# Warping functions
## Warping Function and Torsion of Non-Circular Cylinders

The displacements are given by
    $u_{1}=-\alpha x_{2}x_{3}~;~~u_{2}=\alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})$
where $\alpha$ is the twist per unit length, and $\psi$ is the warping function.
The stresses are given by
    $\sigma _{13}=\mu \alpha (\psi _{,1}-x_{2})~;~~\sigma _{23}=\mu \alpha (\psi _{,2}+x_{1})$
where $\mu$ is the shear modulus.
The projected shear traction is
    $\tau ={\sqrt {(\sigma _{13}^{2}+\sigma _{23}^{2})}}$
Equilibrium is satisfied if
    $\nabla ^{2}{\psi }=0~~~~\forall (x_{1},x_{2})\in {\text{S}}$
Traction-free lateral BCs are satisfied if
    $(\psi _{,1}-x_{2}){\frac {dx_{2}}{ds}}-(\psi _{,2}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
or,
    $(\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
The twist per unit length is given by
    $\alpha ={\frac {T}{\mu {\tilde {J}}}}$
where the torsion constant
    ${\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2}+x_{1}\psi _{,2}-x_{2}\psi _{,1})dA$

### Example 1: Circular Cylinder

Choose warping function
    $\psi (x_{1},x_{2})=0$
Equilibrium ($\nabla ^{2}{\psi }=0$) is trivially satisfied.
The traction free BC is
    $(0-x_{2}){\frac {dx_{2}}{ds}}-(0+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
Integrating,
    $x_{2}^{2}+x_{1}^{2}=c^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
where $c$ is a constant.
Hence, a circle satisfies traction-free BCs.
There is no warping of cross sections for circular cylinders
The torsion constant is
    ${\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2})dA=\int _{S}r^{2}dA=J$
The twist per unit length is
    $\alpha ={\frac {T}{\mu J}}$
The non-zero stresses are
    $\sigma _{13}=-\mu \alpha x_{2}~;~~\sigma _{23}=\mu \alpha x_{1}$
The projected shear traction is
    $\tau =\mu \alpha {\sqrt {(x_{1}^{2}+x_{2}^{2})}}=\mu \alpha r$
Compare results from Mechanics of Materials solution
    $\phi ={\frac {TL}{GJ}}~~\Rightarrow ~~\alpha ={\frac {T}{GJ}}$
and
    $\tau ={\frac {Tr}{J}}~~\Rightarrow ~~\tau =G\alpha r$

### Example 2: Elliptical Cylinder

Choose warping function
    $\psi (x_{1},x_{2})=kx_{1}x_{2}\,$
where $k$ is a constant.
Equilibrium ($\nabla ^{2}{\psi }=0$) is satisfied.
The traction free BC is
    $(kx_{2}-x_{2}){\frac {dx_{2}}{ds}}-(kx_{1}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
Integrating,
    $x_{1}^{2}+{\frac {1-k}{1+k}}x_{2}^{2}=a^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
where $a$ is a constant.
This is the equation for an ellipse with major and minor axes $a$ and $b$, where
    $b^{2}=\left({\frac {1+k}{1-k}}\right)a^{2}$
The warping function is
    $\psi =-\left({\frac {a^{2}-b^{2}}{a^{2}+b^{2}}}\right)x_{1}x_{2}$
The torsion constant is
    ${\tilde {J}}={\frac {2b^{2}}{a^{2}+b^{2}}}I_{2}+{\frac {2a^{2}}{a^{2}+b^{2}}}I_{1}={\frac {\pi a^{3}b^{3}}{a^{2}+b^{2}}}$
where
    $I_{1}=\int _{S}x_{1}^{2}dA={\frac {\pi ab^{3}}{4}}~~;~~I_{2}=\int _{S}x_{2}^{2}dA={\frac {\pi a^{3}b}{4}}$
If you compare ${\tilde {J}}$ and $J$ for the ellipse, you will find that ${\tilde {J}}<J$. 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
    $\alpha ={\frac {(a^{2}+b^{2})T}{\mu \pi a^{3}b^{3}}}$
The non-zero stresses are
    $\sigma _{13}=-{\frac {2\mu \alpha a^{2}x_{2}}{a^{2}+b^{2}}}~~;~~\sigma _{23}=-{\frac {2\mu \alpha b^{2}x_{1}}{a^{2}+b^{2}}}$
The projected shear traction is
    $\tau ={\frac {2\mu \alpha }{a^{2}+b^{2}}}{\sqrt {b^{4}x_{1}^{2}+a^{4}x_{2}^{2}}}~~\Rightarrow ~~\tau _{\text{max}}={\frac {2\mu \alpha a^{2}b}{a^{2}+b^{2}}}~~(b<a)$
* Shear stresses in the cross section of an elliptical cylinder under torsion
For any torsion problem where $\partial$S is convex, the maximum projected shear traction occurs at the point on $\partial$S that is nearest the centroid of S
The displacement $u_{3}$ is
    $u_{3}=-{\frac {(a^{2}-b^{2})Tx_{1}x_{2}}{\mu \pi a^{3}b^{3}}}$
* Displacements ($u_{3}$) in the cross section of an elliptical cylinder under torsion

