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Chapter 7 Balancing of Reciprocating Machines We have already discussed how to obtain inertia forces of linkages (the four-bar and reciprocating engine mechanisms). Such inertia forces cannot be avoided and we have to consider these forces in designing of linkages. (These inertia forces increase stresses in members and loads in bearing). Inertia force 2 αω , where ω is the angular velocity of links. For linkages with very high speed, inertia forces (or loading) results in vibrations, noise emission and sometimes machine failure due to fatigue. Inertia force of a reciprocating mass: y ω 2 A l 3 B x x r+l O 2 r φ Figure 1 A reciprocating mechanism The displacement of the piston is given as 1/2 2 2 sin cos 1 x r l n θ θ = + - with n = l/r (1) The velocity of the piston can be obtained as 0.5 2 2 1 sin 2 sin 2 1 sin x r n n θ ω θ θ - =- + - (2) The acceleration of the piston can be obtained as 2 4 2 1.5 3 2 2 ( 1) cos 2 cos cos 1 sin n x r n n θ θ ω θ θ - - + =- + - (3) In equation (1) the term 0.5 2 2 1 (sin )/ n θ - is expanded bi-nominally to give a harmonic analysis. i.e , xx and x can be represented as the cosine and sine terms of harmonics. The frequencies representing integral multiples of the fundamental frequency are called harmonics, with the fundamental frequency being the first harmonic. For example:
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Feb 07, 2017

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  • Chapter 7 Balancing of Reciprocating Machines

    We have already discussed how to obtain inertia forces of linkages (the four-bar and reciprocating

    engine mechanisms). Such inertia forces cannot be avoided and we have to consider these forces in

    designing of linkages. (These inertia forces increase stresses in members and loads in bearing). Inertia

    force 2 , where is the angular velocity of links. For linkages with very high speed, inertia

    forces (or loading) results in vibrations, noise emission and sometimes machine failure due to fatigue.

    Inertia force of a reciprocating mass:

    y

    2A

    l

    3B

    x

    x

    r+l

    O2

    r

    Figure 1 A reciprocating mechanism

    The displacement of the piston is given as

    1/ 2

    2

    2

    sincos 1x r l

    n

    = +

    with n = l/r (1)

    The velocity of the piston can be obtained as

    0 .5

    2 2

    1 sin 2sin

    2 1 sinx r

    n n

    = +

    (2)

    The acceleration of the piston can be obtained as

    2 42

    1.53 2 2

    ( 1)cos2 coscos

    1 sin

    nx r

    n n

    + = +

    (3)

    In equation (1) the term 0.5

    2 21 (sin ) / n is expanded bi-nominally to give a harmonic analysis. i.e

    ,x x and x can be represented as the cosine and sine terms of harmonics. The frequencies

    representing integral multiples of the fundamental frequency are called harmonics, with the

    fundamental frequency being the first harmonic. For example:

  • 122

    0 02 0n

    fundamental harmonic second harmonic thn harmonic

    A series of harmonic functions are known as the Fourier series. A harmonic function of any arbitrary

    frequency is periodic i.e. it repeats itself at equal intervals of time /2=T , where T is the

    period of excitation. If a function is periodic, it is not necessary that it will be harmonic.Let

    tAx cos= is a harmonic function with period /2 or frequency . Any arbitrary periodic

    functions can be represented by a convergent series of harmonic functions whose frequencies are

    integral multiples of certain fundamental frequency ( )T/20 = .

    Fourier Series:

    1

    0 02 1( ) ( cos sin )

    n

    r rrf t a a r t b r t

    == + + (4)

    with

    / 2

    0

    / 2

    2( )cos

    T

    r

    T

    a f t r t dtT

    = and / 2

    0

    / 2

    2( )sin

    T

    r

    T

    b f t r tT

    =

    T/20 = ; 1,2,3,r =

    where T is the period of function f(t).

    Shaking force: Bearing loads due to inertia forces of rotating/reciprocating masses of mechanism

    causes shaking of the mechanism due to its dynamic nature i.e. it changes with time. This bearing

    loads due to inertia forces of mechanism are called shaking forces. Shaking force can be minimised by

    balancing the inertia forces in opposition to each other such that little or no force is transmitted to

    machine supports. The degree to which this shaking force is undesirable depends upon the frequency,

    , of shaking force and the natural frequency, nf , of flexible members (such as shafts, bearings,

    supports, etc.) through which the force is transmitted. For nf = we have undesirable resonance

    condition.

