Question

Problem Use the following information to answer questions 22-23. The cutting force and thrust force in an orthogonal cutting operation are 1470 N and 1589 N, respectively. The rake angle = 5, the width of the cut=5.0 mm, the chip thickness before the cut = 0.6 mm, and the chip thickness ratio = 0.38. Question 22 Homework – Unanswered What is the shear plane angle in degreesty? B 2133 5. OD 3363 Unanswered Sub Question 23 Home Ungered What is the shear stress for shear strength) in MPa)? A 93,940,320 O B 95.9 O C 43.3 O D 19,262,750 Unanswered

Answer 1

Solution 3 Given fc = 2400 M (Catting force) F = 2589H (throof force) Ralee angle de 50 width of auf b = 5.0mm uncet chip thickness t= 0.6 mm chip thickness reasio r=0.38 Shear angle (0) 0.38a Cos(5) 1-(0.38) (sin (59) tanda rasa i-o sing tand = 0.39 15 T= 24.38° co - 23 from force analysis in mercheut circle diagram, shear force fs = fc .csd -Frasing fs = 1470. COS(21-38°) - 1589«sin (27:38') fs=1368.82–573.308 [fs= 789.52m] now we know, shear streso = . Zc = Shear shear fs a force area fsisind 마하 zg= sing 289.52 - sin (21:38) 590-6 sa gs.ga mpa

co by If Fra goom, fr=goon d=300 Chip sheer force fs= fcceso frosina fs= good (6 30 - 600.sin 36 fs = 729.422 - 300 fs=479.42 M