ME-222 Mechanics Manufacturing Lab-I

8/19/2019 ME-222 Mechanics Manufacturing Lab-I

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ME-222 MECHANICAS AND MANUFACTURING LAB-ITable of Contents

1. Introduction to Lab Equipment and Safety Precautions (Verbal)

2. Time Period of a Simple PendulumTo find time period of a simple pendulum.

3. Fundamental of Statics- Part (a)a. To verify the validation of parallelogram law of forces

b. Resolving forces into their components

4. Fundamental of Statics- Part (b)c. Investigation of equilibrium of momentsd. To find out Torque of non-parallel forces

5. Fundamental of Statics- Part (c)To develop an understanding of levers and to find out mechanical advantage and leverage of all 3 three

classes of levers.

6. Center of gravityTo find the center of gravity of regular and irregular shapes.

7. Reaction Forces in Beams

To find out the support reactions of a simple supported beam.

8. Rolling Disc on an Inclined PlaneTo determine experimentally the moment of inertia of different disc assemblies and compare the resultswith the theoretical values obtained from the mass and the physical dimensions of disc assembly.

9. Friction on a Flat PlaneTo determine the coefficient of friction between various materials and a steel plane.

10. Friction on an Inclined Planea. Find the angle of friction of various materials on a steel plane

b. Verify that the force required parallel to an inclined plane to move a body up the plane corresponds tothe friction coefficient (or angle) already found.

11. Winch

To find out Mechanical advantage, velocity ratio and efficiency of winch.12. Worm and WheelTo find out Mechanical advantage, velocity ratio and efficiency of wheel and worm gear.

13. Toggle JointTo find out Mechanical advantage of toggle joint.

14. Slider Crank Mechanisma. To analyze the variation of displacement of piston in relationship with crank angle & to calculate

velocity and acceleration of piston b. To draw graph between crank angle & piston position.c. Also draw displacement, velocity and acceleration graphs verses time.

15. Whitworth Quick Return Mechanism

a. To analyze the variation of displacement of oscillating rocker in relationship with crank

rotation & to calculate velocity and acceleration of rocker b. To draw graph between crank angle & rocker position.

c. Also draw displacement, velocity and acceleration graphs verses time.

16. Experiments Practice/Viva-Voce


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List of FiguresFigure No. 1

………………………………………………………………………………………..……...5Simple pendulum

Figure No. 2

………………………………………………………………………………………………. 6Simple pendulum Apparatus

Figure No. 3

………………………………………………………………………………………………. 8Law Of Parallelogram Of Forces

Figure No. 4

……………………………………………………………………………………………… .9

Finding the Equilibrant

Figure No. 5……………………………………………………………………………………………… 10

Equipment Setup For Investigation Of Components Of A Force

Figure No. 6……………………………………………………………………………………………… 11

Vector Components

Figure No. 7……………………………………………………………………………………………… 13

Equipment Setup For Equilibrium Of Moments

Figure No. 8………………………………………………………………………………………………. 14

Torque

Figure No. 9

………………………………………………………………………………………………. 15

Equipment Setup Torque Of Nonparallel Forces Figure No. 10………………………………………………………………………………………………. 16Class 1 lever

Figure No. 11

………………………………………………………………………………………………. 17

Class 2 lever

Figure No. 12

………………………………………………………………………………………………. 17

Class 3 lever

Figure No. 13

………………………………………………………………………………………………. 19Regular Shapes

Figure No. 14………………………………………………………………………………………………. 19

Irregular Shapes

Figure No. 15………………………………………………………………………………………………. 21


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Beams

Figure No. 16

………………………………………………………………………………………………. 22Reaction Forces in Beams

Figure No. 17

………………………………………………………………………………………………. 25Spindle

Figure No. 18………………………………………………………………………………………………. 26

Rolling Disc On An Inclined Plane

Figure No. 19………………………………………………………………………………………………. 29

Friction Between Two Surfaces

Figure No. 20………………………………………………………………………………………………. 29

Friction On A Flat Plane Apparatus

Figure No. 21………………………………………………………………………………………………. 31

Frictional On An Inclined Plane

Figure No. 22

………………………………………………………………………………………………. 35Winch

Figure No. 23

………………………………………………………………………………………………. 37Worm And Wheel Arrangement

Figure No. 24

………………………………………………………………………………………………. 38

Worm And Wheel Description

Figure No. 25………………………………………………………………………………………………. 38

Worm And Wheel Apparatus

Figure No. 26………………………………………………………………………………………………. 40

Toggle mechanism. (a) Simple structure. (b) Traditional configuration. (c) Typical application.

Figure No. 27………………………………………………………………………………………………. 41

Toggle Apparatus Description

Figure No. 28………………………………………………………………………………………………. 41

Toggle Apparatus

Figure No. 29

………………………………………………………………………………………………. 44Crank And Connecting Rod

Figure No. 30

………………………………………………………………………………………………. 47


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Crank and Connecting Rod Mechanism

Figure No. 31

………………………………………………………………………………………………. 49Witworth Quick Return Apparatus


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EXPERIMENT # 1 Time Period of a Simple Pendulum

OBJECTIVE

To calculate the time period of simple pendulum and compare it with the theoretical values.