### Example 3: Rectangular Cylinder

In this case, the form of $\psi$ is not obvious and has to be derived from the traction-free BCs
    $(\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
Suppose that $2a$ and $2b$ are the two sides of the rectangle, and $a>b$. Also $a$ is the side parallel to $x_{1}$ and $b$ is the side parallel to $x_{2}$. Then, the traction-free BCs are
    $\psi _{,1}=x_{2}~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~\psi _{,2}=-x_{1}~~{\text{on}}~~x_{2}=\pm b$
A suitable $\psi$ must satisfy these BCs and $\nabla ^{2}{\psi }=0$.
We can simplify the problem by a change of variable
    ${\bar {\psi }}=x_{1}x_{2}-\psi$
Then the equilibrium condition becomes
    $\nabla ^{2}{\bar {\psi }}=0$
The traction-free BCs become
    ${\bar {\psi }}_{,1}=0~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~{\bar {\psi }}_{,2}=2x_{1}~~{\text{on}}~~x_{2}=\pm b$
Let us assume that
    ${\bar {\psi }}(x_{1},x_{2})=f(x_{1})g(x_{2})$
Then,
    $\nabla ^{2}{\bar {\psi }}={\bar {\psi }}_{,11}+{\bar {\psi }}_{,22}=f^{''}(x_{1})g(x_{2})+g^{''}(x_{2})f(x_{1})=0$
or,
    ${\frac {f^{''}(x_{1})}{f(x_{1})}}=-{\frac {g^{''}(x_{2})}{g(x_{2})}}=\eta$

### Case 1: η > 0 or η = 0

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

#### Case 2: η < 0

Let
    $\eta =-k^{2}~~~;~~k>0$
Then,
    ${\begin{aligned}f^{''}(x_{1})+k^{2}f(x_{1})=0~~\Rightarrow &~~f(x_{1})=C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\\g^{''}(x_{2})-k^{2}g(x_{2})=0~~\Rightarrow &~~g(x_{2})=C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\end{aligned}}$
Therefore,
    ${\bar {\psi }}(x_{1},x_{2})=\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\right]$
Apply the BCs at $x_{2}=\pm b$ ~~ (${\bar {\psi }}_{,2}=2x_{1}$), to get
    ${\begin{aligned}\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\\\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[-C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\end{aligned}}$
or,
    $F(x_{1})G^{'}(b)=2x_{1}~~;~~~F(x_{1})G^{'}(-b)=2x_{1}$
The RHS of both equations are odd. Therefore, $F(x_{1})$ is odd. Since, $\cos(kx_{1})$ is an even function, we must have $C_{1}=0$.
Also,
    $F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0$
Hence, $G'(b)$ is even. Since $\sinh(kb)$ is an odd function, we must have $C_{3}=0$.
Therefore,
    ${\bar {\psi }}(x_{1},x_{2})=C_{2}C_{4}\sin(kx_{1})\sinh(kx_{2})=A\sin(kx_{1})\sinh(kx_{2})$
Apply BCs at $x_{1}=\pm a$ ~~ (${\bar {\psi }}_{,1}=0$), to get
    $Ak\cos(ka)\sinh(kx_{2})=0$
The only nontrivial solution is obtained when $\cos(ka)=0$, which means that
    $k_{n}={\frac {(2n+1)\pi }{2a}}~~,~~~n=0,1,2,...$
The BCs at $x_{1}=\pm a$ are satisfied by every terms of the series
    ${\bar {\psi }}(x_{1},x_{2})=\sum _{n=0}^{\infty }A_{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})$
Applying the BCs at $x_{1}=\pm b$ again, we get
    $\sum _{n=0}^{\infty }A_{n}k_{n}\sin(k_{n}x_{1})\cosh(k_{n}b)=2x_{1}~~\Rightarrow ~~~\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})=2x_{1}$
Using the orthogonality of terms of the sine series,
    $\int _{-a}^{a}\sin(k_{n}x_{1})\sin(k_{m}x_{1})dx_{1}={\begin{cases}0&{\rm {if}}~m\neq n\\a&{\rm {if}}~m=n\end{cases}}$
we have
    $\int _{-a}^{a}\left[\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})\right]\sin(k_{m}x_{1})dx_{1}=\int _{-a}^{a}\left[2x_{1}\right]\sin(k_{m}x_{1})dx_{1}$
or,
    $B_{m}a={\frac {4}{k_{m}^{2}}}\sin(k_{m}a)$
Now,
    $\sin(k_{m}a)=\sin \left({\frac {(2m+1)\pi }{2}}\right)=(-1)^{m}$
Therefore,
    $A_{m}={\frac {B_{m}}{k_{m}\cosh(k_{m}b)}}={\frac {(-1)^{m}32a^{2}}{(2m+1)^{3}\pi ^{3}\cosh(k_{m}b)}}$
The warping function is
    $\psi =x_{1}x_{2}-{\frac {32a^{2}}{\pi ^{3}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}{(2n+1)^{3}\cosh(k_{n}b)}}$
The torsion constant and the stresses can be calculated from $\psi$.

