## 5 Net Flexural Shear Flow  
The net flexural shear flow is the sum of the shear flows due to transverse forces \(F_{x}\) and \(F_{y}\) taken separately. The following sum follows from Equation (2a):  
\[q_{x}+q_{y}=(F_{x}/I_{y})D_{y}+(F_{y}/I_{x})D_{x}\] (9a) When the forces and second moments of area are written as ratios, the two flexural shear flows may be added to include the product of these ratios: \[(q_{x}+q_{y})(I_{y}/F_{x})=D_{y}+(F_{y}/F_{x})(I_{y}/I_{x})D_{x} \tag{9b}\]  
The \(I_{x}\) and \(I_{y}\) values for the three channels, A, B and C, are given in Table 3 with their respective geometrical ratios.  
Here, in general, the following expressions for each section's geometric properties apply:  
\begin{table}
\begin{tabular}{l} \hline Section A: \(a=1/2^{\prime\prime}\), \(d=1^{\prime\prime}\), \(t=1/16^{\prime\prime}\), \(L=300\) mm (\(11.81^{\prime\prime}\)) \\ \(a/d=1/2\), \(a/t=8\), \(t/d=1/16\), \(X^{\prime}/a=1/4\), \(X^{\prime}/d=1/8\), \(e_{x}/a=3/8\) \\ \(I_{x}=0.020554^{4}\); \(I_{y}=0.003284^{4}\), \(I_{y}/I_{x}=0.1571\), \(\varnothing/a^{2}=27/32\) \\ \hline Section B: \(a=1^{\prime\prime}\), \(d=13/4^{\prime\prime}\), \(t=1/8^{\prime\prime}\), \(L=1\) m (\(39.37^{\prime\prime}\)) \\ \(a/d=4/7\), \(a/t=8\), \(t/d=1/14\), \(X^{\prime}/a=4/15\), \(X^{\prime}/d=16/105\), \(e_{x}/a=12/31\) \\ \(I_{x}=0.0263754^{4}\); \(I_{y}=0.00536154^{4}\), \(I_{y}/I_{x}=0.20315\), \(\varnothing/a^{2}=0.7207\) \\ \hline Section C: \(a=5/8^{\prime\prime}\), \(d=1^{7}/8^{\prime\prime}\), \(t=3/64^{\prime\prime}\), \(L=340\) mm (\(13.39^{\prime\prime}\)) \\ \(a/d=1/3\), \(a/t=40/3\), \(t/d=1/40\), \(X^{\prime}/a=1/5\), \(X^{\prime}/d=1/15\), \(e_{x}/a=1/3\) \\ \(I_{x}=0.0062514^{4}\); \(I_{y}=0.00043344^{4}\), \(I_{y}/I_{x}=0.06934\), \(\varnothing/a^{2}=27/20\) \\ \hline \end{tabular}
\end{table}
Table 3: Geometric properties of three \(x\)-symmetric channel sections.  
\(X^{\prime}/a=(a/d)/[1+2(a/d)]\)  
\(e_{x}/a=3(a/d)/[1+6(a/d)]\)  
\(\alpha/a^{2}=(1/4)(d/a)[2(1+d/a)+(e/a)(d/a)]/(2+d/a)\)  
\(I_{x}=2[a(d/2)^{2}+at^{3}/12]+td^{3}/12=atd^{2}/2+at^{3}/6+td^{3}/12\)  
\(I_{x}/d^{4}=(1/2)(a/d)(t/d)+(1/6)(a/d)(t/d)^{3}+(1/12)(t/d)\)  
\(I_{y}=dt^{3}/12+dtX^{\prime 2}+2[a^{3}/12+at(a/2-X^{\prime})^{2}]\)  
\(I_{y}/d^{4}=(1/12)(t/d)^{3}+(t/d)(X^{\prime}/d)^{2}+2[(1/12)(t/d)(a/d)^{3}+(a/ d)(t/d)[(1/2)(a/d)-(X^{\prime}/d)]^{2}]\)  
The individual shear flows, \(q_{x}\) and \(q_{y}\), for each channel section A, B and C are as shown in Figure 5a,b when, respectively, \(F_{x}\) and \(F_{y}\) are taken separately to act at the channel section's centroid G. As an example of applying Equation (9b), if the force ratio \(F_{y}/F_{x}\) is unity, the net flexural shear flow at the N.A. (\(x\)-axis) for section A follows from a weighted addition of \(D_{x}\) and \(D_{y}\) between Figure 5a,b:  
\((I_{y}/F_{x})q_{net}=D_{y}+(F_{y}/F_{x})(I_{y}/I_{x})D_{x}\)  
\(=[a^{2}t/2-X^{\prime}\ t\ (a+d/2)]+(1\ \times\ 0.1571)(td^{2}/8+tda/2)\)  
\(=[(1/2)(a/d)^{2}(t/d)-(X^{\prime}/d)(t/d)(a/d+1/2)]d^{3}+0.1571[(1/8)(t/d)+(1/2 )(t/d)(a/d)]d^{3}\)  
\(=[(1/2)(1/2)^{2}(1/16)-(1/8)(1/16)(1/2+1/2)]d^{3}+0.1571[(1/8)(1/16)+(1/2)(1/16) (1/2)]d^{3}\)  
\(=0+(3.682\ \times\ 10^{-3})\ d^{3}\)  
in which \(D_{y}=0\). Hence, \(D_{x}\) only is responsible for the net flexural shear flow at NA:  
\(q_{net}=(F_{x}/I_{y})\ (3.682\ \times\ 10^{-3})d^{3}=F_{x}(3.682\ \times\ 10^{-3})d^{3}/(3.28\ \times\ 10^{-3})d^{4}=1.1226F_{x}/d\)  
which, for shear forces of \(F_{y}=F_{x}=100\) N (say) applied to G, with \(d=25.4\) mm, gives:  
\(q_{net}=1.1226\ \times\ 100/25.4=4.42\) N/mm  
\(\therefore\tau=q_{net}/t=16\ \times\ 4.42=70.7\) MPa