# The Approximate Solution of Nonlinear Flexure of a Cantilever Beam with the Galerkin Method
## 3 Equation of Flexure of an Elastic Beam and Its Solution
 \tag{21}\]  
Then, from Equation (21) with an approximate procedure for an explicit expression, the convergent solution is  
\[A_{1}=\frac{\frac{2\alpha}{3\pi}+\frac{2\alpha A_{0}^{2}(189-27A_{0}^{2}+A_{0} ^{4})}{2835\pi}}{\frac{9\pi^{2}}{8}+\frac{4\alpha A_{0}\big{(}13365-1430A_{0} ^{2}+59A_{0}^{4}\big{)}}{51975\pi}}\, \tag{22}\]  
noting the next approximate solution is  
\[\theta=A_{0}\sin\frac{\pi}{2}x+A_{1}\sin\frac{3\pi}{2}x+A_{2}\sin\frac{5\pi}{2 }x, \tag{23}\]  
and the algebraic equation for \(A_{2}\) is \[ \begin{array}{rl}104187267600\alpha A_{0}^{6}+403446387600\alpha A_{1}^{6}\\ -3205762080\alpha A_{0}^{5}(535A_{1}-751A_{2})+193391815584A_{1}^{5}A_{2}\\ +2042887600\alpha A_{1}^{4}(-6561+2437A_{2}^{2})\\ +93000000\alpha A_{1}^{3}A_{2}(-124729+11855A_{2}^{2})\\ +24812400\alpha A_{0}^{4}(-46189+132753A_{1}^{2}-206834A_{1}A_{2}+17313A_{2}^{ 2})\\ +1708100000\alpha A_{1}A_{2}(116127-14850A_{2}^{2}+625A_{2}^{4})\\ +35377680\alpha A_{1}^{2}(5009445-1890690A_{2}^{2}+131671A_{2}^{4})\\ +8595449577(-25200\alpha+196875\pi^{3}A_{2}\\ +8400\alpha A_{2}^{2}-560\alpha A_{2}^{4}+16\alpha A_{2}^{6})\\ -4315200\alpha A_{3}^{3}[1099929A_{3}^{3}-3598787A_{1}^{2}A_{2}\\ -625A_{2}\big{(}-22287+2785A_{2}^{2}\big{)}+23A_{1}\big{(}-365313+137021A_{2}^{ 2}\big{)}]\\ +71920\alpha A_{2}^{2}[50284785A_{1}^{4}-115877220A_{1}^{3}A_{2}\\ +90A_{1}^{2}(-6589523+2468591A_{2}^{2})\\ +A_{1}\big{(}512155260A_{2}-64424868A_{2}^{2}\big{)}\\ -209\big{(}686205-254610A_{2}^{2}+17599A_{2}^{2}\big{)}]\\ -160\alpha A_{0}[5067002235A_{1}^{5}-55775187135A_{1}^{4}A_{2}\\ -41354A_{1}^{2}A_{2}\big{(}-20529405+2545379A_{2}^{2}\big{)}\\ +16182A_{1}^{3}\big{(}-7763305+2927733A_{2}^{2}\big{)}\\ -9530625A_{2}(358701-44950A_{2}^{2}+1875A_{2}^{2})\\ +3875A_{1}\big{(}277272567-103551750A_{2}^{2}+7178125A_{2}^{2}\big{)}]=0,\end{array}  \tag{24}\]  
with the explicit expression of \(A_{2}\) as  
\[ \begin{array}{rl}A_{2}=&-(-216605329340400\alpha-10314539492400\alpha A_{0}^ {2}\\ -1146059943600\alpha A_{0}^{4}+104187267600\alpha A_{0}^{6}\\ -171908991540000\alpha A_{0}A_{1}+36257169124800\alpha A_{0}^{3}A_{1}\\ -1715082712800\alpha A_{0}^{5}A_{1}+177222542187600\alpha A_{1}^{2}-\\ +42652664474400\alpha A_{0}^{2}A_{1}^{2}+3293920537200\alpha A_{1}^{4}A_{1}^{ 2}\\ +20100128241600\alpha A_{0}A_{1}^{3}-4746413620800\alpha A_{0}^{3}A_{1}^{3}\\ -13403385543600\alpha A_{1}^{4}+3616481737200\alpha A_{0}^{2}A_{1}^{4}\\ -810720357600\alpha A_{0}A_{1}^{5}\\ +403446387600\alpha A_{1}^{6})/(1692229135471875\pi^{3}\\ +546983154900000\alpha A_{0}-60108039000000\alpha A_{0}^{3}\\ +2407527322080\alpha A_{0}^{5}+198356528700000\alpha A_{1}\\ +36834206299200\alpha A_{0}^{2}A_{1}-5132047941600\alpha A_{0}^{4}A_{1}\\ -135835682299200\alpha A_{0}A_{1}^{2}+15529485662400\alpha A_{0}^{3}A_{1}^{2} \\ -115997900000\alpha A_{1}^{3}-8333889662400\alpha A_{0}^{2}A_{1}^{3}\\ +8924029941600\alpha A_{0}A_{1}^{4}+193391815584\alpha A_{1}^{5}).\end{array}  \tag{25}\]  
Now with all three coefficients known, the approximate solution is given as in Equation (23), which is quite different from the power series solution produced by other methods [15]. To verify this solution and make comparisons with other known results, the rotation angle at the free end is obtained by setting \(x=1\) in Equation (24) as  
\[\theta_{B}=A_{0}-A_{1}+A_{2}. \tag{26}\]  
For comparison, the exact solution from Equation (9) is [15]  
\[\sqrt{\alpha}=K(\mu^{2})-F(\phi,\mu),\ \ \mu=\sqrt{\frac{1+\sin\theta_{B}}{2}}\ \ \,\ \ \phi=\arcsin \left(\frac{1}{\sqrt{2}\mu}\right). \tag{27}\]  
Here, \(K(\mu)\) is the complete elliptic function of the first kind, and \(F(\phi,\mu)\) is the elliptic function of the second kind. Then, the relationship between \(\alpha\) and \(\theta_{B}\) can be analytically obtained from these equations. The optimal approximate solution by the HAM is given as [15]  
\[\theta_{B}=\frac{\alpha}{2}\frac{f(\alpha)}{g(\alpha)}, \tag{28}\]  
with  
\[