Dynamics of Machine Manual

8/9/2019 Dynamics of Machine Manual

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EXPERIMENT NO : 5

GEAR TOOTH PROFILE

AIM : (1). To draw tooth profile of gear using gear generation principle.

(2). To determine the dimension of the gear to be produced by the gear

generation principle.

REQUIREMENTS: Model of gear generation (Rack cutter and gear blank in the

form of sheet), Drawing paper , pencil , protector etc ...

THEORY: Gear terminology:

(1) Pitch circle diameter (D): It is the diameter of the circle which by a pure

rolling action would transmit the same motion as the actual gear wheel.

(2) PITCH :

(a) CIRCULAR PITCH (Pc): It is the distance measured along the pitch

circle circumference from a point on one tooth to the corresponding point on

next tooth Pc = D/T, where T = Number of teeth on the gear wheel.

(b). DIAMETERAL PITCH (Pd): It is defined as the number of teeth – per

unit dimension of pitch diameter.

... Pd =

(c). Module (m):

It is defined as the pitch diameter divided by t, number of teeth. Generally it isspecified in mm.

.‟. m =

(d). Base pitch (Pd) : It is the distance as the measured along the base circle

circumstances from the point on one tooth to the corresponding point on next

tooth.

(3). ADDENDUM: IT is the radial distance from the pitch circle to the top of the

tooth.

(4). DEDENDUM: It is the radial distance between the pitch circle and the bottom

of the tooth space.


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PROCEDURE: A blank drawing paper in attached on to the circular board which

represent the gear blank. The circular board is rooted each time by 2 and

correspondingly in to same direction the rack is moved by a distance equal to a

circumferential distance moved by the circular board on the pitch circle. Each time the

position of the rack edge is traced on the paper attached on the circular board. Thecurve should be drawn in such a way that the straight lines are tangent to it. This gives

on involute profile of the gear tooth.

OBSERVATION AND CALCULATION:

Blank Diameter = Addendum diameter Dq = _______cm.

Pitch on the rack Pc of the gear to be generated = _______ cm to be measured

on rack.

As Pc =

= πm , .

.. =

=__________

... m =Module = ______

Pressure Angle = = 20 (Given),

As R = R a – m =

New m = D/T , ... T = D/m =,

D = T.m =

R =

Radius of base circle = R b = R cos = __________ cm.

∴ Base pitch P b =

b D

T = ____________cm

Measure the angle substanded at centre by circle pitch are on the tooth profile drawn

on paper on pitch circle and check whether;

= 18 i.e Q should be 20.

QUESTION

1. State the law of gearing?

2. What are the form of teeth which satisfy the low of gearing?

3. What do you mean by interference in toothed gearing?

4. What are the methods of avoiding interference?

5. What the various type of gear? What will be positions of the shaft axis in each

case?


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REFERENCE:

1. Theory of Machines by Thomes Bevan.

2. Theory of Machines by J.R Shigley.

3. Theory of Machines by J.P Ballaney.


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EXPERIMENT NO: 6

CAM- PROFILE

AIM: (1). To draw the cam profile from the performance of a given cam follower

assembly, when (a) Follower axis passes through the cam shaft centre, (b). When

follower axis is having eccentricity = 1 cm

(2). To check the displacement of follower during rise (uniform acceleration

and retardation on) and during return (SHM) analytically for radial follower case.

(3). To find the velocity of the follower at the given values of (rotation ofcam) for radial follower analytically and graphically.

REQUIREMENTS: Cam follower assembly, scale, graph paper, drawing paper etc.

THEORY:

FOR RISE: During 120 of cam rotation with uniform acceleration and retardation

(parabolic);

X = 2.h (

for 0 ≤ θ ≤ /2............................ (1)

X = h [1-2 (1 -

)

2] for /2 ......................... (2)

where x = displacement of follower at any angle

h = lift of the follower (4 cm for our case )

= total angle turned through during the rise i.e 120˚ in our case.


