Contoh Soal Modal2

CONTOH SOAL 1

STRUKTUR 3 (TIGA) LANTAI

q = 2 t/m

 

Ø  INPUT DATA

Struktur 3 tingkat

3  0  981  0.07   1   3   3  0.05

1  24000  2587.5    300

2  24000  2587.5    600

3  24000  2587.5    900

KETERANGAN

Ø  BARIS 1

3              = ZONA GEMPA

1              = TANAH KERAS

981         = GRAVITASI

0.07        = SKALA GEMPA

1              = I.K

3              = BANYAKNYA TINGKAT

3              = RAGAM GETAR

0,05        = FAKTOR REDAMAN

Ø  MENGHITUNG NILAI W

W           = q x l

·         W1=W2=W3      = 2 x 12

= 24 t = 24000 kg

Ø  PERIODE GETAR ALAMI

------------------------------------------------------------------------

Mode       Eigenvalue      Omega (rad)     Time  Period    Spectral Acc.

------------------------------------------------------------------------

  1   2.0947870E+0001  4.5768843E+0000  1.3728084E+0000  5.6951705E-0002

  2   1.6445869E+0002  1.2824145E+0001  4.8994966E-0001  7.0000000E-0002

  3   3.4341375E+0002  1.8531426E+0001  3.3905568E-0001  7.0000000E-0002

v  Mode 1 (T1)       = 1,37 detik

v  Mode 2(T2)        = 0,49detik

v  Mode 3(T3)        = 0,34 detik

Ø  POLA RAGAM GETAR

Relative displacement U

            Mode-  1            Mode-  2            Mode-  3

 -1.4485811672E+0000 -1.4584733014E-0001 -2.1510874284E-0002

 -2.6102530681E+0000 -6.4908168206E-0002  2.6823621495E-0002

 -3.2549323361E+0000  1.1696047771E-0001 -1.1937634616E-0002

Ø  DISTRIBUSI GAYA GESER GEMPA

Complete Quadratic Combination (CQC) of  3 Modes

Shear coeff = storey shear / cum. storey weight

Floor      Rel.Displacement    Eqv.Lateral Forces          Storey Shear    Shear Coeff

   1     1.4573742762E+0000    9.7457946061E+0002    3.7709559396E+0003       0.052374

   2     2.6115492649E+0000    1.3797786784E+0003    3.0142845343E+0003       0.062798

   3     3.2561868260E+0000    1.7312163587E+0003    1.7312163587E+0003       0.072134

Ø  DISTRIBUSI BEBAN GEMPA

Complete Quadratic Combination (CQC) of  3 Modes

Shear coeff = storey shear / cum. storey weight

Floor      Rel.Displacement    Eqv.Lateral Forces          Storey Shear    Shear Coeff

   1     1.4573742762E+0000    9.7457946061E+0002    3.7709559396E+0003       0.052374

   2     2.6115492649E+0000    1.3797786784E+0003    3.0142845343E+0003       0.062798

   3     3.2561868260E+0000    1.7312163587E+0003    1.7312163587E+0003       0.072134

Analisa Dinamika Struktur

Single Degree of Freedom System

A single degree of freedom system consists of a mass, a spring, and a damper if the system is modeled as a damped system. The spring is modeled as a linear spring, which provides a restoring force. The damper is modeled as a viscous damper, which provides a damping force proportional to a relative displacement and acting in the direction against a velocity vector. If there is a driving force acting on the mass, the system vibrates under the driving force, which is called forced vibration. Otherwise, the system may vibrate under initial displacement and/or initial velocity, which is called free vibration.