livre dynamique des structures

1. Structural Analysis
Superposition
II
Eng. Khaled El-Aswany

2. Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(1.Introduction)

3. 12 t 12 t
12 t
4 t
4 t
12t
32mt
S.F.D
B.M.D
8t
2 t/m
4 m
A B
4 m
2 t/m
4 m
A B
4 m
8t
4 m
A B
4 m
8 t
8t 4t
8 t 8 t
16mt
4 t 4 t
4 t 4 t
4 t
𝒘𝑳𝟐
𝟖
𝑷𝑳
𝟒
=
𝟖𝒙𝟖
𝟒
=16
16mt
2x8=16

4. Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(2.Rules)

5. S.F.D.
𝑷
𝟐
𝑷
𝟐
𝑷
𝟐
𝑷
𝟐
𝑷𝒃
𝑳
𝑷𝒃
𝑳
𝑷𝒂
𝑳
𝑷𝒂
𝑳
𝒘𝑳
𝟐
𝒘𝑳
𝟐
A B
a b
L
𝑷𝑳
𝟒
L
A B
w t/m
𝒘𝑳𝟐
𝟖
B.M.D
P t
𝑷𝒂𝒃
𝑳
P t
L/2
A B
L/2
𝑷
𝟐
𝑷
𝟐
𝒘𝒍
𝟐
𝒘𝒍
𝟐
𝑷𝒃
𝑳
𝑷𝒂
𝑳
wL
𝑷𝒂
B.M.D
b a
P 𝑷
P t
A B
P t
a
𝑷𝒂
a
L
A B
𝑷𝒂
L
L/2
A B
M
L/2
𝑴
𝟐
𝑴
𝟐
P t
P t
𝑴
𝑳
𝑴
𝑳

6. Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(3.Examples)

7. 2 t/m
4 t
3 m
A
3 m
2 t/m
3 m
A
3 m
A B
6m 2m
2m
4 t
𝟔
𝟖
A B
8 mt
6m 2m
2m
2 t/m
4 t
A B
8 mt
6m 2m
2m
A B
6m 2m
2m
2 t/m
𝟖 𝟒
𝟖
𝟏𝟐
𝟐 𝟏
𝟑
4 t
3 m
A
3 m
𝟏𝟐
36
36
𝟗
=

8. A B
4m 2m
2m
8 t 4 t
A B
4m 2m
2m
8 t
A B
4m 2m
2m
4 t
A B
4m 2m
2m
8 t 4 t
A B
4m 2m
2m
4 t 4 t
A B
4m 2m
2m
4 t
𝟏𝟐 𝟔
𝟔
𝟖 𝟖
𝟐
𝟒
𝟏𝟒
𝟏𝟎
𝟐

9. Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(4.Frames)

10. 4t
A D
B C
2m 2m
1m
2m
2m
6t
2t/m
8t
2m 2m
0
2m 2m
0
2t/m
𝑷𝑳
𝟒
=8
𝒘𝑳𝟐
𝟖
=4
3t
6t 𝟔
2m 2m
1
0
4t
2m
2m
4t
𝟏𝟔
𝟖
𝟏𝟔
𝟏𝟔
𝟖
𝟖
𝟏𝟔
𝟖
2m
2m
3t
3t
𝟏𝟐
𝟏𝟐
𝟏𝟐
𝟏𝟐

11. 8mt
A D
B C
8m
3m
3m
2t
4t
2m
8mt
2t
4t
0
𝟖
𝟖
𝟖
4t
𝟐𝟒
𝟐𝟒
0
𝟒
𝟒
𝟒
𝟒
𝟒
𝟒

12. Structural Analysis
Three Moment Equation
II
Eng. Khaled El-Aswany

13. Structural Analysis
Three Moment Equation
II
(Introduction)
Eng. Khaled El-Aswany

14. 2 t/m
6m
A B
6t
2m
C
2m
I1 I2
MA (
𝐋𝟏
𝐈
) + 2 MB (
𝐋𝟏
𝐈
+
𝐋𝟐
𝐈
) + MC (
𝐋𝟐
𝐈
) = -6 (
𝐑𝐛𝟏
𝐈
+
𝐑𝐛𝟐
𝐈
)
MA = MB = MC =
MA (
𝐋𝟏
𝐈𝟏
) + 2 MB (
𝐋𝟏
𝐈𝟏
+
𝐋𝟐
𝐈𝟐
) + MC (
𝐋𝟐
𝐈𝟐
) = -6 (
𝐑𝐛𝟏
𝐈𝟏
+
𝐑𝐛𝟐
𝐈𝟐
)
0 ?? 0
IF ( I ) is constant: 2 t/m
6m
A B
6t
2m
C
2m
I I
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (𝐑𝐛𝟏 + 𝐑𝐛𝟐)

15. Structural Analysis
Three Moment Equation
II
(Elastic Reactions)
Eng. Khaled El-Aswany

16. 2 t/m
6m
A B
6t
2m
C
2m
I1 I2
MA (
𝐋𝟏
𝐈
) + 2 MB (
𝐋𝟏
𝐈
+
𝐋𝟐
𝐈
) + MC (
𝐋𝟐
𝐈
) = -6 (
𝐑𝐛𝟏
𝐈
+
𝐑𝐛𝟐
𝐈
)
MA = MB = MC =
MA (
𝐋𝟏
𝐈𝟏
) + 2 MB (
𝐋𝟏
𝐈𝟏
+
𝐋𝟐
𝐈𝟐
) + MC (
𝐋𝟐
𝐈𝟐
) = -6 (
𝐑𝐛𝟏
𝐈𝟏
+
𝐑𝐛𝟐
𝐈𝟐
)
0 ?? 0
IF ( I ) is constant:
2 t/m
6m
A B
6t
2m
C
2m
I I
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (𝐑𝐛𝟏 + 𝐑𝐛𝟐)
𝑀
6
(2𝑥‫البعيدة‬ + ‫)القريبة‬
A B
a b
L
𝑷𝑳
𝟒
L
A B
w t/m
𝒘𝑳𝟐
𝟖
B.M.D
P t
P t
L/2
A B
L/2
Elastic
Reactions
𝐴 =
1
2
𝑴𝑳
A=
2
3
𝑴𝑳 𝐴2 =
1
2
𝑴𝑏
𝟐
𝟑
b
𝟏
𝟑
b
𝟐
𝟑
a
𝟏
𝟑
a
𝐴1 =
1
2
𝑴𝑎
Pa
B.M.D
Elastic
Reactions
P t
a
A B
b
P t
a
Pa
A B
a b
L
M
𝑀𝑏
𝐿
𝑀𝑎
𝐿
𝐴2𝑥
2
3
𝑏
1
2
𝐴
1
2
𝐴
1
2
𝐴
1
2
𝐴
1
2
𝐴
1
2
𝐴
+𝐴1𝑥(𝑏 +
1
3
𝑎)
L
𝑀
6
(2𝑥‫البعيدة‬ + ‫)القريبة‬
𝐴1𝑥
2
3
𝑎 +𝐴2𝑥(𝑎 +
1
3
𝑏)
L
𝑷𝒂𝒃
𝑳
A=(
𝑏+𝐿
2
)𝑴
𝑀
𝐿
𝑀
𝐿
𝐴1
𝐴2
𝟐
𝟑
a
𝟏
𝟑
a 𝟐
𝟑
b
𝟏
𝟑
b
A2x
2
3
b A1x(b +
1
3
a)
L
𝐴2 − 𝐴1 − 𝑅1
𝑶𝑹 𝑶𝑹

17. Structural Analysis
Three Moment Equation
II
(Example1)
Eng. Khaled El-Aswany

18. 𝟐
𝟑
𝑴𝑳 = 𝟑𝟔
6t
2m 2m
B C
2 t/m
6m
A B
𝒘𝑳𝟐
𝟖
=9
𝑷𝑳
𝟒
= 𝟔
6t
18 t
18 t 1
2
𝑴𝑳 = 12
6t
MA = 0
I 2I
2 t/m
6m
A B
6t
2m
C
2m
MB = MC =
?? 0
6t
2m 2m
B C
2 t/m
6m
A B
MA (
𝐋𝟏
𝐈𝟏
) + 2 MB (
𝐋𝟏
𝐈𝟏
+
𝐋𝟐
𝐈𝟐
) + MC (
𝐋𝟐
𝐈𝟐
) = -6 (
𝐑𝐛𝟏
𝐈𝟏
+
𝐑𝐛𝟐
𝐈𝟐
)
0 + 2 MB (
𝟔
𝟏
+
𝟒
𝟐
) + 0 = -6 (
𝟏𝟖
𝟏
+
𝟔
𝟐
)
16 MB = -126
MB = -7.875 mt
2x6-7.3125
=4.6875
𝟐𝒙𝟔𝒙𝟑 + 𝟕. 𝟖𝟕𝟓
𝟔
𝟔𝒙𝟐 + 𝟕. 𝟖𝟕𝟓 6-4.97
=1.03
7.875
7.875
S.F.D
B.M.D
7.3125
4.6875
4.97
4.97
1.03 1.03
7.875
2.06
S.F.D
B.M.D
4.97
4.97
1.03 1.03
7.875
2.06
7.875
= 7.3125
𝟒
= 𝟒. 𝟗𝟕
2x6
3m

19. Structural Analysis
Three Moment Equation
II
(Example 2)
Eng. Khaled El-Aswany

20. MA = MB = MC =
0 ?? ?? MD = 0
2 t/m
A B
9t
C
6t 6t
9mt
2m 2m 4m
4m 2m 1.5 3m
D
6t 6t
2 t/m
A
2m 4m 2m
B B
9t
C
2m 4m
𝐰𝐋𝟐
𝟖
=16
2
3
𝑴𝑳=85.33
Pa=12 Pa=12
(
4+8
2
)𝑥12=72
𝟗𝒙𝟐𝒙𝟒
𝟔
=12 𝟑
𝟔 𝟗
2.25
2m
1m
0.5
1m
42.67 42.67
36
36
20 16
C
9mt
1.5 3m
D
−2.25 −4.5
𝟐
𝟐
1 2
0 + 2 MB (8 + 6) + MC (6) = -6 (42.67+36 + 20)
28 MB + 6 MC = - 592
MB (6) +2 MC (6 + 4.5) + 0 = -6 (16- 2.25)
6 MB + 21 MC = - 82.5
1
2
MB = -21.625 mt
MC = 2.25 mt
6t 6t
2 t/m
A
2m 4m 2m
B B
9t
C
2m 4m
C
9mt
1.5 3m
D
21.625
21.625
2.25
2.25
16.7
11.3 10 1 1.5 1.5
1
10 10
21.625
2.25
1.75
16.7
1.3
11.3
7.3
6.7
12.7
18.6
7.775
1.5 1.5
4.5
4.5

21. Structural Analysis
Three Moment Equation
II
(Symmetry)
Eng. Khaled El-Aswany

22. A
W t/m
B C
L L L
D
L
E A
W t/m
B C
L L L
D
𝑀𝑎=?? 𝑀𝑏=?? 𝑀𝑐=?? 𝑀𝑎=?? 𝑀𝑏=??
𝑀𝑒= 𝑀𝑎
𝑀𝑑= 𝑀𝑏
3 Unknowns 2 Unknowns
A
W t/m
B C
L L
𝑀𝑎=?? 𝑀𝑏=?? 𝑀𝑐= 𝑀𝑎
2 Unknowns
A
W t/m
B
L
𝑀𝑎=?? 𝑀𝑏= 𝑀𝑎
1 Unknown
𝑀𝑐= 𝑀𝑏 𝑀𝑑= 𝑀𝑎
A
W t/m
B
L L/2
F
𝑀𝑎=?? 𝑀𝑏=?? 𝑀𝐹=??
3 Unknowns

23. A
W t/m
B C
L L L
D
L
E
𝑀𝑎=?? 𝑀𝑏=?? 𝑀𝑐=?? 𝑀𝑒= 𝑀𝑎
𝑀𝑑= 𝑀𝑏
3 Unknowns
A
W t/m
B C
L L
𝑀𝑎=?? 𝑀𝑏=?? 𝑀𝑐= 𝑀𝑎
2 Unknowns
A
W t/m
B
L
𝑀𝑎=?? 𝑀𝑏= 𝑀𝑎
1 Unknown
A
W t/m
B
L
𝑀𝑎=?? 𝑀𝑏= 𝑀𝑎
L
0 0
𝒘𝑳𝟐
𝟖
𝒘𝑳𝟑
𝟏𝟐
𝒘𝑳𝟑
𝟐𝟒
𝒘𝑳𝟑
𝟐𝟒
W t/m
A B
A’ B’
A B
MA’ (L1) + 2 MA (L1 + L2) + MB (L2) = -6 (𝐑𝐚𝟏 + 𝐑𝐚𝟐)
0 + 2 MA (0 + L) + MA (L) = -6 ( 0 +
𝒘𝑳𝟑
𝟐𝟒
)
3 MA L = -
wL3
4
𝑀𝑎=?? 𝑀𝑏= 𝑀𝑎
MA = -
wL2
12
0
0
MB =MA
wL2
12
wL2
12
wL2
12
wL2
12
wL2
12

24. A
W t/m
B C
L L L
D
𝑀𝑎=?? 𝑀𝑏=??
2 Unknowns
𝑀𝑐= 𝑀𝑏 𝑀𝑑= 𝑀𝑎
A
W t/m
B
L L/2
F
𝑀𝑎=?? 𝑀𝑏=?? 𝑀𝐹=??
3 Unknowns
W t/m
𝑀𝑎=?? 𝑀𝑏=?? 𝑀𝑐= 𝑀𝑏 𝑀𝑑= 𝑀𝑎
0
A’
0
C D
L
B
L
D’
A
L
B C D
A
W t/m
W t/m
𝒘𝑳𝟐
𝟖
𝒘𝑳𝟑
𝟏𝟐
𝒘𝑳𝟑
𝟐𝟒
𝒘𝑳𝟑
𝟐𝟒
MA’ (L1) + 2 MA (L1 + L2) + MB (L2) = -6 (𝐑𝐚𝟏 + 𝐑𝐚𝟐)
0 + 2 MA (0 + L) + MB (L) = -6 ( 0 +
𝒘𝑳𝟑
𝟐𝟒
)
0
0
MC =MB
𝒘𝑳𝟐
𝟖
𝒘𝑳𝟑
𝟏𝟐
𝒘𝑳𝟑
𝟐𝟒
𝒘𝑳𝟑
𝟐𝟒
2 MA + MB = -
𝟏
4
wL2 → 𝟏
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (𝐑𝐚𝟏 + 𝐑𝐚𝟐)
MA (L) + 2 MB (L+ L) + MB (L) = -6 (
𝒘𝑳𝟑
𝟐𝟒
+
𝒘𝑳𝟑
𝟐𝟒
)
MA + 5 MB = -
𝟏
2
wL2 → 𝟐
MA = -
wL2
12
MB = -
wL2
12
wL2
12
wL2
12
wL2
12
wL2
12
𝑀𝑎′=0 𝑀𝑑′=0

