ME2307 DYNAMICS LABORATORY MANUAL

LIST OF EXPERIMENTSEX. NODATETITLEMARKSignature

1. a. Study of kinematic gear and pairsb. Study of static and dynamic balancing

2. Spring mass system

3. Bifilar suspension system

4. Flywheel and axle system

5. Compound pendulum

6. Motorized Gyroscope

7. Universal governor apparatus

8. Whirling of shaft apparatus

9. Turn table apparatus

10. a.Vibration table setupb. Transverse vibration free-free beam set upc. Transverse vibration fixed-free beam set up

11. Torsional vibration - Two rotor system

12. Cam analysis Machine

Ex No:1aDate:STUDY OF KINEMATIC GEARS & PAIRSAIM:To study various types of kinematics gears,links, pairs, chains & MechanismsAPPARATUS USED:Arrangement of gear system,Kinematics links, pairs, chains & Mechanisms.GEAR:Gears are used to transmit motion from one shaft to another shaft or between a shaft or slide. This is accomplished by successively engaging teeth.CLASSIFICATION OF GEAR:Gears can be classified according to the relative position of their shaft axis are follows:A: PARALLEL SHAFT(i) Spur gear(ii) Spur rack and pinion(iii) Helical gears or Helical spur gear(iv) Double- helical and Herringbone gearB: INTER SECTING SHAFT(i) Straight bevel gear(ii) Spiral bevel gear(iii) Zerol bevel gearC: SKEW SHAFT(i) Crossed- helical gear(ii) Worm gears ( Non-throated, Single throated, Double throated)

Spur gear:They have straight teeth parallel to the axes and thus are not subjected to axial thrust due to teeth load.Helical gears:In helical gears, the teeth are curved, each being helical in shape. Two mating gears have the same helix angle, but have teeth of opposite hands. At the beginning of engagement, contact occurs only at the point of leading edge of the curved teeth. As the gears rotate, the contact extends along a diagonal line across the teeth. Thus the load application is gradual which result in now impact stresses and reduction in noise. Therefore, the helical gears can be used at higher velocities then the spur gears and have greater load carrying capacity.WORM GEARWorm gear is a special case of spiral gear in which the larger wheel, usually, has a hollow or concave shape such that a portion of the pitch diameter is the other gear is enveloped on it. The smaller of two wheels is called the worm which also has larger spiral angle.BEVEL GEARKinematically, the motion between two intersecting shafts is equivalent to the rolling of two cones, assuming no slipping. The gears, in general, are known as bevel gear. When teeth formed on the cones are straight, the gear are known as straight bevel and when inclined, they are known as spiral or helical bevel.APPLICATION:1. Bevel gears are used for the drive to the differential of automobiles.2. Spur rack and pinion are used in a lathe3. Helical gears are used for greater load at higher velocities4. Gears are used in different machinery.

KINEMATIC LINK:A mechanism is made of a number of resistant bodies out of which some may have motions relative to the others. A resistant body or a group of resistant bodies with rigid connections preventing their relative movement is known as a link. A link also known as kinematic link or element.Examples:A slider-crank mechanism consists of four links: frame and guides, crank connecting rod and slider, the crank link may have crankshaft and flywheel also, forming one link having no relative motion of these.

CLASSIFICATIONS OF LINKS:1. Binary link2. Ternary link3. Quarternary link

KINEMATIC PAIR:A kinematic pair or simply a pair is a joint of two links having relative motion between them.CLASSIFICATIONS OF PAIRS:1. Kinematics pairs according to nature of contact(i) Lower pair (links having surface or area contact)Examples: Nut turning on a screw, shaft rotating in a bearing, universal joint etc.(ii) Higher pair (Point or line contact between the links)Examples: when rolling on a surface, cam and follower pair, tooth gears, ball and roller bearings etc.2. Kinematics pairs according to nature of Mechanical Constraint(a) Closed pair (when the elements of a pair are held together mechanically)Examples: all the lower pairs and some of the higher pair.(b) Unclosed pair (when two links of a pair are in contact either due to force of gravity or some spring action), Example: cam and follower pair.3. Kinematics pairs according to nature of relative motion(i) Sliding pair(ii) Turning pair(iii) Rolling Pair(iv) Screw pair (Helical pair)(v) Spherical pair

KINEMATIC CHAIN:A kinematic chain is an assembly of links in which the relative motions of the links is possible and the motion of each relative to the others is definite. If indefinite motions of other links, it is a non-kinematic chain.Types of kinematics chains (i) Four bar chain or quadric cycle chain(ii) Single slider crank chain(iii) Double slider crank chainMECHANISM:A linkage is obtained if one of the links of a kinematics chain is fixed to the ground. If motion of each link results in definite motions of the others, the linkage is known as a mechanism. If one of the links of a redundant chain is fixed, it is known as a structure. The degree of freedom of a structure is zero or less. A structure with negative degree of freedom is known as a superstructure.OBSERVATION & CONCLUSION1. Comparison between kinematics links, Pairs, chains & Mechanisms.2. Type of Motion to be named

RESULT:Thus the types of kinematics gears, links, pairs, chains & Mechanisms and their applications have been studied.

