11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
21 MACHINING OPERATIONS AND MACHINE TOOLS
Review Questions
21.1
What are the geometric differences between rotational parts and prismatic parts in
machining?
Answer. Rotational parts are cylindrical or disk-shaped and are machined on a turning
machine (e.g., a lathe); prismatic parts are block-shaped or flat and are generally produced
on a milling machine, shaper, or planer.
21.2
Distinguish between generating and forming when machining part geometries.
Answer. Generating refers to creating part geometry using the feed trajectory of the cutting
tool; examples include straight turning, taper turning, and profile milling. Forming involves
creating part geometry using the shape of the cutting tool; common examples include form
turning and drilling.
21.3
Describe the turning process.
Answer. Turning is a machining process in which a single-point tool removes material
from the surface of a rotating cylindrical workpiece, the tool being fed in a direction
parallel to the axis of work rotation.
21.4
What is the difference between threading and tapping?
Answer. A threading operation is performed on a turning machine and produces an
external thread, while tapping is normally performed on a drilling machine and produces an
internal thread.
21.5
How does a boring operation differ from a turning operation?
Answer. Boring produces an internal cylindrical shape from an existing hole, while turning
produces an external cylindrical shape.
21.6
What is meant by the designation 30 cm  90 cm (12 in  36 in) lathe?
Answer. A 30 cm  90 cm (12 in  36 in) lathe has a 30 cm (12 in) swing (maximum work
diameter that can be accommodated) and a 90 cm (36 in) distance between centers
(indicating the maximum work length that can be held between centers).
21.7
Name the various ways in which a work part can be held in a lathe.
Answer. Methods of holding the work in a lathe include: (1) between centers, (2) chuck,
(3) collet, and (4) face plate.
21.8
What is the difference between a live center and a dead center, when these terms are used
in the context of workholding in a lathe?
Answer. A center holds the workpiece at the tailstock end of the lathe. A live center is
mounted in bearings and rotates with the work, while a dead center does not rotate - the
work rotates about it.
21.9
How does a turret lathe differ from an engine lathe?
21-1
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
Answer. A turret lathe has a tool-holding turret in place of a tailstock; the tools in the turret
can be indexed to perform a sequence of different cutting operations on the work without
the need to change tools as in operating a conventional engine lathe.
21.10 What is a blind hole?
Answer. A blind hole does not exit the work; by comparison, a through hole exits the
opposite side of the work part.
21.11 What is the distinguishing feature of a radial drill press?
Answer. A radial drill has a long radial arm along which the drill head can be positioned to
allow the drilling of large work parts. The radial arm can also be swiveled about the
column to drill parts on either side of the worktable.
21.12 What is the difference between peripheral milling and face milling?
Answer. In peripheral milling, the axis of the tool is parallel to the surface being
machined, and the operation is performed by cutting edges on the outside periphery of the
cutter. In face milling, the axis of the cutter is perpendicular to the surface being milled,
and machining is performed by cutting edges on both the end and outside periphery of the
cutter.
21.13 Describe profile milling.
Answer. Profile milling generally involves the milling of the outside periphery of a flat
part.
21.14 What is pocket milling?
Answer. Pocket milling uses an end milling cutter to machine a shallow cavity (a “pocket”)
into a flat work part.
21.15 Describe the difference between up milling and down milling.
Answer. In up milling, the cutter speed direction is opposite the feed direction; in down
milling, the direction of cutter rotation is the same as the feed direction.
21.16 How does a universal milling machine differ from a conventional knee-and-column
machine?
Answer. The universal milling machine has a worktable that can be rotated about a vertical
axis to present the part at any specified angle to the cutter spindle.
21.17 What is a machining center?
Answer. A machining center is a CNC machine tool capable of performing multiple types
of cutting operations involving rotating spindles (e.g., milling, drilling); the machine is
typically equipped with automatic tool-changing, pallet shuttles to speed work part
changing, and automatic work part positioning.
21.18 What is the difference between a machining center and a turning center?
Answer. A machining center is generally confined to rotating spindle operations (e.g.,
milling, drilling); while a turning center performs turning type operations, generally with
single-point tools.
21-2
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
21.19 What can a mill-turn center do that a conventional turning center cannot do?
Answer. A mill-turn center has the capacity to position a rotational work part at a specified
angular location, permitting milling or drilling to be performed at a location on the
periphery of the part.
