Designing Capable and Reliable Products Episode 2 Part 3 docx
Designing Capable and Reliable Products Episode 2 Part 3 docx
... R
A
R
B
30 Â60 Â 5 30 5 .2 45.1 34 6.9 50.8 0.648 724 0 .33 039 1
30 Â60 Â 6 26 4 .2 39 .2 29 7.0 43. 6 0.918951 0.71 531 9
50 Â65 Â 5 21 1.1 31 .2 26 4.8 38 .8 0.99 929 9 0.918648
50 Â65 Â 6 1 83. 0 27 .0 22 5.9 32 . 9 ... ÿ0.041
150 .3 16 0.5161 0.041
150 .3 17 0.5484 0. 121
150 .3 18 0.5806 0 .20 4
150 .3 19 0.6 129 0 .28 7
150 .3 20 0.64 52 0 .3 72
151 .3 21 0.677...
Designing Capable and Reliable Products Episode 2 Part 1 ppsx
... below and involve all three principal stresses (Dowling, 19 93) :
C
1
1
2
s
1
ÿ s
2
jj
Su
c
ÿ 2Su
Su
c
s
1
s
2
!
4:59
C
2
1
2
s
2
ÿ s
3
jj
Su
c
ÿ 2Su
Su
c
s
2
s
3
!
4:60
C
3
1
2
s
3
ÿ ... stress system.
2Sy
2
s
1
ÿ s
2
2
s
2
ÿ s
3
2
s
3
ÿ s
1
2
4:57
where s
1
, s
2
, s
3
are the principal stresses.
Elements...
Designing Capable and Reliable Products Episode 2 Part 2 pptx
... 5 .2 52 8 02 0.8014 6.1 420 0.48 02
495 3. 5 35 837 0. 836 4 6 .20 46 0.5 935
525 2. 9 29 866 0.8654 6 .26 34 0.6959
555 2. 7 27 8 93 0.8 9 23 6 .31 90 0.80 13
585 2. 2 22 915 0.91 43 6 .37 16 0.8989
615 1.8 18 933 ...
L
(MPa)
14 22 5.0 98 .2
15 196.0 85.7
16 1 72. 2 75.1
17 1 52. 6 66.6
18 136 .1 59.4
19 122 .1 53. 3
20 110 .2 48.1
21 100.0 43. 6
22 91.1 39 .7
23...
Designing Capable and Reliable Products Episode 2 Part 4 ppt
... set at the 3
standard deviations limit, therefore:
emin
e
ÿ 3
e
136 :6 ÿ 3 11; 11 03: 6MPa
Therefore:
d
0: 636 62 28 0 Â10
3
1 03: 6 Â10
6
s
... 3
Su
505:9 ÿ 3 25 :3 430 MPa
A
F
Su
min
0:05D ÿ0:0 52
where:
A area of the section
F tensile force:
Solving for D gives:
D
1 :21 Â10
6
0:05 430 Â10
6
0:0 52...
Designing Capable and Reliable Products Episode 2 Part 5 doc
... gives:
407, 415, 420 , 424 , 426 , 428 , 433 , 433 , 436 , 438 , 440, 4 42, 4 43, 444, 445, 447, 447, 449,
450, 450, 451, 4 52, 4 52, 4 53, 4 53, 454, 455, 456, 459, 460, 4 62, 466, 466, 469, 470, 471,
4 72, 474, 477, ... 450, 415, 433 , 420 , 459, 424 , 444, 455, 4 62,
495, 426 , 5 02, 447, 438 , 436 , 498, 474, 515, 4 83, 454, 4 53, 447, 485, 456, 4 43, 449, 466,
445, 4 52, 478, 407, 433 ,...
Designing Capable and Reliable Products Episode 2 Part 6 ppsx
... Occurrence, Severity and Detectability
29 6 Appendix III
distribution for comparison:
y
Nw
2
p
exp
ÿ
x ÿ
2
2
2
50 Â 18
23 :45
2
p
exp
ÿ
x ÿ 457:76
2 23 :45
2
2
To ®nd ... ÿ49:98
2
7 Â49:97 ÿ49:98
2
10 Â49:98 ÿ49:98
2
8 Â49:99 ÿ 49:98
2
6 Â50 ÿ49:98
2
2 Â50:01 ÿ49:98
2
1 Â50: 02 ÿ49:98
2
40
v
u
u
u
u
u
t
...
Designing Capable and Reliable Products Episode 2 Part 7 pdf
... processes
Sand casting ± Steel and iron
Sand casting ± Copper alloys
Sand casting ± Aluminium and magnesium alloys
Shell moulding ± Steel, iron and copper alloys
Shell moulding ± Aluminium and magnesium ... gives a Handling Ratio Total Ratings/Essential Parts 2. 5
. Fitting analysis gives a Fitting Ratio Total Ratings/Essential Parts 2. 5.
Bene®ts
The potential bene®ts of DFA/DFM...
Designing Capable and Reliable Products Episode 2 Part 8 pdf
... Calculation of q
m
Drawing number FQ2186 m
p
1 Â 1 :2 1 :2
g
p
1 Â1 Â 1:7 Â 1 Â1 Â1 :2 2: 04
Adjusted tolerance
Design tolerance
m
p
 g
p
Æ0 :25
1 :2 Â 2: 04
Æ0:1 02
t
p
9 Â 1 9
s
p
1 Â 1 ... 4 :31
Process capability index C
p
4 :33
Manufacturing variability risk q
m
1: 02
33 6 Appendix V
Component name Lamp holder bracket Calculation of q
m
Drawing number...
Designing Capable and Reliable Products Episode 2 Part 9 ppsx
...
2
F
ÿ
F
2
a
b
2
Á
2
a
ÿ
F
a
2
b
2
Á
2
b
!
0:5
Substituting in values for and for each variable gives:
L
1
0: 03 Â 0:05
2
 10 000
2
ÿ
100 000
0: 03
2
 0:05
2
 0:00 03
2
ÿ
100 ... 1:5Â1:05
0:0 32 mm
The standard deviation for one half of the tolerance can be estimated by:
H
b
%
T
2
Á q
2
m
12...
Designing Capable and Reliable Products Episode 2 Part 10 ppsx
... equations:
xo
L
35 0 ÿ 3: 139 44 73 40 22 4: 42 MPa
L
35 0 0 :35 3018440 36 4: 12 MPa
L
3: 44
The reliability equation then becomes:
R
I
22 4: 42
1 ÿexp
ÿ
x 22 4: 42
36 4: 12 22 4: 42
3: 44
!
Á
...
1
0: 03 Â0:05
2
 10 000
2
ÿ
100 000
0: 03
2
 0:05
2
 0:00 03
2
ÿ
100 000
0: 03 Â0:05
2
2
 0:0004
2
!
0:5
2
L
4...