# Ötelenen Elemanlar m F(t) x(t) TEMEL MEKANİK SİSTEM

28.10.2013
TEMEL MEKANİK SİSTEM ELEMANLARI
&Ouml;telenen Elemanlar
x(t)
x(t)
k
F(t)
m
F  mx
D&ouml;nel Elemanlar
1
T  I G 
E1  I G  2
2
IG 
F(t)
1 2
E 2  kx
2
1
F  kx
E1  mx 2
2
 W  F x
1
mL2
12
1
E2  K r2
2
W  T
T  K rθ
1
I G  mR 2
2
x(t)
c
F(t)
F  cx
W   cx x
T  Cr 
W  Cr  
CAD : Katı Modelleme
&Ouml;teleme ve D&ouml;nme Hareketi yapan K&uuml;tle:
1
1
E1  mv G2  I G  2
2
2
1. Kinetik enerji, potansiyel enerji, sanal iş
1
O

2m m  m
m4  m
3
2
3
L
3L
L4  L
R3 
OA 
3
4
9L

OD 
 
16
x
Girdiler: f, T, x4  B 
&Ouml;rnek 1.1:
m2 
2
D
f
A
c 4
k
E1 
E2 
3c
2k
3
B
xB
3c
T 2k
  1 sin   
x4
cos   1
c
k
2
2
2
2
1 2m  3L  1 1 2m  3L   2 1 m 2 1 1 m  L   3( x 4  x B )  1
2
x B 


  

  
  mx 4
22
22 2 3 
L
2 3  8  2 12 3  4 
2
2
1 
3L  1
1
1
2
   2kxB  kx42  kx42
2k  x B 
2
2 
4 
2
2
1
28.10.2013
1
2m m  m
m4  m
3
2
3
L
3L
L4  L
R3 
OA 
3
4
9L

OD 
 
16
x
Girdiler: f, T, x4  B 
m2 
O

2
D
f
A
c 4
3c
xB
3
B
3c
T 2k
2k
  1 sin   
x4
cos   1
c
k
3L   
3L 
 3( x 4  x B ) 

 9L 
 3c x B 
  x B 
   3cx B x B
W  f 
   T 

4
4 
L
16







 cx 4 x 4  cx 4 x 4
0
0
0
3
3
9cL
9cL 
27cL2 
9L
x B  
 x B 

W 
f  Tx 4  Tx B  3cx B x B 
L
L
4
4
16
16
 3cx B x B
 9L
9cL
27cL2  
9cL  
 3
W  
f
x B 
     T  6cx B 
  x B
4
16
4 
 L

 16
k
Q XB
Q
&Ouml;RNEK PROBLEM İ&Ccedil;İN BULUNAN KİNETİK ENERJİ,
POTANSİYEL ENERJİ VE SANAL İŞ:
2
E1 
2
1 2m  3L  1 1 2m  3L   2 1 m 2 1 1 m  L 
x B 


   
 
22
22 2 3
2 3  8  2 12 3  4 
E2 
2
2
 3( x 4  x B )  1
2

  mx 4
L
2
2
1 
3L  1
1
1
2
2k  x B 
   2kxB  kx42  kx42
4  2
2 
2
2
 9L
9cL
27cL2  
9cL  
 3
f
x B 
W  
     T  6cx B 
  x B
16
4
16
L
4 



Q
Q XB
2
28.10.2013
LAGRANGE DENKLEMİ
d  L  L


 Qi
dt  x i  x i
L  E1  E 2
xi : Genel koordinat
Qi : Genel kuvvet
i=1,2,....,n
J.H. Williams, Jr., Fundamentals of Applied Dynamics, John Wiley and
Sons,Inc.,
S
I
1996
2m m  m
m4  m
3
2
3
L
3L
L4  L
R3 
OA 
3
4
9L

OD 
 
16
x
Girdiler: f, T, x4  B 
&Ouml;rnek 1.1 (Devam)
1
m2 
O

2
D
f
c
A
4
3
B
3c
xB
3c
T 2k
2k
k
2
2
1 2m  3L  1 1 2m  3L   2 1 m 2 1 1 m  L 
x B 


  

 
22
22 2 3
2 3  8  2 12 3  4 
E2 
cos   1
c
k
E1 
  1 sin   
x4
2
2
 3( x 4  x B )  1
2
  mx 4
L

2
2
1 
3L  1
1
1
2
2k  x B 
   2kxB  kx42  kx42
2
2 
4 
2
2
 9L
9cL
27cL2  
9cL  
 3
f
x B 
W  
     T  6cx B 
  x B
16
4
16
L
4 



Q
L  E1  E 2
d  L  L
 Q


dt    
Q XB
d  E 1  E 2
 Q


dt    
d   E 1  E 2


 Q XB
dt  x B  x B
3L  9L
9cL
27cL2 
d  18mL2  18mL2   6kL 

 
f
x B 


  
 xB 
4 
4  16
4
16
dt  192
576 

d m
9mL2
3L 
3
9cL 

 x B 
( x 4  x B )   2k  x B 
   2kx B   T  6cx B 

2
dt  2
36L
4 
4
L


mL2  27cL2  9cL
9kL2
3kL
9L


x B 

xB 
f
8
16
4
8
2
16
3m
9cL 
3kL
3
m
  6cx B 
  4kx B   T  x 4
x B 
4
4
2
L
4
3
28.10.2013
mL2  27cL2  9cL
9kL2
3kL
9L


x B 

xB 
f
8
16
4
8
2
16
3m
9cL 
3kL
3
m
  6cx B 
  4kx B   T  x 4
x B 
4
4
2
L
4
 mL2

 8
 0

 
 27cL2
0     
     16
3m      9cL
x
4   B   4

X
M
 9cL      9kL2


4     8
 
 3kL
6c   x B  

 2

X
C

 3kL     
9L
f


2    
16
  

m 
3







x

T
4k  x B 
4
 L
4 

K
F
X
  CX
  KX  F
MX
Doğrusal diferansiyel denklem takımı
&Ccedil;ok serbestlik dereceli titreşim denklemi
&Ouml;ZDEĞER DENKLEMİ
F  0 , serbest titreşim
  CX
  KX  F
MX
X  Aest
[s 2 M  sC  K ]Aest  0
[s 2 M  sC  K ]A  0
det[s 2 M  sC  K ]  0
&Ouml;
&Ouml;zdeğer
denklemi
&Ouml;rnek 1.1 (Devam):
 mL2
 8

 0

    27cL2
0  
  
     16
3m      9cL
x
4   B   4
M

X
 9cL     9kL2
4     8
  
 3kL
6c   x B  
 2

C

X
mL2 2 27cL2
9kL2
s 
s
8
16
8
&Ouml;zdeğer Denklemi:
9cL
3kL
s

4
2
 3kL   
2    0
 
4k   x B 

K
X
9cL
3kL
s
4
2
0
3m 2
s  6cs  4k
4

4
28.10.2013
mL2 2 27cL2
9kL2
s 
s
8
16
8
9cL
3kL

s
4
2
&Ouml;zdeğer Denklemi:
9cL
3kL
s
4
2
0
3m 2
s  6cs  4k
4

m=0.85 kg, L =0.24 m, k=1200 N/m, c=38 Ns/m
0.0061s 2  3.7s  622.08
 20.5s  432
 20.5s  432
0.6375s 2  228s  4800
0
0.0039s 4  3.75s 3  847s 2  141834s  2799360  0
MatLAB kodu:
a=[0.0039,3.75,847,141834,2799360];p=roots(a);vpa(p,4)
clc;clear
m0=0.85;l0=0.24;k0=1200;c0=38;
0 0 85 l0 0 24 k0 1200 0 38
m=[m0*l0^2/8,0;0,3*m0/4];
c=[27*c0*l0^2/16,-9*c0*l0/4;-9*c0*l0/4,6*c0];
k=[9*k0*l0^2,-3*k0*l0/2;-3*k0*l0/2,4*k0];
syms s;p=solve(det(m*s^2+c*s+k));vpa(p,4)
&Ouml;zdeğerler: -104.3+181.4i, -104.3-181.4i, -22.4, -730.2
&Ouml;zdeğerler: -104.3+181.4i, -104.3-181.4i, -22.4, -730.2
Serbest titreşim cevabının formu:
(t )  A 1e 104.3 t cos(181.4t  1 )  A 2 e 22.4 t  A 3e 730.2 t
A1, φ1, A2 ve A3 değerlerini ilk şartlar belirler.
t sonsuza yaklaştığında cevap sıfıra yaklaşır. D&uuml;zg&uuml;n rejim cevabı
olur.
sıfır
&Ouml;zdeğerinin t&uuml;m&uuml;n&uuml;n reel kısımlarının negatif olduğu sistem kararlıdır.
0   2  2
p=-σ+iω
iω
ω0
  0
φ
-σ
p=-σ
  cos 
0 T0  2

1
t 


t   2

 0  104.3 2  181.4 2  209.25 rad/s
104.3
T  t  0.0015

 0.5
t  0
20
209.25
  0 1   2
 T0  0.03
t 
T0

 t   0.06
f0=1/T0
-22.4 i&ccedil;in ∆t=0.0142 ,
t∞=0.28
-730.2 i&ccedil;in ∆t=0.000436 , t∞=0.0086
Sistem i&ccedil;in ∆t=0.000436 , t∞=0.28
5
28.10.2013
1
x( t )  e

t

x
x(t
)
x((t
)
x( t )  e t cos( 1   2 t )
0.
5
ξ=0.1
  10
0.2
t
3
5
t
Bu slaytlar Prof. Dr. Hira Karag&uuml;lle nin sitesinden alınmıştır.
http://kisi.deu.edu.tr/hira.karagulle/
6