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Modelling buckling of long thin carbon fibre tubes

  1. Oct 1, 2013 #1
    Hi all, I´m a researcher in computational biology thinking on topics quite far from my field.

    I want to model long thin-walled tubes of dimensions like those treated here, for example: ≈2 m long, radius ≈50mm and thickness ≈1mm. The tubes I´m interested in should be made of pulltruded/pullwinded carbon fibre. The material properties would be like those shown here:
    http://www.cstsales.com/rod_comp.html
    http://carbonfibretubes.co.uk/technology.html
    http://www.exelcomposites.com/Englis...oundTubes.aspx [Broken]

    Unfortunately, after having a quick look at the excellent documents suggested by AlephZero and Enthalpy in a related thread (https://www.physicsforums.com/showthread.php?t=662216), the formulas given therein seem either appropriate for isotropic materials, like those found in Timoshenko's theory for shell buckling, or too complex, like the NASA design codes for composites.

    1. Is there any simple way to model the local buckling of thin-walled tubes made of unidirectional pulltruded/pullwinded carbon fibres?

    2. If possible, what Poisson coefficient (μ) would be used in the formula σ=(e/R)*E/(3(1-μ2))0.5 for this kind of composites?
    (where "σ" is the stress, "e" thickness, "R" radius, and "E" Young´s modulus)

    Thanks a lot in advance!
    Any help would be really appreciated!
     
    Last edited by a moderator: May 6, 2017
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  3. Oct 9, 2013 #2

    nvn

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    Vigardo: I would like to change the name of your parameter e to t, so it would not be confused with the Napierian constant e = 2.7183.

    Even though your pultruded round tube lay-up contains quite a few longitudinal rovings, it also contains cloth, to increase the tube transverse strength. Therefore, for the purposes of your current analysis, you could assume the laminate properties are quasi-isotropic, unless you find more exact information.

    You can use nu = 0.35, in the tube lengthwise (axial) direction.

    You aptly mentioned local buckling. Your round tube thinness ratio, d/t, is rather high, d/t = 100, where d = tube outside diameter (OD) = 100 mm, and t = tube wall thickness = 1.00 mm. Therefore, you indeed would want to check local buckling (as well as global buckling).

    To compute local buckling critical stress, I currently do not know of a better formula, for pultruded round tubes, than the one you listed in post 1 for a thin-walled cylinder, provided you modify the formula, as follows.

    sigma_cr2 = (k2*E*t/d0)/(1 - nu^2)^0.5, where d0 = thin-walled cylinder mean diameter = d - t, k2 = 0.0580 + 0.522*exp(-beta), beta = 0.0442(d0/t)^0.5, and exp(x) means e^x.

    Keep in mind, you need to use the laminate properties, which are quite different from the fiber properties. For your composite tube, I currently would say, use the following pultruded round tube laminate properties, unless you or anyone can find more accurate laminate properties for your specific tube and lay-up.

    E = compressive modulus of elasticity in the lengthwise (axial) direction = 20 700 MPa; Scu = compressive ultimate strength in the lengthwise direction = 205 MPa at room temperature (25 C).

    Even though I listed Scu, above, it turns out you do not need it, for your particular tube.

    Using the above information, you can now compute the local buckling critical stress, sigma_cr2.

    Next, you need to compute the global buckling critical stress, sigma_cr1. Using the above data, it turns out, your current tube is in the long column range. Therefore, the Euler global buckling formula, for a simply-supported column, applies. Hence, the global buckling critical stress is as follows.

    sigma_cr1 = (E*I/A)(pi/L)^2, where A = 0.25*pi*[d^2 - (d - 2*t)^2], I = (pi/64)[d^4 - (d - 2*t)^4], L = column effective length = 2000 mm, and pi = 3.1416.

    Next, compute the buckling critical stress, sigma_cr = min(sigma_cr1, sigma_cr2).

    sigma_cr, above, applies to room temperature (25 C). If your tube is instead at 40 C, multiply sigma_cr by 0.85. If your tube is instead at 55 C, multiply sigma_cr by 0.70.

    Next, compute the tube allowable axial compressive stress, sigma_allow = sigma_cr/FSu, where FSu = ultimate factor of safety = 2.0.

