## 3 Modeling of the Metallic Braided Tube
### Equivalent Theory Based on Axial Deformation  
The key to generating the equivalent strand model is to obtain the axial deformation of the braided tubes. Therefore, it assumes one strand as one wire for force analysis and the section shape of the wire is still a circle. Figure 5a shows the helix wire, where \(r\) is the radius of rotation. For the sake of analysis, the unfolded helix wire is used, as Figure 5b shows, where OD is the length of the projection of the wire on the XY plane, which can be expressed as \(2\pi r\); DA is the pitch, which can be expressed as \(P\); OA is the length of wire, which can be expressed in term of \(S\); and \(\alpha\) is the initial braiding angle. When the metallic braided tube is stretched by axial force, \(F_{axis}\) is the force along the axial direction, \(F_{wire}\) is the force along the metallic wire's direction, and the relationship between \(S\), \(P\), \(r\), and \(\alpha\) based on the triangle can be expressed as:  
\[S=2\pi r\cdot\text{sin}\alpha \tag{5}\]  
\[P=\frac{2\pi r}{\tan\alpha} \tag{6}\]  
\[F_{axis}=F_{wire}\cdot\text{cos}\alpha \tag{7}\]  
When the metallic braided tube is subjected to axial force, the braiding angle changes from \(\alpha\) to \(\alpha_{0}\), the axial length changes from \(P\) to \(P_{0}\), and the length of the wire changes from  
Figure 4: Modified helix line. Figure 3: The trajectory of the braiding wire. \(S\) to \(S_{0}\). Since both ends of the metallic braided tube are welded to the joint, the projection length of the wire on the XY plane remains unchanged. Figure 5c shows the detailed force analysis of the wire. Supposing that the wire OA is fixed at O, the force \(F_{axis}\) acts on the point A in the direction of AC, and the initial braiding angle is \(\alpha\). In the initial stage, \(F_{axis}\) can be decomposed into two separate forces \(F_{n}\) and \(F_{t}\) along different directions, \(F_{n}\) along the normal direction and \(F_{t}\) along the wire direction. Therefore, the axial displacement \(\Delta P\) is a combination of displacement in two directions, \(w_{n}\) caused by normal force \(F_{n}\), and the other is \(\Delta S\), which is caused by wire force \(F_{t}\), where \(w_{n}\) and \(\Delta S\) can be obtained by deflection calculation of a cantilever beam loaded with a concentrated force according to the laws of material mechanics [13]. \[w_{n}=\frac{F_{n}S^{3}}{3 EI} \tag{8}\]  
where \(S\) is the length of wire, \(E\) is the elastic modulus, and \(I\) is the cross-sectional moment of inertia. When the cross section is a solid circular section, the Moment of Inertia is \(I=\frac{\pi d^{4}}{6\Gamma}\), where \(d\) is the diameter of the section. \[\Delta S=S\cdot\epsilon_{t}=\frac{F_{t}S}{EA} \tag{9}\]  
\[\epsilon_{t}=\frac{F_{t}}{EA} \tag{10}\]  
where \(A\) is the cross-sectional area, \(\Delta S\) is the axial elongation, and \(\epsilon_{t}\) is the strain along the wire. According to the Pythagorean theorem, the elongation of the braided tube \(\Delta P\) can be expressed as:  
\[\Delta P=\sqrt{{w_{n}}^{2}+\Delta S^{2}} \tag{11}\]  
\(\Delta P\) is the axial displacement generated by the wire under axial force. Then, the axial stiffness curve of the metallic braided tube can be obtained by comparing F and \(\Delta P\). The above derived axial displacement is based on the equivalence of treating one strand as a single wire for force analysis, while in the actual metallic braided tube, each wire in the strand is suffering axial force. For force analysis of multiple wires, it is assumed that each wire in the strand is independent and will not be disturbed by other wires when suffering axial force. Figure 6 shows the axial force diagram, in which the blue lines are the wires' deformations after suffering axial force.