MEKANIKA GETARAN

MEKANIKA GETARAN

JATI SUNARYATI

OBJECTS

1. SINGLE DEGREE OF FREEDOM (SDOF) 1.1 Free Vibration 1.2 Force Vibration 2. MULTY DEGREE OF FREEDOM (MDOF) 1.1 Modal Superpotition 1.2 Numerical Solutions

Single Degree of Freedom

)(tPkuucum

Free Vibration

P(t) = 0

Undamped , c = 0

mü + ku = 0

Damped

mü + cú + ku = 0

0c

Undamped , c = 0

mü + ku = 0 ……………………………….. (1)

The solution (linier, homogeneous, second order differential equation) is :

steu ……………………………….. (2)

Substitution into Eq. (1) gives

0)( 2 stekms ………………….. (3)

The characteristic equation is

niskms 2,1

2 0)( ………….. (4)

The general solution of Eq (1) is :

……………….. (5) tsts

eAeAtu21

21)(

Undamped , c = 0

mü + ku = 0

Remember : de Moivre’s theorem

i

eex

eex

ixixixix

2sin

2cos

Substitution Eq (4) to Eq (5) gives :

titi nn eAeAtu

21)( ……………….. (6)

Eq (6) can be written as

and

tBtAtu nn sincos)( ………………… (7)

tBtAtu nnnn cossin)( ………………… (8)

At the time zero,

u(0) = A and ú(0) = ωn B ………………… (9)

The solution is

tnn

utnutu

sin

)0(cos)0()(

……………… (10)

-2

0

2

0 5 10 15

t

u

Amplitude,uo

Tn = 2/n

u(0)

ú(0)

a

b

c

d

e

Undamped , c = 0

mü + ku = 0

Damped , c 0

mü + cú + ku = 0

………………. (11 a)

By dividing by m

022

uuu nn ………………. (11 b)

where

mkkmnmcrc

crcc

nmc

222

2

………………. (12)

………………. (13)

Critical damping coefficient

Damping ratio

Type of motion

1. c = ccr ξ = 1 critical damped system

2. c > ccr ξ > 1 overdamped system

3. c < ccr ξ < 1 underdamped system

The solution of Eq (11 b) is Eq (2), and substitution into Eq (11 b) gives

0)2(22 st

nn ess ………………. (14)

Which is satisfied for all values of t if

0222 nnss

………………. (15)

That has two roots :

2

2,1 1 is n………………. (16)

titi nn eAeAtu

21)(

Hence the general solution is

………………. (17)

Undamped , c ≠ 0

mü + ku = 0

Substitution Eq (16) into Eq (17) gives

)()( 21

titit DDn eAeAetu

………………. (18)

If the initials condition at the time zero u(0) and ú(0)

The equation becomes

where

21 nD………………. (18)

Eq (16) can be written as trigonometric function as

)sincos()( tBtAetu DD

tn

………………. (20)

Undamped , c ≠ 0

mü + ku = 0

t

uutuetu D

D

nD

tn

sin)0()0(

cos)0()(

…. (21)

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8

TD = 2/D

u

t

Undamped , c ≠ 0

mü + ku = 0

Eq(21) indicates that the displacement amplitude decays exponentially with time. The envelope curves en

t, where

and

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1

(D/n)2+2=1

Range of damping for structure

D

n

n

D

T

T

2

2 )0()0()0(

D

nuuu

………………. (22)

21

TnTD

………………. (23)

Decay of motion

A relation between the ratio of two successive peaks of damped free vibration given by

Called logarithmic decrement

21 1

2ln

i

i

u

u………………. (24)

If is small, 11 2 and this gives an approximate equation

2 …………………. (25)

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8

TD = 2/D

u

t

u1 u2 u3

Over j cycles the motion decreases from u1 to uj+1, the ratio is given by

j

j

j

j

eu

u

u

u

u

u

u

u

u

u

14

3

3

2

2

1

1

1 ... ……………….. (26)

therefore

2)/ln()/1 11 juuj ……………….. (27)

Damping ratio

0

2

4 6 8

10

0.2 0.4 0.6 0.8 1

=2

21

2

Loga

rith

mic

dec

rem

ent

Decay of motion