The Slope Deflection Method: Beams - 2 Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : The Slope Deflection Method: Beams - 2 Civil Engineering (CE) Notes | EduRev

 Page 1


0 = +
BC BA
M M    (6) 
 
Substituting the values of  and  in equation (6), 
BA
M
BC
M
 
  0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying, 
 
         (7) 
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
 
Also, the support C is simply supported and hence, 0 =
CB
M 
 
    EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
 
       (8) 
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
 
We have two unknowns 
B
? and 
C
? and there are two equations in 
B
? and
C
? . 
Solving equations (7) and (8) 
 
    radians 
3
10 4286 . 0
-
× - =
B
?
 
      radians    (9)  
3
10 7143 . 1
-
× =
C
?
 
Substituting the values of 
B
? ,
C
? and EI in slope-deflection equations, 
 
    kN.m 285 . 82 =
AB
M 
 
    kN.m 570 . 68 =
BA
M 
 
    kN.m 573 . 68 - =
BC
M 
 
    kN.m 0 =
CB
M    (10) 
 
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d) 
 
Page 2


0 = +
BC BA
M M    (6) 
 
Substituting the values of  and  in equation (6), 
BA
M
BC
M
 
  0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying, 
 
         (7) 
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
 
Also, the support C is simply supported and hence, 0 =
CB
M 
 
    EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
 
       (8) 
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
 
We have two unknowns 
B
? and 
C
? and there are two equations in 
B
? and
C
? . 
Solving equations (7) and (8) 
 
    radians 
3
10 4286 . 0
-
× - =
B
?
 
      radians    (9)  
3
10 7143 . 1
-
× =
C
?
 
Substituting the values of 
B
? ,
C
? and EI in slope-deflection equations, 
 
    kN.m 285 . 82 =
AB
M 
 
    kN.m 570 . 68 =
BA
M 
 
    kN.m 573 . 68 - =
BC
M 
 
    kN.m 0 =
CB
M    (10) 
 
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d) 
 
 
 
In beamAB , 
 
    ,  0 =
? B
M ) ( kN 171 . 30 ? =
A
R
 
     ) ( kN 171 . 30 ? - =
BL
R
 
     ) ( kN 714 . 13 ? - =
BR
R
 
     ) ( kN 714 . 13 ? =
C
R
 
The shear force and bending moment diagram is shown in Fig.15.2e and elastic 
curve is shown in Fig.15.2f. 
 
 
 
 
Page 3


0 = +
BC BA
M M    (6) 
 
Substituting the values of  and  in equation (6), 
BA
M
BC
M
 
  0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying, 
 
         (7) 
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
 
Also, the support C is simply supported and hence, 0 =
CB
M 
 
    EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
 
       (8) 
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
 
We have two unknowns 
B
? and 
C
? and there are two equations in 
B
? and
C
? . 
Solving equations (7) and (8) 
 
    radians 
3
10 4286 . 0
-
× - =
B
?
 
      radians    (9)  
3
10 7143 . 1
-
× =
C
?
 
Substituting the values of 
B
? ,
C
? and EI in slope-deflection equations, 
 
    kN.m 285 . 82 =
AB
M 
 
    kN.m 570 . 68 =
BA
M 
 
    kN.m 573 . 68 - =
BC
M 
 
    kN.m 0 =
CB
M    (10) 
 
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d) 
 
 
 
In beamAB , 
 
    ,  0 =
? B
M ) ( kN 171 . 30 ? =
A
R
 
     ) ( kN 171 . 30 ? - =
BL
R
 
     ) ( kN 714 . 13 ? - =
BR
R
 
     ) ( kN 714 . 13 ? =
C
R
 
The shear force and bending moment diagram is shown in Fig.15.2e and elastic 
curve is shown in Fig.15.2f. 
 
 
 
 
 
 
 
 
Example 15.2 
A continuous beam  is carrying a uniformly distributed load of 5 kN/m as 
shown in Fig.15.3a. Compute reactions and draw shear force and bending 
moment diagram due to following support settlements. 
ABCD
 
 
Page 4


0 = +
BC BA
M M    (6) 
 
Substituting the values of  and  in equation (6), 
BA
M
BC
M
 
  0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying, 
 
         (7) 
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
 
Also, the support C is simply supported and hence, 0 =
CB
M 
 
    EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
 
       (8) 
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
 
We have two unknowns 
B
? and 
C
? and there are two equations in 
B
? and
C
? . 
Solving equations (7) and (8) 
 
    radians 
3
10 4286 . 0
-
× - =
B
?
 
      radians    (9)  
3
10 7143 . 1
-
× =
C
?
 
Substituting the values of 
B
? ,
C
? and EI in slope-deflection equations, 
 
    kN.m 285 . 82 =
AB
M 
 
    kN.m 570 . 68 =
BA
M 
 
    kN.m 573 . 68 - =
BC
M 
 
    kN.m 0 =
CB
M    (10) 
 
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d) 
 
 
 
In beamAB , 
 
    ,  0 =
? B
M ) ( kN 171 . 30 ? =
A
R
 
     ) ( kN 171 . 30 ? - =
BL
R
 
     ) ( kN 714 . 13 ? - =
BR
R
 
     ) ( kN 714 . 13 ? =
C
R
 
The shear force and bending moment diagram is shown in Fig.15.2e and elastic 
curve is shown in Fig.15.2f. 
 
