Vibration and Machine Dynamics Tutorial Problems with Solutions

This article contains solved tutorial problems on Vibration and Machine Dynamics. The problems cover topics such as viscous damper, Newton's second law of motion, moment of inertia, hoisting system, railway buffer, and door motion. The solutions are provided with step-by-step explanations and relevant equations. The subject code, course name, and university are not mentioned.

VIBRATION AND MACHINE DYNAMICS
ENMEC4060
ASSIGNMENT TASK I
TUTORIAL PROBLEMS
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TUTORIAL PROBLEMS
QUESTION 1
A viscous damper consists of a piston in a cylinder containing oil of viscosity 0.38 Ns/m
as a damping fluid. The length of the piston is 100 mm and it has four 6 mm holes between its
faces. If the damping constant is required to be 26 kNs/m, find the piston diameter.
Solution
Damping constant =26kNs/m
viscosity = 0.38 Ns/m
piston length = 100 mm
four 6 mm holes between its faces
piston diameter, D
To obtain the diameter of the piston, we find the area of the holes and relate it to the piston.
d= 0.006 m
The volume displaced by the piston,
Using the velocity of laminar flow in the cylinder,
v= d2 ∆ P
32 μL ... (Equation 1)
Replacing the velocity component in the relationship on equation 1
C ˙x= 32 vμL
d2 A … (Equation 2)
The relationship is further simplified so that the result is given as,
A ( A−4 a ) ˙x= 4 av d2 C ˙x
32 vμL
A ( A−4 a ) ˙x= 4 av d2 C ˙x
32 vμL
A ( A−4 a ) VA = ( A−4 a ) a ( d2 F
8 μL )= ( A−4 a ) A= 1
V a ( d2 F
8 μL )=a ( d2 c
8 μL )= a2 c
2 μπL

Solving the quadratic equation formed, it is possible to obtain the relationship between the area
of the piston and its diameter alongside the velocity.
A2−4 Aa− a2 c
2 μπL =0 → A=D=V
Replacing with values, het roots of A are obtained as,
A=2 a ±0.5 ( √16 a2−( aC d2
2 μL ) )
Fitting the values as provided in the equation as C=100, d=0.006, L=0.1 and a is the area of the
small hole,
A=6.4678e5 m
QUESTION 2
Solutions
The Newton’s second law of motion states that the sum of forces (external and internal) is equal
to the product of mass and the acceleration. There is an assumption that all the conditions are
zero.
Momentum is a product of mass and velocity.
To obtain force from momentum, differentiate the momentum with respect to time,
Simplifying further, considering that mass is a constant,
F=m ( d ( velocity )
dt )→ mdv=Fdt
Integrating further,
The velocity is given as,
v= 400∗0.003
4700 =0.25532 m/s

The maximum deflection and acceleration is given as,
,
For maximum acceleration, assuming that a=0 rad/s2,
Acceleration is achieved by differentiating with respect to time,
The acceleration is obtained as,
acc= 400
4700 =85.1e3 m/s2
The maximum deflection is given at half the velocity,
∆ L
∆ t = v
2 =255.32e-3
2 =12.766e-3 m
QUESTION 3
Moment of inertia of a two-blade aircraft propeller about its rotational axis from the observation
of natural frequency of the free oscillation of the propeller suspended from two light wires
attached to the tips of the blades.
The aircraft propeller is based on yaw and pitch testing the aerodynamic, inertial, and gyroscopic
parameters involved in the rotational motion of the propeller. There are several equations used to
address the rotation of the propeller. Focusing on the moment of inertia,

To find if the number of wires attached to the propeller affect the rotary motion,
for n terms,
Simplifying further,
D
2 θ=hsin β
The n terms,
I ˙θ=−W
n β D
2 + W
n β D
2 −W
n β D
2 −…
I ˙θ=−W
n β D
2 n
I ˙θ=−W
n β D
2
I ˙θ=−Wβ D
2
I ˙θ+Wβ D
2 =0
I ˙θ+W ( D2
4 h θ )=0
The natural frequency based on the moment of inertia is given as,
ωn
2= W D2
4 hI
From the relationship, the number of wires, n, does not affect the inertial status of the motion of
the aircraft propeller.

QUESTION 4
A steel ball of mass supported by undamped springs whose equivalent stiffness. Ball is left to fall
from rest on to tabletop.There are undamped springs attached to the steel ball. When the steel
ball is allowed to drop from the tabletop, it changes from potential energy to kinetic energy.
Assuming that there ar e no losses, PE=KE,
A= √δ2 + V 2
ωx
X ( t ) = Asin ( ωx t−ϕ )
The difference in energy distribution based on the PE and KE levels in the motion,
mgh=m ( ωx
2 δ2 +v2 )
2
kδ=mg
X ( t )= √δ 2+ V 2
ωx
sin ( ωx t−ϕ )
Relating to previous equations, while regarding the spring stiffness,
mg ∆ h=1
2 k δ2
∆ h=
1
2 k δ2
mg

The m, g, k are constant and the change in height is proportional to so as to demonstrate the
energy loss. Therefore, the steel ball may bounce back to its original height in ideal situations but
due to energy loss, there is a change in height.
QUESTION 5
A hoisting system consists of a light 3 m bar hanging vertically of a small servo-driven trolley.
The bar is allowed to swing on its hinge which is controlled by a spring of stiffness 10 kNm/rad
and a damper whose damping coefficient is 800 Nms/rad. The trolley travels on a track built 4 m
about the floor level at the speed profile shown in the figure. The total mass of the basket and the
slewing mechanism (which maintains its horizontal position) is equivalent to 70 kg acting at the
lower end of the bar. The system is designed to handle a payload of 800 kg. Operators from the
loading and unloading stations when the system is in operation.
M L2 ˙θ+C ˙θ+ ( K+ MgL ) θ=−ML ˙y
˙θ=¿ ML ˙y−C ˙θ− ( K + MgL ) θ
M L2
The damping equation is obtained as,
From the Lagrange equation is given as,
The motion is given such that,

QUESTION 6
The structure of the railway buffer when placed on each side,
Resistance= 50kN at a velocity of 0.25m/s
negligible compression, stiffness= 65kN/m
Rigid train of mass 250 tons at a moving speed of 1.6m/s. It collides with the buffer.
Distance taken by train before it stops?
x ( t ) =v ωx sin ωx t
X ( t )=X 0 cos ωx t+ V
ωx
sin ωx t

M ¨xc=Mg sinθ−Ff −kc ( xc−xp )−c ( ˙xc− ˙x p )
mp ¨x p =mp g sin θ+k c ( xc−x p +c ( ˙xc− ˙x p )−Bp ˙xp −P A p )
Ap ˙x p−( V p
β ) dP
dt =Q
m ¨x =PA−B ˙x−K ( x0+x )
m ¨x = ( 50∗0.25∗103 ) −1.6 ˙x−65∗103 ( x0 −x )
25000 ¨x=12500−1.6 ˙x−65000 ( x0 −x )
x=63.45 m towardsthe stop point
QUESTION 7
Moment of the inertia of the door 90kg/m2. Motion of the door if it was opened 1100 and
released. To obtain the time taken to close the door with the indicated conditions above,
From the second order system, time is obtained as,

To obtain the value of natural frequency,
ωx= √ 50
3.567
t= √ 12.52
100 +( 50
3.56 )
t=2.03043 sec