1. Structural Analysis II
Course Code: CIVL322
Dr. Aeid A. Abdulrazeg

2. Moment-Distribution Method
Contents:
introduction
the general procedure.
absolute stiffness of members.
relative stiffness of member.
distribution factor.
carry over factor.
beam with support movement.
moment distribution for frames: no sidesway.
moment distribution for frames: with sidesway

3. Moment-Distribution Method
Introduction
The moment distribution method was developed by Hardy Cross, in 1930.
It has been recognized as one of the most notable advances in structural analysis during the twentieth century.
It provides a convenient means of analyzing statically indeterminate beams and frames to any desired degree of accuracy .
The moment-distribution method is a displacement method of analysis that is easy to apply once certain elastic constants have been determined.

4. The General Procedure:
Artificially restrain temporarily all the joints against rotation and writing down the fixed end moments for all the members.
The joints are then released one by one in succession at each released joint the unbalanced moment are distributed to all ends of the members meeting at that joint.
The released joint is again restrained temporarily before proceeding to next joint.
The same set of operations are carried out at each till all joints are completed.
This completes one cycle of operations. The process is iterated for a number of cycles till the values obtained are within the desired accuracy.

5. General Principles and Definitions:
In order to understand the five steps mentioned, some words need to be defined and relevant derivations made.
Sign Convention: We will establish the same sign convention as that established for the slope-deflection equations: Clockwise moments that act on the member are considered positive, whereas counter clockwise moments are negative.
Fixed-End Moments (FEMs): The moments at the “walls” or fixed joints of a loaded member are called fixed-end moments. These moments can be determined from the table given on the Previous chapter, depending upon the type of loading on the member.

6. General Principles and Definitions:
Absolute Stiffness of Memebrs
Stiffness = Resistance offered by member to a unit displacement or rotation at a point, for given support constraint conditions MB MA
A clockwise moment MA is applied at A to produce a +ve bending in beam AB. Find A and MB. A B A
A RA RB L
E, I – Member properties

7. Using method of slope deflectionMB MA A B A
A RA RB L
E, I – Member properties
Stiffness factor = k = 4EI/L

8. Relative Stiffness of Members
The stiffness factor changes when the far end of the beam is simply-supported. MB MA A B A
A RA RB L
E, I – Member properties

9. General Principles and Definitions:
Joint Stiffness Factor
If several members are fixed connected to a joint and each of their far ends is fixed, then by the principle of superposition, the total stiffness factor at the joint is the sum of the member stiffness factors at the joint, that is,
Distribution Factor
Distribution factor is the ratio according to which an externally applied unbalanced moment M at a joint is apportioned to the various members mating at the joint.

10. Distribution Factor + ve moment M B C A D M I2 I1 L2 L1 I3 L3 A B MAC
MAB C A I2 L2 I1 L1
MAD I3 L3 D D

12. Distribution Factor (DF):That fraction of the total resisting moment supplied by the member is called the distribution factor (DF)

13. Carry Over Factor
Consider again the beam shown in figure, using the slope- deflection method:
Solving for  and equating these equation,
The moment M at the pin induces a moment of M’ = 0.5M at the wall In the case of a beam with the far end fixed, the CO factor is +0.5

14. Member Relative-Stiffness Factor
Quite often a continuous beam or a frame will be made from the same material so its modulus of elasticity E will be the same for all the members.
If this is the case, the common factor 4E will cancel from the numerator and denominator , when the distribution factor for a joint is determined. Hence, it is easier just to determine the member’s relative-stiffness factor

15. (Moment Distribution for Beams)

16. Example 1 (Moment Distribution for Beams)
Determine the internal moments at the supports for the beam shown Fig.
50 kN I I
10 m
10 m
15 m

17. 50 kN I I
10 m
10 m
15 m
0.0
0.43
0.57
0.0
D.F
-125
125
0.00
0.00
F.E.M
unbalanced moment
= 125
U.B. M
D. M
-53.75
-71.2
C.O. M
-26.9
-35.6
71.2
-71.2
-151.9
-35.6
Final M

18. Example 1 (Moment Distribution for Beams)
100 kN
20 kN per m I B 3I C A
2.5 m
2.5 m
5.0 m
7.5 m

19. D.F
0.0
0.33
0.67
1.0
F.E.M
-62.5
62.5
-93.75
93.75
unbalanced moment =
26.04
52.09
13.02
26.05
-26.05
2.149
4.298
8.726
4.363
-2.18
-4.363
0.35
0.72
1.46
0.73
-0.365
-0.73
0.06
0.12
0.244
0.122
-0.06
-0.122
0.01
0.02
0.04
0.02
-46.9
93.69
-93.69
0.00

20. Relative Stiffness of Members
0.0
0.4
0.6
1.0
D.F
-62.5
62.5
-93.75
93.75
F.E.M
unbalanced moment =
31.25
46.88
15.63
-46.87
93.75
-93.75
0.00

21. Example 3 (Moment Distribution for Beams)
Determine the internal moments at the supports for the beam shown Fig.

22. Beam with Support Movement s∆
In Case , one of the support has been rotated
In Case , simply supported beam

23. Example 4 (Moment Distribution for Beams)
Determine the internal moments at the supports for the continuous beam shown Fig. under:
Support (A) sinks by 0.3 cm
Support (A) rotated in the clockwise direction (ƟA = rad)
Support (B) sinks by 1.2 cm
Support (C) sinks by 0.6 cm
EI is the same throughout = 5000 m2.t
7.5 I B I C A
6.0 m
6.0 m

24. 0.0
4/7
3/7
1.0
D.F
-4.17
-5.83
2.5
F.E.M
1.9
1.43
0.95
-3.22
-3.93
3.93
0.00

25. Group Work (Moment Distribution for Beams)
Determine the support moments for the continuous beam shown in figure. EI is constant.
20 k
2k/ft A B C D
25 ft
15 ft
25 ft
10 ft

26. 20 k
2k/ft A B C D
0.00
0.428
0.572
0.572
0.428
0.00

27. Group Work (Moment Distribution for Beams)
Determine the support moments for the continuous beam shown in figure. EI is constant.

28. 0.00
0.33
0.67
0.5
0.5
1.00
150
106.66
0.00
0.00
-53.33
53.33