npr college of engineering & technology

NPR COLLEGE OF ENGINEERING & TECHNOLOGY

1

Subject : Dynamics of Machinery

M.Mathan RajAssistant Professor / Mech

NPRCET

UNIT 2

BALANCING

DYNAMICS OF MACHINERYSTATIC AND DYNAMIC BALANCING

What are the reasons for unbalanced forces?

Unbalanced forces are set up in high speed engines due to the following reasons:

(a) Rotating masses: Rotating masses if not balanced, produce centrifugal forces which act as unbalanced forces and cause undesirable vibrations and noise.

(b) Reciprocating masses: reciprocating masses if not balanced, produce inertia forces which act as unbalanced forces and cause vibrations and noise.

What is balancing?

Balancing is the process of designing or modifying machinery, so that the unbalance in the machinery is reduced to an acceptable level and if possible completely eliminated.

Why balancing is necessary?

If rotating and reciprocating masses are not balanced, they produce centrifugal / inertia forces which will cause excessive vibration, noise, wear and tear of the system. Hence balancing is very essential.

What are the types of balancing?

Balancing

Rotating masses Reciprocating masses

SingleDifferent Primary force Secondary force

Combined rotating and

reciprocating masses


Define static balancing.

A system is said to be in static balance if the net dynamic force acting on the system is zero. Σ F = 0.

Define dynamic balancing.

A system is said to be in dynamic balance if the net dynamic forces as well as net dynamic couples are equal to zero respectively.

What are the various types of balancing of rotating masses?

(a) Balancing of a single rotating mass by a single mass rotating in the same plane (static balancing)

(b) Balancing of a single rotating mass by two masses rotating in different planes (dynamic balancing)

(c) Balancing of several masses rotating in the same plane (static balancing)

(d) Balancing of several masses rotating in different planes (dynamic balancing)


What is rocking?

Sometimes it may not be possible to introduce a single balancing mass in the same plane of rotation of the disturbing mass. Hence the balancing mass is provided in a plane parallel to the plane of rotation of the disturbing mass. This balances the forces, but produces an unbalanced couple. This unbalanced couple tends to rock the shaft in its bearings. The shaft is subjected to bending. This phenomenon is called rocking of the shaft.

Define reference plane.

Reference plane is a plane passing through a point on the axis of rotation and perpendicular to it. It is used as a reference for analysis and balancing.

What are the conditions to be satisfied in reference plane for complete balance in the case of balancing of several masses in different planes?

(a) Forces in the reference plane must balance i.e. resultant force must be zero.

(b) Couples about the reference plane must balance i.e. resultant couple must be zero.


How will you balance several masses rotating in different planes?

When several masses rotate in different planes, they may be transferred to a plane known as reference plane (R P) and hence the problem is reduced to that of several masses rotating in the same plane.

This transfer will introduce the following effects in the reference plane:

(a) An unbalanced force equal and parallel to the centrifugal force produced by the rotating mass (F)

(b) An unbalanced couple whose magnitude is equal to the product of the centrifugal force and the distance between the two planes (F l)

Why are the projection of hands in opposite direction in watches and clocks?

Projections are made such that mass occupied in the projected length is same as mass occupied in the needle from the pivot point. Hans are thus balanced.

DYNAMICS OF MACHINERYBALANCING OF ROTATING MASSES

How would you balance several rotating masses in the same plane by analytical method?

Σ H = Σ mi ri cos θi

Σ V = Σ mi ri sin θi

Resultant force magnitude , R = [(Σ H)2 +(Σ V)2 ]

Direction θ’ = Σ V / Σ H

Inclination of balancing mass with horizontal θ R = 180 + θ’

Let Value of radius of rotation of balancing mass be “ r “

Balancing mass be “ m “

The relation used to find “ m “ or “ r “ is “ m = R / r “

DYNAMICS OF MACHINERYBALANCING OF ROTATING MASSES

DYNAMICS OF MACHINERYBALANCING OF RECIPROCATING MASSES

What do you mean by balancing of reciprocating masses?

