MEKANIKA GETARAN
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Transcript of 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