The Slope Deflection Method: Frames with Sidesway - 4 Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : The Slope Deflection Method: Frames with Sidesway - 4 Civil Engineering (CE) Notes | EduRev

 Page 1


curve " AB rotates by an angle 
AB
? ,  rotates by " "C B
BC
? and  rotates by DC
CD
? as shown in figure. Due to symmetry, 
AB CD
? ? = . From the geometry of the 
figure, 
 
AB AB
AB
L L
BB
1
" ?
- = = ? 
 
But  
a cos
1
?
= ? 
Thus,  
5 cos
?
- =
?
- =
a
?
AB
AB
L
      
 
5
?
- =
CD
? 
  
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
   (1) 
 
We have three independent unknowns for this problem 
C B
? ? , and . The ends ?
A and  are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are, 
 
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M   
 
Now, writing the slope-deflection equations for the six beam end moments, 
 
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- = 
 
? + = EI EI M
B AB
471 . 0 784 . 0 ? 
? + = EI EI M
B BA
471 . 0 568 . 1 ? 
 
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
   
? + = EI EI M
C CD
471 . 0 568 . 1 ? 
? + = EI EI M
C DC
471 . 0 784 . 0 ?   (2) 
 
Now, considering the joint equilibrium of B andC , yields 
 
 
Page 2


curve " AB rotates by an angle 
AB
? ,  rotates by " "C B
BC
? and  rotates by DC
CD
? as shown in figure. Due to symmetry, 
AB CD
? ? = . From the geometry of the 
figure, 
 
AB AB
AB
L L
BB
1
" ?
- = = ? 
 
But  
a cos
1
?
= ? 
Thus,  
5 cos
?
- =
?
- =
a
?
AB
AB
L
      
 
5
?
- =
CD
? 
  
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
   (1) 
 
We have three independent unknowns for this problem 
C B
? ? , and . The ends ?
A and  are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are, 
 
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M   
 
Now, writing the slope-deflection equations for the six beam end moments, 
 
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- = 
 
? + = EI EI M
B AB
471 . 0 784 . 0 ? 
? + = EI EI M
B BA
471 . 0 568 . 1 ? 
 
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
   
? + = EI EI M
C CD
471 . 0 568 . 1 ? 
? + = EI EI M
C DC
471 . 0 784 . 0 ?   (2) 
 
Now, considering the joint equilibrium of B andC , yields 
 
 
0 0 = + ? =
?
BC
BA B
M M M       
     
5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI
C B
? ?     (3) 
 
0 0 = + ? =
?
CD
CB C
M M M              
  
5 . 2 129 . 0 568 . 3 = ? - + EI EI EI
B C
? ?      (4) 
 
 
 
Shear equation for 
Column AB 
0 ) 1 ( 5
1 1
= + - - V M M H
BA AB
    (5) 
  
Column  CD
0 ) 1 ( 5
2 2
= + - - V M M H
DC CD
     (6) 
           
Beam  BC
0 10 2 0
1
= - - - =
?
CB BC C
M M V M     (7) 
 
Page 3


curve " AB rotates by an angle 
AB
? ,  rotates by " "C B
BC
? and  rotates by DC
CD
? as shown in figure. Due to symmetry, 
AB CD
? ? = . From the geometry of the 
figure, 
 
AB AB
AB
L L
BB
1
" ?
- = = ? 
 
But  
a cos
1
?
= ? 
Thus,  
5 cos
?
- =
?
- =
a
?
AB
AB
L
      
 
5
?
- =
CD
? 
  
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
   (1) 
 
We have three independent unknowns for this problem 
C B
? ? , and . The ends ?
A and  are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are, 
 
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M   
 
Now, writing the slope-deflection equations for the six beam end moments, 
 
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- = 
 
? + = EI EI M
B AB
471 . 0 784 . 0 ? 
? + = EI EI M
B BA
471 . 0 568 . 1 ? 
 
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
   
? + = EI EI M
C CD
471 . 0 568 . 1 ? 
? + = EI EI M
C DC
471 . 0 784 . 0 ?   (2) 
 
Now, considering the joint equilibrium of B andC , yields 
 
 
0 0 = + ? =
?
BC
BA B
M M M       
     
5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI
C B
? ?     (3) 
 
0 0 = + ? =
?
CD
CB C
M M M              
  
5 . 2 129 . 0 568 . 3 = ? - + EI EI EI
B C
? ?      (4) 
 
 
 
Shear equation for 
Column AB 
0 ) 1 ( 5
1 1
= + - - V M M H
BA AB
    (5) 
  
Column  CD
0 ) 1 ( 5
2 2
= + - - V M M H
DC CD
     (6) 
           
Beam  BC
0 10 2 0
1
= - - - =
?
CB BC C
M M V M     (7) 
 
?
= + = 5 0
2 1
H H F
X
   (8) 
 
?
= - - = 0 10 0
2 1
V V F
Y
   (9) 
 
 
From equation (7),  
2
10
1
+ +
=
CB BC
M M
V       
 
From equation (8),  
2 1
5 H H - =
 
From equation (9), 10
2
10
10
1 2
-
+ +
= - =
CB BC
M M
V V 
 
Substituting the values of and in equations (5) and (6), 
1 1
,H V
2
V
 
0 2 2 10 60
2
= + + - - -
CB BC BA AB
M M M M H   (10) 
0 2 2 10 10
2
= + + - - + -
CB BC DC CD
M M M M H   (11) 
 
