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I'm building a plane, and one of the solutions I've thought of for connecting the wing spars to the fuselage involves connecting carbon fiber parts. As mentioned in this answer, I found this explanation for how to calculate the scarf ratio:

The scarf ratio is determined by the ratio of fiber tensile strength and the shear strength of the resin. If your fiber takes 3000 N/mm², the resin transmits 20 N/mm² and the fiber content is 60%, the ratio is 1:90, so a 1 mm thick composite layer would need to be joined over a length of 90 mm.

I have tried to find more detailed explanations, but have not yet succeeded. I am interested in knowing where this information comes from. I have at least two specific questions:

  1. does the answer depend on how the composite layer was prepared?
    For example, does the answer change if the composite layer is a pultruded carbon-fiber strip? The fiber ratio is similar to what is claimed above as ideal, but the curing temperature is higher.

  2. It is proposed above that the resin transmits 20N/mm^2 - this square mm, which is a surface, how is it oriented?
    If this was 20 N/mm^2 per mm of overlap of two composite layers, it would seem to make more sense. That's what seems to be being calculated. So is it a typo, or is there something more fundamental that I'm missing?

I think what I'm missing is that I'm thinking of lap-joints, also mentioned. I think that the orientation of the 20 N/mm^2 surface in the above discussion is parallel to the scarf surface, the surface along which the composite layers are joined. In this case, is the 20 N/mm^2 a tensional strength, or shear strength? looks like it's a combination of both? it would be a shear strength in the case of a lap joint, a tensional strength in the case of a but joint.

A similar question about lap joints is here.

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  • $\begingroup$ You should also consider how the strength varies according to the direction of the fibres ie linear , perpendicular or some angle as the effective strength changes. $\endgroup$
    – Solar Mike
    Sep 28 '17 at 14:46
  • $\begingroup$ You seem to have about 6 questions in your essay - start with one and once you have an answer for that your other questions may change... $\endgroup$
    – Solar Mike
    Sep 28 '17 at 14:47
  • $\begingroup$ @SolarMike -point. however, an answer has now been provided. maybe better to stick with the flow now.. $\endgroup$
    – juggler
    Sep 28 '17 at 17:49
  • $\begingroup$ I'm going to propose to edit out all questions on lap joints, as those should really be a separate question. If there aren't any objections, I'll move by this time tomorrow. $\endgroup$
    – Mark
    Sep 28 '17 at 20:14
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    $\begingroup$ @Mark and Mike, -I've done my part to try to follow your suggestions. thanks both of you for the feedback! $\endgroup$
    – juggler
    Sep 28 '17 at 21:14
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Formulas from "The Behavior of Structures Composed of Composite Materials". Consider this source material for a much deeper understanding of composites.

Does the answer depend on how the composite layer was prepared? For example, does the answer change if the composite layer is a pultruded carbon-fiber strip? Looks like the fiber ratio is the same as what is claimed above as ideal, but the curing temperature is higher.

Yes, and no. What you're looking for is the stress present in the mid plane. With higher fiber content and different preparation, you can have lower or higher stresses in the layer. As such, if you have a 1mm thick layer of composite, and you are pulling with 1,000 N, then for a 1mm wide section of the plane, you will have approximately 1,000 N/mm^2. The amount of "pull" that goes through your shell will be based upon the loads you are putting it through, which can depend on how the layer was prepared. This is discussed more in the text below.

It is proposed above that the resin transmits 20N/mm^2 - this square mm, which is a surface, how is it oriented? Is there something more fundamental that I'm missing?

A significant question. We are orienting ourselves so the loads are being transmitted in the x-direction, the joint is in the y-direction, and the thickness of the beam is in the z-direction. Image shown for clarification.

enter image description here

We then look at a 1-mm wide "slice" of the joint - taken in the y-direction, and review. This slice may transmit 3,000 N through the 1-mm wide slice of the joint, and is 1mm thick. If so, we define the adhesive tensile stress / shear stress as follows:

$$ \sigma_{adh} = E_{adh} \frac{\delta_{z1} - \delta_{z2}}{t} $$ $$ \tau_{adh} = G_{adh} \frac{\delta_{x1} - \delta_{x2}}{t} $$

Where,

  • $\sigma_{adh}$ is the tensile stress developed in the adhesive
  • $\tau_{adh}$ is the shear stress developed in the adhesive
  • $E_{adh}$ is the young's modulus of the adhesive
  • $G_{adh}$ is the shear modulus of the adhesive
  • $\delta_{z1}, \delta_{z2}$ are the displacements in z of the pieces, respectively
  • $\delta_{x1}, \delta_{x2}$ are the displacements in x of the pieces, respectively
  • t is the thickness of adhesive perpendicular to the angle of the joint. (Approximately in z).

The adhesive may be only able to take 20 N of force of shear stress in a lab specimen. This documented tensile test is then 20 N/mm^2. Per Vinson:

$$\frac{\tau_{adh}}{\sigma_{fiber}} = -\frac{h}{l}$$ $$\frac{\sigma_{adh}}{\sigma_{fiber}} = \frac{h^2}{l^2}$$

The joint is then analyzed carefully for full analysis. In this example, since the beam is only being pulled, not bent, the shear stress is to be reviewed. Since the thickness of the consists ENTIRELY of adhesive and does not contain any fibers, there is no fiber content to be considered, we simply utilize the shear strength of the adhesive.

As such, the answer before is incorrect. To ensure the adhesive has only 20 N/mm^2 of stress (the maximum the laboratory says it can handle), we apply the formula:

$$\frac{20\frac{N}{mm^2}}{3000\frac{N}{mm^2}} = -\frac{1}{150} = \frac{h}{l}$$ $$l/h = 150$$

However, most joints have minimal tensile load, and typically have the load in bending. This will cause significant z-direction displacements, which are highlighted into the two formulas. Most joint designs lie between 7-9 degrees, depending on the bending loading, tensile loading, and displacement offsets expected due to loading at the joint. Fiberglass is typically done with a 4:1 - 6:1 length ratio, while carbon-epoxy will reach 12:1 ratios. 90:1 ratios are unheard of in practice,

Lap Joints, as Solar Mike Mentioned, are a completely separate topic with significantly more complicated formulas (Also detailed in Vinson's Text).

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  • $\begingroup$ To clarify, the original is also correct to an extent. The squaring in the formula above is based upon normal stress, but it's linear based upon the allowable shear stress. The allowable shear stress is conservatively less than the allowable normal stress by a factor of sqrt(3). en.wikipedia.org/wiki/… $\endgroup$
    – Mark
    Sep 28 '17 at 16:13
  • $\begingroup$ -in your calculation you seem to be ignoring the 60% fiber ratio that solar Mike mentioned. also, you are orienting us such that the force transmission at the joint is normal to the surface of the joint. this is not the same direction as the direction of force transmission in the whole piece. it's like your saying that the force is transmitted along one piece, then changes direction as it crosses the joint, then changes direction again to continue being transmitted along the other piece. is this indeed how you visualise it? $\endgroup$
    – juggler
    Sep 28 '17 at 19:02
  • $\begingroup$ there seems to be a confusion of definitions here between tensile strength and shear strength. sounds like you're describing tensile strength. maybe the two are equal in this situation? $\endgroup$
    – juggler
    Sep 28 '17 at 19:03
  • $\begingroup$ -wow.. you've added to your answer! thanks! I'll accept it soon, have to absorb the info first! $\endgroup$
    – juggler
    Sep 28 '17 at 21:16
  • $\begingroup$ -would not the tensile stress be related to the displacement in y, not z? -the orientation of your axes is not clear. might I suggest you adopt chemical nomenclature where a dashed line indicates pointing into the page, and a solid triangle indicates coming out of the page? another option is a 3d-rendering, but you've put a ton of effort into this already. maybe you're using a right-hand rule for your axes, and I've just forgot the convention.. $\endgroup$
    – juggler
    Sep 29 '17 at 19:57

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