![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/(Ï€D) = (2.30 m/s)/0.200Ï€ = 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)Ï€/(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/Ï€D = 25(103) / (12.7Ï€) = 626.6 rev/min
fr = Nf = 626.6(0.30) = 188 mm/min
Tm = L/fr = 60/188 = 0.319 min
(b) RMR = 0.25Ï€D2fr = 0.25Ï€(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](https://image.slidesharecdn.com/assignment4machining-withsolutions-121213110841-phpapp02/85/Assignment-4-machining-with-solutions-1-320.jpg)
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