It is the dynamic balancing of inertia force and couple (shaking couple) due to reciprocating masses to avoid vibrations of reciprocating engines.

What will happen if reciprocating masses are not balanced?

As the stroke of the reciprocating parts is proportional to radius of crank, the radius of crank acts as an eccentricity of reciprocating masses. Due to this eccentricity inertia force is produced causing vibrations of the engine. Even if there is no unbalanced force due to inertia effect, there will be unbalanced couple called shaking couple causing vibrations of the engine.

DYNAMICS OF MACHINERYBALANCING OF RECIPROCATING MASSES

Why complete balancing of reciprocating mass is not possible?

To balance the rotating mass at crank pin, balancing mass is connected dynamicallyopposite to the crank pin. But the vertical component of the force produced by thebalancing mass is unbalanced and produces hammer blow. In reciprocatingengines, unbalanced forces in the direction of line of stroke are more dangerousthan the forces perpendicular to the line of stroke. Therefore reciprocating massesbalanced only to an extent (fraction “ c “ of reciprocating masses) to avoidhammer blow. Complete balancing of reciprocating masses are not possible.

Distinguish between the nature of unbalanced force due to reciprocating mass and that due to rotating mass.

Inertia forces ( both primary and secondary) due to reciprocating masses are constant in direction but vary in magnitudes.

Centrifugal forces due to rotating masses are constant in magnitude, but vary in direction.

DYNAMICS OF MACHINERYBALANCING A SINGLE CYLINDER ENGINE

DYNAMICS OF MACHINERYPRIMARY AND SECONDARY UNBALANCED FORCES

I

DYNAMICS OF MACHINERYBALANCING MULTI CYLINDER ENGINES

Why are the cranks of a twin cylinder locomotives placed at right angles to each other?

The cranks of a twin cylinder locomotives placed at right angles to each other because

(a) More uniform turning moment is obtained.

(b) Engine can be started easily after stopping in any position.

What do you mean by outside cylinder locomotive?

In outside cylinder locomotive, two cylinders are placed outside the driving wheels, one on each side of the driving wheels to obtain uniform turning moment.

Define tractive force.

Tractive force is the resultant unbalanced primary force along the line of stroke.

Variation of tractive force (effort) of an engine is caused by the resultant unbalanced primary force due to two cylinders along the line of stroke of the locomotive engine.


Define tractive force.

Tractive force is the resultant unbalanced primary force along the line of stroke.

Variation of tractive force (effort) of an engine is caused by the resultant unbalanced primary force due to two cylinders along the line of stroke of the locomotive engine.

Mention the expression for tractive force. At what angles of inclination of the crank with the line of stroke, the tractive force reaches a maximum value?

Tractive force FT = (1-c) m ω2 r (cos θ – sin θ)

It is maximum at θ =1350 and θ = 315 0

FT max = ±√2 (1-c) m ω2 r


Define swaying couple.

The unbalanced primary forces along the line of stroke for two cylinders which are separated by a distance and thus constitute a couple about the centre line of the locomotive engine between the two cylinders. This couple tends to sway the engine alternately in clockwise and anticlockwise direction. Hence this couple is called swaying couple.

Mention the expression for swaying couple. At what angles of inclination of the crank with the line of stroke, the swaying couple reaches a maximum value?

Swaying couple CS = (1-c) m ω2 r (a/2) (cos θ + sin θ)

It is maximum at θ =450 and θ = 225 0


Define hammer blow.

The unbalanced force perpendicular to the line of stroke produces variation in pressure on the rails which results in hammering acting action on the rails. The maximum value of this vertical unbalanced force is called hammer blow. This is caused by the mass provided to balance the reciprocating mass.

Mention the expression for hammer blow. At what angles of inclination of the crank with the line of stroke, the hammer blow reaches a maximum value?

Hammer Blow, FH = B ω2 b sin θ; where B is balancing mass, b is radius of rotation of balancing mass.

