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I am modeling a beam made of an orthotropic material. The beam's length is aligned with the x-direction, and the force is applied in the y-direction (it is a 4 point bend experiment that I am modeling), along the thickness of the beam.

Intuition tells me that the elastic modulus in the y-direction would be the material property that governs the vertical deflection.

Is this actually correct, or do other properties such as shear modulus (i the x-y, x-z, y-z directions) and elastic modulus in the x-direction also have a significant influence over the vertical deflection?

I ran some simulations and I am actually seeing that varying the shear modulus in the (x-y) direction influences the deflection MUCH more than varying the elastic modulus in the y-direction. Could someone explain why this is the case?

2 Answers2

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You've asked several questions in the original question and subsequent comments, so I'll try to address the key points.

  1. For 4-point bending of a long beam aligned in the x direction and deflecting in the y direction, the most important elastic parameter, all else kept equal, is the Young's modulus in the x direction. This parameter dominates because bending relies on induced x-direction stresses causing x-direction contraction (at the top of the beam) and elongation (at the bottom of the beam). The coupling between x-direction normal stress and x-direction strain is governed predominantly by the x-direction Young's modulus.

  2. The y-direction Young's modulus is not particularly important because the y-direction normal stresses are relatively small. This result might seem counterintuitive because—after all—the four-point bending configuration applies downward forces to the beam. But the the x-direction stresses scale up with the beam length (which is large) and inversely to the cube of the beam height (which is small) and predominate over the y-direction stresses.

  3. What about the x-y shear modulus? As you indicate, it seems counterintuitive that increasing any component of stiffness would increase the deflection. A possible explanation comes from the fact that you're adjusting an anisotropic stiffness tensor one element at a time. By increasing a shear modulus without altering any of the other moduli, you're also altering the Poisson ratio for a pair of axes (specifically, you're reducing the Poisson ratio). A reduction in the Poisson ratio implies that the beam doesn't need to laterally contract or expand as much; depending on the constraints, this change might allow additional y-direction deflection. So your change of a shear modulus can have more far-reaching consequences that just changing the constant of proportionality between the shear stress and the shear strain.

  4. Deflection, deformation, displacement all mean whatever you define them to mean; it's typical but not universal to describe the translation of a point on the beam as a deflection or a displacement and the strain as a deformation (to match the other answer in this thread), and that's the convention I use in this answer.

  • Thanks for the answer. I tried to investigate 3 in more detail, and I am very puzzled by what I am seeing. Would you expect a monotonic relationship between shear modulus and the deflection? i.e., increasing x-y shear modulus will increase deflection, or increasing x-y shear modulus will decrease deflection? I would at least expect some monoticity. – anonuser01 Feb 27 '19 at 18:40
  • What I am seeing is definitely not monotonic. In a particular case that I ran, I increased x-y shear modulus starting at 2E+4, to 2E+12. The deflection decreased monotonically until 2E+10. However, when I bumpted it up to 2E+11, the deflection increased. I am very puzzled by this result. – anonuser01 Feb 27 '19 at 18:40
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Since the material is orthotropic the modeling of the deflection may be complex. For an isotropic material (same properties in all directions) the elastic modulus, $E$, is the same in all directions and is the material property that governs deflection according to

$$ρ=\frac{EI}{M}$$

Where $ρ$ is the deflection radius of curvature, $I$ is the moment of inertia about the centroidal axis and $M$ is the bending moment.

For an orthotropic material I think (but I am not sure since I have not analyzed bending in orthotropic materials) that it would be the elastic modulus in the direction of the $x$-axis (along the length of the beam) that would govern the vertical deflection. The reason I think this is because vertical deflection is due to the tensile and compressive stresses (and strains) occurring along the length of the beam, at least for small deflections.

I think the modulus of elasticity in the $y$ direction would govern deformations of the beam along the $y$-axis. I wouldn't consider vertical deflection as the same as vertical deformation. I think any vertical deformation would be small except perhaps at locations of concentrated vertical loads and would be essentially zero for pure bending (again assuming small vertical deflections).

ADDENDUM

In response to your follow up questions:

(1) I edited my OP about 5 minutes ago, where I modified the shear modulus in the x-y direction, and I saw that this significantly changed the vertical displacement. Specifically, increasing this shear modulus increased the vertical displacement. Isn't this a little bit counter-intuitive?

For an isotropic material the shear modulus $G$ is related to the modulus of elasticity $E$ by

$$E=2(1+ν)G$$

Where $ν$ is Poisson’s ratio, which equals –(lateral strain)/(longitudinal strain). Since under axial loading lateral strain is negative (width of beam contracts) with positive longitudinal strain, $E$ will generally be greater than $G$. However, if you substitute $G$ for $E$ in the previous equation, the radius of deflection becomes

$$ρ=\frac{2(1+ν)GI}{M}$$

So increasing $G$ (or $E$) will increase the radius of curvature. The larger the radius of curvature the less the vertical deflection. This agrees with intuition that the stiffer the material (greater $E$ or $G$), the less the deflection. So your observation that an increase in shear modulus increased vertical displacement would not be consistent if the material were isotropic. Perhaps, however, your observation is correct for an orthotropic material, but as I said I am not familiar with the analysis of orthotropic materials.

(2) What is the difference between vertical deflection, vertical deformation, and vertical displacement? I thought the former 2 were the same, and the latter is a quantitative measurement of the former 2?

I should think vertical deflection and vertical displacement (first and third) are the same if it means a change in the position of a point on the beam in the y direction due to a vertical load. Furthermore, I should think that vertical deformation (the second) would be a change in the thickness of the beam due to loading.

Hope this helps.

Bob D
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  • Thank you for the answer. A couple of questions: (1) I edited my OP about 5 minutes ago, where I modified the shear modulus in the x-y direction, and I saw that this significantly changed the vertical displacement. Specifically, increasing this shear modulus increased the vertical displacement. Isn't this a little bit counter-intuitive? (2) What is the difference between vertical deflection, vertical deformation, and vertical displacement? I thought the former 2 were the same, and the latter is a quantitative measurement of the former 2? – anonuser01 Feb 25 '19 at 16:30
  • @Iamanon See the addendum to my answer in response to your follow up questions. Hope it helps. – Bob D Feb 25 '19 at 17:52
  • Thanks again. That helps a lot. I also varied modulus in the x-direction. As you said, this significantly influences the vertical deflection. I also varied modulus in the y-direction -- the influence is minimal. Increasing shear modulus in x-y direction decreased the deflection, which is going against what you said as well as my intuition. Also, when you said that the vertical deflection is due to tensile/compressive stresses along the length of the beam -- Are you referring to compressive and tensile stresses in the y-direction (along thickness of beam) or in x-direction (along length)? – anonuser01 Feb 25 '19 at 18:38
  • I think @chemomechanics below answered the above question. – anonuser01 Feb 25 '19 at 18:41
  • @Iamanon I'm referring to tensile/compressive stresses in the x direction. Visualize the deflected beam. Assume deflection is small (which is normally the case).The extreme fiber at the bottom of the beam undergoes tension and positive strain along the x-axis. The extreme fiber at the top of the beam undergoes compression and negative strain along the x-axis. Along the neutral axis of the beam (again in the direction of the x-axis) the stress-strain is zero. The combination of stretching at the bottom and shrinking at the top along the x-axis is what gives the beam its curvature (deflection) – Bob D Feb 25 '19 at 18:47
  • @Iamanon I think Chemomechanics has said essentially the same thing as I said. – Bob D Feb 25 '19 at 18:50