25. Structural Analysis
Three Moment Equation
II
(Settlement)
Eng. Khaled El-Aswany

26. 2 cm
2 cm
1 cm
𝑀𝑎= 0 𝑀𝑏=?? 𝑀𝑐= 0
A B C
𝑳𝟏 𝑳𝟐
𝑀𝑎=0 𝑀𝑏=?? 𝑀𝑐=?? 𝑀𝑑=0
A B C D
𝑳𝟏 𝑳𝟐 𝑳𝟑
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
∆𝐵−∆𝐴
L1
+
∆𝐵−∆𝐶
L2
)
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
∆𝐵−∆𝐴
L1
+
∆𝐵−∆𝐶
L2
)
MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI (
∆𝐶−∆𝐵
L2
+
∆𝐶−∆𝐷
L3
)
If I constant:
EI : Given
I I
If I variable: MA (
L1
I1
) + 2 MB (
L1
I1
+
L2
I2
) + MC (
L2
I2
) = 6EI (
∆𝐵−∆𝐴
L1
+
∆𝐵−∆𝐶
L2
)
2 Equations

27. 𝑀𝑎= 0 𝑀𝑏=?? 𝑀𝑐=?? 𝑀𝑑= 0
2 cm
𝟔𝒎 𝟒𝒎 𝟔𝒎
A B C D
2 t/m 2 t/m
A B C D
𝟔𝒎 𝟒𝒎 𝟔𝒎
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
∆𝐵−∆𝐴
L1
+
∆𝐵−∆𝐶
L2
)
MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI (
∆𝐶−∆𝐵
L2
+
∆𝐶−∆𝐷
L3
)
EI = 5000 tm2
2 MB (6 + 4) + MC (4) = 6x5000 (
0.02
6
+
0.02
4
)
MB (4) + 2 MC (4 + 6) + MD (6) = 6x5000 (
0−0.02
4
+ 0)
20 MB + 4 MC = 250 → 1
4 MB + 20 MC = -150 → 2
MB = 14.58 mt
MC = -10.42 mt
𝑀𝑎= 0 𝑀𝑏=?? 𝑀𝑐= 𝑀𝑏 𝑀𝑑= 0
2 t/m
A B
𝟔𝒎
C
𝟒𝒎
B
𝒘𝑳𝟐
𝟖
= 𝟗
2
3
𝑴𝑳=36 18
18 0
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (𝑅𝑏1+𝑅𝑏2)
2 MB (6 + 4) + MC (4) = -6 (18 + 0)
2 MB (6 + 4) + MB (4) = -6 (18 + 0)
MB = Mc = - 4.5 mt
MB (tot) = - 4.5 +14.58 = 10.08 mt
Mc (tot) = - 4.5 -10.42 = -14.92 mt
2 t/m
A B
𝟔𝒎
C
𝟒𝒎
B
2 t/m
A B
𝟔𝒎
10.08 14.92 14.92
10.08
4.32
7.68 6.25
6.25 8.49 3.51

28. 10.08
14.92
𝑀𝑎= 0 𝑀𝑏=?? 𝑀𝑐=?? 𝑀𝑑= 0
2 cm
𝟔𝒎 𝟒𝒎 𝟔𝒎
A B C D
2 t/m 2 t/m
A B C D
𝟔𝒎 𝟒𝒎 𝟔𝒎
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
∆𝐵−∆𝐴
L1
+
∆𝐵−∆𝐶
L2
)
MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI (
∆𝐶−∆𝐵
L2
+
∆𝐶−∆𝐷
L3
)
EI = 5000 tm2
2 MB (6 + 4) + MC (4) = 6x5000 (
0.02
6
+
0.02
4
)
MB (4) + 2 MC (4 + 6) + MD (6) = 6x5000 (
0−0.02
4
+ 0)
20 MB + 4 MC = 250 → 1
4 MB + 20 MC = -150 → 2
MB = 14.58 mt
MC = -10.42 mt
𝑀𝑎= 0 𝑀𝑏=?? 𝑀𝑐= 𝑀𝑏 𝑀𝑑= 0
2 t/m
A B
𝟔𝒎
C
𝟒𝒎
B
𝒘𝑳𝟐
𝟖
= 𝟗
2
3
𝑴𝑳=36 18
18 0
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (𝑅𝑏1+𝑅𝑏2)
2 MB (6 + 4) + MC (4) = -6 (18 + 0)
2 MB (6 + 4) + MB (4) = -6 (18 + 0)
MB = Mc = - 4.5 mt
MB (tot) = - 4.5 +14.58 = 10.08 mt
Mc (tot) = - 4.5 -10.42 = -14.92 mt
A B C D

29. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany

30. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Once)

31. A
2 t/m
B
6t
2m
6m
MA
XA
YA YB
A
2 t/m
B
6t
2m
6m
12 mt
9 mt
A
2 t/m
B
6t
2m
6m
M.S.
A B
1 mt
Mo
1 mt
M1
δ10 + X δ11 = 0 X =
−δ10
δ11
δ10 = ʃ
𝑴𝑶
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍
δ11 = ʃ
𝑴𝟏
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
ʃ 𝑴𝟏𝑴𝟏𝒅𝒍
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍
9 mt
6m
1 mt
= Area x Yc
=
2
3
𝑥6𝑥9 x 0.5
Area
Yc
0.5x1=0.5
(Linear)
-
‫مستقيم‬ ‫خط‬
‫منكسر‬ ‫غير‬
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍 = Area x Yc
=
1
2
𝑥6𝑥8 x 3
Area
6m
6 mt
8 mt
Yc
0.5x6=3
(Linear)
-
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍 = Area x Yc
=
1
2
𝑥6𝑥8
6 mt
6m
8 mt
x 4
Area
Yc
𝟐
𝟑
𝐱𝟔 = 𝟒
+
𝟐
𝟑
𝑳
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍 = Area x Yc
=
1
2
𝑥3𝑥8
8 mt
6 mt
3m 3m
+
Area
Yc
𝟐
𝟑
𝑳
𝟏
𝟑
𝑳
𝟐
𝟑
𝐱𝟔 = 𝟒
x 4
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍
x 2
=
L
3
(𝐚𝐜 + 𝐛𝐝+
bc
2
+
ad
2
)
8
4
2
6
8m
8x6
=
8
3 ( + 4x2 +
𝟒𝒙𝟔
𝟐
+
𝟖𝒙𝟐
𝟐
)
8 mt
6 mt
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍 =
6
3
( 8x6 + 0 + 0 + 0 )

32. A
2 t/m
B
6t
2m
6m
MA
XA
YA YB
A
2 t/m
B
6t
2m
6m
12 mt
9 mt
A
2 t/m
B
6t
2m
6m
M.S.
A B
1 mt
Mo
1 mt
M1
δ10 + X δ11 = 0
X =
−δ10
δ11
δ10 = ʃ
𝑴𝑶
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
ʃ 𝑴𝑶𝑴𝟏𝒅𝒍
δ11 = ʃ
𝑴𝟏
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
ʃ 𝑴𝟏𝑴𝟏𝒅𝒍
=
𝟏
𝑬𝑰
[-
𝟐
𝟑
x 6 x 9 x0.5 +
𝟔
𝟑
(
𝟏𝟐𝒙𝟏
𝟐
)] =
−𝟔
𝑬𝑰
=
𝟏
𝑬𝑰
[ 𝟔
𝟑
(1x1)] =
𝟐
𝑬𝑰
=
−( ൗ
−𝟔
EI)
ൗ
𝟐
EI
= 𝟑

33. A
2 t/m
B
6t
2m
6m
3
0
4.5 13.5
A
2 t/m
B
6t
2m
6m
12 mt
9 mt
M.S.
A B
1 mt
Mo
1 mt
M1
δ10 + X δ11 = 0
X =
−δ10
δ11
δ10
δ11
=
𝟏
𝑬𝑰
[-
𝟐
𝟑
x 6 x 9 x0.5 +
𝟔
𝟑
(
𝟏𝟐𝒙𝟏
𝟐
)] =
−𝟔
𝑬𝑰
=
𝟏
𝑬𝑰
[ 𝟔
𝟑
(1x1)] =
𝟐
𝑬𝑰
=
−( ൗ
−𝟔
EI)
ൗ
𝟐
EI
= 𝟑
A B
3
12

34. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Twice)

35. A
4 t/m
D
4t
2m 4m
2m
2m
4m
2m
6t 6t
B C
I I 2I
M.S
4 t/m
4t
6t 6t
A D
B
2m 4m
2m
2m
4m
2m
C
I I 2I
𝑀𝑜
12 12 4 8
𝑀1
1
𝑀2
1
1mt
1mt
δ10 + 𝑿𝟏 δ11 + 𝑿𝟐 δ12 = 0
δ20 + 𝑿𝟏 δ21 + 𝑿𝟐 δ22 = 0
δ10 = ʃ
𝑴𝑶
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[-(
4+8
2
)𝒙𝟏𝟐 x0.5 - 𝟎. 𝟓𝒙𝟒𝒙𝟒 x0.5] = −𝟒𝟎
EI
δ11 = ʃ
𝑴𝟏
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[ 8
3
𝒙(𝟏𝒙𝟏) = 4
EI
+
4
3
𝒙(𝟏𝒙𝟏) ]
δ12 = ʃ
𝑴𝟏
𝑴𝟐
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[4
3
x(
1X1
2
) ] =
൘
2
3
EI
δ02 = ʃ
𝑴𝑶
𝑴𝟐
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[ - 𝟎. 𝟓𝒙𝟒𝒙𝟒 -
𝟏
𝟐
x0.5 𝒙
2
3
𝒙𝟒𝒙𝟖 x0.5 ]=
൘
−2𝟖
3
EI
δ22 = ʃ
𝑴𝟐
𝑴𝟐
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[ 𝒙
4
3
𝒙(𝟏𝒙𝟏) ]
4
3
𝒙(𝟏𝒙𝟏) +
1
2 = 𝟐
EI
−40
EI
+
4
EI
𝑋1 +
൘
2
3
EI
𝑋2 = 0
−40+ 4 𝑋1 +
2
3 𝑋2 = 0
4 𝑋1 +
2
3
𝑋2 = 40
൘
−28
3
EI
+ 𝑋1
൘
2
3
EI
+ 𝑋2
2
EI
= 0
−28
3
+
2
3
𝑋1+2𝑋2 = 0
2
3
𝑋1 + 2 𝑋2 =
28
3
→ 1
→ 2
𝑿𝟏 = 9.76 mt
𝑿𝟐 = 𝟏. 𝟒𝟏 𝒎𝒕

36. A
4 t/m
D
4t
2m 4m
2m
2m
4m
2m
6t 6t
B C
I I 2I
M.S
4 t/m
4t
6t 6t
A D
B
2m 4m
2m
2m
4m
2m
C
I I 2I
𝑀𝑜
12 12 4 8
𝑀1
1
𝑀2
1
1mt
1mt
δ10 + 𝑿𝟏 δ11 + 𝑿𝟐 δ12 = 0
δ20 + 𝑿𝟏 δ21 + 𝑿𝟐 δ22 = 0
δ10 = ʃ
𝑴𝑶
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[-(
4+8
2
)𝒙𝟏𝟐 x0.5 - 𝟎. 𝟓𝒙𝟒𝒙𝟒 x0.5] = −𝟒𝟎
EI
δ11 = ʃ
𝑴𝟏
𝑴𝟏
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[ 8
3
𝒙(𝟏𝒙𝟏) = 4
EI
+
4
3
𝒙(𝟏𝒙𝟏) ]
δ12 = ʃ
𝑴𝟏
𝑴𝟐
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[4
3
x(
1X1
2
) ] =
൘
2
3
EI
δ02 = ʃ
𝑴𝑶
𝑴𝟐
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[ - 𝟎. 𝟓𝒙𝟒𝒙𝟒 -
𝟏
𝟐
x0.5 𝒙
2
3
𝒙𝟒𝒙𝟖 x0.5 ]=
൘
−2𝟖
3
EI
δ22 = ʃ
𝑴𝟐
𝑴𝟐
𝒅𝒍
𝑬𝑰
=
𝟏
𝑬𝑰
[ 𝒙
4
3
𝒙(𝟏𝒙𝟏) ]
4
3
𝒙(𝟏𝒙𝟏) +
1
2 = 𝟐
EI
4 𝑋1 +
2
3
𝑋2 = 40
2
3
𝑋1 + 2 𝑋2 =
28
3
→ 1
→ 2
9.76
1.41 𝑤𝑙2
8
=8
D
A B C
9.76+1.41
2
=5.58
𝑃𝐿
4
=4
1.58
1
4
𝑋9.76=2.44
3
4
𝑋9.76=7.32
12
9.56
12
4.68
𝑿𝟏 = 9.76 mt
𝑿𝟐 = 𝟏. 𝟒𝟏 𝒎𝒕

37. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Support movement)

38. δ10
𝑋1
δ11
δ10 + X1 δ11 = 0
δ10
𝑋1
δ11
δ10 + X1 δ11 = ±∆
Δ

39. 1 𝑡
2 𝑐𝑚
3
X1 δ11 = −𝟎. 𝟎𝟐
δ11 =
𝟏
𝑬𝑰
ʃ 𝑴𝟏𝑴𝟏𝒅𝒍 =
𝟏
𝑬𝑰
(
𝟔
𝟑
( 3x3)x 2 ) =
𝟑𝟔
𝑬𝑰
X1 δ11 = ±∆
2.5𝑡 1.25 𝑡
1.25 𝑡
7.5
B.M.D.
Reactions
X1
36
𝐸𝐼
= −0.02
X1
36
4500
= −0.02
X1 = −2.5 𝑡
EI = 4500 m2t

40. 1 𝑡
2 𝑐𝑚
3
EI = 4500 m2t
δ11 =
𝟏
𝑬𝑰
ʃ 𝑴𝟏𝑴𝟏𝒅𝒍 =
𝟏
𝑬𝑰
(
𝟔
𝟑
( 3x3)x 2 ) =
𝟑𝟔
𝑬𝑰
δ10 + X1 δ11 = ±∆
Reactions
−𝟐𝟕𝟎𝟎
𝑬𝑰
+ X1
36
𝐸𝐼
= −0.02
X1 = 72.5 𝑡
180
M.S.
M0
M1
δ10 =
𝟏
𝑬𝑰
(
−𝟐
𝟑
𝒙𝟔𝒙𝟏𝟖𝟎 x(
𝟓
𝟖
𝒙𝟑) x 2 ) =
−𝟐𝟕𝟎𝟎
𝑬𝑰
5
8
𝐿
3
8
𝐿
5
8
𝑥3
−𝟐𝟕𝟎𝟎
𝟒𝟓𝟎𝟎
+ X1
36
4500
= −0.02
72.5 𝑡 23.75𝑡
23.75𝑡

41. 1 𝑡
2 𝑐𝑚
3
EI = 4500 m2t
δ11 =
𝟏
𝑬𝑰
ʃ 𝑴𝟏𝑴𝟏𝒅𝒍 =
𝟏
𝑬𝑰
(
𝟔
𝟑
( 3x3)x 2 ) =
𝟑𝟔
𝑬𝑰
δ10 + X1 δ11 = ±∆
B.M.D.
−𝟐𝟕𝟎𝟎
𝑬𝑰
+ X1
36
𝐸𝐼
= −0.02
X1 = 72.5 𝑡
180
M.S.
M0
M1
δ10 =
𝟏
𝑬𝑰
(
−𝟐
𝟑
𝒙𝟔𝒙𝟏𝟖𝟎 x(
𝟓
𝟖
𝒙𝟑) x 2 ) =
−𝟐𝟕𝟎𝟎
𝑬𝑰
5
8
𝐿
3
8
𝐿
5
8
𝑥3
−𝟐𝟕𝟎𝟎
𝟒𝟓𝟎𝟎
+ X1
36
4500
= −0.02
37.5

42. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Extra 1)

43. A
2 t/m
B
6t
2m 2m 2m 2m
6t
2m 2m 2m
8mt
2m
6t
4t 4t 8mt
C
1m
3I 2I
M.S.
9
8
8
4
4
𝟗
2mt
2
Mo
𝟗
1mt
1
M1
1
M2
δ11 = 1
𝐸𝐼
(
1
3
*
6
3
(1*1)) =
2
3𝐸𝐼
δ22 = 1
𝐸𝐼
(
1
3
*
6
3
(1*1)) =
2
𝐸𝐼
*
8
3
(1*1))
+(
1
2
δ12 = 1
𝐸𝐼
(
1
3
*
6
3
(
1∗1
2
)) =
1
3𝐸𝐼
δ1o = 1
𝐸𝐼
(
1
3
(-
2
3
x6x9x0.5 =
−34
3𝐸𝐼
0.5
-
2+6
2
x8x0.5)
δ2o = 1
𝐸𝐼
(
1
3
(-
2
3
x6x9x0.5-
2+6
2
x8x0.5)
+(
1
2
(
8
3
X
1𝑋2
2
+
6
3
x9x0.75
0.75
+
2
3
x(9x0.75+
9𝑋1
2
)
0.5
+
4
3
x(4x0.5+
4𝑋1
2
)
-
4
3
x(4x0.5)
0.25
-
2
3
x(9x0.25)
-
6
3
x(9x0.25+
9𝑋1
2
)) =
−17
3𝐸𝐼
δ10 + 𝒙𝟏 δ11 + 𝒙𝟐 δ12= 0
−𝟑𝟒
𝟑
+
𝟐
𝟑
𝒙𝟏 +
𝟏
𝟑
𝒙𝟐 = 0
𝟐 𝒙𝟏 + 𝒙𝟐 = 34
𝒙𝟏 + 6𝒙𝟐 = 17
𝒙𝟏= 17 𝒙𝟐 = 0
δ20 + 𝒙𝟏 δ21 + 𝒙𝟐 δ22= 0
1mt
1mt

44. A
2 t/m
B
6t
2m 2m 2m 2m
6t
2m 2m 2m
8mt
2m
6t
4t 4t 8mt
C
1m
3I 2I
M.S.
9
8
8
4
4
𝟗
2mt
2
Mo
𝟗
1mt
1
M1
1
M2
0.5
0.75 0.5
0.25
0
A 3I B
6t
2m 2m 2m 2m
6t
2m 2m 2m
8mt
2m
6t
4t 4t 8mt
C
1m
2I
2 t/m
A 3I
2m 2m 2m
4t 4t
2 t/m
17
8mt
B
6t
2m 2m 2m 2m
6t
2m
6t
C
1m
2I
8mt
0
𝟏𝟐. 𝟖𝟑 𝟕. 𝟏𝟕 𝟒. 𝟐𝟓
𝟒. 𝟐𝟓
17
4.66 10.34
8
2
2
6.5
3
5
8.5
1mt
1mt

45. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Once)

46. 4t
A
2 t/m
D
B C
I
I
2I
4m
3m
3m
1.19
8
= 4 – 3 = 1
R = U – E
1.19
8
4t
2 t/m
A D
B C
I
I
2I
4m
3m
3m
M.S.
9
6
Mo
1
1t
4
4
M1
δ10
δ11
=
𝟏
𝑬𝑰
ʃ 𝑴𝟎𝑴𝟏𝒅𝒍
=
𝟏
𝑬𝑰
ʃ 𝑴𝟏𝑴𝟏𝒅𝒍
=
𝟏
𝑬𝑰
[- x
𝟐
𝟑
x 6 x 9 x4
- =
−𝟏𝟎𝟖
𝑬𝑰
4 4
𝟏
𝟐
x
𝟏
𝟐
x 6 x 6 x4 ]
𝟏
𝟐
=
𝟏
𝑬𝑰
[
𝟒
𝟑
x (4x4)
+ =
𝟗𝟎.𝟔𝟕
𝑬𝑰
x 𝟒 x 6 x4 ]
𝟏
𝟐
x2
𝑿𝟏 =
− δ10
δ11
=
−( − 108/𝐸𝐼)
90.67/𝐸𝐼
= 1.19
𝑀𝑓 = 𝑀0 + 𝑋1 𝑀1
Mf
𝑀𝑓 = 0 + 1.19 x (-4) = -4.76
4.76 4.76
4.76
4.76
10.24

47. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Twice)

48. 8t
A
2 t/m
D
B C
2I
3I
I
6m
3m
3m
2m
4t
3m
XA
YA
XD
YD
=5– 3 = 2
R = U – E
MA
8t
A
2 t/m
D
B C
2I
3I
I
6m
3m
3m
2m
4t
3m
M.S.
M0
12
9
16
12
M1
1t
1t
3
3
6
6
M2
1mt
0
δ10 =
𝟏
𝑬𝑰
[ x
3
3
x( 12 x 3)
+
=
𝟐𝟒𝟎
𝑬𝑰
𝟏
𝟐
6
3
x(16x6+12x3+
16x3
2
+
12x6
2
)
X4.5 ]
- 2
3
x6x9
δ11 =
𝟏
𝑬𝑰
[ x
3
3
x( 3 x 3)
+
=
𝟏𝟓𝟒.𝟓
𝑬𝑰
𝟏
𝟐
6
3
x(6x6+3x3+
6x3
2
𝐱𝟐)
+ x
6
3
x( 6 x 6) ]
𝟏
𝟑
δ12 =
𝟏
𝑬𝑰
[ x
1
2
x6x6
+ =
𝟐𝟏
𝑬𝑰
𝟏
𝟑
X 1
6
3
x(6x1+
3x1
2
) ]
1
1
1
δ20 =
𝟏
𝑬𝑰
[
=
𝟐𝟔
𝑬𝑰
6
3
x(16x1+12x0+
16x0
2
+
12x1
2
)
X0.5 ]
- 2
3
x6x9
δ22 =
𝟏
𝑬𝑰
[ x (𝟔𝒙𝟏)
=
𝟒
𝑬𝑰
𝟏
𝟑
+
6
3
x( 1x1) ]
x 𝟏

49. 8t
A
2 t/m
D
B C
2I
3I
I
6m
3m
3m
2m
4t
3m
YA
XD
YD
=5– 3 = 2
R = U – E
8t
A
2 t/m
D
B C
2I
3I
I
6m
3m
3m
2m
4t
3m
M.S.
M0
12
9
16
12
M1
1t
1t
3
3
6
6
M2
1mt
0
δ10 =
𝟐𝟒𝟎
𝑬𝑰
δ11
=
𝟏𝟓𝟒.𝟓
𝑬𝑰
δ12
=
𝟐𝟏
𝑬𝑰
1
1
1
δ20 =
𝟐𝟔
𝑬𝑰
δ22
=
𝟒
𝑬𝑰
δ10 + 𝑿𝟏 δ11 + 𝑿𝟐 δ12 = 0
δ20 + 𝑿𝟏 δ21 + 𝑿𝟐 δ22 = 0
𝑿𝟏 = -2.34 mt
𝑿𝟐 = 𝟓. 𝟕𝟖 𝒎𝒕
154.5 𝑋1 + 21 𝑋2 = -240
21 𝑋1 + 4 𝑋2 = −26
→ 1
→ 2
MF
𝑀𝑓 = 𝑀0 + 𝑋1 𝑀1 + 𝑋2 𝑀2
𝑀𝑓 = 𝑀0 -2.34 𝑀1 + 5.78 𝑀2
𝑀𝑓 = (−16) -2.34 (−6) + 5.78 (−1) = -7.74
4.98
8.26
5.78
16
4.98
7.74
2.34
5.78

50. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Symmetry)

51. = 5 – 3 = 2
R = U – E
2 t/m
A
B
D
C
4m
4m
2 t/m
4m
0
8 8
8
= 6 – 3 = 3
R = U – E
2 t/m
A
B
D
C
8m
4m

52. = 5 – 3 = 2
R = U – E
2 t/m
A
B
4m
4m
2 t/m
A
B
4m
4m
M.S.
𝑀𝑜
16
4
16
16
𝑀1
1mt
1
1
1
1
𝑀2
1t
4
δ11 =
𝟏
𝑬𝑰
[4x1x1 + 4x1x1]=
𝟖
𝑬𝑰
δ22 =
𝟏
𝑬𝑰
[𝟒
𝟑
𝒙(𝟒𝒙𝟒)]=
𝟐𝟏.𝟑𝟑
𝑬𝑰
δ12 =
𝟏
𝑬𝑰
[−𝟏
𝟐
𝒙𝟒𝒙𝟒 𝒙𝟏]=
−𝟖
𝑬𝑰
δ20 =
𝟏
𝑬𝑰
[−𝟏
𝟐
𝒙𝟒𝒙𝟒 𝒙𝟏𝟔]=
−𝟏𝟐𝟖
𝑬𝑰
δ10 + 𝑿𝟏 δ11 + 𝑿𝟐 δ12 = 0
δ20 + 𝑿𝟏 δ21 + 𝑿𝟐 δ22 = 0
𝑿𝟏 = -7.47 mt
𝑿𝟐 = 𝟑. 𝟐 𝒎𝒕
8 𝑋1 - 8 𝑋2 = -85.33 → 1
-8 𝑋1 + 21.33 𝑋2 = 128 → 2
𝑀𝑓 = 𝑀0 + 𝑋1 𝑀1 + 𝑋2 𝑀2
𝑀𝑓2 = 16 -7.47 (1) + 3.2 (0) = 8.53
𝑀𝑓 = 𝑀0 -7.47 𝑀1 + 3.2 𝑀2
𝑀𝑓3 = 16 -7.47 (1) + 3.2 (-4) = -4.27
1
2
3 𝑀𝐹
δ10=
𝟏
𝑬𝑰
[(
1
2
x4 x 16) 𝐱𝟏 −
𝟐
𝟑
𝒙𝟒𝒙𝟒 𝒙𝟏 +𝟒𝒙𝟏𝟔 𝒙𝟏 ] =
𝟖𝟓.𝟑𝟑
𝑬𝑰
7.47
8.53
8.53
4.27

53. = 5 – 3 = 2
R = U – E
2 t/m
A
B
4m
4m
2 t/m
A
B
4m
4m
M.S.
𝑀𝑜
16
4
16
16
𝑀1
1mt
1
1
1
1
𝑀2
1t
4
δ11 =
𝟏
𝑬𝑰
[4x1x1 + 4x1x1]=
𝟖
𝑬𝑰
δ22 =
𝟏
𝑬𝑰
[𝟒
𝟑
𝒙(𝟒𝒙𝟒)]=
𝟐𝟏.𝟑𝟑
𝑬𝑰
δ12 =
𝟏
𝑬𝑰
[−𝟏
𝟐
𝒙𝟒𝒙𝟒 𝒙𝟏]=
−𝟖
𝑬𝑰
δ20 =
𝟏
𝑬𝑰
[−𝟏
𝟐
𝒙𝟒𝒙𝟒 𝒙𝟏𝟔]=
−𝟏𝟐𝟖
𝑬𝑰
δ10 + 𝑿𝟏 δ11 + 𝑿𝟐 δ12 = 0
δ20 + 𝑿𝟏 δ21 + 𝑿𝟐 δ22 = 0
𝑿𝟏 = -7.47 mt
𝑿𝟐 = 𝟑. 𝟐 𝒎𝒕
8 𝑋1 - 8 𝑋2 = -85.33 → 1
-8 𝑋1 + 21.33 𝑋2 = 128 → 2
𝑀𝐹
δ10=
𝟏
𝑬𝑰
[(
1
2
x4 x 16) 𝐱𝟏 −
𝟐
𝟑
𝒙𝟒𝒙𝟒 𝒙𝟏 +𝟒𝒙𝟏𝟔 𝒙𝟏 ] =
𝟖𝟓.𝟑𝟑
𝑬𝑰
7.47
8.53
8.53
4.27
8.53
8.53
4.27

54. 1
Structural Fantasy
3
Using the consistent deformation method
Draw the B.M.D. for the shown frames.
2 t/m
A
B
D
C
3m 3m 3m 2m
2m
H
E
J
I
3m 3m 3m 2m
2m
4m
F G
8t 8t 4t
8t 8t
4t
6m
2 t/m
A
B
E
C D
4t 2t
2t
6m
3m 3m 3m
2m 3m 2m
4t
2 t/m
A
B
E
C
D
6t
6t
6m
3m 3m 3m
2m 3m 2m
3m
2
Exercise

55. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Support movement)

56. Draw B.M.D. due to the given loads
and right horizontal movement of
support A by 2 cm. EI = 20000 m2t.
6t
A
2 t/m
D
B C
4m
6m
0.92
YA
= 4 – 3 = 1
R = U – E
6.92
YD
6t
2 t/m
A D
B C
4m
6m
M.S.
Mo
24
24
9
M1
1t 1t
4
4
4
4
δ10 =
𝟏
𝑬𝑰
ʃ 𝑴𝟎𝑴𝟏𝒅𝒍
=
𝟏
𝑬𝑰
[
𝟒
𝟑
x (24x4) -
𝟐
𝟑
x 6 x 9 x4
+ =
𝟐𝟕𝟐
𝑬𝑰
𝟏
𝟐
x 6 x 24 x4 ]
δ11 =
𝟏
𝑬𝑰
ʃ 𝑴𝟏𝑴𝟏𝒅𝒍
=
𝟏
𝑬𝑰
[
𝟒
𝟑
x (4x4) + =
𝟏𝟑𝟖.𝟔𝟕
𝑬𝑰
𝟒 x 6 x4 ]
x2
δ10 + X1 δ11 = 0.02
𝟐𝟕𝟐
𝟐𝟎𝟎𝟎𝟎
+ X1
138.67
20000
= 0.02
X1 = 0.92 𝑡
Mf
3.68 27.68
27.68
3.68
15.68

57. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)

58. R = U - E
R = ( m + r ) - 2j
1 2 3 4
5 6 7 8 9 10 11
12 13
R = ( 13+ 3) - 2x8 = 0 (Determinate)
1 2 3 4
5 6 7 8 9 10 11
12 13
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
1 2 3 4
5 6
7
8 9 10 11
12 13
R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
14
1 2 3 4
5 6
7
8 9 10 11
12 13
R = ( m + r ) - 2j
R = ( 14+ 4) - 2x8 = 2 (Twice indeterminate)
(External & Internal)
14

59. R = U - E
R = ( m + r ) - 2j
1 2 3 4
5 6 7 8 9 10 11
12 13
R = ( 13+ 3) - 2x8 = 0 (Determinate)
1 2 3 4
5 6 7 8 9 10 11
12 13
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
1 2 3 4
5 6
7
8 9 10 11
12 13
R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
14
1 2 3 4
5 6
7
8 9 10 11
12 13
R = ( m + r ) - 2j
R = ( 14+ 4) - 2x8 = 2 (Twice indeterminate)
(External & Internal)
14
δ10 + 𝑿𝟏 δ11 = 0
A B
N0
6 6
A B
N1
1 1
δ10 =
𝟏
𝑬𝑨
ʃ 𝑵𝑶𝑵𝟏𝒅𝒍
4m
=
𝟏
𝑬𝑨
( -6 x 4 x 1
N0 N1
L
C D
C D
+ ……… + ...……)
N0 N1
L N0 N1
L
δ10 =
ΣNON1L
EA
δ11 =
ΣN1N1L
EA
Zero force members
2 members 3 members
P

60. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Example 1)

61. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
0
M.S.

62. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
4
4
No

63. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
𝐹1
𝐹2
𝐹1sin 𝜃
𝐹1cos 𝜃
4
4
Σ𝐹𝑦 = 0
𝐹1sin 𝜃 + 6 = 0
𝐹1=
−6
sin θ
=
−6
0.6
= -10 t (𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑖𝑜𝑛)
No
Σ𝐹𝑥 = 0
𝐹1cos 𝜃 + 𝐹2 = 0
(𝑇𝑒𝑛𝑠𝑖𝑜𝑛)
−10 x0.8+ 𝐹2 = 0 𝐹2 = 8t

64. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
4
4
No
10
8
8

65. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
4
4
No
10
8
8
𝐹3
𝐹4 10cos𝜃
10sin𝜃
𝐹3 cos𝜃
𝐹3 sin𝜃
Σ𝐹𝑦 = 0
10sin 𝜃 - 4 - 𝐹3sin 𝜃 = 0
(𝑇𝑒𝑛𝑠𝑖𝑜𝑛)
Σ𝐹𝑥 = 0
𝐹3cos 𝜃 + 𝐹4 + 10cos 𝜃 = 0
(𝐶𝑜𝑚𝑝𝑒𝑟𝑠𝑖𝑜𝑛)
𝐹4 = -10.67t
𝐹3 = 3.33t
3.33cos 𝜃 + 𝐹4 + 10cos 𝜃 = 0

66. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
4
4
No
10
8
8
3.33
10.67

67. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
4
4
No
10
8
8
3.33
10.67
3.33
10.67
8
8
10

68. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
4
4
No
8
8
3.33
3.33
8
8
4m 4m
4m 4m
3m
0
0
1t
1t
N1
10
10.67
10.67
10
1
1
1
1

69. R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0 𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4𝑚
3𝑚
5𝑚
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
𝜃
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
4
4
No
-10
8
8
3.33
-10.67
3.33
−10.67
8
8
-10
4m 4m
4m 4m
3m
1t
1t
-1
-1
-1
-1
N1
Mem. L No N1 NoN1L N1N1L Nf
1 4 8 -1 -32 4 0
2 4 8 -1 -32 4 0
3 4 8 -1 -32 4 0
4 4 8 -1 -32 4 0
5 5 -10 0 0 0 -10
6 3 4 0 0 0 4
7 5 3.33 0 0 0 3.33
8 3 0 0 0 0 0
9 5 3.33 0 0 0 3.33
10 3 4 0 0 0 4
11 5 -10 0 0 0 -10
12 4 -10.67 0 0 0 -10.67
13 4 -10.67 0 0 0 -1067
-128 16
δ10 =
𝛴N
ON1L
EA
=
−128
EA
δ11 =
𝛴N1N1L
EA
=
16
EA
X1 =
−δ10
δ11
=
−(−128)/𝐸𝐴
16/EA
= 8t
Nf = N0 + X1 N1

70. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Example 2)

71. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No 0
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃

72. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No 0
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
4
10 6
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6

73. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10 6 4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6

74. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10 6
4
𝐹1
𝐹2
𝐹1sin 𝜃
𝐹1cos 𝜃
Σ𝐹𝑦 = 0
𝐹1sin 𝜃 -4 +10 = 0
𝐹1= -10 t (𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑖𝑜𝑛)
Σ𝐹𝑥 = 0
𝐹1cos 𝜃 + 𝐹2 = 0
(𝑇𝑒𝑛𝑠𝑖𝑜𝑛)
−10 x0.8+ 𝐹2 = 0 𝐹2 = 8t
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6

75. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10 6 4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
10
8

76. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10 6 4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
10
8
8

77. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t
6t
4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10 6 4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
10
8
8
𝐹3
𝐹4 10cos𝜃
10sin𝜃
𝐹3 cos𝜃
𝐹3 sin𝜃
Σ𝐹𝑦 = 0
10sin 𝜃 - 6 - 𝐹3sin 𝜃 = 0
Σ𝐹𝑥 = 0
𝐹3cos 𝜃 + 𝐹4 + 10cos 𝜃 = 0
(𝐶𝑜𝑚𝑝𝑒𝑟𝑠𝑖𝑜𝑛)
𝐹4 = - 8t
𝐹3 = 0
0 + 𝐹4 + 10cos 𝜃 = 0

78. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10 6 4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
10
8
8
8
8

79. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10 6 4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
10
8
8
8
8

80. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t
6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
10
6 4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
10
8
8
8
8
𝐹5
𝐹5 cos𝜃
𝐹5 sin𝜃
Σ𝐹𝑦 = 0
10 - 4 - 𝐹5sin 𝜃 = 0
Σ𝐹𝑥 = 0
𝐹5cos 𝜃 - 8 = 0
(𝑇𝑒𝑛𝑠𝑖𝑜𝑛)
𝐹5 = 10t
𝐹5 = 10t

81. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
-10 -6 -4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
-10
8
8
−8
−8
10

82. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
-10 -6 -4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
-10
8
8
−8
−8
10
3m
4m 4m
4m
0
N1
0
1t
1t
𝜃
-0.8
-0.6
𝜃
-0.8
-0.6
𝜃
1

83. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
-10 -6 -4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
-10
8
8
−8
−8
10
3m
4m 4m
4m
0
N1
0
𝜃
𝜃
𝜃
1
δ10 =
𝛴N
ON1L
EA
δ11 =
𝛴N
1N1L
EA
X1 =
−δ10
δ11
=
−(−2)/𝐸𝐴
16/EA
= 0.125t
δ10 + X1 δ11 = 0
=
1
EA
[
1t
1t
-0.8
-0.6
-0.8
-0.6
8x-0.8x4 -6x-0.6x3 +
1
2
x-8x-0.8x4]
=
−2
EA
=
1
EA
[ )0.8(2x4 +)0.6(2x3 x2+)1(2x5 x2
+
1
2
)0.8(2x4] =
16
EA

84. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
-10 -6 -4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
-10
8
8
−8
−8
10
3m
4m 4m
4m
0
N1
0
𝜃
𝜃
𝜃
1
1t
1t
-0.8
-0.6
-0.8
-0.6
Mem. L No N1 A
1 4 0 0 1
2 4 8 -0.8 1
3 4 8 0 1
4 3 -10 0 1
5 5 10 0 1
6 3 -6 -0.6 1
7 5 0 1 1
8 5 0 1 1
9 3 0 -0.6 1
10 5 -10 0 1
11 3 -4 0 1
12 4 -8 0 2
13 4 -8 -0.8 2
14 4 0 0 2

85. R = ( m + r ) - 2j
R = ( 14+ 3) - 2x8 = 1 (Once indeterminate)
(Internal)
No
3m
4m 4m
4m
4t 6t 6t 4t
10t
10t
3m
4m 4m
4m
4t 6t 6t 4t
1 2 3
4 5 6
7 8
13
9 10 11
12 14
2A 2A 2A
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
-10 -6 -4
𝜃
4𝑚
3𝑚
5𝑚
cos 𝜃 =
4
5
= 0.8
sin 𝜃 =
3
5
= 0.6
-10
8
8
−8
−8
10
3m
4m 4m
4m
0
N1
0
𝜃
𝜃
𝜃
1
1t
1t
-0.8
-0.6
-0.8
-0.6
Mem. L No N1 A
𝑁𝑜𝑁1𝐿
𝑨
𝑁𝟏𝑁1𝐿
𝑨
Nf
1 4 0 0 1 0 0 0
2 4 8 -0.8 1 -25.6 2.56 7.9
3 4 8 0 1 0 0 8
4 3 -10 0 1 0 0 -10
5 5 10 0 1 0 0 10
6 3 -6 -0.6 1 10.8 1.08 -6.075
7 5 0 1 1 0 5 0.125
8 5 0 1 1 0 5 0.125
9 3 0 -0.6 1 0 1.08 -0.075
10 5 -10 0 1 0 0 -10
11 3 -4 0 1 0 0 -4
12 4 -8 0 2 0 0 -8
13 4 -8 -0.8 2 12.8 1.28 -8.1
14 4 0 0 2 0 0 0
-2 16
δ10 =
−2
EA
δ11 =
16
EA
X1 = 0.125t

86. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Example 3)

87. B
3m
4t 6t 4t
1 2
3 4 5
6 7
10
R = ( m + r ) - 2j
R = ( 10+ 4) - 2x6 = 2
(Twice indeterminate)
8
9
3m
3m
A A
3m
3m
3m
4t 6t 4t
B
7𝑡
7𝑡
0
No
-7
-7 -6 45°
-3
-3
A
3m
3m
3m
B
N1
1𝑡
0
0
1𝑡
-1
-1 A
3m
3m
3m
B
N2
1𝑡
1𝑡
0 0
0
−1
2
−1
2
−1
2
−1
2
1𝑡
δ10 + X1 δ11 + X2 δ12 = 0
δ20 + X1 δ21 + X2 δ22 = 0
δ10 = 0
δ20 =
51.94
EA
δ11 =
6
EA
δ22 =
14.49
EA
δ12 =
2.12
EA
X1 = 1.34 𝑡
X2 = −3.78 𝑡
Nf = N0 + X1 N1 + X2 N2
Nf = N0 + 1.34 N1 -3.78 N2
6 X1 + 2.12 X2 = 0 → 1
2.12 X1 + 14.49 X2 = -51.94 → 2
Mem. L No N1 N2 NoN1L NoN2L N1N1L N2N2L N1N2L Nf
1 3 0 -1 0 0 0 3 0 0 -1.34
2 3 0 -1 −1
/ 2
0 0 3 1.5 𝟑/ 𝟐 1.34
3 3 -7 0 0 0 0 0 0 0 -7
4 3 2 3 2 0 0 0 0 0 0 0 4.24
5 3 -6 0 −1
/ 2
0 𝟏𝟖/ 𝟐 0 1.5 0 -3.33
6 3 2 0 0 1 0 0 0 𝟑 𝟐 0 -3.78
7 3 2 3 2 0 1 0 𝟏𝟖 0 𝟑 𝟐 0 0.46
8 3 -7 0 −1
/ 2
0 𝟐𝟏/ 𝟐 0 1.5 0 -4.33
9 3 -3 0 0 0 0 0 0 0 -3
51.94
0 6 14.49 2.12

88. 1
Structural Fantasy
3
2
8t
6m
A
B
𝑭𝟏 𝑭𝟐
𝑭𝟖 𝑭𝟗
𝑭𝟑 𝑭𝟒 𝑭𝟓 𝑭𝟔 𝑭𝟕
6m
6m
8t
[𝒀𝑨 = 𝟖𝒕, 𝑭𝟖 = 𝟏𝟔 𝒕]
[𝑭𝟏 = −𝟒. 𝟖𝟏𝒕, 𝑭𝟏𝟎 = 𝟑. 𝟗𝟖 𝒕]
3t
4m
A
B
𝑭𝟏 𝑭𝟐
𝑭𝟖 𝑭𝟗
𝑭𝟑 𝑭𝟒 𝑭𝟓 𝑭𝟔 𝑭𝟕
4m
3m
3m
𝑭𝟏𝟎
C
[𝑿𝑨 = 𝟒. 𝟐𝟓𝒕, 𝑭𝟔 = −𝟒. 𝟓 𝒕]
A
𝑭𝟏
𝑭𝟖
𝑭𝟏𝟔 𝑭𝟏𝟕
𝑭𝟐
𝑭𝟓
𝑭𝟗
𝑭𝟔
3m 3m
8t
4A
𝑭𝟏𝟖 𝑭𝟏𝟗
B
𝑭𝟑 𝑭𝟒
𝑭𝟕
𝑭𝟏𝟎 𝑭𝟏𝟏 𝑭𝟏𝟐 𝑭𝟏𝟑 𝑭𝟏𝟒 𝑭𝟏𝟓
3m
3m
8t
4t 4t 4t
3m 3m
[𝑭𝟏 = 𝟒. 𝟖𝟓𝒕, 𝑭𝟔 = −𝟓. 𝟎𝟓 𝒕]
2m
A B
𝑭𝟏
𝑭𝟐
𝑭𝟖
𝑭𝟔 𝑭𝟕
4m
10t
𝑭𝟑 𝑭𝟒 𝑭𝟓
1.5m
1.5m
4
Using the consistent deformation method
Find the forces in all members.
Exercise

89. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Support
movement)

90. No load Effect
Support movements
Temperature Effects
Fabrication Errors

91. Support movements
A
w t/m
B
No movement:
w t/m
A
B
A B
δ10
1t
δ11
δ10 + X δ11 = 0
A
w t/m
B
Movement:
A
w t/m
B
A B
δ10
X
δ11
δ10 + X δ11 = ±∆
∆
Determine the internal forces in members for the shown truss
due to the external loads and settlement of 2 cm at support B.
3m
4m 4m
4t
A B C
3m
4m 4m
A B C
3m
4m 4m
4t
A B C
M.S.
δ10 + X δ11 = 𝟎. 𝟎𝟐
𝑬𝑨
+ X
𝑬𝑨
= 𝟎. 𝟎𝟐
EA = 10000 t
Answer
X

92. 4t
4t 4t
4m 4m
4m 4m
6t
6t
0 𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
𝜃
3m
4
4
No -10
8
8
3.33
-10.67
3.33
−10.67
8
8
-10
4m 4m
4m 4m
3m
1t
1t
-1
-1
-1
-1
N1
Mem. L No N1 NoN1L N1N1L
1 4 8 -1 -32 4
2 4 8 -1 -32 4
3 4 8 -1 -32 4
4 4 8 -1 -32 4
5 5 -10 0 0 0
6 3 4 0 0 0
7 5 3.33 0 0 0
8 3 0 0 0 0
9 5 3.33 0 0 0
10 3 4 0 0 0
11 5 -10 0 0 0
12 4 -10.67 0 0 0
13 4 -10.67 0 0 0
-128 16
δ10 =
𝛴N
ON1L
EA
=
−128
EA
δ11 =
𝛴N1N1L
=
16
X1 =
−δ10
δ11
=
−(−128)/𝐸𝐴
16/EA
= 8t
Nf = N0 + X1 N1
Determine the internal forces in members for the shown truss due to
the external loads and right horizontal movement of 2 cm at support B.
EA = 10000 t
3m
A B
4t
4t 4t
4m 4m
4m 4m

93. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Temperature)

94. Loads
δ10 + X1 δ11 = 0
Temperature
Temperature Effects
No load Effect
Loads & Temperature
8t
δ1t + X1 δ11 = 0 δ1t + δ10 + X1 δ11 = 0
Loads: δ10 + X1 δ11 = 0
Temp.: δ1t + 𝑿𝟏
′
δ11 = 0
Total: Xtot = X1 +𝑿𝟏
′
δ1t = Σ𝜶. ∆𝒕. 𝑵𝟏. 𝑳
δ1t = 𝜶. ∆𝒕. Σ𝑵𝟏. 𝑳
OR
δ10 + X1 δ11 + X2 δ12 = 0
δ20 + X1 δ21 + X2 δ22 = 0
δ1t + X1 δ11 + X2 δ12 = 0
δ2t + X1 δ21 + X2 δ22 = 0
δ1t + δ10 +X1 δ11 + X2 δ12 = 0
δ2t + δ20 +X1 δ21 + X2 δ22 = 0
8t
10t 10t
δ1t = Σ𝜶. ∆𝒕. 𝑵𝟏. 𝑳
δ2t = Σ𝜶. ∆𝒕. 𝑵𝟐. 𝑳
10−5
Rise (+)
Drop (−)
Tension (+)
compression (−)
ONCE
TWICE

