1) A tapered surface is to be turned on an automatic lathe.

The workpiece is 750 mm long with minimum

and maximum diameters of 100 mm and 200 mm at opposite ends. The 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. (5 p.)

2) In the taper turning job of Problem 1, 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 be required to
complete the job in exactly the same time as your answer to part (a) of that problem. (5 p)
3) A peripheral milling operation is performed on the top surface of a rectangular workpart which is 400
mm long by 60 mm wide. The milling cutter, which is 80 mm in diameter and has five teeth, overhangs
the width of the part on both sides. Cutting speed = 70 m/min, chip load = 0.25mm/tooth, and depth of cut
= 5.0 mm. Determine (a) the actual machining time to make one pass across the surface and (b) the
maximum material removal rate during the cut.
Solution: (a) N = v/πD = 70,000 mm/80 = 279 rev/min
fr = Nntf = 279(5)(0.25) = 348 mm/min
A = (d(D-d))0.5 = (5(80-5))0.5 = 19.4 mm
Tm = (400 + 19.4)/348 = 1.20 min
(b) RMR = wdfr = 60(5)(348) = 104,400 mm3/min

4) A compound die will be used to blank and punch a large washer out of 6061ST aluminum alloy sheet
stock 3.50 mm thick. The outside diameter of the washer is 50.0 mm and the inside diameter is 15.0 mm.
Determine (a) the punch and die sizes for the blanking operation, and (b) the punch and die sizes for the
punching operation (Ac = 0.060). (5 p.)
5) Determine the minimum tonnage press to perform the blanking and punching operation in Problem 4.
The aluminum sheet metal has a tensile strength = 310 MPa, a strength coefficient of 350 MPa, and a
strain-hardening exponent of 0.12. (a) Assume that blanking and punching occur simultaneously. (b)
Assume the punches are staggered so that punching occurs first, then blanking.
Solution: (a) F = 0.7(TS)tL t =3.5 mm from Problem 14.3. L = 50 π + 15 π = 65 π = 204.2 mm F =
0.7(310)(3.5)(235.6) = 155,100 N = 155,1000/(4.4482*2000) tons = 17.4 tons => 18 ton press (b)
Longest length will determine the minimum tonnage press required (b) Longest length will determine the
minimum tonnage press required. Punching length of cut, L= 15 π, blanking length of cut, L=50 π = 157.1
mm (use blanking) F = 0.7(310)(3.5)(157.1) = 119,300 N = 119,3000/(4.4482*2000) tons = 13.4 tons =>
14 ton press

1) A bending operation is to be performed on 5.00 mm thick cold-rolled steel. The part drawing is given
in Figure 1. Determine the blank size required.
Solution: (a) From part drawing in Figure P19.7, ’ = 40, R = 8.50 mm
 = 180  ’ = 140
Ab = 2(/360)(R + Kbat)
R/t = (8.5)/(5.00) = 1.7, which is less than 2.0; therefore, Kba = 0.333
Ab = 2(140/360)(8.5 + 0.33  5.0) = 24.80 mm
Dimensions of starting blank: w = 35 mm, L = 58 + 24.80 + 46.5 = 129.30 mm

