PES INSTITUTE OF TECHNOLOGY BANGALORE SOUTH CAMPUS Hosur Road, (1K.M. Before Electronic City), Bangalore DEPARTMENT OF MECHANICAL ENGINEERING

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1 PES INSTITUTE OF TECHNOLOGY BANGALORE SOUTH CAMPUS Hosur Road, (1K.M. Before Electronic City), Bangalore DEPARTMENT OF MECHANICAL ENGINEERING SOLUTION 3 rd INTERNAL TEST Subject : Machine Tools and Operations Semester: IV Sub. Code : 15ME45B Section: A/B/C Name of the faculty : SJ/MKM/HKJ Q.No 1. a. PART-1 The life of the cutting tool is affected by the following factors: Cutting speed. Feed and depth of cut. Tool geometry. Tool material. Cutting fluid. Work piece material. The properties of the tool material which enhance the tool life include the following Hardness to resist deformation Toughness to resist sudden load in interrupted cuttings Wear resistance High thermal conductivity Low coefficient of friction, at the chip-tool interface, so that the surface finish is good and the wear is minimum The micro structural properties of the work piece material have a significant effect on the life of the cutting tool. Spheriodized pearlite is favorable to tool life whereas laminar pearlite is harmful. Similarly, in cast iron that contains large amount of free graphite and ferrite lead to the increase in tool life than containing free iron. The increase carbide cutting temperature and power consumption vary directly as the hardness of the work piece material. Due to the tendency of pure metals to stick on to tool face of cutting tool, these would adversely affect the tool life. The properties of the work material that tends to increase the tool life include the following Softness Absence of abrasive constituents Lack of work hardening tendency The cutting fluids cool the chip and work piece and may even reduce the frictional stress at tool-work and tool chip interfaces to some extent. This reduces the cutting temperature. If the tool material has the low amount of temperature then there is an appreciable increase in tool life. Marks 4

2 b. Solution: t 2 = 0.62 mm t 1 = 0.2 mm α = 15ᵒ Chip reduction co-efficient (k): k = 1 / r where r = t 1 / t 2 = 0.2/0.62 = k = 1 / r = 1/ = 3.1 Shear angle (ø ): Ø = tan -1 [(r cosα) / ( 1-r sinα)] = tan -1 [(0.322 cos 15ᵒ) / ( sin 15ᵒ)] = 1.74ᵒ 4 2.

3 3. PART-2 The machinability characteristics and their criteria, i.e., the magnitude of cutting forces and temperature, tool life and surface finish are governed or influenced more or less by all the variables and factors involved in machining such as, (a) properties of the work material (b) cutting tool; material and geometry (c) levels of the process parameters (d) machining environments (cutting fluid application etc) Machinability characteristics of any work tool pair may also be further affected by, strength, rigidity and stability of the machine kind of machining operations done in a given machine tool functional aspects of the special techniques, if employed.

4 Role of the properties of the work material on machinability. The work material properties that generally govern machinability in varying extent are: the basic nature brittleness or ductility etc. microstructure mechanical strength fracture or yield hardness hot strength and hot hardness work harden ability thermal conductivity chemical reactivity Role of cutting tool material and geometry on machinability of any work material. Role of tool materials In machining a given material, the tool life is governed mainly by the tool material which also influences cutting forces and temperature as well as accuracy and finish of the machined surface. The composition, microstructure, strength, hardness, toughness, wear resistance, chemical stability and thermal conductivity of the tool material play significant roles on the machinability characteristics though in different degree depending upon the properties of the work material. Role of the process parameters on machinability Proper selection of the levels of the process parameters (V C, S o and t) can provide better machinability characteristics of a given work tool pair even without sacrificing productivity or MRR. Amongst the process parameters, depth of cut, t plays least significant role and is almost invariable. Compared to feed (S o ) variation of cutting velocity (V C ) governs machinability more predominantly. Increase in VC, in general, reduces tool life but it also reduces cutting forces or specific energy requirement and improves surface finish through favorable chip-tool interaction. Some cutting tools, specially ceramic tools perform better and last. Effects of machining environment (cutting fluids) on machinability The basic purpose of employing cutting fluid is to improve machinability characteristics of any work tool pair through : improving tool life by cooling and lubrication reducing cutting forces and specific energy consumption improving surface integrity by cooling, lubricating and cleaning at the cutting zone The favorable roles of cutting fluid application depend not only on its proper selection based on the work and tool materials and the type of the machining process but also on its rate of flow, direction and location of application. 4. Crater wear: consists of a concave section on the tool face formed by the action of the chip sliding on the surface. Crater wear affects the mechanics of the process increasing the actual rake angle of the cutting tool and consequently, making cutting easier. At the same time, the crater wear weakens the tool wedge and increases the possibility for tool breakage. In general, crater wear is of a relatively small concern.

