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Multi Degree of Freedom (MDOF)
24
m3 m2 m1 1 MULTI DEGREE OF FREEDOM (MDOF) TUGAS 5 Persamaan Umum SDOF m ¨ Χ ( t )+ C ¨ Χ ( t )+ Kx ( t )=F ( t )=−m ¨ Χg Persamaan Umum MDOF [ M 1 M 2 M 3 ] { ¨ Χ ( t ) ¨ Χ ( t ) ¨ Χ ( t ) } + [ C 1 C 2 C 3 ] { ¨ Χ ( t ) ¨ Χ ( t ) ¨ Χ( t ) } + [ ( K 1 +K 2 ) −K 2 0 −K 2 K 2 +K 3 −K 3 0 −K 3 K 3 ] { X 1 ( t ) X 2 ( t ) X 3 ( t ) } =− [ M 1 M 2 M 3 ] { ¨ Χg 1 ¨ Χg 2 ¨ Χg 3 } = { X 1 ( t ) X 2 ( t ) X 3 ( t ) } Contoh Kasus 2
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Sekedar tugas kuliah Multi Degree Of Freedom (MDOF)
Transcript of Multi Degree of Freedom (MDOF)
K
1
MULTI DEGREE OF FREEDOM (MDOF)
TUGAS 5
Persamaan Umum SDOF
m Χ̈ ( t )+C Χ̈ ( t )+Kx ( t )=F ( t )=−m Χ̈ gPersamaan Umum MDOF
[M 1 M 2M 3 ]{Χ̈ ( t )
Χ̈ ( t )Χ̈ ( t )}+[C 1 C2
C3 ]{Χ̈ ( t )Χ̈ ( t )Χ̈ ( t )}+[(K1+K 2 ) −K2 0
−K2 K2+K 3 −K 30 −K3 K 3 ]{X 1( t )X 2( t )
X3 ( t )}=−[M 1 M 2
M 3 ]{Χ̈ g1Χ̈ g2Χ̈ g3 }={X 1( t )X 2( t )
X 3( t )}Contoh Kasus
m1
m2
m3
K
2
K3
M1 = 1
M2 = 2
M3 = 3
K1 = 1500.00
K2 = 1000.00
K3 = 500.00
3 0 0[M] = 0 2 0
0 0 1
2500.00 -1000.00 0[K] = -1000.00 1500.00 -500.00
0 -500.00 500.00
|[K ]−ω2 [M ]|=0
|[2500 −1000 0−1000 1500 −5000 −500 500 ]−ω2 [3 0 0
0 2 00 0 1 ]|=0
500|[ 5 −2 0−2 3 −10 −1 1 ]−ω2[3 0 0
0 2 00 0 1 ]|=0
500|[(5−3ω2 ) −2 0
−2 (3−2ω2 ) −10 −1 (1−ω2 ) ]|=0
Di asumsikan
λ= ω2
500
|[(5−3 λ ) −2 0−2 (3−2 λ ) −10 −1 (1− λ) ]|=0
(5−3 λ )[3−2λ −1−1 (1−λ ) ]+2 [−2 −1
0 (1−λ ) ]+0=(5−3 λ ) {(3−2λ )(1−λ )−1 }+2 [−2(1−λ )]=
(5−3 λ ) {(3−2λ )(1−λ )−1 }−4+4 λ=
(5−3 λ )(3−3 λ−2λ+2 λ2−1)−4+4 λ=
{5(3−3 λ−2 λ+2 λ2−1 )−3 λ(3−3 λ−2 λ+2 λ2−1)}−4+4 λ=15−15 λ−10 λ+10 λ2−5−9 λ+9 λ2+6 λ2−6 λ3+3 λ−4+4 λ=−6 λ3+10 λ2+9 λ2+6 λ2−15 λ−10 λ−9 λ+3 λ+4 λ+15−5−4=−6 λ3+25 λ2−34 λ+3 λ+4 λ+6=
−6 λ3+25 λ2−34 λ+3 λ+4 λ+6=
−6 λ3+25 λ2−27 λ+6=
(−6 x3)+25 x2−27 x+6=0
Di dapat: (diurutkan dari nilai yang terkecil)
X1= λ1= 0,299
X2= λ2= 1,304
X3= λ3= 2,563
Masukkan nilai –nilai di atas ke dalam persamaan
|[ [K ]−ω2 [M ] ]|{φ}=0
Untuk λ1= 0,299 Ф31 = 1