Ex No:1bDate:BALANCING OF ROTATING MASSESAIM:This apparatus has been designed to allow the student to check experimentally the normal method of calculating the position of counter balancing weight in rotating mass systems.

DESCRIPTION:The apparatus basically consists of a steel shaft mounted in ball bearings in a stiff rectangular main frame. A set of six blocks of different weights is provided and may be clamped in any position on the shaft, and also be easily detached from the shaft.A disc carrying a circular protractor scale is fitted to one side of the rectangular frame. Shaft carries a disc and rim of this disc is grooved to take a light cold provided with the cylindrical metal containers of exactly the same weight.A scale is fitted to the lower member of the main frame and when used in conjunction with the circular protractor scale, allows the exact longitudinal and angular position of each adjustable block to be determined.The shaft is driven by a 230v single-phase 50hz electric motor, mounted under the main frame, through a belt.For static balancing of individual weights the main frame is suspended to the support frame by chains and in this position the motor driving belt is removed.For dynamic balancing of the rotating mass system the main frame is suspended from the support frame by two short links such that the main frame and the supporting frame are in the same plane.PROCEDURE:STATIC BALANCINGRemove the drive belt. The value of Wr. For each block is determined by clamping each block in turn on the shaft and with the cord and container system suspended over the protractor disc, the number of steel balls, which are of equal weight, are placed into one of the containers to exactly balance the blocks on the shaft. When the block becomes horizontal, the number of balls N` will give the value of Wr. for the block.For finding out Wr during static balancing proceed as follows:Remove the belt. Screw the combined hook to the pulley with groove. (This pulley is different than the belt pulley.) Attach the cord ends of the pans to the above combined hook. Attach the block no. 1 to the shaft at any convenient position and in vertical downward direction. Put steel balls in one of the pans till the block starts moving up. (up to horizontal position.) Number of balls gives the Wr value of block 1. Repeat this for 2-3 times and find the average no of balls. Repeat the procedure for other blocks. DYNAMIC BALANCING: It is necessary to leave the machine before the experiments. Using the values of Wr obtained as above, and if the angular positions and planes of rotation of three of four blocks are known, the student can calculate the positions of other blocks(s) for balancing of the complete system. From the calculations, the student finally clamps all the blocks on the shaft in there appropriate positions. Replace the motor belt; transfer the main frame to its hanging position and then by running the motor, one can verify that these calculations are correct and the blocks are perfectly balanced.

DYNAMIC BALANCING OF 4 BLOCKSObtain dynamic balance on a set of four blocks with unbalance as shown, by properly positioning them in angular and lateral position on the shaft.No.UNBALANCE (W.r product)

10.220 kg

20.210 kg

30.230 kg

40.230 kg

Distance between each block is 3cm. The arrangement is as shown in fig bellow

3 4 1 2

04 3cm 3cm 3cm (Planes 4 and 1 are unbalance planes 3 and 2 are balancing planes)

First of all assume that reference plane is 3. Then find out the couples for blocks 4, 1, &2 with respect to 3 and then draw couple polygon.PlaneWrDistance from N0.3CoupleAngle

3m 3003

40.210306.3135

10.2306013.8240

20.2309020.74

RESULT:Thus the given unbalanced masses are balanced by using rotating mass system.

Ex No:2Date:SPRING MASS SYSTEM (SDF)AIM:To study the longitudinal vibrations of a helical spring and to determine the Natural frequency of vibration theoretically and actually.DESCRIPTIONOne end of open coil spring is fixed to the screw, which is firmly fixed to the upper bracket of vertical frame. The spring is properly gripped by means of lock nut to the screw. Lower end of the spring is attached to the platform carrying the weights. PROCEDURE1.Fix one end of the helical spring to the screw provided to the horizontal bracket.2.Determine free length of the spring.3.Put some weight on the platform and note down the deflection.4.Stretch the spring through some distance and release.5.Count the time required (in sec.) for some, say 10, 20 oscillations.6.Determine the actual period.7.Repeat the procedure for different weights.CALCULATIONSObserved Data:Mass suspendedm = __________ kgWeight suspendedW = mg =__________ NFree length of spring, h1 = ________cmDeflected lengthh2 = ________cmDeflection = h1 h2 = _________cm =____________ mNumber of oscillationsn =____________Time for n oscillationst =_____________ sec.

