Analysis of beams And Frames Using the moment Distribution Method ,, Depending on the Geometry and the material of the beam in consideration.

1. DISPLACEMENT MEDTHOD OF ANALYSIS:
MOMENT DISTRIBUTION
! Member Stiffness Factor (K)
! Distribution Factor (DF)
! Carry-Over Factor
! Distribution of Couple at Node
! Moment Distribution for Beams
! General Beams
! Symmetric Beams
! Moment Distribution for Frames: No Sidesway
! Moment Distribution for Frames: Sidesway
1

2. Member Stiffness Factor (K) & Carry-Over Factor (COF)
CB
P w
EI
A B C D
L1 /2 L1/2 L2 L3
Internal members and far-end member fixed at end support:
4 EI COF = 0.5 2 EI
kCC = k DC =
4 EI L L
K=
L 1
C D
COF = 0.5
B C
K(BC) = 4EI/L2, K(CD) = 4EI/L3
2

3. CB
P w
EI
A B C D
L1 /2 L1/2 L2 L3
Far-end member pinned or roller end support:
COF = 0
3EI
3EI k AA = 0
K= L
L 1
C D
K(AB) = 3EI/L1, K(BC) = 4EI/L2, K(CD) = 4EI/L3
3

4. Joint Stiffness Factor (K)
CB
P w
EI
A B C D
L1 /2 L1/2 L2 L3
K(AB) = 3EI/L1 K(BC) = 4EI/L2, K(CD) = 4EI/L3
Kjoint = KT = ΣKmember
4

5. Distribution Factor (DF)
CB
P w
EI
A B C D
L1 /2 L1/2 L2 L3
K
DF =
ΣK
Notes:
- far-end pined (DF = 1)
- far-end fixed (DF = 0)
A KAB/(K(AB) + K(BC) ) B KBC/(K(AB) + K(BC) ) C K(CD)/(K(BC) + K(CD) ) D
DF 1 DFBA) DFBC DFCB DFCD 0
K(BC)/(K(BC) + K(CD) )
5

6. Distribution of Couple at Node
CB
(EI)1 (EI)2 (EI)3
A B C D
L1 /2 L1/2 L2 L3
A B C D
DF 1 DFBA DFBC DFCA DFCD 0
CB
CO=0 CB(DFBC) CO=0.5
CB( DFBC)
CB(DFBC) B CB( DFBC)
6

7. 50 kN•m
2EI 3EI 3EI
A B C D
4m 4m 8m 8m
L1= L2 = L3
A B C D
DF 1 0.333 0.667 0.5 0.5 0
50 kN•m
CO=0 CO=0.5
50(.667) 50(.667) 16.67
50(.333)
50(.333) B
7

8. Distribution of Fixed-End Moments
P w
(EI)1 M (EI)2 (EI)13
A F B C D
L1 /2 L1/2 L2 L3
L1= L2 = 8 m, L3 = 10 m
A B C D
DF 1 DFBA DFBC DFCB DFCD 0
MF MF B
MF
0 MF(DFBC) MF( DFBC) 0.5
0 MF(DFBC) B MF( DFBC)
8

9. P w
EI
A 1.5PL1/8=30 B wL22/12=16 C wL32/12=25 D
L1 /2 L1/2 L2 L3
L1= L2 = 8 m, L3 = 10 m
A B C D
DF 1 0.4 0.6 0.5 0.5 0
14
30 30 B 16 16
14
0 5.6 5.6 0.5 4.2
0 B 8.4 8.4
9

10. Moment Distribution for Beams
10

11. Example 1
The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ).
Use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams,and qualitative
deflected shape.
20 kN 3 kN/m
2EI 3EI
A B C
4m 4m 8m
11

12. 20 kN 3 kN/m
2EI 3EI
30 16 16
A B C
CO 0.5 0 0.5 0
4m 4m 8m
K1 = 3(2EI)/8 K2 = 4(3EI)/8
K1/(K1+ K2) K2/(K1+ K2)
DF 1 0.333 0.667 0
[FEM]load -30 16 -16
Dist. 4.662 9.338
CO 4.669
Σ -25.34 25.34 -11.33
20 kN 24 kN
A B B C
25.34 kN•m 11.33 kN•m
6.83 kN 13.17 kN 13.75 kN 10.25 kN 12

13. Note : Using the Slope 20 kN 3 kN/m
Deflection
2EI 3EI
20+10 16 16
A B C
4m 4m 8m
3(2 EI )
M BA = θ B − 30 − − − (1)
8
4(3EI )
M BC = θ B + 16 − − − (2)
8
MBA MBC
+ ΣMB = 0: -MBA - MBC = 0
B
(0.75 + 1.5)EIθB - 30 + 16 = 0
θB = 6.22/EI
MBA = -25.33 kN•m,
MBC = 25.33 kN•m
2(3EI )
M CB = θ B − 16 = −11.33 kN • m
8 13

14. 20 kN 3 kN/m
A C
11.33
B
6.83 kN
13.17 + 13.75 = 26.92 kN 10.25 kN
4m 4m 8m
13.75
6.83
V (kN)
+ + x (m)
- 4.58 m - -10.25
-13.17
27.32
6.13
M + +
(kN•m)
- - x (m)
-11.33
-25.33
Deflected
shape x (m)
14