## Prandtl Stress Function (φ)

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 $\phi (x_{1},x_{2})$ using
    ${\sigma _{13}=\phi _{,2}~~;~~\sigma _{23}=-\phi _{,1}}$
You can easily check that this definition satisfies equilibrium.
It can easily be shown that the traction-free BCs are satisfied if
    ${{\frac {d\phi }{ds}}=0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}$
where $s$ is a coordinate system that is tangent to the boundary.
If the cross section is simply connected, then the BCs are even simpler:
    ${\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}$
From the compatibility condition, we get a restriction on $\phi$
    ${\nabla ^{2}{\phi }=C~~\forall ~(x_{1},x_{2})\in {\text{S}}}$
where $C$ is a constant.
Using relations for stress in terms of the warping function $\psi$, we get
    ${\nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}}$
Therefore, the twist per unit length is
    ${\alpha =-{\frac {1}{2\mu }}\nabla ^{2}{\phi }}$
The applied torque is given by
    ${T=-\int _{S}(x_{1}\phi _{,1}+x_{2}\phi _{,2})dA}$
For a simply connected cylinder,
    ${T=2\int _{S}\phi dA}$
The projected shear traction is given by
    ${\tau ={\sqrt {(\phi _{,1})^{2}+(\phi _{,2})^{2}}}}$
The projected shear traction at any point on the cross-section is tangent to the contour of constant $\phi$ at that point.
The relation between the warping function $\psi$ and the Prandtl stress function $\phi$ is
    ${\psi _{,1}={\frac {1}{\mu \alpha }}\phi _{,2}+x2~;~~\psi _{,2}=-{\frac {1}{\mu \alpha }}\phi _{,1}-x1}$

### Membrane Analogy

The equations
    $\nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}~~;~~~\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}$
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 $\nabla ^{2}{\phi }=-2\mu \alpha$ is a Poisson equation. Since the equation is inhomogeneous, the solution can be written as
    $\phi =\phi _{p}+\phi _{h}$
where $\phi _{p}$ is a particular solution and $\phi _{h}$ is the solution of the homogeneous equation.
Examples of particular solutions are, in rectangular coordinates,
    $\phi _{p}=-\mu \alpha x_{1}^{2}~~;~~\phi _{p}=-\mu \alpha x_{2}^{2}$
and, in cylindrical co-ordinates,
    $\phi _{p}=-{\frac {\mu \alpha r^{2}}{2}}$
The homogeneous equation is the Laplace equation $\nabla ^{2}{\phi }=0$, which is satisfied by both the real and the imaginary parts of any { analytic} function ($f(z)$) of the complex variable
    $z=x_{1}+ix_{2}=re^{i\theta }$
Thus,
    $\phi _{h}={\text{Re}}(f(z))~~{\text{or}}~~\phi _{h}={\text{Im}}(f(z))$
Suppose $f(z)=z^{n}$.
Then, examples of $\phi _{h}$ are
    $\phi _{h}=C_{1}r^{n}\cos(n\theta )~~;~~\phi _{h}=C_{2}r^{n}\sin(n\theta )~~;~~\phi _{h}=C_{3}r^{-n}\cos(n\theta )~~;~~\phi _{h}=C_{4}r^{-n}\sin(n\theta )$
where $C_{1}$, $C_{2}$, $C_{3}$, $C_{4}$ are constants.
Each of the above can be expressed as polynomial expansions in the $x_{1}$ and $x_{2}$ 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.

