Milling Dynamics

Abstract

This chapter describes regenerative chatter in milling. Similar to Chap. 3, both analytical and numerical analyses are presented to predict the process behavior. The analytical, frequency domain stability lobe diagram is adapted to milling. Two algorithms are described: average tooth angle and Fourier series. Time domain simulations for square (straight teeth and helical teeth) and ball endmills (helical teeth) are detailed. The cutting force model is then extended and an experimental procedure for identifying the cutting force coefficients is derived for two different fitting approaches. Finally, the effect of process damping on milling stability at low speeds is presented.

If we knew what it was we were doing, it would not be called research, would it?

—Albert Einstein

Appendices

Exercises

• 1. Compute the start and exit angles for the following milling cases.

  • (a) Up milling, 30% radial immersion

  • (b) Down milling, 40% radial immersion

• 2. Determine the tooth passing frequency for a cutter with three teeth rotating at 10,000 rpm.

• 3. Calculate the maximum y direction force for a 40% radial immersion down milling cut carried out using a three tooth end mill; see Fig. 4.82. The material-tool combination gives: k_t = 720 N/mm² and k_n = 200 N/mm² (aluminum alloy). Also, b = 2.5 mm and f_t = 0.2 mm/tooth. Assume a rigid cutting tool and workpiece.

• 4. For the average tooth angle milling stability analysis, complete parts (a) through (d). A 35% radial immersion up milling cut is to be performed using a square end mill with four teeth and the force angle, β, is 68 deg. See Fig. 4.83.

  • (a) Determine the average angle of a tooth in the cut.

  • (b) Calculate the directional orientation factors.

  • (c) Compute the oriented frequency response function and identify the valid chatter frequency range(s) in Hz. The x direction dynamics are given by: f_nx = 1000 Hz, k_x = 7 × 10⁶ N/m, and ζ_x = 0.03 and the y direction dynamics are: f_ny = 1200 Hz, k_y = 6 × 10⁷ N/m, and ζ_y = 0.04. You may assume that these single degree of freedom parameters were obtained from a modal fit to frequency responses measured in the x and y directions with bandwidths of 0–2500 Hz.

  • (d) Plot the first five stability lobes, N = 0 to 4. The workpiece material is 1020 carbon steel; see Table 3.1.

• 5. For the cut described in Exercise 4, use the Fourier approach to obtain the stability lobe diagram (j = 0 to 4).

• 6. Calculate the lowest axial depth for which a constant cutting force is obtained independent of the selected radial depth of cut. The helical square end mill has a diameter of 10 mm, six teeth, and a 42 deg helix angle.

• 7. For a particular milling application, the spindle speed is 6000 rpm. The x and y direction cutting forces are shown in Fig. 4.84. The coordinate system is defined in Fig. 4.85. Only one tooth is engaged in the cut at any time and the cut is stable. The cutter teeth are straight.

  • (a) Is this cut up or down milling?

  • (b) How many teeth are on the milling cutter?

  • (c) If the Fourier transform for either cutting force signal (x or y direction) were computed, at what frequencies (in Hz) would peaks be expected?

  • (d) Determine k_t and k_n, where the tangential and normal force components are expressed as: F_t = k_tbh and F_n = k_nbh, where b = 5 mm, f_t = 0.2 mm/tooth, and the circular tool path approximation may be applied.

• 8. An endmill with four teeth is being used in a cut with an entry (starting) angle of 15 deg and an exit angle of 75 deg.

  • (a) Determine the average number of teeth in the cut (average tooth angle approach).

  • (b) What is the average angle of the surface normal for a tooth engaged in the cut (average tooth angle approach)? Sketch the cut and identify the surface normal and angle on your sketch.

    

  • (c) If β = 60 deg (average tooth angle approach), determine the directional orientation factors.