    Primary and secondary forces can be replaced by rotating masses as:

    1. A mass bm at radius r at A and rotating with crank with the component in horizontal

    direction will be same as by reciprocating mass bm of piston i.e. primary force.

  • 123

    2. A, mass bm at radius )4/( nr and rotating at twice the crank angular velocity . The

    component in the horizontal direction will be same as that due to piston bm i.e. secondary

    force.

    The acceleration of piston can be obtained up to second harmonics as:

    )2cos

    (cos2

    nrAp

    += (5)

    where, n=l/r

    F32

    F32Sin

    O2

    A

    mb2bsin

    mb

    2b

    mb

    mb2cos

    F32 cos

    primary force

    =mbr2cos

    Figure 2 A crack with a counter balance mass

    0cos)sin(cos)sin( 122

    32 =++ TbbmrF b

    with cossincossin 223212 bmrFT b=

    For a single cylinder engine the inertia force due to the reciprocating mass, 43 mmm BB += , will be

    2 cos2cosBF m rn

    = +

    (6)

    where Bm3 is one of the equivalent mass of connecting rod at piston and 4m is the mass of piston.

    For F ve + the inertia force will be directed away from main bearing and for F ve the inertia

  • 124

    force towards the main bearing. Bearing load due to this inertia force will be in the same direction as

    the inertia force.

    In equation (6), various terms can be defined as follows:

    First term:

    cos2rmb (7)

    is called primary force. It reaches its maximum value of 2rmb twice per revolution, when

    0= and .

    Second term:

    nrmB

    2cos2 (8)

    is called secondary force. It reaches its maximum value of 2 /Bm r n four times per revolution of

    crank when 0, / 2, and 3 / 2 = .

    (i) Harmonic motion (Fundamental) is given as tAx cos= , where A is the amplitude and is the

    frequency so that sinx A t = and 2 cosx A t = , where 2 /T = is the period and

    t = .

    The primary inertia force tAmF pi cos2+=

    (ii) Second harmonic motion is given as tAx 2cos= , so that 2 sin 2x A t = and

    24 cos2x A t = .

    The secondary inertia force tAmtAmF si 2cos)4(2cos422 ==

    Both the primary and secondary inertia forces represent a force, which is constant in direction but

    varying in magnitude, and hence cannot be completely balanced by a rotating mass (or which inertia

    force is varying in the direction but is constant in the magnitude).

    Assumption: (i) The replacement of the connecting rod by two masses at its ends maintains

    complete dynamic equivalence (ii) The crank is so designed that the combined centre of mass of

    the crank and the apportioned connecting rod mass, 3m , at the crank pin lies on the axis of rotation,

    and (iii) the engine speed is constant. For multi-cylinder engines, we shall apply these assumptions to

    each constituent cylinder.

  • 125

    Bearing load: In reciprocating engine the bearing load due to the piston inertia can be obtained with

    help of following free body diagram, which is self-explanatory.

    Figure 3 (a) A reciprocating engine

    From ABO2

    sinxq =

    xFT sin34=

    qF .34= 2112 . TqF ==

    21TT =

    Figure 3(b) Free body diagram of the connecting rod

    Figure 3 (c) Free body diagram of the piston

    Figure 3 (d) Free body diagram of the crank

    Figure 3 (e) Free body diagram of bed

    Figure 3 (f) Equivalent system of Figure (e)

  • 126

    The full effect on the engine frame of the inertia of the reciprocating mass is equivalent to the force,

    2 1(cos cos2 )m r n + , along the line of the stroke at O.

    Figure 4 (a) (top) Bearing load due to rotating

    mass (bottom) reaction force on the engine

    frame

    Figure 4 (b) (top) Centrifugal forces on disc (bottom)

    Free body diagram of disc

    From Figure 4, it can be seen that the frame will experience forces: (i) horizontal direction

    cos2em and (ii) vertical direction sin2em . The primary inertia force can be balanced by a

    counter mass bm at radius b such that b Bm b m r= .

