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I have a physics background so apologize my ignorance.

I'm confused by the constants $D$, $\kappa$, and $J$ when it comes to torsion of beams. From what I understand $\kappa$ and $J$ are the same thing, $J$ is (usually) for round beams with isotropic materials while $\kappa$ is for beams with unusual shapes and/or material properties.

What I really don't understand is how you get from $D$ to $\kappa$ though. In a paper, call it paper 1, I found, in the case of an orthotropic beam with rectangular cross section, they derive equations for torsional rigidity $D$ then mention equations such as

$$ \kappa=\frac{D}{2ba^3G^0_x} $$

where $b$ and $a$ are the height and width of the beam while $G^0_x$ is the shear modulus. Where did this equation come from? Is it a standard result? Did they derive the equation for $\kappa$ from FEM simulations?

Then in this paper (call it 2), which deals with torsion of arbitrarily shaped orthotropic composites, they talk about the torsional rigidity $GJ$, which makes me doubt my previous statement about $J$. One of the equations that they give is that the torsional rigidity factor $\beta$ (which is $\kappa$, right?) is

$$ \beta=\frac{GJ}{ab^3G_{zx}} $$

From what I can tell $GJ=D$ from paper 1 and like I said $\beta=\kappa$. So once again I wonder how $D$ (or $GJ$) and $\kappa$ (or $\beta$) are related.

I should explain that the reason why I'm looking to understand better these relationships is because I'm 3D printing springs. I'm looking to find a relationship between the shear moduli of an orthotropic material and the spring constant of a coil spring.

One method that I found is to relate the energy stored in a spring to the energy stored by a beam under torsion $\frac{1}{2}kx^2=\frac{1}{2}\kappa \theta^2$, where $k$ and $x$ are the spring constant and deflection, and $\kappa$ and $\theta$ are the torsional constant and angle of twist of the material.

The two big problems that I have are that my springs have square cross sections (because they're easier to print) and are orthotropic. Hence why I'm having so much trouble.

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  • $\begingroup$ The torsional rigidity measures the resistance of a bar to couples applied at the two ends. If $M$ is the applied torque and $\theta$ is the angle of twist (per unit length of the bar), then the torsional rigidity is defined as $C := M/\theta$. Various symbols have been used to represent the factor $C$. $\endgroup$ – Biswajit Banerjee Jan 11 at 23:22
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Different texts use different notation.

$K$ is sometimes called $J$, torsional stiffness constant, units of $L^4$.

$G$ is shear modulus of rigidity which is sometimes assumed to be 40% of $E$.

$GJ$ or $JK$ is called section modulus of torsion rigidity as $EI$ is in a beam bending moment.

The frequently used equations can be found in Roark's Formulas for Stress and Strain, 7th edition, chapter 10, pp 404.

I am using my phone app, but if you need more detail I responded on my computer using $\LaTeX$.

The equations for torsion in a rectangular solid section:

$$\theta =\frac { TL }{ KG }$$

$${ \tau }_{ max }=\frac { 3T }{ 8a{ b }^{ 2 } } \left[ 1+0.6095\left(\frac { b }{ a }\right) + 0.8865{ \left(\frac { b }{ a } \right) }^{ 2 }-1.8023{ \left(\frac { b }{ a } \right) }^{ 3 }+0.9100{ \left(\frac { b }{ a } \right) }^{ 4 } \right]$$

$$K=a{ b }^{ 3 }\left[ \frac { 16 }{ 3 } -3.36\frac { b }{ a } \left(1-\frac { { b }^{ 4 } }{ { 12a }^{ 4 } } \right) \right] \text{ for } a\ge b $$

$\theta = \text{angle of rotation [rad]} \\ \tau = \text{torsion stress} \\ K = \text{torsional stiffness constant}$

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  • $\begingroup$ Thanks for your answer, the problem is that these equations assume an isotropic material whereas I'm using 3D printing which produces orthotropic materials. Therefore I'm trying to understand better the process of how the equations are derived so that I can produce a solution more suited to my needs. If you could provide more detail, it might help. $\endgroup$ – enea19 Jan 13 at 2:15
  • $\begingroup$ @enea19 : Have you looked at the solutions in Lekhnitskii's book? I would also suggest "On the torsion of functionally graded anisotropic linearly elastic bars" by Cornelius Horgan (2007) because it has a few solutions and a good list of references. $\endgroup$ – Biswajit Banerjee Jan 13 at 2:40
  • $\begingroup$ @enea19, The equations in my answer are based on Saint Venant's formulation for isotropic materials. I'm rusty with tensor analyses. But a google search on 2 or 3 dimensional anisotropic elasticity brings up many free to download articles. One very rough estimate for orthotropic materials is to scale the geometry of your cross section by multiplying the short side of the cross section by ratio of Gy/Gx and test the results for convergence, in a series of iterations using the same equations. $\endgroup$ – kamran Jan 13 at 18:19
  • $\begingroup$ Thanks, also the two papers that I originally quoted did something like that. I edited the original question to better explain my intent in asking the question. Would you be able to help with that? $\endgroup$ – enea19 Jan 14 at 5:31

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