It is maximum at θ = 900 and θ = 270 0

FH max = ± B ω2 b

Effect of hammer blow: Causes variation in pressure between the wheel and the rail.


Mention the expression for permissible value of angular speed ω of the wheel to avoid lifting from the track.

Permissible value of angular speed ω of the wheel to avoid lifting of the wheel from the track , ω = [P/(Bb)] 1/2

Where B is balancing mass, b is radius of rotation of the balancing mass.

What do you mean by multi cylinder engine?

When two or more cylinders are there in an engine, then it is called multi cylinder engine. These engines are broadly classified according to the arrangement of cylinders with respect to each other and crank shaft.

Mention the important types of multi cylinder engines.

Multi cylinder engine

Inline Radial V engine


What do you mean by radial engine? Why are they preferred?

Radial engine is one which has pistons arranged in a circle about the crank centre.

In radial engines, connecting rods (equal to number of cylinders) are connected to a common crank. Unlike in inline engines, the plane of rotation of crank (common crank) is same. Therefore there is no unbalanced primary or secondary couple.

What is V-12 engine? Why is it used in some luxury cars?

The engine has two banks of 6 cylinders each, the banks inclined to each other at an angle. We know that in 6 cylinder inline engine, each bank of cylinders is self balancing and hence the whole engine is balanced. The engine has two banks of 6 cylinders each, the banks inclined to each at an angle. Each bank with 6 cylinders 4 stroke inline engines is self balanced. Hence the whole engine is balanced and hence v-12 engine is used in some luxury cars.


What are the various methods of force balancing of linkages?

(a) Method of static balancing

(b) Method of principle vectors

(c) Method of linearly independent vectors.

(d) Use of cam driven masses.

(e) Addition of an axially symmetric duplicate mechanism

Mention the types of balancing machine.

(a) Static balancing machines: These machines measure the static unbalance only.

(b) Dynamic balancing machines: These machines measure the dynamic unbalance only.

(c) Universal balancing machines: These machines are capable of measuring both static and dynamic unbalance.


How are the cylinders arranged in uncoupled three cylinder locomotives?

In an uncoupled three cylinder locomotive, there are three cylinders out of which two are outside cylinders and one is inside cylinder. On each side of the two driving wheels beyond the wheel, one outside cylinder lies. The inside cylinder lies in between the centre of the two driving wheels to obtain uniformity in turning moment. The angle between the cranks of each cylinder is 1200 .

FIG.


What do you mean by inline engine?

In line engine is one in which the piston axes form a single plane coincident with the crank shaft and in which pistons are all on the same side of the crank shaft.

What are the conditions to be satisfied for the complete balance of multi cylinder in line engines?

(a) Primary forces must balance i.e. primary force polygon must close.

(b) Primary couples must balance i.e. primary couple polygon must close.

(c) Secondary forces must balance i.e. secondary force polygon must close.

(d) Secondary couples must balance i.e. secondary couple polygon must close.

DYNAMICS OF MACHINERYFIRING ORDER

What do you mean by firing order?

The sequence in which charge is ignited inside the the engine cylinders is called firingorder. There are different firing orders for multi cylinder engines

For two stroke engines, interval between the cranks = 360 / n

For four stroke engines, interval between the cranks = 720 / n

The cranks are assumed to rotate in clockwise direction and hence, they arenumbered in anticlockwise as per firing order.

DYNAMICS OF MACHINERYFIRING ORDER

Draw primary and secondary crank positions of four cylinder two stroke in line engine for the firing order 1-5-2-3-4.

1

5

23

4 2

54

31

Primary Crank positions Secondary Crank positions

DYNAMICS OF MACHINERY

STATIC BALANCING DYNAMIC BALANCING

DYNAMICS OF MACHINERYPIVOTED CRADLE BALANCING MACHINES

PIVOTED CRADLE BALANCING MACHINE: This machine measures both static and dynamic unbalance of rotating parts. It is similar to dynamic balancing machine except that in this no preliminary static balancing of the machine part is required.

