me2307dynamics lab manual 2

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IDEAL INSTITUTE OF TECHNOLOGYIDEAL INSTITUTE OF TECHNOLOGYMECHANICAL ENGINEERING DEPARTMENTMECHANICAL ENGINEERING DEPARTMENT

V Semester Dynamics of Machine Lab

DYNAMICS OF MACHINES LABDYNAMICS OF MACHINES LAB

(TME 553)(TME 553)VV Semester

NAMENAME

UNIVERSITY ROLL NOCLASS ROLL NO

BRANCH

BATCH

Department of Mechanical Engineering Ideal Institute Of Technology, Ghaziabad


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IDEAL INSTITUTE OF TECHNOLOGY GHAZIABADIDEAL INSTITUTE OF TECHNOLOGY GHAZIABAD

DEPARTMENT OF MECHANICAL ENGINEERINGDEPARTMENT OF MECHANICAL ENGINEERING

INDEXINDEXEXP.EXP.

NONOOBJECTIVEOBJECTIVE DATEDATE GRADEGRADE REMARKSREMARKS

1

2

3

4

5

6

7

8

9

10

CONTENTS



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Exp.No Name of the Experiment Page No

1. Slider Crank Mechanism 1

2. Cam 4

3. Governor 9

4. Gyroscope 14

5. Whirling Speed of Shaft 99

6. Balancing (Static & Dynamic) 19

7. Vibration (Longitudinal) 24

8. Vibration (Torsional) 29

9. Gear Train 77

10. Gears 88

Experiment. No: 1 Slider Crank Mechanism



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1.1 Objective, 1.2 Apparatus, 1.3 Theory, 1.4 Description of Apparatus, 1.5 Procedure, 1.6 Specification,

1.7 Observation Table 1.8 Calculation, 1.9 Graph, 1.10 Result & Discussion, 1.11 Precautions, 1.12

Sources of error, 1.13 Viva-voce questions

1.1 Objective:To draw the slider displacement versus crank angle and time

versus velocity curve for a slider crank mechanism (reciprocating engine

mechanism) and compare the results with theoretical values.

1.2 Apparatus:Slider crank mechanism, graph sheet.

1.3 Theory: Fig. 1.1 shows the line diagram of a slider crank mechanism.

Fig.1.1, Slider Crank Mechanism

When the crank OC has moved through an angle from IDC ( Inner Dead

Centre), slider has moved from G to F so that the displacement of the slider

FG = x

Let, crank radius = OC = r,

Length of connecting rod = CS = l

If is the angular speed of the crank, it is found that:-

Displacement, x= r. [ (1-cos ) + n - (n2 sin2 )] --- --- --- (1)



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Velocity of slider or piston, vpo= vp = dx / dt = (dx / d)*( d / dt) = (dx /d).

vp = r [ sin + sin 2/ 2 (n2 - sin2 )]

Acceleration of slider or piston ,

ap = d2x / dt2 = dv / dt = (dv / d)*( d / dt) = .(dv / d)

= 2 r [ cos + (cos 2) / n ]

1.4 Description of Apparatus:

The apparatus consists of a slider, which reciprocates inside the cylinder as

the crank rotates. A graduated scale is provided to read the displacement of

the slider corresponding to the crank rotation. When crank is rotated the slider

slides to and fro in a linear motion. The motion of the slider can be read on ascale attached to the frame. A graduated wheel is provided to read the crank

rotation.

1.5 Procedure:-

1. Bring the wheel and the slider to respective reference marks.2. For a given angle of rotation of the crank note down the displacement

of the slider.3. Plot a graph between the slider displacement and the crank rotation.4. Assume that crank is rotating with a uniform angular speed of one rad

per sec (1 rad /sec).5. Convert the crank rotation angle into time and plot the slider

displacement versus time.6. By graphical differentiation determine the velocity time graph.7. By differentiation twice determine the acceleration graph.8. Calculate values of displacement, velocity and acceleration from

equation.9. Compare the results.

1.6 Specification:Length of connecting rod, l = 120 mm.



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Crank radius, r = 50 mm.

1.7 Observation table:

S.No.Crank

rotation

Time

( Sec.)

Slider

displacement

(mm)

Slider

Velocity (m/s)

Slider Acceleration

(m/s2)Remark

Theor. Pract. Theor. Pract. Theor. Pract.1.2.3.4.5.6.7.

