Degarmo's Materials And Processes In Manufacturing
13th Edition
ISBN: 9781119492825
Author: Black, J. Temple, Kohser, Ronald A., Author.
Publisher: Wiley,
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Textbook Question
Chapter 27, Problem 2RQ
Explain why tooth spacing (pitch) is important in sawing.
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Students have asked these similar questions
17) A50-mm-dia endmill cuts 6061 aluminum with 30% tool engagement at a depth
of 6 mm. The tool rotates at 200 rpm and is feeding at 250 mm/min. There are 2
teeth set at a rake angle of 10°.
a. Calculate the required power of cutting.
b. Determine the spindle torque.
c. Determine the cutting force on each tooth.
d. If the thrust force is measured to be 350N, estimate the tool-chip interface
friction coefficient.
How could you physically verify your power estimate?
A turning operation is performed on C1008 steel (a ductile steel) using a tool with a nose radius= 1.3 mm. Cutting speed = 61 m/min and feed = 0.27 mm/rev. Compute an estimate of the surface roughness in this operation. (Hint: the ratio of actual to ideal roughness
can be read on the figure below)
Equations used;
Ra
Ri
Actual
Ratio Theoretical
32NR
= rai Ri
2.4
2.2
2.0
1.8
1.6
1.4
1.2
1.0
0
Ductile metals
Cast irons
Free machining alloys.
100
30.5
200
Cutting speed-ft/min
61
Cutting speed - m/min
300
91.5
400
122
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.
Chapter 27 Solutions
Degarmo's Materials And Processes In Manufacturing
Ch. 27 - Why is sawing one of the most efficient of the...Ch. 27 - Explain why tooth spacing (pitch) is important in...Ch. 27 - What is the tooth gullet used for on a saw blade?Ch. 27 - Explain what is meant by the set of the teeth on a...Ch. 27 - How is tooth set related to saw kerf?Ch. 27 - Why can a bandsaw blade not be hardened throughout...Ch. 27 - What are the advantages of using circular saws?Ch. 27 - Why have bandsawing machines largely replaced...Ch. 27 - Explain how the hole in Figure 27.7 is made on a...Ch. 27 - How would you calculate or estimate Tm for a...
Ch. 27 - What is the disadvantage of using gravity to feed...Ch. 27 - What is unique about the broaching process...Ch. 27 - Can a thick saw blade be used as a broach? Why or...Ch. 27 - Broaching machines are simpler in a basic design...Ch. 27 - Why is broaching particularly well-suited for mass...Ch. 27 - In designing a broach, what would be the first...Ch. 27 - Why is it necessary to relate the design of a...Ch. 27 - What two methods can be utilized to reduce the...Ch. 27 - For a given job, how would a broach having...Ch. 27 - Why are the pitch and radius of the gullet between...Ch. 27 - Why are broaching speeds usually relatively low,...Ch. 27 - What are the advantages of shell-type broach...Ch. 27 - Why are most broaches made from alloy or...Ch. 27 - What are the advantages of TiN-coated broaching...Ch. 27 - For mass-production operations, which process is...Ch. 27 - What is the difference between the roughing teeth...Ch. 27 - The sides of a square, blind hole must be machined...Ch. 27 - The interior, flat surfaces of socket wrenches,...Ch. 27 - Prob. 29RQCh. 27 - What are some ways to improve the efficiency of a...Ch. 27 - To what extent is filing different from sawing?Ch. 27 - Prob. 32RQCh. 27 - Prob. 33RQCh. 27 - Prob. 34RQCh. 27 - A surface 12 in. long is to be machined with a...Ch. 27 - The pitch of the teeth on a simple surface broach...Ch. 27 - Estimate the (approximate maximum) horsepower...Ch. 27 - Estimate the approximate force acting in the...Ch. 27 - In cutting a 6-in.-long slot in a piece of AISI...Ch. 27 - The strength of a pull broach is determined by its...Ch. 27 - Suppose you want to shape a block of metal 7 in....Ch. 27 - Could you have saved any time in Problem 7 by...Ch. 27 - Derive the equation for shaping cutting speed.Ch. 27 - How many strokes per minute would be required to...Ch. 27 - How much time would be required to shape a flat...Ch. 27 - What is the metal removal rate in Problem 11 if...Ch. 27 - Suppose you decide to mill the flat surface...Ch. 27 - A planer has a 10-hp motor, and 75% of the motor...
