applied mechanics group 5

8/2/2019 Applied Mechanics Group 5

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Objective:To study the equilibrium of a particle under the action of forces in a plane.

Apparatus:Hangers, slotted weights, thread, circular table, pulley, rings etc.

Theory:

FORCE AND SYSTEM OF FORCESNewton defined as the external cause which either changes the state of a body (static or dynamic) or tends tochange it. It a body is acted upon by a number of forces, then it constitutes a system of forces. This systemcan be represented by a resultant force, the effect of which is equivalent to the combined effect of all theforces of the system. Forces are said to be coplanar, if they are acting in the same plane. Forces are said to be concurrent, if they are acting at the same point or are passing through it.

Forces are said to be collinear, if they are acting in one direction.

Conditions for a body to be in equilibrium:-

A particle is said to be in equilibrium, when the vector sum of all the forces acting on it is zero. but this

condition does not speak about the rotation of the body. So the second condition of equilibrium is that thesum of the torques about any point must be zero.Mathematically, the first condition of equilibrium is put as,

Fx = 0, Fy = 0 and Fz = 0and, the second condition of equilibrium as

M = 0i.e. the algebraic sum of the moments due to all external forces acting on the body, with respect to anyspecific point, must be zero.The body may be in a state of equilibrium or dynamic equilibrium.If all the forces are assumed to be acting in a single plane, say xy plane, the equations of equilibrium arereduced to:

Fx = 0 and Fy = 0M = 0In the current experiment all the forces are coplanar and concurrent.The ring remains at the center when all the four forces balance each other.


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Static Equilibrium: It describes the condition in which a body is at rest with respect to a frame ofreference.

In static equilibrium:-(i) The vector polygon of all forces acting on body is closed.

(ii) The sum of components of force along any axis is zero.Polygon Method: When adding vectors by polygon method, move the vectors parallel to themselvesto form a polygon in any order, since vector addition is commutative. In case of static equilibrium thevectors very likely add to zero.

Equilibrant: The equilibrant of any number of forces is the single force that is required to produceequilibrium.

Lami's Law: If a particle is in equilibrium under the action of three forces, each force must bear thesame proportionality with the sine of angle between the other two forces.It states that if three coplanar forces are acting on a same point and keep it stationary, then it alwaysobeys the relation:

Sin

F

Sin

F

Sin

F321 ==

A+B+C

A C

B


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Analysis of body at equilibrium:

When all forces which act upon an object are balanced, the object is said to be in a state of

equilibrium. This does not mean that the forces are equal. The net force is 0 and the acceleration is 0m/s2.An object at equilibrium is either:

(i) At rest and staying at rest.Or

(ii) In motion and continuing in motion with the same speed and direction.

If an object is at rest and in a state of equilibrium, then we would say that it is at static equilibrium.

If a particle is in equilibrium under the action of n coplanar forces, the forces must be represented inmagnitude, direction and sense by n sides of polygon taken in order in the same sense.

A particle cannot be in equilibrium when a single force is applied on it. It would be in equilibriumunder the action of two or more forces if the vectorial summation of force is zero.

F4

F3 F5

F2 F6

F1


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= 0F

In particular if a particle is subjected to only two forces, the forces must be equal and opposite, i.e.,equal in magnitude, in the same line of action but opposite in sense in order to keep it in equilibrium.If it is subjected to only three forces, the three forces must be coplanar forces.

This fact can be appreciated by recognizing that the resultant of any two must be equal and oppositeto the third force for equilibrium, and this cannot happen unless the three force lie in one and the sameplane. If a particle is in equilibrium under the action of four or more forces, the forces may be spatial,i.e., not necessarily confined to act along the same line or in the same plane.

When a particle is in equilibrium under the action of three forces 321 &, FFF as shown in figure, thecondition of equilibrium is, i.e., 0321 =++ FFF may alternatively be expressed as Lami's theorem

or Triangle Law of Equilibrium.

Triangle Law of Equilibrium: If a particle is in equilibrium under the action of three forces, theforces must be represented in magnitude, direction and sense by the sides of a triangle, taken in order,in the same sense.

When a particle is in equilibrium, under the action of more that three coplanar forces, the condition ofequilibrium, i.e.

= 0F , 0...4321 =++++ FFFFmay alternatively be stated in forms of the polygon law of equilibrium.

If a particle is in equilibrium under the action of n coplanar forces, the forces must be represented inmagnitude, direction and sense by the n sides of a polygon taken in order in the same sense.

Procedure:(i) Fix the threads at one end to ring and at other end to the hangers.(ii) Pass the threads over the pulleys.(iii) Fix the position of pulleys by taking some suitable angles.


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(iv) Hang the weight at all the hangers.(v) Lift the ring up to some height and leave it, check that the ring must be at the centre.(vi) If the ring is not at centre, then again adjust the weight to bring the ring at centre.

(vii) For the next set of reading set the angles and repeat the above (iv, v, vi) steps.(viii) Draw the vector diagram with suitable scale moving in one direction for all the set ofreading.(ix) Find out the error in magnitude and direction of last side of polygon using the formulae:% error in Magnitude = (| F4-F4 | / F4) x 100% error in Direction = (| 4-4 | / 4) x 100

Observations

S.N

o.

Magnitude of forces

(gram weight)

Angular position of forces

(degrees)

ActualMagnitude ofClosingSideF4(gmwt.)

Observed

Magnitudeof ClosingSideF4(gm.wt.)

ActualPosition

4

ObservedPosition

4

Percentage

Error

F1 F2 F3 F4 1 2 3 4 Mag Dir

1 105 110 110 110 90 90 90 90 109 110 93 90 .91 3.33

2 105 115 175 130 90 90 120 60 134 130 122 120 3.46 8.33

3 105 130 275 80 30 150 120 60 87 80 128 120 8.75 6.67

4 145 110 130 180 60 90 90 120 201 180 87 90 11.67 3.33

5 105 85 255 165 90 90 150 30 177 165 151 150 7.27 0.67

Average error in magnitude = 6.412%Average error in direction = 4.466%

Result:Within the limits of experimental error the observed values and the actual values are found to be inapproximation with each other and hence law of polygon of forces for equilibrium is verified.

% Error in Magnitude = 6.412%% Error in Angle = 4.466%

Precautions:

(i) Weights/Pans should not touch the table.(ii) The weight of hangers should be added to the weights put into pan to find thetotal force in each case.


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(iii) Before taking the reading of forces see that all the hangers are at rest.(iv) See that pulleys must be free from friction.

Sources of Error:

(i) The slotted weights may not be calibrated properly.(ii) There may be friction in pulleys.(iii) Error in measurements of angles of thread.(iv) The threads may not be perfectly radial.(v) The threads are assumed to be light and in extensible.(vi) The muscular force used to put the weights might create an error.

Reference:

www.wikipedia.org

Write-ups
http://www.wikipedia.org/http://www.wikipedia.org/


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TO VERIFY THE LAW OF MOMENT BY USING BELL CRANK

LEVER


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MOMENTS

A moment is a turning force

Law of Moments or Law of Levers

To be totally in balance ( in equilibrium )

both sides of the lever will be still.

Then the clock wise movement will equal the anticlockwise movement


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OBJECTIVE:

To verify the law of moment by using bell crank lever

APPARATUS:

Bell crank lever, slotted weight, spring balance etc.

THEORY:

According to law of moments the moment of a force about an axis is equal to the sum of itscomponents about same axis.

