## 2 Materials and Methods
### Calculation Method of Flexural Rigidity  
In order to solve the calculation of Equation (5), it is indispensable to acquire the value of _EI_. Considering the large deflection of the stalk, and to avoid the calculation of elliptical integration, [PERSON]'s formula was used to compute the value of _EI_. The force analysis is shown in Figure 2. Stalk AO is deformed to A'O under the action of the horizontal force \(F\), the longitudinal displacement, and the deflection of point A are \(\delta_{A}\), \(\Delta_{A}\), respectively. The deformed stalk is then divided into eight equal segments along the longitudinal direction in order to apply [PERSON]'s formula [21] to the calculation.  
Equation (4) can be derived as  
\[\frac{\Delta^{\prime\prime}}{\sqrt{1+\Delta^{\prime 2}}^{1.5}}=\cos\theta\frac{ \mathrm{d}\theta}{\mathrm{d}y}=-\frac{M(y)}{EI} \tag{6}\]  
and the sum of the bending moment of point \(y\) is  
\[M(\Delta)=-F(l-\delta_{A}-y) \tag{7}\]  
The following equation can be obtained by integrating Equation (6) with Equation (7), and substituted into  
\[\sin\theta=-\frac{F(l-\delta_{A}-y)^{2}}{2 EI}+\frac{F(l-\delta_{A})^{2}}{2 EI} \tag{8}\]  
According to the boundary conditions that \(y\) = \(l\) - \(\delta_{A}\) and _th_ = _th_A\({}^{\prime}\), Equation (9) could be obtained:  
\[\sin\theta_{A^{\prime}}=\frac{F(l-\delta_{A})^{2}}{2 EI} \tag{9}\]  
Figure 2: Deflection model of stalk AO. \(\delta_{A}\) vertical displacement of point A, mm; \(\Delta_{A}\) deflection of point A, (mm); \(l\), length of the stalk AO, mm; \(\theta_{A^{\prime}}\), angle between the tangent line of point A and the axis \(Y\), (°); \(\theta_{i}\), angle between _A\({}^{{}^{\prime}}\)Y\({}_{i}\)_ and the axis \(Y\), (°).  
On the basis of Equation (9), the vertical displacement of point A could be acquired:  
\[\delta_{A}=l-\sqrt{\frac{2 EI}{F}\sin\theta_{A^{\prime}}} \tag{10}\]  
\[S=\int_{s}ds=\int_{0}^{l-\delta_{A}}\sqrt{1+\Delta^{\prime 2}}dy=\int_{0}^{l- \delta_{A}}\frac{1}{\sqrt{1-Sin^{2}\theta}}dy \tag{11}\]  
where \(S\) is the length of arc A\({}^{\prime}\)O, mm.  
\[\Delta_{A}=\int_{s}Sin\theta ds=\int_{0}^{l-\delta_{A}}\frac{Sin\theta}{\sqrt{ 1-Sin^{2}\theta}}dy \tag{12}\]  
where \(\Delta_{A}\) is the deflection of point A, mm.  
[PERSON]'s formula can be presented as  
\[\int_{b}^{a}f(x)dx=\frac{h}{3}\Bigg{[}f(a)+4\sum_{k=1}^{n}f(x_{2k-1})+2\sum_{k =1}^{n-1}f(x_{2k})+f(b)\Bigg{]} \tag{13}\]  
In order to apply Equation (13), A\({}^{\prime}\)O is divided into eight segments, that is, the area between [0, \(l_{A}-\delta_{A}\)] is partitioned into eight equal segments, where each length of the segment is \(h=(l-\delta_{A})/8\). Then Equations (11) and (12) could be modified as  
\[\begin{array}{c}S=\int_{0}^{l-\delta_{A}}\frac{1}{\sqrt{1-Sin^{2}\theta}}dy =\\ \frac{l-\delta_{A}}{24}\Bigg{[}1+4\sum_{k=1}^{n}\frac{1}{\sqrt{1-Sin^{2}\theta _{2k-1}}}+2\sum_{k=1}^{n-1}\frac{1}{\sqrt{1-Sin^{2}\theta_{2k}}}+\frac{1}{ \sqrt{1-Sin^{2}\theta_{A^{\prime}}}}\Bigg{]}=l\end{array} \tag{14}\]  
\[\begin{array}{c}\Delta_{A}=\int_{0}^{l-\delta_{A}}\frac{Sin\theta}{\sqrt{1- Sin^{2}\theta}}dy=\\ \frac{l-\delta_{A}}{24}\Bigg{[}4\sum_{k=1}^{n}\frac{Sin\theta_{2k-1}}{\sqrt{1- Sin^{2}\theta_{2k-1}}}+2\sum_{k=1}^{n-1}\frac{Sin\theta_{2k}}{\sqrt{1-Sin^{2} \theta_{2k}}}+\frac{Sin\theta_{A^{\prime}}}{\sqrt{1-Sin^{2}\theta_{A^{\prime }}}}\Bigg{]}\end{array} \tag{15}\]  
where \(Sin\theta\) is defined by Equation (8). Considering the \(y\) value of point \(i\):  
\[y_{i}=0+ih=\frac{i}{2n}(l-\delta_{A}) \tag{16}\]  
\[\sin\theta_{i}=-\frac{F(l-\delta_{A}-y_{i})}{2 EI}+\frac{F(l-\delta_{A})^{2}}{ 2 EI}(i=1,2,\ldots,8) \tag{17}\]  
Combine Equations (9), (16) and (17), \(Sin\theta_{i}\) can be presented as:  
\[Sin\theta_{i}=\Bigg{[}1-\bigg{(}1-\frac{i}{2n}\bigg{)}^{2}\Bigg{]}Sin\theta_{ A^{\prime}} \tag{18}\]  
\(EI\) could be solved by substituting Equations (10) and (17) into Equations (14) and (15), and the deflection curve of stalk AO can be calculated by inputting the value of \(EI\) into Equation (5).