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

## Civil Engineering (CE) : The Slope Deflection 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. Derive slope-deflection equations for the frames undergoing sidesway.
2. Analyse plane frames undergoing sidesway.
3, Draw shear force and bending moment diagrams.
4. Sketch deflected shape of the plane frame not restrained against sidesway.

17.1 Introduction
In this lesson, slope-deflection equations are applied to analyse statically
indeterminate frames undergoing sidesway. As stated earlier, the axial
deformation of beams and columns are small and are neglected in the analysis.
In the previous lesson, it was observed that sidesway in a frame will not occur if

1. They are restrained against sidesway.

In general loading will never be symmetrical. Hence one could not avoid
sidesway in frames.

For example, consider the frame of Fig. 17.1. In this case the frame is
moments  and  are not equal. If is greater than , then . In
BC
M
CB
M b a
CB BC
M M >

Page 2

Instructional Objectives
After reading this chapter the student will be able to
1. Derive slope-deflection equations for the frames undergoing sidesway.
2. Analyse plane frames undergoing sidesway.
3, Draw shear force and bending moment diagrams.
4. Sketch deflected shape of the plane frame not restrained against sidesway.

17.1 Introduction
In this lesson, slope-deflection equations are applied to analyse statically
indeterminate frames undergoing sidesway. As stated earlier, the axial
deformation of beams and columns are small and are neglected in the analysis.
In the previous lesson, it was observed that sidesway in a frame will not occur if

1. They are restrained against sidesway.

In general loading will never be symmetrical. Hence one could not avoid
sidesway in frames.

For example, consider the frame of Fig. 17.1. In this case the frame is
moments  and  are not equal. If is greater than , then . In
BC
M
CB
M b a
CB BC
M M >

such a case joint B andC are displaced toward right as shown in the figure by an
unknown amount ? . Hence we have three unknown displacements in this frame:
rotations
C B
? ? , and the linear displacement ?. The unknown joint rotations
B
? and
C
? are related to joint moments by the moment equilibrium equations.
Similarly, when unknown linear displacement occurs, one needs to consider
force-equilibrium equations. While applying slope-deflection equation to columns
in the above frame, one must consider the column rotation
?
?
?
?
?
? ?
=
h
? as
unknowns. It is observed that in the columnAB , the end B undergoes a linear
displacement with respect to end ? A. Hence the slope-deflection equation for
column AB is similar to the one for beam undergoing support settlement.
However, in this case is unknown. For each of the members we can write the
following slope-deflection equations.
?

[
AB B A
F
AB AB
] M M ? ? ? 3 2
2
- + + =
h
EI
where
h
AB
?
- = ?

AB
? is assumed to be negative as the chord to the elastic curve rotates in the
clockwise directions.

[]
AB A B
F
BA BA
h
EI
M M ? ? ? 3 2
2
- + + =
[]
C B
F
BC BC
h
EI
M M ? ? + + = 2
2

[]
B C
F
CB CB
h
EI
M M ? ? + + = 2
2

[]
CD D C
F
CD CD
h
EI
M M ? ? ? 3 2
2
- + + =
h
CD
?
- = ?
[
CD C D
F
DC DC
h
EI
M M ? ? ? 3 2
2
- + + = ]   (17.1)

As there are three unknowns (
C B
? ? , and ?), three equations are required to
evaluate them. Two equations are obtained by considering the moment
equilibrium of joint B and C respectively.

0 0 = + ? =
?
BC
BA B
M M M   (17.2a)

0 0 = + ? =
?
CD
CB C
M M M  (17.2b)

Now consider free body diagram of the frame as shown in Fig. 17.2. The
horizontal shear force acting at A and B of the column AB is given by

Page 3

Instructional Objectives
After reading this chapter the student will be able to
1. Derive slope-deflection equations for the frames undergoing sidesway.
2. Analyse plane frames undergoing sidesway.
3, Draw shear force and bending moment diagrams.
4. Sketch deflected shape of the plane frame not restrained against sidesway.