Here in this experiment we have to check the moment of the force about the various pointon the lever and that moment must be equal to the spring force multiplied by the fixed distance d. Thedistance d in this experiment is fixed and is equal to 7 inch.

So we have to verify

MOMENT M=W*D=S*d

Where W =force applied on leverD =varying distance on leverS =spring forced =fixed distance (7*2.54=17.78 cm)

PROCEDURE:

1. Engage the chain of spring balance with the lever.2. Hang the weight on the end point marked on the lever.3. Check the pointer to match with the mark made on the lever.4. If point is not matching then adjust the weight to get the correct reading.

5. Note down the spring reading.6. change the position of weight to be hanged on the lever and repeat the above steps.7. Take at least 6 readings.

OBSERVATIONS:

WEIGHT OF HANGER = 258.30 grams


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WEIGHT OF THE SUPPORT = 35 grams

OBSERVATION TABLE:

S NO. WEIGHTW g

DISTANCED cm

MOMENTM=W*D

SPRING FORCES

CALCULATEDVALUE OF S

S1= W*Dd

% ERROR

S1-S *100S1

1 379 30.48 11.551 0.65 .6496 .061

2 419 27.94 11.706 0.65 .6564 1.275

3 444 25.4 11.277 0.65 .6342 2.49

4 494 22.86 11.292 0.65 .6350 2.36

5 549 20.32 11.155 0.65 .6274 3.60

6 644 17.78 11.450 0.65 .6440 .93

CALCULATIONS:

Mean percentage error = .061+1.275+2.49+2.36+3.60+.936

= 1.775 %

RESULT:

As the observed value of force is nearby equal to the calculated value within theexperimental error of 1.775 % hence the law of moment has been verified.

DISCUSSION:

Main aspect: HANDLING OF MOMEMTS DUE TO OTHER PARTS


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ANDACCURACY V/S USE OF HEAVY LOADS

The important point to be noted in this experiment is that we take the readings of spring balanceonly when the pointer becomes parallel to the vertical part of hook on which the spring is attached ,even though the for the weight and the spring is balanced at any moment .

This is because the when the perpendicular hook becomes vertical the moment due to itsweight is 0, as the perpendicular distance of the weight from the central axis is zero, this is not in caseof any other angle.

Now, as far as the moment of horizontal part of hook(where the weight is suspended) isconcerned it is automatically balanced by the special hook given on its opposite side.

In this experiment also the much heavier weights should not be used so that spring does notcross its elastic limit. But still using heavier weights to a limit should be used in order to have moreaccurate results, as the smallerweights of any other part of the apparatus, if involved in the momentscalculation will be negligible, in presence of heavy weights.


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To find the Mechanical advantage, velocity ratio and efficiency of a simple screw jack


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A jackscrew mechanism uses a long threaded rod used to

position a matching nut at any position along its length. Unlike

hydraulic actuators that require continual pressure to remain in

a locked position, forces acting on the nut along the length ofthe rod do not appreciably affect the rotation of the rod so the

mechanism self-locks.

Jackscrews commonly use an Acme thread along the threaded

rod. This pattern is very strong and can resist the large loads
http://en.wikipedia.org/wiki/Nut_(hardware)http://en.wikipedia.org/wiki/Hydraulichttp://en.wikipedia.org/wiki/Acme_thread_formhttp://en.wikipedia.org/w/index.php?title=Threaded_rod&action=edithttp://en.wikipedia.org/w/index.php?title=Threaded_rod&action=edithttp://en.wikipedia.org/wiki/Nut_(hardware)http://en.wikipedia.org/wiki/Hydraulichttp://en.wikipedia.org/wiki/Acme_thread_formhttp://en.wikipedia.org/w/index.php?title=Threaded_rod&action=edithttp://en.wikipedia.org/w/index.php?title=Threaded_rod&action=edit


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imposed on most jackscrews while not being dramatically

weakened by wear over many rotations. Most jackscrews are

lubricated with grease. More-sophisticated screw mechanismsmay use a recirculating-ball nut to minimize friction and prolong

the life of the screw threads, but such jackscrews are usually

not considered self-locking.

As shown in the photo, jackscrews are commonly used in car-

jacks.

The jackscrew figured prominently in the classic Robinson

Crusoe. It was also featured in a recent show [History channel]

as thesaving tool of the Pilgrims' voyage; the main cross beam

(a key to the entire structural integrity) of their small ship,

cracked during a severe storm. A farmer's jackscrew secured

the damage until landfall.
http://en.wikipedia.org/wiki/Wearhttp://en.wikipedia.org/wiki/Grease_(lubricant)http://en.wikipedia.org/wiki/Ball_screwhttp://en.wikipedia.org/wiki/Wearhttp://en.wikipedia.org/wiki/Grease_(lubricant)http://en.wikipedia.org/wiki/Ball_screw


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OBJECTIVE

To find the Mechanical advantage, velocity ratio and efficiency of a simple screw

jack and plot the graph between

Efficiency v/s Load

Effort v/s Load

APPARATUS

Screw jack apparatus,Slotted weights,String,Outside caliper

THEORY

A screw jackis a mechanical lifting device that is used to lift heavy loads with theapplication of a small effort.

Screw Jacks have long been accepted as reliable tools for a variety of jobs. Their

sturdy, simple design makes them ideal for supporting machinery, construction,

structural moving and general maintenance.

PARTS: A screw jack consists of the following parts :

1. Screw threads : It consists of threads that may be square or V in shape.However square threads are more efficient than V threads and are used for

power transmission.2. Head : The screw has a head on its upper end on which the load W rests.


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3. Nut : In the case of a screw jack, the threads of the screw will slide aroundupon the fixed threads of the nut fixed in the frame and which generally formsa part of the body of the screw jack.

PITCH : The axial distance between the corresponding points on two consecutivethreads is known as pitch of the screw. Let this pitch be p.

Weight on the table = W g

Let D be the diameter of the flanged table on which the load W isto be placed and lifted.

Number of starts of the screw = n = 2

P= Total effortin the two hanger including the weight of the hangers.= P1 + P2

Let the table turn through one revolution

LEAD : The lead is the axial distance through which the screw advances inone turn.Load risen in one revolution = Lead of the screw jack = l = n * pitch

l = 2 * p

Effort moved in one revolution = D

V.R. = velocity ratio

V.R. = Distance moved by the effort = DDistance moved by the load l


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V.R. = D2 * p

M.A. = Mechanical Advantage

M.A. = W/P

Percent efficiency = %

% = M.A. * 100V.R.

PROCEDURE


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1. Wrap the string round the circumference of the flanged table and passit over one

pulley. Similarly wrap another string over the flanged table and take itover the second pulley. The free ends of both the strings are tied to twohangers where the weights are to be hanged.

2. Measure the pitch of the thread with the help of vernier caliper.3. Place the load W on the screw head and some weight on the hangers

so that the load W is just lifted, the effort P is equal to the sum of theweights hanged in two hangers.

4. Increase the loads and find the corresponding efforts applied for theconsecutive readings. Take at least six to seven readings.

