2. Illustrate the influences of tool wear on the workpiece and the overall machining operation.
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- 3) Turning: A turning operation is to be adopted under the following machining conditions: Cutting speed: V = 75 m/min Feed rate: f= 0.25 mm/rev Depth of cut: a = 1.75 mm External workpiece diameter: Do = 50 mm Determine: a. The final workpiece diameter b. The rotational speed of the workpiece c. The tool feed rate expressed in m/min d. The material removal rate mm³/min e. The specific energy of the material assuming that the measured cutting force is F= 750 N. Knowing the specific energy, which of materials from the table below could the workpiece be composed? Approximate Specific-Energy Requirements in Cutting Operations SPECIFIC ENERGY* MATERIAL Aluminum alloys Cast irons Copper alloys High-temperature alloys W-s/mm³ 0.4-1.1 1.6-5.5 1.4-3.3 3.3-8.5 0.4-0.6 4.9-6.8 3.8-9.6 3.0-5.2 2.7-9.3 3.0-4.1 hp-min/in³ 0.15-0.4 0.6-2.0 0.5-1.2 1.2-3.1 0.15-0.2 1.8-2.5 1.1-3.5 1.1-1.9 1.0-3.4 1.1-1.5 Magnesium alloys Nickel alloys Refractory alloys Stainless steels Steels Titanium alloys * At…Q1:- Put a mark of True or false for below:- (a):- The milling cutting tool has a single cutting edge? (b):- At the break-even point, the total revenue equals the total cost? (c):- The right-hand rule is used to define machine axes? (d):- Design error is a simple mistake? (e):- The quality of products is poor in mass production? (f):- Allowances must be left before turning? (g):- Up-ward milling is better than down-ward milling?QUESTION 2 Tool life can be defined as the actual machining time of successive surface sharpening between raw material and cutting tool. Tool life is commonly presented in the form of a curve in'which the log of the cutting speed is plotted against the log of the tool life measured in minutes. Justify the action that should be taken to remove maximum material per minute with the same tool life and at the same time keep the good surface finish.
- Question 3. Three tool materials are to be compared for the same finish turning operation on a batch of 100 steel parts: high speed steel, cemented carbide, and ceramic. For the high speed steel tool, the 170 Taylor equation parameters are: n= 0.125 and C= 70. The price of the HSS tool is $15.00 and it is estimated that it can be ground and reground 15 times at a cost of $1.50. Tool change time = 3 min. Both carbide and ceramic tools are in insert form and can be held in the same mechanical toolholder. The Taylor equation parameters for the cemented carbide are: n = 0.25 and C = 500; and for the ceramic: n = 0.6 and C = 3,000. The cost per insert for the carbide = $6.00 and for the ceramic = $8.00. Number of cutting edges per insert in both cases = 6. Tool change time = 1.0 min for both tools. Time to change parts = 2.0 min. Feed = 0.25 mm/rev, and depth = 3.0 mm. The cost of machine time = $30/hr. The part dimensions are: diameter = 56.0 mm and length = 290 mm. Setup time for the…Question 1 By using a labelled diagram describe the function of a horizontal milling machine. Describe two methods of producing screw threads in the workshop. Describe the various tool life criteria that are used to estimate the life of a cutting tool. Describe the various milling cutters that are used on a vertical and horizontal milling machine and state for what purpose they are used. Describe 2 methods of producing tapers on a lathe machine. Your answers to each question should be supported by suitable diagrams for clarity.Question: Determine the optimum cutting speed for an operation on a Lathe machine using the following information: Tool change time: Tool regrinds time: 3 min Machine running cost Re.0.50 per min Depreciation of tool regrinds Rs. 5.0 The constants in the tool life equation are 60 and 0.2 3 min
- Question 3. A cemented carbide tool is used to turn a part with length = 18.0 in and diameter = 3.0 in. The parameters in the Taylor equation are: n= 0.27 and C = 1200. The rate for the operator and machine tool = $33.00/hr, and the tooling cost per cutting edge = $2.00. It takes 3.0 min to load and unload the workpart and 1.50 min to change tools. The feed = 0.013 in/rev. Determine: a) Cutting speed for maximum production rate, b) Tool life in min of cutting, and c) Cycle time and cost per unit of product.2. The following data was obtained from an orthogonal cutting test: Rake angle = 20° Cutting speed = 100 m/min Chip length before cutting = 29.4 mm Chip length after cutting = 12.9 mm Vertical cutting force 1050 N Horizontal cutting force = 630 N Using Merchant's analysis, calculate (a) resultant force (c) friction force and friction angle (b) shear plane angle (d) total work doneWrite TRUE is the statement/s is True. If the statement is false, identify the word/s that makes the sentence incorrect and write its corresponding correct answer. Alterations on the independent factors during cutting process can cause tool wear. Surface finish and surface integrity of the workpiece are affected by the tool shape and its sharpness.
- 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 030mmIn machining operation, if velocity doubled then the tool life reduced to 1/16 th of initial value. What is the Taylor's tool life index (n) for a machining operation?Make a comprehensive write up on tool-wear and tool life with due reference to cutting speed and tool life. ( Manufacturing process)