Presentation on theme: "Metal Cutting and Cutting Tools"— Presentation transcript:
1 Metal Cutting and Cutting Tools CHAPTER (3)Metal Cutting and Cutting ToolsDr. Ahmed Abou El-Wafa
2 Clearance necessary to allow cutting to take place. 3.1 Clearance Angle:The clearance angle may be defined as the angle between the flank face of the tool and a tangent to the work surface originating at the cutting edge.FLeading edge(a) Heel fouls work & prevents leading edge from entering.P(b) Heel cleared a way allowing leading edge to enter the work.FPClearance angle: 6-8ºClearance necessary to allow cutting to take place.
3 Effect of workpiece shape on the clearance angle. Primary Clearance angleSecondary Clearance angle(b) Internal cylinder(a) External cylinderPFClearance angle: 5-7ºEffect of workpiece shape on the clearance angle.An excessive clearance angle will not increase efficiency and will merely weaken the tool.
4 Tool angles showing the rake angle. The rake angle is the angle between the top face of the tool and the normal to the work surface at the cutting edge.ChipRake anglepositiveTool angles showing the rake angle.Although increasing the rake angle improves the cutting action, it tends to weaken the tool. Therefore, the choice of a suitable rake angle becomes a compromise between adequate strength of tool and good cutting action.- Generally, for work materials that give a continuous chip, the greater the work-material strength, the smaller the rake angle.
5 3.3 Chip Formation 3.3.1 Types of chips Ductile materials: e.g. st. Brittle materials: e.g. C IWith BUE-Tool-chip friction causes portions of chip to adhere to rake face- BUE forms, then breaks off, cyclicallyAssociated with difficult-to-machine metals at high cutting speedsSerrated chip
7 Features of a single point cutting tool. 3.4 Single Point Cutting Tools3.4.1 The straight-edge cutting toolSection B-B at 90º to cutting edgeFeatures of a single point cutting tool.
8 3.4.2 The parting-off toolSuch a tool must be set on or very slightly above the center, but never below center, or the work will climb on top of the tool just before it is parted off. The tool is naturally rather weak and can easily be broken if this happens.
9 3.4.3 Tool setting Effect of tool setting. (b) Tool below center. Work tends to climb over tool.(a) Tool above center. Tool tends to rub.(c) Tool on center but inclined upward at front. Tool rubs.(d) Tool on center but inclined downward at front. Work tends to drag tool in.Effect of tool setting.
10 3.5 Multi-Point Cutting Tools 3.5.1 ReamersThe clearance angle of reamer is very small. A reamer must obviously not be ground on its clearance faces or its size will be destroyed, and it is ground on its rake faces, i.e., along its flutes.
11 Calculation of drill size required for producing M48x3 thread; 3.5.2 TapsExample 3.1:Calculation of drill size required for producing M48x3 thread;Depth of thread = 0.54*pitch = 0.54*3 = 1.62 mm.90% depth of thread = 1.62*0.9 = 1.46 mm.Tapping size = 48-(2*1.46) = mm diameter.- The next convenient drill size above this should be used.A tap.
12 3.6 Copying and Generating Processes - Copying process: in this process the surface is dependent for its shape on the shape of the cutter, (as an example form tools).W.PTool- Generating process: in this process the surface is independent on the shape of the tool, but depends on the relative motions of work and tool.
13 3.7 Cutting Fluids 3.7.1 Types of cutting fluids Water A poor lubricant, has little wetting action, it also encourages rusting and is suitable only as a coolant during tool grinding.(2) Soluble OilsWith water, they form an intimate mixture, or emulsion, by adding emulsifying agents. While the oil prevents rusting, it is suitable for light cutting operations, i.e. low metal removal rates.(3) Mineral OilsThey are suitable for heavier cutting operations, i.e. high metal removal rates, because of their much better lubricating properties. Sulfur compounds are added to prevent the chip welding to the rake face and forming BUE. They should not be used on Cu and its alloys since they have corrosive effect on them, but are suitable for steels.(4) Vegetable OilsThey are little used since they are liable to decompose and smell, and may become a health hazard.
14 3.7.2 Functions or uses of coolants or cutting fluids The important functions of cutting fluids are given as follows;(i) Cutting fluid washes away the chips and hence keeps the cutting region free.(ii) It helps in keeping freshly machined surface bright by giving a protective coating against atmospheric, oxygen and thus protects the finished surface from corrosion.(iii) It decreases wear and tear of cutting tool and hence increases tool life.(iv) It improves machinability and reduces power requirements.(v) It prevents expansion of workpieces.(vi) It cools the tool and workpiece and removes the generated heat from the cutting zone.(vii) It decreases adhesion between chip and tool, provides lower friction and wear, and a smaller built-up edge.
