The Slope Deflection Method: Frames with Sidesway - 3 Civil Engineering (CE) Notes | EduRev

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Civil Engineering (CE) : The Slope Deflection Method: Frames with Sidesway - 3 Civil Engineering (CE) Notes | EduRev

 Page 1


Solution: 
 
In this problem rotations and translations at joints B and need to be evaluated. 
Hence, in this problem we have three unknown displacements: two rotations and 
one translation. Fixed end moments are  
C
 
. 0 ; 0 ; 0 ; 0
; . 9 ; . 9
36
9 3 12
= = = =
- = =
× ×
=
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M M M
m kN M m kN M
       (1)   
 
The joints B and  translate by the same amount C ?. Hence, the chord to the 
elastic curve rotates in the clockwise direction as shown in Fig. 17.3b. 
 
6
?
- =
AB
?     
and     
3
?
- =
CD
?       (2) 
 
 
 
Now, writing the slope-deflection equations for six beam end moments, 
 
?
?
?
?
?
? ?
+ + =
2 6
) 2 ( 2
9
B AB
EI
M ? 
 
? + + = EI EI M
B AB
333 . 0 667 . 0 9 ? 
 
? + + - = EI EI M
B BA
333 . 0 333 . 1 9 ? 
 
 
Page 2


Solution: 
 
In this problem rotations and translations at joints B and need to be evaluated. 
Hence, in this problem we have three unknown displacements: two rotations and 
one translation. Fixed end moments are  
C
 
. 0 ; 0 ; 0 ; 0
; . 9 ; . 9
36
9 3 12
= = = =
- = =
× ×
=
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M M M
m kN M m kN M
       (1)   
 
The joints B and  translate by the same amount C ?. Hence, the chord to the 
elastic curve rotates in the clockwise direction as shown in Fig. 17.3b. 
 
6
?
- =
AB
?     
and     
3
?
- =
CD
?       (2) 
 
 
 
Now, writing the slope-deflection equations for six beam end moments, 
 
?
?
?
?
?
? ?
+ + =
2 6
) 2 ( 2
9
B AB
EI
M ? 
 
? + + = EI EI M
B AB
333 . 0 667 . 0 9 ? 
 
? + + - = EI EI M
B BA
333 . 0 333 . 1 9 ? 
 
 
C B BC
EI EI M ? ? 5 . 0 + = 
 
C B CB
EI EI M ? ? + = 5 . 0 
   
? + = EI EI M
C CD
667 . 0 333 . 1 ? 
 
? + = EI EI M
C DC
667 . 0 667 . 0 ?    (3) 
 
Now, consider the joint equilibrium of B andC . 
 
0 0 = + ? =
?
BC
BA B
M M M        (4) 
 
0 0 = + ? =
?
CD
CB C
M M M         (5) 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame. Considering the free body diagram of the member (vide Fig. 
17.4c), 
BC
       
0
2 1
= +H H .   
        (6) 
 
 
 
Page 3


Solution: 
 
In this problem rotations and translations at joints B and need to be evaluated. 
Hence, in this problem we have three unknown displacements: two rotations and 
one translation. Fixed end moments are  
C
 
. 0 ; 0 ; 0 ; 0
; . 9 ; . 9
36
9 3 12
= = = =
- = =
× ×
=
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M M M
m kN M m kN M
       (1)   
 
The joints B and  translate by the same amount C ?. Hence, the chord to the 
elastic curve rotates in the clockwise direction as shown in Fig. 17.3b. 
 
6
?
- =
AB
?     
and     
3
?
- =
CD
?       (2) 
 
 
 
Now, writing the slope-deflection equations for six beam end moments, 
 
?
?
?
?
?
? ?
+ + =
2 6
) 2 ( 2
9
B AB
EI
M ? 
 
? + + = EI EI M
B AB
333 . 0 667 . 0 9 ? 
 
? + + - = EI EI M
B BA
333 . 0 333 . 1 9 ? 
 
 
C B BC
EI EI M ? ? 5 . 0 + = 
 
C B CB
EI EI M ? ? + = 5 . 0 
   
? + = EI EI M
C CD
667 . 0 333 . 1 ? 
 
