IME 258 Manufacturing Methods and Materials I
Homework 3
From Fundamentals of Modern Manufacturing, 5e (published by Wiley)  MPGroover 2012, with selected
additional exercises.
Machining, Parts I & II
Turning and Related Operations
1.
(SI units) An engine lathe is used to turn a cylindrical work part 150 mm in diameter by 500 mm
long. Cutting speed = 2.50 m/s, feed = 0.30 mm/rev, and depth of cut = 3.0 mm. Determine (a) cutting
time and (b) metal removal rate.
2.
(SI units) In a production turning operation, the cylindrical workpiece is 375 mm long and 150 mm
in diameter. Feed = 0.30 mm/rev, and depth of cut = 4.0 mm. What cutting speed must be used to
achieve a machining time of 5.0 min? If the ideal arithmetic surface roughness should be less than 3
micrometers, what minimum tool nose radius would be required? What workpiece or machine tool
factors may affect this surface finish?
3.
(USCS units) A facing operation is performed on a cylindrical part with diameter = 6 in and length
= 15 in. The engine lathe spindle rotates at 200 rev/min. Depth of cut = 0.110 in, and feed = 0.008
in/rev. Assuming the cutting tool moves from the outer diameter of the workpiece to exactly the center
at a constant velocity, determine (a) the velocity of the tool as it moves from the outer diameter toward
the center and (b) the cutting time.
4.
(SI units) A tapered surface is turned on an automatic lathe. The workpiece is 550 mm long with
minimum and maximum diameters of 100 mm and 200 mm at opposite ends. Automatic controls on the
lathe permit the surface speed to be maintained at a constant value of 200 m/min by adjusting the
rotational speed as a function of workpiece diameter. Feed = 0.25 mm/rev, and depth of cut = 3.0 mm.
The rough geometry of the piece has already been formed, and this operation will be the final cut.
Determine (a) the time required to turn the taper and (b) the rotational speeds at the beginning and end
of the cut (assume cutting starts at the smaller diameter).
5.
(SI units) In the taper turning job of Problem 4, suppose that the automatic lathe with surface speed
control is not available and a conventional lathe must be used. Determine the rotational speed that
would complete the job in exactly the same time as in part (a) of that problem.
6.
(SI units) In problems 4 and 5, how many parts per unit time can be made? If the tool nose radius is
1.25 mm, what would be the ideal surface roughness? What workpiece or machine tool factors may
affect this surface finish?
Drilling
7.
(SI units) A drilling operation is performed on a steel part using a 12.7-mm-diameter twist drill with
point angle = 118. The hole is a blind hole with a depth of 60 mm. Cutting speed = 15 m/min, and feed
= 0.20 mm/rev. Determine (a) cutting time of the operation and (b) metal removal rate after the drill bit
reaches full diameter.
8. (USCS units) A two-spindle drill simultaneously drills a ½-in hole and a ¾-in hole through a workpiece
that is 1.0 in thick. Both drills are twist drills with point angles of 118°. Cutting speed for the material is
300 ft/min. The rotational speed of each spindle can be set individually. The feed rate for both holes
must be set to the same value because the two spindles feed at the same rate. The feed rate is set so the
total metal removal rate of both drills does not exceed 1.50 in3/min. Determine (a) maximum feed rate
(in/min) that can be used, (b) individual feeds (in/rev) for each hole, and (c) cutting time to drill the
holes.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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IME 258 Manufacturing Methods and Materials I
Homework 3
From Fundamentals of Modern Manufacturing, 5e (published by Wiley)  MPGroover 2012, with selected
additional exercises.
Milling
9.
(SI units) Peripheral milling is performed on the top surface of a rectangular work part that is 400
mm long by 50 mm wide. The milling cutter is 70 mm in diameter and has five teeth. It overhangs the
width of the part on both sides. Cutting speed = 60 m/min, chip load = 0.25 mm/tooth, and depth of cut
= 6.5 mm. Determine (a) machining time of the operation and (b) maximum material removal rate
during the cut.
10.
(SI units) A face milling operation removes 6.0 mm from the top surface of a rectangular piece of
aluminum that is 300 mm long by 90 mm wide by 75 mm thick. The cutter follows a path that is
centered over the workpiece. It has four teeth and is 100 mm in diameter. Cutting speed = 2.0 m/s, and
chip load = 0.27 mm/tooth. Determine (a) machining time and (b) maximum metal removal rate during
cutting.
11.
(USCS units) Slab milling is performed on the top surface of a rectangular workpiece that is 12.0 in
long by 2.5 in wide by 4.0 in thick. The helical milling cutter, which has a 3.0-in diameter and eight
teeth overhangs the width of the part on both sides. Cutting speed = 125 ft/min, feed = 0.006 in/tooth,
and depth of cut = 0.300 in. Determine (a) machining time and (b) maximum metal removal rate during
the cut. (c) If an additional approach distance of 0.5 in is provided at the beginning of the pass (before
cutting begins), and an overtravel distance is provided at the end of the pass equal to the cutter radius
plus 0.5 in, what is the duration of the feed motion.
12.
(SI units) A high-speed steel tool is used to turn a steel work part that is 350 mm long and 75 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 = $36.00/hr, and the tooling cost per cutting edge = $4.25.
