Torsion constant

There is no exact closed form method for calculating the torsion constant, \(J\), for an arbitrary section. Instead, various approximations can be used.

Simplified method for open sections

Break the section down into a series of rectangles with sides \(b\) and \(t\), with \(t < b\).

\[J = \sum_{}^{}{k\frac{bt^{3}}{3}}\]

The value k can be approximated using the following formula:

\[k = 1 - 0.63\frac{t}{b} + 0.052\left( \frac{t}{b} \right)^{2}\]

Alternative, simpler, but slightly less accurate:

\[J = \sum_{}^{}\left( \frac{bt^{3}}{3} - \frac{t^{4}}{5} \right)\]

Grillage Models

The torsion constant for a b × t section of a slab for use in a grillage model is not the same as the torsion constant for a b × t rectangle. Instead use:

\[J = \frac{bt^{3}}{6}\]

Example – 305 x 127 x 42 kg/m Universal Beam

Figure 1: Open section
Figure 1: Open section
Element \(b\) \(t\) \(k\) \(\frac{kbt^{3}}{3}\)
Top Flange 124.3 12.1 0.939 68936
Web 283.0 8.0 0.982 47441
Bottom Flange 124.3 12.1 0.939 68936

\[\begin{align} \sum \frac{kbt^{3}}{3} & = 68936 + 47441 + 68936 \\ & = 185313 \end{align}\]

Calculated J value of \(185313\textrm{ mm}^{4}\) compares to a data book value of \(211000\textrm{ mm}^{4}\).

Simplified method for closed sections

Break the section down into a series of strips forming a loop where the centreline of the loop encloses an area Ae, with each strip of length s and thickness t. J is given by the formula:

\[J = \frac{4{A_{e}}^{2}}{\sum_{}^{}\frac{s}{t}}\]

Example – 160 x 80 x 6mm Rectangular Hollow Section

Trace an idealised loop (shown as a dashed line) around the centre of the walls of the section:

Figure 2: Closed section
Figure 2: Closed section

Calculate \(A_{e}\), the area enclosed by the dashed line:

\[\begin{align} A_{e} & = 154 \times 74 \\ & = 11396 \end{align}\]

Calculate \(\sum \frac{s}{t}\), the sum of the lengths of each dashed line section divided by the wall thickness along that line:

\[\begin{align} \sum\frac{s}{t} = \ \frac{74}{6} + \frac{154}{6} + \frac{74}{6} + \frac{154}{6} & = 76 \end{align}\]

Finally calculate \(J\):

\[\begin{align} J & = \frac{4{A_{e}}^{2}}{\sum_{}^{}\frac{s}{t}} \\ & = \frac{4 \times 11396^{2}}{76} \\ & = 6835201 \end{align}\]

The calculated answer of 6835201 compares with a data book value of 7020000.

Exact formula

To calculate the torsional stiffness, \(J\), of an arbitrary section, area \(A\), find a function \(\varphi\), such that everywhere:

\[ \frac{\partial^{2}\varphi}{\partial x^{2}} + \frac{\partial^{2}\varphi}{\partial y^{2}} + 1 = 0\label{a}\tag{Eq.1} \]

and on the external boundary:

\[ \varphi = 0\label{b}\tag{Eq.2} \]

and on every internal boundary:

\[\oint_{}^{}{\frac{\partial\varphi}{\partial n}\operatorname{ds} = A_{e}}\]

where \(n\) and \(s\) are co-ordinate axes normal and tangent to the boundary respectively, and \(A_{e}\) is the area enclosed by the internal boundary, then:

\[ J = 4\int_{A}^{}{\varphi\operatorname{dA}}\label{c}\tag{Eq.3} \]

Worked example – 1x1 square (Finite difference method)

As there are very few closed solutions to the above equations, we will find an approximate solution using the finite difference method.

Figure 3: Finite difference method grid
Figure 3: Finite difference method grid

Consider a square with sides 1, broken down into a 10x10 grid. The second derivative at each point can be approximated in terms of the value at that point and the four surrounding points by using the finite difference method:

\[ \frac{\partial^{2}\varphi}{\partial x^{2}} \approx \frac{\varphi_{1} - {2\varphi}_{0} + \varphi_{2}}{\delta^{2}} \]

Thus, \(\ref{a}\) can be rewritten as follows:

\[\frac{\varphi_{1} - {2\varphi}_{0} + \varphi_{2}}{\delta^{2}} + \frac{\varphi_{3} - {2\varphi}_{0} + \varphi_{4}}{\delta^{2}} + 1 = 0\]

Rearrange in terms of \(\varphi_{0}\)

\[\begin{align} \frac{1}{\delta^{2}}\left( \varphi_{1} + \varphi_{2} + \varphi_{3} + \varphi_{4} \right) + 1 & = \frac{4}{\delta^{2}}\varphi_{0} \\ \varphi_{0} & = \frac{\varphi_{1} + \varphi_{2} + \varphi_{3} + \varphi_{4} + \delta^{2}}{4} \end{align}\]

Knowing from \(\ref{b}\) that the values of \(\varphi\) around the perimeter are all zero, we can iterate over all the internal grid intersections using this expression until all the values converge. One tool you can use for this is the iterative solver in Excel. If we set up our grid with the above formula (setting \(\delta = 0.1\)) in every internal cell, and set the perimeter cells to zero:

Figure 4: Finite difference method in Excel
Figure 4: Finite difference method in Excel

You will likely get a circular reference warning, in which case go to File > Options > Formula and tick "Enable iterative calculations". You can press F9 to keep iterating further.

\(\ref{c}\) tells us that \(J\) is equal to the product of \(4\delta^{2}\) and the sum of all the values in the table or 0.136, compared to a data book value of 0.141. This method would converge on the correct value as you refine the grid.

This method is easily extended for sections that can be overlaid with a perfect square grid, but for other shapes you will need to either approximate the section using squares or reformulate the finite difference approximation for non-uniform \(\delta\) values.

Further reading

This document is based on the principles of Chapter 10 and the Appendix of Theory of Elasticity by Timoshenko.