95. 𝜃
-1.67
Determine the internal forces in
members for the shown truss
due to the external loads and
rise of temperature 100oC.
𝛼 = 1𝑥10−5
/𝑜𝐶
8m
B
6t
6t
6m
A
Example 1
EA = 45000 t
R = ( m + r ) - 2j
R = ( 5+ 4) - 2x4= 1
(Once indeterminate)
6𝑥8 − 𝑌𝑏𝑥6 = 0
𝑌𝑏 = 8𝑡
Ʃ𝑀𝐴 = 0
𝜃
𝑆𝑖𝑛𝜃 =
8
10
𝑐𝑜𝑠𝜃 =
6
10
6m
8m
10m
8m
B
6t
6t
6m
A
6t
8t
2t
𝜃
𝜃
𝜃 𝜃
-8
-6
-6
10
No
8m
6m
N1
B
A
0
0
1t
1t
𝜃
𝜃
1.33
-1.67
1.33
1
𝜃
𝐹1 𝐹2 𝐹3
𝐹4
𝐹5
Mem. L No N1 NoN1L N1N1L N1L Nf
1 8 -6 1.333 -64 14.22 10.67 3.21
2 10 10 -1.667 -166.7 27.78 -16.67 -1.52
3 10 0 -1.667 0 27.78 -16.67 -11.52
4 8 -8 1.333 -85.33 14.22 10.67 1.21
5 6 -6 1 -36 6 6 0.91
-352 90 -6
Due to loads
δ10 + X1 δ11 = 0
δ1t + X1
’ δ11 = 0
X1 = 3.91 t
δ1t = 𝛼. ∆𝑡. 𝛴𝑁1. 𝐿
Due to temperature
δ1t = 1𝑥10−5
𝑥100𝑥 − 6 = −0.006
X1
’ = 3t
-0.006 + X1
’ 𝟗𝟎
𝟒𝟓𝟎𝟎𝟎
= 0
Xtot = X1 +𝑿𝟏
′
= 𝟔. 𝟗𝟏𝒕
Total
Nf = N0 + Xtot N1

96. B
3m
4t 6t 4t
1 2
3 4 5
6 7
10
R = ( m + r ) - 2j
R = ( 10+ 4) - 2x6 = 2
(Twice indeterminate)
8
9
3m
3m
A
A
3m
3m
3m
4t 6t 4t
B
7𝑡
7𝑡
0
No
-7
-7 -6 45°
-3
-3
1𝑡
A
3m
3m
3m
B
N1 0
0
1𝑡
-1
-1 A
3m
3m
3m
B
N2
1𝑡
1𝑡
0 0
0
−1
2
−1
2
−1
2
−1
2
1𝑡
δ1t + 𝑋1
′
δ11 + 𝑋2
′
δ12 = 0
δ2t + 𝑋1
′
δ21 + 𝑋2
′
δ22 = 0
δ10 = 0
δ20 =
51.94
EA
δ11 =
6
EA
δ22 =
14.49
EA
δ12 =
2.12
EA
X1 = 1.34 𝑡
X2 = −3.78 𝑡
Nf = N0 + X1t N1 + X2t N2
Nf = N0 + 1.34 N1 -3.78 N2
6
45000
𝑋1
′
+
2.12
45000
𝑋2
′
= 0.003
2.12 𝑋2
′
+ 14.49 𝑋2
′
= 0
Determine the internal forces in
members for the shown truss due
to the external loads and rise of
temperature 50oC.
𝛼 = 1𝑥10−5
/𝑜𝐶
Example 2
EA = 45000 t
Due to loads
Due to temperature
X1t = X1 +𝑿𝟏
′
= 𝟐𝟓. 𝟎𝟒𝒕
Total
δ1t = 𝛼. ∆𝑡. 𝛴𝑁1. 𝐿
δ1t = 1𝑥10−5𝑥50𝑥(−1𝑥3𝑥2) = −0.003
δ2t = 1𝑥10−5
𝑥50𝑥
−1
2
𝑥3𝑥4 + 1𝑥3 2𝑥2 = 0
X2t = X2 +𝑿𝟐
′
= −𝟕. 𝟐𝟓𝒕
𝑋1
′
= 23.7 𝑡
𝑋2
′
= −3.47𝑡
Mem. Nf
1 -25.04
2 -19.9
3 -7
4 4.24
5 -0.87
6 -7.25
7 -3
8 -1.87
9 -3
10 2.12

97. Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Fabrication error)

98. Loads
δ10 + X1 δ11 = 0
Fabricated
Fabrication errors
No load Effect
Loads & fabrication
8t
δ1f + X1 δ11 = 0 δ1f + δ10 + X1 δ11 = 0
Loads: δ10 + X1 δ11 = 0
fabrication: δ1f + 𝑿𝟏
′
δ11 = 0
Total: Xtot = X1 +𝑿𝟏
′
δ1f = Σ∆. 𝑵𝟏
OR
δ10 + X1 δ11 + X2 δ12 = 0
δ20 + X1 δ21 + X2 δ22 = 0
δ1f + X1 δ11 + X2 δ12 = 0
δ2f + X1 δ21 + X2 δ22 = 0
δ1f + δ10 +X1 δ11 + X2 δ12 = 0
δ2f + δ20 +X1 δ21 + X2 δ22 = 0
10t
Long (+)
short (−)
Tension (+)
compression (−)
ONCE
TWICE
δ1f = Σ∆. 𝑵𝟏
δ2f = Σ∆. 𝑵𝟐
8t
10t

99. Determine the internal forces
in members for the shown
truss due to the external loads
and fabrication error in
members F3 by (+0.5cm) and
F5 by (-1cm)..
8m
B
6t
6t
6m
A
Example 1
EA = 45000 t
R = ( m + r ) - 2j
R = ( 5+ 4) - 2x4= 1
(Once indeterminate)
𝐹1 𝐹2 𝐹3
𝐹4
𝐹5
8m
B
6t
6t
6m
A
6t
8t
2t
𝜃
𝜃
𝜃 𝜃
-8
-6
-6
10
No
𝜃
-1.67
8m
6m
N1
B
A
0
0
1t
1t
𝜃
𝜃
1.33
-1.67
1.33
1
𝜃
-352 90
Due to loads
δ10 + X1 δ11 = 0
X1 = 3.91 t
Due to Fabrication
δ1f + X1
’ δ11 = 0
X1
’ = 9.2t
-0.018 + X1
’ 𝟗𝟎
𝟒𝟓𝟎𝟎𝟎
= 0
Xtot = X1 +𝑿𝟏
′
= 𝟏𝟑. 𝟏𝒕
Total
Nf = N0 + Xtot N1
Mem. L No N1 NoN1L N1N1L Δ ΔN1 Nf
1 8 -6 1.333 -64 14.22 0 0 11.47
2 10 10 -1.667 -166.7 27.78 0 0 -11.8
3 10 0 -1.667 0 27.78 0.005 -0.008 -21.8
4 8 -8 1.333 -85.33 14.22 0 0 9.47
5 6 -6 1 -36 6 -0.01 -0.01 7.1
-0.018

100. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany

101. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Introduction)

102. C
3 t/m
2I I
6m
A B
6t
2m 2m
Degree of Freedom
C
A B
Rotation (𝛼) OR (𝜃) Translation (∆)
𝛼 = 0
𝛼 ≠ 0
𝛼 = 0
∆𝑥 = 0
∆𝑦 = 0
𝛼 = 
∆𝑥 = 0
∆𝑦 = 0
𝛼 = 
∆𝑥 = 
∆𝑦 = 0
∆= 0
Beams:
[ Except settlement cases]

103. C
3 t/m
2I I
6m
A B
6t
2m 2m
3 t/m
6m
A B C
6t
2m 2m
B
Fixed End Moment
Rotation Moment
Sway Moment
L
A B C
L
B
L
A B C
L
B
4EI
L
αA +
2EI
L
αB
4EI
L
αB +
2EI
L
αA
𝑀𝐴𝐵
𝐹
−
6EI
L2
∆
𝑀𝐵𝐴
𝐹
MAB = MAB
F +
2EI
L
(2αA+ αB -3
∆
L
)
MAB = MAB
F +
4EI
L
αA +
2EI
L
αB −
6EI
L2 ∆
αA
αB
−
6EI
L2
∆
‫الساعة‬ ‫عقارب‬ ‫مع‬
𝑀𝐵𝐶
𝐹
𝑀𝐶𝐵
𝐹
αB
αC
4EI
L
αB +
2EI
L
αC
4EI
L
αC +
2EI
L
αB
∆
−
6EI
L2
∆ −
6EI
L2
∆
∆
Ψ
𝐾
Slope Deflection Equations:
MAB = MAB
F +
2EI
L
(2αA+ αB -3 Ψ )
MBA = MBA
F +
2EI
L
(2αB+ αA -3 Ψ )
MBC = MBC
F +
2EI
L
(2αB+ αC -3 Ψ )
MCB = MCB
F +
2EI
L
(2αC+ αB -3 Ψ )

104. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Fixed End Moment)

105. w t/m
L
A B
B
Pt
L/2 L/2
A
−
𝑤𝐿2
12
𝑤𝐿2
12
𝑤𝐿2
12
𝑤𝐿2
12
𝑃𝐿
8
𝑃𝐿
8
−
𝑃𝐿
8
𝑃𝐿
8
−
Pa(L−a)
L
Pa(L−a)
L
Pa(L−a)
L
Pa(L−a)
L
L
A B
Pt Pt
a
a b
−
2
9
PL
2
9
PL
2
9
PL 2
9
PL
L
A B
Pt Pt
a a a
−
Pa𝑏2
𝐿2
Pb𝑎2
𝐿2
Pa𝑏2
𝐿2
P𝑏𝑎2
𝐿2
L
A B
Pt
a b
−
𝑤𝑙2
20
𝑤𝑙2
30
𝑤𝑙2
30
L
A B
w t/m
w𝑙2
20
Mb(2a−b)
L2
Ma(2b−a)
L2
L
A B
M
a b
−
Mb(2a−b)
L2
−
Ma(2b−a)
L2
L
A B
M
a b
6x4(2x1−4)
52
6x1(2x4−1)
52
5m
A B
6mt
1m 4m

106. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 1)

107. C
3 t/m
2I I
6m
A B
6t
2m 2m
3 t/m
6m
A B C
6t
2m 2m
B
𝛼𝐴 = 0 𝛼𝐵 = ?? 𝛼𝐶 = 0
−9 9 −3 3
𝑀𝐴𝐵
𝐹
= -9 mt, 𝑀𝐵𝐴
𝐹
= 9 mt
𝑀𝐵𝐶
𝐹
= -3 mt, 𝑀𝐶𝐵
𝐹
= 3 mt
1) Unknowns
𝛼𝐵
2) Fixed End Moment
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶
2𝑥2𝐸𝐼
6
:
2𝑥𝐸𝐼
4
2 ∶ 1.5
2𝐸𝐼
𝐿 𝐴𝐵
:
2𝐸𝐼
𝐿 𝐵𝐶
2
6
∶
1
4
× 6
× 2
4 ∶ 3
4) Slope deflection equations
MAB = MAB
F +
2EI
L
(2αA+ αB -3 Ψ )
MBA = MBA
F +
2EI
L
(2αB+ αA -3 Ψ )
MBC = MBC
F +
2EI
L
(2αB+ αC -3 Ψ )
MCB = MCB
F +
2EI
L
(2αC+ αB -3 Ψ )
= −9 + 4 ( 0 + αB - 0 )
= 9 + 4 ( 2αB + 0 - 0 )
= −3 + 3 ( 2αB + 0 - 0 )
= 3 + 3 ( 0 + αB - 0 )
= −9 + 4αB
= 9 + 8αB
= −3 + 6αB
= 3 + 3αB
5) Compatibility eq.
ΣMb = 0
𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0
9 + 8αB +−3 + 6αB = 0
αB = −0.4286
𝑀𝐴𝐵 = -10.71
𝑀𝐵𝐴 = 5.57
𝑀𝐵𝐶 = -5.57
𝑀𝐶𝐵 = 1.71
6) Moment values
3 t/m
6m
A B C
6t
2m 2m
B
10.71 5.57 5.57 1.71
Reactions
8.16
9.86 3.96 2.04
F.E.M.
10.71
5.57
1.71
5.57+1.71
2
= 3.64
2.36
B.M.D. 𝑤𝐿2
8
𝑃𝐿
4
=6

108. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 2)

109. 3 t/m
6m
A B C
6t
2m 2m
B
𝛼𝐴 = 0 𝛼𝐵 = ??
−9 9 −3 3
𝑀𝐴𝐵
𝐹
= -9 mt, 𝑀𝐵𝐴
𝐹
= 9 mt
𝑀𝐵𝐶
𝐹
= -3 mt, 𝑀𝐶𝐵
𝐹
= 3 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶
2𝑥2𝐸𝐼
6
:
2𝑥𝐸𝐼
4
2 ∶ 1.5
2𝐸𝐼
𝐿 𝐴𝐵
:
2𝐸𝐼
𝐿 𝐵𝐶
2
6
∶
1
4
× 6
× 2
4 ∶ 3
4) Slope deflection equations
MAB = MAB
F +
2EI
L
(2αA+ αB -3 Ψ )
MBA = MBA
F +
2EI
L
(2αB+ αA -3 Ψ )
MBC = MBC
F +
2EI
L
(2αB+ αC -3 Ψ )
MCB = MCB
F +
2EI
L
(2αC+ αB -3 Ψ )
= −9 + 4 ( 0 + αB - 0 )
= 9 + 4 ( 2αB + 0 - 0 )
= −3 + 3 ( 2αB + 0 - 0 )
= 3 + 3 ( 0 + αB - 0 )
= −9 + 4αB
= 9 + 8αB
= −3 + 6αB + 3αC
= 3 + 3αB + 6αC
5) Compatibility eq.
ΣMb = 0
𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0
9 + 8αB −3 + 6αB+3αC = 0
αB = −0.36
αC = −0.32
𝑀𝐴𝐵 = -10.44
𝑀𝐵𝐴 = 6.12
𝑀𝐵𝐶 = -6.12
𝑀𝐶𝐵 = 0
6) Moment values
3 t/m
6m
A B
10.44 6.12 6.12
Reactions
F.E.M.
𝛼𝐶 = ??
C
3 t/m
2I I
6m
A B
6t
2m 2m
𝛼𝐵 , 𝛼𝐶
𝑀𝐶𝐵 = 0
3 + 3αB+6αC = 0
3αB+6αC = -3 → 2
14αB +3αC = -6 → 1
C
6t
2m 2m
B
8.28
9.72 4.53 1.47
10.44
6.12
6.12
2
=
3.06
2.94
B.M.D. 𝑃𝐿
4
=6
𝑤𝐿2
8

110. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Modified)

111. 3 t/m
6m
A B
Fixed End Moment
Rotation Moment
Sway Moment
L
A B
L
A B C
L
B
4EI
L
αA +
2EI
L
αB
4EI
L
αB +
2EI
L
αA
𝑀𝐴𝐵
𝐹
−
6EI
L2
∆
𝑀𝐵𝐴
𝐹
MBC =𝑀𝐵𝐶
𝐹′
+
3EI
L
(αB −
∆
L
)
MBC = 𝑀𝐵𝐶
𝐹′
+
3EI
L
αB −
3EI
L2 ∆
αA
αB
−
6EI
L2
∆
‫الساعة‬ ‫عقارب‬ ‫مع‬
𝑀𝐵𝐶
𝐹′
3EI
L
αB
∆
−
3EI
L2
∆
∆
Ψ
𝐾
Slope Deflection Equations:
MAB =𝑀𝐴𝐵
𝐹
+
2EI
L
(2αA+ αB -3 Ψ )
MBA = 𝑀𝐵𝐴
𝐹
+
2EI
L
(2αB+ αA -3 Ψ )
MBC = 𝑀𝐵𝐶
𝐹′
+
3EI
L
(αB - Ψ )
C
3 t/m
2I I
6m
A B
6t
2m 2m
C
6t
2m 2m
B
αB
C
L
B
𝑀𝐴 = ?? 𝑀𝐶 = 0
𝑀𝐵 = ??

