# Experimental and Numerical Study of the Jib Connection Frame of a Wheeled Crane
## 4 Calculation Method  
The preceding analysis revealed that in addition to chord 2, the transverse connection of the slewing plane is also susceptible to instability in this design scheme. This study proposes a structural treatment for the structure that is analogous to that described in Figure 18, which is simplified into the form shown in Figure 19. The direct connection is reduced to a hinged connection at both extremities, while the intermediate diagonal connection is transformed into an elastic support. Figure 20 illustrates the local deformation. The modulus of elasticity of the chord is designated as E, while the moment of inertia is represented by I. The chord is then supported in a straightforward manner.  
Figure 19: Simplified model.  
Figure 20: Local simplified model.  
The chord may be simplified as a compression rod with an elastic support in the middle. If the chord exhibits a buckling behavior under pressure, the reaction force \(Q\), generated by the support at hinge point C, will be proportional to the deflection \(\delta\), thereby representing the spring constant.  
\[Q=\alpha\delta \tag{1}\]  
To obtain the reaction forces, both spans are considered to be simply supported beams.  
\[Q=\frac{P\delta}{l_{1}}+\frac{P\delta}{l_{2}}+\frac{M_{1}}{l_{1}}+\frac{M_{2}} {l_{2}} \tag{2}\]  
We define \(k^{2}=\frac{P}{EI}\), which is substituted into Equation (2), yielding:  
\[Q=\frac{k^{2}EI\delta}{l_{1}}+\frac{k^{2}EI\delta}{l_{2}}+\frac{M_{1}}{l_{1}}+ \frac{M_{2}}{l_{2}} \tag{3}\]  
According to [PERSON] [30], the deflection curve of two force couples acting on both ends of the rod is:  
\[y=\frac{M_{2}}{P}\left(\frac{\sin kx}{\sin kl}-\frac{x}{l}\right)+\frac{M_{1}} {P}\left(\frac{\sin k(l-x)}{\sin kl}-\frac{l-x}{l}\right) \tag{4}\]  
The deflection curve equation for a transverse load \(Q\) is  
\[y=\left\{\begin{array}{l}\frac{Q\sin kl_{2}}{Pk\sin kl}\sin kx-\frac{Ql_{1} }{P}x,(0\leq x\leq l_{1})\\ \frac{Q\sin kl_{1}}{Pk\sin kl}\sin k(l-x)-\frac{Ql_{1}}{Pl}(l-x),(l_{1}\leq x \leq l)\end{array}\right. \tag{5}\]  
The deflection equation of the simplified model presented in this article can be obtained by employing Formulas (4) and (5), as illustrated in Formula (6):  
\[y=\frac{M_{2}}{P}\left(\frac{\sin kx}{\sin kl}-\frac{x}{l}\right)+\frac{M_{1}} {P}\left(\frac{\sin k(l-x)}{\sin kl}-\frac{l-x}{l}\right)-\frac{Q\sin kl_{2}}{ Pk\sin kl}\sin kx+\frac{Ql_{2}}{Pl}x \tag{6}\]  
At the elastic support \(y_{(x=l_{1})}=\delta\)  
\[\frac{M_{2}}{P}\left(\frac{\sin kl_{1}}{\sin kl}-\frac{l_{1}}{l}\right)+\frac {M_{1}}{P}\left(\frac{\sin k(l-l_{1})}{\sin kl}-\frac{l-l_{1}}{l}\right)- \frac{Q\sin kl_{2}}{Pk\sin kl}\sin kl_{1}+\frac{Ql_{2}}{Pl}l_{1}=\frac{Q}{\alpha} \tag{7}\]  
To proceed, it is necessary to substitute the spring constant \(\alpha=\frac{48 EI}{P}\).  
The force point C in this article is the midpoint of the chord, and the special case \(l_{1}=l_{2}=\frac{l}{2}\) is substituted into Formulas (3) and (7):  
\[\left\{\begin{array}{l}(M_{1}+M_{2})+\left(\frac{k^{2}l^{3}}{24}-\frac{l}{ 2}\right)Q=0\\ \left(\frac{\sin kl/2}{\sin kl}-\frac{1}{2}\right)\frac{(M_{1}+M_{2})}{k^{2} EI}-\left(\frac{\sin^{2}kl/2}{k^{3}\sin kl}-\frac{l}{4k^{2}}+\frac{l^{3}}{48} \right)\frac{Q}{EI}=0\end{array}\right. \tag{8}\]  
For the two simultaneous formulations in Equation (8) to hold true, it is necessary that their determinant is equal to zero.  
\[\left|\begin{array}{l}1\\ \left(\frac{\sin kl/2}{\sin kl}-\frac{1}{2}\right)/k^{2}EI\end{array}\right. \left(\frac{\sin^{2}kl/2}{k^{3}\sin kl}-\frac{l}{4k^{2}}+\frac{l^{3}}{48} \right)/EI\right|=0 \tag{9}\]  
The stable equation is obtained as:  
\[\left(\frac{\sin^{2}kl/2}{k^{3}\sin kl}-\frac{l}{4k^{2}}+\frac{l^{3}}{48} \right)/EI-\left(\frac{k^{2}l^{3}}{24}-\frac{l}{2}\right)\left(\frac{\sin kl/2 }{\sin kl}-\frac{1}{2}\right)/k^{2}EI=0 \tag{10}\]Introducing symbolom \(u=kl\), it can be determined that:  
\[\frac{\sin^{2}(u/2)}{u\sin u}-\frac{1}{4}+\frac{u^{2}}{48}-\left(\frac{u^{2}}{24}- \frac{1}{2}\right)\left(\frac{\sin(u/2)}{\sin u}-\frac{1}{2}\right)=0 \tag{11}\]