Unit-I: Theory of Metal Cutting Type-I (Cutting Forces Analysis) 1. In orthogonal cutting of a 60mm diameter MS bar on lathe, the following data was obtained, Rake angle = 15 0, Cutting Speed = 100 m/min, Cutting Force = 200 N, Feed Force = 70 N, Chip thickness = 0.3 mm, feed rate = 0.2 mm/rev.: feed force=60n Calculate: i. Shear plane angle ii. Coefficient of friction iii. Chip flow velocity iv. Friction Angle 2. In orthogonal turning of MS bar of 65mm diameter on a lathe a feed of 0.8 mm was used. A continuous chip of 1.4mm thickness was removed at a rotational speed of 85 rpm of work. Calculate: i. Chip thickness ratio ii. Chip reduction ratio iii. Total length of the chip removed in one min. [Ans: 0.57, 1.75, 9917.96 mm/min] 3. In an orthogonal turning operation on a Lathe, the following data was obtained, Back Rake angle = 15 0, Cutting Speed = 100 m/min, Cutting Force = 120 N, Feed Force = 30 N, Cutting thickness = 0.3 mm, feed rate = 0.2 mm/rev, Work piece diameter =120mm, Depth of Cut=0.4mm. Calculate: i. Shear Plane Angle ii. Coefficient of friction iii. Chip flow velocity iv. Friction angle v. chip thickness ratio vi. Strain energy vii. Shear strain, viii. Shear stress [Ans: 37.90 0, 0.555, 1.11 m/sec, 29.03 0, 0.667, 999.56N/mm 2, 1.706, 585.59 N/mm 2 ] 4. A carbide tipped tool of designation 0-10-5-5-9-90-1 mm (ORS) is used to turn a steel work piece of 50 mm diameter with a cutting speed of 240m/min and feed 0.25 mm/rev. Refer following data: Cutting force: 180N, Feed Force: 100N, chip thickness: 0.32mm. Calculate: i. Shear angle ii. Shear force iii. Friction force iv. Normal force acting on shear plane v. Friction angle [Ans: 41.673 0, 67.96N, 129.73N, 194.373, 39 0 ] 1 Department of Mechanical Engineering (T.E.)
5. In an orthogonal cutting operation following observation were made: Cutting speed=25m/min, width of cut=2.5 mm, feed =0.24mm/rev. chip thickness=0.4mm, Cutting force= 1400N, Thrust force=400 N. Tool rack angle= 5 0 Calculate: i. Shear Plane angle: ii. Friction angle iii. Chip Velocity iv. Shear strain v. Power [Ans: 34.24 0, 20.94 0,15 m/min, 2.1, 0.17 kw] 6. A seamless tube of 50mm outside diameter is turned on a lathe with a cutting speed of 20m/min. the tool rack angle is 15 0 and feed rate is 0.2 mm/rev. the length of continuous chip in one revolution measures 80 mm. Calculate: i. Chip thickness ratio ii. Shear plane angle iii. Shear flow speed iv. Shear strain [Ans: 0.509, 29.52 0, 10.18m/min, 2.025] 7. In orthogonal cutting operation, the following observations are obtained: cutting speed:120m/min. uncut chip thickness 0.127 mm, rack angle:10 0,width of cut 6.35 mm, Cutting force 567 mm, Thrust force 227N, Chip thickness:0.228mm. Calculate the following: Shear angle, friction angle, shear stress along the shear plane, chip velocity, Shear strain, and cutting power. [Ans: 31.27 0, 31.82 0, 236.09N/mm 2, 66.84m/min, 2.04, 1.134kW] 8. During machining of C-25 steel with 0-10-6-6-90-1 mm (ORS) shaped carbide cutting tool, the following observations have been made. Cutting speed of 200m/min and feed 0.2mm/rev. tangential Cutting force: 1600N, Feed thrust Force: 800N, chip thickness: 0.39mm. Calculate: i. kinetic coefficient of friction ii. Shear force: iii Friction force iv. Shear strain v. shear strain rate vi. Specific power consumption [Ans: 0.71, 1011.5N, 1065.68N, 2.148, 10.74rev/mm, 0.066kw-min/cm 3 ] 2 Department of Mechanical Engineering (T.E.)