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FOR RETURN : during 120˚ of cam rotation with SHM (i.e. from 150˚ to 270˚ of

cam rotation )

x= (h/2) * 1-Cos(πθ / α )

where α = total angle turned through by cam during return stroke.

PROCEDURE : Take the readings on x scale for the various angular position of cam

at the interval of 10° in both the case (a) and (b). Convert the readings on x scale such

that for = 0

X = 0. Draw the displacement diagram x v/s and from this draw the cam profile in both the cases (a) and (b). Check the displacement of the follower at various angular

position of cam analytically for the known rise and return.

OBSERVATION AND TABULATION:

Minimum radius of cam = 30 mm.

Diameter of the roller follower = 35 mm.

Angular velocity of cam= 1 rad/s

TABLE A : RADIAL FOLLOWER

Θ XS X = XS-XMIN θ XS X = XS-XMIN


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TABLE B: OFFSET FOLLOWER

Θ XS X = XS-XMIN θ XS X = XS-XMIN

CALCULATIONS:

(1). Calculation of x for various value of .

(2). Calculate of – velocity (x) for various value of

GRAPHS:

Draw the graph of x v/s θ for both the cases and from this draw the graph of (dx/dθ)

V/s θ for radial follower only.

RESULTS :

Sr.

No

θ X(Experimentally)

X(Analytically)

Sr.

No

θ X(Experimentally)

X(Analytically)


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QUESTION:

1. What are the various type cam and follower?

2. What are the various type of applications of Cam?

3. Why are the spring used in cam follower system?

REFERENCE:





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EXPERIMENT NO: 7

HOOK’S JOINT

AIM: To study the performance of a hook‟s joint and to draw displacement diagram

(ϕ Vs θ) ,considering w=angular velocity of the driver shaft = 1 rad/s

REQUIREMENT: Hook‟s Joint, scales, Graph paper(full size) etc.

THEORY:


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This joint is used to connect two non-parallel intersecting shafts. The end of each shaft

is forked and each fork provides two bearings for the arms of a cross. The arms of the

cross are at right angles and the cross serves to transmit motion from the driving to the

driven shaft. The applications of the Hook‟s joint are found in the transmission from

the gear box to the back axle of automobiles and in the transmission of the drive to thespindles of the multiple drilling machines.

In fig.2 an end elevation looking along the axis of the driving shaft and a top view are

shown. The planes or rotation of the cross are represented by traces PP and QQ in top

view.

Let the initial position of the cross be such that both the arms lie in the place of paper

in elevation, while the arm is attached to the driving shaft lies in the plane containing

the axis of the two shafts. Let the driving shaft turn through the angle θ so that the arm

AB is displaced to A1B1. Then the projection C1D1 of the other arm CD must also turn

through the angle θ. But the true length of the C1O is given by C2O and therefore the

angle through which the driven shaft has turned is given by φ;

From figure, tan φ =

and tan φ =

=

=

If α is the inclination of the driven shaft to the driving shaft, then OM/ON = cos α

= 1/cos α ; tan θ =cos α.tan φ ………………..(1)

Let w = Angular velocity driving shaft =

and w1 = Angular velocity of driven shaft =

Differentiating both sides of equation (1)

Sec2 φ

= cos α sec2 φ

=

= cos α cos

2 θ sec

2 φ

But, sec2φ =1 +

=

=

α . …………………………(2)

For a given values of α this expression is a maximum and cos θ =0


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i.e. when θ = π/2, 3 π/2 etc. and it is minimum when cos θ = ±1 when θ = 0, π etc.

If the speed ω of the driving shaft is constant , the maximum speed w1 of the driven

shaft is given by

= 1/cos α & minimum speed of the driven shaft w1/ w = cos α,

The value of θ for which the speeds of the driving and driven shafts are equal w= w1

may be found by equating equation (2) to unity.

