, and then it is constant across to the other support.This wing with two supports and equal shear and bending from the tips to the supports works the same way between the supports;

While the only failure seen was in the top cap, we do not know that other regions of the wing will not fail in future tests. The author did not report on how long the wing held the load prior to failure. For all we know, it broke while the jacks were still being unloaded, in which case, he never developed full load in the wing. Knowing the actual load on the spar when it broke is important to knowing how much it needs to be beefed up;

The fact that the cap buckled into deformation along the fore-aft direction indicates that the top portion of the spar did not have sufficient bending stiffness in the fore-aft direction. Bending stiffness (EI in the fore-aft direction) is the thing that allows something to carry a given column load without buckling. The correct beefup would be to add cap width from outboard of the first rib through the center section. How much to increase EI in that direction would be entirely a function of how much more load you want to carry than what took the spar into elastic instability. See Euler's rule on columns. Buckling load moves linearly with EI, and I (in this case) increases with the third power of cap longitudenal dimension - for instance, he could double the buckling load by making the cap 26% wider, triple its buckling load by making it 44% wider;

His beefup of the spar with plates bolted to the shear web seems like the heavy way to do it, but if enough material is used, he can increase EI enough to carry the intended load. I am NOT saying that his proposed change will or will not be adequate. Right now I think he is guessing, and I HATE guessing, particularly when guessing is so easily replaced with a little math...

His assumption that he only needs to repair the area where the spar collapsed may be in error. The innermost ribs attached to a very firm structure, and the buckled spar may have locally overloaded the ribs, caps, and web around those mounts. They need to be very carefully examined for delamination and other issues as large deformations definitely occurred near these ribs while the spar was buckling;

Adding the plates by bolting along the web center may cause issues. Web fabric will be cut at each of the illustrated holes, the effective cross section of the web will be reduced by these holes, with the web shear strength being reduced in proportion to remaining uncut fibers in a section taken through the holes. The spar will also see an abrupt change in stiffness at the ends of the plates, just where the spar also sees a a change from zero shear to the largest shear carried in the wing. In general, I do not like this mod;

His addition of aluminum in contact with carbon-epoxy composite is particularly ill-advised and scary. Unless the aluminum plate and carbon fiber are electrically insulated from each other, galvanic corrosion can be expected. If they are electrically insulated from each other, I have to question if they will elastically deform with each other, reinforcing each other against buckling. If they are bonded with epoxy to connect them together, the aluminum will eventually separate from the epoxy due to corrosion and again, the buckling strength will drop when that happens. Other home shop applicable adhesive methods have similar problems in contact with aluminum alloys.

In total, I think that you brother in arms on this project should rethink his spar repair, and consider a new spar beefed up appropriately.

Billski
 

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
5,256
Location
NJ
At McDonnel (now Boeing) they insulated CF to aluminum with a layer of Glass Cloth on the AV8-B
 

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
5,256
Location
NJ
The horizontal root shear might be high when testing a single wing with a weak support setup.
But those two wings together the shear cancels to zero at center.
I love it when others say things I am afraid to say. I think we both know the shear and moment is not zero there (or it would not have broken) although looking (perhaps incorrectly) at pieces of the theory can suggest that.
 

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
5,256
Location
NJ
To the OP's question, nope, the moment coming in from one wing and the moment from the other wing are NOT additive. See comments below on four-point bending, then go look up the topic. Have confidence...

Billski
I have problems getting my head around this one...(no confidence in "my" conclusions). I will assume also if the more simple case is examined as diagrammed (not 4 point bending, not existing failure), the moments still do not add.

Just for reference sake (which is where I have my problem as regards logic) when we examine a wing joined at the center with a single bolt in top and bottom caps, ....the cap thickness is double at that point where the two wings are pinned together.
 
Last edited:

Autodidact

Well-Known Member
Joined
Oct 21, 2009
Messages
4,513
Location
Oklahoma
Compare the failed spar with this double shear spar connection of the original design which would have much greater resistance to lateral buckling than what the failed spar appears to have:
dblshr.jpgIMG_0110.jpg

I think we both know the shear and moment is not zero there (or it would not have broken)
This is not correct; the moment is not zero between the shear fittings (they attach the wing to the fuselage), but the shear most certainly is zero in that region - there is no air load there and no attachment to the fuselage either, therefore there is no shear, and it is called pure bending and happens when you apply only a moment to the ends of a beam segment. The failure is lateral buckling and is caused by compression loads only.
 

mcrae0104

Well-Known Member
HBA Supporter
Log Member
Joined
Oct 27, 2009
Messages
4,074
Location
KBJC
Doing the shear and moment diagram the moments shifts from high positive to high negative at the exact center. ( I think)
That's what the shear will do (and actually since the attach fittings are not both at the center, the shear will be zero between them).

For the moment diagram, it will increase from zero at one wingtip up to its maximum at the middle, then fall to zero at the other wingtip.

Just a thought--maybe it would be a good idea to brush up on beam theory and then you'll know if your program is telling you the truth. Peery and Bruhn each have chapters on this (I'm citing them because those are the two I have on hand) but I bet there's some stuff on YouTube that's more basic.