    Therefore, the tube allowable axial compressive load is P_allow = sigma_allow*A.

    It would be good if you could test a few columns, to see if your test results correspond to these analysis results.
     
    Last edited: Oct 9, 2013
  4. Oct 9, 2013 #3
    First, thank you for your great and detailed response!

    Before continuing, I want to share two relevant documents I found after posting:

    1. A recent NASA technical report about "Simple Formulas and Results for Buckling- Resistance and Stiffness Design of Compression-Loaded Laminated-Composite Cylinders" Nemeth and Mikulas (2009) http://shellbuckling.com/papers/classicNASAReports/2009NASA-TP-2009-215778.pdf
    (found in a web about shell/tube-buckling which contains more relevant bibliography:
    http://shellbuckling.com/bibliography.php)

    2. A research paper where the author determines, using quite simple load vs. shape factor diagrams, the predicted failure mechanism for advanced composite tubes: Weaver (2000) http://www.sciencedirect.com/science/article/pii/S1359836800000299

    The pultruded round tubes I´m talking about are mainly unidirectional (95%, see: https://www.acpsales.com/Carbon-Tubes.html), so I think we can´t assume they are quasi-isotropic. However, in the paper Weaver (2000) it is said that a quasi-isotropic ply-layout is the most efficient preventing local buckling under compressive loads. Perhaps, a quasi-isotropic layout may be what I need since, as you noted, such thin-walled cylinders are likely locally buckling.

    Are so temperature dependant these critical loads?

    To check the different formulas (Timoshenko, NASA tech. report by Nemeth & Mikulas (2009), and yours), I computed the critical stresses (σ) using a cylinder of 1m, r=50cm and t=1mm under axial load. For the material properties I employed those used as example in NASA´s technical report by Nemeth & Mikulas (2009): E1=137.9 GPa, E2= 15.9 GPa, G12=4.1 GPa and v12= 0.3.

    Table 1:
    ---------------------------------
    σ[GPa] Formula
    ---------------------------------
    1.67 Timosheko
    0.24 NASA-report (eq.41a)*
    0.56 nvn
    1.70 Euler-buckling**
    ---------------------------------
    *I used equation 41a for a homogeneous orthotropic material, and v21 = v12*E2/E1 (http://en.wikipedia.org/wiki/Poisson's_ratio)
    **both ends pinned, just for checking

    According to Table 1, and if I didn´t make any mistake, it seems that the critical stress computed with your fomula (nvn) is the half of that obtained with Timoshenko´s theory for isotropic tubes and twice than that computed considering the composite as an homogeneous orthotropic material (NASA´s report).

    I´m a bit confused... Am I doing something wrong? Or critical stress predictions are so indeterminated for advanced composites tubes?

    By the way, do you remember where you found your fomula?

    Thanks a lot four your time and help!
     
  5. Oct 9, 2013 #4

    nvn

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    Just to ensure I understand, here you are saying, for this calculation, you used a cylinder of dimensions L = 1000 mm, r = 500 mm, and t = 1 mm, right? And is r mean radius, or outside radius?

    By the way, always leave a space between a numeric value and its following unit symbol. E.g., 1 mm, not 1mm. See the international standard for writing units (ISO 31-0).
     
    Last edited: Oct 9, 2013
  6. Oct 10, 2013 #5
    I'm really sorry nvn, I meant r = 50 mm instead of r = 50 cm. All calculations were made with 5 cm radius, 1 m length and 1 mm thickness. Yesterday I was really tired... Should I edit my post to fix it?
     
  7. Oct 10, 2013 #6

    nvn

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    Vigardo: Thanks for the clarification regarding mean (wall midsurface) radius r = 50.0 mm.

    In table 1 of post 3, I was able to match your 1.67 GPa sigma value for Timoshenko, and your 0.24 GPa for Nemeth (2009).

    In Nemeth (2009), which equation number did you use to compute ncr? And which equation number did you use to compute epsilon?

    I could not yet match your 0.56 GPa for nvn. I instead got 0.57 GPa (569 MPa). Did you perhaps make a typographic mistake? (And the nvn value is currently one third of the Timoshenko value, not one half.)

