Homework #4: Due on November 3rd at midnight through the course web site.
The
solution will be posted the day after so that you can check your answers before the
exam.
•
Chapter 23 problems : 23.5, 23.6, 23.8, 23.12, 23.13, 23.17, 23.22, 23.24
•
Chapter 24 problems: 24.1, 24.3, 24.5, 24.7, 24.15, 24.17, 24.21
Chapter 23 problems
23.5 Tool life tests on a lathe have resulted in the following data: (1) at a cutting speed of 375 ft/
min, the tool life was 5.5 min; (2) at a cutting speed of 275 ft/min, the tool life was 53
min. (a) Determine the parameters
n
and
C
in the Taylor tool life equation. (b) Based on
the
n
and
C
values, what is the likely tool material used in this operation? (c) Using your
equation, compute the tool life that corresponds to a cutting speed of 300 ft/min. (d)
Compute the cutting speed that corresponds to a tool life
T
= 10 min.
Solution
: (a)
VT
n
=
C
; Two equations: (1) 375(5.5)
n
=
C
and (2) 275(53)
n
=
C
375(5.5)
n
= 275(53)
n
375/275 = (53/5.5)
n
1.364 = (9.636)
n
ln 1.364 =
n
ln 9.636
0.3102 = 2.2655
n
n
= 0.137
C
= 375(5.5)
0.137
= 375(1.2629)
C
= 474
Check:
C
= 275(53)
0.137
= 275(1.7221) = 474
(b) Comparing these values of
n
and
C
with those in Table 23.2, the likely tool material is
high speed steel.
(c) At
v
= 300 ft/min,
T
= (
C
/
v
)
1/
n
= (474/300)
1/0.137
= (1.579)
7.305
=
28.1 min
(d) For
T
= 10 min,
v
=
C
/
T
n
= 474/10
0.137
= 474/1.371 =
346 ft/min
23.6 Tool life tests in turning yield the following data: (1) when cutting speed is 100 m/min, tool
life is 10 min; (2) when cutting speed is 75 m/min, tool life is 30 min. (a) Determine the
n
and
C
values in the Taylor tool life equation. Based on your equation, compute (b) the
tool life for a speed of 110 m/min, and (c) the speed corresponding to a tool life of 15
min.
Solution
: (a) Two equations: (1) 100(10)
n
=
C
and (2) 75(30)
n
=
C
.
100(10)
n
= 75(30)
n
ln 100 +
n
ln 10 = ln 75 +
n
ln 30
4.6052 + 2.3026
n
= 4.3174 + 3.4012
n
4.6052 - 4.3174 = (3.4012 – 2.3026)
n
0.2877 = 1.0986
n
n
= 0.2619
C
= 100(10)
0.2619
= 100(1.8275)
C
= 182.75
(b) 110
T
0.2619
= 182.75
T
0.2619
= 182.75/110 = 1.661
T
= 1.661
1/0.2619
= 1.661
3.819
=
6.95 min
(c)
v
(15)
0.2619
= 182.75
v
= 182.75/(15)
0.2619
= 182.75/2.0322 =
89.9 m/min
23.8 A 15.0 in by 2.0 in workpart is machined in a face milling operation using a 2.5 in diameter
fly cutter with a single carbide insert. The machine is set for a feed of 0.010 in/tooth and