Surface Location Error in Milling

Abstract

Chapter 5 studies the influence of forced vibrations on part geometric accuracy in milling. The associated errors are referred to as surface location errors, or SLE. After describing SLE, an analytical, frequency domain solution is first presented which includes a Fourier series force model and SLE variation with axial location along the tool axis. A time domain simulation is then detailed which models the cycloidal motion of the cutter teeth and predicts SLE.

No amount of experimentation can ever prove me right; a single experiment can prove me wrong.

—Albert Einstein

Appendices

Exercises

• 1. Determine the value of the mean y direction cutting force for the following cuts using Eq. 5.3. The aluminum alloy-four tooth cutter combination gives: k_t = 790 N/mm² and k_n = 190 N/mm², k_te = 8 N/mm, and k_ne = 4 N/mm. Also, b = 5 mm and f_t = 0.15 mm/tooth. Assume a rigid cutting tool and workpiece.

  • (a) Up milling, 30% radial immersion

  • (b) Down milling, 40% radial immersion

• 2. Plot the y direction force over one cutter revolution for: down milling, 50% radial immersion, N_t = 2, d = 19 mm, γ = 30 deg, b = 2 mm, f_t = 0.2 mm/tooth, k_t = 730 N/mm², k_n = 205 N/mm², k_te = k_ne = 0 N/mm, and Ω = 10,000 rpm. Use the Fourier series approach and show results for both five and 25 terms.

• 3. Calculate the surface location error for the following conditions: 25% radial immersion down milling, spindle speeds from 11,000 rpm to 13,000 rpm, N_t = 4, γ = 30 deg, d = 12.7 mm diameter, f_t = 0.15 mm/tooth, b = 2 mm, k_t = 700 N/mm², k_n = 210 N/mm², k_te = k_ne = 2 N/mm, and symmetric structural dynamics with a stiffness of 8 × 10⁶ N/m, ζ = 0.02, and 800 Hz natural frequency. For the Fourier computations, use 15 terms, five axial steps, and a spindle speed resolution of 20 rpm. At a spindle speed of 12,140 rpm is the surface overcut or undercut?

• 4. Using time domain simulation, determine the surface location error at the free end of the cutter for the same conditions described in Exercise 3. Use a spindle speed equal to the best speed calculated from Eq. 4.29 for the N = 0 (rightmost) lobe. Carry out your simulation for 40 revolutions with 500 steps per tooth.

• 5. Plot the y direction force over one cutter revolution for: up milling, 50% radial immersion, N_t = 4, d = 12.7 mm, γ = 30 deg, b = 2 mm, f_t = 0.1 mm/tooth, k_t = 700 N/mm², k_n = 200 N/mm², k_te = k_ne = 0 N/mm, and Ω = 12,000 rpm. Use the Fourier series approach and show results for both five and 25 terms. Compare these results to the y direction force for straight teeth with all other parameters identical.

Appendix: Fourier Force Series Coefficients

The Fourier coefficients, a_n and b_n, for the y direction force series:

$$ {F}_y\left(\phi \right)=\sum \limits_{j=1}^A\sum \limits_{i=1}^{N_{\mathrm{t}}}\left({a}_0+\sum \limits_{n=1}^{\infty}\left({a}_n\cos \left(n{\phi}_i\right)+{b}_n\sin \left(n{\phi}_i\right)\right)\right), $$

where ${\phi}_i=\omega t+\frac{2\pi }{N_{\mathrm{t}}}\left(i-1\right)-\chi \left(j-1\right)$, are provided here [18]. The terms in Eqs. 5.18 through 5.23 were determined using Eqs. 5.5 and 5.6. The integration limits for down milling (ϕ₁ to π) are shown. For up milling, the limits are modified to be zero to ϕ₁.

$$ {a}_1=-\frac{bN_{\mathrm{t}}}{\pi }{\left[\begin{array}{l}{k}_{\mathrm{t}}c\left(-\frac{1}{4}\sin \phi +\frac{1}{12}\sin 3\phi \right)+{k}_{\mathrm{n}}c\left(-\frac{1}{4}\cos \phi -\frac{1}{12}\cos 3\phi \right)\\ {}+{k}_{\mathrm{t}\mathrm{e}}\left(\frac{1}{4}\cos 2\phi \right)+{k}_{\mathrm{n}\mathrm{e}}\left(\frac{1}{2}\phi +\frac{1}{4}\sin 2\phi \right)\end{array}\right]}_{\phi_1}^{\pi } $$