    I could not yet match your 1.70 GPa for Euler buckling. Do you mean you used my Euler buckling formula in post 2? If so, did you make a mistake?

    My formula in post 2 is just the Timoshenko formula, except algebraically simplified, and then multiplied by equation (eq.) 50 in Nemeth (2009), except I currently multiplied eq. 50 by an arbitrary additional strength reduction factor of rf1 = ~0.58. In other words, my k2 is k2 = rf1*gamma, where gamma is eq. 50.
     
  8. Oct 10, 2013 #7
    I really appreciate the high attention you´ve paid to my numbers and I´m happy that our results are very close. Thanks a lot again!

    Just to clarify. For me, "r" is the outher radius. If you want to use the mean (wall midsurface) radius I recomend to name it "r0" (following your nomenclature for "d0"). However, it´s worth noting that both, r and r0 (or d and d0), are going to be very similar in the very thin walled tubes we´re talking about here. In any case, I used d0 in your local buckling formula, as you required.

    I used eq. 31 to compute ncr and eq. 34b for ε. The β, μ and ρ values required by eq. 31 were computed according to eqs. 37, 38 and 39, respectively.


    The value I see today in my excel file is 0.57569244 GPa, and yes, Timoshenko value is one third... Yesterday I was really tired, sorry again.

    Sorry, I did not use yours, I used the eq. 49b from Nemeth (2009) instead. I know that yours can be more accurate, but since our tubes are very thin I considered that the approximations made in eq 49b were accurate enougth to obtain valid results. If my numbers are not wrong we should obtain very similar results, shouldn´t we?

    Thanks to point out the reduction in loadcarrying capacity associated with the presence imperfections, it is something that I was not considering yet. In Nemeth (2009) they say that gamma factor is used to account for initial geometric imperfections (according to some NASA code). Should I understand that your "rf1" factor is intended to account for some other kind of imperfections related with the fabrication process of carbon fibre tubes or it is just a more conservative factor? How "arbitrary" the reduction factor is?


    In any case, I´m still trying to figure out what is going on with the low critical stress obtained with eq. 47 from Nemeth (2009). May this low value be related to the fact that I´m considering the tube as mainly unidirectional (axially oriented)?

    In my design structural efficiency is critical (in terms of weight). Perhaps unidirectional pultruded tubes are not those what I need given they seem to have quite low local buckling stress when they are very thin. I asked about pultruded tubes because I thought that they were going to be the most efficient choice, but now it seems that a more quasi-isotropic layout may be more appropriate. What do you think/recomend about this?
     
  9. Oct 11, 2013 #8

    nvn

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    (1) I thought R in eq. 49b means mean radius (r0 = 49.5 mm), instead of outside radius. Therefore, I currently got 1.67 GPa, instead of your 1.70 GPa value in table 1 of post 3. If I use outside radius (r = 50.0 mm), then I get 1.70 GPa. And, should eq. 49a be used, instead of eq. 49b?

    However, if Ex for the laminate is Ex = 76.9 GPa, as explained in item 5 below, then one would expect the Euler buckling stress would be approximately 0.5*Ex*(pi*r0/L)^2 = 930 MPa (?). But eq. 49a instead gives 1081 MPa, which is much higher. Why would an orthotropic laminate having Ex = Ey = 76.9 GPa give a much higher Euler buckling stress than an isotropic material having E = 76.9 GPa? Hence, eq. 49a is not making sense to me, yet.

    (2) I am not familiar with eq. 47. And eq. 47 is not critical stress. I am not yet understanding what you mean here, nor how eq. 47 is used. Could you explain?

    (3) Aside from my above question, I am struggling to understand the significance of n = 0, and why (or when, or if) we should use n = 0. Could you shed any light on this?

    (4) E1 = 137.9 GPa, E2 = 15.9 GPa, etc., on p. 18 in Nemeth appears to be the tensile properties of a single ply of unidirectional tape (perhaps Hercules AS or AS1 carbon fibre unidirectional tape?), with perhaps something like 3501 or 3501-6 epoxy resin. For this Nemeth data set, his E_bar = 54.4 GPa, and nu_bar = 0.34, on p. 20 are correct.

    I would tend to think this is a higher-grade ply, perhaps used in aerospace, whereas plies for commercial or pultruded products might not be this stiff (?).