15. Example 2
From the beam shown use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams,
and qualitative deflected shape.
10 kN 50 kN•m 5 kN/m
2EI 3EI 3EI
A B C D
4m 4m 8m 8m
15

16. 10 kN 50 kN•m 5 kN/m
2EI 15
26.67 26.67 40 3EI
A 0.5 B 0.5 C 0.5 D
4m 4m 8m 8m
K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8
K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3)
DF 1 0.333 0.667 0.571 0.428 1
Joint couple -16.65 -33.35
50 kN•m
CO=0 CO=0.5
50(.667) 50(.667)
50(.333)
50(.333) B
16

17. 10 kN 50 kN•m 5 kN/m
2EI 15
26.67 26.67 40 3EI
A 0.5 B 0.5 C 0.5 D
4m 4m 8m 8m
K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8
K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3)
DF 1 0.333 0.667 0.571 0.429 1
Joint couple -16.65 -33.35
CO -16.675
FEM -15 26.667 -26.667 40
Dist. -3.885 -7.782 1.905 1.437
CO 0.953 -3.891
Dist. -0.317 -0.636 2.218 1.673
CO 1.109 -0.318
Dist. -0.369 -0.740 0.181 0.137
Σ -36.22 -13.78 -43.28 43.25
17

18. 10 kN 50 kN•m 5 kN/m
2EI 36.22
13.78 43.25 43.25 3EI
A B C D
4m 4m 8m 8m
10 kN
40 kN
36.22 kN•m
A B 43.25 kN•m
C D
Ay = 0.47 kN ByL = 9.53 kN CyR = 25.41 kN Dy = 14.59 kN
40 kN
13.78 kN•m 43.25 kN•m
B C
ByR = 12.87 kN CyL= 27.13 kN
18

19. 10 kN 50 kN•m 5 kN/m
A D
2EI B C 3EI
0.47 kN 14.59 kN
9.53+12.87=22.4 kN 27.13+25.41=52.54 kN
4m 4m 8m 8m
12.87 25.41 2.92 m
V (kN) 0.47
x (m)
2.57 m
-9.53 -14.59
30.32 -27.13 21.29
13.78
M(kN•m) 1.88
x (m)
-36.22
-43.25
Deflected
x (m)
shape
19

20. Example 3
From the beam shown use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams, and
qualitative deflected shape.
2EI 40 kN 50 kN•m 10 kN/m 40 kN
A D
EI
B C
3m 3m 9m 3m
20

21. 120 kN•m 40 kN
2EI 40 kN 50 kN•m 10 kN/m
A D
30 30 101.25 EI
B C
0.5 0.5
3m 3m 9m 3m
K1 = 4(2EI)/6 K2 = 3(EI)/9
K1/(K1+ K2) K2/(K1+ K2)
DF 0 0.80 0.20 1
Joint couple 40 10 -120
CO 20 -60
FEM 30 -30 101.25
Dist.
Dist. -9 -2.25
CO -4.5
Σ 45.5 1 49 -120
21

22. 2EI 40 kN 50 kN•m 10 kN/m 40 kN
1
A D
49 120 120
45.5 EI
B C
3m 3m 9m 3m
40 kN
40 kN
45.5 kN•m 1 kN•m
A B 120 kN•m
C D
Ay = 27.75 kN ByL = 12.25 kN CyR = 40 kN
90 kN
49 kN•m 120 kN•m
B C
ByR = 37.11 kN CyL = 52.89 kN
22

23. 2EI 40 kN 50 kN•m 10 kN/m 40 kN
A D
EI
B C
27.75 kN
12.25+37.11 = 49.36 kN
52.89+40 = 92.89 kN
3m 3m 9m 3m
37.11 40
27.75
V (kN) + + +
- x (m)
-12.25 3.71 m -
37.75 -52.89
19.84
M(kN•m) + 1 + x (m)
- -
-
-45.5 -49
-120
Deflected x (m)
shape
23

24. Example 4
The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ) settles 10 mm.
Use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams,
and qualitative deflected shape.
20 kN 15 kN•m
12 kN•m 3 kN/m
2EI 3EI
A B C
4m 4m 8m
24

25. 20 kN 15 kN•m
12 kN•m 3 kN/m
B
A C
2EI 30 16 3EI
0.5 4 m 16
4m 10 mm 8 m
0.5
6(2 EI )∆ 6(2 EI )∆ 6(3EI ) ∆
A − = 9.375 6(3EI ) ∆ = 28.125
L 2
2L 2 = 28.125 L2
L2
∆ [FEM]∆ ∆
B B
6(2 EI )∆ C
L2 K1 = 3(2EI)/8 K2 = 4(3EI)/8
K1/(K1+ K2) K2/(K1+ K2)
DF 1 0.333 0.667 0
Joint couple -12 5 10
CO -6 5
[FEM]load -30 16 -16
[FEM]∆ 9.375 -28.125 -28.125
Dist. 12.90 25.85
CO 12.92
Σ -12 -8.72 23.72 -26.20 25