Fig. 4.82

Down milling force geometry

Fig. 4.83

35% radial immersion up milling geometry

Fig. 4.84

Time domain forces in the x and y directions

Fig. 4.85

Coordinate system for forces

Appendix: Reformulation of Fourier Series Eigenvalue Problem

In Sect. 4.3.3, we detailed the truncated Fourier series approach to obtaining the analytical stability limit for milling [5]. The dynamic milling equation (Eq. 4.49) was presented as:

$$ \left(\begin{array}{c}{F}_x\\ {}{F}_y\end{array}\right){\mathrm{e}}^{i{\omega}_{\mathrm{c}}t}=\frac{1}{2}{bK}_{\mathrm{t}}\left[{A}_0\right]\left(1-{\mathrm{e}}^{i{\omega}_{\mathrm{c}}\tau}\right)\left[\begin{array}{cc}{\mathrm{FRF}}_{xx}& 0\\ {}0& {\mathrm{FRF}}_{yy}\end{array}\right]\left(\begin{array}{c}{F}_x\\ {}{F}_y\end{array}\right){\mathrm{e}}^{i{\omega}_{\mathrm{c}}t}, $$

(4.117)

which is true if $\left(\begin{array}{c}{F}_x\\ {}{F}_y\end{array}\right){\mathrm{e}}^{i{\omega}_{\mathrm{c}}t}=\left[\begin{array}{cc}1& 0\\ {}0& 1\end{array}\right]\left(\begin{array}{c}{F}_x\\ {}{F}_y\end{array}\right){\mathrm{e}}^{i{\omega}_{\mathrm{c}}t}=\left[I\right]\left(\begin{array}{c}{F}_x\\ {}{F}_y\end{array}\right){\mathrm{e}}^{i{\omega}_{\mathrm{c}}t}$. This enables us to write $\frac{1}{2}{bK}_{\mathrm{t}}\left[{A}_0\right]\left(1-{\mathrm{e}}^{i{\omega}_{\mathrm{c}}\tau}\right)\left[\begin{array}{cc}{\mathrm{FRF}}_{xx}& 0\\ {}0& {\mathrm{FRF}}_{yy}\end{array}\right]=\left[I\right]$ or $\frac{N_{\mathrm{t}}}{4\pi }{bK}_{\mathrm{t}}\left(1-{\mathrm{e}}^{-i{\omega}_{\mathrm{c}}\tau}\right)\left[{\mathrm{FRF}}_{\mathrm{or}}\right]=\left[I\right]$. The eigenvalue problem was then posed in [5] as:

$$ \det \left(\left[I\right]+\Lambda \left[{\mathrm{FRF}}_{\mathrm{or}}\right]\right)=0, $$

(4.118)

where $\Lambda =-\frac{N_{\mathrm{t}}}{4\pi }{bK}_{\mathrm{t}}\left(1-{\mathrm{e}}^{-i{\omega}_{\mathrm{c}}\tau}\right)$ gives the eigenvalues. To apply the Matlab^® eig function, we must restate the eigenvalue problem as det([FRF_or] − λ[I]) = 0. The new complex eigenvalues are therefore $\lambda ={\lambda}_{\mathrm{Re}}+i{\lambda}_{\mathrm{Im}}=\frac{4\pi }{N_{\mathrm{t}}}\frac{1}{bK_{\mathrm{t}}\left(1-{\mathrm{e}}^{-i{\omega}_{\mathrm{c}}\tau}\right)}$. We determine the corresponding stability limit by solving this expression for b, rationalizing the result, and then substituting for ${\mathrm{e}}^{-i{\omega}_{\mathrm{c}}\tau }$ using the Euler identity ${\mathrm{e}}^{-i{\omega}_{\mathrm{c}}\tau }=\cos \left({\omega}_{\mathrm{c}}\tau \right)-i\sin \left({\omega}_{\mathrm{c}}\tau \right)$.

$$ {\displaystyle \begin{array}{l}{\tilde{b}}_{\mathrm{lim}}=\frac{4\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda}_{\mathrm{Re}}+i{\lambda}_{\mathrm{Im}}\right)\left(1-{\mathrm{e}}^{-i{\omega}_{\mathrm{c}}\tau}\right)}\\ {}{\tilde{b}}_{\mathrm{lim}}=\frac{4\pi \left({\lambda}_{\mathrm{Re}}-i{\lambda}_{\mathrm{Im}}\right)}{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda}_{\mathrm{Re}}+i{\lambda}_{\mathrm{Im}}\right)\left({\lambda}_{\mathrm{Re}}-i{\lambda}_{\mathrm{Im}}\right)\left(1-{\mathrm{e}}^{-i{\omega}_{\mathrm{c}}\tau}\right)}\\ {}{\tilde{b}}_{\mathrm{lim}}=\frac{4\pi \left({\lambda}_{\mathrm{Re}}-i{\lambda}_{\mathrm{Im}}\right)}{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)+i\sin \left({\omega}_{\mathrm{c}}\tau \right)\right)}\end{array}} $$