1. To Find Stiffness of the spring (S)Stiffness of the springS = W / =__________N/mOBSERVATIONSSl. No.

MassWeight 'W'DeflectionStiffness of SpringNo. of OscillationsTime for 'n' oscillationsPeriodic Time'Tp'Natural Frequencyfn (Experimental)Natural Frequency fn (Theoretical)

m'W = mgFree length h1Deflected length h2Deflection h1 - h2S =ntTp = t / nfn = 1/ Tpfn =

kgNcmcmcmmN/msecsecHzHz

1

2

3

4

5

6

MEAN VALUE Smean

2. To find Periodic Time of oscillations (Tp)Periodic TimeTp = t/nseconds= __________ seconds3. To Find Natural Frequency Experimentally (fn) ex.Natural Frequencyfn=1/TpHz= _______ Hz.4. To Find Natural Frequency Theoretically (fn) ThNatural Frequencyfn=Hz= _______ Hz

Result:Thus the longitudinal vibrations of a helical spring were studied and the Natural frequency was determined both theoretically and experimentally.EX NO:3 BIFILAR SUSPENSION

Aim: To determine the radius of gyration and the moment of Inertia of a given rectangular plate.Apparatus required: Main frame, bifilar plate, weights, stopwatch, threadFormula used:Time period T=t/NNatural frequency fn = 1/T hzRadius of gyration k =(Tb/2)(g/L) (mm)Where, b=distance of string from centre of gravity, T= time periodL= length of the string, N= number of oscillationst= time taken for N oscillationsProcedure:1. Select the bifilar plate2. With the help of chuck tighten the string at the top.3. Adjust the length of string to desired value.4. Give a small horizontal displacement about vertical axis.5. Start the stop watch and note down the time required for N oscillation.6. Repeat the experiment by adding weights and also by changing the length of the strings.7. Do the model calculation

Graph:A graph is plotted between weights added and radius of gyration

Calculations:

Observation:Type of suspension = bifilar suspensionNumber of oscillation n=10Tabulation:

Sl.No.Mass of the bar(kg)Mass added(kg)Total mass (kg)Length of string L (m)Time taken for N osc.T secNo of oscillationsPeriodic timeNatural frequencyfn (Hz)Radius of gyration k (mm)Mom-ent of inertia

Result:

Ex No:4Date:FLY WHEELAIM:To determine the torsinol frequency of fly wheel and moment of inertia of fly wheel.DESCRIPTION:The arrangement consist of a long elastic shaft gripped at the upper end by the chuck in the bracket. The bracket is clamped to the upper beam of the main frame. A heavy steel fly wheel clamped at the shaft suspends from the bracket.

PROCEDURE:1. Support the fly wheel2. Allow the fly wheel to oscillate then determine torsional frequency the time for say 10 oscillation without mass.3. Complete the observation table given below.4. Determine the torsional frequency of fly wheel.

SI NoLength of suspension of fly wheelCGTime (t) for n oscillationNo of oscillation nPeriodic time tp = t/nTorsional frequency

1

2

3

4

5

Calculation:1. Torsional stiffness kt kt = GIp /LG = modulus of rigidity = 0.8 * 106 Kg/cm2 Ip = polar moment of inertia of fly wheel = d4/32 d = dia of fly wheell= length of the fly wheel 2. Moment of inertia of fly wheel W/g * D2/8 W = weight of the disc = 7.8 kgD = dia of disc = 25 cm

Torsional frequency = 2 *

RESULT:Thus the torsional frequency and moment of inertia was determined and tabulated. Ex No:5Date:COMPOUND PENDULUMAim:1. To determine the radius of gyration KG of given compound pendulum.2. . To verify the relationTp=2KG2+ (OG)2

g (OG)

Where,Tp=Periodic time in sec.

K=Radius of gyration about C.G. m.

OG =Distance of the C.G. of rod from support. m

L=Free Length of suspended pendulum - m.

g=9.81 m/s2

DESCRIPTIONThe compound pendulum consists of steel bar. The bar is supported in the hole by the knife edge.PROCEDURE1.Support the rod on knife-edge.2.Note the length of suspended pendulum and determine OG.3.Allow the bar to oscillate and determine T by knowing the time for say 10 oscillations without attaching lumped masses.4.Complete the observation table given below.5.Attach Lumped mass of desired wt. and carry out the above procedure and complete the observation table.CALCULATION:Observed Data:Length of the barL= ____________ mDistance of G from supportO OG= _____________ mNumber of oscillationsn =____________Time for n oscillationst =_____________ sec.1. To find Periodic Time - TpTp = t / nsec= ________ Sec2. To Find radius of gyration experimentally - KG Tp=2 KG2 + (OG)2

g . (OG)2

OBSERVATION TABLETable - 1.Without Lumped Mass - Only barSl. No.Length of the barDistance OGNo. of OscillationsTime for 'n' oscillationsPeriodic TimeRadius of Gyration