### Example: Equilateral Triangle

* Torsion of a cylinder with a triangular cross section
The equations of the three sides are
    ${\begin{aligned}{\text{side}}~\partial S^{(1)}~:~~&f_{1}(x_{1},x_{2})=x_{1}-{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(2)}~:~~&f_{2}(x_{1},x_{2})=x_{1}+{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(3)}~:~~&f_{3}(x_{1},x_{2})=x_{1}-a=0\end{aligned}}$
Let the Prandtl stress function be
    $\phi =Cf_{1}f_{2}f_{3}$
Clearly, $\phi =0$ 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 $\phi$, all we have to do is satisfy the compatibility condition to get the value of $C$. If we can get a closed for solution for $C$, then the stresses derived from $\phi$ will satisfy equilibrium.
Expanding $\phi$ out,
    $\phi =C(x_{1}-{\sqrt {3}}x_{2}+2a)(x_{1}+{\sqrt {3}}x_{2}+2a)(x_{1}-a)$
Plugging into the compatibility condition
    $\nabla ^{2}{\phi }=12Ca=-2\mu \alpha$
Therefore,
    $C=-{\frac {\mu \alpha }{6a}}$
and the Prandtl stress function can be written as
    $\phi =-{\frac {\mu \alpha }{6a}}(x_{1}^{3}+3ax_{1}^{2}+3ax_{2}^{2}-3x_{1}x_{2}^{2}-4a^{3})$
The torque is given by
    $T=2\int _{S}\phi dA=2\int _{-2a}^{a}\int _{-(x_{1}+2a)/{\sqrt {3}}}^{(x_{1}+2a)/{\sqrt {3}}}\phi dx_{2}dx_{1}={\frac {27}{5{\sqrt {3}}}}\mu \alpha a^{4}$
Therefore, the torsion constant is
    ${\tilde {J}}={\frac {27a^{4}}{5{\sqrt {3}}}}$
The non-zero components of stress are
    ${\begin{aligned}\sigma _{13}=\phi _{,2}&={\frac {\mu \alpha }{a}}(x_{1}-a)x_{2}\\\sigma _{23}=-\phi _{,1}&={\frac {\mu \alpha }{2a}}(x_{1}^{2}+2ax_{1}-x_{2}^{2})\end{aligned}}$
The projected shear stress
    $\tau ={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}$
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 $(a,0)$,
    $\tau _{\text{max}}={\frac {3\mu \alpha a}{2}}$
The out-of-plane displacements can be obtained by solving for the warping function $\psi$. For the equilateral triangle, after some algebra, we get
    $u_{3}={\frac {\alpha x_{2}}{6a}}(3x_{1}^{2}-x_{2}^{2})$
The displacement field is plotted below
* Displacements $u_{3}$ 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 $b$ is much larger than the thickness $t$, and that $t$ does not vary rapidly with change along the length axis $\xi$.
Using the membrane analogy, we can neglect the curvature of the membrane in the $\xi$ direction, and the Poisson equation reduces to
    ${\frac {d^{\phi }}{d\eta ^{2}}}=-2\mu \alpha$
which has the solution
    $\phi =\mu \alpha \left({\frac {t^{2}}{4}}-\eta ^{2}\right)$
where $\eta$ is the coordinate along the thickness direction.
The stress field is
    $\sigma _{3\xi }={\frac {\partial }{\partial }}{\phi }{\eta }=-e\mu \beta \eta ~~;~~~\sigma _{3\eta }=0$
Thus, the maximum shear stress is
    $\tau _{\text{max}}=\mu \beta t_{\text{max}}$

### Thin-walled Closed Sections

The Prandtl stress function $\phi$ can be approximated as a linear function between $\phi _{1}$ and $0$ on the two adjacent boundaries.
The local shear stress is, therefore,
    $\sigma _{3s}={\frac {\phi _{1}}{t}}$
where $s$ is the parameterizing coordinate of the boundary curve of the cross-section and $t$ is the local wall thickness.
The value of $\phi _{1}$ can determined using
    $\phi _{1}={\frac {2\mu \alpha A}{\oint _{S}{\frac {dS}{t}}}}$
where $A$ is the area enclosed by the mean line between the inner and outer boundary.
The torque is approximately
    $T=2A\phi _{1}$

## Related content

Introduction to Elasticity
* ↑ 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
* ↑ Simplified approach for torsional analysis of non-homogenous tubes with non-circular cross-sections, B Golami Bazehhour, J Rezaeepazhand, Date: 2012: Journal: International Journal of Engineering, Volume: 25, Issue: 3,
* ↑ Torsion of tubes with quasi-polygonal holes using complex variable method, Gholami Bazehhour, Benyamin, Rezaeepazhand, J. Journal: Mathematics and mechanics of solids ISSN: 1081-2865 Date: 05/2014 Volume: 19 Issue: 3 Page: 260-276 DOI: 10.1177/1081286512462836