Vibration and Machine Dynamics Tutorial Problems with Solutions

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