BK70A0500 Machine Dynamics (6 ECTC), Intermediate test 19.10.2015
Professor Jussi Sopanen, LUT School of Energy Systems / Mechanical Engineering
1(3)
No material allowed in the exam! Pocket, scientific or programmable calculator can be used.
1. Derive the equations of motion for the system shown in Figure below using Lagrange’s method.
(6 p)
2. a) Two systems are shown in Figures I and II. Four equations of motions (1, 2, 3 and 4) are given
below. Which set of equations is correct for the system’s shown in Figures I and II? (2 p)
b) Select one system (either Figure I or Figure II) and derive the equation of motions using
Newton’s method. (4 p)
Systems:
Figure I
Figure II
Equations of motion:
1.
2.
3.
4.
m
0
x1
c
0
x1
k
0
x1
0
0
m
x2
0 2c
x2
0 2k
x2
0
m
0
x1
c
c
x1
k
k
x1
0
0
m
x2
c 2c
x2
k
2k
x2
0
m 0
0 m
x1
c
c
c
c
x1
k
k
k
k
x1
0
0
2c
c
2k
k
c
c
k
k
m
0
0
m
x2
x1
x2
x2
x1
x2
x2
x1
x2
0
0
Continues on next page…
BK70A0500 Machine Dynamics (6 ECTC), Intermediate test 19.10.2015
Professor Jussi Sopanen, LUT School of Energy Systems / Mechanical Engineering
2(3)
3. The process of modal analysis in the calculation of a 2DOF system response to initial conditions is
partly shown below. However, there are three sections where the analysis has produced multiple
solutions, but only one of them is correct. Your job is to identify the correct results and explain
why they are correct. (6 p)
Mass and stiffness matrices of the system and the initial conditions are:
M
1 0
0 4
, K
12
2
x0
2 12
1
1
,
0
x0
0
Part I: Three options for the mass normalized stiffness matrix are:
12
A) K
1
1
3
B) K
6
0
0 12
C) K
12
2
2
3
The eigenvalues and the matrix of eigenvectors obtained by computer solution are:
1
2.8902 ,
2
12.1098 ,
P
0.1091
0.9940
0.9940
0.1091
PartII: Three options for the modal matrix are:
A) S
0.062
0.998
0.333
0.021
B) S
0.109
0.944
0.497
0.055
C) S
0.29
0.957
0.497
0.145
Part III: Three options for the modal initial conditions are:
A) r0
0
0
, r0
1.624
1.537
B) r0
0.290
0.957
, r0
0
0
C) r0
2.204
0.378
, r0
0
0
Finally, explain the remaining steps needed to obtain the physical displacements x (t ) . Note
that is not required to calculate the physical displacements.
Continues on next page…
BK70A0500 Machine Dynamics (6 ECTC), Intermediate test 19.10.2015
Professor Jussi Sopanen, LUT School of Energy Systems / Mechanical Engineering
3(3)
4. Consider the torsional vibration analysis of the geared system shown in Figure below.
50J
Gear ratio n = n2/n1
4k
10J
20J
2k
20J
k
5J
a) Draw the equivalent shaft system (for torsional analysis) and derive the stiffness and mass
matrices. (4 p)
b) Computer solution of the natural frequencies ( ) and mode shapes ( u ) of the system gives
the following results. Draw the mode shapes and identify the rigid body mode (2 p).
0
17.006
24.018
50.338
rad/s
u
1
1
1
0.277
1
0.442
0.187
1
1
1
0.067
0.43
0.607
0.464
0.285
0.031
5. In a damped 2 DOF system, the mass and stiffness matrices are M and K as given below.
M
9 0
0 1
, K
27
3
3
3
The damping matrix C is unknown, but it is proportional to the mass and stiffness matrices.