    Secondly, I currently do not think you can get away with using 100 % longitudinal (0 deg) unidirectional tape plies in your tube. I would tend to think 100 % longitudinal unidirectional-tape ply tubes might be acceptable for only tensile force, but not bending moment, shear force, nor compressive force (?). Of course, I could be wrong.

    Therefore, I think you might need, as a bare minimum, at least a [0, 90, 0, 90, 0] unidirectional tape laminate for your tube. If we use the Nemeth data set for this cross-ply laminate, and use the rule-of-mixtures, then I think we would get, for the laminate, Ex = 0.60*137.9 + 0.40*15.9 = 89.1 GPa (?), in the tube axial direction.

    However, one thing they seem to fail to mention in most references is, E1 = 137.9 GPa and E2 = 15.9 GPa (in Nemeth and other sources) are tensile moduli of elasticity. The authors or companies are often failing to mention, or characterize, the compressive modulus of elasticity (and compressive strength) of the laminate, which is very important. I think most polymer-matrix composite materials are strong in tension, but significantly weaker in compression.

    Therefore, what you need are realistic values for the compressive modulus, of the laminate (not a single ply), in the tube axial direction. And you need the laminate compressive strength in the tube axial direction. You are a good researcher; therefore, perhaps you can find some credible data for these values.

    We see that as you approach a quasi-isotropic lay-up of the above unidirectional tape, the tensile modulus approaches E_bar = 54.4 GPa. But compressive modulus would be significantly less. Plus, the Nemeth ply is probably an aerospace-grade ply. Furthermore, commercial tubes might not be 100 % unidirectional tape, especially pultruded tubes. Therefore, using E1 = 137.9 GPa, or even 89.1 GPa, which are tensile moduli, and aerospace grade, instead of compressive properties, for your calculations, is probably completely unrealistic.

    I am not nearly convinced that we have credible compressive moduli (and compressive strength) properties for any applicable tubes yet.

    (5) Furthermore, I am not yet quite understanding how Nemeth takes into account the lay-up of the laminate. Nemeth does not contain a clear description of what he means by "homogenous, specifically orthotropic laminate." Do you think he means the following? Fifty percent of the unidirectional tape plies are oriented at 0 deg; i.e., along the tube axial (x) direction. And 50 % of the unidirectional tape plies are oriented at 90 deg.

    If so, then it sounds like the laminate tensile moduli would be, Ex = Ey = 0.50*137.9 + 0.50*15.9 = 76.9 GPa, right?

    The above is for the main body of the document. Then he briefly mentions theta, in figure 2, in eqs. 51 and 52. Therefore, it seems the main body of the document uses theta = 0 deg, right? How to use theta > 0 deg is not clear to me, yet.
     
    Last edited: Oct 11, 2013
  10. Oct 11, 2013 #9
    Acording to figure 2 and symbols legend from Nemeth (2009), "R" seems to be the inner radius. And yes, now I think eq.49a should be used instead of 49b because the authors say that it includes the transverse shear flexibility. I´ll include this quantity in the table later...

    I´m not now at my computer to repeat this calculations, but it seems not possible. Since you´re dividing by a number bigger than 1, it should be smaller. Did you noticed the small "-1" at the top right side of the brackets in eq. 49a? (it is very tiny).

    I typed 47 instead of 41a, the formula used to compute the local buckling stress, sorry.


    The n = 0 refers to the axisymmetric local buckling mode, while n ≠ 0 refers to all the asymmetric ones (read the paragraph between eqs. 12 and 13 in page 8).
    To the best of my knowledge, I think that the authors are just considering all posibilities. To determine the local buckling loads, they first assume some buckling mode shapes (eq.12) and then compute the critical loads associated to them. The minimum value will be the buckling load. To find such value, they assume that eq. 24 is diferentiable and, after some change of variable (λ), they obtain the minimum value of "n" directly (eq.31). However, for some reason I don´t know yet, when n = 0 (symmetric mode) it should be treated different than when n ≠ 0 (asymmetric modes), leading to different formulas: 32b and 32a, respectively. In any case, the authors say in the paragraph after eq.32c that when β>μ that the n = 0 solution should be the lower one. So, depending on the relative value of β and μ the final formulas for the critical stress to be used here should be 41a or 41b. I did not check eq.41b yet, I´ll do it when I get home. Thanks a lot to highlight this point, I did not notice this.

    I´ll comment as soon as possible the rest of your very interesting post. Thanks again!
     
  11. Oct 11, 2013 #10
    I´ll continue my last post here...

    For the material properties taken from Nemeth (2009), β = 0.277 and μ = 5.55, so the critical stress should be computed with eq.41a (as I did) and it corresponds to the asymmetrical local buckling mode (n ≈ 5, in this case, computed using eq.31). I´ll update the Table 1 here and include the results for the modified Euler buckling formula (eq.49a):

    Table 1 (updated)
    ----------------------------------------
    σ[GPa] Formula
    ----------------------------------------
    1.67 Timosheko (eq.42b)
    0.24 NASA-report asymmetric (eq.41a)*
    0.54 NASA-report axisymmetric (eq.41b)*
    0.56 nvn
    1.70 Euler-buckling** (eq.49b)
    1.10 Euler-buckling** + shear (eq.49a)
    -----------------------------------------
    *v21 = v12*E2/E1 (http://en.wikipedia.org/wiki/Poisson's_ratio)
    **Both ends pinned

    Surprisingly, now your formula for local buckling seems very close to the axisymmetric result from NASA´s report.

    Yes, it´s hard, but I did! Some of the links below have both kinds of properties.

    I don´t know how "high-grade" are those plies, nor how expensive they are, but I´ve found may similar or even higher specifications in internet. Here there are some examples (including compresive strength, modulus, Poisson ratio, etc):
    - Composites with Tensile modulus = 240 GPa and compressive strength = 1000 MPa (see last page) http://www.dpp-pultrusion.com/wp-content/uploads/2012/12/DPP-specificationsheet_feb_2013.pdf [Broken]
    - http://www.performance-composites.com/carbonfibre/mechanicalproperties_2.asp
    - http://www.goodfellow.com/E/Carbon-Epoxy-Composite.html
    Do you think I should not trust these properties? Are not commercial products?

    Do the links above convince you?

    Now I think you´re right. Thin walled tubes made from 100% 0º unidirectional tape plies seem not to be as efficient as other "more quasi-isotropic" layups. The former layups, perhaps, are efficient under compressive forces when they are designed as solid and relatively short bars, or if they are thick tubes. However I´m not sure about this, perhaps someone else that is familiar with this kind of advanced composites can illuminate us...

    I´ll try to repeat the computations to obtain the bar-like properties from the single ply ones later, it´s very interesting for my project. I need to think about these two later points carefully during weekend.

    Thank´s again for this very interesting discussion and have a nice weekend!
     
    Last edited by a moderator: May 6, 2017
  12. Oct 11, 2013 #11

    nvn

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    Vigardo: All stress values in your updated table 1 in post 10 look correct. "Cylinder radius" in Nemeth is mean radius (r0). Figure 2 does not make this clear, but other publications treating thin-walled cylinders make it clear that "cylinder radius" is mean radius (r0), only for thin-walled cylinders.

    No, I am still not convinced regarding the tube laminate properties. We have virtually no laminate compressive moduli for specific carbon fibre tubes you want to purchase or use for analysis. The only link, so far, that sounds semi-credible is your "performance-composites" link, perhaps under the "Std CF fabric" heading; it might be usable ... but it has no compressive modulus listed.

    Keep in mind, Nemeth eq. 49a cannot exceed the tube material axial compressive strength, or bearing strength; otherwise, the tube will rupture (fail). But instead of using the published compressive strength, you might take the minimum axial compressive strength of the composite tube, then multiply it by 0.90, and say this is its axial compressive strength, Scu. Therefore, the global buckling critical stress would then be, sigma_cr1 = min(Scu, sigma_49a), where sigma_49a is the eq. 49a stress. The local buckling critical stress, sigma_cr2, would be the eq. 41a (or 41b) stress. Then, as explained in post 2, compute sigma_cr = min(sigma_cr1, sigma_cr2). Finally, the tube allowable axial compressive stress would be, sigma_allow = sigma_cr/FSu, where FSu = ultimate factor of safety = 2.0.
     