26. Note : Using the 20 kN 18.75-9.375 15 kN•m
3 kN/m
slope deflection 12 kN•m (9.375 )∆
B (28.125)∆
A C
2EI 3EI 28.125
(20+10)load (16)load 16
4m 4m 10 mm 8m
15 kN•m
-12 MBA MBC
4(2 EI ) 2(2 EI )
M AB = θA + θ B + 20 − 18.75 − − − (1)
8 8
B
2(2 EI ) 4(2 EI )
M BA = θA + θ B − 20 + 18.75 − − − (2)
8 8 + ΣMB = 0: - MBA - MBC + 15 = 0
( 2) − (1) 3(2 EI )
: M BA = θ B − 30 + 9.375 − 12 / 2 − − − (2a)
2 8 (0.75 + 1.5)EIθB - 38.75 - 15 = 0
4(3EI ) θB = 23.9/EI
M BC = θ B + 16 − 28.125 − − − (3)
8
MBA = -8.7 kN•m,
2(3EI )
M CB = θ B − 16 − 28.125 − − − (4)
8 MBC = 23.72 kN•m
2(3EI )
M CB = θ B − 16 − 28.125
8
= −26.2 kN • m 26

27. 20 kN 15 kN•m
12 kN•m 3 kN/m
2EI 8.725 23.7253EI 26.205
A B C
4m 4m 8m
20 kN 24 kN
12 kN•m
8.725 kN•m 26.205 kN•m
A B B C
23.725 kN•m
Ay = 7.41 kN ByL = 12.59 kN ByR = 11.69 kN Cy = 12.31 kN
27

28. 20 kN 15 kN•m
12 kN•m 3 kN/m
A C
2EI B 3EI 26.205
7.41 kN
12.59+11.69 = 24.28 kN 12.31 kN
4m 4m 8m
7.41 11.69
V (kN)
+ + x (m)
- 3.9 m - -12.31
-12.59
41.64
12
+
M
(kN•m) x (m)
-8.725 - -0.93 -
-23.725 -26.205
Deflected 10 mm θB = 23.9/EI
shape x (m)
28

29. Example 5
For the beam shown, support A settles 10 mm downward, use the moment
distribution method to
(a)Determine all the reactions at supports
(b)Draw its quantitative shear, bending moment diagrams, and qualitative
deflected shape.
Take E= 200 GPa, I = 50(106) mm4.
12 kN•m
6 kN/m
B
A
2EI C 1.5EI 10 mm
3m 3m
29

30. 12 kN•m
4.5+(4.5/2) 6 kN/m
= 6.75
B
A
4.5 10 mm
2EI C 1.5EI
3m 0.5 3 m 0.5
100-(100/2) = 50 6(1.5 × 200 × 50)(0.01)
0.01 m
C 32
MF∆
A = 100 kN • m
K1 = 4(2EI)/3 K2 = 3(1.5EI)/3
K1/(K1+ K2) K2/(K1+ K2)
DF 0 0.64 0.36 1
Joint couple 12
CO 6
[FEM]load 6.75
[FEM]∆ 50
Dist. -40.16 -22.59
CO -20.08
Σ -20.08 -40.16 40.16 12 30

31. 12 kN•m
6 kN/m
B
A
2EI C 1.5EI
3m 3m
ΣΜ -20.08 -10.16 40.16 12
20.08 kN•m 40.16 kN•m
B C
40.16 + 20.08
20.08 kN = 20.08 kN
3
18 kN
12 kN•m
6 kN/m
40.16 kN•m
C A
26.39 kN 8.39 kN
31

32. 12 kN•m
6 kN/m
B
A
2EI C 1.5EI 10 mm
3m 3m
20.08 kN•m 40.16 kN•m
B C
20.08 kN 20.08 kN 12 kN•m
6 kN/m
40.16 kN•m
C A
26.39 kN 8.39 kN
V (kN) 26.39
+ 8.39
x (m)
-
M (kN•m) -20.08
20.08 12
x (m)
Deflected shape -40.16
x (m)
32

33. Example 6
For the beam shown, support A settles 10 mm downward, use the moment
distribution method to
(a)Determine all the reactions at supports
(b)Draw its quantitative shear, bending moment diagrams, and qualitative
deflected shape.
Take E= 200 GPa, I = 50(106) mm4.
12 kN•m
6 kN/m
B
A
2EI C 1.5EI 10 mm
3m 3m
33

34. • Overview 12 kN•m
6 kN/m
B
A
C 10 mm
2EI 1.5EI
3m 3m
12 kN•m
6 kN/m
B
A
10 mm
C
R
∆
B ×C
A
R´
R + R ' C = 0 − − − (1*)
34