Furthermore, the results of the Damped Eigenvalue Problem are available and a computer
solution yields the following eigenvalues:
1,2
0.0707 1.4124 j
3,4
0.2000 1.9900 j
Determine the proportional damping matrix C . (6 p)
BK70A0500 Machine Dynamics (6 ECTC), Intermediate test
Appendix: Equations
1 DOF Undamped Free Vibration:
mx kx
Natural frequencies (rad/s or Hz):
1 k
2
2
m
A sin( n t ) B cos( n t ) , where A v0 and B
n
Solution with Initial Conditions (ICs):
0
k
m
x (t )
1
T
fn
n
x0
n
or alternatively
x (t )
C sin(
n
2 2
n 0
v02 and
x
) , where C
t
tan
x
1
n 0
v0
n
Velocity and acceleration: x(t )
x(t )
n C cos( nt
2
n
C sin( nt
1 DOF Damped Free Vibration:
Critical damping:
ccr
2m
Damped natural frequency:
Logarithmic decrement:
Solution with ICs:
x(t ) e
)
mx cx kx
0
2 km , damping ratio:
n
d
ln
)
x(t )
x(t T )
nt
A sin
c
2m
c
2 km
n
2
1
n
c
ccr
2 , damping ratio:
T
n d
, n periods:
n
2
4
x and B
d
, where A
dt
B cos
dt
v0
1
x( t )
ln
n
x(t nT )
2
x0
n 0
d
Equations of motion:
Newton’s II law:
M
J
T
qi
U
qi
D
qi
generalized non-conservation forces Q
i
W
qi
Lagrange’s equation:
F
d
dt
mx,
T
qi
Kinetic energy (rectilinear motion): T
Rayleigh dissipation function D
n
1
2
i 1, 2, ...., n ,
Qi
xj
m
Fj
qi
j
1 2 , potential energy of spring:
mv
U
2
1 2
kx
2
n
cij qi q j
i 1 j 1
Modal Analysis
1
Cholesky decomposition in a special case of diagonal mass matrix:
m1
1/ 2
L
M
0
The mass normalized stiffness matrix: K
Eigenproblem:
Kv i
Matrix of eigenvectors:
P
Modal (spectral) matrix:
Km
i
v1
vi ,
M
1/ 2
M
M
m1
1/ 2
0
0
1
m2
1/ 2
i
v2
vn
2
i
P KP diag ( i ) diag (
q Pr ,
M 1/ 2 q ,
1/ 2
P
x
S
r0
2
i
)
1
Sr
T
P L PT M1/ 2
1
S x0
i ri (t )
K , where relations 2
Direct solution method for harmonic forced response
Mx Cx Kx Fs sin f t Fc cos f t
x (t ) a sin f t b cos f t ,
x (t )
f a cos f t
f b sin
Solution of coefficient vectors a and b
2
Fs
a
K
fM
fC
2
Fc
b
K
fC
fM
KM
m2
L1
2
i
T
Coordinate transformations: x
Modal matrix:
S L 1P M
Modal ICs:
r0 S 1x 0
Modal Damping in modal coordinates: ri t
Proportional Damping: C
L 1K L
1 T
,
0
2
i i
r t
i
i
f
t,
0
i 1, 2,
2
i
,n
and
x(t )
i
i
2
f
2
a sin
i 1, 2,
2
i
f
t
2
f
b cos
f
, n must hold
t
Tuning of undamped vibration absorber k2
m2
2
f
BK70A0500 Machine Dynamics (6 ECTC), Intermediate test
Appendix: Equations
Numerical Integration
Equation of motion:
Initial conditions:
Mx(t ) Cx(t ) Kx(t ) F(t )
y1 (0)
x(0)
y 2 (0)
State space equation: y(t ) Ay (t ) f (t ) , where A
x(0)
0
M 1K
I
M 1C
, y(t )
y2
z 2 dA
y1 (t )
x (t )
x (t )
y 2 (t )
and f (t )
0
M 1F (t )
Torsional vibrations
Natural frequencies and mode shapes of two inertia system:
0
1
A1
A2
0
k ( J1 J 2 )
J1 J 2
1
0
A1
A2
J2
J1
1
Polar area moment of inertia for circular cross-section: I
p
A
d4
32
Rules for geared system’s reduction to equivalent system (n is gear ratio):
Damped eigenvalue problem
Equation of motion Mx Cx Kx
State space format: A
0
0
i
Eigenvalues and vectors:
eigenvalue problem Az
M 1C
j
i
i
i
ui
i ui
zi =
where
,
I
M 1K
ni
i
j
i
zi =
i
i
ui
i ui
i 1
z
i
ni
i
ni
2
1
ni
ni
j
i
2
1
i
j
is natural frequency of mode i
is damping ratio of mode i
Mathematics
Determinant for characteristic equation (for solution of natural frequencies): K
2x2 matrix determinant: A
a
b
c
d
A
ad
cb
2x2 matrix inverse: A
2
M
0
a
c
b
d
A
1
1
A
d
c
b
a