ME 482 - Exam #2
Oct. 31, 2005
Name____________________
Problems
(40)
1. Consider a turning operation on a part of diameter 6”. The part material has shear strength of 40,000 psi
and a tool rake angle of 10 deg. In addition, the cutting conditions are rotational speed = 500 rpm, feed =
0.0075 in/rev, and depth = 0.075 in. The chip thickness after the cut is 0.015 in. Determine the
horsepower required in this operation.
Solution
Power from P (hp) = Fc v/33,000; thus need Fc and v.
r = 0.075/0.015 = 0.5
 = 2(45°) +  – 2 = 2 (45°) + 10° – 2(28.3°) = 43.4°
 = tan-1 (0.5 cos10°/(1 – 0.5 sin10°)) = 28.3°
As = 0.0075(0.075)/sin 28.3° = 0.00119 in2
Fs = 40,000 (0.00119) = 47.5 lb
Fc = 47.5 cos(43.4 – 10)/cos(28.3 + 43.4 – 10) = 83.6 lb
v = 500 rpm (6 in/rev)/(12 in/ft) = 785 fpm
HP = 83.6(785)/33,000 = 1.99 hp
Grading:
1.
2.
3.
4.
(20)
2.
Student determines chip thickness ratio and equations to determine shear plane angle.
Student applies Merchant relation to get friction angle.
Student determines shearing force and applies force vector equations to determine cutting force.
Student determines cutting horsepower.
10 pts
10 pts
10 pts
10 pts
Answer the following T/F questions:
a. The value of the Merchant Eqn. is that it relates rake angle, shear angle and chip thickness.
T__ F_X_
b. In the machining optimization paper, the cost minimized cutting speed is constant over the
planning period over any time of day or week.
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c. In the high speed machining of titanium alloys paper, it was shown that at a lower feed
rate the tool life can actually be higher at a higher cutting speed.
T_X_ F__
d. Flank wear causes a surface crater caused by by the chip sliding against the surface.
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e. WC-Co should not be used for machining steel alloys.
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f. In optimized machining the production rate is improved by decreasing the tool change time.
T_X_ F__
g. Titanium is more machinable than copper.
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h. Cast cobalt alloys have a better hardness than high-speed steels at all temperatures.
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i. Taylor’s tool life equation was shown to apply to high speed machining in the text.
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k. Active lubrication of the tool is very effective in high speed machining.
T__ F_X_
Grading:
2 pts each correct answer
2
(40) 3.
A vertical boring mill is used to bore the inside diameter of a large batch of tube-shaped parts. The
diameter = 28 in. and the length of bore = 14 in. Current cutting conditions are speed = 200 fpm, feed =
0.015 in./rev, and depth = 0.125 in. The Taylor equation parameters are n = 0.23 and C = 850. Tool
change time = 3 min., and tooling cost = $3.50 per cutting edge. The total time to load and unload the
part is 12 min. The cost of machine time on the mill is $42/hr.
Management has decreed that the production rate for this job must be increased by 25%. Is this possible?
Assume that the feed must remain unchanged to achieve the required surface finish. What are the current
production rate and the maximum possible production rate for this job?
Solution:
At current operating speed v = 200 fpm:
From Taylor’s eqn: T = (850/200)1/0.23 = 540 min
Machining time = Tm =  (28)(14)/[(200)(12)(0.015)] = 34.2 min/pc
np = 540/34.2 = 15 pc/tool life
Tc = 12 + 34.2 + 3/15 = 46.4 min/pc
Production rate = Rc = 60/46.4 = 1.293 pc/hr current production rate
Find vmax for optimum production rate:
vmax = 850/[(1/0.23 – 1)(3.0)]0.23 = 500 fpm
Tmax = (1/0.23 -1)(3.0)) = 10 min
Machining time = Tm =  (28)(14)/[(500)(12)(0.015)] = 13.7 min/pc
np = 10/13.7 = 0.73 pc/tool life
Tc = 12 + 13.7 + 3/0.73 = 29.8 min/pc
Production rate = Rc = 60/29.8 = 2.01 pc/hr max production rate
Since this represents a 56% increase in production rate, a 25% increase is possible.
Grading:
1. Student determines machining time and tool life time at current speed.
2. Student determines optimum machining speed and tool life time.
3. Student determines current and maximum production rate (or cycle time)
4. Student notes that 25% increase possible.
10 pts
10 pts
10 pts
10 pts