Fundamentals of Structural Analysis Episode 1 Part 1 pptx
Fundamentals of Structural Analysis Episode 2 Part 1 ppsx
... +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 0.00
SUM 2. 46 −4. 92 +4. 92 +14 .27 +15 .73 +7.87
In ...
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...
Fundamentals of Structural Analysis Episode 2 Part 1 pdf
... 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 0.00
SUM 2. 46 −4. 92 +4. 92 +14 .27 +15 .73 +7.87
In the ... = 2 kN-m
M
F
bc
= 2 kN-m
2 m
2 m
4 m
a
d
b
4 kN
c
2 m2 m
4 kN
1. 33
1. 33
1. 33
2. 67
2. 67
I
nflection point
Beam and Frame Analysis: Displacement Method, Part...
Fundamentals of Structural Analysis Episode 2 Part 2 pptx
... 0 .22 0.45 0 1 0 .23 0.46 0.31 0
MEM M
ab
M
ba
M
bc
M
be
M
eb
M
ab
M
ba
M
bc
M
be
M
eb
EAM
FEM
− 62. 5
62. 5
− 62. 5
62. 5
DM
62. 5 62. 5
COM
31.3 31.3
DM
−31.0 20 .6 − 42. 2 21 .6 −43 .2 29 .0
COM
0.0
21 .1
0.0
−14.5
SUM
0.0 ... (2)
(3)
(4) (5)
(6) (7)
Problem 1.
2 m
2 m
3 m
a
c
b
3 kN/m
4 kN
2EI
E
I
2 m
2 m
3 m
a
cb
8 kN
2EI
E
I
2 kN-m
50 kN
ab
cd
50 kN
2 m 2 m
2 m 2...
Fundamentals of Structural Analysis Episode 2 Part 2 ppt
... M
F
−
8
PL
8
PL
− ab
2
PL − a
2
bPL
− a(1-a)PL a(1-a)PL
−(6−8a+3a
2
)
12
22
wLa
(4-3a)
12
23
wLa
−
12
2
wL
12
2
wL
−
20
2
wL
12
2
wL
−
96
5
2
wL
96
5
2
wL
− b(2a-b)M a(2b-a)M
Note: Positive ... 0 .22 0.45 0 1 0 .23 0.46 0.31 0
MEM M
ab
M
ba
M
bc
M
be
M
eb
M
ab
M
ba
M
bc
M
be
M
eb
EAM
FEM
− 62. 5
62. 5
− 62. 5
62. 5
DM
62. 5 62. 5
COM
31.3 31.3
DM
−31.0 20 .6 − 4...
Fundamentals of Structural Analysis Episode 2 Part 3 potx
... (6)
Problem 2.
2 m
2 m
3 m
a
c
b
3 kN/m
4 kN
2EI
E
I
2 m
2 m
3 m
a
cb
8 kN
2EI
E
I
2 kN-m
50 kN
a
b
c
4 m 4 m
4 m
8 m
E
I
2EI2EI
50 kN
ab
c
4 m 4 m
4 m
8 m
E
I
2EI2EI
2 m
2 m
2 m
4 kN
a
b
c
2 m
2 m
2 m
4 ... kN
a
b
c
d
2. 77 kN
3. 36 kN-m
1 . 23 kN
0 .27 kN-m
0 .27 kN-m
1.64 kN-m
0.69 kN
0.69 kN
c
b
1.64 kN-m
0. 82 kN-m
1 . 23 kN
1 . 23 kN
1 . 23 kN
0.69 kN
0.6...
Fundamentals of Structural Analysis Episode 2 Part 3 pps
... kN
a
b
c
d
2. 77 kN
3. 36 kN-m
1 . 23 kN
0 .27 kN-m
0 .27 kN-m
1.64 kN-m
0.69 kN
0.69 kN
c
b
1.64 kN-m
0. 82 kN-m
1 . 23 kN
1 . 23 kN
1 . 23 kN
0.69 kN
0.69 kN
0.69 kN
0.69 kN
0.69 kN
1 . 23 kN
c
1 . 23 kN
1 . 23 kN
0.69 ... (6)
Problem 2.
2 m
2 m
3 m
a
c
b
3 kN/m
4 kN
2EI
E
I
2 m
2 m
3 m
a
cb
8 kN
2EI
E
I
2 kN-m
50 kN
a
b
c
4 m 4 m
4 m
8 m
E
I
2EI2EI
50 kN
ab
c
4 m 4 m
4...
