This document provides information about analyzing structural frames using the moment distribution method with sway. It discusses analyzing portal frames where the amount of sway is unknown by assuming initial fixed end moments. It provides equations to calculate the ratio of fixed end moments at column heads for different end conditions. As an example, it analyzes a portal frame with given properties and loadings using the moment distribution method accounting for sway. It calculates the fixed end moments, distribution factors, moment distribution table, reactions, and final bending moments at the joints.

1. Structural Analysis II
(CE-403)
By
Iqbal Hafeez Khan
Assistant Professor
Department of Civil Engineering
SISTec GN 1

2. References:
• C S Reddy, Basic Structural Analysis, Tata McGraw Hill
• C.K. Wang, Indeterminate Structural Analysis, Addison Wesley.
• B.C. Punmia , Theory of Structures,Laxmi Publication
• S. Ramamrutham, Theory of Structures,Dhanpat Rai Publication
• R C Hibbeler, Structural Analysis,Pearson
2

3. Moment Distribution with Sway Analysis
3
B
A
C
D
W

4. Moment Distribution with Sway Analysis
• In the case of continuous beam the effect of yielding or settlement of
support was taken into account by introducing initial fixed end
moments
6𝐸𝐼𝛿
𝐿2
• In the case of portal frame the amount of sway or joint movements is
not known and the analysis is done by assuming some arbitary
movement
• These assumed fixed moments due to side sway are then distributed
and reactions are found.
4

5. Reason of side sway
• Unsymmetrical loading on the portal frame
5
a b
W

6. • Unsymmetrical outline of the portal frame
6
a
b

7. Different end conditions of the column of the portal frame
7

8. Non uniform section of the vertical member
8
𝑰 𝟏 𝑰 𝟐

9. Horizontal Loading on column of the frame
9
W

10. Settlement of the support of the frame
10
𝜹

11. Ratio of Sway moments at Column Heads
Both ends fixed:
𝑀𝐴𝐵 = 𝑀 𝐵𝐴 =
6𝐸𝐼1δ
𝐿1
2
𝑀 𝐷𝐶 = 𝑀 𝐶𝐷 =
6𝐸𝐼2δ
𝐿2
2
𝑀 𝐴𝐵
𝑀 𝐷𝐶
=
𝑀 𝐵𝐴
𝑀 𝐶𝐷
=
൘
𝑰 𝟏
𝑳 𝟏
𝟐
൘
𝑰 𝟐
𝑳 𝟐
𝟐
11
B
A
C
D
W
𝜹 𝜹
𝑰 𝟏 𝑰 𝟐𝑳 𝟏 𝑳 𝟐

12. Both ends hinged:
𝑀 𝐵𝐴 =
3𝐸𝐼1δ
𝐿1
2
𝑀 𝐶𝐷 =
3𝐸𝐼2δ
𝐿2
2
𝑀 𝐵𝐴
𝑀 𝐶𝐷
=
൘
𝑰 𝟏
𝑳 𝟏
𝟐
൘
𝑰 𝟐
𝑳 𝟐
𝟐
12
B
A
C
D
W
𝜹 𝜹
𝑰 𝟏 𝑰 𝟐𝑳 𝟏 𝑳 𝟐

13. One end fixed and other is hinged:
𝑀𝐴𝐵 = 𝑀 𝐵𝐴 =
6𝐸𝐼1δ
𝐿1
2
𝑀 𝐶𝐷 =
3𝐸𝐼2δ
𝐿2
2
𝑀 𝐴𝐵
𝑀 𝐶𝐷
=
𝑀 𝐵𝐴
𝑀 𝐶𝐷
=
൘
𝟐𝑰 𝟏
𝑳 𝟏
𝟐
൘
𝑰 𝟐
𝑳 𝟐
𝟐
13
B
A
C
D
W
𝜹 𝜹
𝑰 𝟏 𝑰 𝟐𝑳 𝟏 𝑳 𝟐

14. Analyse the frame by moment distribution method
14
6 kN/m
A
B C
D
2 m
3 m
2 m
I
2 I
I

15. Solution:
Maximum Bending Moment:
For Span AB
= 0
For Span BC
For Span CD
= 0
15
=
𝑤𝐿2
8
=
6 X 22
8
= 𝟑 𝐤𝐍𝐦

16. Fixed End Moments:
16
𝑀 𝐹𝐴𝐵 = 𝟎
𝑀 𝐹𝐵𝐴 = 𝟎
𝑀 𝐹𝐵𝐶 = −
𝑤𝑙²
12
= −
6 X 2²
12
=−2 kNm
𝑀 𝐹𝐶𝐵 =
𝑤𝑙²
12
=
6 X 2²
12
= 2 kNm
𝑀 𝐹𝐶𝐷 = 𝟎
𝑀 𝐹𝐷𝐶 = 𝟎