9. A pipe of 38mm is being turned on a lathe with a tool having a rack of 33 0 and a feed of 0.15mm/rev. The length of chip over one revolution of work piece is 72 mm. The cutting speed is 12.5m/min. The tangential force is 410 N. Calculate: i. Coefficient of friction on the rake force. ii. Thickness of chip. iii. Angle of shear iv. Velocity of chip along the tool face. v. Velocity of shear. [Ans: 1.456, 0.248, 36.98 o, 7.54m/min, 10.51m/min] 10. During machining of alloy steel with tool having geometry 0-5-6-6-8-75-1mm ORS the tangential force and axial force measured by dynamometer to be 280N and 130N respectively. Calculate the radial force, friction force and coefficient of friction at the chip tool interface. [Ans: 267.6N, 153.9N, 0.575] 10. Identify the following tool signature 8-12-6-6-6-10-1 in ASA System. 11. A cutting tool in ORS system in designated as a0-10-5-5-8-80-2. Identify the tool. 12. Describe a tool with 8-14-6-6-20-15-4 signature 13. An orthogonal cut 2.5 mm wide is made at a speed of 0.5m/s and feed of 0.26mm with a HSS tool having a 20 0 rake angle. The chip thickness ratio is to be found 0.58, the cutting force is 1400N and the feed thrust force is 360N. Find: i. Chip thickness ii. Shear angle. iii. Resultant force iv. Coefficient of friction on face of tool. v. Shear force & Normal force. Vi. Specific Energy [Ans: 0.448, 34.211 o, 1445.5N, 0.685, 955.35N & 1084.8N, 0.0358kw-min/cm 3 ] 14. A steel tube 42mm outside diameter is turned on a lathe. The following data was obtained, Rack angle 32 o, cutting speed 18m /min. Feed=0.12mm/rev. Length of continuous chip in one revolution =52mm, cutting force=180kg, feed force=60kg. Determine: i. Chip velocity ratio, ii. Chip thickness iii. Shear plane angle iv. Velocity of chip along tool face. v. Coefficient of friction. [Ans: 0.39, 0.30, 22.87 o, 7.09, 1.21] 15. In a orthogonal cutting if the feed is 1.25mm/rev and chip thickness after cutting is 2mm, Determine: i. Chip thickness ratio ii. Shear angle. For rake angle 10 0 [Ans: 0.625, 34.62 0,] 3 Department of Mechanical Engineering (T.E.)
Type-II (Tool Life) 1. The following equation for tool life has been obtained for H.S.S tool. VT 0.13 f 0.6 d 0.3 = C. A 60 min tool life is obtained using the following cutting conditions: v o =40m/min; t o = 0.25mm; d o = 2mm. Calculate the effect on the tool life if speed, feed and depth of cut are together increased by 25% and also they are individually increased by 25% [Ans: Together increased T=2.3min, S increased 10.78min, f increased 21.38min, d increased 35.76min] 2. In a certain tool test, a single point cutting tool had a life of 10 minutes when operating at 240 m/min at what speed should the tool have to be operated in order to have tool life of 3 hours? Take n= 0.2 [Ans:134.6m/min] 3. The following cutting speeds of cutting time observation have been noted in a machining process. Calculate: i. n and C ii. Recommend the cutting speed for a desired tool life of 60 minutes. Cutting Speed V 25m/min 35m/min Cutting Time 90 min 20 min 4. A tool life of 70 min is to be obtaining at a cutting speed of 15m/min and 7 min at 50m/min. Determine: i. Taylor s tool life equation. ii. Calculate percentage increased in tool life when cutting speed is reduced by 50% [Ans: VT0.522= 138.26, 229.45%.] 5. A tool life of 80min is obtained at a speed of 30mpm and 8min at 60mpm. Determine the following i. Tool life equation ii. Cutting speed for 4min tool life. [Ans: VT 0.301 =112 18, V=73.91mpm] 6. The following equation of room life is given for a turning operation. VT 0.13 f 0.77 d 0.37 = C. A 60min tool life is obtained while cutting operation at V=30m/min, f=0.30mm/rev. and depth of cut d=2.5mm. Calculate the change in tool life if the cutting speed, feed, depth of cut is increased by 25% individually ad also taken together. What will be their effect on tool life? [Ans: v increased T 10.77min, f increased T15.98min. d increased 31.77 min. Together1.522min.] Conclusion: Maximum reduction in tool life is due to increases in cutting speed. Minimum effect on the tool life is due to increases is depth of cut. The total reduction in the life is due to the combined effect of all the parameters. 4 Department of Mechanical Engineering (T.E.)
