ss

1. 1
Lecture No. : 5 ‫المحاضرة‬
‫الخامسة‬

2. 2
Fixed End Moment
Relative Stiffness
Distribution Factor
Table :
FEM
RS
DF
Fixed End Moment
Distribution Moment
Carry Over Moment
Distribution Moment
Solution Steps :

3. 3
Fixed End Moment FEM
L
P
L
w
8
L
P
12
L
w 2
L
w
L
P
a b
w L2
20
w L2
30
P a b2
L2
P b a2
L2
Solution Steps :

4. 4
Relative Stiffness RS
K =
L
3 EI
K = 3/4 Ko
K = 1/2 Ko
Sym
K =
L
2 EI
K =
L
6 EI
Anti-Sym
K = 3/2 Ko
K =
L
4 EI
K = Ko
Solution Steps :

5. 5
Distribution Factor DF
1
3
2
1
1+2+3
2
1+2+3
3
1+2+3
1
6
1
3
1
2
Solution Steps :

6. 6
Table :
Fixed End Moment
‫تمهيدية‬ ‫مرحلة‬
Solution Steps :

7. 7
Table :
Fixed End Moment
Distribution Moment
Carry Over Moment
Distribution Moment
Carry Over Moment
Distribution Moment
Carry Over Moment
Distribution Moment
‫مرحلة‬
‫أولية‬
‫مرحلة‬
‫متكررة‬
‫مرحلة‬
‫متكررة‬
‫مرحلة‬
‫متكررة‬
Solution Steps :

8. 8
Table :
DF
FEM
DM
COM
DM
MFinal
COM
DM
COM
DM
Solution Steps :

9. 9
Frames
Without Sway With Sway
Symmetry With Support
Solution Steps :

10. 10
A D
2 t/m
8 t
8 t
B C
12
I I
4I
6
6
Solution Steps :

11. 11
Take it
In first DM
only
Solution Steps :

12. 12
Solution Steps :

13. 13
Advanced Example
in beams

14. 14
Beams with settlement
Fixed End Moment
D

15. 15
Using 3M equation method
Apply 3M eqn. at A :
L
EI
MA
0+ ( )
0+
2
L
EI
MB
+ ( ) = 6
D
L
6EI
MA
2 MB
+ =
D
L2
D
A B

16. 16
Apply 3M eqn. at B :
L
EI
MB
+ ( )
0+
2 0
+
L
EI
MA ( ) = 6
D
L
-6EI
MA 2MB
+ =
D
L2
D
A B

17. 17
6EI
MA
2 MB
+ =
D
L2
By addition
MA 0
MB
+ =
MA
MB
-
=
MA 2MB
+ =
-6EI D
L2
6EI
MA=
D
L2
MB =
-6EI D
L2

18. 18
D
A B
6EI
MA =
D
L2
MB =
-6EI D
L2
6EI D
L2
6EI D
L2
+
_

19. 19

20. 20
6
I
A
B
8
2I
C
EI=4800 t.m2
2 cm
FEM :
Example 16 :
6EI D
L2
6x4800x0.02
62
6x2x4800x0.02
82
-16 18
Draw B.M.D for the shown beam

21. 21
RS :
KAB : KBC
1
6
2
8
:
1
6
1
4
:
:
2 3
6
I
A
B
8
2I
C

22. 22
DF :
DFAB : DFBC
2
2+3
3
2+3
:
:
0.4 0.6
KAB : KBC
:
2 3
6
I
A
B
8
2I
C

23. 23
A B C
DF 0.4 0.6
FEM - 16 -16 +18 +18
+2
DM 0 0
-
-0.8 1.2
COM -0.4 -0.6
0 0
DM 0 0
0 0
MFinal -16.4 -16.8 16.8 17.4
16.4 16.8 17.4
16.8

24. 24
16.4
16.8
17.4

25. 25
Frames
with Sway

26. 26
A D
2 t/m
8 t
B C
12
I I
2 I
6
Example 17 :
Draw B.M.D & N.F.D and S.F.D for the shown frame

