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

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

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

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

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

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