EXP #2 TORSIONAL VIBRATION
CALCULATIONS
1) Torsional Vibration without damping :
𝐺𝐽 𝜏 2
𝐼= 2 ( )
4𝜋
𝐿
𝐼 = 𝑆𝑒𝑐𝑜𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎
𝜋
𝐽 = 𝑃𝑜𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑑 4 , d = 6.2 mm
32
𝐺 = 𝑅𝑖𝑔𝑖𝑑𝑖𝑡𝑦 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 = 80 𝐺𝑁/𝑚2
From fig.1 we can find
𝜏2
𝐿
= 0.0005 which represent the slope of
the curve.
𝐽=
𝐼=
𝜋
∗ 6.24 = 145.066 𝑚𝑚4
32
80 ∗ 145.066
(0.0005) = 0.147 𝑚𝑚4
4𝜋 2
length
(mm)
0
480
502
601
730
830
10t (sec)
0
4.57
4.67
5.37
5.8
6.3
t
0
0.457
0.467
0.536
0.58
0.63
t^2
0
0.208
0.218
0.287
0.336
0.396
EXP #2 TORSIONAL VIBRATION
fig.1 Torsional Vibration without Damping
0,45
0,4
period 𝜏^2 (sec)
0,35
0,3
0,25
y = 0,0005x - 0,0077
R² = 0,9941
0,2
0,15
0,1
0,05
0
-0,05
0
100
200
300
400
500
600
700
800
900
length (mm)
2) Torsional Vibration with damping:
To find the logarithm decrement we use the next equation
𝛿=
2𝜋𝜉
√1 − 𝜉 2
=
1 𝑥𝜊
𝑥1
𝑙𝑛
= 𝑙𝑛
𝑛 𝑥𝑛
𝑥2
𝜉 = 𝑇ℎ𝑒 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑟𝑎𝑡𝑖𝑜
Experimentally we found the following data when the
term √1 − 𝜉 2 ≅ 1
Levels
𝒙𝝄
𝒙𝒏
𝒏
𝝃
1
15
3
24
0.01067
2
11
4
15
0.01073
3
19
6
14
0.01310
EXP #2 TORSIONAL VIBRATION
DISCUSSTION
1- Discuss the source of inaccuracies for both parts of the experiment?
- The machine itself may has a small deviation from the correct way of
installation and fixing in its jaws over the years.
- Human errors will cause a source of inaccuracies such as not taking
the time properly, not holding the wheel in the right way can make
inaccuracy in length readings.
- For the second part the installation of the oil container can be
unstable which therefore will cause the wheel to touch the container
and add an extra unwanted damping.
2- Mention another method to estimate the moment of inertia for a
wheel then compare with the method of this experiment?
- We can divide the wheel into equal parts and calculate the moment
for each part as the following then adding them together total kinetic
energy is
K = ½ (m1r12 + m2r22 + …)Ω2 = ½ (∑miri2)Ω2
The sum of the products of the masses of the particles by the squares of
their distances from the axis is
I = ∑miri2
So the kinetic energy of the entire rigid body can now be written as,
K = ½ I Ω2
We integrate I for a rigid body with a continuous distribution of
matter
EXP #2 TORSIONAL VIBRATION
I = ∫r2dm
And calculating it for a uniform solid cylinder rotating about a central
diameter we find:
I = MR2/4 + ML2/12
3-
Give some practical applications for the torsional vibration?
Torsional Model of a Compressor Drive Train
Turbomachinery drive train
Synchronous Electric Motor Drive Trains
4- Discuss the effect of increasing oil viscosity on the damping factor?
- The resistance offered by the fluid to the moving body it what leads
to the absorption of energy and that depends on many factors
including the viscosity, when the viscosity increases the damping
ratio will increase along with the change of harmonic motion.