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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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FAQs on The Slope Deflection Method: Frames with Sidesway - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the slope deflection method? |  |
Ans. The slope deflection method is a structural analysis technique used to determine the bending moments and rotations at the ends of individual members in a structural frame. It is based on the principles of equilibrium and compatibility and is often used to analyze frames with sidesway.
| 2. How does the slope deflection method handle frames with sidesway? | |
Ans. The slope deflection method can handle frames with sidesway by considering the additional lateral displacements and rotations that occur due to sidesway. These additional displacements and rotations are incorporated into the analysis by introducing additional unknowns, known as sidesway factors, which represent the relative lateral displacements and rotations of the ends of the members.
| 3. What are the advantages of using the slope deflection method for analyzing frames with sidesway? | |
Ans. The slope deflection method offers several advantages for analyzing frames with sidesway. Firstly, it provides a more accurate representation of the actual behavior of the structure compared to other simpler methods. Secondly, it allows for the consideration of lateral displacements and rotations, which are important in determining the structural response under lateral loads. Lastly, it can handle frames with complex geometries and loadings, making it a versatile method for structural analysis.
| 4. Are there any limitations or assumptions associated with the slope deflection method for frames with sidesway? | |
Ans. Yes, there are some limitations and assumptions associated with the slope deflection method for frames with sidesway. One major assumption is that the structure remains within the elastic range and that the behavior of the members is linearly elastic. Additionally, the method assumes that the lateral displacements and rotations are small, so it may not be suitable for highly flexible or highly deforming structures. It also assumes that the frame is statically determinate, meaning that the number of unknowns can be determined solely from the number of equilibrium equations.
| 5. How can the slope deflection method be applied to solve practical engineering problems? | |
Ans. The slope deflection method can be applied to solve practical engineering problems by following a systematic step-by-step approach. Firstly, the structural frame is divided into individual members, and the member properties such as length, material properties, and cross-sectional properties are determined. Secondly, the external loads and support conditions are identified. Then, the equilibrium equations are set up for each joint, taking into account the additional unknowns for sidesway. Finally, the unknowns, including the bending moments and rotations, are solved using the slope deflection equations and boundary conditions. The resulting values can be used to assess the structural behavior, design members, and ensure structural stability.
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