35. • Artificial joint applied 12 kN•m
4.5+(4.5/2) 6 kN/m
= 6.75
B
A
4.5 10 mm
2EI C 1.5EI
3m 0.5 3 m 0.5
100-(100/2) = 50 6(1.5 × 200 × 50)(0.01)
0.01 m
C 32
MF∆
A = 100 kN • m
K1 = 4(2EI)/3 K2 = 3(1.5EI)/3
K1/(K1+ K2) K2/(K1+ K2)
DF 0 0.64 0.36 1
Joint couple 12
CO 6
[FEM]load 6.75
[FEM]∆ 50
Dist. -40.16 -22.59
CO -20.08
Σ -20.08 -40.16 40.16 12 35

36. 12 kN•m
6 kN/m
B
A
2EI C 1.5EI
3m 3m
ΣΜ -20.08 -40.16 40.16 12
20.08 kN•m 40.16 kN•m
B C
40.16 + 20.08 18 kN
20.08 kN = 20.08 kN 12 kN•m
3 6 kN/m
40.16 kN•m
C A
26.39 kN 8.39 kN
40.16 kN•m
40.16 kN•m
20.08 C 26.39 kN + ↑ ΣFy = 0 : − 20.08 − 26.39 + R = 0
R
R = 46.47 kN 36

37. • Artificial joint removed
∆
B
A
C
2EI 0.5 1.5EI 0.5
3m R´ 3m
C C
75
B ∆ ∆ 6(1.5EI )( )
100 EI = 75
75-(75/2) A 32
6(2 EI ) ∆ C 75
= 100 → ∆ = = 37.5
32 EI
DF 0 0.64 0.36 1
[FEM]∆ -100 -100 +37.5
Dist. 40 22.5
CO 20
Σ -80 -60 60
80 kN•m 60 kN•m 60 kN•m
B C C A
46.67 kN 46.67 kN 20 kN 20 kN
C
46.67 20 kN + ↑ ΣFy = 0 : R' = 66.67 kN 37
R´

38. • Solve equation
Substitute R = 46.47 kN and R ' = 66.67 kN in (1*)
46.47 + 66.67C = 0
C = −0.6970
12 kN•m
6 kN/m
B
20.08 kN•m A
10 mm
20.08 kN C
R = 46.47 kN 8.39 kN
∆
B
80 kN•m A × C = −0.6970
46.67 kN 20 kN
R´ = 66.67 kN
12 kN•m
6 kN/m
35.68 kN•m B
A
2EI C 1.5EI
12.45 kN 5.55 kN 38

39. 12 kN•m
6 kN/m
35.68 kN•m B
A
2EI C 1.5EI 10 mm
12.45 kN 5.55 kN
3m 3m
V (kN) 12.45 0.925 m
+
x (m)
-5.55
M (kN•m) 14.57 12
1.67 +
x (m)
-
-35.68
Deflected shape
x (m)
39

40. Symmetric Beam
• Symmetric Beam and Loading
P P
θ θ
A B C D
L´ L L´
real beam
V´B L L V´C
M L
2 2 + ΣMC´ = 0: − VB ' ( L ) + ( L)( ) = 0
EI 2
B´ C´
ML
M VB ' = θ =
M 2 EI
L EI
EI
2 EI
conjugate beam M= θ
L
The stiffness factor for the center span is, therefore,
2 EI
K=
L
40

41. • Symmetric Beam with Antisymmetric Loading
P
θ
A D
B θ C
P
L´ L L´
real beam
1 M L M 1 M L 2L
( )( ) + ΣMC´ = 0: − VB ' ( L) + ( )( )( ) = 0
2 EI 2 EI 2 EI 2 3
V´B 2 ML
L VB ' = θ =
B´ 3 6 EI
C´
6 EI
1 M L V´C M= θ
M ( )( ) L
2 EI 2
EI The stiffness factor for the center span is, therefore,
conjugate beam
6 EI
K=
L 41

42. Example 5a
Determine all the reactions at supports for the beam below. EI is constant.
15 kN/m
A D
B C
4m 6m 4m
42

43. 15 kN/m
A D
wL2/15 = 16 B wL2/12 = 45 C wL2/15 = 16
4m 6m 4m
3EI 3EI 2 EI 2 EI
K ( AB ) = = , K ( BC ) = =
L 4 L 6
K ( AB ) K ( AB ) (3EI / 4)
( DF ) AB = = 1,( DF ) BA = = = 0.692,
K ( AB ) K ( AB ) + K ( BC ) (3EI / 4) + ( 2 EI / 6)
K ( BC ) (2 EI / 6)
( DF ) BC = = = 0.308
K ( AB ) + K ( BC ) (3EI / 4) + (2 EI / 6)
DF 1.0 0.692 0.308
[FEM]load 0 -16 +45
Dist. -20.07 -8.93
ΣΜ -36.07 +36.07
30 kN 30 kN
8 90 kN 8
36.07 kN•m 36.07 kN•m
3 3
A B B C C D
4m 3m 3m 4m
0.98 kN 29.02 kN 29.02 kN 0.98 kN 43
45 kN 45 kN

44. 30 kN 30 kN
8 90 kN 8
36.07 kN•m 36.07 kN•m
3 3
A B B C C D
4m 3m 3m 4m
0.98 kN 29.02 kN 29.02 kN 0.98 kN
45 kN 45 kN
15 kN/m
A D
B 74.02 kN C 74.02 kN
0.98 kN 0.98 kN
4m 6m 4m
45
29.02
V 0.98
(kN) x (m)
-0.98
-29.02 -45
31.42
M +
(kN•m) x (m)
- -
-36.07 -36.07
Deflected
shape 44

