UNIVERSITAS NEGERI MALANGJURUSAN TEKNIK SIPILPRODI S1 TEKNIK SIPIL

NAMA = DESY KRISNA CAHYA

NIM = 120523400098OFF = D

Kekakuank1 = 29596.7 lb/inchk2 = 42694.4 lb/inch

m = 136 00 66

k = 72291.15 -42694.444444444

W2=50 lb/ft

W10x21

20 psf

W1=100 lb/ft²

W10x4520 psf

30'

15'

10'

u1

u2

k =-42694.4444444444 42694.4444444444

Persamaan Eigen Problem72291.15 -136 λ -42694.4444444444

-42694.4444444444 42694.4444444444

Penyederhanaan persamaan 372291.15 -136 λ ɑ1

-42694.4444444 ɑ1 + 42694.4444444444

= 1> Persamaan 4

72291.15 ɑ1 -136 λ72291.15 -136 λ +

> Subtitusi ersamaan 6 ke Persamaan 5-42694.4444444 + 42694.4 -66-42694.4444444 + 72291.1522633745 -111.8578985672529596.70781893 -247.86826126155 λ +

Coba 1Persamaan = -237Akar = 136.09Pendekatan = 1

Coba 2Persamaan = 22966Akar = 1150.4Pendekatan = 1000

λ1 = 136.09λ2 = 1150.40

Mode 1 = 136,09 Mode 2 = 1150,4ɑ1 = 1 1 1ɑ2 = 1.26 -1.97

ω1 = 172.085 rad/sω2 = 862.252 rad/s

ɑ11 = 1

Asumsi ɑ1

Mencari nilai λ

Nilai λ

Nilai Fungsi ɑNilai ɑ

1,62-0,00286λ

ɑ12 = 1ɑ21 = 1.26ɑ22 = -1.97

F1(t) = 10000 (1-t/td)lbF2(t) = 20000 (1-t/td)lb

td = 0.1 sec

F1(t) = F2(t) =

= P1(t) …Persamaan 82+ω2²z2z̈� = P2(t)

Dimana:P1 = …Persamaan 9P2 = φ12F1(t) + φ22F2(t)

Subtitusi Persamaan 9 ke 8

1z̈� + 140 z1

2+ω2²z2z̈�2+ω2²z2z̈�

2z̈� + 1082 z2

Periode NaturalT1 =

= 0.531 sec

T2 == 0.191 sec

Dimana: f(t) = 1f(t) = 0

td/T1 = 0.188td/T2 = 0.523

(DLF)1max = 0.59(DLF)2max = 1.22

z1 st == 16.20

z�1+ω1²z1

φ11F1(t) + φ21F2(t)

z�1+ω1²z1

z�1+ω1²z1

2π/ω1

2π/ω2

F01/ω1²

z2 st == -1.37

z1max = 9.6z2max = 1.67

u1max = φ11 z1max

u2max = φ21 z1max

u1max = 0.70u2max = 0.61

u1max = 0.62u2max = 0.79

vu =

k1 = 14798.35V11 = 9.11664324049997V12 = 1.25050412678155

V1max = 9202

Untuk kolom di lantai duak2 = 21347.2222222222

V21 = 3.41526218133146V22 = -5.3604363580586

V2max = 6356

F02/ω2²

z jmax(φij-φi-1j)

UNIVERSITAS NEGERI MALANGJURUSAN TEKNIK SIPILPRODI S1 TEKNIK SIPIL

Diketahui :E =

I W10X45 =I W10X22 =

L =h1 =h2 =W1 =W2 =w1 =w2 =

=g =

Massam1 = 136 lb.sec2/inchm2 = 66 lb.sec2/inch

… Persamaan 1… Persamaan 2

-42694.4444444444 x ɑ1 =-66 λ ɑ2

+ -42694.4444444444 ɑ2 = 0-66 λ ɑ2 = 0

ɑ1 + -42694.4444444 ɑ2 =-42694.4444444444 ɑ2 = 0

ɑ2 = 1.69 +

λ 1.69 + -0.00319 λλ -136.010362694301 λ + 0.21045221826

0.210452218259391 λ² = 0 …Persamaan 7

ω1 = 11.83 rad/sω2 = 32.89 rad/s

φ11 = 0.064 φ

φ12 = 0.050φ21 = 0.081φ22 = -0.099

0 for t > 0,1 sec

= P1(t)== 2268 f(t)

= P2(t)= φ12F1(t) + φ22F2(t)= -1485 f(t)

untuk t>0,1

for t ≤0,1 sec

φ11F1(t) + φ21F2(t)

t/td untuk t≤0,1

+ φ12 z2max…Persamaan 10

+ φ22 z2max

inchinch

inchinch

…Persamaan 11

= 15 lb/inch= 9117 lb= 1251 lblb

= 21.347 lb= 3415 lb= -5360 lblb

Diketahui :29000000 lb/inc2 Dari SAP

248 inch4 Dari SAP106 inch4 Dari SAP

30 ft = 360 inc15 ft = 180 inc10 ft = 120 inch

100 lb/ft = 8.3333333333 lb/inch50 lb/ft = 4.1666666667 lb/inch

52500 lb25500 lb

20 psf386 inch/s2

0 …Persamaan 30

…Persamaan 4…Persamaan 5

0

-0.00319 λ …Persamaan 6

= 0λ² = 0

= 0.064 0.050

0.081 -0.099