Hence it is also called Universal Balancing Machine.

FIG

DYNAMICS OF MACHINERYPIVOTED CRADLE BALANCING MACHINES

PIVOTED CRADLE BALANCING MACHINE: This machine measures both static and dynamic unbalance of rotating parts. It is similar to dynamic balancing machine except that in this no preliminary static balancing of the machine part is required.

Hence it is also called Universal Balancing Machine.

It works on the principle that a given system of revolving masses can be balanced by introducing two balancing masses in arbitrarily chosen planes of motion. First two planes say L and M are selected. By using these planes, the balancing effects are to be measured and necessary corrections are applied.

The rotating part is mounted on the cradle in such a manner that the axis of the pivots P-P lies on one of the chosen planes say L. The purpose of this that the out of balance effect in this plane can not cause oscillations of the cradle. The out of balance effect in the other plane M produces a moment (Fm a sin θ) in the plane of oscillation and rocks the cradle about the axis P-P. The maximum value of this moment Fm a can be measured so that the amount of unbalance (wm r) in plane M can be obtained. Thereafter the cradle is slided along the guide, so that the axis of oscillation lies in plane M and the amount and angular position of the unbalance (wl r) in plane L can be obtained.

Problem 1: Given: Four masses attached to a shaft in the same plane

m1 m2 m3 m4 Unit

Mass 200 300 240 260 kg

Radius 0.2 0.15 0.25 0.3 m

Angle between successive masses

m1-m2 m2-m3 m3-m4

θ 450 75 0 135 0

Balancing mass is provided at radius 0.2 m

Find the magnitude and angular position of the balancing mass

Problem 2Four masses A,B,C,D are attached to a rotating shaft

A B C D UnitMass 20 10 8 kg

Radius 50 62.5 100 75 mmDistance between planes of successive masses

A and B B and C C and D600 600 600 mm

Shaft is in complete balanceFind (a) the mass of A; (b) Angular positions of A, B, C, and D


A B C D UnitMass 200 300 400 200 kg

Radius 80 70 60 80 mmDistance

from Plane A 0 300 400 700 mmAngle between cranks measured anticlockwise

A and B B and C C and Dθ 450 70 0 120 0

Balancing masses in planes X and YDistance between planes

A and X X and Y Y and D100 400 200 mm

Balancing masses at radius 100 mmFind (a) the balancing masses; (b) Angular positions of the balancing masses.

Problem 4Four masses A,B,C,D are to be balanced.


Radius 180 240 120 150 mmAngle between planes

B and C B and D C and D BD and CD areθ 90 0 210 0 120 0 in same sense

Planes B and C are 300 mm apartFind (a) the mass of rotor A; (b) angular position of A; (c ) Position of plane A and D .

Problem 5Four masses A,B,C,D are in order along the axis. Shaft and masses are in complete balance.


Radius 180 240 120 150 mmAngular spacing

C and B D and Bθ 90 0 210 0

Planes B and C are 300 mm apartFind (a) the mass of rotor A; (b) angular position of rotor A; (c ) Position of plane A and D .


A B C D UnitMass 200 300 400 200 kg

Radius 80 70 60 80 mmDistance

from Plane A 0 300 400 700 mmAngle between cranks measured anticlockwise

A and B B and C C and Dθ 450 70 0 120 0

Balancing masses in planes X and YDistance between planes

A and X X and Y Y and D100 400 200 mm

Balancing masses at radius 100 mmFind (a) the balancing masses; (b) Angular positions of the balancing masses.

Problem 7A shaft carries Four masses A,B,C,D in parallel planes.Shaft is in complete balance

A B C D UnitMass 18 21.5 kg

Eccentricity 80 60 60 80 mm

Angle between massesB and C B and A

θ 100 0 190 0 in same senseDistance between planes

A and B B and C100 200 mm

Find (a) mass of A and D; (b) Distance between A and D;(c ) Angular position of mass at D.

