STUDENT NAME MD ATIQUR RAHMAN FAISAL

STUDENT ID SCM-012154

COURSE BACHELOR IN MECHANICAL ENGINEERING

LECTURER DR. CHIA

SUBMISSION DATE 19th April 2012

SUBJECT ENGINEERING MECHANICS (EAT 227)

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 o θ̈ …………………. ( Equation 1 )

where Io = mass moment of inertia of the disk

Hence, −kθ=I o θ̈ ……..……... ( Equation 2 )

where k = torsional stiffness of the shaft

Rearrange Equation 2

θ̈+ω

n2θ=0

.………..……... ( Equation 3 )

where natural frequency of the system,

ωn=√ kI o …..…….…..……... ( Equation 4 )

From Simple Theory of Torsion,

TJ= τR

=GθL

where T = Applied torque J = Polar second moment of area

τ = Shear stress R = Radius of shaft

G = Shear modulus θ = Angle of twist

L = Length of shaft

As torsional stiffness k=T

θ , it can be determined through k=GJ

L ………….. ( Equation 5 )

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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.

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TORSION VIBRATION

Results :

Time, (s)Length 0.3m 0.4m 0.5m 0.3m(With

Annular disk)0.4m(With Annular disk)

0.5m(With Annular disk)

Number of cycles10 4.74 4.89 5.08 4.53 5.55 7.2420 8.93 10.08 10.55 11.25 11.62 13.3330 13.27 14.64 15.7 16.77 17.77 19.5540 18.33 19.67 21.14 22.08 24.24 25.450 23.05 25.73 26.45 27.49 30.37 31.71

Given Information :

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) = π (r 4 )2

= π ¿¿ = 2.51327×10-11

We know, I 0 (moment of inertia) = 12mr2 =

12 (4.536) (0.075)2 = 0.0127575 (circular disk)

We know, I 0 (moment of inertia) for annular disk = 12mr 2+ 1

2m ¿ =

12

(4.536 )¿ =0.28728.

We know, K (torsion stiffness of the shaft) = GJL = (80×109 )×(2.51327×10−11)

0.3 = 6.70205

We know, ɷn(angular speed) = √ KI 0 = √ 6.702050.0127575

= 22.97 rad/s

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We know, τ (shear stress) = 2×πɷn

= 2× π22.97 = 0.2735

Comparing experimental and theoretical value of ɷn,

ɷn = 2×π×f, where f is collected from graph.

ɷn = 2×π×2.17 = 13.63rad/s

Percentage errors :

ɷn ( practical)−ɷn(theoritical)ɷn (practical )

×100%,

22.97−13.6313.63

×100%

= 68.525%

Plotted Graph :

0 5 10 15 20 250

10

20

30

40

50

60

f(x) = 2.17001445544614 x + 0.348922480783916

N vs T ,L=0.30m (without annular disk)

N vs T ,L=0.30m (without annular disk)Linear (N vs T ,L=0.30m (without annular disk))

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TORSION VIBRATION

0 5 10 15 20 25 300

10

20

30

40

50

60

f(x) = 1.94580528102256 x + 0.809029174099596

N vs T ,L=0.35m ( without annular disk )

N vs T ,L=0.35m ( without annular disk )Linear (N vs T ,L=0.35m ( without annular disk ))

0 5 10 15 20 25 300

10

20

30

40

50

60

f(x) = 1.87499305623235 x + 0.405109600428538

N vs T ,L=0.4m ( without annular disk )

N vs T ,L=0.4m ( without annular disk )Linear (N vs T ,L=0.4m ( without annular disk ))

0 5 10 15 20 25 300

10

20

30

40

50

60

f(x) = 1.75803317382267 x + 1.12606315313642

N vs T, L=0.30m ( with annular disk )

N vs T, L=0.30m ( with annular disk )Linear (N vs T, L=0.30m ( with annular disk ))

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TORSION VIBRATION

0 5 10 15 20 25 30 350

10

20

30

40

50

60

f(x) = 1.60598936528597 x + 1.23673046772828

N vs T, L=0.35m ( with annular disk )

N vs T, L=0.35m ( with annular disk )Linear (N vs T, L=0.35m ( with annular disk ))

5 10 15 20 25 30 350

10

20

30

40

50

60

f(x) = 1.63891480576455 x − 1.8703373128975

N vs T, L = 0.40m ( with annular disk )

N vs T, L = 0.40m ( with annular disk )Linear (N vs T, L = 0.40m ( with annular disk ))

Discussion :

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.

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