## 7 Net Axial Stress
### Constrained Axial Stress Distribution
Substituting into Equation (13a) leads to a non-dimensional form of \(\sigma_{z}\):  
\[(at^{2}/t)\sigma_{z}=2.2172(0.7206-\omega/a^{2}) \tag{13c}\]  
Taking \(\omega/a^{2}\) ordinates from Figure 5a, Equation (13c) applies to points 1, 3 and 4 (NA) as follows:  
\(\mathrm{at}\) 1, \(\omega_{1}=0\), \(\therefore(at^{2}/T)\sigma_{z1}=2.2172(0.7206-0)=1.5977\)  
\(\mathrm{at}\) 3, \(\omega_{3}=ad/2\), \(\therefore(at^{2}/T)\sigma_{z3}=2.2172[0.7206-(1/2)(d/a)]=-0.3423\)  
\(\mathrm{at}\) 4, \(\omega_{4}=(a+e)d/2\), \(\therefore(at^{2}/T)\sigma_{z4}=2.2172[0.7206-(1/2)(d/a)(1+e/a)]=-1.0933\)  
With symmetry evident in the swept area plot, it follows that \(\sigma_{z1}=\sigma_{z7}\) and \(\sigma_{z3}=\sigma_{z5}\). Torque \(T\) corresponding to transverse shear forces \(F_{x}=F_{y}=100\) N applied to E follows as  
\(T\)/\(a=F_{y}(e_{x}/a+X^{\prime}/a)=100\) (\(12/31+4/15)=65.38\) N  
Hence, the constrained stresses at these positions within the channel fixing are as follows:  
\(\sigma_{z1}=1.5977\) (\(T/at^{2}\)) = 1.5977 \(\times\) 65.38/3.1752 = 10.36 MPa  
\(\sigma_{z3}=0.2834\) (\(T/at^{2}\)) = \(-0.3423\) \(\times\) 65.38/3.1752 = \(-2.22\) MPa  
\(\sigma_{z4}=-1.0933\) (\(T/at^{2}\)) = \(-1.0933\) \(\times\) 65.38/3.1752 = \(-7.09\) MPa  
These three stress calculations are sufficient to establish the spread in stress that arises over the section at the fixing where there is full constraint (see Figure 9). Channel C:  
\(d/a=3\), \(e_{x}/a=1/3\), \(L=340\) mm, \(a=15.875\) mm, \(t=1.19063\) mm,  
\(\Gamma_{1}/a^{5}t=1.6375\), \(\varpi/a^{2}=1.35\), \(X^{\prime}/a=1/5\), \(\mu=0.5825t/a^{2}\),  
\(\mu L=0.5825(t/a)(L/a)=0.9356\), \((1-\mathrm{e}^{-2\mu L})/(1+\mathrm{e}^{-2\mu L})=0.7332\). Substituting into Equation (13a) leads to a non-dimensional form of \(\sigma_{z}\):  
\[(at^{2}/t)\sigma_{z}=0.7687(1.35-\omega/a^{2}) \tag{13d}\]  
Taking \(\omega/a^{2}\) ordinates from Figure 2b, Equation (13d) applies to perimeter points 1, 3 and 4 (NA) as  
\(\mathrm{at}\) 1 \(\omega_{1}=0\), \(\therefore(at^{2}/T)\sigma_{z1}=0.7687\) (\(1.35-0\)) = 1.0377  
\(\mathrm{at}\) 3, \(\omega_{3}=ad/2\), \(\therefore(at^{2}/T)\sigma_{z3}=0.7687\) [\(1.35-(1/2)(d/a)\)] = \(-0.1153\)  
\(\mathrm{at}\) 4, \(\omega_{4}=(a+e)d/2\), \(\therefore(at^{2}/T)\sigma_{z4}=0.7687\) [\(1.35-(1/2)(d/a)(1+e/a)\)] = \(-0.4996\)  
With symmetry evident in the swept area plot, it follows that \(\sigma_{z1}=\sigma_{z7}\) and \(\sigma_{z3}=\sigma_{z5}\). Here, torque \(T\) corresponding to transverse shear forces \(F_{x}=F_{y}=100\) N at E follows from  
\(T\)/\(a=F_{y}\) (\(e_{x}/a+X^{\prime}/a\)) = 100 (\(12/31+4/15\)) = 53.33 N Hence, the stresses at these positions within the channel C fixing are as follows:  
\(\sigma_{z1}=1.5977\) (\(T/at^{2}\)) = 1.0377 \(\times\) 53.33/1.190632 = 39.04 MPa  
\(\sigma_{z3}=0.2834\) (\(T/at^{2}\)) = \(-0.1153\)\(\times\) 53.33/1.190632 = \(-4.34\) MPa  
\(\sigma_{z4}=-1.0933\) (\(T/at^{2}\)) = \(-0.49966\)\(\times\) 53.33/1.190632 = \(-18.80\) MPa  
These three stresses appear in their respective positions 1, 3 and 4 in the distribution shown in Figure 9.  
Table 8 reveals the influence of channel section geometry and beam length within which are the three essential calculations given above along with others required to provide the constrained axial stress distribution for each fixed-end section. Note: Points 1 and 7 and points 3 and 5 are'mirrored' within Table 8. Stresses for points 2 and 6, upon the \(y\)-axis, are found from linear interpolation. It appears from Table 8 that the maximum axial stresses that arise from constraining warping displacement within the fixing are all well within the elastic range. However, the possibility of compressive elastic buckling in the web (for points 3, 4 and 5) should not be overlooked. For this, the net axial stress distribution is required as follows.