1. ## Wing Spar Calculation

-This is a procedure I learned about in the book: Composite Basics by Andrew Marshall for sizing a composite spar. It seems to give too light structures by a factor of 10 but their could be errors in my calculations. The units seem to work out but are these relatively simple calculations nonsense as I suspect?

-800lbs gross, designed for 10Gs, 22' span, 2.75' cord, .4125'=4.95" thickness

-weight center at 5' from wing root

-M=5' * 800lbs/2 * 10Gs = 20kft.lbs

-estimate thickness of cap to calculate average spar depth: about .017' or .204" for one side: .4125'-.017'=.3955'

-20kft.lbs / .3955' = 50.6klbs / 2 since each cap takes 1/2 load = 25.3klbs

-Rutan lay-up gives 48ksi in tension and 40ksi in compression

-layer of Rutan 7715UNI or 7725BID is .0075" thick

-multiply by 1 for root, by .56 for 3/4 from tip, .25 for 1/2, .06 for 1/4

-25.3klbs/48ksi=.53sq.in for lower cap at root/3" wide cap/.0075"/layer=24 layers; 14 at 3/4(from tip), 6 at 1/2; 2 layers at 1/4

-length from root to determine weight of lower spar cap = 2layer*11' + 4layers*5.5'+ 6layers*2.75' + 10layers*1.25' = 22'+22'+16.5'+12.5'=73'*12"/'* 3" * .0075"= 19.7cu.in * 1.08oz/cu.in=22.3oz for lower cap

-25.3/40=.63.sq.in for upper cap at root=29 layers; 16 layers at 3/4; 8 layers at 1/2; 2 layers at 1/4

-length from root to determine weight of top spar cap = 2layer*11' + 6layers*5.5'+ 8layers*2.75' + 13layers*1.25' = 22'+33'+22'+16.25'=93.25'*12"/'* 3" * .0075"= 25.2cu.in * 1.08oz/cu.in=27.2oz for upper cap

-shear stress (15ksi, about half tensile strength)

-4000lbs/15000psi=.267sq.in

-.267sq.in/4.95"(spar depth=total wing thickness since spar caps carry some shear)=.054"/.0075"/layer=7.2 layers so 8 total layers at root, 4layers at 3/4 from tip, 2 layers at 1/2

-length from root to determine weight of shear web = 2layer*11' + 2layers*2.75' + 4layers*1.25' = 22'+5.5'+5'=32.5'*12"/'* 3" * .0075"= 2.9cu.in * 1.08oz/cu.in=3.16oz for shear web

Brock

2. ## Re: Wing Spar Calculation

I have read Marshall, have engineering degrees, and prefer real engineering texts. The methods he uses are intuitive, but a little less than thorough.

The next thing is that we are all reluctant to do someone elses' engineering...

First error I see is that composites are designed with a factor of safety of 2.0 or more. Maybe you already included that, but it was not explicitly covered.

Next is that the shear web has to extend onto the flanges to tie things together, so that will add some weight.

Then come the big issue - when you make up a spar and apply shear and bending loads to it, the flanges lengthen and contract, and the shear web moves with it, as well as just carrying the shear load. So a little square of the shear web, not only tries to become a parrallelogram, you also superimpose deformation like a trapazoid on it. In airplane wing spars, you can get nearly as much deformation from the caps extending and compressing than from shear... And you have to enlarge both the caps and the web to make it the shear web carry the load...

The only way I know to do that is to set everything up in Excel, including the failure checks for the caps and web, and then iterate the number of plies until everything passes. Then do it again with a different number of cap plies, and find the combination with the lease cross sectional area. This is optimization. The rest of the analysis is in both Jones and in Tsai and Hahn. They are listed in the books part of the Composites FAQ's. Sorry, it takes studying, but I don't know any other way around composites.

Billski

3. ## Re: Wing Spar Calculation

Originally Posted by wsimpso1
I have read Marshall, have engineering degrees, and prefer real engineering texts. The methods he uses are intuitive, but a little less than thorough.

