Fundamentals Of Structural Analysis Episode 1 Part 5 pptx
Fundamentals Of Structural Analysis Episode 1 Part 5 pptx
... -0. 71 -1. 00 -3.40 -4.80
6
-56 .56 17 ,680 -3.20 1. 00 0.94 -3.20 -3.00
7
40.00 25, 000 1. 60 -0. 71 0.33 -1. 14 0 .53
8
-56 .56 17 ,680 -3.20 0.00 -0.47 0.00 1. 50
9
-40.00 25, 000 -1. 60 -0. 71 -0.33 1. 14 ... (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
-11 3 .13 17 ,6...
Fundamentals of Structural Analysis Episode 1 Part 5 pdf
... -0. 71 -1. 00 -3.40 -4.80
6
-56 .56 17 ,680 -3.20 1. 00 0.94 -3.20 -3.00
7
40.00 25, 000 1. 60 -0. 71 0.33 -1. 14 0 .53
8
-56 .56 17 ,680 -3.20 0.00 -0.47 0.00 1. 50
9
-40.00 25, 000 -1. 60 -0. 71 -0.33 1. 14 ... (mm/kN)
1 -0.33 25, 000 -0. 013 -0.8 0. 010 -0.032 0.026
2 0 33,333 0 -0.6 0 -0. 018 0. 011
3 0 25, 000 0 -0.8 0 -0.032 0.026
4 0 .50 33,333 0. 0 15 -0.6 -0.009 -0...
Fundamentals of Structural Analysis Episode 1 Part 5 pps
... -0. 71 -1. 00 -3.40 -4.80
6
-56 .56 17 ,680 -3.20 1. 00 0.94 -3.20 -3.00
7
40.00 25, 000 1. 60 -0. 71 0.33 -1. 14 0 .53
8
-56 .56 17 ,680 -3.20 0.00 -0.47 0.00 1. 50
9
-40.00 25, 000 -1. 60 -0. 71 -0.33 1. 14 ... 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 =12 m
1
2
3
4
5
6
1
2
3
4
5...
Fundamentals of Structural Analysis Episode 1 Part 4 pptx
... form:
x
y
1
2
4m
3m
3
3m
1
2
3
x
y
1
2
4m
3m
3
3m
1
2
3
x
y
1
2
4m
3m
3
3m
1
2
3
x
y
1
2
4m
3m
3
3m
1
2
3
x
y
1
2
4m
3m
3
3m
1
2
3
x
y
1
2
4m
3m
3
3m
1
2
3
1 kN
1 kN
1 kN
1 kN
1 kN
1 kN
Truss Analysis: Force ... 6
−0.8
3 31
1. 0
2 0.0 5 1. 0 6
−0.0
Contribution of Reaction Forces
Equation Number and Value of EntryReaction
Number
Force
Number
2i -1...
Fundamentals of Structural Analysis Episode 2 Part 5 pptx
...
=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. 25
0.6 25
0. 41
F
CJ
6 m
9 m
1. 18 1. 82
Influence Lines by S. T. Mau
256
Solution. ... location of the group load.
Placing the group load to maximize F
CJ
.
1kN 1kN
2 kN
1m
2m
2 kN
10 kN/m 10 kN/m
x...
Fundamentals of Structural Analysis Episode 1 Part 1 docx
... as
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
2
2
1
1
y
x
y
x
F
F
F
F
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
444342 41
343332 31
242322 21
1 413 1 211
kkkk
kkkk
kkkk
kkkk
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
2
2
1
1
v
u
v
u
(12 )
2
1
1
2
x
x
Truss Analysis: Matrix ... form:
1
2
3
1
2
3
1
2
1
2
3
2
1
3
3
P
y2
P
x2
(F
y2
)
1
+(F
y2
)
2
(F
x2
)
1
+
(F
x2
)
2
P
y3
P
x3
(F
y3
)
2
+(F
y3
)
3
(F
x3
)
2
+
(F
x...
Fundamentals of Structural Analysis Episode 1 Part 1 pptx
... as
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
2
2
1
1
y
x
y
x
F
F
F
F
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
444342 41
343332 31
242322 21
1 413 1 211
kkkk
kkkk
kkkk
kkkk
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
2
2
1
1
v
u
v
u
(12 )
2
1
1
2
x
x
Truss Analysis: Matrix ... and Frames 15 3
Table: Beam Deflection Formulas 16 8
Problem 6. 17 3
Beam and Frame Analysis: Displacement Method, Part I 17 5
1. Introducti...
Fundamentals of Structural Analysis Episode 1 Part 2 pot
... equation:
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
66 656 46362 61
56 555 453 52 51
46 454 44342 41
36 353 43332 31
26 252 42322 21
1 6 15 1 413 1 211
KKKKKK
KKKKKK
KKKKKK
KKKKKK
KKKKKK
KKKKKK
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
3
3
2
2
1
1
v
u
v
u
v
u
=
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
3
3
2
2
1
1
y
x
y
x
y
x
P
P
P
P
P
P
( 15 )
where ... solved by the method o...
Fundamentals of Structural Analysis Episode 1 Part 3 pps
... F
4
(3 /5) –F
1
= 0,
F
1
= –6 kN.
2
1
2
3
6 kN
4
5
6
R
y1
R
x1
1
3
4
5
R
x5
R
y5
F
5
3
4
5
6 kN
3
F
6
3
4
5
2
F
5
F
4
F
1
3
5
4
3
4
5
Truss Analysis: Force Method, Part I by S. T. Mau
51
In this ... joint.
4
F
6
F
3
F
2
3
5
4
3
4
5
1
F
3
F
1
R
y1
R
x1
3
4
5
F
4
F
2
R
y5
5
R
x5
3
5
4
10 kN
3@2m=6m
1
3
2
4
5
6
7
1
2 3
5
6
7
8
10
11
2m
9...
Fundamentals of Structural Analysis Episode 1 Part 3 pps
... F
4
(3 /5) –F
1
= 0,
F
1
= –6 kN.
2
1
2
3
6 kN
4
5
6
R
y1
R
x1
1
3
4
5
R
x5
R
y5
F
5
3
4
5
6 kN
3
F
6
3
4
5
2
F
5
F
4
F
1
3
5
4
3
4
5
Truss Analysis: Force Method, Part I by S. T. Mau
49
This particular ... joint.
4
F
6
F
3
F
2
3
5
4
3
4
5
1
F
3
F
1
R
y1
R
x1
3
4
5
F
4
F
2
R
y5
5
R
x5
3
5
4
10 kN
3@2m=6m
1
3
2
4
5
6
7
1
2 3
5
6
7
8
10
11...