... shown.
(1- a) (1- b)
(2)
(3)
(4- a) (4- b)
Problem 2.
4m
4@ 3m =12 m
1 kN
a
b
4m
4@ 3m =12 m
1 kN
a
b
12 kN
4 @ 4m =16 m
2m
3m
a
b
c
6@3m =18 m
4m
4m
12 kN
a
b
c
3@3m=9m
4m
4m
15 kN
a
AB
b
c
3@3m=9m
4m
4m
15 kN
a
AB
b
... 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...
... shown.
(1- a) (1- b)
(2)
(3)
(4- a) (4- b)
Problem 2.
4m
4@ 3m =12 m
1 kN
a
b
4m
4@ 3m =12 m
1 kN
a
b
12 kN
4 @ 4m =16 m
2m
3m
a
b
c
6@3m =18 m
4m
4m
12 kN
a
b
c
3@3m=9m
4m
4m
15 kN
a
AB
b
c
3@3m=9m
4m
4m
15 kN
a
AB
b
... 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...
... Frame Analysis: Force Method, Part I by S. T. Mau
11 9
(11 ) (12 )
(13 ) (14 )
(15 ) (16 )
Problem 2. Frame problems.
5 m
5 m
10 kN
5 m
5 m
10 kN-m
5 m
5 m
10 kN
5 m
5 m
10 kN-m
5 m
5 m
10 kN
5 m
5 m
10 ... Beam
P
a/2EI
P
a/EI
2aaa
R
eactions
P
a/2EI
P
a/EI
11 Pa
2
/12 EI
5Pa
2
/12 EI
Shear(Rotation)Diagram
( Unit: Pa
2
/EI )
1
1/ 12
5 /12
1/ 6
M
oment(Deflection) Diagram...
... displaced
configuration of the frame as shown below.
Displaced configuration.
P
2P
2P
P
1
2
2
1
1
1/ L1/L
1
1/L
1/ L
1
1
2PL
2L
1
1
2L
2L
P
Beam and Frame Analysis: Force Method, Part II by S. T. Mau
14 0
The ...
∆
d
Load
Diagram
Moment
Diagram
(M)(m)(m)(m)(m)
a~b
EI
1
(
3
1
)
(2PL)(2L)(2L)
=
EI
PL
3
8
3
00
EI
1
(
3
1
)
(2PL)(2L)(2L)
=
EI
PL
3
8
3
b~c
EI2
1
(...
... solutions.
P
1
2
3
θ
2
’
M
1
M
1
θ
11
θ
1
’
θ
3
’
M
1
θ
21
M
1
θ
31
M
1
M
2
θ
21
M
2
θ
22
M
2
θ
32
M
3
M
3
θ
31
M
3
θ
23
M
3
θ
33
θ
2
=0
θ
1
=0
θ
3
=0
P
Beam and Frame Analysis: Force Method, Part ... self-evident.
L
L
w
P
a
b
5 P /16
11 P /16
3PL /16
L
/2
L
/2
Beam and Frame Analysis: Force Method, Part III by S. T. Mau
16 3
⎥
⎦
⎤
⎢
⎣
⎡
2 212
12 1...
... 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...
... shown.
(1- a) (1- b)
(2)
(3)
(4- a) (4- b)
Problem 2.
4m
4@ 3m =12 m
1 kN
a
b
4m
4@ 3m =12 m
1 kN
a
b
12 kN
4 @ 4m =16 m
2m
3m
a
b
c
6@3m =18 m
4m
4m
12 kN
a
b
c
3@3m=9m
4m
4m
15 kN
a
AB
b
c
3@3m=9m
4m
4m
15 kN
a
AB
b
... 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...
... redundant
force.
P
a
b
5P /16
11 P /16
3PL /16
V
M
5PL/32
∆
11 P /16
−5P /16
−3PL /16
I
nflection point
L
/2
L
/2
L
/2
L
/2
P
a
b
Beam and Frame Analysis: Force Method, Part III by S. T. Mau
17 4
Beam and Frame Analysis: ... good
approximation to the correct solutions.
P
1
2
3
θ
2
’
M
1
M
1
θ
11
θ
1
’
θ
3
’
M
1
θ
21
M
1
θ
31
M
1
M
2
θ
21
M
2
θ
22
M
2
θ
3...