    Figure 5 A reciprocating machine with a counter balance mass

    The component of inertia force bm along the line of stroke is cos2bmb . But a force

    sin2bmb is then introduced in a direction (perpendicular) to line of stroke. In most practical

    applications, the reciprocating mass is partially balanced by a rotating mass bm , reducing the inertia

    force in the line of stroke and direction, such that

    bmrcm bB =

  • 127

    where c is the fraction of reciprocating mass to be balanced and mB is the mass of the slider. The

    unbalanced primary force along the line of action is given by

    )coscos( 22 bmrm bB = or coscos.22 rcmrm Bb

    or cos)1( 2rmc B (9)

    The unbalance force in the vertical direction

    sin2bmb = sin2rcmB (10)

    Resultant unbalance force at any instant

    = 22222 sincos)1( ccrmB + (11)

    If the revolving and reciprocating (primary force) masses to be balanced simultaneously then,

    rcmmrcmrmbm BABAb )( +=+= (12)

    where Am is the mass of rotating components. The direction of resultant force at any instant is given

    as

    tan

    1cos)1(tan

    2

    2

    c

    c

    rmc

    rcm

    B

    B

    =

    = (13)

    (a)

    (b)

    (c) Equivalent of Figure (c)

    Figure 6 Frame reaction in the horizontal direction

    Resultant R will be rotating in opposite direction to crank (i.e. cw). For 1

    2c = ; R is constant and

    rotating with in cw direction. So primary force can be balanced as shown in Figure 7.

  • 128

    Figure 7

    Conditions of the balance in a multi cylinder engine:

    c2 c1

    c3 c4

    m2m3 m4

    m1

    stroke

    12

    3

    4

    2

    3

    m4

    m3

    m2m1

    Figure 8 Multi-cylinder in-line engine

    In Figure 8 1 , 2 and 3 are 4 are angular positions of the multi cylinder engine (Four engines in

    Figure 8) and 2 , 3 and 4 are the phase difference of cranks 2, 3, 4 with respect to 1. The line of

    stroke of each cylinder is assumed to in the horizontal plane through the crankshaft axis. The inertia

    force of reciprocating masses will be situated in same horizontal plane. For each cylinder: r is the

    crank radius, l is the connecting rod length, m is the total reciprocating mass (the piston and the

    connecting rod) and i is the angle between cranks, which depends upon the firing order. Subscript i

    represent the cylinder number (i = 1, 2, 3, 4).

    Equilibrium of Primary forces:

    Conditions for the complete equilibrium of primary forces are:

    1. The algebraic sum of primary forces is zero and

    2. The algebraic sum of their moments with respect to any of the transverse plane ( to

    crankshaft) is equal to zero.

  • 129

    The resultant primary shaking force will be

    nnnp rmrmrmrmF coscoscoscos2

    32

    3322

    2212

    11 ++++=

    ==

    n

    iiiirm

    1

    2 cos (14)

    with ii += 1 ni = 3,2,1 (15)

    where 1 is the variable and i are constants and 01 = . Substituting (14) into (15), we get

    ( ) += iiip rmF 12 cos (16)

    or ===

    n

    iiii

    n

    iiii rmrm

    11

    2

    11

    2 sinsincos.cos

    or = =

    n

    i

    n

    iiiiiii rmrm

    1 11

    21

    2 )sin(sin)cos(cos (17)

    The primary force, pF , will be maximum, when 01

    =

    pF i.e.

    =

    cos

    sintan

    11

    mr

    mr (18)

    For pF to be zero for all values of 1 , then from eqn. (17), we get

    0cos1

    =

    =

    n

    i

    mr (19)

    and

    ==

    n

    i

    mr1

    0sin (20)

    Similarly, for moments, we can write

    0cos1

    =

    =

    n

    i

    mra (21)

    and

    0sin1

    =

    =

    n

    i

    mra (22)

    where a is the distance from the center line (line of stroke) of any cylinder to an arbitrarily selected

    reference plane ( to the crank shaft axis). Thus, for complete balancing of primary forces and

    moments, eqns. (19) to (22) should be satisfied.