? + = EI EI M
C DC
667 . 0 667 . 0 ?    (3) 
 
Now, consider the joint equilibrium of B andC . 
 
0 0 = + ? =
?
BC
BA B
M M M        (4) 
 
0 0 = + ? =
?
CD
CB C
M M M         (5) 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame. Considering the free body diagram of the member (vide Fig. 
17.4c), 
BC
       
0
2 1
= +H H .   
        (6) 
 
 
 
The forces  and  are calculated from the free body diagram of column 
1
H
2
H
AB andCD . Thus, 
 
6
6
1
AB BA
M M
H
+
+ - =      
 
and  
 
3
2
DC CD
M M
H
+
=     (7) 
 
Substituting the values of  and  into equation (6) yields, 
1
H
2
H
 
36 2 2 = + + +
DC CD AB BA
M M M M      (8) 
 
 
 
Substituting the beam end moments from equation (3) in equations (4), (5) and 
(8), yields 
 
9 333 . 0 5 . 0 333 . 2 = ? + + EI EI EI
C B
? ?     
        
 
0 667 . 0 5 . 0 333 . 2 = ? + + EI EI EI
B C
? ?     
      
  
36 333 . 3 4 2 = ? + + EI EI EI
C B
? ?      (9) 
 
Solving equations (9), (10) and (11), 
 
88 . 4 ; 76 . 2 - = =
C B
EI EI ? ?     and    00 . 15 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
 
15.835 kN.m (counterclockwise)
AB
M = 
0.325 kN.m(clockwise)
BA
M =- 
0.32 kN.m
BC
M = 
3.50 kN.m
CB
M =-   
3.50 kN.m
CD
M = 
6.75 kN.m
DC
M = . 
 
The bending moment diagram for the frame is shown in Fig. 17.4 d.  
 
Page 4


Solution: 
 
In this problem rotations and translations at joints B and need to be evaluated. 
Hence, in this problem we have three unknown displacements: two rotations and 
one translation. Fixed end moments are  
C
 
. 0 ; 0 ; 0 ; 0
; . 9 ; . 9
36
9 3 12
= = = =
- = =
× ×
=
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M M M
m kN M m kN M
       (1)   
 
The joints B and  translate by the same amount C ?. Hence, the chord to the 
elastic curve rotates in the clockwise direction as shown in Fig. 17.3b. 
 
6
?
- =
AB
?     
and     
3
?
- =
CD
?       (2) 
 
 
 
Now, writing the slope-deflection equations for six beam end moments, 
 
?
?
?
?
?
? ?
+ + =
2 6
) 2 ( 2
9
B AB
EI
M ? 
 
? + + = EI EI M
B AB
333 . 0 667 . 0 9 ? 
 
? + + - = EI EI M
B BA
333 . 0 333 . 1 9 ? 
 
 
C B BC
EI EI M ? ? 5 . 0 + = 
 
C B CB
EI EI M ? ? + = 5 . 0 
   
? + = EI EI M
C CD
667 . 0 333 . 1 ? 
 
? + = EI EI M
C DC
667 . 0 667 . 0 ?    (3) 
 
Now, consider the joint equilibrium of B andC . 
 
0 0 = + ? =
?
BC
BA B
M M M        (4) 
 
0 0 = + ? =
?
CD
CB C
M M M         (5) 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame. Considering the free body diagram of the member (vide Fig. 
17.4c), 
BC
       
0
2 1
= +H H .   
        (6) 
 
 
 
The forces  and  are calculated from the free body diagram of column 
1
H
2
H
AB andCD . Thus, 
 
6
6
1
AB BA
M M
H
+
+ - =      
 
and  
 
3
2
DC CD
M M
H
+
=     (7) 
 
Substituting the values of  and  into equation (6) yields, 
1
H
2
H
 
36 2 2 = + + +
DC CD AB BA
M M M M      (8) 
 
 
 
Substituting the beam end moments from equation (3) in equations (4), (5) and 
(8), yields 
 
9 333 . 0 5 . 0 333 . 2 = ? + + EI EI EI
C B
? ?     
        