17. Distribution factor
Joint Member Stiffness Total Stiffness Distribution Factor
B
BA
BC
4𝐸(2𝐼)
3
4𝐸𝐼
2
28 𝐸𝐼
6
0.57
0.43
C
CB
CD
4𝐸𝐼
2
4𝐸𝐼
2
8𝐸𝐼
2
0.5
0.5
17

18. Moment Distribution Table:
0.57 0.43 0.5 0.5
FEMs 0 0 -2 2 0 0
Balance - 1.14 0.86 -1 -1 -
Carry 0.57 - -0.5 0.43 - -0.5
Balance - 0.29 0.21 -0.22 -0.22 -
Carry 0.15 - -0.11 0.11 - -0.11
Balance - 0.06 0.05 -0.06 -0.06 -
0.72 1.49 -1.49 1.26 -1.26 -0.61FM
A B C D
18

19. Horizontal Reactions:
Taking moment about B
-𝐻𝐴 x 3 +𝑀𝐴 + 𝑀 𝐵 = 0
-𝐻𝐴 x 3 + 0.72 + 1.49= 0
𝐻𝐴= 0.74 kN
Taking moment about C
𝐻 𝐷 x 2 +𝑀 𝐷 + 𝑀𝑐 = 0
𝐻 𝐷 x 2 - 0.61 - 1.26 = 0
𝐻 𝐷= 0.94 kN
𝑯 𝑨
𝑯 𝑫
𝑽 𝑨
𝑽 𝑫
A
B C
D
19
2 m
3 m
2 m
6 kN/m

20. Summation of Horizontal Force
= 𝐻𝐴 − 𝐻 𝐷
= 0.74– (0.94)
= -0.2 kN( )
= 0.2 kN( )
This is the unbalanced force due to which frame will sway.
We have to apply S force on C for sway analysis due to which clockwise
moment will be induced at columns head.
20

21. Sway Analysis
Both ends are fixed
𝑀 𝐴𝐵
𝑀 𝐷𝐶
=
𝑀 𝐵𝐴
𝑀 𝐶𝐷
=
൘
𝑰 𝟏
𝑳 𝟏
𝟐
൘
𝑰 𝟐
𝑳 𝟐
𝟐
𝑀 𝐴𝐵
𝑀 𝐷𝐶
=
𝑀 𝐵𝐴
𝑀 𝐶𝐷
=
ൗ𝟐𝑰
𝟑 𝟐
ൗ𝑰
𝟐 𝟐
𝑀 𝐴𝐵
𝑀 𝐷𝐶
=
𝑀 𝐵𝐴
𝑀 𝐶𝐷
=
𝟖
𝟗
𝑀𝐴𝐵=𝑀 𝐵𝐴= + 8 kNm
𝑀 𝐷𝐶=𝑀 𝐶𝐷 = + 9 kNm
21
2 m
3 m
2 m
I
2 I
I
S

22. Moment Distribution Table:
0.57 0.43 0.5 0.5
FEMs 8 8 0 0 9 9
Balance - -4.57 -3.43 -4.5 -4.5 -
Carry -2.29 - -2.25 -1.72 - -2.25
Balance - 1.29 0.96 0.86 0.86 -
Carry 0.64 - 0.43 0.48 - 0.43
Balance - -0.25 -0.18 -0.24 -0.24 -
Carry -0.13 - -0.09-0.12 -0.12
-
Balance - 0.07 0.05 0.05 0.05 -
6.22 4.54 -4.54 -5.16 5.16 7.06FM
A B C D
22

23. Horizontal Reactions:
Taking moment about B
-𝐻𝐴 x 3 +𝑀𝐴 + 𝑀 𝐵 = 0
-𝐻𝐴 x 3 + 6.22 + 4.54= 0
𝐻𝐴= 3.59 kN
Taking moment about C
𝐻 𝐷 x 2 +𝑀 𝐶 + 𝑀 𝐷 = 0
𝐻 𝐷 x 2 +5.16 + 7.06 = 0
𝐻 𝐷= -6.11 kN
23
2 m
3 m
2 m
I
2 I
I
S
𝑯 𝑨
𝑯 𝑫
𝑽 𝑨
𝑽 𝑫
A
B
D
C

24. Now Summation of horizontal forces
𝐻𝐴 -𝐻 𝐷- S =0
3.59 – (-6.11) – S =0
S = 9.7 kN
24
2 m
3 m
2 m
I
2 I
I
S
𝑯 𝑨
𝑯 𝑫
𝑽 𝑨
𝑽 𝑫

25. 25
A B C D
Sway force = 9.70 kN 6.22 4.54 -4.54 -5.16 5.16 7.06
Sway force = 0.20 kN 0.13 0.09 -0.09 -0.10 0.10 0.15
Non Sway moment 0.72 1.49 -1.49 1.26 -1.26 -0.61
Final Moments 0.85 1.58 -1.58 1.16 -1.16 -0.46
for
9.70 kN = 6.22
1 kN =
6.22
9.70
0.20 kN =
6.22
9.70
x 0.20 =0.13

26. 26
A
B
D
C
0.86
1.6
1.6
1.18
1.18
0.49