  • 130

    Equilibrium of Secondary Forces: Conditions for complete equilibrium of the secondary forces

    are:

    1. The algebraic sum of the secondary forces is zero and

    2. The algebraic sum of their moments with respect of any transverse plane is also zero.

    Thus the resultant of secondary forces:

    2

    1

    cos2n

    s i i

    i i

    rF m r

    i

    =

    with 1 i

    i i

    r

    n l

    =

    (23)

    or ( )2 11

    cos2n

    s i i i

    i i

    rF m r

    l

    = +

    or 2 2

    2 2

    1 1cos2 cos2 sin 2 sin 2smr mr

    Fl l

    = (24)

    (i) sF is maximum, when

    1

    0sF

    =

    which gives

    2

    1

    1 2

    sin 21tan

    2cos2

    mr

    l

    mr

    l

    =

    (25)

    (ii) For 0=sF i.e. criteria for the balancing of the secondary force is (from eqn. 24)

    = 02cos2

    l

    mr (26)

    and

    = 02sin2

    l

    mr (27)

    (iii) For balancing of moments, we have:

    = 02sin.2

    al

    mr (28)

    and

    = 02cos.2

    al

    mr (29)

  • 131

    Thus for complete secondary balancing eqns. (26)-(29) has to be satisfied. Fundamental eqns. (19) to

    (22) and eqns. (26) to (29) express completely the primary and secondary balances for reciprocating

    engine with n-cylinders arranged in a line on one side of the crank shaft with the cylinder centre lies

    all in the plane containing the crankshaft axis. Firing order affects the balance of multi cylinder

    engine: For example Two-stroke engine: One cycle of operation in one revolution of crankshaft.

    Interval between the crank= n/2 . Four- stroke engine: One cycle of operation in two revolutions of

    crankshaft. Interval between the crank= n/4 , where n is numbers of cylinder in the multi-cylinder

    engine.

    (A) Crank arrangement for two-stroke engine:

    (1) The crank arrangement can be so chosen to make the cylinder firing not in a sequential order

    but distribute this on either side of the crankshaft with respect to its middle plane.

    (2) Once crank arrangement is fixed there may be several choice on firing order itself (see

    specially four- stroke engine).

    Cranks are rotating in the cw direction. Firing will start say from engine 1. Next cylinder, which is

    2/ in phase in ccw, will be ready for firing (because of completion of compression stroke). Next

    firing can be done in engine 4 then in 2 and finally in 3. Hence, the firing order will be: 1-4-2-3.

    Another version can be 1-3-2-4. For given crank arrangement (see in Figure 9 for 4, 5 and 6 cylinders)

    for two- stroke engine only unique firing order is possible i.e. for 4 cylinder: 1-4-2-3, 5 cylinder: 1-5-

    2-4-3 and 6 cylinder: 1-6-2-5-3-4.

    1

    2

    3490

    Firing order 1-4-2-3

    1 2 3 4

    (a) Four cylinder engine (interval=2

    4 2

    = )

  • 132

    1

    5

    24

    372

    1 2 3 4 5

    Firing order 1-5-2-4-3

    (b) Five cylinder engine: interval=2

    725

    o =

    1 2 3 4 5 61

    4

    3

    5

    2

    6

    60

    Firing order 1-6-2-5-3-4

    (c) Six cylinder engine: interval=2

    606

    o =

    Figure 9 Two-stroke engine

    (B) Four stroke engine: For a given crank arrangement there may be several firing order possible.

    Refer to Figure 10. For 4- cylinder the firing order could be: 1-4-2-3 or 1-3-2-4; for 5-cylinder the

    firing order could be: 1-5-3-4-2 (only one) and for 6- cylinder the firing order could be: 1-6-2-5-3-4,

    1-6-4-5-3-4 or 1-3-4-5-6-2.

    1 2

    4 3

    1 2 3 4

    Firing order 1-4-2-3

    (a) Four cylinder engine: internal 4

    4

    = =

  • 133

    1

    4

    32

    3

    1 2 3 4 5

    Firing order 1-5-2-4-3

    (b) Five cylinder: interval 4

    1445

    = =

    4

    53

    2

    144

    1-0

    5-144

    2-288

    4-432 72

    3-576 216

    1 5

    63

    4

    2

    Firing order 1-6-2-5-3-4

    1 2 3 4 5 6

    (c) Six cylinder: interval4

    1206

    = =

    Figure 10 Four-stroke engine

    Balancing of Radial Engines (Direct and reverse crank method)

    B

    A

    O2

    Figure 11(a) Slider-crack mechanism

    A

    A

    O

    Direct crank

    Reverse crank

    Fig. 1(a)

    Figure 11(b) Primary direct and reverse crank

    Reverse crank OA is an image of direct crank OA. Direct crank and reverse crank rotates in opposite

    direction. Let rm is the reciprocating mass at B, then primary force will be cos2rmr . This is

    2

    rm

    2

    rm

  • 134

    equivalent to the component of the centrifugal force produced by mass rm at A. Let total

    reciprocating mass rm is divided into two parts i.e. 2/rm for each in the direct crank and in the

    reverse crank as shown in Fig. 11(a). The centrifugal force acting on primary direct and reverse crank

    is 2

    2

    rm r . Total component in the direction of line of stroke 22 cos2

    rm r =

    2 cosrm r = . The

    component in vertical direction cancel each other i.e. 2 sin2

    rm r .

