ME 3162
Tutorial 1 & 2
(Metal cutting)
Tutorial 1
1. Two thousand bars 80mm diameter and 300mm long must be turned down to 65mm
diameter for 150mm of their length. The surface finish and accuracy requirements are
such that a heavy roughing cut (removing most of the material) followed by a light
finishing cut are needed. The roughing cut is to be taken at maximum power. The light
finishing cut is to be taken at a feed of 0.13mm/rev and a cutting speed of 1.5m/s.
If the lathe has 2kW motor and an efficiency of 50 percent, calculate the total production
time in kiloseconds (ks) for the batch of work. Assume that the specific cutting energy for
the work material is 2.73GJ/m3, the time taken to return the tool to the beginning of the
cut is 15s, and the time taken to load and unload a workpiece is 120s.
Summary:
• 2000 pcs
• dw=80 mm
• dm=65 mm
• lw=150 mm
• Finish cut: f=0.13 mm/rev
V=1.5m/s
• Pe=2 kW
• ηm=0.5
• ps=2.73 GJ/m3
• Tool return time = 15 s
• Load/Unload time =120 s
• Power available:
Pm = Pe ×η m = 2000 × 0.5 = 1000(W )
• Maximum removal rate:
Pm = ps × Z w
⇒ Zw =
Pm
1000
=
= 3.663 ×10 −7 (m 3 / s )
9
ps
2.73 ×10
(
• For finishing cut:
)
Zw
3.663 × 10−7
=
= 1.88 ~ 2( mm )
ap =
fV (0.13 × 10−3 ) × 1.5
Summary:
• 2000pcs
• dw=80 mm
• dm=65 mm
• lw=150 mm
• Finish cut:
f=0.13 mm/rev
V=1.5m/s
• Pe=2 kW
• ηm=0.5
• ps=2.73 GJ/m3
• Tool return time = 15 s
• Load/Unload time =120 s
lw
lw  π × d w  150  π × (69 × 10−3 ) 
 = 167( s )
= ×
×
tmf =
=
f × nw
f  V  0.13 
1.5

• For rough cut:
2
π
× 80 ×10 −3 − 69 ×10 −3
volume − of − removed − material 4
=
t mr =
Zw
3.663 ×10 −7
[(
) (
(
) ]× (150 ×10 )
= 532( s )
)
2
−3
• Total production time:
= no. of pieces x (loading time + tool return time + rough cut time + finish cut time)
∑ time = 2000 × (120 + 15 × 2 + 532 + 167) = 1698(ks)
Note: Depth-of-cut for finish cut should be ≤ 2mm
Tutorial 1 - Problem 1
2. In a drilling operation using a twist drill, the rotational frequency of a drill is 5
revolutions per second, the feed 0.25mm per revolution, the major cutting-edge angle
60°, and the drill diameter 12mm. Assuming that the specific cutting energy for the work
material is 2GJ/m3, calculate:
a. The maximum metal removal rate, in µm3/s
b. The undeformed chip thickness, in mm
c. The drill torque, in Nm
Summary:
• nt= 5 rev/s
• f= 0.25 mm/rev
• kr= 60°
• dm = 12mm
• ps = 2 GJ/m3
a. Maximum metal removal rate:
πd m2
Zw =
Vf
4
Zw =
πd m2
π × (12 ×10 −3 )
2
fnt =
4
4
Z w = 1.414 ×10 −7 (m 3 / s )
(
)
× 0.25 ×10 −3 × 5
Z w = 1.414 ×1011 ( µm 3 / s )
b. Undeformed chip thickness:
ac =
f
0.25
sin k r =
sin 60° = 0.108(mm)
2
2
c. Torque:
P
T= m
ω
(
) (
)
ps Z w
2 ×109 × 1.414 ×10 −7
T=
=
= 9( Nm)
2πnt
2×π × 5
Tutorial 1 - Problem 2
Summary:
• nt= 5 rev/s
• f= 0.25 mm/rev
• kr= 60°
• dm = 12mm
• ps = 2 GJ/m3
3. In a face milling operation, the depth of cut is 5mm, the work feed speed is 0.65mm/s,
and the width of the workpiece is 50mm. The cutter has a diameter of 100mm and has 20
teeth. If the cutting speed is to be 1m/s, calculate:
a. The rotational frequency of the cutter in revolutions per second.
b. The maximum metal removal rate, in µm3/s.
c. The time taken to machine 1000 work pieces of length 150mm if it takes 180s to load
and unload a workpiece and return the cutter to the beginning of the cut.
d. The maximum undeformed chip thickness in mm.
Summary:
• ap= 5 mm
• Vf= 0.65 mm/s
• ae=50 mm
• dt=100 mm
• N=20 teeth
• V=1 m/s
• 1000 pieces
• lw= 150 mm
• Loading time + tool return time = 180s
a. Rotational speed:
V
1
nt =
=
= 3.18(rev / s )
πd t π × (100 ×10 −3 )
b. Maximum material removal rate:
Z w = a p aeV f = (5 × 10−3 )(50 × 10−3 )(0.65 × 10−3 )
Z w = 1.63 × 10−7 ( m 3 / s )
Z w = 1.63 × 1011 ( µm 3 / s )
c. Time to machine each piece:
l + d t 150 + 100
tm = w
=
= 385( s )
Vf
0.65
Summary:
• ap= 5 mm
• Vf= 0.65 mm/s
• ae=50 mm
• dt=100 mm
• N=20 teeth
• V=1 m/s
• 1000 pieces
• lw= 150mm
• Loading time + tool return time = 180s
Production time:
= No. of pieces x (machining time + [loading time + tool return time])
∑ time = 1000 × (385 + 180) = 565(ks)
d. Undeformed chip thickness: (tool axis passes over the workpiece)
acmax = a f
acmax =
Vf
Nnt
=
0.65
= 0.0102(mm)
20 × 3.18
Tutorial 1 - Problem 3
Two situations of vertical-milling:
1. Tool axis passes over the workpiece (in face-milling)
ap: depth of cut
ae: width of workpiece
Maximum undeformed chip thickness:
Machining time:
Two situations of vertical-milling:
2. Tool axis does not pass over the workpiece
ap: depth of cut
ae: width of slot
Maximum undeformed chip thickness:
Machining time:
Tutorial 2
In an orthogonal cutting test on mild steel, the following results were obtained:
Calculate the shear angle φ and mean shear strength of the work material τs, in MN/m2
a. Cutting ratio:
a
0.25
= 0.333
rc = c =
ao 0.75
Shear angle:
r cos γ ne
0.333 × cos 0°
tan φ = c
=
= 0.333
1 − rc sin γ ne 1 − 0.333 × sin 0°
⇒ φ = 18.435°
b. Shear plane area:
A
a ×a
As = c = c w
sin φ
sin φ
Mean shear strength
F [(F cos φ − Ft sin φ )]sin φ
τs = s = c
As
ac × a w
[(900 × cos18.435° − 450 × sin 18.435°)]× sin 18.435°
τs =
(0.25 ×10−3 )× (2.5 ×10−3 )
τ s = 360 ×106 ( N / m 2 )
ac = 0.25mm
ao = 0.75mm
aw = 2.5mm
l f = 0.5mm
Fc = 900 N
Ft = 450 N
τ s = 360( MN / m 2 )
Tutorial 2
γ ne = 0°