Problem 8Four Cylinder Vertical EngineCrank radius r = 225 mmPlane of rotation of 1st crank from 3rd crank at 600 mmPlane of rotation of 2nd crank from 3rd crank at 300 mmPlane of rotation of 4th crank from 3rd crank at 300 mmReciprocating masses of 1st cylinder 100 kg.Reciprocating masses of 2nd cylinder 120 kg.Reciprocating masses of 4th cylinder 100 kg.Find (a) Reciprocating mass for 3rd cylinder;(b) Angular positions of 1st, 2nd,3rd, and 4th cylinders for complete balance.

Problem 9

Four Cylinder Vertical EngineCrank radius r = 150 mmPlane of rotation of 1st crank from 3rd crank at 400 mmPlane of rotation of 2nd crank from 3rd crank at 200 mmPlane of rotation of 4th crank from 3rd crank at 200 mmReciprocating masses of 1st cylinder 50 kg.Reciprocating masses of 2nd cylinder 60 kg.Reciprocating masses of 4th cylinder 50 kg.Find (a) Reciprocating mass for 3rd cylinder;(b) Angular positions of 1st, 2nd,3rd, and 4th cylinders for complete balance.

Problem10Six Cylinder Vertical 4 stroke inline engineFiring order 1-4-2-6-3-5Stroke length L = 2 r = 0.1 mConnecting rod length l = 0.2 mPitch distance between cylinder centre lines 100, 100, 150, 100, 100 mmReciprocating mass per cylinder mR = 1 kg.Engine speed N = 3000 rpmFind Unbalanced primary and secondary forces and couples. Consider plane midway between cylinders 3 and 4 as reference.

Problem11Three Cylinder inline IC engineCranks are set at angle = 120 0

Pitch of the cylinders = 1 mStroke length of the piston L = 0.6 mReciprocating mass for inside cylinder = 300 kgReciprocating mass for outside cylinder = 260 kgDistance between plane of rotation of balancing mass from inside crank = 0.8 mPercentage of reciprocating masses to be balanced, c = 0.4Radius of balancing masses = 0.6 mFind (a) magnitudes and (b) angular positions of balancing masses; (c ) Hammer blow/wheel at axle speed = 6 rps.

Problem12Uncoupled outside cylinderMass of rotating parts per cylinder = 360 kg.Mass of reciprocating parts per cylinder m R= 300 kg.Angle between cranks 90 0

Crank radius r = 0.3 mCylinder centres = 1.75 mRadius of balancing masses = 0.75 mWheel centres = 1.45 mAll the rotating masses and 2/3 of reciprocating masses are balanced in planes of driving wheels.Find (a) magnitudes and (b) angular positions of balancing masses; (c )Limiting speed in kph ( for the wheel to lift off the rails) if load on each driving wheel is 30 kN and diameter of the wheel is 1.8 m(d) Swaying couple at the above speed.

Problem13Two cylinder inside uncoupled locomotiveMass of rotating parts per cylinder = 250 kg.Mass of reciprocating parts per cylinder m R= 300 kg.Angle between cranks 90 0

Crank radius r = 0.3 mDistance between cylinders = 0.6 mRadius of balancing masses = 1 mDistance between driving wheels = 1.5 mDiameter of driving wheels = 2 mSpeed of locomotive = 80 km/hrAll the rotating masses and 2/3 of reciprocating masses are balanced in planes of driving wheels.Find (a) Hammer blow; (b ) Maximum variation in tractive effort; (c ) Maximum swaying couple.

Problem14Two cylinder inside uncoupled locomotiveMass of reciprocating parts per cylinder m R= 300 kg.Angle between cranks 90 0

Crank radius r = 0.3 mDistance between cylinders = 0.65 mDistance between driving wheels = 1.6 mDiameter of driving wheels = 1.8 mMaximum hammer blow = 45000 N at Speed of locomotive = 100 km/hrAll the rotating masses and 2/3 of reciprocating masses are balanced in planes of driving wheels.Find (a) Fraction of reciprocating masses to be balanced c; (b) Variation in tractive effort; (c ) Maximum swaying couple.

