The Moment Distribution Method: Frames with Sidesway - 2 Civil Engineering (CE) Notes | EduRev

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Civil Engineering (CE) : The Moment Distribution Method: Frames with Sidesway - 2 Civil Engineering (CE) Notes | EduRev

 Page 1


429 . 0 =
CB
DF ; 571 . 0 =
CD
DF .    (1) 
   
b) Calculate fixed end moment due to applied loading. 
 
0 =
F
AB
M ;          kN.m 0 =
F
BA
M
 
 ;     kN.m 0 1 + =
F
BC
M kN.m 0 1 - =
F
CB
M
        
kN.m 0 =
F
CD
M    ;     .   (2) kN.m 0 =
F
DC
M
 
 
 
Now the frame is prevented from sidesway by providing a support at C as shown 
in Fig 21.4b (ii). The moment-distribution for this frame is shown in Fig 21.4c.  Let 
, and be the balanced end moments. Now calculate 
horizontal reactions at A and D from equations of statics. 
BA AB
M M ' , '
CD
M '
DC
M '
 
3
' '
1
BA AB
A
M M
H
+
= 
       
        = 
3
268 . 7 635 . 3 + -
 
        . ) ( 635 . 3 ? - = KN
 
) ( kN 635 . 3
3
269 . 17 636 . 3
1
? =
-
=
D
H . 
 
 
Page 2


429 . 0 =
CB
DF ; 571 . 0 =
CD
DF .    (1) 
   
b) Calculate fixed end moment due to applied loading. 
 
0 =
F
AB
M ;          kN.m 0 =
F
BA
M
 
 ;     kN.m 0 1 + =
F
BC
M kN.m 0 1 - =
F
CB
M
        
kN.m 0 =
F
CD
M    ;     .   (2) kN.m 0 =
F
DC
M
 
 
 
Now the frame is prevented from sidesway by providing a support at C as shown 
in Fig 21.4b (ii). The moment-distribution for this frame is shown in Fig 21.4c.  Let 
, and be the balanced end moments. Now calculate 
horizontal reactions at A and D from equations of statics. 
BA AB
M M ' , '
CD
M '
DC
M '
 
3
' '
1
BA AB
A
M M
H
+
= 
       
        = 
3
268 . 7 635 . 3 + -
 
        . ) ( 635 . 3 ? - = KN
 
) ( kN 635 . 3
3
269 . 17 636 . 3
1
? =
-
=
D
H . 
 
 
) ( kN 10 ) 635 . 3 635 . 3 ( 10 ? - = + - - = R   (3)  
 
           
 
d) Moment-distribution for arbitrary known sidesway  ' ? . 
 
Since  is arbitrary, Choose any convenient value. Let  ' ? ' ? = 
EI
150
   Now calculate 
fixed end beam moments for this arbitrary sidesway. 
 
L
EI
M
F
AB
? 6
- =  )
3
150
(
3
6
EI
EI
- × - = =   kN.m 100
 
kN.m 100 =
F
BA
M 
 
kN.m 100 + = =
F
DC
F
CD
M M     (4) 
 
 
Page 3


429 . 0 =
CB
DF ; 571 . 0 =
CD
DF .    (1) 
   
b) Calculate fixed end moment due to applied loading. 
 
0 =
F
AB
M ;          kN.m 0 =
F
BA
M
 
 ;     kN.m 0 1 + =
F
BC
M kN.m 0 1 - =
F
CB
M
        
kN.m 0 =
F
CD
M    ;     .   (2) kN.m 0 =
F
DC
M
 
 
 
Now the frame is prevented from sidesway by providing a support at C as shown 
in Fig 21.4b (ii). The moment-distribution for this frame is shown in Fig 21.4c.  Let 
, and be the balanced end moments. Now calculate 
horizontal reactions at A and D from equations of statics. 
BA AB
M M ' , '
CD
M '
DC
M '
 
3
' '
1
BA AB
A
M M
H
+
= 
       
        = 
3
268 . 7 635 . 3 + -
 
        . ) ( 635 . 3 ? - = KN
 
) ( kN 635 . 3
3
269 . 17 636 . 3
1
? =
-
=
D
H . 
 
 
) ( kN 10 ) 635 . 3 635 . 3 ( 10 ? - = + - - = R   (3)  
 
           
 
d) Moment-distribution for arbitrary known sidesway  ' ? . 
 