21.20 How do shaping and planing differ?
Answer. In shaping, the work is stationary during the cut, and the speed motion is
performed by the cutting tool; whereas in planing, the cutting tool is stationary, and the
work part is moved past the tool in the speed motion.
21.21 Broaching is performed using a rotating multi-tooth cutting tool: (a) true or (b) false?
Answer. (b) Broaching uses a multiple-tooth tool to take multiple cuts by moving the tool
linearly relative to the work in the direction of the tool axis.
21.22 What is the difference between internal broaching and external broaching?
Answer. Internal broaching is accomplished on the inside surface (hole) of a work part;
while external broaching is performed on one of the outside surfaces of the part.
21.23 Identify the three basic forms of sawing operation?
Answer. The three forms of sawing are: (1) hacksawing, (2) bandsawing, and (3) circular
sawing.
21.24 What does the term thread chasing refer to?
Answer. The term thread chasing is applied to production thread-making operations that
utilize self-opening dies which are designed with an automatic device that opens the cutting
teeth at the end of the cut.
21.25 What is the name of the most common process for cutting internal threads?
Answer. Tapping.
21.26 Gear hobbing is basically which of the following types of machining operations , (a)
grinding, (b) milling, or (c) shaping?
Answer. (b). Gear hobbing is basically a milling operation, but the cutter, called a hob, is
much more complex than a conventional form milling cutter. In addition, special milling
machines (called hobbing machines) are required to accomplish the relative speed and feed
motions between the cutter and the gear blank.
Problems
Answers to problems labeled (A) are listed in an Appendix at the back of the book.
Turning and Related Operations
21.1
(A) (SI units) An engine lathe is used to turn a cylindrical work part 125 mm in diameter
by 400 mm long. Cutting speed = 2.50 m/s, feed = 0.40 mm/rev, and depth of cut = 3.0
mm. Determine (a) cutting time and (b) metal removal rate.
Solution: (a) N = v/(πD) = (2.50 m/s)/0.125 = 6.366 rev/s
fr = Nf = 6.366(0.40) = 2.55 mm/s
Tm = L/fr = 400/2.55 = 157 s = 2.62 min
21-3
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
Alternative method using Equation (21.5),
Tm = 125(400)π/(2500  0.40) = 157 s = 2.62 min
(b) RMR = vfd = (2500 m/s)(0.40 mm)(3.00 mm) = 3000 mm3/s
21.2
(USCS units) A cylindrical work bar with diameter = 4.5 in and length = 52 in is chucked
in an engine lathe and supported at the opposite end using a live center. A 46.0 in portion of
the length is to be turned to a diameter of 4.25 in one pass at a speed of 450 ft/min. The
metal removal rate should be 6.75 in3/min. Determine (a) the required depth of cut, (b) the
required feed, and (c) the cutting time.
Solution: (a) depth d = (4.50  4.25)/2 = 0.125 in
(b) RMR = vfd; f = RMR/(12vd) = 6.75/(12  450  0.125) = 0.010 in
f = 0.010 in/rev
(c) N = v/πD = 450  12/4.5 = 382 rev/min
fr = 382(0.010) = 3.82 in/min
Tm = 46/3.82 = 12.04 min
21.3
(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.
Solution: (a) fr = fN = (0.008 in/rev)(200 rev/min) = 1.60 in/min
(b) L = distance from outside to center of part = 0.5D = 0.5(15 – 6) = 3.0 in
Tm = L/fr = 0.5D/fr = 3/(1.6) = 1.875 min
21.4
(A) (SI units) A tapered surface is turned on an automatic lathe. The workpiece is 550 mm
long with minimum and maximum diameters of 120 mm and 200 mm at opposite ends.
Automatic controls on the lathe permit the surface speed to be maintained at a constant
value of 150 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 approximate 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).