17.1 Introduction
In this lesson, slope-deflection equations are applied to analyse statically
indeterminate frames undergoing sidesway. As stated earlier, the axial
deformation of beams and columns are small and are neglected in the analysis.
In the previous lesson, it was observed that sidesway in a frame will not occur if

1. They are restrained against sidesway.

In general loading will never be symmetrical. Hence one could not avoid
sidesway in frames.

For example, consider the frame of Fig. 17.1. In this case the frame is
moments  and  are not equal. If is greater than , then . In
BC
M
CB
M b a
CB BC
M M >

such a case joint B andC are displaced toward right as shown in the figure by an
unknown amount ? . Hence we have three unknown displacements in this frame:
rotations
C B
? ? , and the linear displacement ?. The unknown joint rotations
B
? and
C
? are related to joint moments by the moment equilibrium equations.
Similarly, when unknown linear displacement occurs, one needs to consider
force-equilibrium equations. While applying slope-deflection equation to columns
in the above frame, one must consider the column rotation
?
?
?
?
?
? ?
=
h
? as
unknowns. It is observed that in the columnAB , the end B undergoes a linear
displacement with respect to end ? A. Hence the slope-deflection equation for
column AB is similar to the one for beam undergoing support settlement.
However, in this case is unknown. For each of the members we can write the
following slope-deflection equations.
?

[
AB B A
F
AB AB
] M M ? ? ? 3 2
2
- + + =
h
EI
where
h
AB
?
- = ?

AB
? is assumed to be negative as the chord to the elastic curve rotates in the
clockwise directions.

[]
AB A B
F
BA BA
h
EI
M M ? ? ? 3 2
2
- + + =
[]
C B
F
BC BC
h
EI
M M ? ? + + = 2
2

[]
B C
F
CB CB
h
EI
M M ? ? + + = 2
2

[]
CD D C
F
CD CD
h
EI
M M ? ? ? 3 2
2
- + + =
h
CD
?
- = ?
[
CD C D
F
DC DC
h
EI
M M ? ? ? 3 2
2
- + + = ]   (17.1)

As there are three unknowns (
C B
? ? , and ?), three equations are required to
evaluate them. Two equations are obtained by considering the moment
equilibrium of joint B and C respectively.

0 0 = + ? =
?
BC
BA B
M M M   (17.2a)

0 0 = + ? =
?
CD
CB C
M M M  (17.2b)

Now consider free body diagram of the frame as shown in Fig. 17.2. The
horizontal shear force acting at A and B of the column AB is given by

h
M M
H
AB BA
+
=
1
(17.3a)

Similarly for memberCD , the shear force is given by
3
H

3
CD DC
MM
H
h
+
=    (17.3b)

Now, the required third equation is obtained by considering the equilibrium of
member , BC

13
00
X
FH =? +
?
H=

0 =
+
+
+
h
M M
h
M M
DC CD AB BA
(17.4)

Substituting the values of beam end moments from equation (17.1) in equations
(17.2a), (17.2b) and (17.4), we get three simultaneous equations in three
unknowns
C B
? ? , and , solving which joint rotations and translations are
evaluated.
?

Page 4

Instructional Objectives
After reading this chapter the student will be able to
1. Derive slope-deflection equations for the frames undergoing sidesway.
2. Analyse plane frames undergoing sidesway.
3, Draw shear force and bending moment diagrams.
4. Sketch deflected shape of the plane frame not restrained against sidesway.

17.1 Introduction
In this lesson, slope-deflection equations are applied to analyse statically
indeterminate frames undergoing sidesway. As stated earlier, the axial
deformation of beams and columns are small and are neglected in the analysis.
In the previous lesson, it was observed that sidesway in a frame will not occur if

1. They are restrained against sidesway.

In general loading will never be symmetrical. Hence one could not avoid
sidesway in frames.

For example, consider the frame of Fig. 17.1. In this case the frame is
moments  and  are not equal. If is greater than , then . In
BC
M
CB
M b a
CB BC
M M >

such a case joint B andC are displaced toward right as shown in the figure by an
unknown amount ? . Hence we have three unknown displacements in this frame:
rotations
C B
? ? , and the linear displacement ?. The unknown joint rotations
B
? and
C
? are related to joint moments by the moment equilibrium equations.
Similarly, when unknown linear displacement occurs, one needs to consider
force-equilibrium equations. While applying slope-deflection equation to columns
in the above frame, one must consider the column rotation
?
?
?
?
?
? ?
=
h
? as
unknowns. It is observed that in the columnAB , the end B undergoes a linear
displacement with respect to end ? A. Hence the slope-deflection equation for
column AB is similar to the one for beam undergoing support settlement.
However, in this case is unknown. For each of the members we can write the
following slope-deflection equations.
?

