Fast prediction of chatter stability in milling process based on an updated numerical solution scheme

Abstract

The stability prediction is an effective approach to suppress the unstable milling process caused by the regenerative chatter. Through updating a numerical solution scheme (NSS), a fast prediction method is proposed in this paper. Firstly, a NSS is constructed by combining the Lagrange polynomial and Euler’s method. Then, the solution of the differential equations described the milling process is transformed into the initial-value problem of the equation which is expressed by the built NSS over each small discrete time interval. Subsequently, the state transition matrix over one tooth passing period is established to predict the stability lobe diagram (SLD) via the Floquet theory. Finally, the computational performances of the proposed method are verified by the single and two DOF dynamic models. Compared with the complete discretization scheme, the computational accuracy and efficiency of the proposed method are increased by 2.69 times and 2.01 times, respectively. The proposed method reduces the computational accuracy by 2.16 times in comparison to the first-order full-discretization method, but it increases the computational efficiency by 2.2 times.

Appendix 1 Coefficient matrixes for single and two DOF milling systems

The coefficient matrixes for the single DOF milling system can be described as

$$\begin{array}{l}\mathbf{A}=\left[\begin{array}{cc}-\xi {\omega }_{n}& \frac{1}{{m}_{t}}\\ \left({\xi }^{2}-1\right){m}_{t}{\omega }_{n}^{2}& -\xi {\omega }_{n}\end{array}\right]\\ \mathbf{B}\left(t\right)=\left[\begin{array}{cc}0& 0\\ wh\left(t\right)& 0\end{array}\right]\end{array}$$

where h(t) is the cutting force coefficient.

The coefficient matrixes for the two DOF milling system can be described as

$$\begin{array}{l}\mathbf{A}=\left[\begin{array}{cc}-{\mathbf{M}}^{-1}\mathbf{C}/2& {\mathbf{M}}^{-1}\\ {\mathbf{C}\mathbf{M}}^{-1}\mathbf{C}/4-\mathbf{K}& -{\mathbf{C}\mathbf{M}}^{-1}/2\end{array}\right]\\ \mathbf{B}\left(t\right)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ {wh}_{xx}\left(t\right)& {wh}_{xy}\left(t\right)& 0& 0\\ {wh}_{yx}\left(t\right)& {wh}_{yy}\left(t\right)& 0& 0\end{array}\right]\end{array}$$

with

$$\mathbf{M}=\begin{array}{cc}\left[\begin{array}{l}{m}_{t}\\ \begin{array}{cc}& {m}_{t}\end{array}\end{array}\right]& \mathbf{C}=\end{array}\begin{array}{cc}\left[\begin{array}{l}{2m}_{t}\xi \\ \begin{array}{cc}& {2m}_{t}\xi \end{array}\end{array}\right]& \mathbf{K}=\end{array}\left[\begin{array}{l}{m}_{t}{\omega }_{n}^{2}\\ \begin{array}{cc}& {m}_{t}{\omega }_{n}^{2}\end{array}\end{array}\right]$$