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) = k x_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 a b^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) $$

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_1x_2}{\mu\pi a^3b^3} $$

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_1x_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{align}

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{align}$$

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{align}

\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{align}$$ or,

F(x_1) G^{'}(b) = 2 x_1 ; F(x_1) G^{'}(-b) = 2 x_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

A k \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) = 2 x_1 \Rightarrow \sum_{n=0}^{\infty} B_n \sin(k_n x_1) = 2 x_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 \ne 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[2 x_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 + i x_2 = r e^{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 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.
 * Circular cross-section.
 * Elliptical cross-section.
 * Circle with semicircular groove.
 * Equilateral triangle.

Example: Equilateral Triangle
The equations of the three sides are
 * $$\begin{align}

\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{align}$$

Let the Prandtl stress function be

\phi = C f_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} = 12 C a = -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_1x_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{27 a^4}{5\sqrt{3}} $$

The non-zero components of stress are
 * $$\begin{align}

\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{align}$$

The projected shear stress

\tau = \sqrt{\sigma_{13}^2+ \sigma_{23}^2} $$ is plotted below The maximum value occurs at the middle of the sides. For example, at $$(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

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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