(4.119)

After the Euler identity substitution, we again rationalize to obtain Eq. 4.120.

$$ {\displaystyle \begin{array}{l}{\tilde{b}}_{\mathrm{lim}}=\frac{4\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)}\frac{\left({\lambda}_{\mathrm{Re}}-i{\lambda}_{\mathrm{Im}}\right)\left(1+\cos \left({\omega}_{\mathrm{c}}\tau \right)-i\sin \left({\omega}_{\mathrm{c}}\tau \right)\right)}{\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)+i\sin \left({\omega}_{\mathrm{c}}\tau \right)\right)\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)-i\sin \left({\omega}_{\mathrm{c}}\tau \right)\right)}\\ {}{\tilde{b}}_{\mathrm{lim}}=\frac{4\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)}\frac{\left({\lambda}_{\mathrm{Re}}-i{\lambda}_{\mathrm{Im}}\right)\left(1+\cos \left({\omega}_{\mathrm{c}}\tau \right)-i\sin \left({\omega}_{\mathrm{c}}\tau \right)\right)}{\left(2-2\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)}\\ {}{\tilde{b}}_{\mathrm{lim}}=\frac{2\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)}\frac{\left(\begin{array}{l}\left({\lambda}_{\mathrm{Re}}\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)+{\lambda}_{\mathrm{Im}}\sin \left({\omega}_{\mathrm{c}}\tau \right)\right)\\ {}+i\left({\lambda}_{\mathrm{Im}}\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)-{\lambda}_{\mathrm{Re}}\sin \left({\omega}_{\mathrm{c}}\tau \right)\right)\end{array}\right)}{\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)}\end{array}} $$

(4.120)

Because ${\tilde{b}}_{\mathrm{lim}}$ must be real valued, the imaginary part from the parenthetical portion of the numerator in the final line in Eq. 4.120 must be equal to zero:

$$ {\lambda}_{\mathrm{Im}}\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)-{\lambda}_{\mathrm{Re}}\sin \left({\omega}_{\mathrm{c}}\tau \right)=0. $$

(4.121)

This gives $\frac{\lambda_{\mathrm{Im}}}{\lambda_{\mathrm{Re}}}=\frac{\sin \left({\omega}_{\mathrm{c}}\tau \right)}{1-\cos \left({\omega}_{\mathrm{c}}\tau \right)}=\tilde{\kappa}$. Substitution in Eq. 4.120 yields Eq. 4.122.

$$ {\displaystyle \begin{array}{l}{\tilde{b}}_{\mathrm{lim}}=\frac{2\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)}\left(\frac{\lambda_{\mathrm{Re}}\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)}{\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)}+\frac{\lambda_{\mathrm{Im}}\sin \left({\omega}_{\mathrm{c}}\tau \right)}{\left(1-\cos \left({\omega}_{\mathrm{c}}\tau \right)\right)}\right)\\ {}{\tilde{b}}_{\mathrm{lim}}=\frac{2\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)}\left({\lambda}_{\mathrm{Re}}+{\lambda}_{\mathrm{Im}}\tilde{\kappa}\right)\\ {}{\tilde{b}}_{\mathrm{lim}}=\frac{2\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)}{\lambda}_{\mathrm{Re}}\left(1+\frac{\lambda_{\mathrm{Im}}}{\lambda_{\mathrm{Re}}}\overset{\sim }{\kappa}\right)=\frac{2\pi }{N_{\mathrm{t}}{K}_{\mathrm{t}}\left({\lambda_{\mathrm{Re}}}^2+{\lambda_{\mathrm{Im}}}^2\right)}{\lambda}_{\mathrm{Re}}\left(1+{\overset{\sim }{\kappa}}^2\right)\end{array}} $$

(4.122)