I don't see much of a need for FEA on a light plane, but maybe I'm just a Luddite. Use the tool you like, just be sure you know what it's telling you. You'll get there!
 

BBerson

Light Plane Philosopher
HBA Supporter
Joined
Dec 16, 2007
Messages
15,268
Location
Port Townsend WA
I love it when others say things I am afraid to say. I think we both know the shear and moment is not zero there (or it would not have broken) although looking (perhaps incorrectly) at pieces of the theory can suggest that.
This graph isn't exact because wings are never supported at tips only, but shows how horizontal shear goes to zero at center.
Bending moment (compression in upper cap) is of course maximum.
image.jpg
Don't forget about vertical shear also.
 

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
5,256
Location
NJ
it is called pure bending and happens when you apply only a moment to the ends of a beam segment. The failure is lateral buckling and is caused by compression loads only.
I hate being stupid, I have the books, I have not spent hours and hours in a statics class doing exercises as should properly have been done. Your posts illustrate the double up of the caps at the center I was referring to......Now that could be because there is a big old bolt in the center and the Lugs needed the extra for tear-out/failure on bearing.

View attachment Peery Wing Structures.pdf

This shows 1/2 the wing, where the 2 wing halves join.....there is double the load......what am I missing?
 
Last edited:

Autodidact

Well-Known Member
Joined
Oct 21, 2009
Messages
4,513
Location
Oklahoma
You're not stupid, but I am impatient and that's a fault I have. I did not spend hours in a statics class either, I just bought the books and read them (much more efficient than driving b&f to a class as long as you don't crave a degree), and there is no substitute for doing that IMO. There are as many subtle differences in the analysis of structural configurations as there are different configurations, but there is a set of basic principles that can be applied to them all and they must be learned or the task is impossible without prohibitive amounts of trial and error.

The bolt hole is exactly the reason for the extra cap depth; it does need the bearing strength, but more importantly cutting a section of the cap at the bolt hole reveals the reduced section area; since the diameter of the bolt is somewhat similar to the depth of the spar cap (a coincidence, more or less, for example an aluminum bolt would have required a much larger hole), the cap depth needs to be increased to regain the section area. The taper down to the cap section w/o holes is to avoid a stress riser.
 

Autodidact

Well-Known Member
Joined
Oct 21, 2009
Messages
4,513
Location
Oklahoma
View attachment 63227

This shows 1/2 the wing, where the 2 wing halves join.....there is double the load......what am I missing?
distrib.jpg

That's 10lb per inch, IOW there is 10 lb acting on each inch if the wing span, not counting the triangular part of the load distribution, so for that rectangular portion of the load distribution the total load (with its centroid at the mid-span, incidentally) is (10lb/inch)*100inch=1000lb. The triangular portion starts at 0lb per inch and increases linearly to 10lb/inch. Since the area of a triangle is half the area of a rectangle of same height/width, the load from the triangular portion of the load distribution is only half as much as the rectangle, ie, 500lb and its centroid is, IIRC from geometry, 1/3 of the way from the root to the tip. So the total load on that half wing is actually 1500lb. To really understand the load distribution formula requires using calculus:

∫(10+0.1x)dx = ∫10dx+∫0.1xdx

Integral of dx is x so ∫10dx=10x and the integral of xdx is 1/2x^2, so

∫10dx+∫0.1xdx = 10x+0.05x^2 = 10(100)+0.05(100^2) = 1000+500 = 1500.

This is just an idealized representation meant to illustrate the method of creating these diagrams, and the real load distribution is of course elliptical. An exact method would use Calculus, but a close and conservative approximation can be done by slicing up the elliptical distribution into a number of pieces (the more pieces, the more accurate) and you can then use each slice to calculate the moment of that slice of load distribution by multiplying it times its distance from the root, add all of those moments together, and that gives you the total moment at the root.

What you're missing is that you are not reading the book. The text explains exactly how to do these things and it just requires patience. Now, if you have difficulty reading, I can understand not being able to easily (it's never easy) master these subjects, but you don't appear to have any trouble at all writing things and that leads me to believe that you have no trouble reading either, you just don't want to, and neither do I, but it is just the only way. There is no other way. I don't mean to be overly critical of you, but I see someone who could do this and I think you are missing a critical step and it could get you hurt - sorry!:ermm:
 
Last edited:

mcrae0104

Well-Known Member
HBA Supporter
Log Member
Joined
Oct 27, 2009
Messages
4,074
Location
KBJC
You beat me to it, Mad MAC!

diagrams.jpg
shear.jpg
moment.jpg

PropP, I think the usual presentation of only one wing might be unhelpful. It might lead one to think there's X in-lb at the left root and X in-lb at the right root, therefore there's 2X in-lb in the middle, but that's not right. Hopefully the moment diagram of the full spar is helpful in visualizing the way the moment builds up from the wingtip to a maximum value, and then falls off back to zero at the other wingtip.
 
Top