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dinamika
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kuum 0=+
nn tqt )()(u =
Tnnn )( 21 =
tBtAtq nnnnn sincos)( +=
)sincos()( tBtAt nnnnn +=u
kuum 0=+
0 ]km =+ )([ 2 tqnnnn
00=
0m]k = nn 2[
0m]k = 2det[ n
}]{[}]{[}]{[}]{[
2
2
rrr
nnn
MKMK
=
=
( )Tnnn MK }]{[}]{[ 2 = simetrismatriksAAT ][][ =
][}{][}{ 2 MK TnnTn =
][}{][}{ 2 MK TnnTn =}]{[}]{[ 2 rrr MK =
}{ r
Tn}{
0}]{[}){(}]{[}{}]{[}{}]{[}{}]{[}{
22
2
2
=
=
=
rT
nrn
rT
nrrT
n
rT
nnrT
n
MMKMK
0}]{[}{ 0}]{[}{ == rTnrTn KjugaM
nnn mk 2=
22
22212
12
122121
2
12
2
21
2
1
22
221
)(0
0
mkkmkkk
mm
kkkkk
=+
=+
=
+
2121211211 mm , 2111 ,
2222212
221 mm ,
2212 ,
21222221112
2212221
2121211
211 mmmm +=+
0))(( 22212121112221 =+ mm
02221212111 =+ mm
00 == rTnrTn m k
}]{[}{ 1}]{[}{ IMatauMM TnTnn ===
33243
111324
002
1002
hEIkkk
kkhEI
mmm cc =
=
=
=
= k m
0252
023det
det[
2242
2
2
2
=+
=
=
kkmmmkkkmk
n
n
n
0m]k
mk
mk
mk
mk 22
22 222211
====
05.02
2023
21
11
2
21
112
1
21
1
=
==
kkkk
mk
mkkkmk
5.0 nilaidiperoleh sehingga ,1 1121 ==
1 nilaidiperoleh sehingga ,1 1222 ==
=
=
=
= 11
121
22
122
21
111
{ } 011
111312/1
021=
=
k
T k
{ } 011
100212/1
021=
=
m
T m
)()(u tqt nn=
=
=
+=
+=
n
nnnnnnn
n
nnnnnn
tBtAt
tBtAt
1
1
)cossin()(
)sincos()(
u
u
)0()(
)0()0(
1
1
=
=
=
=
n
nnn
n
nnn
qt
q
u
u
n
nn
n
nnnn
n
n
nnnn
qBBt
qAA
)0()(
)0()0(
1
1
==
==
=
=
u
u
)0()(
)0()0(
1
1
=
=
=
=
n
nnn
n
nnn
qt
q
u
u
n
nn
n
nnnn
n
n
nnnn
qBBt
qAA
)0()(
)0()0(
1
1
==
==
=
=
u
u
)()sin)0(cos)0(()(11
tqtqtqtn
nnn
n
nn
n
nnnn
==
=+=
u
0)0(,0)0(1)0(,1)0( 2121 ==== qqdanqq
=
=
=
= 11
121
22
122
21
111
)()sin)0(cos)0(()(11
tqtqtqtn
nnn
n
nn
n
nnnn
==
=+=
u
0)0(,0)0(1)0(,1)0( 2121 ==== qqdanqq ttqttq 2211 cos1)(cos1)( ==
tqtqtq nn
nnnn
sin)0(cos)0()( +=
tttutu
212
1 cos11cos1
2/1)()(
+
=
0kuucum =++
qu =
0qkqcqm =++ Tn
0=++ qkqcqm TnTnTn
0=++ qKqCqM nnn
02 2 =++ nnnnnn qqq
,nnM mTn=nnn
Tnn MK 2== k
nnnnTnn MC 2== c
=
=
=
= 11
121
22
122
21
111
Solusi:Menentukan matriks massa, matriks kekakuan dan matriks redamansebagai berikut
m
=
= 10
020
02
1 mmm
1113 k
22
221
=
+= kkk
kkk
2226 c
22
221
=
+= ccc
ccc
nTnnK k=
{ }
{ }
{ }
{ }
=
=
===
=
==
=
==
=
===
60075.0
611
111311
015.0
111311
011
111315.0
75.0150
111315.0
22222
1221
2112
11111
k
kkKK
kK
kK
k.kKK
T
T
T
T
K
k
k
k
k
C M
=
= 120
05.13005.1 cm
=
+
+
00
60075.0
12005.1
3005.1
2
1
2
1
2
1
02 2 =++ nnnnnn qqq
++= tqqtqetq dn
dn
onnodnn
tn
nn
sincos)(
21 nndn =
=
++=
n
ndn
dn
onnodno
tn t
qqtqet nn1
sincos)
u(
200/km mk 2/1 =
0)0(,0)0(0)0(,1)0( 2121 ==== qqdanqq
=
=121
21
111
=
++=
n
ndn
dn
onnodno
tn t
qqtqet nn1
sincos)
u( 21 nndn =
+
=
ttetu
tudd
t12
1
11
2
1 sin1
cos15.0
)()(
11
200/km mk /21 =
0)0( ,0)0( 0)0( ,0)0( 2121 ==== qqdanqq
=
= 11
22
122
=
++=
n
ndn
dn
onnodno
tn t
qqtqet nn1
sincos)
u( 21 nndn =
+
=
ttetu
tudd
t22
2
22
2
1 sin1
cos11
)()(
22
}}{{}{}]{[}]{[}]{[}]{[
0}]{[}]{[}]{[
ii
g
g
AuuMuKuCuM
uKuCuuM
=
=++
=+++
][}}{{ }}{]{[}{ }}{]{[}{ }}{]{[}{ MuAKACAM TigiiTiiiTiiiTi =++
Ti}{
][][][2][}{
}]{[}{
2 MKMCMRMM
n
n
Tii
iT
ii
=
=
=
=
i
igiiiii
igiiiiiiii
MRuAAA
RuAMAMAM
=++
=++
2
2 2
2
i
igiiiii MRuAAA =++ 2 2
giiiii uDDD =++ 2 2
ntdispleceme pseudo === Dii
ii SDDM
RA
[ ] 5.02max )}({= ii Au
w1 = 2943 kg, K1 = 1600 kg/cmw2 = 4414 kg, K2 = 2000 kg/cmw3 = 4414 kg, K3 = 2400 kg/cm
Hitung :1. Frekuensi alami dan waktu getar alami dari sistem struktur di atas.2. Gambar mode shape dari masing-masing waktu getar alami yang terjadi.3. Hitung gaya gempa disetiap lantai dari sistem struktur tersebut jika berada
di wilayah gempa 3 dengan jenis tanah lunak SNI .
