DeGarmo's Materials and Processes in Manufacturing
12th Edition
ISBN: 9781118987674
Author: J. T. Black, Ronald A. Kohser
Publisher: WILEY
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Chapter 21, Problem 7P
The power required to machine metal is related to the cutting force
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The orthogonal cutting of steel is done with 10° rake tool with a depth of cut of 4mm
feed rate of 0.4mm/rev. The cutting speed is 200m/min. The chip thickness ratio is
0.35. The vertical cutting force is 1200N and the horizontal cutting force is 700N.
Calculate from the Merchant's theory, the various work done in metal cutting and shear
stress.
In 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:
Answer
Calculate the r/min required for finish turning of a 45mm diameter piece of machine steel (The cutting speed of the machine steel is 30m/min).
Chapter 21 Solutions
DeGarmo's Materials and Processes in Manufacturing
Ch. 21 - Why has the metal-cutting process resisted...Ch. 21 - What variables must be considered in understanding...Ch. 21 - Which of the seven basic chip formation processes...Ch. 21 - How is feed related to speed in the machining...Ch. 21 - Before you select speed and feed for a machining...Ch. 21 - Milling has two feeds. What are they, and which...Ch. 21 - What is the fundamental mechanism of chip...Ch. 21 - What is the difference between oblique machining...Ch. 21 - What are the implications of Figure 21.13, given...Ch. 21 - Note that the units for the approximate equation...
Ch. 21 - For orthogonal machining, the cutting edge radius...Ch. 21 - How do the magnitude of the strain and strain rate...Ch. 21 - Why is titanium such a difficult metal to machine?...Ch. 21 - Explain why you get segmented or discontinuous...Ch. 21 - Why is metal cutting shear stress such an...Ch. 21 - Which of the three cutting forces in oblique...Ch. 21 - How is the energy in a machining process typically...Ch. 21 - Where does the energy consumed in metal cutting...Ch. 21 - What are two ways of estimating the primary...Ch. 21 - What are the three different ways to perform...Ch. 21 - Why does the cutting force Fc increase with...Ch. 21 - Why doesnt the cutting force Fc increase with...Ch. 21 - Prob. 23RQCh. 21 - How does the selection of the machining parameters...Ch. 21 - Suppose you had a machining operation (boring)...Ch. 21 - Make a sketch like that shown in Figure 21.1 with...Ch. 21 - Show how you would do near orthogonal machining in...Ch. 21 - Can you do orthogonal machining on a shaper or...Ch. 21 - What process and material combination would yield...Ch. 21 - What is meant by the statement that machining...Ch. 21 - Prob. 31RQCh. 21 - Figure 21.4 provides suggested cutting speeds and...Ch. 21 - For problem 1, suppose you selected a speed of 145...Ch. 21 - If the cutting forces is 1000 lb calculate the...Ch. 21 - Explain how you would estimate the cutting force...Ch. 21 - For a turning operation, you have selected a...Ch. 21 - For a slab milling operation using a...Ch. 21 - The power required to machine metal is related to...Ch. 21 - In order to drill a hole in the material described...Ch. 21 - Suppose you have the data in Table 21.A obtained...Ch. 21 - Calculate the horsepower that a process is going...Ch. 21 - Explain how you would estimate the cutting force...Ch. 21 - Derive equations for F and N using the circular...Ch. 21 - Prob. 14PCh. 21 - Prob. 15PCh. 21 - A manufacturing engineer needs an estimate of the...Ch. 21 - Using Figure 21.4 for input data, determine the...Ch. 21 - Estimate the horsepower needed to remove metal at...Ch. 21 - For a turning process, the horsepower required was...
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- 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_forwardIn turning of stales steel alloy, 1100 mm length and 400 mm diameter, the Feed was 0.35 mm/rev, and depth of cut = 2.5 mm. The tool used in this cutting is cemented carbide tool where Taylor tool life parameters are n = 0.24 and C = 450 (tool life (min) and cutting speed (m/min). Compute the cutting speed that will allow the tool life to be 10% longer than the machining time for this part.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
- Note: 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_forwardIn an orthogonal cutting operation an 8 mm deep groove is to be turned on a 50 mm diameter steel bar. Spindle speed is 300 rpm and a feed rate of 0.25 mm/rev is given to the tool. Produced chips have a width of 2 mm. Calculate the material removal rate at the beginning and at the end of the cut. Can u help me please?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
- Find 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_forwardWhat’s the answer for this pleasearrow_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_forward
- In 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_forward3. On an upright drilling machine, a 20 mm diameter hole is to be produced in a plate of SAEE112 steel of 30 mm thickness. The cutting speed selected is 10 m/min, and the cutting torque measured is 20 N.m. Calculate the spindle speed, the depth of cut, the main cutting force, and the cutting power.arrow_forwardA motorised metal guillotine machine is required to cut 45 mm diameter hole in a plate of 20 mm thickness at rate of 35 holes per minute. It requires a torque of 7 Nm for an area of hole in mm². If the cutting takes 1/10 of a second and the speed of it's flywheel varies from 165 rpm to 145 rpm, calculate 4.1)Energy required to cut a hole 4.2)Energy required for cutting work per second. 4.3)Maximum fluctuation of energy of the flywheel 4.4)Mass of the flywheel having radius of gyrations of 1,5 marrow_forward
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