112. w t/m
L
A B
−
𝑤𝐿2
12
𝑤𝐿2
12
L
A B
𝑀𝐴𝐵
𝐹
𝑀𝐵𝐴
𝐹
𝑀𝐴𝐵
𝐹′
L
A B
𝑀𝐴𝐵
𝐹 𝑀𝐵𝐴
𝐹
= − 0.5
𝑀𝐴𝐵
𝐹′
L
A B
w t/m
= −1.5
wL2
12
B
Pt
L/2 L/2
A
−
𝑃𝐿
8
𝑃𝐿
8
𝑀𝐴𝐵
𝐹′
−
𝑃𝐿
8
= 𝑥 1.5
B
Pt
L/2 L/2
A
−
Pa𝑏2
𝐿2
Pb𝑎2
𝐿2
L
A B
Pt
a b
L
A B
Pt
a b
𝑀𝐴𝐵
𝐹′
−
Pa𝑏2
𝐿2
Pb𝑎2
𝐿2
= − 0.5
𝑀𝐴𝐵
𝐹′
=
L
A B
4EI
L
αA +
2EI
L
αB
4EI
L
αB +
2EI
L
αA
αA
αB
3EI
L
αA
𝑀𝐴𝐵
𝐹′
= − 0.5
L
A B
−
6EI
L2
∆ (−
6EI
L2 ∆)
−
6EI
L2
∆ −
6EI
L2
∆
∆
L
A B
L
A B

113. w t/m
L
A B
−
𝑤𝐿2
12
𝑤𝐿2
12
L
A B
𝑀𝐴𝐵
𝐹
𝑀𝐵𝐴
𝐹
𝑀𝐴𝐵
𝐹′
L
A B
𝑀𝐴𝐵
𝐹 𝑀𝐵𝐴
𝐹
= − 0.5
𝑀𝐴𝐵
𝐹′
L
A B
w t/m
= −1.5
wL2
12
B
Pt
L/2 L/2
A
−
𝑃𝐿
8
𝑃𝐿
8
𝑀𝐴𝐵
𝐹′
−
𝑃𝐿
8
= 𝑥 1.5
B
Pt
L/2 L/2
A
−
Pa𝑏2
𝐿2
Pb𝑎2
𝐿2
L
A B
Pt
a b
L
A B
Pt
a b
𝑀𝐴𝐵
𝐹′
−
Pa𝑏2
𝐿2
Pb𝑎2
𝐿2
= − 0.5
6m
A B
2 t/m 6t
2m
12
−6 6 -12
6m
A B
2 t/m 6t
2m
𝑀𝐴𝐵
𝐹′
= −6 6 -12
− 0.5 ( + ) = -3
B

114. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Modified)
(Redo Example 2)

115. 𝑀𝐴𝐵
𝐹
= -9 mt, 𝑀𝐵𝐴
𝐹
= 9 mt
𝑀𝐵𝐶
𝐹
= -4.5 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶
4 ∶
9
2
2𝐸𝐼
𝐿 𝐴𝐵
:
3𝐸𝐼
𝐿 𝐵𝐶
2𝑥2
6
∶
3𝑥1
4
𝛼𝐵
3 t/m
6m
A B C
6t
2m 2m
B
𝛼𝐴 = 0 𝛼𝐵 = ?? 𝛼𝐶 = ??
−9 9 −4.5
F.E.M.
C
3 t/m
2I I
6m
A B
6t
2m 2m
𝑀𝐶 = 0
𝑀𝐴 = ?? 𝑀𝐵 = ??
:
8 ∶ 9
4) Slope deflection equations
= −9 + 8 ( 0 + αB - 0 )
= 9 + 8 ( 2αB + 0 - 0 )
= −4.5 + 9 ( αB - 0)
MAB = MAB
F +
2EI
L
(2αA+ αB -3 Ψ )
MBA = MBA
F +
2EI
L
(2αB+ αA -3 Ψ )
MBC = 𝑀𝐵𝐶
𝐹′
+
3EI
L
(αB - Ψ )
= −9 + 8αB
= 9 + 16αB
= −4.5 + 9αB
5) Compatibility eq.
ΣMb = 0
𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0
9 + 16αB −4.5 + 9αB = 0
αB = −0.18
𝑀𝐴𝐵 = -10.44
𝑀𝐵𝐴 = 6.12
𝑀𝐵𝐶 = -6.12
6) Moment values
6.12
3 t/m
6m
A B
10.44 6.12
Reactions C
6t
2m 2m
B
8.28
9.72 4.53 1.47
10.44
6.12
6.12
2 =
3.06
2.94
B.M.D. 𝑃𝐿
4
=6
𝑤𝐿2
8

116. 𝑀𝐴𝐵
𝐹
= -9 mt, 𝑀𝐵𝐴
𝐹
= 9 mt
𝑀𝐵𝐶
𝐹
= -4.5 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶
4 ∶
9
2
2𝐸𝐼
𝐿 𝐴𝐵
:
3𝐸𝐼
𝐿 𝐵𝐶
2𝑥2
6
∶
3𝑥1
4
𝛼𝐵
3 t/m
6m
A B C
6t
2m 2m
B
𝛼𝐴 = 0 𝛼𝐵 = ?? 𝛼𝐶 = ??
−9 9 −4.5
F.E.M.
C
3 t/m
2I I
6m
A B
6t
2m 2m
𝑀𝐶 = 0
𝑀𝐴 = ?? 𝑀𝐵 = ??
:
8 ∶ 9
4) Slope deflection equations
= −9 + 8 ( 0 + αB - 0 )
= 9 + 8 ( 2αB + 0 - 0 )
= −4.5 + 9 ( αB - 0)
MAB = MAB
F +
2EI
L
(2αA+ αB -3 Ψ )
MBA = MBA
F +
2EI
L
(2αB+ αA -3 Ψ )
MBC = 𝑀𝐵𝐶
𝐹′
+
3EI
L
(αB - Ψ )
= −9 + 8αB
= 9 + 16αB
= −4.5 + 9αB
5) Compatibility eq.
ΣMb = 0
𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0
9 + 16αB −4.5 + 9αB = 0
αB = −0.18
𝑀𝐴𝐵 = -10.44
𝑀𝐵𝐴 = 6.12
𝑀𝐵𝐶 = -6.12
6) Moment values
6.12
3 t/m
6m
A B
10.44 6.12
Reactions C
6t
2m 2m
B
8.28
9.72 4.53 1.47
10.44
6.12
6.12
2 =
3.06
2.94
B.M.D. 𝑃𝐿
4
=6
𝑤𝐿2
8
Type
Equation
D.O.F αA, αB , Ψ αA, Ψ
Stiffness 2EI
L
3EI
L
F.E.M.
𝑀𝐴𝐵
𝐹′
= 𝑀𝐴𝐵
𝐹
-
1
2
𝑀𝐵𝐴
𝐹
Summary
MAB = MAB
F +
2EI
L
(2αA+ αB -3 Ψ )
MBA = MBA
F +
2EI
L
(2αB+ αA -3 Ψ )
MAB = 𝑀𝐴𝐵
𝐹′
+
3EI
L
(αA - Ψ )
A B B
A
B
A
𝑀𝐴𝐵
𝐹′
A B
𝑀𝐴𝐵
𝐹
𝑀𝐵𝐴
𝐹

117. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 3)

118. 𝑀𝐴𝐵
𝐹
= -9 mt, 𝑀𝐵𝐴
𝐹
= 9 mt
𝑀𝐵𝐶
𝐹
= -3 mt, 𝑀𝐶𝐵
𝐹
= 3 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶
2𝑥2𝐸𝐼
6
:
2𝑥𝐸𝐼
4
2 ∶ 1.5
2𝐸𝐼
𝐿 𝐴𝐵
:
2𝐸𝐼
𝐿 𝐵𝐶
2
6
∶
1
4
× 6
× 2
4 ∶ 3
4) Slope deflection equations
MAB = MAB
F +
2EI
L
(2αA+ αB -3 Ψ )
MBA = MBA
F +
2EI
L
(2αB+ αA -3 Ψ )
MBC = MBC
F +
2EI
L
(2αB+ αC -3 Ψ )
MCB = MCB
F +
2EI
L
(2αC+ αB -3 Ψ )
= −9 + 4 ( 0 + αB - 0 )
= 9 + 4 ( 2αB + 0 - 0 )
= −3 + 3 ( 2αB + 0 - 0 )
= 3 + 3 ( 0 + αB - 0 )
= −9 + 4αB
= 9 + 8αB
= −3 + 6αB + 3αC
= 3 + 3αB + 6αC
5) Compatibility eq.
ΣMb = 0
𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0
9 + 8αB −3 + 6αB+3αC = 0
αB = −0.84
αC = 1.92
𝑀𝐴𝐵 = -12.36
𝑀𝐵𝐴 = 2.28
𝑀𝐵𝐶 = -2.28
𝑀𝐶𝐵 = 12
6) Moment values
12.36 2.28 2.28
Reactions
𝛼𝐵 , 𝛼𝐶
𝑀𝐶𝐵 + 𝑀𝐶𝐷 = 0
3 + 3αB+6αC −12 = 0
3αB+6αC = 9 → 2
14αB +3αC = -6 → 1
𝛼𝐴 = 0 𝛼𝐵 = ?? 𝛼𝐶 = ??
C
3 t/m
2I I
6m
A B
6t
2m 2m
6t
2m
D
3 t/m
6m
A B C
6t
2m 2m
B
−9 9 −3 3
12
−12
F.E.M.
ΣMc = 0
3 t/m
6m
A B C
6t
2m 2m
B
12 6t
2m
C D
12
6t
2m
C D
7.32
10.68 0.57 5.43
12.36
2.28
1.14
B.M.D. 𝑤𝐿2
8
6
12

119. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 4)

120. A
9t
2m
4m
2m
6t
B
1) Unknowns
𝛼𝐵 , 𝛼𝐶
2) Fixed End Moment 𝛼𝐴 = ?? 𝛼𝐵 = ?? 𝛼𝐶 = ??
C
3 t/m
B
8m
D
3I 2I
2m
2m
2m
6t
6t
2m
4m
A I
9t
2m
6t
I
3 t/m
D
C
8m
𝛼𝑑 = ??
𝑀𝐴 = 12 𝑀𝐵 = ?? 𝑀𝐶 = ?? 𝑀𝑑 = 0
−24
4 6t
6t
C
B
2m
2m
2m
3 t/m
17
−17
𝑀𝐵𝐴
𝐹′
= 4 mt
𝑀𝐵𝐶
𝐹
= -17 mt, 𝑀𝐶𝐵
𝐹
= 17 mt
𝑀𝐶𝐷
𝐹′
= -24 mt
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶 ∶ 𝐾𝐶𝐷
3𝐸𝐼
𝐿 𝐴𝐵
:
2𝐸𝐼
𝐿 𝐵𝐶
:
3𝐸𝐼
𝐿 𝐶𝐷
× 4
:
:
3𝑥1
6
2𝑥3
6
3𝑥2
8
:
:
1
2
1
3
4
:
:
2 4 3
:
:
4) Slope deflection equations
= 4 + 2 ( αB - 0 )
= −17 + 4 ( 2αB + αC - 0) )
= 17 + 4 ( 2αC + αB - 0)
= −24 + 3 (αC - 0 )
MBC = 𝑀𝐵𝐶
𝐹
+
2EI
L
(2αB+ αC -3 Ψ )
MCB = 𝑀𝐶𝐵
𝐹
+
2EI
L
(2αC+ αB -3 Ψ )
MCD = 𝑀𝐶𝐷
𝐹′
+
3EI
L
(αC - Ψ)
MBA = 𝑀𝐵𝐴
𝐹′
+
3EI
L
(αB - Ψ)
= 4 + 2αB
= −17 + 8αB + 4αC
= 17 + 4αB + 8αC
= −24 + 3αC
5) Compatibility eq.
ΣMb = 0
𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0
αB = 1.2234
αC = 0.1915
𝑀𝐶𝐵 + 𝑀𝐶𝐷 = 0
4αB+11αC = 7 → 2
10αB +4αC = 13 → 1
ΣMc = 0
6) Moment Values
MBC = −6.45 𝑚𝑡
MCB = 23.43 𝑚𝑡
MCD = −23.43 𝑚𝑡
MBA = 6.45 𝑚𝑡
A
9t
2m
4m
2m
6t
B
3 t/m
D
C
8m
6.45 6t
6t
C
B
2m
2m
2m
3 t/m
23.43
6.45 23.43
Reactions
17.83
12.17
5.075
9.925 14.93 9.07
12.28
6.23
11.89
3.7
6.45
23.43
12
B.M.D.

121. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(Sway or without sway)

122. Delete
✓
Can translate ?
X
Symmetry ?
Ψ
Ψ
Frame
Without Sway
Symmetry No translation
With Sway
Δ Δ

123. Without Sway With Sway
Delete

124. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(Without Sway)

125. Delete
C
1.5 t/m
6I
I
A B
6t
3m
2m
4I
1.5I 4t
6t
F
E
D
3m
4m
1) Unknowns
𝛼𝐵 , 𝛼𝐶
12m 8m

126. Delete
1) Unknowns
𝛼𝐵 , 𝛼𝐶
2) Fixed End Moment I
3m
1.5I 4t
6t
E
D
3m
4m
12m 8m 2m
C
1.5 t/m
6I
A B
6t
4I F
C
B
1.5 t/m
-12
8
18
-18 -8
2m
-4.5
B C
-5.33
2.67
𝑀𝐴𝐵
𝐹
= -18 mt, 𝑀𝐵𝐴
𝐹
= 18 mt
𝑀𝐵𝐶
𝐹
= -8 mt, 𝑀𝐶𝐵
𝐹
= 8 mt
𝑀𝐵𝐷
𝐹
= -5.33 mt, 𝑀𝐷𝐵
𝐹
= 2.67 mt
ഥ
𝑀𝐶𝐸
𝐹
= -4.5 mt

127. Delete
I
3m
1.5I 4t
6t
E
D
3m
4m
12m 8m 2m
C
1.5 t/m
6I
A B
6t
4I F
C
B
1.5 t/m
-12
8
18
-18 -8
2m
-4.5
B C
-5.33
2.67
1) Unknowns
𝛼𝐵 , 𝛼𝐶
2) Fixed End Moment
𝑀𝐴𝐵
𝐹
= -18 mt, 𝑀𝐵𝐴
𝐹
= 18 mt
𝑀𝐵𝐶
𝐹
= -8 mt, 𝑀𝐶𝐵
𝐹
= 8 mt
𝑀𝐵𝐷
𝐹
= -5.33 mt, 𝑀𝐷𝐵
𝐹
= 2.67 mt
ഥ
𝑀𝐶𝐸
𝐹
= -4.5 mt
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶 ∶ 𝐾𝐵𝐷 ∶ 𝐾𝐶𝐸
2𝑋6
12
:
2𝑋4
8
:
2𝑋1.5
6
:
3𝑋1
6
1 ∶ 1 ∶
1
2
:
1
2
2 ∶ 2 ∶ 1 : 1

128. Delete
I
3m
1.5I 4t
6t
E
D
3m
4m
12m 8m 2m
C
1.5 t/m
6I
A B
6t
4I F
C
B
1.5 t/m
-12
8
18
-18 -8
2m
-4.5
B C
-5.33
2.67
1) Unknowns
𝛼𝐵 , 𝛼𝐶
2) Fixed End Moment
𝑀𝐴𝐵
𝐹
= -18 mt, 𝑀𝐵𝐴
𝐹
= 18 mt
𝑀𝐵𝐶
𝐹
= -8 mt, 𝑀𝐶𝐵
𝐹
= 8 mt
𝑀𝐵𝐷
𝐹
= -5.33 mt, 𝑀𝐷𝐵
𝐹
= 2.67 mt
ഥ
𝑀𝐶𝐸
𝐹
= -4.5 mt
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶 ∶ 𝐾𝐵𝐷 ∶ 𝐾𝐶𝐸
2𝑋6
12
:
2𝑋4
8
:
2𝑋1.5
6
:
3𝑋1
6
1 ∶ 1 ∶
1
2
:
1
2
2 ∶ 2 ∶ 1 : 1
4) Slope Deflection Equation
𝑀𝐴𝐵 = -18 + 2 (2𝛼𝐴 + 𝛼𝐵)
𝑀𝐵𝐴 = 18 + 2 (2𝛼𝐵 + 𝛼𝐴)
𝑀𝐵𝐶 = -8 + 2 (2𝛼𝐵 + 𝛼𝐶)
𝑀𝐶𝐵 = 8 + 2 (2𝛼𝐶 + 𝛼𝐵)
𝑀𝐵𝐷 = -5.33 + (2𝛼𝐵 + 𝛼𝐷)
𝑀𝐷𝐵 = 2.67 + (2𝛼𝐷 + 𝛼𝐵)
𝑀𝐶𝐸 = -4.5 + (𝛼𝐶)
= -18 + 2 𝛼𝐵
= 18 + 4𝛼𝐵
= -8 + 4𝛼𝐵 + 2𝛼𝐶
= 8 + 4𝛼𝐶 + 2𝛼𝐵
= -5.33 + 2𝛼𝐵
= 2.67 + 𝛼𝐵
= -4.5 + 𝛼𝐶

129. Delete
1) Unknowns
𝛼𝐵 , 𝛼𝐶
2) Fixed End Moment
𝑀𝐴𝐵
𝐹
= -18 mt, 𝑀𝐵𝐴
𝐹
= 18 mt
𝑀𝐵𝐶
𝐹
= -8 mt, 𝑀𝐶𝐵
𝐹
= 8 mt
𝑀𝐵𝐷
𝐹
= -5.33 mt, 𝑀𝐷𝐵
𝐹
= 2.67 mt
ഥ
𝑀𝐶𝐸
𝐹
= -4.5 mt
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶 ∶ 𝐾𝐵𝐷 ∶ 𝐾𝐶𝐸
2𝑋6
12
:
2𝑋4
8
:
2𝑋1.5
6
:
3𝑋1
6
1 ∶ 1 ∶
1
2
:
1
2
2 ∶ 2 ∶ 1 : 1
4) Slope Deflection Equation
𝑀𝐴𝐵 = -18 + 2 (2𝛼𝐴 + 𝛼𝐵)
𝑀𝐵𝐴 = 18 + 2 (2𝛼𝐵 + 𝛼𝐴)
𝑀𝐵𝐶 = -8 + 2 (2𝛼𝐵 + 𝛼𝐶)
𝑀𝐶𝐵 = 8 + 2 (2𝛼𝐶 + 𝛼𝐵)
𝑀𝐵𝐷 = -5.33 + (2𝛼𝐵 + 𝛼𝐷)
𝑀𝐷𝐵 = 2.67 + (2𝛼𝐷 + 𝛼𝐵)
𝑀𝐶𝐸 = -4.5 + (𝛼𝐶)
= -18 + 2 𝛼𝐵
= 18 + 4𝛼𝐵
= -8 + 4𝛼𝐵 + 2𝛼𝐶
= 8 + 4𝛼𝐶 + 2𝛼𝐵
= -5.33 + 2𝛼𝐵
= 2.67 + 𝛼𝐵
= -4.5 + 𝛼𝐶
I
3m
1.5I 4t
6t
E
D
3m
4m
12m 8m 2m
C
1.5 t/m
6I
A B
6t
4I F
C
B
1.5 t/m
-12
8
18
-18 -8
2m
-4.5
B C
-5.33
2.67
5) Comp. equations
ΣMb = 0
𝑀𝐵𝐴 + 𝑀𝐵𝐶 + 𝑀𝐵𝐷 = 0
ΣMc = 0
𝑀𝐶𝐵 + 𝑀𝐶𝐸 + 𝑀𝐶𝐹 = 0
10𝛼𝐵 + 2𝛼𝐶 = -4.67
2𝛼𝐵 + 5𝛼𝐶 = 8.5
𝛼𝐵 = -0.87717
𝛼𝐶 = 2.05087
𝑀𝐴𝐵 = -19.75
𝑀𝐵𝐴 = 14.49
𝑀𝐵𝐶 = -7.4
𝑀𝐶𝐵 = 14.45
𝑀𝐵𝐷 = -7.09
𝑀𝐷𝐵 = 1.80
𝑀𝐶𝐸 = -2.45
6) Moments
-12
12
14.45
14.49
19.75 7.4
2.45
7.09
1.80
3m
4t
6t
E
D
3m
4m
12m 8m 2m
C
1.5 t/m
A B
6t
F
C
B
1.5 t/m
2mB C
9.44 8.56 5.12 6.88 6
1.12
4.88 2.4
1.6
19.75
14.49
7.4
14.4512
1.8
7.09
2.7
2.45
4.8

130. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(Without Sway)
(Symmetry)

131. Delete 1) Unknowns
𝛼𝐵 , 𝛼𝐶 , 𝛼𝐷 , 𝛼𝐸
2) Fixed End Moment
𝛼𝐷 = - 𝛼𝐶 𝛼𝐸 = - 𝛼𝐵
3) Relative Stiffness
𝐾𝐴𝐵 ∶ 𝐾𝐵𝐶 ∶ 𝐾𝐵𝐸 ∶ 𝐾𝐶𝐷
6 ∶ 2 ∶ 2.5 : 3.75
C
3 t/m
A
B
5m
8m
2I
2I
F
E
D
5m
6 t/m
2I
I I
3I
B E
B E
C D
8m
5m
32
16
-32
-16
5m
4) Slope Deflection Equations
𝑀𝐵𝐴 = 0 + 6 (𝛼𝐵)
𝑀𝐵𝐶 = 0 + 2 (2𝛼𝐵 + 𝛼𝐶)
𝑀𝐶𝐵 = 0 + 2 (2𝛼𝐶 + 𝛼𝐵)
𝑀𝐵𝐸 = -16 + 2.5(2𝛼𝐵 + 𝛼𝐸)
𝑀𝐶𝐷 = -32 + 3.75(2𝛼𝑐 + 𝛼𝐷)
= 6𝛼𝐵
= 4𝛼𝐵 + 2𝛼𝐶
= 4𝛼𝐶 + 2𝛼𝐵
= -16 + 2.5(2𝛼𝐵 - 𝛼𝐵)
= -32 + 3.75(2𝛼𝑐 - 𝛼𝑐)
= -16 + 2.5𝛼𝐵
= -32 + 3.75𝛼𝑐
5) Compatibility Equations
ΣMb = 0 𝑀𝐵𝐴 + 𝑀𝐵𝐶 + 𝑀𝐵𝐸 = 0
ΣMc = 0 𝑀𝐶𝐵 + 𝑀𝐶𝐷 = 0
12.5𝛼𝐵 + 2𝛼𝐶 = 16
2𝛼𝐵 + 7.75𝛼𝐶 = 32
𝛼𝐵 = 0.664603
𝛼𝐶 = 3.962315
6) Moment values
𝑀𝐵𝐴 = 3.88
𝑀𝐵𝐶 = 10.51
𝑀𝐶𝐵 = 17.14
𝑀𝐵𝐸 = -14.38
𝑀𝐶𝐷 = -17.14
C
3 t/m
A
B
5m
8m
2I
2I
F
E
D
5m
𝜶𝑩 =??
𝜶𝑪 =??
6 t/m
2I
I I
3I
𝜶𝑬 =??
𝜶𝑫 =??
C
A
B
F
E
D

132. Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(With Sway)

139. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany

140. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Introduction)

141. Introduction
1- Fixed End Moment
2- Distribution Moment
(Distribution Factor)
3- Carry over
C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
A
6m
3 t/m
B
3
9 -3
-6
-9
Carry
over
Carry
over
D.M.
D.M.

142. Fixed end moment
Distribution Factor Carry over
w t/m
L
𝑤𝐿2
12
−
𝑤𝐿2
12
𝑃𝐿
8
−
𝑃𝐿
8
−
𝑃𝑎𝑏2
𝐿2
𝑃𝑏𝑎2
𝐿2
L/2
Pt
L/2
𝑤𝐿2
30
−
𝑤𝐿2
20
𝑃𝑎(𝐿 − 𝑎)
𝐿
−
𝑃𝑎(𝐿 − 𝑎)
𝐿
L
𝑀𝑎(2𝑏 − 𝑎)
𝐿2
𝑀𝑏(2𝑎 − 𝑏)
𝐿2
w t/m
L
−
𝑤𝐿2
12
𝒙𝟏. 𝟓
−
𝑃𝐿
8
𝒙𝟏. 𝟓
−
𝑃𝑎𝑏2
𝐿2
L/2
Pt
L/2
a
Pt
b
L
L
Pt Pt
a a
b
M
a b
L
Pt
a b
L
− 𝟎. 𝟓 𝐱
𝑷𝒃𝒂𝟐
𝑳𝟐

143. Fixed end moment
Distribution Factor
Carry over
D.F. for Joint B
KBA : KBC
C
3 t/m
6m
A B
6t
2m 2m
6t
4m
2m
D
3I 2I I
3 t/m
6m
A 3I B C
B
6t
2m 2m
2I
6t
4m
2m
D
I
C
D.F. for Joint C
KCB : KCD
KBA + KBC KBA + KBC KCB + KCD KCB + KCD

144. Fixed end moment
Distribution Factor
Carry over
Joint B
KBA : KBC
C
3 t/m
6m
A B
6t
2m 2m
6t
4m
2m
D
3I 2I I
3 t/m
6m
A 3I B C
B
6t
2m 2m
2I
6t
4m
2m
D
I
C
Joint C
KCB : KCD
4𝑥𝐸𝑥3
6
:
4𝑥𝐸𝑥2
4
4𝑥𝐸𝑥2
4
:
4𝑥𝐸𝑥1
6
1
2
:
1
2
1
2
:
1
6
6 : 2
𝟑
𝟒
:
𝟏
𝟒
3 : 1
1
2
1
2
+
1
2
:
1
2
1
2
+
1
2
𝟏
𝟐
:
𝟏
𝟐
L
L
K =
𝑰
𝑳
K =
𝑰
𝑳
𝒙
𝟑
𝟒
C
3 t/m
6m
A B
6t
2m 2m
3 t/m
6m
D
3I 2I 3I
K =
𝑰
𝑳
𝒙
𝟏
𝟐
L
𝑲𝑨𝑩=
𝑰
𝑳
𝑲𝑩𝑪=
𝑰
𝑳
𝒙
𝟏
𝟐
Example for Symmetry Beam:
(Symmetry)

145. Fixed end moment Distribution Factor
Carry over
L
A B
L
B C
D.M. D.M.
𝟏
𝟐
D.M. 𝟎
x
𝟏
𝟐

146. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 1)

147. C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
A
6m
3 t/m
B
3
9 -3
-9
FEM
1- Fixed End Moment
−
𝑤𝐿2
12
=
𝑤𝐿2
12
=
PL
8
=
-
PL
8
=

148. C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
A
6m
3 t/m
B
3
9 -3
-9
3
9 -3
-9
FEM
A B C
𝟐/𝟓 𝟑/𝟓
D.M.
-6
-2.4
D.F.
FEM
1
-
‫نجمع‬
2
-
‫االشارة‬ ‫نغير‬
3
-
‫في‬ ‫نضرب‬
‫التوزيع‬ ‫نسب‬
Joint B
KBA : KBC
1
6
:
1
4
4 : 6
2 : 3
2
5
:
3
5
I
L
:
I
L
Distribution factors:
2- Distribution Moment

149. C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
A
6m
3 t/m
B
3
9 -3
-9
3
9 -3
-9
FEM
A B C
𝟐/𝟓 𝟑/𝟓
D.M.
-6
-2.4 -3.6
D.F.
1
-
‫نجمع‬
2
-
‫االشارة‬ ‫نغير‬
3
-
‫في‬ ‫نضرب‬
‫التوزيع‬ ‫نسب‬
FEM
Joint B
KBA : KBC
1
6
:
1
4
4 : 6
2 : 3
2
5
:
3
5
I
L
:
I
L
Distribution factors:
2- Distribution Moment

150. C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
A
6m
3 t/m
B
3
9 -3
-9
3
9 -3
-9
FEM
A B C
𝟐/𝟓 𝟑/𝟓
D.M. -2.4 -3.6
C.O. -1.2 -1.8
D.F.
Joint B
KBA : KBC
1
6
:
1
4
4 : 6
2 : 3
2
5
:
3
5
I
L
:
I
L
Distribution factors:
FEM
3- Carry Over

151. C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
A
6m
3 t/m
B
3
9 -3
-9
3
9 -3
-9
FEM
A B C
𝟐/𝟓 𝟑/𝟓
D.M. -2.4 -3.6
C.O. -1.2 -1.8
F.M. -10.2 6.6 -6.6 1.2
D.F.
A 6m
3 t/m
B B
6t
2m 2m
C
10.2 6.6 6.6 1.2
‫سالب‬
(
‫الساعة‬ ‫عقارب‬ ‫عكس‬
)
‫موجب‬
(
‫الساعة‬ ‫عقارب‬ ‫مع‬
)
Joint B
KBA : KBC
1
6
:
1
4
4 : 6
2 : 3
2
5
:
3
5
I
L
:
I
L
Distribution factors:

152. C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
A
6m
3 t/m
B
3
9 -3
-9
3
9 -3
-9
FEM
A B C
𝟐/𝟓 𝟑/𝟓
D.M. -2.4 -3.6
C.O. -1.2 -1.8
F.M. -10.2 6.6 -6.6 1.2
D.F.
A 6m
3 t/m
B B
6t
2m 2m
C
10.2 6.6 6.6 1.2
= 8.4
9.6
=4.35
1.65
C
A B
10.2
6.6
1.2
2.1
3x6x3 + 6.6 -10.2
6
6x2+6.6-1.2
4
FEM
Joint B
KBA : KBC
1
6
:
1
4
4 : 6
2 : 3
2
5
:
3
5
I
L
:
I
L
Distribution factors:
B.M.D.

153. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 2)

154. A
6m
3 t/m
B
9 -4.5
-9
C
3 t/m
6m
A B
6t
2m 2m
B
6t
2m 2m
C
0
9 -4.5
-9
FEM
𝟖/𝟏𝟕 9/𝟏𝟕
D.F.
A B C
Joint B
KBA : KBC
1
6
:
3
16
16 : 18
8 : 9
8
17
:
9
17
1
6
:
1
4
𝑥
3
4
D.M. -2.12 -2.38
C.O. -1.06
F.M. -10.06 6.88 -6.88 0
A 6m
3 t/m
B
10.06 6.88 6.88
B
6t
2m 2m
C
9.53 8.47 1.28
4.72
C
A B
10.06
6.88
2.56
B.M.D.
Distribution factors:
I
L
:
I
L
𝑥
3
4

155. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 3)

156. C
2 t/m
A B
8t 1t
6m
2m 2m 2m
I 2I
A
8t
2m 2m
I B C
2 t/m
B
1t
6m 2m
2I
−4 4
1t
2m
C
2 t/m
B
6m
6
−6 −2
−0.5 ( )= -8
FEM
−6 −2
6

157. C
2 t/m
A B
8t 1t
6m
2m 2m 2m
I 2I
A
8t
2m 2m
I B C
2 t/m
B
1t
6m 2m
2I
−4 4
C
2 t/m
B
6m 2m
C
B
1t
6m 2m
−9
2
1
1
−8
FEM

158. C
2 t/m
A B
8t 1t
6m
2m 2m 2m
I 2I
A
8t
2m 2m
I B C
2 t/m
B
1t
6m 2m
2I
−4 4 −8
4 -8
-4
FEM
𝟎. 𝟓 0.5
D.M.
D.F.
A B C
Joint B
KBA : KBC
1
4
:
1
4
1 : 1
1
2
:
1
2
1
4
:
2
6 𝑥
3
4
I
L
:
I
L
𝑥
3
4
2 2
C.O. 1
F.M. -3 6 -6
3 6 6
3.25 4.75 6.33
6.67
B.M.D.
A
8t
2m 2m
B C
2 t/m
B
1t
6m 2m
3.5
C
A B
3
6
2
Distribution factors:
FEM

159. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 4)

160. C
3 t/m
10m
A B
8t
7.5m 7.5m 10m
D
I 2I I
3 t/m
10m
A I B C
B
8t
7.5m 7.5m
2I D
I
C
10m
25
−25 −15 15 0 0
Joint B
KBA : KBC
Joint C
KCB : KCD
𝐼
𝐿
:
𝐼
𝐿
1
10
:
2
15
15 : 20
𝟑
𝟕
:
𝟒
𝟕 25 −15 15 0 0
−25
15
35
:
20
35
𝐼
𝐿
:
𝐼
𝐿
2
15
:
1
10
20 : 15
𝟒
𝟕
:
𝟑
𝟕
20
35
:
15
35
3/7 4/7 4/7 3/7
−5.714 −8.55 −6.45
A B C D
FEM
D.M.1
D.F.
C.O.1
−4.286
D.M.2 2.443 1.629 1.229
1.832
-2.143 -4.275 -2.857 -3.225
C.O.2 0.916 0.814 1.221 0.614
D.M.3 −0.465
−0.349 −0.525
−0.696
C.O.3 -0.174 -0.348 -0.233 -0.263
D.M.4 0.199
0.149 0.133
C.O.4 0.075 0.066 0.099 0.05
0.1
D.M.5 −0.038
−0.028 -0.057 -0.043
F.M. -26.33 -22.32 5.69 -2.82
-5.69
22.32
22.32
26.33 5.69 2.82
3 t/m
10m
A B
8t
7.5m
B C
7.5m 10m
C D
5.69
22.32
0.851 0.851
5.11 2.89
15.4 14.6
C
A B
B.M.D.
26.33
22.32
5.69
2.82
𝑤𝑙2
8
26.33 + 22.32
2
13.175
=37.5
=24.32
𝑃𝐿
4
22.32 + 5.69
2
16
=30
=14
Distribution factors:
Reactions

161. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 5 - Symmetry)

162. C
1 t/m
4m
A B
12m
D
9 mt
2m 4m
2m
9 mt
A B
9 mt
4m 2m
C
B
12m
1 t/m
C D
9 mt
4m
2m
−3 0
FEM
−12 12 0 3
0 -12
-3
FEM
0.8 0.2
D.F.
A B C
D.M. 9.6 2.4
Joint B
KBA : KBC
1
6
:
1
24
24 : 6
4
5
:
1
5
1
6
:
1
12
𝑥
1
2
I
L
:
I
L
𝑥
1
2
4 : 1
C.O. 4.8
F.M. 1.8 9.6 -9.6
A B
9 mt
4m 2m
C
B
12m
1 t/m
1.8 9.6 9.6 9.6
0.4
0.4 6 6
8.8
0.2
1.8
8.8
0.2
1.8
C
A B
9.6
9.6
8.4
B.M.D.

163. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 6)

164. 3 t/m
D
C
8m
-16 16
C
3 t/m
B
8m
D
3I 2I
2m
2m
2m
6t
6t
3m
3m
A I
8t
2m
6t
I
3
3m
3m
A
8t
2m
6t
B
FEM
𝑃𝐿
8
𝑥1.5 = 9
3m
3m
A
8t
2m
B
6t
6t
C
B
2m
2m
2m
−6
3m
3m
A
2m
6t
B
12
6
6t
6t
C
B
−17 17
2m
2m
2m
3 t/m
C
B
2m
2m
2m
3 t/m
𝑤𝐿2
12
= 9
−
𝑤𝐿2
12
= -9
−
𝑃𝑎(𝐿−𝑎)
𝐿
= -8
𝑃𝑎(𝐿−𝑎)
𝐿
= 8

165. 3 t/m
6t
6t
C
3 t/m
B
8m
D
3I 2I
C
B 3I D
2I
C
8m
3 −17 17 -16 16
Joint B Joint C
KCB : KCD
𝐼
𝐿
𝑥
3
4
:
𝐼
𝐿
3 -17 17 -16 16
𝐼
𝐿
:
𝐼
𝐿
1
2
:
1
4
2
𝟎. 𝟐 𝟎. 𝟖 2/𝟑 1/𝟑
11.2 -0.667 -0.333
A B C D
FEM
D.M.1
D.F.
C.O.1
2.8
D.M.2 0.267 -3.733 -1.867
0.667
-0.333 5.6 -0.167
C.O.2 -1.867 0.133 -0.933
D.M.3 1.493
0.373 -0.044
-0.089
C.O.3 -0.044 -0.747 -0.022
D.M.4 0.036
0.009 -0.498
C.O.4 -0.249 0.018 -0.124
-0.249
D.M.5 0.199
0.05 -0.012 -0.006
F.M. -6.3 18.499 14.75
-18.499
6.3
C
A B
B.M.D.
12
6.3
18.5
14.75
2.85
13.64
Distribution factors:
2m
2m
2m
6t
6t
3m
3m
A I
8t
2m
6t
I
2m
2m
2m
3 t/m
3m
3m
A I
8t
2m
6t
I B
3
6
:
2
8
:
4
1
:
2
𝟏
𝟑
:
𝟐
𝟑
KBA : KBC
1
6
𝑥
3
4
:
3
6
1
8
:
1
2
8
:
2
4
:
1
𝟒
𝟓
:
𝟏
𝟓
6.3 18.5 14.75
18.5
6.3
12.47 11.53
12.97 17.03
3.05
8m
C D
3 t/m
6t
B C
2m
3 t/m
6t
2m
2m
3m
8t
A B
3m
2m
6t
9.56
7.375
Reactions
D
FEM
10.95

166. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Settlement)

167. D
I
C
10m
1 cm
Determinate structures
Indeterminate structures
−12 −12
−3𝐸𝐼∆
𝐿2
∆
∆
∆
C
3 t/m
10m
A B
8t
7.5m 7.5m 10m
D
I 2I I
2 cm 1 cm
EI = 10000 m2t
3 t/m
10m
A I B
2 cm
C
B
8t
7.5m 7.5m
2I
2 cm 1 cm
10m
A B
2 cm
I
−6𝐸𝐼∆
𝐿2
−6𝐸𝐼∆
𝐿2 3 t/m
10m
A I B
25
−25
13
−37
C
B
8t
7.5m 7.5m
2I
C
B
15m
2I
2 cm
1 cm
Δ
5.33
15
−15
5.33
20.33
−9.67 3

168. 36.18
14.93
5.79
19.64
11.95
D
I
C
10m
1 cm
C
3 t/m
10m
A B
8t
7.5m 7.5m 10m
D
I 2I I
2 cm 1 cm
EI = 10000 m2t
3 t/m
10m
A I B
2 cm
C
B
8t
7.5m 7.5m
2I
2 cm 1 cm
13
−37 20.33
−9.67 3
Joint B
KBA : KBC
Joint C
KCB : KCD
𝐼
𝐿
:
𝐼
𝐿
1
10
:
2
15
15 : 20
𝟑
𝟕
:
𝟒
𝟕
15
35
:
20
35
𝐼
𝐿
:
𝐼
𝐿
𝑥
3
4
2
15
:
1
10
𝑥
3
4
80 : 45
𝟎. 𝟔𝟒 : 𝟎. 𝟑𝟔
80
125
:
45
125
13 −9.67 20.33 3 0
−37
3/7 4/7 𝟎. 𝟔𝟒 𝟎. 𝟑𝟔
−1.903 −14.93 −8.399
A B C D
FEM
D.M.1
D.F.
C.O.1
−1.428
D.M.2 4.266 0.609 0.342
3.199
-0.714 -7.466 -0.951
C.O.2 1.6 0.304 2.133
D.M.3 −0.174
−0.13 −0.768
−1.365
C.O.3 -0.065 -0.682 -0.087
D.M.4 0.39
0.292 0.056 0.031
F.M. -36.18 -14.93 5.79 0
-5.79
14.93
C
A B D
B.M.D.

169. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 1)

170. 5t
1.5 t/m
B C
2I 4m
4m
5t
A
1.5 t/m
D
B C
I
I
2I
5m
4m
4m
−13 13
A D
I
I C
B
5m
0
0
5m
0 0
0 -13
0
FEM
8/13 5/13
D.F.
A B C D
D.M.
Joint B
KBA : KBC
1
5
:
1
8
8 : 5
8
13
:
5
13
1
5
:
2
8
𝑥
1
2
I
L
:
I
L 𝑥
1
2
Distribution Factors
8 5
C.O. 4
F.M. 4 8 -8
5t
1.5 t/m
B C
4m
4m
8
4 4
8
8
8
2.4
A
B
5
m
2.4
8.5 8.5
2.4
C
D
5
m
2.4
8
4 4
8
8
8
B.M.D.
14
𝑤𝑙2
8
+
𝑃𝐿
4
= 22

171. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 2)

172. 4t
A
3 t/m
B C
8m
−16 16
4t
A
3 t/m
B
C
3m
8m
3m
B
−4.5
3m 3m
16 -4.5
-16
FEM
0.5 0.5
D.F.
A B C
D.M. -5.75 -5.75
C.O. -2.875
F.M. -18.875 10.25 -10.25
Joint B
KBA : KBC
1
8
:
1
8
1 : 1
𝟎. 𝟓 : 𝟎. 𝟓
1
8
:
1
6 𝑥
3
4
I
L
:
I
L
𝑥
3
4
Distribution Factors
:
A
3 t/m
B
8m
18.875 10.25
10.25
13.08 10.92
B
4t
C
3m
3m
3.71 A B
C
10.25
10.25
18.875
0.29
0.875

173. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 3)

174. A
3 t/m
B
8m
−16 16
4t
C
B
−4.5
3t
B
−6
3m 3m
16 -4.5
-16
FEM
0.5 0.5
D.F.
A B C
D.M.
Joint B
KBA : KBC
1
8
:
1
8
1 : 1
𝟎. 𝟓 : 𝟎. 𝟓
1
8
:
1
6 𝑥
3
4
I
L
:
I
L
𝑥
3
4
:
-6
4t
A
3 t/m
B
C
3m
8m 2m
3m
3t
0
-2.75 -2.75
C.O. -1.375
F.M. -17.375 13.25 -6 -7.25 0
A
3 t/m
B
8m
17.375 13.25
7.25
B
4t
C
3m
3m
A B
C
7.25
13.25
17.375
2.375
3t
2m
B
6
6
𝑝𝑙
4
B.M.D.

175. Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 4)

176. I
3m
4t
E
DB
3m
12m
2m
1.5 t/m
6I
A B
C
-12
-4.5
BD
C
-5.33
2.67
A B
-18 18
C
-5.33 -8 8
8m
C
4I
B
1.5 t/m
-8 8
E
-12 -4.5 0
D
18
-18
6t
2.67
C
1.5 t/m
6I
I
A B
6t
3m
12m 8m 2m
4I
1.5I 4t
6t
F
E
D
3m
4m
FEM
D.F.
Joint B
KBA : KBD : KBC
I
L
:
Distribution Factors
: :
I
L
I
L
:
6
12
: 1.5
6
4
8
:
6 : 3 6
:
0.4 : 0.2 0.4
:
Joint C
KCB : KCE
:
I
L
: I
L x
3
4
4
8
: 1
6
x
3
4
4
8
: 1
8
4 : 1
0.8 : 0.2
0.4 0.2 0.4 0.8 0.2

177. I
3m
4t
E
DB
3m
12m
2m
1.5 t/m
6I
A B
C
-12
-4.5
BD
C
-5.33
2.67
A B
-18 18
C
-5.33 -8 8
8m
C
4I
B
1.5 t/m
-8 8
E
-12 -4.5 0
D
18
-18
6t
2.67
FEM
D.F.
Joint B
KBA : KBD : KBC
I
L
:
Distribution Factors
: :
I
L
I
L
:
6
12
: 1.5
6
4
8
:
6 : 3 6
:
0.4 : 0.2 0.4
:
Joint C
KCB : KCE
:
I
L
: I
L x
3
4
4
8
: 1
6
x
3
4
4
8
: 1
8
4 : 1
0.8 : 0.2
0.4 0.2 0.4 0.8 0.2
C
1.5 t/m
6I
I
A B
6t
3m
12m 8m 2m
4I
1.5I
4t
6t
F
E
D
3m
4m
D.M.1 -1.867 -0.933 6.8 1.7
C.O.1 -0.933 -0.467
3.4 -0.933
D.M.2
C.O.2
-1.36 -0.68 -1.36 0.747 0.187
0.373 -0.68
-0.68 -0.34
D.M.3
C.O.3
-0.149 -0.075 -0.149
0.272
0.544
-0.075
0.136
-0.075 -0.037
D.M.4 -0.109 -0.054 -0.109 0.06 0.015
F.M. -19.69 14.52 -7.08 -7.44 14.46 -12 -2.46 0 1.82
12m
1.5 t/m
A B
19.69 14.52
8m
1.5 t/m
B C
7.44 14.46
6t
D
4m
2m B
1.82
7.08 2.46
6t
E
C
3m
3m
2m
C
6t
12 14.52
19.69
7.44
14.46
12
1.82
7.08
2.7
2.46
4.8
B.M.D.
-1.867