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

${\mathrm{\nabla }}^2\psi =0\mathrm{\forall }(x_1,x_2)\in \text{S}$

Traction-free lateral BCs are satisfied if

$({\psi }_{,1}-x_2)\frac{d{x}_2}{ds}-({\psi }_{,2}+x_1)\frac{d{x}_1}{ds}=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

or,

$({\psi }_{,1}-x_2){\hat{n}}_1+({\psi }_{,2}+x_1){\hat{n}}_2=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

The twist per unit length is given by

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

where the torsion constant

$\tilde{J}=\underset{S}{\int }(x_1^2+x_2^2+x_1{\psi }_{,2}-x_2{\psi }_{,1})dA$

Example 1: Circular Cylinder

Choose warping function

$\psi (x_1,x_2)=0$

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

The traction free BC is

$(0-x_2)\frac{d{x}_2}{ds}-(0+x_1)\frac{d{x}_1}{ds}=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

Integrating,

$x_2^2+x_1^2=c^2\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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

$\tilde{J}=\underset{S}{\int }(x_1^2+x_2^2)dA=\underset{S}{\int }{r}^{2}dA=J$

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

$\varphi =\frac{TL}{GJ}\Rightarrow \alpha =\frac{T}{GJ}$

and

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

Example 2: Elliptical Cylinder

Choose warping function

$\psi (x_1,x_2)=k{x}_1{x}_2$

where $k$ is a constant.

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

The traction free BC is

$(k{x}_2-x_2)\frac{d{x}_2}{ds}-(k{x}_1+x_1)\frac{d{x}_1}{ds}=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

Integrating,

$x_1^2+\frac{1-k}{1+k}{x}_2^2=a^2\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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

$I_1=\underset{S}{\int }{x}_1^{2}dA=\frac{\pi a{b}^3}{4};I_2=\underset{S}{\int }{x}_2^{2}dA=\frac{\pi a^{3}b}{4}$

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

$\tau =\frac{2\mu \alpha }{a^2+b^2}\sqrt{b^4{x}_1^2+a^4{x}_2^2}\Rightarrow {\tau }_{\text{max}}=\frac{2\mu \alpha a^{2}b}{a^2+b^2}(b<a)$

File:Torsion elliptic cs stress.png 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

$u_3=-\frac{(a^2-b^2)T{x}_1{x}_2}{\mu \pi a^3{b}^3}$

File:Torsion elliptic cs disp.png 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

$({\psi }_{,1}-x_2){\hat{n}}_1+({\psi }_{,2}+x_1){\hat{n}}_2=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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 ${\mathrm{\nabla }}^2\psi =0$.

We can simplify the problem by a change of variable

$\overline{\psi }=x_1{x}_2-\psi$

Then the equilibrium condition becomes

${\mathrm{\nabla }}^2\overline{\psi }=0$

The traction-free BCs become

${\overline{\psi }}_{,1}=0\text{on}{x}_1=\pm a,\text{and}{\overline{\psi }}_{,2}=2x_1\text{on}{x}_2=\pm b$

Let us assume that

$\overline{\psi }(x_1,x_2)=f(x_1)g(x_2)$

Then,

${\mathrm{\nabla }}^2\overline{\psi }={\overline{\psi }}_{,11}+{\overline{\psi }}_{,22}=f^{{\text{'}}^{\prime }}(x_1)g(x_2)+g^{{\text{'}}^{\prime }}(x_2)f(x_1)=0$

or,

$\frac{f^{{\text{'}}^{\prime }}(x_1)}{f(x_1)}=-\frac{g^{{\text{'}}^{\prime }}(x_2)}{g(x_2)}=\eta$

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,

$\begin{array}{cc}{f}^{{\text{'}}^{\prime }}(x_1)+k^{2}f(x_1)=0\Rightarrow & f(x_1)=C_1\cos (k{x}_1)+C_2\sin (k{x}_1)\\ g^{{\text{'}}^{\prime }}(x_2)-k^{2}g(x_2)=0\Rightarrow & g(x_2)=C_3\cosh (k{x}_2)+C_4\sinh (k{x}_2)\end{array}$

