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

Structural Analysis

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
XAD
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
XAD
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
XAD
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= ?
 
 
 
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Free

,

past year papers

,

Objective type Questions

,

shortcuts and tricks

,

Summary

,

ppt

,

Sample Paper

,

practice quizzes

,

Exam

,

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

,

Semester Notes

,

study material

,

Viva Questions

,

mock tests for examination

,

Previous Year Questions with Solutions

,

Extra Questions

,

pdf

,

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

,

MCQs

,

video lectures

,

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

,

Important questions

;