The document provides steps to solve two examples using the moment distribution method for beams. It first divides each beam into sections and calculates the fixed end moment for each section. It then calculates the stiffness and distribution factor for each section. Moment distribution tables are constructed and the distributing moment and carryover moment are calculated in cycles until equilibrium is reached. Finally, the bending moment diagram and shear force diagram are drawn.

1. 1
Eng. Abdulrahman Shaaban
Solved Examples at Moment Distribution method for Beams
Example (1)
Draw B.M.D. & S.F.D. for the following Beams:
Step 1: Coding joint and Dived the beam into individual parts then Calculate Fixed End
Moment (F.E.M) for each part
Step 2: Calculate Stiffness for each member then calculate its Distribution Factor (D.F).
For member AB: 𝐾𝐴𝐵 =
4𝐸𝐼
𝐿
=
4 ∗ 3𝐸𝐼
6
= 2𝐸𝐼
For member BC: 𝐾 𝐵𝐶 =
4𝐸𝐼
𝐿
=
4 ∗ 4𝐸𝐼
10
= 1.6𝐸𝐼
For member CD: 𝐾𝐶𝐷 =
4𝐸𝐼
𝐿
=
4 ∗ 2𝐸𝐼
6
= 4 3⁄ 𝐸𝐼
For joint B:
𝐷𝐹𝐵𝐴 =
𝐾 𝐵𝐴
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
=
2
2 + 1.6
= 5 9⁄
𝐷𝐹𝐵𝑐 =
𝐾 𝐵𝑐
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
=
1.6
2 + 1.6
= 4 9⁄
A
B C D
A B C D
6 -6 5 -5 4 -8

2. 2
Eng. Abdulrahman Shaaban
For joint C:
𝐷𝐹𝐶𝐵 =
𝐾𝐶𝐵
𝐾𝐶𝐵 + 𝐾𝐶𝐷
=
1.6
1.6 + 4 3⁄
= 6 11⁄
𝐷𝐹𝐶𝐷 =
𝐾𝐶𝐷
𝐾𝐶𝐵 + 𝐾𝐶𝐷
=
4 3⁄
1.6 + 4 3⁄
= 5 11⁄
Step 3: Construct a table to summarize results and facilitation of calculation procedures.
Joint A B C D
Member AB BA BC CB CD DC
D.F 0 5/9 4/9 6/11 5/11 0
F.E.M 6 -6 5 -5 4 -8
Step 4: Calculate the Distributing Moment (Ds.M) for each member by sum FEM of its
linked joint then revers the value sign then multiply with its D.F.
For example:
For joint B:
𝐷𝑠. 𝑀 𝐵𝐴 = −(𝑀 𝐵𝐴 + 𝑀 𝐵𝐶) ∗ 𝐷𝐹𝐵𝐴 = −(−6 + 5) ∗
5
9
= 0.556
𝐷𝑠. 𝑀 𝐵𝐶 = −(𝑀 𝐵𝐴 + 𝑀 𝐵𝐶) ∗ 𝐷𝐹𝐵𝐶 = −(−6 + 5) ∗
4
9
= 0.444
Joint A B C D
Member AB BA BC CB CD DC
D.F 0 5/9 4/9 6/11 5/11 0
F.E.M 6 -6 5 -5 4 -8
Ds.M 0.000 0.556 0.444 0.545 0.455 0.000
Step 5: Calculate the Carry Over Moment (C.O.M) between end of each member.
Note that we don’t distribute COM to the pure hinged of free ended joints.
Joint A B C D
Member AB BA BC CB CD DC
D.F 0 5/9 4/9 6/11 5/11 0
F.E.M 6 -6 5 -5 4 -8
Ds.M 0.000 0.556 0.444 0.545 0.455 0.000
C.O.M 0.278 0.000 0.273 0.222 0.000 0.227

3. 3
Eng. Abdulrahman Shaaban
Step 6: Repeat step 4&5 by making many cycles if needed to reach to equilibrium of
joints (3 cycles seems to be good).
Joint A B C D
Member AB BA BC CB CD DC
D.F 0 5/9 4/9 6/11 5/11 0
F.E.M 6 -6 5 -5 4 -8
Ds.M 0.000 0.556 0.444 0.545 0.455 0.000
C.O.M 0.278 0.000 0.273 0.222 0.000 0.227
Ds.M 0.000 -0.152 -0.121 -0.121 -0.101 0.000
C.O.M -0.076 0.000 -0.061 -0.061 0.000 -0.051
Ds.M 0.000 0.034 0.027 0.033 0.028 0.000
C.O.M 0.017 0.000 0.017 0.013 0.000 0.014
Step 7: Calculate Final Moment (F.M) by Sum all previous moments for each member.
Joint A B C D
Member AB BA BC CB CD DC
D.F 0 5/9 4/9 6/11 5/11 0
F.E.M 6 -6 5 -5 4 -8
Ds.M 0.000 0.556 0.444 0.545 0.455 0.000
C.O.M 0.278 0.000 0.273 0.222 0.000 0.227
Ds.M 0.000 -0.152 -0.121 -0.121 -0.101 0.000
C.O.M -0.076 0.000 -0.061 -0.061 0.000 -0.051
Ds.M 0.000 0.034 0.027 0.033 0.028 0.000
C.O.M 0.017 0.000 0.017 0.013 0.000 0.014
F.M 6.22 -5.56 5.58 -4.37 4.38 -7.81
Step 7: Draw B.M.D
6.22 5.57
4.37
7.81
5.03
5.33

