Dinamika Struktur

Wednesday, March 06th 2019

Dinamika Struktur dan Rekayasa Gempa (CIV-308) ASSIGMENT 4 Kelompok : 1. Bagas Makarim S 2016091008 2. Diza Roshaunda 2016091013 3. Raka Maulana P 2016091037 4. Shafira Khalisha 2016091045 5. Alfan Safari 2016091056 6. Elysa Damayanti 2016091062

Assignment 4 An SDF system has the following properties :  m = 4.500 kg-sec2/m  k = 178.400 kgf/m  Tn = 1 sec (wn = 6,283 rad/sec)  x = 0,05. Determine the response u(t) of this system due to El-Centro 1940 N-S, using : a. Interpolation b. Central Difference c. Newmark Average Acceleration d. Newmark Linear Acceleration

Find the 𝑼𝒎𝒂𝒙 ! m k ωn ξ ωd pt Δt

Diketahui 4500 178400 6.296383442 0.05 6.288508037 470.5667186 0.02

Diketahui 𝐾𝑔. 𝑠 2 /𝑚 𝐾𝑔/𝑚 𝑟𝑎𝑑/𝑠𝑒𝑐 𝑟𝑎𝑑/𝑠𝑒𝑐 𝑠𝑒𝑐

A B C D A' B' C' D'

0.992114709 0.019822113 2.94435E-08 1.47566E-08 -0.785836666 0.979633946 2.19491E-06 2.21E-06

𝐴 = 𝑒 −ξ𝜔𝐷Δt (

ξ √1−ξ2

𝑆𝑖𝑛𝜔𝐷 Δt + Cos𝜔𝐷 Δt) 0.05

𝐴 = 𝑒 −0.05 𝑥 6.28 𝑥 0.02 (

√1−0.052

𝑆𝑖𝑛(6.28 𝑥 0.02) + Cos (6.28 x 0.02)

𝐴 = 0.992114709

1

𝐵 = 𝑒 −ξ𝜔𝐷 Δt (𝜔 𝑆𝑖𝑛𝜔𝐷 Δt) 𝐷

1

𝐵 = 𝑒 −0.05 x 6.28 x 0.02 (6.28 𝑆𝑖𝑛 (6.28 𝑥 0.02)) 𝐵 = 0.019822113

𝐶=

𝐶=

1 2ξ + { 𝑘 𝜔𝑛 Δt

2

1−ξ

𝑒 −ξ𝜔𝐷 Δt [(𝜔

𝐷 Δt

1 2 𝑥 0.05 + { 178400 (6.29 𝑥 0.02)

ξ

−

√1−ξ

2

2ξ ) Cos𝜔𝑛 Δt]} 𝑛 Δt

) 𝑆𝑖𝑛𝜔𝐷 Δt − (1 + 𝜔 2

1−0.05

𝑒 −0.05 x 6.28 x 0.02 [((6.28 𝑥 0.02) −

0.05 √1−0.052

) 𝑆𝑖𝑛 (6.28 𝑥 0.02) −

2 x 0.05

(1 + (6.29 𝑥 0.02)) Cos(6.29 𝑥 0.02)]} 𝐶 = 2.944 𝑥 10−8

2

1 2ξ 2ξ − 1 2ξ 𝐷 = [1 − + 𝑒 −ξ𝜔𝐷 Δt ( 𝑆𝑖𝑛𝜔𝐷 Δt + )] 𝑘 𝜔𝑛 Δt 𝜔𝐷 Δt 𝜔𝑛 Δt 𝐷=

1 2 x 0.05 [1 − 178400 (6.29 𝑥 0.02) 2

+ 𝑒 −0.05 x 6.28 x 0.02 ( 𝐷 = 1.47 𝑥 10−8

2 x 0.05 − 1 2 x 0.05 𝑆𝑖𝑛(6.28 𝑥 0.02) + )] (6.28 𝑥 0.02) (6.29 𝑥 0.02)

𝐴′ = 𝑒 −ξ𝜔𝐷 Δt (

𝜔𝑛 √1−ξ2

𝑆𝑖𝑛𝜔𝐷 Δt)

𝐴′ = 𝑒 −0.05 x 6.28 x 0.02 (

6.29

√1−0.052

𝑆𝑖𝑛(6.28 𝑥 0.02))

𝐴′ = −0.785836666

𝐵′ = 𝑒 −ξ𝜔𝐷 Δt (Cos𝜔𝑛 Δt −

ξ √1−ξ2

𝑆𝑖𝑛𝜔𝐷 Δ)

𝐵′ = 𝑒 −0.05 x 6.28 x 0.02 (Cos(6.29 𝑥 0.02) −

0.05 √1−0.052

𝑆𝑖𝑛(6.28 𝑥 0.02))

𝐵′ = 0.979633946

𝐶′ = 1

0.02

1 1 {− 0.02 178400

+ 𝑒 −0.05 x 6.28 x 0.02 [(

6.29

√1−0.052

+

0.05 √1−0.052

) 𝑆𝑖𝑛(6.28 𝑥 0.02) +

Cos(6.28 𝑥 0.02)]}

𝐶′ =

1 1 {− Δt + 𝑘

𝑒 −ξ𝜔𝐷 Δt [(

𝜔𝑛 √1−ξ

2

+

ξ √1−ξ

2

) 𝑆𝑖𝑛𝜔𝐷 Δt +

1

Δt

Cos𝜔𝐷 Δt]}

𝐶′ = 2.194 𝑥 10−6

𝐷′ =

𝐷′ =

1 ξ 1 − 𝑒 −ξ𝜔𝐷 Δt 𝑆𝑖𝑛𝜔𝐷 Δt + Cos𝜔𝐷 Δt 𝑘Δt 2 √ [ ( 1−ξ )] 1 0.05 [1 − 𝑒 −0.05 x 6.28 x 0.02 ( 𝑆𝑖𝑛(6.28 𝑥 0.02) + Cos(6.28 𝑥 0.02))] 178400 𝑥 0.02 √1 − 0.052

𝐷′ = 2.21 𝑥 10−6

Chart Title 0.00003 0.00002

Axis Title

0.00001 0 0

5

10

15

-0.00001 -0.00002 -0.00003

Axis Title

20

25

30