ME 362 VIBRATIONS I
Lecture 3
VIBRATION OF TWO-DEGREE-OF-FREEDOM SYSTEMS
Faisal Wahib Adam
Mechanical Engineering Department, KNUST
Jan 2014
Models
1
Systems with more than one degree of
freedom
• Many real systems can be represented by a single degree of
freedom model.
• However, most actual systems have several bodies and several
restraints and therefore several degrees of freedom.
• Since no body is completely rigid, and no spring is without
mass, every real system has more than one degree of
freedom, and sometimes it is not sufficiently realistic to
approximate a system by a single degree of freedom model.
• Thus, it is necessary to study the vibration of systems with more
than one degree of freedom.
General Solution-Free Undamped Systems
 .. 
m x   k x  0
 
x  A sin t
Mode Shapes  Principal Coordinates
D   I  xi   0

let    2
k / m   x  0
let D  k/m
D    I   0
D 
1
0
0
0
0
0
D 
2
.
.
.
.
0
.
.
.
.
.
0
.
.
0
.
0
0
0
D 
n
3
i

 x   x   x  
 1   1   1  
 x2   x2   x2  
p      ...  
.
.
.





 

 xn   xn   xn  
1
2
n

Normalised Weighed Modal Matrix
1
p
p
xT mx
General Solution-Forced Damped Systems
 .. 
.
m x  c x  k x  F 
 
 
x  p y
 
 
 

T
T
T
 .. 
.
p m p  y   p c  p  y   p k  p y  p F 
 
 
c   M    K ; Rayleigh Proportional Damping
T
 a1 0 . . 0 
b1 0 . . 0 
 c1 
1 0 . . 0
0

0

 
0 1 0 . . 
0
.
.
0
.
.
a
b
2
2
 .  

c 2 

  ..  
0 . . 0 0  y    0 . . 0 0   y    0 . . 0 0 y   . 
  


  
.
0
.
.
.
0
0
.
.
.
0
0
.
.
.
0






 
 0 . . . a n 
 0 . . . b n 
0 . . . 1
c n 
4
General Solution


y1  a1 y1  b1 y1  c1





y n  an y n  bn yn  cn
x  p  y
1
5
Vibration of two-degree-of-freedom systems
𝑚1 𝑥1 + 𝑘1 + 𝑘 𝑥1 − 𝑘𝑥2 = 0
𝑚2 𝑥2 − 𝑘𝑥1 + 𝑘 + 𝑘2 𝑥2 = 0
𝑥1 = 𝐴𝑠𝑖𝑛(𝜔𝑡 + ∅); 𝑥2 = 𝐵𝑠𝑖𝑛(𝜔𝑡 + ∅)
𝑥1 = −𝜔2 𝑥1 ; 𝑥2 = −𝜔2 𝑥2
Solution cont’d
Algebraic Method
2
𝑘1 + 𝑘 − 𝑚1 𝜔 𝐴 − 𝑘𝐵 = 0
𝑘2 + 𝑘 − 𝑚2 𝜔2 𝐵 − 𝑘𝐴 = 0
𝐵
𝑘1 + 𝑘 − 𝑚1 𝜔2
𝑘
=
=
𝐴
𝑘
𝑘2 + 𝑘 − 𝑚2 𝜔 2
4
𝑚1 𝑚2 𝜔 + 𝑘1 + 𝑘 𝑘2 + 𝑘
− 𝑚1 𝑘1 + 𝑘 + 𝑚2 𝑘1 + 𝑘 𝜔2 − 𝑘 2 = 0
Consider the case where 𝑘1 = 𝑘2 = 𝑘, and 𝑚1 =
𝑚2 = m
2 4
𝑚 𝜔 + 2𝑘 2𝑘 − 2𝑚 2𝑘 𝜔2 − 𝑘 2 = 0
𝑚2 𝜔4 − 4𝑚𝑘𝜔2 + 3𝑘 2 = 0
𝑘
𝐿𝑒𝑡 𝑎 =
𝑎𝑛𝑑 λ = 𝜔2
𝑚
λ2 − 4𝑎λ + 3𝑎2 = 0
λ = 𝑎, 3𝑎
Natural frequencies
𝜔1 =
𝑘
, 𝜔2 =
𝑚
𝑘
3
𝑚
Mode shapes
𝐵1
2𝑘 − 𝑚𝜔1 2
=
= 𝑣1 = 1
𝐴1
𝑘
𝐵2
2𝑘 − 𝑚𝜔2 2
=
= 𝑣2 = −1
𝐴2
𝑘
Solution cont’d
Matrix Method
𝑚1
0
Let
0 𝑥1
𝑘 +𝑘
+ 1
𝑚2 𝑥2
𝑘
λ=𝜔2
(𝑘1 +𝑘) − λ𝑚1
𝑘
−𝑘
𝑘 + 𝑘2
𝑘
2𝑎 − λ
−𝑎
= 0; a =
−𝑎
2𝑎 − λ
𝑚
λ2 − 4𝑎λ + 3𝑎2 = 0; λ = a, 3a
𝑥1
𝑥2 = 0
−𝑘
𝑘 + 𝑘2 − λ𝑚2
𝑥1
𝑥2 = 0
𝑘1 + 𝑘
−𝑘
−λ
𝑚1
𝑚1
=0
−𝑘
𝑘2 + 𝑘
−λ
𝑚2
𝑚2
Consider the case when 𝑘1 = 𝑘2 = 𝑘, and 𝑚1 = 𝑚2 = m
2𝑘
−λ
𝑚
−𝑘
𝑚
−𝑘
𝑚
=0
2𝑘
−λ
𝑚
𝜔1 = 𝑎 =
𝑘
𝑟𝑎𝑑/𝑠
𝑚
𝜔2 = 3𝑎 =
3𝑘
𝑟𝑎𝑑/𝑠
𝑚
Eigen vectors/Mode shapes
λ = λ1 = 𝑎
𝑎
−𝑎
−𝑎 𝐴1
= 0;
𝑎 𝐵1
𝐵1
= 𝑣1 = 1
𝐴1
λ = λ2 = 3𝑎
−𝑎
−𝑎
−𝑎 𝐴′′
= 0;
−𝑎 𝐵′′
𝐵2
= 𝑣2 = −1
𝐴2
Solution cont’d
General solutions
𝑥1 = 𝐴1 𝑠𝑖𝑛 𝜔1 𝑡 + ∅1 + 𝐴2 𝑠𝑖𝑛(𝜔2 𝑡 + ∅2 )
𝑥2 = 𝐵1 𝑠𝑖𝑛 𝜔1 𝑡 + ∅1 + 𝐵2 𝑠𝑖𝑛(𝜔2 𝑡 + ∅2 )
𝐵1 / 𝐴1 = 1; 𝐵2 / 𝐴2 = −1;
𝑥1 = 𝐴1 𝑠𝑖𝑛 𝜔1 𝑡 + ∅1 + 𝐴2 𝑠𝑖𝑛(𝜔2 𝑡 + ∅2 )
𝑥2 = 𝐴1 𝑠𝑖𝑛 𝜔1 𝑡 + ∅1 − 𝐴2 𝑠𝑖𝑛 𝜔2 𝑡 + ∅2
The constants 𝐴1 , 𝐴2 , ∅1 , ∅2 could be determined from the initial conditions.
Example 2
Consider the simple two-degree-of-freedom system with a harmonic force applied to one mass as indicated
in the Figure. For this example, let m1 = 9 kg, m2 = 1 kg, k1 = 24 N/m, k2 = 3 N/m, c1 = 2.4 N· s/m and c2 = 0.3
N·s/m. Calculate
1
0
i. natural frequency ii. Free response ii. Steady- State Response
x(0) =
, 𝑥(0) =
0
0
F.B.D ?
Equations of motion ?
9
0
0
2.7 −0.3
27 −3
𝐱+
𝐱+
𝐱=
1
−0.3 0.3
−3 3
0
3𝑐𝑜𝑠2𝑡
i.
9 0
27 −3
𝐱+
𝐱=𝟎
0 1
−3 3
27 − 9λ −3
= 0;
−3
3−λ
λ=2, 4
Natural frequencies
𝜔1 = 2 = 1.414 𝑟𝑎𝑑/𝑠
𝜔2 = 4 = 2 𝑟𝑎𝑑/𝑠
Free- Response
First mode, λ=2
27 − 9(2)
−3
𝐶1
= 0;
−3
3 − (2) 𝐶2
𝐶2 3
=
𝐶1 1
Second mode
27 − 9(4)
−3
𝐶1
= 0;
−3
3 − (4) 𝐶2
𝐶2
3
=−
𝐶1
1
Modal Matrix, p
11
1  1
p

3
3


 x1  1
 1
     A sin(1t  1 )    B sin(2t  2 )
3
 x2  3
x1  A sin(1t  1 )  B sin(2t  2 )
x2  3 A sin(1t  1 )  3B sin(2t  2 )
Applying Initial Conditions
1
x(0)    and
0
 .  0
 x(0)   

 0
1  2   / 2
A  1 / 2, B  1 / 2
x1  0.5 sin( 2t   / 2)  0.5 sin(2t   / 2)
x2  1.5 sin( 2t   / 2)  1.5 sin(2t   / 2)
Steady- State Response
Modal Matrix, p
1  1
p

3
3


 
9 0 1
x ' m x  1 3
 18



1
1
0 1 3
27 3 1
x ' k  x  1 3
 18



1
1
 3 3 3
 
Normalised Weighed Modal Matrix
1
p
p
T
x mx
p
12
1
18
1  1
3 3 


Solution
 
0.2
p c  p  
 0
T
 .. 
.
m x  c x  k x  F 
 
 

1
p F   2.12 cos 2t  
1
T
x  p y
 
 .. 
p m p  y  
 
T
1
p m p  
0
T
 
 
 
2 0
p k  p  

0
4


T
13
 

T
T
.
p c  p  y   p k  p y  p F 
 
0
1
T
0 
0.4




y1  0.2 y1  2 y1  2.12 cos 2t
y1  0.4 y1  4 y1  2.12 cos 2t
y1 

1     2 
2 2
2
y1  1.04 cos(2t  0.198)
y2  2.65 cos(2t   / 2)
Solution
x  p y
1 1  1 1.04 cos(2t  0.198)
x 
3 3   2.65 cos(2t   / 2) 
18 


x1  0.245 cos(2t  0.198)  0.625 sin 2t
x2  0.736 cos(2t  0.198)  1.875 sin 2t
14
Assignment 1
The Figure shown, shows a system consisting
of two bodies and three springs with a
harmonic force
F (t) = 15 cos 3t kN, applied to one mass.
Determine the natural frequencies of this
system and calculate its steady-state
response. Obtain the solution in the physical
coordinate system x(t). Work from first
principles using matrix methods.
Take m1 =7 kg, m2 =17kg, k1 = 30 kN/m, k2 = 20
kN/m, and k3 = 16 kN/m, where k denotes
spring
constant.
Assume
zero
initial
conditions.
Assignment 2
Find the natural frequency and the Free response at zero initial conditions
of the system shown below
Assignment 3
The figure shows a trolley of mass M, which runs on a
frictionless horizontal plane. A t O the trolley carries a
simple pendulum of length l with a body of mass m at
its end.
Two equal springs, each of stiffness k, are attached to
the trolley and to the fixed walls. By using the
independent co-ordinates x and 𝜑 as shown in the
figure, determine, for the small free oscillations, the
equations of motion.
𝑀 + 𝑚 𝑥 + 2𝑘𝑥 + 𝑚𝑙 𝜑 = 0
𝑥 + 𝑙 𝜑 + 𝑔𝜑 = 0
If M=100 kg, m=10 kg, k= 2 kN/m, l= 2 m. Determine
the natural frequencies of the system.