8.9.10.11.12.

1.8 Calculations:

1.9 Graph: Plot a graph between the slider displacement and the crank rotation

1.10 Result & Discussion:

1.11 Precautions:

1. Displacement of slider should be measured at equal interval of crank rotation.

2. Smooth curves should be drawn in plotting the graph.

1.12 Sources of Errors:

1. Clearances in the joints.

2. Inaccurate graduation.

3. Inaccuracy in performing experiments.

1.13 Viva-voce questions



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V max = 2 S / t= 2 S / o during ascent=V 2 S/2 during descent

Fig.2.1, Cam & Follower Apparatus


The apparatus is shown fig. 2.1. It consists of a cam with flat-faced follower.

The angle of rotation of the cam and follower displacement can be read from

the graduation marked on cam and follower scale.

1.5 Procedure:-

1. Bring the cam & following to zero position.

2. Rotate cam slowly and note down the angle of rotation of the cam at

regular interval and the corresponding displacement of the follower.



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3. Plot a graph between displacement of the follower and the angle of

rotation of the cam.

4. Plot the velocity and acceleration diagram.

5. Determine the maximum velocity and acceleration during ascent and

descent.


S.NO. Angle of rotation

( 0 )

Displacement of follower

(cm)

1.2.

3.4.5.

1.7 Calculations:

1.8 Graph:Plot a graph between displacement of the follower and the angle of rotation

of the cam. Plot the velocity and acceleration diagram


1.10 Precautions:

1. Cam should be rotate lowly and continuously.

2. Lubricant the can the roller bearing to decrease friction.


1. Effect of clearance in the roller and cam spindle.2. Effect of the elasticity of the links.

3. Lateral shift in the roller follower and cam.



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Experiment No: 3 Governor


1.7 Observation Table 1.8 Calculation, 1.9 Graph, 1.10 Result & Discussion, 1.11 Precautions, 1.12 Viva-

voce questions.

1.1 Objective:To find the controlling force (Fc) for porter governor and proell

governor.

1.2 Apparatus:Governor Arrangements, vary volt, tachometer,

1.3 Theory:Definitions of

Sensitivity:

Stability:

Hunting:

Isochronisms:

Effort & power:

Insensitiveness:



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Fig.3.1 Porter Governor

Fig.3.2 Proell Governor



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1.7 Observation table:Porter Governor:

Weight placed on the sleeve-----------------Kg.

S. No. Speed, N(rpm)

Angularspeed (rad

/sec)

Sleevedisplacement, x

in mm

Height ofGovernor

h=hi-x/2

Radius ofrotation r=12- h2

ControllingForce,

Fc=m 2r

Remark

1.2.3.4.5.

Proell Governor:Weight placed on the sleeve-----------------Kg.

S. No. Speed, N

(rpm)

Angular

speed (rad

/sec)

Sleeve

displacement, x

in mm

Height of

Governor

h=hi-x/2

Radius ofrotation r=12- h2

Controlling

Force,

Fc=m 2r

Remark

1.2.

3.4.5.

1.8 Calculations:

1.9 Graph:Plot a graph between the angular speed and sleeve displacement for both thegovernors.

Plot a graph between the controlling force and radius of rotation for both thegovernors.


1.11 Precautions:



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1. Reading should be taken carefully.2. Speed should be increased gradually and slowly noting that sleeve may not

come out.

Experiment No.: 4 Gyroscope



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1.7 Observation Table 1.8 Calculation, 1.9 Graph, 1.10 Result & Discussion, 1.11 Precautions, 1.12

Sources of error, 1.13 Viva-voce questions.

1.1 Objective:To verify the law of gyroscopic couple, C=I p with the help

of motorized Gyroscope.

1.2 Apparatus:Motorized Gyroscope, weights, stopwatch & tachometer

1.3 Theory: Fig. 4.1 shows motorized gyroscope.

Fig.4.1 Gyroscope

The various terms involved are:

GYROSCOPE: It is rotating body, which processes perpendicular to plate of

rotation, i.e. axis of rotation also changes its direction under the action of

external forces.Axis of Spin: Is the axis about which a disc/rotor rotates as shown in figure

Precession: It means the rotation of axes in other plane or about other axis

(axis of precession) which is perpendicular to both the axis i.e. axis of spin

and axis of couple.