Since  is arbitrary, Choose any convenient value. Let  ' ? ' ? = 
EI
150
   Now calculate 
fixed end beam moments for this arbitrary sidesway. 
 
L
EI
M
F
AB
? 6
- =  )
3
150
(
3
6
EI
EI
- × - = =   kN.m 100
 
kN.m 100 =
F
BA
M 
 
kN.m 100 + = =
F
DC
F
CD
M M     (4) 
 
 
 
 
 
The moment-distribution for this case is shown in Fig 24.4d. Now calculate 
horizontal reactions  and . 
2 A
H
2 D
H
 
 = 
2 A
H ) ( kN 15 . 43
3
48 . 76 98 . 52
? =
+
 
2 D
H = ) ( kN 15 . 43
3
49 . 76 97 . 52
? =
+
  
 
) ( kN 30 . 86 ? - = F 
 
Page 4


429 . 0 =
CB
DF ; 571 . 0 =
CD
DF .    (1) 
   
b) Calculate fixed end moment due to applied loading. 
 
0 =
F
AB
M ;          kN.m 0 =
F
BA
M
 
 ;     kN.m 0 1 + =
F
BC
M kN.m 0 1 - =
F
CB
M
        
kN.m 0 =
F
CD
M    ;     .   (2) kN.m 0 =
F
DC
M
 
 
 
Now the frame is prevented from sidesway by providing a support at C as shown 
in Fig 21.4b (ii). The moment-distribution for this frame is shown in Fig 21.4c.  Let 
, and be the balanced end moments. Now calculate 
horizontal reactions at A and D from equations of statics. 
BA AB
M M ' , '
CD
M '
DC
M '
 
3
' '
1
BA AB
A
M M
H
+
= 
       
        = 
3
268 . 7 635 . 3 + -
 
        . ) ( 635 . 3 ? - = KN
 
) ( kN 635 . 3
3
269 . 17 636 . 3
1
? =
-
=
D
H . 
 
 
) ( kN 10 ) 635 . 3 635 . 3 ( 10 ? - = + - - = R   (3)  
 
           
 
d) Moment-distribution for arbitrary known sidesway  ' ? . 
 
Since  is arbitrary, Choose any convenient value. Let  ' ? ' ? = 
EI
150
   Now calculate 
fixed end beam moments for this arbitrary sidesway. 
 
L
EI
M
F
AB
? 6
- =  )
3
150
(
3
6
EI
EI
- × - = =   kN.m 100
 
kN.m 100 =
F
BA
M 
 
kN.m 100 + = =
F
DC
F
CD
M M     (4) 
 
 
 
 
 
The moment-distribution for this case is shown in Fig 24.4d. Now calculate 
horizontal reactions  and . 
2 A
H
2 D
H
 
 = 
2 A
H ) ( kN 15 . 43
3
48 . 76 98 . 52
? =
+
 
2 D
H = ) ( kN 15 . 43
3
49 . 76 97 . 52
? =
+
  
 
) ( kN 30 . 86 ? - = F 
 
 
Let  be a factor by which the solution of case (ii k i ) needs to be multiplied. Now 
actual moments in the frame is obtained by superposing the solution ( ) on the 
solution obtained by multiplying case (ii
ii
i ) by . Thus cancel out the holding 
force R such that final result is for the frame without holding force. 
k kF
  
Thus, .   R F k =
 
1161 . 0
13 . 86
10
=
-
-
= k     (5) 
 
 
Now the actual end moments in the frame are, 
 
AB AB AB
M k M M ' ' ' + = 
3.635 0.1161( 76.48) 5.244 kN.m
AB
M =- + + =+ 
7.268 0.1161( 52.98) 1.117 kN.m
BA
M =- + + =- 
7.268 0.1161( 52.98) 1.117 kN.m
BC
M =+ + - =+ 
7.269 0.1161( 52.97) 13.419 kN.m
CB
M =- + - =- 
7.268 0.1161( 52.97) 13.418 kN.m
CD
M =+ + + =+ 
3.636 0.1161( 76.49) 12.517 kN.m
DC
M =+ + + =+ 
 
The actual sway is computed as,  
EI
k
150
1161 . 0 ' × = ? = ? 
 
    
EI
415 . 17
= 
  
The joint rotations can be calculated using slope-deflection equations. 
 
