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# The Moment Distribution Method: Frames with Sidesway - 1 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : The Moment Distribution Method: Frames with Sidesway - 1 Civil Engineering (CE) Notes | EduRev

``` Page 1

Instructional Objectives
After reading this chapter the student will be able to
1. Extend moment-distribution method for frames undergoing sidesway.
2. Draw free-body diagrams of plane frame.
3. Analyse plane frames undergoing sidesway by the moment-distribution
method.
4. Draw shear force and bending moment diagrams.
5. Sketch deflected shape of the plane frame not restrained against sidesway.

21.1 Introduction
In the previous lesson, rigid frames restrained against sidesway are analyzed
using moment-distribution method. It has been pointed in lesson 17, that frames
which are unsymmetrical or frames which are loaded unsymmetrically usually get
displaced either to the right or to the left. In other words, in such frames apart
from evaluating joint rotations, one also needs to evaluate joint translations
(sidesway). For example in frame shown in Fig 21.1, the loading is symmetrical
but the geometry of frame is unsymmetrical and hence sidesway needs to be
considered in the analysis. The number of unknowns is this case are: joint
rotations
B
? and
C
? and member rotation ? . Joint B and C get translated by the
same amount as axial deformations are not considered and hence only one
independent member rotation need to be considered. The procedure to analyze
rigid frames undergoing lateral displacement using moment-distribution method
is explained in section 21.2 using an example.

Page 2

Instructional Objectives
After reading this chapter the student will be able to
1. Extend moment-distribution method for frames undergoing sidesway.
2. Draw free-body diagrams of plane frame.
3. Analyse plane frames undergoing sidesway by the moment-distribution
method.
4. Draw shear force and bending moment diagrams.
5. Sketch deflected shape of the plane frame not restrained against sidesway.

21.1 Introduction
In the previous lesson, rigid frames restrained against sidesway are analyzed
using moment-distribution method. It has been pointed in lesson 17, that frames
which are unsymmetrical or frames which are loaded unsymmetrically usually get
displaced either to the right or to the left. In other words, in such frames apart
from evaluating joint rotations, one also needs to evaluate joint translations
(sidesway). For example in frame shown in Fig 21.1, the loading is symmetrical
but the geometry of frame is unsymmetrical and hence sidesway needs to be
considered in the analysis. The number of unknowns is this case are: joint
rotations
B
? and
C
? and member rotation ? . Joint B and C get translated by the
same amount as axial deformations are not considered and hence only one
independent member rotation need to be considered. The procedure to analyze
rigid frames undergoing lateral displacement using moment-distribution method
is explained in section 21.2 using an example.

21.2 Procedure
A special procedure is required to analyze frames with sidesway using moment-
distribution method. In the first step, identify the number of independent rotations
( ?) in the structure. The procedure to calculate independent rotations is
explained in lesson 22. For analyzing frames with sidesway, the method of
superposition is used. The structure shown in Fig. 21.2a is expressed as the
sum of two systems: Fig. 21.2b and Fig. 21.2c. The systems shown in figures
21.2b and 21.2c are analyzed separately and superposed to obtain the final
answer. In system 21.2b, sidesway is prevented by artificial support atC . Apply
all the external loads on frame shown in Fig. 21.2b. Since for the frame,
sidesway is prevented, moment-distribution method as discussed in the previous
lesson is applied and beam end moments are calculated.
Let and  be the balanced moments obtained by
distributing fixed end moments due to applied loads while allowing only joint
rotations  (
' ' ' ' '
, , , ,
CD CB BC BA AB
M M M M M
'
DC
M
B
? and
C
? ) and preventing sidesway.
Now, calculate reactions  and (ref. Fig 21.3a).they are ,
1 A
H
1 D
H

Page 3

Instructional Objectives
After reading this chapter the student will be able to
1. Extend moment-distribution method for frames undergoing sidesway.
2. Draw free-body diagrams of plane frame.
3. Analyse plane frames undergoing sidesway by the moment-distribution
method.
4. Draw shear force and bending moment diagrams.
5. Sketch deflected shape of the plane frame not restrained against sidesway.

21.1 Introduction
In the previous lesson, rigid frames restrained against sidesway are analyzed
using moment-distribution method. It has been pointed in lesson 17, that frames
which are unsymmetrical or frames which are loaded unsymmetrically usually get
displaced either to the right or to the left. In other words, in such frames apart
from evaluating joint rotations, one also needs to evaluate joint translations
(sidesway). For example in frame shown in Fig 21.1, the loading is symmetrical
but the geometry of frame is unsymmetrical and hence sidesway needs to be
considered in the analysis. The number of unknowns is this case are: joint
rotations
B
? and
C
? and member rotation ? . Joint B and C get translated by the
same amount as axial deformations are not considered and hence only one
independent member rotation need to be considered. The procedure to analyze
rigid frames undergoing lateral displacement using moment-distribution method
is explained in section 21.2 using an example.