5. Calculate mechanical advantage, velocity ratio and efficiency in eachcase.

6. Plot the graph between efficiency v/s load, effort v/s load.

OBSERVATION

Pitch = 6 mmWeight of screw jack system = 5.875 kgDiameter = 127 mmCircumference of the table = 2 * * radius = D = 398.976 mmLead of the screw = l = 2 * p = 2 * 6 = 12 mm


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V.R. = velocity ratio = 33.23

S.No. LOAD

(W)

(in kg)

EFFORT

(P)

(P1+P2)

(in kg)

MECHANICAL

ADVANTAGE=(w/p

)

%=(M.A/V.R)*100

1. 5.875 .503 11.67 35.112. 6.833 .563 12.13 36.52

3. 7.322 .583 12.55 37.794. 7.835 .613 12.78 38.465. 8.271 .633 13.06 39.316. 8.808 .658 13.38 40.26

Mean 12.59537.90


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PRECAUTIONS

1. Use both the pulleys to find the values of effort P to avoid the side thrust.2. The load and effort should move slowly.3. Add weights in hangers gently.4. Lubricate the screw to decrease friction.5. The string should not overlap.6. There should be no knot in the string.7. See that both the pans should move downwards.

RESULT

The mechanical advantage of the screw jack system is : 12.595The velocity ratio of the screw jack system is : 33.23The percent efficiency of the screw jack system is : 37.90%

DISCUSSIONS

If we increase the diameter D of the flanged table we would require less effort toraise the same load by the same distance .

It is because torque required to raise the load will be constant. ButTORQUE = FORCE * DISTANCE

So if we increase the diameter(i.e. distance) so we have to apply less force to raise thesame load by the same distance. But we cant increase the diameter to greater extent

because we cant neglect the weight of the flanged table and that will give inaccurateresults.


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6.1 Objective:

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

wheel and plot a graph of:1. Efficiency Vs Load2. Effort Vs Load

6.2 Apparatus:

Worm and worm wheel apparatus, string, meter rod, outside caliper, pan etc.

6.3 Theory:


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The worm is shown with the worm above the worm wheel. The gear set can also bearranged with the worm below the worm wheel. Other alignments are used lessfrequently.

A worm gear is used when a large speed reduction ratio is required between crossedaxis shafts which do not intersect. A basic helical gear can be used but the power,which can be transmitted, is low. A worm drive consists of a large diameter wormwheel with a worm screw meshing with teeth on the periphery of the worm wheel.The worm is similar to a screw and the worm wheel is similar to a section of a nut.As the worm is rotated the worm wheel is caused to rotate due to the screw likeaction of the worm. The size of the worm gear set is generally based on the centredistance between the worm and the worm wheel.

If the worm gears are machined basically as crossed helical gears the result is ahighly stress point contact gear. However normally the worm wheel is cut with aconcave as opposed to a straight width. This is called a single envelope worm gearset. If the worm is machined with a concave profile to effectively wrap around theworm wheel the gear set is called a double enveloping worm gear set and has thehighest power capacity for the size. Single enveloping gear sets require accuratealignment of the worm wheel to ensure full line tooth contact. Double envelopinggear sets require accurate alignment of both the worm and the worm wheel to obtainmaximum face contact.

Worm teeth ProfileThe sketch below shows the normal (not axial) worm tooth profile as indicated in BS721-2 for unit module (m = 1mm) other module teeth are in proportion e.g. 2mmmodule teeth are 2 times larger


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Initial sizing of worm gear... (Thermal)

Worm gears are often limited not by the strength of the teeth but by the heatgenerated by the low efficiency. It is necessary therefore to determine the heatgenerated by the gears = (Input power - Output power). The worm gearbox must havelubricant to remove the heat from the teeth in contact and sufficient area on theexternal surfaces to distribute the generated heat to the local environment. Thisrequires completing an approximate heat transfer calculation. If the heat lost to theenvironment is insufficient then the gears should be adjusted (more starts, largergears) or the box geometry should be adjusted, or the worm shaft could include a fanto induced forced air flow heat loss.Formulae

Friction Coefficient

Cast Iron and Phosphor Bronze... Table x 1, 15Cast Iron and Cast Iron... Table x 1, 33Quenched Steel and Aluminum Alloy. Table x 1, 33Steel and Steel. Table x 2


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Friction coefficients - For Case Hardened Steel Worm / PHs Bros WheelSliding

Speed

Friction

Coefficient

Sliding

Speed

Friction

Coefficientm/s m/s

0 0,145 1,5 0,0380,001 0,12 2 0,0330,01 0,11 5 0,0230,05 0,09 8 0,020,1 0,08 10 0,0180,2 0,07 15 0,0170,5 0,055 20 0,0161 0,044 30 0,016

The picture to the right is a typical set-up for a motor and worm gear system. As theworm revolves the worm wheel (spur gear) also revolves but the rotary motion istransmitted through a ninety-degree angle.The gear ratio of a worm gear is worked out through the following formula:

Number of teeth on worm wheelNumber of teeth on wormThe worm acts as a single toothed gear so the ratio is;

Number of teeth on worm wheel
http://images.google.com/imgres?imgurl=http://www.cardenal-newman.edu/staff/wormwheel.jpg&imgrefurl=http://www.cardenal-newman.edu/staff/cortina.htm&h=290&w=217&sz=8&hl=en&start=2&um=1&tbnid=LbweQh2wKE4FIM:&tbnh=115&tbnw=86&prev=/images%3Fq%3Dwormwheel%26svnum%3D10%26um%3D1%26hl%3Den%26rlz%3D1T4RNWN_enIN213IN218


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1A double-action worm gear set - so called because the worm has both rotational and

longitudinal degrees of freedom - has been invented primarily for use in a motorizedspacecraft hinge. This paper presents an analysis of the mechanical efficiency of thedouble-action worm gear set, together with preliminary test results. Results ofanalysis and test show that the double-action feature of the worm makes it possible tocontrol the forward and backward driving efficiency of the gear set. Compared to astandard worm gear set, the double-action worm gear set has the advantages of theability to absorb small applied displacements to the worm wheel shaft in an unlocked

position, the ability to be non-back drivable with a high degree of certainty whenlocked against an end stop and the ability to maintain a specific preload in a locked

position.

In working out the problems on levers, belts and pulleys, inclined planes and so forth,we have not taken account of friction or other sources of energy loss. In other words,we have supposed them to be perfect, when in fact they are not. To measure the

performance of a machine, we often find its efficiency, which is defined as

(2-4)WhereThe efficiency of a machine,

Win = the input work to a machine, andOut = the output work of a machine.Gear Ratios:It is important when working with gears to know what number of teeth the gearsshould have so that they can mesh properly in a gear train. The size of the teeth forconnecting gears must be match properly.The Mechanical Advantage of a machine is the ratio of Force being moved W to theEffort F

Mechanical Advantage = W /F

The Velocity Ratio of a machine is the ratio of the distance moved by the Effort andthe distance moved by the Force being overcome.Velocity Ratio = Distance moved by Effort/ distance moved by forceIn the ideal frictionless/weightless machineVelocity Ratio = Mechanical AdvantageThe efficiency of a simple machineEfficiency = Work done by the machine / Work supplied to machine


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The only parameter that can be determined from the machine dimensions is thevelocity ratio.

Machines generally follow the linear rule...F = a + b. WThe worm gear combines the inclined plane wrapped around a cylinder with the leverin the form of a gear. The linear pitch of the worm gear, the distance between teeth,must match the linear pitch of the spur gear. The teeth on the spur gear must beinclined at the pitch angle of the worm. Each turn of the worm advances the gear byone tooth, so the velocity ratio is N: 1. Worm gears can be made with multiplethreads, so that in one revolution of the worm the gear moves a space of n teeth,giving a velocity ratio of N: n. Worm gears are a good way to get large velocity

ratios, and also prevent the mechanism from driving in the reverse direction,something desired in steering mechanisms,

Mechanical Advantage:The ratio, load/effort is called mechanical advantage of a machine. The mechanicaladvantage should be greater than one. If, in a machine, the ratio is less than one, itwould be more accurate to call is mechanical disadvantage.

