## 9 Non-Symmetric Section
### Centroid Forces \(F_{x}\) and \(F_{y}\)
This stress, which diminishes with the length to attain zero at the free end, depends upon the primary warping constant for the section, which is found from  
\[\Gamma_{1}=\int_{s}\omega^{2}t\mathrm{d}s-\varpi^{2}\int_{s}t\ \mathrm{d}s \tag{22}\]  
With \(t\) constant, the first integral in Equation (22) is found from the square of each line equation in Figure 15.  
1-3: \(t\int_{s}\omega^{2}\mathrm{d}s=t\int_{0}^{d/3}(0.8966 ds)^{2}\mathrm{d}s=0.8039 d^{2}t\mid s^{3}/3\mid_{0}^{d/3}=0.009925d^{5}t\)  
3-5: \(t\int_{s}\omega^{2}\mathrm{d}s=t\int_{d/3}^{dd/3}(0.3447d^{2}-0.1373 ds)^{2} \mathrm{d}s\)  
\(=\ t\int_{d/3}^{d/3}\left(0.1182d^{4}-0.09466d^{3}s+0.01885d^{2}s^{2}\right) \mathrm{d}s\)  
\(=\ t\mid 0.1182d^{4}s-0.04733d^{3}s^{2}+0.006284d^{2}s^{3}\mid_{d/3}^{4d/3}\)  
\(=\ t(0.08835d^{5}-0.03437d^{5})=0.05398d^{5}t\)  
5-7: \(t\int_{s}\omega^{2}\mathrm{d}s=t\int_{d/3}^{d/3}\left(0.0236d^{2}+0.1035 ds \right)^{2}\mathrm{d}s\)  
\(=\ t\int_{4d/3}^{2d}\left(0.000557d^{4}+0.004885d^{3}s+0.01071d^{2}s^{2}\right) \mathrm{d}s\)  
\(=\ t\mid 0.000557d^{4}s+0.002443d^{3}s^{2}+0.00357d^{2}s^{3}\mid_{4d/3}^{2d}\)  
\(=\ t(0.03945d^{5}-0.01355d^{5})=0.025904d^{5}t\)  
Substituting these into Equation (22) with \(\varpi\) and \(\int t\ \mathrm{d}s=t(2d)\) within the second term:  
\(\Gamma_{1}=(0.009925d^{5}t+0.05398d^{5}t+0.025904d^{5}t)-(0.2049d^{2})^{2}t( 2d)\)  
\(=(0.089809\,-0.083968)\,d^{5}t=0.00584d^{5}t\)  
The length coefficient \(\mu\) follows:  
\[\mu=\sqrt{(\mathrm{G}J/E\Gamma_{1})}=\sqrt{(\mathrm{J}/8\Gamma_{1})}=6.5428t/d ^{2}\]  
in which \(E=2G(1+\
u)=8G/3\) for \(\
u=1/3\) and \(J=2 dt^{3}/3\). Then, \(\mu L=6.5428(t/d)(L/d)=1.7172\) and \(\exp(-2\ \mu L)=0.03225\). These constants provide the axial stress, from constrained warping at the fixing, as  
\[\sigma_{z}=-(\omega-\varpi)T/\mu\Gamma_{1}\times(1-\mathrm{e}^{-2\mu L})/(1+ \mathrm{e}^{-2\mu L}) \tag{23a}\]  
Equation (23a) is applied in a convenient form to the end section's median points 1, 2, 3... 7 as follows:  
\[ \begin{array}{l}(d^{3}t^{2}/T)\sigma_{z}=24.536(\varpi-\omega)\\ =24.536(0.2049d^{2}-\omega)\end{array}  \tag{23b}\]  
Figure 16: Unconstrained axial warping displacement \(w\) distribution. \[\omega_{1} = 0,\,(d^{3}t^{2}/T)\sigma_{z1}=5.027d^{2}\] \[\omega_{2} = 0.1743d^{2},\,(d^{3}t^{2}/T)\sigma_{z2}=24.536(0.2049d^{2}-0.1743d^{ 2})=0.751d^{2}\] \[\omega_{3} = 0.2988d^{2},\,(d^{3}t^{2}/T)\sigma_{z3}=24.536(0.2049d^{2}-0.2988d ^{2})=-2.304d^{2}\] \[\omega_{4} = 0.2188d^{2},\,(d^{3}t^{2}/T)\sigma_{z4}=24.536(0.2049d^{2}-0.2188 d^{2})=-0.341d^{2}\] \[\omega_{5} = 0.1616d^{2},\,(d^{3}t^{2}/T)\sigma_{z5}=24.536(0.2049d^{2}-0.1616 d^{2})=1.062d^{2}\] \[\omega_{6} = 0.1760d^{2},\,(d^{3}t^{2}/T)\sigma_{z3}=24.536(0.2049d^{2}-0.1760 d^{2})=0.709d^{2}\] \[\omega_{7} = 0.2306d^{2},\,(d^{3}t^{2}/T)\sigma_{z7}=24.536(0.2049d^{2}-0.2306 d^{2})=-0.631d^{2}\]  
The corresponding _constrained_ axial stress distribution, \(\sigma_{z}\), in Figure 17 is proportional to the _unconstrained_ warping displacement (see Figure 16). Linear variations between tension and compression apply to each limb. A conversion of each ordinate \((d^{3}t^{2}/T)\sigma_{z}\) into stress requires a multiplication factor, \(T/d^{3}t^{2}\), so that for median point 3, for example, Equation (23b) yields  
\[\sigma_{z3}=-2.304d^{2}\times T/d^{3}t^{2}=-2.304T/dt^{2}\]  
which, for a unit torque (\(T=1\) N m), a maximum elastic compressive stress is seen to occur at point 3 in the fixed-end cross-section:  
\[\sigma_{z3}=-2.304\times 1000/[(25.4)^{3}\times 15/8\times(3/64)^{2}]=-34.13 \,\mathrm{MPa}\]  
The greatest tensile stress applies to point 1. This converts to \(\sigma_{z1}=74.47\) MPa, which is also elastic under the stated torque. The constrained axial stress, \(\sigma_{z}\), in the beam diminishes with the length according to the first term in Equation (3a).