|[ [K ]−ω2 [M ] ]|{φ}=0
|[(5−3 λ ) −2 0−2 (3−2 λ ) −10 −1 (1− λ) ]|{φ
11
φ21
φ31}+{000 }=0
|[ ((5−3 λ )×φ11 )+(−2×φ
21)+(0×φ
31 )
(−2×φ11 )+((3−2 λ )×φ
21 )+(−1×φ
31 )
(0×φ11 )+(−1×φ
21 )+((1−λ )×φ
31 )
]|=0|[ ((5−3 λ )×φ
11 )+(−2×φ
21 )
(−2×φ11 )+((3−2 λ )×φ
21 )+(−1×1)
(−1×φ21 )+((1−λ )×1 )
]|=0|[ ((5−3×0 ,299)×φ
11 )+(−2×φ
21)
(−2×φ11 )+((3−2×0 ,299 )×φ
21 )+(−1×1 )
(−1×φ21 )+((1−0 ,299)×1)
]|=0|[ ( 4 ,103φ
11 )+(−2φ
21 )
(−2φ11 )+(2 ,402φ
21 )+(−1)
(−φ21 )+(0 ,701)
]|=0
Dengan subtitusi mundur di dapat nilai:
(−φ21)+(0 ,701 )=0
φ21=0 ,701
(−2φ11)+(2 ,402φ
21 )+(−1)=0
(−2φ11)+(2 ,402×0 ,701)+(−1 )=0
(−2φ11)+(1 ,684−1 )=0
(−2φ11)+0 ,684=0
−2φ11=−0 ,684
2φ11=0 ,684
φ11=0 ,6842
=0 ,342
φ31=1
Untuk λ2= 1,304 Ф32 = 1
|[ [K ]−ω2 [M ] ]|{φ}=0
|[(5−3 λ ) −2 0−2 (3−2 λ ) −10 −1 (1− λ) ]|{φ
12
φ22
φ32}+{000 }=0
|[ ((5−3 λ )×φ12 )+(−2×φ
22)+(0×φ
32)
(−2×φ12 )+((3−2 λ )×φ
22 )+(−1×φ
32 )
(0×φ12 )+(−1×φ
22 )+((1−λ )×φ
32 )
]|=0|[ ((5−3 λ )×φ
12 )+(−2×φ
22 )
(−2×φ12 )+((3−2 λ )×φ
22 )+(−1×1 )
(−1×φ22 )+((1−λ )×1 )
]|=0|[ ((5−3×1,304 )×φ
12 )+(−2×φ
22)
(−2×φ12 )+((3−2×1 ,304 )×φ
22 )+(−1×1 )
(−1×φ22 )+((1−1 ,304 )×1)
]|=0|[ (1,088φ
12 )+(−2φ
22 )
(−2φ12 )+(0 ,392φ
22 )+(−1)
(−φ22 )+(−0 ,304 )
]|=0Dengan subtitusi mundur di dapat nilai:
(−φ22)+(−0 ,304 )=0
φ22=−0 ,304
(−2φ12)+(0 ,392φ
22 )+(−1)=0
(−2φ12)+(0 ,392×−0 ,304 )+(−1)=0
(−2φ12)+(−0 ,119−1 )=0
(−2φ12)+(−1 ,119 )=0
−2φ12=1 ,119
φ12=1,119
−2=−0 ,56
φ32=1
Untuk λ3= 2,563 Ф33 = 1
|[ [K ]−ω2 [M ] ]|{φ}=0
|[(5−3 λ ) −2 0−2 (3−2 λ ) −10 −1 (1− λ) ]|{φ
13
φ23
φ33}+{000 }=0
|[ ((5−3 λ )×φ13 )+(−2×φ
23 )+(0×φ
33 )
(−2×φ13 )+((3−2 λ )×φ
23 )+(−1×φ
33 )
(0×φ13 )+(−1×φ
23 )+((1−λ )×φ
33)
]|=0|[ ((5−3 λ )×φ
13 )+(−2×φ
23 )
(−2×φ13 )+((3−2 λ )×φ
23 )+(−1×1 )
(−1×φ23 )+((1−λ )×1 )
]|=0|[ ((5−3×2 ,563 )×φ
13 )+(−2×φ
23 )
(−2×φ13 )+((3−2×2,563 )×φ
23 )+(−1×1 )
(−1×φ23 )+((1−2 ,563)×1)
]|=0
|[ (−2 ,689φ13 )+(−2φ
23 )
(−2φ13 )+(−2 ,126φ
23)+(−1 )
(−φ23 )+(−1 ,563 )
]|=0
Dengan subtitusi mundur di dapat nilai:
(−φ23 )+(−1,563 )=0
φ23=−1 ,563
(−2φ13 )+(−2 ,126φ