Velocity Ratio:The ratio

Is called the velocity ratio of a machine. The two distances are moved in the sameinterval of time, so they are proportional to the velocities of the effort and load.

Pd1 = Wd2 in a perfect machine.Or

Where d1 and are d2 are the distances moved by effort P and load W respectively.Hence, in a perfect machine the mechanical advantage is equal to velocity ratio.


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TopEfficiency of Machine:In all machines some work is always wasted friction. The result of it is that the workdone by the effort in a given time, called total work or work input (= p d1) isalways greater than the work done on the load (= W d2) called useful work or workoutput. The difference of the latter from the former = lost work (Pd1-Wd2).

The ratioIs called efficiency of machine. It is also defined as the ratio

In any actual machine the efficiency is always less than one but in a perfect or idealmachine in where there is no friction at all the efficiency is equal to unity.

Or Mechanical advantage = Efficiency velocity ratioI.e. M.A. =Velocity Ratio

The apparatus consists of toothed wheel fixed with a drum on it. The worm mesheswith the toothed wheel. The worm is fixed on a metallic spindle. The spindle carries a

pulley from which a string hangs for application of effort. Another string passes onthe drum for carrying the weight to be lifted.

D = diameter of the pulley to worm.
http://www.eformulae.com/engineering/machines.php#top%23tophttp://www.eformulae.com/engineering/machines.php#top%23top


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D = diameter of drum fixed on the wheel.T = number of teeth on the worm.

If one revolution is given to the pulley two teeth of the worm wheel moves of theworm threads are of single start.

Displacement of effort p = adDisplacement of load w = 2d

TU.R = dot = DT

2d 2d

M.A = W/p = M.A/U.R = 2wd x 100

PetDiameter of effort pulley=122mm

Diameter of load pulley=186mm

No. of teeth=95

Hook weight(load side)=30gm

Hook weight(effort side)=25gm

6.4 Procedure:

(1) Measure the circumference of the drum and the pulley with the help ofoutside

Calipers.

(2) Wrap the string around the pulley for effort and also wrap another stringaroundThe drum to carry the load.

(3) Suspend some load to the string and go on adding weights. In the panhanging

(4) From the pulley till the load w just starts moving upwards.

(5) Note down the weights in the effort arm.


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(6) Repeat the experiment with different loads.

6.5 OBSERVATION TABLE :

Snow. Loadsuspendedw (g )

Totaleffort p(g)

M.A.=w/p V.R.=dt/2d %=M.A.*100

1 80 60 1.33 62.31 2.132 120 65 1.85 62.31 2.973 170 70 2.43 62.31 3.904 200 75 2.67 62.31 4.295 275 85 3.235 62.31 5.19

6.6 Result:

The mechanical advantage, velocity ratio and efficiency of worm and worm wheelhave been found out and graph of:

1 Efficiency Vs Load2 Effort Vs Load

Has been plotted.Mechanical advantage of worm wheel =2.303Velocity ratio of worm wheel =62.31Efficiency of worm wheel=3.696%

6.7 Precaution:


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(1) Lubricate the bearing of the worm and teeth of the worm wheel to decreaseFriction.

(2) Weight should be gently put in the effort arm.(3) The string should not overlap on the pulley or on the drum.(4) The pan weighing the weights shouldnt touch the wall.(5) Load and effort should move slowly.(6) There should be no knot in the string.

6.8 Improvements on the experiment:

(1) The string sometimes overlaps on the pulley or on the drum.(2) The weights used in the experiment may not be exact due to mishandling by students.(3) The weights should have been put gently on the pan.

6.9 CONCLUSION:1. The worm and worm wheel set up is an apparatus used to lift a given load by using effort

having a value much less than the load.2. The worm wheel apparatus is used to calculate efficiency, mechanical advantage and velocity

ratio of a pulley system. It is a better method than screw jack apparatus as there is only oneeffort point. However, if used after a long interval or not lubricated properly, extra effort is

required due to internal stiffness.3. The mechanical advantage, i.e., the ratio of load to effort remains almost constant with loadapplied. The graph between effort and load is found to be a straight line. The graph betweenload and efficiency is a carve that becomes constant after a certain value of load.

6.10 References-

(1) www.wikipedia.org(2) www.cs.cmu.edu(3) www.eformulae.com/engineering/machines.php - 100k

(4) www.du.edu/~jcalvert/tech/machines/machines.(5) www.roymech.co.uk/Useful_Tables/Drive/Worm_Gears.html(6) www.technologystudent.com/gears1
http://www.wikipedia.org/http://www.du.edu/~jcalvert/tech/machines/machineshttp://www.du.edu/~jcalvert/tech/machines/machineshttp://www.du.edu/~jcalvert/tech/machines/machineshttp://www.du.edu/~jcalvert/tech/machines/machineshttp://www.du.edu/~jcalvert/tech/machines/machineshttp://www.roymech.co.uk/Useful_Tables/Drive/Worm_Gears.htmlhttp://www.wikipedia.org/http://www.du.edu/~jcalvert/tech/machines/machineshttp://www.roymech.co.uk/Useful_Tables/Drive/Worm_Gears.html


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Types - There are several types of axle

Transaxle - The most common type, which encases an automatic or manual

transmission and a differential. The main purpose is the gear changingcapabilities of a transmission.

Drive Axle - Also called live axle, which contains the differentials. It providesthe torque transfer capabilities of a differential.

Straight Axle - A straight beam that connects the wheels together, but hasno differentials. It helps to support the weight of the vehicle and serves asan attachment point for the wheels.

Wheel: A wheel is a circular device capable of rotating on its axis, facilitating

movement or transportation or performing labour in machines. A wheeltogether with an axle overcomes friction by facilitating motion by rolling.A differential is a device that sends power from a drive shaft to both sides ofan axle. The rotational torque from a drive shaft is generally laid outhorizontally down the length of the vehicle (typical RWD). But the wheelsneed to be turned at a 90-degree angle from the position of the drive shaft.The axle driving the wheels is split into two parts and the inner ends of bothsides are connected to the differential. The differential is made up of a seriesof gears that can direct the rotational power from the drive shaft to 90-degreeangles and turn both sides of the axle. Differential can also be used betweenthe front and rear axles on a four-wheel-drive or all-wheel-system by splitting

power between both axles and ultimately driving all four wheels.A wheel and differential axle is a simple lifting machine, which is employed tolift a larger load W at a point by employing a smaller force P at some otherpoint. During the process the distance y moved by the effort may be muchmore than the distance x moved by the load.

Work input in the machine = P * yWork output in the machine = W * xEfficiency of the machine = = W * x = W / P = Mechanical AdvantageP * y y / x Velocity ratio

Mechanical advantage (MA) - is the factor by which a mechanismmultiplies the force put into it. It is the ratio of the force exerted by a machine(the output) to the force exerted on the machine, usually by an operator (theinput). The theoretical mechanical advantage of a system is the ratio of theforce that performs the useful work to the force applied, assuming there is nofriction in the system. In practice, the actual mechanical advantage will beless than the theoretical value by an amount determined by the amount offriction.


