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Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(1.Introduction)
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
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
๐‘ท๐‘ณ
๐Ÿ’
=
๐Ÿ–๐’™๐Ÿ–
๐Ÿ’
=16
16mt
2x8=16
Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(2.Rules)
S.F.D.
๐‘ท
๐Ÿ
๐‘ท
๐Ÿ
๐‘ท
๐Ÿ
๐‘ท
๐Ÿ
๐‘ท๐’ƒ
๐‘ณ
๐‘ท๐’ƒ
๐‘ณ
๐‘ท๐’‚
๐‘ณ
๐‘ท๐’‚
๐‘ณ
๐’˜๐‘ณ
๐Ÿ
๐’˜๐‘ณ
๐Ÿ
A B
a b
L
๐‘ท๐‘ณ
๐Ÿ’
L
A B
w t/m
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
B.M.D
P t
๐‘ท๐’‚๐’ƒ
๐‘ณ
P t
L/2
A B
L/2
๐‘ท
๐Ÿ
๐‘ท
๐Ÿ
๐’˜๐’
๐Ÿ
๐’˜๐’
๐Ÿ
๐‘ท๐’ƒ
๐‘ณ
๐‘ท๐’‚
๐‘ณ
wL
๐‘ท๐’‚
B.M.D
b a
P ๐‘ท
P t
A B
P t
a
๐‘ท๐’‚
a
L
A B
๐‘ท๐’‚
L
L/2
A B
M
L/2
๐‘ด
๐Ÿ
๐‘ด
๐Ÿ
P t
P t
๐‘ด
๐‘ณ
๐‘ด
๐‘ณ
Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(3.Examples)
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
๐Ÿ”
๐Ÿ–
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
๐Ÿ– ๐Ÿ’
๐Ÿ–
๐Ÿ๐Ÿ
๐Ÿ ๐Ÿ
๐Ÿ‘
4 t
3 m
A
3 m
๐Ÿ๐Ÿ
36
36
๐Ÿ—
=
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
๐Ÿ๐Ÿ ๐Ÿ”
๐Ÿ”
๐Ÿ– ๐Ÿ–
๐Ÿ
๐Ÿ’
๐Ÿ๐Ÿ’
๐Ÿ๐ŸŽ
๐Ÿ
Structural Analysis
Superposition
II
Eng. Khaled El-Aswany
(4.Frames)
4t
A D
B C
2m 2m
1m
2m
2m
6t
2t/m
8t
2m 2m
0
2m 2m
0
2t/m
๐‘ท๐‘ณ
๐Ÿ’
=8
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
=4
3t
6t ๐Ÿ”
2m 2m
1
0
4t
2m
2m
4t
๐Ÿ๐Ÿ”
๐Ÿ–
๐Ÿ๐Ÿ”
๐Ÿ๐Ÿ”
๐Ÿ–
๐Ÿ–
๐Ÿ๐Ÿ”
๐Ÿ–
2m
2m
3t
3t
๐Ÿ๐Ÿ
๐Ÿ๐Ÿ
๐Ÿ๐Ÿ
๐Ÿ๐Ÿ
8mt
A D
B C
8m
3m
3m
2t
4t
2m
8mt
2t
4t
0
๐Ÿ–
๐Ÿ–
๐Ÿ–
4t
๐Ÿ๐Ÿ’
๐Ÿ๐Ÿ’
0
๐Ÿ’
๐Ÿ’
๐Ÿ’
๐Ÿ’
๐Ÿ’
๐Ÿ’
Structural Analysis
Three Moment Equation
II
Eng. Khaled El-Aswany
Structural Analysis
Three Moment Equation
II
(Introduction)
Eng. Khaled El-Aswany
2 t/m
6m
A B
6t
2m
C
2m
I1 I2
MA (
๐‹๐Ÿ
๐ˆ
) + 2 MB (
๐‹๐Ÿ
๐ˆ
+
๐‹๐Ÿ
๐ˆ
) + MC (
๐‹๐Ÿ
๐ˆ
) = -6 (
๐‘๐›๐Ÿ
๐ˆ
+
๐‘๐›๐Ÿ
๐ˆ
)
MA = MB = MC =
MA (
๐‹๐Ÿ
๐ˆ๐Ÿ
) + 2 MB (
๐‹๐Ÿ
๐ˆ๐Ÿ
+
๐‹๐Ÿ
๐ˆ๐Ÿ
) + MC (
๐‹๐Ÿ
๐ˆ๐Ÿ
) = -6 (
๐‘๐›๐Ÿ
๐ˆ๐Ÿ
+
๐‘๐›๐Ÿ
๐ˆ๐Ÿ
)
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 (๐‘๐›๐Ÿ + ๐‘๐›๐Ÿ)
Structural Analysis
Three Moment Equation
II
(Elastic Reactions)
Eng. Khaled El-Aswany
2 t/m
6m
A B
6t
2m
C
2m
I1 I2
MA (
๐‹๐Ÿ
๐ˆ
) + 2 MB (
๐‹๐Ÿ
๐ˆ
+
๐‹๐Ÿ
๐ˆ
) + MC (
๐‹๐Ÿ
๐ˆ
) = -6 (
๐‘๐›๐Ÿ
๐ˆ
+
๐‘๐›๐Ÿ
๐ˆ
)
MA = MB = MC =
MA (
๐‹๐Ÿ
๐ˆ๐Ÿ
) + 2 MB (
๐‹๐Ÿ
๐ˆ๐Ÿ
+
๐‹๐Ÿ
๐ˆ๐Ÿ
) + MC (
๐‹๐Ÿ
๐ˆ๐Ÿ
) = -6 (
๐‘๐›๐Ÿ
๐ˆ๐Ÿ
+
๐‘๐›๐Ÿ
๐ˆ๐Ÿ
)
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 (๐‘๐›๐Ÿ + ๐‘๐›๐Ÿ)
๐‘€
6
(2๐‘ฅโ€ซุงู„ุจุนูŠุฏุฉโ€ฌ + โ€ซ)ุงู„ู‚ุฑูŠุจุฉโ€ฌ
A B
a b
L
๐‘ท๐‘ณ
๐Ÿ’
L
A B
w t/m
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
B.M.D
P t
P t
L/2
A B
L/2
Elastic
Reactions
๐ด =
1
2
๐‘ด๐‘ณ
A=
2
3
๐‘ด๐‘ณ ๐ด2 =
1
2
๐‘ด๐‘
๐Ÿ
๐Ÿ‘
b
๐Ÿ
๐Ÿ‘
b
๐Ÿ
๐Ÿ‘
a
๐Ÿ
๐Ÿ‘
a
๐ด1 =
1
2
๐‘ด๐‘Ž
Pa
B.M.D
Elastic
Reactions
P t
a
A B
b
P t
a
Pa
A B
a b
L
M
๐‘€๐‘
๐ฟ
๐‘€๐‘Ž
๐ฟ
๐ด2๐‘ฅ
2
3
๐‘
1
2
๐ด
1
2
๐ด
1
2
๐ด
1
2
๐ด
1
2
๐ด
1
2
๐ด
+๐ด1๐‘ฅ(๐‘ +
1
3
๐‘Ž)
L
๐‘€
6
(2๐‘ฅโ€ซุงู„ุจุนูŠุฏุฉโ€ฌ + โ€ซ)ุงู„ู‚ุฑูŠุจุฉโ€ฌ
๐ด1๐‘ฅ
2
3
๐‘Ž +๐ด2๐‘ฅ(๐‘Ž +
1
3
๐‘)
L
๐‘ท๐’‚๐’ƒ
๐‘ณ
A=(
๐‘+๐ฟ
2
)๐‘ด
๐‘€
๐ฟ
๐‘€
๐ฟ
๐ด1
๐ด2
๐Ÿ
๐Ÿ‘
a
๐Ÿ
๐Ÿ‘
a ๐Ÿ
๐Ÿ‘
b
๐Ÿ
๐Ÿ‘
b
A2x
2
3
b A1x(b +
1
3
a)
L
๐ด2 โˆ’ ๐ด1 โˆ’ ๐‘…1
๐‘ถ๐‘น ๐‘ถ๐‘น
Structural Analysis
Three Moment Equation
II
(Example1)
Eng. Khaled El-Aswany
๐Ÿ
๐Ÿ‘
๐‘ด๐‘ณ = ๐Ÿ‘๐Ÿ”
6t
2m 2m
B C
2 t/m
6m
A B
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
=9
๐‘ท๐‘ณ
๐Ÿ’
= ๐Ÿ”
6t
18 t
18 t 1
2
๐‘ด๐‘ณ = 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 (
๐‹๐Ÿ
๐ˆ๐Ÿ
) + 2 MB (
๐‹๐Ÿ
๐ˆ๐Ÿ
+
๐‹๐Ÿ
๐ˆ๐Ÿ
) + MC (
๐‹๐Ÿ
๐ˆ๐Ÿ
) = -6 (
๐‘๐›๐Ÿ
๐ˆ๐Ÿ
+
๐‘๐›๐Ÿ
๐ˆ๐Ÿ
)
0 + 2 MB (
๐Ÿ”
๐Ÿ
+
๐Ÿ’
๐Ÿ
) + 0 = -6 (
๐Ÿ๐Ÿ–
๐Ÿ
+
๐Ÿ”
๐Ÿ
)
16 MB = -126
MB = -7.875 mt
2x6-7.3125
=4.6875
๐Ÿ๐’™๐Ÿ”๐’™๐Ÿ‘ + ๐Ÿ•. ๐Ÿ–๐Ÿ•๐Ÿ“
๐Ÿ”
๐Ÿ”๐’™๐Ÿ + ๐Ÿ•. ๐Ÿ–๐Ÿ•๐Ÿ“ 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
๐Ÿ’
= ๐Ÿ’. ๐Ÿ—๐Ÿ•
2x6
3m
Structural Analysis
Three Moment Equation
II
(Example 2)
Eng. Khaled El-Aswany
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
๐ฐ๐‹๐Ÿ
๐Ÿ–
=16
2
3
๐‘ด๐‘ณ=85.33
Pa=12 Pa=12
(
4+8
2
)๐‘ฅ12=72
๐Ÿ—๐’™๐Ÿ๐’™๐Ÿ’
๐Ÿ”
=12 ๐Ÿ‘
๐Ÿ” ๐Ÿ—
2.25
2m
1m
0.5
1m
42.67 42.67
36
36
20 16
C
9mt
1.5 3m
D
โˆ’2.25 โˆ’4.5
๐Ÿ
๐Ÿ
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
Structural Analysis
Three Moment Equation
II
(Symmetry)
Eng. Khaled El-Aswany
A
W t/m
B C
L L L
D
L
E A
W t/m
B C
L L L
D
๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘Ž=?? ๐‘€๐‘=??
๐‘€๐‘’= ๐‘€๐‘Ž
๐‘€๐‘‘= ๐‘€๐‘
3 Unknowns 2 Unknowns
A
W t/m
B C
L L
๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘Ž
2 Unknowns
A
W t/m
B
L
๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž
1 Unknown
๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= ๐‘€๐‘Ž
A
W t/m
B
L L/2
F
๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐น=??
3 Unknowns
A
W t/m
B C
L L L
D
L
E
๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘’= ๐‘€๐‘Ž
๐‘€๐‘‘= ๐‘€๐‘
3 Unknowns
A
W t/m
B C
L L
๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘Ž
2 Unknowns
A
W t/m
B
L
๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž
1 Unknown
A
W t/m
B
L
๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž
L
0 0
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
W t/m
A B
Aโ€™ Bโ€™
A B
MAโ€™ (L1) + 2 MA (L1 + L2) + MB (L2) = -6 (๐‘๐š๐Ÿ + ๐‘๐š๐Ÿ)
0 + 2 MA (0 + L) + MA (L) = -6 ( 0 +
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
)
3 MA L = -
wL3
4
๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž
MA = -
wL2
12
0
0
MB =MA
wL2
12
wL2
12
wL2
12
wL2
12
wL2
12
A
W t/m
B C
L L L
D
๐‘€๐‘Ž=?? ๐‘€๐‘=??
2 Unknowns
๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= ๐‘€๐‘Ž
A
W t/m
B
L L/2
F
๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐น=??
3 Unknowns
W t/m
๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= ๐‘€๐‘Ž
0
Aโ€™
0
C D
L
B
L
Dโ€™
A
L
B C D
A
W t/m
W t/m
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
MAโ€™ (L1) + 2 MA (L1 + L2) + MB (L2) = -6 (๐‘๐š๐Ÿ + ๐‘๐š๐Ÿ)
0 + 2 MA (0 + L) + MB (L) = -6 ( 0 +
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
)
0
0
MC =MB
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
2 MA + MB = -
๐Ÿ
4
wL2 โ†’ ๐Ÿ
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (๐‘๐š๐Ÿ + ๐‘๐š๐Ÿ)
MA (L) + 2 MB (L+ L) + MB (L) = -6 (
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
+
๐’˜๐‘ณ๐Ÿ‘
๐Ÿ๐Ÿ’
)
MA + 5 MB = -
๐Ÿ
2
wL2 โ†’ ๐Ÿ
MA = -
wL2
12
MB = -
wL2
12
wL2
12
wL2
12
wL2
12
wL2
12
๐‘€๐‘Žโ€ฒ=0 ๐‘€๐‘‘โ€ฒ=0
Structural Analysis
Three Moment Equation
II
(Settlement)
Eng. Khaled El-Aswany
2 cm
2 cm
1 cm
๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘= 0
A B C
๐‘ณ๐Ÿ ๐‘ณ๐Ÿ
๐‘€๐‘Ž=0 ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘‘=0
A B C D
๐‘ณ๐Ÿ ๐‘ณ๐Ÿ ๐‘ณ๐Ÿ‘
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
โˆ†๐ตโˆ’โˆ†๐ด
L1
+
โˆ†๐ตโˆ’โˆ†๐ถ
L2
)
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
โˆ†๐ตโˆ’โˆ†๐ด
L1
+
โˆ†๐ตโˆ’โˆ†๐ถ
L2
)
MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI (
โˆ†๐ถโˆ’โˆ†๐ต
L2
+
โˆ†๐ถโˆ’โˆ†๐ท
L3
)
If I constant:
EI : Given
I I
If I variable: MA (
L1
I1
) + 2 MB (
L1
I1
+
L2
I2
) + MC (
L2
I2
) = 6EI (
โˆ†๐ตโˆ’โˆ†๐ด
L1
+
โˆ†๐ตโˆ’โˆ†๐ถ
L2
)
2 Equations
๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘‘= 0
2 cm
๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž
A B C D
2 t/m 2 t/m
A B C D
๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
โˆ†๐ตโˆ’โˆ†๐ด
L1
+
โˆ†๐ตโˆ’โˆ†๐ถ
L2
)
MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI (
โˆ†๐ถโˆ’โˆ†๐ต
L2
+
โˆ†๐ถโˆ’โˆ†๐ท
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 (
0โˆ’0.02
4
+ 0)
20 MB + 4 MC = 250 โ†’ 1
4 MB + 20 MC = -150 โ†’ 2
MB = 14.58 mt
MC = -10.42 mt
๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= 0
2 t/m
A B
๐Ÿ”๐’Ž
C
๐Ÿ’๐’Ž
B
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
= ๐Ÿ—
2
3
๐‘ด๐‘ณ=36 18
18 0
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (๐‘…๐‘1+๐‘…๐‘2)
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
๐Ÿ”๐’Ž
C
๐Ÿ’๐’Ž
B
2 t/m
A B
๐Ÿ”๐’Ž
10.08 14.92 14.92
10.08
4.32
7.68 6.25
6.25 8.49 3.51
10.08
14.92
๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘‘= 0
2 cm
๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž
A B C D
2 t/m 2 t/m
A B C D
๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž
MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI (
โˆ†๐ตโˆ’โˆ†๐ด
L1
+
โˆ†๐ตโˆ’โˆ†๐ถ
L2
)
MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI (
โˆ†๐ถโˆ’โˆ†๐ต
L2
+
โˆ†๐ถโˆ’โˆ†๐ท
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 (
0โˆ’0.02
4
+ 0)
20 MB + 4 MC = 250 โ†’ 1
4 MB + 20 MC = -150 โ†’ 2
MB = 14.58 mt
MC = -10.42 mt
๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= 0
2 t/m
A B
๐Ÿ”๐’Ž
C
๐Ÿ’๐’Ž
B
๐’˜๐‘ณ๐Ÿ
๐Ÿ–
= ๐Ÿ—
2
3
๐‘ด๐‘ณ=36 18
18 0
MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (๐‘…๐‘1+๐‘…๐‘2)
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
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Once)
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
ฮด10 + X ฮด11 = 0 X =
โˆ’ฮด10
ฮด11
ฮด10 = สƒ
๐‘ด๐‘ถ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’
ฮด11 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’
9 mt
6m
1 mt
= Area x Yc
=
2
3
๐‘ฅ6๐‘ฅ9 x 0.5
Area
Yc
0.5x1=0.5
(Linear)
-
โ€ซู…ุณุชู‚ูŠู…โ€ฌ โ€ซุฎุทโ€ฌ
โ€ซู…ู†ูƒุณุฑโ€ฌ โ€ซุบูŠุฑโ€ฌ
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ = Area x Yc
=
1
2
๐‘ฅ6๐‘ฅ8 x 3
Area
6m
6 mt
8 mt
Yc
0.5x6=3
(Linear)
-
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ = Area x Yc
=
1
2
๐‘ฅ6๐‘ฅ8
6 mt
6m
8 mt
x 4
Area
Yc
๐Ÿ
๐Ÿ‘
๐ฑ๐Ÿ” = ๐Ÿ’
+
๐Ÿ
๐Ÿ‘
๐‘ณ
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ = Area x Yc
=
1
2
๐‘ฅ3๐‘ฅ8
8 mt
6 mt
3m 3m
+
Area
Yc
๐Ÿ
๐Ÿ‘
๐‘ณ
๐Ÿ
๐Ÿ‘
๐‘ณ
๐Ÿ
๐Ÿ‘
๐ฑ๐Ÿ” = ๐Ÿ’
x 4
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’
x 2
=
L
3
(๐š๐œ + ๐›๐+
bc
2
+
ad
2
)
8
4
2
6
8m
8x6
=
8
3 ( + 4x2 +
๐Ÿ’๐’™๐Ÿ”
๐Ÿ
+
๐Ÿ–๐’™๐Ÿ
๐Ÿ
)
8 mt
6 mt
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ =
6
3
( 8x6 + 0 + 0 + 0 )
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
ฮด10 + X ฮด11 = 0
X =
โˆ’ฮด10
ฮด11
ฮด10 = สƒ
๐‘ด๐‘ถ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’
ฮด11 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’
=
๐Ÿ
๐‘ฌ๐‘ฐ
[-
๐Ÿ
๐Ÿ‘
x 6 x 9 x0.5 +
๐Ÿ”
๐Ÿ‘
(
๐Ÿ๐Ÿ๐’™๐Ÿ
๐Ÿ
)] =
โˆ’๐Ÿ”
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ ๐Ÿ”
๐Ÿ‘
(1x1)] =
๐Ÿ
๐‘ฌ๐‘ฐ
=
โˆ’( เต—
โˆ’๐Ÿ”
EI)
เต—
๐Ÿ
EI
= ๐Ÿ‘
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
ฮด10 + X ฮด11 = 0
X =
โˆ’ฮด10
ฮด11
ฮด10
ฮด11
=
๐Ÿ
๐‘ฌ๐‘ฐ
[-
๐Ÿ
๐Ÿ‘
x 6 x 9 x0.5 +
๐Ÿ”
๐Ÿ‘
(
๐Ÿ๐Ÿ๐’™๐Ÿ
๐Ÿ
)] =
โˆ’๐Ÿ”
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ ๐Ÿ”
๐Ÿ‘
(1x1)] =
๐Ÿ
๐‘ฌ๐‘ฐ
=
โˆ’( เต—
โˆ’๐Ÿ”
EI)
เต—
๐Ÿ
EI
= ๐Ÿ‘
A B
3
12
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Twice)
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
๐‘€๐‘œ
12 12 4 8
๐‘€1
1
๐‘€2
1
1mt
1mt
ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0
ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0
ฮด10 = สƒ
๐‘ด๐‘ถ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[-(
4+8
2
)๐’™๐Ÿ๐Ÿ x0.5 - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ x0.5] = โˆ’๐Ÿ’๐ŸŽ
EI
ฮด11 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ 8
3
๐’™(๐Ÿ๐’™๐Ÿ) = 4
EI
+
4
3
๐’™(๐Ÿ๐’™๐Ÿ) ]
ฮด12 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[4
3
x(
1X1
2
) ] =
เต˜
2
3
EI
ฮด02 = สƒ
๐‘ด๐‘ถ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ -
๐Ÿ
๐Ÿ
x0.5 ๐’™
2
3
๐’™๐Ÿ’๐’™๐Ÿ– x0.5 ]=
เต˜
โˆ’2๐Ÿ–
3
EI
ฮด22 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ ๐’™
4
3
๐’™(๐Ÿ๐’™๐Ÿ) ]
4
3
๐’™(๐Ÿ๐’™๐Ÿ) +
1
2 = ๐Ÿ
EI
โˆ’40
EI
+
4
EI
๐‘‹1 +
เต˜
2
3
EI
๐‘‹2 = 0
โˆ’40+ 4 ๐‘‹1 +
2
3 ๐‘‹2 = 0
4 ๐‘‹1 +
2
3
๐‘‹2 = 40
เต˜
โˆ’28
3
EI
+ ๐‘‹1
เต˜
2
3
EI
+ ๐‘‹2
2
EI
= 0
โˆ’28
3
+
2
3
๐‘‹1+2๐‘‹2 = 0
2
3
๐‘‹1 + 2 ๐‘‹2 =
28
3
โ†’ 1
โ†’ 2
๐‘ฟ๐Ÿ = 9.76 mt
๐‘ฟ๐Ÿ = ๐Ÿ. ๐Ÿ’๐Ÿ ๐’Ž๐’•
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
๐‘€๐‘œ
12 12 4 8
๐‘€1
1
๐‘€2
1
1mt
1mt
ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0
ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0
ฮด10 = สƒ
๐‘ด๐‘ถ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[-(
4+8
2
)๐’™๐Ÿ๐Ÿ x0.5 - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ x0.5] = โˆ’๐Ÿ’๐ŸŽ
EI
ฮด11 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ 8
3
๐’™(๐Ÿ๐’™๐Ÿ) = 4
EI
+
4
3
๐’™(๐Ÿ๐’™๐Ÿ) ]
ฮด12 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[4
3
x(
1X1
2
) ] =
เต˜
2
3
EI
ฮด02 = สƒ
๐‘ด๐‘ถ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ -
๐Ÿ
๐Ÿ
x0.5 ๐’™
2
3
๐’™๐Ÿ’๐’™๐Ÿ– x0.5 ]=
เต˜
โˆ’2๐Ÿ–
3
EI
ฮด22 = สƒ
๐‘ด๐Ÿ
๐‘ด๐Ÿ
๐’…๐’
๐‘ฌ๐‘ฐ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[ ๐’™
4
3
๐’™(๐Ÿ๐’™๐Ÿ) ]
4
3
๐’™(๐Ÿ๐’™๐Ÿ) +
1
2 = ๐Ÿ
EI
4 ๐‘‹1 +
2
3
๐‘‹2 = 40
2
3
๐‘‹1 + 2 ๐‘‹2 =
28
3
โ†’ 1
โ†’ 2
9.76
1.41 ๐‘ค๐‘™2
8
=8
D
A B C
9.76+1.41
2
=5.58
๐‘ƒ๐ฟ
4
=4
1.58
1
4
๐‘‹9.76=2.44
3
4
๐‘‹9.76=7.32
12
9.56
12
4.68
๐‘ฟ๐Ÿ = 9.76 mt
๐‘ฟ๐Ÿ = ๐Ÿ. ๐Ÿ’๐Ÿ ๐’Ž๐’•
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Support movement)
ฮด10
๐‘‹1
ฮด11
ฮด10 + X1 ฮด11 = 0
ฮด10
๐‘‹1
ฮด11
ฮด10 + X1 ฮด11 = ยฑโˆ†
ฮ”
1 ๐‘ก
2 ๐‘๐‘š
3
X1 ฮด11 = โˆ’๐ŸŽ. ๐ŸŽ๐Ÿ
ฮด11 =
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ =
๐Ÿ
๐‘ฌ๐‘ฐ
(
๐Ÿ”
๐Ÿ‘
( 3x3)x 2 ) =
๐Ÿ‘๐Ÿ”
๐‘ฌ๐‘ฐ
X1 ฮด11 = ยฑโˆ†
2.5๐‘ก 1.25 ๐‘ก
1.25 ๐‘ก
7.5
B.M.D.
Reactions
X1
36
๐ธ๐ผ
= โˆ’0.02
X1
36
4500
= โˆ’0.02
X1 = โˆ’2.5 ๐‘ก
EI = 4500 m2t
1 ๐‘ก
2 ๐‘๐‘š
3
EI = 4500 m2t
ฮด11 =
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ =
๐Ÿ
๐‘ฌ๐‘ฐ
(
๐Ÿ”
๐Ÿ‘
( 3x3)x 2 ) =
๐Ÿ‘๐Ÿ”
๐‘ฌ๐‘ฐ
ฮด10 + X1 ฮด11 = ยฑโˆ†
Reactions
โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ
๐‘ฌ๐‘ฐ
+ X1
36
๐ธ๐ผ
= โˆ’0.02
X1 = 72.5 ๐‘ก
180
M.S.