23 )+(−1)=0
(−2φ13 )+(−2 ,126×−1 ,563 )+(−1)=0
(−2φ13 )+(3 ,323 )+(−1)=0
(−2φ13 )+(2 ,323 )=0
(−2φ13 )=−2,323
φ13=−2 ,323
−2=1 ,1615
φ33=1
φ31=1
φ21=0 ,701
φ11=0 ,304
Ragam 1
Untuk λ1= 0,299
λ= ω2
500 ⇒ω=√ λ×500=√0 ,299×500=12 ,227T=2π
ω= 2π12 ,227
=0 ,514
φ22=−0 ,304
φ12=−0 ,560
φ32=1
F= 1T
= 10 ,514
=1 ,945
Ragam 2
Untuk λ2= 1,304
λ= ω2
500 ⇒ω=√ λ×500=√1 ,304×500=25 ,534
φ23=−1 ,563
φ13=1 ,1615
φ33=1
T=2πω
= 2π25 ,534
=0 ,246
F= 1T
= 10 ,246
=4 ,065
Ragam 3
Untuk λ3= 2,563
K
1
λ= ω2
500 ⇒ω=√ λ×500=√2 ,563×500=35 ,798
T=2πω
= 2π35 ,798
=0 ,175
F= 1T
= 10 ,175
=5 ,714
Kasus 2
m1
m2
m3
K
2
K3
M1 = 2
M2 = 2
M3 = 2
K1 = 1500.00
K2 = 1000.00
K3 = 500.00
2 0 0[M] = 0 2 0
0 0 2
2500.00 -1000.00 0[K] = -1000.00 1500.00 -500.00
0 -500.00 500.00
|[K ]−ω2 [M ]|=0
|[2500 −1000 0−1000 1500 −5000 −500 500 ]−ω2 [2 0 0
0 2 00 0 2 ]|=0
500|[ 5 −2 0−2 3 −10 −1 1 ]−ω2[2 0 0
0 2 00 0 2 ]|=0
500|[(5−2ω2) −2 0
−2 (3−2ω2) −10 −1 (2−ω2 ) ]|=0
Di asumsikan
λ= ω2
500
|[(5−2 λ ) −2 0−2 (3−2 λ ) −10 −1 (2−λ ) ]|=0
(5−3 λ )[3−2λ −1−1 (2−λ ) ]+2 [−2 −1
0 (2−λ ) ]−0=(5−3 λ ) {(3−2λ )(2−λ )−1 }+2 [−2(2−λ )]=
(5−3 λ ) {(3−2λ )(2−λ )−1 }−8+8 λ=
(5−3 λ )(6−3 λ−4 λ+2 λ2−1)−8+8 λ=
{5(6−3 λ−4 λ+2λ2−1 )−3 λ (6−3 λ−4 λ+2 λ2−1 )}−8+8 λ=30−15 λ−20 λ+10 λ2−5−18 λ+9 λ2+12λ2−6 λ3+3 λ−8+8 λ=−6 λ3+10 λ2+9 λ2+12 λ2−15 λ−20 λ−18 λ+3 λ+8 λ+30−5−8=−6 λ3+31 λ2−53 λ+3 λ+8 λ+17=
−6 λ3+31 λ2−53 λ+11 λ+17=
−6 λ3+31 λ2−42λ+17=
(−6 x3)+31 x2−42 x+17=0
Di dapat: (diurutkan dari nilai yang terkecil)
X1= λ1= 0,856
X2= λ2= 1
X3= λ3= 3,311
Masukkan nilai –nilai di atas ke dalam persamaan
|[ [K ]−ω2 [M ] ]|{φ}=0
Untuk λ1= 0,856 Ф31 = 1
|[ [K ]−ω2 [M ] ]|{φ}=0
|[(5−3 λ ) −2 0−2 (3−2 λ ) −10 −1 (2− λ) ]|{φ
11
φ21
φ31}+{000 }=0
|[ ((5−3 λ )×φ11 )+(−2×φ
21)+(0×φ
31 )
(−2×φ11 )+((3−2 λ )×φ
21 )+(−1×φ
31 )
(0×φ11 )+(−1×φ
21 )+((2−λ )×φ
31 )
]|=0|[ ((5−3 λ )×φ
11 )+(−2×φ
21 )
(−2×φ11 )+((3−2 λ )×φ
21 )+(−1×1)
(−1×φ21 )+((2−λ )×1 )
]|=0|[ ((5−3×0 ,856)×φ
11)+(−2×φ
21 )
(−2×φ11 )+((3−2×0 ,856 )×φ
21 )+(−1×1 )
(−1×φ21 )+((2−0 ,856 )×1)
]|=0|[ (2 ,432φ
11 )+(−2φ