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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 on C1008 steel (a ductile steel) using a tool with a nose radius = 1.3 mm. Cutting speed = 61 m/min and feed = 0.27 mm/rev. Compute an estimate of the surface roughness in this operation. (Hint: the ratio of actual to ideal roughness can be read on the figure below) Equations used R₁ = f² 32NR Ra = rai Ri Actual Theoretical Ratio= 2.4 2.2 2.0 1.8 1.4 1.2 1.0 0 Ductile metals Cast irons Free machining alloys 100 I 30.5 200 Cutting speed-ft/min I 61 300 91.5 Cutting speed - m/min T 400 I 122arrow_forwardThe vertical band saw is particularly useful, as it can be used to make straight cuts as well as ______________.arrow_forwardIn orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool. the cutting velocity is 90 m/min The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm If the orthogonal rake angle is zero and the principal cutting edge angle is 90° Calculate the shear angle in degree.arrow_forward
- .cutting speed is 9m/min. The return time to cutting time is 1:4 and the feed is 3mm Q.4 (i) Find the time required for taking a cemplete cut on a plate 600 × 900 mm", if the cutting speed is 9m/min. The return time to cutting time is 1:4 and the feed is 3mm for the shaper. The clearance at each end is 75 mm. (11) A hole of 30 mm diameter and 75 mm depth is to be drilled. The feed is 1.3 mm/rev and the cutting speed is 62 m/min. Assuming tool approach and tool over travel as 6 mm. Calculate : (a) Cutting time and (b) Material removal rate Q.5 Write short note on the following - (i) Up milling and down milling processes (ii) Balancing of grinding wheels (iii) Gear hobbing ittp3.7 w w w.ItuommC.comarrow_forwardQuestion 4. The following data are available from orthogonal cutting experiments. The depth of cut (feed) to = 0.13 mm, width of cut b = 2.5 mm, rake angle a = - 5°, and cutting speed V = 2 m/s. Chip thickness, t. (mm) = 0.23 Cutting force, F. (N) = 430 Thrust force, F; (N) = 280 Determine the shear angle ð, friction coefficient u , shear stress t, shear strain y on the shear plane, chip velocity Ve, and shear velocity Vs, as well as energies uf ,Ug, and uarrow_forward3) The following data are available from orthogonal cutting experiment, Depth of cut t, = 0.13 mm, width of cut w = 2.5 mm, rake angle a = -5°, cutting speed v = 2 m/s, Chip thickness, t= 0.23 mm, cutting force, F. = 430 N, thrust force, F = 280 N. Determine the following: Shear angle Friction Coefficient u (using F= µ N) Shear Stress t,S Shear strain y on the shear plane. Power required to perform the operation. Gross power required if the efficiency of the machine is 85%. Specific Energy, Utarrow_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_forwardHow much machining time required to turn a mild steel rod from 65mm to 58 mm over a length of 100 mm by turning using a carbide insert. If the approach length and over run length is = 5 mm, Cutting speed is 20 m/min and feed is =0.2 mm/rev, and the depth of cut is 0.5mm Darrow_forwardFind the machining time required to turn a mild steel rod from 65mm to 58 mm over a length of 100 mm by using a carbide insert. If the approach length and over run length is taken as 5 mm, Cutting speed as 20 m/min and feed is =0.2 mm/rev, and the depth of cut is 0.5mmarrow_forward
- Assuming that the coefficient of friction is 0.25, calculate the maximum depth of cut forturning a hard aluminium alloy on a 15-kW lathe (with a mechanical efficieny of 80%) at awidth of cut of 6 mm, rake angle of 0˚, and a cutting speed of 90 m/min, shear strength of 150 MPa.arrow_forwardPlease answer it correctly. Thank youarrow_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
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