INTRODUCTION

A simple pendulum in its simplest form consists of heavy bob suspended at the end of lightextensible and flexible string. The other end of the string is fixed

Figure No. 1 Simple pendulum

L = Length of the string

M = Mass of the bob in kg

W = Weight of the bob in Newtonθ = Angle through which the string is displaced

When the bob is at A the pendulum is in equilibrium position. If the bob is brought to B or C andreleased, it will start oscillating between the two positions B and C with A as mean position. It

has been observed that if the angle θ is very small then the bob will have simple harmonic

motion. Now the couple tending to restore the bob to the equilibrium position or restoring torque,

T = m g Sin θ * L

Sin θ is very small therefore, sin θ ≈ θ radians

T = m g L θ


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We know that mass moment of inertia of the bob about an axis through the point of suspension,

I = Mass * (Length)2 = m L

2

Angular acceleration of the string,

We know that, periodic time

From above eq we see that the periodic time of a simple pendulum depends only upon its length

and acceleration due to gravity. The mass of the bob has no effect on it.

APPARATUS

Figure No. 2 Simple pendulum Apparatus

PROCEDURE

1. Take a long string and tight the bob on its one end.

2. Then tight the string with pin on wall mounted pendulum apparatus.3.

Deflect the bob from its original position by keeping string tight.

4. Take a stop watch to note the time.

5. Release the bob and instantly start the stop watch.6. Note the time of 20 oscillations and calculate the time period by using the formula.7. Repeat the procedure to 2-3 times and take the average time period.

8. Reduce the length of the string and repeat the same procedure.


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9. Compare it with theoretical value.

OBSERVATION AND CALCULATIONS

Mass of the bob = ____________________

Sr. # Length

(m)

Time for 20 oscillations

(sec)

Time Period

(Theoratical)(sec)

Time Period

(Practical)(sec)

Error

(sec)


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EXPERIMENT # 2 Fundamental of Statics – Part (a)

EXPERIMENT a

OBJECTIVE

To verify the validation of parallelogram law of forces.

THEORY

In Figure 1, spaceships x and y are pulling on an asteroid with forces indicated by vectors Fx and

Fy. Since these forces are acting on the same point of the asteroid, they are called concurrent

forces. As with any vector quantity, each force is defined both by its direction, the direction of

the arrow, and by its magnitude, which is proportional to the length of the arrow. (The magnitude

of the force is independent of the length of the tow rope.)

Figure No. 3 Law Of Parallelogram Of Forces

The total force on the asteroid can be determined by adding vectors Fx and Fy. In the illustration,

the parallelogram method is used. The diagonal of the parallelogram defined by Fx and Fy is Fr,

the vector indicating the magnitude and direction of the total force acting on the asteroid. Fr is

called the resultant of Fx and Fy.

PROCEDURE

1. Setup the apparatus as shown in figure 4.

2. Use the holding pin to hold the force ring in place.

3. Add weights in the respective hangers and make sure that the ring is centralized.

4. Note down the forces and their respective angles.


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5. Find out the resultant by using parallelogram law of forces. Compare the graphical

resultant with the experimental resultant.

Figure No. 4 Finding the Equilibrant


Sr. # F1

(N)

θ 1

(Degree)

F2

(N)

θ 2

(Degree)

Fr

(N)

θ r

(Degree)


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EXPERIMENT b

OBJECTIVE

Resolving Forces into their components.

THEORY

When forces are resolved in their X and Y components then graphically and experimentally they

fgive the same result . for graphical method force on an angle is drawn and is then resolved into

its components but analytically the formula Fx=FCosӨ and Fy=FSinӨ are used respectively.

PROCEDURE

1. Set up the equipment as shown in Figure 5.

2. As shown, determine a force vector, F, by hanging a mass from the Force Ring over a

pulley.

3.

Use the Holding Pin to hold the Force Ring in place.

4. Set up the Force Balance and a pulley so the string from the balance runs horizontally

from the bottom of the pulley to the Force Ring. Hang a second Mass Hanger directly

from the Force Ring.

5. Now pull the Force Balance toward or away from the pulley to adjust the horizontal, or

“x-component” of the force. Adjust the mass on the vertical Mass Hanger to adjus t the

vertical or “y-component” of the force. Adjust the x and y components in this way until

the Holding Pin is centered in the Force Ring. (Notice that these x and y components are

actually the x and y components of the equilibrant of F, rather than of F itself.)

Figure No. 5 Equipment Setup For Investigation Of Components Of A Force


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Figure No. 6 Vector Components


Sr. # F(N)

θ(degree)

Fx = F Cos θ (N)

Fy= F Sin θ (N)


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EXPERIMENT # 3 Fundamental of Statics – Part (b)

EXPERIMENT c

OBJECTIVE

Investigation of the equilibrium of moments.

THEORY

The Law of Moments allows us to determine when an object is balanced. It has important

applications in aviation because pilots need to know if their aircraft will fly straight and level.

The law of moments says that an object such as a scale will be in equilibrium (will not tip in

either direction) when the Counterclockwise Moment is equal to the Clockwise Moment .

Following are some basic principles used in study of Equilibrium.