Solution: (a) RMR = vfd = (150 m/min)(103 mm/m)(0.25 mm)(3.0 mm) = 112,500 mm3/min
Area of frustrum of cone A = (R1 + R2){h2 + (R1 – R2)2}0.5
Given R1 = 100 mm, R2 = 60 mm, and h = 550 mm
A = (100 + 60){5502 + (100 – 60)2}0.5 = 160(304,100)0.5 = 277,190 mm2
Depth of cut d = 3.0 mm, volume cut V = Ad = (277,190 mm2)(3.0 mm) = 831,571 mm3
Tm = V/RMR = (831,571 mm3)/(112,500 mm3/min) = 7.39 min
(b) At beginning of cut (D1 = 200 mm), N = 150,000/200 = 238.7 rev/min
At end of cut (D2 = 120 mm), N = v/D = 150,000/120 = 397.9 rev/min
21.5
(SI units) In the taper turning job of the previous problem, 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 8.0 min.
21-4
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
Solution: At a constant rotational speed and feed, feed rate fr is constant and Equations
(21.3) and (21.4) can be used. Combining, Tm = L/Nf and rearranging, N = L/fTm
Given L = 550 mm and f = 0.25 mm/rev from the previous problem,
N = 550/(0.25  8.0) = 275 rev/min
21.6
(USCS units) A 4.00-in-diameter workpiece that is 30 in long is to be turned down to a
diameter of 3.50 in, using two passes on an engine lathe at a cutting speed = 350 ft/min,
feed = 0.015 in/rev, and depth of cut = 0.125 in. The bar will be held in a chuck and
supported on the opposite end in a live center. With this workholding setup, one end must
be turned to diameter; then the bar must be reversed to turn the other end. Using an
overhead crane available at the lathe, the total time to load and unload the bar is 6.3 min,
and the time to reverse the bar is 4.2 min. For each turning cut, an approach allowance of
0.25 in must be added to the cut length. Determine the total cycle time to complete this job.
Solution: First end: cut 15 in of 30 in length
N = 350  12/4.0 = 334.2 rev/min, fr = 334.2(0.015) = 5.013 in/min
Tm = (15 + 0.25)/5.013 = 3.04 min; this reduces diameter to 3.75 in
N = 350  12/3.75 = 356.5 rev/min, fr = 356.5(0.015) = 5.348 in/min
Tm = 15.25/5.348 = 2.85 min to reduce the diameter to 3.50 in
Reverse the bar, which takes 4.2 min and cut the remaining 15 in of 30 in length. The
machining times will be the same as before because the length and diameters are the same.
Total cycle time = 6.3 + 3.04 + 2.85 + 4.2 + 3.04 + 2.85 = 22.28 min
21.7
(USCS units) The end of a large tubular work part is faced on a vertical boring mill. The
part outside diameter = 35.0 in, inside diameter = 20.0 in, and height = 18.0 in. If the facing
operation is performed at a rotational speed = 40 rev/min, feed = 0.015 in/rev, and depth of
cut = 0.180 in, determine (a) cutting time to complete the facing operation and (b) cutting
speeds and metal removal rates at the beginning and end of the cut.
Solution: (a) Distance traveled L = (Do  Di)/2 = (35  20)/2 = 7.5 in
fr = (40 rev/min)(0.015 in/rev) = 0.60 in/min
Tm = 7.5/0.6 = 12.5 min
(b) At Do = 35 in, N = v/πD, v = NπD = (40 rev/min)(35/12) = 366.5 ft/min
RMR = vfd = (366.5  12)(0.015)(0.18) = 11.88 in3/min
At Di = 20 in, v = NπD = (40 rev/min)(20/12) = 209.4 ft/min
RMR = vfd = (209.4  12)(0.015)(0.18) = 6.79 in3/min
Drilling
21.8
(A) (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 = 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.
Solution: (a) N = v/πD = 15(103)/(12.7) = 376 rev/min
fr = Nf = 376(0.20) = 75.2 mm/min
A = 0.5D tan (90 – θ/2) = 0.5(12.7)tan(90 – 118/2) = 3.82 mm
Tm = (d + A)/fr = (60 + 3.82)/75.2 = 0.845 min
21-5
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
(b) RMR = 0.25πD2fr = 0.25(12.7)2(75.2) = 9,526 mm3/min
21.9
(USCS units) A CNC drill press performs a series of through-hole drilling operations on a
1.75-in thick aluminum plate that is a component in a heat exchanger. Each hole diameter =
3/4 in. There are 100 holes in all, arranged in a 10 by 10 matrix pattern, and the distance
between adjacent hole centers (along the square) = 1.5 in. Cutting speed = 300 ft/min,
penetration feed (z-direction) = 0.015 in/rev, and traverse rate between holes (x-y plane) =
15.0 in/min. Assume that x-y moves are made at a distance of 0.50 in above the work
surface, and that this distance must be included in the penetration feed rate for each hole.