 
 
 
 
 
 
 
Example 15.2 
A continuous beam  is carrying a uniformly distributed load of 5 kN/m as 
shown in Fig.15.3a. Compute reactions and draw shear force and bending 
moment diagram due to following support settlements. 
ABCD
 
 
Support B   0.005m vertically downwards 
Support C    0.01 m vertically downwards 
Assume E =200 GPa,  
4 3
10 35 . 1 m I
-
× =
 
 
 
In the above continuous beam, four rotations
A
? ,
B
? ,
C
? and 
D
? are to be 
evaluated. One equilibrium equation can be written at each support.Hence, 
solving the four equilibrium equations, the rotations are evaluated and hence the 
moments from slope-deflection equations. Now consider the kinematically 
restrained beam as shown in Fig.15.3b. 
 
Referring to standard tables the fixed end moments may be evaluated .Otherwise 
one could obtain fixed end moments from force method of analysis. The fixed 
end moments in the present case are (vide fig.15.3b) 
 
 
 
    kN.m 667 . 41 =
F
AB
M
 
   (clockwise) kN.m 667 . 41 - =
F
BA
M
 
   (counterclockwise) kN.m 667 . 41 =
F
BC
M
 
    (clockwise) kN.m 667 . 41 - =
F
CB
M
 
 
Page 5


0 = +
BC BA
M M    (6) 
 
Substituting the values of  and  in equation (6), 
BA
M
BC
M
 
  0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying, 
 
         (7) 
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
 
Also, the support C is simply supported and hence, 0 =
CB
M 
 
    EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
 
       (8) 
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
 
We have two unknowns 
B
? and 
C
? and there are two equations in 
B
? and
C
? . 
Solving equations (7) and (8) 
 
    radians 
3
10 4286 . 0
-
× - =
B
?
 
      radians    (9)  
3
10 7143 . 1
-
× =
C
?
 
Substituting the values of 
B
? ,
C
? and EI in slope-deflection equations, 
 
    kN.m 285 . 82 =
AB
M 
 
    kN.m 570 . 68 =
BA
M 
 
    kN.m 573 . 68 - =
BC
M 
 
    kN.m 0 =
CB
M    (10) 
 
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d) 
 
 
 
In beamAB , 
 
    ,  0 =
? B
M ) ( kN 171 . 30 ? =
A
R
 
     ) ( kN 171 . 30 ? - =
BL
R
 
     ) ( kN 714 . 13 ? - =
BR
R
 
     ) ( kN 714 . 13 ? =
C
R
 
The shear force and bending moment diagram is shown in Fig.15.2e and elastic 
curve is shown in Fig.15.2f. 
 
 
 
 
 
 
 
 
Example 15.2 
A continuous beam  is carrying a uniformly distributed load of 5 kN/m as 
shown in Fig.15.3a. Compute reactions and draw shear force and bending 
moment diagram due to following support settlements. 
ABCD
 
 
Support B   0.005m vertically downwards 
Support C    0.01 m vertically downwards 
Assume E =200 GPa,  
4 3
10 35 . 1 m I
-
× =
 
 
 
In the above continuous beam, four rotations
A
? ,
B
? ,
C
? and 
D
? are to be 
evaluated. One equilibrium equation can be written at each support.Hence, 
solving the four equilibrium equations, the rotations are evaluated and hence the 
moments from slope-deflection equations. Now consider the kinematically 
restrained beam as shown in Fig.15.3b. 
 
Referring to standard tables the fixed end moments may be evaluated .Otherwise 
one could obtain fixed end moments from force method of analysis. The fixed 
end moments in the present case are (vide fig.15.3b) 
 
 
 
    kN.m 667 . 41 =
F
AB
M
 
   (clockwise) kN.m 667 . 41 - =
F
BA
M
 
   (counterclockwise) kN.m 667 . 41 =
F
BC
M
 
    (clockwise) kN.m 667 . 41 - =
F
CB
M
 
 
    (counterclockwise) kN.m 667 . 41 =
F
CD
M
 
   (clockwise)    (1) kN.m 667 . 41 - =
F
DC
M
 
In the next step, write slope-deflection equations for each span. In the 
spanAB ,B is below A and hence the chord joining B A ' rotates in the clockwise 
direction (see Fig.15.3c) 
 
 
 
0005 . 0
10
005 . 0 0
- =
-
=
AB
? radians (negative as the chord B A ' rotates in the 
clockwise direction from the original direction) 
 
0005 . 0 - =
BC
? radians (negative as the chord C B ' ' rotates in the clockwise 
direction) 
 
001 . 0
10
01 . 0
= =
CD
? radians (positive as the chord D C ' rotates in the counter 
clockwise direction from the original direction)     (2) 
 
Now, writing the expressions for the span end moments, for each of the spans,  
 
  ) 0005 . 0 2 ( 2 . 0 667 . 41 + + + =
B A AB
EI M ? ? 
  ) 0005 . 0 2 ( 2 . 0 667 . 41 + + + - =
A B BA
EI M ? ? 
 
  ) 0005 . 0 2 ( 2 . 0 667 . 41 + + + =
C B BC
EI M ? ? 
  ) 0005 . 0 2 ( 2 . 0 667 . 41 + + + - =
B C CB
EI M ? ? 
 
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