(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 5 ppsx
Machinery''''s Handbook 27th Episode 2 Part 5 potx
... = 150 5 Constant C = 1010
V
T
SMRR
V
T
SMRR
V
T
SMRR
V
T
SMRR
100 6 85 1 15 5 95 1 95 455 22 5 3 05 23 0
10 1 25 0 21 0 1080 355 8 25 4 15 555 4 15
1 22 70 3 85 1970 650 150 5 750 1010 760
Fig. 2a. T–V Fig. 2b. ... C = 53 85 Constant C = 38 85
V
T
SMRR
V
T
SMRR
V
T
SMRR
V
T
SMRR
100 26 95 460 21 05 6 95 1 6 25 8 15 11 75 880
10 49 05 8 35 3830 126 5 29 60 1480 21...
Bearing Design in Machinery Episode 2 Part 5 potx
... is 20 ,000 N, or 50 00 N on each pad. The total manufac-
tured clearance between the two pads ðh
1
þ h
2
Þ is 0.4 mm. Each
circular pad is of 100-mm diameter and recess diameter of 50 mm.
R ¼ 50 ... 50 mm and R
0
¼ 25 mm. The oil viscosity is 0:01 N-s=m
2
.In
order to minimize vertical displacement, the slider plate is
prestressed. The reaction force at the top is W
1
¼ 50 00 N, an...
Fundamentals of Structural Analysis Episode 2 Part 1 ppsx
... us:
M
F
bc
= −
12
)( )(
2
Lengthw
= −
12
(4) (3)
2
= − 4 kN-m
M
F
cb
=
12
)( )(
2
Lengthw
=
12
(4) (3)
2
= 4 kN-m
(d) Compute DF at b:
2 m
2 m
4 m
a
c
b
3 kN/m
4 kN
2 m
2 kN
2 m
2 m
4 m
a
c
b
3 ... +7 .50
DM −4.69 2. 81
COM 2. 35 −1.41
DM +0.71 +0.70
COM +0.36 0. 35
DM −0 .22 −0.14
COM −0.11 −0.07
DM +0.04 +0.03
COM +0. 02 +0. 02
DM −0.01 −0.01
COM 0.00...
Fundamentals of Structural Analysis Episode 2 Part 4 ppsx
... =
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
+−−−−−
−−+−−−−−
−−
−−−−−−+−
−−−−−+
EK
L
EK
C
L
EK
S-EK
L
EK
C
L
EK
S
L
EK
C
L
EK
C
L
EA
S
L
EK
L
EA
CS
L
EK
C
L
EK
C
L
EA
S
L
EK
L
EA
CS
L
EK
S
L
EK
L
EA
CS
L
EK
S
L
EA
C
L
EK
S
L
EK
L
EA
CS
L
EK
S
L
EA
C
EK
L
EK
C
L
EK
SEK
L
EK
C
L
EK
S
L
EK
C
L
EK
C
L
EA
S
L
EK
L
EA
CS
L
EK
C
L
EK
C
L
EA
S
L
EK
L
EA
CS
L
EK
S
L...
Fundamentals of Structural Analysis Episode 2 Part 5 pptx
... maximize F
CJ
.
1kN 1kN
2 kN
1m
2m
2 kN
10 kN/m 10 kN/m
x
6 m
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
2 kN
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
2 kN
1kN 1kN
1m2m
7 m
9 m
Influence Lines by S. T. Mau
25 9
Example 6. Find the ... Mau
27 0
Placing finite length uniform load for maximum compression in member CJ.
(F
CJ
)
max
= −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )-0 .5( 4. 82) (0. 6...
Fundamentals of Structural Analysis Episode 2 Part 5 pdf
... below.
a
b
c
d
5 m 5 m
5 m
a
b
cd
V
c
5 m
5 m 5 m
1 /2
1 /2
1 /2
a
b
c
d
5 m 5 m
5 m
10 kN-m 10 kN-m
Influence Lines by S. T. Mau
26 8
(F
CJ
)
max
= − [2( 0. 6 25 )+1(0. 6 25 )(7/9)+1(0. 6 25 )(6/9)]= 2. 15 kN
This ... maximize F
CJ
.
1kN 1kN
2 kN
1m
2m
2 kN
10 kN/m 10 kN/m
x
6 m
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
2 kN
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
2 kN
1kN...
Fundamentals Of Structural Analysis Episode 2 Part 5 pot
... maximize F
CJ
.
1kN 1kN
2 kN
1m
2m
2 kN
10 kN/m 10 kN/m
x
6 m
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
2 kN
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
2 kN
1kN 1kN
1m2m
7 m
9 m
Influence Lines by S. T. Mau
26 2
Problem 1.
(1) Construct ... CJ.
(F
CJ
)
max
= −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )-0 .5( 4. 82) (0. 6 25 )(4. 82/ 9)]
= −33.0 kN
10 kN/m
6 m
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
6...
Industrial Machinery Repair Part Episode 2 Part 5 doc
... areas and are screwed
directly into the bearing or into a short pipe connection to the bearing.
3 52 Machinery Installation
Development of your preventive maintenance tasks is highly recommended
before ... “polar”—that is, they carry an electric charge that causes them to be
attracted to any electric field extending out a few molecule lengths from
most metallic bearing surfaces. This el...
Machinery Components Maintenance And Repair Episode 2 Part 5 ppt
... life.
Figure 7- 52 depicts a shaft equipped with MRC’s “PumPac.” The two
bearings are mounted back-to-back, with the apex of the etched “V” point-
ing in the direction of predominant thrust.
422 Machinery ... other characteristics
424 Machinery Component Maintenance and Repair
Figure 7 -53 . Check internal contact surfaces by turning outer ring slowly while holding inner
ring.
430 Mac...
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