Also, the rate at which the drill is retracted from each hole is twice the penetration feed
rate. The drill has a point angle = 100. Determine the time required from the beginning of
the first hole to the completion of the last hole, assuming the most efficient drilling
sequence is used to accomplish the job.
Solution: Time to drill each hole:
N = 300  12/0.75 = 1527.7 rev/min
fr = 1527.7(0.015) = 22.916 in/min
Distance per hole = 0.5 + A + 1.75
A = 0.5(0.75) tan(90 – 100/2) = 0.315 in
Tm = (0.5 + 0.315 + 1.75)/22.916 = 0.112 min
Time to retract drill from hole = 0.112/2 = 0.056 min
All moves between holes are at a distance = 1.5 in using a back and forth path between
rows of holes. Time to move between holes = 1.5/15 = 0.1 min. With 100 holes, the
number of moves between holes = 99.
Total cycle time to drill 100 holes = 100(0.112 + 0.056) + 99(0.1) = 26.7 min
21.10 (USCS units) A gun-drilling operation is used to drill a 9/64-in diameter hole. It takes 2.5
min to perform the operation using high pressure fluid delivery of coolant to the drill point.
The current spindle speed = 4000 rev/min, and feed = 0.0017 in/rev. In order to improve
surface finish in the hole, it has been decided to increase the speed by 20% and decrease the
feed by 25%. How long will it take to perform the operation at the new cutting conditions?
Solution: fr = Nf = 4000 rev/min (0.0017 in/rev) = 6.8 in/min
Hole depth d = 2.5 min(6.8 in/min) = 17.0 in
New speed v = 4000(1 + 0.20) = 4800 rev/min
New feed f = 0.0017(1  0.25) = 0.001275 in/min
New feed rate fr = 4800(0.001275) = 6.12 in/min
New drilling time Tm = 17.0/6.12 in/min = 2.78 min
Milling
21.11 (A) (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.
Solution: (a) N = v/πD = 60(103) mm/70 = 273 rev/min
fr = Nntf = 273(5)(0.25) = 341 mm/min
21-6
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
A = (d(D  d))0.5 = (6.5(70  6.5))0.5 = 20.3 mm
Tm = (400 + 20.3)/341 = 1.23 min
(b) RMR = wdfr = 50(6.5)(341) = 110,825 mm3/min
21.12 (SI units) A face milling operation removes 6.0 mm from the top surface of a rectangular
piece of aluminum that is 250 mm long by 75 mm wide by 50 mm thick. Thickness after
the cut = 44.0 mm. The cutter follows a path that is centered over the workpiece. It has four
teeth and an 85-mm diameter. Cutting speed = 2.0 m/s, and chip load = 0.18 mm/tooth.
Determine (a) machining time and (b) maximum metal removal rate during cutting.
Solution: (a) N = v/πD = 2.0(103) mm/s)/85 = 7.49 rev/s
fr = Nntf = 7.49(4)(0.18) = 5.39 mm/s

A  0.5 D  D 2  w2
 = 0.585 

852  752 = 0.5(85 – 40) = 22.5 mm
Tm = (L + A)/fr = (250 + 22.5))/5.39 = 50.6 s = 0.84 min
(b) RMR = wdfr = 75(6)(5.39) = 2426 mm3/s
21.13 (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.
Solution: (a) N= v/πD = 125(12)/3 = 159.15 rev/min
fr = Nntf = 159.15(8)(0.006) = 7.64 in/min
A = (d(D  d))0.5 = (0.30(3.0  0.30))0.5 = 0.90 in
Tm = (L + A)/fr = (12.0 + 0.9)/7.64 = 1.69 min
(b) RMR = wdfr = 2.5(0.30)(7.64) = 5.73 in3/min
(c) The cutter travels 0.5 in before making contact with the work. It moves 0.90 in before
reaching full depth of cut. It then feeds the length of the work (12.0 in). The overtravel
consists of the cutter radius (1.5 in) plus an additional 0.5 in. Thus,
Tf = (0.5 + 0.9 + 12.0 + 1.5 + 0.5)/7.64 = 2.02 min
21.14 (USCS units) A face milling operation is performed on the top surface of a steel block that
is 12.0 in long by 2.0 in wide by 2.5 in thick. The milling cutter follows a path that is
centered over the width of the block. It has five teeth and a 3.0 in diameter. Cutting speed =
250 ft/min, feed = 0.006 in/tooth, and depth of cut = 0.150 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?