M0
M1
ฮด10 =
๐Ÿ
๐‘ฌ๐‘ฐ
(
โˆ’๐Ÿ
๐Ÿ‘
๐’™๐Ÿ”๐’™๐Ÿ๐Ÿ–๐ŸŽ x(
๐Ÿ“
๐Ÿ–
๐’™๐Ÿ‘) x 2 ) =
โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ
๐‘ฌ๐‘ฐ
5
8
๐ฟ
3
8
๐ฟ
5
8
๐‘ฅ3
โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ
๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ
+ X1
36
4500
= โˆ’0.02
72.5 ๐‘ก 23.75๐‘ก
23.75๐‘ก
1 ๐‘ก
2 ๐‘๐‘š
3
EI = 4500 m2t
ฮด11 =
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ =
๐Ÿ
๐‘ฌ๐‘ฐ
(
๐Ÿ”
๐Ÿ‘
( 3x3)x 2 ) =
๐Ÿ‘๐Ÿ”
๐‘ฌ๐‘ฐ
ฮด10 + X1 ฮด11 = ยฑโˆ†
B.M.D.
โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ
๐‘ฌ๐‘ฐ
+ X1
36
๐ธ๐ผ
= โˆ’0.02
X1 = 72.5 ๐‘ก
180
M.S.
M0
M1
ฮด10 =
๐Ÿ
๐‘ฌ๐‘ฐ
(
โˆ’๐Ÿ
๐Ÿ‘
๐’™๐Ÿ”๐’™๐Ÿ๐Ÿ–๐ŸŽ x(
๐Ÿ“
๐Ÿ–
๐’™๐Ÿ‘) x 2 ) =
โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ
๐‘ฌ๐‘ฐ
5
8
๐ฟ
3
8
๐ฟ
5
8
๐‘ฅ3
โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ
๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ
+ X1
36
4500
= โˆ’0.02
37.5
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Beams)
(Extra 1)
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
๐Ÿ—
2mt
2
Mo
๐Ÿ—
1mt
1
M1
1
M2
ฮด11 = 1
๐ธ๐ผ
(
1
3
*
6
3
(1*1)) =
2
3๐ธ๐ผ
ฮด22 = 1
๐ธ๐ผ
(
1
3
*
6
3
(1*1)) =
2
๐ธ๐ผ
*
8
3
(1*1))
+(
1
2
ฮด12 = 1
๐ธ๐ผ
(
1
3
*
6
3
(
1โˆ—1
2
)) =
1
3๐ธ๐ผ
ฮด1o = 1
๐ธ๐ผ
(
1
3
(-
2
3
x6x9x0.5 =
โˆ’34
3๐ธ๐ผ
0.5
-
2+6
2
x8x0.5)
ฮด2o = 1
๐ธ๐ผ
(
1
3
(-
2
3
x6x9x0.5-
2+6
2
x8x0.5)
+(
1
2
(
8
3
X
1๐‘‹2
2
+
6
3
x9x0.75
0.75
+
2
3
x(9x0.75+
9๐‘‹1
2
)
0.5
+
4
3
x(4x0.5+
4๐‘‹1
2
)
-
4
3
x(4x0.5)
0.25
-
2
3
x(9x0.25)
-
6
3
x(9x0.25+
9๐‘‹1
2
)) =
โˆ’17
3๐ธ๐ผ
ฮด10 + ๐’™๐Ÿ ฮด11 + ๐’™๐Ÿ ฮด12= 0
โˆ’๐Ÿ‘๐Ÿ’
๐Ÿ‘
+
๐Ÿ
๐Ÿ‘
๐’™๐Ÿ +
๐Ÿ
๐Ÿ‘
๐’™๐Ÿ = 0
๐Ÿ ๐’™๐Ÿ + ๐’™๐Ÿ = 34
๐’™๐Ÿ + 6๐’™๐Ÿ = 17
๐’™๐Ÿ= 17 ๐’™๐Ÿ = 0
ฮด20 + ๐’™๐Ÿ ฮด21 + ๐’™๐Ÿ ฮด22= 0
1mt
1mt
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
๐Ÿ—
2mt
2
Mo
๐Ÿ—
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
๐Ÿ๐Ÿ. ๐Ÿ–๐Ÿ‘ ๐Ÿ•. ๐Ÿ๐Ÿ• ๐Ÿ’. ๐Ÿ๐Ÿ“
๐Ÿ’. ๐Ÿ๐Ÿ“
17
4.66 10.34
8
2
2
6.5
3
5
8.5
1mt
1mt
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Once)
4t
A
2 t/m
D
B C
I
I
2I
4m
3m
3m
1.19
8
= 4 โ€“ 3 = 1
R = U โ€“ 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
ฮด10
ฮด11
=
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐ŸŽ๐‘ด๐Ÿ๐’…๐’
=
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’
=
๐Ÿ
๐‘ฌ๐‘ฐ
[- x
๐Ÿ
๐Ÿ‘
x 6 x 9 x4
- =
โˆ’๐Ÿ๐ŸŽ๐Ÿ–
๐‘ฌ๐‘ฐ
4 4
๐Ÿ
๐Ÿ
x
๐Ÿ
๐Ÿ
x 6 x 6 x4 ]
๐Ÿ
๐Ÿ
=
๐Ÿ
๐‘ฌ๐‘ฐ
[
๐Ÿ’
๐Ÿ‘
x (4x4)
+ =
๐Ÿ—๐ŸŽ.๐Ÿ”๐Ÿ•
๐‘ฌ๐‘ฐ
x ๐Ÿ’ x 6 x4 ]
๐Ÿ
๐Ÿ
x2
๐‘ฟ๐Ÿ =
โˆ’ ฮด10
ฮด11
=
โˆ’( โˆ’ 108/๐ธ๐ผ)
90.67/๐ธ๐ผ
= 1.19
๐‘€๐‘“ = ๐‘€0 + ๐‘‹1 ๐‘€1
Mf
๐‘€๐‘“ = 0 + 1.19 x (-4) = -4.76
4.76 4.76
4.76
4.76
10.24
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Twice)
8t
A
2 t/m
D
B C
2I
3I
I
6m
3m
3m
2m
4t
3m
XA
YA
XD
YD
=5โ€“ 3 = 2
R = U โ€“ 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
ฮด10 =
๐Ÿ
๐‘ฌ๐‘ฐ
[ x
3
3
x( 12 x 3)
+
=
๐Ÿ๐Ÿ’๐ŸŽ
๐‘ฌ๐‘ฐ
๐Ÿ
๐Ÿ
6
3
x(16x6+12x3+
16x3
2
+
12x6
2
)
X4.5 ]
- 2
3
x6x9
ฮด11 =
๐Ÿ
๐‘ฌ๐‘ฐ
[ x
3
3
x( 3 x 3)
+
=
๐Ÿ๐Ÿ“๐Ÿ’.๐Ÿ“
๐‘ฌ๐‘ฐ
๐Ÿ
๐Ÿ
6
3
x(6x6+3x3+
6x3
2
๐ฑ๐Ÿ)
+ x
6
3
x( 6 x 6) ]
๐Ÿ
๐Ÿ‘
ฮด12 =
๐Ÿ
๐‘ฌ๐‘ฐ
[ x
1
2
x6x6
+ =
๐Ÿ๐Ÿ
๐‘ฌ๐‘ฐ
๐Ÿ
๐Ÿ‘
X 1
6
3
x(6x1+
3x1
2
) ]
1
1
1
ฮด20 =
๐Ÿ
๐‘ฌ๐‘ฐ
[
=
๐Ÿ๐Ÿ”
๐‘ฌ๐‘ฐ
6
3
x(16x1+12x0+
16x0
2
+
12x1
2
)
X0.5 ]
- 2
3
x6x9
ฮด22 =
๐Ÿ
๐‘ฌ๐‘ฐ
[ x (๐Ÿ”๐’™๐Ÿ)
=
๐Ÿ’
๐‘ฌ๐‘ฐ
๐Ÿ
๐Ÿ‘
+
6
3
x( 1x1) ]
x ๐Ÿ
8t
A
2 t/m
D
B C
2I
3I
I
6m
3m
3m
2m
4t
3m
YA
XD
YD
=5โ€“ 3 = 2
R = U โ€“ 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
ฮด10 =
๐Ÿ๐Ÿ’๐ŸŽ
๐‘ฌ๐‘ฐ
ฮด11
=
๐Ÿ๐Ÿ“๐Ÿ’.๐Ÿ“
๐‘ฌ๐‘ฐ
ฮด12
=
๐Ÿ๐Ÿ
๐‘ฌ๐‘ฐ
1
1
1
ฮด20 =
๐Ÿ๐Ÿ”
๐‘ฌ๐‘ฐ
ฮด22
=
๐Ÿ’
๐‘ฌ๐‘ฐ
ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0
ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0
๐‘ฟ๐Ÿ = -2.34 mt
๐‘ฟ๐Ÿ = ๐Ÿ“. ๐Ÿ•๐Ÿ– ๐’Ž๐’•
154.5 ๐‘‹1 + 21 ๐‘‹2 = -240
21 ๐‘‹1 + 4 ๐‘‹2 = โˆ’26
โ†’ 1
โ†’ 2
MF
๐‘€๐‘“ = ๐‘€0 + ๐‘‹1 ๐‘€1 + ๐‘‹2 ๐‘€2
๐‘€๐‘“ = ๐‘€0 -2.34 ๐‘€1 + 5.78 ๐‘€2
๐‘€๐‘“ = (โˆ’16) -2.34 (โˆ’6) + 5.78 (โˆ’1) = -7.74
4.98
8.26
5.78
16
4.98
7.74
2.34
5.78
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Symmetry)
= 5 โ€“ 3 = 2
R = U โ€“ E
2 t/m
A
B
D
C
4m
4m
2 t/m
4m
0
8 8
8
= 6 โ€“ 3 = 3
R = U โ€“ E
2 t/m
A
B
D
C
8m
4m
= 5 โ€“ 3 = 2
R = U โ€“ E
2 t/m
A
B
4m
4m
2 t/m
A
B
4m
4m
M.S.
๐‘€๐‘œ
16
4
16
16
๐‘€1
1mt
1
1
1
1
๐‘€2
1t
4
ฮด11 =
๐Ÿ
๐‘ฌ๐‘ฐ
[4x1x1 + 4x1x1]=
๐Ÿ–
๐‘ฌ๐‘ฐ
ฮด22 =
๐Ÿ
๐‘ฌ๐‘ฐ
[๐Ÿ’
๐Ÿ‘
๐’™(๐Ÿ’๐’™๐Ÿ’)]=
๐Ÿ๐Ÿ.๐Ÿ‘๐Ÿ‘
๐‘ฌ๐‘ฐ
ฮด12 =
๐Ÿ
๐‘ฌ๐‘ฐ
[โˆ’๐Ÿ
๐Ÿ
๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ]=
โˆ’๐Ÿ–
๐‘ฌ๐‘ฐ
ฮด20 =
๐Ÿ
๐‘ฌ๐‘ฐ
[โˆ’๐Ÿ
๐Ÿ
๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ๐Ÿ”]=
โˆ’๐Ÿ๐Ÿ๐Ÿ–
๐‘ฌ๐‘ฐ
ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0
ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0
๐‘ฟ๐Ÿ = -7.47 mt
๐‘ฟ๐Ÿ = ๐Ÿ‘. ๐Ÿ ๐’Ž๐’•
8 ๐‘‹1 - 8 ๐‘‹2 = -85.33 โ†’ 1
-8 ๐‘‹1 + 21.33 ๐‘‹2 = 128 โ†’ 2
๐‘€๐‘“ = ๐‘€0 + ๐‘‹1 ๐‘€1 + ๐‘‹2 ๐‘€2
๐‘€๐‘“2 = 16 -7.47 (1) + 3.2 (0) = 8.53
๐‘€๐‘“ = ๐‘€0 -7.47 ๐‘€1 + 3.2 ๐‘€2
๐‘€๐‘“3 = 16 -7.47 (1) + 3.2 (-4) = -4.27
1
2
3 ๐‘€๐น
ฮด10=
๐Ÿ
๐‘ฌ๐‘ฐ
[(
1
2
x4 x 16) ๐ฑ๐Ÿ โˆ’
๐Ÿ
๐Ÿ‘
๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ +๐Ÿ’๐’™๐Ÿ๐Ÿ” ๐’™๐Ÿ ] =
๐Ÿ–๐Ÿ“.๐Ÿ‘๐Ÿ‘
๐‘ฌ๐‘ฐ
7.47
8.53
8.53
4.27
= 5 โ€“ 3 = 2
R = U โ€“ E
2 t/m
A
B
4m
4m
2 t/m
A
B
4m
4m
M.S.
๐‘€๐‘œ
16
4
16
16
๐‘€1
1mt
1
1
1
1
๐‘€2
1t
4
ฮด11 =
๐Ÿ
๐‘ฌ๐‘ฐ
[4x1x1 + 4x1x1]=
๐Ÿ–
๐‘ฌ๐‘ฐ
ฮด22 =
๐Ÿ
๐‘ฌ๐‘ฐ
[๐Ÿ’
๐Ÿ‘
๐’™(๐Ÿ’๐’™๐Ÿ’)]=
๐Ÿ๐Ÿ.๐Ÿ‘๐Ÿ‘
๐‘ฌ๐‘ฐ
ฮด12 =
๐Ÿ
๐‘ฌ๐‘ฐ
[โˆ’๐Ÿ
๐Ÿ
๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ]=
โˆ’๐Ÿ–
๐‘ฌ๐‘ฐ
ฮด20 =
๐Ÿ
๐‘ฌ๐‘ฐ
[โˆ’๐Ÿ
๐Ÿ
๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ๐Ÿ”]=
โˆ’๐Ÿ๐Ÿ๐Ÿ–
๐‘ฌ๐‘ฐ
ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0
ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0
๐‘ฟ๐Ÿ = -7.47 mt
๐‘ฟ๐Ÿ = ๐Ÿ‘. ๐Ÿ ๐’Ž๐’•
8 ๐‘‹1 - 8 ๐‘‹2 = -85.33 โ†’ 1
-8 ๐‘‹1 + 21.33 ๐‘‹2 = 128 โ†’ 2
๐‘€๐น
ฮด10=
๐Ÿ
๐‘ฌ๐‘ฐ
[(
1
2
x4 x 16) ๐ฑ๐Ÿ โˆ’
๐Ÿ
๐Ÿ‘
๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ +๐Ÿ’๐’™๐Ÿ๐Ÿ” ๐’™๐Ÿ ] =
๐Ÿ–๐Ÿ“.๐Ÿ‘๐Ÿ‘
๐‘ฌ๐‘ฐ
7.47
8.53
8.53
4.27
8.53
8.53
4.27
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
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Frames)
(Support movement)
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 โ€“ 3 = 1
R = U โ€“ 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
ฮด10 =
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐ŸŽ๐‘ด๐Ÿ๐’…๐’
=
๐Ÿ
๐‘ฌ๐‘ฐ
[
๐Ÿ’
๐Ÿ‘
x (24x4) -
๐Ÿ
๐Ÿ‘
x 6 x 9 x4
+ =
๐Ÿ๐Ÿ•๐Ÿ
๐‘ฌ๐‘ฐ
๐Ÿ
๐Ÿ
x 6 x 24 x4 ]
ฮด11 =
๐Ÿ
๐‘ฌ๐‘ฐ
สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’
=
๐Ÿ
๐‘ฌ๐‘ฐ
[
๐Ÿ’
๐Ÿ‘
x (4x4) + =
๐Ÿ๐Ÿ‘๐Ÿ–.๐Ÿ”๐Ÿ•
๐‘ฌ๐‘ฐ
๐Ÿ’ x 6 x4 ]
x2
ฮด10 + X1 ฮด11 = 0.02
๐Ÿ๐Ÿ•๐Ÿ
๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ๐ŸŽ
+ X1
138.67
20000
= 0.02
X1 = 0.92 ๐‘ก
Mf
3.68 27.68
27.68
3.68
15.68
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
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
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
ฮด10 + ๐‘ฟ๐Ÿ ฮด11 = 0
A B
N0
6 6
A B
N1
1 1
ฮด10 =
๐Ÿ
๐‘ฌ๐‘จ
สƒ ๐‘ต๐‘ถ๐‘ต๐Ÿ๐’…๐’
4m
=
๐Ÿ
๐‘ฌ๐‘จ
( -6 x 4 x 1
N0 N1
L
C D
C D
+ โ€ฆโ€ฆโ€ฆ + ...โ€ฆโ€ฆ)
N0 N1
L N0 N1
L
ฮด10 =
ฮฃNON1L
EA
ฮด11 =
ฮฃN1N1L
EA
Zero force members
2 members 3 members
P
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Example 1)
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
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.
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
4
4
No
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
๐น1
๐น2
๐น1sin ๐œƒ
๐น1cos ๐œƒ
4
4
ฮฃ๐น๐‘ฆ = 0
๐น1sin ๐œƒ + 6 = 0
๐น1=
โˆ’6
sin ฮธ
=
โˆ’6
0.6
= -10 t (๐ถ๐‘œ๐‘š๐‘๐‘Ÿ๐‘’๐‘ ๐‘–๐‘œ๐‘›)
No
ฮฃ๐น๐‘ฅ = 0
๐น1cos ๐œƒ + ๐น2 = 0
(๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›)
โˆ’10 x0.8+ ๐น2 = 0 ๐น2 = 8t
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
4
4
No
10
8
8
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
4
4
No
10
8
8
๐น3
๐น4 10cos๐œƒ
10sin๐œƒ
๐น3 cos๐œƒ
๐น3 sin๐œƒ
ฮฃ๐น๐‘ฆ = 0
10sin ๐œƒ - 4 - ๐น3sin ๐œƒ = 0
(๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›)
ฮฃ๐น๐‘ฅ = 0
๐น3cos ๐œƒ + ๐น4 + 10cos ๐œƒ = 0
(๐ถ๐‘œ๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘–๐‘œ๐‘›)
๐น4 = -10.67t
๐น3 = 3.33t
3.33cos ๐œƒ + ๐น4 + 10cos ๐œƒ = 0
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
4
4
No
10
8
8
3.33
10.67
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
4
4
No
10
8
8
3.33
10.67
3.33
10.67
8
8
10
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0
M.S.
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
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
R = ( m + r ) - 2j
R = ( 13+ 4) - 2x8 = 1 (Once indeterminate)
(External)
4t
4t 4t
4m 4m
4m 4m
6t
6t
0 ๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4๐‘š
3๐‘š
5๐‘š
3m
4t
4t 4t
4m 4m
4m 4m
1 2 3 4
5 6 7 8 9 10 11
12 13
3m
๐œƒ
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
4
4
No
-10
8
8
3.33
-10.67
3.33
โˆ’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
ฮด10 =
๐›ดN
ON1L
EA
=
โˆ’128
EA
ฮด11 =
๐›ดN1N1L
EA
=
16
EA
X1 =
โˆ’ฮด10
ฮด11
=
โˆ’(โˆ’128)/๐ธ๐ด
16/EA
= 8t
Nf = N0 + X1 N1
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Example 2)
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
4
10 6
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10 6 4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10 6
4
๐น1
๐น2
๐น1sin ๐œƒ
๐น1cos ๐œƒ
ฮฃ๐น๐‘ฆ = 0
๐น1sin ๐œƒ -4 +10 = 0
๐น1= -10 t (๐ถ๐‘œ๐‘š๐‘๐‘Ÿ๐‘’๐‘ ๐‘–๐‘œ๐‘›)
ฮฃ๐น๐‘ฅ = 0
๐น1cos ๐œƒ + ๐น2 = 0
(๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›)
โˆ’10 x0.8+ ๐น2 = 0 ๐น2 = 8t
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10 6 4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
10
8
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10 6 4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
10
8
8
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10 6 4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
10
8
8
๐น3
๐น4 10cos๐œƒ
10sin๐œƒ
๐น3 cos๐œƒ
๐น3 sin๐œƒ
ฮฃ๐น๐‘ฆ = 0
10sin ๐œƒ - 6 - ๐น3sin ๐œƒ = 0
ฮฃ๐น๐‘ฅ = 0
๐น3cos ๐œƒ + ๐น4 + 10cos ๐œƒ = 0
(๐ถ๐‘œ๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘–๐‘œ๐‘›)
๐น4 = - 8t
๐น3 = 0
0 + ๐น4 + 10cos ๐œƒ = 0
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10 6 4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
10
8
8
8
8
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10 6 4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
10
8
8
8
8
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
10
6 4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
10
8
8
8
8
๐น5
๐น5 cos๐œƒ
๐น5 sin๐œƒ
ฮฃ๐น๐‘ฆ = 0
10 - 4 - ๐น5sin ๐œƒ = 0
ฮฃ๐น๐‘ฅ = 0
๐น5cos ๐œƒ - 8 = 0
(๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›)
๐น5 = 10t
๐น5 = 10t
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
-10 -6 -4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
-10
8
8
โˆ’8
โˆ’8
10
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
-10 -6 -4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
-10
8
8
โˆ’8
โˆ’8
10
3m
4m 4m
4m
0
N1
0
1t
1t
๐œƒ
-0.8
-0.6
๐œƒ
-0.8
-0.6
๐œƒ
1
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
-10 -6 -4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
-10
8
8
โˆ’8
โˆ’8
10
3m
4m 4m
4m
0
N1
0
๐œƒ
๐œƒ
๐œƒ
1
ฮด10 =
๐›ดN
ON1L
EA
ฮด11 =
๐›ดN
1N1L
EA
X1 =
โˆ’ฮด10
ฮด11
=
โˆ’(โˆ’2)/๐ธ๐ด
16/EA
= 0.125t
ฮด10 + X1 ฮด11 = 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]
=
โˆ’2
EA
=
1
EA
[ )0.8(2x4 +)0.6(2x3 x2+)1(2x5 x2
+
1
2
)0.8(2x4] =
16
EA
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
-10 -6 -4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
-10
8
8
โˆ’8
โˆ’8
10
3m
4m 4m
4m
0
N1
0
๐œƒ
๐œƒ
๐œƒ
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
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
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
-10 -6 -4
๐œƒ
4๐‘š
3๐‘š
5๐‘š
cos ๐œƒ =
4
5
= 0.8
sin ๐œƒ =
3
5
= 0.6
-10
8
8
โˆ’8
โˆ’8
10
3m
4m 4m
4m
0
N1
0
๐œƒ
๐œƒ
๐œƒ
1
1t
1t
-0.8
-0.6
-0.8
-0.6
Mem. L No N1 A
๐‘๐‘œ๐‘1๐ฟ
๐‘จ
๐‘๐Ÿ๐‘1๐ฟ
๐‘จ
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
ฮด10 =
โˆ’2
EA
ฮด11 =
16
EA
X1 = 0.125t
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Example 3)
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
7๐‘ก
7๐‘ก
0
No
-7
-7 -6 45ยฐ
-3
-3
A
3m
3m
3m
B
N1
1๐‘ก
0
0
1๐‘ก
-1
-1 A
3m
3m
3m
B
N2
1๐‘ก
1๐‘ก
0 0
0
โˆ’1
2
โˆ’1
2
โˆ’1
2
โˆ’1
2
1๐‘ก
ฮด10 + X1 ฮด11 + X2 ฮด12 = 0
ฮด20 + X1 ฮด21 + X2 ฮด22 = 0
ฮด10 = 0
ฮด20 =
51.94
EA
ฮด11 =
6
EA
ฮด22 =
14.49
EA
ฮด12 =
2.12
EA
X1 = 1.34 ๐‘ก
X2 = โˆ’3.78 ๐‘ก
Nf = N0 + X1 N1 + X2 N2
Nf = N0 + 1.34 N1 -3.78 N2
6 X1 + 2.12 X2 = 0 โ†’ 1
2.12 X1 + 14.49 X2 = -51.94 โ†’ 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 โˆ’1
/ 2
0 0 3 1.5 ๐Ÿ‘/ ๐Ÿ 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 โˆ’1
/ 2
0 ๐Ÿ๐Ÿ–/ ๐Ÿ 0 1.5 0 -3.33
6 3 2 0 0 1 0 0 0 ๐Ÿ‘ ๐Ÿ 0 -3.78
7 3 2 3 2 0 1 0 ๐Ÿ๐Ÿ– 0 ๐Ÿ‘ ๐Ÿ 0 0.46
8 3 -7 0 โˆ’1
/ 2
0 ๐Ÿ๐Ÿ/ ๐Ÿ 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
1
Structural Fantasy
3
2
8t
6m
A
B
๐‘ญ๐Ÿ ๐‘ญ๐Ÿ
๐‘ญ๐Ÿ– ๐‘ญ๐Ÿ—
๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’ ๐‘ญ๐Ÿ“ ๐‘ญ๐Ÿ” ๐‘ญ๐Ÿ•
6m
6m
8t
[๐’€๐‘จ = ๐Ÿ–๐’•, ๐‘ญ๐Ÿ– = ๐Ÿ๐Ÿ” ๐’•]
[๐‘ญ๐Ÿ = โˆ’๐Ÿ’. ๐Ÿ–๐Ÿ๐’•, ๐‘ญ๐Ÿ๐ŸŽ = ๐Ÿ‘. ๐Ÿ—๐Ÿ– ๐’•]
3t
4m
A
B
๐‘ญ๐Ÿ ๐‘ญ๐Ÿ
๐‘ญ๐Ÿ– ๐‘ญ๐Ÿ—
๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’ ๐‘ญ๐Ÿ“ ๐‘ญ๐Ÿ” ๐‘ญ๐Ÿ•
4m
3m
3m
๐‘ญ๐Ÿ๐ŸŽ
C
[๐‘ฟ๐‘จ = ๐Ÿ’. ๐Ÿ๐Ÿ“๐’•, ๐‘ญ๐Ÿ” = โˆ’๐Ÿ’. ๐Ÿ“ ๐’•]
A
๐‘ญ๐Ÿ
๐‘ญ๐Ÿ–
๐‘ญ๐Ÿ๐Ÿ” ๐‘ญ๐Ÿ๐Ÿ•
๐‘ญ๐Ÿ
๐‘ญ๐Ÿ“
๐‘ญ๐Ÿ—
๐‘ญ๐Ÿ”
3m 3m
8t
4A
๐‘ญ๐Ÿ๐Ÿ– ๐‘ญ๐Ÿ๐Ÿ—
B
๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’
๐‘ญ๐Ÿ•
๐‘ญ๐Ÿ๐ŸŽ ๐‘ญ๐Ÿ๐Ÿ ๐‘ญ๐Ÿ๐Ÿ ๐‘ญ๐Ÿ๐Ÿ‘ ๐‘ญ๐Ÿ๐Ÿ’ ๐‘ญ๐Ÿ๐Ÿ“
3m
3m
8t
4t 4t 4t
3m 3m
[๐‘ญ๐Ÿ = ๐Ÿ’. ๐Ÿ–๐Ÿ“๐’•, ๐‘ญ๐Ÿ” = โˆ’๐Ÿ“. ๐ŸŽ๐Ÿ“ ๐’•]
2m
A B
๐‘ญ๐Ÿ
๐‘ญ๐Ÿ
๐‘ญ๐Ÿ–
๐‘ญ๐Ÿ” ๐‘ญ๐Ÿ•
4m
10t
๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’ ๐‘ญ๐Ÿ“
1.5m
1.5m
4
Using the consistent deformation method
Find the forces in all members.