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Gyroscopic Couple: it is applied couple needed to change the angular

momentum vector of rotating disc/Gyroscope when it processes. It acts in

the plane of coupe which is perpendicular to both the other planes (plane of

spin and plane of precession) it is given as:-

C= I p

Where,

I = Moment of inertia of rotor.

= Angular velocity of rotor.

p= Angular velocity of precession.


1.5 Procedure:-

1. Balance the initial horizontal position of the rotor.

2. Start the motor by increasing the voltage with the transformer & watch

until it attains a constant speed.

3. Process the yoke frame no.2 about vertical axis by applying necessary

force by hand to the same.

4. It will be observed that the rotor frame swing about the horizontal axis Y-

Y. Motor side is seen coming upward and the weight pan side doing

downwards.

5. Rotate the vertical Yoke axis in the anti-clock wise direction seen from

above & observe that the rotor frame swing in opposite sense.

6. Balance the rotor position on the horizontal frame.

7. Start the motor by measuring the voltage with the autotransformer & waittill it attains constant speed.

8. Put weight in the weight pan & start the stopwatch to note the time in sec

required

9. Speed may be measured by the tachometer provided on control panel.



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10 Enter the observation in the table.

1.6 Specification:

1. Weight of rotor - 6.25 kg

2. Rotor diameter - 301 mm

3. Rotor thickness - 100.45 mm


Speed of

discC for 90o

precession

0.5 1 1.5 2 2.5

LoadTime

1.8 Calculations:1. I=

2. =

3. p= d / dt = /2/E 2 S / 22

1.9 Graph:1.10 Result & Discussion:

1.11 Precautions:

1. At starting the pointer should be at zero mark.

2. For comparison of Gyroscopic couple angular displacement for different

loads should be insured before conducting the experiment.

3. Proper lubrication should be placed gently and without impact.


1. Rotor should run at a steady speed.



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2. Rotor should rotate in a vertical plane.

Experiment No: 5 Whirling Speed of Shaft


1.7 Observation Table 1.8 Calculation, 1.9 Graph, 1.10 Result & Discussion, 1.11 Precautions, 1.12 Viva-

voce questions.



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1.1 Objective: Determine the whirling speed of various shafts

1.2 Apparatus:Whirling of shaft apparatus, auto-transformer, various shafts,

tachometer.

1.3 Theory:Describewhirling of shaft and effects of whirling.

Deflection due to mass of shaft.

5 w L4

=384 E I

Critical speed or whirling of speed

Nc= (1/2 ) ( g /) rps.

Where,

L= Length of the shaft

W= weight of the shaft= mass of the shaft x 9.81

I=Moment of inertia in mm4

E= Youngs modulus of elasticity= 210 N/M2


Fig. 5.1 shows whirling of shaft apparatus.



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Fig.4.1 whirling of shaft apparatus

1.5 Procedure: -

1. Fix the shaft properly at both the ends.

2. Check the whole apparatus for tightening of screws etc.

3. First increases the voltage slowly for maximum level and then start

slowing down step by step.4. Observe the loops appearing on the shaft and note down the number of

loops and the speed at which they are appearing.

5. Slowly bring the shaft to rest and switch off the supply.

6. Repeat the same procedure for different shaft.

1.6 Specification:L=

W=

I=

E= Youngs modulus of elasticity= 210 N/M2


S.No Shaft diameter (cm)

Moment ofinertia (cm4)

Weight(Kg./cm)

Length(cm)

123



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Critical speed

Shaft-1 Shaft-2 Shaft-3First NodeSecond NodeThird Node

1.8 Calculations:

1.9 Graph:


1.11 Precautions:

1. The shaft should be straight

2. The shaft should be properly tightened.3. Voltage should not be very high.

4. Reading should be taken properly.


3. Rotor should run at a steady speed.

4. Rotor should rotate in a vertical plane.

Experiment No: 6 Balancing (Static & Dynamic)


1.7 Result & Discussion, 1.8 Viva-voce questions.

1.1 Objective: To verify the fundamental laws of balancing by using rotating

masses.

1.2 Apparatus:Balancing apparatus, steel shaft, weights etc.

1.3 Theory: Fig. 6.1 shows balancing apparatus.



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Fig.6.1 Balancing Apparatus

When a disc is rotating along its centre of gravity with uniform speed, inertia

forces and torques will be zero if the matter is uniformly distributed about its

C.G. but if the centre of rotation and the geometrical centre of G are

different the inertia force and inertia torque will have some finite values.