Solution: (a) N = v/πD = 250(12)/3 = 318.3 rev/min
fr = 318.3(5)(0.006) = 9.55 in/min
21-7
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015

A  0.5 D  D 2  w2
 = 0.53 

32  2.02  0.382 in
Tm = (12.0 + 0.382)/9.55 = 1.30 min
(b) RMR = 2.0(0.150)(9.55) = 2.87 in3/min
(c) The cutter travels 0.5 in before making contact with the work. It moves 1.50 in before
its center is aligned with the starting edge of the 12.0 in workpiece. It then feeds the length
of the work (12.0 in). The overtravel consists of the cutter radius (1.5 in) plus an additional
0.5 in. Thus,
Tf = (0.5 + 1.5 + 12.0 + 1.5 + 0.5)/9.55 = 1.68 min
21.15 (USCS units) Solve the previous problem except that the workpiece is 5.0 in wide and the
cutter is offset to one side so that the swath cut by the cutter = 1.0 in wide. This is called
partial face milling, Figure 21.20(b).
Solution: (a) N = 250(12)/3 = 318.3 rev/min
fr = 318.3(5)(0.006) = 9.55 in/min
A  w  D  w  1.0  3.0  1.0  = 1.414 in
Tm = (12.0 + 1.414)/9.55 = 1.405 min
(b) RMR = 1.0(0.150)(9.55) = 1.43 in3/min
(c) The cutter travels 0.5 in before making contact with the work. It moves 1.414 in before
reaching full width of cut. It then feeds the length of the work (12.0 in). The overtravel
consists of the cutter radius (1.5 in) plus an additional 0.5 in. Thus,
Tf = (0.5 + 1.414 + 12.0 + 1.5 + 0.5)/9.55 = 1.67 min
Other Operations
21.16 (A) A three-axis computer numerical control machining center is tended by a worker who
loads and unloads parts between machining cycles. The machining cycle takes 7.8 min, and
the worker takes 4.2 min using a hoist to unload the part just completed and load the next
part into the fixture on the machine worktable. A proposal has been made to install a twoposition pallet shuttle at the machine so that the worker and the machine tool can perform
their respective tasks simultaneously rather than sequentially. The pallet shuttle would
transfer the parts between the machine worktable and the load/unload station in 15 sec.
Determine (a) the current cycle time for the operation and (b) the cycle time if the proposal
is implemented. (c) What is the percentage increase in hourly production rate that would
result from using the pallet shuttle?
Solution: (a) The current cycle time is the machine cycle time plus the load unload time.
Tc = 7.8 + 4.2 = 12.0 min
(b) The cycle time under the proposal is Tc = Max{7.8, 4.2} + 0.25 = 8.05 min
(c) The current hourly production rate Rp = 60/12 = 5.0 pc/hr
The production rate under the proposal Rp = 60/8.05 = 7.45 pc/hr
This is an increase of (7.45 – 5.0)/5.0 = 0.49 = 49%
21.17 (SI units) A shaper is used to reduce the thickness of a 50-mm cast iron part to 45 mm. The
top surface of the part is 600 mm long by 200 mm wide. Cutting speed = 0.12 m/sec, and
feed = 0.50 mm/pass. The ram is hydraulically driven and has a return stroke that takes
21-8
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
50% of the forward stroke time. An extra 2 sec must be added at the end of each forward
and reverse stroke to account for acceleration and deceleration of the ram. The ram moves
parallel to the long dimension of the part. How long will it take to perform the operation?