1-cos2 θ sin

2α= cos α

cos2 θ =

=

………………………..(2)

sin2 θ = 1 - cos

2 θ =

And , tan2 θ = cos α ; tan θ =±√ ………………..(3)

Equation (3) gives four values of θ for a particular value of which gives w= w1

The angular acceleration of the driven shaft is given by

…………………………. (4)

The value of the θ for which the acceleration is maximum may be found by,

Cos 2θ =

……………………………. (5)

PROCEDURE:

Adjust the pointers on both the scales alongwith both the shafts so that when θ=0,ϕ=0.

Rotate the driven shaft through 200 each time and take the reading of φ for driven

shaft. Draw the displacement diagram φ and θ. Calculate the values of θ at which w =

w1 and find maximum and minimum value of w1 for a given value of α and w(=1

rad/s).

OBSERVATIONS AND CALCULATION:

α = Angle between the shaft axis.

W = Angular velocity of driver shaft = 1 rad/s (given)


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TABLE :

Θ Φ Θ Φ

CALCULATIONS:

1. Find θ @ which w1=w

2. Find θ @ which (w1)max & (w1)min

GRAPH :

Draw a graph of Φ vs θ.

RESULT :

QUESTIONS:

1 What is other name of hook joint ?

2 How can you find percentage fluctuation of speed of driven shaft for given value of

α and ω ?

3 What is "Double Hook joint? why it is Required?

4 State the condition under which two shafts connected together by a Double Hooks

joint shall have the same angular velocities.

REFERANCE :

1.

Theory of Machines by Thomes Bevan.

2.

Theory of Machines by J.R Shigley.

3.

Theory of Machines by J.P Ballaney.


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EXPRIMENT NO : 08

BALANCING OF FOUR ROTOR SYSTEM

AIM:

To study the balancing of several masses revolving in different planes

theoretically and experimentally.

REQUIREMENTS : Scale, Masses

Theory:

The apparatus has been designed to carry out experiment on primary balancingin the laboratory and verify the theory of balancing.

It consists of a rectangular steel frame suspended by four spring from a strong

steel stand as shown in fig.1. One rectangular frame, two bearings are mounted which

support a steel shaft carrying four balanced dies equally spaced. One of the dies is

grooved and this is connected to a balanced 220 V A.C Electrical motor by v-belt so

that the whole system can be rotated.

In all the four dies circumferential slot are provided at four different road.

Number of steel Places to act as balancing masses are included and these places can

be attached in the slots of the dies by means of screwed rod and nuts. The discs are

marked with radial lines and numbered, to read the angular position of the balancingmasses. The balancing weight attached under the left hand corner of the frame ensures


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that the spring tensions are identical and the spirit level. By attaching the balancing

weight to the circumferential slots of the discs at different positions, various

combination of “ out of balance condition “ can be obtained either in one plane or two

plates . By switching on the motor and making the system to rotate, stage of “out of

balance” can be clearly observed due to vibration and oscillation set up in the system.

Now the distance between discs positions of balancing weights,

The standard equipment includes about 40 weights and 5 screwed rods with

nuts, washers etc. This screwed rod with nuts and washers form one standard weight,

and the weights can be attached to these rods according to the requirements.

The equipment is complete with strong iron stand for floor mounting, set of tools and

a scale.

The unbalanced masses are placed in the planes 1 and 2 e.g. 200 gms at 180

and 80 respectively. Now in order to find the balancing masses required to be placed

in the reference planes L & R, we use the following table.

PROCEDURE:

For the giving four rotor system and the unbalanced masses placed in two rotors in the

slots provides, use the following method to balance the system.

PLANE Mass, m

(kg.)

Radius,

(cm)

Cont.force/W

(kg.cm)

Distance

From ref.

Plane(m)

Couple / w

1

2

3

4

The two equation Σ F = 0 and Σ m = 0 are used to determine the amount and

location of the corrections.

The above two vector equation can be solved graphically or analytically. Using both

these methods determine the location and amount of masses to be placed in theremaining two rotors.


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CONCLUSION:-

QUESTION :

(1). Explain the term „Static Balancing‟ and Dynamic Balancing‟. State the necessary

condition to achieve them.

(2). Why balancing of rotating plates necessary for high speed engines ?

REFERANCE :




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Dynamics of Machine Manual

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