1. The document contains 5 problems related to machining operations including turning, drilling, and tool life calculations.
2. Problem 1 involves determining the cutting time and metal removal rate for turning a cylindrical workpiece.
3. Problem 2 calculates the required cutting speed to complete a turning operation in 5 minutes given specifications for the workpiece, feed rate, and depth of cut.
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MILLING – Cutting parameters, machine time calculation
Milling operation – Plain milling, side & face milling, form milling, gang milling, end milling, face milling, T slot milling, slitting
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This lecture note is basically designed for mechanical Engineering Manufacturing stream students and Instructors.
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The following presentation consists of a brief introduction to power screw that we use in our day to day life, its types, analysis of load, efficiency, application and examples with images.
MILLING – Cutting parameters, machine time calculation
Milling operation – Plain milling, side & face milling, form milling, gang milling, end milling, face milling, T slot milling, slitting
GEAR CUTTING – Gear cutting on milling machine – dividing head and indexing method, gear hobbing, principle of operation, advantages & limitation, hobbing tech, gear shaping, gear finishing process
education is a key for everything so the objective of this slide is to share knowledge to the glob in my area of specialization.
This lecture note is basically designed for mechanical Engineering Manufacturing stream students and Instructors.
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Assignment [4] machining with solutions
1. ASSIENMENT [4]
MACHINING
PROBLEMS
1- A cylindrical workpart 200 mm in diameter and 700 mm long is to be turned in an engine
lathe. Cutting conditions are as follows: cutting speed is 2.30 m/s, feed is 0.32 mm/rev, and
depth of cut is 1.80 mm. Determine (a) cutting time, and (b) metal removal rate.
Solution: (a) N = v/(πD) = (2.30 m/s)/0.200π = 3.66 rev/s
fr = Nf = 6.366(.3) = 1.17 mm/s
Tm = L/fr = 700/1.17 = 598 s = 9.96 min
Alternative calculation using Eq. (22.5), Tm = 200(700)π/(2,300 x 0.32) = 597.6 sec = 9.96
min
(b) RMR = vfd = (2.30 m/s)(103)(0.32 mm)(1.80 mm) = 1320 mm3/s .
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?
Solution: Starting with Eq. (22.5): Tm = πDoL/vf.
Rearranging to determine cutting speed: v = πDoL/fTm
v = π(0.4)(0.15)/(0.30)(10-3)(5.0) = 0.1257(103) m/min = 125.7 m/min
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.
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
14- 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) 120(7)n = C and (2) 80(28)n = C.
120(7)n = 80(28)n
ln 120 + n ln 7 = ln 80 + n ln 28
4.7875 + 1.9459 n = 4.3820 + 3.3322 n
4.7875 - 4.3820 = (3.3322 – 1.9459) n
0.4055 = 1.3863 n n = 0.2925
C = 120(7)0.2925 = 120(1.7668) C = 212.0
Check: C = 80(28)0.2925 = 80(2.6503) = 212.0
(b) 110 T0.2925 = 212.0
2. T0.2925 = 212.0/110 = 1.927
T = 1.9271/0.2925 = 1.9273.419 = 9.42 min
(c) v (15)0.2925 = 212.0
v = 212.0/(15)0.2925 = 212.0/2.2080 = 96.0 m/min .
5- Turning tests have resulted in 1-min tool life at a cutting speed = 4.0 m/s and a 20-min tool
life at a speed = 2.0 m/s. (a) Find the n and C values in the Taylor tool life equation. (b)
Project how long the tool would last at a speed of 1.0 m/s.
Solution: (a) For data (1) T = 1.0 min, then C = 4.0 m/s = 240 m/min
For data (2) v = 2 m/s = 120 m/min
120(20)n = 240
20n = 240/120 = 2.0
n ln 20 = ln 2.0
2.9957 n = 0.6931 n = 0.2314
(b) At v = 1.0 m/s = 60 m/min
60(T)0.2314 = 240
(T)0.2314 = 240/60 = 4.0
T = (4.0)1/0.2314 = (4)4.3215 = 400 min