EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 9780100793439
Author: KALPAKJIAN
Publisher: YUZU
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Chapter 21, Problem 60QTP
For a turning operation using a ceramic cutting tool, if the speed is increased by 50%, by what factor must the feed rate be modified to obtain a constant tool life? Use n = 0.5 and y = 0.6.
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The outside diameter of a cylinder made of steel is to be turned. The starting diameter is 120 mm and the length is 1400 mm. The feed is 0.3 mm/rev and the depth of cut is 2.5mm. The cut will be made with a cemented carbide cutting tool whose Taylor tool life parameters are: n= 0.33 and C=500. Units for the Taylor equation are min for tool life and m/min for cutting speed. Compute the cutting speed that will allow the tool life to be just equal to the cutting time required to complete this turning operation.
In a production turning operation, the workpart is 60 mm in diameter and 500 mm long. A feed of 0.75 mm/rev is used in the operation. If cutting speed=9 m/s, the tool must be changed every 4 workparts; But if cutting speed=5 m/s, the tool can be used to produce 50 pieces between tool changes. Determine the Taylor tool life equation for this job. (use the equations given below for solution)
In a production turning operation, the workpart is 60 mm in diameter and 500 mm long. A feed of 0.75 mm/rev is used in the operation. If cutting speed=9 m/s, the tool must be changed every 4 workparts; But if cutting speed=5 m/s, the tool can be used to produce 50 pieces between tool changes. Determine the Taylor tool life equation for this job. (use the equations given below for solution)
Chapter 21 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 21 - Explain why continuous chips are not necessarily...Ch. 21 - Name the factors that contribute to the formation...Ch. 21 - What is the cutting ratio? Is it always less than...Ch. 21 - Explain the difference between positive and...Ch. 21 - Explain how a dull tool can lead to negative rake...Ch. 21 - Comment on the role and importance relief angle.Ch. 21 - Explain the difference between discontinuous chips...Ch. 21 - Why should we be interested in the magnitude of...Ch. 21 - What are the differences between orthogonal and...Ch. 21 - What is a BUE? Why does it form?
Ch. 21 - Is there any advantage to having a built-up edge...Ch. 21 - What is the function of chip breakers? How do they...Ch. 21 - Identify the forces involved in a cutting...Ch. 21 - Explain the characteristics of different types of...Ch. 21 - List the factors that contribute to poor surface...Ch. 21 - Explain what is meant by the term machinability...Ch. 21 - What is shaving in machining? When would it be...Ch. 21 - List reasons that machining operations may be...Ch. 21 - Are the locations of maximum temperature and...Ch. 21 - Is material ductility important for machinability?...Ch. 21 - Explain why studying the types of chips produced...Ch. 21 - Prob. 22QLPCh. 21 - Tool life can be almost infinite at low cutting...Ch. 21 - Explain the consequences of allowing temperatures...Ch. 21 - The cutting force increases with the depth of cut...Ch. 21 - Why is it not always advisable to increase the...Ch. 21 - What are the consequences if a cutting tool chips?Ch. 21 - What are the effects of performing a cutting...Ch. 21 - Prob. 29QLPCh. 21 - Prob. 30QLPCh. 21 - Prob. 31QLPCh. 21 - Prob. 32QLPCh. 21 - Comment on your observations regarding Figs. 21.1...Ch. 21 - Prob. 34QLPCh. 21 - Comment on your observations regarding the...Ch. 21 - Why does the temperature in cutting depend on the...Ch. 21 - You will note that the values of a and b in Eq....Ch. 21 - Prob. 38QLPCh. 21 - Prob. 39QLPCh. 21 - Explain whether it is desirable to have a high or...Ch. 21 - The Taylor tool-life equation is directly...Ch. 21 - Prob. 42QLPCh. 21 - Why are tool temperatures low at low cutting...Ch. 21 - Can high-speed machining be performed without the...Ch. 21 - Prob. 45QLPCh. 21 - Prob. 46QLPCh. 21 - State whether or not the following statements are...Ch. 21 - Let n = 0.5 and C = 400 in the Taylor equation for...Ch. 21 - Assume that, in orthogonal cutting, the rake angle...Ch. 21 - Prob. 50QTPCh. 21 - Prob. 51QTPCh. 21 - Using trigonometric relationships, derive an...Ch. 21 - An orthogonal cutting operation is being carried...Ch. 21 - Prob. 54QTPCh. 21 - Prob. 55QTPCh. 21 - Prob. 56QTPCh. 21 - Show that, for the same shear angle, there are two...Ch. 21 - With appropriate diagrams, show how the use of a...Ch. 21 - In a cutting operation using a 5 rake angle, the...Ch. 21 - For a turning operation using a ceramic cutting...Ch. 21 - In Example 21.3, if the cutting speed V is...Ch. 21 - Using Eq. (21.30), select an appropriate feed for...Ch. 21 - With a carbide tool, the temperature in a cutting...Ch. 21 - The following flank wear data were collected in a...Ch. 21 - The following data are available from orthogonal...Ch. 21 - Prob. 66QTPCh. 21 - Design an experimental setup whereby orthogonal...Ch. 21 - Describe your thoughts on whether chips produced...Ch. 21 - Recall that cutting tools can be designed so that...Ch. 21 - Recall that the chip-formation mechanism also can...Ch. 21 - Prob. 73SDPCh. 21 - Describe your thoughts regarding the recycling of...Ch. 21 - List products that can be directly produced from...Ch. 21 - Obtain a wood planer and some wood specimens. Show...Ch. 21 - It has been noted that the chips from certain...Ch. 21 - As we have seen, chips carry away the majority of...