The Moment of a force is the turning effect about a pivot point. To develop a moment, the force

must act upon the body to attempt to rotate it. A moment is can occur when forces are equal and

opposite but not directly in line with each other.

The Moment of force acting about a point or axis is found by multiplying the Force (F) by the

perpendicular distance from the axis (d), called the lever arm.

Moment = Force x Perpendicular Distance

M = F x d

(N m) = (N) x (m)

PROCEDURE

1. Setup the apparatus as shown in figure 7.

2. Pass the rod from the pivot and screw it tightly.

3. Apply load on each side and note down its value

4. Measure the distance from center when rod is got balanced.

5. Verify that the clockwise moment is equal to anticlockwise moment.


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Figure No. 7 Equipment Setup For Equilibrium Of Moments


Sr.

#

F1

(N)

d1

(mm)

t1= (F1 x d1)

(N-mm)

F2

(N)

d2

(mm)

t2 = ( F2 x d2 )

(N-mm)


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EXPERIMENT d

OBJECTIVE

To find out torque of nonparallel forces.

THEORY

In Experiment 3, you investigated torques applied to the balance beam, and discovered that when

the torques about the point of rotation are balanced, the beam remains balanced. However, all the

forces in that experiment were parallel to each other and perpendicular to the balance beam.

What happens when one or more of the forces is not perpendicular to the beam.

Fortunately, it turns out that the formula for torque that you determined in Experiment 3 (τ = F d)

can be generalized to account for this more general case.

The generalized formula is:

τ = F d sinθ;

where F is the magnitude of the applied force, d is the distance from the pivot point to the point

at which the force is applied, and θ is the angle between F and d (see Figure 1).

Figure No. 8 Torque

In this experiment, you will investigate the validity of this definition for torque.

PROCEDURE

1. Set up the equipment as shown in Figure 9.

2.

First balance the beam without any applied forces.3. Then use a hanging mass and the force Balance to apply forces F1 and F2 as in Figure 2.

4.

Attach the degree dial by masking tape on the board at the centre of application of force

F1.

5. Note down F1,F2,d1,d2 and Ө then check the validity of formula τ =F.d with τ = F d sinθ


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Figure No. 9 Equipment Setup Torque Of Nonparallel Forces


Sr.

#

F1

(N)

θ

(degree)

d1

(mm)

τ1 = F1 d1Sinθ

(N-mm)

F2

(N)

d2

(mm)

τ2 = F2 .d2(N-mm)


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EXPERIMENT # 4 Fundamental of Statics – Part (c)

OBJECTIVE

To develop an understanding of levers and to find out mechanical advantage and leverage of all 3classes of levers.

THEORY

A lever is a rigid rod or bar capable of turning about a fixed point called fulcrum. It is used as a

machine to lift a load by the application of a small effort. The ratio of the load lifted to the effort

applied is called the mechanical advantage. The perpendicular distance between the load point

and fulcrum is known as load arm and the perpendicular distance between the effort point and

fulcrum is called effort arm. The ratio of the effort arm to the load arm is called leverage.

The levers may be of first type, second type and third type. In the first type of levers, the fulcrumis in between the load and effort. Since the effort arm is equal to load arm, therefore, the

mechanical advantage is equal to one. Such type of levers are commonly found in bell cranked

levers used in railway signaling arrangement, rocker arm in internal combustion engines, handle

of a hand pump, hand wheel of a punching press, beam of a balance, foot lever etc.

In the second type of levers, the load is in between the fulcrum and effort. In this case, the effort

arm is more than load arm, therefore, the mechanical advantage is more than one. The

application of such type of levers is found in levers of loaded safety valves.

In the third type of levers, the effort in between the fulcrum and load. Since the effort arm, in thiscase, is less than the load arm, therefore, the mechanical advantage is less than one. A pair of

tongs, the treadle of a sewing machine etc. are examples of type of lever.

Figure No. 10 Class 1 lever


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PROCEDURE

1.

Set up the equipment as shown in Figure 1.

2. Measure length d1 which is the effort arm

3. Measure length d2 which is the load arm

4. Measure the load and effort applied.


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5. Calculate the mechanical advantage and leverage.

OBSERVATIONS AND CALCULATIONS

Sr.

#

d1

(mm)

d2

(mm)

Load W

(N)

Effort P

(N)

Leverage = d1/d2 M.A = W/P


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EXPERIMENT # 5 Center Of Gravity

OBJECTIVE

To find the Center of Gravity of Regular and Irregular shapes

THEORY

Locating the center of gravity of an object is very important in our daily lives. The earth pulls

down on each particle of an object with a gravitational force that we call weight.

Although individual particles throughout an object all contribute weight in this way, the net

effect is as if the total weight of the object were concentrated in a single point - the object's

center of gravity.

In general, determining the center of gravity (cg) is a complicated procedure because the mass

(and weight) may not be uniformly distributed throughout the object. The general case requiresthe use of calculus. If the mass is uniformly distributed, the problem is greatly simplified. If the

object has a line (or plane) of symmetry, the cg lies on the line of symmetry. For a solid block of

uniform material, the center of gravity is simply at the average location of the physical

dimensions.