Exercise
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Support
movement)
No load Effect
Support movements
Temperature Effects
Fabrication Errors
Support movements
A
w t/m
B
No movement:
w t/m
A
B
A B
ฮด10
1t
ฮด11
ฮด10 + X ฮด11 = 0
A
w t/m
B
Movement:
A
w t/m
B
A B
ฮด10
X
ฮด11
ฮด10 + X ฮด11 = ยฑโˆ†
โˆ†
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.
ฮด10 + X ฮด11 = ๐ŸŽ. ๐ŸŽ๐Ÿ
๐‘ฌ๐‘จ
+ X
๐‘ฌ๐‘จ
= ๐ŸŽ. ๐ŸŽ๐Ÿ
EA = 10000 t
Answer
X
4t
4t 4t
4m 4m
4m 4m
6t
6t
0 ๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
๐œƒ
3m
4
4
No -10
8
8
3.33
-10.67
3.33
โˆ’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
ฮด10 =
๐›ดN
ON1L
EA
=
โˆ’128
EA
ฮด11 =
๐›ดN1N1L
=
16
X1 =
โˆ’ฮด10
ฮด11
=
โˆ’(โˆ’128)/๐ธ๐ด
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
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Temperature)
Loads
ฮด10 + X1 ฮด11 = 0
Temperature
Temperature Effects
No load Effect
Loads & Temperature
8t
ฮด1t + X1 ฮด11 = 0 ฮด1t + ฮด10 + X1 ฮด11 = 0
Loads: ฮด10 + X1 ฮด11 = 0
Temp.: ฮด1t + ๐‘ฟ๐Ÿ
โ€ฒ
ฮด11 = 0
Total: Xtot = X1 +๐‘ฟ๐Ÿ
โ€ฒ
ฮด1t = ฮฃ๐œถ. โˆ†๐’•. ๐‘ต๐Ÿ. ๐‘ณ
ฮด1t = ๐œถ. โˆ†๐’•. ฮฃ๐‘ต๐Ÿ. ๐‘ณ
OR
ฮด10 + X1 ฮด11 + X2 ฮด12 = 0
ฮด20 + X1 ฮด21 + X2 ฮด22 = 0
ฮด1t + X1 ฮด11 + X2 ฮด12 = 0
ฮด2t + X1 ฮด21 + X2 ฮด22 = 0
ฮด1t + ฮด10 +X1 ฮด11 + X2 ฮด12 = 0
ฮด2t + ฮด20 +X1 ฮด21 + X2 ฮด22 = 0
8t
10t 10t
ฮด1t = ฮฃ๐œถ. โˆ†๐’•. ๐‘ต๐Ÿ. ๐‘ณ
ฮด2t = ฮฃ๐œถ. โˆ†๐’•. ๐‘ต๐Ÿ. ๐‘ณ
10โˆ’5
Rise (+)
Drop (โˆ’)
Tension (+)
compression (โˆ’)
ONCE
TWICE
๐œƒ
-1.67
Determine the internal forces in
members for the shown truss
due to the external loads and
rise of temperature 100oC.
๐›ผ = 1๐‘ฅ10โˆ’5
/๐‘œ๐ถ
8m
B
6t
6t
6m
A
Example 1
EA = 45000 t
R = ( m + r ) - 2j
R = ( 5+ 4) - 2x4= 1
(Once indeterminate)
6๐‘ฅ8 โˆ’ ๐‘Œ๐‘๐‘ฅ6 = 0
๐‘Œ๐‘ = 8๐‘ก
ฦฉ๐‘€๐ด = 0
๐œƒ
๐‘†๐‘–๐‘›๐œƒ =
8
10
๐‘๐‘œ๐‘ ๐œƒ =
6
10
6m
8m
10m
8m
B
6t
6t
6m
A
6t
8t
2t
๐œƒ
๐œƒ
๐œƒ ๐œƒ
-8
-6
-6
10
No
8m
6m
N1
B
A
0
0
1t
1t
๐œƒ
๐œƒ
1.33
-1.67
1.33
1
๐œƒ
๐น1 ๐น2 ๐น3
๐น4
๐น5
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
ฮด10 + X1 ฮด11 = 0
ฮด1t + X1
โ€™ ฮด11 = 0
X1 = 3.91 t
ฮด1t = ๐›ผ. โˆ†๐‘ก. ๐›ด๐‘1. ๐ฟ
Due to temperature
ฮด1t = 1๐‘ฅ10โˆ’5
๐‘ฅ100๐‘ฅ โˆ’ 6 = โˆ’0.006
X1
โ€™ = 3t
-0.006 + X1
โ€™ ๐Ÿ—๐ŸŽ
๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ๐ŸŽ
= 0
Xtot = X1 +๐‘ฟ๐Ÿ
โ€ฒ
= ๐Ÿ”. ๐Ÿ—๐Ÿ๐’•
Total
Nf = N0 + Xtot N1
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
7๐‘ก
7๐‘ก
0
No
-7
-7 -6 45ยฐ
-3
-3
1๐‘ก
A
3m
3m
3m
B
N1 0
0
1๐‘ก
-1
-1 A
3m
3m
3m
B
N2
1๐‘ก
1๐‘ก
0 0
0
โˆ’1
2
โˆ’1
2
โˆ’1
2
โˆ’1
2
1๐‘ก
ฮด1t + ๐‘‹1
โ€ฒ
ฮด11 + ๐‘‹2
โ€ฒ
ฮด12 = 0
ฮด2t + ๐‘‹1
โ€ฒ
ฮด21 + ๐‘‹2
โ€ฒ
ฮด22 = 0
ฮด10 = 0
ฮด20 =
51.94
EA
ฮด11 =
6
EA
ฮด22 =
14.49
EA
ฮด12 =
2.12
EA
X1 = 1.34 ๐‘ก
X2 = โˆ’3.78 ๐‘ก
Nf = N0 + X1t N1 + X2t N2
Nf = N0 + 1.34 N1 -3.78 N2
6
45000
๐‘‹1
โ€ฒ
+
2.12
45000
๐‘‹2
โ€ฒ
= 0.003
2.12 ๐‘‹2
โ€ฒ
+ 14.49 ๐‘‹2
โ€ฒ
= 0
Determine the internal forces in
members for the shown truss due
to the external loads and rise of
temperature 50oC.
๐›ผ = 1๐‘ฅ10โˆ’5
/๐‘œ๐ถ
Example 2
EA = 45000 t
Due to loads
Due to temperature
X1t = X1 +๐‘ฟ๐Ÿ
โ€ฒ
= ๐Ÿ๐Ÿ“. ๐ŸŽ๐Ÿ’๐’•
Total
ฮด1t = ๐›ผ. โˆ†๐‘ก. ๐›ด๐‘1. ๐ฟ
ฮด1t = 1๐‘ฅ10โˆ’5๐‘ฅ50๐‘ฅ(โˆ’1๐‘ฅ3๐‘ฅ2) = โˆ’0.003
ฮด2t = 1๐‘ฅ10โˆ’5
๐‘ฅ50๐‘ฅ
โˆ’1
2
๐‘ฅ3๐‘ฅ4 + 1๐‘ฅ3 2๐‘ฅ2 = 0
X2t = X2 +๐‘ฟ๐Ÿ
โ€ฒ
= โˆ’๐Ÿ•. ๐Ÿ๐Ÿ“๐’•
๐‘‹1
โ€ฒ
= 23.7 ๐‘ก
๐‘‹2
โ€ฒ
= โˆ’3.47๐‘ก
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
Structural Analysis
Consistent Deformations
II
Eng. Khaled El-Aswany
(Trusses)
(Fabrication error)
Loads
ฮด10 + X1 ฮด11 = 0
Fabricated
Fabrication errors
No load Effect
Loads & fabrication
8t
ฮด1f + X1 ฮด11 = 0 ฮด1f + ฮด10 + X1 ฮด11 = 0
Loads: ฮด10 + X1 ฮด11 = 0
fabrication: ฮด1f + ๐‘ฟ๐Ÿ
โ€ฒ
ฮด11 = 0
Total: Xtot = X1 +๐‘ฟ๐Ÿ
โ€ฒ
ฮด1f = ฮฃโˆ†. ๐‘ต๐Ÿ
OR
ฮด10 + X1 ฮด11 + X2 ฮด12 = 0
ฮด20 + X1 ฮด21 + X2 ฮด22 = 0
ฮด1f + X1 ฮด11 + X2 ฮด12 = 0
ฮด2f + X1 ฮด21 + X2 ฮด22 = 0
ฮด1f + ฮด10 +X1 ฮด11 + X2 ฮด12 = 0
ฮด2f + ฮด20 +X1 ฮด21 + X2 ฮด22 = 0
10t
Long (+)
short (โˆ’)
Tension (+)
compression (โˆ’)
ONCE
TWICE
ฮด1f = ฮฃโˆ†. ๐‘ต๐Ÿ
ฮด2f = ฮฃโˆ†. ๐‘ต๐Ÿ
8t
10t
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)
๐น1 ๐น2 ๐น3
๐น4
๐น5
8m
B
6t
6t
6m
A
6t
8t
2t
๐œƒ
๐œƒ
๐œƒ ๐œƒ
-8
-6
-6
10
No
๐œƒ
-1.67
8m
6m
N1
B
A
0
0
1t
1t
๐œƒ
๐œƒ
1.33
-1.67
1.33
1
๐œƒ
-352 90
Due to loads
ฮด10 + X1 ฮด11 = 0
X1 = 3.91 t
Due to Fabrication
ฮด1f + X1
โ€™ ฮด11 = 0
X1
โ€™ = 9.2t
-0.018 + X1
โ€™ ๐Ÿ—๐ŸŽ
๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ๐ŸŽ
= 0
Xtot = X1 +๐‘ฟ๐Ÿ
โ€ฒ
= ๐Ÿ๐Ÿ‘. ๐Ÿ๐’•
Total
Nf = N0 + Xtot N1
Mem. L No N1 NoN1L N1N1L ฮ” ฮ”N1 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
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Introduction)
C
3 t/m
2I I
6m
A B
6t
2m 2m
Degree of Freedom
C
A B
Rotation (๐›ผ) OR (๐œƒ) Translation (โˆ†)
๐›ผ = 0
๐›ผ โ‰  0
๐›ผ = 0
โˆ†๐‘ฅ = 0
โˆ†๐‘ฆ = 0
๐›ผ = ๏๏
โˆ†๐‘ฅ = 0
โˆ†๐‘ฆ = 0
๐›ผ = ๏๏
โˆ†๐‘ฅ = ๏๏
โˆ†๐‘ฆ = 0
โˆ†= 0
Beams:
[ Except settlement cases]
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
ฮฑA +
2EI
L
ฮฑB
4EI
L
ฮฑB +
2EI
L
ฮฑA
๐‘€๐ด๐ต
๐น
โˆ’
6EI
L2
โˆ†
๐‘€๐ต๐ด
๐น
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3
โˆ†
L
)
MAB = MAB
F +
4EI
L
ฮฑA +
2EI
L
ฮฑB โˆ’
6EI
L2 โˆ†
ฮฑA
ฮฑB
โˆ’
6EI
L2
โˆ†
โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซู…ุนโ€ฌ
๐‘€๐ต๐ถ
๐น
๐‘€๐ถ๐ต
๐น
ฮฑB
ฮฑC
4EI
L
ฮฑB +
2EI
L
ฮฑC
4EI
L
ฮฑC +
2EI
L
ฮฑB
โˆ†
โˆ’
6EI
L2
โˆ† โˆ’
6EI
L2
โˆ†
โˆ†
ฮจ
๐พ
Slope Deflection Equations:
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = MBA
F +
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MBC = MBC
F +
2EI
L
(2ฮฑB+ ฮฑC -3 ฮจ )
MCB = MCB
F +
2EI
L
(2ฮฑC+ ฮฑB -3 ฮจ )
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Fixed End Moment)
w t/m
L
A B
B
Pt
L/2 L/2
A
โˆ’
๐‘ค๐ฟ2
12
๐‘ค๐ฟ2
12
๐‘ค๐ฟ2
12
๐‘ค๐ฟ2
12
๐‘ƒ๐ฟ
8
๐‘ƒ๐ฟ
8
โˆ’
๐‘ƒ๐ฟ
8
๐‘ƒ๐ฟ
8
โˆ’
Pa(Lโˆ’a)
L
Pa(Lโˆ’a)
L
Pa(Lโˆ’a)
L
Pa(Lโˆ’a)
L
L
A B
Pt Pt
a
a b
โˆ’
2
9
PL
2
9
PL
2
9
PL 2
9
PL
L
A B
Pt Pt
a a a
โˆ’
Pa๐‘2
๐ฟ2
Pb๐‘Ž2
๐ฟ2
Pa๐‘2
๐ฟ2
P๐‘๐‘Ž2
๐ฟ2
L
A B
Pt
a b
โˆ’
๐‘ค๐‘™2
20
๐‘ค๐‘™2
30
๐‘ค๐‘™2
30
L
A B
w t/m
w๐‘™2
20
Mb(2aโˆ’b)
L2
Ma(2bโˆ’a)
L2
L
A B
M
a b
โˆ’
Mb(2aโˆ’b)
L2
โˆ’
Ma(2bโˆ’a)
L2
L
A B
M
a b
6x4(2x1โˆ’4)
52
6x1(2x4โˆ’1)
52
5m
A B
6mt
1m 4m
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 1)
C
3 t/m
2I I
6m
A B
6t
2m 2m
3 t/m
6m
A B C
6t
2m 2m
B
๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = 0
โˆ’9 9 โˆ’3 3
๐‘€๐ด๐ต
๐น
= -9 mt, ๐‘€๐ต๐ด
๐น
= 9 mt
๐‘€๐ต๐ถ
๐น
= -3 mt, ๐‘€๐ถ๐ต
๐น
= 3 mt
1) Unknowns
๐›ผ๐ต
2) Fixed End Moment
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ
2๐‘ฅ2๐ธ๐ผ
6
:
2๐‘ฅ๐ธ๐ผ
4
2 โˆถ 1.5
2๐ธ๐ผ
๐ฟ ๐ด๐ต
:
2๐ธ๐ผ
๐ฟ ๐ต๐ถ
2
6
โˆถ
1
4
ร— 6
ร— 2
4 โˆถ 3
4) Slope deflection equations
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = MBA
F +
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MBC = MBC
F +
2EI
L
(2ฮฑB+ ฮฑC -3 ฮจ )
MCB = MCB
F +
2EI
L
(2ฮฑC+ ฮฑB -3 ฮจ )
= โˆ’9 + 4 ( 0 + ฮฑB - 0 )
= 9 + 4 ( 2ฮฑB + 0 - 0 )
= โˆ’3 + 3 ( 2ฮฑB + 0 - 0 )
= 3 + 3 ( 0 + ฮฑB - 0 )
= โˆ’9 + 4ฮฑB
= 9 + 8ฮฑB
= โˆ’3 + 6ฮฑB
= 3 + 3ฮฑB
5) Compatibility eq.
ฮฃMb = 0
๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0
9 + 8ฮฑB +โˆ’3 + 6ฮฑB = 0
ฮฑB = โˆ’0.4286
๐‘€๐ด๐ต = -10.71
๐‘€๐ต๐ด = 5.57
๐‘€๐ต๐ถ = -5.57
๐‘€๐ถ๐ต = 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. ๐‘ค๐ฟ2
8
๐‘ƒ๐ฟ
4
=6
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 2)
3 t/m
6m
A B C
6t
2m 2m
B
๐›ผ๐ด = 0 ๐›ผ๐ต = ??
โˆ’9 9 โˆ’3 3
๐‘€๐ด๐ต
๐น
= -9 mt, ๐‘€๐ต๐ด
๐น
= 9 mt
๐‘€๐ต๐ถ
๐น
= -3 mt, ๐‘€๐ถ๐ต
๐น
= 3 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ
2๐‘ฅ2๐ธ๐ผ
6
:
2๐‘ฅ๐ธ๐ผ
4
2 โˆถ 1.5
2๐ธ๐ผ
๐ฟ ๐ด๐ต
:
2๐ธ๐ผ
๐ฟ ๐ต๐ถ
2
6
โˆถ
1
4
ร— 6
ร— 2
4 โˆถ 3
4) Slope deflection equations
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = MBA
F +
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MBC = MBC
F +
2EI
L
(2ฮฑB+ ฮฑC -3 ฮจ )
MCB = MCB
F +
2EI
L
(2ฮฑC+ ฮฑB -3 ฮจ )
= โˆ’9 + 4 ( 0 + ฮฑB - 0 )
= 9 + 4 ( 2ฮฑB + 0 - 0 )
= โˆ’3 + 3 ( 2ฮฑB + 0 - 0 )
= 3 + 3 ( 0 + ฮฑB - 0 )
= โˆ’9 + 4ฮฑB
= 9 + 8ฮฑB
= โˆ’3 + 6ฮฑB + 3ฮฑC
= 3 + 3ฮฑB + 6ฮฑC
5) Compatibility eq.
ฮฃMb = 0
๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0
9 + 8ฮฑB โˆ’3 + 6ฮฑB+3ฮฑC = 0
ฮฑB = โˆ’0.36
ฮฑC = โˆ’0.32
๐‘€๐ด๐ต = -10.44
๐‘€๐ต๐ด = 6.12
๐‘€๐ต๐ถ = -6.12
๐‘€๐ถ๐ต = 0
6) Moment values
3 t/m
6m
A B
10.44 6.12 6.12
Reactions
F.E.M.
๐›ผ๐ถ = ??
C
3 t/m
2I I
6m
A B
6t
2m 2m
๐›ผ๐ต , ๐›ผ๐ถ
๐‘€๐ถ๐ต = 0
3 + 3ฮฑB+6ฮฑC = 0
3ฮฑB+6ฮฑC = -3 โ†’ 2
14ฮฑB +3ฮฑC = -6 โ†’ 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. ๐‘ƒ๐ฟ
4
=6
๐‘ค๐ฟ2
8
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Modified)
3 t/m
6m
A B
Fixed End Moment
Rotation Moment
Sway Moment
L
A B
L
A B C
L
B
4EI
L
ฮฑA +
2EI
L
ฮฑB
4EI
L
ฮฑB +
2EI
L
ฮฑA
๐‘€๐ด๐ต
๐น
โˆ’
6EI
L2
โˆ†
๐‘€๐ต๐ด
๐น
MBC =๐‘€๐ต๐ถ
๐นโ€ฒ
+
3EI
L
(ฮฑB โˆ’
โˆ†
L
)
MBC = ๐‘€๐ต๐ถ
๐นโ€ฒ
+
3EI
L
ฮฑB โˆ’
3EI
L2 โˆ†
ฮฑA
ฮฑB
โˆ’
6EI
L2
โˆ†
โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซู…ุนโ€ฌ
๐‘€๐ต๐ถ
๐นโ€ฒ
3EI
L
ฮฑB
โˆ†
โˆ’
3EI
L2
โˆ†
โˆ†
ฮจ
๐พ
Slope Deflection Equations:
MAB =๐‘€๐ด๐ต
๐น
+
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = ๐‘€๐ต๐ด
๐น
+
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MBC = ๐‘€๐ต๐ถ
๐นโ€ฒ
+
3EI
L
(ฮฑB - ฮจ )
C
3 t/m
2I I
6m
A B
6t
2m 2m
C
6t
2m 2m
B
ฮฑB
C
L
B
๐‘€๐ด = ?? ๐‘€๐ถ = 0
๐‘€๐ต = ??
w t/m
L
A B
โˆ’
๐‘ค๐ฟ2
12
๐‘ค๐ฟ2
12
L
A B
๐‘€๐ด๐ต
๐น
๐‘€๐ต๐ด
๐น
๐‘€๐ด๐ต
๐นโ€ฒ
L
A B
๐‘€๐ด๐ต
๐น ๐‘€๐ต๐ด
๐น
= โˆ’ 0.5
๐‘€๐ด๐ต
๐นโ€ฒ
L
A B
w t/m
= โˆ’1.5
wL2
12
B
Pt
L/2 L/2
A
โˆ’
๐‘ƒ๐ฟ
8
๐‘ƒ๐ฟ
8
๐‘€๐ด๐ต
๐นโ€ฒ
โˆ’
๐‘ƒ๐ฟ
8
= ๐‘ฅ 1.5
B
Pt
L/2 L/2
A
โˆ’
Pa๐‘2
๐ฟ2
Pb๐‘Ž2
๐ฟ2
L
A B
Pt
a b
L
A B
Pt
a b
๐‘€๐ด๐ต
๐นโ€ฒ
โˆ’
Pa๐‘2
๐ฟ2
Pb๐‘Ž2
๐ฟ2
= โˆ’ 0.5
๐‘€๐ด๐ต
๐นโ€ฒ
=
L
A B
4EI
L
ฮฑA +
2EI
L
ฮฑB
4EI
L
ฮฑB +
2EI
L
ฮฑA
ฮฑA
ฮฑB
3EI
L
ฮฑA
๐‘€๐ด๐ต
๐นโ€ฒ
= โˆ’ 0.5
L
A B
โˆ’
6EI
L2
โˆ† (โˆ’
6EI
L2 โˆ†)
โˆ’
6EI
L2
โˆ† โˆ’
6EI
L2
โˆ†
โˆ†
L
A B
L
A B
w t/m
L
A B
โˆ’
๐‘ค๐ฟ2
12
๐‘ค๐ฟ2
12
L
A B
๐‘€๐ด๐ต
๐น
๐‘€๐ต๐ด
๐น
๐‘€๐ด๐ต
๐นโ€ฒ
L
A B
๐‘€๐ด๐ต
๐น ๐‘€๐ต๐ด
๐น
= โˆ’ 0.5
๐‘€๐ด๐ต
๐นโ€ฒ
L
A B
w t/m
= โˆ’1.5
wL2
12
B
Pt
L/2 L/2
A
โˆ’
๐‘ƒ๐ฟ
8
๐‘ƒ๐ฟ
8
๐‘€๐ด๐ต
๐นโ€ฒ
โˆ’
๐‘ƒ๐ฟ
8
= ๐‘ฅ 1.5
B
Pt
L/2 L/2
A
โˆ’
Pa๐‘2
๐ฟ2
Pb๐‘Ž2
๐ฟ2
L
A B
Pt
a b
L
A B
Pt
a b
๐‘€๐ด๐ต
๐นโ€ฒ
โˆ’
Pa๐‘2
๐ฟ2
Pb๐‘Ž2
๐ฟ2
= โˆ’ 0.5
6m
A B
2 t/m 6t
2m
12
โˆ’6 6 -12
6m
A B
2 t/m 6t
2m
๐‘€๐ด๐ต
๐นโ€ฒ
= โˆ’6 6 -12
โˆ’ 0.5 ( + ) = -3
B
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Modified)
(Redo Example 2)
๐‘€๐ด๐ต
๐น
= -9 mt, ๐‘€๐ต๐ด
๐น
= 9 mt
๐‘€๐ต๐ถ
๐น
= -4.5 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ
4 โˆถ
9
2
2๐ธ๐ผ
๐ฟ ๐ด๐ต
:
3๐ธ๐ผ
๐ฟ ๐ต๐ถ
2๐‘ฅ2
6
โˆถ
3๐‘ฅ1
4
๐›ผ๐ต
3 t/m
6m
A B C
6t
2m 2m
B
๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = ??
โˆ’9 9 โˆ’4.5
F.E.M.
C
3 t/m
2I I
6m
A B
6t
2m 2m
๐‘€๐ถ = 0
๐‘€๐ด = ?? ๐‘€๐ต = ??
:
8 โˆถ 9
4) Slope deflection equations
= โˆ’9 + 8 ( 0 + ฮฑB - 0 )
= 9 + 8 ( 2ฮฑB + 0 - 0 )
= โˆ’4.5 + 9 ( ฮฑB - 0)
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = MBA
F +
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MBC = ๐‘€๐ต๐ถ
๐นโ€ฒ
+
3EI
L
(ฮฑB - ฮจ )
= โˆ’9 + 8ฮฑB
= 9 + 16ฮฑB
= โˆ’4.5 + 9ฮฑB
5) Compatibility eq.
ฮฃMb = 0
๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0
9 + 16ฮฑB โˆ’4.5 + 9ฮฑB = 0
ฮฑB = โˆ’0.18
๐‘€๐ด๐ต = -10.44
๐‘€๐ต๐ด = 6.12
๐‘€๐ต๐ถ = -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. ๐‘ƒ๐ฟ
4
=6
๐‘ค๐ฟ2
8
๐‘€๐ด๐ต
๐น
= -9 mt, ๐‘€๐ต๐ด
๐น
= 9 mt
๐‘€๐ต๐ถ
๐น
= -4.5 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ
4 โˆถ
9
2
2๐ธ๐ผ
๐ฟ ๐ด๐ต
:
3๐ธ๐ผ
๐ฟ ๐ต๐ถ
2๐‘ฅ2
6
โˆถ
3๐‘ฅ1
4
๐›ผ๐ต
3 t/m
6m
A B C
6t
2m 2m
B
๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = ??