It takes 3.0 min to load and unload the work part and 4.0 min to change tools. Determine (a) cutting
speed for maximum production rate, (b) tool life, and (c) cycle time and cost per unit of product.
13.
(SI units) Solve Problem 12 except that in part (a), determine cutting speed for minimum cost.
14.
(USCS units) Cemented carbide inserts are used to turn a part with a length of 21.0 in and diameter
= 4.0 in. The parameters in the Taylor equation are: n = 0.25 and C = 800 (ft/min). The operator and
machine tool rate = $45.00/hr, and the tooling cost per cutting edge = $2.00. It takes 2.5 min to load and
unload the work part and 1.50 min to change inserts. Feed = 0.015 in/rev. Determine (a) cutting speed
for maximum production rate, (b) tool life, and (c) cycle time and cost per unit of product.
15.
(USCS units) Solve Problem 14 except that in part (a) determine cutting speed for minimum cost.
16.
(SI units) The same grade of cemented carbide tooling is available in two forms for turning
operations in the machine shop: disposable inserts and brazed inserts that must be reground. The Taylor
equation parameters for this grade are n = 0.25 and C = 300 (m/min) under the cutting conditions here.
For the disposable inserts, the price of each insert = $6.00, there are four cutting edges per insert, and
the average tool change time = 1.0 min. For the brazed insert, the price of the tool = $30.00, and it can
be used a total of 15 times before it must be scrapped. The tool change time for the regrindable tooling =
3.0 min. The standard time to grind or regrind the cutting edge is 5.0 min, and the grinder is paid at a
rate = $20.00/hr. Machine time on the lathe costs $24.00/hr. The work part to be used in the comparison
is 375 mm long and 62.5 mm in diameter, and it takes 2.0 min to load and unload the work. Feed = 0.30
mm/rev. For the two cases, compare (a) cutting speeds for minimum cost, (b) tool lives, and (c) cycle
time and cost per unit of production. (d) Which tool would you recommend?
17.
(SI units) Solve Problem 16 except that in part (a), determine the cutting speeds for maximum
production rate.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
21-1
IME 258 Manufacturing Methods and Materials I
Homework 3
From Fundamentals of Modern Manufacturing, 5e (published by Wiley)  MPGroover 2012, with selected
additional exercises.
18.
(SI units) Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be
compared for the same turning operation on a batch of 50 steel parts. For the high-speed steel tool, the
Taylor equation parameters are: n = 0.130 and C = 80 (m/min). The price of the HSS tool is $20.00, and
it is estimated that it can be ground and reground 15 times at a cost of $2.00 per grind. Tool change time
is 3 min. Both carbide and ceramic tools are inserts and can be held in the same mechanical toolholder.
The Taylor equation parameters for the cemented carbide are n = 0.30 and C = 650 (m/min), and for the
ceramic: n = 0.6 and C = 3500 (m/min). The cost per insert for the carbide is $8.00, and for the ceramic
is $10.00. There are six cutting edges per insert in both cases. Tool change time = 1.0 min for both tools.
The time to change a part = 2.5 min. Feed = 0.30 mm/rev, and depth of cut = 3.5 mm. Cost of operator
and machine time = $40/hr. Part diameter = 73 mm, and length = 250 mm. Setup time for the batch =
2.0 hr. For the three tooling cases, compare (a) cutting speeds for minimum cost, (b) tool lives, (c) cycle
time, (d) cost per production unit, and (e) total time to complete the batch and production rate. (f) What
is the proportion of time spent actually cutting metal for each tool material? Use of a spreadsheet
calculator is recommended.
19.
(SI units) Solve Problem 18 except that in parts (a) and (b) determine the cutting speeds and tool
lives for maximum production rate. Use of a spreadsheet calculator is recommended.
20.
(USCS units) A vertical boring mill is used to bore the inside diameter of a large batch of
tube-shaped parts. The diameter = 28.0 in, and the length of the bore = 14.0 in. Cutting speed = 150
ft/min, feed = 0.015 in/rev, and depth of cut = 0.125 in. The parameters of the Taylor equation for the
cutting tool in the operation are n = 0.25 and C = 1100 (ft/min). Tool change time = 3.0 min and tooling
cost = $3.50 per cutting edge. Time to load and unload the parts = 14.0 min, and the cost of machine
time = $42.00/hr. Management wants to increase production rate by 50%. Is that possible? Assume that
feed is unchanged to achieve the required surface finish. What is the current production rate and the
maximum possible production rate for this job?
Additional problems:
21.
Using the chart given in the class Power Point slides, determine the spindle speed (rpm), the feed
rate (in/min) and the cutting power required for turning a 3” dia aluminum rod (6061T6 alloy, BHN =
150) with a depth of cut of 0.1” using a M2 HSS single point tool.
22.
It is required to face mill a low carbon (annealed) steel block (BHN 100) to size it down from
2.1 to 2.0 thick using a 5 diameter 16-toothed carbide cutting tool. The block is 4.0 wide and 8
long. Determine the following using data given in the class Power Point slides:
a)
The recommended spindle speed in revolutions per minute.
b)
The feed per tooth.
c)
The cutting power required.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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