    The secondary force is ( )22 cos24

    r

    rm

    n

    , where rm is replaced by the mass of 2/rm each

    placed at c and c as shown in Figure 11(c) with 4r

    oc ocn

    = = =

    crank radius of secondary crank.

    Secondary crank oc rotates at 2 rad/sec . Secondary crank co rotates at 2 rad/sec .

    c

    o

    2

    2

    reversecrank

    c

    directcrank

    Figure 11(b) Secondary direct & reverse crank

    Three cylinders radial engine: Line of stroke is 120. All three lines of stroke are in one plane.

    Shaking forces from reciprocating parts of each cylinder will be acting along the line of strokes in a

    single plane. Let rm be the mass of reciprocating part of each cylinder.

    Cylinder no. Direction of rotation Crank angle Direction of rotation Crank angle

    1 ccw 0 cw 0

    2 ccw 240 cw 120

    3 ccw 120 cw 240

    Primary unbalance: mass3

    2

    rm= ; crank radius = r; speed= cw; 232

    rp

    mF r= . It can be balanced

    by a balancing mass bm at radius b i.e. 2

    3 rb

    mbm = and placed at opposite to the crank.

  • 135

    1

    Line of stroke

    connecting rod

    for cylinder 1

    connecting rod for cylinder 3

    connectingrod forcylinder2

    120

    120120

    32lineofstroke

    Figure 12(a) Three cylinders radial engine

    1,2,3 1

    3 2

    Direct primarycranks

    Reverse primarycranks

    Figure 12(b)

    1

    3 2

    1,2,3

    Direct Secodary cranks(Balanced by itself)

    Reverse secondarycranks

    22

    Figure 12(c)

    Secondary unbalance:mass2

    3 rm= , radius n

    r

    4= ,

    r

    ln = and speed 2 ccw= + . Hence,

    ( )23 22 4

    rs

    m rF

    n =

    ; by gear train we have to obtain secondary balancing.

    rm

    rm rm

    2

    rm

    2

    rm

    23 r

    m

    2

    rm

    2

    rm

    23 r

    m

    2

    rm

    2

    rm

  • 136

    Tutorial Problems:

    (1) A single cylinder horizontal oil engine has a crank 190.5 mm long and a connecting rod 838.2 mm

    long. The revolving parts are equivalent to 489.3 N at crank radius and the piston and gudgeon pin

    weigh 400.32 N. The connecting rod weighs 511.52 N and its center of gravity is 266.7 mm from the

    crank pin centre. Revolving balance weights are introduced at a radius of 215.9 mm on extensions of

    the crank webs in order to balance all the revolving parts and one-half of the reciprocating parts. Find

    the magnitude of the total balance weight and neglecting the obliquity of the connecting rod.

    ( ) . . / 0i e l r , the nature and magnitude of the residual unbalanced force on the engine. RPM=300 cw.

    (Answer: 987.456 N, 5390.976 N and revolving at 300 rpm ccw).

    (2) A four cylinder Marin oil engine has the cranks arranged at angular interval of 90. The inner

    cranks are 1.22 m apart and are placed symmetrically between the out cranks, which are 3.05 m apart.

    Each crank is 457.2 mm long, the engine runs at 90 rpm; and the weight of the reciprocating parts for

    each cylinder is 8006.4 N. In which order should the cranks be arranged for the best balance of the

    reciprocating masses, and what will then be the magnitude of the unbalanced primary couple?

