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Page 1
571 . 0 ; 429 . 0 = =
CD CB
DF DF
b) Calculate fixed end moments due to applied loading.
2
2
12 3 3
9.0 kN.m
6
F
AB
M
××
== ; 9.0 kN.m
F
BA
M =-
0kN.m
F
BC
M = ; 0kN.m
F
CB
M =
0kN.m
F
CD
M = ; 0kN.m
F
DC
M =
c) Prevent sidesway by providing artificial support at C. Carry out moment-
distribution ( Case . .e i A in Fig. 21.5b). The moment-distribution for this case is
shown in Fig. 21.5c.
Page 2
571 . 0 ; 429 . 0 = =
CD CB
DF DF
b) Calculate fixed end moments due to applied loading.
2
2
12 3 3
9.0 kN.m
6
F
AB
M
××
== ; 9.0 kN.m
F
BA
M =-
0kN.m
F
BC
M = ; 0kN.m
F
CB
M =
0kN.m
F
CD
M = ; 0kN.m
F
DC
M =
c) Prevent sidesway by providing artificial support at C. Carry out moment-
distribution ( Case . .e i A in Fig. 21.5b). The moment-distribution for this case is
shown in Fig. 21.5c.
Now calculate horizontal reaction at A and from equations of statics. D
()
1
11.694 3.614
6 7.347 kN
6
A
H
-
=+= ?
()
1
1.154 0.578
0.577 kN
3
D
H
--
==- ?
( ) 12 (7.347 0.577) 5.23 kN R=- - =- ?
d) Moment-distribution for arbitrary sidesway ' ? (case B, Fig. 21.5c)
Calculate fixed end moments for the arbitrary sidesway of
EI
150
' = ? .
6 (2 ) 12 150
( ) 50 kN.m ; 50 kN.m ;
66
F F
AB BA
EI EI
MM
L EI
? =- = × - =+ =+
Page 3
571 . 0 ; 429 . 0 = =
CD CB
DF DF
b) Calculate fixed end moments due to applied loading.
2
2
12 3 3
9.0 kN.m
6
F
AB
M
××
== ; 9.0 kN.m
F
BA
M =-
0kN.m
F
BC
M = ; 0kN.m
F
CB
M =
0kN.m
F
CD
M = ; 0kN.m
F
DC
M =
c) Prevent sidesway by providing artificial support at C. Carry out moment-
distribution ( Case . .e i A in Fig. 21.5b). The moment-distribution for this case is
shown in Fig. 21.5c.
Now calculate horizontal reaction at A and from equations of statics. D
()
1
11.694 3.614
6 7.347 kN
6
A
H
-
=+= ?
()
1
1.154 0.578
0.577 kN
3
D
H
--
==- ?
( ) 12 (7.347 0.577) 5.23 kN R=- - =- ?
d) Moment-distribution for arbitrary sidesway ' ? (case B, Fig. 21.5c)
Calculate fixed end moments for the arbitrary sidesway of
EI
150
' = ? .
6 (2 ) 12 150
( ) 50 kN.m ; 50 kN.m ;
66
F F
AB BA
EI EI
MM
L EI
? =- = × - =+ =+
6( ) 6 150
( ) 100 kN.m ; 100 kN.m ;
33
F F
CD DC
EI EI
MM
L EI
? =- =- × - =+ =+
The moment-distribution for this case is shown in Fig. 21.5d. Using equations of
static equilibrium, calculate reactions and .
2 A
H
2 D
H
) ( 395 . 12
6
457 . 41 911 . 32
2
? =
+
= kN H
A
) ( 952 . 39
3
285 . 73 57 . 46
2
? =
+
= kN H
D
) ( 347 . 52 ) 952 . 39 395 . 12 ( ? - = + - = kN F
e) Final results
Now, the shear condition for the frame is (vide Fig. 21.5b)
Page 4
571 . 0 ; 429 . 0 = =
CD CB
DF DF
b) Calculate fixed end moments due to applied loading.
2
2
12 3 3
9.0 kN.m
6
F
AB
M
××
== ; 9.0 kN.m
F
BA
M =-
0kN.m
F
BC
M = ; 0kN.m
F
CB
M =
0kN.m
F
CD
M = ; 0kN.m
F
DC
M =
c) Prevent sidesway by providing artificial support at C. Carry out moment-
distribution ( Case . .e i A in Fig. 21.5b). The moment-distribution for this case is
shown in Fig. 21.5c.
Now calculate horizontal reaction at A and from equations of statics. D
()
1
11.694 3.614
6 7.347 kN
6
A
H
-
=+= ?
()
1
1.154 0.578
0.577 kN
3
D
H
--
==- ?
