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# The Moment Distribution Method: Frames with Sidesway - 3 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : The Moment Distribution Method: Frames with Sidesway - 3 Civil Engineering (CE) Notes | EduRev

``` Page 1

571 . 0 ; 429 . 0 = =
CD CB
DF DF

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

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

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

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

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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## Structural Analysis

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