5 Flank wear: occurs on the tool flank as a result of friction between the machined surface of the work piece and the tool flank. Flank wear appears in the form of so-called wear land and is measured by the width of this wear land, VB, Flank wear affects to the great extend the mechanics of cutting. Cutting forces increase significantly with flank wear. If the amount of flank wear exceeds some critical value (VB > 0.5~0.6 mm), the excessive cutting force may cause tool failure. 5. PART-3 Different types of chips of various shape, size, colour etc. are produced by machining depending upon: Type of cut, i.e., continuous (turning, boring etc) or intermittent cut (milling). Work material (brittle or ductile etc.). Cutting tool geometry (rake, cutting angles etc.). Levels of the cutting velocity and feed (low, medium or high). Cutting fluid (type of fluid and method of application). The basic major types of chips and the conditions generally under which such types of chips form are given below: Continuous chips without BUE When the cutting tool moves towards the work piece, there occurs a plastic deformation of the work piece and the metal is separated without any discontinuity and it moves like a ribbon. The chip moves along the face of the tool. This mostly occurs while cutting a ductile material. It is desirable to have smaller chip thickness and higher cutting speed in order to get continuous chips. Lesser power is consumed while continuous chips are produced. Total life is also mortised in this process. The formation of continuous chips is schematically shown in Fig.

6 The following condition favors the formation of continuous chips without BUE chips: Work material - ductile. Cutting velocity - high. Feed - low. Rake angle - positive and large. Cutting fluid - both cooling and lubricating. Discontinuous chips This is also called as segmental chips. This mostly occurs while cutting brittle material such as cast iron or low ductile materials. Instead of shearing the metal as it happens in the previous process, the metal is being fractured like segments of fragments and they pass over the tool faces. Tool life can also be more in this process. Power consumption as in the previous case is also low. The formation of continuous chips is schematically shown in Fig. The following condition favors the formation of discontinuous chips: Of irregular size and shape: - work material - brittle like grey cast iron. Of regular size and shape: - work material ductile but hard and work hardenable. Feed rate - large. Tool rake - negative. Cutting fluid - absent or inadequate. Continuous chips with BUE When cutting a ductile metal, the compression of the metal is followed by the high heat at tool face. This in turns enables part of the removed metal to be welded into the tool. This is known as built up edge, a very hardened layer of work material attached to the tool face, which tends to act as a cutting edge itself replacing the real cutting tool edge. The built-up edge tends to grow until it reaches a critical size (~0.3 mm) and then passes off with the chip, leaving small fragments on the machining surface. Chip will break free and cutting forces are smaller, but the effect is a rough machined surface. The built-up edge disappears at high cutting speeds. The weld metal is work hardened or strain hardened. While the cutting process is continued, some of built up edge may be combined with the chip and pass along the tool face. Some of the built up edge may be permanently fixed on the tool face. This produces a rough surface finish and the tool life may be reduced. The formation of continuous chips with BUE is schematically shown in Fig.

7 6. Orthogonal cutting The cutting edge of the tool is perpendicular to the direction of feed motion. Chip flow is expected to in a direction perpendicular to the cutting edge. There are only two components of force; these components are mutually perpendicular. The cutting edge is larger than cutting width. Chips are in the form of a spiral coil. High heat concentration at cutting region. For a given feed and depth of cutting, the force acts on a small area as compared with oblique cutting, so tool life is less. Surface finish is poor. Used in grooving, parting, slotting, pipe cutting. Oblique Cutting The cutting edge of the tool is inclined to the direction of feed motion. The chip flow angle is more than zero. There are three mutually perpendicular forces acting while cutting process. The cutting edge may or may not be larger than cutting width. Chip flow is in a sideways direction. Less concentration of heat at cutting region compared to orthogonal cutting. Force is acting on a large area, results in more tool life. Good surface finish obtained. Used almost all industrial cutting, used in drilling, grinding, milling.