Therefore,

$\overline{\psi }(x_1,x_2)=[C_1\cos (k{x}_1)+C_2\sin (k{x}_1)][C_3\cosh (k{x}_2)+C_4\sinh (k{x}_2)]$

Apply the BCs at $x_2=\pm b$ ~~ (${\overline{\psi }}_{,2}=2x_1$), to get

$\begin{array}{cc}[C_1\cos (k{x}_1)+C_2\sin (k{x}_1)][C_3\sinh (kb)+C_4\cosh (kb)]& =2x_1\\ [C_1\cos (k{x}_1)+C_2\sin (k{x}_1)][-C_3\sinh (kb)+C_4\cosh (kb)]& =2x_1\end{array}$

or,

$F(x_1)G^{\text{'}}(b)=2x_1;F(x_1)G^{\text{'}}(-b)=2x_1$

The RHS of both equations are odd. Therefore, $F(x_1)$ is odd. Since, $\cos (k{x}_1)$ is an even function, we must have $C_1=0$.

Also,

$F(x_1)[G^{\text{'}}(b)-G^{\text{'}}(-b)]=0$

Hence, $G^{\prime }(b)$ is even. Since $\sinh (kb)$ is an odd function, we must have $C_3=0$.

Therefore,

$\overline{\psi }(x_1,x_2)=C_2{C}_4\sin (k{x}_1)\sinh (k{x}_2)=A\sin (k{x}_1)\sinh (k{x}_2)$

Apply BCs at $x_1=\pm a$ ~~ (${\overline{\psi }}_{,1}=0$), to get

$Ak\cos (ka)\sinh (k{x}_2)=0$

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

$\overline{\psi }(x_1,x_2)=\sum _{n=0}^{\mathrm{\infty }}{A}_n\sin (k_n{x}_1)\sinh (k_n{x}_2)$

Applying the BCs at $x_1=\pm b$ again, we get

$\sum _{n=0}^{\mathrm{\infty }}{A}_n{k}_n\sin (k_n{x}_1)\cosh (k_{n}b)=2x_1\Rightarrow \sum _{n=0}^{\mathrm{\infty }}{B}_n\sin (k_n{x}_1)=2x_1$

Using the orthogonality of terms of the sine series,

$\underset{-a}{\overset{a}{\int }}\sin (k_n{x}_1)\sin (k_m{x}_1)d{x}_1=\{\begin{array}{cc}0& \mathrm{i}\mathrm{f}m\ne n\\ a& \mathrm{i}\mathrm{f}m=n\end{array}$

we have

$\underset{-a}{\overset{a}{\int }}[\sum _{n=0}^{\mathrm{\infty }}{B}_n\sin (k_n{x}_1)]\sin (k_m{x}_1)d{x}_1=\underset{-a}{\overset{a}{\int }}\left[2x_1\right]\sin (k_m{x}_1)d{x}_1$

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

$\psi =x_1{x}_2-\frac{32a^2}{{\pi }^3}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n\sin (k_n{x}_1)\sinh (k_n{x}_2)}{(2n+1{)}^3\cosh (k_{n}b)}$

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 $\varphi (x_1,x_2)$ using

${\sigma }_{13}={\varphi }_{,2};{\sigma }_{23}=-{\varphi }_{,1}$

You can easily check that this definition satisfies equilibrium.

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

$\frac{d\varphi }{ds}=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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:

$\varphi =0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

From the compatibility condition, we get a restriction on $\varphi$

${\mathrm{\nabla }}^2\varphi =C\mathrm{\forall }(x_1,x_2)\in \text{S}$

where $C$ is a constant.

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

${\mathrm{\nabla }}^2\varphi =-2\mu \alpha \mathrm{\forall }(x_1,x_2)\in \text{S}$

Therefore, the twist per unit length is

$\alpha =-\frac{1}{2\mu }{\mathrm{\nabla }}^2\varphi$

The applied torque is given by

$T=-\underset{S}{\int }(x_1{\varphi }_{,1}+x_2{\varphi }_{,2})dA$

For a simply connected cylinder,

$T=2\underset{S}{\int }\varphi dA$

The projected shear traction is given by

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

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

The relation between the warping function $\psi$ and the Prandtl stress function $\varphi$ is

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

Membrane Analogy

The equations

${\mathrm{\nabla }}^2\varphi =-2\mu \alpha \mathrm{\forall }(x_1,x_2)\in \text{S};\varphi =0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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

$\varphi ={\varphi }_p+{\varphi }_h$

where ${\varphi }_p$ is a particular solution and ${\varphi }_h$ is the solution of the homogeneous equation.

