# Warping functions

## Warping Function and Torsion of Non-Circular Cylinders

The displacements are given by

${\displaystyle u_{1}=-\alpha x_{2}x_{3}~;~~u_{2}=\alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})}$

where ${\displaystyle \alpha }$ is the twist per unit length, and ${\displaystyle \psi }$ is the warping function.

The stresses are given by

${\displaystyle \sigma _{13}=\mu \alpha (\psi _{,1}-x_{2})~;~~\sigma _{23}=\mu \alpha (\psi _{,2}+x_{1})}$

where ${\displaystyle \mu }$ is the shear modulus.

The projected shear traction is

${\displaystyle \tau ={\sqrt {(\sigma _{13}^{2}+\sigma _{23}^{2})}}}$

Equilibrium is satisfied if

${\displaystyle \nabla ^{2}{\psi }=0~~~~\forall (x_{1},x_{2})\in {\text{S}}}$

Traction-free lateral BCs are satisfied if

${\displaystyle (\psi _{,1}-x_{2}){\frac {dx_{2}}{ds}}-(\psi _{,2}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

or,

${\displaystyle (\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

The twist per unit length is given by

${\displaystyle \alpha ={\frac {T}{\mu {\tilde {J}}}}}$

where the torsion constant

${\displaystyle {\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2}+x_{1}\psi _{,2}-x_{2}\psi _{,1})dA}$

### Example 1: Circular Cylinder

Choose warping function

${\displaystyle \psi (x_{1},x_{2})=0}$

Equilibrium (${\displaystyle \nabla ^{2}{\psi }=0}$) is trivially satisfied.

The traction free BC is

${\displaystyle (0-x_{2}){\frac {dx_{2}}{ds}}-(0+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

Integrating,

${\displaystyle x_{2}^{2}+x_{1}^{2}=c^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

where ${\displaystyle c}$ is a constant.

Hence, a circle satisfies traction-free BCs.

There is no warping of cross sections for circular cylinders

The torsion constant is

${\displaystyle {\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2})dA=\int _{S}r^{2}dA=J}$

The twist per unit length is

${\displaystyle \alpha ={\frac {T}{\mu J}}}$

The non-zero stresses are

${\displaystyle \sigma _{13}=-\mu \alpha x_{2}~;~~\sigma _{23}=\mu \alpha x_{1}}$

The projected shear traction is

${\displaystyle \tau =\mu \alpha {\sqrt {(x_{1}^{2}+x_{2}^{2})}}=\mu \alpha r}$

Compare results from Mechanics of Materials solution

${\displaystyle \phi ={\frac {TL}{GJ}}~~\Rightarrow ~~\alpha ={\frac {T}{GJ}}}$

and

${\displaystyle \tau ={\frac {Tr}{J}}~~\Rightarrow ~~\tau =G\alpha r}$

### Example 2: Elliptical Cylinder

Choose warping function

${\displaystyle \psi (x_{1},x_{2})=kx_{1}x_{2}\,}$

where ${\displaystyle k}$ is a constant.

Equilibrium (${\displaystyle \nabla ^{2}{\psi }=0}$) is satisfied.

The traction free BC is

${\displaystyle (kx_{2}-x_{2}){\frac {dx_{2}}{ds}}-(kx_{1}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

Integrating,

${\displaystyle x_{1}^{2}+{\frac {1-k}{1+k}}x_{2}^{2}=a^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

where ${\displaystyle a}$ is a constant.

This is the equation for an ellipse with major and minor axes ${\displaystyle a}$ and ${\displaystyle b}$, where

${\displaystyle b^{2}=\left({\frac {1+k}{1-k}}\right)a^{2}}$

The warping function is

${\displaystyle \psi =-\left({\frac {a^{2}-b^{2}}{a^{2}+b^{2}}}\right)x_{1}x_{2}}$

The torsion constant is

${\displaystyle {\tilde {J}}={\frac {2b^{2}}{a^{2}+b^{2}}}I_{2}+{\frac {2a^{2}}{a^{2}+b^{2}}}I_{1}={\frac {\pi a^{3}b^{3}}{a^{2}+b^{2}}}}$

where

${\displaystyle I_{1}=\int _{S}x_{1}^{2}dA={\frac {\pi ab^{3}}{4}}~~;~~I_{2}=\int _{S}x_{2}^{2}dA={\frac {\pi a^{3}b}{4}}}$

If you compare ${\displaystyle {\tilde {J}}}$ and ${\displaystyle J}$ for the ellipse, you will find that ${\displaystyle {\tilde {J}}. This implies that the torsional rigidity is less than that predicted with the assumption that plane sections remain plane.

The twist per unit length is

${\displaystyle \alpha ={\frac {(a^{2}+b^{2})T}{\mu \pi a^{3}b^{3}}}}$

The non-zero stresses are

${\displaystyle \sigma _{13}=-{\frac {2\mu \alpha a^{2}x_{2}}{a^{2}+b^{2}}}~~;~~\sigma _{23}=-{\frac {2\mu \alpha b^{2}x_{1}}{a^{2}+b^{2}}}}$

The projected shear traction is

${\displaystyle \tau ={\frac {2\mu \alpha }{a^{2}+b^{2}}}{\sqrt {b^{4}x_{1}^{2}+a^{4}x_{2}^{2}}}~~\Rightarrow ~~\tau _{\text{max}}={\frac {2\mu \alpha a^{2}b}{a^{2}+b^{2}}}~~(b
 Shear stresses in the cross section of an elliptical cylinder under torsion

For any torsion problem where ${\displaystyle \partial }$S is convex, the maximum projected shear traction occurs at the point on ${\displaystyle \partial }$S that is nearest the centroid of S

The displacement ${\displaystyle u_{3}}$ is

${\displaystyle u_{3}=-{\frac {(a^{2}-b^{2})Tx_{1}x_{2}}{\mu \pi a^{3}b^{3}}}}$
 Displacements (${\displaystyle u_{3}}$) in the cross section of an elliptical cylinder under torsion

### Example 3: Rectangular Cylinder

In this case, the form of ${\displaystyle \psi }$ is not obvious and has to be derived from the traction-free BCs

${\displaystyle (\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

Suppose that ${\displaystyle 2a}$ and ${\displaystyle 2b}$ are the two sides of the rectangle, and ${\displaystyle a>b}$. Also ${\displaystyle a}$ is the side parallel to ${\displaystyle x_{1}}$ and ${\displaystyle b}$ is the side parallel to ${\displaystyle x_{2}}$. Then, the traction-free BCs are

${\displaystyle \psi _{,1}=x_{2}~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~\psi _{,2}=-x_{1}~~{\text{on}}~~x_{2}=\pm b}$

A suitable ${\displaystyle \psi }$ must satisfy these BCs and ${\displaystyle \nabla ^{2}{\psi }=0}$.

We can simplify the problem by a change of variable

${\displaystyle {\bar {\psi }}=x_{1}x_{2}-\psi }$

Then the equilibrium condition becomes

${\displaystyle \nabla ^{2}{\bar {\psi }}=0}$

The traction-free BCs become

${\displaystyle {\bar {\psi }}_{,1}=0~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~{\bar {\psi }}_{,2}=2x_{1}~~{\text{on}}~~x_{2}=\pm b}$

Let us assume that

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=f(x_{1})g(x_{2})}$

Then,

${\displaystyle \nabla ^{2}{\bar {\psi }}={\bar {\psi }}_{,11}+{\bar {\psi }}_{,22}=f^{''}(x_{1})g(x_{2})+g^{''}(x_{2})f(x_{1})=0}$

or,

${\displaystyle {\frac {f^{''}(x_{1})}{f(x_{1})}}=-{\frac {g^{''}(x_{2})}{g(x_{2})}}=\eta }$

### Case 1: ${\displaystyle \eta >0}$ or ${\displaystyle \eta =0}$

In both these cases, we get trivial values of ${\displaystyle C_{1}=C_{2}=0}$.

#### Case 2: ${\displaystyle \eta <0}$

Let

${\displaystyle \eta =-k^{2}~~~;~~k>0}$

Then,

{\displaystyle {\begin{aligned}f^{''}(x_{1})+k^{2}f(x_{1})=0~~\Rightarrow &~~f(x_{1})=C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\\g^{''}(x_{2})-k^{2}g(x_{2})=0~~\Rightarrow &~~g(x_{2})=C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\end{aligned}}}

Therefore,

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\right]}$

Apply the BCs at ${\displaystyle x_{2}=\pm b}$ ~~ (${\displaystyle {\bar {\psi }}_{,2}=2x_{1}}$), to get

{\displaystyle {\begin{aligned}\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\\\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[-C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\end{aligned}}}

or,

${\displaystyle F(x_{1})G^{'}(b)=2x_{1}~~;~~~F(x_{1})G^{'}(-b)=2x_{1}}$

The RHS of both equations are odd. Therefore, ${\displaystyle F(x_{1})}$ is odd. Since, ${\displaystyle \cos(kx_{1})}$ is an even function, we must have ${\displaystyle C_{1}=0}$.