    Last edited: Oct 11, 2013
  13. Oct 12, 2013 #12

    nvn

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    Vigardo: Corrections to post 11, paragraph 3: Scu = tube laminate axial compressive ultimate strength. Compute the global buckling critical stress, sigma_cr1 = sigma_49a. If sigma_cr1 > 0.5*Scu, then compute sigma_cr1 = Scu*[1 - 0.25(Scu/sigma_49a)]. This latter equation is merely where I am drawing a smooth, parabolic arc to transition from sigma_49a up to Scu, instead of making an erroneous, sharp transition at Scu. Next, compute the local buckling critical stress, sigma_cr2 = gamma_50*sigma_41a (or 41b), where gamma_50 is eq. 50. Then, as explained in post 2, compute sigma_cr = min(sigma_cr1, sigma_cr2). Finally, compute the tube allowable axial compressive stress, sigma_allow = sigma_cr/FSu, where FSu = 2.0.

    For Scu in paragraph 1 above, unless you find something better, we currently perhaps could use "performance-composites, Std CF fabric"; therefore, we currently could assume Scu = 570 MPa (?). But no axial compressive modulus is listed. I vaguely seem to recall, carbon fibre reinforced polymers are significantly less stiff in compression than tension (?), perhaps anywhere from 75 to 90 % of the tensile modulus. Therefore, perhaps currently assume the tube axial compressive modulus is Ec1 = 0.85*E1 = 0.85*70.0 = 60.0 GPa, where E1 is E1 from "performance-composites, Std CF fabric." But continue to try to find actual axial compressive modulus Ec1 (and compressive ultimate strength Scu) test results for actual carbon fibre tubes you want to purchase or use, in case Ec1 or Scu is lower than I assumed here. I am very skeptical about using any Ec1 or Scu value with no credible compression test results for the actual tubes and lay-up you want to purchase.
     
    Last edited by a moderator: Oct 15, 2013
  14. Oct 12, 2013 #13

    nvn

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    Vigardo: I now found some credible carbon fibre, epoxy, ply properties in some notes I have.

    (1) T300 carbon fibre, F934 epoxy resin, cloth (not unidirectional tape), ply thickness ho = 0.175 mm, E1 = 66.0 GPa, E2 = 66.0 GPa, G12 = 4.10 GPa, nu12 = 0.04, Stu1 = 375 MPa, Scu1 = 279 MPa, Stu2 = 368 MPa, Scu2 = 278 MPa.

    But if the cloth were instead something like Hercules AS1 carbon fibre cloth, with 3501 epoxy resin, which is slightly weaker than T300/F934, then I think E1 and E2 in item 1 would decrease to 0.932*66.0 = 61.5 GPa.

    If all plies in a laminate were item 1 plies at 0 deg, then the tensile (and compressive) moduli of the laminate would be Ex = Ey = 66.0 GPa. And the strength would be as shown in item 1. Or we perhaps could multiply each value by 0.932 for AS1 cloth, with 3501 epoxy.

    A quasi-isotropic laminate of item 1 plies would be E = 47.1 GPa, nu = 0.315. Or for AS1 cloth, 3501 epoxy, perhaps E = 43.9 GPa, nu = 0.315. All of these values assume room temperature.

    Cloth is slightly weaker than unidirectional tape. If you are sure your tube would consist of only unidirectional tape, and you do not want to use cloth values, let me know.

    Unlike E-glass/epoxy, I did not find evidence that the compressive modulus of carbon fibre reinforced polymers is less than tensile modulus -- although compressive ultimate strength (Scu) is sometimes less than tensile ultimate strength (Stu), but not always.
     
    Last edited: Oct 13, 2013
  15. Oct 13, 2013 #14

    nvn

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    Vigardo: Here are some additional, credible carbon fibre, epoxy, cloth ply properties. Subscript t means tensile, and c means compressive.

    (2) T300 carbon fibre, F934 epoxy resin, cloth (not unidirectional tape), ply thickness ho = 0.325 mm, E1 = 74.0 GPa, E2 = 74.0 GPa, G12 = 4.55 GPa, nu12 = 0.05, Stu1 = 499 MPa, Scu1 = 352 MPa, Stu2 = 458 MPa, Scu2 = 352 MPa.