2331
2222
2111
/.5,49814414kg4414
/.5,49814414kg4414
/.39812943kg2943
scmkggWmW
scmkggWmW
scmkggWmW
====
====
====
Menyusun matriks kekakuan [K]
[ ]( ) ( )( ) ( ) ( )
( ) ( )
( )( )
=
+
+
=
+
+
=
=
440020000200036001600
0160016002400200020000
2000200016001600016001600
0
0
322
2211
11
333231
232221
131211
KKKKKKK
KK
KKKKKKKKK
K
Menyusun matriks massa [M]
[ ]
=
=
5,40005,40003
000000
3
2
1
mm
mM
[ ] [ ]( ) { }{ } [ ] [ ]( ) 0det0
02
2
=
=
MKMK
n
n
( )( )
( )( )
( )( )
( ) ( )( ){ } ( ) ( )( ){ }( ) ( )( ){ }
( ) 010264,11343210231102000200031600
5,44400160016005,444005,4360031600
05,4440020000
20005,4360016000160031600
det
05,4440020000
20005,4360016000160031600
05,400
05,40003
440020000200036001600016001600
8246
2
2222
2
2
2
2
2
2
2
=+
=
=
=
=
nnn
n
nnnn
n
n
n
n
n
n
n
( ) 010264,113432102311 8246 =+ nnn Misalkan 2n =
( ) 010264,113432102311 823 =+
det17,092,3722T
rad/det92,371438/detrad1438
det23,05,2722T
rad/det5,27757/detrad757
det58,08,1022T
rad/det8,10116/detrad116
33
31222
33
22
22222
22
11
11222
11
===
=====
===
=====
===
=====
Bentuk mode 1 (vektor shape) relatif displacement. { } 11 [ ] [ ]( ) { } [ ] [ ]( ) { } 00 1212 == MKMK n
( )( )( )
( )( )
0387820000200030781600
016001252
0387820000200030781600016001252
01165,4440020000
20001165,4360016000160011631600
05,400
05,40003
440020000200036001600016001600
1312
131211
1211
13
12
11
13
12
11
13
12
112
=+
=+
=
=
=
=
n
Bila harga , dan harga-harga yang lainnya dinyatakanterhadap harga , maka diperoleh:
0,111 =11
( )
( )( ) 4036,03878
7825,02000038787825,02000
0387820007825,01600
12520160011252
13
13
1312
12
12
==
=+
=+
=
=
=
{ }
=
=
4036,07825,01
13
12
11
1
2 3
[ ] [ ]{ }{ } 02 = MK n
{ }
=
=
9844,0419,01
23
22
21
2
{ }
=
=
634,1696,11
33
32
31
3
rad/det92,37rad/det5,27rad/det8,10det17,0Tdet23,0Tdet58,0T
321
321
===
===
{ } [ ]{ }
[ ] 0984,0419,00,1
5,40005,40003
4036,07825,01
021
= MT
{ } [ ]{ }
[ ] 0634,1696,10,1
5,40005,40003
984.0419,01
032
= MT
{ }
55.0det17,092,3775.0det23,05,2775.0det58,08,10
33
22
11
===
===
===
d
d
d
ccc
3
2
1
Trad/detTrad/detTrad/det
2
2
dimana
.
i
dD
i
ii
Diii
dii
clacementpseudodispS
MRipartisipasfaktor
SMcRA
==
==
==
[ ] { }
[ ]
/cmkg.det91,2/cmkg.det65,2
/cmkg.det33,85,45,4
3403,0782,01
23
22
2
11
=
=
=
=
=
RR
MR T
[ ] { }{ }
[ ] kg.det/cm97,7403,0782,01
5,40005,40003
403,0782,01
111
=
=
= MM T
/cmkg.det03,28/cmkg.det00,72
3
22
=
=
MM
( )( )( )( )
( )( )
cm036,003.287,3998155,091.2.
cm368,075,2798175,065,2.
cm316,097,78,1098175,033,8.
23
23
333
22
22
222
21
21
111
===
=
==
===
McRA
McRA
McRA
d
d
d
cmuuu
u
u
AAAu
=
=
+
+
=
++=
388.0298.0486.0
634.1696.11
036.09844.0419.0
1368.0
403.0782.01
316.0
)]}{()}{()}{[(
3
2
1
max
5.0222
max
5.0233
222
211max
kgFFF
F
F
uKF
=
=
=
=
0659.1114797.483
061.302388.0298.0486.0
440020000200036001600016001600
}]{[
3
2
1
max