EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 9780100793439
Author: KALPAKJIAN
Publisher: YUZU
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Chapter 23, Problem 18QLP
Explain the reasoning behind the various design guide-lines for turning.
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Using the orthogonal model as an approximation of turning, you have been asked to determine:
(a) Friction angle.
(b) Friction coefficient.
(c) Discuss the effect of increasing shear angle on shear force and shear stress.
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…
Discuss the Applications and Importance of knurling and Taper turning Operations.
Chapter 23 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 23 - Describe the types of machining operations that...Ch. 23 - What is turning? What kind of chips are produced...Ch. 23 - What is the thrust force in turning? What is the...Ch. 23 - What are the components of a lathe?Ch. 23 - (a) What is a tracer lathe? (b) What is an...Ch. 23 - Describe the operations that can be performed on a...Ch. 23 - Why were power chucks developed?Ch. 23 - Explain why operations such as boring on a lathe...Ch. 23 - Why are turret lathes typically equipped with more...Ch. 23 - Describe the differences between boring a...
Ch. 23 - How is drill life determined?Ch. 23 - What is the difference between a conventional...Ch. 23 - Why are reaming operations performed?Ch. 23 - Explain the functions of the saddle on a lathe.Ch. 23 - Describe the relative advantages of (a)...Ch. 23 - Explain how external threads are cut on a lathe.Ch. 23 - Prob. 17RQCh. 23 - Explain the reasoning behind the various design...Ch. 23 - Note that both the terms tool strength and...Ch. 23 - (a) List and explain the factors that contribute...Ch. 23 - Explain why the sequence of drilling, boring, and...Ch. 23 - Why would machining operations be necessary even...Ch. 23 - A highly oxidized and uneven round bar is being...Ch. 23 - Describe the difficulties that may be encountered...Ch. 23 - (a) Does the force or torque in drilling change as...Ch. 23 - Explain the similarities and differences in the...Ch. 23 - Describe the advantages and applications of having...Ch. 23 - Assume that you are asked to perform a boring...Ch. 23 - Explain the reasons for the major trend that has...Ch. 23 - Describe your observations concerning the contents...Ch. 23 - The footnote to Table 23.12 states that as the...Ch. 23 - In modern manufacturing, which types of metal...Ch. 23 - Sketch the tooling marks you would expect if a...Ch. 23 - What concerns would you have in turning a powder...Ch. 23 - The operational severity for reaming is much lower...Ch. 23 - Review Fig. 23.6, and comment on the factors...Ch. 23 - Explain how gun drills remain centered during...Ch. 23 - Comment on the magnitude of the wedge angle on the...Ch. 23 - If inserts are used in a drill bit (see Fig....Ch. 23 - Refer to Fig. 23.11b, and in addition to the tools...Ch. 23 - Calculate the same quantities as in Example 23.1...Ch. 23 - Estimate the machining time required to rough turn...Ch. 23 - A high-strength cast-iron bar 8 in. in diameter is...Ch. 23 - A 0.30-in.-diameter drill is used on a drill press...Ch. 23 - In Example 23.4, assume that the workpiece...Ch. 23 - For the data in Problem 23.45, calculate the power...Ch. 23 - A 6-in.-diameter aluminum cylinder 10 in. in...Ch. 23 - A lathe is set up to machine a taper on a bar...Ch. 23 - Assuming that the coefficient of friction is 0.25,...Ch. 23 - A 3-in.-diameter, gray cast iron cylindrical part...Ch. 23 - Would you consider the machining processes...Ch. 23 - Would it be difficult to use the machining...Ch. 23 - If a bolt breaks in a hole, it typically is...Ch. 23 - An important trend in machining operations is the...Ch. 23 - Review Fig. 23.8d, and explain if it would be...Ch. 23 - Boring bars can be designed with internal damping...Ch. 23 - A large bolt is to be produced from extruded...Ch. 23 - Make a comprehensive table of the process...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- A turning operation is performed with HSS tooling on mild steel, with Taylor tool life parameters n = 0.12, C = 60 m/min. Work part length = 450 mm and diameter = 80 mm. Feed = 0.20 mm/rev. Handling time per piece = 4.0 min, and tool change time = 1.5 min. Cost of machine and operator = $27/hr, and tooling cost = $2 per cutting edge. Find the a. cutting speed for maximum production rate and b. cutting speed for minimum cost Equations used n *-=c(")* Vmax = C 1-n Tt Vmin = C =c(₁" n 1 n Co n CoTt + Ct narrow_forwardExplain any one of the turning operations in terms of tool shape,tool path, final and generated geometry, possible tolerances. Then list possible applications or products that are made using that operation.arrow_forwardDifferentiate between (i) Plain Turning and Facing operation (ii)Taper Turning and Knurling operationarrow_forward
- A 200 mm long magnesium alloy bar, 63 mm in diameter is turned on a lathe using a high speed steel cutter travelling at 180 mm/min. The spindle rotates at 450 rpm and lathe is equipped with a 10 kW motor, operating at a mechanical efficiency of 92%. The final diameter of the magnesium alloy bar is 59,5 mm. Indicate with a sketch the recommend size and location of the following tool angles: back rake, side rake, end relief, side relief and side and end cutting edge. Calculate the cutting time for the machining process.Calculate the required cutting force.arrow_forwardQuestion 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…arrow_forwardIn 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_forward
- Explain the difference between gear hobbing and gear shaping with referenceto various relative motions and applicationsarrow_forwardIn 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)arrow_forwardIn a straight turning operations the diameter of a workpiece is reduced from 50 mm 40 mm using a feed rate of 0.1 mm/rev. The cutting tool used for the process has a normal working rake angle of 15°, clearance angle of 4° and a lead angle of 90°. The cutting force of 1200N and a thrust force of 660 N is measured during cutting operations. The workpiece material was stainless steel with a density of 7850 Kg/m³, thermal conductivity is 15 J/smK, specific heat is 480 J/kgK. The width of the secondary deformation zone divided by the chip thickness wo is assumed to be 0.2 for the above mentioned cutting conditions. The chip/tool contact length is measured as 1.5 mm. (a) Calculate the temperature rise in the primary deformation zone (b) Calculate the temperature rise in the secondary deformation zone (c) Calculate the maximum temperature if the room temperature is 24°C If the tool life for the average cutting velocity of 90 m/min is measured as 10 min. and for a cutting velocity of 160 m/min…arrow_forward
- Tool life tests in turning yield the following data: (1) v = 100 m/min, T = 10 min; (2) v = 75 m/min, T = 30 min. (a) Determine the n and C values in the Taylor tool life equation. Based on your equation, compute (b) the tool life for a speed of 90 m/min, and (c) the speed corresponding to a tool life of 20 minarrow_forwardIn 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)arrow_forwardIn a turning operation, the workpiece diameter is Dm=44.00 mm and the diameter after the operation should be 22.00 mm. The cutting speed is set to 105.00 m/min and the federate is 0.03 mm/rev. Calculate the material 3 removal rate (Cm²Imin) for this operation (Do not input units). Your Answer: Answerarrow_forward
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