## 9 Non-Symmetric Section
### Torsional Shear Flow  
The corresponding torsional shear flow (see Section 2) is found from applying Equation (5b) to the swept area diagram as follows:  
\[q=Et\mathrm{d}^{3}\theta/\mathrm{d}z^{3}\times Q\] (24a) where \[Q=\int_{s}\left(\omega-\varpi\right)\mathrm{d}s \tag{24b}\]  
The integration for \(Q\) in Equation (24b) gives the change in the shear flow between points  
1-3 : \(\omega-\varpi\ =\ 0.8966 ds\ -\ 0.2049d^{2}\)  
\(Q_{1-3}\ =\ \int_{0}^{d/3}\left(0.8966 ds\ -\ 0.2049d^{2}\right)\mathrm{d}s\)  
\(=\ |0.4483 ds^{2}-0.2049d^{2}s\ |_{0}^{d/3}\ =\ 0.0498d^{3}\ -\ 0.0683d^{3}\ =\ -0.0185d^{3}\)  
3-5 : \(\omega\ -\ \varpi\ =\ 0.3447d^{2}-0.1373 ds\ -\ 0.2049d^{2}\ =\ 0.1398d^{2}\ -\ 0.137 3 ds\)  
\(Q_{3-5}=\int_{d/3}^{4d/3}\left(0.1398d^{2}\ -\ 0.1373 ds\right)\mathrm{d}s\ =\ |\ 0.1398d^{2}s\ -\ 0.06865 ds^{2}|_{d/3}^{4d/3}\)  
\(=\ \left(0.1864d^{3}\ -\ 0.12204d^{3}\right)\ -\ \left(0.0466d^{3}\ -\ 0.00763d^{3}\right)\ =\ 0.02539d^{3}\)  
5-7 : \(\omega\ -\ \varpi\ =\ 0.0236d^{2}\ +\ 0.1035 ds\ -\ 0.2049d^{2}\ =\ -0.1813d^{2}\ +\ 0.10 35 ds\)  
\(Q_{5-7}\ =\ \int_{4d/3}^{2d}\left(-0.1813d^{2}\ +\ 0.1035 ds\right)\mathrm{d}s\ =\ |\ -\ 0.1813d^{2}s\ +\ 0.05175 ds^{2}|_{4d/3}^{2d}\)  
\(=\ \left(-0.3626d^{3}\ +\ 0.2070d^{3}\right)\ -\ \left(-0.2417d^{3}\ +\ 0.0920d^{3}\right)\ =\ -0.0059d^{3}\)  
Intermediate positions are taken to find the shape of the shear flow distribution between points 3 and 4 and 5 and 6 more accurately as follows:  
1-2 : \(Q_{1-2}\ =\ |0.4483 ds^{2}-0.2049d^{2}s|_{0}^{7d/36}=0.01695d^{3}-0.03984d^{3} =\ -0.0229d^{3}\)  
3-4 : \(Q_{3-4}\ =\ |0.1398d^{2}s\ -\ 0.06865 ds^{2}|_{4/3}^{11d/12}\)  
\(=\ \left(0.1282d^{3}\ -\ 0.05769d^{3}\right)\left(-\ 0.0466d^{3}\ \right)\left(The conversion to \(q\) (N/mm) from Equation (24a) applies \(q=Et\) d\({}^{3}\theta/\)d\({}^{3}\times Q\). For example, at point 3 within the fixed-end cross-section, Figure 18b shows \(Q=-0.0185d^{3}\), which is converted to \(q\) for a unit torque of 1 Nm applied to section C, with \(d=17/s^{\prime\prime}\), as follows:  
\[q=Et\) d\({}^{3}\theta/\)d\({}^{3}\times Q\]  
At the fixing where \(z=0\), the third derivative in Equation (6c) is reduced to  
\[ \begin{array}{l}\text{d}^{3}\theta/\text{d}z^{3}=-(\mu^{2}T/|G)=-T/E\Gamma_{ 1}\\ \therefore\,q=-Tt/\Gamma_{1}\times Q=(-Tt/0.00584d^{5}t)\,\times(-0.0185d^{3})=3.16 78T/d^{2}\\ =(3.1678\times 1000)/(1.875\times 25.4)^{2}=1.398\text{ N/mm}\end{array} \]  
Note the sign change that appears for \(q\). Here, a full conversion from \(Q\) to \(q\) reverses the distribution given in Figure 18b. Equation (6c) is required in full when conversions are applied to intermediate span positions \(z=L/4\), \(L/2\), etc. (see Table 1). Variation in the torsional stiffness with the length follows from the twist rate [1]:  
\[\text{d}\theta/\text{d}z=(T/G)\left[1-\cosh\mu(L-z)/\cosh\mu L\right]\]  
Integrating for \(\theta\),  
\[\theta=(T/G)\left[z+\sinh\mu(L-z)/\mu\cosh\mu L\right]+C\]  
where \(C=-(T/G)(\sinh\mu L)/(\mu\cosh\mu L)\), is required to provide zero twist at the fixed end (\(z=0\)). Hence,  
\[ \begin{array}{l}\theta=(T/G)[z+sinh\mu(L-z)/\mu cosh\mu L]-(T/G)(sinh\mu L) /(\mu cosh\mu L)\\ =(T/G)\{z+[sinh\mu(L\text{-}z)-sinh\mu L]/\mu cosh\mu L\}\end{array} \] (25a) Stiffness \[T/\theta\] at position \[z\] is found from Equation ( 25a ): \[T/\theta=G/\{z+[\sinh\mu(L-z)-\sinh\mu L]/\mu\cosh\mu L\} \tag{25b}\]  
For example, at the first quarter position \(z=L/4\), substituting \(\mu L=1.7172\) for \(L=500\) mm, \(G=70\times 10^{3}\) MPa and \(J=53.588\) mm\({}^{4}\), found previously, into Equation (25b) provides the following:  
\[ \begin{array}{l}T/\theta=G/\{L/4+[\sinh(3\ \mu L/4)-\sinh\mu L]/\mu\cosh\mu L \}\\ =(70\times 10^{3}\times 53.588)/[125+(1.675-2.695)/(0.003434\times 2.874]\\ =173.27\times 10^{3}\text{ N mm/c}^{\text{c}}=173.27\text{ N m/c}^{\text{c}}\end{array} \]  
Figure 18: Shear flow measure (**a**) q/d\({}^{3}\) for channel median line in (**b**).