45. Example 5b
Determine all the reactions at supports for the beam below. EI is constant.
15 kN/m
A C D
B
15 kN/m
4m 3m 3m 4m
45

46. Fixed End Moment 15 kN/m
A C D
wL2/15 = 16 B 11wL2/192 5wL2/192
= 30.938 = 14.063
5wL2/192 11wL2/192
= 14.063 = 30.938 wL2/15 = 16
A C D
B
15 kN/m
15 kN/m
16.875 16
A C D
16 B 16.875
15 kN/m
46

47. 15 kN/m
16.875 16
A C D
16 B 16.875
15 kN/m
4m 3m 3m 4m
3EI 3EI 6 EI 6 EI
K ( AB ) = = = 0.75EI , K ( BC ) = = = EI
L 4 L 6
0.75 1
( DF ) AB = 1, ( DF ) BA = = 0.429, ( DF ) BC = = 0.571
0.75 + 1 0.75 + 1
DF 1.0 0.429 0.571
[FEM]load 0 -16 16.875
Dist. -0.375 -0.50
ΣΜ -16.375 16.375
8
30 kN
8 45 kN 3
16.375 kN•m 16.375 kN•m 4m
3
A B B C C D
4m
5.91 kN 24.09 kN 27.96 kN 5.91 kN
27.96 kN 45 kN 24.09 kN 30 kN 47

48. 8
30 kN
8 45 kN 3
16.375 kN•m 16.375 kN•m 4m
3
A B B C C D
4m
5.91 kN 24.09 kN 27.96 kN 5.91 kN
27.96 kN 45 kN 24.09 kN 30 kN
15 kN/m
A C D
B
15 kN/m
5.91 kN 52.05 kN 52.05 kN 5.91 kN
27.96 27.96
V 5.91 5.91
(kN) x (m)
-24.09 -24.09
M 16.375
(kN•m) x (m)
-16.375
Deflected shape
48

49. Moment Distribution Frames: No Sidesway
49

50. Example 6
From the frame shown use the moment distribution method to:
(a) Determine all the reactions at supports
(b) Draw its quantitative shear and bending moment diagrams,
and qualitative deflected shape.
40 kN•m
48 kN 8 kN/m
B C
A
2.5EI 2.5EI
4m
3EI
5m D 5m
50

51. 40 kN•m
48 kN 8 kN/m
B C
A
2.5EI 45 25 2.5EI
0.5 0.5
4m
KAB = KBC = 3(2.5EI)/5 = 1.5 EI 3EI 0.5
KBD = 4(3EI)/4 = 3EI
5m D 5m
A B D C
Member AB BA BC BD DB CB
DF 1 0.25 0.25 0.5 0 1
Joint load -10 -10 -20
CO -10
FEM -45 25
Dist. 5 5 10
CO 5
Σ 0 -50 20 -10 -5 0 51

52. 40 kN•m
48 kN 8 kN/m
B C
A
20
2.5EI 50 2.5EI
14 kN 10 16 kN
3EI
5 Member AB BA BC BD DB CB
D Σ 0 -50 20 -10 -5 0
40 kN•m
58 kN 34 24 3.75
50 20
48 kN 3.75 40 kN
A B 0 50 kN•m 58
10 B C 3.75
58 20 3.75
14 kN 34 kN 10 kN•m 16 kN
24 kN
3.75 kN
5
3.75 kN
58
52

53. 40 kN•m
48 kN 8 kN/m
B C
A
2.5EI 20 2.5EI
14 kN 50 16 kN
3EI 10 4m
5
D
5m 5m
58 kN
35
16
10
20
50 Moment diagram Deflected shape
5
53

54. Moment Distribution for Frames: Sidesway
54

55. Single Frames
P P
∆ ∆
C C R´
B C B R B
x C1
A D A D A D
Artificial joint applied Artificial joint removed
(no sidesway) (sidesway)
0 = R + C1R´
55

56. Multistory Frames ∆2 P3
P2
P4
P1 ∆1
P3
∆´´ ∆´´
P2 R2´´
R2 R2´
∆´ ∆´
P4
x C1 x C2
P1 R1´
R1 R1´´
0 =R2 + C1R2´ + C2R2´´
0 =R1 + C1R1´ + C2R2´´
56

57. Example 7
From the frame shown use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams, and
qualitative deflected shape.
EI is constant.
16 kN
1m 4m
B C
5m
A D
57

58. • Overview
16 kN 16 kN
1m 4m C C R´
B C B R B
5m 5m = + x C1
A D A D A D
artificial joint applied artificial joint removed
(no sidesway) ( sidesway)
R + C1R´ = 0 ---------(1)
58

59. • Artificial joint applied (no sidesway)
Fixed end moment:
16 kN
b=1m a=4m C
B R
Pa2b/L2 Pb2a/L2
= 10.24 = 2.56
A D
Equilibrium condition :
+ ΣF = 0: A + D + R = 0
x x x
59