Eliminating  in equation (10) and (11), 
2
H
 
25 = - - + + +
CB BC DC CD BA AB
M M M M M M   (12) 
 
Substituting the values of  in (12) we get the required third 
equation. Thus, 
DC CD BA AB
M M M M , , ,
 
+ ? + EI EI
B
471 . 0 784 . 0 ? + ? + EI EI
B
471 . 0 568 . 1 ? + ? + EI EI
C
471 . 0 568 . 1 ? 
? + EI EI
C
471 . 0 784 . 0 ? -( ? - + + EI EI EI
C B
6 . 0 2 5 . 2 ? ? )-
( ? - + + - EI EI EI
C B
6 . 0 2 5 . 2 ? ? ) 25 = 
 
Simplifying, 
 
25 084 . 3 648 . 0 648 . 0 = ? + - - EI EI EI
B C
? ?   (13) 
 
Solving simultaneously equations (3) (4) and (13), yields 
 
205 . 1 ; 741 . 0 = - =
C B
EI EI ? ?     and    204 . 8 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
  
3.28 kN.m
AB
M = 
 
Page 4


curve " AB rotates by an angle 
AB
? ,  rotates by " "C B
BC
? and  rotates by DC
CD
? as shown in figure. Due to symmetry, 
AB CD
? ? = . From the geometry of the 
figure, 
 
AB AB
AB
L L
BB
1
" ?
- = = ? 
 
But  
a cos
1
?
= ? 
Thus,  
5 cos
?
- =
?
- =
a
?
AB
AB
L
      
 
5
?
- =
CD
? 
  
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
   (1) 
 
We have three independent unknowns for this problem 
C B
? ? , and . The ends ?
A and  are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are, 
 
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M   
 
Now, writing the slope-deflection equations for the six beam end moments, 
 
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- = 
 
? + = EI EI M
B AB
471 . 0 784 . 0 ? 
? + = EI EI M
B BA
471 . 0 568 . 1 ? 
 
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ? 
   
? + = EI EI M
C CD
471 . 0 568 . 1 ? 
? + = EI EI M
C DC
471 . 0 784 . 0 ?   (2) 
 
Now, considering the joint equilibrium of B andC , yields 
 
 
0 0 = + ? =
?
BC
BA B
M M M       
     
5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI
C B
? ?     (3) 
 
0 0 = + ? =
?
CD
CB C
M M M              
  
5 . 2 129 . 0 568 . 3 = ? - + EI EI EI
B C
? ?      (4) 
 
 
 
Shear equation for 
Column AB 
0 ) 1 ( 5
1 1
= + - - V M M H
BA AB
    (5) 
  
Column  CD
0 ) 1 ( 5
2 2
= + - - V M M H
DC CD
     (6) 
           
Beam  BC
0 10 2 0
1
= - - - =
?
CB BC C
M M V M     (7) 
 
?
= + = 5 0
2 1
H H F
X
   (8) 
 
?
= - - = 0 10 0
2 1
V V F
Y
   (9) 
 
 
From equation (7),  
2
10
1
+ +
=
CB BC
M M
V       
 
From equation (8),  
2 1
5 H H - =
 
From equation (9), 10
2
10
10
1 2
-
+ +
= - =
CB BC
M M
V V 
 
Substituting the values of and in equations (5) and (6), 
1 1
,H V
2
V
 
0 2 2 10 60
2
= + + - - -
CB BC BA AB
M M M M H   (10) 
0 2 2 10 10
2
= + + - - + -
CB BC DC CD
M M M M H   (11) 
 
Eliminating  in equation (10) and (11), 
2
H
 
25 = - - + + +
CB BC DC CD BA AB
M M M M M M   (12) 
 
Substituting the values of  in (12) we get the required third 
equation. Thus, 
DC CD BA AB
M M M M , , ,
 
+ ? + EI EI
B
471 . 0 784 . 0 ? + ? + EI EI
B
471 . 0 568 . 1 ? + ? + EI EI
C
471 . 0 568 . 1 ? 
? + EI EI
C
471 . 0 784 . 0 ? -( ? - + + EI EI EI
C B
6 . 0 2 5 . 2 ? ? )-
( ? - + + - EI EI EI
C B
6 . 0 2 5 . 2 ? ? ) 25 = 
 
Simplifying, 
 
25 084 . 3 648 . 0 648 . 0 = ? + - - EI EI EI
B C
? ?   (13) 
 
Solving simultaneously equations (3) (4) and (13), yields 
 
205 . 1 ; 741 . 0 = - =
C B
EI EI ? ?     and    204 . 8 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
  
3.28 kN.m
AB
M = 
 
2.70 kN.m
BA
M = 
 
2.70 kN.m
BC
M = - 
 
5.75 kN.m
CB
M = -  
  
5.75 kN.m
CD
M = 
 
4.81 kN.m
DC
M = .      (14) 
 
The bending moment diagram for the frame is shown in Fig. 17.5 d.  
 
 
 
 
Summary 
In this lesson, slope-deflection equations are derived for the plane frame 
undergoing sidesway. Using these equations, plane frames with sidesway are 
analysed. The reactions are calculated from static equilibrium equations. A 
couple of problems are solved to make things clear. In each numerical example, 
the bending moment diagram is drawn and deflected shape is sketched for the 
plane frame. 
 
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