APPARATUS

Figure No. 13 Regular Shapes

Figure No. 14 Irregular Shapes


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PROCEDURE

The Plumb Line Method

1. You have received different shapes of materials. The shapes of regular object, irregularobject and letter were cut out from the acrylic pieces.

2.

Small holes at non-collinear points were punched on each sample.

3. The sample need to be suspended on the board supplied with the apparatus at thesuspending pin at the top center of the body

4. The sample should hang loosely from the support and it should not touch any surface.

5. A plumb bob was suspended from the support with the cord extending down in front ofthe main body and suspending sample.

6. A line need to be drawn on the sample along the path of the cord.

7. The sample has to be removed and suspended again through another hole. The line has to

be drawn again.8. The intersection of the two lines was marked as C (the center of gravity).

9. Repeat the above procedure for the other shapes to get the center of gravity.


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EXPERIMENT # 6 Reaction of Simply Supported Beam

OBJECTIVE

To find out the support reactions of a simply supported beam.

THEORY

General Beams Theory

A bar subject to forces or couples that lie in a plane containing the longitudinal axis of the bar is

called a beam. The forces are understood to act perpendicular to the longitudinal axis.

Simple Beams

A beam that is freely supported at both ends is called a simple beam. The term "freely supported"

implies that the end supports are capable of exerting only forces upon the bar and are not capable

of exerting any moments. Thus there is no restraint offered to the angular rotation of the ends of

the bar at the supports as the bar deflects under the loads. Two simple beams are sketched in Fig.

1

Figure No. 15 Beams

It is to be observed that at least one of the supports must be capable of undergoing horizontal

movement so that no force will exist in the direction of the axis of the beam. It neither end were

free to move horizontally, then some axial force would arise in the beam as it deforms under load

The beam of Fig. 1(a) is said to be subject to a concentrated force; that of Fig. 1(b) is loaded by a

uniformly distributed load as well as a couple.

Statically Determinate Beams

The beams considered above, are ones in which the reactions of the supports may be determined

by use of the equations of static equilibrium. The values of these reactions are independent of the

deformations of the beam. Such beams are said to be statically determinate.


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Statically Indeterminate Beams

If the number of reactions exerted upon the beam exceeds the number of equations of static

equilibrium, then the statics equations must be supplemented by equations based upon thedeformations of the beam. In this case the beam is said to be statically indeterminate.

Internal Forces and Moments in Beams

When a beam is loaded by forces and couples, internal stresses arise in the bar. In general, both

normal and shearing stresses will occur. In order to determine the magnitude of these stresses at

any section of the beam, it is necessary to know the resultant force and moment acting at that

section. These may be found by applying the equations of static equilibrium.

The transverse applied load on a beam can be in two forms, either concentrated or distributed. A

distributed load occupies a length of the beam surface. It is taken as being constant across the beam width but it can be uniformly or non-uniformly distributed over part or the whole length of

the beam. Two simplified forms of support are used for the ease of analysis. These are termed

simply supported and built in or fixed. The number and type of supports can make a beam either

statically determinate or statically indeterminate. In -the former a solution can be obtained by

considering force and moment equilibrium. If the beam is statically indeterminate we also have

to consider the deflection of the beam in order to obtain a solution.

Both stresses and deflections during bending are related to the shear force and the bending

'moment. Thus it is essential to know how these quantities vary along a beam and where the

maximal and minimal are located.

APPARATUS

Figure No. 16 Reaction Forces in Beams


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PROCEDURE

1. Assemble the apparatus as shown in previous figure.

2. Attach force gauges to their holders and tighten the screws.

3.

Hang beam on force gauges with the help of thread.

4. Place hanger to the desired slot of beam.

5. Measure the distance of applied load from reference „A‟ with a measuring tape.

6. Make the force gauge display zero by revolving aluminum dial on the gauge.

7. Now add weight on hanger.

8. Compare the theoretical reaction forces with the reaction forces displayed on force gauge.


Case

#

W1

(N)

W2

(N)

W3

(N)

U

(cm)

X

(cm)

Y

(cm)

Z

(cm)

L

(cm)

R Ath

(N)

R Bth

(N)

R Aexp

(N)

R Bexp

(N)


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EXPERIMENT # 7 Rolling Disc on an Inclined Plane

OBJECTIVE

To determine experimentally the moment of inertia of different disc assemblies and compare the

results with the theoretical values obtained from the mass and the physical dimensions of discassembly.

THEORY

A disc with mass m and radius R, rolls from rest at top position and takes time t(s), to reach

bottom position.

Let the linear velocity of the disc centre at the bottom position = v (m/s)

Then, the angular velocity of the disc at this position = ω (rad/sec) = v/r (rad/sec)

Average linear velocity = ½ v (m/s) = L/t (m/s)

Where L is the linear distance travelled

From conservation of energy,

Potential energy (at highest position) = Kinetic energy (at lowest position)

Therefore, moment of inertia of disc,


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Where

m: Mass of disc assembly

r: Radius of spindle

Figure No. 17 Spindle

Volume of disc, VD = π R 2l1

Volume of the spindle, VS = π r 2

(l2+ l3)

Theoretically value of „I‟ can be calculated from the mass and physical di mensions of disc

assembly; determine the volume of disc VD and the volume of the spindle VS, which may be

considered as a single cylinder.

