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# The Slope Deflection Method: Frames Without Sidesway - 1 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : The Slope Deflection Method: Frames Without Sidesway - 1 Civil Engineering (CE) Notes | EduRev

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Instructional Objectives
After reading this chapter the student will be able to
1. State whether plane frames are restrained against sidesway or not.
2. Able to analyse plane frames restrained against sidesway by slope-deflection
equations.
3. Draw bending moment and shear force diagrams for the plane frame.
4. Sketch the deflected shape of the plane frame.

16.1 Introduction
In this lesson, slope deflection equations are applied to solve the statically
indeterminate frames without sidesway. In frames axial deformations are much
smaller than the bending deformations and are neglected in the analysis. With
this assumption the frames shown in Fig 16.1 will not sidesway. i.e. the frames
will not be displaced to the right or left. The frames shown in Fig 16.1(a) and Fig
16.1(b) are properly restrained against sidesway. For example in Fig 16.1(a) the
joint can’t move to the right or left without support A also moving .This is true
also for joint . Frames shown in Fig 16.1 (c) and (d) are not restrained against
sidesway. However the frames are symmetrical in geometry and in loading and
hence these will not sidesway. In general, frames do not sidesway if
D

1) They are restrained against sidesway.

Page 2

Instructional Objectives
After reading this chapter the student will be able to
1. State whether plane frames are restrained against sidesway or not.
2. Able to analyse plane frames restrained against sidesway by slope-deflection
equations.
3. Draw bending moment and shear force diagrams for the plane frame.
4. Sketch the deflected shape of the plane frame.

16.1 Introduction
In this lesson, slope deflection equations are applied to solve the statically
indeterminate frames without sidesway. In frames axial deformations are much
smaller than the bending deformations and are neglected in the analysis. With
this assumption the frames shown in Fig 16.1 will not sidesway. i.e. the frames
will not be displaced to the right or left. The frames shown in Fig 16.1(a) and Fig
16.1(b) are properly restrained against sidesway. For example in Fig 16.1(a) the
joint can’t move to the right or left without support A also moving .This is true
also for joint . Frames shown in Fig 16.1 (c) and (d) are not restrained against
sidesway. However the frames are symmetrical in geometry and in loading and
hence these will not sidesway. In general, frames do not sidesway if
D

1) They are restrained against sidesway.

Page 3

Instructional Objectives
After reading this chapter the student will be able to
1. State whether plane frames are restrained against sidesway or not.
2. Able to analyse plane frames restrained against sidesway by slope-deflection
equations.
3. Draw bending moment and shear force diagrams for the plane frame.
4. Sketch the deflected shape of the plane frame.

16.1 Introduction
In this lesson, slope deflection equations are applied to solve the statically
indeterminate frames without sidesway. In frames axial deformations are much
smaller than the bending deformations and are neglected in the analysis. With
this assumption the frames shown in Fig 16.1 will not sidesway. i.e. the frames
will not be displaced to the right or left. The frames shown in Fig 16.1(a) and Fig
16.1(b) are properly restrained against sidesway. For example in Fig 16.1(a) the
joint can’t move to the right or left without support A also moving .This is true
also for joint . Frames shown in Fig 16.1 (c) and (d) are not restrained against
sidesway. However the frames are symmetrical in geometry and in loading and
hence these will not sidesway. In general, frames do not sidesway if
D

1) They are restrained against sidesway.

For the frames shown in Fig 16.1, the angle ? in slope-deflection equation is
zero. Hence the analysis of such rigid frames by slope deflection equation
essentially follows the same steps as that of continuous beams without support
settlements. However, there is a small difference. In the case of continuous
beam, at a joint only two members meet. Whereas in the case of rigid frames two
or more than two members meet at a joint. At joint  in the frame shown in Fig
16.1(d) three members meet. Now consider the free body diagram of joint C as
shown in fig 16.2 .The equilibrium equation at joint C is
C

?
= 0
C
M ?         0 = + +
CD CE CB
M M M

Page 4

Instructional Objectives
After reading this chapter the student will be able to
1. State whether plane frames are restrained against sidesway or not.
2. Able to analyse plane frames restrained against sidesway by slope-deflection
equations.
3. Draw bending moment and shear force diagrams for the plane frame.
4. Sketch the deflected shape of the plane frame.

16.1 Introduction
In this lesson, slope deflection equations are applied to solve the statically
indeterminate frames without sidesway. In frames axial deformations are much
smaller than the bending deformations and are neglected in the analysis. With
this assumption the frames shown in Fig 16.1 will not sidesway. i.e. the frames
will not be displaced to the right or left. The frames shown in Fig 16.1(a) and Fig
16.1(b) are properly restrained against sidesway. For example in Fig 16.1(a) the
joint can’t move to the right or left without support A also moving .This is true
also for joint . Frames shown in Fig 16.1 (c) and (d) are not restrained against
sidesway. However the frames are symmetrical in geometry and in loading and
hence these will not sidesway. In general, frames do not sidesway if
D

1) They are restrained against sidesway.

For the frames shown in Fig 16.1, the angle ? in slope-deflection equation is
zero. Hence the analysis of such rigid frames by slope deflection equation
essentially follows the same steps as that of continuous beams without support
settlements. However, there is a small difference. In the case of continuous
beam, at a joint only two members meet. Whereas in the case of rigid frames two
or more than two members meet at a joint. At joint  in the frame shown in Fig
16.1(d) three members meet. Now consider the free body diagram of joint C as
shown in fig 16.2 .The equilibrium equation at joint C is
C

?
= 0
C
M ?         0 = + +
CD CE CB
M M M

At each joint there is only one unknown as all the ends of members meeting at a
joint rotate by the same amount. One would write as many equilibrium equations
as the no of unknowns, and solving these equations joint rotations are evaluated.
Substituting joint rotations in the slope–deflection equations member end
moments are calculated. The whole procedure is illustrated by few examples.
Frames undergoing sidesway will be considered in next lesson.

Example 16.1
Analyse the rigid frame shown in Fig 16.3 (a). Assume EI to be constant for all
the members. Draw bending moment diagram and also sketch the elastic curve.

Solution
In this problem only one rotation needs to be determined i. e. .
B
? Thus the
required equations to evaluate
B
? is obtained by considering the equilibrium of
joint B . The moment in the cantilever portion is known. Hence this moment is
applied on frame as shown in Fig 16.3 (b).  Now, calculate the fixed-end
moments by fixing the support B (vide Fig 16.3 c). Thus

Page 5

Instructional Objectives
After reading this chapter the student will be able to
1. State whether plane frames are restrained against sidesway or not.
2. Able to analyse plane frames restrained against sidesway by slope-deflection
equations.
3. Draw bending moment and shear force diagrams for the plane frame.
4. Sketch the deflected shape of the plane frame.

16.1 Introduction
In this lesson, slope deflection equations are applied to solve the statically
indeterminate frames without sidesway. In frames axial deformations are much
smaller than the bending deformations and are neglected in the analysis. With
this assumption the frames shown in Fig 16.1 will not sidesway. i.e. the frames
will not be displaced to the right or left. The frames shown in Fig 16.1(a) and Fig
16.1(b) are properly restrained against sidesway. For example in Fig 16.1(a) the
joint can’t move to the right or left without support A also moving .This is true
also for joint . Frames shown in Fig 16.1 (c) and (d) are not restrained against
sidesway. However the frames are symmetrical in geometry and in loading and
hence these will not sidesway. In general, frames do not sidesway if
D

1) They are restrained against sidesway.

For the frames shown in Fig 16.1, the angle ? in slope-deflection equation is
zero. Hence the analysis of such rigid frames by slope deflection equation
essentially follows the same steps as that of continuous beams without support
settlements. However, there is a small difference. In the case of continuous
beam, at a joint only two members meet. Whereas in the case of rigid frames two
or more than two members meet at a joint. At joint  in the frame shown in Fig
16.1(d) three members meet. Now consider the free body diagram of joint C as
shown in fig 16.2 .The equilibrium equation at joint C is
C

?
= 0
C
M ?         0 = + +
CD CE CB
M M M

At each joint there is only one unknown as all the ends of members meeting at a
joint rotate by the same amount. One would write as many equilibrium equations
as the no of unknowns, and solving these equations joint rotations are evaluated.
Substituting joint rotations in the slope–deflection equations member end
moments are calculated. The whole procedure is illustrated by few examples.
Frames undergoing sidesway will be considered in next lesson.

Example 16.1
Analyse the rigid frame shown in Fig 16.3 (a). Assume EI to be constant for all
the members. Draw bending moment diagram and also sketch the elastic curve.

Solution
In this problem only one rotation needs to be determined i. e. .
B
? Thus the
required equations to evaluate
B
? is obtained by considering the equilibrium of
joint B . The moment in the cantilever portion is known. Hence this moment is
applied on frame as shown in Fig 16.3 (b).  Now, calculate the fixed-end
moments by fixing the support B (vide Fig 16.3 c). Thus

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## Structural Analysis

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