2) Determine the bending force required in Problem 1 if the bend is to be performed in a Vdie with a die
opening dimension of 40 mm. The material has a tensile strength of 600 MPa and a shear strength of 430
MPa.
NUMUNE SI units) Determine the bending force in Problem 19.9 if the bend is performed in a V-die
witha die opening dimension = 40 mm. The steel has a tensile strength = 350 MPa and shear strength =
250 MPa. Solution:
For V-bending, Kbf = 1.33.F = Kbf(TS)wt2/D = 1.33(350)(35)(5.0)2/40 = 10,183 N
3) A deep drawing operation is performed in which the inside of the cylindrical cup has a diameter of
4.25 in and a height = 2.65 in. The stock thickness = 1/8 in, and the starting blank diameter = 7.7 in.
Punch and die radii = 5/32 in. The metal has a tensile strength = 65,000 lb/in2 , a yield strength = 32,000
lb/in2 , and a shear strength of 40,000 lb/in2 . Determine (a) drawing ratio, (b) reduction, (c) drawing
force, and (d) blankholder force.
Solution: (a) DR = 7.7/4.25 = 1.81(b) r = (Db – Dp)/Db = (7.7 – 4.25)/7.7 = 3.45/7.70 = 0.448 = 44.8%(c)
F = Dpt(TS)(Db/Dp - 0.7) = (4.25)(0.1875)(65,000)(7.7/4.25 - 0.7) = 180,900 lb(d) Fh = 0.015Y(Db2 -
(Dp + 2.2t + 2Rd)2)Fh = 0.015(32,000)(7.72 - (4.25 + 2.2 x 0.1875 + 2 x 0.15625)2) =
0.015(32,000)(7.72 - 4.9752)
Fh=52.100lb
4) A drawing operation is performed on 3.0 mm stock. The part is a cylindrical cup with height = 50 mm
and inside diameter = 70 mm. Assume the corner radius on the punch is zero. (a) Find the required
starting blank size Db. (b) Is the drawing operation feasible?
Solution: Use surface area computation, assuming thickness t remains constant. Cup area = wall area +
base area = πDph + πDp 2 /4 = π(70)(50) + 0.25π(70)2 = 14,846 mm2 . Blank area = πDb 2 /4 =
0.7855Db 2 Setting blank area = cup area: 0.7855Db 2 = 14,846 Db 2 = 14,846/0.7855 = 18,900 Db =
137.48 mm Test for feasibility: DR = Db/Dp = 137.48/70 = 1.964; t/Db = 3/137.48 = 0.0218 = 2.18%.
These criteria values indicate that the operation is feasible; however, with a punch radius Rp = 0, this
shape would be difficult to draw because the drawing punch would act on the metal like a blanking
punch.
5) A two-spindle drill simultaneously drills a ½ in hole and a ¾ in hole through a workpiece that is 1.0
inch thick. Both drills are twist drills with point angles of 118°. Cutting speed for the material is 230
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 2 spindles lower at the same rate. The feed rate is set so the total metal
removal rate does not exceed 1.50 in3 /min. Determine (a) the maximum feed rate (in/min) that can be
used, (b) the individual feeds (in/rev) that result for each hole, and (c) the time required to drill the holes.
1) A blanking operation is to be performed on 2.0 mm thick cold-rolled steel (half hard). The part is
circular with diameter = 75.0 mm. Determine the appropriate punch and die sizes for this operation.
Solution: (a) Since a = 0.075, the clearance is given by, c = 0.075 (2) = 0.15 mm. Thus the Punch
diameter Dh is calculated as Dh = Db - 2c = 75.0 - 2(0.15) = 74.70 mm. and the Die diameter is Db = 75
mm.
2) Determine the blanking force required in Problem 1, if the shear strength of the steel = 325 MPa and
the tensile strength is 450 MPa.
Solution the blanking force is given by F = StL The thick of the metal stock t is given by the problem as t
= 2 mm The length of cut edge is calculated as: L = πD = 75π = 235.65 mm Thus the blanking force is F
= 325 (2) (235.65) = 153,200 N
3) The top surface of a rectangular workpart is machined using a peripheral milling operation. The
workpart is 735 mm long by 50 mm wide by 95 mm thick. The milling cutter, which is 60 mm in
diameter and has five teeth, overhangs the width of the part equally on both sides. Cutting speed = 80
m/min, chip load = 0.30 mm/tooth, and depth of cut = 7.5 mm. (a) Determine the time required to make
one pass across the surface, given that the setup and machine settings provide an approach distance of 5
mm before actual cutting begins and an overtravel distance of 25 mm after actual cutting has finished. (b)
What is the maximum material removal rate during the cut? (5p.)
4) A shaper is used to reduce the thickness of a 50 mm part to 45 mm. The part is made of cast iron and
has a tensile strength of 270 MPa and a Brinell hardness of 165 HB. The starting dimensions of the part
are 750 mm x 450 mm x 50 mm. The cutting speed is 0.125 m/sec and the feed is 0.40 mm/pass. The
shaper ram is hydraulically driven and has a return stroke time that is 50% of the cutting stroke time. An
extra 150 mm must be added before and after the part for acceleration and deceleration to take place.
Assuming the ram moves parallel to the long dimension of the part, how long will it take to machine?
Solution: Time per forward stroke = (150 + 750 + 150)/(0.125 x 1000) = 8.4 sec Time per reverse stroke
= 0.50(8.4) = 4.2 sec Total time per pass = 8.4 + 4.2 = 12.6 sec = 0.21 min Number of passes = 450/0.40
= 1125 passes Total time T m = 1125(0.21) = 236 min
5) A drilling operation is to be performed with a 12.7 mm diameter twist drill in a steel workpart. The
hole is a blind hole at a depth of 60 mm and the point angle is 118°. The cutting speed is 25 m/min and
the feed is 0.30 mm/rev. Determine (a) the cutting time to complete the drilling operation, and (b) metal
removal rate during the operation, after the drill bit reaches full diameter.
Solution: (a) N = v/πD = 25(103 ) / (12.7π) = 626.6 rev/min fr = Nf = 626.6(0.30) = 188 mm/min A =
0.5D tan (90 – /2) = 0.5(12.7)tan(90 – 118/2) = 3.82 mm Tm = (d + A)/fr = (60 + 3.82)/188 = 0.339 min
(b) RMR = 0.25πD2 fr = 0.25π(12.7)2 (188) = 23,800 mm3 /min
2) A blanking die is to be designed to blank the part outline shown in Figure 1. The material is 4 mm thick
stainless steel (half hard). Determine the dimensions of the blanking punch and the die opening ( Ac =
0.075). (5 p.)
Solution: Since a = 0.075, the clearance is given by, c = 0.075 (4) = 0.3 mm. Blanking die dimensions:
the same as for the part in the figure: L = 85 mm w = 50 mm t = 25 mm s = 25 mm Blanking punch
dimensions: Length L = 85 - 2(0.3) = 84.4 mm Width w = 50 - 2(0.3) = 49.4 mm Top and bottom t widths
= 25 - 2(0.3) = 24.4 mm The s = 25 mm inset dimension remains the same.