ExperimentalTheoretical

LntTp = t/nKGKG

mcmmsecsecmm

1

2

3

Where,Tp=Periodic time.3.To find radius of gyration theoretically - KGKG theoretical=L

2 3

Table - 2. With Lumped Mass - Bar with attached massSl. No.Length of the barLumpedMassDistance OGNo. of OscillationsTime for 'n' oscillationsPeriodic TimeRadius of Gyration

Exp.Th

LMOGOGntTp = t/nKGKG

mkgcmmsecsecmm

1

2

3

4

CALCULATIONSObserved Data:Length of the barL= ____________ mDistance of G from supportO OG= _____________ mNumber of oscillationsn =____________Time for n oscillationst =_____________ sec.1. To find Periodic Time - TpTp = t / nsec= ________ Sec2. To Find radius of gyration experimentally - KG Tp=2 KG2+ (OG)2

g . (OG)2

Where, Tp=Periodic time.

RESULT:Thus the radius of Gyration for the bar and for various lumped masses were determined using Compound Pendulum apparatus.

Ex No:6Date:MOTORISED GYROSCOPE

AIM:To study the Gyroscopic Principles and to determine the Gyroscopic effect.INTRODUCTIONA)AXIS OF SPINIf a body is revolving about an axis the latter is known as axis of spin OXB)PRECESSIONPrecession means the rotation about the third axis OZ, which is perpendicular to both the axis of spin OX and that of couple OY.C)AXIS OF PRECESSIONThe third axis OZ is perpendicular to both the axis of spin OX and that of couple OY is known as axis of precession.D)GYROSCOPEIt is a body, which, while spinning about an axis, is free to rotate in either direction under the action of external forces.Examples : Locomotive, automobile and aero-plane making a turn. In certain cases the gyroscopic forces are undesirable whereas in other cases the gyroscopic effect may be utilized in developing desirable forces.

E)GYROSCOPIC EFFECTTo a body, revolving (or spinning) about an axis say OX if a couple represented by a vector OY perpendicular to OX is applied, the body tries to precess about an axis OZ which is perpendicular both to OX and OY. Thus the plane of spin, plane of precession and plane of gyroscopic couple are mutually perpendicular.The above combined effect is known as precession or gyroscopic effect.RULE NO.1The spinning body exerts a torque or couple in such a direction which tends to make the axis of spin coincide with that of the precession.To study the rule of gyroscopic behaviour following procedure may be adopted.

a)Balance the initial horizontal position of the rotor.b)Start the motor by increasing the voltage with the dimmer, and wait until it attains constant speed.Process the yoke frame No.2 about vertical axis by applying necessary force by hand to the same ( in the clockwise sense seen from above).It will be observed that the rotor frame swings about the horizontal axis YY. Motor side is seen coming upward and the weight pan side going downward.Rotate the vertical yoke axis in the anti-clockwise direction seen from above and observe that the rotor frame swing in opposite sense (as compared to that in previous case following the above rule).RULE NO.2The spinning body precesses in such a way as to make the axis of spin coincide with that of the couple applied, through 90o turn axis.a)Balance the rotor position on the horizontal frame.b)Start the motor by increasing the voltage with the dimmer and wait till the disc attains constant speed.c)Put weight ( 0.5 Kg., 1.0 Kg. or 2 Kg) in the weight pan, and start the stop watch to note the time in seconds required for precession, through 90o or 180o etc.d)The vertical yoke precesses about OZ axis as per the rule No.2.e)Speed may be measured by the tachometerf)Enter the observation in the table.

OBSERVATION TABLEOBSERVED DATA1)Mass of Rotor(M): 6.5 Kgs.2)Rotor Diameter (D): 300 mm ( 0.30 m. )3)Rotor Thickness: 010mm ( 0.01 m )4)Moment of inertia of the I =M x D2/ 8 kg.m2 disc, coupling and motor : rotor about central axis (I) 5)Distance of bolt of: L = 18.0 cm = 0.18 mweight pan from disccenter (L)Sl. No.Speed of the RotorLoadAngle TurnedTime takenPrecessional Angular velocityAngular velocity of the rotorGyroscopic Couple - TheoreticalGyroscopic Couple - Actual

NMass - mWeight - Wdd x /180dtp = d/dt = 2N/60C = IpC = W.L

rpmkgNDegreeRadianssecrad/secrad/secNmNm

Calculation:

Cth=I x x p, where (Nm)=

Cact=W x LNm.

