Warping Function and Torsion of Non-Circular Cylinders[edit | edit source]
The displacements are given by

where
is the twist per unit length, and
is the
warping function.
The stresses are given by

where
is the shear modulus.
The projected shear traction is

Equilibrium is satisfied if

Traction-free lateral BCs are satisfied if

or,

The twist per unit length is given by

where the torsion constant

Example 1: Circular Cylinder[edit | edit source]
Choose warping function

Equilibrium (
) is trivially satisfied.
The traction free BC is

Integrating,

where
is a constant.
Hence, a circle satisfies traction-free BCs.
There is no warping of cross sections for circular cylinders
The torsion constant is

The twist per unit length is

The non-zero stresses are

The projected shear traction is

Compare results from Mechanics of Materials solution

and

Example 2: Elliptical Cylinder[edit | edit source]
Choose warping function

where
is a constant.
Equilibrium (
) is satisfied.
The traction free BC is

Integrating,

where
is a constant.
This is the equation for an ellipse with major and minor axes
and
,
where

The warping function is

The torsion constant is

where

If you compare
and
for the ellipse, you will find that
. 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

The non-zero stresses are

The projected shear traction is

Shear stresses in the cross section of an elliptical cylinder under torsion
|
For any torsion problem where
S is convex, the maximum projected shear traction occurs at the point on
S that is nearest the centroid of S
The displacement
is

Displacements (  ) in the cross section of an elliptical cylinder under torsion
|
Example 3: Rectangular Cylinder[edit | edit source]
In this case, the form of
is not obvious and has to be
derived from the traction-free BCs

Suppose that
and
are the two sides of the rectangle, and
.
Also
is the side parallel to
and
is the side parallel to
.
Then, the traction-free BCs are

A suitable
must satisfy these BCs and
.
We can simplify the problem by a change of variable

Then the equilibrium condition becomes

The traction-free BCs become

Let us assume that

Then,

or,

In both these cases, we get trivial values of
.
Let

Then,

Therefore,
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bcb2287f2c30081bbf32884dfdc0a1b041a4235)
Apply the BCs at
~~ (
), to get
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8f3546d1357cc29b7f3b8e541db9f4809ba2b02)
or,

The RHS of both equations are odd. Therefore,
is odd. Since,
is an even function, we must have
.
Also,
![{\displaystyle F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7d9023ae4253379b8f94da24302aff5143a00b4)
Hence,
is even. Since
is an odd function, we must
have
.
Therefore,

Apply BCs at
~~ (
), to get

The only nontrivial solution is obtained when
, which means that

The BCs at
are satisfied by every terms of the series

Applying the BCs at
again, we get

Using the orthogonality of terms of the sine series,

we have
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a7507d4b2ef54e1539e8202728c76e9fef29c6e)
or,

Now,

Therefore,

The warping function is

The torsion constant and the stresses can be calculated from
.
Prandtl Stress Function (φ)[edit | edit source]
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
using

You can easily check that this definition satisfies equilibrium.
It can easily be shown that the traction-free BCs are satisfied if

where
is a coordinate system that is tangent to the boundary.
If the cross section is simply connected, then the BCs are even simpler:

From the compatibility condition, we get a restriction on

where
is a constant.
Using relations for stress in terms of the warping function
, we
get

Therefore, the twist per unit length is

The applied torque is given by

For a simply connected cylinder,

The projected shear traction is given by

The projected shear traction at any point on the cross-section is tangent to the contour of constant
at that point.
The relation between the warping function
and the Prandtl stress
function
is

The equations

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.
The equation
is a Poisson equation. Since the
equation is inhomogeneous, the solution can be written as

where
is a particular solution and
is the solution
of the homogeneous equation.
Examples of particular solutions are, in rectangular coordinates,

and, in cylindrical co-ordinates,

The homogeneous equation is the Laplace equation
,
which is satisfied by both the real and the imaginary parts of
any { analytic} function (
) of the complex variable

Thus,

Suppose
.
Then, examples of
are

where
,
,
,
are constants.
Each of the above can be expressed as polynomial expansions in the
and
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 [1][2][3].
Example: Equilateral Triangle[edit | edit source]
Torsion of a cylinder with a triangular cross section
|
The equations of the three sides are

Let the Prandtl stress function be

Clearly,
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
, all
we have to do is satisfy the compatibility condition to get the value of
. If we can get a closed for solution for
, then the stresses
derived from
will satisfy equilibrium.
Expanding
out,

Plugging into the compatibility condition

Therefore,

and the Prandtl stress function can be written as

The torque is given by

Therefore, the torsion constant is

The non-zero components of stress are

The projected shear stress

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
,

The out-of-plane displacements can be obtained by solving for the
warping function
. For the equilateral triangle, after some
algebra, we get

The displacement field is plotted below
Displacements  in a cylinder with a triangular cross section.
|
Examples are I-beams, channel sections and turbine blades.
We assume that the length
is much larger than the thickness
,
and that
does not vary rapidly with change along the length axis
.
Using the membrane analogy, we can neglect the curvature of the membrane
in the
direction, and the Poisson equation reduces to

which has the solution

where
is the coordinate along the thickness direction.
The stress field is

Thus, the maximum shear stress is

Thin-walled Closed Sections[edit | edit source]
The Prandtl stress function
can be approximated as a linear
function between
and
on the two adjacent boundaries.
The local shear stress is, therefore,

where
is the parameterizing coordinate of the boundary curve of
the cross-section and
is the local wall thickness.
The value of
can determined using

where
is the area enclosed by the mean line between the inner and
outer boundary.
The torque is approximately

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