AssignmentQ.N.1. In a turning operation, it was observed that the tool life was 150 minutes when the cutting speed was 20 meters per minute. As the speed was increased to 25 meters per minute, the tool life dropped to 25.2 minutes. If the time required to change the tool was 2 minutes and if the cost of regrinding the tool was ten times the cost of turning per minute, calculate: (i) The most economical cutting speed, (ii) Tool life for maximum production. Q.N.2. A tool with 18o rake angle is making an orthogonal cut, 3 mm wide, at a speed of 31 m/min and feed of 0.25 mm. The chip thickness ratio is 0.55, cutting force is 1392 N and feed force is 363 N. Find: (1) Chip thickness, (3) Co-efficient of friction on tool face, (2) Shear plane angle, (4) Shear force on shear plane, (5) Energy consumed in KW-min per cubic centimeter of metal removed. Q.N.3. The following data refers to an orthogonal cutting test: (1) Depth of cut: 1.5 mm, (3) Rake angle: 15o (5) Chip thickness ratio: 0.4, (6) Shear strength at zero compressive stress i.e., So =150 N/mm2 (7) Material constant, K = 0.2 in the relation S= So + KSn Calculate: (1) Shear plane angle, (3) Chip flow speed (5) Power at tool. (2) Fraction angle, (4) Shear force on shear plane, (2) Width of cut: 5 mm (4) Cutting speed = 25 m/min 5 mm. friction force. Determine: (1) Chip thickness ratio. Q. specific cutting energy.N.5. (4) Tool feed rate = 80 mm/min. (iii) Shear strain. normal force at shear plane. the following observations were made Feed = 0. kinetic co-efficient of friction. = 60 mm. the following observations made : Depth of cut = 1. Cutting speed = 80 m/min.5 mm. (2) Shear plane angle. (5) Cut chip thickness = 0. (3) Dynamic shear strain. During turning of a metallic workpiece with a single point having tool signature of 0-8-67-10-10-60-1 as per (ORS) system. Q.Q. Feed thrust force = 800 N Chip thickness = 0. (2) Work dia. Chip thickness = 0.40 mm. Speed = 200 m/min. Tangential cutting force = 1500 N.35.15 mm/rev. . Depth of cut = 1 mm.6. During machining of C-45 steel with 0-10-6-6-8-90-1 mm (ORS) shape carbide cutting tool.N.N. (1) Spindle speed = 400 rpm (3) Depth of cut = 2. Calculate the following: (i) Chip reduction co-efficient. The following observations are made while machining Mild steel with a tool of (0-10-66-75-1 mm ORS) shape.15 mm/rev.4. Calculate shear force.45. Feed = 0. (ii) Shear angle. Cutting force Thrust force. shear stress along shear plane. shear strain in chip. friction angle. Ft = 225 N = 0. frictional angle. a chip thickness of 0. Q. Q. Width of cut Chip thickness ratio Rake angle. power for cutting operation.45 is obtained. In an orthogonal turning of M. (ii) Shear angle.N.2 mm/rev. = 4. Determine shear angle. shear strain in chip and shear strain rate.N. Uncut chip thickness. v = 2 m/s. = 0.Q. Chip thickness. = 128 N Determine shear angle. Uncut chip thickness.7. Cutting force Thrust force.13 mm = 0.228 mm. = 0. α = 15o Fc = 500 N. = 20o = 363 N. In an orthogonal cutting operation the following data have been observed: Cutting speed. = 0. shear stress along the shear plane and power for the cutting operation and also find chip velocity. If the rake angle of tool is 0o.61.S part using a depth of cut 1.9. Rake angle. . Calculate the following: (i) Chip thickness ratio (iii) Chip contraction co-efficient. chip velocity.24 m/s. shear strain rate.5 mm and feed 0.8.08 mm.129 mm. In an orthogonal cutting operation. the following data have been observed: Cutting speed.N. = 0. Depth of cut Feed Cutting force = 0. Fx = 800 N Feed = 0. the following data was observed: Cutting speed.10. Q.5 mm/rev. Calculate co-efficient of friction. the following forces have been recorded on a two dimensional dynamometer. Find cutting power.S. It is found that cutting force = 2500 N and feed force = 1000 N. Calculate the following: (i) (ii) (iii) Radial component of force fy Frictional force P and normal force N Kinematic co-efficient of friction µ. motor power.15 mm/rev. A 50 mm outside diameter tubing is turned on a lathe with cutting speed of 25 rpm. tool for M. = 200 N.N.Q.S. shear angle.N.12. = 3 mm = 0. While Machining C20 steel with a triple carbide tool of shape 