3) Determine the tonnage requirement for the blanking operation in Problem 2, given that the stainless
steel has a yield strength = 500 MPa, a shear strength = 600 MPa, and a tensile strength = 700 MPa.
Numune Determine the tonnage requirement for the blanking operation in Problem 19.4, given that the
stainless steel has a yield strength = 275 MPa, shear strength = 450 MPa, and tensile strength = 650 MPa.
Solution: F = StL t = 4 mm from Problem 19.4. L = 85 + 50 + 25 + 25 + 35 + 25 + 25 + 50 = 320 mm F =
450(4.0)(320) = 576,000 N
4) The top surface of a rectangular workpart is machined using a peripheral milling operation. The
workpart is 735 mm long by 50 mm wide by 95 mm thick. The milling cutter, which is 60 mm in
diameter and has five teeth, overhangs the width of the part equally on both sides. Cutting speed = 80
m/min, chip load = 0.30 mm/tooth, and depth of cut = 7.5 mm. (a) Determine the time required to make
one pass across the surface, given that the setup and machine settings provide an approach distance of 5
mm before actual cutting begins and an overtravel distance of 25 mm after actual cutting has finished. (b)
What is the maximum material removal rate during the cut?
1) A facing operation is performed on an engine lathe. The diameter of the cylindrical part is 6 in and the
length is 15 in. The spindle rotates at a speed of 180 rev/min. Depth of cut = 0.110 in, and feed = 0.008
in/rev. Assume 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 towards the
center and (b) the cutting time.

