Make a neat schematic drawing of the required surface finish for roughness 1.6 achieved by milling.
Q: How is chamfering done in a lathe? Make an illustration
A: Chamfering - it is a process done on lathe machine.
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A: For solution refer below images.
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A: Solution: The following changes need to be done to obtain the required surface finish:
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- Expert Q&A Done The top surface of a rectangular workpart is machined using a peripheral milling operation. The workpart is 735 mm long by 50 mm wide by 95 mm thick. The milling cutter, which is 60 mm in diameter and has five teeth, overhangs the width of the part equally on both sides. Cutting speed = 80 m/min, chip load = 0.30 mm/tooth, and depth of cut = 7.5 mm. (a) Determine the time required to make one pass across the surface, given that the setup and machine settings provide an approach distance of 5 mm before actual cutting begins and an overtravel distance of 25 mm after 1.2 9. 20 65 73 actual cutting has finished v in seconds. (b) What is the maximum material 3.9 239 removal rate during the cut v in mm3/sec? 0.127 5A mild steel specimen of Initial diameter of 30.5 mm is turned to final diameter of 27 mm for an initial length of 153 mm on a lathe machine. Using the given data find the following. (i) Feed of 0.4 mm/rev & Depth of cut is 0.5 mm (ii)During machining the tool's approach length is 5 mm, over run length is 2mm (iii) Time required to complete one single turning operation is 3.5 minutes (a)Find number of passes to finish the entire turning operation (b) Total time required to complete the turning operation in minutesA mild steel specimen of Initial diameter of 30.5 mm is turned to final diameter of 27 mm for an initial length of 153 mm on a lathe machine. Using the given data find the following. (i) Feed of 0.4 mm/rev & Depth of cut is 0.5 mm (ii)During machining the tool's approach length is 5 mm, over run length is 2mm (iii) Time required to complete one single turning operation is 3.5 minutes (c) Spindle speed in in rpm (= (d) Cutting speed in m/minutes(=
- A mild steel specimen of Initial diameter of 32.5 mm is turned to final diameter of 25 mm for an initial length of 154 mm on a lathe machine. Using the given data find the following. (i) Feed of 0.3 mm/rev & Depth of cut is 0.5 mm (ii)During machining the tool's approach length is 6 mm, over run length is 3mm (iii) Time required to complete one single turning operation is 2.3 minutes (a)Find number of passes to finish the entire turning operation (b) Total time required to complete the turning operation in minutes (c) Spindle speed in in rpm (d) Cutting speed in m/minutesA mild steel specimen of Initial diameter of 53.5 mm is turned to final diameter of 47 mm for an initial length of 154 mm on a lathe machine. Using the given data find the following. (i) Feed of 0.4 mm/rev & Depth of cut is 0.5 mm (ii) During machining the tool's approach length is 5 mm, over run length is 2mm (iii) Total time required to complete the turning operation is 59.6 minutesA mild steel specimen of Initial diameter of 53.5 mm is turned to final diameter of 47 mm for an initial length of 154 mm on a lathe machine. Using the given data find the following. (i) Feed of 0.4 mm/rev & Depth of cut is 0.5 mm (ii) During machining the tool's approach length is 5 mm, over run length is 2mm (iii) Total time required to complete the turning operation is 59.6 minutes (a) Find number of passes to finish the entire turning operation ( (b) The actual length of the turning operation in mm ('
- A mild steel specimen of Initial diameter of 53.5 mm is turned to final diameter of 49 mm for an initial length of 151 mm on a lathe machine. Using the given data find the following. (i) Feed of 0.2 mm/rev & Depth of cut is 0.5 mm (ii) During machining the tool's approach length is 7 mm, over run length is 4mm (iii) Total time required to complete the turning operation is 52.6 minutes (a) Find number of passes to finish the entire turning operation (b) The actual length of the turning operation in mm (c)The time required to complete one single turning operation in minutes (d) Spindle speed in rpmQ1. Calculate the time required to machine a workpiece 170 mm long, 60 mm diameter to 165 mm long 50 mm diameter. The workpiece rotates at 440 rpm, feed is 0.3 mm/rev and maximum depth of cut is 2 mm. Assume total approach and overtravel distance as 5 mm for turning operation. Answer Q2. A gray cast iron surface 280 wide and 540mm long may be machined either on a vertical milling machine, using a 100mm - diameter face mill having eight inserted HSS teeth, or on a horizontal milling machine using an HSS slab mill with eight teeth on a 200-mm. diameter. Which machine has the faster cutting time? The values of feed per tooth and cutting speed for both processes are 0.4mm/tooth and 80m/min, respectively. The depth of cut = 3.0 mm and assume A and O equal to 5. AnswerENT 152 Computer Aided Manufacturing I Milling Worksheet-SHOW ALL WORK A milling operation is to be performed at 100 sfpm at .005 ipt. A 500" diameter, 4 flute HSs end mill is to be used. The operation is a slot .100" deep across the diagonal of a rectangular part 4 inches wide and 6 inches long. The material is 4140 medium carbon steel (BHN-250). Use a specific HP of 1.2 HP/in/min. Calculate: RPM feed (ipm) feed (ipr) Length of cut Cut time MRR a. c. b. f. HP required at the spindle HP required at the motor (assume the drive system is 809% efficient) C. d. h.
- operation, the shear IS cIose to, In turning operation, spindle speed is set to provide a cutting speed of 2.2 m/: The feed and depth of cut are 0.32 mm and 2.1 mm, respectively. The tool rake angle is 15°. After the cut, the deformed chip thickness is measured to be 0.43 mm. Using the Orthogonal model as an approximation of turning operation, the shear strain is close to, In turning operation, spindle speed is set to provide a cutting speed of 2.4 m/Task 4 Here students need to calculate (a) The cutting speed (b) Turning time (c) Amount of material removed. For a specimen as shown below, if spindle rotates at N = 360 rpm, number of pass = 03, feed rate= 0.2mm/rev, if density of material=7.85× 10-6 Kg/mm³. Appendix Volume for Cylinder = rr?h Volume of frustum= nh (a?+b²+ab)/3 radii 'a' & b' height(Length) 'h'. b-a Taper angle : = tan20 h TDN Cutting speed:V : meters/min 1000 Where: V = Cutting speed in Meter/minute. T= 3.14 D = Diameter of work piece in mm N = rpm Cutting Time Length of the job to be turned NO.of cuts Time to turn mins. Feed/Rev rpm Amount of Material Removed = Original Mass – Machined Mass Original Size of Component 380 mm Machined Component 80 mm 100 mm 120 mm 6 20mm + 25mm + 40mm + 25mm 030mmSee image and answer pls