There is no exact closed form method for calculating the torsion constant, \(J\), for an arbitrary section. Instead, various approximations can be used.
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:
\[J = \sum_{}^{}\left( \frac{bt^{3}}{3} - \frac{t^{4}}{5} \right)\]
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:
\[J = \frac{bt^{3}}{6}\]
| 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 \(211000\textrm{ mm}^{4}\).
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:
\[J = \frac{4{A_{e}}^{2}}{\sum_{}^{}\frac{s}{t}}\]
Trace an idealised loop (shown as a dashed line) around the centre of the walls of the 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.
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:
\[ J = 4\int_{A}^{}{\varphi\operatorname{dA}}\label{c}\tag{Eq.3} \]
As there are very few closed solutions to the above equations, we will find an approximate solution using the finite difference method.
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:
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.
This document is based on the principles of Chapter 10 and the Appendix of Theory of Elasticity by Timoshenko.