The inertia force in this case will be balanced by the input torque but inertiaforce will cause deformation of the shaft in radial direction i.e. along the line

joining the center of rotation and C.G. if the disc is allowed to move in one

plane and is suspended by a spring to provide a restoring force the disc will

oscillate due to the fact that a force of type Fsinwt, Fcoswt will act upon it.

In the apparatus the C.G. is made to change from C.G. of rotation by adding

some weight at a certain distance from the C.G. of rotation of disc. The

unbalance added will depend upon the product weight added and the

distance at which at which it is added.

The balancing law can be written by applying condition of equilibrium to the

system.



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The apparatus basically consists of a steel shaft mounted in ball bearing in

a stiff rectangular main frame. A set of six blocks of different weights is

provided & may by clamped in any position on the shaft, and also be easily

detached from the shaft.

The disc caring circular protector scale is fitted in the side of the rectangular

frame. Shaft carried a disc & rim of this disc is grooved to take a tight hold

provided with two cylinder metal containers of exactly the same weight. The

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 &

angular position of each angular block to be determined.A 230 V drives the shaft, single phase 50 cycles 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 chain & in

this position 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 as that the mainframe & the supporting frame are in the same frame.

1.5 Procedure:-

Static Balancing: Remove the drive belt, the value of wrN for each block is

determined by clamping each block in turn on the shaft & with a cord &

container system suspended over the protector disc, the no. of steel balls,

which are of equal weights, are placed into one of the container to exactly

balance the block on the shaft. When the block becomes horizontal, the no. of

balls N will give the value of weight for the block.

For finding our wr during static balancing proceed as follows:



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1. Remove the belt.

2. Screw the combine hook to the pulley with the groove (this pulley is

different than the belt pulley)

3. Attach the cord ends of the pass to the above combined hooks.

4. Attach the block no.1 to the shaft at any convenient position & in

vertical downward direction.

5. Put steel balls in one of the pan till the block starts moving up. (upto

horizontal position)

6. No. of balls gives the wr value of block 1 repeat this for 2-3 times &

find the avg. no. of balls.

7. Repeat the procedure for the other blocks.Dynamic Balancing :

It is necessary to leave the machine before the experiment. Using the value of

wr obtained as above & if the angular position & planes of rotation of three of

four blocks are known, the students can calculate the position of other blocks,

(s) for balancing of the complete system, from the calculations, the students

finally clamps all the blocks on the haft in their appropriate position & then byrunning the one can verify that these calculations are correct & the blocks are

perfectly balanced.

1.6 Specification:




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Experiment No: 7 Vibration (Longitudinal)


1.7 Observation Table 1.8 Calculation, 1.9 Result & Discussion, 1.10 Precautions.

1.1 Objective: To study the longitudinal vibration of helical spring and todetermine the frequency of period of vibrator theoretically & actually byexperiment.

1.2 Apparatus:Vibration apparatus, stopwatch, weights, stand scale etc.



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1.3 Theory:

Longitudinal vibration:

Spring stiffness:


Fig. 7.1 shows the line diagram of vibration apparatus.

Fig.7.1, Vibration apparatus

1.5 Procedure:-

1. Fix one end of helical spring by upper screw.2. Determine the free length.3. Put some weight on platform & note down the deflection.4. Stretch spring length some distance & release.

5. Count the time required in sec. for say 10,20 oscillations.6. Determine the actual period.7. Repeat the procedure for different weights.

1.6 Specification:



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Axial length of spring =Mean diameter of spring =Wire diameter =

1.7 Observation table: For Mean Stiffness

S.No. Wt. attachedW= (m x 9.81) N

Deflection ofspring (cm)

Stiffness (k)(N/cm)

1.2.3.4.

For Mean Period

S.No. Wt. attachedW= (m x 9.81)

N

No.ofoscillations

(n)

Time foroscillations

(t)

Period (t/n)

1.2.3.4.

1.8 Calculations:


1.10 Precautions:

1. Note down the time correctly.



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2. Note down the oscillations properly.3. Dont stretch spring very much.