15 3.8 The Mechanics of Metal Cutting 3.8.1 Introduction
16 Terminology in orthogonal cutting. wwctctoTerminology in orthogonal cutting.
20 3.8.3 Force relationshipWhere: r = chip thickness ratio (or cutting ratio); to = thickness of the chip prior to chip formation; and tc = chip thickness after separationChip thickness after cut always greater than before, so chip ratio always less than 1.0Where: = rake angle
21 τ = - Coefficient of friction between tool and chip: where β is: the friction angle- Shear stress acting along the shear plane:where S is: the shear strengthwhere As is: the shear plane areato is: cut depthw is: cutting edge width,Φ is: shear plane angleτ =- Of all the possible angles at which shear deformation can occur, the work material will select a shear plane angle that minimizes energy, given by;
22 Fc = Fs cos(β - α) / [cos( + β - α)] F, N, Fs, and Fn cannot be directly measuredThe only forces that can be measured are the forces acting on the tool:Cutting force FcThrust force Ft- Then;F = Fc sin + Ft cosN = Fc cos ‑ Ft sinFs = Fc cos ‑ Ft sinFn = Fc sin + Ft cos- From Merchant circle:Fc = Fs cos(β - α) / [cos( + β - α)]Ft = Fs sin(β - α) / [cos( + β - α)]Shear Strain: = tan( - ) + cot -
23 Power calculations - Power is force times speed: Pc = Fc v (hp or kW) Where: v is the cutting speed- The cutting horsepower ishpc = Fc v / 33, (hp)1 kW = 1.34 hp- Due to efficiency losses (η about 90%), the gross horsepower (hpg) required is; hpg = hpc / η- The unit horsepower is;hpu = hpc / MRR (hp.min/mm3)Unit power is also known as the specific energy UMRR- Units for specific energy are typically N.m/mm3 or J/mm3 (in.lb/in3)
24 Example 3.2In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine;(a) the machining shear strain, (b) shear stress, and (c) cutting horsepower.SOLUTION(a) (Shear strain):r = 0.02/0.045 = 0.444Shear plane angle: tan = r cos / [1 – r sin ]tan = cos 10 /[1 – sin 10] = 25.4°Shear strain: = tan( - ) + cot = tan( ) + cot 25.4 = 2.386
25 SOLUTION (cont.)(b) (Shear stress):Shear force: Fs = Fc cos - Ft sin Fs = 350 cos sin 25.4 = 194 lbShear plane area:As = to w / sin As = (0.02) (0.125)/sin 25.4= in2The shear stress isτ = 194/ = 33,276 lb/in2(c) (Cutting horsepower):Cutting hp: hpc = Fc v / 33,000hpc = (350) (200) / 33,000 = 2.12 hp
26 Tool wear (a) Crater wear and (b) Flank wear. 3.9 Tool Failure and Tool Life- Tool wearTool wear (a) Crater wear and (b) Flank wear.
27 2. VBB,max = 0.6 mm, if the flank is irregularly worn, or; I: Initial wearII: Steady state wearIII: Rapid wearThe criteria recommended by ISO3685:1993 to define the effective tool life for cemented carbides tools, high-speed steels (HSS) and ceramics are:Cemented carbides:1. VBB = 0.3 mm, or2. VBB,max = 0.6 mm, if the flank is irregularly worn, or;HSS and ceramics:1. Catastrophic failure, or;2. VBB = 0.3 mm, if the flank is regularly in region B; or3. VBB,max = 0.6mm, if the flank is irregularly in region B.
28 - Tool life Taylor’s equation: Another form of Taylor’s equation: Where:V: Cutting speed (m/min.)TL: Tool life (min.)n, C : Constants depend on machining conditions, e.g., feed, depth of cut, coolant, etc.
29 Example 3.3The outside diameter of a cylinder made of titanium alloy is to be turned. The starting diameter = 500 mm and the length = 1000 mm. Cutting conditions are feed = 0.4 mm/rev and depth of cut = 3.0 mm. The cut will be made with a cemented carbide cutting tool whose Taylor tool life parameter n = 0.23 and C = 400 (m/min). Compute the cutting speed that will allow the tool life to be just equal the cutting time for this part.SOLUTIONv TLn = C(1)Cutting time (Tm) =, where: N = (1000 * v) / D
30 (2)For Tool life (TL) = Cutting time (Tm) i.e., Equating (1) & (2):V = m/min.
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