## 9 Non-Symmetric Section
### Centroid Forces \(F_{x}\) and \(F_{y}\)
\(q_{2}t/F=0\)  
\(q_{2}t/F=(8.6646\ \times\ 0.002836)+(34.0\ \times\ 0.0004726)=0.04064\)  
\(q_{3}t/F=(8.6646\ \times\ 0.004861)+(34.0\ \times\ 0.0002315)=0.12083\)  
\(q_{4}t/F=(8.6646\ \times\ 0.009115)-(34.0\ \times\ 0.001794)=0.01798\)  
\(q_{5}t/F=(8.6646\ \times\ 0.006944)-(34.0\ \times\ 0.003241)=-0.05003\)  
\(q_{6}t/F=(8.6646\ \times\ 0.005498)-(34.0\ \times\ 0.00348)=-0.07068\)  
\(q_{7}t/F=0\)  
(3a) \(F_{x}/F_{y}=1\), \((F_{x}=F_{y}=F)\)  
\(qt/F=15.109(D_{x}/d^{3})+42.775(D_{y}/d^{3})\)  
Taking \(D_{x}/d^{3}\) and \(D_{y}/d^{3}\) from Table 11 provides the shear flows at median points 1, 2, 3... 7 under a force ratio of unity:  
\(q_{1}t/F=0\)  
\(q_{2}t/F=(15.109\ \times\ 0.002836)+(42.775\ \times\ 0.0004726)=0.2450\)  
\(q_{3}t/F=(15.109\ \times\ 0.004861)+(42.775\ \times\ 0.0002315)=0.08335\)  
\(q_{4}t/F=(15.109\ \times\ 0.009115)-(42.775\ \times\ 0.001794)=0.06098\)  
\(q_{5}t/F=(15.109\ \times\ 0.006944)-(42.775\ \times\ 0.003241)=-0.0337\)  
\(q_{6}t/F=(15.109\ \times\ 0.005498)-(42.775\ \times\ 0.00348)=-0.06578\)  
(3b) \(F_{x}/F_{y}=+1\), \((F_{x}=F_{y}=-F)\)  
All the calculations for 3a) above apply with a sign change. (4a) \(F_{x}/F_{y}=-1\), \((F_{x}=-F,F_{y}=F)\)  
\(qt/F=(6.494\ -\ 8.6646)(D_{x}/d^{3})+(-34.0+8.775)(D_{y}/d^{3})\)  
\(=-2.2206(D_{x}/d^{3})-(25.225(D_{y}/d^{3})\)  
Taking \(D_{x}/d^{3}\) and \(D_{y}/d^{3}\) from Table 11 provides the shear flows at median points 1, 2, 3... 7, under a force ratio of minus unity:  
\(q_{1}t/F=0\)  
\(q_{2}t/F=-(2.2206\ \times\ 0.002836)-(25.225\ \times\ 0.0004726)=-0.01822\)  
\(q_{3}t/F=-(2.2206\ \times\ 0.004861)-(25.225\ \times\ 0.0002315)=-0.01663\)  
\(q_{4}t/F=-(2.2206\ \times\ 0.009115)+(25.225\ \times\ 0.001794)=-0.0250\)  
\(q_{5}t/F=-(2.2206\ \times\ 0.006944)+(25.225\ \times\ 0.003241)=0.06633\)  
\(q_{6}t/F=-(2.2206\ \times\ 0.005498)+(25.225\ \times\ 0.00348)=0.07557\)  
(4b) \(F_{x}/F_{y}=-1\), \((F_{x}=F,F_{y}=-F)\)  
All the calculations for 4a) above apply with a sign change. (5) \(F_{x}/F_{y}=2\), \((F_{x}=2F,F_{y}=F)\)  
\(qt/F=(6.444+2\ \times\ 8.6646)(D_{x}/d^{3})+(34.0\ \times\ 2+8.775)(D_{y}/d^{3})\)  
\(=23.7732(D_{x}/d^{3})+76.775(D_{y}/d^{3})\)  
which gives the shear flow distribution under a force ratio of 2:  
\(q_{1}t/F=0\)  
\(q_{2}t/F=(23.7732\ \times\ 0.002836)+(76.775\ \times\ 0.0004726)=0.1037\)  
\(q_{3}t/F=(23.7732\ \times\ 0.004861)+(76.775\ \times\ 0.0002315)=0.1333\)  
\(q_{4}t/F=(23.7732\ \times\ 0.009115)-(76.775\ \times\ 0.001794)=0.07896\)  
\(q_{5}t/F=(23.7732\ \times\ 0.006944)-(76.775\ \times\ 0.003241)=-0.08375\)  
\(q_{6}t/F=(23.7732\ \times\ 0.005498)-(76.775\ \times\ 0.00348)=-0.13647\)  
(6) \(F_{x}/F_{y}=1/2\), \((F_{x}=F,F_{y}=2F)\)  
\(qt/F=(2\ \times\ 6.444+8.6646)(D_{x}/d^{3})+(34.0+2\ \times\ 8.775)(D_{y}/d^{3})\)  
\(=21.5526(D_{x}/d^{3})+31.55(D_{y}/d^{3})\)which gives the shear flow distribution under a force ratio of \(\
icefrac{{1}}{{2}}\):  
\[q_{1}t/F = 0\] \[q_{2}t/F = (21.5526\ \times\ 0.002836)+(51.55\ \times\ 0.0004726)=0.08548\] \[q_{3}t/F = (21.5526\ \times\ 0.004861)+(51.55\ \times\ 0.0002315)=0.11673\] \[q_{4}t/F = (21.5526\ \times\ 0.009115)-(51.55\ \times\ 0.001794)=0.10397\] \[q_{5}t/F = (21.5526\ \times\ 0.006944)-(51.55\ \times\ 0.003241)=-0.01741\] \[q_{6}t/F = (21.5526\ \times\ 0.005498)-(51.55\ \times\ 0.00348)=-0.0609\]  
Figure 14a-f show the normalised shear flow distributions, \(qt/F\), at each of these force ratios. The conversion to \(q_{net}\) in N/mm units required a multiplication factor, \(F/t\), which, for a force of 100 N (say) and a section thickness of \(3/64^{\prime\prime}\), gives \(F/t=84\) N/mm. The swept area diagram for this non-symmetrical section is constructed in Figure 15. The area swept (enclosed) between the top flange 1-2-3 and E is counter-clockwise. The area swept between web 3-4-5 and E is clockwise and the area swept between the bottom flange and E is counter-clockwise.