27. 27
A D
2 t/m
8 t
B C
12
6
FEM :
M FBC
12
L
w 2
=
12
12
2 2

= = - 24 tm
8
L
P
M FAB = =
8
6
8
= - 6 tm M FBA
= 6 tm
M FCB
= 24 tm

28. 28
A D
B C
12
I I
2 I
6
RS :
KAB : KBC
1
6
2
12
:
:
1 1
: KCD
1
6
:
: 1
DF :
DFBA : DFBC
1
2
1
2
:
DFCB : DFCD
1
2
1
2
:

29. 29
6
I
A B D
2 t/m
2I I
12 6
C
8 t
DF 0.5 0.5 0.5 0.5
FEM
DM
COM
DM
- 6 +6 -24 +24 0 0
0 0
+9 +9 -12 -12
+4.5 0 +4.5
-6 -6
0
0 0
+3 +3 -2.25 -2.25
COM
DM
+1.5 0 +1.5
-1.13 -1.13
0
0 0
+0.57 +0.56 -0.75 -0.75

30. 30
DF 0.5 0.5 0.5 0.5
FEM
DM
COM
DM
- 6 +6 -24 +24 0 0
0 0
+9 +9 -12 -12
+4.5 0 +4.5
-6 -6
0
0 0
+3 +3 -2.25 -2.25
COM
DM
+1.5 0 +1.5
-1.13 -1.13
0
0 0
+0.57 +0.56 -0.75 -0.75
MFinal
+0.28 +18.76 +15.14 -15.14 -7.51
COM
DM
+0.28 0 +0.28
-0.38 -0.38
0
0 0
+0.19 +0.19 -0.14 -0.14
-18.76

31. 31
MFinal
+0.28 +18.76 +15.14 -15.14 -7.51
-18.76
0.28
18.76
18.76
15.14
15.14
7.51

32. 32
2 t/m
12
6
18.76
0.28
7.51
15.14
18.76 15.14
8 t
4 t
4 t
19.04/6
19.04/6
7.17 t
0.83 t
3.17
12+? t 12-? t
3.78 t
3.78 t
Free Body Diagram 24 t

33. 33
2 t/m
7.17 t
0.83 t
? t ? t
3.78 t
3.78 t
8 t
7.17 t 3.78 t
S Fx ≠ 0
Free Body Diagram

34. 34
2 t/m
? t ? t
7.17 t 3.78 t
S Fx ≠ 0
Reason
Sway do not included in the solution
Solution
Take Sway into consideration
Sway is unknown ?

35. 35
A D
2 t/m
8 t
B C
Ro
A D
B C
R1
d
Mo
M1
First Stage
Second Stage
Solution Stages

36. 36
First Stage
Second Stage
Sway Moment Reaction
Mo
Ro
0
d R1
M1
2 d 2 R1
2 M1
3 d 3 R1
3 M1
D d D R1
D M1
.
.
.
.
.
.
.
.
.
.
.
.
Final Stage D d Mo + D M1
Ro
- D R1

37. 37
Sway Moment Reaction
Final Case D d Mo + D M1
Ro - D R1
RFinal
No support
In original structure
Ro - D R1 = 0
Ro = D R1
D
R1
Ro
=
MFinal= Mo+ D M1
MFinal= Mo+ M1
R1
Ro
= 0

38. 38
Return to our problem

39. 39
2 t/m
7.17 t
0.83 t
? t ? t
3.78 t
3.78 t
8 t
7.17 t 3.78 t
S Fx ≠ 0
First Stage :

40. 40
2 t/m
7.17 t
3.78 t
S Fx = 0
Ro
7.17 - 3.78 - Ro = 0
Ro = 3.39

41. 41
Second Stage :
A D
B C
d
FEM :
6EI d
L2
A D
B C
12
I I
2 I
6
6EI d
62

42. 42
6EI d
L2
6EI d
62
EI is unknown
Assume :
EI d = 60
6 x 60
62
= 10
M FAB
= - 10 tm M FBA
= - 10 tm
M FCD
= - 10 tm M FDC
= - 10 tm
d is unknown

43. 43
Second Stage can be solved as previous
DF 0.5 0.5 0.5 0.5
FEM
DM
- 10 - 10 0 0 - 10 - 10