Nümunə :(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

2) The following stress and strain values were measured in the plastic region during a tensile test carried
out on a new experimental metal: (1) true stress = 43,608 lb/in2 and true strain = 0.27 in/in, and (2) true
stress = 52,048 lb/in2 and true strain = 0.85 in/in. Based on these data points, determine the strength
coefficient and strain-hardening exponent.
3) A hot working operation is carried out at various speeds. The strength constant = 30,000 lb/in2 and the
strain-rate sensitivity exponent = 0.15. Determine the flow stress if the strain rate is (a) 0.01/sec (b)
1.0/sec, (c) 100/sec.
4) A face milling operation removes 0.32 in depth of cut from the end of a cylinder that has a diameter of
3.90 in. The cutter has a 4-in diameter with 4 teeth, and its feed trajectory is centered over the circular
face of the work. The cutting speed is 375 ft/min and the chip load is 0.006 in/tooth. Determine (a) the
time to machine, (b) the average metal removal rate (considering the entire machining time), and (c) the
maximum metal removal rate.
5) A tapered surface is to be turned on an automatic lathe. The workpiece is 750 mm long with minimum
and maximum diameters of 100 mm and 200 mm at opposite ends. The 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.

1) A specimen with 6.0 in starting gage length is subjected to a tensile test in which the grips holding the
end of the test specimen are moved with a relative velocity = 1.0 in/sec. Construct a plot of the strain rate
as a function of length as the specimen is pulled to a length = 8.0 in.

2) A sheet-metal part 3.0 mm thick and 20.0 mm long is bent to an included angle = 60° and a bend radius
= 7.5 mm in a V-die. The metal has a yield strength = 220 MPa and a tensile strength = 340 MPa.
Calculate the required force to bend the part, given that the die opening dimension = 15 mm.
Solution: For V-bending, Kbf = 1.33. F = Kbf(TS)wt2 /D = 1.33(340)(20)(3)2 /15 = 5426 N
3) In a tensile test, two pairs of values of stress and strain were measured for the specimen metal after it
had yielded: (1) true stress = 217 MPa and true strain = 0.35, and (2) true stress = 259 MPa and true strain
= 0.68. Based on these data points, determine the strength coefficient and strain-hardening exponent.
Solution: Solve two equations, two unknowns: ln K = ln σ - n ln ε (1) ln K = ln 217 – n ln 0.35 (2) ln K =
ln 259 – n ln 0.68 (3) ln K = 5.3799 – (-1.0498)n = 5.3799 + 1.0498 n (4) ln K = 5.5568 – (-0.3857)n =
5.5568 + 0.3857 n 5.3799 + 1.0498 n = 5.5568 + 0.3857 n 1.0498 n – 0.3857 n = 5.5568 – 5.3799 0.6641
n = 0.1769 n = 0.2664 ln K = 5.3799 + 1.0498 (0.2664) = 5.6596 K = 287 MPa
4) A shaper is used to reduce the thickness of a 50 mm part to 45 mm. The part is made of cast iron and
has a tensile strength of 270 MPa and a Brinell hardness of 165 HB. The starting dimensions of the part
are 750 mm x 450 mm x 50 mm. The cutting speed is 0.125 m/sec and the feed is 0.40 mm/pass. The
shaper ram is hydraulically driven and has a return stroke time that is 50% of the cutting stroke time. An
extra 150 mm must be added before and after the part for acceleration and deceleration to take place.
Assuming the ram moves parallel to the long dimension of the part, how long will it take to machine? (5
p.)
Solution: Time per forward stroke = (150 + 750 + 150)/(0.125 x 1000) = 8.4 sec
Time per reverse stroke = 0.50(8.4) = 4.2 sec
Total time per pass = 8.4 + 4.2 = 12.6 sec = 0.21 min
Number of passes = 450/0.40 = 1125 passes
Total time T m = 1125(0.21) = 236 min

5) A cylindrical workpart 200 mm in diameter and 700 mm long is to be turned in an engine lathe. Cutting
speed = 2.30 m/s, feed = 0.32 mm/rev, and depth of cut = 1.80 mm. Determine (a) cutting time, and (b)
metal removal rate.

