Lathe final.ppt
- 1. © rkm2003 Introduction Lathe is a machine, which removes the metal from a piece of work to the required shape & size “Mother of Machine Tools “ Cylindrical and flat surfaces Turning Operations: Machine Tool – LATHE Job (work piece) – rotary motion Tool – linear motions
- 2. © rkm2003 The Lathe Head stock Spindle Feed rod Bed Compound rest and slide(swivels) Carriage Apron selector Feed change gearbox Spindle speed Lead screw Cross slide Dead center Tool post Guide ways Tailstock quill Tailstock Handle
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- 5. © rkm2003 Lathe parts Toolpost Fitted on top slide and carries the cutting tool or the cutting tool holder
- 6. © rkm2003 Lathe parts Tool holders Used for holding cutting tool bits Available in Right hand, left hand and straight
- 7. © rkm2003 Cutting Tools Can be High Speed Steel held in tool holders Can be also Ceramic (Tungsten carbide) bits held directly in toolpost
- 8. © rkm2003 Chucks Three jaw Four Jaw
- 9. © rkm2003 Centers Workpiece Headstock center (Live Centre) Tailstock center (Dead Centre)
- 10. © rkm2003 Turning Parameters 1. Cutting Speed v 2. Feed f 3. Depth of Cut d
- 11. © rkm2003 Cutting speed: The Peripheral Speed of Work piece past the Cutting Tool Feed: The distance the tool advances for every rotation of work piece (mm/rev) Depth of Cut: perpendicular distance between machined surface and uncut surface of the Work piece d = (D1 – D2)/2 (mm) m/min 1000 N D v D – Diameter (mm) N – Revolutions per Minute (rpm)
- 12. © rkm2003 Material Removal Rate MRR Volume of material removed in one revolution MRR = D d f mm3 • Job makes N revolutions/min MRR = D d f N (mm3/min) In terms of v MRR is given by MRR = 1000 v d f (mm3/min)
- 13. © rkm2003 Operations on Lathe Turning Facing Grooving Parting Chamfering Taper turning Drilling Threading
- 14. © rkm2003 Turning .. Cylindrical job: 1)Excess Material is removed to reduce Diameter 2) a depth of cut of 1 mm will b reduce diameter by 2 mm Cutting speed Chip Workpiece Depth of cut (d) Depth of cut Tool Feed Chuck N Machined surface
- 15. © rkm2003 Facing .. machine end of job Flat surface or to Reduce Length of Job Turning Tool Feed: in direction perpendicular to workpiece axis Length of Tool Travel = radius of workpiece Depth of Cut: in direction parallel to workpiece axis Depth of cut Feed Workpiece Chuck Cutting speed Tool d Machined Face
- 16. © rkm2003 Grooving Produces a Groove on workpiece Shape of tool shape of groove Carried out using Grooving Tool Shape produced by form tool Groove Grooving tool Feed or depth of cut Form tool
- 17. © rkm2003 Parting Tool is fed radially into rotating work at some location to cut off end of part, or provide a groove
- 18. © rkm2003 Chamfering Cutting edge cuts an angle on the corner of the cylinder, forming a "chamfer"
- 19. © rkm2003 Taper Turning Taper: Methods Form Tool Swiveling Compound Rest Taper Turning Attachment Simultaneous Longitudinal and Cross Feeds C B A L D 90° 2 D1 L D D 2 tan 2 1
- 20. © rkm2003 Taper Turning .. By Form Tool Taper Workpiece Straight cutting edge Direction of feed Form tool
- 21. © rkm2003 Taper Turning ,, By Compound Rest Face plate Dog Tail stock quill Tail stock Mandrel Direction of feed Compound rest Slide Compound rest Hand crank Tool post & Tool holder Cross slide
- 22. © rkm2003 Threading Pointed form tool is fed linearly across surface of rotating workpart parallel to axis of rotation at a large feed rate, thus creating threads
- 23. © rkm2003 Drilling Drill – cutting tool – held in TS – feed from TS Feed Drill Quill clamp moving quill Tail stock clamp Tail stock
- 24. © rkm2003 Example A mild steel rod having 50 mm diameter and 500 mm length is to be turned on a lathe. Determine the machining time to reduce the rod to 45 mm in one pass when cutting speed is 30 m/min and a feed of 0.7 mm/rev is used. Given data: D = 50 mm, Lj = 500 mm v = 30 m/min, f = 0.7 mm/rev Substituting the values of v and D in Machining time: t = 500 / (0.7191) = 3.74 minutes m/min 1000 N D v N = 191 rpm min N f L t j
- 25. © rkm2003 Example A 150 mm long 12 mm diameter stainless steel rod is to be reduced in diameter to 10 mm by turning on a lathe in one pass. The spindle rotates at 500 rpm, and feed is 0.4 mm/rev. Calculate the cutting speed, material removal rate and the time required for machining the steel rod. Given data: Lj = 150 mm, D1 = 12 mm, D2 = 10 mm, N = 500 rpm Using Equation (1) v = 12500 / 1000 = 18.85 m/min. depth of cut = d = (12 – 10)/2 = 1 mm From Equation : MRR = D d f N MRR = 3.142120.41500 = 7538.4 mm3/min t = 150/(0.4500) = 0.75 min. min N f L t j
- 26. © rkm2003 Example Determine the angle at which the compound rest would be swiveled for cutting a taper on a workpiece having a length of 150 mm and outside diameter 80 mm. The smallest diameter on the tapered end of the rod should be 50 mm and the required length of the tapered portion is 80 mm. Given data: D1 = 80 mm, D2 = 50 mm, Lj = 80 mm (with usual notations) tan = (80-50) / 280 or = 10.620 The compound rest should be swiveled at 10.62o
- 27. © rkm2003 Process sequence From raw material of 175 mm length and 60 mm diameter a workpiece is to be made which has been specified in the given figure 40 Dia 50 50 40 20 50 20 Dia Threading
- 28. © rkm2003 Process sequence A possible sequence for producing the component would be: Turning (reducing completely to 50 mm) Facing (to reduce the length to 160 mm) Step turning (reducing from 50 mm to 40 mm) Thread cutting. Grooving