โˆ’9 9 โˆ’4.5
F.E.M.
C
3 t/m
2I I
6m
A B
6t
2m 2m
๐‘€๐ถ = 0
๐‘€๐ด = ?? ๐‘€๐ต = ??
:
8 โˆถ 9
4) Slope deflection equations
= โˆ’9 + 8 ( 0 + ฮฑB - 0 )
= 9 + 8 ( 2ฮฑB + 0 - 0 )
= โˆ’4.5 + 9 ( ฮฑB - 0)
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = MBA
F +
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MBC = ๐‘€๐ต๐ถ
๐นโ€ฒ
+
3EI
L
(ฮฑB - ฮจ )
= โˆ’9 + 8ฮฑB
= 9 + 16ฮฑB
= โˆ’4.5 + 9ฮฑB
5) Compatibility eq.
ฮฃMb = 0
๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0
9 + 16ฮฑB โˆ’4.5 + 9ฮฑB = 0
ฮฑB = โˆ’0.18
๐‘€๐ด๐ต = -10.44
๐‘€๐ต๐ด = 6.12
๐‘€๐ต๐ถ = -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. ๐‘ƒ๐ฟ
4
=6
๐‘ค๐ฟ2
8
Type
Equation
D.O.F ฮฑA, ฮฑB , ฮจ ฮฑA, ฮจ
Stiffness 2EI
L
3EI
L
F.E.M.
๐‘€๐ด๐ต
๐นโ€ฒ
= ๐‘€๐ด๐ต
๐น
-
1
2
๐‘€๐ต๐ด
๐น
Summary
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = MBA
F +
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MAB = ๐‘€๐ด๐ต
๐นโ€ฒ
+
3EI
L
(ฮฑA - ฮจ )
A B B
A
B
A
๐‘€๐ด๐ต
๐นโ€ฒ
A B
๐‘€๐ด๐ต
๐น
๐‘€๐ต๐ด
๐น
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 3)
๐‘€๐ด๐ต
๐น
= -9 mt, ๐‘€๐ต๐ด
๐น
= 9 mt
๐‘€๐ต๐ถ
๐น
= -3 mt, ๐‘€๐ถ๐ต
๐น
= 3 mt
1) Unknowns
2) Fixed End Moment
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ
2๐‘ฅ2๐ธ๐ผ
6
:
2๐‘ฅ๐ธ๐ผ
4
2 โˆถ 1.5
2๐ธ๐ผ
๐ฟ ๐ด๐ต
:
2๐ธ๐ผ
๐ฟ ๐ต๐ถ
2
6
โˆถ
1
4
ร— 6
ร— 2
4 โˆถ 3
4) Slope deflection equations
MAB = MAB
F +
2EI
L
(2ฮฑA+ ฮฑB -3 ฮจ )
MBA = MBA
F +
2EI
L
(2ฮฑB+ ฮฑA -3 ฮจ )
MBC = MBC
F +
2EI
L
(2ฮฑB+ ฮฑC -3 ฮจ )
MCB = MCB
F +
2EI
L
(2ฮฑC+ ฮฑB -3 ฮจ )
= โˆ’9 + 4 ( 0 + ฮฑB - 0 )
= 9 + 4 ( 2ฮฑB + 0 - 0 )
= โˆ’3 + 3 ( 2ฮฑB + 0 - 0 )
= 3 + 3 ( 0 + ฮฑB - 0 )
= โˆ’9 + 4ฮฑB
= 9 + 8ฮฑB
= โˆ’3 + 6ฮฑB + 3ฮฑC
= 3 + 3ฮฑB + 6ฮฑC
5) Compatibility eq.
ฮฃMb = 0
๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0
9 + 8ฮฑB โˆ’3 + 6ฮฑB+3ฮฑC = 0
ฮฑB = โˆ’0.84
ฮฑC = 1.92
๐‘€๐ด๐ต = -12.36
๐‘€๐ต๐ด = 2.28
๐‘€๐ต๐ถ = -2.28
๐‘€๐ถ๐ต = 12
6) Moment values
12.36 2.28 2.28
Reactions
๐›ผ๐ต , ๐›ผ๐ถ
๐‘€๐ถ๐ต + ๐‘€๐ถ๐ท = 0
3 + 3ฮฑB+6ฮฑC โˆ’12 = 0
3ฮฑB+6ฮฑC = 9 โ†’ 2
14ฮฑB +3ฮฑC = -6 โ†’ 1
๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = ??
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
โˆ’9 9 โˆ’3 3
12
โˆ’12
F.E.M.
ฮฃMc = 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. ๐‘ค๐ฟ2
8
6
12
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Example 4)
A
9t
2m
4m
2m
6t
B
1) Unknowns
๐›ผ๐ต , ๐›ผ๐ถ
2) Fixed End Moment ๐›ผ๐ด = ?? ๐›ผ๐ต = ?? ๐›ผ๐ถ = ??
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
๐›ผ๐‘‘ = ??
๐‘€๐ด = 12 ๐‘€๐ต = ?? ๐‘€๐ถ = ?? ๐‘€๐‘‘ = 0
โˆ’24
4 6t
6t
C
B
2m
2m
2m
3 t/m
17
โˆ’17
๐‘€๐ต๐ด
๐นโ€ฒ
= 4 mt
๐‘€๐ต๐ถ
๐น
= -17 mt, ๐‘€๐ถ๐ต
๐น
= 17 mt
๐‘€๐ถ๐ท
๐นโ€ฒ
= -24 mt
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ถ๐ท
3๐ธ๐ผ
๐ฟ ๐ด๐ต
:
2๐ธ๐ผ
๐ฟ ๐ต๐ถ
:
3๐ธ๐ผ
๐ฟ ๐ถ๐ท
ร— 4
:
:
3๐‘ฅ1
6
2๐‘ฅ3
6
3๐‘ฅ2
8
:
:
1
2
1
3
4
:
:
2 4 3
:
:
4) Slope deflection equations
= 4 + 2 ( ฮฑB - 0 )
= โˆ’17 + 4 ( 2ฮฑB + ฮฑC - 0) )
= 17 + 4 ( 2ฮฑC + ฮฑB - 0)
= โˆ’24 + 3 (ฮฑC - 0 )
MBC = ๐‘€๐ต๐ถ
๐น
+
2EI
L
(2ฮฑB+ ฮฑC -3 ฮจ )
MCB = ๐‘€๐ถ๐ต
๐น
+
2EI
L
(2ฮฑC+ ฮฑB -3 ฮจ )
MCD = ๐‘€๐ถ๐ท
๐นโ€ฒ
+
3EI
L
(ฮฑC - ฮจ)
MBA = ๐‘€๐ต๐ด
๐นโ€ฒ
+
3EI
L
(ฮฑB - ฮจ)
= 4 + 2ฮฑB
= โˆ’17 + 8ฮฑB + 4ฮฑC
= 17 + 4ฮฑB + 8ฮฑC
= โˆ’24 + 3ฮฑC
5) Compatibility eq.
ฮฃMb = 0
๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0
ฮฑB = 1.2234
ฮฑC = 0.1915
๐‘€๐ถ๐ต + ๐‘€๐ถ๐ท = 0
4ฮฑB+11ฮฑC = 7 โ†’ 2
10ฮฑB +4ฮฑC = 13 โ†’ 1
ฮฃMc = 0
6) Moment Values
MBC = โˆ’6.45 ๐‘š๐‘ก
MCB = 23.43 ๐‘š๐‘ก
MCD = โˆ’23.43 ๐‘š๐‘ก
MBA = 6.45 ๐‘š๐‘ก
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.
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(Sway or without sway)
Delete
โœ“
Can translate ?
X
Symmetry ?
ฮจ
ฮจ
Frame
Without Sway
Symmetry No translation
With Sway
ฮ” ฮ”
Without Sway With Sway
Delete
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(Without Sway)
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
๐›ผ๐ต , ๐›ผ๐ถ
12m 8m
Delete
1) Unknowns
๐›ผ๐ต , ๐›ผ๐ถ
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
๐‘€๐ด๐ต
๐น
= -18 mt, ๐‘€๐ต๐ด
๐น
= 18 mt
๐‘€๐ต๐ถ
๐น
= -8 mt, ๐‘€๐ถ๐ต
๐น
= 8 mt
๐‘€๐ต๐ท
๐น
= -5.33 mt, ๐‘€๐ท๐ต
๐น
= 2.67 mt
เดฅ
๐‘€๐ถ๐ธ
๐น
= -4.5 mt
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
๐›ผ๐ต , ๐›ผ๐ถ
2) Fixed End Moment
๐‘€๐ด๐ต
๐น
= -18 mt, ๐‘€๐ต๐ด
๐น
= 18 mt
๐‘€๐ต๐ถ
๐น
= -8 mt, ๐‘€๐ถ๐ต
๐น
= 8 mt
๐‘€๐ต๐ท
๐น
= -5.33 mt, ๐‘€๐ท๐ต
๐น
= 2.67 mt
เดฅ
๐‘€๐ถ๐ธ
๐น
= -4.5 mt
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ท โˆถ ๐พ๐ถ๐ธ
2๐‘‹6
12
:
2๐‘‹4
8
:
2๐‘‹1.5
6
:
3๐‘‹1
6
1 โˆถ 1 โˆถ
1
2
:
1
2
2 โˆถ 2 โˆถ 1 : 1
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
๐›ผ๐ต , ๐›ผ๐ถ
2) Fixed End Moment
๐‘€๐ด๐ต
๐น
= -18 mt, ๐‘€๐ต๐ด
๐น
= 18 mt
๐‘€๐ต๐ถ
๐น
= -8 mt, ๐‘€๐ถ๐ต
๐น
= 8 mt
๐‘€๐ต๐ท
๐น
= -5.33 mt, ๐‘€๐ท๐ต
๐น
= 2.67 mt
เดฅ
๐‘€๐ถ๐ธ
๐น
= -4.5 mt
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ท โˆถ ๐พ๐ถ๐ธ
2๐‘‹6
12
:
2๐‘‹4
8
:
2๐‘‹1.5
6
:
3๐‘‹1
6
1 โˆถ 1 โˆถ
1
2
:
1
2
2 โˆถ 2 โˆถ 1 : 1
4) Slope Deflection Equation
๐‘€๐ด๐ต = -18 + 2 (2๐›ผ๐ด + ๐›ผ๐ต)
๐‘€๐ต๐ด = 18 + 2 (2๐›ผ๐ต + ๐›ผ๐ด)
๐‘€๐ต๐ถ = -8 + 2 (2๐›ผ๐ต + ๐›ผ๐ถ)
๐‘€๐ถ๐ต = 8 + 2 (2๐›ผ๐ถ + ๐›ผ๐ต)
๐‘€๐ต๐ท = -5.33 + (2๐›ผ๐ต + ๐›ผ๐ท)
๐‘€๐ท๐ต = 2.67 + (2๐›ผ๐ท + ๐›ผ๐ต)
๐‘€๐ถ๐ธ = -4.5 + (๐›ผ๐ถ)
= -18 + 2 ๐›ผ๐ต
= 18 + 4๐›ผ๐ต
= -8 + 4๐›ผ๐ต + 2๐›ผ๐ถ
= 8 + 4๐›ผ๐ถ + 2๐›ผ๐ต
= -5.33 + 2๐›ผ๐ต
= 2.67 + ๐›ผ๐ต
= -4.5 + ๐›ผ๐ถ
Delete
1) Unknowns
๐›ผ๐ต , ๐›ผ๐ถ
2) Fixed End Moment
๐‘€๐ด๐ต
๐น
= -18 mt, ๐‘€๐ต๐ด
๐น
= 18 mt
๐‘€๐ต๐ถ
๐น
= -8 mt, ๐‘€๐ถ๐ต
๐น
= 8 mt
๐‘€๐ต๐ท
๐น
= -5.33 mt, ๐‘€๐ท๐ต
๐น
= 2.67 mt
เดฅ
๐‘€๐ถ๐ธ
๐น
= -4.5 mt
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ท โˆถ ๐พ๐ถ๐ธ
2๐‘‹6
12
:
2๐‘‹4
8
:
2๐‘‹1.5
6
:
3๐‘‹1
6
1 โˆถ 1 โˆถ
1
2
:
1
2
2 โˆถ 2 โˆถ 1 : 1
4) Slope Deflection Equation
๐‘€๐ด๐ต = -18 + 2 (2๐›ผ๐ด + ๐›ผ๐ต)
๐‘€๐ต๐ด = 18 + 2 (2๐›ผ๐ต + ๐›ผ๐ด)
๐‘€๐ต๐ถ = -8 + 2 (2๐›ผ๐ต + ๐›ผ๐ถ)
๐‘€๐ถ๐ต = 8 + 2 (2๐›ผ๐ถ + ๐›ผ๐ต)
๐‘€๐ต๐ท = -5.33 + (2๐›ผ๐ต + ๐›ผ๐ท)
๐‘€๐ท๐ต = 2.67 + (2๐›ผ๐ท + ๐›ผ๐ต)
๐‘€๐ถ๐ธ = -4.5 + (๐›ผ๐ถ)
= -18 + 2 ๐›ผ๐ต
= 18 + 4๐›ผ๐ต
= -8 + 4๐›ผ๐ต + 2๐›ผ๐ถ
= 8 + 4๐›ผ๐ถ + 2๐›ผ๐ต
= -5.33 + 2๐›ผ๐ต
= 2.67 + ๐›ผ๐ต
= -4.5 + ๐›ผ๐ถ
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
ฮฃMb = 0
๐‘€๐ต๐ด + ๐‘€๐ต๐ถ + ๐‘€๐ต๐ท = 0
ฮฃMc = 0
๐‘€๐ถ๐ต + ๐‘€๐ถ๐ธ + ๐‘€๐ถ๐น = 0
10๐›ผ๐ต + 2๐›ผ๐ถ = -4.67
2๐›ผ๐ต + 5๐›ผ๐ถ = 8.5
๐›ผ๐ต = -0.87717
๐›ผ๐ถ = 2.05087
๐‘€๐ด๐ต = -19.75
๐‘€๐ต๐ด = 14.49
๐‘€๐ต๐ถ = -7.4
๐‘€๐ถ๐ต = 14.45
๐‘€๐ต๐ท = -7.09
๐‘€๐ท๐ต = 1.80
๐‘€๐ถ๐ธ = -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
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(Without Sway)
(Symmetry)
Delete 1) Unknowns
๐›ผ๐ต , ๐›ผ๐ถ , ๐›ผ๐ท , ๐›ผ๐ธ
2) Fixed End Moment
๐›ผ๐ท = - ๐›ผ๐ถ ๐›ผ๐ธ = - ๐›ผ๐ต
3) Relative Stiffness
๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ธ โˆถ ๐พ๐ถ๐ท
6 โˆถ 2 โˆถ 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
๐‘€๐ต๐ด = 0 + 6 (๐›ผ๐ต)
๐‘€๐ต๐ถ = 0 + 2 (2๐›ผ๐ต + ๐›ผ๐ถ)
๐‘€๐ถ๐ต = 0 + 2 (2๐›ผ๐ถ + ๐›ผ๐ต)
๐‘€๐ต๐ธ = -16 + 2.5(2๐›ผ๐ต + ๐›ผ๐ธ)
๐‘€๐ถ๐ท = -32 + 3.75(2๐›ผ๐‘ + ๐›ผ๐ท)
= 6๐›ผ๐ต
= 4๐›ผ๐ต + 2๐›ผ๐ถ
= 4๐›ผ๐ถ + 2๐›ผ๐ต
= -16 + 2.5(2๐›ผ๐ต - ๐›ผ๐ต)
= -32 + 3.75(2๐›ผ๐‘ - ๐›ผ๐‘)
= -16 + 2.5๐›ผ๐ต
= -32 + 3.75๐›ผ๐‘
5) Compatibility Equations
ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ + ๐‘€๐ต๐ธ = 0
ฮฃMc = 0 ๐‘€๐ถ๐ต + ๐‘€๐ถ๐ท = 0
12.5๐›ผ๐ต + 2๐›ผ๐ถ = 16
2๐›ผ๐ต + 7.75๐›ผ๐ถ = 32
๐›ผ๐ต = 0.664603
๐›ผ๐ถ = 3.962315
6) Moment values
๐‘€๐ต๐ด = 3.88
๐‘€๐ต๐ถ = 10.51
๐‘€๐ถ๐ต = 17.14
๐‘€๐ต๐ธ = -14.38
๐‘€๐ถ๐ท = -17.14
C
3 t/m
A
B
5m
8m
2I
2I
F
E
D
5m
๐œถ๐‘ฉ =??
๐œถ๐‘ช =??
6 t/m
2I
I I
3I
๐œถ๐‘ฌ =??
๐œถ๐‘ซ =??
C
A
B
F
E
D
Structural Analysis
Slope deflection Method
II
Eng. Khaled El-Aswany
(Frames)
(With Sway)
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Introduction)
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.
Fixed end moment
Distribution Factor Carry over
w t/m
L
๐‘ค๐ฟ2
12
โˆ’
๐‘ค๐ฟ2
12
๐‘ƒ๐ฟ
8
โˆ’
๐‘ƒ๐ฟ
8
โˆ’
๐‘ƒ๐‘Ž๐‘2
๐ฟ2
๐‘ƒ๐‘๐‘Ž2
๐ฟ2
L/2
Pt
L/2
๐‘ค๐ฟ2
30
โˆ’
๐‘ค๐ฟ2
20
๐‘ƒ๐‘Ž(๐ฟ โˆ’ ๐‘Ž)
๐ฟ
โˆ’
๐‘ƒ๐‘Ž(๐ฟ โˆ’ ๐‘Ž)
๐ฟ
L
๐‘€๐‘Ž(2๐‘ โˆ’ ๐‘Ž)
๐ฟ2
๐‘€๐‘(2๐‘Ž โˆ’ ๐‘)
๐ฟ2
w t/m
L
โˆ’
๐‘ค๐ฟ2
12
๐’™๐Ÿ. ๐Ÿ“
โˆ’
๐‘ƒ๐ฟ
8
๐’™๐Ÿ. ๐Ÿ“
โˆ’
๐‘ƒ๐‘Ž๐‘2
๐ฟ2
L/2
Pt
L/2
a
Pt
b
L
L
Pt Pt
a a
b
M
a b
L
Pt
a b
L
โˆ’ ๐ŸŽ. ๐Ÿ“ ๐ฑ
๐‘ท๐’ƒ๐’‚๐Ÿ
๐‘ณ๐Ÿ
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
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
4๐‘ฅ๐ธ๐‘ฅ3
6
:
4๐‘ฅ๐ธ๐‘ฅ2
4
4๐‘ฅ๐ธ๐‘ฅ2
4
:
4๐‘ฅ๐ธ๐‘ฅ1
6
1
2
:
1
2
1
2
:
1
6
6 : 2
๐Ÿ‘
๐Ÿ’
:
๐Ÿ
๐Ÿ’
3 : 1
1
2
1
2
+
1
2
:
1
2
1
2
+
1
2
๐Ÿ
๐Ÿ
:
๐Ÿ
๐Ÿ
L
L
K =
๐‘ฐ
๐‘ณ
K =
๐‘ฐ
๐‘ณ
๐’™
๐Ÿ‘
๐Ÿ’
C
3 t/m
6m
A B
6t
2m 2m
3 t/m
6m
D
3I 2I 3I
K =
๐‘ฐ
๐‘ณ
๐’™
๐Ÿ
๐Ÿ
L
๐‘ฒ๐‘จ๐‘ฉ=
๐‘ฐ
๐‘ณ
๐‘ฒ๐‘ฉ๐‘ช=
๐‘ฐ
๐‘ณ
๐’™
๐Ÿ
๐Ÿ
Example for Symmetry Beam:
(Symmetry)
Fixed end moment Distribution Factor
Carry over
L
A B
L
B C
D.M. D.M.
๐Ÿ
๐Ÿ
D.M. ๐ŸŽ
x
๐Ÿ
๐Ÿ
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 1)
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
โˆ’
๐‘ค๐ฟ2
12
=
๐‘ค๐ฟ2
12
=
PL
8
=
-
PL
8
=
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
๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“
D.M.
-6
-2.4
D.F.
FEM
1
-
โ€ซู†ุฌู…ุนโ€ฌ
2
-
โ€ซุงุงู„ุดุงุฑุฉโ€ฌ โ€ซู†ุบูŠุฑโ€ฌ
3
-
โ€ซููŠโ€ฌ โ€ซู†ุถุฑุจโ€ฌ
โ€ซุงู„ุชูˆุฒูŠุนโ€ฌ โ€ซู†ุณุจโ€ฌ
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
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
๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“
D.M.
-6
-2.4 -3.6
D.F.
1
-
โ€ซู†ุฌู…ุนโ€ฌ
2
-
โ€ซุงุงู„ุดุงุฑุฉโ€ฌ โ€ซู†ุบูŠุฑโ€ฌ
3
-
โ€ซููŠโ€ฌ โ€ซู†ุถุฑุจโ€ฌ
โ€ซุงู„ุชูˆุฒูŠุนโ€ฌ โ€ซู†ุณุจโ€ฌ
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
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
๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“
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
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
๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“
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
โ€ซุณุงู„ุจโ€ฌ
(
โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซุนูƒุณโ€ฌ
)
โ€ซู…ูˆุฌุจโ€ฌ
(
โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซู…ุนโ€ฌ
)
Joint B
KBA : KBC
1
6
:
1
4
4 : 6
2 : 3
2
5
:
3
5
I
L
:
I
L
Distribution factors:
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
๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“
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.
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 2)
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
๐Ÿ–/๐Ÿ๐Ÿ• 9/๐Ÿ๐Ÿ•
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
๐‘ฅ
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
๐‘ฅ
3
4
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 3)
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
โˆ’4 4
1t
2m
C
2 t/m
B
6m
6
โˆ’6 โˆ’2
โˆ’0.5 ( )= -8
FEM
โˆ’6 โˆ’2
6
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
โˆ’4 4
C
2 t/m
B
6m 2m
C
B
1t
6m 2m
โˆ’9
2
1
1
โˆ’8
FEM
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
โˆ’4 4 โˆ’8
4 -8
-4
FEM
๐ŸŽ. ๐Ÿ“ 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 ๐‘ฅ
3
4
I
L
:
I
L
๐‘ฅ
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
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 4)
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
โˆ’25 โˆ’15 15 0 0
Joint B
KBA : KBC
Joint C
KCB : KCD
๐ผ
๐ฟ
:
๐ผ
๐ฟ
1
10
:
2
15
15 : 20
๐Ÿ‘
๐Ÿ•
:
๐Ÿ’
๐Ÿ• 25 โˆ’15 15 0 0
โˆ’25
15
35
:
20
35
๐ผ
๐ฟ
:
๐ผ
๐ฟ
2
15
:
1
10
20 : 15
๐Ÿ’
๐Ÿ•
:
๐Ÿ‘
๐Ÿ•
20
35
:
15
35
3/7 4/7 4/7 3/7
โˆ’5.714 โˆ’8.55 โˆ’6.45
A B C D
FEM
D.M.1
D.F.
C.O.1
โˆ’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 โˆ’0.465
โˆ’0.349 โˆ’0.525
โˆ’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 โˆ’0.038
โˆ’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
๐‘ค๐‘™2
8
26.33 + 22.32
2
13.175
=37.5
=24.32
๐‘ƒ๐ฟ
4
22.32 + 5.69
2
16
=30
=14
Distribution factors:
Reactions
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 5 - Symmetry)
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
โˆ’3 0
FEM
โˆ’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
๐‘ฅ
1
2
I
L
:
I
L
๐‘ฅ
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.
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Example 6)
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
๐‘ƒ๐ฟ
8
๐‘ฅ1.5 = 9
3m
3m
A
8t
2m
B
6t
6t
C
B
2m
2m
2m
โˆ’6
3m
3m
A
2m
6t
B
12
6
6t
6t
C
B
โˆ’17 17
2m
2m
2m
3 t/m
C
B
2m
2m
2m
3 t/m
๐‘ค๐ฟ2
12
= 9
โˆ’
๐‘ค๐ฟ2
12
= -9
โˆ’
๐‘ƒ๐‘Ž(๐ฟโˆ’๐‘Ž)
๐ฟ
= -8
๐‘ƒ๐‘Ž(๐ฟโˆ’๐‘Ž)
๐ฟ
= 8
3 t/m
6t
6t
C
3 t/m
B
8m
D
3I 2I
C
B 3I D
2I
C
8m
3 โˆ’17 17 -16 16
Joint B Joint C
KCB : KCD
๐ผ
๐ฟ
๐‘ฅ
3
4
:
๐ผ
๐ฟ
3 -17 17 -16 16
๐ผ
๐ฟ
:
๐ผ
๐ฟ
1
2
:
1
4
2
๐ŸŽ. ๐Ÿ ๐ŸŽ. ๐Ÿ– 2/๐Ÿ‘ 1/๐Ÿ‘
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
๐Ÿ
๐Ÿ‘
:
๐Ÿ
๐Ÿ‘
KBA : KBC
1
6
๐‘ฅ
3
4
:
3
6
1
8
:
1
2
8
:
2
4
:
1
๐Ÿ’
๐Ÿ“
:
๐Ÿ
๐Ÿ“
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
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Beams)
(Settlement)
D
I
C
10m
1 cm
Determinate structures
Indeterminate structures
โˆ’12 โˆ’12
โˆ’3๐ธ๐ผโˆ†
๐ฟ2
โˆ†
โˆ†
โˆ†
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
โˆ’6๐ธ๐ผโˆ†
๐ฟ2
โˆ’6๐ธ๐ผโˆ†
๐ฟ2 3 t/m
10m
A I B
25
โˆ’25
13
โˆ’37
C
B
8t
7.5m 7.5m
2I
C
B
15m
2I
2 cm
1 cm
ฮ”
5.33
15
โˆ’15
5.33
20.33
โˆ’9.67 3
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
โˆ’37 20.33
โˆ’9.67 3
Joint B
KBA : KBC
Joint C
KCB : KCD
๐ผ
๐ฟ
:
๐ผ
๐ฟ
1
10
:
2
15
15 : 20
๐Ÿ‘
๐Ÿ•
:
๐Ÿ’
๐Ÿ•
15
35
:
20
35
๐ผ
๐ฟ
:
๐ผ
๐ฟ
๐‘ฅ
3
4
2
15
:
1
10
๐‘ฅ
3
4
80 : 45
๐ŸŽ. ๐Ÿ”๐Ÿ’ : ๐ŸŽ. ๐Ÿ‘๐Ÿ”
80
125
:
45
125
13 โˆ’9.67 20.33 3 0
โˆ’37
3/7 4/7 ๐ŸŽ. ๐Ÿ”๐Ÿ’ ๐ŸŽ. ๐Ÿ‘๐Ÿ”
โˆ’1.903 โˆ’14.93 โˆ’8.399
A B C D
FEM
D.M.1
D.F.
