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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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FAQs on The Moment Distribution Method: Frames with Sidesway - 2 - Structural Analysis - Civil Engineering (CE)
| 1. What is the Moment Distribution Method? |  |
Ans. The Moment Distribution Method is a structural analysis technique used to analyze frames with sidesway. It involves the distribution of moments at the joints of the frame until equilibrium is achieved. This method is commonly used to determine the moments and rotations at each joint of the frame, allowing for the calculation of member forces and displacements.
| 2. How does the Moment Distribution Method handle frames with sidesway? | |
Ans. The Moment Distribution Method handles frames with sidesway by considering the lateral displacements of the frame. Sidesway refers to the lateral movement or rotation of a frame caused by horizontal loads or lack of lateral stability. The method incorporates these displacements into the analysis, allowing for the determination of member forces and rotations that consider the effects of sidesway.
| 3. What are the advantages of using the Moment Distribution Method for frames with sidesway? | |
Ans. The Moment Distribution Method offers several advantages for analyzing frames with sidesway. Firstly, it provides a relatively simple and intuitive approach to analyze the behavior of such frames. Additionally, it allows for the consideration of sidesway effects, which is crucial for accurate analysis and design. The method also provides a systematic procedure for distributing moments and determining member forces, making it a practical and efficient analysis technique.
| 4. Are there any limitations to using the Moment Distribution Method for frames with sidesway? | |
Ans. Yes, the Moment Distribution Method has some limitations when it comes to frames with sidesway. One major limitation is that it assumes that the frame remains elastic and that the member stiffness does not change significantly during loading. This assumption may not hold true for frames with significant sidesway, as the lateral displacements can lead to changes in member stiffness. Additionally, the method may become more complex and time-consuming for frames with highly irregular geometry or complex loadings.
| 5. How can the Moment Distribution Method be applied in practice for frames with sidesway? | |
Ans. To apply the Moment Distribution Method in practice for frames with sidesway, the following steps are typically followed:
1. Determine the stiffness matrix of the frame, considering the stiffness of each member and joint.
2. Apply the external loads to the frame.
3. Distribute the moments at each joint using the moment distribution method, taking into account the sidesway displacements.
4. Repeat the distribution process until convergence is achieved, ensuring that the moments at each joint remain constant.
5. Calculate the member forces and rotations based on the distributed moments.
6. Check the frame's stability and assess its structural adequacy based on the calculated member forces and displacements.
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Nov 21, 2024 Last updated |
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