Examples of particular solutions are, in rectangular coordinates,

${\varphi }_p=-\mu \alpha x_1^2;{\varphi }_p=-\mu \alpha x_2^2$

and, in cylindrical co-ordinates,

${\varphi }_p=-\frac{\mu \alpha r^2}{2}$

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

$z=x_1+i{x}_2=r{e}^{i\theta }$

Thus,

${\varphi }_h=\text{Re}(f(z))\text{or}{\varphi }_h=\text{Im}(f(z))$

Suppose $f(z)=z^n$.

Then, examples of ${\varphi }_h$ are

${\varphi }_h=C_1{r}^n\cos (n\theta );{\varphi }_h=C_2{r}^n\sin (n\theta );{\varphi }_h=C_3{r}^{-n}\cos (n\theta );{\varphi }_h=C_4{r}^{-n}\sin (n\theta )$

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

File:Torsion of triangle cylinder.png Torsion of a cylinder with a triangular cross section

The equations of the three sides are

$\begin{array}{cc}\text{side}\partial S^{(1)}:& f_1(x_1,x_2)=x_1-\sqrt{3}{x}_2+2a=0\\ \text{side}\partial S^{(2)}:& f_2(x_1,x_2)=x_1+\sqrt{3}{x}_2+2a=0\\ \text{side}\partial S^{(3)}:& f_3(x_1,x_2)=x_1-a=0\end{array}$

Let the Prandtl stress function be

$\varphi =C{f}_1{f}_2{f}_3$

Clearly, $\varphi =0$ 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 $\varphi$, 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 $\varphi$ will satisfy equilibrium.

Expanding $\varphi$ out,

$\varphi =C(x_1-\sqrt{3}{x}_2+2a)(x_1+\sqrt{3}{x}_2+2a)(x_1-a)$

Plugging into the compatibility condition

${\mathrm{\nabla }}^2\varphi =12Ca=-2\mu \alpha$

Therefore,

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

and the Prandtl stress function can be written as

$\varphi =-\frac{\mu \alpha }{6a}(x_1^3+3a{x}_1^2+3a{x}_2^2-3x_1{x}_2^2-4a^3)$

The torque is given by

$T=2\underset{S}{\int }\varphi dA=2\underset{-2a}{\overset{a}{\int }}\underset{-(x_1+2a)/\sqrt{3}}{\overset{(x_1+2a)/\sqrt{3}}{\int }}\varphi d{x}_{2}d{x}_1=\frac{27}{5\sqrt{3}}\mu \alpha a^4$

Therefore, the torsion constant is

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

The non-zero components of stress are

$\begin{array}{cc}{\sigma }_{13}={\varphi }_{,2}& =\frac{\mu \alpha }{a}(x_1-a)x_2\\ {\sigma }_{23}=-{\varphi }_{,1}& =\frac{\mu \alpha }{2a}(x_1^2+2a{x}_1-x_2^2)\end{array}$

The projected shear stress

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

is plotted below

File:Torsion triangle cs stress.png 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

File:Torsion triangle cs disp.png 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

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

which has the solution

$\varphi =\mu \alpha (\frac{t^2}{4}-{\eta }^2)$

where $\eta$ is the coordinate along the thickness direction.

The stress field is

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

Thus, the maximum shear stress is

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

Thin-walled Closed Sections

The Prandtl stress function $\varphi$ can be approximated as a linear function between ${\varphi }_1$ and $0$ on the two adjacent boundaries.

The local shear stress is, therefore,

${\sigma }_{3s}=\frac{{\varphi }_1}{t}$

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

The value of ${\varphi }_1$ can determined using

${\varphi }_1=\frac{2\mu \alpha A}{{\oint }_S\frac{dS}{t}}$

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

The torque is approximately

$T=2A{\varphi }_1$

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