Experiment No: 8 Vibration (Torsional)


1.7 Observation Table 1.8 Calculation, 1.9 Result & Discussion, 1.10 Precautions.

1.1 Objective:To study the torsional vibration (undamped) of single rotor shaftsystem.

1.2 Apparatus:Torsional vibration apparatus, stopwatch etc.

1.3 Theory:

Torsional vibration:

Modulus of rigidity:Polar moment of inertia:

Fig. 8.1 shows the line diagram of a torsional vibration apparatus.



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Fig.8.1, Torsional vibration apparatus


One end of the shaft is gripped in the chuck and heavy flywheel free torotate in ball bearing is fixed at the other end of the haft. The bracket withfixed end of the shaft can be clamped at any convenient position alonglower beam. Thus length of the shaft can be varied during theexperiments.The ball bearing housing is fixed to side member of the main

frame.

1.5 Procedure:-

1. Fix the bracket at convenient position along the lower beam.2. Grip one end of the shaft at bracket by chuck.3. Fix the rotor on other end of the shaft.4. Twist the rotor through some angle and release.

5. Note down the time required for 10,20 oscillation.6. Repeat the procedure in different length of the shaft.

1.6 Specification:(a) Shaft diameter=



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(b) Diameter of disc=(c) Weight if the disc=(d) Modulus of rigidity for shaft= 0.8*106 Kg/cm2

1.7 Observation table:S.No. Length of shaft (L) No. of

Oscillations (n)Time taken forn oscillations

(t)

Periodictime(T=t/n)

1.2.3.4.5.

1.8 Calculations:i. Find the torsional stiffness KtKt= GIP/L Where L= length of shaft

D= Diameter of shaftIp= P.I. of shaftG= Modulus of rigidity

ii Theoretical

T=2 I/kt

Where, I= M.I. of disc=

iii Experimental

Time of oscillating

T=No. of oscillation




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1.10 Precautions:

1. The chuck should properly tighten the shaft.2. Note down the time correctly .

Experiment No: 9 Gear Train

Aim: To study the different types of gear train.

Gear Train:Sometime two or more gears are made to mesh with each other to transmitpower from one shaft to another such combination is called gear train.Following are the different types of gear train, depending upon thearrangement of wheels.

1. Simple gear train.2. Compound gear train.3. Reverted gear train4. Epicyclic gear train.

1. Simple gear Train:-When there is only one gear on each shaft is Knownas simple gear train.

Since circumferential velocity of meshing gear are same. (fig. a)



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d1 N1 d2 N2=

60 60

d1 N1 = d2 N2

N1 Z2... = .

N2 Z1 Where: d1 = P.C.D. of driver geard2 = P.C.D of driven gear Z1 = no. of

Teeth on Driverm = module Z2 = no. of Teeth on Driven

= P.C.D./ z Z = no. of Teeth on gear N1 = Speed of driver ( r.p.m.)N2 = Speed of drive ( r.p.m.)

The ratio of N1 and N2 is known as speed ratio.Train value is reciprocal of speed ratio i.e. speed ratio of driven gear to drivergear.

N2 Z1=

N1 Z2

It may be noted (from fig. ) that when the number of intermediate gear are oddthe motion of driven and driver are same and if number of intermediate gearare even the motion of driver & driven is opposite direction from fig. (b)

Let N1= Speed of driver gear 1 Z1 = No. of teeth on driver gearN2= Speed of intermediate gear2 Z2 = No of teeth on intermediate gearN3= Speed of driven gear Z3 = No. of teeth on driven gear

Since gear 1 and gear 2 are in meshing.

N1 Z2= --- --- --- (i) and similarly gear 2,3 are in meshing .

N2 Z1

N2 Z3= --- --- --- (ii)

N3 Z2



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Multiply both equations

N1 N2 Z2 Z3 =

N2 N3 Z1 Z2N1 Z3

=N3 Z1

Speed of driver No. of teeth on driveni .e. speed ratio =

Speed of driven No. of teeth on driver

Speed of driven No. of teeth driverand train value =

Speed of driver No. of teeth on driven

From above we see that the speed ratio and train value in a simple train ofgear is independent of the size and no. of intermediates gears. These

intermediates gears are called Idler gear.Idler gear does not effect on the train value and speed ratio.

COMPUND GEAR TRAIN: In compound gear train there are more then onegear on a shaft.

