Contoh Pada Continous Beam
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Transcript of Contoh Pada Continous Beam
![Page 1: Contoh Pada Continous Beam](https://reader036.fdokumen.com/reader036/viewer/2022082501/547693e9b4af9f71588b4621/html5/thumbnails/1.jpg)
PENYELESAIAN DISPLACEMENT PADA BEAMDENGAN METODE FLEXIBILITAS DAN METODE KEKAKUAN
1. Diselesaikan dengan metode Gaya (Flexibilitas).
12 Ton 2 Ton/m’
2 A
B C
24m 4m 4m
2 t/m 12 T
96 Tm 96 Tm 12 Tm 12 Tm
24 T 24 T 6 T 6 T
WL
2 = = 24 T
P
2 =
12
2 = 6 T.
WL2
12 =
2 24
12
2. = 96 Tm.
PL
8 =
12 8
8
. = 12 Tm.
24 T 30 T 6 T
96 Tm 84 Tm 12 Tm
3 L = 24 m L = 8 m
Karena berupa ‘Continous Beam’, maka translasi aksial dapat diabaikan, sehingga
didapat Matrix Flexibilitas sebagai berikut :
Batang 1 (panjang 3L).
FM1 =
L
EI
L
EIL
EI
L
EI
3 2
23 2
2
= 54 27
27 18
2L L
L
L
EI6
Batang 2 (panjang L)
FM2 =
L
EI
L
EIL
EI
L
EI
3 2
23 2
2
= 2 3
3 6
2L L
L
L
EI6
Matrix Flexibilitas terhadap sumbu global
![Page 2: Contoh Pada Continous Beam](https://reader036.fdokumen.com/reader036/viewer/2022082501/547693e9b4af9f71588b4621/html5/thumbnails/2.jpg)
FM =
54 27 0 0
27 18 0 0
0 0 2 3
0 0 3 6
2
2
L L
L
L L
L
L
EI6
Matrix kesetimbangan (Matrix transformasi).
BMS = B BMJ MQ =
1 0 1 1
0 0
0 1 0 1
0 0 0
L L
L
1
L
Matrix Reaksi Tumpuan.
BRS = B BRJ RQ = 1 0 1 1
1 1 1 2
0 1 0 1
1
L
Matrix Flexibilitas FS = BMS
T . FM . BMS
FS =
1 0 0
0 0 1
1 0 0 0
1 1 0
L
L
L
1
L
54 27 0 0
27 18 0 0
0 0 2 3
0 0 3 6
2
2
L L
L
L L
L
L
EI6
1 0 1 1
0 0
0 1 0 1
0 0 0
L L
L
1
L
=
18 0 27 18
0 2 0 1
27 0 54 27
18 1 27 20
L
EI6= F F
F FJJ JQ
QJ QQ
AQ = FQQ-1 (DQ - FQJ . AJ)
AQ = - 6
351
EI
L
20 27
27 54
27 0
18 1 L
EI6
84
12
=
12
60
Reaksi Tumpuan
AR = - ARC + BRJ . AJ + BRQ . AQ
AR = -
24
30
6 +
1 0
1 1
0 1
1
L
84
12
+
1 1
1 2
0 1
1
L
12
60
=24
30
6
+
10 5
9
1 5
,
, +
6
13 5
7 5
,
, =
28 5
34 5
3
,
,
(Ton)
(AQ)akhir = AQF + AQ =
96
12 +
12
60
=
84
72
Perpindahan titik buhul
DJ = FJJ AJ + FJQ AQ = 18 0
0 2
L
EI6
82
12
+
27 18
0 1
L
EI6
84
72 =
0 000946
0 0000183
,
,
Gaya Ujung Batang
AM = AMF + BMJ AJ + BMQ AQ
=
24
96
6
12
+
1 0
0
0 1
0
L
L
1
L
82
12
+
1 1
0 0
0 1
0 0
1
L
12
60
![Page 3: Contoh Pada Continous Beam](https://reader036.fdokumen.com/reader036/viewer/2022082501/547693e9b4af9f71588b4621/html5/thumbnails/3.jpg)
=
24
96
6
12
+
10 25
82
1 5
12
,
, +
9
0
7 5
0
,
=
22 72
14
12
0
,
2. Diselesaikan dengan Metode Kekakuan.
Kekakuan Batang 1 terhadap sumbu lokal dan global (panjang = 3L)
KL =
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
3 2 3 2
2 2
3 2 3 2
2 2
L L L L
L L L L
L L L L
L L L L
EI =
4 6 4 6
6 12 6 6
4 6 4 6
6 6 6 12
2 2
2 2
L L
L L L L
L L
L L L L
EI
L9 3
T = Matrix Identitas 4 x 4
T T = Matrix Identitas 4 x 4
Kg = T T KL T = KL
Kekakuan Batang 2 terhadap sumbu lokal dan global (panjang = L)
KL =
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
3 2 3 2
2 2
3 2 3 2
2 2
L L L L
L L L L
L L L L
L L L L
EI = EI
L9 3
T = Matrix Identitas 4 x 4
T T = Matrix Identitas 4 x 4
Kg = T T KL T = KL
Overall matrix Kekakuan
d1 d2 d3
108 54 108 54
54 36 54 18
108 54 108 54
54 18 54 36
2 2
2 2
L L
L L L L
L L
L L L L
4 6 4 6
6 12 6 6
4 6 4 6
6 6 6 12
2 2
2 2
L L
L L L L
L L
L L L L
![Page 4: Contoh Pada Continous Beam](https://reader036.fdokumen.com/reader036/viewer/2022082501/547693e9b4af9f71588b4621/html5/thumbnails/4.jpg)
g
m
g
g
1
1
2
3
84
12
=
4 6 4 6 0 0
6 12 6 6 0 0
4 6 112 48 108 54
6 6 48 48 54 18
0 0 108 54 108 54
0 0 54 18 54 36
2 2
2 2 2
2 2
L L
L L L L
L L L
L L L L L L
L L
L L L L
EI
L9 3
0
0
0
02
3
Matrix kekakuan yang sudah ditata.
d1 d2 d3
84
12
1
1
2
3
g
m
g
g
=
48 18 6 6 48 54
18 36 0 0 54 54
6 0 4 6 4 0
6 0 6 12 6 0
48 54 4 6 112 108
54 54 0 0 108 108
2 2 2
2 2
2 2
L L L L L L
L L L L
L L
L L L L
L L L
L L
EI
L9 3
2
3
0
0
0
0
Displacement dan Reaction
2
3
=
EI
L9 3
48 18
18 36
2 2
2 2
1L L
L L
84
12
=
EI
L9
48 18
18 36
1
84
12
2
3
=
9
1404
L
EI
36 18
18 48
84
12
=
0 006428
0 002142
,
,