## 3 Torsional Shear Flow Calculations  
This analysis begins with the application of Equation (5b) to an \(x\) axially symmetric channel section of uniform thickness. At a given position \(z\) along the beam's length, the torsional shear flow in the cross-section is controlled by the integral \(Q\), with torque \(T\) applied to shear centre E (see Figure 2a). The swept area diagram (see Figure 2b) provides \(Q\) at points 1, 2, 3 and 4 within each straight length 1-2, 2-3, etc., of the channel section. To permit different channel geometries, each \(Q\) appears finally in a non-dimensional form involving the ratios between channel dimensions \(a\), \(d\), \(e\) and \(t\) as follows:  
1-3: \(0\leq s\leq a\); \(\omega=ds/2\); \(Q_{1-3}=\int_{s}\,(ds/2-\omega)\mathrm{d}s=ds^{2}/4-\omega s\)  
\(Q_{1}=0\), \(Q_{3}=da^{2}/4-\varpi a\); \(1\)\(Q_{3}/a^{3}=(1/4)(d/a)-\varpi/a^{2}\)  
3-4: \(a\leq s\leq a+d/2\); \(\omega=es+(ad/2-\varpi a)\); \(Q_{3-4}=\int_{s}\,[es+(ad/2-\varpi a)-\varpi]\,\mathrm{d}s\)  
\(Q_{4}/a^{3}=(1/2)(e/a)[1+(1/2)(d/a)]^{2}+(k_{o}/a^{2}-\varpi/a^{2})[1+(1/2)( d/a)]+k_{1}/a^{3}\)  
4-5: \(a+d/2\leq s\leq a+d\); \(y=-se+(ad/2+e\varpi+\varpi)\); \(Q_{4-5}=\int_{s}\,[-se+(ad/2+e\varpi+\varpi a)-\varpi]\mathrm{d}s\)  
\(=\int_{s}\,(k_{3}-se-\varpi)\,\mathrm{d}s\)  
\(Q_{5}/a^{3}=-(1/2)(e/a)[1+d/a)^{2}-(\varpi/a^{2}-k_{3}/a^{2})(1+d/a)+k_{4}/a^{3}\)  
5-7: \(a+d\leq s\leq 2a+d\); \(\omega=-ds/2+d(a+d/2)\); \(Q_{5-7}=\int_{s}\,[-ds/2+d(a+d/2)-\varpi]\mathrm{d}s\)  
\(=\int_{s}\,(k_{5}-sd/2-\varpi)\,\mathrm{d}s\)  
\(Q_{7}/a^{3}=-(1/4)(d/a)(2+d/a)^{2}-(2+d/a)(\varpi/a^{2}-k_{5}/a^{2})+k_{6}/a^{3}\)  
where  
\(k_{o}/a^{2}=(1/2)(d/a)-(e/a)\)  
\(k_{1}/a^{3}=-(1/4)(d/a)+(1/2)(e/a)\)  
\(k_{3}/a^{2}=(e/a)(d/a)+e/a+(1/2)(d/a)\)  
\(k_{4}/a^{3}=(e/a)[1+(1/2)(d/a)]^{2}+[1+(1/2)(d/a)](k_{o}/a^{2}-k_{3}/a^{2})+ k_{1}/a^{3}\)  
\(k_{5}/a^{2}=(d/a)[1+(1/2)(d/a)]\)  
\(k_{6}/a^{3}=(1+d/a)^{2}[(1/4)(d/a)-(e/a)]+(1+d/a)(k_{3}/a^{2}-k_{5}/a^{2})+k_{4}/a^ {3}\)  
\(\exists/a^{2}=(1/4)(d/a)[2(1+d/a)+(e/a)(d/a)]/(2+d/a)\)  
\(e/a=3(a/d)/[1+6(a/d)]\)  
The \(Q\) coefficients listed in Table 1 above are proportional to the shear flow within each channel section. Shear flow \(q\) (N/mm) at positions 1, 2,...7 in the channel wall are found from multiplying \(Q\) (mm\({}^{3}\)) by \(Et\) (d\({}^{3}\theta/\)d\({}^{2}\)) (N/mm\({}^{4}\)). Then, the shear stress at these positions follows from \(\tau=q/t=E(\mathrm{d}^{3}\theta/\mathrm{d}z^{3})\). Equation (6c) shows that the third derivative depends upon \(z\) and \(q=\mathrm{c}Q\) within the proportionality referred to. Table 2 shows how \(\mathrm{d}^{3}\theta/\mathrm{d}z^{3}\) varies with \(z/L\) in each channel section for a unit torque, using constants \(J\) and \(\mu\). From Equation (6c),  
\[\mathrm{d}^{3}\theta/\mathrm{d}z^{3}=-(T/G)\;[\mu^{2}\mathrm{cosh}\mu(L-z)/ \mathrm{cosh}\;\mu L] \tag{8a}\] \[(G/T)\;\mathrm{d}^{3}\theta/\mathrm{d}z^{3}=-(\mu^{2}/J)\mathrm{cosh}[\mu L(1 -z/L)]/\mathrm{cosh}\;\mu L \tag{8b}\]  
A: \(J=67.74\;\mathrm{mm}^{4}\), \(\mu=\sur(GJ/E\Gamma_{1})=6.562\times 10^{-3}\;\mathrm{mm}^{-1}\), \(\mu L=6.562\times 10^{-3}\times 300=1.9685\); B: \(J=1016.8\;\mathrm{mm}^{4}\), \(\mu=\sur(GJ/E\Gamma_{1})=4.55\times 10^{-3}\;\mathrm{mm}^{-1}\), \(\mu L=4.55\times 10^{-3}\times 10^{3}=4.55\); C: \(J=44.656\;\mathrm{mm}^{4}\), \(\mu=\sur(GJ/E\Gamma_{1})=2.73\times 10^{-3}\;\mathrm{mm}^{-1}\), \(\mu L=2.73\times 10^{-3}\times 340=0.936\).