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Let W be the resistance overcome through distance d. P be the effortapplied through a distance D.Losses due to friction = Input Output

= P * D W * dWhen the machine is doing work under the action of load W above (ie P = 0)

then,W * D > frictional lossesW * D > P * D W * dW * D > 1P * D 2

Or Mechanical Advantage > 1Velocity Ratio 2

Thus for a machine to be reversible, its efficiency should be more than 50%.The efficiency of a self locking machine will be less than 50%. A wheel anddifferential axle is a simple machine and is used to lift heavy loads. It has awheel of larger radius (R) and an axle of smaller radius (r) fixed on the sameshaft. Wheel and axle are free to rotate about its shaft.

Mechanical advantage of wheel and differential axle: -The effort is applied to the rim of the wheel and the load is raised by a rope

wound around the axle.In one rotation wheel covers a distance of2pRIn one rotation load is raised by a distance of2prIf we neglect force of friction,

Output = inputW x 2pr = P x 2pRW/P = 2pR/2prW/P = R/r

Since [W/P = M.A.]M.A. = R/rOr

M.A. = radius of wheel / radius of axleThis expression indicates that in order to increase the mechanicaladvantageRadius of :wheel must have a large value.Radius of axle must be smaller than that of wheel.


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PROCEDURE:

1. Wrap the string around the wheel and the axle.

2. Put same weight in one of the pulleys (denoted as load) and for each valueof load put weights in other axles such that it starts moving.

3. Change the value of load and repeat step -2.

4. Note down the diameters of wheel and axles and calculate the velocityratio.

5. Calculate the mechanical advantage.

6. A plot of Load Vs Effort and Effort Vs efficiency is drawn

OBSERVATIONS:- Diameter of pulley and axle :

D = 194mm

d1 = 91 mm

d2 = 51 mm

Weight of hook on effort side = 50 gms

Weight of hook on load side = 0.5 gms

OBSERVATION TABLE:-

S.No. Load(W)(gm)

Effort (P)(gm)

W/P(M.A.)

Distancemoved

byeffort

(y)(cm)

Distance

movedby load

(x)(cm)

VelocityRatio(VR) =y / x

Efficiency

=(M.A)V.R

1) 175 25.5 7.058 48.1 5.1 9.43 0.727


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2) 215 30.5 7.049 48.1 5.1 9.43 0.7263) 255 35.5 7.283 48.1 5.1 9.43 0.7514) 285 40.5 7.037 48.1 5.1 9.43 0.725

5) 355 45.5 7.802 48.1 5.1 9.43 0.8046) 450 55.5 8.108 48.1 5.1 9.43 0.836

CALCULATIONS:-

Mean of Velocity Ratio = (9.43+9.43+9.43+9.43+9.43+9.43)/6

= 9.43Mean of efficiency = (0.727+0.726+0.751+0.725+0.804+0.836)/6= 4.569/6= 0.762

RESULT:-

The velocity ratio found out is 9.43

The percent efficiency is 76.20%.

CONCLUSIONS:-

The wheel and differential axle apparatus shows that for different loads andefforts that the graph between effort and load is a straight line whereasthat between efficiency v/s load is a parabola that becomes constant for aparticular load.

The efficiency of the wheel and differential apparatus is greater than 50% asit is a reversible machine.

PRECAUTIONS:-

1. The readings must be taken and noted down carefully.2. The load and effort should move slowly.


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3. Weights must be added gently.4. Any overlapping of the string must be avoided.5. There should be no knot in the string.

6. This is a light-loading machine and thus only lightweights must be usedduring the course of experiment.

SOURCES OF ERROR:-

1. The friction offered by wheel is bound to introduce some error in thereadings. No matter what amount of lubrication is done, wheel can neverbe made completely frictionless.2. Friction due to the wheel may also introduce error.3. If the string is not inextensible or if it overlaps or if there is a knot in thestring, then it will result in error.4. Weights in the hook may be added forcefully. This should not happen, asthis will introduce the impulse causing an error.5. Human error may creep in while performing the experiment.

REFERENCES:-

www.google.com

www.wikipedia.org

Applied Mechanics A.K.Tayal

Write-ups
http://www.google.com/http://www.google.com/


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AIM:To verify the reactions at the supports of a simply supported beam.

APPARATUS:A graduated wooden beam, two compression spring balance, slotted weights etc.

THEORY:

Simply supported beam

A horizontal bar supported at the two ends by two spring-loaded vertical supports can be idealized asa simply supported beam.


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A simply supported beam has only two supports and supports multiple loads either applied externalloads or loads from other beams. A horizontal bar supported at the two ends by two spring loadedvertical supports can be idealized as a simply supported beam.

The distance b/w the two supports is said to be the span of the beam.The specific characteristics of an individual beam are:1) length of the beam2) number of point loads on the beam.3) the positions and values of the reaction.

One end of the beam can rotate slightly around a support with a hinge joint, called the fixed node.The other end rests on some sort of roller or bearing surface, called the rolling node. Although thefixed node allows that end of the beam to rotate slightly it cannot move vertically or horizontally. Therolling node allows the other end to move horizontally and rotate slightly but it cannot movevertically. Examples of simply supported beams include beam bridges, truss bridges, gangways, etc.The load is seen to impart a bending moment to the beam that subjects its upper edge to compressive

stresses that tend to shorten it and subjects its lower edge to tensile stresses that tend to lengthen it. Inbetween there is a gradual shift from compressive to tensile stresses. In figure, the purple line passingthrough the middle of the beam, called the neutral axis, remains the same length, which indicates theabsence of these stresses there. The shift in stresses from compression to tension causes the uppersection of the beam to tend to slide past the lower section. This induces another simply supportedbeam stress in the beam called shear. This horizontal shear stress is greatest along the neutral axis.Solid beams have very high shear strengths. Therefore shear stress is not a major factor in beamdesign except for very tall, thin beams.The importance of providing a rolling node for one end of the beam as that end of the beam moves inand out when the load is applied and removed. If that end is not allowed to move additional stressesare placed on the beam and it will no longer behave as a simply supported beam.

Moment of ForceThe SI unit for moment is the newton meter (Nm).Moment = Magnitude of force X perpendicular distance to the pivotWe have to make use of law of equilibrium, which states that:

1. The algebraic sum of vertical forces must be zero.i.e. FY = 0

2. The algebraic sum of horizontal forces must be zero.i.e. FX = 0

3. The algebraic sum of moments about a point must be zero.i.e. MA = 0


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Free Body Diagram of Beam

Therefore, total load applied = total reaction at the ends of the beam

321WWWRR

BA++=+ - (i)

By taking moment about A

- (ii)

From the selected values of321

&, lll &321

&, WWW determine21

&RR .Percentage error :

100'

'

A

AA

R

RR & 100'

'

B

BB

R

RR

Experimentally and compare these values with exact value given by above relation.PROCEDURE:1) The two springs were calibrated, as they are not accurate. On each of these spring dead weight of1 , 2 , 3 , 4 kgf was put and corresponding to each load the reading at all the points were taken.2) The beam was placed and the initial reading of both the spring balances was taken, as this is to bededucted from all the readings to take care of mass-less beam.3) The weights were suspended at different points of the beam.4) The spring reading was noted and the distance of weights from one end of the beam was measured.5) At least five readings were taken by keeping the weight at different points of the beam.6) Equation 1 & 2 were used to calculate RA & RB.7) From the calibration curve the actual load was found out for the observed load reading at all thepoints.8) Finally the %error in RA & RB, between the calculated load and calibrated load was calculated.

l

lWlWlWR

B

332211++

=


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ASSUMPTIONS :The following are the key assumptions that are used to simplify the experiment.1) The centre of gravity of the beam coincides with the geometric centre.