 
0 667 . 0 5 . 0 333 . 2 = ? + + EI EI EI
B C
? ?     
      
  
36 333 . 3 4 2 = ? + + EI EI EI
C B
? ?      (9) 
 
Solving equations (9), (10) and (11), 
 
88 . 4 ; 76 . 2 - = =
C B
EI EI ? ?     and    00 . 15 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
 
15.835 kN.m (counterclockwise)
AB
M = 
0.325 kN.m(clockwise)
BA
M =- 
0.32 kN.m
BC
M = 
3.50 kN.m
CB
M =-   
3.50 kN.m
CD
M = 
6.75 kN.m
DC
M = . 
 
The bending moment diagram for the frame is shown in Fig. 17.4 d.  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Page 5


Solution: 
 
In this problem rotations and translations at joints B and need to be evaluated. 
Hence, in this problem we have three unknown displacements: two rotations and 
one translation. Fixed end moments are  
C
 
. 0 ; 0 ; 0 ; 0
; . 9 ; . 9
36
9 3 12
= = = =
- = =
× ×
=
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M M M
m kN M m kN M
       (1)   
 
The joints B and  translate by the same amount C ?. Hence, the chord to the 
elastic curve rotates in the clockwise direction as shown in Fig. 17.3b. 
 
6
?
- =
AB
?     
and     
3
?
- =
CD
?       (2) 
 
 
 
Now, writing the slope-deflection equations for six beam end moments, 
 
?
?
?
?
?
? ?
+ + =
2 6
) 2 ( 2
9
B AB
EI
M ? 
 
? + + = EI EI M
B AB
333 . 0 667 . 0 9 ? 
 
? + + - = EI EI M
B BA
333 . 0 333 . 1 9 ? 
 
 
C B BC
EI EI M ? ? 5 . 0 + = 
 
C B CB
EI EI M ? ? + = 5 . 0 
   
? + = EI EI M
C CD
667 . 0 333 . 1 ? 
 
? + = EI EI M
C DC
667 . 0 667 . 0 ?    (3) 
 
Now, consider the joint equilibrium of B andC . 
 
0 0 = + ? =
?
BC
BA B
M M M        (4) 
 
0 0 = + ? =
?
CD
CB C
M M M         (5) 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame. Considering the free body diagram of the member (vide Fig. 
17.4c), 
BC
       
0
2 1
= +H H .   
        (6) 
 
 
 
The forces  and  are calculated from the free body diagram of column 
1
H
2
H
AB andCD . Thus, 
 
6
6
1
AB BA
M M
H
+
+ - =      
 
and  
 
3
2
DC CD
M M
H
+
=     (7) 
 
Substituting the values of  and  into equation (6) yields, 
1
H
2
H
 
36 2 2 = + + +
DC CD AB BA
M M M M      (8) 
 
 
 
Substituting the beam end moments from equation (3) in equations (4), (5) and 
(8), yields 
 
9 333 . 0 5 . 0 333 . 2 = ? + + EI EI EI
C B
? ?     
        
 
0 667 . 0 5 . 0 333 . 2 = ? + + EI EI EI
B C
? ?     
      
  
36 333 . 3 4 2 = ? + + EI EI EI
C B
? ?      (9) 
 
Solving equations (9), (10) and (11), 
 
88 . 4 ; 76 . 2 - = =
C B
EI EI ? ?     and    00 . 15 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
 
15.835 kN.m (counterclockwise)
AB
M = 
0.325 kN.m(clockwise)
BA
M =- 
0.32 kN.m
BC
M = 
3.50 kN.m
CB
M =-   
3.50 kN.m
CD
M = 
6.75 kN.m
DC
M = . 
 
The bending moment diagram for the frame is shown in Fig. 17.4 d.  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Example 17.3 
Analyse the rigid frame shown in Fig. 17.5 a. Moment of inertia of all the 
members are shown in the figure. Draw bending moment diagram. 
 
 
 
Under the action of external forces, the frame gets deformed as shown in Fig. 
17.5b. In this figure, chord to the elastic curve are shown by dotted line. ' BB is 
perpendicular to AB and is perpendicular to . The chords to the elastic " CC DC
 
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