Mass of the disc MD:

Mass of the spindle MS:


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Theoretical moment of inertia of disc ID,

Theoretical moment of inertia of disc Is,

Thus, theoretical total moment of inertia of the disc assembly,

APPARATUS

Figure No. 18 Rolling Disc On An Inclined Plane

PROCEDURE

1. Refer to the technical data for physical dimensions and weights of discs.

2. Place the inclined plane apparatus on a level surface and ensure that the top surfaces of the

two rails are at the same level. Wipe off any grease and dirt, which may be on the tops of rails.

3. Set one end of the two flanking rails of apparatus at a level above that of the other end. Set a

distance of L(m) along the length of the plane (ex:!m) and at height h=100mm between theextremities of the distance traversed by the centre of the disc.

4.Allow the spindle of the small disc assembly to rest on the two flanking rails and release it so

that it starts rolling unaided down the incline, ensuring that the disc not rub against the rails

during its motion. Note time t(sec) taken for the disc to traverse the distance L(m).


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5. Carry out the procedures three times to get average time taken.

6. Repeat procedure for the other disc.


Large Disc Small Disc

Diameter of disc, DD (mm)

Diameter of Spindle, DS

Thickness of disc, l1 (mm)

Length of spindle, l2 , l3 (mm)

Mass of disc, m (kg) 635g 375g

Time, t (sec) Large Disc Small Disc

t1

t2

t3

Average t


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EXPERIMENT # 8 Friction on an Flat Plane

OBJECTIVE

The objective of this experiment is to determine the coefficient of friction between various

materials and a steel plane.

THEORY

Friction is the resistive force that impedes the motion of a body when one tries to slide the object

along a surface. The friction force acts parallel to the surfaces in contact, opposes the relative

velocity of the body with respect to the surface, and its magnitude depends on the nature of the

particular materials that are rubbing together, but not on other variables, such as the area of

contact. This will be varied experimentally, and is true only in the macroscopic sense, since on

the molecular level things are much more complicated. For the case where the surfaces are in

motion relative to each other, the force is called the force of kinetic friction, and is found to be

proportional to the normal force acting at the region of contact, and always in opposition to the

velocity of the body relative to the surface of contact;

Thus the magnitude of the friction force can be written as

where the constant of proportionality, μk is the coefficient of kinetic friction.

If the two bodies in contact have no relative velocity, an even larger static frictional force must

be overcome in order to initiate slipping. This is of the same form

Only now Fe is the externally applied force that is attempting to cause to bodies to slip. This

static friction only acts to cancel out the external forces to prevent relative motion, and has a

maximum magnitude

where μs is called the coefficient of static friction. As indicated above, for most surfaces we find

that


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We can investigate kinetic friction by observing the motion of a block along a level surface

under the influence of an applied force. The block has a mass m 0, and extra masses m can be

added to it. A second mass M, hanging at the end of a string passing over a pulley, applies a

constant force to the block with its added masses, causing the system to move. As the mass M

falls, the block slides toward the right, and its motion is retarded by the friction force pointing

toward the left. If the mass M is chosen so that its weight just balances the friction force, then the

masses move at constant speed. Under this condition, the equations describing the motion of the

masses are

When T, the string tension, is eliminated from these equations we get

Figure No. 19 Frictional Force Between Two Surfaces

APPARATUS

Figure No. 20 Friction On A Flat Plane Apparatus


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PROCEDURE

1. Clamp the plane in the 0o position and use a spirit level to ensure the whole apparatus is level.

Place the sample tray on the horizontal steel channel at the end remote from the pulley.

2. Attach the towing cord and arrange it over the pulley with the load hanger suspended.

3. Add load to the hanger until the tray will continue to slide at roughly constant velocity after being

given a slight push to start it moving. Record this load in table.

4. You may find that you need to lightly tap the bench which the unit is on or the apparatus itself to

induce movement in the tray.

5.

Also ensure that the hanger is not swaying before loading.


Mass of hanger = ____________

Sr.

#

Mass of

Sample

Tray

(Kg)

Slide

Load (g)

Normal Force R N

(g) (mass of

hanger + Slide

Load)

Load on

hanger (g)

Sliding Force F(g)

(Hanger +load on

hanger)

Coefficient of

friction

μ= F/R N


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EXPERIMENT # 9 Friction on an Inclined Plane

OBJECTIVE

The object of this experiment is first to finder the angle of friction of various materials on a steel

plane. The second object is to verify that the force required parallel to an inclined plane to move

a body up the plane corresponds to the friction coefficient (or angle) already found.

THEORY

Suppose motion of a block along an inclined surface under the influence of an applied force. The

block has a mass m0. A second mass M, hanging at the end of a string passing over a pulley,

applies a constant force to the block, causing the system to move. As the mass M falls, the block

slides upward, and its motion is retarded by the friction force.