44. 44
solved as anti-symmetry problem
A D
B C
Or

45. 45
A D
B C
12
I I
2 I
6
RS :
KAB : KBC
1
6
2
12
:
:
2 3
DF :
DFBA : DFBC
2
5
3
5
:
3
2
x

46. 46
DF
A B D
C
0.4 0.6
FEM
DM
COM
DM
- 10 -10 0
0 +4 +6
+2 0 0
0 0 0
MFinal - 8 - 6 + 6 + 6 - 6 - 8

47. 47
MFinal
6
6
8
- 8 - 6 + 6 + 6 - 6 - 8
6
8
6
M1

48. 48
12
6
6
8
14/6
14/6
6
14/6
14/6
14/6
14/6
R1
Free Body Diagram
8

49. 49
14/6
14/6
R1
S Fx = 0
14/6 + 14/6 - R1 = 0
R1 = 14/3

50. 50
Ro
Ro = 3.39
R1
R1 =14 / 3
RFinal = Ro - D R1 = 0
Ro = D R1
D
R1
Ro
=
14 / 3
3.39
= 0.726
=
MFinal= Mo+ 0.726 M1

51. 51
0.28
18.76
18.76
15.14
15.14
7.51
6
6
8
6
8
6
M1
Mo
MFinal= Mo+ 0.726 M1
-
+
-
-
-
+
+ +
-
-
+.28+.726x-8
+7.51+.726x+8
-18.76+.726x6
-15.14+.726x-6
- 5.53
13.32
- 14.40 - 19.50

52. 52
- 5.53 13.32
- 14.40 - 19.50
5.53
14.40
13.32
19.50
14.40
19.50

53. 53
A D
2 t/m
8 t
B C
12
6
5.53
14.40
13.32
19.50
14.40
19.50
9.97
16.95
12
2.03
36
19.05
+
+
+
-
-
- -
-

54. 54
2 t/m
8 t
12
6
14.40
5.53 13.32
19.50
14.40 19.50
4 t
4 t
8.87/6
8.87/6
5.48 t
2.52 t
1.47
5.47 t
32.82/6
32.82/6
5.48 t
Free Body Diagram
- 5.53 13.32
- 14.40 - 19.50

55. 55
2 t/m
14.40 19.50
24 t
12 t 12 t
5.1/12 t
5.1/12 t 0.425 t
11.575 t 12.425 t

56. 56
2 t/m
8 t
5.48 t
2.52 t
5.48 t
5.48 t
24 t
11.575 t 12.425 t
5.48 t
5.48 t
11.575 t
12.425 t

57. 57
2.52
5.48
5.48
11.575
12.425
SFD
- -
+
+
+
2 t/m
8 t
5.48 t
2.52 t
5.48 t
5.48 t
24 t
11.575 t 12.425 t
5.48 t
5.48 t
11.575 t
12.425 t

58. 58
2 t/m
8 t
5.48 t
2.52 t
5.48 t
5.48 t
24 t
11.575 t 12.425 t
5.48 t
5.48 t
11.575 t
12.425 t
11.575
5.48
NFD
-
-
-
12.425

59. 59
A D
16 t
B C
12
I 2 I
2 I
6
Example 18 :
Draw B.M.D for the shown frame
3

60. 60
A D
16 t
B C
12
6
3
First Stage :

61. 61
FEM :
M FBC= -27 tm
A
D
16 t
B C
12
6
3
B C
16 t
9
3
16 x 3 x 92
122
16 x 9 x 32
122
- 27 + 9
M FCB= +9 tm

62. 62
RS :
KAB : KBC
1
6
2
12
:
:
2 2
: KCD
2
6
:
: 3
DF :
DFBA : DFBC
1
2
1
2
:
DFCB : DFCD
2
5
3
5
:
A D
B C
12
I 2 I
2 I
6
3
4
x

63. 63
DF 0.5 0.5 0.4 0.6
FEM
DM
COM
DM
0 0 -27 +9 0 0
0 0
+13.5 +13.5 -3.6 -5.4
+6.75 0 +6.75
- 1.8 0
0
0 0
+0.9 +0.9 -2.7 -4.05
COM
DM
+0.45 0 +0.45
-1.35 0
0
0 0
+0.67 +0.68 -0.18 -0.27
6
I
A B D
2 t/m
2I 2I
12 6
C
8 t