21 )
(−2φ11 )+(1 ,288φ
21 )+(−1 )
(−φ21)+(1 ,144 )
]|=0Dengan subtitusi mundur di dapat nilai:
(−φ21)+(1,144 )=0
φ21=1 ,144
(−2φ11)+(1 ,288φ
21 )+(−1)=0
(−2φ11)+(2 ,402×1 ,144 )+(−1 )=0
(−2φ11)+(2 ,748−1)=0
(−2φ11)+1 ,748=0
−2φ11=−1 ,748
2φ11=1 ,748
φ11=1,7482
=0 ,874
φ31=1
Untuk λ2= 1 Ф32 = 1
|[ [K ]−ω2 [M ] ]|{φ}=0
|[(5−3 λ ) −2 0−2 (3−2 λ ) −10 −1 (2− λ) ]|{φ
12
φ22
φ32}+{000 }=0
|[ ((5−3 λ )×φ12 )+(−2×φ
22)+(0×φ
32)
(−2×φ12 )+((3−2 λ )×φ
22 )+(−1×φ
32 )
(0×φ12 )+(−1×φ
22 )+((2−λ )×φ
32 )
]|=0|[ ((5−3×1)×φ
12 )+(−2×φ
22)
(−2×φ12 )+((3−2×1 )×φ
22 )+(−1×1 )
(−1×φ22 )+((2−1)×1)
]|=0|[ (2×φ
12)+(−2×φ
22 )
(−2×φ12 )+(1×φ
22 )+(−1×1)
(−1×φ22 )+(1 )
]|=0|[ (2φ
12 )+(−2φ
22)
(−2φ12 )+(1φ
22 )+(−1)
(−φ22 )+(1)
]|=0Dengan subtitusi mundur di dapat nilai:
(−φ22)+(1)=0
φ22=−1
(−2φ12)+(1φ
22 )+(−1 )=0
(−2φ12)+(1×−1 )+(−1)=0
(−2φ12)+(−1−1)=0
(−2φ12)+(−2)=0
−2φ12=2
φ12=2
−2=−1
φ32=1
Untuk λ3= 3,311 Ф33 = 1
|[ [K ]−ω2 [M ] ]|{φ}=0
|[(5−3 λ ) −2 0−2 (3−2 λ ) −10 −1 (2− λ) ]|{φ
13
φ23
φ33}+{000 }=0
|[ ((5−3 λ )×φ13 )+(−2×φ
23 )+(0×φ
33 )
(−2×φ13 )+((3−2 λ )×φ
23 )+(−1×φ
33 )
(0×φ13 )+(−1×φ
23 )+((2−λ )×φ
33 )
]|=0|[ ((5−3 λ )×φ
13 )+(−2×φ
23 )
(−2×φ13 )+((3−2 λ )×φ
23 )+(−1×1 )
(−1×φ23 )+((2−λ )×1)
]|=0
|[ ((5−3×3,311 )×φ13 )+(−2×φ
23 )
(−2×φ13 )+((3−2×3,311)×φ
23 )+(−1×1)
(−1×φ23 )+((2−3,311)×1 )
]|=0|[ (−4 ,933 φ
13 )+(−2φ
23 )
(−2φ13 )+(−3 ,622φ
23)+(−1 )
(−φ23 )+(−1 ,311)
]|=0Dengan subtitusi mundur di dapat nilai:
(−φ23 )+(−1,311 )=0
φ23=−1 ,311
(−2φ13 )+(−3 ,622φ
23 )+(−1 )=0
(−2φ13 )+(−2 ,126×−1 ,311)+(−1)=0
(−2φ13 )+(2 ,787 )+(−1)=0
(−2φ13 )+(1 ,787 )=0
(−2φ13 )=−1,787
φ13=−1 ,787
−2=0 ,894
φ33=1
φ31=1
φ21=1 ,144
φ11=0 ,874
Ragam 1
Untuk λ1= 0,856
λ= ω2
500 ⇒ω=√ λ×500=√0 ,856×500=20 ,688
T=2πω
= 2π20 ,688
=0 ,304
F= 1T
= 10 ,304
=3 ,293
φ22=−1
φ12=−1
φ32=1
Ragam 2
Untuk λ2= 1
λ= ω2
500 ⇒ω=√ λ×500=√1×500=22 ,361
T=2πω
= 2π2 ,3612
=0 ,281
F= 1T
= 10 ,281
=3 ,559
φ23=−1 ,311
φ13=0 ,894
φ33=1
Ragam 3
Untuk λ3= 3,311
λ= ω2
500 ⇒ω=√ λ×500=√3 ,311×500=40 ,688T=2π
ω= 2 π40 ,688
=0 ,154
F= 1T
= 10 ,154
=6 ,476