Figure No. 21 Frictional On An Inclined Plane

To study static friction, we can use an inclined plane. As the angle of inclination is increased

from zero, the component of the block's weight pointing down the plane increases. Because of

the variable nature of static friction, the magnitude of the friction force keeps increasing as the

ramp is raised. At a certain critical angle, however, the friction force reaches its maximum value,

and any further increase in the angle will cause the block to begin sliding down the ramp. At that

critical angle, the forces on the block are described by

from which we find

Thus, by measuring the angle of inclination at which the block just begins to slide, we can determine the

coefficient of static friction.


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PROCEDURE

1.

Clamp the plane in the 10o inclination.

2. Place the sample tray on the horizontal steel channel at the end remote from the pulley.

3.

Attach the towing cord and arrange it over the pulley with the load hanger suspended.

4. Add load to the hanger until the tray will continue to slide at roughly constant velocity after being

given a slight push to start sliding slowly up the plane. Record this load in table.

5. You may find that you need to lightly tap the bench which the unit is on or the apparatus itself to

induce movement in the tray.

6. Repeat the procedure with increased inclination.

7. Also ensure that the hanger is not swaying before loading.


Mass of hanger = ____________

Mass of

Sample

Tray

(Kg)

Angle of

inclination

θ (degree)

Mass of

block +

added

mass W

(g)

Towing

Force

(hanger +

Weight on

hanger) P

(g)

Normal

Force

WCosθ

Sliding

Force

P-WSinθ

Friction

Coefficient

μ = P-Wsinθ

WCosθ

Friction

angle

tan-1

µ


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EXPERIMENT # 10 Winch

OBJECTIVE

To find the Mechanical Advantage, Velocity Ratio and Efficiency of Winch.

THEORY

Winches are lifting, hauling or holding devices in which a tensioned rope is wound round a

rotating drum. They are extensively used for transporting people or goods, and they can be found

especially in mines and in marine applications. Winches are the fundamental elements, for

example, in crane and mooring systems, for activating cable cars, lifts and as a matter of fact,

whenever a dynamic pull is required from a flexible rope. Throughout history winches have been

used and probably the earliest illustration of a directly coupled winch is the mechanism used at a

well-head for lifting water containers.

An indirect coupling would be to use a clutch or gear and the intermediate of both components.

Most systems are gear coupled when the power source is not capable of producing adequate

torque, but when it can be used, the direct coupling system is mechanically better. It eliminates

gearing, reduces the number of bearings and simplifies the overall design.


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APPARATUS

Figure No. 22 Winch

PROCEDURE

1.

Firstly stabilize the single purchase crab machine and wrap the cord around the load drum and the effort wheel.

2. Put some weight on the load drum and add some effort to the effort wheel via hanger.

3. Stop adding effort until both the load and effort got stabilized.

4. Write down the reading in the observation table.

5. After this apply the above procedure, four to five times with gradually increasing the load

as well as effort to the load drum and effort wheel respectively.

6. Write down all the readings in the given observation table.

7. Measure the Diameter of load drum and effort wheel.

8. Calculate M.A, V.R and Efficiency of machine.


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T1 = Number of teeth on the pinion

T2 = Number of teeth on the spur wheel

D = Diameter of the effort wheel

d = Diameter of the load drum

V.R = Distance moved by effort / Distance moved by load

= πD / ( πd / (T1/T2) )

M.A = Load / Effort

Efficiency = M.A / V.R

Sr. # Load W

(N)

Effort P (N) Mechanical

Advantage

Velocity Ratio Efficiency


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EXPERIMENT # 10 Worm and Wheel

OBJECTIVE

To find the mechanical advantage, velocity ratio, and efficiency of worm and worm wheel and

plot a graph of1. Efficiency v/s load and;

2. Effort v/s load.

THEORY

A worm wheel is a simple lifting machine. The basic motion of the Lifting Machines is the rotary

motion. This is usually achieved by the use of pulleys and belts. However, in those machines

where a positive drive (i.e. no slip drive) is essential and no slip between belt and pulleys can be

accepted, a toothed belt and pulley is used. A gear is a wheel with accurately machined teeth

round its edge. One type of gear is the WORM and the WORM WHEEL.

The worm and worm wheel arrangement is widely used for performing mechanical jobs. As in

screw jack this arrangement also fundamentally provides some mechanical advantage & this is

used to lift the loads. The concept that rolling friction is less than sliding friction is used in this

experiment. At the point of release the string is in a state of pure rolling with respect to the

drums.

A worm gear, or worm wheel, is a type of gear that engages with a worm to greatly reduce

rotational speed or to allow higher torque to be transmitted. The image below shows a section of

a gear box with bronze worm gear being driven by a worm. A worm gear is an example of a

screw.

Figure No. 23 Worm And Wheel Arrangement


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The arrangement of gears seen above is called a worm and worm wheel. The worm, which in this

example is brown in colour, only has one tooth but it is like a screw thread. The worm wheel,

coloured yellow, is like a normal gear wheel or spur gear. The worm always drives the worm

wheel round, it is never the opposite way round as the system tends to lock and jam.

Figure No. 24 Worm And Wheel Description

APPARATUS

The apparatus consists of a toothed wheel fixed with a drum on the wheel meshes with the

toothed wheel. The worm is fixed on a metallic spindle. The spindle carries a pulley from which

hangs for application of effort. Another string also passes on the drum for carrying the weight to

be lifted.

