EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 8220100793431
Author: KALPAKJIAN
Publisher: PEARSON
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Chapter 24, Problem 10RQ
Why is sawing a commonly used process? Why do some saw blades have staggered teeth? Explain.
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A student is using a lathe with 80-hp and 80% efficiency to fabricate a copper alloy with Sy = 1200ksi If the width of cut is 0.30 inand the student set a rake angle of 0and a cutting speed of 200fl / min while she assumed a coefficient of friction to be 0.5. What is the maximum depth of cut the student can achieve?
In orthogonal turning of low carbon steel pipe with
principal cutting edge angle of 90°, the main
cutting force is 1000 N and the feed force is
800 N. The shear angle is 25° and orthogonal rake
angle is zero. Employing Merchant's theory, the
ratio of friction force to normal force acting on the
cutting tool is
(a) 1.56
(b) 1.25
(d) 0.64
(c) 0.80
During orthogonal cutting operation of material has shear strength 95.5 Mpa.
The cutting force is more than thrust force by 10%. The rake angle = 5°, the width of
the cut = 5.0 mm, the chip thickness before the cut = 0.6, and the chip thickness ratio
= 0.38. Determine (a) both cutting force and thrust force and (b) the coefficient of
friction in the operation.
Chapter 24 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 24 - Explain why milling is such a versatile machining...Ch. 24 - Describe a milling machine. How is it different...Ch. 24 - Describe the different types of cutters used in...Ch. 24 - Define the following: face milling, peripheral...Ch. 24 - Can threads be machined on a mill? Explain.Ch. 24 - What is the difference between feed and feed per...Ch. 24 - Explain the relative characteristics of climb...Ch. 24 - Describe the geometric features of a broach and...Ch. 24 - What is a pull broach? A push broach?Ch. 24 - Why is sawing a commonly used process? Why do some...
Ch. 24 - What advantages do bed-type milling machines have...Ch. 24 - Explain why the axis of a hob is tilted with...Ch. 24 - What is a shell mill? Why is it used?Ch. 24 - Why is it difficult to saw thin sheet metals?Ch. 24 - Of the processes depicted in Fig. 24.2, which is...Ch. 24 - Describe the tool motion during gear shaping.Ch. 24 - When is filing necessary?Ch. 24 - Would you consider the machining processes...Ch. 24 - Why is end milling such an important versatile...Ch. 24 - List and explain factors that contribute to poor...Ch. 24 - Explain why broaching crankshaft bearings is an...Ch. 24 - Several guidelines are presented in this chapter...Ch. 24 - What are the advantages of helical teeth over...Ch. 24 - Explain why hacksaws are not as productive as band...Ch. 24 - What similarities and differences are there in...Ch. 24 - Why do machined gears have to be subjected to...Ch. 24 - How would you reduce the surface roughness shown...Ch. 24 - Why are machines such as the one shown in Fig....Ch. 24 - Comment on your observations concerning the...Ch. 24 - Explain how contour cutting could be started in a...Ch. 24 - Prob. 32QLPCh. 24 - Describe the parts and conditions under which...Ch. 24 - Explain the reason that it is difficult to use...Ch. 24 - Would you recommend broaching a keyway on a gear...Ch. 24 - Prob. 37QTPCh. 24 - A slab-milling operation is being performed at a...Ch. 24 - Show that the distance lc in slab milling is...Ch. 24 - Prob. 40QTPCh. 24 - Calculate the chip depth of cut, tc, and the...Ch. 24 - Estimate the time required to face mill a...Ch. 24 - A 12-in.-long, 1-in.-thick plate is being cut on a...Ch. 24 - A single-thread hob is used to cut 40 teeth on a...Ch. 24 - Assume that m the face-milling operation shown in...Ch. 24 - A slab-milling operation will take place on a part...Ch. 24 - Prob. 47QTPCh. 24 - In describing the broaching operations and the...Ch. 24 - The parts shown in Fig. 24.1 are to be machined...Ch. 24 - Would you prefer to machine the part in Fig. 24....Ch. 24 - Prob. 51SDPCh. 24 - Suggest methods whereby milling cutters of various...Ch. 24 - Prepare a comprehensive table of the process...Ch. 24 - Prob. 55SDPCh. 24 - Make a list of all the processes that can be used...
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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 1 mm thick cylindrical tube, 100 mm in diameter, is orthogonally turned such that the entire wall thickness of the tube is cut in a single pass. The axial feed of the tool is 1 m/minute and the specific cutting energy (u) of the tube material is 6 J/mm³. Neglect contribution of feed force towards power. The power required to carry out this operation isarrow_forwardParvinbhaiarrow_forward(e) Briefly describe types of chips that occur in metal cutting. (f) For orthogonal cutting, the tool rake angle =15°. The chip thickness before the cut is 0.30mm and the cut yields a deformed chip thickness = 0.65mm. Calculate the shear plane angle and shear strain.arrow_forward
- (11,00 Puanlar) 39 An orthogonal cutting operation is being carried out under the following conditions: t0=0,38 mm, tc=0,65 mm, width of the cut= 2.5 mm, V= 3.5 m/s, rake angle=D 6 , Fc= 515 N, and Ft= 210 N. Calculate the percentage of the total energy that is dissipated in the shear plane. The power input in cutting= F.V Power for shearing=F,V, t I'c tan ø = Iccos a tc 1-re sin a cin -1-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_forwardExample 3.18 A low carbon steel bar of 147 mm diameter with length of 630 mm is being turned with uncoated carbide insert. The observed tool life are 24 and 12 for cutting velocities of 90 m/min and 120 m/min respectively. The feed and depth of cut are 0.2 mm./rev and 2 mm respectively. Use the unmachined to calculate the cutting velocity (i) When tool life is 20 min. the cutting velocity in m/min is (a) 87 (b) 97 (c) 107 (d) 114 (ii) Neglect over travel or approach of the tool. When tool life is 20 min, the machining time in remain for a single pass is (a) 5 (b) 10 (c) 15 (d) 20arrow_forward
- 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.arrow_forwardEstimate the machining time required in rough turning a 2.0-m-long, annealed aluminum-alloy round bar that is 75 mm in diameter, using (a) a high-speed steel tool; and (b) a carbine tool. Use a feed of 2 mm/rev. Assume max cutting speed for high-speed tools is moving 4 m/s and for carbide tools is moving 7 m/s.arrow_forward2 1.46 Explain why the power requirements in cutting depend on the cutting force but not the thrust force.arrow_forward
- (b) A 400 mm long, 19.5 mm diameter of 304 stainless steel rod (assume specific energy of steel is 4 W.s/mm³) is being reduced in diameter to 17 mm by turning on a lathe machine. The spindle rotates at N = 700 rpm, and the tool is traveling at an axial speed of 300 mm/min. Calculate: (i) (ii) (iii) (iv) (v) cutting speed material removal rate cutting time power dissipated cutting forcearrow_forward2 2 . 16 Explain why so many different types of cutting-tool materials have been developed over the years. Why are they still being developed further?arrow_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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