Also,

${\displaystyle F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0}$

Hence, ${\displaystyle G'(b)}$ is even. Since ${\displaystyle \sinh(kb)}$ is an odd function, we must have ${\displaystyle C_{3}=0}$.

Therefore,

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=C_{2}C_{4}\sin(kx_{1})\sinh(kx_{2})=A\sin(kx_{1})\sinh(kx_{2})}$

Apply BCs at ${\displaystyle x_{1}=\pm a}$ ~~ (${\displaystyle {\bar {\psi }}_{,1}=0}$), to get

${\displaystyle Ak\cos(ka)\sinh(kx_{2})=0}$

The only nontrivial solution is obtained when ${\displaystyle \cos(ka)=0}$, which means that

${\displaystyle k_{n}={\frac {(2n+1)\pi }{2a}}~~,~~~n=0,1,2,...}$

The BCs at ${\displaystyle x_{1}=\pm a}$ are satisfied by every terms of the series

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=\sum _{n=0}^{\infty }A_{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}$

Applying the BCs at ${\displaystyle x_{1}=\pm b}$ again, we get

${\displaystyle \sum _{n=0}^{\infty }A_{n}k_{n}\sin(k_{n}x_{1})\cosh(k_{n}b)=2x_{1}~~\Rightarrow ~~~\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})=2x_{1}}$

Using the orthogonality of terms of the sine series,

${\displaystyle \int _{-a}^{a}\sin(k_{n}x_{1})\sin(k_{m}x_{1})dx_{1}={\begin{cases}0&{\rm {if}}~m\neq n\\a&{\rm {if}}~m=n\end{cases}}}$

we have

${\displaystyle \int _{-a}^{a}\left[\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})\right]\sin(k_{m}x_{1})dx_{1}=\int _{-a}^{a}\left[2x_{1}\right]\sin(k_{m}x_{1})dx_{1}}$

or,

${\displaystyle B_{m}a={\frac {4}{k_{m}^{2}}}\sin(k_{m}a)}$

Now,

${\displaystyle \sin(k_{m}a)=\sin \left({\frac {(2m+1)\pi }{2}}\right)=(-1)^{m}}$

Therefore,

${\displaystyle A_{m}={\frac {B_{m}}{k_{m}\cosh(k_{m}b)}}={\frac {(-1)^{m}32a^{2}}{(2m+1)^{3}\pi ^{3}\cosh(k_{m}b)}}}$

The warping function is

${\displaystyle \psi =x_{1}x_{2}-{\frac {32a^{2}}{\pi ^{3}}}\sum _{n=0}^{\infty }{\frac {(-1)^{m}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}{(2n+1)^{3}\cosh(k_{n}b)}}}$

The torsion constant and the stresses can be calculated from ${\displaystyle \psi }$.

## Prandtl Stress Function (${\displaystyle \phi }$)

The traction free BC is obviously difficult to satisfy if the cross-section is not a circle or an ellipse.

To simplify matters, we define the Prandtl stress function ${\displaystyle \phi (x_{1},x_{2})}$ using

${\displaystyle {\sigma _{13}=\phi _{,2}~~;~~\sigma _{23}=-\phi _{,1}}}$

You can easily check that this definition satisfies equilibrium.

It can easily be shown that the traction-free BCs are satisfied if

${\displaystyle {{\frac {d\phi }{ds}}=0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}}$

where ${\displaystyle s}$ is a coordinate system that is tangent to the boundary.

If the cross section is simply connected, then the BCs are even simpler:

${\displaystyle {\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}}$

From the compatibility condition, we get a restriction on ${\displaystyle \phi }$

${\displaystyle {\nabla ^{2}{\phi }=C~~\forall ~(x_{1},x_{2})\in {\text{S}}}}$

where ${\displaystyle C}$ is a constant.

Using relations for stress in terms of the warping function ${\displaystyle \psi }$, we get

${\displaystyle {\nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}}}$

Therefore, the twist per unit length is

${\displaystyle {\alpha =-{\frac {1}{2\mu }}\nabla ^{2}{\phi }}}$

The applied torque is given by

${\displaystyle {T=-\int _{S}(x_{1}\phi _{,1}+x_{2}\phi _{,2})dA}}$

For a simply connected cylinder,

${\displaystyle {T=2\int _{S}\phi dA}}$

The projected shear traction is given by

${\displaystyle {\tau ={\sqrt {(\phi _{,1})^{2}+(\phi _{,2})^{2}}}}}$

The projected shear traction at any point on the cross-section is tangent to the contour of constant ${\displaystyle \phi }$ at that point.