RESULTIt is found that the equation T = I x x p is verified.

Ex No:7Date:GOVERNORSAIM:To study the characteristics of various types of Governors and to draw their characteristic curves.a) Watt Governor b) Porter Governor c) Proell Governor PROCEDURE:1. The knob of the dimmerstat must be kept in zero position before switching on the main supply.2.The main supply is switched on and gradually the speed of the motor is increased. Due to this the center sleeve rises from the lower stop aligning with the marking on the scale. This is initial lift of the sleeve.3.The readings of the sleeve position and speed for the initial lift are to be noted. Speed of the motor is to be measured by hand tachometer, from the counter hole provided on the spindle.4.Then the speed is increased in steps to give suitable sleeve movement and the corresponding sleeve displacement and the speed are noted.

A) WATT GOVERNOR: 1. Watt Governor set up is arranged as shown in Fig. by using the proper linkages provided.2. The speed is increased gradually and the speed of rotation `N` and corresponding sleeve displacement `X` are noted down.

CALCULATION:Observed Data:Length of each link L =125 mm.=________ mInitial height of Governor ho=094 mm.=________ mInitial radius of rotation ro=136 mm.=_______ mMass of each ball m=700 gms. =________ kgTo Find Radius of rotation `r` at any position: 1)To Find height h = ho X/2 mtr.=__________ m2) To find Cos =h / L=_____________ 3)Then r =0.05 + L Sin mtr.=______________ m4)Angular Velocity = 2N/60 rad/sec =___________ rad / sec 5) Centrifugal force Fc = m 2r =____________ NOBSERVATIONS: a)Length of each link L =125 mm.=________ mb)Initial height of Governor ho=094 mm.=________ mc)Initial radius of rotation ro=136 mm.=_______ md)Mass of each ball m=700 gms. =________ kg

Sl. No.Speed of the GovernorSleeve DisplacementAngular velocity of the Governor'Height of the governorRadius of rotationCentrifugal Force

NX = 2 N/60h = ho x/2

rFc = m 2r

rpmmrad / secmmN

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6

Graphs to be plotted:a)Force Vs Radius of rotation.b)Speed Vs Sleeve Displacement.____________________________________________________________B)PORTER GOVERNOR1. Arrange the set-up as shown in Fig. by using the proper linkages & weights provided. 2. The speed is increased gradually and the speed of rotation `N` and corresponding sleeve displacement `X` are noted down.OBSERVATIONS: a)Length of each link L =125 mm.=________ mb)Initial height of Governor ho=094 mm.=________ mc)Initial radius of rotation ro=136 mm.=_______ md)Mass of each ball m=700 gms. =________ kge)Mass of Sleeve weightM=500 gms.=_________ kg OBSERVATION TABLESl. No.Speed of the GovernorSleeve DisplacementAngular velocity of the Governor'Height of the governorRadius of rotationCentrifugal Force

NX = 2 N/60h = ho x/2

rFc = m 2r

rpmmrad / secmmN

1

2

3

4

5

CALCULATION:Observed Data:Length of each link L =125 mm.=________ mInitial height of Governor ho=094 mm.=________ mInitial radius of rotation ro=136 mm.=_______ mMass of each ball m=700 gms. =________ kgMass of Sleeve weightM=500 gms.=_________ kg To Find Radius of rotation `r` at any position: 1)To Find height h = ho X/2 mtr. =__________ m2) To find Cos =h / L=_____________ 3)Then r =0.05 + L Sin mtr.=______________ m4)Angular Velocity = 2N/60 rad/sec =_______ rad / sec 5) Centrifugal force Fc = m 2r =____________ NGraphs to be plotted:a)Force Vs Radius of rotation.b)Speed Vs Sleeve Displacement.________________________________________________________________C)PROELL GOVERNOR1. Arrange the set-up for Proell Governor as shown in Fig. by using the proper linkages & weights provided. 2. The speed is increased gradually and the speed of rotation `N` and corresponding sleeve displacement `X` are noted down.

a)Length of each link L =125 mm.=________ mb)Initial height of Governor ho=094 mm.=________ mc)Initial radius of rotation ro=141.5 mm.=_______ md)Mass of each ball m=700 gms. =________ kge)Mass of Sleeve weightM=500 gms.=_________ kg f) Extensiion of length BG=075 mm.=_________ mOBSERVATION TABLE

Sl. No.Speed of the GovernorSleeve DisplacementAngular velocity of the Governor'Height of the governorRadius of rotationCentrifugal Force