C.O.1
โˆ’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 โˆ’0.174
โˆ’0.13 โˆ’0.768
โˆ’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.
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 1)
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
โˆ’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
๐‘ฅ
1
2
I
L
:
I
L ๐‘ฅ
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
๐‘ค๐‘™2
8
+
๐‘ƒ๐ฟ
4
= 22
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 2)
4t
A
3 t/m
B C
8m
โˆ’16 16
4t
A
3 t/m
B
C
3m
8m
3m
B
โˆ’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
๐ŸŽ. ๐Ÿ“ : ๐ŸŽ. ๐Ÿ“
1
8
:
1
6 ๐‘ฅ
3
4
I
L
:
I
L
๐‘ฅ
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
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 3)
A
3 t/m
B
8m
โˆ’16 16
4t
C
B
โˆ’4.5
3t
B
โˆ’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
๐ŸŽ. ๐Ÿ“ : ๐ŸŽ. ๐Ÿ“
1
8
:
1
6 ๐‘ฅ
3
4
I
L
:
I
L
๐‘ฅ
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
๐‘๐‘™
4
B.M.D.
Structural Analysis
Moment Distribution
II
Eng. Khaled El-Aswany
(Frames)
(Example 4)
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
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

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  • 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 ๐’˜๐‘ณ๐Ÿ ๐Ÿ– ๐‘ท๐‘ณ ๐Ÿ’ = ๐Ÿ–๐’™๐Ÿ– ๐Ÿ’ =16 16mt 2x8=16
  • 5. S.F.D. ๐‘ท ๐Ÿ ๐‘ท ๐Ÿ ๐‘ท ๐Ÿ ๐‘ท ๐Ÿ ๐‘ท๐’ƒ ๐‘ณ ๐‘ท๐’ƒ ๐‘ณ ๐‘ท๐’‚ ๐‘ณ ๐‘ท๐’‚ ๐‘ณ ๐’˜๐‘ณ ๐Ÿ ๐’˜๐‘ณ ๐Ÿ A B a b L ๐‘ท๐‘ณ ๐Ÿ’ L A B w t/m ๐’˜๐‘ณ๐Ÿ ๐Ÿ– B.M.D P t ๐‘ท๐’‚๐’ƒ ๐‘ณ P t L/2 A B L/2 ๐‘ท ๐Ÿ ๐‘ท ๐Ÿ ๐’˜๐’ ๐Ÿ ๐’˜๐’ ๐Ÿ ๐‘ท๐’ƒ ๐‘ณ ๐‘ท๐’‚ ๐‘ณ wL ๐‘ท๐’‚ B.M.D b a P ๐‘ท P t A B P t a ๐‘ท๐’‚ a L A B ๐‘ท๐’‚ L L/2 A B M L/2 ๐‘ด ๐Ÿ ๐‘ด ๐Ÿ P t P t ๐‘ด ๐‘ณ ๐‘ด ๐‘ณ
  • 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 ๐Ÿ” ๐Ÿ– 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 ๐Ÿ– ๐Ÿ’ ๐Ÿ– ๐Ÿ๐Ÿ ๐Ÿ ๐Ÿ ๐Ÿ‘ 4 t 3 m A 3 m ๐Ÿ๐Ÿ 36 36 ๐Ÿ— =
  • 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 ๐Ÿ๐Ÿ ๐Ÿ” ๐Ÿ” ๐Ÿ– ๐Ÿ– ๐Ÿ ๐Ÿ’ ๐Ÿ๐Ÿ’ ๐Ÿ๐ŸŽ ๐Ÿ
  • 10. 4t A D B C 2m 2m 1m 2m 2m 6t 2t/m 8t 2m 2m 0 2m 2m 0 2t/m ๐‘ท๐‘ณ ๐Ÿ’ =8 ๐’˜๐‘ณ๐Ÿ ๐Ÿ– =4 3t 6t ๐Ÿ” 2m 2m 1 0 4t 2m 2m 4t ๐Ÿ๐Ÿ” ๐Ÿ– ๐Ÿ๐Ÿ” ๐Ÿ๐Ÿ” ๐Ÿ– ๐Ÿ– ๐Ÿ๐Ÿ” ๐Ÿ– 2m 2m 3t 3t ๐Ÿ๐Ÿ ๐Ÿ๐Ÿ ๐Ÿ๐Ÿ ๐Ÿ๐Ÿ
  • 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 ( ๐‹๐Ÿ ๐ˆ ) + 2 MB ( ๐‹๐Ÿ ๐ˆ + ๐‹๐Ÿ ๐ˆ ) + MC ( ๐‹๐Ÿ ๐ˆ ) = -6 ( ๐‘๐›๐Ÿ ๐ˆ + ๐‘๐›๐Ÿ ๐ˆ ) MA = MB = MC = MA ( ๐‹๐Ÿ ๐ˆ๐Ÿ ) + 2 MB ( ๐‹๐Ÿ ๐ˆ๐Ÿ + ๐‹๐Ÿ ๐ˆ๐Ÿ ) + MC ( ๐‹๐Ÿ ๐ˆ๐Ÿ ) = -6 ( ๐‘๐›๐Ÿ ๐ˆ๐Ÿ + ๐‘๐›๐Ÿ ๐ˆ๐Ÿ ) 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 (๐‘๐›๐Ÿ + ๐‘๐›๐Ÿ)
  • 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 ( ๐‹๐Ÿ ๐ˆ ) + 2 MB ( ๐‹๐Ÿ ๐ˆ + ๐‹๐Ÿ ๐ˆ ) + MC ( ๐‹๐Ÿ ๐ˆ ) = -6 ( ๐‘๐›๐Ÿ ๐ˆ + ๐‘๐›๐Ÿ ๐ˆ ) MA = MB = MC = MA ( ๐‹๐Ÿ ๐ˆ๐Ÿ ) + 2 MB ( ๐‹๐Ÿ ๐ˆ๐Ÿ + ๐‹๐Ÿ ๐ˆ๐Ÿ ) + MC ( ๐‹๐Ÿ ๐ˆ๐Ÿ ) = -6 ( ๐‘๐›๐Ÿ ๐ˆ๐Ÿ + ๐‘๐›๐Ÿ ๐ˆ๐Ÿ ) 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 (๐‘๐›๐Ÿ + ๐‘๐›๐Ÿ) ๐‘€ 6 (2๐‘ฅโ€ซุงู„ุจุนูŠุฏุฉโ€ฌ + โ€ซ)ุงู„ู‚ุฑูŠุจุฉโ€ฌ A B a b L ๐‘ท๐‘ณ ๐Ÿ’ L A B w t/m ๐’˜๐‘ณ๐Ÿ ๐Ÿ– B.M.D P t P t L/2 A B L/2 Elastic Reactions ๐ด = 1 2 ๐‘ด๐‘ณ A= 2 3 ๐‘ด๐‘ณ ๐ด2 = 1 2 ๐‘ด๐‘ ๐Ÿ ๐Ÿ‘ b ๐Ÿ ๐Ÿ‘ b ๐Ÿ ๐Ÿ‘ a ๐Ÿ ๐Ÿ‘ a ๐ด1 = 1 2 ๐‘ด๐‘Ž Pa B.M.D Elastic Reactions P t a A B b P t a Pa A B a b L M ๐‘€๐‘ ๐ฟ ๐‘€๐‘Ž ๐ฟ ๐ด2๐‘ฅ 2 3 ๐‘ 1 2 ๐ด 1 2 ๐ด 1 2 ๐ด 1 2 ๐ด 1 2 ๐ด 1 2 ๐ด +๐ด1๐‘ฅ(๐‘ + 1 3 ๐‘Ž) L ๐‘€ 6 (2๐‘ฅโ€ซุงู„ุจุนูŠุฏุฉโ€ฌ + โ€ซ)ุงู„ู‚ุฑูŠุจุฉโ€ฌ ๐ด1๐‘ฅ 2 3 ๐‘Ž +๐ด2๐‘ฅ(๐‘Ž + 1 3 ๐‘) L ๐‘ท๐’‚๐’ƒ ๐‘ณ A=( ๐‘+๐ฟ 2 )๐‘ด ๐‘€ ๐ฟ ๐‘€ ๐ฟ ๐ด1 ๐ด2 ๐Ÿ ๐Ÿ‘ a ๐Ÿ ๐Ÿ‘ a ๐Ÿ ๐Ÿ‘ b ๐Ÿ ๐Ÿ‘ b A2x 2 3 b A1x(b + 1 3 a) L ๐ด2 โˆ’ ๐ด1 โˆ’ ๐‘…1 ๐‘ถ๐‘น ๐‘ถ๐‘น
  • 17. Structural Analysis Three Moment Equation II (Example1) Eng. Khaled El-Aswany
  • 18. ๐Ÿ ๐Ÿ‘ ๐‘ด๐‘ณ = ๐Ÿ‘๐Ÿ” 6t 2m 2m B C 2 t/m 6m A B ๐’˜๐‘ณ๐Ÿ ๐Ÿ– =9 ๐‘ท๐‘ณ ๐Ÿ’ = ๐Ÿ” 6t 18 t 18 t 1 2 ๐‘ด๐‘ณ = 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 ( ๐‹๐Ÿ ๐ˆ๐Ÿ ) + 2 MB ( ๐‹๐Ÿ ๐ˆ๐Ÿ + ๐‹๐Ÿ ๐ˆ๐Ÿ ) + MC ( ๐‹๐Ÿ ๐ˆ๐Ÿ ) = -6 ( ๐‘๐›๐Ÿ ๐ˆ๐Ÿ + ๐‘๐›๐Ÿ ๐ˆ๐Ÿ ) 0 + 2 MB ( ๐Ÿ” ๐Ÿ + ๐Ÿ’ ๐Ÿ ) + 0 = -6 ( ๐Ÿ๐Ÿ– ๐Ÿ + ๐Ÿ” ๐Ÿ ) 16 MB = -126 MB = -7.875 mt 2x6-7.3125 =4.6875 ๐Ÿ๐’™๐Ÿ”๐’™๐Ÿ‘ + ๐Ÿ•. ๐Ÿ–๐Ÿ•๐Ÿ“ ๐Ÿ” ๐Ÿ”๐’™๐Ÿ + ๐Ÿ•. ๐Ÿ–๐Ÿ•๐Ÿ“ 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 ๐Ÿ’ = ๐Ÿ’. ๐Ÿ—๐Ÿ• 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 ๐ฐ๐‹๐Ÿ ๐Ÿ– =16 2 3 ๐‘ด๐‘ณ=85.33 Pa=12 Pa=12 ( 4+8 2 )๐‘ฅ12=72 ๐Ÿ—๐’™๐Ÿ๐’™๐Ÿ’ ๐Ÿ” =12 ๐Ÿ‘ ๐Ÿ” ๐Ÿ— 2.25 2m 1m 0.5 1m 42.67 42.67 36 36 20 16 C 9mt 1.5 3m D โˆ’2.25 โˆ’4.5 ๐Ÿ ๐Ÿ 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 ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘’= ๐‘€๐‘Ž ๐‘€๐‘‘= ๐‘€๐‘ 3 Unknowns 2 Unknowns A W t/m B C L L ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘Ž 2 Unknowns A W t/m B L ๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž 1 Unknown ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= ๐‘€๐‘Ž A W t/m B L L/2 F ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐น=?? 3 Unknowns
  • 23. A W t/m B C L L L D L E ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘’= ๐‘€๐‘Ž ๐‘€๐‘‘= ๐‘€๐‘ 3 Unknowns A W t/m B C L L ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘Ž 2 Unknowns A W t/m B L ๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž 1 Unknown A W t/m B L ๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž L 0 0 ๐’˜๐‘ณ๐Ÿ ๐Ÿ– ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ W t/m A B Aโ€™ Bโ€™ A B MAโ€™ (L1) + 2 MA (L1 + L2) + MB (L2) = -6 (๐‘๐š๐Ÿ + ๐‘๐š๐Ÿ) 0 + 2 MA (0 + L) + MA (L) = -6 ( 0 + ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ ) 3 MA L = - wL3 4 ๐‘€๐‘Ž=?? ๐‘€๐‘= ๐‘€๐‘Ž 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 ๐‘€๐‘Ž=?? ๐‘€๐‘=?? 2 Unknowns ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= ๐‘€๐‘Ž A W t/m B L L/2 F ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐น=?? 3 Unknowns W t/m ๐‘€๐‘Ž=?? ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= ๐‘€๐‘Ž 0 Aโ€™ 0 C D L B L Dโ€™ A L B C D A W t/m W t/m ๐’˜๐‘ณ๐Ÿ ๐Ÿ– ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ MAโ€™ (L1) + 2 MA (L1 + L2) + MB (L2) = -6 (๐‘๐š๐Ÿ + ๐‘๐š๐Ÿ) 0 + 2 MA (0 + L) + MB (L) = -6 ( 0 + ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ ) 0 0 MC =MB ๐’˜๐‘ณ๐Ÿ ๐Ÿ– ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ 2 MA + MB = - ๐Ÿ 4 wL2 โ†’ ๐Ÿ MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (๐‘๐š๐Ÿ + ๐‘๐š๐Ÿ) MA (L) + 2 MB (L+ L) + MB (L) = -6 ( ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ + ๐’˜๐‘ณ๐Ÿ‘ ๐Ÿ๐Ÿ’ ) MA + 5 MB = - ๐Ÿ 2 wL2 โ†’ ๐Ÿ MA = - wL2 12 MB = - wL2 12 wL2 12 wL2 12 wL2 12 wL2 12 ๐‘€๐‘Žโ€ฒ=0 ๐‘€๐‘‘โ€ฒ=0
  • 25. Structural Analysis Three Moment Equation II (Settlement) Eng. Khaled El-Aswany
  • 26. 2 cm 2 cm 1 cm ๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘= 0 A B C ๐‘ณ๐Ÿ ๐‘ณ๐Ÿ ๐‘€๐‘Ž=0 ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘‘=0 A B C D ๐‘ณ๐Ÿ ๐‘ณ๐Ÿ ๐‘ณ๐Ÿ‘ MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI ( โˆ†๐ตโˆ’โˆ†๐ด L1 + โˆ†๐ตโˆ’โˆ†๐ถ L2 ) MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI ( โˆ†๐ตโˆ’โˆ†๐ด L1 + โˆ†๐ตโˆ’โˆ†๐ถ L2 ) MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI ( โˆ†๐ถโˆ’โˆ†๐ต L2 + โˆ†๐ถโˆ’โˆ†๐ท L3 ) If I constant: EI : Given I I If I variable: MA ( L1 I1 ) + 2 MB ( L1 I1 + L2 I2 ) + MC ( L2 I2 ) = 6EI ( โˆ†๐ตโˆ’โˆ†๐ด L1 + โˆ†๐ตโˆ’โˆ†๐ถ L2 ) 2 Equations
  • 27. ๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘‘= 0 2 cm ๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž A B C D 2 t/m 2 t/m A B C D ๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI ( โˆ†๐ตโˆ’โˆ†๐ด L1 + โˆ†๐ตโˆ’โˆ†๐ถ L2 ) MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI ( โˆ†๐ถโˆ’โˆ†๐ต L2 + โˆ†๐ถโˆ’โˆ†๐ท 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 ( 0โˆ’0.02 4 + 0) 20 MB + 4 MC = 250 โ†’ 1 4 MB + 20 MC = -150 โ†’ 2 MB = 14.58 mt MC = -10.42 mt ๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= 0 2 t/m A B ๐Ÿ”๐’Ž C ๐Ÿ’๐’Ž B ๐’˜๐‘ณ๐Ÿ ๐Ÿ– = ๐Ÿ— 2 3 ๐‘ด๐‘ณ=36 18 18 0 MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (๐‘…๐‘1+๐‘…๐‘2) 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 ๐Ÿ”๐’Ž C ๐Ÿ’๐’Ž B 2 t/m A B ๐Ÿ”๐’Ž 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 ๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘=?? ๐‘€๐‘‘= 0 2 cm ๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž A B C D 2 t/m 2 t/m A B C D ๐Ÿ”๐’Ž ๐Ÿ’๐’Ž ๐Ÿ”๐’Ž MA (L1) + 2 MB (L1 + L2) + MC (L2) = 6EI ( โˆ†๐ตโˆ’โˆ†๐ด L1 + โˆ†๐ตโˆ’โˆ†๐ถ L2 ) MB (L2) + 2 MC (L2 + L3) + MD (L3) = 6EI ( โˆ†๐ถโˆ’โˆ†๐ต L2 + โˆ†๐ถโˆ’โˆ†๐ท 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 ( 0โˆ’0.02 4 + 0) 20 MB + 4 MC = 250 โ†’ 1 4 MB + 20 MC = -150 โ†’ 2 MB = 14.58 mt MC = -10.42 mt ๐‘€๐‘Ž= 0 ๐‘€๐‘=?? ๐‘€๐‘= ๐‘€๐‘ ๐‘€๐‘‘= 0 2 t/m A B ๐Ÿ”๐’Ž C ๐Ÿ’๐’Ž B ๐’˜๐‘ณ๐Ÿ ๐Ÿ– = ๐Ÿ— 2 3 ๐‘ด๐‘ณ=36 18 18 0 MA (L1) + 2 MB (L1 + L2) + MC (L2) = -6 (๐‘…๐‘1+๐‘…๐‘2) 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
  • 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 ฮด10 + X ฮด11 = 0 X = โˆ’ฮด10 ฮด11 ฮด10 = สƒ ๐‘ด๐‘ถ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ ฮด11 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ 9 mt 6m 1 mt = Area x Yc = 2 3 ๐‘ฅ6๐‘ฅ9 x 0.5 Area Yc 0.5x1=0.5 (Linear) - โ€ซู…ุณุชู‚ูŠู…โ€ฌ โ€ซุฎุทโ€ฌ โ€ซู…ู†ูƒุณุฑโ€ฌ โ€ซุบูŠุฑโ€ฌ สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ = Area x Yc = 1 2 ๐‘ฅ6๐‘ฅ8 x 3 Area 6m 6 mt 8 mt Yc 0.5x6=3 (Linear) - สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ = Area x Yc = 1 2 ๐‘ฅ6๐‘ฅ8 6 mt 6m 8 mt x 4 Area Yc ๐Ÿ ๐Ÿ‘ ๐ฑ๐Ÿ” = ๐Ÿ’ + ๐Ÿ ๐Ÿ‘ ๐‘ณ สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ = Area x Yc = 1 2 ๐‘ฅ3๐‘ฅ8 8 mt 6 mt 3m 3m + Area Yc ๐Ÿ ๐Ÿ‘ ๐‘ณ ๐Ÿ ๐Ÿ‘ ๐‘ณ ๐Ÿ ๐Ÿ‘ ๐ฑ๐Ÿ” = ๐Ÿ’ x 4 สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ x 2 = L 3 (๐š๐œ + ๐›๐+ bc 2 + ad 2 ) 8 4 2 6 8m 8x6 = 8 3 ( + 4x2 + ๐Ÿ’๐’™๐Ÿ” ๐Ÿ + ๐Ÿ–๐’™๐Ÿ ๐Ÿ ) 8 mt 6 mt สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ = 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 ฮด10 + X ฮด11 = 0 X = โˆ’ฮด10 ฮด11 ฮด10 = สƒ ๐‘ด๐‘ถ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐‘ถ๐‘ด๐Ÿ๐’…๐’ ฮด11 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ [- ๐Ÿ ๐Ÿ‘ x 6 x 9 x0.5 + ๐Ÿ” ๐Ÿ‘ ( ๐Ÿ๐Ÿ๐’™๐Ÿ ๐Ÿ )] = โˆ’๐Ÿ” ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ ๐Ÿ” ๐Ÿ‘ (1x1)] = ๐Ÿ ๐‘ฌ๐‘ฐ = โˆ’( เต— โˆ’๐Ÿ” EI) เต— ๐Ÿ EI = ๐Ÿ‘
  • 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 ฮด10 + X ฮด11 = 0 X = โˆ’ฮด10 ฮด11 ฮด10 ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ [- ๐Ÿ ๐Ÿ‘ x 6 x 9 x0.5 + ๐Ÿ” ๐Ÿ‘ ( ๐Ÿ๐Ÿ๐’™๐Ÿ ๐Ÿ )] = โˆ’๐Ÿ” ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ ๐Ÿ” ๐Ÿ‘ (1x1)] = ๐Ÿ ๐‘ฌ๐‘ฐ = โˆ’( เต— โˆ’๐Ÿ” EI) เต— ๐Ÿ EI = ๐Ÿ‘ 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 ๐‘€๐‘œ 12 12 4 8 ๐‘€1 1 ๐‘€2 1 1mt 1mt ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0 ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0 ฮด10 = สƒ ๐‘ด๐‘ถ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [-( 4+8 2 )๐’™๐Ÿ๐Ÿ x0.5 - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ x0.5] = โˆ’๐Ÿ’๐ŸŽ EI ฮด11 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ 8 3 ๐’™(๐Ÿ๐’™๐Ÿ) = 4 EI + 4 3 ๐’™(๐Ÿ๐’™๐Ÿ) ] ฮด12 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [4 3 x( 1X1 2 ) ] = เต˜ 2 3 EI ฮด02 = สƒ ๐‘ด๐‘ถ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ - ๐Ÿ ๐Ÿ x0.5 ๐’™ 2 3 ๐’™๐Ÿ’๐’™๐Ÿ– x0.5 ]= เต˜ โˆ’2๐Ÿ– 3 EI ฮด22 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ ๐’™ 4 3 ๐’™(๐Ÿ๐’™๐Ÿ) ] 4 3 ๐’™(๐Ÿ๐’™๐Ÿ) + 1 2 = ๐Ÿ EI โˆ’40 EI + 4 EI ๐‘‹1 + เต˜ 2 3 EI ๐‘‹2 = 0 โˆ’40+ 4 ๐‘‹1 + 2 3 ๐‘‹2 = 0 4 ๐‘‹1 + 2 3 ๐‘‹2 = 40 เต˜ โˆ’28 3 EI + ๐‘‹1 เต˜ 2 3 EI + ๐‘‹2 2 EI = 0 โˆ’28 3 + 2 3 ๐‘‹1+2๐‘‹2 = 0 2 3 ๐‘‹1 + 2 ๐‘‹2 = 28 3 โ†’ 1 โ†’ 2 ๐‘ฟ๐Ÿ = 9.76 mt ๐‘ฟ๐Ÿ = ๐Ÿ. ๐Ÿ’๐Ÿ ๐’Ž๐’•