Solution: Time per forward stroke = 0.600/0.12 = 5.0 sec
Time for reverse stroke = 0.50(5.0) = 2.5 sec
Total time per pass = 5.0 + 2 + 2.5 + 2 = 11.5 sec
Number of passes = 200/0.50 = 500 passes
Total time Tm = 500(11.5) = 5750 sec = 95.83 min
21.18 (A) (USCS units) An open-side planer is used to plane the top surface of a rectangular part
that is 20.0 in by 45.0 in by 5 in thick. The part is made of carbon steel with a tensile
strength of 50,000 lb/in2 and a Brinell hardness of 110 HB. Cutting speed = 30 ft/min, feed
= 0.016 in/pass, and depth of cut = 0.250 in. The length of the stroke across the work must
be set up so that 10 in are allowed at both the beginning and end of the stroke for approach
and overtravel. The return stroke takes 60% of the time for the forward stroke, which
includes time losses for acceleration and deceleration during each pass. How long will it
take to complete the job, assuming that the part is oriented on the planer table so as to
minimize the total time?
Solution: Orient the part so that its length (L = 45 in) is in the direction of the stroke. This
will minimize the number of passes required which will minimize time in this case. Time
per forward stroke = (10 + 45 + 10)/(30  12) = 0.18 min
Time per reverse stroke = 0.60(.18) = 0.11 min
Total time per pass = 0.18 + 0.11 = 0.29 min
Number of passes = 20.0/0.016 = 1250 passes
Total time Tm = 1250(0.29) = 362.5 min
Check: Orient work so that its width (w = 20 in) is in direction of stroke.
Time per forward stroke = (10 + 20 + 10)/(30  12) = 0.11 min
Time per reverse stroke = 0.60(.11) = 0.067 min
Total time per pass = 0.11 + 0.067 = 0.177 min
Number of passes = 45.0/0.016 = 2812.5 rounded up to 2813 passes
Total time = 2813(0.177) = 497.9 min
21.19 (SI units) A series of end-milling cuts is currently used to produce an aluminum part that is
an aircraft component. The purpose of the machining operation is to remove 95% of the
part weight to create a structural frame. A total of 4.0 min is lost during the milling cycle
due to tool repositioning. The part has a length = 1.6 m, width = 0.5 m, and thickness =
140 mm. The operation uses a four-tooth indexable-insert end mill with diameter = 32
mm at a cutting speed = 600 m/min, chip load = 0.15 mm/tooth, and average crosssectional area of cut = 240 mm2. High-speed machining has been proposed to replace the
conventional milling process. The same chip load and average area of cut will be used,
but the cutting speed will be increased to 3600 m/min (Table 21.1), and the time lost for
tool repositioning will be reduced to 2.0 min. Determine (a) the cycle time of the current
milling operation and (b) the cycle time of the proposed HSM operation. (c) Is this part a
good candidate for high-speed machining? Explain.
Solution: Assume the same indexable-insert end mill (with appropriate inserts for HSM)
will be used in the new operation. To calculate the two cycle times, the total volume of
21-9
11-17, 11-18-2014
Solutions for Fundamentals of Modern Manufacturing, 6e (published by Wiley)  MPGroover 2015
material removed will be determined and then divided by the material removal rates for
the two operations.
Total volume of starting aluminum block = 1600(500)(140) = 112,000,000 mm3
Total volume of metal removed = 0.95(112,000,000) = 106,400,000 mm3
(a) End milling at 600 m/min: N = v/πD = 600(103)/32 = 5,968 rev/min
fr = Nntf = 5,968(4)(0.15) = 3,581 mm/min
RMR = fr  cross-sectional area of cut = (3581 mm/min)(240 mm2) = 859,436.7 mm3/min
Tm = 4.0 + 106,400,000/859,436.7 = 127.8 min
(b) End milling at 3600 m/min: N = v/πD = 3600(103)/32 = 35,810 rev/min
fr = Nntf = 35,810(4)(0.15) = 21,486 mm/min
RMR = fr(cross-sectional area of cut) = (21,486 mm/min)(240 mm2) = 5,156,620 mm3/min
Tm = 2.0 + 106,400,000/5,156,620 = 22.6 min
(c) This is a good candidate for HSM because the conventional machining time is so
long. The HSM time saving = 127.8 – 22.6 = 105.2 min or 82%. Generally, HSM is
justified by at least one of the following conditions: (1) large volumes of metal removed
from large parts, (2) multiple cutting operations requiring many different tools, and (3)
complex shapes.
21-10