60. 16 kN
1m 4m C
R A B C D
B
0.5 DF 0 0.50 0.50 0.50 0.50
10.24 2.56 0
FEM 10.24 -2.56
5m 5m Dist. -5.12 -5.12 1.28 1.28
0.5 0.5
CO -2.56 0.64 -2.56 0.64
A D Dist. -0.32 -0.32 1.28 1.28
CO -0.16 0.64 -0.16 0.64
5.78 kN•m Dist. -0.32 -0.32 0.08 0.08
2.72 kN•m
CO -0.16 0.04 -0.16 0.04
Dist. -0.02 -0.02 0.08 0.08
B C Σ -2.88 -5.78 5.78 -2.72 2.72 1.32
5m 5m Equilibrium condition :
+ ΣF = 0: 1.73 - 0.81 + R = 0
x
A Ax = 1.73 kN D
Dx = 0.81 kN R = - 0.92 kN
2.88 kN•m 1.32 kN•m
60

61. • Artificial joint removed ( sidesway)
Fixed end moment:
Since both B and C happen to be displaced the same amount ∆, and AB and DC
have the same E, I, and L so we will assume fixed-end moment to be 100 kN•m.
∆ ∆
B 5m C
100 kN•m R´
100 kN•m
5m 5m
100 kN•m 100 kN•m
A
D
Equilibrium condition :
+ ΣF = 0: A + D + R´ = 0
x x x
61

62. ∆ ∆
B 5m C
R´ A B C D
0.5 0.50 0.50
100 kN•m 100 kN•m DF 0 0.50 0.50 0
FEM 100 100 100 100
5m 5m Dist. -50 -50 -50 -50
0.5 0.5
100 kN•m 100 kN•m CO -25.0 -25.0 -25.0 -25.0
A Dist. 12.5 12.5 12.5 12.5
D
CO 6.5 6.5 6.5 6.5
Dist. -3.125 -3.125 -3.125 -3.125
60 kN•m 60 kN•m CO -1.56 -1.56 -1.56 -1.56
Dist. 0.78 0.78 0.78 0.78
CO 0.39 0.39 0.39 0.39
B C
Dist. -0.195 -0.195 -0.195 -0.195
Σ 80 60 -60 -60 60 80
5m 5m
Equilibrium condition: + ΣFx = 0:
A Ax = 28 kN D
Dx = 28 kN -28 - 28 + R´ = 0
80 kN•m 80 kN•m R´ = 56 kN
62

63. Substitute R = -0.92 and R´= 56 in (1) :
R + C1R´ = 0
-0.92 + C1(56) = 0
0.92
C1 =
56
16 kN 16 kN
B C C R´ 1m 4m C
R B B
5.78 2.72 60 60 4.79 3.71
2.72 60 60 3.71
5.78 4.79
+ x C1 = 5m 5m
2.88 80 80 1.57 2.63
1.32
1.73 kN 0.81 28 28 1.27 kN
A D A D A D 1.27
63

64. 8.22
16 kN
3.71
1m 4m C
B 4.79
4.79 3.71 3.71
4.79
4.79 3.71
5m 5m
1.57 2.63
1.27 kN 1.27 kN 1.57 2.63
A D 2.99 kN Bending moment
13.01 kN diagram (kN•m)
∆ ∆
Deflected shape 64

65. Example 8
From the frame shown use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams, and
qualitative deflected shape.
20 kN/m 3m
B pin
C
3EI
3m 2EI
4EI 4m
A
D
65

66. • Overview
20 kN/m
B 3m, 3EI B C B C
C
R R´
3 m , 2EI
4m , 4EI
= + x C1
A A A
D D D
artificial joint applied artificial joint removed
(no sidesway) (sidesway)
R + C1R´ = 0 ----------(1)
66

67. • Artificial joint applied (no sidesway)
20 kN/m A B C D
B 3m , 3EI C R 0.471 0.529
DF 0 1.00 1.00 0
0.5
15 FEM 15.00 -15.00
0.5 Dist. 7.065 7.935
0.5
CO 3.533
3 m , 2EI 4m, 4EI
15 Σ 18.53 -7.94 7.94
A
D 7.94 kN•m
KBA = 4(2EI)/3 = 2.667EI
B C + ΣF = 0:
x
KBC = 3(3EI)/3 = 3EI
60 - 33.53 - 0 + R = 0
60
KCD = 3(4EI)/4 = 3EI 3m 4m
R = - 26.47 kN
A Ax = 33.53 kN D Dx = 0
18.53 kN•m 0
67

68. • Artificial joint removed ( sidesway)
• Fixed end moment
∆ ∆ ∆ ∆
B 3m, 3EI C B 3m, 3EI C
R´ R´
100 kN•m
6(2EI∆)/(3) 2 6(4EI)∆/(4) 2 100 kN•m
3 m, 2EI
100 kN•m 4m, 4EI 3 m, 2EI 4m, 4EI
6(2EI∆)/(3) 2 100 kN•m
3(4EI∆)/(4) 2 3(4EI)(75/EI)/(4) 2
A A = 56.25 kN•m
D D
Assign a value of (FEM)AB = (FEM)BA = 100 kN•m
6(2 EI ) ∆
2
= 100
3
∆AB = 75/EI
68