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 4 doc
... =
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
+−−−−−
−−+−−−−−
−−
−−−−−−+−
−−−−−+
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
... −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 m
9 m
1.18 1. 82
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)]= ... 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 2...
Fundamentals of Structural Analysis Episode 2 Part 5 pdf
... 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 4
S = (
L
aL −
) S
i
...
=10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)−0 .5( 1.18)(0.41)(1.18/6)] = 14. 65 kN
10 kN/m
6 m
1 . 25
1 . 25
0. 6 25
0.41
F
CJ
10 kN/m
6 m
9 m
1.18 1. 82
1 . 25
1 ....
Fundamentals of Structural Analysis Episode 2 Part 6 doc
... 17 .63 0 42. 98
σ
due to axial force (kN/cm
2
)
0.0 72 0.0 72 0.088 0.088 0. 125 0. 124
σ
due to moment (kN/cm
2
)
0 0.0 06 0 0.013 0 0.0 32
Total σ (kN/cm
2
)
0.0 72 0.078 0.088 0.101 0. 125 0.1 56
Error ... depth of the beam, h, is the same for both members.
The fixed-end moments are obtained from the above table:
P
L
/2
L
/2
w
L
L
/2
L
/2
L
0.2L0.2L 0.6L
h
h
b
a
0....
Fundamentals of Structural Analysis Episode 2 Part 7 potx
... Solutions
Member
Force
(MN)
Elongation
(m)
Member
Force
(MN)
Elongation
(m)
1 0. 375 0.011 8 0 0
2 0. 375 0.011 9 0. 625 0.031
3 0. 375 0.011 10 0 0
4 0. 375 0.011 11 -0. 625 -0.031
5 -0. 625 -0.031 12 -0 .75 0 -0. 023
6 0 0 13 -0 .75 0 -0. 023
7 0. 625 -0.031
Case (a): ... kN
0.5 kN 1.5 kN
−1. 12 kN −1. 12
0 kN
2 kN
2. 24
−1 kN
0 kN
2. 24
0 kN
1 kN
1. 12
2 .24
−1. 12
−1 kN
1 k...
Fundamentals Of Structural Analysis Episode 2 Part 1 pps
... +15 +15
COM +7.50 +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 0.00
SUM 2. 46 −4. 92 ...
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...
Fundamentals Of Structural Analysis Episode 2 Part 2 pptx
... a(1-a)PL
−(6−8a+3a
2
)
12
22
wLa
(4-3a)
12
23
wLa
−
12
2
wL
12
2
wL
−
20
2
wL
12
2
wL
−
96
5
2
wL
96
5
2
wL
− b(2a-b)M a(2b-a)M
Note: Positive moment acts clockwise.
aL
w
L
w
L
w
L
w
L
/2
L
/2
P
aL
aL
P
P
aL
bL
PP
M
bL
aL
... 0 .22 0.45 0 1 0 .23 0.46 0.31 0
MEM M
ab
M
ba
M
bc
M
be
M
eb
M
ab
M
ba
M
bc
M
be
M
eb
EAM
FEM
− 62. 5
62. 5
− 62. 5
62. 5
DM
62. 5 62. 5...
Fundamentals Of Structural Analysis Episode 2 Part 8 pptx
... kN-m
Problem 2.
(1) M
ba
= − 525 kN-m, M
bc
= 525 kN-m, M
ab
= 26 3 kN-m, M
cb
= 2, 363 kN-m.
(2) M
ba
= 28 .8 kN-m, M
bc
= 28 .8 kN-m, M
ab
= 28 .8 kN-m, M
cb
= 28 .8 kN-m
12 Matrix Algebra ... 29 6
Gasset plate, 28 4
Gaussian elimination method, 29 3
Geometric non-linearity, 28 8
Global Coordinate, 4, 5, 23 8
Hardening, 28 8
Hinge, 93
Influence lines, 24 9...
Từ khóa:
- virtual work method of structural analysis
- fundamentals of data analysis in access
- the picture of dorian gray analysis chapter 1
- the moaning of life karl pilkington episode 1
- fundamentals of the java programming language java se 6 ed 1
- the grammar of the english tense system a comprehensive analysis volume 1
- the moaning of life karl pilkington stream episode 1
- the moaning of life karl pilkington episode 1 online free
- the moaning of life karl pilkington episode 1 free