# Slope Deflection Method in Nonlocal Axially Functionally Graded Tapered Beams
## 6 Examples
### A Frame Subjected to Horizontal Triangular Load  
In this section, a frame system subjected to non-symmetrical load is given as shown in Figure 14. Equilibrium equations can be written in nodes \(D\) and \(C\) and as follows [64] (see Equations (34), (35), (63) and (77)).  
\[\begin{array}{l}\left(\frac{4 EI}{L_{1}}+\frac{4 gEI}{L_{2}}\right)\theta_{D}+ \frac{2 gEI}{L_{2}}\theta_{C}+\frac{2 gEI\theta_{C}}{L_{2}}+\frac{6 EI\delta}{L_ {1}^{2}}-\frac{PI_{1}^{2}}{30}=0\\ \left(\frac{4 EI}{L_{1}}+\frac{4 gEI}{L_{2}}\right)\theta_{C}+\frac{2 gEI\theta_{ D}}{L_{2}}+\frac{6 EI\delta}{L_{1}^{2}}=0\end{array} \tag{94}\]  
The equilibrium condition in the \(AB\) direction is  
\[\left(\frac{12 EI}{L_{1}^{3}}+\frac{12 EI}{L_{1}^{3}}\right)\delta+\frac{\left( \frac{4 EI}{L_{1}}+\frac{2 EI}{L_{1}}\right)\theta_{C}}{L_{1}}+\frac{\left( \frac{4 EI}{L_{1}}+\frac{2 EI}{L_{1}}\right)\theta_{D}}{L_{1}}-\frac{pL_{1}}{2} +\frac{7 pL_{1}}{20}+\frac{\gamma^{2}p}{L_{1}}=0 \tag{95}\]  
\(\theta_{C}\), \(\theta_{D}\), and \(\delta\) are found by solving Equations (94) and (95) as follows.  
\[\begin{array}{l}\theta_{C}=\frac{pL_{1}L_{2}\left(60\beta\gamma^{2}L_{1}-17 gL_{1}^{3}+120\gamma^{2}L_{2}-12\gamma^{2}L_{2}\right)}{240 EI\left(61L_{1}+L_{2} \right)\left(gL_{1}+2L_{2}\right)}\\ \theta_{D}=\frac{pL_{1}L_{2}\left(60\beta\gamma^{2}L_{1}+7 gL_{1}^{3}+120 \gamma^{2}L_{2}-8L_{1}^{2}L_{2}\right)}{240 EI\left(61L_{1}+L_{2}\right)\left( gL_{1}+2L_{2}\right)}\\ \delta=\frac{pL_{1}^{2}\left(-60\beta\gamma^{2}L_{1}+9 gL_{1}^{3}-40\gamma^{2}L _{2}+44L_{1}^{2}L_{2}\right)}{240 EI\left(61L_{1}+L_{2}\right)}\end{array} \tag{96}\]  
The unit displacements obtained as a result of the present work are validated by the results given in [64] for \(L_{1}=3\) m, \(L_{2}=3\) m, \(\beta=1\), \(\gamma=0\), and \(p=30\) kN/m, for \(\theta_{C}\), \(\theta_{D}\), and \(\delta\) (see Table 2).  
The end moments can be written as below.  
\[M_{AD}=-\left(\frac{2 EI}{L_{1}}\theta_{D}+\frac{6 EI}{L_{1}^{2}}\delta+\frac{1} {20}p\left(20\gamma^{2}+L_{1}^{2}\right)\right) \tag{97}\]  
\[M_{DA}=\frac{4 EI}{L}\theta_{D}+\frac{6 EI}{L_{1}^{2}}\delta-\frac{pL_{1}^{2}}{30} \tag{98}\]  
\begin{table}
\begin{tabular}{c c c} \hline \hline
**Unit Displacements** & **Reference Study (\(\gamma\)/L = 0)**[64] & **Present (\(\gamma\)/L = 0)** \\ \hline \(\theta_{C}\) & \(-\frac{4.66}{EI}\) & \(-\frac{4.66}{EI}\) \\ \(\theta_{D}\) & \(-\frac{0.161}{EI}\) & \(-\frac{0.161}{EI}\) \\ \(\delta\) & \(\frac{18.8}{EI}\) & \(\frac{18.8}{EI}\) \\ \hline \hline \end{tabular}
\end{table}
Table 2: Unit displacements for a frame system subjected to horizontal triangular load [64].  
Figure 14: Frame subjected to non-symmetrical loading (\(p\)).  
\[M_{DC}=-\bigg{(}\frac{2 EI}{L_{2}}\theta_{C}+\frac{4 EI}{L_{2}}\theta_{D}\bigg{)} \tag{99}\]  
\[M_{CD}=\frac{2 EI}{L_{2}}\theta_{D}+\frac{4 EI}{L_{2}}\theta_{C} \tag{100}\]  
\[M_{BC}=\frac{2 EI}{L_{1}}\theta_{C}+\frac{6 EI}{L_{1}^{2}}\delta \tag{101}\]  
\[M_{CB}=-\bigg{(}\frac{4 EI}{L_{1}}\theta_{C}+\frac{6 EI}{L_{1}^{2}}\delta\bigg{)} \tag{102}\]  
In the limit \(\gamma\) goes to zero, the end moments (see Table 3) revert to the moments given in [64].  
The variation of the rotation at the joint \(D\) in classical and nonlocal theory with respect to the \(\gamma/L\) ratio is given in Figure 15.