Question

QUESTION 5 During the orthogonal cutting of medium carbon steel rod with diameter of 75 mm and yield strength of 240 N/mm², the following machining data were used: Rake angle = 12°C Spindle speed = 950 rev/min Chip thickness ratio = 0.25 mm Depth of cut = 0.8 mm Width of cut = 5.0 mm Horizontal component of the cutting force = 1000 N The vertical component of the cutting force = 1400 N

Answer 1

Dat given ane diameted, Da 75 mm yield Strength, a= 240 N/mma Rake angle, a 120 Spindle speed, N= 950 rpm Chip thickness cratio or= 0. 25 mm depth ot cut, d= Oo8mm Width of cut, Wi=5 mm horizontal component of cutting force fra TOOON Ventical componene -d fore culting Fc = 1400 N cas $ Shear plane angle tand rcos Q 1- Sina

Merchant's Circle; mi 10 establish relationship the and actual forces measurable blw ir SA ho? Fs Tool Fc Brad R= Resultant SERIER circle. Dia of R Sin in 8 = R by property of triangle. fs Fan R= Sin (+B-2) sing cosß Fc Cos(Bra) cos(& + 8-a) fr Sin (B-d)

tand 0.25 x cos (120) 0.25x Sin (120) 14. 46° isht o Circle b) from merchanit's tan (B-V ) = FT fc 5 1000 1400 tan (B-12) 35.54 gil tant (5/7) 8-12 47.540 friction angle, 1000+14009 R= Eat fy? Per Resultant, R= 1720. 46 N fe frictional force on tool chip interface - 1720.46x Sin (47.54) 1269.27 N F= R sine)

Ne Normal fome to Friction . Na Rcos (8) 1720.46 x cos (47.54) = 1161.44 N fs Resistance to Shear the metal fs Rcos ( $+ -d) 1720 Cos(14.46+47.54-10) 110506 N fone Normal Esn to Shear plane Esn R sin ( $+B-Q) 1720 sin(14,46 +47.57-12) 1317.6 N velocity @ analyss 90-100) Vf V ia Ve= Cutting velocity Vs= shear velocity VF = chip flow velocity Bao-a < Vc

Sin rale, According to Vc Sin (90-(0-6) Vs VF Sin (4) Sid (90-x) TX 75x950 Gow 60 3.73 m/s Vc = 1000 x 60 Vs Sin (90-0) Sin (90-(6-1)) Vs 3.73 sin (90-12) Sin (90-(14.46-12) Vs= 3.652 m/s