NX = 2 N/60h = ho x/2

rFc = m 2r

rpmmrad / secmmN

1

2

3

4

5

CALCULATION:Observed Data:Length of each link L =125 mm.=________ mInitial height of Governor ho=094 mm.=________ mInitial radius of rotation ro=136 mm.=_______ mMass of each ball m=700 gms. =________ kgMass of Sleeve weightM=500 gms.=_________ kg To Find Radius of rotation `r` at any position: 1)To Find height h = ho X/2 mtr. =__________ m2)To find radius of rotation, lift manually at different heights.3)Angular Velocity = 2N/60 rad/sec =___________ rad / sec 4) Centrifugal force Fc = m 2r =____________ NFollowing graphs to be plotted:a)Sleeve Disp. 'X' Vs 'r' Radius of rotation.To draw this graph proceed as follows :1)Keep the Governor in static position.2)By lifting the Governor Mechanism manually measure the sleeve displacement and corresponding radius of rotation r of balls.a)Force Vs Radius of rotation 'r'.b)Speed Vs Sleeve Displacement.

Results:

Thus the characteristic of various types of governors are studied and their characteristic curves were drawn.Ex No:8Date:WHIRLING OF SHAFTSAIM:To study the vibrations of the whirling shafts and to determine the frequency of vibrations.PROCEDURE:1. Fix the given shaft for suitable end condition. (Free-Free or Fixed Free).2. Slowly increase the speed of rotation and find the minimum speed where the amplitude is maximum for the first mode.3. Note down the speed.4. Again slowly increase the speed of rotation and find the minimum speed where the amplitude is maximum for the second mode.5. Note down the speed.6. Repeat the procedure for variuos sizes of the shafts and for various end conditions.

WHIRLING OF ELASTIC SHAFTSLetL=Length of the shaft in m.E=Youngs Modulus = 2.060 x 106 Kg/cm2I=2nd moment of inertia(Area) of the shaft m4=(/64) x d4W=Weight of the shaft per unit length N/m.g=Acceleration due to gravity in m/sec2 = 9.81Then the frequency of vibration for the various modes is given by the equation

f=KxE.I.g

W.L (C.P.S.)

Sl.No.Dia. Of ShaftConditionK' valueSpeed of rotationI - MI of shaftW - Weight per unit lengthFrequency of vibration

1

2

3

The various values for K are given below :End ConditionValue of K

1st Mode2nd mode

Supported, Supported1.576.28

Fixed, Supported2.459.80

Fixed, fixed3.5614.24

DATAShaft Dia.I = m4W = Kg/m

5 mm6.14 x 10-11 m40.12 kg/m

6 mm1.275 x 10-10 m40.26 x 10 kg/m

8 mm4.02 x 10-10 m40.38 x 10 kg/m

TYPICAL TEST OBSERVATIONS1)Both ends of shaft free (supported) 1st and 2nd mode of vibration can be observed on shafts with 3/16 dia and dia.2)One end of shaft fixed and the other free; 1st and 2nd mode of vibration can be observed on shaft with 3/16 dia.3)Both ends of shaft fixed 2nd mode of vibration can not be observed on any of the shafts as the speeds are very high and hence beyond the range of the apparatus.4)There is a difference between theoretical speed of whirling and actual speed observed, due to following reasons .

RESULT:Thus the vibrations of whirling shafts are studied and the frequency of vibration is determined.Ex No:9Date:TURN TABLE APPARATUSAIM:To determine the moment of inertia of the given disc.INTRODUCTION:The moment of inertia of a body depends upon its mass distribution and shape.For a symmetrical, homogeneous object, the MI can be derived theoretically and expressed in terms of the objects total mass and dimension.However the the MI of any object can be determined experimentally from the dynamical equation of torque = I by measuring & , or from energy considerations, KE of rotation = I2/2.PROCEDURE:1. A wire cord is attached to the weight pan.2. Weight is added in the weight pan.3. The turn table is rotated in such a fashion that the pan moves upward and hold the table when the point of weight pan reaches at the mark made on the angle frame.4. Now the turn table is released and the weight pan will move downwards.5. When turn table is released and the weight pan starts to move downwards, the stopwatch is started to note down the time taken.6. Time taken is noted until the pan reaches the bottom position.7. This procedure is repeated with the turn table be rotated in same direction for 3 to 5 times with varying weights.8. Moment of inertia of the turn table is found out from the calculations.

