The Machining Process

This document discusses the machining process, including the analysis of parts and machining parameters. It also covers manufacturing processes, quality control, and sampling acceptance plans.

The Machining Process
Name
Institutional Affiliation

1. Analysis of the parts of an air conditioning pipe
Pipe frame (1), an extension pipes between two pipes
Two coaxial holes (2)
Bushings (3)
Top hole covered with a metal sheet (4) attached to (1)
Mounting plate (8)
Sub-assembly composed of (7) and (8) in the frame pipe
Control stick (5)
Control handle (6)
The following components would be manufactured/designed together
 Attach the mounting plate and valve together, since they seal the same hole
 Attach control stick, metal tower and handle together, they are on the same shaft
 The bushings can be fixed together in the coaxial holes
Screen shots of a gas-flow meter are as follows:
Figure 1:The sections of the machine

Figure 2:The whole machine assembled
Design efficiency is the number of essential parts divided by the number of parts
Expressed as A/(A+B) 5/(5+12) = 29.4%
2. Question 2.1 (Machining Parameters)
The machining parameters for an end-milling operation for the profile shown in fig. 2 listed in
the table below
Feed rate
(mm/min
)
Cutting
depth
(mm)
Spindle
speed
(RPM)
Cutter size
(diameter
in mm)
No. of teeth
(flutes) for
the cutter
Radial cutting
depth
K=1 or 2 300 2 6000 10 2 40% of cutter size
K=3 or 4 350 2.5 5000 12 2 30% of cutter size
K=5 or 6 400 1.5 6000 15 4 25% of cutter size
K=7 or 8 300 3 6500 20 2 35% of cutter size
K=9 or 0 600 2.5 6000 25 2 20% of cutter size
Take the value of K for your work to be the last digital number of your student ID. For example,
1234567, then K=7 so the machining parameters for your analysis would be: Feed rate =300mm/min;

cutting depth=3; spindle speed =6500; cutter size (diameter)=20 number of teeth/flutes =2 and radial
cutting depth=35% of cutter size. Based on the table 1 and K (the last digit of number of your student
ID), calculate the following
i. The chip load (mm/tooth/revolution)
ii. The surface speed of the end mill (m/min)
iii. The material removal rate (mm3/min)
iv. Assume the specific power of the part is 15mJ/kg, and the density of the part to be machined is
2700kg/m3, calculate the power needed for the machining.
v. For the machining from A to B in figure 2;
a. What are the minimal leading in and leading out distances?
b. What is the machining time from A to B taking into consideration of leading in and
leading out times (assuming the s=distance between A and B is 100mm?
Solution
Your k=3, parameters highlighted in yellow:
Chip load = feed rate/ (RPM X No. teeth) = 350/ (5000 X 2) = 0.035mm/tooth/rev
Surface speed = 350/1000 = 0.35m/min
Removal rate = (Area X cutting depth) X speed = (π X 12/4) X 2.5 = 31.42mm3 X 5000 =
157,079.63mm3/min
Power = 15mJ/kg, density = 2700kg/m3
The volume per milli-joule would be 15 X 2700 = 40,500mJ/m3
If S = 100mm, the minimal leading in and out distances is 30% cutter size, 30mm and 70mm distances
T m= L+2 A
F r
Where Tm is the machining time, L =length of cut, A approach distance, Fr feed rate
T m= 30+70 x 2
350 =0.49 min
3. Question 2.2 (Manufacturing Processes)