69. ∆ ∆
A B C D
B 3m C R´
0.5 DF 0 0.471 0.529 1.00 1.00 0
100
FEM 100 100 56.25
0.5
Dist. -47.1 -52.9 0
0.5
3m 4m CO -28.55
100
Σ 76.45 52.9 -52.9 56.25
56.25
A
D
52.9. kN•m
+ ΣF = 0:
C x
B -43.12 - 14.06 + R´ = 0
3m 4m R´ = 57.18 kN
A Ax = 43.12kN
D 14.06 kN
76.45 kN•m
56.25 kN•m 69

70. Substitute R = -26.37 and R´= 57.18 in (1) :
R + C1R´ = 0
-26.47 + C1(57.18) = 0
26.47
C1 =
57.18
20 kN/m
B B 52.9 kN•m B
C C 3m
R C
7.94 kN•m R´ 16.55 kN•m
7.94 kN•m 52.9
= 3m
+ 76.45 kN•m
x C1 4m
18.53 kN•m 53.92 kN•m
33.53 kN 43.12 kN 53.49 kN
A A 56.25
0 A 26.04 kN•m
5.52 kN kN
6.51
0
D D 14.06 D
5.52 kN
70

71. 20 kN/m
16.55
B 3m 16.55
C B C
16.55 kN•m
3m
53.92 kN•m 4m
53.92
53.49 kN
A 26.04 kN•m A
5.52 kN kN
6.51 26.04
Moment diagram
∆ ∆ D
D
5.52 kN
B C
A
Deflected shape
71
D

72. Example 8
From the frame shown use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams,
and qualitative deflected shape.
EI is constant.
10 kN B
C
4m
A D
4m 3m
72

73. • Overview
10 kN B
C
4m
A D
4m 3m
R + C1R´ = 0 ---------(1)
=
10 kN B C B C
R R´
+ x C1
A D A D
artificial joint applied artificial joint removed
(no sidesway) (sidesway)
73

74. • Artificial joint applied (no sidesway)
10 kN B C
R
0 0
A D
Equilibrium condition : + ΣF = 0:
x
10 + R = 0
R = - 10 kN
74

75. • Artificial joint removed (sidesway)
• Fixed end moment 6EI∆BC/(4) 2
B
R´
6EI∆AB/(4) 2
C 3EI∆CD/(5) 2
100 kN•m 100 kN•m
6EI∆CD/(5) 2 4 m
6EI∆AB/(4) 2
A D
4m 3m
Assign a value of (FEM)AB = (FEM)BA = 100 kN•m : 6 EI∆ AB = 100
42
∆AB = 266.667/EI
∆ CD ∆BC
B∆ C´
R´
∆ ∆CD = ∆ / cos 36.87° = 1.25 ∆ = 1.25(266.667/EI)
B´ C = 333.334/EI
36.87° C´
4m ∆CD
∆BC = ∆ tan 36.87° = 0.75 ∆
36.87° = 0.75(266.667/EI)
A D C ∆AB = ∆ = 200/EI
4m 3m
75

76. ∆BC= 200/EI, ∆CD = 333.334/EI
6EI∆BC/(4) 2 = 6(200)/42 = 75 kN•m
B
R´
100 kN•m
C 3EI∆CD/(5) 2 = 3(333.334)/52 = 40 kN•m
100 kN•m
A D
Equilibrium condition :
+ ΣF = 0: A + D + R´ = 0
x x x
76

77. 75 A B C D
B 75
R´ DF 0 0.50 0.50 0.625 0.375 1
0.5 C
100 FEM 100 100 -75 -75 40
40
4 m Dist. -12.5 -12.5 21.875 13.125
0.5 0.5
100 CO -6.25 10.938 -6.25
A D Dist. -5.469 -5.469 3.906 2.344
4m 3m CO -2.735 1.953 -2.735
Dist. -0.977 -0.977 1.709 1.026
KBA = 4EI/4 = EI, KBC = 4EI/4 = EI,
KCD = 3EI/5 = 0.6EI Σ 91.02 81.05 -81.05 -56.48 56.48
34.38 kN 34.38 kN
81.05 81.05 B C 56.48
B 56.48 C
34.38 kN 34.38 kN
+ ΣF = 0: 39.91 kN
A 43.02 kN x D
91.02 -43.02 - 39.91 + R´ = 0
34.38 kN R´ = 82.93 kN 34.38 kN
77

78. Substitute R = -10 kN and R´= 82.93 kN in (1) : -10 + C1(82.93) = 0
R + C1R´ = 0 ---------(1) C1 = 10/82.93
81.05
10 kN B C B C
R R´
81.05 56.48
x C1= 10/82.93
+ 56.48
0 0 91.02 39.91 kN
A 0 D A 43.02 kN D
0
=
34.38 kN
0 34.38 kN
9.77
10 kN B
C
9.77 6.81
6.81 4m
10.98 4.81 kN
A 5.19 kN
D
4.15 kN 4.15 kN
4m 3m
78

