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

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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. 
2. If the frame geometry and the loading are symmetrical. 
 
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 
symmetrical but not the loading. Due to unsymmetrical loading the beam end 
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. 
2. If the frame geometry and the loading are symmetrical. 
 
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 
symmetrical but not the loading. Due to unsymmetrical loading the beam end 
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. 
2. If the frame geometry and the loading are symmetrical. 
 
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 
symmetrical but not the loading. Due to unsymmetrical loading the beam end 
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. 
2. If the frame geometry and the loading are symmetrical. 
 
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 
symmetrical but not the loading. Due to unsymmetrical loading the beam end 
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. 
2. If the frame geometry and the loading are symmetrical. 
 
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 
symmetrical but not the loading. Due to unsymmetrical loading the beam end 
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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