(1). Yes, when the axis spindle is on a vertical position, it can be used to bore the hole and with
the section A B, which needs precision and when well set for drilling. The spindle (whose size is
Wmm) have a rectangular surface which will ensure that precision is achieved (height H and
length AB) to drill/mill the hole.
(2). There are three processes that is:
a. Milling b. Drilling and c. turning
(3) yes, to some extent. only by being careful, can the part of the metal be gorged out using chemical
processes, use of acids. The process is very difficult and needs an accuracy of a high level, which might
not be attained. I would not recommend.
(4) Material extrusion could be one of them, Vat photopolymerization might be used in this instance as
well. I would not use any other method, since not many would gorge out the hole with the precision
required.
Now power shall be 40,500/31.42 = 1288.99mJ
4. Assume during the machining, 9 cylinders are randomly collected as the measurement samples
and the dimensions of the 9 samples are inspected as shown in Table 2.
i. Calculate the mean value and standard deviation
ii. Create a control chart indicating the mean, UCL, and LCL and plot the data in table
within the control chart.

iii. Suggest whether or not the quality of the machined parts satisfy the tolerance
requirements
iv. Calculate the process of capability Cp and Cpk
v. Recommend the new tolerance for the diameter based on measurement data in table
assuming Cp>=1.3 and Cpk> =1
Solution
Mean calculated and results shown in the table
Diameter Observation 1 Observation 2 Observation 3 Mean SD
Sample 1 20.032 19.971 19.999
20.0
01
0.03
05
Sample 2 19.979 19.979 19.993
19.9
84
0.00
81
Sample 3 20.015 20.013 19.973
20.0
00
0.02
37
Sample 4 19.985 19.978 20.011
19.9
91
0.01
74
Sample 5 20.031 19.973 20.047
20.0
17
0.03
89
Sample 6 19.987 19.969 19.982
19.9
79
0.00
93
Sample 7 20.038 20.033 19.997
20.0
23
0.02
24
Sample 8 20.012 19.965 20.021
19.9
99
0.03
01
Sample 9 19.995 19.979 20.043
20.0
06
0.03
33
The control chart
Diameter Observation 1 Observation 2 Observation 3 Mean Mean (2) UCL LCL
Sample 1 20.032 19.971 19.999
20.0
01 20.000 20.04 19.96
Sample 2 19.979 19.979 19.993
19.9
84 20.000 20.04 19.96
Sample 3 20.015 20.013 19.973
20.0
00 20.000 20.04 19.96
Sample 4 19.985 19.978 20.011
19.9
91 20.000 20.04 19.96
Sample 5 20.031 19.973 20.047
20.0
17 20.000 20.04 19.96
Sample 6 19.987 19.969 19.982
19.9
79 20.000 20.04 19.96
Sample 7 20.038 20.033 19.997
20.0
23 20.000 20.04 19.96

Sample 8 20.012 19.965 20.021
19.9
99 20.000 20.04 19.96
Sample 9 19.995 19.979 20.043
20.0
06 20.000 20.04 19.96
The chart below is the control chart for the data on the table above:
1 2 3 4 5 6 7 8 9
19.900
19.920
19.940
19.960
19.980
20.000
20.020
20.040
20.060
20.000
20.04
19.96
The tolerance requirement of a machine
In the chart above, is represented by three lines, the mean, the UCL and the LCL. The UCL region
represents the better tolerance of a machine, and the LCL region shows the worst of tolerant a machine
can be. From the curve we observe that the curve is more on the lower side than the upper, that means
the machine is below the tolerance level and therefore has a quality lower that cannot satisfy the
tolerance level.
The CP and Cpk are calculated, with the help of the mean, UCL, LCL, and Standard deviation as shown in
the table below, the figures are derived from the tables above.
UCL 20.04
LCL 19.96
Mean 20.000
Standard Deviation 0.014164
Upper six limit 20.08
Lower six limit 19.92
Cpk (Process capability Index) - 468.80
CP (Max. possible process capability) 0.94
The new tolerance limits for the diameter is that which is assumed to be confined when the Cpk>=1.3,
and CP>=1. Between sample 1 and sample 4