Figure No. 25 Worm And Wheel Apparatus


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This lifting machine consists of the following parts:

1. Worm: It is a gear with just one tooth. The tooth is in the form of a screw thread.

2. Worm Wheel: This is like a normal gear wheel.

3. Load Drum: This is mounted on the worm wheel and rotates when load to be lifted isapplied to it.

4.

Metallic Spindle: This is attached to the worm. It is attached to a pulley where the effort

is applied.

PROCEDURE

1. Measure the circumference of drum and pulley with the help of outside caliper.2. Wrap the string round the pulley of the worm for effort and also wrap string round the

drum to carry the load.

3. Suspend some load to the string passing round the drum and go on adding weights.

4. Add weight to the effort hanger until it move down.5. Repeat the experiment with different loads.


V.R. = D T/2d

M.A.= W/P

η = M.A / V.R = (W/P) / (D*T/2*d)

Diameter of Drum fixed on wheel = d = _____________

Diameter of Pulley attached to worm =D = _____________

Number of teeth on worm wheel = ____________

Sr. # Load

W(N)

Effort

P(N)

Mechanical

Advantage

Velocity Ratio Efficiency


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EXPERIMENT # 11 Toggle Joint

OBJECTIVE

To determine the Mechanical Advantage of a Toggle Joint.

THEORY

Toggle mechanism is combination of solid, usually metallic links (bars), connected by pin

(hinge) joints that are so arranged that a small force applied at one point can create a much larger

force at another point.

The basic action of a toggle mechanism is shown in illustration a. When α = 90° the forces P and

Q are independent of each other. Again when α = 0° the forces are isolated, force Q being

sustained entirely by the frame, and force P serving only to hold the link in position. At α = 45°

from the symmetry | P | = |Q|, the mechanism serves to transfer the direction of forces to achieveequilibrium.

Because the simple configuration of illustration a requires low-friction sliders, it is impractical.

A more useful structure replaces the vertical slider with a second link pinned to the frame

(illustration b), in which case input P sets up forces in both links. A further modification

(illustration c) replaces the other slider with a link.

Figure No. 26 Toggle mechanism. (a) Simple structure. (b) Traditional configuration. (c)Typical application.

http://www.answers.com/topic/toggle-mechanismhttp://www.answers.com/topic/impracticalhttp://www.answers.com/topic/sliderhttp://www.answers.com/topic/sliderhttp://www.answers.com/topic/impracticalhttp://www.answers.com/topic/toggle-mechanism


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APPARATUS

This apparatus is designed to evaluate forces within a toggle mechanism. Load is applied to the

two pairs of links by a hanger suspended from their connecting pivot. One end of the links is

pivoted to a base, and the other end is able to move sideways on low friction ball bearing wheels.The moving links are restrained by a horizontal spring balance, which measures the horizontal

reaction directly. The angle of the toggle can be varied.

Figure No. 27 Toggle Apparatus Description

Figure No. 28 Toggle Apparatus


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PROCEDURE

1. By means of a measuring tape, measure the vertical height (h) of the apparatus and the

horizontal length (D) with some loads attached (as a reference). 2. Now add a known weight to the hanger. This is the effort P. 3. Note down the reading from the spring balance. This is the load W. each division on the

spring balance equals to 0.5 kg. 4. The actual value of Mechanical Advantage is calculated by dividing load (W) by effort

(P) whereas the theoretical value is given by D/4h. 5. The experiment is repeated for different values of P.


D = ____________

P = ____________

H = ____________

W = ____________

S# Load

W

(kg)

Effort

P

(kg)

Height

h

(mm)

Length

D

(mm)

Mechanical

Advantage

%age

errror

Experimental

(F/P)

Theoratical

(D/4h)

1

2

3


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EXPERIMENT # 12 Slider Crank Mechanism

OBJECTIVE

a. To analyze the variation of displacement of piston in relationship with crank angle & to

calculate velocity and acceleration of piston b. To draw graph between crank angle & piston displacement.


THEORY

In order to simplify the study of mechanisms, it is necessary to understand some definitions and

the basic knowledge as follows:

MECHANISM is defined as combinations of rigid bodies formed and connected to each other

and transmit relative motion to each other such as crankshaft connecting rod and piston of an

engine.

MACHINE is defined as a combination of a mechanism or more to transmit force and motion

from the source of power to another resisting element, for example: an operation of an internal

combustion engine.

The motion of a mechanism, in which each point of the element moves in parallel planes, is

called "PLANE MOTION".

If each point moves in straight line and parallel to each other, the motion is known as

"TRANSLATION".

If each point moves with a constant distance from its axis, this motion is known as"ROTATION".

The movement of a point of a mechanism may also be in translation, rotation or both.

However, there are some other types of movements which the position of moving points

may not be in the same plane for example: THREAD MOTION, HELICAL MOTION

etc.

When an element of a mechanism moves through all the possible positions and returns to its

original position, it is said to have completed a cycle of motion and the amount of time required

for this completed a cycle is called "PERIOD".

Crankshaft&

connecting rod

The main driving shaft of an engine that receives reciprocating motion from the piston and

converts it to rotary motion, is called crank shaft. Together, the crankshaft and the connecting

rods transform the pistons' reciprocating motion into rotary motion.