( ) 12 (7.347 0.577) 5.23 kN R=- - =- ?
d) Moment-distribution for arbitrary sidesway ' ? (case B, Fig. 21.5c)
Calculate fixed end moments for the arbitrary sidesway of
EI
150
' = ? .
6 (2 ) 12 150
( ) 50 kN.m ; 50 kN.m ;
66
F F
AB BA
EI EI
MM
L EI
? =- = × - =+ =+
6( ) 6 150
( ) 100 kN.m ; 100 kN.m ;
33
F F
CD DC
EI EI
MM
L EI
? =- =- × - =+ =+
The moment-distribution for this case is shown in Fig. 21.5d. Using equations of
static equilibrium, calculate reactions and .
2 A
H
2 D
H
) ( 395 . 12
6
457 . 41 911 . 32
2
? =
+
= kN H
A
) ( 952 . 39
3
285 . 73 57 . 46
2
? =
+
= kN H
D
) ( 347 . 52 ) 952 . 39 395 . 12 ( ? - = + - = kN F
e) Final results
Now, the shear condition for the frame is (vide Fig. 21.5b)
129 . 0
12 ) 952 . 39 395 . 12 ( ) 577 . 0 344 . 7 (
12 ) ( ) (
2 2 1 1
=
= + + -
= + + +
k
k
H H k H H
D A D A
Now the actual end moments in the frame are,
AB AB AB
M k M M ' ' ' + =
11.694 0.129( 41.457) 17.039 kN.m
AB
M=+ + =+
3.614 0.129( 32.911) 0.629 kN.m
BA
M =- + + =
3.614 0.129( 32.911) 0.629 kN.m
BC
M=+ - =-
1.154 0.129( 46.457) 4.853 kN.m
CB
M =- + - =-
1.154 0.129( 46.457) 4.853 kN.m
CD
M =- + + =+
0.578 0.129( 73.285) 8.876 kN.m
DC
M =- + + =+
The actual sway
EI
k
150
129 . 0 ' × = ? = ?
EI
35 . 19
=
The joint rotations can be calculated using slope-deflection equations.
[ ] ? ? ? 3 2
) 2 ( 2
- + + = -
B A
F
AB AB
L
I E
M M
or
[]
?
?
?
?
?
?
?
?
?
?
?
?
- - =
?
?
?
?
?
?
+ - = +
L
EI
M M
EI
L
L
EI
M M
EI
L
F
AB AB
F
AB AB B A
? ?
? ?
12
4
12
4
2
[]
?
?
?
?
?
?
?
?
?
?
?
?
- - =
?
?
?
?
?
?
+ - = +
L
EI
M M
EI
L
L
EI
M M
EI
L
F
BA BA
F
BA BA A B
? ?
? ?
12
4
12
4
2
17.039 kN.m
AB
M =+
Page 5
571 . 0 ; 429 . 0 = =
CD CB
DF DF
b) Calculate fixed end moments due to applied loading.
2
2
12 3 3
9.0 kN.m
6
F
AB
M
××
== ; 9.0 kN.m
F
BA
M =-
0kN.m
F
BC
M = ; 0kN.m
F
CB
M =
0kN.m
F
CD
M = ; 0kN.m
F
DC
M =
c) Prevent sidesway by providing artificial support at C. Carry out moment-
distribution ( Case . .e i A in Fig. 21.5b). The moment-distribution for this case is
shown in Fig. 21.5c.
Now calculate horizontal reaction at A and from equations of statics. D
()
1
11.694 3.614
6 7.347 kN
6
A
H
-
=+= ?
()
1
1.154 0.578
0.577 kN
3
D
H
--
==- ?
( ) 12 (7.347 0.577) 5.23 kN R=- - =- ?
d) Moment-distribution for arbitrary sidesway ' ? (case B, Fig. 21.5c)
Calculate fixed end moments for the arbitrary sidesway of
EI
150
' = ? .
6 (2 ) 12 150
( ) 50 kN.m ; 50 kN.m ;
66
F F
AB BA
EI EI
MM
L EI
? =- = × - =+ =+
6( ) 6 150
( ) 100 kN.m ; 100 kN.m ;
33
F F
CD DC
EI EI
MM
L EI
? =- =- × - =+ =+
The moment-distribution for this case is shown in Fig. 21.5d. Using equations of
static equilibrium, calculate reactions and .