0-8-6-7-10-70-1 mm (ORS).08 mm.84). For facing operation with H. α = 15o . Q.3 m/s. bar. specific cutting resistance and unit power (assume lathe efficiency 0.S. velocity of chip along tool face. Chip thickness = 0. feed = 0. Depth of cut = 2 mm.4 mm. The length of continuous chip is 80 mm in one revolution. Uncut chip thickness.11. Fz = 1300 N. .N.18 mm/rev. 25 mm/rev and chip thickness after cutting is 2 mm. v = 30 m/min.N.N. the following observations were made: α = 15o.Q. During machining of SAE (1040 steel) with a double carbide cutting tool 0-8-6-7-10-701 mm (ORS) shape.14. shear strength = 600 MPa. µ = 0. In orthogonal turning if the feed is 1. Find out the following: (i) (ii) (iii) (iv) (v) Chip thickness ratio and chip reduction ratio Shear angle Shear force Shear stress and shear strain Work done in shear. During turning of metallic workpiece with a single point orthogonal cutting tool. Find following: Chip thickness ratio. Depth of cut = 2 mm.N. Fv = 80 N.15. v = 60 m/min.4 mm. Fc = 200 N.4 mm.13. Chip thickness = 0.25 mm/rev. Q.5 mm (iv) Work diameter =60 mm (v) Chip thickness = 0. shear plane angle. Q. α =10o. Determine the following: (i) (ii) (iii) (iv) (v) (vi) Chip thickness ratio Shear angle Shear force Frictional angle Cutting force Power at the cutting tool. the following observations were made: (i) Spindle speed = 400 rpm (ii) Tool feed rate = 80 mm/min (iii) Depth of cut = 2. . Feed = 0. shear strain.9. b = 10 mm. Tool changing time is 1 min. 1. Select the speed that minimizes the machining cost and calculate the corresponding tool life. Determine the following: (i) Tool life equation. bar is turned on a lathe with feed of 0. . Rs. 5 per hr.25 per edge 15 sec 5 min 0. Compare the life of two tools at a speed of 30 m/min. A tool life of 80 minutes is obtained at a speed of 50 m/min and 8 minutes at 100 m/min during a cutting test on a single point cutting tool.N. Reconditioning cost of tool edge is Rs. An automatic lathe is to be used to machine brass components 75 mm long X 50 mm diameter using a depth of cut of 1. the tool life is 128 min. A 200 mm long and 30 mm dia.18. Q. 1 per edge.25 mm. = = = = = Rs. if at a cutting speed of 24 m/min. the operating cost is 5 paise per min.16.N. while the tool cost is Rs. Q. VTn = 300.Q. 0. Compare the machining cost per component while operating under most economical conditions for the two materials. The tool life for a HSS tool is given by VT1/7 = C1 and for carbide tool is VT1/5 = C2.2 mm/rev Q.25 mm/rev. Assume that: Labour + overhead rate Reconditioning cost of tool edge Loading-unloading time of the workpiece Tool change time Feed Tool life relation. The requirements are such that either of the two tool materials X or Y can be used.N.19.N.17. (ii) Cutting speed for 10 minutes tool life. Also estimate the cutting speed for minimum time of production. 23.22. A tool life of 80 mm is obtained at a speed of 300 m/min and 8 mm at 60 m/min. VTn = C.28.N. Q.21. Q.N.2 and VTn = C. Tool regrind time = 3 min. n = 0. Determine optimum cutting speed for an operation carried on a lathe machine using following operations: Tool changing time = 4 min. A carbide tool while machining a MS workpiece was found to have a life of 1 hour and 40 minutes when the cutting at 50 m/min. .Q. Find tool life if tool is to operate t speed 30% higher than previous one.N. Let. Let n= 0. to reduce tool life to 2/5 of its former value). Also calculate cutting speed if tool is required to have a life of 2 hours and 45 minutes. Determine the percentage change in cutting speed required to give 60% reduction in tool life (i. Determine the following: (a) Tool life equation. Assume n = 0. (b) Cutting speed for 4 mm tool life. Find optimum cutting speed and tool life for minimum cost for machining medium carbon steel under the following conditions: Cost of operating machine Total cost of tool change = = 20 paise per min. The life relation of tool is VTn = C. Let C = 60..N.e. n = 1/5 Q. The cutting speed is 40 m/min when tool life is 50 min for the depth of cut and feed employed. The tool life equation is given by.24. Depreciation tool regrind = Rs. Q. Machine running cost = 20 paise per min.2. 1. 8.N.20. Rs.