The next thing is that we are all reluctant to do someone elses' engineering...

First error I see is that composites are designed with a factor of safety of 2.0 or more. Maybe you already included that, but it was not explicitly covered.

Next is that the shear web has to extend onto the flanges to tie things together, so that will add some weight.

Then come the big issue - when you make up a spar and apply shear and bending loads to it, the flanges lengthen and contract, and the shear web moves with it, as well as just carrying the shear load. So a little square of the shear web, not only tries to become a parrallelogram, you also superimpose deformation like a trapazoid on it. In airplane wing spars, you can get nearly as much deformation from the caps extending and compressing than from shear... And you have to enlarge both the caps and the web to make it the shear web carry the load...

The only way I know to do that is to set everything up in Excel, including the failure checks for the caps and web, and then iterate the number of plies until everything passes. Then do it again with a different number of cap plies, and find the combination with the lease cross sectional area. This is optimization. The rest of the analysis is in both Jones and in Tsai and Hahn. They are listed in the books part of the Composites FAQ's. Sorry, it takes studying, but I don't know any other way around composites.

Billski
That was better than anything I have ever read in any engineering text. Compression in the spar cap is the short side of the parallelogram and tension is the long side of the parallelogram of the shear web. Impose that on the spar caps and iterate. Clear as polished glass. Thanks for that.

4. ## Re: Wing Spar Calculation

I have read Tsai & Hahn along with Jones and bruhne? Not a one of those books calculated loads, deformation, or resistance for me. Still uncertain if it is better to reread or destructively test..... Probably should do both. It's possible to fail @ either one.
Billiski, this is the second time I've read it stated here that composites are designed with an ignorance factor of 2. The fibers follow near linear elasticity and the resin properties are equally well known, in addition the lamina analysis methods have been around since I was born. Abbrieviated analysis is foolhardy at best, but why is the increased factor inherint?

Oh, and to be more specific, the methods you should use include the loss of potential strength when your wingskin panels start to buckle. To me it seems the lack of simplistic analysis for composites, and stressed skins has detured many of better sense than myself...

Anyone designing in composites should read about jointery. Otherwise your 10g structure could be held on by a 5g tab, or almost as bad you're weight could climb pointlessly high as you eyeball and overengineer every connection. The technical nature is no less than any of the other text, but it should at least give you an idea of no-nos.

Wonderous Mountain

5. ## Re: Wing Spar Calculation

If one understands the behavior of composites and has a good grasp of physical properties of the laminate in question, many conventional forms of analysis (such as those covered in Bruhn's work) are still applicable and can be used quite successfully for first cut analysis/design. And the more familiar one is with the material, the closer the results of conventional methods will be.

The design safety factor of 2.0 is a requirement of the FARs. It came into being due to the variability of material properties that was so common ten or twenty years ago, and still is when design and analysis is being done at an amateur level. I think it is that reason that it is still being recommended today, despite there being much better databases of properties available to the designer.

6. ## Re: Wing Spar Calculation

The reason I got to use the FoS of 2.0 with composites was that water infiltration will cost any structure a fair chunk of its strength over time. Yeah, the FAR's used 2.0 as a factor of ignorance, but hygroscopiscity and hygrothermal behaviour is the reason to keep it. Yeah, some places have less water than others, but we travel in these things too...

Billski

7. ## Re: Wing Spar Calculation

I was aware that water entered and left composites; However, it's ability to reduce composite/fiber strength over time wasn't expressed clearly in any of the classic text I found at Embry Riddle. At worst I heard it increased weight by a few percent and decreased strength by just over ten, and regained it once the composite had dried out. Thinking now that appeared an irrational view given waters role as a universal solvent. Although long term slow chemical reactions aren't understood well by many.

If things continue to go as they have for me I will have more understanding of women than composites-though I may always feel safer with the composites.

