... 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
(...
... 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...