64. 64
DF 0.5 0.5 0.4 0.6
MFinal
+7.54 +15.11 +9.92 -9.92 0
COM
DM
+0.34 0 +0.34
-0.09 0
0
0 0
+0.04 +0.05 -0.14 -0.20
-15.11
FEM
DM
COM
DM
0 0 -27 +9 0 0
0 0
+13.5 +13.5 -3.6 -5.4
+6.75 0 +6.75
- 1.8 0
0
0 0
+0.9 +0.9 -2.7 -4.05
COM
DM
+0.45 0 +0.45
-1.35 0
0
0 0
+0.67 +0.68 -0.18 -0.27

65. 65
MFinal
7.54
15.11
15.11
9.92
9.92
+7.54 +15.11 +9.92 -9.92 0
-15.11

66. 66
12
6
15.11
7.54
9.92
15.11 9.92
22.65/6
22.65/6
3.78
9.92/6
1.65 t
Free Body Diagram
9.92/6

67. 67
3.78 t
3.78 t
? t ? t
1.65 t
1.65 t
3.78 t 1.65 t
First Stage :

68. 68
3.78 t
1.65 t
S Fx = 0
Ro
3.78 - 1.65 - Ro = 0
Ro = 2.13

69. 69
Second Stage :
A D
B C
d
FEM :
6EI d
L2
6EI d
62
A D
B C
12
I I
2 I
6

70. 70
6EI d
L2
6EI d
62
EI is unknown
Assume :
EI d = 60
6 x 60
62
= 10
M FAB
= - 10 tm M FBA
= - 10 tm
M FCD
= - 10 tm M FDC
= - 10 tm
d is unknown

71. 71
DF 0.5 0.5 0.4 0.6
FEM
DM
COM
DM
-10 -10 0 0 -5 0
0 0
+5 +5 +2 +3
+2.5 0 +2.5
+1 0
0
0 0
-0.5-0.5 -1 -1.5
COM
DM
-0.25 0 -0.25
-0.5 0
0
0 0
+0.25 +0.25 +0.1 +0.15
A B D
C
-10 -10 0 0 -10 -10
+10
+5
FEM

72. 72
DF 0.5 0.5 0.4 0.6
MFinal
- 7.62 - 5.27 +3.43 - 3.43 0
COM
DM
+0.13 0 +0.13
+0.05 0
0
0 0
-0.02 -0.03 -0.05 -0.08
5.27
FEM
DM
COM
DM
-10 -10 0 0 -5 0
0 0
+5 +5 +2 +3
+2.5 0 +2.5
+1 0
0
0 0
-0.5-0.5 -1 -1.5
COM
DM
-0.25 0 -0.25
-0.5 0
0
0 0
+0.25 +0.25 +0.1 +0.15

73. 73
MFinal
3.43
3.43
5.27
7.62
5.27
M1
- 7.62 - 5.27 +3.43 - 3.43 0
5.27

74. 74
12
6
5.27
7.62
12.89/6
12.89/6
3.43
3.43/6
3.43/6
12.89/6
3.43/6
R1
Free Body Diagram

75. 75
R1
S Fx = 0
12.89/6 + 3.43/6 - R1 = 0
R1 = 2.72
12.89/6
3.43/6

76. 76
Ro
Ro = 2.13
R1
R1 =2.72
RFinal = Ro - D R1 = 0
Ro = D R1
D
R1
Ro
=
2.72
2.13
= 0.783
=
MFinal= Mo+ 0.783 M1

77. 77
M1
Mo
MFinal= Mo+ 0.783 M1
-
-
-
-
+
+
-
-
+7.54+.783x-7.62
-15.11+.783x5.27
-9.92+.783x-3.43
+1.57
- 10.98 - 12.61
7.54
15.11
15.11
9.92
9.92
3.43
3.43
5.27
7.62
5.27

78. 78
+1.57
- 10.98 - 12.61
1.57
10.98
12.61
10.98
12.61

79. 79
1.57
10.98
12.61
10.98
12.61
A
D
16 t
B C
12
6
3
?

