## 2 Study of the Axial Strength of Windings under Short-Circuit Force  
In order to study the axial mechanical stress on windings under short-circuit forces, a mechanical stress calculation model for windings was established, as shown in Figure 1.  
For transformer windings, the axial electromagnetic forces generated by radial leakage magnetic fields cause axial bending deformations in the windings between two sets of cushion blocks. It is considered that the windings and the two sets of cushion blocks constitute a fixed pivot beam structure at both ends [24]. The axial force load on the winding is uniformly distributed. In Figure 1, \(q_{1}\) represents the force per unit length acting downward along the winding, while \(R_{1}\) and \(R_{2}\) are the support reactions at both ends. \(M_{1}\) and \(M_{2}\) denote the bending moments, \(w\) stands for the height of the winding, and \(L\) represents the length of the winding. To determine the value of the bending stress, we must first derive the force per unit length \(q_{\text{st}}\) acting on the wires between the two sets of cushion blocks:  
\[q_{\text{st}}=\frac{F_{\text{Y}}}{N_{\text{s}}\pi D} \tag{1}\]  
where \(N_{\text{s}}\) denotes the number of strands of the winding, \(D\) denotes the average diameter of the winding, and \(F_{\text{Y}}\) denotes the maximum axial force applied to the wire cake. There are six unknown quantities in the problem, two vertical forces, two horizontal forces, and two bending moments, but only three static equilibrium equations are presented, making it a hyperstatic problem. The two horizontal forces in opposite directions are of equal magnitude, exerting mutual tensile forces on each other. Due to the small magnitude of these forces, they can be disregarded [24]. The two vertical forces are of equal magnitude and act downward in the same direction. Therefore, the support reactions happen at both ends, \(R_{1}=R_{2}=\)\(q_{1}L/2\), where \(q_{1}\) is the force per unit length acting downward along the beam. Given that this structure is subjected to uniformly distributed axial force loads, \(q_{1}\) is equal to \(q_{\text{st}}\). The moment of inertia about the z-axis can be expressed as follows:  
\[I_{\text{z}}=\frac{tw^{3}}{12} \tag{2}\]  
where \(t\) denotes the radial width of the winding. Figure 1: Winding mechanical stress calculation model. Cushion block constraints can be regarded as fixed-end constraints. Because the force of the winding is symmetrical and the constraint form at both ends is the same, the bending moment constrained by the winding is equal, that is, \(M_{1}\) is equal to \(M_{2}\). The total bending moment at position \(x\) can be expressed as follows:  
\[M(x)=\frac{q_{1}Lx}{2}-\frac{q_{1}x^{2}}{2}-M_{1} \tag{3}\]  
The deflection curve equation of the beam should be calculated firstly to obtain the bending moment \(M_{1}\):  
\[\frac{d^{2}y}{dx^{2}}=-\frac{M(x)}{EI_{z}} \tag{4}\]  
where \(E\) is the elastic modulus of the beam. Substituting (3) into (4), (4) can then be re-written as follows:  
\[\frac{d^{2}y}{dx^{2}}=-\frac{1}{EI_{z}}\bigg{(}M_{1}-\frac{q_{1}Lx}{2}+\frac{q _{1}x^{2}}{2}\bigg{)} \tag{5}\]  
Integrating (5) once gives the following:  
\[\frac{dy}{dx}=\frac{1}{EI_{z}}\bigg{(}M_{1}x-\frac{q_{1}Lx^{2}}{2}+\frac{q_{1 }x^{3}}{2}\bigg{)}+C \tag{6}\]  
where \(C\) is the integration constant. Since we are investigating a beam structure with fixed supports, it follows that \(dy/dx=0\) at \(x=0\) and \(x=L\).