Dinamika Struktur - Getaran

Laporan Tugas

Nama : Fajrin Siddiq NIM : 0310610029

Fakultas Teknik Jurusan Sipil Universitas Brawijaya Malang 2008 Tugas Dinamika

(Sebagai prasyarat mengikuti kuis)

Sistem seperti terlihat pada gambar di atas mempunyai harga k = 40 lbin, berat = 38.6 lb. Pada kondisi awal, U0 = U = 0 . Sketsalah o

hasil getarannya jika diberi gaya eksitasi P( t ) = 10 cos(10t ) dan rasio redaman = 0.29 !  Penyelesaian :  Solusi umum : u = Ucos( Ωt −α ) + e−ζωnt (A1 cosωdt + A2 sinωdt)

ωn = Uo = r=

k kg 40(386) = = = 20 raddtk m W 38.6 Po 10 = = 0.25 in k 40

Ω 10 = = 0.5 ω n 20

ζωn = (0.29)(20) = 5.8 rad/s U= U=

[(1 − r )

2 2

U0 + (2ζ r)2

[(1 − 0.5 )

U = 0.311

2 2

]

1 2

0.25 + (2(0.29)(0.5))2

]

tanα =

2ζ r 2(0.29)(0.5) = = 0.387 1 − r2 1 − (0.5)2

α = 0.37 ωd = ωn 1 − ζ

2

= 20 1 − (0.29)2 = 19.14 rad/s

• Mencari nilai A1 u = Ucos(Ω t − α ) + e−ζωnt (A1 cosωdt + A2 sinωdt) u(0) = 0 = 0.311cos(−0.37) + A1 A1 = −0.311cos(−0.37) A1 = −0.29 in •

Mencari Nilai A2

u = −ΩUsin(Ω t − α ) + e−ζω nt [ ( A2ωd − A1ζωn ) cosωdt − ( A1ωd − A2ζω n ) sinωdt] u (0) = 0 = −ΩUsin(−α ) + [ ( A2ωd − A1ζωn ) ] u (0) = 0 = −10(0.311)sin(−0.37 ) + [ ( A2 (19.14) − (−0.29)(0.29)(20)) ] u (0) = 0 = (−3.11)(−0.362) + [ ( A2 (19.14) + 1.682 ) ] u (0) = 0 = 1.126 + A2 (19.14) + 1.682 A2 = −

•

2.808 = −0.147 in 19.14

Maka persamaannya akan menjadi : u = 0.311cos( 10 t − 0.37 ) + e−(5.8)t (−0.29 cos( 19.14t) − 0.147 sin( 19.14t) in

•

Sketsa hasil getaran : 0.7

U(t)

0.6 0.5 0.4 0.3 0.2

t

0.1 0 -0.1 0 -0.2 -0.3 -0.4

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5