## 9 Non-Symmetric Section
### Trans-Moments  
The trans-moments, which arise in the transfer of \(M_{x}\) and \(M_{y}\) from centroid G to shear centre E, are given as  
\[B_{x}=M_{x}\;e_{y} \tag{28a}\] \[B_{y}=M_{y}\;e_{x} \tag{28b}\]  
where \(e_{x}\) and \(e_{y}\) are the perpendicular distances between rectangular axes \(x\) and \(y\) at G and E. Equation (28a,b) are taken to contribute, respectively, to the first and second terms in the axial stress Equation (26a):  
\[\sigma_{z1}=B_{x}/I_{Ex}=M_{x}\;E_{y}/I_{Ex}\] (29a) and \[\sigma_{z2}=B_{y}/I_{Ey}=M_{y}\;E_{x}/I_{Ey} \tag{29b}\]  
Combining Equation (26a) with Equation (29a,b) gives the total axial stress from the two bending sources:  
\[\sigma_{z}=[(M_{x}^{\prime}y/)/I_{x}+(M_{x}\;E_{y})/I_{Ex}]+[(M_{y}^{\prime}x/ )/I_{x}+(M_{y}\;E_{x})/I_{Ey}]\] (30a) in which the parallel axis theorem provides for the change in \[I_{x}\] and \[I_{y}\] accompanying a separation between the respective parallel axes for G and E: (i) \[E_{x}=e_{x}+X^{\prime}\] between \[I_{y}\] and (ii) \[E_{y}=Y^{\prime}-e_{y}\] between \[B_{y}=M_{y}\;e_{x} \tag{28b}\]0 and \[B_{y}=M_{y}\;e_{x} \tag{28b}\]1 (see Figure 20a,b) \[B_{y}=M_{y}\;e_{x} \tag{28b}\]2 (30b) \[B_{y}=M_{y}\;e_{x} \tag{28b}\]3 (30c) in which \[B_{y}=M_{y}\;e_{x} \tag{28b}\]4 is the channel section area. To apply Equation ( 30a - c ), recall that the following section properties are required: \[B_{y}=M_{y}\;e_{x} \tag{28b}\]5 and \[B_{y}=M_{y}\;e_{x} \tag{28b}\]6 which provide the three ratios between second moments of area appearing in Equation ( 26b,c) \[B_{y}=M_{y}\;e_{x} \tag{28b}\]7 Initial calculations are made in the absence of trans-moments, substituting into Equation ( 26a,c): \[B_{y}=M_{y}\;e_{x} \tag{28b}\]8 (31a) \[B_{y}=M_{y}\;e_{x} \tag{28b}\]9 This shows that the greatest stress occurs at the fixing ( \[\sigma_{z1}=B_{x}/I_{Ex}=M_{x}\;E_{y}/I_{Ex}\]0 ) when Equation ( 31a ) simplifies to  
\[\sigma_{z1}=B_{x}/I_{Ex}=M_{x}\;E_{y}/I_{Ex}\]1 Unsymmetrical channel showing geometry for (**a**) shear centre and (**b**) centroid G.  
\[ \begin{array}{l}\sigma_{z}=(761F_{y}+1023F_{x})(y/d)/0.2361d^{3}t+(761F_{x}+193.95F_ {y})(x/d)/0.04475d^{3}t\\ \sigma_{z}\ d^{2}t/10^{3}=(3.223F_{y}+4.333F_{x})(y/d)+(17.005F_{x}+4.334F_{y})(x/ d)\end{array}  \tag{31b}\]  
with \(\sigma_{z}\) in MPa, \(d\) and \(t\) in mm. Equation (31b) is reduced for the following load combinations:  
\[ \begin{array}{l}(1)\quad F_{x}=F;F_{y}=0,\ \mid\sigma_{z}\ d^{2}t/10^{3}=4.333(y/d) +17.005(x/d);\\ (2)\quad F_{y}=F;F_{x}=0,\ \mid\sigma_{z}\ d^{2}t/10^{3}=3.223(y/d)+4.334(x/d);\\ (3)\quad F_{x}=F_{y}=F,\ \mid\sigma_{z}\ d^{2}t/10^{3}=7.556(y/d)+21.339(x/d); \\ (4)\quad F_{x}=-F;F_{y}=F,\ \mid\sigma_{z}\ d^{2}t/10^{3}=-1.11(y/d)-12.671(x/d). \end{array} \]  
Table 12 is constructed for the axial stresses that apply to median points 1, 2, 3... 7. The first two rows provide the co-ordinates of each point, which apply to each of (1)-(4) above. Figure 21a-d shows the elastic bending stress distributions for the four load combinations where \(F=100\) N. With equal forces applied to G in the negative \(x\) and \(y\) directions, Figure 21c becomes the sum of the stresses with each of these forces applied separately (i.e., Figure 21a,b). When the direction of \(F_{x}\) is reversed, this has the effect of raising compression in each flange while lowering tension in the web (see Figure 21d). Linear stress distributions apply within each limb, but intersections giving zero stress do not correspond to 'neutral axis' points 2, 4 and 6 in Figure 20a. In this author's interpretation of trans-moment stress distributions, the four force combinations are treated separately and then superimposed in Figure 21a-d as required.