(5.18)

$$ {a}_2=-\frac{bN_{\mathrm{t}}}{\pi }{\left[\begin{array}{l}{k}_{\mathrm{t}}c\left(\frac{1}{4}\phi -\frac{1}{4}\sin 2\phi +\frac{1}{16}\sin 4\phi \right)+{k}_{\mathrm{n}}c\left(-\frac{1}{16}\cos 4\phi \right)\\ {}+{k}_{\mathrm{t}\mathrm{e}}\left(-\frac{1}{2}\cos \phi +\frac{1}{6}\cos 3\phi \right)+{k}_{\mathrm{n}\mathrm{e}}\left(\frac{1}{2}\sin \phi +\frac{1}{6}\sin 3\phi \right)\end{array}\right]}_{\phi_1}^{\pi } $$

(5.19)

$$ {a}_n=-\frac{bN_{\mathrm{t}}}{\pi }{\left[\begin{array}{l}{k}_{\mathrm{t}}c\left(-\frac{1}{2n}\sin n\phi +\frac{1}{4\left(n-2\right)}\sin \left(n-2\right)\phi +\frac{1}{4\left(n+2\right)}\sin \left(n+2\right)\phi \right)\\ {}+{k}_{\mathrm{n}}c\left(\frac{1}{4\left(n-2\right)}\cos \left(n-2\right)\phi -\frac{1}{4\left(n+2\right)}\cos \left(n+2\right)\phi \right)\\ {}+{k}_{\mathrm{t}\mathrm{e}}\left(-\frac{1}{2\left(n-1\right)}\cos \left(n-1\right)\phi +\frac{1}{2\left(n+1\right)}\cos \left(n+1\right)\phi \right)\\ {}+{k}_{\mathrm{n}\mathrm{e}}\left(\frac{1}{2\left(n-1\right)}\sin \left(n-1\right)\phi +\frac{1}{2\left(n+1\right)}\sin \left(n+1\right)\phi \right)\end{array}\right]}_{\phi_1}^{\pi },n=3,4,\dots $$

(5.20)

$$ {b}_1=-\frac{bN_{\mathrm{t}}}{\pi }{\left[\begin{array}{l}{k}_{\mathrm{t}}c\left(\frac{3}{4}\cos \phi -\frac{1}{12}\cos 3\phi \right)+{k}_{\mathrm{n}}c\left(\frac{1}{4}\sin \phi -\frac{1}{12}\sin 3\phi \right)\\ {}+{k}_{\mathrm{t}\mathrm{e}}\left(-\frac{1}{2}\phi +\frac{1}{4}\sin 2\phi \right)+{k}_{\mathrm{n}\mathrm{e}}\left(-\frac{1}{4}\cos 2\phi \right)\end{array}\right]}_{\phi_1}^{\pi } $$

(5.21)

$$ {b}_2=-\frac{bN_{\mathrm{t}}}{\pi }{\left[\begin{array}{l}{k}_{\mathrm{t}}c\left(\frac{1}{4}\cos 2\phi -\frac{1}{16}\cos 4\phi \right)+{k}_{\mathrm{n}}c\left(\frac{1}{4}\phi -\frac{1}{16}\sin 4\phi \right)\\ {}+{k}_{\mathrm{t}\mathrm{e}}\left(-\frac{1}{2}\sin \phi +\frac{1}{6}\sin 3\phi \right)+{k}_{\mathrm{n}\mathrm{e}}\left(-\frac{1}{2}\cos \phi -\frac{1}{6}\cos 3\phi \right)\end{array}\right]}_{\phi_1}^{\pi } $$

(5.22)

$$ {b}_n=-\frac{bN_{\mathrm{t}}}{\pi }{\left[\begin{array}{l}{k}_{\mathrm{t}}c\left(\frac{1}{2n}\cos n\phi -\frac{1}{4\left(n-2\right)}\cos \left(n-2\right)\phi -\frac{1}{4\left(n+2\right)}\cos \left(n+2\right)\phi \right)\\ {}+{k}_{\mathrm{n}}c\left(\frac{1}{4\left(n-2\right)}\sin \left(n-2\right)\phi -\frac{1}{4\left(n+2\right)}\sin \left(n+2\right)\phi \right)\\ {}+{k}_{\mathrm{t}\mathrm{e}}c\left(-\frac{1}{2\left(n-1\right)}\sin \left(n-1\right)\phi +\frac{1}{2\left(n+1\right)}\sin \left(n+1\right)\phi \right)\\ {}+{k}_{\mathrm{n}\mathrm{e}}c\left(-\frac{1}{2\left(n-1\right)}\cos \left(n-1\right)\phi -\frac{1}{2\left(n+1\right)}\cos \left(n+1\right)\phi \right)\end{array}\right]}_{\phi_1}^{\pi },n=3,4,\dots $$

(5.23)