Best regards,

Wonderous Mountain

8. ## Re: Wing Spar Calculation

K, thanks for the information. I had forgotten to add the safety factor of 2. I guess I really need someone to do the engineering for me as I doubt even if I have a spreadsheet for the calculations that I would be able to perform/understand them. I do understand that it is too much to ask for someone to do the engineering for me (for free). I am also considering pultruded carbon rods:

-18Gs, 800lbs gross, AR=8, 22' span, 2.75' cord, .4125'=4.95" thickness
-weight center at 5' from root
-M=5' * 800lbs/2 * 18Gs = 36kft.lbs
-estimate thickness of cap to calculate average spar depth: about .017' or .204" for one side: .4125'-.017'=.3955'
-36kft.lbs / .3955' = 91klbs / 2 since each cap takes 1/2 load = 45.5klbs
-150ksi compression, 175ksi tension
-.07x.437 Graphlite solid rectangle = .0306sq.in
-lower cap: 45.5klbs/175ksi=.26sq.in /.0306sq.in/layer=8.5=9 layers at root, 5 at 3/4(from tip), 3 at 1/2; 1 layers at 1/4
-length from root to determine weight of top spar cap = 1layer*11' + 2layers*5.5'+ 2layers*2.75' + 4layers*1.25' = 11'+11'+5.5'+5'=32.5'*12"/'*.0306sq.in = 11.9cu.in
-upper cap: 45.5/150=.303sq.in at root=9.9=10 layers, 5.5=6 at 3/4; 2.5=3 layers at 1/2; 1 layers at 1/4
-length from root to determine weight of top spar cap = 1layer*11' + 2layers*5.5'+ 3layers*2.75' + 4layers*1.25' = 11'+11'+8.25'+5'=35.25'*12"/'*.0306sq.in = 12.9cu.in
-multiply by 1 for root, by .56 for 3/4 from tip, .25 for 1/2, .06 for 1/4
-shear stress (15ksi, about half tensile strength)
-7200lbs/15000psi=.48sq.in
-.48sq.in/4.95"(spar depth=total wing thickness since spar caps carry some shear)=.097"/.0075"/layer=12.9, 14 total layers at root, 8 layer at 3/4 from tip, 4 layers at 1/2, 1 layer at 1/4,
-length from root to determine weight shear web = 1layer*11' + 3layers*5.5'+ 4layers*2.75' + 6layers*1.25' = 11'+16.5'+11'+7.5'=46'*12"/' * 4.95" * .0075" = 20.5cu.in * 1.08oz/cu.in=22oz

Brock

9. ## Re: Wing Spar Calculation

Some of the numbers didn't sound right so I did a quick check - the axial load in each cap is 91,914 pounds - not half that. This is a simple load resolution of a couple where if you look at a simple free-body diagram you'll be able to see that the moment resolution is either done about the overall neutral axis or about the neutral axis of either cap. If you do it about the neutral axis then yes, the load gets divided by two due to the two caps but the arm from the neutral axis to the cap is half the height of the cross section (this assumes that both caps are identical in cross section and stiffness of course). Normally you resolve the moment of a couple about one of the reactants so you use the whole section height but in this case you do not divide by two. Either way the result is the same.

Then, for the pultruded uni, the property values are highly dependent on tooling and setup, the resin you use and of course the fibers themselves. In normal homebuilding where wet layup is used, I'd recommend not using property values above about 125ksi.

10. ## Re: Wing Spar Calculation

Using 20n*400#*60"/4.95"=~96,960

Eghh, just curious.

11. ## Re: Wing Spar Calculation

Thanks Orion for checking the number for me. Using half the axial load was my mistake, not the book I got the information from. I was working from too sparse notes rather than the original source and that seemed right to me. Once again I've come to the conclusion that I'm only wrong when I think.

In the above calculations calculating the moment seems to make sense to me, but dividing the moment by the spar depth to get the load on the spar caps is where the calculation seems to break down. I think one needs to use beam theory instead.

