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

## 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)

3
10 4286 . 0
-
× - =
B
?

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)

3
10 4286 . 0
-
× - =
B
?

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)

3
10 4286 . 0
-
× - =
B
?

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)

3
10 4286 . 0
-
× - =
B
?

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)

3
10 4286 . 0
-
× - =
B
?

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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## Structural Analysis

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