[
AB B A
F
AB AB
] M M ? ? ? 3 2
2
- + + =
h
EI
where
h
AB
?
- = ?

AB
? is assumed to be negative as the chord to the elastic curve rotates in the
clockwise directions.

[]
AB A B
F
BA BA
h
EI
M M ? ? ? 3 2
2
- + + =
[]
C B
F
BC BC
h
EI
M M ? ? + + = 2
2

[]
B C
F
CB CB
h
EI
M M ? ? + + = 2
2

[]
CD D C
F
CD CD
h
EI
M M ? ? ? 3 2
2
- + + =
h
CD
?
- = ?
[
CD C D
F
DC DC
h
EI
M M ? ? ? 3 2
2
- + + = ]   (17.1)

As there are three unknowns (
C B
? ? , and ?), three equations are required to
evaluate them. Two equations are obtained by considering the moment
equilibrium of joint B and C respectively.

0 0 = + ? =
?
BC
BA B
M M M   (17.2a)

0 0 = + ? =
?
CD
CB C
M M M  (17.2b)

Now consider free body diagram of the frame as shown in Fig. 17.2. The
horizontal shear force acting at A and B of the column AB is given by

h
M M
H
AB BA
+
=
1
(17.3a)

Similarly for memberCD , the shear force is given by
3
H

3
CD DC
MM
H
h
+
=    (17.3b)

Now, the required third equation is obtained by considering the equilibrium of
member , BC

13
00
X
FH =? +
?
H=

0 =
+
+
+
h
M M
h
M M
DC CD AB BA
(17.4)

Substituting the values of beam end moments from equation (17.1) in equations
(17.2a), (17.2b) and (17.4), we get three simultaneous equations in three
unknowns
C B
? ? , and , solving which joint rotations and translations are
evaluated.
?

Knowing joint rotations and translations, beam end moments are calculated from
slope-deflection equations. The complete procedure is explained with a few
numerical examples.
Example 17.1
Analyse the rigid frame as shown in Fig. 17.3a. Assume EI to be constant for all
members. Draw bending moment diagram and sketch qualitative elastic curve.

Solution
In the given problem, joints B and  rotate and also translate by an amount C ? .
Hence, in this problem we have three unknown displacements (two rotations and
one translation) to be evaluated. Considering the kinematically determinate
structure, fixed end moments are evaluated. Thus,

. 0 ; 0 ; . 10 ; . 10 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M        (1)

The ends A and  are fixed. Hence, D . 0 = =
D A
? ? Joints B and  translate by
the same amount . Hence, chord to the elastic curve
C
? ' AB and ' rotates by an
amount (see Fig. 17.3b)
DC

3
?
- = =
CD AB
? ?       (2)

Chords of the elastic curve ' AB and ' rotate in the clockwise direction;
hence
DC
AB
? and
CD
? are taken as negative.

Page 5

Instructional Objectives
After reading this chapter the student will be able to
1. Derive slope-deflection equations for the frames undergoing sidesway.
2. Analyse plane frames undergoing sidesway.
3, Draw shear force and bending moment diagrams.
4. Sketch deflected shape of the plane frame not restrained against sidesway.

17.1 Introduction
In this lesson, slope-deflection equations are applied to analyse statically
indeterminate frames undergoing sidesway. As stated earlier, the axial
deformation of beams and columns are small and are neglected in the analysis.
In the previous lesson, it was observed that sidesway in a frame will not occur if

1. They are restrained against sidesway.

In general loading will never be symmetrical. Hence one could not avoid
sidesway in frames.