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- In a production turning operation, the workpart is 60 mm in diameter and 500 mm long. A feed of 0.75 mm/rev is used in the operation. If cutting speed-9 m/s, the tool must be changed every 4 workparts; But if cutting speed=5 m/s, the tool can be used to produce 50 pieces between tool changes. Determine the Taylor tool life equation for this job. (use the equations given below for solution) L Tm- 1,= Nf N AD, vT" = C %3| AD,L Tm fvarrow_forwardQuestion 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.arrow_forwardNote: Read the question carefully and give me right solutions according to the question. In orthogonal cutting of steel tube of 150 mm diameter and 2 mm thick, the cutting force was 130 kg and feed force was 35 kg for chip thickness of 0.3mm. The orthogonal cut was taken at 60 meter per minute with a feed of 0.14 mm/rev. If the back rack angle of the cutting tool was - 8 o (minus 8 degree), then calculate the shear strain and strain energy per unit volume.arrow_forward
- In a turning operation, cutting speed =200 m/min; feed = 0.25mm mm/rev, and depth of cut = 4.00mm Thermal diffusivity of the work material = 20m mm^2/s and volumetric specific heat =3.5(10^ -3 )J/mm^ 3 -C If the temperature increase above ambient temperature (20degreesC) is the angle measured by a tool-chip thermocouple to be 700degreesC, determine the specific energy for the work material in this operation.arrow_forwardIn an orthogonal cutting test with a bar of 75 mm diameter is reduced to 73 mm by using a HSS tool with arake angle = 10o, following observations were made: length of the chip, lc = 69.44 mm, cutting ratio r =0.3, the horizontal component of the cutting force, FH = 1450 N, and the vertical component of the cuttingforce, FV = 850 N. The various parameters recorded in this cutting operation are: depth of cut, d = 2 mm;feed rate, f = 0.3 mm/rev, cutting speed, V = 60 m/min. Using Merchant’s theory calculate the following:1) Friction force along rake face2) Normal force acting on rake face3) Shear force along the shear plane4) Normal force acting on shear plane5) The percentage error in shear angle predicted by Merchant’s theory6) Shear velocity7) Chip velocity8) Total work done9) The shear work proportion out of the total work done10) The friction work proportion out of the total work donearrow_forwardPuanlar) 16 The outside diameter of a cylinder made of titanium alloy is to be turned. The starting diameter is 400 mm and the length is 1100 mm. The feed is 0.35 mm/rev and the depth of cut is 2.5 mm. The cut will be made with a cemented carbide cutting tool whose Taylor tool life parameters are: n= 0.24 and C=450. Units for the Taylor equation are min for tool life and m/min for cutting speed. Compute the cutting speed that will allow the tool life to be just equal to the cutting time for this part. V vT" = C. %3D Ce Tm 1. %3D 19 fv 25 31 37 43 49 187.9 m/min b) 325.8 m /min 275.8 m/min d) 226.6 m/minarrow_forward
- 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 batch is 2.0…arrow_forwardLow carbon steel having a tensile strength = 300 MPa and a shear strength = 220 MPa is turned at a cutting speed = 2.5 m/s, feed = 0.20 mm/rev and depth of cut = 3.0 mm. The rake angle = 5° in the direction of chip flow. The resulting chip ratio = 0.45. Using the orthogonal model to approximate turning, determine the cutting force and feed force.arrow_forward2) In an orthogonal cutting operation, the cutting tool has a rake angle = 20°. The chip thickness before the cut = 0.25 mm. If the cutting force and thrust force are measured during an orthogonal cutting operation with values, Fc = 1450 N and Ft = 1990 N. The width of the orthogonal cutting operation w = 2.5 mm. the friction angle B is = 79.91° Determine: The Shear Angle The Shear Strain The Shear Strength of the material The Force of Frictionarrow_forward
- calculate the time taken to complete a 300 mm long cut on an aluminium plate using a 75 mm diameter slab mill with 6 teeth.arrow_forwardA number of through holes with 10-mm-diameter have been drilled through 30-mm thick cast iron plate. At a cutting speed of 25 m/min, the high-speed steel drill tool lasted for 44 holes. But, when the cutting speed increased to 35 m/min, the drill tool lasted for only five holes. The feed used in the both cases is 0.08 mm/rev. Determine the values of n and Cin the Taylor tool life equation for the data, where cutting speed v is expressed in m/min, and tool life T is expressed in min.arrow_forwardA 16mm Solid carbide tool with two teeth is used to machine Aluminium using the cutting speed of 275.00 m/min. The feed per tooth is 0.28 mm. What is the right feed rate to use to program the CNC machine? Your Answer: Answerarrow_forward
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