80. 80
10.98
12.61
16 t
B C
9
3
1.63
1.63/4=0.41
11.39
16 x 3 x 9
12
= 36
24.61

81. 81
1.57
10.98
12.61
10.98
12.61
11.39
24.61
+
+
-
- -
-
BMD

82. 82
A
D
16 t
B C
12
I 2 I
2 I
6
Example 19 :
Draw B.M.D for the shown frame
8
4 t

83. 83
First Stage :
A
D
16 t
B C
12
6
8
4 t

84. 84
FEM :
A
D
16 t
B C
12
6 8
4 t
L
P
8
L
P
8
L
P
M FBC = = - 24 tm
M FCB = 24 tm
16 x 12
8
=

85. 85
RS :
KAB : KBC
1
6
2
12
:
:
2 2
: KCD
2
8
:
: 3
DF :
DFBA : DFBC
1
2
1
2
:
DFCB : DFCD
2
5
3
5
:
A
D
B C
12
I
2 I
2 I
6 8

86. 86
DF 0.5 0.5 0.4 0.6
FEM
DM
COM
DM
0 0 -24 +24 0 0
0 0
+12 +12 -9.6 -14.4
+6 0 +6
- 4.8 -7.2
0
0 0
+2.4 +2.4 -2.4 -3.6
COM
DM
+1.2 0 +1.2
-1.2 -1.8
0
0 0
+0.6 +0.6 -0.48 -0.72
6
I
A B D
2I 2I
12 8
C

87. 87
DF 0.5 0.5 0.4 0.6
MFinal
+7.5 +15.12 +18.9 -18.9 9.36
COM
DM
+0.3 0 +0.3
-0.24 -0.36
0
0 0
+0.12 +0.12 -0.12 -0.18
-15.12
FEM
DM
COM
DM
0 0 -24 +24 0 0
0 0
+12 +12 -9.6 -14.4
+6 0 +6
- 4.8 -7.2
0
0 0
+2.4 +2.4 -2.4 -3.6
COM
DM
+1.2 0 +1.2
-1.2 -1.8
0
0 0
+0.6 +0.6 -0.48 -0.72

88. 88
7.5
15.12
15.12
18.9
18.9
MFinal
+7.5 +15.12 +18.9 -18.9
-15.12 9.36
9.36

89. 89
12
6
15.12
7.5
18.9
15.12 18.9
22.62/6
22.62/6
3.77
28.26/8
3.53 t
Free Body Diagram
28.86/8
9.36
8

90. 90
3.77 t
3.77 t
? t ? t
3.53 t
3.53 t
3.77 t
3.53 t
First Stage :
A
D
16 t
B C
4 t
4 t

91. 91
3.77 t
3.53 t
S Fx = 0
Ro
3.77 - 3.53 - 4 - Ro = 0
Ro = -3.76
4 t

92. 92
Second Stage :
A
D
B C
d
FEM :
6EI d
L2
6EI d
62
6x2EI d
82
A
D
B C
12
I
2 I
2 I
6 8

93. 93
6EI d
L2
6EI d
62
EI is unknown
Assume :
EI d = 48
6 x 48
62
= 8
M FAB
= - 8 tm M FBA
= - 8 tm
M FCD
= - 9 tm M FDC
= - 9 tm
d is unknown
12 x 48
82
= 9
6x2EI d
82

94. 94
DF 0.5 0.5 0.4 0.6
FEM
DM
COM
DM
- 8 - 8 0 0 - 9 - 9
0 0
+4 +4 +3.6 +5.4
+2 0 +2
+1.8 +2.7
0
0 0
-0.9 -0.9 -0.8 -1.2
COM
DM
-0.45 0 -0.45
-0.4 -0.6
0
0 0
+0.2 +0.2 +0.18 +0.27
A B D
C