1) A compound die will be used to blank and punch a large washer out of 6061ST aluminum alloy sheet
stock 3.50 mm thick. The outside diameter of the washer is 50.0 mm and the inside diameter is 15.0 mm.
Determine (a) the punch and die sizes for the blanking operation, and (b) the punch and die sizes for the
punching operation.
2) A sheet-metal part 3.0 mm thick and 20.0 mm long is bent to an included angle = 60° and a bend radius
= 7.5 mm in a V-die. The metal has a yield strength = 220 MPa and a tensile strength = 340 MPa.
Calculate the required force to bend the part, given that the die opening dimension = 15 mm.
Solution: For V-bending, Kbf = 1.33. F = Kbf(TS)wt2 /D = 1.33(340)(20)(3)2 /15 = 5426 N
3) The following stress and strain values were measured in the plastic region during a tensile test carried
out on a new experimental metal: (1) true stress = 43,608 lb/in2 and true strain = 0.27 in/in, and (2) true
stress = 52,048 lb/in2 and true strain = 0.85 in/in. Based on these data points, determine the strength
coefficient and strain-hardening exponent.
4) For a certain metal, the strength coefficient = 700 MPa and strain-hardening exponent = 0.27.
Determine the average flow stress that the metal experiences if it is subjected to a stress that is equal to its
strength coefficient K, ε must be equal to 1.0 .
Solution: Yf = K = 700 = Kε n = 700ε .27 ε must be equal to 1.0. Y f = 700(1.0).27/1.27 = 700/1.27 =
551.2 MPa
5) A particular metal has a flow curve with strength coefficient = 35,000 lb/in2 and strainhardening
exponent = 0.26. A tensile specimen of the metal with gage length = 2.0 in is stretched to a length = 3.3
in. Determine the flow stress at this new length and the average flow stress that the metal has been
subjected to during deformation.
Solution: ε = ln (3.3/2.0) = ln 1.65 = 0.501 Flow stress Yf = 35,000(0.501)0.26 = 29,240 lb/in2 . Average
flow stress Y f = 35,000(0.501)0.26/1.26 = 23,206 lb/in2 .
1) A specimen with 6.0 in starting gage length is subjected to a tensile test in which the grips holding the
end of the test specimen are moved with a relative velocity = 1.0 in/sec. Construct a plot of the strain rate
as a function of length as the specimen is pulled to a length = 8.0 in.

2) The strength coefficient = 550 MPa and strain-hardening exponent = 0.22 for a certain metal. During a
forming operation, the final true strain that the metal experiences = 0.85. Determine the flow stress at this
strain and the average flow stress that the metal experienced during the operation.
Solution: Flow stress Yf = 550(0.85)0.22 = 531 MPa. Average flow stress Y f = 550(0.85)0.22/1.22 = 435
MPa.
3) A power shears is used to cut soft cold-rolled steel that is 4.75 mm thick. At what clearance should the
shears be set to yield an optimum cut ( Ac = 0.060)?
Solution: From Table 14.1, Ac = 0.060. Thus, c = Act = 0.060(4.75) = 0.285 mm
4) A drawing operation is performed on 3.0 mm stock. The part is a cylindrical cup with height = 50 mm
and inside diameter = 70 mm. Assume the corner radius on the punch is zero. (a) Find the required st
Solution: Use surface area computation, assuming thickness t remains constant. Cup area = wall area +
base area = πDph + πDp 2 /4 = π(70)(50) + 0.25π(70)2 = 14,846 mm2 . Blank area = πDb 2 /4 =
0.7855Db 2 Setting blank area = cup area: 0.7855Db 2 = 14,846 Db 2 = 14,846/0.7855 = 18,900 Db =
137.48 mm Test for feasibility: DR = Db/Dp = 137.48/70 = 1.964; t/Db = 3/137.48 = 0.0218 = 2.18%.
These criteria values indicate that the operation is feasible; however, with a punch radius Rp = 0, this
shape would be difficult to draw because the drawing punch would act on the metal like a blanking
punch.arting blank size Db. (b) Is the drawing operation feasible?