21.2 Procedure
A special procedure is required to analyze frames with sidesway using moment-
distribution method. In the first step, identify the number of independent rotations
( ?) in the structure. The procedure to calculate independent rotations is
explained in lesson 22. For analyzing frames with sidesway, the method of
superposition is used. The structure shown in Fig. 21.2a is expressed as the
sum of two systems: Fig. 21.2b and Fig. 21.2c. The systems shown in figures
21.2b and 21.2c are analyzed separately and superposed to obtain the final
answer. In system 21.2b, sidesway is prevented by artificial support atC . Apply
all the external loads on frame shown in Fig. 21.2b. Since for the frame,
sidesway is prevented, moment-distribution method as discussed in the previous
lesson is applied and beam end moments are calculated.
Let and  be the balanced moments obtained by
distributing fixed end moments due to applied loads while allowing only joint
rotations  (
' ' ' ' '
, , , ,
CD CB BC BA AB
M M M M M
'
DC
M
B
? and
C
? ) and preventing sidesway.
Now, calculate reactions  and (ref. Fig 21.3a).they are ,
1 A
H
1 D
H

2 2
' '
1
h
Pa
h
M M
H
BA AB
A
+
+
=

1
' '
1
h
M M
H
DC CD
D
+
=       (21.1)

again,            (21.2) ) (
1 1 D A
H H P R + - =

Page 4

Instructional Objectives
After reading this chapter the student will be able to
1. Extend moment-distribution method for frames undergoing sidesway.
2. Draw free-body diagrams of plane frame.
3. Analyse plane frames undergoing sidesway by the moment-distribution
method.
4. Draw shear force and bending moment diagrams.
5. Sketch deflected shape of the plane frame not restrained against sidesway.

21.1 Introduction
In the previous lesson, rigid frames restrained against sidesway are analyzed
using moment-distribution method. It has been pointed in lesson 17, that frames
which are unsymmetrical or frames which are loaded unsymmetrically usually get
displaced either to the right or to the left. In other words, in such frames apart
from evaluating joint rotations, one also needs to evaluate joint translations
(sidesway). For example in frame shown in Fig 21.1, the loading is symmetrical
but the geometry of frame is unsymmetrical and hence sidesway needs to be
considered in the analysis. The number of unknowns is this case are: joint
rotations
B
? and
C
? and member rotation ? . Joint B and C get translated by the
same amount as axial deformations are not considered and hence only one
independent member rotation need to be considered. The procedure to analyze
rigid frames undergoing lateral displacement using moment-distribution method
is explained in section 21.2 using an example.

21.2 Procedure
A special procedure is required to analyze frames with sidesway using moment-
distribution method. In the first step, identify the number of independent rotations
( ?) in the structure. The procedure to calculate independent rotations is
explained in lesson 22. For analyzing frames with sidesway, the method of
superposition is used. The structure shown in Fig. 21.2a is expressed as the
sum of two systems: Fig. 21.2b and Fig. 21.2c. The systems shown in figures
21.2b and 21.2c are analyzed separately and superposed to obtain the final
answer. In system 21.2b, sidesway is prevented by artificial support atC . Apply
all the external loads on frame shown in Fig. 21.2b. Since for the frame,
sidesway is prevented, moment-distribution method as discussed in the previous
lesson is applied and beam end moments are calculated.
Let and  be the balanced moments obtained by
distributing fixed end moments due to applied loads while allowing only joint
rotations  (
' ' ' ' '
, , , ,
CD CB BC BA AB
M M M M M
'
DC
M
B
? and
C
? ) and preventing sidesway.
Now, calculate reactions  and (ref. Fig 21.3a).they are ,
1 A
H
1 D
H

2 2
' '
1
h
Pa
h
M M
H
BA AB
A
+
+
=

1
' '
1
h
M M
H
DC CD
D
+
=       (21.1)

again,            (21.2) ) (
1 1 D A
H H P R + - =

In Fig 21.2c apply a horizontal force  in the opposite direction of F R. Now
, then the superposition of beam end moments of system (b) and times
(c) gives the results for the original structure. However, there is no way one could
analyze the frame for horizontal force , by moment-distribution method as sway
comes in to picture. Instead of applying , apply arbitrary known displacement /
sidesway ' as shown in the figure. Calculate the fixed end beam moments in
the column
R F k = k
F
F
?
AB and CD for the imposed horizontal displacement. Since joint
displacement is known beforehand, one could use moment-distribution method to
analyse this frame. In this case, member rotations  ? are related to joint
translation which is known. Let  and are the
balanced moment obtained by distributing the fixed end moments due to
assumed sidesway  at joints
' ' ' ' ' ' ' ' ' '
, , , ,
CD CB BC BA AB
M M M M M
' '
DC
M
' ? B and . Now, from statics calculate horizontal
force due to arbitrary sidesway
C
F ' ? .