    (3) CFS003 (Amoco T300) carbon fibre, LTM25 epoxy resin, cloth (not unidirectional tape), ply thickness ho = 0.230 mm, Et1 = 53.6 GPa, Ec1 = 47.2 GPa, Et2 = 55.2 GPa, Ec2 = 48.6 GPa, G12 = 2.85 GPa, nut12 = 0.042, nuc12 = 0.033, Stu1 = 618 MPa, Scu1 = 643 MPa, Stu2 = 652 MPa, Scu2 = 556 MPa.​

    If all plies in a laminate are oriented at 0 deg, the lowest property values we have seen, so far, for carbon fibre, epoxy, cloth, are, from item 3, Ec1 = 47.2 GPa, Ec2 = 48.6 GPa, G12 = 2.85 GPa, nuc12 = 0.033, and from item 1, Scu1 = 279 MPa (or from item 2, Scu1 = 352 MPa; or from item 3, Scu1 = 643 MPa; or from post 12, Scu1 = 570 MPa).

    A quasi-isotropic laminate of item 3 plies would be Ec = 34.1 GPa, nuc = 0.31.

    I currently do not know why Scu1 of items 1 and 2 is so much lower than Scu1 of item 3 or post 12.
     
    Last edited by a moderator: Oct 15, 2013
  16. Oct 15, 2013 #15
    Great!

    Ok I´ll take this into account. Thanks for clarifying.

    Then I prefer unidirectional tape.

    Thanks a lot for the material properties you´ve provided (items 1-3), it seems they are representative comercial carbon fibre composites. Unfortunately, I´ll need plies with higher modulus and strength specifications; even if they are very expensive or aerospace advanced composites. Just to give an idea of what I´m looking for, the quasi-isotropic properties of the final tube should be like Ec = 150 GPa and Scu = 1200 MPa. I think these properties would be attained using some kind of quasi-isotropic layout made from 0º, 45º and 90º plies of Ec1 ≈ 250 GPa and Scu ≈ 1600 MPa. Am I dreaming or this may be feasible?
    (At this moment I´m only working in paper, cost should not be a problem :-)

    Now I pay more atention to compressive properties. From what I´ve seen so far, I would say that the compressive E and Su for carbon fibre composites are ≈20% smaller their tensile counterparts. Please, check these links:
    http://www.hexcel.com/Resources/DataSheets/Brochure-Data-Sheets/Prepreg_Technology.pdf [Broken] (in page 28 there are some prepreg compressive properities)
    http://www.torayca.com/en/techref/index.html [Broken] (clicking in "compressive properties" link there is a comparative chart evidencing the differences between tensile and compressive strengths)

    As I said, I´ve been studying this weekend some key things about orthotropic materials... From what it is said in Nemeth 2009 and here (http://www.aeroway.ca/Laminatetheory.html), to obtain the tube properties from single ply ones it´s necessary:

    1. Obtain the "reduced stiffness constants" for a single ply: Q11, Q22, Q12 and Q66. (Q16 and Q26 can be neglected under the assumption of plane stress)

    2. Compute the "transformed reduced stiffness constants" (Q11', Q22', Q12' and Q66') using the eq.6. found in http://www.aeroway.ca/Laminatetheory.html (they name these constants with an horizontal line over the Q, but I do Q').

    3. Apply the rule of mixtures indicated in the eq.52 of Nemeth 2009.

    4. Introduce the "homogenized stiffness constants" (Qij) obtained in 3 into the eqs.53a-e to get the corresponding material properties of the tube.

    This way, for the layout you´ve proposed (50% of ud. 0º and 50% of ud. 90º plies) one gets:
    Ex = Ey = 77.4 GPa, Gxy = 4.1 GPa and nu_xy = nu_yx = 0.06; which is slightly different from appliying rule of mixtures directly (76.9 GPa).