LetN1= Speed of the driving gear, N2, N3, N4, N5, N6 speed of respectivegears.Z1= No. of teeth on driving gear Z2, Z3, Z4, Z5, Z6 no. of teeth onrespective gears.

Since gear 1 in mesh with gear 2.N1 Z2



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Speed ratio = = --- --- --- (i) similarlyN2 Z1

N3 Z4= = ---- --- --- (ii)

N4 Z3

N5 Z6= = --- --- --- (iii)

N6 Z5Speed ratio of compound gear train.Multiplying equation (i), (ii) and (iii) we get.

N1 N3 N5 Z2 Z4 Z6= =

N2 N4 N6 Z1 Z3 Z5

N2 = N3, N5 = N4

N1 Z2 Z4 Z6=

N6 Z2 Z4 Z6

Speed of the first driver Product of the no. of teeth on driven

Speed ratio: = ---------------------------- = -----------------------------------Speed of the last driven product of the no. of teeth on drivers

Speed of the last driven Product of the no. of teeth on driversTrain ratio: = ---------------------------- = ---------------------------------

Speed of the first driven product of the no. of teeth on driven

The advantage of compound train over a simple gear train is that a muchlarger speed reduction from first shaft to the last shaft can be obtain with smallgears.Reverted Gear Train: When the axis of the first gear and last gear are co-axial, then the gear train is known as reverted gear train. In reverted gear motion offirst and last gear is in same direction.



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Let Z1 = no. of teeth on gear1 Z2, Z3, Z4, no. of teeth on respective gears.d1 = P.C.D. of gear d2, d3, d4 P.C.D. of respective gearsN1= speed of gear 1 in (r. p .m).

If a is the distance between the centre of shaft. (It is assume module of allgears are same)

d1+d2 d3+d4a = =

2 2

or mZ1 + mZ2 mZ3 + mZ3a = --------------- = -----------------

2 2

a = Z1 +Z2 = Z3 +Z4

or Z1 + Z2=Z3 + Z4Epicyclic Gear Train:

In an epicyelic gear train, the axis of the shaft, over which the gear aremounted , may move relative to a fixed axis. A simple epicyclic gear train isshown in fig. where gear A and the arm C have a common axis at O1 about

which they can rotate. Gear B meshes with gear A and has its axis on the armat O2, about which the gear B can rotate, if the arm is fixed , the gear train issimple and gear a can drive gear B or vice versa, but if gear A is fixed and thearm is rotated about the axis of gear A ( i.e. O1).then the gear B is forced torotate upon and around gear A . Such a motion is called epicyclic and the geartrains arranged in such a manner that one or ore of their members move uponand around another member are known as epicyclic gear trains (epi. Meansupon and cyclic mean around). The epicyclic gear trains my be simple orcompound.The epicyclic gear trains are useful for transmitting high velocity ratio with

gears of moderate size in a comparatively lesser space. The epicyclic geartrains are used in the back gear of lathe, differential gears of the automobiles.Hoists, pulley blocks. Wrist watches etc.

Velocity Ratio of Epicyclic Gear Train:The following two methods may be used for finding out the velocity ratio of anepicyclic gear train.



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Experiment No: 10 Gears

Aim : To study the Gears.

Gear: Gear are defined as toothed wheels or multilobed cams which transmitpower and motion from one shaft to another by means of successiveengagement of teeth .

The motion and power transmitted by gears is kinematically equivalent to thattransmitted by friction wheels or discs. In order to understand how the motioncan be transmitted by two toothed wheels, consider two plain circular wheels

A and B mounted on shafts, having sufficient rough surfaces and pressingagainst each other as shown in fig. 10.1 (a).Let the wheel A be keyed to the rotating shaft and the wheel B to the shaft, tobe rotated. A little consideration will show, that when the wheel A is rotated bya rotating shaft, it will rotate the wheel B in the opposite direction as shown inFig. 10.1 (a).If P>F sleeping will takes place, P= is tangential forceIf P< F sleeping not occurs, F= is frictional force

In order to avoid sleeping a number of projection (called teeth) are provided on

the periphery of wheel.

TERMINOLOGY:

Pitch Circle: - It is an imaginary circle which by pure rolling action would givethe same motion as actual gear.



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Pitch circle diameter (P.C.D.): It is the diameter of pitch circle. The size ofgear is usually specified by the P.C.D.

Pitch Point: It is common point of contact between two pitch circles.

