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

Structural Analysis

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 
 
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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