Brock

12. ## Re: Wing Spar Calculation

Originally Posted by durabol
In the above calculations calculating the moment seems to make sense to me, but dividing the moment by the spar depth to get the load on the spar caps is where the calculation seems to break down. I think one needs to use beam theory instead.
Depends on application and what you're after. If you're after an approximation of structure, weight distribution and material needs, then the simple methodology works fine. But keep in mind that in practical design, beam theory can be accurately and safely applied using simplified terms - after all, the simplicity of looking at spar caps as a moment couple is actually derived from beam theory as long as deflections are small and material properties are linear.

13. ## Re: Wing Spar Calculation

Originally Posted by orion
Depends on application and what you're after. If you're after an approximation of structure, weight distribution and material needs, then the simple methodology works fine. But keep in mind that in practical design, beam theory can be accurately and safely applied using simplified terms - after all, the simplicity of looking at spar caps as a moment couple is actually derived from beam theory as long as deflections are small and material properties are linear.
Okay, thanks once again.
Brock

14. ## Re: Wing Spar Calculation

-I've done some more number crunching, I derated the pultruded material to 100ksi in compression and 125ksi in tension, I designed the wing for 10Gs times a safety factor of 2. I still come out with spar caps that weigh about 2lbs each (8lbs total) which seem too low and too good to be true. I worry that the wing would be too flexible due to such a small amount of material. I can't find any errors in my calculations.
-10Gs x safety factor of 2=20Gs, 800lbs gross, AR=8, 22' span, 2.75' cord, .494'=5.94" thickness
-weight center at 5' from root
-M=5' * 800lbs/2 * 20Gs = 40kft.lbs
-estimate thickness of cap to calculate average spar depth: about .01' for one side: .494'-.01'=.484'
-
-40kft.lbs / .484' = 83klbs
-100ksi compression, 125ksi tension
-multiply by 1 for root, by .56 for 3/4 from tip, .25 for 1/2, .06 for 1/4
-
-rectangular carbon rod:
-.07x.437=.03059sq.in cross section area, 20.2lbs/1000'=.0202lbs/'=.202lbs/ft/12"/' =.00168lbs/in, .00168/.03059sq.in=.055lb/cu.in, .88oz/cu.in
-.092x.22=.02024sq.in cross section area, 13.3lbs/1000'=.0133lbs/'=.00111lb/in/.0224, .79oz/cu.in
-
-upper cap: 83klbs/100ksi=.83sq.in/layer/.0306sq.in=27.1=28 layers at root, 16 at 3/4(from tip), 7 at 1/2; 2 layers at 1/4
-length from root to determine weight of top spar cap = 2layer*11' + 5layers*5.5'+ 7layers*2.75' + 12layers*1.25' = 22'+27.5'+19.25'+15'=83.25'*12"/'*.0306sq.in = 30.6cu.in, 30.6*.88oz/cu.in=27oz
-
-lower cap: 83/125=.664sq.in at root=22 layers, 13 at 3/4; 6 layers at 1/2; 2 layers at 1/4
-length from root to determine weight of lower spar cap = 2layer*11' + 4layers*5.5'+ 7 layers*2.75' + 9layers*1.25' = 22'+22'+11.25'+15=70.25'*12"/'*.0306sq.in = 25.8cu.in, 25.8*.88=22.7oz

Brock

15. ## Re: Wing Spar Calculation

One of the nice things about using light and strong materials is simply that you don't need that much of them to satisfy strength. And it might satisfy stiffness since the modulus of the carbon pulltrusion will realistically be somewhere between 22 Msi and about 28 Msi. But if you go ahead and do the deflection calculations, the old rule of thumb is that on a composite cantilever wing of aspect ratio between about 6.5 and about 7.5, ideally you want to limit deflection to about 1" per G. Personally I try to shoot for less - maybe around .5" per G. These are old rules of thumb and quite frankly I can't tell you the source. I only know that I've been given the numbers many years ago from old-timers who have been in the light plane industry for much of their professional lives. But I've seen as much as about 1.5" per G with no significant issues.

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