EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 9780100793439
Author: KALPAKJIAN
Publisher: YUZU
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Chapter 26, Problem 40QTP
Assume that the energy cost for grinding an aluminum part with a specific energy requirement of 8 W-s/mm3 is $1.50 per piece. What would be the energy cost of carrying out the same operation if the workpiece material were T15 tool steel?
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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.
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 26 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 26 - What is an abrasive? What are superabrasives?Ch. 26 - How is the size of an abrasive grain related to...Ch. 26 - Why are most abrasives made synthetically?Ch. 26 - Describe the structure of a grinding wheel and its...Ch. 26 - Explain the characteristics of each type of bond...Ch. 26 - What causes grinding sparks in grinding? Is it...Ch. 26 - Define metallurgical burn.Ch. 26 - Define (a) friability, (b) wear flat, (c) grinding...Ch. 26 - What is creep-feed grinding and what are its...Ch. 26 - How is centerless grinding different from...
Ch. 26 - What are the differences between coated and bonded...Ch. 26 - What is the purpose of the slurry in...Ch. 26 - Explain why grinding operations may be necessary...Ch. 26 - Why is there such a wide variety of types, shapes,...Ch. 26 - Prob. 15QLPCh. 26 - The grinding ratio, G, depends on the type of...Ch. 26 - What are the consequences of allowing the...Ch. 26 - Explain why speeds are much higher in grinding...Ch. 26 - Prob. 19QLPCh. 26 - Prob. 20QLPCh. 26 - Prob. 21QLPCh. 26 - Referring to the preceding chapters on processing...Ch. 26 - Explain the reasons that so many deburring...Ch. 26 - What precautions should you take when grinding...Ch. 26 - Prob. 25QLPCh. 26 - What factors could contribute to chatter in...Ch. 26 - Prob. 27QLPCh. 26 - Prob. 28QLPCh. 26 - Describe the effects of a wear flat on the overall...Ch. 26 - What difficulties, if any, could you encounter in...Ch. 26 - Prob. 31QLPCh. 26 - Prob. 32QLPCh. 26 - Prob. 33QLPCh. 26 - Jewelry applications require the grinding of...Ch. 26 - List and explain factors that contribute to poor...Ch. 26 - Calculate the chip dimensions in surface grinding...Ch. 26 - If the strength of the workpiece material is...Ch. 26 - Assume that a surface-grinding operation is being...Ch. 26 - Estimate the percent increase in the cost of the...Ch. 26 - Assume that the energy cost for grinding an...Ch. 26 - It is known that, in grinding, heat checking...Ch. 26 - Prob. 45QTPCh. 26 - With appropriate sketches, describe the principles...Ch. 26 - Prob. 47SDPCh. 26 - Vitrified grinding wheels (also called ceramic...Ch. 26 - Conduct a literature search, and explain how...Ch. 26 - Visit a large hardware store and inspect the...Ch. 26 - Obtain a small grinding wheel or a piece of a...Ch. 26 - In reviewing the abrasive machining processes in...Ch. 26 - Obtain pieces of sandpaper and emery cloth of...Ch. 26 - On the basis of the contents of this chapter,...
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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_forwardA 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_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_forward
- A slab-milling operation is carried out on a 200 mm long, 80-mm-wide annealed mild-steel workpiece having a feedrate of 0.1 mm/tooth and a depth of cut of 4.0 mm. The cutter of 50 mm diameter has 18 straight teeth and rotates at 135 rpm. The given specific energy for this material is 3.5 W s/mm3 and the slab mill is wider than the workpiece to be machined. Calculate:‧ the material-removal rate;‧ the power and torque required for this operation;‧ the cutting time. (243 mm/min, 77760 mm3/min, 4.5 kW, 52.8 s)arrow_forwardIn machining a mild steel work piece with carbide tool, the life of the tool was found to be 1 hour and 40 minutes, at a spindle speed of 30 m/min. Calculate the tool life if it has to be operated at a speed of 40% higher than the initial cutting speed. Also calculate the cutting speed if the tool is required to have a life of 2 hours and 45 minutes. Assume Taylor's exponent valuen is 0.28.arrow_forwardA student is performing a turning operation with a workpiece with an initial diameter of 40 mm to produce a 30 mm diameter rod that is 100 mm long. The lathe power is 20 kW and is operating on 85% mechanical efficiency. If the student set the cutting speed to 0.5 m/min and the cutting tool is set to have a rake angle of 5 degrees: a.) What material can we choose for the rod is the coefficient of friction is 0.5? b.) If we select 4130 normalized heat-treated steel for the rod, and coefficient of friction is 0.5, what will the maximum depth of cut we can achieve?arrow_forward
- A 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_forward11) An internal bore is ground to 100 mm, removing 0.3 mm stock. To accomplish this, a grinding wheel 80% of the finished size is used. The wheel is dressed every 15 cycles because it is observed that the radial wheel wear is 0.05 mm on average. Assume wheel and part widths are equal. a. What dressing depth would you recommend and why? b. Calculate the G-ratio of this operation.arrow_forwardHello, in the second case it was said that 80% of Ore is grinded to a particle size of below 0.4mm. I would like to know why 0.9 mm was chosen as the feed size in the second case.arrow_forward
- Solve the math in a detailed wayarrow_forwardManufacturers are designing grinding spindles to turn at higher and higher speeds. What is the main reason for this, and what is one drawback of increasing speed? Name 2 reasons why cost increases as finish allowances decrease. Tool temperatures are low at low cutting speeds and high at high cutting speeds, but low again at even higher cutting speeds. Explain why.arrow_forwardA 600mm*30mm flat surface of a plate is to be finish machined on a shaper .The plate has been fixed with 600 mm side along the tool travel direction. If the tool over-travel at each end of the plate is 20 mm, average cutting speed is 8 m/min, feed rate is 0.3 mm/stroke and the ratio of return time to cutting time of the tool is 1:2 Determine time required for machining?arrow_forward
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