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# The Moment Distribution Method: Frames with Sidesway - 4 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : The Moment Distribution Method: Frames with Sidesway - 4 Civil Engineering (CE) Notes | EduRev

``` Page 1

Now calculate reactions from free body diagram shown in Fig. 21.5d.

Page 2

Now calculate reactions from free body diagram shown in Fig. 21.5d.

Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?

11
5 2.29
A
HV +=-      (3)

Column  CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2

(4)
12
5 2.284
D
HV -=

Beam  BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?

( )
1
5.002 kN V=?
( )
2
4.998 kN V= ?      (5)

Thus from (3) ( )
1
1.458 kN
A
H =- ?

from (4) ( )
1
1.456 kN
D
H= ?      (6)

Page 3

Now calculate reactions from free body diagram shown in Fig. 21.5d.

Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?

11
5 2.29
A
HV +=-      (3)

Column  CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2

(4)
12
5 2.284
D
HV -=

Beam  BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?

( )
1
5.002 kN V=?
( )
2
4.998 kN V= ?      (5)

Thus from (3) ( )
1
1.458 kN
A
H =- ?

from (4) ( )
1
1.456 kN
D
H= ?      (6)

Version 2 CE IIT, Kharagpur

0=
()
11
05
5.002 kN
FHHR
R
=++-
=+?
?
(7)

d) Moment-distribution for arbitrary sidesway ' ? .

Calculate fixed end beam moments for arbitrary sidesway of

EI
75 . 12
' = ?

The member rotations for this arbitrary sidesway is shown in Fig. 21.6e.

Page 4

Now calculate reactions from free body diagram shown in Fig. 21.5d.

Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?

11
5 2.29
A
HV +=-      (3)

Column  CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2

(4)
12
5 2.284
D
HV -=

Beam  BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?

( )
1
5.002 kN V=?
( )
2
4.998 kN V= ?      (5)

Thus from (3) ( )
1
1.458 kN
A
H =- ?

from (4) ( )
1
1.456 kN
D
H= ?      (6)

Version 2 CE IIT, Kharagpur

0=
()
11
05
5.002 kN
FHHR
R
=++-
=+?
?
(7)

d) Moment-distribution for arbitrary sidesway ' ? .

Calculate fixed end beam moments for arbitrary sidesway of

EI
75 . 12
' = ?

The member rotations for this arbitrary sidesway is shown in Fig. 21.6e.

1
1
"'
;
cos 5
AB
AB AB
BB
LL
5.1'
?
a
? ? ?
==- ?= =
2
2'
0.4 '
5
?
?= = ?
'
()
5
AB
clockwise ?
?
=- ;
'
()
5
CD
clockwise ?
?
=-
2
2'tan '
()
22 5
BC
counterclockwise
a
?
? ? ?
== =

6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F AB
AB AB
AB
EI EI
M
LEI
?
??
=- =- - = +
??
??

6.0 kN.m
F
BA
M =+
6 6 ( ) 12.75
7.65 kN.m
25
F BC
BC BC
BC
EI EI
M
LEI
?
??
=- =- =-
??
??

7.65 kN.m
F
CB
M =-
6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F CD
CD CD
CD
EI EI
M
LEI
?
??
=- =- - =+
??
??

6.0 kN.m
F
DC
M =+

The moment-distribution for the arbitrary sway is shown in Fig. 21.6f. Now
reactions can be calculated from statics.

Page 5

Now calculate reactions from free body diagram shown in Fig. 21.5d.

Column AB
11
0 5 1.526 0.764 0
AA
MH V =? + + + =
?

11
5 2.29
A
HV +=-      (3)

Column  CD
1
0 5 1.522 0.762 0
DD
MH V =? - - - =
? 2

(4)
12
5 2.284
D
HV -=

Beam  BC
1
0 2 1.522 1.526 10 1 0
C
MV =? + - - × =
?

( )
1
5.002 kN V=?
( )
2
4.998 kN V= ?      (5)

Thus from (3) ( )
1
1.458 kN
A
H =- ?

from (4) ( )
1
1.456 kN
D
H= ?      (6)

Version 2 CE IIT, Kharagpur

0=
()
11
05
5.002 kN
FHHR
R
=++-
=+?
?
(7)

d) Moment-distribution for arbitrary sidesway ' ? .

Calculate fixed end beam moments for arbitrary sidesway of

EI
75 . 12
' = ?

The member rotations for this arbitrary sidesway is shown in Fig. 21.6e.

1
1
"'
;
cos 5
AB
AB AB
BB
LL
5.1'
?
a
? ? ?
==- ?= =
2
2'
0.4 '
5
?
?= = ?
'
()
5
AB
clockwise ?
?
=- ;
'
()
5
CD
clockwise ?
?
=-
2
2'tan '
()
22 5
BC
counterclockwise
a
?
? ? ?
== =

6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F AB
AB AB
AB
EI EI
M
LEI
?
??
=- =- - = +
??
??

6.0 kN.m
F
BA
M =+
6 6 ( ) 12.75
7.65 kN.m
25
F BC
BC BC
BC
EI EI
M
LEI
?
??
=- =- =-
??
??

7.65 kN.m
F
CB
M =-
6 6 (2 ) 12.75
6.0 kN.m
5.1 5
F CD
CD CD
CD
EI EI
M
LEI
?
??
=- =- - =+
??
??

6.0 kN.m
F
DC
M =+

The moment-distribution for the arbitrary sway is shown in Fig. 21.6f. Now
reactions can be calculated from statics.

Column AB

21
0 5 6.283 6.567 0
AA
MH V =? - - + =
?

11
5 12.85
A
HV +=      (3)

Column  CD

22
0 5 6.567 6.283 0
DD
MH V =? - - - =
?

12
5 12.85
D
HV -=      (4)

Beam  BC
1
0 2 6.567 6.567 0
C
MV =? + + =
?

( )
1
6.567 VkN
( )
2
6.567 kN V =- ? ;+?   (5) =
Thus from 3 ( )
2
3.883 kN
A
H =+ ?

from 4        (6) ()
2
3.883 kN
D
H= ?

```
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## Structural Analysis

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