--- title: "Orthogonal cutting forces and power numerical" subject: "Manufacturing and Production Processes" chapter: 7 key: "semester_4/manufacturing_and_production_processes:orthogonal-cutting-forces-power-numerical" source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/orthogonal-cutting-forces-power-numerical site: BCT Engineering --- # Orthogonal cutting forces and power numerical Manufacturing and Production Processes · Chapter 7: Material Removal Processes: “ Chip-forming” - Asked in: year not given, 6 marks - Video: [Metal cutting solved example problem | orthogonal metal cutting examples](https://www.youtube.com/watch?v=ExMp6FOTCjg) — Top Mech Edu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/orthogonal-cutting-forces-power-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-7 ## Question In an orthogonal cutting test, the uncut chip thickness is 0.25 mm, the chip thickness is 0.75 mm, the width of cut is 3 mm and the rake angle is 10 degree. The cutting force is 900 N and the thrust force is 450 N at a cutting speed of 90 m/min. Calculate the shear angle, friction force, normal force on the tool face, coefficient of friction, shear force, shear stress, cutting power and the specific cutting energy. ## Short answer Chip thickness ratio r = 0.333 yields shear angle φ = 19.2° via tan φ = r cos α/(1 - r sin α). Forces: friction 599 N, normal 808 N, coefficient 0.74. Shear force 702 N, shear stress 308 MPa. Cutting power 1.35 kW, specific energy 1.2 J/mm³. ## Answer **Given:** $t_0 = 0.25$ mm, $t_c = 0.75$ mm, $b = 3$ mm, $\alpha = 10^\circ$, $F_c = 900$ N, $F_t = 450$ N, $V = 90$ m/min $= 1.5$ m/s. #### Shear angle Chip thickness ratio $r = t_0/t_c = 0.25/0.75 = 0.3333$. $$ \tan\phi = \frac{r\cos\alpha}{1 - r\sin\alpha} = \frac{0.3333\times0.9848}{1 - 0.3333\times0.1736} = \frac{0.3283}{0.9421} = 0.3485 $$ $$ \phi = 19.2^\circ $$ #### Forces on the tool face $$ F = F_c\sin\alpha + F_t\cos\alpha = 900(0.1736) + 450(0.9848) = 156.3 + 443.2 = 599.4\ \text{N} $$ $$ N = F_c\cos\alpha - F_t\sin\alpha = 900(0.9848) - 450(0.1736) = 886.3 - 78.1 = 808.2\ \text{N} $$ $$ \mu = \frac{F}{N} = \frac{599.4}{808.2} = 0.742,\qquad \beta = \tan^{-1}0.742 = 36.6^\circ $$ #### Shear force and shear stress $$ F_s = F_c\cos\phi - F_t\sin\phi = 900(0.9444) - 450(0.3289) = 850.0 - 148.0 = 702\ \text{N} $$ Shear plane area: $$ A_s = \frac{b\,t_0}{\sin\phi} = \frac{3\times0.25}{0.3289} = 2.28\ \text{mm}^2 $$ $$ \tau = \frac{F_s}{A_s} = \frac{702}{2.28} = 308\ \text{MPa} $$ #### Power and specific energy $$ P_c = F_c V = 900\times1.5 = 1350\ \text{W} = 1.35\ \text{kW} $$ Metal removal rate $= b\,t_0\,V = 3\times0.25\times1500 = 1125\ \text{mm}^3/\text{s}$. $$ u = \frac{P_c}{MRR} = \frac{1350}{1125} = 1.2\ \text{J/mm}^3 $$ | Quantity | Result | |---|---| | Shear angle | $19.2^\circ$ | | Friction force | 599 N | | Normal force on rake face | 808 N | | Coefficient of friction | 0.742 | | Shear force | 702 N | | Shear stress | 308 MPa | | Power | 1.35 kW | | Specific energy | 1.2 J/mm$^3$ | **Answer:** $\phi = 19.2^\circ$, $F = 599$ N, $N = 808$ N, $\mu = 0.74$, $F_s = 702$ N, $\tau = 308$ MPa, $P = 1.35$ kW, $u = 1.2$ J/mm$^3$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the shear angle found from the chip thickness ratio? A. tan(phi) = r sin(alpha) / (1 - r cos(alpha)), giving about 3 deg B. tan(phi) = tc/t0 = 3, giving 71.6 deg C. phi = 45 deg + alpha/2 = 50 deg D. tan(phi) = r cos(alpha) / (1 - r sin(alpha)), with r = t0/tc = 0.333, giving 19.2 deg Answer: D. Using r = 0.25/0.75 and alpha = 10 deg gives tan(phi) = 0.3485 and phi = 19.2 deg. (Key point: Shear angle) 2. What are the friction force F and normal force N on the tool face? A. F = 599 N and N = 808 N B. F = 808 N and N = 599 N C. F = 900 N and N = 450 N D. F = 702 N and N = 599 N Answer: A. F = Fc sin(a) + Ft cos(a) = 599 N; N = Fc cos(a) - Ft sin(a) = 808 N. (Key point: Friction and normal force) 3. How is shear stress found? A. Cutting force 900 N divided by b t0 = 0.75 mm^2, giving 1200 MPa B. Shear force 702 N divided by shear plane area b t0/sin(phi) = 2.28 mm^2, giving 308 MPa C. Shear force divided by b tc, giving about 312 MPa D. Shear force 702 N divided by 3 mm^2 area Answer: B. The shear plane area is b t0/sin(phi), and tau = Fs/As = 308 MPa. (Key point: Shear stress) 4. How is the specific cutting energy obtained? A. Power divided by chip thickness B. Cutting force times speed in m/min = 81 000 C. Power 1350 W divided by MRR 1125 mm^3/s = 1.2 J/mm^3 D. Power 1350 W divided by 90 mm^3/s = 15 J/mm^3 Answer: C. Power = Fc V = 900 x 1.5; MRR = b t0 V = 3 x 0.25 x 1500 = 1125 mm^3/s; u = 1.2 J/mm^3. 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