This lecture includes analysis of frame by moment distribution method

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