Problem15

Four cylinder inline engine

Speed N = 1800 rpm

Crank radius r = 0.06m

Connecting rod length l = 0.24 m

Spacing between cylinders = 150 mm

Mass of reciprocating parts per cylinder m R= 1.5 kg.

Angular positions of cranks 1-4-2-3 at 90 o intervals in end view.

Cylinders are numbered in sequence 1-2-3-4 from one end.Find Unbalanced primary and secondary forces and couples with reference to central plane of engine.

Unit 2: Problem 1- A 270 cm long shaft carries three pulleys, two atits ends and the third at its mid point. The two end pulleys havemasses 48 kg and 20 kg and their centre of gravities are 1.5 cm and1.25 cm respectively from the axis of the shaft, the middle pulley hasa mass of 56 kg and its CG is 1.5 cm from the shaft axis. The pulleysare so keyed to the shaft that the assembly is in static balance. Theshaft rotates at 300 rpm in two bearings 180 cm apart with equalover hangings on either side. Determine the relative angular positionof the pulleys.Given Data: Length of shaft L = 2.7 mLet A, C, and E be the pulleys at left end, middle and right endrespectively.Let B and D be the bearings; Length between bearings = 1.8 m

Mass (kg) Radius (m) CentrifugalForce /ω2

A 48 0.015 0.72

C 56 0.015 0.84

E 20 0.0125 0.25

To find : relative angular position of the pulleys51

Plane Diagram Space Diagram

2.70

1.35

0.45

A C EB D

C

Table

Mass m (kg)

Radius r (m)

Centrifugal Force /ω2

=mr

Angleθ

mr cos θ ‘ mr sin θ

a 48 0.015 0.72 θA 0.72cosθA 0.72sinθA

c 56 0.015 0.84 0 0.84 0

e 20 0.0125 0.25 θE 0.25cosθE 0.25sinθE

Sum = 0 0

Solving the two equations, 0.72cosθA+0.84+0.25cosθE = 0, and 0.72sinθA+0.25sinθE = 0 you may find it difficult!!!

Solving by familiar equation for a triangle, 52

C

E

A

EA

‘ a = 72‘ e = 25

‘ c =84

2.70

1.35

0.45

A C EB D

C

E

A

a c eMagnitude 72 84 25

A C E Total

cos 0.595 -0.346 0.96

53.5 110.2 16.3 180

180+53.5 180-16.3

Angle deg 233.5 163.7

53

Angular position of mass A= 163.7 o

Angular position of mass E= 233.5 o

Unit 2: Problem 2- A, B, C, and D are four masses carried by a rotating shaft at radii100 mm, 150 mm, 150 mm, and 200 mm respectively. The planes in which the massesrotate are spaced at 500 mm apart and the magnitude of the masses B, C, and D are 9kg, 5 kg and 4 kg respectively. Find the required mass A and the relative angularsettings of the four masses so that the shaft shall be in complete balance.

Given Data A B C D

Mass m (kg) ma 9 5 4

Radius r (m) 0.1 0.15 0.15 0.2

Distance between planes = 0.5 m


B

1.50

1.00

0.50

A(RP) CB D

54

D

C

B

A

DC

‘ c = 0.75

‘ d = 1.20

‘ b =0.675

B

C

D

Table:

Plane Mass Radius Force/ω2

Distance from RP Couple/ω2 Angle

m (kg) r (m)mr (kg

m) l (m)mrl (kg

m2) θ (deg)