5) A three-axis CNC machining center is tended by a worker who loads and unloads parts between
machining cycles. The machining cycle takes 5.75 min, and the worker takes 2.80 min using a hoist to
unload the part just completed and load and fixture the next part onto the machine worktable. A proposal
has been made to install a two-position 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. What is the
percentage increase in hourly production rate that would result from using the pallet shuttle? (5 p.
1) If the part illustrated in Figure 1 were made by shear spinning, determine (a) the wall thickness along
the cone-shaped portion, and (b) the spinning reduction r.
Solution: (a) tf = t sin  = (2.4)sin 30 = 2.4(0.5) = 1.2 mm
(b) r = (t  tf)/t = (2.4 – 1.2)/2.4 = 0.50 = 50%
(c) Based on sidewise displacement of metal through a shear angle of 30,
Shear strain  = cot 30 = 1.732.
2) Determine the shear strain that is experienced by the material that is shear spun in Problem 1. (5 p)
3) Derive an expression for the reduction r in drawing as a function of drawing ratio DR. (5 p.)
4) A blanking die is to be designed to blank the part outline shown in Figure 2. The material is 4 mm thick
stainless steel (half hard). Determine the dimensions of the blanking punch and the die opening (Ac =
0.075). (5 p.)
Solution: Since a = 0.075, the clearance is given by, c = 0.075 (4) = 0.3 mm. Blanking die dimensions:
the same as for the part in the figure: L = 85 mm w = 50 mm t = 25 mm s = 25 mm Blanking punch
dimensions: Length L = 85 - 2(0.3) = 84.4 mm Width w = 50 - 2(0.3) = 49.4 mm Top and bottom t widths
= 25 - 2(0.3) = 24.4 mm The s = 25 mm inset dimension remains the same.
5) Determine the tonnage requirement for the blanking operation in Problem 4, given that the stainless
steel has a yield strength = 500 MPa, a shear strength = 600 MPa, and a tensile strength = 700 MPa.
Solution: F= StLt= 4 mm from Problem 20.4. L= 85 + 50 + 25 + 25 + 35 + 25 + 25 + 50 = 320 mm F=
600(4.0)(320) = 768,000 N. This is about86.3 tons
1) A peripheral milling operation is performed on the top surface of a rectangular workpart which is 400
mm long 60 mm wide. The milling cutter, which is 80 mm in diameter and has five teeth, overhangs the
width of the part on both sides. Cutting speed = 70 m/min, chip load =0.25 mm/tooth, and depth of cut =
5.0 mm. Determine (a) the actual machining time to make one pass across the surface and (b) the
maximum material removal rate during the cut
Solution: (a) N = v/πD = 70,000 mm/80π = 279 rev/min fr = Nntf = 279(5)(0.25) = 348 mm/min A =
(d(D-d))0.5 = (5(80-5))0.5 = 19.4 mm Tm = (400 + 19.4)/348 = 1.20 min
(b) RMR = wdfr = 60(5)(348) = 104,400 mm3 /min
2) The strength coefficient = 550 MPa and strain-hardening exponent = 0.22 for a certain metal. During a
forming operation, the final true strain that the metal experiences = 0.85. Determine the flow stress at this
strain and the average flow stress that the metal experienced during the operation.
Solution: Flow stress Yf = 550(0.85)0.22 = 531 MPa. Average flow stress Y f = 550(0.85)0.22/1.22 = 435
MPa.
3) A blanking operation is to be performed on 2.0 mm thick cold-rolled steel (half hard). The part is
circular with diameter = 75.0 mm. Determine the appropriate punch and die sizes for this operation, Ac =
0.075
Solution: Since a = 0.075, the clearance is given by, c = 0.075 (2) = 0.15 mm. Thus the Punch diameter
Dh is calculated as Dh = Db - 2c = 75.0 - 2(0.15) = 74.70 mm. and the Die diameter is Db = 75 mm.
4) A 42.0-mm-thick plate made of low carbon steel is to be reduced to 34.0 mm in one pass in a rolling
operation. As the thickness is reduced, the plate widens by 4%. The yield strength of the steel plate is 174
MPa and the tensile strength is 290 MPa. The entrance speed of the plate is 15.0 m/min. The roll radius is
325 mm and the rotational speed is 49.0 rev/min. Determine (a) the minimum required coefficient of
friction that would make this rolling operation possible, (b) exit velocity of the plate, and (c) forward slip.
Solution: (a) Maximum draft d max = µ 2 R Given that d = t o - t f = 42 - 34 = 8.0 mm, µ 2 = 8/325 =
0.0246 µ = (0.0246) 0.5 = 0.157
(b) Plate widens by 4%. t o w o v o = t f wf v f wf = 1.04 w o 42(w o )(15) = 34(1.04w o )v f v f = 42(w o
)(15)/ 34(1.04w o ) = 630/35.4 = 17.8 m/min
(c) v r = π r 2 N= π(0.325) 2 (49.0) = 16.26 m/min
5) In a turning operation, spindle speed is set to provide a cutting speed of 1.8 m/s. The feed and depth of
cut of cut are 0.30 mm and 2.6 mm, respectively. The tool rake angle is 8°. After the cut, the deformed
chip thickness is measured to be 0.49 mm. Determine (a) shear plane angle, (b) shear strain, and (c) ma
Solution: (a) r = to/tc = 0.30/0.49 = 0.612 φ = tan-1 (0.612 cos 8/(1 – 0.612 sin 8)) = tan-1 (0.6628) =
33.6° (b) γ = cot 33.6 + tan (33.6 - 8) = 1.509 + 0.478 = 1.987
(c) RMR = (1.8 m/s x 103 mm/m)(0.3)(2.6) = 1404 mm3 /sterial removal rate. Use the orthogonal cutting
model as an approximation of the turning process.