Unit-II: Machine tool and their application Type: A. Numerical on Indexing Simple Indexing: 1. Divide the periphery of job into 60 equal divisions and the crank movement. [Ans: it indicates 12 hole on 18 hole plate] 2. Divide the plate in to 35 divisions. Find the indexing movement. [Ans: It indicates 1 full turn and 3 holes on 21 holes circle 2] 3. It is required to divide the periphery of a job in 28 equal divisions. Find the indexing arrangement. [Ans: It indicates 1 full turn and 9 holes on21 holes circle, plate number 2] Compound Indexing: 4. Index for 87 divisions by compound indexing. [Ans: The work piece will be indexed through 1/87 of a revolution each time as the crank is moved forward 23 holes circle and the plate and crank backward 11 holes on 33 hole circle.] 5. Index for 51 divisions by using compound indexing method. [Ans: Crank is moved 2 holes on 17 hole circle and the plate and 12 holes on 18 holes circle, in the same direction] 6. Index 69 divisions by using compound indexing method. [Ans: Crank is moved 21 holes on 23 hole circle and the plate and 6 holes on 18 holes circle, in the same direction ] Type: B. Numerical on Milling Machine 7. A 63.5 mm of diameter plain milling cutter having 6 teeth is used for face milling a block of aluminum 18 cm long and 3 cm wide. The spindle speed is 1500 rpm and feed is 0.125mm/tooth. Determine: i. Table feed in mm/min ii. Cutting time [Ans:1125mm/min, 0.167 minutes] 8. A face milling cutter has 100 mm diameter has 10 teeth. The cutting speed is 22 m/min, if the feed /tooth/rev is 1 mm. what time will take for one cut along a work piece of CI 1400 mm long. [Ans:2 min] 9. A plain surface 60mm wide and 230 mm long is to be milled on a horizontal milling machine with cutter diameter 80 mm and speed 50m/min. take feed per tooth is 0.11 mm and number of teeth on cutter is 12. Calculate machining time [Ans: 0.946 minutes] 5 Department of Mechanical Engineering (T.E.)
Type: C. Numerical on Drilling Machine 1. Calculate the time in seconds for high speed steel drill 10mm diameter to penetrate a 18mm thick steel plate. Assume a feed of 0.2mm/rev. And cutting speed for steel as a 20 m/min. [Ans: 9.89sec] 2. Calculate the spindle speed in RPM. For a HSS drill 10mm diameter, cutting a MS plate of 20mm thickness. Also calculate the machining time if feed is 0.18 mm/rev and cutting speed of 35mpm(meter per minute) [Ans: 1114.08rpm, 6.881seconds] 3. Calculate the cutting speed for drilling with a 12.7 mm diameter drill at 688rpm. Also find a machining time in seconds if a drill is used to penetrate a hole in 16mm thick plate a feed of 2.25mm/rev. [Ans: 27.45m/min, 6.91 seconds] 4. At what speed a 20mm drill will run for cutting steel plate at 25mpm cutting speed? Also find machining time in minutes to penetrate a hole up to 18mm depth at a feed 0.28mm/rev. [Ans: 397.88rpm, 0.215minutes] 5. Calculate the machining time for drill 4 hole of 16mm diameter each on a flange from the following data: Flange thickness=32mm, cutting speed=24mpm, feed =0.2mm/rev 6. A hole of 25mm diameter and 62.5mm depth is to be drilled. Take feed as 1.25mm/rev. And cutting speed as 60mpm. What are the feed speed, spindle rpm and cutting time. Assume clearance height (approach length) as 5mm. [Ans:954.9mm/min,763.94rpm, 0.071minutes] 7. A hole of 30mm diameter and 75mm depth is to be drilled. The suggested feed 1.3 mm per rev and the cutting speed 62m/min. Assuming tool approach and tool over travel as 6mm Calculate i.spindle rpm. ii. Feed speed. iii. Cutting time. [Ans: 657.84rpm, 855.19mm/min, 0.094minutes] 8. Calculate the machining time required for producing 20holes on an MS plate pf 40mm thickness with the following data. Drill diameter:30mm, cutting speed:25m/min, feed:0.1mm/rev, over run: 0.5 x drill diameter(mm) [Ans: 41.46minutes] 6 Department of Mechanical Engineering (T.E.)
Unit-III: Grinding & Finishing Process Grinding Wheel Signature: 1. Explain the meaning of following letter printed on the grinding wheel: W-A-46-K-5-V-17 2. Explain the meaning of following letter printed on the grinding wheel W-C-80-T-9-S-0 explain the meaning of each letter mentioned in the above specification of grinding wheel. 3. Explain the meaning of following letter printed on the grinding wheel: W-C-500-H-4-V-17 Describe the meaning of any four letters (except first and last letter) mentioned in the above specification of grinding wheel. 4. On the grinding wheel it is printed as under W-S-30-R-7-V-17 Explain in each case, what each letter and number indicate. 5. Explain the meaning of grinding wheel signature: 26-C-60-M-7-V-28 6. Explain the meaning of following letters printed on grinding wheel: W-A-40-J-6-V-17 7 Department of Mechanical Engineering (T.E.)