Download - Lanny Nattasha Tonak
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TUGAS 2
DINAMIKA STRUKTUR
Oleh
NAMA : LANNY NATTASHA TONAK
NIM : 0806012895
JURUSAN TEKNIK SIPIL
FAKULTAS SAINS DAN TEKNIK
UNIVERSITAS NUSA CENDANA
KUPANG
2011
SOAL !
Tentukan ragam getar !m1=1
m2=4
m3=4
k1=4
k2=4
k3=4
m1
m2
m3
k1
k2
k3
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JAWAB :
-m3 y3 - k3 y3 +k2( y2 – y1 ) = 0
-m2 y2 + k2 ( y2 – y3 ) - k1 ( y2 – y1 ) = 0
-m1 y1 + k1( y1 – y2 ) = 0
m3 0 0 y3 k3+k2 -k2 0 y3
0 m2 0 y2 + -k2 k2+k1 -k2 y2 = 0
0 0 m1 y1 0 -k1 k1 y1
4m3 0 0 8 -4 0
m = 0 4m2 0 k = -4 8 -4
m1=1
m2=4
m3=4
k1=4
k2=4
k3=4
m1
m2
m3
k1
k2
k3
m 3 m 1m 2
k3 k2k1
Y3 Y2 Y1
m 3 m 2 m1k3y3k2(y2-y3) k2
k1(y2-y1) k1k1(y1-y2)
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0 0 4m3 0 -4 4
Agent value :
8k – m3ω2 -4k 0 Ф3
-4k 8k-m2ω2 -4k Ф2 = 0
0 -4k 4k-m1ω2 Ф1
8k – 4m3ω2 -4k 0 Ф3 k
-4k 8k-m2ω2 -4k Ф2 = 0 K
0 -4k 4k-m1ω2 Ф1 k
Misalkan : 8-4λ -4 0 Ф3
-4 8-4λ -4 Ф2 = 0
0 -4 4-λ Ф1
(8 - 4λ)Ф3 - 4Ф2 = 0
-4Ф3 + ( 8 - 4λ )Ф2 - 4Ф1= 0
-4Ф2 + ( 4 – λ )Ф1 = 0
Misalkan Ф3 =1, maka:
( 8 -4λ ) (1) - 4Ф2 = 0
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8 - 4λ = 4Ф2
Ф2 = 2 - λ
Ф2 substitusi ke persamaan (2):
-4(1) + ( 8 - 4λ ) ( 2-λ ) - 4Ф1 = 0
-4 + (16 - 8λ - 8λ + 4λ2 ) = 4Ф1
4λ² - 16λ + 12 = 4Ф1
Φ1 = λ² - 4λ + 3
Φ2 dan Φ1 substitusi ke persamaan 1
-4 ( 2 – λ ) + ( 4 – λ ) ( λ² - 4λ + 3 ) = 0
-8 + 4λ + ( 4λ² - 16λ + 12 - λ³ + 4λ² - 3λ ) = 0
-8 + 4λ + 8λ² - 19λ - λ³ + 12 = 0
-λ³ + 8λ² - 15λ + 4 = 0
λ³ - 8λ² + 15λ – 4 = 0
Ragam getar tiap mode di tabulasikan dalam table berikut:
Mode 1 Mode 2 Mode3
λ 3 = 0.319 λ 2 = 2.358 λ 1 = 5.323
Φ3 =1 Φ3 = 1 Φ3 = 1 Φ3 = 1
Φ2 =2-λ Φ2 = 1.681 Φ2 = -0.358 Φ2 = -3.323
Φ1 = λ²-4λ+3 Φ1 = 1.826 Φ1 = -0.872 Φ1 = 10.042
10.04
-3.32
1.00
-0.87
1.00
MODE 1
1.83
1.68
1.00
MODE 3
-0.36
MODE 2