    Althoug in this 0º/90º layout differences are small, when the angles are not orthogonal they are much bigger. To both, illustrate this and cross-validate the procedure, I compared the predicted buckling loads computed using Nemeth´s eq. 41a or 41b (and the knocdown factor from eq.50) with some experimental data (from Priyadarsini et al. 2012: http://www.worldscientific.com/doi/pdf/10.1142/S0219455412500289)

    Ply properties: See Table 2 from Priyadarsini et al. 2012.
    Tube dimensions: L = 0.4 m, R = 0.15 m, t = 1 mm, R/t = 150
    Layout: 50% 0º + 50% 45º
    Predicted tube properties: Ex = 83.3 GPa, Ey = 22.8 GPa, Gxy = 21.6 GPa, nu_xy = 0.73 and nu_yx = 0.20
    Predicted local buckling (axisymmetric mode): σ_pred = 0.0941 GPa
    Experimental stress at buckling: σ_exp = 0.082 GPa

    This means that the experimental value is only 15% lower than predicted!
    This is great, don´t you think? Please, let me know if I´m doing something wrong.
     
    Last edited by a moderator: May 6, 2017
  17. Oct 16, 2013 #16
    I´ve compared carefully the formulas I introduced in my spreadsheet (Excel) with those obtained from the web http://www.aeroway.ca/Laminatetheory.html (Laminatetheory). They look right. The only thing I found is that the first parenthesis in the formula for Q22' (eq.6) seems inconsistent. To fix this I just removed it. Is this correct?

    In the following I pasted the input and output figures copied directly from the spreadsheet:

    Ply material properties: E1 = 148.5, E2 = 9.9, G12 = 4.8, nu_12 = 0.286.

    Reduced stiffnesses: Q11 = 149.3100857, Q22 = 9.90373296, Q12 = 2.832467627, Q66 = 4.8
    (obtained using eqs.1-3 from Laminatetheory)

    Transformed reduced stiffnesses:
    0º ply: Q11' = 149.3100857, Q22' = 9.90373296, Q12' = 2.832467627, Q66' = 4.8
    45º ply: Q11' = 46.01968849, Q22' = 43.61968849, Q12' = 36.41968849, Q66' = 38.38722086

    Laminate reduced stiffnesses: Q11 = 97.66488711, Q22 = 26.76171072, Q12 = 19.62607806, Q66 = 21.59361043

    Laminate material properties: E1 = 83.27182593, E2 = 22.81778624, G12 = 21.59361043, nu_12 = 0.733364106, nu_21 = 0.200953266.

    I hope this helps to clarify, and again, thanks a lot for your cross-checking efforts and time!
     
  18. Oct 16, 2013 #17
    I did not see your new post while typing my answer...
    Ok, don´t worry, take your time (always very appreciated!).
     
  19. Oct 16, 2013 #18
    I did not see your new post while typing my answer...
    Ok, don´t worry, take your time (always very appreciated!).
     
  20. Oct 16, 2013 #19

    nvn

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    Vigardo: First, change your E2 = 9.9 GPa to E2 = 9.85 GPa, to match table 2.

    The Q22' equation you used appears to be wrong. You were correct to delete the extraneous, erroneous, leading parenthesis. Next, in the Q22' equation, change "Q66" to "2*Q66", without changing any other character in the equation. All other equations, besides Q22', are correct. Please definitely let me know if you disagree. (We need a third, independent verification, to prove or disprove this typographic mistake. Can you find a verification?)

    In addition to the above, I had a minor typographic mistake in a nu equation, which I now corrected.

    Using the table 2 ply properties (without rounding), the correct answer, for the tube laminate properties, for your stated lay-up, currently appears to be as follows.

    Ex = 83.89 GPa, Ey = 24.02 GPa, Gxy = 21.59 GPa, nu_xy = 0.7019, nu_yx = 0.2010.

    After our laminate properties match, then I later plan to start looking at the other parts of your post, as time permits. It might take me a few days thereafter to reply.
     
  21. Oct 17, 2013 #20

    nvn

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    Vigardo: I could not match your above sigma_pred value. Not only is your current Q22bar equation incorrect (as mentioned in post 19), but also it appears you are currently using an incorrect cylinder diameter. The document states the cylinder inside diameter is 300 mm. Therefore, the cylinder mean diameter is 301 mm, not 300 mm. Your cylinder currently appears to be too small.

    Using my laminate properties at the bottom of post 19, I currently got sigma_pred = 96.16 MPa (axisymmetric mode).
     
    Last edited: Oct 18, 2013
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