The relation between the warping function ${\displaystyle \psi }$ and the Prandtl stress function ${\displaystyle \phi }$ is

${\displaystyle {\psi _{,1}={\frac {1}{\mu \alpha }}\phi _{,2}+x2~;~~\psi _{,2}=-{\frac {1}{\mu \alpha }}\phi _{,1}-x1}}$

### Membrane Analogy

The equations

${\displaystyle \nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}~~;~~~\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}$

are similar to the equations that govern the displacement of a membrane that is stretched between the boundaries of the cross-sectional curve and loaded by an uniform normal pressure.

This analogy can be useful in estimating the location of the maximum shear stress and the torsional rigidity of a bar.

• The stress function is proportional to the displacement of the membrane from the plane of the cross-section.
• The stiffest cross-sections are those that allow the maximum volume to be developed between the deformed membrane and the plane of the cross-section for a given pressure.
• The shear stress is proportional to the slope of the membrane.

## Solution Strategy

The equation ${\displaystyle \nabla ^{2}{\phi }=-2\mu \alpha }$ is a Poisson equation. Since the equation is inhomogeneous, the solution can be written as

${\displaystyle \phi =\phi _{p}+\phi _{h}}$

where ${\displaystyle \phi _{p}}$ is a particular solution and ${\displaystyle \phi _{h}}$ is the solution of the homogeneous equation.

Examples of particular solutions are, in rectangular coordinates,

${\displaystyle \phi _{p}=-\mu \alpha x_{1}^{2}~~;~~\phi _{p}=-\mu \alpha x_{2}^{2}}$

and, in cylindrical co-ordinates,

${\displaystyle \phi _{p}=-{\frac {\mu \alpha r^{2}}{2}}}$

The homogeneous equation is the Laplace equation ${\displaystyle \nabla ^{2}{\phi }=0}$, which is satisfied by both the real and the imaginary parts of any { analytic} function (${\displaystyle f(z)}$) of the complex variable

${\displaystyle z=x_{1}+ix_{2}=re^{i\theta }}$

Thus,

${\displaystyle \phi _{h}={\text{Re}}(f(z))~~{\text{or}}~~\phi _{h}={\text{Im}}(f(z))}$

Suppose ${\displaystyle f(z)=z^{n}}$.

Then, examples of ${\displaystyle \phi _{h}}$ are

${\displaystyle \phi _{h}=C_{1}r^{n}\cos(n\theta )~~;~~\phi _{h}=C_{2}r^{n}\sin(n\theta )~~;~~\phi _{h}=C_{3}r^{-n}\cos(n\theta )~~;~~\phi _{h}=C_{4}r^{-n}\sin(n\theta )}$

where ${\displaystyle C_{1}}$, ${\displaystyle C_{2}}$, ${\displaystyle C_{3}}$, ${\displaystyle C_{4}}$ are constants.

Each of the above can be expressed as polynomial expansions in the ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ coordinates.

Approximate solutions of the torsion problem for a particular cross-section can be obtained by combining the particular and homogeneous solutions and adjusting the constants so as to match the required shape.

Only a few shapes allow closed-form solutions. Examples are

• Circular cross-section.
• Elliptical cross-section.
• Circle with semicircular groove.
• Equilateral triangle.

There are a few other papers which propose closed-form or semi-closed-form solutions to the torsion problem for cross-sections with irregular shapes [1][2][3].

### Example: Equilateral Triangle

 Torsion of a cylinder with a triangular cross section

The equations of the three sides are

{\displaystyle {\begin{aligned}{\text{side}}~\partial S^{(1)}~:~~&f_{1}(x_{1},x_{2})=x_{1}-{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(2)}~:~~&f_{2}(x_{1},x_{2})=x_{1}+{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(3)}~:~~&f_{3}(x_{1},x_{2})=x_{1}-a=0\end{aligned}}}

Let the Prandtl stress function be

${\displaystyle \phi =Cf_{1}f_{2}f_{3}}$

Clearly, ${\displaystyle \phi =0}$ at the boundary of the cross-section (which is what we need for solid cross sections).

Since, the traction-free boundary conditions are satisfied by ${\displaystyle \phi }$, all we have to do is satisfy the compatibility condition to get the value of ${\displaystyle C}$. If we can get a closed for solution for ${\displaystyle C}$, then the stresses derived from ${\displaystyle \phi }$ will satisfy equilibrium.