1) A metal has a flow curve with strength coefficient = 850 MPa and strain-hardening exponent = 0.30. A
tensile specimen of the metal with gage length = 100 mm is stretched to a length = 157 mm. Determine
the flow stress at the new length and the average flow stress that the metal has been subjected to during
the deformation. (5 p.)

2) A drilling operation is to be performed with a 12.7 mm diameter twist drill in a steel workpart. The
hole is a blind hole at a depth of 60 mm and the point angle is 118°. The cutting speed is 25 m/min and
the feed is 0.30 mm/rev. Determine (a) the cutting time to complete the drilling operation, and (b) metal
removal rate during the operation, after the drill bit reaches full diameter. (5p.)
Solution: (a) N = v/πD = 25(103) / (12.7π) = 626.6 rev/min fr = Nf = 626.6(0.30) = 188 mm/min Tm =
L/fr = 60/188 = 0.319 min
(b) RMR = 0.25πD2fr = 0.25π(12.7)2(188) = 23,800 mm3/min
3) A compound die will be used to blank and punch a large washer out of 6061ST aluminum alloy sheet
stock 3.50 mm thick. The outside diameter of the washer is 50.0 mm and the inside diameter is 15.0 mm.
Determine (a) the punch and die sizes for the blanking operation, and (b) the punch and die sizes for the
punching operation, Ac = 0.060. (5 p.)

