Theory of Structures II Moment Distribution Beams Reference: Structural Analysis 10th Edition in SI Units by R.C. Hibbeler

1. CED 426
Structural Theory II
Lecture 17
Displacement Method of Analysis:
Moment Distribution for Beams
Mary Joanne C. Aniñon
Instructor

2. Procedure For Analysis
• The following procedure provides a general method for determining
the end moments on the beam spans using moment distribution.
1. Determine the Distribution Factors and Fixed-End Moments
2. Perform Moment-Distribution Process

3. Distribution Factors and Fixed-End Moments
1.a. Identify the joints and spans on the beam.
1.b. Calculate member and joint stiffness factors for each span.
1.c. Determine the distribution factors (DF). Remember that DF = 0 for
fixed end and DF = 1 for an end pin or end roller support
1.d. Determine the fixed-end moments

4. Moment-Distribution Process
Assume all joints are initially locked. Then:
2.a. Determine the moment that is needed to put each joint in
equilibrium
2.b. Release or unlock the joints and distribute the counterbalancing
moments into the members at each joint
2.c. Carry these moments over to the other end of the member by
multiplying each moment by the carry-over factor +
1
2

5. Moment-Distribution Process
• By repeating this cycle of locking and unlocking the joints, it will be
found that the moment corrections will diminish since the beam
tends to achieve its final deflected shape
• When a small enough value for the corrections is obtained, the
process should be stopped.
• Each column of FEMs, distributed moments, and carry-over moments
should then be added. If this is done correctly, moment equilibrium at
the joints will be achieved

6. Example 1
• Determine the internal moments at each support of the beam shown
in Fig.11-7a. EI is constant

7. Example 1
Step 1.a. Identify the joints and spans in the
beam: A, B, C, D and span AB, DC, CD
Step 1.b. Calculate member and joint stiffness
factors for each span.
𝐾𝑇 = 𝐾
Step 1.c. Determine the distribution factors (DF).
𝐷𝐹 =
𝐾
𝐾
K =
4𝐸𝐼
𝐿

8. Example 1
1.d. Determine the fixed-end moments

9. Example 1
2.a. Determine the moment that is
needed to put each joint in
equilibrium
Joint A B C D
Member AB BA BC CB CD DC
DF 0 0.5 0.5 0.4 0.6 0
FEM
Dist.
0
0
0
120
-240
120
240
4
-250
6
250
0
CO
Dist.
60
0
0
-1
2
-1
60
-24
0
-36
3
0
CO
Dist.
-0.5
0
0
6
-12
6
-0.5
0.2
0
0.3
-18
0
CO
Dist.
3
0
0
-0.05
0.1
-0.05
3
-1.2
0
-1.8
0.15
0
CO
Dist.
-0.025
0
0
0.3
-0.6
0.3
-0.025
0.01
0
0.015
-0.9
0
∑M 62.475 125.25 -125.25 281.485 -281.485 234.25
2.b. Release or unlock the joints
and distribute the
counterbalancing moments into
the members at each joint
2.c. Carry these moments over to
the other end of the member by
multiplying each moment by the
carry-over factor +
1
2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Blue – 2.b. (Joint) – Add, Opposite in
Sign-Multiply to DF
Green – 2.c. (Span) – Divided by 2
(don’t change the sign) – distribute to
the other side
Stop on Color Blue!
Stop when the values in Blue Rows is
<= 0.05

10. Example 1
• Starting with the FEMs, line 4, the moments at joints B and C are
distributed simultaneously, line 5.
• These moments are then carried over simultaneously to the
respective ends at each span, line 6.
• The resulting moments are again simultaneously distributed and
carried over, lines 7 and 8.
• The process is continued until the resulting moments are diminished
an appropriate amount, line 13
• The resulting moments are found by summation, line 14

11. Example 1
Joint A B C D
Member AB BA BC CB CD DC
∑M 62.475 125.25 -125.25 281.485 -281.485 234.25
Initial Assumption: All CLOCKWISE
3.a. Compute the shear
Placing these moments on each span and applying the equations of equilibrium yields the end shears.

12. Example 1
3.a. Compute the shear and moment diagrams