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Page 1
Now calculate reactions from free body diagram shown in Fig. 21.5d.
Page 2
Now calculate reactions from free body diagram shown in Fig. 21.5d.
Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?
11
5 2.29
A
HV +=- (3)
Column CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2
(4)
12
5 2.284
D
HV -=
Beam BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?
( )
1
5.002 kN V=?
( )
2
4.998 kN V= ? (5)
Thus from (3) ( )
1
1.458 kN
A
H =- ?
from (4) ( )
1
1.456 kN
D
H= ? (6)
Page 3
Now calculate reactions from free body diagram shown in Fig. 21.5d.
Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?
11
5 2.29
A
HV +=- (3)
Column CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2
(4)
12
5 2.284
D
HV -=
Beam BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?
( )
1
5.002 kN V=?
( )
2
4.998 kN V= ? (5)
Thus from (3) ( )
1
1.458 kN
A
H =- ?
from (4) ( )
1
1.456 kN
D
H= ? (6)
Version 2 CE IIT, Kharagpur
0=
()
11
05
5.002 kN
XAD
FHHR
R
=++-
=+?
?
(7)
d) Moment-distribution for arbitrary sidesway ' ? .
Calculate fixed end beam moments for arbitrary sidesway of
EI
75 . 12
' = ?
The member rotations for this arbitrary sidesway is shown in Fig. 21.6e.
Page 4
Now calculate reactions from free body diagram shown in Fig. 21.5d.
Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?
11
5 2.29
A
HV +=- (3)
Column CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2
(4)
12
5 2.284
D
HV -=
Beam BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?
( )
1
5.002 kN V=?
( )
2
4.998 kN V= ? (5)
Thus from (3) ( )
1
1.458 kN
A
H =- ?
from (4) ( )
1
1.456 kN
D
H= ? (6)
Version 2 CE IIT, Kharagpur
0=
()
11
05
5.002 kN
XAD
FHHR
R
=++-
=+?
?
(7)
d) Moment-distribution for arbitrary sidesway ' ? .
Calculate fixed end beam moments for arbitrary sidesway of
EI
75 . 12
' = ?
The member rotations for this arbitrary sidesway is shown in Fig. 21.6e.
1
1
"'
;
cos 5
AB
AB AB
BB
LL
5.1'
?
a
? ? ?
==- ?= =
2
2'
0.4 '
5
?
?= = ?
'
()
5
AB
clockwise ?
?
=- ;
'
()
5
CD
clockwise ?
?
=-
2
2'tan '
()
22 5
BC
counterclockwise
a
?
? ? ?
== =
6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F AB
AB AB
AB
EI EI
M
LEI
?
??
=- =- - = +
??
??
6.0 kN.m
F
BA
M =+
6 6 ( ) 12.75
7.65 kN.m
25
F BC
BC BC
BC
EI EI
M
LEI
?
??
=- =- =-
??
??
7.65 kN.m
F
CB
M =-
6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F CD
CD CD
CD
EI EI
M
LEI
?
??
=- =- - =+
??
??
6.0 kN.m
F
DC
M =+
The moment-distribution for the arbitrary sway is shown in Fig. 21.6f. Now
reactions can be calculated from statics.
Page 5
Now calculate reactions from free body diagram shown in Fig. 21.5d.
Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?
11
5 2.29
A
HV +=- (3)
Column CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2
(4)
12
5 2.284
D
HV -=
Beam BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?
( )
1
5.002 kN V=?
( )
2
4.998 kN V= ? (5)
Thus from (3) ( )
1
1.458 kN
A
H =- ?
from (4) ( )
1
1.456 kN
D
H= ? (6)
Version 2 CE IIT, Kharagpur
0=
()
11
05
5.002 kN
XAD
FHHR
R
=++-
=+?
?
(7)
d) Moment-distribution for arbitrary sidesway ' ? .
Calculate fixed end beam moments for arbitrary sidesway of
EI
75 . 12
' = ?
The member rotations for this arbitrary sidesway is shown in Fig. 21.6e.
1
1
"'
;
cos 5
AB
AB AB
BB
LL
5.1'
?
a
? ? ?
==- ?= =
2
2'
0.4 '
5
?
?= = ?
'
()
5
AB
clockwise ?
?
=- ;
'
()
5
CD
clockwise ?
?
=-
2
2'tan '
()
22 5
BC
counterclockwise
a
?
? ? ?
== =
6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F AB
AB AB
AB
EI EI
M
LEI
?
??
=- =- - = +
??
??
6.0 kN.m
F
BA
M =+
6 6 ( ) 12.75
7.65 kN.m
25
F BC
BC BC
BC
EI EI
M
LEI
?
??
=- =- =-
??
??
7.65 kN.m
F
CB
M =-
6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F CD
CD CD
CD
EI EI
M
LEI
?
??
=- =- - =+
??
??
6.0 kN.m
F
DC
M =+
The moment-distribution for the arbitrary sway is shown in Fig. 21.6f. Now
reactions can be calculated from statics.
Column AB
21
0 5 6.283 6.567 0
AA
MH V =? - - + =
?
11
5 12.85
A
HV += (3)
Column CD
22
0 5 6.567 6.283 0
DD
MH V =? - - - =
?
12
5 12.85
D
HV -= (4)
Beam BC
1
0 2 6.567 6.567 0
C
MV =? + + =
?
( )
1
6.567 VkN
( )
2
6.567 kN V =- ? ;+? (5) =
Thus from 3 ( )
2
3.883 kN
A
H =+ ?
from 4 (6) ()
2
3.883 kN
D
H= ?
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FAQs on The Moment Distribution Method: Frames with Sidesway - 4 - Structural Analysis - Civil Engineering (CE)
| 1. What is the moment distribution method? |  |
Ans. The moment distribution method is a structural analysis technique used to determine the moments and rotations at each joint of a frame structure. It is based on the principle of stiffness distribution, where the moments are distributed among the members based on their relative stiffness.
| 2. How does the moment distribution method handle frames with sidesway? | |
Ans. The moment distribution method can handle frames with sidesway by considering the sidesway effects during the analysis. Sidesway refers to the lateral movement or displacement of a frame due to horizontal loads. The method incorporates the sidesway effects by including additional stiffness terms in the analysis equations.
| 3. What are the advantages of using the moment distribution method for frames with sidesway? | |
Ans. The moment distribution method offers several advantages when analyzing frames with sidesway. Firstly, it provides an accurate representation of the distribution of moments and rotations at each joint, considering the effects of sidesway. Secondly, it allows for the determination of the final moments and rotations after the frame has stabilized under load. Lastly, it offers a simplified and systematic approach to analyzing complex frame structures.
| 4. Are there any limitations to using the moment distribution method for frames with sidesway? | |
Ans. Yes, there are some limitations to using the moment distribution method for frames with sidesway. One limitation is that the method assumes that the frame remains elastic and does not undergo significant plastic deformation. Additionally, it assumes that the frame members do not experience significant axial deformations. If these assumptions are violated, the accuracy of the analysis results may be compromised.
| 5. How can the moment distribution method be applied to practical civil engineering problems? | |
Ans. The moment distribution method can be applied to practical civil engineering problems by following a step-by-step procedure. First, the frame geometry and member properties are defined. Then, the fixed-end moments are calculated based on the applied loads. The distribution factors and carryover factors are determined, and an iterative process is used to distribute the moments and rotations at each joint. Finally, the final moments and rotations are obtained, and the frame's stability and strength can be evaluated.
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