  • 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 ๐‘€๐‘œ 12 12 4 8 ๐‘€1 1 ๐‘€2 1 1mt 1mt ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0 ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0 ฮด10 = สƒ ๐‘ด๐‘ถ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [-( 4+8 2 )๐’™๐Ÿ๐Ÿ x0.5 - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ x0.5] = โˆ’๐Ÿ’๐ŸŽ EI ฮด11 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ 8 3 ๐’™(๐Ÿ๐’™๐Ÿ) = 4 EI + 4 3 ๐’™(๐Ÿ๐’™๐Ÿ) ] ฮด12 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [4 3 x( 1X1 2 ) ] = เต˜ 2 3 EI ฮด02 = สƒ ๐‘ด๐‘ถ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ - ๐ŸŽ. ๐Ÿ“๐’™๐Ÿ’๐’™๐Ÿ’ - ๐Ÿ ๐Ÿ x0.5 ๐’™ 2 3 ๐’™๐Ÿ’๐’™๐Ÿ– x0.5 ]= เต˜ โˆ’2๐Ÿ– 3 EI ฮด22 = สƒ ๐‘ด๐Ÿ ๐‘ด๐Ÿ ๐’…๐’ ๐‘ฌ๐‘ฐ = ๐Ÿ ๐‘ฌ๐‘ฐ [ ๐’™ 4 3 ๐’™(๐Ÿ๐’™๐Ÿ) ] 4 3 ๐’™(๐Ÿ๐’™๐Ÿ) + 1 2 = ๐Ÿ EI 4 ๐‘‹1 + 2 3 ๐‘‹2 = 40 2 3 ๐‘‹1 + 2 ๐‘‹2 = 28 3 โ†’ 1 โ†’ 2 9.76 1.41 ๐‘ค๐‘™2 8 =8 D A B C 9.76+1.41 2 =5.58 ๐‘ƒ๐ฟ 4 =4 1.58 1 4 ๐‘‹9.76=2.44 3 4 ๐‘‹9.76=7.32 12 9.56 12 4.68 ๐‘ฟ๐Ÿ = 9.76 mt ๐‘ฟ๐Ÿ = ๐Ÿ. ๐Ÿ’๐Ÿ ๐’Ž๐’•
  • 37. Structural Analysis Consistent Deformations II Eng. Khaled El-Aswany (Beams) (Support movement)
  • 38. ฮด10 ๐‘‹1 ฮด11 ฮด10 + X1 ฮด11 = 0 ฮด10 ๐‘‹1 ฮด11 ฮด10 + X1 ฮด11 = ยฑโˆ† ฮ”
  • 39. 1 ๐‘ก 2 ๐‘๐‘š 3 X1 ฮด11 = โˆ’๐ŸŽ. ๐ŸŽ๐Ÿ ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ ( ๐Ÿ” ๐Ÿ‘ ( 3x3)x 2 ) = ๐Ÿ‘๐Ÿ” ๐‘ฌ๐‘ฐ X1 ฮด11 = ยฑโˆ† 2.5๐‘ก 1.25 ๐‘ก 1.25 ๐‘ก 7.5 B.M.D. Reactions X1 36 ๐ธ๐ผ = โˆ’0.02 X1 36 4500 = โˆ’0.02 X1 = โˆ’2.5 ๐‘ก EI = 4500 m2t
  • 40. 1 ๐‘ก 2 ๐‘๐‘š 3 EI = 4500 m2t ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ ( ๐Ÿ” ๐Ÿ‘ ( 3x3)x 2 ) = ๐Ÿ‘๐Ÿ” ๐‘ฌ๐‘ฐ ฮด10 + X1 ฮด11 = ยฑโˆ† Reactions โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ ๐‘ฌ๐‘ฐ + X1 36 ๐ธ๐ผ = โˆ’0.02 X1 = 72.5 ๐‘ก 180 M.S. M0 M1 ฮด10 = ๐Ÿ ๐‘ฌ๐‘ฐ ( โˆ’๐Ÿ ๐Ÿ‘ ๐’™๐Ÿ”๐’™๐Ÿ๐Ÿ–๐ŸŽ x( ๐Ÿ“ ๐Ÿ– ๐’™๐Ÿ‘) x 2 ) = โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ ๐‘ฌ๐‘ฐ 5 8 ๐ฟ 3 8 ๐ฟ 5 8 ๐‘ฅ3 โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ ๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ + X1 36 4500 = โˆ’0.02 72.5 ๐‘ก 23.75๐‘ก 23.75๐‘ก
  • 41. 1 ๐‘ก 2 ๐‘๐‘š 3 EI = 4500 m2t ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ ( ๐Ÿ” ๐Ÿ‘ ( 3x3)x 2 ) = ๐Ÿ‘๐Ÿ” ๐‘ฌ๐‘ฐ ฮด10 + X1 ฮด11 = ยฑโˆ† B.M.D. โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ ๐‘ฌ๐‘ฐ + X1 36 ๐ธ๐ผ = โˆ’0.02 X1 = 72.5 ๐‘ก 180 M.S. M0 M1 ฮด10 = ๐Ÿ ๐‘ฌ๐‘ฐ ( โˆ’๐Ÿ ๐Ÿ‘ ๐’™๐Ÿ”๐’™๐Ÿ๐Ÿ–๐ŸŽ x( ๐Ÿ“ ๐Ÿ– ๐’™๐Ÿ‘) x 2 ) = โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ ๐‘ฌ๐‘ฐ 5 8 ๐ฟ 3 8 ๐ฟ 5 8 ๐‘ฅ3 โˆ’๐Ÿ๐Ÿ•๐ŸŽ๐ŸŽ ๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ + X1 36 4500 = โˆ’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 ๐Ÿ— 2mt 2 Mo ๐Ÿ— 1mt 1 M1 1 M2 ฮด11 = 1 ๐ธ๐ผ ( 1 3 * 6 3 (1*1)) = 2 3๐ธ๐ผ ฮด22 = 1 ๐ธ๐ผ ( 1 3 * 6 3 (1*1)) = 2 ๐ธ๐ผ * 8 3 (1*1)) +( 1 2 ฮด12 = 1 ๐ธ๐ผ ( 1 3 * 6 3 ( 1โˆ—1 2 )) = 1 3๐ธ๐ผ ฮด1o = 1 ๐ธ๐ผ ( 1 3 (- 2 3 x6x9x0.5 = โˆ’34 3๐ธ๐ผ 0.5 - 2+6 2 x8x0.5) ฮด2o = 1 ๐ธ๐ผ ( 1 3 (- 2 3 x6x9x0.5- 2+6 2 x8x0.5) +( 1 2 ( 8 3 X 1๐‘‹2 2 + 6 3 x9x0.75 0.75 + 2 3 x(9x0.75+ 9๐‘‹1 2 ) 0.5 + 4 3 x(4x0.5+ 4๐‘‹1 2 ) - 4 3 x(4x0.5) 0.25 - 2 3 x(9x0.25) - 6 3 x(9x0.25+ 9๐‘‹1 2 )) = โˆ’17 3๐ธ๐ผ ฮด10 + ๐’™๐Ÿ ฮด11 + ๐’™๐Ÿ ฮด12= 0 โˆ’๐Ÿ‘๐Ÿ’ ๐Ÿ‘ + ๐Ÿ ๐Ÿ‘ ๐’™๐Ÿ + ๐Ÿ ๐Ÿ‘ ๐’™๐Ÿ = 0 ๐Ÿ ๐’™๐Ÿ + ๐’™๐Ÿ = 34 ๐’™๐Ÿ + 6๐’™๐Ÿ = 17 ๐’™๐Ÿ= 17 ๐’™๐Ÿ = 0 ฮด20 + ๐’™๐Ÿ ฮด21 + ๐’™๐Ÿ ฮด22= 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 ๐Ÿ— 2mt 2 Mo ๐Ÿ— 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 ๐Ÿ๐Ÿ. ๐Ÿ–๐Ÿ‘ ๐Ÿ•. ๐Ÿ๐Ÿ• ๐Ÿ’. ๐Ÿ๐Ÿ“ ๐Ÿ’. ๐Ÿ๐Ÿ“ 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 โ€“ 3 = 1 R = U โ€“ 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 ฮด10 ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐ŸŽ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ [- x ๐Ÿ ๐Ÿ‘ x 6 x 9 x4 - = โˆ’๐Ÿ๐ŸŽ๐Ÿ– ๐‘ฌ๐‘ฐ 4 4 ๐Ÿ ๐Ÿ x ๐Ÿ ๐Ÿ x 6 x 6 x4 ] ๐Ÿ ๐Ÿ = ๐Ÿ ๐‘ฌ๐‘ฐ [ ๐Ÿ’ ๐Ÿ‘ x (4x4) + = ๐Ÿ—๐ŸŽ.๐Ÿ”๐Ÿ• ๐‘ฌ๐‘ฐ x ๐Ÿ’ x 6 x4 ] ๐Ÿ ๐Ÿ x2 ๐‘ฟ๐Ÿ = โˆ’ ฮด10 ฮด11 = โˆ’( โˆ’ 108/๐ธ๐ผ) 90.67/๐ธ๐ผ = 1.19 ๐‘€๐‘“ = ๐‘€0 + ๐‘‹1 ๐‘€1 Mf ๐‘€๐‘“ = 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 =5โ€“ 3 = 2 R = U โ€“ 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 ฮด10 = ๐Ÿ ๐‘ฌ๐‘ฐ [ x 3 3 x( 12 x 3) + = ๐Ÿ๐Ÿ’๐ŸŽ ๐‘ฌ๐‘ฐ ๐Ÿ ๐Ÿ 6 3 x(16x6+12x3+ 16x3 2 + 12x6 2 ) X4.5 ] - 2 3 x6x9 ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ [ x 3 3 x( 3 x 3) + = ๐Ÿ๐Ÿ“๐Ÿ’.๐Ÿ“ ๐‘ฌ๐‘ฐ ๐Ÿ ๐Ÿ 6 3 x(6x6+3x3+ 6x3 2 ๐ฑ๐Ÿ) + x 6 3 x( 6 x 6) ] ๐Ÿ ๐Ÿ‘ ฮด12 = ๐Ÿ ๐‘ฌ๐‘ฐ [ x 1 2 x6x6 + = ๐Ÿ๐Ÿ ๐‘ฌ๐‘ฐ ๐Ÿ ๐Ÿ‘ X 1 6 3 x(6x1+ 3x1 2 ) ] 1 1 1 ฮด20 = ๐Ÿ ๐‘ฌ๐‘ฐ [ = ๐Ÿ๐Ÿ” ๐‘ฌ๐‘ฐ 6 3 x(16x1+12x0+ 16x0 2 + 12x1 2 ) X0.5 ] - 2 3 x6x9 ฮด22 = ๐Ÿ ๐‘ฌ๐‘ฐ [ x (๐Ÿ”๐’™๐Ÿ) = ๐Ÿ’ ๐‘ฌ๐‘ฐ ๐Ÿ ๐Ÿ‘ + 6 3 x( 1x1) ] x ๐Ÿ
  • 49. 8t A 2 t/m D B C 2I 3I I 6m 3m 3m 2m 4t 3m YA XD YD =5โ€“ 3 = 2 R = U โ€“ 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 ฮด10 = ๐Ÿ๐Ÿ’๐ŸŽ ๐‘ฌ๐‘ฐ ฮด11 = ๐Ÿ๐Ÿ“๐Ÿ’.๐Ÿ“ ๐‘ฌ๐‘ฐ ฮด12 = ๐Ÿ๐Ÿ ๐‘ฌ๐‘ฐ 1 1 1 ฮด20 = ๐Ÿ๐Ÿ” ๐‘ฌ๐‘ฐ ฮด22 = ๐Ÿ’ ๐‘ฌ๐‘ฐ ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0 ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0 ๐‘ฟ๐Ÿ = -2.34 mt ๐‘ฟ๐Ÿ = ๐Ÿ“. ๐Ÿ•๐Ÿ– ๐’Ž๐’• 154.5 ๐‘‹1 + 21 ๐‘‹2 = -240 21 ๐‘‹1 + 4 ๐‘‹2 = โˆ’26 โ†’ 1 โ†’ 2 MF ๐‘€๐‘“ = ๐‘€0 + ๐‘‹1 ๐‘€1 + ๐‘‹2 ๐‘€2 ๐‘€๐‘“ = ๐‘€0 -2.34 ๐‘€1 + 5.78 ๐‘€2 ๐‘€๐‘“ = (โˆ’16) -2.34 (โˆ’6) + 5.78 (โˆ’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 โ€“ 3 = 2 R = U โ€“ E 2 t/m A B D C 4m 4m 2 t/m 4m 0 8 8 8 = 6 โ€“ 3 = 3 R = U โ€“ E 2 t/m A B D C 8m 4m
  • 52. = 5 โ€“ 3 = 2 R = U โ€“ E 2 t/m A B 4m 4m 2 t/m A B 4m 4m M.S. ๐‘€๐‘œ 16 4 16 16 ๐‘€1 1mt 1 1 1 1 ๐‘€2 1t 4 ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ [4x1x1 + 4x1x1]= ๐Ÿ– ๐‘ฌ๐‘ฐ ฮด22 = ๐Ÿ ๐‘ฌ๐‘ฐ [๐Ÿ’ ๐Ÿ‘ ๐’™(๐Ÿ’๐’™๐Ÿ’)]= ๐Ÿ๐Ÿ.๐Ÿ‘๐Ÿ‘ ๐‘ฌ๐‘ฐ ฮด12 = ๐Ÿ ๐‘ฌ๐‘ฐ [โˆ’๐Ÿ ๐Ÿ ๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ]= โˆ’๐Ÿ– ๐‘ฌ๐‘ฐ ฮด20 = ๐Ÿ ๐‘ฌ๐‘ฐ [โˆ’๐Ÿ ๐Ÿ ๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ๐Ÿ”]= โˆ’๐Ÿ๐Ÿ๐Ÿ– ๐‘ฌ๐‘ฐ ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0 ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0 ๐‘ฟ๐Ÿ = -7.47 mt ๐‘ฟ๐Ÿ = ๐Ÿ‘. ๐Ÿ ๐’Ž๐’• 8 ๐‘‹1 - 8 ๐‘‹2 = -85.33 โ†’ 1 -8 ๐‘‹1 + 21.33 ๐‘‹2 = 128 โ†’ 2 ๐‘€๐‘“ = ๐‘€0 + ๐‘‹1 ๐‘€1 + ๐‘‹2 ๐‘€2 ๐‘€๐‘“2 = 16 -7.47 (1) + 3.2 (0) = 8.53 ๐‘€๐‘“ = ๐‘€0 -7.47 ๐‘€1 + 3.2 ๐‘€2 ๐‘€๐‘“3 = 16 -7.47 (1) + 3.2 (-4) = -4.27 1 2 3 ๐‘€๐น ฮด10= ๐Ÿ ๐‘ฌ๐‘ฐ [( 1 2 x4 x 16) ๐ฑ๐Ÿ โˆ’ ๐Ÿ ๐Ÿ‘ ๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ +๐Ÿ’๐’™๐Ÿ๐Ÿ” ๐’™๐Ÿ ] = ๐Ÿ–๐Ÿ“.๐Ÿ‘๐Ÿ‘ ๐‘ฌ๐‘ฐ 7.47 8.53 8.53 4.27
  • 53. = 5 โ€“ 3 = 2 R = U โ€“ E 2 t/m A B 4m 4m 2 t/m A B 4m 4m M.S. ๐‘€๐‘œ 16 4 16 16 ๐‘€1 1mt 1 1 1 1 ๐‘€2 1t 4 ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ [4x1x1 + 4x1x1]= ๐Ÿ– ๐‘ฌ๐‘ฐ ฮด22 = ๐Ÿ ๐‘ฌ๐‘ฐ [๐Ÿ’ ๐Ÿ‘ ๐’™(๐Ÿ’๐’™๐Ÿ’)]= ๐Ÿ๐Ÿ.๐Ÿ‘๐Ÿ‘ ๐‘ฌ๐‘ฐ ฮด12 = ๐Ÿ ๐‘ฌ๐‘ฐ [โˆ’๐Ÿ ๐Ÿ ๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ]= โˆ’๐Ÿ– ๐‘ฌ๐‘ฐ ฮด20 = ๐Ÿ ๐‘ฌ๐‘ฐ [โˆ’๐Ÿ ๐Ÿ ๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ๐Ÿ”]= โˆ’๐Ÿ๐Ÿ๐Ÿ– ๐‘ฌ๐‘ฐ ฮด10 + ๐‘ฟ๐Ÿ ฮด11 + ๐‘ฟ๐Ÿ ฮด12 = 0 ฮด20 + ๐‘ฟ๐Ÿ ฮด21 + ๐‘ฟ๐Ÿ ฮด22 = 0 ๐‘ฟ๐Ÿ = -7.47 mt ๐‘ฟ๐Ÿ = ๐Ÿ‘. ๐Ÿ ๐’Ž๐’• 8 ๐‘‹1 - 8 ๐‘‹2 = -85.33 โ†’ 1 -8 ๐‘‹1 + 21.33 ๐‘‹2 = 128 โ†’ 2 ๐‘€๐น ฮด10= ๐Ÿ ๐‘ฌ๐‘ฐ [( 1 2 x4 x 16) ๐ฑ๐Ÿ โˆ’ ๐Ÿ ๐Ÿ‘ ๐’™๐Ÿ’๐’™๐Ÿ’ ๐’™๐Ÿ +๐Ÿ’๐’™๐Ÿ๐Ÿ” ๐’™๐Ÿ ] = ๐Ÿ–๐Ÿ“.๐Ÿ‘๐Ÿ‘ ๐‘ฌ๐‘ฐ 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 โ€“ 3 = 1 R = U โ€“ 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 ฮด10 = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐ŸŽ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ [ ๐Ÿ’ ๐Ÿ‘ x (24x4) - ๐Ÿ ๐Ÿ‘ x 6 x 9 x4 + = ๐Ÿ๐Ÿ•๐Ÿ ๐‘ฌ๐‘ฐ ๐Ÿ ๐Ÿ x 6 x 24 x4 ] ฮด11 = ๐Ÿ ๐‘ฌ๐‘ฐ สƒ ๐‘ด๐Ÿ๐‘ด๐Ÿ๐’…๐’ = ๐Ÿ ๐‘ฌ๐‘ฐ [ ๐Ÿ’ ๐Ÿ‘ x (4x4) + = ๐Ÿ๐Ÿ‘๐Ÿ–.๐Ÿ”๐Ÿ• ๐‘ฌ๐‘ฐ ๐Ÿ’ x 6 x4 ] x2 ฮด10 + X1 ฮด11 = 0.02 ๐Ÿ๐Ÿ•๐Ÿ ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ๐ŸŽ + X1 138.67 20000 = 0.02 X1 = 0.92 ๐‘ก Mf 3.68 27.68 27.68 3.68 15.68
  • 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 ฮด10 + ๐‘ฟ๐Ÿ ฮด11 = 0 A B N0 6 6 A B N1 1 1 ฮด10 = ๐Ÿ ๐‘ฌ๐‘จ สƒ ๐‘ต๐‘ถ๐‘ต๐Ÿ๐’…๐’ 4m = ๐Ÿ ๐‘ฌ๐‘จ ( -6 x 4 x 1 N0 N1 L C D C D + โ€ฆโ€ฆโ€ฆ + ...โ€ฆโ€ฆ) N0 N1 L N0 N1 L ฮด10 = ฮฃNON1L EA ฮด11 = ฮฃN1N1L 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 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. ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 ๐น1 ๐น2 ๐น1sin ๐œƒ ๐น1cos ๐œƒ 4 4 ฮฃ๐น๐‘ฆ = 0 ๐น1sin ๐œƒ + 6 = 0 ๐น1= โˆ’6 sin ฮธ = โˆ’6 0.6 = -10 t (๐ถ๐‘œ๐‘š๐‘๐‘Ÿ๐‘’๐‘ ๐‘–๐‘œ๐‘›) No ฮฃ๐น๐‘ฅ = 0 ๐น1cos ๐œƒ + ๐น2 = 0 (๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›) โˆ’10 x0.8+ ๐น2 = 0 ๐น2 = 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. ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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. ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 4 4 No 10 8 8 ๐น3 ๐น4 10cos๐œƒ 10sin๐œƒ ๐น3 cos๐œƒ ๐น3 sin๐œƒ ฮฃ๐น๐‘ฆ = 0 10sin ๐œƒ - 4 - ๐น3sin ๐œƒ = 0 (๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›) ฮฃ๐น๐‘ฅ = 0 ๐น3cos ๐œƒ + ๐น4 + 10cos ๐œƒ = 0 (๐ถ๐‘œ๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘–๐‘œ๐‘›) ๐น4 = -10.67t ๐น3 = 3.33t 3.33cos ๐œƒ + ๐น4 + 10cos ๐œƒ = 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. ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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. ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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. ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4๐‘š 3๐‘š 5๐‘š 3m 4t 4t 4t 4m 4m 4m 4m 1 2 3 4 5 6 7 8 9 10 11 12 13 3m ๐œƒ cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 4 4 No -10 8 8 3.33 -10.67 3.33 โˆ’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 ฮด10 = ๐›ดN ON1L EA = โˆ’128 EA ฮด11 = ๐›ดN1N1L EA = 16 EA X1 = โˆ’ฮด10 ฮด11 = โˆ’(โˆ’128)/๐ธ๐ด 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ
  • 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 4 10 6 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐น1 ๐น2 ๐น1sin ๐œƒ ๐น1cos ๐œƒ ฮฃ๐น๐‘ฆ = 0 ๐น1sin ๐œƒ -4 +10 = 0 ๐น1= -10 t (๐ถ๐‘œ๐‘š๐‘๐‘Ÿ๐‘’๐‘ ๐‘–๐‘œ๐‘›) ฮฃ๐น๐‘ฅ = 0 ๐น1cos ๐œƒ + ๐น2 = 0 (๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›) โˆ’10 x0.8+ ๐น2 = 0 ๐น2 = 8t ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 10 8 8 ๐น3 ๐น4 10cos๐œƒ 10sin๐œƒ ๐น3 cos๐œƒ ๐น3 sin๐œƒ ฮฃ๐น๐‘ฆ = 0 10sin ๐œƒ - 6 - ๐น3sin ๐œƒ = 0 ฮฃ๐น๐‘ฅ = 0 ๐น3cos ๐œƒ + ๐น4 + 10cos ๐œƒ = 0 (๐ถ๐‘œ๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘–๐‘œ๐‘›) ๐น4 = - 8t ๐น3 = 0 0 + ๐น4 + 10cos ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 10 6 4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 10 8 8 8 8 ๐น5 ๐น5 cos๐œƒ ๐น5 sin๐œƒ ฮฃ๐น๐‘ฆ = 0 10 - 4 - ๐น5sin ๐œƒ = 0 ฮฃ๐น๐‘ฅ = 0 ๐น5cos ๐œƒ - 8 = 0 (๐‘‡๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘›) ๐น5 = 10t ๐น5 = 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ -10 -6 -4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 -10 8 8 โˆ’8 โˆ’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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ -10 -6 -4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 -10 8 8 โˆ’8 โˆ’8 10 3m 4m 4m 4m 0 N1 0 1t 1t ๐œƒ -0.8 -0.6 ๐œƒ -0.8 -0.6 ๐œƒ 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ -10 -6 -4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 -10 8 8 โˆ’8 โˆ’8 10 3m 4m 4m 4m 0 N1 0 ๐œƒ ๐œƒ ๐œƒ 1 ฮด10 = ๐›ดN ON1L EA ฮด11 = ๐›ดN 1N1L EA X1 = โˆ’ฮด10 ฮด11 = โˆ’(โˆ’2)/๐ธ๐ด 16/EA = 0.125t ฮด10 + X1 ฮด11 = 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] = โˆ’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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ -10 -6 -4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 -10 8 8 โˆ’8 โˆ’8 10 3m 4m 4m 4m 0 N1 0 ๐œƒ ๐œƒ ๐œƒ 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 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ -10 -6 -4 ๐œƒ 4๐‘š 3๐‘š 5๐‘š cos ๐œƒ = 4 5 = 0.8 sin ๐œƒ = 3 5 = 0.6 -10 8 8 โˆ’8 โˆ’8 10 3m 4m 4m 4m 0 N1 0 ๐œƒ ๐œƒ ๐œƒ 1 1t 1t -0.8 -0.6 -0.8 -0.6 Mem. L No N1 A ๐‘๐‘œ๐‘1๐ฟ ๐‘จ ๐‘๐Ÿ๐‘1๐ฟ ๐‘จ 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 ฮด10 = โˆ’2 EA ฮด11 = 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 7๐‘ก 7๐‘ก 0 No -7 -7 -6 45ยฐ -3 -3 A 3m 3m 3m B N1 1๐‘ก 0 0 1๐‘ก -1 -1 A 3m 3m 3m B N2 1๐‘ก 1๐‘ก 0 0 0 โˆ’1 2 โˆ’1 2 โˆ’1 2 โˆ’1 2 1๐‘ก ฮด10 + X1 ฮด11 + X2 ฮด12 = 0 ฮด20 + X1 ฮด21 + X2 ฮด22 = 0 ฮด10 = 0 ฮด20 = 51.94 EA ฮด11 = 6 EA ฮด22 = 14.49 EA ฮด12 = 2.12 EA X1 = 1.34 ๐‘ก X2 = โˆ’3.78 ๐‘ก Nf = N0 + X1 N1 + X2 N2 Nf = N0 + 1.34 N1 -3.78 N2 6 X1 + 2.12 X2 = 0 โ†’ 1 2.12 X1 + 14.49 X2 = -51.94 โ†’ 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 โˆ’1 / 2 0 0 3 1.5 ๐Ÿ‘/ ๐Ÿ 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 โˆ’1 / 2 0 ๐Ÿ๐Ÿ–/ ๐Ÿ 0 1.5 0 -3.33 6 3 2 0 0 1 0 0 0 ๐Ÿ‘ ๐Ÿ 0 -3.78 7 3 2 3 2 0 1 0 ๐Ÿ๐Ÿ– 0 ๐Ÿ‘ ๐Ÿ 0 0.46 8 3 -7 0 โˆ’1 / 2 0 ๐Ÿ๐Ÿ/ ๐Ÿ 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 ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ– ๐‘ญ๐Ÿ— ๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’ ๐‘ญ๐Ÿ“ ๐‘ญ๐Ÿ” ๐‘ญ๐Ÿ• 6m 6m 8t [๐’€๐‘จ = ๐Ÿ–๐’•, ๐‘ญ๐Ÿ– = ๐Ÿ๐Ÿ” ๐’•] [๐‘ญ๐Ÿ = โˆ’๐Ÿ’. ๐Ÿ–๐Ÿ๐’•, ๐‘ญ๐Ÿ๐ŸŽ = ๐Ÿ‘. ๐Ÿ—๐Ÿ– ๐’•] 3t 4m A B ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ– ๐‘ญ๐Ÿ— ๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’ ๐‘ญ๐Ÿ“ ๐‘ญ๐Ÿ” ๐‘ญ๐Ÿ• 4m 3m 3m ๐‘ญ๐Ÿ๐ŸŽ C [๐‘ฟ๐‘จ = ๐Ÿ’. ๐Ÿ๐Ÿ“๐’•, ๐‘ญ๐Ÿ” = โˆ’๐Ÿ’. ๐Ÿ“ ๐’•] A ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ– ๐‘ญ๐Ÿ๐Ÿ” ๐‘ญ๐Ÿ๐Ÿ• ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ“ ๐‘ญ๐Ÿ— ๐‘ญ๐Ÿ” 3m 3m 8t 4A ๐‘ญ๐Ÿ๐Ÿ– ๐‘ญ๐Ÿ๐Ÿ— B ๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’ ๐‘ญ๐Ÿ• ๐‘ญ๐Ÿ๐ŸŽ ๐‘ญ๐Ÿ๐Ÿ ๐‘ญ๐Ÿ๐Ÿ ๐‘ญ๐Ÿ๐Ÿ‘ ๐‘ญ๐Ÿ๐Ÿ’ ๐‘ญ๐Ÿ๐Ÿ“ 3m 3m 8t 4t 4t 4t 3m 3m [๐‘ญ๐Ÿ = ๐Ÿ’. ๐Ÿ–๐Ÿ“๐’•, ๐‘ญ๐Ÿ” = โˆ’๐Ÿ“. ๐ŸŽ๐Ÿ“ ๐’•] 2m A B ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ ๐‘ญ๐Ÿ– ๐‘ญ๐Ÿ” ๐‘ญ๐Ÿ• 4m 10t ๐‘ญ๐Ÿ‘ ๐‘ญ๐Ÿ’ ๐‘ญ๐Ÿ“ 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 ฮด10 1t ฮด11 ฮด10 + X ฮด11 = 0 A w t/m B Movement: A w t/m B A B ฮด10 X ฮด11 ฮด10 + X ฮด11 = ยฑโˆ† โˆ† 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. ฮด10 + X ฮด11 = ๐ŸŽ. ๐ŸŽ๐Ÿ ๐‘ฌ๐‘จ + X ๐‘ฌ๐‘จ = ๐ŸŽ. ๐ŸŽ๐Ÿ EA = 10000 t Answer X
  • 92. 4t 4t 4t 4m 4m 4m 4m 6t 6t 0 ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ ๐œƒ 3m 4 4 No -10 8 8 3.33 -10.67 3.33 โˆ’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 ฮด10 = ๐›ดN ON1L EA = โˆ’128 EA ฮด11 = ๐›ดN1N1L = 16 X1 = โˆ’ฮด10 ฮด11 = โˆ’(โˆ’128)/๐ธ๐ด 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 ฮด10 + X1 ฮด11 = 0 Temperature Temperature Effects No load Effect Loads & Temperature 8t ฮด1t + X1 ฮด11 = 0 ฮด1t + ฮด10 + X1 ฮด11 = 0 Loads: ฮด10 + X1 ฮด11 = 0 Temp.: ฮด1t + ๐‘ฟ๐Ÿ โ€ฒ ฮด11 = 0 Total: Xtot = X1 +๐‘ฟ๐Ÿ โ€ฒ ฮด1t = ฮฃ๐œถ. โˆ†๐’•. ๐‘ต๐Ÿ. ๐‘ณ ฮด1t = ๐œถ. โˆ†๐’•. ฮฃ๐‘ต๐Ÿ. ๐‘ณ OR ฮด10 + X1 ฮด11 + X2 ฮด12 = 0 ฮด20 + X1 ฮด21 + X2 ฮด22 = 0 ฮด1t + X1 ฮด11 + X2 ฮด12 = 0 ฮด2t + X1 ฮด21 + X2 ฮด22 = 0 ฮด1t + ฮด10 +X1 ฮด11 + X2 ฮด12 = 0 ฮด2t + ฮด20 +X1 ฮด21 + X2 ฮด22 = 0 8t 10t 10t ฮด1t = ฮฃ๐œถ. โˆ†๐’•. ๐‘ต๐Ÿ. ๐‘ณ ฮด2t = ฮฃ๐œถ. โˆ†๐’•. ๐‘ต๐Ÿ. ๐‘ณ 10โˆ’5 Rise (+) Drop (โˆ’) Tension (+) compression (โˆ’) ONCE TWICE
  • 95. ๐œƒ -1.67 Determine the internal forces in members for the shown truss due to the external loads and rise of temperature 100oC. ๐›ผ = 1๐‘ฅ10โˆ’5 /๐‘œ๐ถ 8m B 6t 6t 6m A Example 1 EA = 45000 t R = ( m + r ) - 2j R = ( 5+ 4) - 2x4= 1 (Once indeterminate) 6๐‘ฅ8 โˆ’ ๐‘Œ๐‘๐‘ฅ6 = 0 ๐‘Œ๐‘ = 8๐‘ก ฦฉ๐‘€๐ด = 0 ๐œƒ ๐‘†๐‘–๐‘›๐œƒ = 8 10 ๐‘๐‘œ๐‘ ๐œƒ = 6 10 6m 8m 10m 8m B 6t 6t 6m A 6t 8t 2t ๐œƒ ๐œƒ ๐œƒ ๐œƒ -8 -6 -6 10 No 8m 6m N1 B A 0 0 1t 1t ๐œƒ ๐œƒ 1.33 -1.67 1.33 1 ๐œƒ ๐น1 ๐น2 ๐น3 ๐น4 ๐น5 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 ฮด10 + X1 ฮด11 = 0 ฮด1t + X1 โ€™ ฮด11 = 0 X1 = 3.91 t ฮด1t = ๐›ผ. โˆ†๐‘ก. ๐›ด๐‘1. ๐ฟ Due to temperature ฮด1t = 1๐‘ฅ10โˆ’5 ๐‘ฅ100๐‘ฅ โˆ’ 6 = โˆ’0.006 X1 โ€™ = 3t -0.006 + X1 โ€™ ๐Ÿ—๐ŸŽ ๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ๐ŸŽ = 0 Xtot = X1 +๐‘ฟ๐Ÿ โ€ฒ = ๐Ÿ”. ๐Ÿ—๐Ÿ๐’• 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 7๐‘ก 7๐‘ก 0 No -7 -7 -6 45ยฐ -3 -3 1๐‘ก A 3m 3m 3m B N1 0 0 1๐‘ก -1 -1 A 3m 3m 3m B N2 1๐‘ก 1๐‘ก 0 0 0 โˆ’1 2 โˆ’1 2 โˆ’1 2 โˆ’1 2 1๐‘ก ฮด1t + ๐‘‹1 โ€ฒ ฮด11 + ๐‘‹2 โ€ฒ ฮด12 = 0 ฮด2t + ๐‘‹1 โ€ฒ ฮด21 + ๐‘‹2 โ€ฒ ฮด22 = 0 ฮด10 = 0 ฮด20 = 51.94 EA ฮด11 = 6 EA ฮด22 = 14.49 EA ฮด12 = 2.12 EA X1 = 1.34 ๐‘ก X2 = โˆ’3.78 ๐‘ก Nf = N0 + X1t N1 + X2t N2 Nf = N0 + 1.34 N1 -3.78 N2 6 45000 ๐‘‹1 โ€ฒ + 2.12 45000 ๐‘‹2 โ€ฒ = 0.003 2.12 ๐‘‹2 โ€ฒ + 14.49 ๐‘‹2 โ€ฒ = 0 Determine the internal forces in members for the shown truss due to the external loads and rise of temperature 50oC. ๐›ผ = 1๐‘ฅ10โˆ’5 /๐‘œ๐ถ Example 2 EA = 45000 t Due to loads Due to temperature X1t = X1 +๐‘ฟ๐Ÿ โ€ฒ = ๐Ÿ๐Ÿ“. ๐ŸŽ๐Ÿ’๐’• Total ฮด1t = ๐›ผ. โˆ†๐‘ก. ๐›ด๐‘1. ๐ฟ ฮด1t = 1๐‘ฅ10โˆ’5๐‘ฅ50๐‘ฅ(โˆ’1๐‘ฅ3๐‘ฅ2) = โˆ’0.003 ฮด2t = 1๐‘ฅ10โˆ’5 ๐‘ฅ50๐‘ฅ โˆ’1 2 ๐‘ฅ3๐‘ฅ4 + 1๐‘ฅ3 2๐‘ฅ2 = 0 X2t = X2 +๐‘ฟ๐Ÿ โ€ฒ = โˆ’๐Ÿ•. ๐Ÿ๐Ÿ“๐’• ๐‘‹1 โ€ฒ = 23.7 ๐‘ก ๐‘‹2 โ€ฒ = โˆ’3.47๐‘ก 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 ฮด10 + X1 ฮด11 = 0 Fabricated Fabrication errors No load Effect Loads & fabrication 8t ฮด1f + X1 ฮด11 = 0 ฮด1f + ฮด10 + X1 ฮด11 = 0 Loads: ฮด10 + X1 ฮด11 = 0 fabrication: ฮด1f + ๐‘ฟ๐Ÿ โ€ฒ ฮด11 = 0 Total: Xtot = X1 +๐‘ฟ๐Ÿ โ€ฒ ฮด1f = ฮฃโˆ†. ๐‘ต๐Ÿ OR ฮด10 + X1 ฮด11 + X2 ฮด12 = 0 ฮด20 + X1 ฮด21 + X2 ฮด22 = 0 ฮด1f + X1 ฮด11 + X2 ฮด12 = 0 ฮด2f + X1 ฮด21 + X2 ฮด22 = 0 ฮด1f + ฮด10 +X1 ฮด11 + X2 ฮด12 = 0 ฮด2f + ฮด20 +X1 ฮด21 + X2 ฮด22 = 0 10t Long (+) short (โˆ’) Tension (+) compression (โˆ’) ONCE TWICE ฮด1f = ฮฃโˆ†. ๐‘ต๐Ÿ ฮด2f = ฮฃโˆ†. ๐‘ต๐Ÿ 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) ๐น1 ๐น2 ๐น3 ๐น4 ๐น5 8m B 6t 6t 6m A 6t 8t 2t ๐œƒ ๐œƒ ๐œƒ ๐œƒ -8 -6 -6 10 No ๐œƒ -1.67 8m 6m N1 B A 0 0 1t 1t ๐œƒ ๐œƒ 1.33 -1.67 1.33 1 ๐œƒ -352 90 Due to loads ฮด10 + X1 ฮด11 = 0 X1 = 3.91 t Due to Fabrication ฮด1f + X1 โ€™ ฮด11 = 0 X1 โ€™ = 9.2t -0.018 + X1 โ€™ ๐Ÿ—๐ŸŽ ๐Ÿ’๐Ÿ“๐ŸŽ๐ŸŽ๐ŸŽ = 0 Xtot = X1 +๐‘ฟ๐Ÿ โ€ฒ = ๐Ÿ๐Ÿ‘. ๐Ÿ๐’• Total Nf = N0 + Xtot N1 Mem. L No N1 NoN1L N1N1L ฮ” ฮ”N1 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 (๐›ผ) OR (๐œƒ) Translation (โˆ†) ๐›ผ = 0 ๐›ผ โ‰  0 ๐›ผ = 0 โˆ†๐‘ฅ = 0 โˆ†๐‘ฆ = 0 ๐›ผ = ๏๏ โˆ†๐‘ฅ = 0 โˆ†๐‘ฆ = 0 ๐›ผ = ๏๏ โˆ†๐‘ฅ = ๏๏ โˆ†๐‘ฆ = 0 โˆ†= 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 ฮฑA + 2EI L ฮฑB 4EI L ฮฑB + 2EI L ฮฑA ๐‘€๐ด๐ต ๐น โˆ’ 6EI L2 โˆ† ๐‘€๐ต๐ด ๐น MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 โˆ† L ) MAB = MAB F + 4EI L ฮฑA + 2EI L ฮฑB โˆ’ 6EI L2 โˆ† ฮฑA ฮฑB โˆ’ 6EI L2 โˆ† โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซู…ุนโ€ฌ ๐‘€๐ต๐ถ ๐น ๐‘€๐ถ๐ต ๐น ฮฑB ฮฑC 4EI L ฮฑB + 2EI L ฮฑC 4EI L ฮฑC + 2EI L ฮฑB โˆ† โˆ’ 6EI L2 โˆ† โˆ’ 6EI L2 โˆ† โˆ† ฮจ ๐พ Slope Deflection Equations: MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = MBA F + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MBC = MBC F + 2EI L (2ฮฑB+ ฮฑC -3 ฮจ ) MCB = MCB F + 2EI L (2ฮฑC+ ฮฑB -3 ฮจ )
  • 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 โˆ’ ๐‘ค๐ฟ2 12 ๐‘ค๐ฟ2 12 ๐‘ค๐ฟ2 12 ๐‘ค๐ฟ2 12 ๐‘ƒ๐ฟ 8 ๐‘ƒ๐ฟ 8 โˆ’ ๐‘ƒ๐ฟ 8 ๐‘ƒ๐ฟ 8 โˆ’ Pa(Lโˆ’a) L Pa(Lโˆ’a) L Pa(Lโˆ’a) L Pa(Lโˆ’a) L L A B Pt Pt a a b โˆ’ 2 9 PL 2 9 PL 2 9 PL 2 9 PL L A B Pt Pt a a a โˆ’ Pa๐‘2 ๐ฟ2 Pb๐‘Ž2 ๐ฟ2 Pa๐‘2 ๐ฟ2 P๐‘๐‘Ž2 ๐ฟ2 L A B Pt a b โˆ’ ๐‘ค๐‘™2 20 ๐‘ค๐‘™2 30 ๐‘ค๐‘™2 30 L A B w t/m w๐‘™2 20 Mb(2aโˆ’b) L2 Ma(2bโˆ’a) L2 L A B M a b โˆ’ Mb(2aโˆ’b) L2 โˆ’ Ma(2bโˆ’a) L2 L A B M a b 6x4(2x1โˆ’4) 52 6x1(2x4โˆ’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 ๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = 0 โˆ’9 9 โˆ’3 3 ๐‘€๐ด๐ต ๐น = -9 mt, ๐‘€๐ต๐ด ๐น = 9 mt ๐‘€๐ต๐ถ ๐น = -3 mt, ๐‘€๐ถ๐ต ๐น = 3 mt 1) Unknowns ๐›ผ๐ต 2) Fixed End Moment 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ 2๐‘ฅ2๐ธ๐ผ 6 : 2๐‘ฅ๐ธ๐ผ 4 2 โˆถ 1.5 2๐ธ๐ผ ๐ฟ ๐ด๐ต : 2๐ธ๐ผ ๐ฟ ๐ต๐ถ 2 6 โˆถ 1 4 ร— 6 ร— 2 4 โˆถ 3 4) Slope deflection equations MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = MBA F + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MBC = MBC F + 2EI L (2ฮฑB+ ฮฑC -3 ฮจ ) MCB = MCB F + 2EI L (2ฮฑC+ ฮฑB -3 ฮจ ) = โˆ’9 + 4 ( 0 + ฮฑB - 0 ) = 9 + 4 ( 2ฮฑB + 0 - 0 ) = โˆ’3 + 3 ( 2ฮฑB + 0 - 0 ) = 3 + 3 ( 0 + ฮฑB - 0 ) = โˆ’9 + 4ฮฑB = 9 + 8ฮฑB = โˆ’3 + 6ฮฑB = 3 + 3ฮฑB 5) Compatibility eq. ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0 9 + 8ฮฑB +โˆ’3 + 6ฮฑB = 0 ฮฑB = โˆ’0.4286 ๐‘€๐ด๐ต = -10.71 ๐‘€๐ต๐ด = 5.57 ๐‘€๐ต๐ถ = -5.57 ๐‘€๐ถ๐ต = 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. ๐‘ค๐ฟ2 8 ๐‘ƒ๐ฟ 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 ๐›ผ๐ด = 0 ๐›ผ๐ต = ?? โˆ’9 9 โˆ’3 3 ๐‘€๐ด๐ต ๐น = -9 mt, ๐‘€๐ต๐ด ๐น = 9 mt ๐‘€๐ต๐ถ ๐น = -3 mt, ๐‘€๐ถ๐ต ๐น = 3 mt 1) Unknowns 2) Fixed End Moment 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ 2๐‘ฅ2๐ธ๐ผ 6 : 2๐‘ฅ๐ธ๐ผ 4 2 โˆถ 1.5 2๐ธ๐ผ ๐ฟ ๐ด๐ต : 2๐ธ๐ผ ๐ฟ ๐ต๐ถ 2 6 โˆถ 1 4 ร— 6 ร— 2 4 โˆถ 3 4) Slope deflection equations MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = MBA F + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MBC = MBC F + 2EI L (2ฮฑB+ ฮฑC -3 ฮจ ) MCB = MCB F + 2EI L (2ฮฑC+ ฮฑB -3 ฮจ ) = โˆ’9 + 4 ( 0 + ฮฑB - 0 ) = 9 + 4 ( 2ฮฑB + 0 - 0 ) = โˆ’3 + 3 ( 2ฮฑB + 0 - 0 ) = 3 + 3 ( 0 + ฮฑB - 0 ) = โˆ’9 + 4ฮฑB = 9 + 8ฮฑB = โˆ’3 + 6ฮฑB + 3ฮฑC = 3 + 3ฮฑB + 6ฮฑC 5) Compatibility eq. ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0 9 + 8ฮฑB โˆ’3 + 6ฮฑB+3ฮฑC = 0 ฮฑB = โˆ’0.36 ฮฑC = โˆ’0.32 ๐‘€๐ด๐ต = -10.44 ๐‘€๐ต๐ด = 6.12 ๐‘€๐ต๐ถ = -6.12 ๐‘€๐ถ๐ต = 0 6) Moment values 3 t/m 6m A B 10.44 6.12 6.12 Reactions F.E.M. ๐›ผ๐ถ = ?? C 3 t/m 2I I 6m A B 6t 2m 2m ๐›ผ๐ต , ๐›ผ๐ถ ๐‘€๐ถ๐ต = 0 3 + 3ฮฑB+6ฮฑC = 0 3ฮฑB+6ฮฑC = -3 โ†’ 2 14ฮฑB +3ฮฑC = -6 โ†’ 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. ๐‘ƒ๐ฟ 4 =6 ๐‘ค๐ฟ2 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 ฮฑA + 2EI L ฮฑB 4EI L ฮฑB + 2EI L ฮฑA ๐‘€๐ด๐ต ๐น โˆ’ 6EI L2 โˆ† ๐‘€๐ต๐ด ๐น MBC =๐‘€๐ต๐ถ ๐นโ€ฒ + 3EI L (ฮฑB โˆ’ โˆ† L ) MBC = ๐‘€๐ต๐ถ ๐นโ€ฒ + 3EI L ฮฑB โˆ’ 3EI L2 โˆ† ฮฑA ฮฑB โˆ’ 6EI L2 โˆ† โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซู…ุนโ€ฌ ๐‘€๐ต๐ถ ๐นโ€ฒ 3EI L ฮฑB โˆ† โˆ’ 3EI L2 โˆ† โˆ† ฮจ ๐พ Slope Deflection Equations: MAB =๐‘€๐ด๐ต ๐น + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = ๐‘€๐ต๐ด ๐น + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MBC = ๐‘€๐ต๐ถ ๐นโ€ฒ + 3EI L (ฮฑB - ฮจ ) C 3 t/m 2I I 6m A B 6t 2m 2m C 6t 2m 2m B ฮฑB C L B ๐‘€๐ด = ?? ๐‘€๐ถ = 0 ๐‘€๐ต = ??
