# Slope Deflection Method in Nonlocal Axially Functionally Graded Tapered Beams
## 6 Examples  
### A Frame Subjected to Triangular Load  
In this example, the moments at each joint of the frame shown in Figure 12a are obtained in the frame of nonlocal elasticity.  
Equilibrium equations can be written in nodes \(B\) and \(C\) as follows [64] (see Equations (43) and (63)).  
\[\left(\frac{4\beta EI}{L_{2}}\ +\ \frac{4 EI}{L_{1}}\right)\theta_{B}+\frac{2 \beta EI}{L_{2}}\ \theta_{C}+\frac{1}{96}p\ (48\gamma^{2}+5L_{2}{}^{2})=0 \tag{84}\]  
\[\frac{2\beta EI}{L_{2}}\ \theta_{B}+\left(\frac{4\beta EI}{L_{2}}\ +\ \frac{4 EI}{L_{1}}\right)\theta_{C}-\frac{1}{96}p\ (48\gamma^{2}+5L_{2}{}^{2})=0 \tag{85}\]  
Solving Equations (84) and (85) simultaneously, \(\theta_{B}\) and \(\theta_{C}\) are found as follows.  
\[\theta_{B}=-\frac{L_{1}\ L_{2}p\ (48\gamma^{2}+5L_{2}{}^{2})}{192 EI\ (\beta\ L_{1}+2L_{2})},\ \theta_{C}=\frac{p\ (48\gamma^{2}L_{1}\ L_{2}+5L_{1}L_{2}{}^{3})}{192 EI\ (\beta\ L_{1}+2L_{2})} \tag{86}\]  
Figure 12: (**a**) The frame subjected to the two triangular loads \((p)\); (**b**) the variation of rotation (\(\theta_{B}\)) at point B.  
In the limit \(\gamma\) goes to zero, \(\theta_{B}\) and \(\theta_{C}\) are  
\[\lim_{\gamma\to 0}\theta_{B}=-\frac{5L_{1}\ L_{2}{}^{3}p}{192\beta EIL_{1}+384 EIL_{2}},\ \lim_{\gamma\to 0}\theta_{C}=\frac{5L_{1}\ L_{2}{}^{3}p}{192\beta EIL_{1}+384 EIL_{2}} \tag{87}\]  
The unit displacements obtained as a result of the present work are validated by the results given in [64] for \(L_{1}\) = 12 m, \(L_{2}\) = 8 m, \(\beta\) = 1, \(\gamma\) = 0, and \(p\) = 24 kN/m, for \(\theta_{B}\) and \(\theta_{C}\) (see Table 1).  
\[\theta_{B}=-\frac{960}{7 EI},\ \theta_{C}=\frac{960}{7 EI} \tag{88}\]  
Special Case: Variable cross section.  
In the frame shown in Figure 12a, the variation of the moment of inertia and modulus of elasticity function of the elements \(AB\) and \(DC\) are (see Figure 13)  
\[I(z)=-0.2\ \frac{I}{L_{1}}\ z+1.2\ I\,\ E(z)=E+\frac{E}{L_{1}}\ z \tag{89}\]  
For \(L_{1}=12\) m, \(m_{i\ \theta_{i}}\) is calculated using Equation (62) as follows.  
\[m_{i\ \theta_{i}}=0.6015\ EI \tag{90}\]  
Equations (84) and (85) take the following form:  
\[\begin{array}{l}\left(\frac{4\beta EI}{L_{2}}+0.6015 EI\right)\theta_{B}+ \frac{2\beta EI}{L_{2}}\ \theta_{C}+\frac{1}{96}p\ (48\gamma^{2}+5L_{2}{}^{2})=0\\ \frac{2\beta EI}{L_{2}}\ \theta_{B}+\left(\frac{4\beta EI}{L_{2}}+0.6015 EI \right)\theta_{C}-\frac{1}{96}p\ (48\gamma^{2}+5L_{2}{}^{2})=0\end{array} \tag{91}\]  
\(\theta_{B}\) and \(\theta_{C}\) are obtained by solving the system given in Equation (91).  
\[\theta_{B}=\frac{L_{2}p\ (-8.29182\beta\gamma^{2}-0.863731\beta L_{2}{}^{2}-0.8 31255\gamma^{2}L_{2}-0.0865891L_{2}{}^{3})}{EI\ (33.1673\beta^{2}+13.3001\beta L_{2}+L_{2}{}^{2})}=- \theta_{C} \tag{92}\]  
For \(L_{1}=12\) m, \(L_{2}=8\) m, \(\beta=1\), \(\gamma=0\), and \(p=24\) kN/m,  
\[\theta_{B}=-\frac{93.9518}{EI}=-\theta_{C} \tag{93}\]  
\begin{table}
\begin{tabular}{c c c} \hline \hline
**Unit Displacements** & **Reference Study (\(\gamma\)/\(L\) = 0)**[64] & **Present (\(\gamma\)/\(L\) = 0)** \\ \hline \(\theta_{B}\) & \(-\frac{960}{7 EI}\) & \(-\frac{960}{7 EI}\) \\ \(\theta_{C}\) & \(\frac{960}{7 EI}\) & \(\frac{960}{7 EI}\) \\ \hline \hline \end{tabular}
\end{table}
Table 1: Unit displacements for a frame system subjected to vertical triangular load [64].  
Figure 13: The variation of the moment of inertia of the elements \(AB\) and \(DC\) (see Figure 12a).