95. 95
DF 0.5 0.5 0.4 0.6
MFinal
- 6.35 - 4.75 +4.59 - 4.59 -6.76
COM
DM
+0.1 0 +0.1
+0.09 +0.14
0
0 0
-0.05 -0.04 -0.04 -0.06
4.75
FEM
DM
COM
DM
- 8 - 8 0 0 - 9 - 9
0 0
+4 +4 +3.6 +5.4
+2 0 +2
+1.8 +2.7
0
0 0
-0.9 -0.9 -0.8 -1.2
COM
DM
-0.45 0 -0.45
-0.4 -0.6
0
0 0
+0.2 +0.2 +0.18 +0.27

96. 96
MFinal
4.59
4.59
4.75
6.35
4.75
M1
- 6.35 - 4.75 +4.59 - 4.59 -6.76
4.75
6.76

97. 97
6
4.75
6.35
11.1/6
11.1/6
4.59
11.35/8
11.35/8
11.1/6
11.35/8
R1
Free Body Diagram
6.76
4.59
4.59
4.75
6.35
4.75
6.76

98. 98
R1
S Fx = 0
11.1/6 + 11.35/8 - R1 = 0
R1 = 3.27
11.1/6
11.35/8

99. 99
Ro
Ro = 3.76
R1
R1 = 3.27
RFinal = Ro + D R1 = 0
Ro = - D R1
D =
R1
- Ro
3.27
-3.76
= -1.15
=
MFinal= Mo-1.15 M1

100. 100
M1
Mo
MFinal= Mo- 1.15 M1
-
-
-
-
+
+
-
-
+7.5-1.15x-6.35
-15.12-1.15x4.75
-18.9-1.15x-4.59
+14.8
- 20.58 - 13.62
4.59
4.59
4.75
6.35
4.75
6.76
7.5
15.12
15.12
18.9
18.9
9.36 +
+
+9.36-1.15x+6.76
1.83

101. 101
14.8
20.58 13.62
20.58
13.62
- 20.58 - 13.62
+1.83
+14.8
1.83

102. 102
14.8
20.58 13.62
20.58
13.62
1.83
8
A
D
16 t
B C
12
6
4 t
17.1
48
30.9
BMD
+
-
+
-

103. 103
Advanced Idea

104. 104
Beams with springs
First Stage :

105. 105
First Stage :
Ro
Second Stage :
R1
1 cm

106. 106
First Stage
Second Stage
Settlement Moment Reaction
Mo
Ro
0
1 cm R1
M1
2 cm 2 R1
2 M1
3 cm 3 R1
3 M1
D cm D R1
D M1
.
.
.
.
.
.
.
.
.
.
.
.
Final Stage Mo + D M1
Ro
+ D R1
D cm

107. 107
Settlement Moment Reaction
Final Stage Mo + D M1
Ro
+ D R1
D cm
F = K d
Ro + D R1 = K
D
MFinal= Mo+ D M1

108. 108
Beams with Intermediate Hinge
Determinate Part Indeterminate Part
Determinate Part
Determinate Part

109. 109
Indeterminate Part
Indeterminate Parts

110. 110
Solution Steps :
Ro
R1
Mo
M1
d

111. 111
First Stage
Second Stage
Moment Reaction
Mo
Ro
0
d R1
M1
2 d 2 R1
2 M1
3 d 3 R1
3 M1
D d D R1
D M1
.
.
.
.
.
.
.
.
.
.
.
.
Final Stage D d Mo + D M1
Ro
- D R1
Settlement

112. 112
Sway Moment Reaction
Final Case D d Mo + D M1
Ro - D R1
RFinal
No support
In original structure
Ro - D R1 = 0
Ro = D R1
D
R1
Ro
=
MFinal= Mo+ D M1
MFinal= Mo+ M1
R1
Ro
= 0

113. 113
Summary
Frames
Without Sway With Sway
Symmetry With Support

114. 114
Frames With Sway
Summary

115. 115
First Stage :
Mo
Ro
S Fx = 0

116. 116
Second Stage :
A D
B C
d
M1
R1
S Fx = 0

117. 117
Sway Moment Reaction
Final Case Mo + D M1
Ro - D R1
RFinal
No support
In original structure
Ro - D R1 = 0
Ro = D R1
D
R1
Ro
=
MFinal= Mo+ D M1
MFinal= Mo+ M1
R1
Ro
= 0

118. 118
Questions