Sl. No.Weight of the panHeight travelled by the weightRebouncing heightTime taken for height h1Acceleration of the panAngular acceleration of the rotating tableTension in the stringTorque produced on the cordFrictional torqueMoment of inertia of the table

MassmWeight W= mghhtaTcfI

kgNcmmcmmsm/s2rad/s2NNmNmKg m2

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5

1. To find acceleration of the pan moving downwards (a)h1 = Ut + (1/2)(at2)U=0 (starting from rest)a = (2h1)/(t2)2. To find the angular acceleration = a / r3. To find tension on the cord (T)T = W ma = mg ma= m(g-a)4. To find the torque produced by the cord on the system (c)c = Tension force x radius of rotation = T x r

5. The average frictional torque during the motion is given by

f = mgr (h1-h2) (h1 + h2)6. To find the moment of inertia of the table experimentally, Net torque (c - f) = I I = _______ kg.m27. Mean value of the moment of inertia of the table = ______________ kg.m2.

Result:

Ex No:10 aDate:VIBRATION TABLE SETUPAIM:To determine the transmissibility ratio by using the vibrating platform.PROCEDURE:1. Starts the units by giving some speed say around 100 150rpm initially. Wait for few seconds for achieving the stability of the system.2. Note down the readings speed, rpm and load on the indicator and enter it in the table.3. By varying the speed take readings and enter it in the observation table.4. See that take at least 5 readings before stopping the unit. Ensure that the dimmer stat is on zero before restarting the unit.

TABULATION:

S.NoSpeedLoadAmplit-udeAngular speedNatural frequen-cyTheoretical amplitudeForce observedTR thTR ex

Calculations:1. Weight of the motor, exciter plate and base plate W 17.3 kg2. Weight of exciter W1 = 0.15 kg- 2 Nos.3. Total spring stiffness (4 Nos) K = 0.4545 kg/mm x 4 = 1818kg/m.4. Radius of eccentric weight attached e = 60mm = 0.065. Natural frequency n = (K/M) Where M = W/g6. Angular speed = 2N/607. Theoretical Amplitude X = 8. F0 = Total spring stiffness x Amplitude = K x X .kg9. TR th = Ft/Fo10. TR ex = Ft / (K.Y)

Result:

Ex No:10 bDate:TRANSVERSE VIBRATION(FREE FREE SIMPLY SUPPORTED BEAM)AIM:To determine the Youngs Modulus of Elasticity of the given material of the Beam and the natural frequency using a Simply supported Beam.TOOLS REQUIRED:1. A Bar of uniform cross section.2. Simple supports setup3. Dial Gauge with magnetic Base4. Point Load setup5. Measuring Tape.PROCEDURE1.The bar is placed on the given supports and it is simply supported at its ends.2.Measure the length of the bar between the supports (L).3.Also measure the lateral dimensions(B & H)4.Apply some point loads at any point on the beam.5.Measure the distances of the point load P from the ends A and B. (a & b)6.Fix the dial gauge on the loading point and measure the deflection of the beam at the loading point ().7. Repeat the experiment by varying the positions and magnitudes of the point load.8.Using the theoretical relation determine the Youngs Modulus of Elasticity of the material of the Bar.9.Also calculate the natural frequency.CALCULATIONS:Observed Data:Length of the bar between supports ABL= ___________mWidth of the barB = ____________mThickness of the barH = ____________mMoment of Inertia of the cross section of the bar (Second moment of Area)I = B x H3 12= __________ m4.Mass loadedm= _________ kgWeight or Load in (N)W= mg= ________ NOBSERVATIONS:Length of the bar between supports ABL = ___________mWidth of the barB = ____________ mThickness of the barH = ____________ m

Sl. No.LoadDistance of Point load P from end 'A'Distance of Point load 'P' from end 'B'Deflection at loaded point - Young's Modulus of Elasticity - 'E'Natural frequency fn

E = Wa2b2 3I L fn = g 2

Mass 'm'WeightW = mgab = L - a

kgNmmmmmN/m2Hz

1

2

3

4

5

6

Distance of Point load P from end 'A'a= __________ m

Distance of Point load P from end 'B'b= __________ mDeflection of beam at the point P= ____________ mTo Find Youngs Modulus E

=Wa2b2 3EILHence E=Wa2b2 3IL= _____________ N/m2

To Find Natural frequency.

fn=g 2= __________ HzResult:

Thus the Youngs Modulus of the material of the bar and the natural frequency of the Simply supported beam were determined.

