## 7 Net Axial Stress
### Net Bending Stress  
When forces \(F_{x}\) and \(F_{y}\) are aligned with the section's principal directions (\(x\) and \(y\)), the combined axial stress arising from \(M_{x}\) and \(M_{y}\) within each quadrant's points in the channel section is as follows:  
\[\sigma_{z}=\pm\,M_{x}\,y/\!M_{x}\pm M_{y}\,x/\!I_{y} \tag{10}\]  
Quadrant 1: points 1 and 2: \(\sigma_{z}=+\,F_{y}(L-z)y/\!I_{x}+F_{x}(L-z)x/\!I_{y}\)  
Quadrant 2: points 2, 3 and 4: \(\sigma_{z}=+\,F_{y}(L-z)y/\!I_{x}-F_{x}(L-z)x/\!I_{y}\)  
Quadrant 3: points 4, 5 and 6: \(\sigma_{z}=-\,F_{y}(L-z)y/\!I_{x}-F_{x}(L-z)x/\!I_{y}\)  
Quadrant 4: points 6 and 7: \(\sigma_{z}=-\,F_{y}(L-z)y/\!I_{x}+F_{x}(L-z)x/\!I_{y}\)  
in which the respective signs are simply those of the co-ordinates (\(x\), \(y\)) across the four quadrants in this convention. It can be seen from Equation (11a-d) that the net bending stress distribution is 3D, for which Figure 7a,b are its 2D projection in the \(z\)-\(y\) plane. To show the 3D distribution in an oblique view, median perimeter points 1, 2, 3... 7 lie along the channel's mid-thickness in which the following co-ordinates (with signs) in Table 7 apply. Substituting from Table 7, Equation (11a-d) are re-written, for \(F_{x}=F_{y}=F\), in non-dimensional form to provide for specific channel constants and length ratios as follows:  
\[(d^{4}/a)(\sigma_{z1}/FL)=(1/2k_{x})(d/a)+(1-X^{\prime}/a)/k_{y} \tag{12a}\]  
\[(d^{4}/a)(\sigma_{z2}/FL)=(1/2k_{x})(d/a) \tag{12b}\]  
\[(d^{4}/a)(\sigma_{z3}/FL)=(1/2k_{x})(d/a)\,-(X^{\prime}/a)/k_{y} \tag{12c}\]  
\[(d^{4}/a)(\sigma_{z4}/FL)=-(X^{\prime}/a)/k_{y} \tag{12d}\]  
\[(d^{4}/a)(\sigma_{z5}/FL)=-(1/2k_{x})(d/a)\,-(X^{\prime}/a)/k_{y} \tag{12e}\]  
\[(d^{4}/a)(\sigma_{z6}/FL)=-(1/2k_{x})(d/a) \tag{12f}\]  
\[(d^{4}/a)(\sigma_{z7}/FL)=-(1/2k_{x})(d/a)+(1-X^{\prime}/a)/k_{y} \tag{12g}\]  
Note: Constants \(k_{x}\) and \(k_{y}\) appear in the second moment of area expressions \(I_{x}=k_{x}d^{4}\) and \(I_{y}=k_{y}d^{4}\) (with units in mm\({}^{4}\)) for each channel as follows. Channel A: \(d/a=2\), \(X^{\prime}/a=1/4\), \(k_{x}=0.02085\), \(k_{y}=0.003276\), \(L=300\) mm and \(d=25.4\) mm. Substituting into Equation (12a-g) leads to the net bending stress (in MPa) at each mid-thickness perimeter position. \[(d^{4}/a)(\sigma_{z1}/FL)=1/(2\,\times\,0.02085)(2)+(1-1/4)/0.003276=276.9,\, |\,\sigma_{z1}=2534.6\,\times\,10^{-3}F\]  
\[(d^{4}/a)(\sigma_{z2}/FL)=1/0.02085=47.96,\,|\,\sigma_{z2}=439\,\times\,10^{- 3}F\]  
\[(d^{4}/a)(\sigma_{z1}/FL)=(1/2k_{x})(d/a)+(1-X^{\prime}/a)/k_{y} \tag{12a}\]0  
\[(d^{4}/a)(\sigma_{z1}/FL)=(1/2k_{x})(d/a)+(1-X^{\prime}/a)/k_{y} \tag{12a}\]1  
\[(d^{4}/a)(\sigma_{z1}/FL)=(1/2k_{x})(d/a)+(1-X^{\prime}/a)/k_{y} \tag{12a}\]2  
\[(d^{4}/a)(\sigma_{z1}/FL)=(1/2k_{x})(d/a)+(1-X^{\prime}/a)/k_{y} \tag{12a}\]3 \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline
**Point** & **1** & **2** & **3** & **4** & **5** & **6** & **7** \\ \hline \(x\) & \(a-X^{\prime}\) & 0 & \(-X^{\prime}\) & \(-X^{\prime}\) & \(-X^{\prime}\) & 0 & \(a-X^{\prime}\) \\ \hline \(y\) & \(d\) /2 & \(d\) /2 & \(d\) /2 & 0 & \(-d/2\) & \(-d/2\) & \(-d/2\) \\ \hline \hline \end{tabular}
\end{table} 
Table 7: Perimeter point co-ordinates with quadrant signs for bending. where it can be seen that with \(F\) = 100 N, these stress levels are elastic, e.g., \(\sigma_{z2}\) = 43.9 MP, (but not so if \(F\) = 1000 N). Figure 8 displays the elastic axial stress levels calculated above as a distribution connecting linear segments between perimeter points 1 and 2, 3 and 5 and 5 and 7. The arrows shown indicate the stress direction and the sense shows where the web and flanges receive their greatest compression and tension with equal forces \(F\) applied at G.  
Channel B:  
\(d/a=7/4\), \(X^{\prime}/a=4/15\), \(k_{x}=0.026395\), \(k_{y}=0.0053615\), \(L\) = 1 m and \(d\) = 44.45 mm, \(a\) = 25.4 mm.