Fundamentals of Structural Analysis Episode 2 Part 5 pptx
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 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 ... 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...
Fundamentals Of Structural Analysis Episode 1 Part 5 pptx
... (kN-mm)
1 80.00 25 ,000 3 .20 0.00 -0.33 0.00 -1.06
2 80.00 25 ,000 3 .20 -0.71 -0.33 -2. 26 -1.06
3 40.00 25 ,000 1.60 0.00 0.33 0.00 0 .53
4
-113.13 17,680 -6.40 0.00 0.47 0.00 -3.00
5
120 .00 25 ,000 4.80 ...
α
=5( 10
-6
)/
o
C.
Problem 5- 3.
1
2
4m
3m
3
3m
1
2
3
1.0 kN
0 .5 kN
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
4m
3@4m=12m
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
4m
3@4m=...
Fundamentals Of Structural Analysis Episode 2 Part 8 pptx
... kN-m, M
db
= − 12. 50 kN-m.
- 32. 8
-14
33.6
26 .6
-1.93
-2. 15
1. 85
1.14
I
nflection point
-2. 30
1.48 m
-5 -2. 0
4 .5
I
nflection point
2. 0
Index by S. T. Mau
324
principle of virtual, 72
shear, 94
thrust, ... 29 2
Single degree of freedom(SDOF), 181,
29 5
Singular, 9
Slope-deflection method, 20 9
treatment of load between nodes, 21 5
treatment of side-sway,...
Fundamentals of Structural Analysis Episode 1 Part 5 pdf
... -0.0 32 0. 026
2 0 33,333 0 -0.6 0 -0.018 0.011
3 0 25 ,000 0 -0.8 0 -0.0 32 0. 026
4 0 .50 33,333 0.0 15 -0.6 -0.009 -0.018 0.011
5 -0.83 20 ,000 -0.0 42 1.0 -0.0 42 0. 050 0. 050
6 0 20 ,000 0 1.0 0 0. 050 ... (kN-mm)
1 80.00 25 ,000 3 .20 0.00 -0.33 0.00 -1.06
2 80.00 25 ,000 3 .20 -0.71 -0.33 -2. 26 -1.06
3 40.00 25 ,000 1.60 0.00 0.33 0.00 0 .53
4
-113.13 17,680 -6.40...
Fundamentals of Structural Analysis Episode 1 Part 5 pps
... is
α
=5( 10
-6
)/
o
C.
Problem 5- 3.
1
2
4m
3m
3
3m
1
2
3
1.0 kN
0 .5 kN
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
4m
3@4m=12m
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
4m
3@4m=12m
Truss Analysis: Force Method, Part II ... -0.0 32 0. 026
2 0 33,333 0 -0.6 0 -0.018 0.011
3 0 25 ,000 0 -0.8 0 -0.0 32 0. 026
4 0 .50 33,333 0.0 15 -0.6 -0.009 -0.018 0.011
5 -0.83 20 ,000 -0.0 42 1.0...
Fundamentals of Structural Analysis Episode 2 Part 1 ppsx
... + 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
2 m...
Fundamentals of Structural Analysis Episode 2 Part 1 pdf
... Distribution for a Two-DOF Problem
Node ab cd
Member ab bc cd
DF 0 0. 6 25 0.3 75 0 .5 0 .5 0
MEM M
ab
M
ba
M
bc
M
cb
M
cd
M
dc
EAM 30
DM + 15 + 15
COM +7 .50 +7 .50
DM −4.69 2. 81
COM 2. 35 −1.41
DM +0.71 +0.70
COM ... below.
FBDs of the two members.
1 .5 kN-m
2 m
2 m
4 kN
3 kN-m
2. 38 kN
1. 62 kN
3 kN-m
4 m
3 kN/m
4 .5 kN-m
6.38 kN
5. 62 kN
Beam and Frame Analysis: Displ...
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 ... kN
ab
cd
50 kN
2 m 2 m
2 m 2 m 4 m
50 kN
ab
c
4 m 4 m
4 m
8 m
2EI
2EI
E
I
E
I
2EI2EI
50 kN
ab
c
4 m 4 m
4 m
8 m
E
I
2EI2EI
2 m
2 m...
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