Ex No:10cDate: TRANSVERSE VIBRATION(FIXED FREE - CANTILEVER BEAM)AIM:To determine the Youngs Modulus of Elasticity of the given material of the Beam and the natural frequency using a Cantilever Beam.TOOLS REQUIRED:1. A Bar of uniform cross section.2. Dial Gauge with magnetic Base3. Point Load setup4. Measuring Tape.PROCEDURE1.The bar is placed on the system such that its one end is fixed by the support and the other end remaining free.2.Measure the length of the bar (L).3.Also measure the lateral dimensions(B & H)4.Apply some point loads at any point on the beam.5.Measure the distance of the point load P from the fixed end A (a)6.Fix the dial gauge on the loading point and measure the deflection of the beam at the loading point ().7. Repeat the experiment by varying the positions and magnitudes of the point load.8.Using the theoretical relation determine the Youngs Modulus of Elasticity of the material of the Bar.9.Also calculate the natural frequency.CALCULATIONS:Observed Data:Length of the bar between supports ABL= ___________mWidth of the barB = ____________mThickness of the barH = ____________mMoment of Inertia of the cross section of the bar (Second moment of Area)I = B x H3 12= __________ m4.Mass loadedm= _________ kgWeight or Load in (N)W= mg= ________ NDistance of Point load P from fixed end 'A' a= __________ m

Deflection of beam at the point P= ____________ m

OBSERVATIONSLength of the bar L = ___________mWidth of the barB = ____________ mThickness of the barH = ____________ m

Sl. No.LoadDistance of Point load P from fixed end 'A'Deflection at loaded point - Young's Modulus of Elasticity - 'E'Natural frequency fn

E = Wa3 3I fn = g 2

Mass 'm'WeightW = mga

kgNmmmmN/m2Hz

1

2

3

4

5

6

To Find Youngs Modulus E=Wa3 3EIHence E=Wa3 3I= _____________ N/m2

To Find Natural frequency.

fn=g 2= __________ Hz

Result:

Thus the Youngs Modulus of the material of the bar and natural frequency of the Cantilever Beam were determined.

TORSIONAL VIBRATION OF TWO ROTOR SYSTEMEX. No: 11Date:Aim:To study the undamped torsional vibration of two rotor shaft system.Apparatus required:1. Rotor assembly2. Weights3. Stop clock4. Measuring tapeProcedure:1. Fix the bracket at any convenient position along the lower beam.2. Fix the rotor on the each end of the shaft.3. Twist the through some angle and release.4. Note down the time required for 10 oscillations.5. Repeat the procedure for different mass attached in disc B.6. Make the following observations:a. Diameter of the disk A= 250 mm.b. Diameter of the disk B= 200 mm.c. Weight of the disk A= 3.8 kg.d. Weight of the disk B= 2.5 kge. Length of cross arm= 150 mm.f. Diameter of the shaft= 3 mm.g. Modulus of rigidity of the shaft (G) = 0.8*106 kg/cm2h. Length of shaft between rotors= 1100 mm.

Tabulation:S.NOMass of disc AMass of disc B + cross armMass attachedTotal massNo. Of oscillations nTime For n Oscillations t secPeriodic Time TexpsecMoment of inertia of disc AMoment of inertia of disc BNatural Frequency FexpHzNatural Frequency FthHz

1

2

3

4

5

Model Calculation:1. Torsional stiffness Kt = G*Ip / LWhere , Ip = polar moment of inertia of the shaft = *d4 / 32 = 2. T theoretical = Ttheoretical = 2*(I/Kt)0.5Where , M . I of the disc Ia = (MA)*DA2/8

=

M . I of the disc Ib = (MB)*DB2/8

= 3. Natural frequency of torsional vibration theoretically

fn = ____________ Hz

4. Natural frequency of torsional vibration Experimentallyfn = 1/Tp Hzfn = ____________ Hz

Result:Thus the undamped vibration of the two rotor system is studied.

Ex No:12Date:CAM ANALYSIS MACHINEAim:To study the profile of given can using cam analysis system and to draw the displacement diagram for the follower and the cam profile. Also to study the jump-speed characteristics of the cam & follower mechanism.Apparatus required: Cam analysis system and Dial gauge

Description:A cam is a machine element such as a cylinder or any other solid with a surface of contact so designed as to give a predetermined motion to another element called the follower.A cam is a rotating body importing oscillating motor to the follower. All cam mechanisms are composed of at least there links viz: 1.Cam, 2. Follower and 3. Frame which guides follower and cam.

Procedure:Cam analysis system consists of cam roller follower, pull rod and guide of pull rod.1. Set the cam at 0 and note down the projected length of the pull rod2. Rotate the can through 10 and note down the projected length of the pull rod above the guide3. Calculate the lift by subtracting each reading with the initial reading.Jump-speed:1. The cam is run at gradually increasing speeds, and the speed at which the follower jumps off is observed.2. This jump-speed is observed for different loads on the follower.Graph:Displacement diagram and also the cam profile is drawn using a polar graph chart.The Force Vs Jump-speed curve is drawn.

Tabulation:1.Cam profileSl.No.Angle of rotation (degrees)Lift in mmLift + base circle radius (mm)

1

2. Jump-speed.Sl.No.Load on theFollower, F (N)Jump-speedN (RPM)

Result.

4