2) The weights and reaction lie in the same vertical plane.3) There are only two supports for the beam.4) There are no external moments.5) There are no external applied loads except the loads we have taken under consideration.6) The deflection of beam is negligibly small.7) The curvature of beam due to loading is negligibly small.

OBSESRVATIONS:1. Initial reading of left spring = 1.4 kgf2. Initial reading of right spring = 1.4 kgf3. Hence weight of the beam = 2.8 kgf4. Total length of the beam = 100cm

WEIGHT R A RB450 0.4 0.4900 0.8 0.81350 1.2 1.21800 1.6 1.6

S.

No

Load in Kg Distanc

e from

A

Observed

Load

Calculated

Reaction

from (i) &

(ii)

Calibrated

Reaction

from

Calibration

Curve

%Error

W1 W2 L1 L2 RA1 RB1 RA RB RA' RB' in R A in RB1. 1.35

01.350 25 75 1.2 1.2 1.35 1.35 1.52 1.52 11.05 11.05

2. 1.750

1.750 15 85 1.6 1.6 1.75 1.75 1.97 1.97 11.57 11.57

3. 1.900 1.900 30 70 1.8 1.8 1.90 1.90 2.14 2.14 11.63 11.634. 2.15

02.150 10 90 2.0 2.0 2.15 2.15 2.42 2.42 11.55 11.55

5. 2.700

2.700 35 65 2.4 2.4 2.70 2.70 3.04 3.04 11.59 11.59

6. 3.100

3.100 40 60 2.8 2.8 3.10 3.10 3.49 3.49 11.58 11.58


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Result:The supports of simply supported beam are nearly equal to the calculated reactions. The

percentage errors are as follows:


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Average % error in RA = 11.495%Average % error in RB = 11.495%

Precautions:1. Note down the distances to the exact mm.

2. Note down the zero error at the spring balances.

3. Wooden beam should be horizontal.

4. Knife edges at the spring balances must be set perpendicular to the axis of

the beam.

Sources of Error:a. Slotted weights may not be calibrated accurately.

b. If the reactions are different, wooden beam may not remain horizontal due

to uneven compression of springs in spring balances.

c. Inaccuracy in the measurement of distances.

d. Inaccuracy in the readings of reactions at spring balance

References: Write-ups

www.wikipedia.com


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3.1 AIM : To determine the coefficient of static friction between two given material surfaces.

3.2 APPARATUS REQUIRED :1- Adjustable inclined plane.2- Frictinless pulley.3- wooden box.

4- inextensible string.5- Hanger with pan .6- Standard weights.


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Friction and Surface Roughness

In general, the coefficients of friction forstatic and kineticfriction are different.Like all simple statements about friction, this picture of friction is too simplistic. Sayingthat rougher surfaces experience more friction sounds safe enough - two pieces of coarsesandpaper will obviously be harder to move relative to each other than two pieces of finesandpaper. But if two pieces of flat metal are made progressively smoother, we will reach apoint where the resistance to relative movement increases. If we make them very flat andsmooth, and remove all surface contaminants in a vacuum, the smooth flat surfaces willactually adhere to each other, making what is called a "cold weld". Once we reach a certaindegree of mechanical smoothness, the frictional resistance is found to depend on the natureof the molecular forces in the area of contact, so that substances of comparable"smoothness" can have significantly different coefficients of friction.An easily observed counterexample to the idea that rougher surfaces exhibit more friction isthat of ground glass versus smooth glass .Smooth glass plates in contact exhibit much morefrictional resistance to relative motion than the rougher ground glass.
http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html#coehttp://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#stahttp://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#stahttp://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kinhttp://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kinhttp://hyperphysics.phy-astr.gsu.edu/hbase/frict.html#coehttp://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#stahttp://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin


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We can build a simple model of the friction force that is useful in many situations. The model friction

force has the following properties:There are two types of frictional force. The force of static friction and the force of kinetic friction.The direction of the static frictional force is along the contact surface and opposite in direction of anyapplied force.The magnitude of the static friction force is given by

The direction of the kinetic frictional force is opposite the direction of motion of the object it acts on.The magnitude of the kinetic friction force is given by

The coefficients of friction depend on the nature of the surface.The frictional force is nearly independent of the contact area between the objects.The kinetic friction force is usually less than the maximum static friction force.

A free-body diagram of a block resting on a rough inclined plane, with its weight (W), normal force(N) and friction (F) shown.

The plot below of the frictional force vs. the applied force illustrates some of the features of the

frictional force.
http://en.wikipedia.org/wiki/Image:Free_body_diagram.pnghttp://en.wikipedia.org/wiki/Image:Free_body_diagram.png


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Note that the frictional force equals the applied force (in magnitude) until it reaches the maximumpossible value usN. Then the object begins to move as the applied force exceeds the maximumfrictional force. When the object is moving the frictional force is kinetic and roughly constant at thevalue ukN which is below the maximum static friction force.The table below summarizes the main characteristics of the frictional force.

Static Friction Kinetic Friction

Symbol f s fk

Directionopposite direction of

applied force

opposite direction of

object's motion

Magnitude


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LubricantsA common way to reduce friction is by using a lubricant, such as oil or water, which is placedbetween the two surfaces, often dramatically lessening the coefficient of friction. The science of

friction and lubrication is called tribology. Lubricant technology is when lubricants are mixed with theapplication of science, especially to industrial or commercial objectives.Superlubricity, a recently-discovered effect, has been observed in graphite: it is the substantialdecrease of friction between two sliding objects, approaching zero levels (a very small amount offrictional energy would still be dissipated).

Lubricants to overcome friction need not always be thin, turbulent fluids or powdery solids such asgraphite andtalc;acoustic lubrication actually uses sound as a lubricant.Energy of frictionAccording to the law ofconservation of energy, no energy is destroyed due to friction, though it maybe lost to the system of concern. Energy is transformed from other forms into heat. A sliding hockeypuck comes to rest due to friction as its kinetic energy changes into heat. Since heat quickly

dissipates, many early philosophers, includingAristotle, wrongly concluded that moving objects loseenergy without a driving force.When an object is pushed along a surface, the energy converted to heat is given by:

whereNis the normal force,k is the coefficient of kinetic friction,x is the coordinate along which the object transverses.Physical deformation is associated with friction. While this can be beneficial, as inpolishing, it isoften a problem, as the materials are worn away, and may no longer hold the specified tolerances.

The work done by friction can translate into deformation and heat that in the long run may affect thesurface's specification and the coefficient of friction itself. Friction can in some cases cause solidmaterials to melt.Summary:Friction is a measure of the force pressing the two objects together.We can measure friction in terms of a coefficient of friction; the ratio of the force needed to

move two objects in contact with one another and the force holding the two objects together. Friction can be usefulFriction can be reduced by using lubricants, ball bearings, air-cushions and by streamliningFriction reduces the efficiency of machinesFriction produces heat

Friction results in wear and tear.

3.4 PROCEDURE:1-A particular angle of inclination was set.2-The box was put on the inclined plane .3-The box was attached to a thread and passed over a
http://en.wikipedia.org/wiki/Lubricanthttp://en.wikipedia.org/wiki/Tribologyhttp://en.wikipedia.org/wiki/Tribologyhttp://en.wikipedia.org/wiki/Superlubricityhttp://en.wikipedia.org/wiki/Superlubricityhttp://en.wikipedia.org/wiki/Graphitehttp://en.wikipedia.org/wiki/Talchttp://en.wikipedia.org/wiki/Talchttp://en.wikipedia.org/wiki/Acoustic_lubricationhttp://en.wikipedia.org/wiki/Acoustic_lubricationhttp://en.wikipedia.org/wiki/Conservation_of_energyhttp://en.wikipedia.org/wiki/Aristotlehttp://en.wikipedia.org/wiki/Aristotlehttp://en.wikipedia.org/wiki/Normal_forcehttp://en.wikipedia.org/wiki/Normal_forcehttp://en.wikipedia.org/wiki/Polishinghttp://en.wikipedia.org/wiki/Polishinghttp://en.wikipedia.org/wiki/Tolerance_(engineering)http://en.wikipedia.org/wiki/Meltinghttp://en.wikipedia.org/wiki/Lubricanthttp://en.wikipedia.org/wiki/Tribologyhttp://en.wikipedia.org/wiki/Superlubricityhttp://en.wikipedia.org/wiki/Graphitehttp://en.wikipedia.org/wiki/Talchttp://en.wikipedia.org/wiki/Acoustic_lubricationhttp://en.wikipedia.org/wiki/Conservation_of_energyhttp://en.wikipedia.org/wiki/Aristotlehttp://en.wikipedia.org/wiki/Normal_forcehttp://en.wikipedia.org/wiki/Polishinghttp://en.wikipedia.org/wiki/Tolerance_(engineering)http://en.wikipedia.org/wiki/Melting


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pulley.4-Some weights were put in a pan and the weight was checked to have the motion of the box.5- For same reading on same angle, put some weight in box and repeat procedure

3 for motion.6-Take atleast three to four readings on same angle.7- Change inclination and repeat above procedure.8- Take average of the values of the friction for same materials.9- Repeat above procedure for different surfaces in cotact and compare their friction with each other.3.5 OBSERVATION TABLE :

TABLE 1: WOOD AND WOOD

Weight of wooden box=84.15 gm

s.no.

Inclination()

m(gm.)

M(wt. In pan)(gm)

Sin Cos

1. 30 84.15 86.17 .5 .866 .6052. 30 89.15 93.17 .5 .866 .6303. 30 94.15 98.17 .5 .866 .6264. 30 99.15 103.17 .5 .866 .624

1. 20 84.15 63.17 .342 .939 .4352. 20 89.15 65.17 .342 .939 .4503. 20 94.15 67.17 .342 .939 .3964. 20 99.15 71.17 .342 .939 .400

mean =.5208

TABLE 2: WOOD AND GI SHEET


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mean =.329

3.6 PRECAUTIONS:1-The pulley should be well lubricated.2-String should be inextensible.3-Angle of inclined plane should be accurately measured and inclined plane should be steady.4-Condition under which the block just slides is arbitrarily determined, hence care should be takenthat it is almost same for all observation.5-The inclined plane may not be smooth all over .Hence coefficient of friction varies from place toplace.

3.7 RESULT :Average value of coefficient of friction for following surfaces are :wood and GI sheet = 0.329wood and wood =0.5208

3.8 CONCLUSION:1-In the experiment ,we are finding coefficient of static friction between two surfaces .when the block justslides ,frictional force acting on it is maxmimum.2-We observe that coefficient of static friction between wood and wood is greater than wood and aluminumsheet.

3-The difference in values of coefficient of static friction for the same material is due to non uniformroughness of the whole surface.

3.9 REFERENCES :1. en.wikipedia.org/wiki/Friction2. www.fearofphysics.com/Friction/frintro.html3. hyperphysics.phy-astr.gsu.edu/hbase/frict.html4. www.thinkingfountain.org/f/friction/friction.html

s.no.

Inclination()

m(gm.)

M(wt. Inpan)(gm)

Sin Cos

1. 30 84.15 68.17 .5 .866 .3582. 30 89.15 75.17 .5 .866 .3963. 30 94.15 79.17 .5 .866 .3934. 30 99.15 83.17 .5 .866 .391

1. 20 84.15 50.17 .342 .939 .2702. 20 89.15 52.17 .342 .939 .2593. 20 94.15 57.17 .342 .939 .2824. 20 99.15 60.17 .342 .939 .282


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5. U.C.JINDAL6. A.K.TAYAL

Aim

Construct the following:1) Simple roof truss2) Roof truss with angled tie rod

And find the forces in each member of the truss.

Apparatus

Stands and cross bars, hinges, hinge holder hooks, member rods,loading rod with platform, spring balances, and slotted weights.


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Theory

A truss is a structure consisting of straight members connected at theirextremities only. The members being slender and unable to supportlateral loads, all the loads must be applied at the joints; a truss maythus be assumed to consist of pins and two-force members.

A truss can be thought of as a Beam (structure) where the web consistsof a series of separate members instead of a continuous plate. In thetruss, the lower horizontal member (the bottom chord) and the upperhorizontal member (the top chord) carry Tension and Physicalcompression, fulfilling the same function as the Flange of an I-beam.Which chord carries tension and which carries compression depends onthe overall direction of Bending. In the truss pictured above right, thebottom chord is in tension, and the top chord in compression.


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RIGID OR PERFECT TRUSS:- If the truss follows the following relation thenit is said to be a rigid truss or a non collapsible truss, i.e. it has a greatstrength.

M=2j-3Where, M= no. of members of the truss

J= no. of joints in the trussIf M>2j-3 then the truss is called to be statically indeterminate truss.


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If M


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three unknown support reactions (A, Dx, Dy), giving a total of 10unknowns to solve for using the 10 equations obtained fromequilibrium.

The method of sections: This method uses free-body-diagrams of sectionsof the truss to obtain unknown forces. For example, if one needs only tofind the force in BC, it is possible to do this by only writing twoequations. First, draw the free body diagram of the full truss and solvefor the reaction at A by taking moments about D. Next draw the freebody diagram of the section shown and take moments about E to findthe force in BC.

In the method of sections one can write three equations for each free-body-diagram (two components of force and one moment equation).

Consider the equilibrium of TRUSS:-F=0,

F1

FEC

F3

BB

E

A

A

FBC

FED

A C

B

W

L1

L2


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-W + R1 + R2 =0

Taking moment about A :

Ma = 0R2 . L1 W .L2 =0R2 = (W . L2) / L1

After solving above equations we can find the value of R1 & R2

Procedure

1)First of all, all the materials required to make the truss were broughttogether.

2) Then the simple roof truss was made by putting the materialtogether.

3) The truss was hanged in such a manner so that the hooks on the barwere parallel to the truss.

4) Spring balances were hanged onto the chain parallel to the truss.

5) Some weight was putted onto the upper part of the truss.

6) The corresponding readings on the two spring balances were noted.

A

B

C

D

L1

L2

R2R1

1 2W2

W1


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7) Then this truss was removed and it was made in the shape of rooftruss with angled tie rod.

8) The above steps were being repeated for this truss.

9) At least four readings were noted for each type of truss.

Observations

For Simple Roof Truss

Initial value of R1= 0.8 kgfInitial value of R2=0.75 kgf

Length of Rod A= 44.5 inchesLength of Rod E= 24 inches

S.No.

Weight

(kg.)(W)

Springreading

Calculated reading

Percentage error

Forces inmembers

R1 R2 R1 R2 R1 R2 AB BC AC

1.

410 1 0.95 0.205

0.205

2.75 2.75 0.54 0.54 0.50

2. 820 1.20 1.15 0.41 0.41 2.5 2.5 1.08 1.08 1.00

3. 1715 1.65 1.6 0.86 0.86 1.3 1.3 2.30 2.30 2.13

For Roof Truss with Angled Rod

Initial Value of R1= 1.15 kgfInitial Value of R2= 1.1 kgfLength of C= 31 inchesLength of D= 26 inchesLength of H= 10 inches


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S.N

o.

Weight(kg)Final

springreading

Calculated

reading

Percentage

errorForces in members

W1upper

W2Lower

R1 R2 R1 R2 R1 R2 AB AD BD BC DC

1. 430 430 1.6 1.55

0.43

0.43

4.4 4.4 0.65

0.65

0.62

0.62

0.61

2. 910 840 2.05

2 0.88

0.88

2.3 2.3 1.45

1.45

1.38

1.38

1.37

3. 1332 1285 2.45

2.4 1.31

1.31

1.0 1.0 2.10

2.10

2.00

2.00

1.99

Result

Hence the simple roof truss and roof truss with angled rod is successivelyconstructed and forces in members was found.

Precautions

1) All connections of the truss should be tight. All the screws should betightened, so that members do not fall off when the load is applied.

2) Enough space should be left for introducing the weights. However thecentral limb ofthe truss (with which it is attached to the upper horizontal rod) should not

be made tohang from its extreme end as this will make it fall when appreciable force

is applied.

3) The truss should be hung parallel to the upper rod.

4) While making the truss, appropriate rods should be taken ( as certain rodsto be put

are of equal length).

5) Readings in spring balance should be measured correctly.

Sources of Error


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1) The screw may not be properly tightened. This may change the anglesbetween different members of the truss when the load is applied. Hence,the internal forces developed in the members would change and cause

error.2) High least count of the spring balance introduces the error in calculation.3) The spring balance used to measure the reactions at the ends of the truss

may not be calibrated properly.4) Human errors may creep in. The readings may not be taken properly or

parallax may occur while taking the readings.

References www.google.com

www.wikipedia.com

Write-ups

OBJECTIVE:To determine the forces in the members of a loaded shear legs space frame experimentally, vectoriallyand graphically.

APPARATUS:Shear legs apparatus consisting of two rigid bars AB and AC and a tie bar AD together with aprovision for loading at A as shown in figure, metre rod, spring balances and standard weights.

APPARATUS DESCRIPTION:The shear consists of two rigid rods joined at a point and a tie-bar joined from the joints to a fixedsupport. The shear legs apparatus provides a method to extensively study the action of various forcesin space, not even coplanar. The apparatus is loaded at the joint. It is used to lift the load. As theweight is distributed in the three arms of the legs, hence, heavy weights can be suspended.

Shear legs are often used to make temporary cranes. In this experiment theideas developed in experiments with several forces in one plane are extended to
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three dimensions. The double shear legs are mounted on rollers, which run on around bar. Compression forces in the legs are measured with integral springbalances, and restraint is by an adjustable tie chain. The backstay is adjustable

and is also fitted with a spring balance. Loading is by a weight hung from theapex. Clarity of thought is encouraged, since there are both compressive andtensile forces present.

THEORY:The forces may be of three types:

1. Compression2. Tension3. Shear Force

Out of the above three forces tension and shear force are more dangerous.

Compression:- Compression has many implications in material science, physicsand structural engineering, for compression yields noticeable amounts ofstressand tension.

By inducing compression, mechanical properties such as compressive strength ormodulus of elasticity, can be measured. Scientists may utilize press machines toinduce compression

Tension:-Tension is a reactionforce applied by a stretched string (rope or asimilar object) on the objects which stretch it. The direction of the force oftension is parallel to the string, towards the string.

Tension exists also inside the string itself: if the string is considered to becomposed of two parts, tension is the force which the two parts of the stringapply on each other

Shear Force:- Shear strength in reference to soil is a term used to describe the maximum strength ofsoil at which point significantplastic deformation oryielding occurs due to an appliedshear stress.
http://en.wikipedia.org/wiki/Material_sciencehttp://en.wikipedia.org/wiki/Physicshttp://en.wikipedia.org/wiki/Structural_engineeringhttp://en.wikipedia.org/wiki/Stress_(physics)http://en.wikipedia.org/wiki/Tension_(mechanics)http://en.wikipedia.org/wiki/Compressive_strengthhttp://en.wikipedia.org/wiki/Modulus_of_elasticityhttp://en.wikipedia.org/wiki/Scientisthttp://en.wikipedia.org/wiki/Reactionhttp://en.wikipedia.org/wiki/Forcehttp://en.wikipedia.org/wiki/Ropehttp://en.wikipedia.org/wiki/Plasticity_(physics)http://en.wikipedia.org/wiki/Yield_(engineering)http://en.wikipedia.org/wiki/Yield_(engineering)http://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Material_sciencehttp://en.wikipedia.org/wiki/Physicshttp://en.wikipedia.org/wiki/Structural_engineeringhttp://en.wikipedia.org/wiki/Stress_(physics)http://en.wikipedia.org/wiki/Tension_(mechanics)http://en.wikipedia.org/wiki/Compressive_strengthhttp://en.wikipedia.org/wiki/Modulus_of_elasticityhttp://en.wikipedia.org/wiki/Scientisthttp://en.wikipedia.org/wiki/Reactionhttp://en.wikipedia.org/wiki/Forcehttp://en.wikipedia.org/wiki/Ropehttp://en.wikipedia.org/wiki/Plasticity_(physics)http://en.wikipedia.org/wiki/Yield_(engineering)http://en.wikipedia.org/wiki/Shear_stress


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There is no definitive 'shear strength' of a soil, as it depends on a number of factors affecting the soilat any given time and on the frame of reference, in particular the rate at which the shearing occurs.Since the frame is in equilibrium, every point and sub part of the frame must also be in equilibrium.

Consider the joint A. The forces that act A to keep it in equilibrium are the known loads actingvertically downward and the forces along AB, AC & AD. The direction of AB, AC & AD aredetermined from co-ordinate geometry only the magnitude of forces are unknown, which may beobtained, experimentally by reading the spring balances installed in the members.Basic principle: Every point on the frame is in equilibrium. Hence the forces acting at a point add tozero. The forces are the load-weight and the forces along the arms AB, AC and AD. The directions ofFAB, FAC and FAD are obtained by measuring the coordinates of A,B,C,D and fixing one point asorigin.

Procedure:


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PRECAUTIONS:1) The compression in the beams should be noted carefully.

2) The load applied should not be too heavy.

3) Before taking the reading of forces see that all beams are properly

fixed in their position.

4) Measurements of the respective lengths of the beams should be

made carefully.

5) The weight hanging should be vertical without oscillating.

SOURCES OF ERROR1) The measurement of different lengths of beam may not be correct.

2) The weights might be oscillating while experiment is being

performed.

3) Some friction may also be present in the beams while they are being

compressed.

4) The weight should be vertical so that the force is acting vertically

downwards.

REFERENCES:- www.google.com

www.wikipedia.org

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