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Mathematical Relation for Piston Motion

I

= rod length (distance between piston pin and crank pin)r = crank r adius (distance between crank pin and crank center, i.e. half stroke)

A = crank angle (from cylinder bore centerline at TDC)

x = piston pin position (upward from crank center along cylinder bore centerline)

v

= piston pin velocity (upward from crank center along cylinder bore centerline)a = piston pin acceleration (upward from crank center along cylinder bore centerline)ω = crank angular velocity in rad/s

Figure No. 29 Crank And Connecting Rod

Angular velocity

The crankshaft angular velocity is r elated to the engine revolutions per minute (RPM):

ω = 2 π N / 60

Triangle relation

As shown in the diagram, the cr ank pin, crank center and piston pin form triangle NOP.

By the cosine law it is seen that:


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Equations with respect to angular position (Angle Domain)

The equations that follow describe the reciprocating motion of the piston with respect tocrank angle. Exam ple graphs of these equations are shown below.

Position

Position with respect to crank angle (by rearranging the triangle relation):

VelocityVelocity with respect to crank angle (take first derivative, using the chain rule):

AccelerationAcceleration with respect to crank angle (take second derivative, using the chain rule and thequotient r ule):


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Equations with respect to time (Time Domain)

Angular velocity derivatives

If angular velocity is constant, then A=ωt

And the f ollowing r elations apply:

dA/dt = ω

d2A / dt

2 = 0

Converting from Angle Domain to Time Domain

The equations that follow describe the reciprocating motion of the piston with respect to

time. If time domain is required instead of angle domain, first replace A with cot in the

equations, and then scale for angular velocity as follows:

Position

Position with respect to time is simply:

x

Velocity

Velocity with respect to time (using the chain rule):

Acceleration

Acceleration with respect to time (using the chain rule and product rule, and the angularvelocity derivatives):


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APPARATUS

Figure No. 30 Crank and Connecting Rod Mechanism

This bench top unit demonstrates the conversion of smooth rotary motion into reciprocating

motion. A millimeter scale is fitted for the outlet stroke. Crank radius can both be adjusted &have three positions.

Technical Specifications are given below

Crank radius (can be adjusted at three points)

R1 =25mm, R2 =37.5mm, R3 =50mm

Connecting rod length

L =140mm

PROCEDURE

1. Bring the wheel and the slider at reference points and mark these points.

2. For a given angle of rotation (fixed), note down the displacement of slider.

3. Plot a graph between the slider displacement and crank angle.

4. Assume that crank is rotating with a unif orm speed.

5.

Replace the crank angle with equal interval of time& draw slider displacement versustime, find slope at each reading. Then draw velocity-time graph.

6. From Velocity -Time graph, take slope of velocity curve& draw acceleration graph.

7. Compare the results of each gr aph and draw them on a single graph with crank

angle along x-axis.


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Sr.

#Crank

Rotation

(degree)

Time

(sec)

Slider Position

(mm)Slider

Displacement

(mm)

Slider

velocity

(m/sec)

Slider

acceleration

(m/sec2)

1

2

3

4

5

6

7

8

9

10

11

12

13


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EXPERIMENT # 13 Whitworth Quick Return Mechanism

OBJECTIVE

a. To analyze the variation of displacement of oscillating rocker in relationship with crank

rotation & draw graph between rocker oscillation and crank rotation. b. To calculate velocity and acceleration of rocker


THEORY

The Whitworth quick return mechanism converts rotary motion into reciprocating motion, but

unlike the crank and slider, the forward reciprocating motion is at a different rate than the backward stroke. At the bottom of the drive arm, the peg only has to move through a few degrees

to sweep the arm from left to right, but it takes the remainder of the revolution to bring the arm

back. This mechanism is most commonly seen as the drive for a shaping machine.

APPARATUS

Whitworth's quick return is used to generate uneven reciprocating motion with slow feed and

quick return. This table model clearly demonstrates the transmission behaviour of such a layout.

The input angle is set by turning the crank. The output stroke is read on a ruler on the slider. The

transmission components are manufactured in aluminium. All axles are equipped with ball bearings. Due to its low weight, the unit is easy to carry using the two handles.

Figure No. 31 Witworth Quick Return Apparatus

PROCEDURE:

1. Bring the crank and the rocker at reference points and mark these points.

2. For a given angle of rotation (fixed), note down the displacement of Rocker.


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3. Plot a graph between the Rocker displacement and Crank rotation.

4. Assume that Crank is rotating with a unif orm speed.

5. Replace the Crank angle with equal interval of time& draw Rocker displacement versus

time, find slope at each reading. Then draw velocity-time graph.

6. From Velocity -Time graph, take slope of velocity curve& draw acceleration graph.

7. Compare the results of each gr aph and draw them on a single graph with time along x-

axis.


Sr.

#Crank

Rotation

(degree)

Time

(sec)

Rocker Position

(mm) Rocker

Displacement

(mm)

Rocker

velocity(m/sec)

Rocker

acceleration(m/sec

2)

1

2

3

4

5

6

7

8

9

10

11

12

13

Documents

ME-222 Mechanics Manufacturing Lab-I