Warping functions

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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


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


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


Example 2: Elliptical Cylinder[edit | edit source]

Choose warping function

where is a constant.

Equilibrium () is satisfied.

The traction free BC is


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


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



Case 1: η > 0 or η = 0[edit | edit source]

In both these cases, we get trivial values of .

Case 2: η < 0[edit | edit source]




Apply the BCs at ~~ (), to get


The RHS of both equations are odd. Therefore, is odd. Since, is an even function, we must have .


Hence, is even. Since is an odd function, we must have .


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




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

Membrane Analogy[edit | edit source]

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.

Solution Strategy[edit | edit source]

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


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


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.

Thin-walled Open Sections[edit | edit source]

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

Related content[edit | edit source]

Introduction to Elasticity

  1. 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
  2. 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,
  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