4) A series of cold rolling operations are to be used to reduce the thickness of a plate from 50 mm down to
25 mm in a reversing two-high mill. Roll diameter = 700 mm and coefficient of friction between rolls and
work = 0.15. The specification is that the draft is to be equal on each pass. Determine (a) minimum
number of passes required, and (b) draft for each pass? (5p.)
5) In an orthogonal cutting operation, the 0.250 in wide tool has a rake angle of 5°. The lathe is set so the
chip thickness before the cut is 0.010 in. After the cut, the deformed chip thickness is measured to be
0.027 in. Calculate (a) the shear plane angle and (b) the shear strain for the operation. (

1) A cylindrical work bar with 4.5 in diameter and 52 in length 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. (5 p.)
Solution: (a) depth d = (4.50 - 4.25)/2 = 0.125 in
(b) RMR = vfd; f = RMR/(12vd) = 6.75/(12 x 450 x 0.125) = 0.010 in
f = 0.010 in/rev
(c) N = v/πD = 450 x 12/4.5 = 382 rev/min
fr = 382(0.010) = 3.82 in/min
Tm = 46/3.82 = 12.04 min

2) In a production turning operation, the foreman has decreed that a single pass must be completed on the
cylindrical workpiece in 5.0 min. The piece is 400 mm long and 150 mm in diameter. Using a feed = 0.30
mm/rev and a depth of cut = 4.0 mm, what cutting speed must be used to meet this machining time
requirement? (5 p)
3) A drilling operation is to be performed with a 12.7 mm diameter twist drill in a steel workpart. The
hole is a blind hole at a depth of 60 mm and the point angle is 118°. The cutting speed is 25 m/min and
the feed is 0.30 mm/rev. Determine (a) the cutting time to complete the drilling operation, and (b) metal
removal rate during the operation, after the drill bit reaches full diameter. (5 p.)
Solution: (a) N = v/πD = 25(103 ) / (12.7π) = 626.6 rev/min fr = Nf = 626.6(0.30) = 188 mm/min A =
0.5D tan (90 – /2) = 0.5(12.7)tan(90 – 118/2) = 3.82 mm Tm = (d + A)/fr = (60 + 3.82)/188 = 0.339 min
(b) RMR = 0.25πD2 fr = 0.25π(12.7)2 (188) = 23,800 mm3 /mi
4) A drawing operation is performed on 3.0 mm stock. The part is a cylindrical cup with height = 50 mm
and inside diameter = 70 mm. Assume the corner radius on the punch is zero. (a) Find the required
starting blank size Db. (b) Is the drawing operation feasible? (5 p.)
Solution: Use surface area computation, assuming thickness t remains constant. Cup area = wall area +
base area = πDph + πDp 2 /4 = π(70)(50) + 0.25π(70)2 = 14,846 mm2 . Blank area = πDb 2 /4 =
0.7855Db 2 Setting blank area = cup area: 0.7855Db 2 = 14,846 Db 2 = 14,846/0.7855 = 18,900 Db =
137.48 mm Test for feasibility: DR = Db/Dp = 137.48/70 = 1.964; t/Db = 3/137.48 = 0.0218 = 2.18%.
These criteria values indicate that the operation is feasible; however, with a punch radius Rp = 0, this
shape would be difficult to draw because the drawing punch would act on the metal like a blanking
punch.
5) A slab milling operation is performed on the top surface of a steel rectangular workpiece 12.0 in long
by 2.5 in wide. The helical milling cutter, which has a 3.0 in diameter and ten teeth, is set up to overhang
the width of the part on both sides. Cutting speed is 125 ft/min, feed is 0.006 in/tooth, and depth of cut =
0.300 in. Determine (a) the actual machining time to make one pass across the surface and (b) the
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(10)(0.006) = 9.55 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)/9.55 = 1.35 min
(b) RMR = wdfr = 2.5(0.30)(9.55) = 7.16 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)/9.55 = 1.56 min