Manufacturing Mechanics & Metal Cutting
Merchant Shear-Plane Angle & Cutting Force Formula
Calculate M.E. Merchant’s Minimum-Power Shear-Plane Angle phi = 45 + alpha/2 - beta/2, Rake-Face Friction Coefficient TAN(beta), Shear-Plane Area, and Tangential Cutting Force F_c.
Predict CNC spindle tangential cutting forces from Merchant’s force circle—in radians!
M. Eugene Merchant (1945, Journal of Applied Physics) minimized plastic shear power in orthogonal chip formation to derive the foundational Shear-Plane Angle equation: phi_deg = 45 + alpha_deg / 2 - beta_deg / 2, where alpha_deg is tool rake angle and beta_deg = ATAN(mu) is rake-face friction angle! Across shear-plane area A_s = (b * t_1) / SIN(phi) (for cut width b and uncut chip feed thickness t_1) at dynamic alloy shear strength tau_s (MPa = N/mm^2), Merchant’s force circle gives Main Tangential Cutting Force F_c (N) = (tau_s * b * t_1 * COS(beta - alpha)) / (SIN(phi) * COS(phi + beta - alpha))! Remember the universal spreadsheet trig rule: convert all degree inputs by PI() / 180 before calling SIN, COS, or TAN!
Verify on Rake_Angle alpha = 10° (B2), Friction_Angle beta = 40° (B3 -> Merchant Shear Angle phi = 45 + 5 - 20 = 30.00° in E2, Friction Coeff mu = TAN(40°) = 0.8391 in E3), Alloy_Shear_Strength tau_s = 200 MPa (B4), Cut_Width b = 3.0 mm (B5), and Uncut_Chip_Feed t_1 = 0.20 mm (B6 -> uncut cross-section = 0.60 mm^2): Shear-Plane Area A_s is 0.60 / SIN(30°) = 1.20 mm^2 (E4), and Tangential Cutting Force F_c is (120 * COS(30°)) / (SIN(30°) * COS(60°)) = (120 * 0.866025) / 0.25 = 415.69 N (E5)! Apply high-pressure flood coolant + sharper positive rake (alpha = 20°, beta = 20°): phi opens to 45.00° and F_c plunges 42% to 240.00 N!
| Merchant Force-Circle Metric (alpha=10°, beta=40°, tau_s=200MPa, b=3mm, t1=0.2mm) | Formula | Verified Output |
|---|---|---|
| Merchant Shear Angle phi (deg) | 45 + alpha_deg / 2 - beta_deg / 2 | 30.00° (45.00° @ alpha=20, beta=20) |
| Rake Friction Coefficient mu | TAN(beta_deg * PI() / 180) | 0.8391 (0.3640 @ beta = 20°) |
| Shear-Plane Area A_s (mm^2) | (b * t_1) / SIN(phi * PI() / 180) | 1.20 mm^2 |
| Tangential Cutting Force F_c (N) | (tau_s*b*t1*COS(b-a))/(SIN(p)*COS(p+b-a)) | 415.69 N (240.00 N Lubricated) |
Merchant Tangential Cutting Force F_c Formula
Validate beta > 0, tau_s > 0, b > 0, t_1 > 0, and kinematic feasibility 0 < phi < 90 and (phi + beta - alpha) < 90!
=LET(al_d, B2, bt_d, B3, ts_m, B4, wd_b, B5, fd_t, B6, ph_d, 45 + al_d / 2 - bt_d / 2, IF(OR(bt_d <= 0, bt_d >= 90, ts_m <= 0, wd_b <= 0, fd_t <= 0, ph_d <= 0, ph_d + bt_d - al_d >= 90), "Invalid Orthogonal Geometry", LET(dg_r, PI() / 180, nm_f, ts_m * wd_b * fd_t * COS((bt_d - al_d) * dg_r), dn_f, SIN(ph_d * dg_r) * COS((ph_d + bt_d - al_d) * dg_r), ROUND(nm_f / dn_f, 2))))Lubricated Positive-Rake Reduction Check (alpha = 20°, beta = 20°)
Set B2 = 20 and B3 = 20: SIN(45°) * COS(45°) = 0.50 exact, dropping F_c cleanly to 240.00 N!
=IF(ISTEXT(E5), E5, ROUND((B5 * B6) / SIN(E2 * PI() / 180), 2))Try the example
Use the original values or change them to test your understanding.