5. Question 3.1
Assume the spindle speed is 2000rpm and the cutting depth is 2mm, assuming the nose radius
of the cutter is 0.7mm, recommend the feed rate (mm/rev) to achieve the surface finish?
Assume three machines shown in table 2 are available for the machining of the cylinder shown
in fig 4
Machine Name Standard deviation
Machine A 0.005
Machine B 0.01
Machine C 0.02
a. Which machines in table 1, could be used to machine the cylinder in figure 4 to achieve the
length tolerance but not the diameter tolerance assuming the process capability Cp>=1;
b. Which of the machine(s) in table 1 could be used to machine the cylinder in figure 4 to
achieve the dimensional tolerance for both the diameter and length assuming the process
capability Cp>=1;
c. Which of the machine(s) in table 1 could be used to machine the cylinder in figure 4 to
achieve the dimensional tolerance for both the diameter and length assuming the process
capability Cp>=1?
Solution
Here we consider:
Inch per minute = 2mm
Revs. Per minute=2000
Cutter diameter = 1.4mm
Cutting speed = (π X D X S)/1000 = (π X 1.4 X 2000)/1000 = 8.796mm/rev
Machine B
Machine C
Machine A
6. Question 3.2
After the producer has manufactured the cylinders shown in the figure 4 in mass quantity, the
producer would like to agree a sampling acceptance plan. For Lot size quantity, the producer
would like to agree a sampling acceptance plan. For Lot size=2000, three sampling plans
proposed:
Plan 1: Sampling size = n = 10, reject if the number of defectives C>=1
Plan 2: Sample size = n = 20, reject if number defectives C>=2
Plan 3: Sample size = n = 20, reject if number defectives C=1
i. Create the operating Characteristic (OC) curves for these three sample plans
ii. Which one is the best for the producer if the producer wants to have <5% producer’s
risk? Why?
iii. Which is the best for the customer if the customer wants 10% customer’s risk? Why

Solution
Consider the table below used to produce the curve
Po npo pa
0 0 1
0.05 0.5 0.909796
0.1 1 0.735759
0.15 1.5 0.557825
0.2 2 0.406006
0.25 2.5 0.287297
0.3 3 0.199148
0.35 3.5 0.135888
0.4 4 0.091578
0.45 4.5 0.061099
0.5 5 0.040428
0.55 5.5 0.026564
0.6 6 0.017351
0.65 6.5 0.011276
0.7 7 0.007295
0.75 7.5 0.004701
0.8 8 0.003019
0.85 8.5 0.001933
0.9 9 0.001234
0.95 9.5 0.000786
1 10 0.000499
1.05 10.5 0.000317
1.1 11 0.0002
1.15 11.5 0.000127
1.2 12 7.99E-05
1.25 12.5 5.03E-05

1.3 13 3.16E-05
1.35 13.5 1.99E-05
1.4 14 1.25E-05
1.45 14.5 7.82E-06
1.5 15 4.89E-06
1.55 15.5 3.06E-06
1.6 16 1.91E-06
1.65 16.5 1.19E-06
1.7 17 7.45E-07
1.75 17.5 4.65E-07
1.8 18 2.89E-07
1.85 18.5 1.8E-07
1.9 19 1.12E-07
1.95 19.5 6.97E-08
2 20 4.33E-08
2.05 20.5 2.69E-08
2.1 21 1.67E-08
0 5 10 15 20 25 30 35 40 45
0
0.2
0.4
0.6
0.8
1
1.2
% non conforming in population
Probability od acceptance
Figure 3:The first scenario

0 5 10 15 20 25 30 35 40 45
0
0.2
0.4
0.6
0.8
1
1.2
% non conforming in population
Probability od acceptance
Figure 4:The second Scenario
0 5 10 15 20 25 30 35 40 45
0
0.2
0.4
0.6
0.8
1
1.2
% non conforming in population
Probability od acceptance
Figure 5:The third scenario
All the scenarios are the best, because they seem the same at 5% and beyond, which would be high risk
if the 5% would have been a significant number.
Non is the best either if the customer wanted 10%, because at 10% the probability of acceptance is 0 or
almost 0