4. 4
Eng. Abdulrahman Shaaban
Step 7: Draw S.F.D
A B C D
6.22
5.57
4.37
7.81
2*6/2 = 6 6
(6.22-5.57)/6 = 0.108
4/2 = 2 2
(5.57-4.37)/10 = 0.12
9*2/6 = 3 9*4/6 = 6
(7.81-4.37)/6 = 0.57
6.108
2.12 2.12 2.43
-6.57
-1.88
-5.892

5. 5
Eng. Abdulrahman Shaaban
Example (2)
Draw B.M.D. & S.F.D. for the following Beams:
Step 1: Coding joint and Dived the beam into individual parts then Calculate Fixed End
Moment (F.E.M) for each part
Step 2: Calculate Stiffness for each member then calculate its Distribution Factor (D.F).
For member AB: 𝐾𝐴𝐵 =
3𝐸𝐼
𝐿
=
3 ∗ 𝐸𝐼
6
= 0.5 𝐸𝐼
For member BC: 𝐾 𝐵𝐶 =
4𝐸𝐼
𝐿
=
4 ∗ 2𝐸𝐼
8
= 1.0 𝐸𝐼
For member CD: 𝐾𝐶𝐷 =
4𝐸𝐼
𝐿
=
4 ∗ 𝐸𝐼
6
= 2 3⁄ 𝐸𝐼
For joint B:
𝐷𝐹𝐵𝐴 =
𝐾 𝐵𝐴
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
=
0.5
1 + 0.5
= 1 3⁄
𝐷𝐹𝐵𝑐 =
𝐾 𝐵𝑐
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
=
1
1 + 0.5
= 2 3⁄
A B C D
-12+6/2 = -9 16 -16 9 -9
A
B C D

6. 6
Eng. Abdulrahman Shaaban
For joint C:
𝐷𝐹𝐶𝐵 =
𝐾𝐶𝐵
𝐾𝐶𝐵 + 𝐾𝐶𝐷
=
1
1 + 2 3⁄
= 0.6
𝐷𝐹𝐶𝐷 =
𝐾𝐶𝐷
𝐾𝐶𝐵 + 𝐾𝐶𝐷
=
2 3⁄
1 + 2 3⁄
= 0.4
Step 3: Construct a table to summarize results and facilitation of calculation procedures.
Joint A B C D
Member AB BA BC CB CD DC
D.F 0.00 0.33 0.67 0.60 0.40 0.00
F.E.M 6 -9 16 -16 9 -9
Step 4: Calculate the Distributing Moment (Ds.M) for each member by sum FEM of its
linked joint then revers the value sign then multiply with its D.F.
Joint A B C D
Member AB BA BC CB CD DC
D.F 0.00 0.33 0.67 0.60 0.40 0.00
F.E.M 6 -9 16 -16 9 -9
DSm 0.00 -2.33 -4.67 4.20 2.80 0.00
Step 5: Calculate the Carry Over Moment (C.O.M) between end of each member.
Note that we don’t distribute COM to the pure hinged of free ended joints.
Joint A B C D
Member AB BA BC CB CD DC
D.F 0.00 0.33 0.67 0.60 0.40 0.00
F.E.M 6 -9 16 -16 9 -9
DSm 0.00 -2.33 -4.67 4.20 2.80 0.00
Com 0.00 0 2.1 -2.33333 0 1.4

7. 7
Eng. Abdulrahman Shaaban
Step 6: Repeat step 4&5 by making many cycles if needed to reach to equilibrium of
joints (3 cycles seems to be good).
Joint A B C D
Member AB BA BC CB CD DC
D.F 0.00 0.33 0.67 0.60 0.40 0.00
F.E.M 6 -9 16 -16 9 -9
DSm 0.00 -2.33 -4.67 4.20 2.80 0.00
Com 0.00 0 2.1 -2.33333 0 1.4
DSm 0.00 -0.70 -1.40 1.40 0.93 0.00
Com 0.00 0 0.7 -0.7 0 0.4667
DSm 0.00 -0.23 -0.47 0.42 0.28 0.00
Com 0.00 0 0.21 -0.23333 0 0.14
DSm 0.00 -0.07 -0.14 0.14 0.09 0.00
Com 0.00 0 0.07 -0.07 0 0.0467
Step 7: Calculate Final Moment (F.M) by Sum all previous moments for each member.
Joint A B C D
Member AB BA BC CB CD DC
D.F 0.00 0.33 0.67 0.60 0.40 0.00
F.E.M 6 -9 16 -16 9 -9
DSm 0.00 -2.33 -4.67 4.20 2.80 0.00
Com 0.00 0 2.1 -2.33333 0 1.4
DSm 0.00 -0.70 -1.40 1.40 0.93 0.00
Com 0.00 0 0.7 -0.7 0 0.4667
DSm 0.00 -0.23 -0.47 0.42 0.28 0.00
Com 0.00 0 0.21 -0.23333 0 0.14
DSm 0.00 -0.07 -0.14 0.14 0.09 0.00
Com 0.00 0 0.07 -0.07 0 0.0467
FM 6 -12.34 12.41 -13.18 13.11 -6.95

8. 8
Eng. Abdulrahman Shaaban
Step 7: Draw B.M.D
Step 7: Draw S.F.D
6
12.37 13.15
6.95
A
B C D
6.0 t.m
12.37 13.15
6.95
2*6/2 = 6 6
(12.37-6)/6 = 1.062
3
3*8/2 = 12 12
(13.15-12.37)/8 = 0.097
12/2 = 6 6
(13.15-6.95)/6 = 1.03
-3
4.938
11.9
7.03
-4.97
-4.09
-1.06
1.75
3.87 7.95
3.0 t
-7.06

9. 9
Eng. Abdulrahman Shaaban
Fixed End Moment Table: