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Q3.(a)

m
l
Assume the rigid bar moves by , then m moves by l. The location at a
moves by a. The equation of motion about  (taking moments about the
pin joint) is
ml 2  [k (a )  c(a)]a
Q5. (P2-11)
Assume the friction force beneath the wheel is
f. Equations of motion for the x and  are
mx  kx  f
J    fr
x
f
kx

o
Merge into a single equation as
J o
mx  kx 
r
0
Using geometric relationship x  r , one gets
( J o  mr 2 )  kr 2  0
and
k
(m  J o / r 2 )
n 
x1
Q6.
f1
f1
m
kx1
Equilibrium about the central support
f1a  f 2 b  0 ( f 2  k2 x2 )
Geometric relationship
f2
x2
x2
x
 1
b
a
Equation of motion
mx1  f1  k1 x1
finally
b2
mx1  ( k1  k2 2 ) x1  0
a
Q7. (P3-20)
The ground excitation is y (t )  Y sin

2 πv
.
L
When  
2πv
k

,
L
m
2 πvt
L
and so the excitation frequency is
there is resonance and the largest
acceleration occurs.
Q8.
fsint
The equations of motion for x and  are
mx  kx  R  f sin  t
m (d  d ) 2   Rd
b
1
2
x
m
1
Using geometric relationship x  d1 , one gets
kx
R
[md12  mb (d1  d 2 ) 2 ]x  kd12 x  fd12 sin  t
mb
R

Q9. (P5-20)
Assume the center of mass moves
downwards by x and the rod rotates clockwise around it by .
x1
The equations of motion for x and  are
mx  kx1  kx2  k ( x1  x2 )
kx2
kx1
J c  kx1
l
l
 kx2
4
2
The geometric relationship between displacements is
l
l
x1  x  
x2  x  
4
2
The final equations of motion become
x
x2
kl 

2
k
m 0   x 
4  x  0

 0 J    kl 5kl 2   
c   

 
 4 16 
Q10.
The matrix equation of motion is
3m 0   x1   2k  k   x1 
 0 m x    k 4k   x   0

 2  
 2 
The corresponding eigenvalue problem is
0   a1 
 2k  k 
2 3m
(



 0 m)a   0

k
4
k



  2
This requires
 2k
det( 
 k
 k
0
2 3m


 0 m )  0
4k 


This leads to
(2k  3m 2 )(4k  m 2 )  k 2  0
or
3m 2 4  14mk 2  7k 2  0
This quadratic equation
14  14 2  4  3  7 k
k
k
 
 0.57 and 4.10
6
m
m
m
2
Substituting each of these  into (2) yields modes as
 1 
 1 
k
k
and 

 for 1  2.025
 for 2  0.755
m
m
0.29
 10.3
2
Q11.
Because the center of gravity is offset from the elastic
center, the wing vibrates vertically and torsionally. The
two degrees-of-freedom are vertical displacement x
measured from the elastic center O and angular
displacement  rotating about O in clockwise direction
respectively.
(1)
(2)
(3)
The equation of motion of the airfoil is
m(x  eg)  kx, J o  meg x  K
mx  meg  kx  0 , meg x  J o  K  0
Assume x(t )  A exp( it ) ,  (t )  B exp( it )
It follows that
k  m 2  meg 2   A
0

2
2  

me

K

J

B
g
o

  

x
G
O
This leads to
m( J o  meg2 ) 4  (kJo  mK ) 2  kK  0.76 4  1800 2  1000000  0
Two natural frequencies can be found as 29.83rad/s and 38.45rad/s (also confirmed
using MATLAB).
The lift force and pitching moment acting at the aerodynamic center are
L  qSa ,
M  qScM,0 c (q 
1
V2 )
2
The moment equilibrium about center O is
Lea  K  M  0
It follows that

qScM,0 c
qSaea  K
Divergence occurs when the denominator of the above equation becomes zero.
This allows the critical speed for divergence to be determined as
Vdiv 
2K

Saeg
2  100
 68.04m/s
1.2  0.3  1  6  0.02
The equation of motion of the wing under air flow is
mx  meg  kx  L , meg x  J o  K  M
There are mainly three reasons why flutter may occur. (1) Inertia coupling: the
terms represented by meg which links the acceleration of x with acceleration of .
(2) Aerodynamic coupling: a wing rotation  causes a lift change which in turn
alters translation x; or a translation x (so a vertical linear velocity) alters effective
incidence angle which brings about a lift change that affects wing rotation . (3)
Stiffness coupling: the force in the vertical spring induced by x produces a torque
about a reference point while a wing rotation about the same reference point
produces a vertical force in the vertical spring. Since the reference point in Figure 4
is the elastic center of the wing, elastic coupling is not present in this question.
Flutter can be prevented in theory by removing the above-mentioned inertia,
aerodynamic and elastic coupling by arranging for the center of gravity, the center
of independence and the elastic center to coincide. Increasing the stiffness of a
wing also helps to raise the flutter speed, though this normally carries a weight
penalty. It is difficult to make the center of independence coincide with the elastic
center in practice.
Q12
M x3 k1x1
k2x2
First, you need to define the degrees-offreedom and then the displacements
forces
concerned in terms of these degrees-offreedom. Assume the center of gravity G

moves upwards by x and the rod rotates
x3 x1 x
x2
anti-clockwise around it by these are the
def chosen degrees-of-freedomDenote x1
displacement profile
and x2 as displacements at k1 and k2
respectively, and x3 as the displacement of
the lumped mass M).
The geometric relationship between displacements is
x1  x  l1
x2  x  l2
x3  x  l3
With displacements defined, forces acting on the rigid bar can be
defined.
The difficult is how to deal with the lumped mass M. A convenient way
is to use D’Alembert principle: a force Mx3 is placed at the location (in
place) of the lumped mass in the direction opposite to x3.
The equations of motion for x and  are
mx  k1 x1  k2 x2  Mx3
J   k x l  k x l  Mx l
G
1 1 1
2
2 2
3 3
Substituting the above relationship into the previous equations of motion
and further mathematical manipulation yields
( m  M ) x  Ml3  ( k1  k2 ) x  ( k2l2  k1l1 )  0
( J G  Ml32 )  Ml3 x  ( k2l2  k1l1 ) x  ( k1l12  k2l22 )  0
or more compactly
m  M
  Ml
3

 Ml3   x  k1  k2
 
J G  Ml32   k2l2  k1l1
k2l2  k1l1   x 
 0
k1l12  k2l22   
These are the (final) equations of motion of the structure in terms of x
and , which are amenable to further mathematical treatment (for
example, for determining the natural frequencies).