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

## 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)

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)

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)

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)

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)

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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## Structural Analysis

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