## 3 Deviation Analysis of New Integrated TBM Disc Cutter
### Analysis of Assembly Deviation of Thread Fastening System
#### 3.1.2 Spinor Representation of Tolerance Zone of the Dimension Chains  
(1) Dimension chain 1  
(1) Between bearing 1 and cutter holder  
There are two parallel sub-connecting pairs between bearing 1 and the cutter holder. The lower end face of bearing 1 and the end face of the hole on the cutter holder form a face-to-face connecting pair, \(T_{1}\). The clearance between the outer ring of bearing 1 and the hole's wall on the cutter holder is matched to form a cylindrical surface to the cylindrical surface connecting pair, \(T_{2}\). Figure 4: Part tolerances of thread fastening system. 1—Fastening bolt, 2—Cutter holder. Figure 3: Dimensional chains of threaded fastening system. According to the accuracy requirements, the tolerance field spinor of \(T_{1}\) is as follows. \[\left\{ \begin{array}{l}u_{C1}=w_{C1}=\frac{0.022+0.015+0.02+0.02}{2}=0.0385\\ v_{C1}=0\\ a_{C1}=\delta_{C1}=\frac{0.022+0.015+0.02+0.02}{2.25}=0.0342\\ \beta_{C1}=0\end{array} \right. \tag{8}\]  
So,  
\[T_{1}=\left[0.0385\quad 0\quad 0.0385\quad 0.0342\quad 0\quad 0.0342\right]^{T} \tag{9}\]  
According to the accuracy requirements, the tolerance field spinor of \(T_{2}\) is as follows. \[\left\{ \begin{array}{l}u_{P2}=w_{P2}=0\\ v_{P2}=0.02+0.02=0.0400\\ a_{P2}=\delta_{P2}=\frac{0.02+0.02}{10.667}=0.00375\\ \beta_{P2}=0\end{array} \right. \tag{10}\]  
So,  
\[T_{2}=\left[0\quad 0.0400\quad 0\quad 0.00375\quad 0\quad 0.00375\right]^{T} \tag{11}\]  
According to the solution method of the local parallel dimension chain [18], the union operation is performed on the three translation parameters of two parallel sub-connecting pairs, and the intersection operation is performed on the three rotation parameters. So,  
\[CFE_{1}=\left[0.0385\quad 0.0400\quad 0.0385\quad 0.00375\quad 0\quad 0.00375 \right]^{T} \tag{12}\]  
2 Between bearing 1 and fastening bolt  
There is an interference fit between bearing 1 and the fastening bolt. Therefore, only the surface connection pair composed of the upper-end face of bearing 1 and the shoulder end face of the fastening bolt is considered, and its tolerance field spinor is as follows. \[\left\{ \begin{array}{l}u_{P2}=w_{P2}=0\\ v_{P1}=0.02+0.2+0.1=0.32\\ a_{P1}=\delta_{P1}=\frac{0.02+0.2+0.1}{8.75}=0.0366\\ \beta_{P1}=0\end{array} \right. \tag{13}\]  
So,  
\[CFE_{2}=\left[0\quad 0.320\quad 0\quad 0.0366\quad 0\quad 0.0366\right]^{T} \tag{14}\]  
Then, the Jacobian-Torsor model of dimension chain 1 is as follows. \[PFE_{1}=\left[ \begin{array}{l}J_{CFE1}\\ J_{IFE3}\\ J_{CFE2}\end{array} \right]^{T}\left[ \begin{array}{l}[CE1]\\ [IFE3]\\ [CFE2]\end{array} \right]=\left[ \begin{array}{l}0.0385\\ 0.36\\ 0.0385\\ 0.0416\\ 0\\ 0.0416\end{array} \right] \tag{15}\]  
where  
\[ \begin{array}{l}J_{CFE1}=\text{diag}(1,1,1,1,1,1)_{CFE1}\\ J_{CFE2}=\text{diag}(1,1,1,1,1,1)_{CFE2}\\ J_{IFE3}=\text{diag}(1,1,1,1,1,1)_{IFE3}\\ [IFE3]=0\end{array} \](2)  
Dimension chain 2  
1 Between bearing 2 and cutter holder  
There is a clearance fit between the outer ring of bearing 2 and the mounting hole of the cutter base, so it forms a cylindrical to cylindrical connecting pair. According to the accuracy requirements, its tolerance field spinor is as follows. \[\left\{ \begin{array}{l}u_{\text{C3}}=w_{\text{C3}}=\frac{0.022+0.015+0.02+0.02} {2.5}=0.0385\\ \alpha_{\text{C3}}=\delta_{\text{C3}}=\frac{0.022+0.015+0.02+0.02}{2.5}=0.0308 \end{array} \right. \tag{16}\]  
So,  
\[CFE_{3}=\left[0.0385\quad 0\quad 0.0385\quad 0.0308\quad 0\quad 0.0308\right]^{T} \tag{17}\]  
There is an interference fit between the inner ring of bearing 2 and the fastening bolt. Therefore, only the surface connection pair composed of the lower end face of bearing 2 and the end face of the shoulder of the fastening bolt is considered. According to the accuracy requirements, its tolerance field spinor is as follows. \[\left\{ \begin{array}{l}u_{\text{C2}}=w_{\text{C2}}=\frac{0.2+0.1+0.02}{2.5}= 0.160\\ \alpha_{\text{C2}}=\delta_{\text{C2}}=\frac{0.2+0.1+0.02}{1.25}=0.256\end{array} \right.