THEORY OF MACHINES AND MECHANISNMS I LAB MANUAL EXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD AIM:
To determine the radius of gyration of given bar using bifilar suspension method.
APPARATUS USED: Uniform steel bar, cylindrical weights, stop watch. DESCRIPTION SET UP
Fig 1 shows the general arrangement for carrying out experiment. A uniform rectangular section bar is suspended from support frame by two parallel cords. Two small chucks fitted in the frame and other ends secured in bifilar bar. Ends of these cords pass through a chuck so as to facilitate for change in length of the cord it is possible to adjust length of chord by loosening the chuck. The suspension may also be used to determine radius of gyration of any body, In this case the body under investigation is bolted to the center. Radius of gyration of combined bar and body is then determined.
PROCEDURE
1) Suspend the bar from chuck and adjust length length of chord ‘ L’ conveniently conveniently note the suspension length of each cord must be same. 2) Allow the bar to oscillate about the vertical axis passing through center of grav gravit ity y and and meas measur ure e peri period odic ic time time ‘T’ ‘T’ by know knowin ing g time time for for 20 oscillations. 3) Repeat the experiment experiment by mounting the weights one equal distance from the center.
THEORY Consider bar AB whose moment of inertia is to be determine is suspended by two parallel strings as shown in figure 2. Let,
L m K I 2r T θ φ α
= length of each string. = mass of bar. = radius of gyration about the vertical axis axis through G. = mass moment of inertia about about the vertical axis through G. = distance between two strings = tension in each string = angle of twist of bar. = angle of twist of string = angular accelaration of the bar. bar.
Initial position of bar is AGB. Now, if bar is set gently in vibration in horizontal plane, it will start oscillating about vertical axis through its mass center G. when bar is twisted about its C.G, through an angle θ, bar is in position A’GB’ as shown. When bar is getting θ as angular displacement, then string is getting φ as angular displacement about vertical, distance moved by point A is arc AA’
arc AA’ = r θ =L φ φ = r θ L
PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT
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THEORY OF MACHINES AND MECHANISNMS I LAB MANUAL EXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD now tension in each string is
T = mg 2
(2)
For twisted position, tension T gives rise to the component T sinφ which Acts at A’ and B’ normal to A’B’ in horizontal plane. It applies moment of bar about vertical axis through G which is of magnitude 2 T sinφ r Now , Acceleration torque = IG α
= (m k2) α Restoring torque = 2 T sin φ . r
=
2 m g sinφ . R
from (2)
2 As, T = m g
and sin φ
= φ
(for value of small φ )
2 Restoring torque =m g φ r Substituting (1) in (4) Restoring torque = m g ( r θ ) r
L = m g r 2 θ
L For static equilibrium of rigid body at any instant, (Accelerating torque) = (restoring torque)
(m k2 ) α
= m g .r 2
θ
L α
=
θ
r 2
g
k2
L
=
constant
Angular acceleration Angular displacement
α
= r 2
g
θ
k2
L
=
constant
Hence, motion of bar is approximately S. H.M Therefore, frequency of Oscillation of bifilar suspension is given by
f n =
1
g r 2
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THEORY OF MACHINES AND MECHANISNMS I LAB MANUAL EXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD
L k2
2π f n
= 1 2π
where,
r
g
k
L
f n
=
cycles/sec
g
=
m/s2
k
=
m
r, l
=
m
The periodic time T =
1 f n
Tp
=
2π
k
L
r
g
OBSERVATIONS: 1) Length of strings
= L =
2) Distance between C. G of bar
= r =
and point of suspension 3) Distance between C.G of cylinder
=
and C.G of bar 4) Length of bar
L1
=
5) Mass of bar
mb
=
6) Cylindrical masses
mc
=
For rectangular bar : Sr.
Time
Time taken for 20 oscillations
No. T1
T2
Period Avg.
T3
Tp = Tp
T(sec)
20
To calculate radius of gyration
Tp
=
2π
k
L
r
g
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THEORY OF MACHINES AND MECHANISNMS I LAB MANUAL EXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD
Kbar
=
Tp .r 2π
g L
To calculate mass moment of inertia, I
Ibar
Theoretically,
Ibar
=
=
mb kbar 2
mb kbar 2= mb L1 2 / 12
OBSERVATIONS AND CALCULATIONS
For combined body i.e (bar + cylinder) Sr.
Time
Time taken for 20 oscillations
No. T1
T2
Period Avg.
T3
Tp = Tp
T(sec)
20
To calculate radius of gyration K2
=
Tp .r
g
2π
L
To calculate M.I of combined body
I2
=
(m2) (k2)2 = (mb + 2mc) (k22)
Calculations for cylinder: To calculate radius of gyration Since, radius of gyration of each cylinder is k c about it own C.G, then mass M.I of Bar and cylinder about axis of oscillation is given by
I2
=
mb kb2 + 2 [ mc kc2 + mc (r 2) ] [ by parallel axis theorem ]
=
mb kb2 + 2 [ mc ( kc2 + r 2 )
To calculate M. I of cylinder,
PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT
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THEORY OF MACHINES AND MECHANISNMS I LAB MANUAL EXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD
I =mc kc2 Theoretically, for cylinder,
I =mc kc2
=
m r 2 2
RESULT Radius of gyration
Mass moment of inertia
Component Theo (m)
Expt (m)
Theor (kgm 2)
Expt (kgm 2)
Rectangular bar Rectangular bar with cylindrical weights
CONCLUSION
CONCLUSION:
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THEORY OF MACHINES AND MECHANISNMS I LAB MANUAL EXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD
1. The suspension of bar is bifilar which performs S.H.M. 2. Bifilar suspension can be used to determine radius of gyration of machine parts having complete regular geometrical shapes 3. Difference in the theoretical values and Experimental values is due to manual error in time measurement.
.
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