EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 8220100793431
Author: KALPAKJIAN
Publisher: PEARSON
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Chapter 26, Problem 38QTP
Assume that a surface-grinding operation is being carried out under the following conditions: D = 200 mm, d = 0.1 mm, v = 40 m/min, and V = 2000 m/min. These conditions are then changed to the following: D = 150 mm, d = 0.1 mm, v = 30 m/min, and V = 2500 m/min. How different is the temperature rise from the rise that occurs with the initial conditions?
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A HSS tool is used to turn a steel workpart that is 300 mm long and 80 mm in diameter. The
parameters in the Taylor equation are: n=0.13 and C= 75 (m/min) for a feed of 0.4 mm/rev.
The operator and machine tool rate = $30.00/hr, and the tooling cost per cutting edge =
$4.00. It takes 2.0 min to load and unload the workpart and 3.50 min to change tools.
Determine:
a. Tutting speed for maximum production rate,
b. Tool life in min of cutting, and
c. Cycle time and cost per unit of product.
A HSS tool is used to turn a steel workpart that is 300 mm long and 80 mm in diameter. The parameters in the Taylor equation are: n = 0.13 and C = 75 (m/min) for a feed of 0.4 mm/rev. The operator and machine tool rate = $30.00/hr, and the tooling cost per cutting edge = $4.00. It takes 2.0 min to load and unload the workpart and 3.50 min to change tools. Determine:
Tutting speed for maximum production rate,
Tool life in min of cutting, and
Cycle time and cost per unit of product.
determine cutting speed for minimum cost.
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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- (Grinding Operations and Grinding Machines) (USCS units) In a centerless grinding operation, the grinding wheel diameter = 8 in, and the regulating wheel diameter = 5.0 in. The grinding wheel rotates at 1500 rev/min, and the regulating wheel rotates at 180 rev/min. The inclination angle of the regulating wheel = 2.3°º. Determine the production rate of cylindrical work parts whose diameter = 0.5 in and length = 5.0 in.arrow_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_forwardI need the answer as soon as possiblearrow_forward
- 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)arrow_forwardIn a surface grinding operation performed on a 6150 steel workpart (annealed, approximately 200 BHN), the designation on the grinding wheel is C-24-D-5-V. Wheel diameter =178 mm and its width= 25 mm. Rotational speed = 3000 rev/min. Infeed = 0.05 mm per pass, crossfeed = 12.5 mm, and workpiece speed = 6 m/min. This operation has been a source of trouble right from the beginning. The surface finish is not as good as the 0.4 µm specified on the part print, and there are signs of metallurgical damage on the surface. In addition, the wheel seems to become clogged almost as soon as the operation begins. In short, nearly everything that can go wrong with the job has gone wrong. (a) Determine the rate of metal removal when the wheel is engaged in the work. (b) If the number of active grits per cm? = 30, determine the average chip length and the number of chips formed per time. (c) What changes would you recommend in the grinding wheel to help solve the problems encountered? Explain why you…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
- Problem 5. (Grinding Operations and Grinding Machines) (USCS units) In a centerless grinding operation, the grinding wheel diameter = 8 in, and the regulating wheel diameter = 5.0 in. The grinding wheel rotates at 1500 rev/min, and the regulating wheel rotates at 180 rev/min. The inclination angle of the regulating wheel = 2.3°. Determine the production rate of cylindrical work parts whose diameter = 0.5 in and length = 5.0 in. Equations used V = TDN RMR lc = = √ Dd rg = vwwd nc = U w' vwC Fev Vwwd 0.25 w d K₁ (TWOW) 0.5 (1) 0² VC D F'c = K₁ Ts = K₂d0.75 GR= Vw Vg 0.5 (TCU) OD fr = TD,N, sin I D0.25arrow_forwardQ1. Calculate the time required to machine a workpiece 170 mm long, 60 mm diameter to 165 mm long 50 mm diameter. The workpiece rotates at 440 rpm, feed is 0.3 mm/rev and maximum depth of cut is 2 mm. Assume total approach and overtravel distance as 5 mm for turning operation. Answer Q2. A gray cast iron surface 280 wide and 540mm long may be machined either on a vertical milling machine, using a 100mm - diameter face mill having eight inserted HSS teeth, or on a horizontal milling machine using an HSS slab mill with eight teeth on a 200-mm. diameter. Which machine has the faster cutting time? The values of feed per tooth and cutting speed for both processes are 0.4mm/tooth and 80m/min, respectively. The depth of cut = 3.0 mm and assume A and O equal to 5. Answerarrow_forwardA 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_forward
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