For example, consider the frame of Fig. 17.1. In this case the frame is
moments  and  are not equal. If is greater than , then . In
BC
M
CB
M b a
CB BC
M M >

such a case joint B andC are displaced toward right as shown in the figure by an
unknown amount ? . Hence we have three unknown displacements in this frame:
rotations
C B
? ? , and the linear displacement ?. The unknown joint rotations
B
? and
C
? are related to joint moments by the moment equilibrium equations.
Similarly, when unknown linear displacement occurs, one needs to consider
force-equilibrium equations. While applying slope-deflection equation to columns
in the above frame, one must consider the column rotation
?
?
?
?
?
? ?
=
h
? as
unknowns. It is observed that in the columnAB , the end B undergoes a linear
displacement with respect to end ? A. Hence the slope-deflection equation for
column AB is similar to the one for beam undergoing support settlement.
However, in this case is unknown. For each of the members we can write the
following slope-deflection equations.
?

[
AB B A
F
AB AB
] M M ? ? ? 3 2
2
- + + =
h
EI
where
h
AB
?
- = ?

AB
? is assumed to be negative as the chord to the elastic curve rotates in the
clockwise directions.

[]
AB A B
F
BA BA
h
EI
M M ? ? ? 3 2
2
- + + =
[]
C B
F
BC BC
h
EI
M M ? ? + + = 2
2

[]
B C
F
CB CB
h
EI
M M ? ? + + = 2
2

[]
CD D C
F
CD CD
h
EI
M M ? ? ? 3 2
2
- + + =
h
CD
?
- = ?
[
CD C D
F
DC DC
h
EI
M M ? ? ? 3 2
2
- + + = ]   (17.1)

As there are three unknowns (
C B
? ? , and ?), three equations are required to
evaluate them. Two equations are obtained by considering the moment
equilibrium of joint B and C respectively.

0 0 = + ? =
?
BC
BA B
M M M   (17.2a)

0 0 = + ? =
?
CD
CB C
M M M  (17.2b)

Now consider free body diagram of the frame as shown in Fig. 17.2. The
horizontal shear force acting at A and B of the column AB is given by

h
M M
H
AB BA
+
=
1
(17.3a)

Similarly for memberCD , the shear force is given by
3
H

3
CD DC
MM
H
h
+
=    (17.3b)

Now, the required third equation is obtained by considering the equilibrium of
member , BC

13
00
X
FH =? +
?
H=

0 =
+
+
+
h
M M
h
M M
DC CD AB BA
(17.4)

Substituting the values of beam end moments from equation (17.1) in equations
(17.2a), (17.2b) and (17.4), we get three simultaneous equations in three
unknowns
C B
? ? , and , solving which joint rotations and translations are
evaluated.
?

Knowing joint rotations and translations, beam end moments are calculated from
slope-deflection equations. The complete procedure is explained with a few
numerical examples.
Example 17.1
Analyse the rigid frame as shown in Fig. 17.3a. Assume EI to be constant for all
members. Draw bending moment diagram and sketch qualitative elastic curve.

Solution
In the given problem, joints B and  rotate and also translate by an amount C ? .
Hence, in this problem we have three unknown displacements (two rotations and
one translation) to be evaluated. Considering the kinematically determinate
structure, fixed end moments are evaluated. Thus,

. 0 ; 0 ; . 10 ; . 10 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M        (1)

The ends A and  are fixed. Hence, D . 0 = =
D A
? ? Joints B and  translate by
the same amount . Hence, chord to the elastic curve
C
? ' AB and ' rotates by an
amount (see Fig. 17.3b)
DC

3
?
- = =
CD AB
? ?       (2)

Chords of the elastic curve ' AB and ' rotate in the clockwise direction;
hence
DC
AB
? and
CD
? are taken as negative.

Now, writing the slope-deflection equations for the six beam end moments,

[]
AB B A
F
AB AB
EI
M M ? ? ? 3 2
3
2
- + + =

.
3
; 0 ; 0
?
- = = =
AB A
F
AB
M ? ?

? + = EI EI M
B AB
3
2
3
2
?

? + = EI EI M
B BA
3
2
3
4
?

C B BC
EI EI M ? ?
2
1
10 + + =

C B CB
EI EI M ? ? + + - =
2
1
10

? + = EI EI M
C CD
3
2
3
4
?

```
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## Structural Analysis

30 videos|122 docs|28 tests

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