## 4 Flexural Shear Flow  
Here, Figure 4a,b apply to each uniformly thick, singly symmetric, channel A, B and C, having \(x\) and \(y\) as its principal axes with aligned forces \(F_{x}\) and \(F_{y}\) applied at centroid G. Working along the mean centre line of an \(x\)-symmetric channel, web \(d\times t\) and flanges \(a\times t\), centroid position \(X^{\prime}\) and the second moments of area for centroidal axes \(x\) and \(y\) in Figure 5a,b appear in their ND forms \(X^{\prime}/a\), \(I_{x}/d^{4}\) and \(I_{y}/d^{4}\):  
\[ \begin{array}{l}X^{\prime}/a=(a/d)/[1+2(a/d)]\\ I_{x}=2[at(d/2)^{2}+at^{3}/12]+td^{3}/12=atd^{2}/2+at^{3}/6+td^{3}/12\\ I_{x}/d^{4}=(1/2)(a/t)(t/d)+(1/6)(a/d)(t/d)^{3}+(1/12)(t/d)\\ I_{y}=dt^{3}/12+dtX^{\prime 2}+2[{a}^{3}/12+at(a/2-X^{\prime})^{2}]\\ I_{y}/d^{4}=(1/12)(t/d)^{3}+(t/d)(X^{\prime}/d)^{2}+2[(1/12)(t/d)(d/d)^{3}+(a/ d)(t/d)[(1/2)(a/d)-(X^{\prime}/d)]^{2}]\end{array} \]  
Figure 4: (**a**) Transverse forces applied at G replaced by (**b**) equivalent force system at E.  
Each ND form employs the dimension ratios \(a/d\), \(t/d\) and \(X^{\prime}/d\) for any \(x\)-symmetric channel geometry. The first moment of area integrals \(D_{x}\) and \(D_{y}\) in Equation (2a) follow separately as  
1-3: \(D_{x}=t\int_{s}y\;\mathrm{d}s=tds/2+D_{x1}\)  
at 1, \(s=0\), \(D_{x1}=0\) (free edge)  
at 2, (\(y\)-axis), \(s=a-X^{\prime}\), \(D_{x2}=td\) (\(a-X^{\prime}\))/2  
at 3. \(s=a\), \(D_{x3}=tda/2\)  
3-5, \(D_{x}=t\int_{s}(d/2-s)\;\mathrm{d}s=t(ds/2-s^{2}/2)+D_{x3}\)  
at 3, \(s=0\), \(D_{x3}=tda/2\)  
at 4, (\(x\)-axis) \(s=d/2\), \(D_{x4}=t(d^{2}/4-d^{2}/8)+tda/2=td^{2}/8+tda/2\) (NA, web centre)  
at 5, \(s=d\), \(D_{x5}=D_{x3}=tda/2\)  
5-7 \(D_{x}=-t\int_{s}(d/2)\;\mathrm{d}s+D_{x5}=-tds/2+D_{x5}\)  
at 5, \(s=0\), \(D_{x}=D_{x5}=tda/2\)  
at 6, (\(y\)-axis), \(s=X^{\prime}\), \(D_{x6}=-tdX^{\prime}/2+tda/2=D_{x2}\)  
at 7, \(s=a\), \(D_{x7}=-tda/2+tda/2=0\) (at the free edge)  
Given that \(D_{x}\) is proportional to \(q_{x}\), Figure 5a shows that the \(q_{x}\) distribution is linear between flange points 1, 2 and 3 and between 5, 6 and 7. Web points, 3, 4 and 5 connect \(q\) with a convex parabola, showing maximum shear flow upon the neutral axis (NA) at point 4 on the section's centroidal axis. To find the \(q_{y}\) distribution (Figure 5b), take \(s\) counter-clockwise direction with the origin initially at free end point 1 on the top flange, where \(D_{y}=0\), and thereafter at median points 2, 3... 7:  
1-3: \(D_{y}=t\int_{s}[a-(s+X^{\prime})]\;\mathrm{d}s=(a-X^{\prime})s-s^{2}/2+D_{y1}\)  
at 1, \(s=0\), \(D_{y1}=0\)  
at 2 (\(y\)-axis, NA), \(s=a-X^{\prime}\), \(D_{y2}=t\;[(a-X^{\prime})^{2}-(a-X^{\prime})^{2}/2]=(t/2)(a-X^{\prime})^{2}\)  
at 3, \(s=a\), \(D_{y3}=t\;[(a-X^{\prime})a-a^{2}/2]=at(a/2-X^{\prime})\)  
3-5: \(D_{y}=-t\int_{s}X^{\prime}\;\mathrm{d}s+D_{y3}=-t\;X^{\prime}\;s+D_{y3}\)  
at 3, \(s=0\), \(D_{y3}=at(a/2-X^{\prime})\)  
at 4 (\(x\)-axis), \(s=d/2\), \(D_{y4}=a^{2}t/2-X^{\prime}\;t\;(a+d/2)\)  
at 5, \(s=d\), \(D_{y5}=-t\;X^{\prime}\;d+at\;(a/2-X^{\prime})=a^{2}t/2-X^{\prime}\;t\;(a+d)\)  
5--7: \(D_{y}=-t\int_{s}(s-X^{\prime})\;\mathrm{d}s+D_{y5}=ts\;(s/2-X^{\prime})+D_{y5}\)  
at 5: \(s=0\), \(D_{y5}=a^{2}t/2-X^{\prime}\;t\;(a+d)\)  
at 6 (\(y\)-axis, NA): \(s=X^{\prime}\), \(D_{y6}=X^{\prime}\;t\;(X^{\prime}/2-X^{\prime})+a^{2}t/2-X^{\prime}\;t\;(a+d)\)  
\(D_{y6}=(a^{2}t/2)-X^{\prime}\;t\;(X^{\prime}/2+d+a)\)  
at 7, \(s=a\), \(D_{y7}=ta\;(a/2-X^{\prime})+a^{2}t/2-X^{\prime}\;t\;(a+d)=a^{2}t-X^{\prime}\;t\;(2a+d)\)  
Figure 5: Shear flow distributions in channel section: (a) under \(F_{y}\) and (b) under \(F_{x}\).