79. 9.77
10 kN B
C
9.77 6.81
6.81 4m
10.98
A 5.19 kN 4.81 kN
D
4.15 kN 4.15 kN
4m 3m
9.77
B B
9.77 C C
6.81
A 10.98 A
D D
Bending moment diagram Deflected shape
(kN•m)
79

80. Example 9
From the frame shown use the moment distribution method to:
(a) Determine all the reactions at supports, and also
(b) Draw its quantitative shear and bending moment diagrams,and
qualitative deflected shape.
EI is constant.
40 kN
20 kN B
C
3EI
3m
4EI
4EI 4m
A
D
2m 3m 2m
80

81. 40 kN
• Overview
20 kN B
C
3EI
3m
4EI
4EI 4m
A
D
2m 3m 2m
R + C1R´ = 0 ----------(1)
40 kN
=
20 kN B C B C
R R´
+ x C1
artificial joint applied artificial joint removed
(no sidesway) (no sidesway)
A A
D D
2m 3m 2m 2m 3m 2m
81

82. • Artificial joint applied (no sidesway)
Fixed end moments:
40 kN
PL/8 = 15
B C
20 kN R
15+(15/2)
= 22.5 kN•m
A
D
2m 3m 2m
Equilibrium condition :
+ ΣF = 0: A + D + R = 0
x x x
82

83. A B C D
40 kN
20 kN B C DF 0 0.60 0.40 1.00 1.00 0
R
22.5 FEM 22.5
15
0.5 Dist. -13.5 -9.0
0.5 0.5 CO -6.75
Σ -6.75 -13.5 13.5
A
D
2m 3m 2m KBA = 4(4EI)/3.6 = 4.444EI, KBC = 3(3EI)/3 = 3EI,
24.5 kN 15.5 kN
13.5 kN•m
C + ΣF = 0:
B x
40 kN 23.08 + 20 -7.75 + R´ = 0
13.5 B C R´ = - 35.33 kN
A 7.75 kN
23.08 kN
24.5 kN 15.5 kN D
6.75 kN•m 0
24.5 kN 15.5 kN
83

84. • Artificial joint removed (sidesway)
Fixed end moments: 3m
B C
R´
m
3EI
06
4. 4
3.6
72
m
4EI
4EI
A
D
3(3EI)∆BC/(3) 2 6(3EI)∆BC/(3) 2
100 kN•m
R´
6(4EI)∆AB/(3.606) 2 B C
6(4EI)∆CD/(4.47) 2
100 kN•m
6(4EI)∆AB/(3.61) 2
3(4EI)∆CD/(4.472) 2
A
D
Assign a value of (FEM)AB = (FEM)BA = 100 kN•m : 6(4 EI ) ∆ AB = 100, ∆AB = 54.18/EI
3.612 84

85. CC´ ∆ = ∆ABcos 33.69° = 45.08/EI
B BB´ C C´
R´ C´
B´ ∆ CD
.69 o C ∆ tan 26.57 = 22.54/EI
26.57°
26.57°
33
B∆ 33.69o
AB
=5 ∆ tan 33.69 = 30.05/EI
4.3
A /E I
D
B´
∆ BC = B' C ' = 22.54 / EI + 30.05 / EI = 52.59 / EI
∆CD = ∆/cos 26.57°= 50.4/EI
3(3EI)∆BC/(3) 2 = 3(3EI)(52.59/EI) /(3) 2 = 52.59 kN•m
R´
100 kN•m B C
100 kN•m
3(4EI)∆CD/(4.472) 2
A = 3(4EI)(50.4/EI)/(4.472) 2
D
= 30.24 kN•m
85

86. A B C D
3EI
B 52.59 C DF 0 0.60 0.40 1.00 1.00 0
R´
100 0.5 FEM 100 100 -52.59 30.24
3m Dist. -28.45 -18.96
100 0.5 0.5 4m CO -14.223
4EI Σ 85.78 71.55 -71.55
4EI 30.24
A 30.24
D
2m 3m 2m
23.85 kN 23.85 kN
71.55 kN•m + ΣF = 0:
x
C
B -68.34 - 19.49 + R´ = 0
71.55 B C R´ = 87.83 kN
A 68.34 kN 23.85 23.85 19.49 kN
D
85.78kN•m 30.24 kN•m
23.85 kN
23.85 kN
86

87. Substitute R = -35.33 and R´= 87.83 in (1) : -35.33 + C1(87.83) = 0
40 kN C1 = 35.33/87.83
20 kN B C B C
35.33 kN 90.59 kN
13.5 kN•m 71.55 kN•m
+85.78 kN•m x C1
6.75 kN•m
23.08 kN 68.34 kN
0 30.24 kN•m
A 24.5 kN A 23.85 kN
7.75 kN 19.485 kN
D D
15.5 kN 23.85 kN
=
40 kN
20 kN B
C
15.28 kN•m
27.76 kN•m
4.41 kN
12.16 kN•m
A 14.91 kN
15.59 kN
D
25.09 kN 87