THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT 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 Fig 1 shows the general arrangement for carrying out experiment. A SET UP 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 of chord ‘ L’ 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 gravity and measure periodic time ‘T’ by knowing time for 20 oscillations. 3) Repeat the 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 = length of each string. m = mass of bar. K = radius of gyration about the vertical axis through G. I = mass moment of inertia about the vertical axis through G. 2r = distance between two strings T = tension in each string

= angle of twist of bar. = angle of twist of string = angular accelaration of the 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 , bar is in position A’GB’ as shown. When bar is getting as angular displacement, then string is getting as angular displacement about vertical, distance moved by point A is arc AA’

arc AA’ = r =L = r L ( 1)

PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT 1

THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD

now tension in each string is T = m g (2) 2 For twisted position, tension T gives rise to the component T sin 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 sin r Now , Acceleration torque = IG = (m k2)

Restoring torque = 2 T sin . r

= 2 m g sin . R from (2)

2

As, T = m g and sin = (for value of small )

2

Restoring torque =m g r

Substituting (1) in (4)

Restoring torque = m g ( r ) r

L

= m g r 2

L

For static equilibrium of rigid body at any instant,

(Accelerating torque) = (restoring torque)

(m k2 ) = m g .r 2

L

= r 2 g = constant

k2 L

Angular acceleration = r 2 g = constant

Angular displacement k2 L

Hence, motion of bar is approximately S. H.M Therefore, frequency of

Oscillation of bifilar suspension is given by

fn = 1 g r 2

2 L k2

fn = 1 r g

PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT 2

THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD

2 k L

where, fn = cycles/sec

g = m/s2

k = m

r, l = m

The periodic time T = 1

fn

Tp = 2 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.

No. Time taken for 20 oscillations

Time

Period

T1 T2 T3

Avg.

T(sec)

Tp = Tp

20

To calculate radius of gyration

Tp = 2 k L

r g

Kbar = Tp .r g

PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT 3

THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD

2 L

To calculate mass moment of inertia, I

Ibar = mb kbar2

Theoretically, Ibar = mb kbar2= mb L1 2 / 12

OBSERVATIONS AND CALCULATIONS

For combined body i.e (bar + cylinder)

Sr.

No. Time taken for 20 oscillations

Time

Period

T1 T2 T3

Avg.

T(sec)

Tp = Tp

20

To calculate radius of gyration K2 = Tp .r g

2 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 kc about it own C.G, then mass M.I of

Bar and cylinder about axis of oscillation is given by I2 = mb kb

2 + 2 [ mc kc2 + mc (r2) ] [ by parallel axis theorem ]

= mb kb2 + 2 [ mc ( kc

2 + r2 )

To calculate M. I of cylinder, I =mc kc

2

PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT 4

THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD

Theoretically, for cylinder, I =mc kc

2 = m r 2 2

RESULT

Component Radius of gyration

Theo (m) Expt (m)

Mass moment of inertia

Theor (kgm2) Expt (kgm2)

Rectangular bar

Rectangular bar with cylindrical weights

CONCLUSION

CONCLUSION:

1. The suspension of bar is bifilar which performs S.H.M.

PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT 5

THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD

2. Bifilar suspension can be used to determine radius of gyration of machine parts having complete regular geometrical shapes3. Difference in the theoretical values and Experimental values is due to manual error in time measurement.

.

PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT 6