Solved Examples on statically indeterminate beams structure by using moment distribution method

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: