slope deflection bergoyang
TRANSCRIPT
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BAN NO.3
CONTOH SOAL Kontr uksi Bergoyang
I
I
1). Hitung Nilai K
EI
15
2 EI
12
EI
10
2). Menentukan Momen Frimer (MF)
12 X 9 X 6 2
15 2
-12 X 9 2 X 6
15 2
24 X 12
8
= - = 0
3). Tentukan nilai R ( ∆ )
∆
15
∆
10
30
30
30
= 17.28
= -25.92
MFDC
=
=
3 R
R BC
R CD
=
MFCD
X =
0
=
MFAB =
2 R
3
K BC =
x 30
R AB = x 30
=
K CB =
SOLUSI
5
K BD = K DB = X =
36 Kip ft
Kip ft
K AB = K BA = X = 2
Kip ft
MFBC = - MFCB = =
MFBA
SLOFE DEFLECTION
2I
10 M
24 k
A
B C
D
6 M 6 M
12 k
6 M
9 M
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Masukkan kembali nilai yang didapat kedalam rumus Slope Deflection
= + K ( -2 ΦA - ΦB + 2 R )
= + 2 { ( -2 x ) - ( + ( 2 x ) }
= + 2 ( + )
= + 2 ( )
= +
=
= + K ( -2 ΦB - ΦA + 2 R )
= + 2 { ( -2 x ) - ( ) + 2 x ) }
= + 2 ( + )
= + 2 ( )
= +
=
= + K ( -2 ΦB - ΦC + 0 R )
= + 5 { ( -2 x ) - ( ) + 0 x ) }
= + 5 ( + )
= + 5 ( )
= +
=
= + K ( -2 ΦC - ΦB + 0 R )
= + 5 { ( -2 x ) - ( ) + 0 x ) }
= + 5 ( + )
= + 5 ( )
= +
=
= + K ( -2 ΦC - ΦD + 3 R )
= + 3 { ( -2 x ) - ( ) + 3 x ) }
= + 3 ( + )= + 3 ( )
= +
=
= + K ( -2 ΦD - ΦC + 3 R )
= + 3 { ( -2 x ) - ( ) + 3 x ) }
= + 3 ( + )
= + 3 ( )
= +
=
-MFDC
57.311
27.28
19.104
0.0 0
0.00
0.0 -54.55
0.0 0.00
57.311
27.275 2.76
-2.76
-1.99
19.104
38.207
19.104
38.207
1.99
73.545
-36 -5.521
2.7604
19.104
0
19.104
0
19.104
1.99
-27.28
-25.92 -3.976
-25.92
-7.51
-36 2.7604
-36
MDC
17.28 -8.64
17.28 -17.28
0.00
-25.92
0.0 -5.521
0.0
-25.92 23.603
-2.32
36
17.28 22.43
1.9881
17.28 -44.86 -1.99
2.76
36 -3.976 -2.76
MCD MFCD
MBC MFBC
1.9881 22.43
-22.43
MCB -MFCB
36 -6.74
36
24.51
73.54
0.0
MAB MFAB
2.32
MBA -MFBA
-33.68
11.80
-36 -37.54
-73.54
0.0
0.0
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8). Lakukaan pemotongan pada masing-masing titik ( Free body )
B 2I C
I
I
a). potongan A & B
Kf
HB = Kip
HA = Kip
1. Reaksi perletakan
= 0
- x 15 + P x 9 + = 0
- 15 HB + 12 x 9 + = 0- 15 HB + + = 0
- 15 HB + = 0
= / -15
= Kip ( )
= 0
HA x 15 - P x 6 + = 0
15 HA - 12 x 6 + = 0
15 HA - +
15 HA = 0
= / 15
= Kip ( )
2). Momen dan Lintang
= x X = x X - P ( X - 9 )
= X = X - X +
= X +
= =
73.542.32
4.65
2.32
7.35
69.7
4.65
2.32
0
ΣHA
HB
0
12
2.32
73.54
-7.35 108
-7.35
108 2.32
110.3
2.32
2.32
72 2.32
-110.3
4.65
-69.7
ΣHB
Mx 4.65
Interval 0 ≤ x ≤ 9
Mx
108
Interval 9 ≤ x ≤ 15
DxDx 4.65
2.32
0
4.65
7.35
4.65
RHA
RHA
RHB
RHB
D
A
24 k
6 6
12 k
6 ft
9 ft
B
C
10 ft
12 k
6 ft
9 ft
A
B
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Tabel Momen dan Lintang pada potongan A & B
b. Potongan B & C
24 K
B C
= K = K
1. Reaksi perletakan
= 0- x 12 + P x 6 + - = 0
- 12 RC + 24 x 6 + - = 0
- 12 RC + + - = 0
- 12 RC + = 0
= / -12
= K ( )
= 0
R B x 12 - P x 6 - + =
12 R B - 24 x 6 - + = 0
12 R B - - + = 0
12 R B = 0
= /
= K ( )
2. Momen dan Lintang
= x X - =
= X -
73.54
73.54
6
2.32
2.32
2.32
Nilai Momen
Kip ft
41.81
Kip ft
23.23
6.06
6
Nilai LintangX Rumus Momen Rumus Lintang
0 0.00
9
Kip
Kip ft
Kip
2 9.29 Kip ft 4.65 Kip
1
Kip ft 4.65 Kip
4.65
13.94
Kip ft
Kip ft 4.65
37.16 Kip ft 4.65
7 32.52
4.65
4.65
4 18.58
Kip
Kip
Kip
4.65
4.65
4.65
Kip
6 27.87 Kip ft
13 12.39 Kip ft
9 41.81 Kip ft
8
10
12
Kip ft
Kip ft
27.10
4.65
Kip
Kip ft
Kip
Kip ft Kip
Kip ft
Kip
-7.35
Kip
-7.35
-2.32
Kip
Kip ft Kip
-7.35
11
14
3
5
-7.35 Kip
-7.35
-7.35
Kip
-7.35
-72.8
Interval 0 ≤ x ≤ 6
RC 73.54
2.32
15
6.06
Dx2.32
17.94
2.32
72.77 12
ΣMC
6.06 2.32
Mx RBV
2.32 73.54
17.94R BV 6.06
R CV
R BV
73.54
2.32 73.54
144
R CV
73.54
2I
34.46
19.75
5.04
ΣMB
R CV -215.23
215.23
144
R BV
7,35 X + 108 7,35
4,65 X 4,65
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= x X - P ( X - 6 ) -
= X - X + -
= X +
=
Tabel Momen dan Lintang pada potongan A & B
C). Potongan C & D
= K
= K
1. Reaksi perletakan
= 0
x 10 - = 0
= / 10
= kip ( )
= 0
- HD x 10 - = 0
HD = / -10
HD = kip ( )
2). Momen dan Lintang
= x X
= X
=
-7.35
Dx 7.35
7.35
Interval 0 ≤ x ≤ 10
Mx 7.35
73.5
Hc
Hc 7.35
73.5
ΣHD
ΣHC
Hc
73.5
73.5
0
73.5
Interval 6 ≤ x ≤ 12
Mx RBV
6.06 24 144
0
1
Dx -17.9
X Rumus Momen
Kip ft 6.06
3.75 Kip ft 6.06
-17.9 141.68
2 9.81 Kip ft 6.06
Kip
Nilai Momen Nilai Lintang
Kip-2.32
4 21.94 Kip ft 6.06
Kip
3 15.88 Kip ft 6.06 Kip
6 34.07 Kip ft 6.06
Kip
5 28.00 Kip ft 6.06 Kip
7 16.13
8 -1.80
-17.94 Kip
Kip
Kip
Kip ft -17.94
11 -55.61 Kip ft
Kip
12 -73.54
10
Kip ft -17.94
6
Kip ft -17.94
9 -19.74
Kip ft -17.94
-37.67
Kip ft
Kip
Kip ft -17.94 Kip
Kip-17.94
34.07
Kip
2.32
2.32
Rumus Lintang
7.35
Hc 7.35
Hc
6,06 X - 2,32 6,06
-17,9X + 141,68 -17,9
C
D
I 10 ft
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Tabel Momen dan Lintang pada potongan A & B
GAMBAR MOMEN, LINTANG DAN NORMAL
Kip ft
Kip ft
Kip ft
Kip ft
0
0
Gambar Momen
-2.32
10
9
1
X
Kip ft 7.35
Nilai Momen Nilai Lintang
Kip
Kip73.54
Kip
66.19 Kip ft 7.35 Kip
8 58.84 Kip ft 7.35
Kip
Kip7 51.48 Kip ft 7.35
6 44.13 Kip ft 7.35
5 36.77 Kip ft
3 22.06 Kip ft
14.71 Kip ft 7.35
7.35
2
Rumus Momen Rumus Lintang
0
7.35 Kip
4
Kip ft 7.35 Kip
Kip ft 7.35
Kip
Kip
Kip
0.00
73.54
34.07
73.54
7.35
29.42 Kip ft 7.35
41.81
-7,35 X -7,35
A
B
C
D
- 2,32 Kip ft
(+
(+
(-)
(-)
(+
(-)
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kip
7.35 kip
4.65 kip -7.35 Kip
0 7.35 kip
kip
0
Gambar Lintang
Kip
kip
Gambar Normal
6.06
17.94
6.06
-17.94
A
B
C
D
(+
(+
(-)(-) (-)
A
B C
D
(-)
-
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Detail Asumsi Pembebanan
ASUMSI PEMBEBANAN
q1 - Beban Mati
Balok Ukuran 30 x 30 cm = x x =
q1 = x = 0.353 T/1.2 0.294
SLOPE DEFLECTION
SO US
0.35 0.35 2.4
4 M
q1
A B
C
D
E
4.5 M 4.5 M
EI EI
0.01 M Keramik
0.02 M Spesi K
0.12 M Plat Lan
0.35 x 0.35 M B
2.25 M
2.25 M
EI EI
q2
2.25 M
4.5 M
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q2 - Beban Hidup + Beban Mati
- Beban Hidup
Untuk Ruang Pertemuan =
Wu1 = x =
- Beban Mati
~ Plat Lantai = x = T/m2
~ Spesi Keramik = x = T/m2
~ Keramik = x = /m +T/m
Wu2 = x =
Wu = Wu1 + Wu2 =
PENYELESAIAN
Dari penyelesaian diatas dapat saya gambarkan sebagai berikut :
q2 = T/m' q2 = Wu x
=
=
q1 = T/m'
1). Menentukan Momen Frimer (MF)
0.353
2.756
1.225
1.2 0.354 0.425
2.756
2 0.021 0.042
1 0.024 0.024
0.5 0.8 T/
T/
1.225 T/m
0.354
500 Kg/m
0.12 2.4 0.288
1.6
4 M
A B
C
D
E
4.5 M 4.5 M
EI EI
EI EI
2.25 M
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1 5
12 96
x 2 x
= +
= TM
1 5
12 96
x 2 x
= +
= TM
2). Rumus Slope Deflection
M AB = 1 EI x θB - MBD = x (
= EI θB - = EI θB
MBA = 1 EI x 2 θB + MDB = x (
4.5
= EI θB + = EI θD
MBC = 1 EI x 2 θB MDE = 1 EI x 2 θD
4 4
= EI θB = EI θD
MCB = 1 EI x θB MED = 1 EI x θD
= EI θB = EI θD
3). Joint Condition
MBA + MBC + MBD = 0
0.5 0.5
4 4
0.25 0.25
3.502 1 EI
4.5
0.44 3.502 0.44
3.502 1 EI
0.22 3.502 0.44
2.
9
0.5954 2.91
4.5 4.5
+
5
3.502
+ qL2
=
0.353 4.5
12
3.502
- MFDB = MFBD = qL2
qL2
+
5 2.
9=
0.353 4.5
12
0.5954 2.91
= qL2- MF AB = MFBA +
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= TM
MBD = x + x -
= - -
= TM
MDB = x + x +
= + +
= TM
MDE = x
= TM
MED = x
= TM
6). Free Body
= T/m1
TM TM TM
= /m1
M
TM
4.5 4.
0.308
-3.85
-0.963
3.365 3.776 4.084
0.3528
2.756
-1.712 0.14 3.50
5.351
0.5 -3.85
-1.926
0.25
0.62 3.502
0.22 -3.85 3.502
-4.084
0.44 -3.85 0.22
0.274 0.86 3.50
0.154
0.44 0.62
q1
q2
A B q1
B
B
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R AV = - = T
-
ΣV = 0R AV + R BV = R 1 + R 2
+ = +=
b). Momen dan Lintang
2. Titik potongan A dan B
= T/m
TM TM
= T/m1
T T
M M
M
~ Menghitung momen dari titik A ke titik BR 1 =
TM = T/m =
=
= T/m
R AV = T R 2 =
M
= X =
==
~ Bidang MomenMx = x X - - R 1 x - R 2 (
Interval 0 ≤ X ≥ 2,25
R AV 3.365 X
2
0.3528
X
2.25q2' 2.76
2.25
3.365 2.756
x3.803 q2'
=2.25 2.76
3.803 R BV= 3.985
4.5
3.365 3.776
0.3528
17.112
2.25 2.25
7.788 7.788
2.756
R AV=
3.803
4.5
3.803 3.985 1.59 6.201
q1
q2
A
q1
q2
A B
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= X - - X2
- X2
(
= X - X2
- X3
-
~ Bidang Lintang
Dx = - X - X2
~ Menghitung momen dari titik B ke titik A
TM
R BV = T
M
~ Bidang Momen
Mx = - x -X - + R 1 x + R 2
= X - - X + X (
= X - X - X -
~ Bidang Lintang Dx = - f' (Mx) / Dx
Dx = + X + X2
3. Titik potongan B dan C
2.25
X Rumus Momen Rumus Lintang
0
-X
3.776
-3.985 0.3528 0.612
2
3.985 0.1764 0.204
x 3.985
2
3.985 3.776 0.353 0.612
Interval 0 ≤ X ≥ 2,25
R BV 3.776
2.25
2.25
3.776
0
3.803 0.3528 0.612
X Rumus Momen Rumus Lintang
0.1764 0.204
3.803 3.365 0.353 0.612
3.365
2
3.803
3.803 X - 0.1764 X2 - 0.204 X3- 3.365 3.803 - 0.3528 X - 0.612 X2
q1
q2
A
3.985 X - 0.1764 X2 - 0.204 X3- 3.776 - 3.985 + 0.3528 X + 0.612 X2
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T
TM
4 M
TM
T
a). Reaksi Perletakan
= 0- x 4 + + = 0
- 4 + = 0
- =
-
= 0
- x 4 + + = 0
- 4 + = 0
= T
b). Momen dan Lintang
T
TM
T
4 M
T
TM
T
7.598
0.308
0.116
0.116
0.154
7.598
0.116
-4
0.462 0.116
4
0.462
= -0.462
ΣMC
0.308 0.154
ΣMB0.154 0.308
0.462
=
7.598
0.308
0.154
7.598
B
C
RCH
RCH
RCH
B
C
RCH
RBH
RBH
RBH
RCH
RBH
RBH
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- R DV = -
R DV = - = T
-
ΣMD = 0
R BV x - R 1 x - R 2 x -R BV - x - x -
R BV - - - + = 0
R BV - = 0
- R BV = -
R BV = - = T
-
ΣV = 0R
BV
+ R DV
= R 1
+ R 2
+ = +
=
b). Momen dan Lintang
= T/m
TM TM
= T/m
T T
M M
M
R 1TM = T/m
1
= T/m1
R BV = T R 23.612 q2'
4.084 2.756
X
0.3528
x
3.612 R DV= 4.176
4.5
16.256 3.612
4.5
2.25 2.25
7.788 7.788
2.756
4.084
R BV =
3.612 4.176 1.59 6.201
5.351
0.3528
5.351
4.5 16.256
4.5 16.256
4.5 3.572 13.95 4.084
4.0844.5 1.59 2.25 6.201 2.25 4.
18.791 4.176
4.5
4.5 2.25 2.25
4.5 18.791
q1
q2
B
q1
q2
B D
q2'
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M= X
~ Bidang Momen
Mx = x X - - R 1 x - R 2
= X - - X2
- X2
= X - X2 - X3 -
~ Bidang Lintang
Dx = - X - X2
~ Menghitung momen dari titik B ke titik D
TM
R DV = T
M
~ Bidang MomenMx = - x -X - + R 1 x + R 2
= X - - X2
+ X2
(
2
4.176 5.351 0.353 0.612
Interval 0 ≤ X ≥ 2,25
R DV 5.351 -X
2.25
2.25
5.351
x 4.176
0
X Rumus Momen Rumus Lintang
0.353 0.612
2
3.612 0.1764 0.204 4.084
Interval 0 ≤ X ≥ 2,25
3.612 0.3528 0.612
2
3.612 4.084
R BV 4.084 X
2.25q2' 2.8
2.25
2.8 2.25
3.612 X - 0.1764 X2 - 0.204 X3- 4.084 3.612 - 0.3528 X - 0.612 X2
q1
q2
D
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= X - X - X -
~ Bidang Lintang Dx = - f' (Mx) / Dx
Dx = + X + X2
5. Titik potongan B dan C
T
TM
4 M
TM
T
a). Reaksi Perletakan
= 0
x 4 - - = 04 Rc - = 0
= T
= 0
x 4 - - = 0
4 - = 0
2.890 0.722
4
2.890
ΣME
1.926 0.963
ΣMD
0.963 1.9262.890
=
4.176
1.926
0.963
4.176
2.25
X Rumus Momen Rumus Lintang
0
5.351
-4.176 0.3528 0.612
2
4.176 0.1764 0.204
4.176 X - 0.3528 X2 - 0.612 X3- 5.351 - 4.176 - 0.3528 X - 0.612 X2
D
E
REH
RDH
REH
RDH
REH
RDH
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= T
b). Momen dan Lintang
TM
TMT
4 M
T
TMTM
~ Bidang Momen
Mx = x X -
= X -
~ Bidang Lintang
Dx =
Rumus Lintang
TM
TM
TM
9) Gambar
4 0.963 0.722
2.25 -0.301 0.722
1.926
0.722 1.926
Nilai Momen Nilai Lintan
0 -1.926 0.722
0.722
X Rumus Momen
4.176
1.9260.722
0.722
0.9634.176
Interval 0 ≤ X ≥ 4
0.722
4=
2.890
0.722 X - 1.926
D
E
0.722
RDH
RDH
REH
RDH
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TM TM
TM
TM
TM
TM
TM
Gambar Momen
T
T
T T
T
T
T
Gambar Lintang
-0.282
-0.116
-3.985
0.826
1.973
-0.154 0.
3.612
-0.091 -0.116
3.803
-3.365 -4.084
-3.776
0.308(+) (+)
(-) (-)(-)
(-)
(+)
(-)
(-)
(+)
A
B
C
A B
C
(+)
8/13/2019 Slope Deflection Bergoyang
http://slidepdf.com/reader/full/slope-deflection-bergoyang 28/44
T
Gambar Normal
7.598
(-)
A B
C
8/13/2019 Slope Deflection Bergoyang
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/m1
0.294
Garis leleh
rmik
tai
alok
q1
q2
4.5 M
4.5 M
8/13/2019 Slope Deflection Bergoyang
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t
x
T/m'
2.25
0
2
8/13/2019 Slope Deflection Bergoyang
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x 2
x 2
θD
2 θB + θD ) -
+ EI θD -
2 θD + θB ) +
+ EI θB +
3.502
3.502
0.22 3.502
0.22 3.502
56 4.5
6
56 4.5
6
8/13/2019 Slope Deflection Bergoyang
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EI θD - ) = 0
……….. Pers ( 1 )
……….. Pers ( 2 )
3.50 2
8/13/2019 Slope Deflection Bergoyang
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8/13/2019 Slope Deflection Bergoyang
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4 M
TM
L
x = T
- = 0
- = 0
+ = 0
+ = 03.77665
3.776
76
4.5 6.201
3.365
3.365
5
0.963
C
8/13/2019 Slope Deflection Bergoyang
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x
x
X
1 x al x t
20.5 x X x X
X x XX
X - 2 X )
3
2.25
0.6120.5 1.225
0.3528
2.76
0.3528 X
q1 X
8/13/2019 Slope Deflection Bergoyang
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1 X )
3
( -X + 2 X )3
-1 X )
3
1.973 TM -0.091
-3.985 TTM
T
Nilai Momen Nilai Lintang
-3.776
1.973 TM -0.091 T
Nilai Momen Nilai Lintang
-3.365 TM 3.803 T
8/13/2019 Slope Deflection Bergoyang
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8/13/2019 Slope Deflection Bergoyang
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x L
x = T
- = 0
- = 04.08451
4.5
4.084
4.5
6 6.201
L
g
8/13/2019 Slope Deflection Bergoyang
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+ = 0+ = 0
= x
= x
= X
= 1 x al x t
0.3528
0.3528 X
q1 X
5.35184 5.351
8/13/2019 Slope Deflection Bergoyang
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2
= 0.5 x X x X
= X x X
= X
( X + 2 X )
3
( 1 X )
3
( -X + 2 X )
3
-1 X )
T
T0.826 TM -0.282
3.612
Nilai Lintang
-4.084 TM
Nilai Momen
0.5 1.225
0.612
2.8
2.25
8/13/2019 Slope Deflection Bergoyang
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3
( )
0.826 TM -0.282
-4.176 TTM
T
Nilai Momen Nilai Lintang
-5.351
8/13/2019 Slope Deflection Bergoyang
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( )
TM-5.351
g
8/13/2019 Slope Deflection Bergoyang
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TM
TM
T
T
TM
0.722
0.722
-4.176
63
-1.926
(+)
(-)
(-)
(+)
(-)
D
E
D
E
8/13/2019 Slope Deflection Bergoyang
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T
11.774
3.803
15.576
7.788
7.788
15.576
4.176
(-)
D
E