A (RP) mA 0.1 0.1mA 0 0

B 9 0.15 1.35 0.5 0.675 0

C 5 0.15 0.75 1 0.75 θ C = 65.4

D 4 0.2 0.8 1.5 1.2 θ D = 214.6

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Plane Mass Radius Force/ω2 Angle

m (kg) r (m) mr (kg m) θ (deg) mr cosθ mr sinθ

A (RP) mA 0.1 0.1mA

B 9 0.15 1.35 0 1.35 0

C 5 0.15 0.75 65.4 0.311897 0.6821

D 4 0.2 0.8 214.6 -0.657824 -0.455

Sum = 1.004074 0.2268

Resultant= 0.1 mA = 1.029

Angle of resultant θ= 12.72

Angle of equilibrant θA =180+12.72 192.72deg

mA= 10.29kg

D

C

B

A

Results

Plane Mass Radius Angle

m (kg) r (m) θ (deg)

A (RP) 10.29 0.1 192.7

B 9 0.15 0

C 5 0.15 65.4

D 4 0.2 214.6

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Unit 2: Problem 3- The following data refer to two cylinder uncoupled locomotiveRotating mass per cylinder = 300 kg;Reciprocating mass per cylinder = 330 kg;Distance between wheels = 1.5 m;Distance between cylinders (pitch) = 600 mm;Diameter of treads of driving wheels = 1.8 m;Crank radius = 325 mm;Radius of centre of balance mass = 650 mm;Locomotive speed = 60 km/hr;Angle between cylinder cranks = 90 o ;Dead load on each wheel = 40 kN; Determine(a) Balancing mass required in the planes of driving wheels if whole of the revolving

mass and 2/3 of the reciprocating mass are to be balanced.(b) (b) Swaying couple;(c) Variation in tractive force;(d) Maximum and minimum pressure on the wheels; and(e) Maximum speed of the locomotive without lifting the wheels from the rails.

57

Given data: mo = 300 kg; mr = 330 kg; mT = mo+ (2/3)mr = 520 kg; Distance between wheels, lW = 1.5 m; Distance between cylinders, lc= a = 0.6 m;D = 1.8 m, R = D/2 = 0.9 m; Velocity, v = ωR = 60x1000/3600= 16.667 m/s; (1 km = 1000 m and 1 hr = 3600 sc); ω = 16.667/0.9 = 18.52 rad/s;

Plane Mass Radius Force/ω2

Distance from RP Couple/ω2 Angle

m (kg) r (m) mr (kg m) l (m) mrl (kg m2) θ (deg)

A (RP) mA 0.65 0.65mA 0 0 -

B 520 0.325 169 0.45 76.05 0

C 520 0.325 169 1.05 177.45 90

D mD 0.65 0.65mD 1.5 0.975mD θ D = 246.8

C

B

1.50

1.05

0.45

A(RP) CB D

Wheel1 Cyl.1 Cyl.2 Wheel2


A D

C

B

58

Plane Couple/ω2 Angle

mrl (kg m2) θ (deg) mrl cosθ mrl sinθ

A (RP) 0 -

B 76.05 0 76.05 0

C 177.45 90 -0.112192 177.45

D 0.975mD θ D

SUM = 75.93781 177.45

θ = Tan -1 (177.45/75.94)= 66.8 deg

θ D 180+66.8= 246.8 deg

R= 193.0156

mD =193.0156/0.975 = 198 kg

Table:

Plane Mass Radius Force/ω2 Angle mr cosθ mr sinθ

m (kg) r (m) mr (kg m) θ (deg)

A (RP) mA 0.65 0.65mA -

B 520 0.325 169 0 169 0

C 520 0.325 169 90 -0.106849 169

D 198 0.65 128.7 246.8 -50.49516 -118.4

SUM= 118.398 50.62

θ= 23.13919

θA = 203.1392deg

R= 128.765kgm

mA= R/0.65= 198kgm59

60

Maximum and Minimum Pressure on wheels:

61

Unit 2: Problem 4- The firing order of a six cylinder vertical four stroke in line engine is1-4-2-6-3-5. The piston stroke is 80 mm and the length of each connecting rod is 180mm. the pitch distance between the cylinder centre lines are 80 mm, 80 mm, 120 mm,8 mm, and 80 mm respectively. The reciprocating mass per cylinder is 1.2 kg and theengine speed is 2400 rpm. Determine the out of balance primary and secondary forcesand couples of the engine taking a plane midway between the cylinder 3 and 4 asreference point.Given Data: n = 6; l = 0.18 m; m r= 1.2 kg; L = 0.08 m;N = 2400 rpm; ω = 2πN/60 = 251.33 rad/s;To find: UPF, UPC, USF, and USC

Interval (angle) between the cranks,θ = 4π/n = 4x180/6 = 120 o

2,5

1,6

4,3

654321

0.120.08 0.08 0.080.08

Space Diagram for Primary Forces

1,6

4,32,5

654321

0.120.08 0.08 0.080.08

Space Diagram for Secondary Forces

62

Table: (for Primary forces and couples)Plane Mass Radius Force/ω2 Distance from RP Couple/ω2 Angle


1 1.2 0.04 0.048 -0.22 -0.01056 902 1.2 0.04 0.048 -0.14 -0.00672 3303 1.2 0.04 0.048 -0.06 -0.00288 5704 1.2 0.04 0.048 0.22 0.00288 2105 1.2 0.04 0.048 0.14 0.00672 6906 1.2 0.04 0.048 0.06 0.01056 450

Table: (for Secondaryforces and couples)Plane Mass Radius Force/ω2 Distance from RP Couple/ω2 Angle


1 1.2 0.04 0.048 -0.22 -0.01056 902 1.2 0.04 0.048 -0.14 -0.00672 5703 1.2 0.04 0.048 -0.06 -0.00288 10504 1.2 0.04 0.048 0.22 0.00288 3305 1.2 0.04 0.048 0.14 0.00672 12906 1.2 0.04 0.048 0.06 0.01056 810

Do graphically and check answers: UPC=0; UPF=0; USC=0; USF=0.See analytical answer also.

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Table: (for Primary couples) Table: (for Primary forces )Plane Couple/ω2 Angle mrl cosθ mrl sinθ Plane Force/ω2 Angle mr cosθ mr sinθ

mrl (kg m2) θ (deg) (kg m2) (kg m2) mr (kg m) θ (deg) (kg m) (kg m)

1 -0.01056 90 0.00000 -0.01056 1 0.048 90 0.00000 0.048002 -0.00672 330 -0.00582 0.00336 2 0.048 330 0.04157 -0.024003 -0.00288 570 0.00249 0.00144 3 0.048 570 -0.04157 -0.024004 0.00288 210 -0.00249 -0.00144 4 0.048 210 -0.04157 -0.024005 0.00672 690 0.00582 -0.00336 5 0.048 690 0.04157 -0.024006 0.01056 450 0.00000 0.01056 6 0.048 450 0.00000 0.04800

SUM= 0.00000 0.00000 SUM= 0.00000 0.00000

Table: (for Secondary couples) Table: (for Secondary forces)PlaneCouple/ω2 Angle mrl cosθ mrl sinθ Plane Force/ω2 Angle mr cosθ mr sinθ

mrl (kg m2)θ (deg) (kg m2) (kg m2) mr (kg m) θ (deg) (kg m) (kg m)

1 -0.01056 90 0.00000 -0.01056 1 0.048 90 0.00000 0.048002 -0.00672 570 0.00582 0.00336 2 0.048 570 -0.04157 -0.024003 -0.00288 1050 -0.00249 0.00144 3 0.048 1050 0.04157 -0.024004 0.00288 330 0.00249 -0.00144 4 0.048 330 0.04157 -0.024005 0.00672 1290 -0.00582 -0.00336 5 0.048 1290 -0.04157 -0.024006 0.01056 810 0.00000 0.01056 6 0.048 810 0.00000 0.04800

SUM= 0.00000 0.00000 SUM= 0.00000 0.00000

Analytical Answer

USC=0; USF=0

UPC=0; UPF=0;

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