8 7. Solution: PART-4 L 2 = 96 mm L 1 = 240 mm α = 20ᵒ d = 0.6 mm F c = 2400 N F t = 240 N Shear plane angle: Ø = tan -1 [(r cosα) / ( 1-r sinα)] Where r = t 1/ t 2 = L 2 / L 1 = 96/240 = 0.4 = tan -1 [(0.4 cos 20ᵒ) / ( sin 20ᵒ)] = 23.5ᵒ Chip thickness: r = t 1 / t = 0.6 / t 2 t 2 = 1.5 mm Friction angle: μ = tan β β = tan -1 (μ) where μ = (F c sinα + F t cosα) / (F c cosα - F t sinα) = (2400 sin20ᵒ cos20ᵒ) / (200 cos20ᵒ sin20ᵒ) = 0.4 β = tan -1 (0.4) = 25.6ᵒ

9 Resultant cutting force: R = F c 2 + F t 2 = =2412 N. Solution: Let V 1 be the initial cutting speed and T 1 be the tool life corresponding to V 1 n = 0.2 Therefore V 1 T n 1 = C V 1 (T 1 ) 0.2 = C (1) If the tool life is reduced to 60% then the cutting speed has been increased Let V 2 be the increase in cutting speed Therefore Tool life T 2 = T T 1 = 0.4 T 1 Therefore V 2 T 2 n = C V 2 (T 2 ) 0.2 = C V 2 (0.4 T 1 ) 0.2 = C (2) Comparing eqn (1) and (2) We have V 1 (T 1 ) 0.2 = V 2 (0.4 T 1 ) 0.2 V 1 = V 2 (0.4) 0.2 or V 2 = V 1 / (0.4) 0.2 V 2 = 1.2 V 1 This means that, when the tool loses 60% of its life, the speed has been increased to 1.2 times the initial speed. Therefore % change in cutting speed = [ (V 2 - V 1 ) / V 1 ] x100 Therefore % change = [ (1.2 V 1 - V 1 ) / V 1 ] x100 = 0.2 x 100 = 20%

10 9. PART-5 Solution: α = 10ᵒ V c = 200 m/min b 1 = 3 mm t 1 = 0.3 mm τ s = 400 N/mm 2 μ = 0.5 Shear angle: Ø = (π/4) - (β/2) + (α/2) μ = tan β Therefore β = tan -1 (μ) = tan -1 (0.5) = 26.56ᵒ Ø = (π/4) - (26.56/2) + (10/2) = 36.72ᵒ Cutting force and Thrust force: where F c = F R cos (β α)..(1) F t = F R sin (β α)..(2) But F R = F s / cos (ø + β α).(3) But F s =? shear stress = τ s = shear force(f s) / shear area ( A s ).(4) τ s = 400 N/mm 2 A s = A 0 / sinø = b 1 t 1 / sinø = (3x0.3) / sin = mm 2 From eqn (4) we have, F s = τ s x A s = 400 x = 602 N Eqn (3) reduces to F R = 602 / cos ( ) = N

11 From eqn (1) F c = cos ( ) = N And From eqn (2) F t = sin ( ) = N Machining constant: 2ø + β α = C 2 ( 36.72) = C C = Solution: At 24 m/min, T = 12 min For HSS VT 1/7 = C 1 (24) (12) 1/7 = C 1 C 1 = 4 For Carbide VT 1/5 = C 2 (24) (12) 1/5 = C 2 C 2 = To calculate T for a cutting speed of 30 m/min For HSS (30) T 1/7 = 4 T 1/7 = 1.6 T 1 = 26.4 min For Carbide (30) T 1/5 = T 1/5 = T 2 = 41.9 min Therefore T 1 / T 2 = 26.4 / 41.9 = 0.64 Therefore T 1 / T 2 = 0.64 or T 1 = 0.64 T 2