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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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FAQs on The Slope Deflection Method: Beams - 2 - Structural Analysis - Civil Engineering (CE)
| 1. What is the slope deflection method in beam analysis? |  |
Ans. The slope deflection method is a structural analysis technique used to determine the bending moments, shear forces, and deflections of beams subjected to loads. It considers the actual rotation at each end of a member, rather than assuming fixed or hinged connections.
| 2. How does the slope deflection method work? | |
Ans. The slope deflection method works by considering the equilibrium of forces and moments at each joint of a beam. It involves calculating the slope (rotation) and deflection at each joint using a set of equilibrium equations derived from the beam's geometry, material properties, and loading conditions. These calculations are repeated for each joint until a system of simultaneous equations is formed and solved to determine the unknown reactions, moments, and shears in the beam.
| 3. What are the advantages of using the slope deflection method? | |
Ans. The slope deflection method offers several advantages in beam analysis. Firstly, it provides a more accurate representation of the actual behavior of the beam by considering the rotational stiffness of connections. Secondly, it allows for the analysis of continuous beams with multiple supports and variable cross-sections. Additionally, the method can handle complex loading conditions and is applicable to both statically determinate and indeterminate structures.
| 4. Are there any limitations or assumptions associated with the slope deflection method? | |
Ans. Yes, there are certain limitations and assumptions associated with the slope deflection method. One major assumption is that the beam is made of a linearly elastic material, meaning it behaves elastically within its proportional limit. Another assumption is that the beam is initially straight and its deflections are small. Additionally, the method assumes that the structure is statically determinate or can be reduced to a determinate system through the use of virtual work principles.
| 5. How is the slope deflection method different from other beam analysis methods? | |
Ans. The slope deflection method differs from other beam analysis methods, such as the moment distribution method or the matrix stiffness method, in its approach to modeling joint rotations. While the moment distribution method assumes fixed or hinged connections, the slope deflection method considers the actual rotation at each joint. This makes it more accurate for analyzing beams with semi-rigid or flexible connections. The matrix stiffness method, on the other hand, uses matrix algebra to solve for unknown displacements and forces, whereas the slope deflection method relies on a series of equilibrium equations.
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