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22.1
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.
Solution
: (a)
N
=
v
/(
π
D
) = (2.30 m/s)/0.200
π
= 3.66 rev/s
f
r
=
Nf
= 6.366(.3) = 1.17 mm/s
T
m
=
L
/
f
r
= 700/1.17 = 598 s =
9.96 min
Alternative calculation using Eq. (22.5),
T
m
= 200(700)
π
/(2,300 x 0.32) = 597.6 sec = 9.96 min
(b)
R
MR
=
vfd
= (2.30 m/s)(10
3
)(0.32 mm)(1.80 mm) =
1320 mm
3
/s
22.3
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.
Solution:
(a)
f
r
=
fN
= (0.008 in/rev)(180 rev/min) =
1.44 in/min
(b)
L
= distance from outside to center of part =
D
/2;
T
m
=
L
/
f
r
=
D
/(2
f
r
) = 6/(2 x 1.44) =
2.083 min
22.6
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
in
3
/min. Determine (a) the required depth of cut, (b) the required feed, and (c) the cutting time.
Solution
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 Fall '08
 Groover

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