Designing Capable and Reliable Products Episode 2 Part 9 ppsx
Designing Capable and Reliable Products pptx
... the
potential risks of failure and their consequences, both from normal use and fore-
seeable misuse, is a key element in designing capable and reliable products (Wright,
198 9).
1.5 Designing for quality
The ... methods for predicting quality and evaluat-
ing the long-term quality of an engineer's design (Mùrup, 199 3; Russell and Taylor,
199 5; Shah, 199 8; Taguch...
Designing Capable and Reliable Products pot
... quality and reliability engineering
Designing Capable and Reliable Products
(Dale, 199 4; Kehoe, 199 6; Maylor, 199 6). This ®gure can be as high as 40% in the
service industry! (Bendell et al., 199 3). ... (Kotz and Lovelace, 199 8;
Vasseur et al., 19 92 ) . Making the product robust to variation is the driving force
behind designing capable and reliable products, les...
Designing Capable and Reliable Products J.D. potx
... (Kotz and Lovelace, 199 8;
Vasseur et al., 19 92 ) . Making the product robust to variation is the driving force
behind designing capable and reliable products, lessens the need for inspection and
can ... GR/J97 92 2 and GR/L 623 13), has funded the work presented in this
book.
J.D. Booker, M. Raines, K.G. Swift
School of Engineering, University of Hull, UK
May 20 00
Preface...
Dimensioning and Tolerancing Handbook Episode 2 Part 9 pps
... Model:
( ) ( ) ( )
2
13
2
12
2
11
2
eScSaST
ASM
δδδ ++=
( )
( )( )( )
( )( )( )
) 790 9 32( )4500 82( 2
) 790 9 32( 2
2
222
627 221 018 696 45008
646 92 0 14 92 2 7 790 93
627 22
)0004)(548310()646 92 ( )017453(
./.
./
da
... 0.0005 0.00 12
0.600-0 .99 9 0.046 821 58 0.5654 92 0.0006 0.0015
1.000-1. 499 0.0 420 49 92 0.6 021 191 0.0008 0.0 02
1.500 -2. 799 0.048 096 84 0....
Dimensioning and Tolerancing Handbook Episode 2 Part 8 ppsx
... )
( )
( ) ( )
46 823 .1/43 899 .1
46 823 .1/1
46537.1/43 899 .1
46537.1/1
46 823 .1/43 899 .1
46 823 .1/1
1 599 7.43 899 .
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1 599 7.43 899 .
125 76.46537.
1 599 7.43 899 .
0 720 2.46 823 .
0 025 .0 025 .0015.015.
BB
B
B
TT
TT
+
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+=−−−
The ... 1.500
2. 800
4.500
7.800
13.600
0. 599
0 .99 9
1. 499
2. 799
4. 499
7. 799
13. 599...
Dimensioning and Tolerancing Handbook Episode 2 Part 5 ppsx
... 2. 8717E- 02 2.8067E- 02 2.7 4 29 E- 02 2.6804E- 02 2.6 190 E- 02 2.5588E- 02 2. 499 8E- 02 2.4419E- 02 2.3852E- 02 2. 3 29 6E- 02
2 2. 2750E- 02 2 .22 16E- 02 2.1 692 E- 02 2.1178E- 02 2.0675E- 02 2.0182E- 02 1 .96 99E- 02 1. 92 2 6E- 02 ...
89. 02
83. 02
1
10
30
50
100
150
20 0
300
400
500
800
120 0 75.63
99 .99 966%
99 .99 66
99 .99
99 .98
99 .96 6
99 .94...
Handbook of Micro and Nano Tribology Episode 2 Part 9 pot
... LLC, 199 9
© 199 9 by CRC Press LLC
© 199 9 by CRC Press LLC
Preface
Second Edition, 199 9
The first edition of the
Handbook of Micro/Nanotribology
was published in the Spring of 199 5. ... 194 9- .
TJ1075.H245 199 9
621 .8
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9 dc21
98 -24 466
CIP
This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with
pe...
Engineering Mechanics - Statics Episode 2 Part 9 ppsx
... F
H
w
2h
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
=
tan
θ
max
()
wa
2F
H
= cos
θ
max
()
2F
H
4
F
H
2
wa()
2
+
=
T
max
F
H
cos
θ
max
()
= F
H
2
wa()
2
4
+=
wa
2
a
2
16h
2
1+=
Guess h 1 m=
Given T
max
wa
2
a
2
16h
2
1+= h Find h()= h 7. 09 m=
Problem ... y−()=
Σ
M
= 0;
Mwy
y
2
⎛
⎜
⎝
⎞
⎟
⎠
+ wb
b
2
⎛
⎜
⎝
⎞
⎟
⎠
+ wa
a
2
⎛
⎜
⎝
⎞
⎟
⎠
+ wby− 0=
My() wby
1
2
wy
2
−
1
2
wb
2
−
1
2...
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