  • 112. w t/m L A B โˆ’ ๐‘ค๐ฟ2 12 ๐‘ค๐ฟ2 12 L A B ๐‘€๐ด๐ต ๐น ๐‘€๐ต๐ด ๐น ๐‘€๐ด๐ต ๐นโ€ฒ L A B ๐‘€๐ด๐ต ๐น ๐‘€๐ต๐ด ๐น = โˆ’ 0.5 ๐‘€๐ด๐ต ๐นโ€ฒ L A B w t/m = โˆ’1.5 wL2 12 B Pt L/2 L/2 A โˆ’ ๐‘ƒ๐ฟ 8 ๐‘ƒ๐ฟ 8 ๐‘€๐ด๐ต ๐นโ€ฒ โˆ’ ๐‘ƒ๐ฟ 8 = ๐‘ฅ 1.5 B Pt L/2 L/2 A โˆ’ Pa๐‘2 ๐ฟ2 Pb๐‘Ž2 ๐ฟ2 L A B Pt a b L A B Pt a b ๐‘€๐ด๐ต ๐นโ€ฒ โˆ’ Pa๐‘2 ๐ฟ2 Pb๐‘Ž2 ๐ฟ2 = โˆ’ 0.5 ๐‘€๐ด๐ต ๐นโ€ฒ = L A B 4EI L ฮฑA + 2EI L ฮฑB 4EI L ฮฑB + 2EI L ฮฑA ฮฑA ฮฑB 3EI L ฮฑA ๐‘€๐ด๐ต ๐นโ€ฒ = โˆ’ 0.5 L A B โˆ’ 6EI L2 โˆ† (โˆ’ 6EI L2 โˆ†) โˆ’ 6EI L2 โˆ† โˆ’ 6EI L2 โˆ† โˆ† L A B L A B
  • 113. w t/m L A B โˆ’ ๐‘ค๐ฟ2 12 ๐‘ค๐ฟ2 12 L A B ๐‘€๐ด๐ต ๐น ๐‘€๐ต๐ด ๐น ๐‘€๐ด๐ต ๐นโ€ฒ L A B ๐‘€๐ด๐ต ๐น ๐‘€๐ต๐ด ๐น = โˆ’ 0.5 ๐‘€๐ด๐ต ๐นโ€ฒ L A B w t/m = โˆ’1.5 wL2 12 B Pt L/2 L/2 A โˆ’ ๐‘ƒ๐ฟ 8 ๐‘ƒ๐ฟ 8 ๐‘€๐ด๐ต ๐นโ€ฒ โˆ’ ๐‘ƒ๐ฟ 8 = ๐‘ฅ 1.5 B Pt L/2 L/2 A โˆ’ Pa๐‘2 ๐ฟ2 Pb๐‘Ž2 ๐ฟ2 L A B Pt a b L A B Pt a b ๐‘€๐ด๐ต ๐นโ€ฒ โˆ’ Pa๐‘2 ๐ฟ2 Pb๐‘Ž2 ๐ฟ2 = โˆ’ 0.5 6m A B 2 t/m 6t 2m 12 โˆ’6 6 -12 6m A B 2 t/m 6t 2m ๐‘€๐ด๐ต ๐นโ€ฒ = โˆ’6 6 -12 โˆ’ 0.5 ( + ) = -3 B
  • 114. Structural Analysis Slope deflection Method II Eng. Khaled El-Aswany (Modified) (Redo Example 2)
  • 115. ๐‘€๐ด๐ต ๐น = -9 mt, ๐‘€๐ต๐ด ๐น = 9 mt ๐‘€๐ต๐ถ ๐น = -4.5 mt 1) Unknowns 2) Fixed End Moment 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ 4 โˆถ 9 2 2๐ธ๐ผ ๐ฟ ๐ด๐ต : 3๐ธ๐ผ ๐ฟ ๐ต๐ถ 2๐‘ฅ2 6 โˆถ 3๐‘ฅ1 4 ๐›ผ๐ต 3 t/m 6m A B C 6t 2m 2m B ๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = ?? โˆ’9 9 โˆ’4.5 F.E.M. C 3 t/m 2I I 6m A B 6t 2m 2m ๐‘€๐ถ = 0 ๐‘€๐ด = ?? ๐‘€๐ต = ?? : 8 โˆถ 9 4) Slope deflection equations = โˆ’9 + 8 ( 0 + ฮฑB - 0 ) = 9 + 8 ( 2ฮฑB + 0 - 0 ) = โˆ’4.5 + 9 ( ฮฑB - 0) MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = MBA F + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MBC = ๐‘€๐ต๐ถ ๐นโ€ฒ + 3EI L (ฮฑB - ฮจ ) = โˆ’9 + 8ฮฑB = 9 + 16ฮฑB = โˆ’4.5 + 9ฮฑB 5) Compatibility eq. ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0 9 + 16ฮฑB โˆ’4.5 + 9ฮฑB = 0 ฮฑB = โˆ’0.18 ๐‘€๐ด๐ต = -10.44 ๐‘€๐ต๐ด = 6.12 ๐‘€๐ต๐ถ = -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. ๐‘ƒ๐ฟ 4 =6 ๐‘ค๐ฟ2 8
  • 116. ๐‘€๐ด๐ต ๐น = -9 mt, ๐‘€๐ต๐ด ๐น = 9 mt ๐‘€๐ต๐ถ ๐น = -4.5 mt 1) Unknowns 2) Fixed End Moment 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ 4 โˆถ 9 2 2๐ธ๐ผ ๐ฟ ๐ด๐ต : 3๐ธ๐ผ ๐ฟ ๐ต๐ถ 2๐‘ฅ2 6 โˆถ 3๐‘ฅ1 4 ๐›ผ๐ต 3 t/m 6m A B C 6t 2m 2m B ๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = ?? โˆ’9 9 โˆ’4.5 F.E.M. C 3 t/m 2I I 6m A B 6t 2m 2m ๐‘€๐ถ = 0 ๐‘€๐ด = ?? ๐‘€๐ต = ?? : 8 โˆถ 9 4) Slope deflection equations = โˆ’9 + 8 ( 0 + ฮฑB - 0 ) = 9 + 8 ( 2ฮฑB + 0 - 0 ) = โˆ’4.5 + 9 ( ฮฑB - 0) MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = MBA F + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MBC = ๐‘€๐ต๐ถ ๐นโ€ฒ + 3EI L (ฮฑB - ฮจ ) = โˆ’9 + 8ฮฑB = 9 + 16ฮฑB = โˆ’4.5 + 9ฮฑB 5) Compatibility eq. ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0 9 + 16ฮฑB โˆ’4.5 + 9ฮฑB = 0 ฮฑB = โˆ’0.18 ๐‘€๐ด๐ต = -10.44 ๐‘€๐ต๐ด = 6.12 ๐‘€๐ต๐ถ = -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. ๐‘ƒ๐ฟ 4 =6 ๐‘ค๐ฟ2 8 Type Equation D.O.F ฮฑA, ฮฑB , ฮจ ฮฑA, ฮจ Stiffness 2EI L 3EI L F.E.M. ๐‘€๐ด๐ต ๐นโ€ฒ = ๐‘€๐ด๐ต ๐น - 1 2 ๐‘€๐ต๐ด ๐น Summary MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = MBA F + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MAB = ๐‘€๐ด๐ต ๐นโ€ฒ + 3EI L (ฮฑA - ฮจ ) A B B A B A ๐‘€๐ด๐ต ๐นโ€ฒ A B ๐‘€๐ด๐ต ๐น ๐‘€๐ต๐ด ๐น
  • 117. Structural Analysis Slope deflection Method II Eng. Khaled El-Aswany (Example 3)
  • 118. ๐‘€๐ด๐ต ๐น = -9 mt, ๐‘€๐ต๐ด ๐น = 9 mt ๐‘€๐ต๐ถ ๐น = -3 mt, ๐‘€๐ถ๐ต ๐น = 3 mt 1) Unknowns 2) Fixed End Moment 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ 2๐‘ฅ2๐ธ๐ผ 6 : 2๐‘ฅ๐ธ๐ผ 4 2 โˆถ 1.5 2๐ธ๐ผ ๐ฟ ๐ด๐ต : 2๐ธ๐ผ ๐ฟ ๐ต๐ถ 2 6 โˆถ 1 4 ร— 6 ร— 2 4 โˆถ 3 4) Slope deflection equations MAB = MAB F + 2EI L (2ฮฑA+ ฮฑB -3 ฮจ ) MBA = MBA F + 2EI L (2ฮฑB+ ฮฑA -3 ฮจ ) MBC = MBC F + 2EI L (2ฮฑB+ ฮฑC -3 ฮจ ) MCB = MCB F + 2EI L (2ฮฑC+ ฮฑB -3 ฮจ ) = โˆ’9 + 4 ( 0 + ฮฑB - 0 ) = 9 + 4 ( 2ฮฑB + 0 - 0 ) = โˆ’3 + 3 ( 2ฮฑB + 0 - 0 ) = 3 + 3 ( 0 + ฮฑB - 0 ) = โˆ’9 + 4ฮฑB = 9 + 8ฮฑB = โˆ’3 + 6ฮฑB + 3ฮฑC = 3 + 3ฮฑB + 6ฮฑC 5) Compatibility eq. ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0 9 + 8ฮฑB โˆ’3 + 6ฮฑB+3ฮฑC = 0 ฮฑB = โˆ’0.84 ฮฑC = 1.92 ๐‘€๐ด๐ต = -12.36 ๐‘€๐ต๐ด = 2.28 ๐‘€๐ต๐ถ = -2.28 ๐‘€๐ถ๐ต = 12 6) Moment values 12.36 2.28 2.28 Reactions ๐›ผ๐ต , ๐›ผ๐ถ ๐‘€๐ถ๐ต + ๐‘€๐ถ๐ท = 0 3 + 3ฮฑB+6ฮฑC โˆ’12 = 0 3ฮฑB+6ฮฑC = 9 โ†’ 2 14ฮฑB +3ฮฑC = -6 โ†’ 1 ๐›ผ๐ด = 0 ๐›ผ๐ต = ?? ๐›ผ๐ถ = ?? 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 โˆ’9 9 โˆ’3 3 12 โˆ’12 F.E.M. ฮฃMc = 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. ๐‘ค๐ฟ2 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 ๐›ผ๐ต , ๐›ผ๐ถ 2) Fixed End Moment ๐›ผ๐ด = ?? ๐›ผ๐ต = ?? ๐›ผ๐ถ = ?? 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 ๐›ผ๐‘‘ = ?? ๐‘€๐ด = 12 ๐‘€๐ต = ?? ๐‘€๐ถ = ?? ๐‘€๐‘‘ = 0 โˆ’24 4 6t 6t C B 2m 2m 2m 3 t/m 17 โˆ’17 ๐‘€๐ต๐ด ๐นโ€ฒ = 4 mt ๐‘€๐ต๐ถ ๐น = -17 mt, ๐‘€๐ถ๐ต ๐น = 17 mt ๐‘€๐ถ๐ท ๐นโ€ฒ = -24 mt 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ถ๐ท 3๐ธ๐ผ ๐ฟ ๐ด๐ต : 2๐ธ๐ผ ๐ฟ ๐ต๐ถ : 3๐ธ๐ผ ๐ฟ ๐ถ๐ท ร— 4 : : 3๐‘ฅ1 6 2๐‘ฅ3 6 3๐‘ฅ2 8 : : 1 2 1 3 4 : : 2 4 3 : : 4) Slope deflection equations = 4 + 2 ( ฮฑB - 0 ) = โˆ’17 + 4 ( 2ฮฑB + ฮฑC - 0) ) = 17 + 4 ( 2ฮฑC + ฮฑB - 0) = โˆ’24 + 3 (ฮฑC - 0 ) MBC = ๐‘€๐ต๐ถ ๐น + 2EI L (2ฮฑB+ ฮฑC -3 ฮจ ) MCB = ๐‘€๐ถ๐ต ๐น + 2EI L (2ฮฑC+ ฮฑB -3 ฮจ ) MCD = ๐‘€๐ถ๐ท ๐นโ€ฒ + 3EI L (ฮฑC - ฮจ) MBA = ๐‘€๐ต๐ด ๐นโ€ฒ + 3EI L (ฮฑB - ฮจ) = 4 + 2ฮฑB = โˆ’17 + 8ฮฑB + 4ฮฑC = 17 + 4ฮฑB + 8ฮฑC = โˆ’24 + 3ฮฑC 5) Compatibility eq. ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ = 0 ฮฑB = 1.2234 ฮฑC = 0.1915 ๐‘€๐ถ๐ต + ๐‘€๐ถ๐ท = 0 4ฮฑB+11ฮฑC = 7 โ†’ 2 10ฮฑB +4ฮฑC = 13 โ†’ 1 ฮฃMc = 0 6) Moment Values MBC = โˆ’6.45 ๐‘š๐‘ก MCB = 23.43 ๐‘š๐‘ก MCD = โˆ’23.43 ๐‘š๐‘ก MBA = 6.45 ๐‘š๐‘ก 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 โœ“ Can translate ? X Symmetry ? ฮจ ฮจ Frame Without Sway Symmetry No translation With Sway ฮ” ฮ”
  • 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 ๐›ผ๐ต , ๐›ผ๐ถ 12m 8m
  • 126. Delete 1) Unknowns ๐›ผ๐ต , ๐›ผ๐ถ 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 ๐‘€๐ด๐ต ๐น = -18 mt, ๐‘€๐ต๐ด ๐น = 18 mt ๐‘€๐ต๐ถ ๐น = -8 mt, ๐‘€๐ถ๐ต ๐น = 8 mt ๐‘€๐ต๐ท ๐น = -5.33 mt, ๐‘€๐ท๐ต ๐น = 2.67 mt เดฅ ๐‘€๐ถ๐ธ ๐น = -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 ๐›ผ๐ต , ๐›ผ๐ถ 2) Fixed End Moment ๐‘€๐ด๐ต ๐น = -18 mt, ๐‘€๐ต๐ด ๐น = 18 mt ๐‘€๐ต๐ถ ๐น = -8 mt, ๐‘€๐ถ๐ต ๐น = 8 mt ๐‘€๐ต๐ท ๐น = -5.33 mt, ๐‘€๐ท๐ต ๐น = 2.67 mt เดฅ ๐‘€๐ถ๐ธ ๐น = -4.5 mt 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ท โˆถ ๐พ๐ถ๐ธ 2๐‘‹6 12 : 2๐‘‹4 8 : 2๐‘‹1.5 6 : 3๐‘‹1 6 1 โˆถ 1 โˆถ 1 2 : 1 2 2 โˆถ 2 โˆถ 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 ๐›ผ๐ต , ๐›ผ๐ถ 2) Fixed End Moment ๐‘€๐ด๐ต ๐น = -18 mt, ๐‘€๐ต๐ด ๐น = 18 mt ๐‘€๐ต๐ถ ๐น = -8 mt, ๐‘€๐ถ๐ต ๐น = 8 mt ๐‘€๐ต๐ท ๐น = -5.33 mt, ๐‘€๐ท๐ต ๐น = 2.67 mt เดฅ ๐‘€๐ถ๐ธ ๐น = -4.5 mt 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ท โˆถ ๐พ๐ถ๐ธ 2๐‘‹6 12 : 2๐‘‹4 8 : 2๐‘‹1.5 6 : 3๐‘‹1 6 1 โˆถ 1 โˆถ 1 2 : 1 2 2 โˆถ 2 โˆถ 1 : 1 4) Slope Deflection Equation ๐‘€๐ด๐ต = -18 + 2 (2๐›ผ๐ด + ๐›ผ๐ต) ๐‘€๐ต๐ด = 18 + 2 (2๐›ผ๐ต + ๐›ผ๐ด) ๐‘€๐ต๐ถ = -8 + 2 (2๐›ผ๐ต + ๐›ผ๐ถ) ๐‘€๐ถ๐ต = 8 + 2 (2๐›ผ๐ถ + ๐›ผ๐ต) ๐‘€๐ต๐ท = -5.33 + (2๐›ผ๐ต + ๐›ผ๐ท) ๐‘€๐ท๐ต = 2.67 + (2๐›ผ๐ท + ๐›ผ๐ต) ๐‘€๐ถ๐ธ = -4.5 + (๐›ผ๐ถ) = -18 + 2 ๐›ผ๐ต = 18 + 4๐›ผ๐ต = -8 + 4๐›ผ๐ต + 2๐›ผ๐ถ = 8 + 4๐›ผ๐ถ + 2๐›ผ๐ต = -5.33 + 2๐›ผ๐ต = 2.67 + ๐›ผ๐ต = -4.5 + ๐›ผ๐ถ
  • 129. Delete 1) Unknowns ๐›ผ๐ต , ๐›ผ๐ถ 2) Fixed End Moment ๐‘€๐ด๐ต ๐น = -18 mt, ๐‘€๐ต๐ด ๐น = 18 mt ๐‘€๐ต๐ถ ๐น = -8 mt, ๐‘€๐ถ๐ต ๐น = 8 mt ๐‘€๐ต๐ท ๐น = -5.33 mt, ๐‘€๐ท๐ต ๐น = 2.67 mt เดฅ ๐‘€๐ถ๐ธ ๐น = -4.5 mt 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ท โˆถ ๐พ๐ถ๐ธ 2๐‘‹6 12 : 2๐‘‹4 8 : 2๐‘‹1.5 6 : 3๐‘‹1 6 1 โˆถ 1 โˆถ 1 2 : 1 2 2 โˆถ 2 โˆถ 1 : 1 4) Slope Deflection Equation ๐‘€๐ด๐ต = -18 + 2 (2๐›ผ๐ด + ๐›ผ๐ต) ๐‘€๐ต๐ด = 18 + 2 (2๐›ผ๐ต + ๐›ผ๐ด) ๐‘€๐ต๐ถ = -8 + 2 (2๐›ผ๐ต + ๐›ผ๐ถ) ๐‘€๐ถ๐ต = 8 + 2 (2๐›ผ๐ถ + ๐›ผ๐ต) ๐‘€๐ต๐ท = -5.33 + (2๐›ผ๐ต + ๐›ผ๐ท) ๐‘€๐ท๐ต = 2.67 + (2๐›ผ๐ท + ๐›ผ๐ต) ๐‘€๐ถ๐ธ = -4.5 + (๐›ผ๐ถ) = -18 + 2 ๐›ผ๐ต = 18 + 4๐›ผ๐ต = -8 + 4๐›ผ๐ต + 2๐›ผ๐ถ = 8 + 4๐›ผ๐ถ + 2๐›ผ๐ต = -5.33 + 2๐›ผ๐ต = 2.67 + ๐›ผ๐ต = -4.5 + ๐›ผ๐ถ 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 ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ + ๐‘€๐ต๐ท = 0 ฮฃMc = 0 ๐‘€๐ถ๐ต + ๐‘€๐ถ๐ธ + ๐‘€๐ถ๐น = 0 10๐›ผ๐ต + 2๐›ผ๐ถ = -4.67 2๐›ผ๐ต + 5๐›ผ๐ถ = 8.5 ๐›ผ๐ต = -0.87717 ๐›ผ๐ถ = 2.05087 ๐‘€๐ด๐ต = -19.75 ๐‘€๐ต๐ด = 14.49 ๐‘€๐ต๐ถ = -7.4 ๐‘€๐ถ๐ต = 14.45 ๐‘€๐ต๐ท = -7.09 ๐‘€๐ท๐ต = 1.80 ๐‘€๐ถ๐ธ = -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 ๐›ผ๐ต , ๐›ผ๐ถ , ๐›ผ๐ท , ๐›ผ๐ธ 2) Fixed End Moment ๐›ผ๐ท = - ๐›ผ๐ถ ๐›ผ๐ธ = - ๐›ผ๐ต 3) Relative Stiffness ๐พ๐ด๐ต โˆถ ๐พ๐ต๐ถ โˆถ ๐พ๐ต๐ธ โˆถ ๐พ๐ถ๐ท 6 โˆถ 2 โˆถ 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 ๐‘€๐ต๐ด = 0 + 6 (๐›ผ๐ต) ๐‘€๐ต๐ถ = 0 + 2 (2๐›ผ๐ต + ๐›ผ๐ถ) ๐‘€๐ถ๐ต = 0 + 2 (2๐›ผ๐ถ + ๐›ผ๐ต) ๐‘€๐ต๐ธ = -16 + 2.5(2๐›ผ๐ต + ๐›ผ๐ธ) ๐‘€๐ถ๐ท = -32 + 3.75(2๐›ผ๐‘ + ๐›ผ๐ท) = 6๐›ผ๐ต = 4๐›ผ๐ต + 2๐›ผ๐ถ = 4๐›ผ๐ถ + 2๐›ผ๐ต = -16 + 2.5(2๐›ผ๐ต - ๐›ผ๐ต) = -32 + 3.75(2๐›ผ๐‘ - ๐›ผ๐‘) = -16 + 2.5๐›ผ๐ต = -32 + 3.75๐›ผ๐‘ 5) Compatibility Equations ฮฃMb = 0 ๐‘€๐ต๐ด + ๐‘€๐ต๐ถ + ๐‘€๐ต๐ธ = 0 ฮฃMc = 0 ๐‘€๐ถ๐ต + ๐‘€๐ถ๐ท = 0 12.5๐›ผ๐ต + 2๐›ผ๐ถ = 16 2๐›ผ๐ต + 7.75๐›ผ๐ถ = 32 ๐›ผ๐ต = 0.664603 ๐›ผ๐ถ = 3.962315 6) Moment values ๐‘€๐ต๐ด = 3.88 ๐‘€๐ต๐ถ = 10.51 ๐‘€๐ถ๐ต = 17.14 ๐‘€๐ต๐ธ = -14.38 ๐‘€๐ถ๐ท = -17.14 C 3 t/m A B 5m 8m 2I 2I F E D 5m ๐œถ๐‘ฉ =?? ๐œถ๐‘ช =?? 6 t/m 2I I I 3I ๐œถ๐‘ฌ =?? ๐œถ๐‘ซ =?? C A B F E D
  • 132. Structural Analysis Slope deflection Method II Eng. Khaled El-Aswany (Frames) (With Sway)
  • 133.
  • 134.
  • 135.
  • 136.
  • 137.
  • 138.
  • 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 ๐‘ค๐ฟ2 12 โˆ’ ๐‘ค๐ฟ2 12 ๐‘ƒ๐ฟ 8 โˆ’ ๐‘ƒ๐ฟ 8 โˆ’ ๐‘ƒ๐‘Ž๐‘2 ๐ฟ2 ๐‘ƒ๐‘๐‘Ž2 ๐ฟ2 L/2 Pt L/2 ๐‘ค๐ฟ2 30 โˆ’ ๐‘ค๐ฟ2 20 ๐‘ƒ๐‘Ž(๐ฟ โˆ’ ๐‘Ž) ๐ฟ โˆ’ ๐‘ƒ๐‘Ž(๐ฟ โˆ’ ๐‘Ž) ๐ฟ L ๐‘€๐‘Ž(2๐‘ โˆ’ ๐‘Ž) ๐ฟ2 ๐‘€๐‘(2๐‘Ž โˆ’ ๐‘) ๐ฟ2 w t/m L โˆ’ ๐‘ค๐ฟ2 12 ๐’™๐Ÿ. ๐Ÿ“ โˆ’ ๐‘ƒ๐ฟ 8 ๐’™๐Ÿ. ๐Ÿ“ โˆ’ ๐‘ƒ๐‘Ž๐‘2 ๐ฟ2 L/2 Pt L/2 a Pt b L L Pt Pt a a b M a b L Pt a b L โˆ’ ๐ŸŽ. ๐Ÿ“ ๐ฑ ๐‘ท๐’ƒ๐’‚๐Ÿ ๐‘ณ๐Ÿ
  • 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 4๐‘ฅ๐ธ๐‘ฅ3 6 : 4๐‘ฅ๐ธ๐‘ฅ2 4 4๐‘ฅ๐ธ๐‘ฅ2 4 : 4๐‘ฅ๐ธ๐‘ฅ1 6 1 2 : 1 2 1 2 : 1 6 6 : 2 ๐Ÿ‘ ๐Ÿ’ : ๐Ÿ ๐Ÿ’ 3 : 1 1 2 1 2 + 1 2 : 1 2 1 2 + 1 2 ๐Ÿ ๐Ÿ : ๐Ÿ ๐Ÿ L L K = ๐‘ฐ ๐‘ณ K = ๐‘ฐ ๐‘ณ ๐’™ ๐Ÿ‘ ๐Ÿ’ C 3 t/m 6m A B 6t 2m 2m 3 t/m 6m D 3I 2I 3I K = ๐‘ฐ ๐‘ณ ๐’™ ๐Ÿ ๐Ÿ L ๐‘ฒ๐‘จ๐‘ฉ= ๐‘ฐ ๐‘ณ ๐‘ฒ๐‘ฉ๐‘ช= ๐‘ฐ ๐‘ณ ๐’™ ๐Ÿ ๐Ÿ Example for Symmetry Beam: (Symmetry)
  • 145. Fixed end moment Distribution Factor Carry over L A B L B C D.M. D.M. ๐Ÿ ๐Ÿ D.M. ๐ŸŽ x ๐Ÿ ๐Ÿ
  • 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 โˆ’ ๐‘ค๐ฟ2 12 = ๐‘ค๐ฟ2 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 ๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“ D.M. -6 -2.4 D.F. FEM 1 - โ€ซู†ุฌู…ุนโ€ฌ 2 - โ€ซุงุงู„ุดุงุฑุฉโ€ฌ โ€ซู†ุบูŠุฑโ€ฌ 3 - โ€ซููŠโ€ฌ โ€ซู†ุถุฑุจโ€ฌ โ€ซุงู„ุชูˆุฒูŠุนโ€ฌ โ€ซู†ุณุจโ€ฌ 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 ๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“ D.M. -6 -2.4 -3.6 D.F. 1 - โ€ซู†ุฌู…ุนโ€ฌ 2 - โ€ซุงุงู„ุดุงุฑุฉโ€ฌ โ€ซู†ุบูŠุฑโ€ฌ 3 - โ€ซููŠโ€ฌ โ€ซู†ุถุฑุจโ€ฌ โ€ซุงู„ุชูˆุฒูŠุนโ€ฌ โ€ซู†ุณุจโ€ฌ 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 ๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“ 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 ๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“ 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 โ€ซุณุงู„ุจโ€ฌ ( โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซุนูƒุณโ€ฌ ) โ€ซู…ูˆุฌุจโ€ฌ ( โ€ซุงู„ุณุงุนุฉโ€ฌ โ€ซุนู‚ุงุฑุจโ€ฌ โ€ซู…ุนโ€ฌ ) 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 ๐Ÿ/๐Ÿ“ ๐Ÿ‘/๐Ÿ“ 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 ๐Ÿ–/๐Ÿ๐Ÿ• 9/๐Ÿ๐Ÿ• 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 ๐‘ฅ 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 ๐‘ฅ 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 โˆ’4 4 1t 2m C 2 t/m B 6m 6 โˆ’6 โˆ’2 โˆ’0.5 ( )= -8 FEM โˆ’6 โˆ’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 โˆ’4 4 C 2 t/m B 6m 2m C B 1t 6m 2m โˆ’9 2 1 1 โˆ’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 โˆ’4 4 โˆ’8 4 -8 -4 FEM ๐ŸŽ. ๐Ÿ“ 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 ๐‘ฅ 3 4 I L : I L ๐‘ฅ 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 โˆ’25 โˆ’15 15 0 0 Joint B KBA : KBC Joint C KCB : KCD ๐ผ ๐ฟ : ๐ผ ๐ฟ 1 10 : 2 15 15 : 20 ๐Ÿ‘ ๐Ÿ• : ๐Ÿ’ ๐Ÿ• 25 โˆ’15 15 0 0 โˆ’25 15 35 : 20 35 ๐ผ ๐ฟ : ๐ผ ๐ฟ 2 15 : 1 10 20 : 15 ๐Ÿ’ ๐Ÿ• : ๐Ÿ‘ ๐Ÿ• 20 35 : 15 35 3/7 4/7 4/7 3/7 โˆ’5.714 โˆ’8.55 โˆ’6.45 A B C D FEM D.M.1 D.F. C.O.1 โˆ’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 โˆ’0.465 โˆ’0.349 โˆ’0.525 โˆ’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 โˆ’0.038 โˆ’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 ๐‘ค๐‘™2 8 26.33 + 22.32 2 13.175 =37.5 =24.32 ๐‘ƒ๐ฟ 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 โˆ’3 0 FEM โˆ’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 ๐‘ฅ 1 2 I L : I L ๐‘ฅ 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 ๐‘ƒ๐ฟ 8 ๐‘ฅ1.5 = 9 3m 3m A 8t 2m B 6t 6t C B 2m 2m 2m โˆ’6 3m 3m A 2m 6t B 12 6 6t 6t C B โˆ’17 17 2m 2m 2m 3 t/m C B 2m 2m 2m 3 t/m ๐‘ค๐ฟ2 12 = 9 โˆ’ ๐‘ค๐ฟ2 12 = -9 โˆ’ ๐‘ƒ๐‘Ž(๐ฟโˆ’๐‘Ž) ๐ฟ = -8 ๐‘ƒ๐‘Ž(๐ฟโˆ’๐‘Ž) ๐ฟ = 8
  • 165. 3 t/m 6t 6t C 3 t/m B 8m D 3I 2I C B 3I D 2I C 8m 3 โˆ’17 17 -16 16 Joint B Joint C KCB : KCD ๐ผ ๐ฟ ๐‘ฅ 3 4 : ๐ผ ๐ฟ 3 -17 17 -16 16 ๐ผ ๐ฟ : ๐ผ ๐ฟ 1 2 : 1 4 2 ๐ŸŽ. ๐Ÿ ๐ŸŽ. ๐Ÿ– 2/๐Ÿ‘ 1/๐Ÿ‘ 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 ๐Ÿ ๐Ÿ‘ : ๐Ÿ ๐Ÿ‘ KBA : KBC 1 6 ๐‘ฅ 3 4 : 3 6 1 8 : 1 2 8 : 2 4 : 1 ๐Ÿ’ ๐Ÿ“ : ๐Ÿ ๐Ÿ“ 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 โˆ’12 โˆ’12 โˆ’3๐ธ๐ผโˆ† ๐ฟ2 โˆ† โˆ† โˆ† 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 โˆ’6๐ธ๐ผโˆ† ๐ฟ2 โˆ’6๐ธ๐ผโˆ† ๐ฟ2 3 t/m 10m A I B 25 โˆ’25 13 โˆ’37 C B 8t 7.5m 7.5m 2I C B 15m 2I 2 cm 1 cm ฮ” 5.33 15 โˆ’15 5.33 20.33 โˆ’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 โˆ’37 20.33 โˆ’9.67 3 Joint B KBA : KBC Joint C KCB : KCD ๐ผ ๐ฟ : ๐ผ ๐ฟ 1 10 : 2 15 15 : 20 ๐Ÿ‘ ๐Ÿ• : ๐Ÿ’ ๐Ÿ• 15 35 : 20 35 ๐ผ ๐ฟ : ๐ผ ๐ฟ ๐‘ฅ 3 4 2 15 : 1 10 ๐‘ฅ 3 4 80 : 45 ๐ŸŽ. ๐Ÿ”๐Ÿ’ : ๐ŸŽ. ๐Ÿ‘๐Ÿ” 80 125 : 45 125 13 โˆ’9.67 20.33 3 0 โˆ’37 3/7 4/7 ๐ŸŽ. ๐Ÿ”๐Ÿ’ ๐ŸŽ. ๐Ÿ‘๐Ÿ” โˆ’1.903 โˆ’14.93 โˆ’8.399 A B C D FEM D.M.1 D.F. C.O.1 โˆ’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 โˆ’0.174 โˆ’0.13 โˆ’0.768 โˆ’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 โˆ’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 ๐‘ฅ 1 2 I L : I L ๐‘ฅ 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 ๐‘ค๐‘™2 8 + ๐‘ƒ๐ฟ 4 = 22
  • 171. Structural Analysis Moment Distribution II Eng. Khaled El-Aswany (Frames) (Example 2)
  • 172. 4t A 3 t/m B C 8m โˆ’16 16 4t A 3 t/m B C 3m 8m 3m B โˆ’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 ๐ŸŽ. ๐Ÿ“ : ๐ŸŽ. ๐Ÿ“ 1 8 : 1 6 ๐‘ฅ 3 4 I L : I L ๐‘ฅ 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 โˆ’16 16 4t C B โˆ’4.5 3t B โˆ’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 ๐ŸŽ. ๐Ÿ“ : ๐ŸŽ. ๐Ÿ“ 1 8 : 1 6 ๐‘ฅ 3 4 I L : I L ๐‘ฅ 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 ๐‘๐‘™ 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