STUDENT NAME
MD ATIQUR RAHMAN FAISAL
STUDENT ID
SCM-012154
COURSE
BACHELOR IN MECHANICAL ENGINEERING
LECTURER
DR. CHIA
SUBMISSION DATE
19 April 2012
SUBJECT
ENGINEERING MECHANICS (EAT 227)
th
TORSION VIBRATION
Experiment Title:
Torsion Vibration.
Introduction
Torsion is the twisting of a metallic rod shaped object, when a torque is applied on
:
two sides’ perpendicular to the radius of a uniform cross -sectional bar.
Objective
:
Determining the natural frequency of a system undergoing tortional vibration.
Theory
:
Using Newton’s second law of tortional system.
T I …………………. ( Equation 1 ) o
where Io = mass moment of inertia of the t he disk
……..……... ( Equation 2 ) Hence, k I o where k = k = torsional stiffness of the shaft
Rearrange Equation 2 2
0 .………..……... ( Equation 3 ) n where natural frequency of the system,
n
k
I o
…..…….…..……... ( Equation 4 )
From Simple Theory of Torsion,
T J
where T = Applied torque
G
R
L
J = Polar second moment of area
= Shear stress
R = Radius of shaft
G = Shear modulus
= Angle of twist
L = Length of shaft As torsional stiffness k
MD. Atiqur Rahman Faisal
T
, it can be determined through k
GJ
L
………….. ( Equation 5 )
Page 2
TORSION VIBRATION
Apparatus
:
One solid circular disk with mass = 4.536kg, diameter = 150mm and thickness = 30mm. One annular circular disk with mass 1.89kg, outer diameter 150mm, inner diameter = 110mm and thickness = 30mm. Two chucks; one steel rod; one stopwatch.
Procedure
:
1. The diameter of the provided rod is measured at three different locations to get the average diameter of the rod. 2. The anchor is chucked tightly to the solid circular disk. 3. The length of the rod or the distance between the two chucks is initially kept 30cm. 4. The disk is displaced slightly, so that the rod can be twisted. 5. The disk is released and the stopwatch is switched on simultaneously. 6. The time taken is recorded according 10, 20, 30, 40 and 50 cycles of the disk. 7. From step 3 to step 6 is repeated by increasing the length between the two chucks from 35 cm to 40 cm. 8. The whole procedure is repeated by attaching the annular circular disk on top of the solid disk.
MD. Atiqur Rahman Faisal
Page 3
TORSION VIBRATION
Results
:
Length
0.3m
0.4m
0.5m
Number of cycles 10 20 30 40 50
4.74 8.93 13.27 18.33 23.05
4.89 10.08 14.64 19.67 25.73
5.08 10.55 15.7 21.14 26.45
Given Information
Time, (s) 0.3m(With Annular disk)
4.53 11.25 16.77 22.08 27.49
0.4m(With Annular disk)
0.5m(With Annular disk)
5.55 11.62 17.77 24.24 30.37
7.24 13.33 19.55 25.4 31.71
:
Mass of the circular disk = 4.536kg. Diameter of the circular disk = 150mm. Thickness of the circular disk = 20mm. Mass of the annular disk = 1.86kg. Outer diameter of the annular disk = 150mm. Inner diameter of the annular disk = 110mm. Thickness of the annular disk = 30mm. G ( Shear module ) = 80GPa. Sample calculation
:
We know, J (polar second moment of area) =
() = = 2.5132710
-11
(moment of inertia) = mr = (4.536) (0.075) = 0.0127575 (circular disk) We know, (moment of inertia) for annular disk = = =0.28728. 2
We know,
2
= () = 6.70205 = 22.97 rad/s We know, (angular speed) = √ = √ = = 0.2735 We know, (shear stress) = We know, K (torsion K (torsion stiffness of the shaft) =
MD. Atiqur Rahman Faisal
Page 4
TORSION VIBRATION
Comparing experimental and theoretical value of
= 2 = 2
,
×π× f, where f is collected collected from graph. ×π×2.17 = 13.63rad/s
Percentage errors
:
×100%, = 68.525%
Plotted Graph
:
N vs T ,L=0.30m (without (without annular disk) 60 y = 2.17x + 0.3489 50 N vs T ,L=0.30m (without annular disk)
40 30
Linear (N vs T ,L=0.30m (without annular disk))
20 10 0 0
5
10
MD. Atiqur Rahman Faisal
15
20
25
Page 5
TORSION VIBRATION
N vs T ,L=0.35m ( without annular disk ) 60 y = 1.9458x + 0.809 50 N vs T ,L=0.35m ( without annular disk )
40 30
Linear (N vs T ,L=0.35m ( without annular disk ))
20 10 0 0
10
20
30
N vs T ,L=0.4m ( without annular disk ) 60
y = 1.875x + 0.4051
50 40
N vs T ,L=0.4m ( without annular disk )
30
Linear (N vs T ,L=0.4m ( without annular disk ))
20 10 0 0
10
20
30
N vs T, T, L=0.30m ( with annular disk ) 60 y = 1.758x + 1.1261 50 N vs T, L=0.30m ( with annular disk )
40 30
Linear (N vs T, L=0.30m ( with annular disk ))
20 10 0 0
10
MD. Atiqur Rahman Faisal
20
30
Page 6
TORSION VIBRATION
N vs T, L=0.35m ( with annular disk ) 60 y = 1.606x + 1.2367 50 40 30
N vs T, L=0.35m ( with annular disk )
20
Linear (N vs T, L=0.35m ( with annular disk ))
10 0 0
10
20
30
40
N vs T, L = 0.40m ( with annular disk ) 60
y = 1.6389x - 1.8703
50 N vs T, L = 0.40m ( with annular disk )
40 30
Linear (N vs T, L = 0.40m ( with annular disk ))
20 10 0 0
10
Discussion
20
30
40
: Comparing results, it is observed that the error percentage is quiet high, this is due to the
high sensitive values that have been collected during the experiment. Possible reason for the errors are, when the disk was rotating, it was quiet difficult to take the exact turn, or rotation of the disk. So there was also some human reflex and visual errors. Even though, when the experiment was performing, there was some unwanted air resistance application running beside the experiment. Some of the measurement was so sensitive, and it was hard to take the measurement. The apparatus tools, and measurement mechines was also not so accurate. The experiment could be repeated for some more time to get some more result to find it more accurate. Considering all the facts, the experiment can be accepted with errors.
MD. Atiqur Rahman Faisal
Page 7
