## 2 Wheel-Track Coupling Model
### Flexible-Body Modelling Theory  
To properly simulate flexible multibody dynamics, the primary issue is to reproduce the finite element model in the multibody dynamic model. This procedure is usually achieved through model reduction on the degrees of freedom (DOFs) of finite element model. Various reduction techniques can be applied to the finite element model, such as the Guyan method, dynamic condensation method, and component mode synthesis method [25; 26; 27; 28; 29]. According to the Guyan method, the undamped system vibration equation takes the following form:  
\[M\ddot{a}+Ka=F \tag{1}\]  
where \(M\) and \(K\) are the \(n\times n\) system mass matrix and stiffness matrix, respectively; \(a\) is the \(n\times 1\) system displacement vector; and \(F\) is the \(n\times 1\) load vector. Dividing the displacement vector \(a\) into blocks of \(a_{m}\) and \(a_{s}\) gives:  
\[a=\left[\begin{array}{c}a_{m}\\ a_{s}\end{array}\right] \tag{2}\]  
where \(a_{m}\) is the \(m\times 1\) displacement vector of the master DOFs and \(a_{s}\) is the \((n\)-\(m)\times 1\) displacement vector of the slave DOFs. Assume \(a_{m}\) and \(a_{s}\) have the following relation:  
\[a_{s}=Ta_{m} \tag{3}\]  
where \(T\) is the \(m\times m\) transforming matrix. The system displacement vector can be written in block form as follows:  
\[a=\left[\begin{array}{c}a_{m}\\ a_{s}\end{array}\right]=\left[\begin{array}{c}I\\ T\end{array}\right]a_{m}=T^{*}a_{m} \tag{4}\]where \(T^{*}\) is the \(n\times m\) transforming matrix. Equation (1) can be rewritten in block matrices as:  
\[\left[\begin{array}{cc}M_{mm}&M_{ms}\\ M_{sm}&M_{ss}\end{array}\right]\left[\begin{array}{c}\ddot{a}_{m}\\ \ddot{a}_{s}\end{array}\right]+\left[\begin{array}{cc}K_{mm}&K_{ms}\\ K_{sm}&K_{ss}\end{array}\right]\left[\begin{array}{c}a_{m}\\ a_{s}\end{array}\right]=\left[\begin{array}{c}F_{m}\\ F_{s}\end{array}\right] \tag{5}\]  
By pre-multiplying Equation (1) with \((T^{*})^{T}\) and eliminating \(a_{\rm s}\) with Equation (4), Equation (1) can be written as:  
\[M^{*}\ddot{a}_{m}+K^{*}a_{m}=(T^{*})^{T}F \tag{6}\]  
where  
\[M^{*}=(T^{*})^{T}MT^{*},K^{*}=(T^{*})^{T}KT^{*} \tag{7}\]  
In Equation (6), \(M^{*}\) and \(K^{*}\) are both \(m\times m\) matrices. Thus, the rank of the system DOFs decreases from \(n\) to \(m\).  
In the static condensation process, the equivalent inertia items are omitted. Taking the system internal force \(F_{s}\) as zero, Equation (6) is converted to:  
\[\left[\begin{array}{cc}K_{mm}&K_{ms}\\ K_{sm}&K_{ss}\end{array}\right]\left[\begin{array}{c}a_{m}\\ a_{s}\end{array}\right]=\left[\begin{array}{c}F_{m}\\ 0\end{array}\right] \tag{8}\]  
Solving Equation (8) for \(a_{\rm s}\) gives:  
\[a_{s}=-{K_{ss}}^{-1}K_{sm}a_{m}=Ta_{m} \tag{9}\]  
where  
\[T=-{K_{ss}}^{-1}K_{sm} \tag{10}\]  
The reduced mass matrix \(M^{*}\) and stiffness matrix \(K^{*}\) can be obtained by substituting Equation (9) into Equation (7):  
\[\begin{array}{l}M^{*}=M_{mm}-{K_{sm}}^{T}{K_{ss}}^{-1}M_{sm}-M_{ms}{K_{ss}}^ {-1}K_{sm}+{K_{sm}}^{T}{K_{ss}}^{-1}M_{ss}{K_{ss}}^{-1}K_{sm},\\ K^{*}=K_{mm}-{K_{sm}}^{T}{K_{ss}}^{-1}K_{sm}\end{array} \tag{11}\]  
From the above equations, it can be seen that the Guyan method is based on Equation (9). The error of this method can be obtained by solving Equation (5):  
\[a_{\rm s}=-{K_{ss}}^{-1}\big{[}M_{sm}\ddot{a}_{m}+M_{ss}\ddot{a}_{s}+K_{sm}a_{ m}\big{]} \tag{12}\]  
Differentiating Equation (12) twice with respect to time and comparing it with Equation (9), it is found that the method has exact solutions whenever:  
\[{K_{ss}}{M_{ss}}^{-1}={K_{sm}}{M_{sm}}^{-1} \tag{13}\]