Page 5

Instructional Objectives
After reading this chapter the student will be able to
1. Extend moment-distribution method for frames undergoing sidesway.
2. Draw free-body diagrams of plane frame.
3. Analyse plane frames undergoing sidesway by the moment-distribution
method.
4. Draw shear force and bending moment diagrams.
5. Sketch deflected shape of the plane frame not restrained against sidesway.

21.1 Introduction
In the previous lesson, rigid frames restrained against sidesway are analyzed
using moment-distribution method. It has been pointed in lesson 17, that frames
which are unsymmetrical or frames which are loaded unsymmetrically usually get
displaced either to the right or to the left. In other words, in such frames apart
from evaluating joint rotations, one also needs to evaluate joint translations
(sidesway). For example in frame shown in Fig 21.1, the loading is symmetrical
but the geometry of frame is unsymmetrical and hence sidesway needs to be
considered in the analysis. The number of unknowns is this case are: joint
rotations
B
? and
C
? and member rotation ? . Joint B and C get translated by the
same amount as axial deformations are not considered and hence only one
independent member rotation need to be considered. The procedure to analyze
rigid frames undergoing lateral displacement using moment-distribution method
is explained in section 21.2 using an example.

21.2 Procedure
A special procedure is required to analyze frames with sidesway using moment-
distribution method. In the first step, identify the number of independent rotations
( ?) in the structure. The procedure to calculate independent rotations is
explained in lesson 22. For analyzing frames with sidesway, the method of
superposition is used. The structure shown in Fig. 21.2a is expressed as the
sum of two systems: Fig. 21.2b and Fig. 21.2c. The systems shown in figures
21.2b and 21.2c are analyzed separately and superposed to obtain the final
answer. In system 21.2b, sidesway is prevented by artificial support atC . Apply
all the external loads on frame shown in Fig. 21.2b. Since for the frame,
sidesway is prevented, moment-distribution method as discussed in the previous
lesson is applied and beam end moments are calculated.
Let and  be the balanced moments obtained by
distributing fixed end moments due to applied loads while allowing only joint
rotations  (
' ' ' ' '
, , , ,
CD CB BC BA AB
M M M M M
'
DC
M
B
? and
C
? ) and preventing sidesway.
Now, calculate reactions  and (ref. Fig 21.3a).they are ,
1 A
H
1 D
H

2 2
' '
1
h
Pa
h
M M
H
BA AB
A
+
+
=

1
' '
1
h
M M
H
DC CD
D
+
=       (21.1)

again,            (21.2) ) (
1 1 D A
H H P R + - =

In Fig 21.2c apply a horizontal force  in the opposite direction of F R. Now
, then the superposition of beam end moments of system (b) and times
(c) gives the results for the original structure. However, there is no way one could
analyze the frame for horizontal force , by moment-distribution method as sway
comes in to picture. Instead of applying , apply arbitrary known displacement /
sidesway ' as shown in the figure. Calculate the fixed end beam moments in
the column
R F k = k
F
F
?
AB and CD for the imposed horizontal displacement. Since joint
displacement is known beforehand, one could use moment-distribution method to
analyse this frame. In this case, member rotations  ? are related to joint
translation which is known. Let  and are the
balanced moment obtained by distributing the fixed end moments due to
assumed sidesway  at joints
' ' ' ' ' ' ' ' ' '
, , , ,
CD CB BC BA AB
M M M M M
' '
DC
M
' ? B and . Now, from statics calculate horizontal
force due to arbitrary sidesway
C
F ' ? .

2
' ' ' '
2
h
M M
H
BA AB
A
+
=

1
' ' ' '
2
h
M M
H
DC CD
D
+
=      (21.3)

) (
2 2 D A
H H F + =         (21.4)

In Fig 21.2, by method of superposition

R kF =  or  F R k / =

Substituting the values of R and from equations (21.2) and (21.4), F

) (
) (
2 2
1 1
D A
D A
H H
H H P
k
+
+ -
=           (21.5)

Now substituting the values of , ,  and  in 21.5,
1 A
H
2 A
H
1 D
H
2 D
H

?
?
?
?
?
?
?
? +
+
+
+
?
?
?
?
?
?
?
?
+
+
-
=
+
1 2
1 2 2
' ' ' '
' ' ' '
' '
' '
h
M M
h
M M
h
M M
h
Pa
h
M M
P
k
DC CD
BA AB
DC CD
BA AB
(21.6)

Hence, beam end moment in the original structure is obtained as,

) ( ) ( c system b system original
kM M M + =

If there is more than one independent member rotation, then the above
procedure needs to be modified and is discussed in the next lesson.

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