Pressure angle or angle of obliquity: it is the angle between commonnormal to two gear teeth at a point of contact and the common tangent at thepitch point . Slandered pressure angle are 14, 20o .

Addendum: It is a radial distance of a tooth from pitch circle to the top of thetooth .

Dedendum: It is a radical distance of a tooth from pitch circle. to the bottom oftooth .

Clearance: Dedendum-Addendum.

Circular Pitch:- Circular pitch is the distance measured along the pitch circlebetween two similar point on adjacent teeth .

P.C.D.Pc = Z = no. of teeth on wheel

ZModule:- is the ratio of P.C.D. to the no of teeth.

P. C. D.m = --------

Z

Diametral Pitch: it is the ratio no of teeth to pitch circle diameter.Z

Pd= -------- P.C.D

P.C.D. Z

PcPd = ----------- ---------Z P.C.D

Addendum (ha) =mDedendum ( Hf ) = 1.25 m.Clearance =( hf ) = (hf-ha) = 0.25 mTooth thickness = 1.5708 m


PcPd=


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TYPES OF GEARGears are broadly classified in to four groups.-Spur gear-Helical gear-Bevel gear

-Worm gear

Spur Gear:-Teeth are cut parallel to the axis of the shaft. Profile of the gear

tooth is in the shape of involute curve and remains identical along entire widthof gear wheel. As the teeth are parallel to the axis of the shaft spur gear areused only when the shaft are parallel. Spur gear impose radical load on theshafts.

Helical Gear:-The teeth of these gears are cut at an angle with the axis of theshaft. Helical gear have an involute profile similar to that of spur gear. Howeverthis involute profile is in a plane which is perpendicular to the tooth elements.The magnitude of helix angle of pinion and gear is same, however the hand ofhelix is opposite. A right hand pinion meshes with left hand gear and viceversa. Helical gear impose radical and thrust load on the shaft.

There is a special types of helical gear consisting a double helical gear withsmall grove between two helices. The grove is required for hobbing andgrinding operation. These gears are called herringbone gear. Theconstruction results in equal and opposite thrust reaction balancing each otherand imposing no thrust load on the shaft .Herringbone gear are used only forparallel shafts.

Bevel gear: - Bevel gear have a shape of truncated cone. The size of geartooth, including the thickness and height, decreases towards the apex of thecone. Bevel gear are normally used for shafts which are right angles toeach other. This however is not rigid condition and the angle can be slightlymore or less then 90degrees. The tooth of the bevel gears can be cut straightor spiral (4) Bevel gear impose radical and thrust load on the shafts.



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Worm gear:-The warm gears consist of a warm and a warm wheel. The warmis in the form of a threaded screw, which meshes with the matching wheel. Thethreads on the warms can be single or multi start and usually have a smalllead. Warm gear drives are used for the shafts, the axis of which do notintersect and are perpendicular or to each other. The warm impose high thrust

load while worm wheel impose high radical load on the shaft. Worm gear driveare characterized by high speed reduction ratio.

Law of gearing:- The common normal at the point of contact between apair of teeth must always pass through a fixed point in order to obtainedconstant velocity ratio. Fixed point is called pitch point.

Forms of Teeth:- Two types of teeth commonly used.

(i) Cycloidal Teeth

(ii) Involute Teeth

Interference :- The phenomenon when tip of tooth undercut the root on itsmating gear is known as interference .Only Involute and cycloidal curves satisfy the fundamental law of gearing. Incase of involute profile the common normal at the point of contact alwayspasses through the pitch point (p) and maintains a constant inclination ( ) withcommon tangent to the pitch circle. The is called pressure angle . In case ofcycloid curves the pitch point is fixed but inclination various , it is due to this

reason cycloidal carves become obsolete . Some time combination ofinvolutes and cycloid carves is used for gear tooth in order to avoidinterference . In this case middle third of the tooth profile has an involuteshape while the remaining profile is cycloidal.The disadvantage of the involute teeth is that the interference occurs withpinion having smaller no. of teeth . This may be avoided by altering theheights of addendum and dedendum of mating teeth or angle of ablightly.Envolute teeth are easy to manufacturer then cycloid teeth .

Cycloidal gears are stronger then the involute gear for the same pitch. Less

wear in cycloidal gear as compared to involute gears. In cycloidal gearinterference does not occur.



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me2307dynamics lab manual 2

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