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

$\tau = \frac{2\mu\alpha}{a^2+b^2}\sqrt{b^4 x_1^2 + a^4 x_2^2} ~~ \Rightarrow ~~ \tau_{\text{max}} = \frac{2\mu\alpha a^2 b}{a^2+b^2} ~~ (b < a)$
 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

$u_3 = -\frac{(a^2-b^2)Tx_1x_2}{\mu\pi a^3b^3}$
 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

$\bar{\psi} = x_1x_2 - \psi$

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: $\eta > 0$ or $\eta = 0$

In both these cases, we get trivial values of $C_1 = C_2 = 0$.

Case 2: $\eta < 0$

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,

$B_m a = \frac{4}{a k_m^2} \sin(k_m a)$

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 ($\phi$)

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 $\phi(x_1,x_2)$ 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 restrictionon $\phi$

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

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

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

Example: Equilateral Triangle

 Torsion of a cylinder with a triangular cross section

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

 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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