Expanding ${\displaystyle \phi }$ out,

${\displaystyle \phi =C(x_{1}-{\sqrt {3}}x_{2}+2a)(x_{1}+{\sqrt {3}}x_{2}+2a)(x_{1}-a)}$

Plugging into the compatibility condition

${\displaystyle \nabla ^{2}{\phi }=12Ca=-2\mu \alpha }$

Therefore,

${\displaystyle C=-{\frac {\mu \alpha }{6a}}}$

and the Prandtl stress function can be written as

${\displaystyle \phi =-{\frac {\mu \alpha }{6a}}(x_{1}^{3}+3ax_{1}^{2}+3ax_{2}^{2}-3x_{1}x_{2}^{2}-4a^{3})}$

The torque is given by

${\displaystyle T=2\int _{S}\phi dA=2\int _{-2a}^{a}\int _{-(x_{1}+2a)/{\sqrt {3}}}^{(x_{1}+2a)/{\sqrt {3}}}\phi dx_{2}dx_{1}={\frac {27}{5{\sqrt {3}}}}\mu \alpha a^{4}}$

Therefore, the torsion constant is

${\displaystyle {\tilde {J}}={\frac {27a^{4}}{5{\sqrt {3}}}}}$

The non-zero components of stress are

{\displaystyle {\begin{aligned}\sigma _{13}=\phi _{,2}&={\frac {\mu \alpha }{a}}(x_{1}-a)x_{2}\\\sigma _{23}=-\phi _{,1}&={\frac {\mu \alpha }{2a}}(x_{1}^{2}+2ax_{1}-x_{2}^{2})\end{aligned}}}

The projected shear stress

${\displaystyle \tau ={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}}$

is plotted below

 Stresses in a cylinder with a triangular cross section under torsion

The maximum value occurs at the middle of the sides. For example, at ${\displaystyle (a,0)}$,

${\displaystyle \tau _{\text{max}}={\frac {3\mu \alpha a}{2}}}$

The out-of-plane displacements can be obtained by solving for the warping function ${\displaystyle \psi }$. For the equilateral triangle, after some algebra, we get

${\displaystyle u_{3}={\frac {\alpha x_{2}}{6a}}(3x_{1}^{2}-x_{2}^{2})}$

The displacement field is plotted below

 Displacements ${\displaystyle u_{3}}$ in a cylinder with a triangular cross section.

### Thin-walled Open Sections

Examples are I-beams, channel sections and turbine blades.

We assume that the length ${\displaystyle b}$ is much larger than the thickness ${\displaystyle t}$, and that ${\displaystyle t}$ does not vary rapidly with change along the length axis ${\displaystyle \xi }$.

Using the membrane analogy, we can neglect the curvature of the membrane in the ${\displaystyle \xi }$ direction, and the Poisson equation reduces to

${\displaystyle {\frac {d^{\phi }}{d\eta ^{2}}}=-2\mu \alpha }$

which has the solution

${\displaystyle \phi =\mu \alpha \left({\frac {t^{2}}{4}}-\eta ^{2}\right)}$

where ${\displaystyle \eta }$ is the coordinate along the thickness direction.

The stress field is

${\displaystyle \sigma _{3\xi }={\frac {\partial }{\partial }}{\phi }{\eta }=-e\mu \beta \eta ~~;~~~\sigma _{3\eta }=0}$

Thus, the maximum shear stress is

${\displaystyle \tau _{\text{max}}=\mu \beta t_{\text{max}}}$

### Thin-walled Closed Sections

The Prandtl stress function ${\displaystyle \phi }$ can be approximated as a linear function between ${\displaystyle \phi _{1}}$ and ${\displaystyle 0}$ on the two adjacent boundaries.

The local shear stress is, therefore,

${\displaystyle \sigma _{3s}={\frac {\phi _{1}}{t}}}$

where ${\displaystyle s}$ is the parameterizing coordinate of the boundary curve of the cross-section and ${\displaystyle t}$ is the local wall thickness.

The value of ${\displaystyle \phi _{1}}$ can determined using

${\displaystyle \phi _{1}={\frac {2\mu \alpha A}{\oint _{S}{\frac {dS}{t}}}}}$

where ${\displaystyle A}$ is the area enclosed by the mean line between the inner and outer boundary.

The torque is approximately

${\displaystyle T=2A\phi _{1}}$

## Related content

1. Approximate Torsional Analysis of Multi-layered Tubes with Non-circular Cross-Sections, Gholami Bazehhour, Benyamin, Rezaeepazhand, Jalil, Journal: Applied composite materials ISSN: 0929-189X Date: 12/2011 Volume: 18 Issue: 6 Page: 485-497 DOI: 10.1007/s10443-011-9213-z
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