2 A
H
2 D
H
) ( 395 . 12
6
457 . 41 911 . 32
2
? =
+
= kN H
A
) ( 952 . 39
3
285 . 73 57 . 46
2
? =
+
= kN H
D
) ( 347 . 52 ) 952 . 39 395 . 12 ( ? - = + - = kN F
e) Final results
Now, the shear condition for the frame is (vide Fig. 21.5b)
129 . 0
12 ) 952 . 39 395 . 12 ( ) 577 . 0 344 . 7 (
12 ) ( ) (
2 2 1 1
=
= + + -
= + + +
k
k
H H k H H
D A D A
Now the actual end moments in the frame are,
AB AB AB
M k M M ' ' ' + =
11.694 0.129( 41.457) 17.039 kN.m
AB
M=+ + =+
3.614 0.129( 32.911) 0.629 kN.m
BA
M =- + + =
3.614 0.129( 32.911) 0.629 kN.m
BC
M=+ - =-
1.154 0.129( 46.457) 4.853 kN.m
CB
M =- + - =-
1.154 0.129( 46.457) 4.853 kN.m
CD
M =- + + =+
0.578 0.129( 73.285) 8.876 kN.m
DC
M =- + + =+
The actual sway
EI
k
150
129 . 0 ' × = ? = ?
EI
35 . 19
=
The joint rotations can be calculated using slope-deflection equations.
[ ] ? ? ? 3 2
) 2 ( 2
- + + = -
B A
F
AB AB
L
I E
M M
or
[]
?
?
?
?
?
?
?
?
?
?
?
?
- - =
?
?
?
?
?
?
+ - = +
L
EI
M M
EI
L
L
EI
M M
EI
L
F
AB AB
F
AB AB B A
? ?
? ?
12
4
12
4
2
[]
?
?
?
?
?
?
?
?
?
?
?
?
- - =
?
?
?
?
?
?
+ - = +
L
EI
M M
EI
L
L
EI
M M
EI
L
F
BA BA
F
BA BA A B
? ?
? ?
12
4
12
4
2
17.039 kN.m
AB
M =+
0.629 kN.m
BA
M =
()
9 0.129(50) 15.45 kN.m
F
AB
M =+ =
()
9 0.129(50) 2.55 kN.m
F
BA
M =- + = -
1
change in near end + - change in far end
2
3
1
(17.039 15.45) (0.629 2.55)
2
0.0
3
6
A
EI
L
EI
?
??
??
??
=
??
-+- +
??
??
==
4.769
B
EI
? =
Example 21.3
Analyse the rigid frame shown in Fig. 21.6a. The moment of inertia of all the
members are shown in the figure.
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FAQs on The Moment Distribution Method: Frames with Sidesway - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the Moment Distribution Method? |  |
Ans. The Moment Distribution Method is a structural analysis technique used in civil engineering to analyze frames with sidesway. It allows engineers to determine the distribution of moments and shears in the members of a frame subjected to lateral loads. The method is based on the principle of successive approximation, where the moments are distributed and redistributed until equilibrium is achieved.
| 2. How does the Moment Distribution Method handle frames with sidesway? | |
Ans. The Moment Distribution Method is specifically designed to handle frames with sidesway, which refers to the lateral deflection of a frame caused by horizontal loads. This method accounts for the flexibility of the frame by taking into consideration the moments and rotations at the joints. By distributing and redistributing the moments, the method accounts for the additional flexibility of the sidesway, providing a more accurate analysis of the frame's behavior.
| 3. What are the advantages of using the Moment Distribution Method for frames with sidesway? | |
Ans. The Moment Distribution Method offers several advantages for analyzing frames with sidesway. Firstly, it provides a more accurate representation of the frame's behavior under lateral loads compared to other methods. Secondly, it allows engineers to easily determine the distribution of moments and shears in the members, enabling them to assess the structural integrity and design of the frame. Lastly, the method is relatively straightforward and computationally efficient, making it a popular choice for structural engineers.
| 4. Are there any limitations to the Moment Distribution Method for frames with sidesway? | |
Ans. While the Moment Distribution Method is a powerful tool for analyzing frames with sidesway, it does have some limitations. One limitation is that it assumes the frame remains within the elastic range, meaning it cannot accurately predict the behavior of frames that experience significant plastic deformation. Another limitation is that it may not provide accurate results for frames with complex geometry or irregular member stiffness. In such cases, more advanced analysis methods may be required.
| 5. What are some practical applications of the Moment Distribution Method for frames with sidesway? | |
Ans. The Moment Distribution Method finds extensive use in the analysis and design of various structural systems in civil engineering. It is commonly applied in the design of buildings, bridges, and other large structures subjected to lateral loads. By accurately predicting the distribution of moments and shears, engineers can ensure the structural integrity and stability of these systems. The method also allows for the optimization of member sizes and reinforcement, resulting in cost-effective and efficient designs.
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