 
[
AB B A
F
AB AB
]
L
EI
M M ? ? ? 3 2
2
- + + =    where 
L
AB
?
- = ?              
 
[]
AB A B
F
BA BA
L
EI
M M ? ? ? 3 2
2
- + + = 
 
Page 5


429 . 0 =
CB
DF ; 571 . 0 =
CD
DF .    (1) 
   
b) Calculate fixed end moment due to applied loading. 
 
0 =
F
AB
M ;          kN.m 0 =
F
BA
M
 
 ;     kN.m 0 1 + =
F
BC
M kN.m 0 1 - =
F
CB
M
        
kN.m 0 =
F
CD
M    ;     .   (2) kN.m 0 =
F
DC
M
 
 
 
Now the frame is prevented from sidesway by providing a support at C as shown 
in Fig 21.4b (ii). The moment-distribution for this frame is shown in Fig 21.4c.  Let 
, and be the balanced end moments. Now calculate 
horizontal reactions at A and D from equations of statics. 
BA AB
M M ' , '
CD
M '
DC
M '
 
3
' '
1
BA AB
A
M M
H
+
= 
       
        = 
3
268 . 7 635 . 3 + -
 
        . ) ( 635 . 3 ? - = KN
 
) ( kN 635 . 3
3
269 . 17 636 . 3
1
? =
-
=
D
H . 
 
 
) ( kN 10 ) 635 . 3 635 . 3 ( 10 ? - = + - - = R   (3)  
 
           
 
d) Moment-distribution for arbitrary known sidesway  ' ? . 
 
Since  is arbitrary, Choose any convenient value. Let  ' ? ' ? = 
EI
150
   Now calculate 
fixed end beam moments for this arbitrary sidesway. 
 
L
EI
M
F
AB
? 6
- =  )
3
150
(
3
6
EI
EI
- × - = =   kN.m 100
 
kN.m 100 =
F
BA
M 
 
kN.m 100 + = =
F
DC
F
CD
M M     (4) 
 
 
 
 
 
The moment-distribution for this case is shown in Fig 24.4d. Now calculate 
horizontal reactions  and . 
2 A
H
2 D
H
 
 = 
2 A
H ) ( kN 15 . 43
3
48 . 76 98 . 52
? =
+
 
2 D
H = ) ( kN 15 . 43
3
49 . 76 97 . 52
? =
+
  
 
) ( kN 30 . 86 ? - = F 
 
 
Let  be a factor by which the solution of case (ii k i ) needs to be multiplied. Now 
actual moments in the frame is obtained by superposing the solution ( ) on the 
solution obtained by multiplying case (ii
ii
i ) by . Thus cancel out the holding 
force R such that final result is for the frame without holding force. 
k kF
  
Thus, .   R F k =
 
1161 . 0
13 . 86
10
=
-
-
= k     (5) 
 
 
Now the actual end moments in the frame are, 
 
AB AB AB
M k M M ' ' ' + = 
3.635 0.1161( 76.48) 5.244 kN.m
AB
M =- + + =+ 
7.268 0.1161( 52.98) 1.117 kN.m
BA
M =- + + =- 
7.268 0.1161( 52.98) 1.117 kN.m
BC
M =+ + - =+ 
7.269 0.1161( 52.97) 13.419 kN.m
CB
M =- + - =- 
7.268 0.1161( 52.97) 13.418 kN.m
CD
M =+ + + =+ 
3.636 0.1161( 76.49) 12.517 kN.m
DC
M =+ + + =+ 
 
The actual sway is computed as,  
EI
k
150
1161 . 0 ' × = ? = ? 
 
    
EI
415 . 17
= 
  
The joint rotations can be calculated using slope-deflection equations. 
 
 
[
AB B A
F
AB AB
]
L
EI
M M ? ? ? 3 2
2
- + + =    where 
L
AB
?
- = ?              
 
[]
AB A B
F
BA BA
L
EI
M M ? ? ? 3 2
2
- + + = 
 
 
 
In the above equation, except 
A
? and 
B
? all other quantities are known. Solving 
for 
A
? and 
B
? , 
EI
B A
55 . 9
; 0
-
= = ? ? . 
 
The elastic curve is shown in Fig. 21.4e. 
 
 
 
 
 
 
 
 
 
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