    (Answer: 1,4,2,3;42.82 kN-m)

    (3) In a three-cylinder radial engine all three connecting rods act on a single crank. The cylinder

    center lines are set at 120 , the weight of the reciprocating parts per line is 22.24 N, the crank length

    is 76.2 mm, the connecting rod length is 279.4 mm and the rpm are 1800 cw. Determine with regard

    to the inertia of the reciprocating parts (a) the balance weight to be fitted at 101.6 mm radius to give

    primary balance (b) the nature and magnitude of the secondary unbalanced force (c) whether the

    fourth and sixth order forces are balanced or unbalanced. (Answer: (a) 25 N (b) Constant magnitude

    2508.67 N, 3600 rpm ccw (c) th4 harmonic is unbalanced, th6 harmonic is balanced).

    Primary Balance of Multi-cylinder In-line Engines:

    The conditions in order to give primary balance of the reciprocating masses are:

    = 0cos. 2 rmrec (i.e. Algebraic sum of primary forces shall be zero) and

    = 0cos. 2. ramrec (i.e. Algebraic sum of the (primary force) couple about any point in the plane of force shall also be zero).

    where a is the distance of the plane of rotation of the crank from a parallel reference plane. Above

    two equations should satisfy for the angular position of the crankshaft relative to the dead center

    (reference line). The primary force due to the reciprocating mass is identical with the component

    parallel to the line of stroke of the centrifugal force produced by an equal mass attached to and

  • 137

    revolving with the crank pin. Let the reciprocating masses are attached to crank pin at A, B, C and D

    are revolving with crank as shown in Figure 13(a). The force polygon (centrifugal forces) may be

    drawn. (i.e. oa, ab, bc and cd). Primary forces of the individual cylinder are equal to the component of

    oa, ab, bc and cd along the line of stroke (say PQ). Now ef, fg ,gh and he are primary forces and since

    algebraic sum of these are equal to zero (for this configuration), the engine is balanced for this

    configuration. But suppose cranks turns clockwise through an angle , then the effect will be same as

    crank shaft is fixed and line of stroke is rotating ccw by angle. Let us say SP is now new line of

    stroke. Now 1, , and kl lm mn nk are primary forces and these algebraic sum is not zero, but it is equal

    to 1kk (primary force is not balanced). The primary force will be only balanced when 1k coincides

    with k or point d with o in force polygon. i.e. the centrifugal force polygon is a closed one.

    A B

    CD

    O

    Four cylinder engineOA,OB,OC,OD are four cranks

    Figure 13(a) Four cylinder engine

    b

    a

    oS

    lm k k

    1 np

    e

    h

    fc

    d

    Qtoline ofstroke

    Force polygon of the centrifugalforces produced when revolvingmasses respectively equal toreciprocating masses at crank pin

    Figure 13(b) Force polygon

    In similar way the primary couple can only be balanced if the couple polygon for the corresponding

    centrifugal forces is closed. Hence, if a system of reciprocating masses is to be primary balance, the

    system of revolving masses, which is obtained by substituting an equal revolving mass at the crank

    pin for each reciprocating mass, must be balanced. The problem of the primary balance of

    reciprocating masses may therefore be solved by using the method for revolving masses.

    Secondary Balance of Multi-cylinder In-line Engine: The conditions for the complete secondary

    balance of an engine are

    ( )22 cos2 04

    rec

    rm

    n =

    and

  • 138

    ( )22 cos2 04

    rec

    rm a

    n =

    for all angular positions of the crank shaft relative to the line of stroke. As for the primary balance,

    these conditions can only be satisfied if the force and couple polygons are closed for the

    corresponding system of revolving masses. If to each imaginary secondary crank (angular

    velocity= 2 and crank length ( )/ 4r n ) a revolving mass is attached, equal to the corresponding

    reciprocating mass, then the system thus obtained must be completely balanced.

    Example 1: A shaft carries four masses in parallel planes A, B, C and D in this order, along it. The

    masses at B and C are 18 kg and 12.5 kg, respectively, and each has an eccentricity of 6 cm. The

    masses at A and D have an eccentricity of 8 cm. The angle between the masses at B and C is 1000,

    and that between the masses at B and A is 1900 (both angles measured in the same direction). The

    axial distance between the planes A and B is 10 cm. And between B and C is 20 cm. If the shaft is in

    complete dynamic balance, determine (i) the masses at and D (ii) the distance between the planes C

    and D, and (iii) the angular position of the mass at D.

    Solution:

    B (Reference)

    ld 1000

    D

    10cm 20cm CDl 1900 0

    A B C D

    A Figure 14 Multi-cylinder engine

    Given data are: 18 6 ; 12.5 6 ; 8 B B C C A Dm kg r cm m kg r cm r r cm= = = = = =

    Aim is to obtain: ? ?; ? ? (with respect to B)A D CD Dm m l = = = =

    Condition of dynamics balancing are

    cos 0; sin 0; cos 0; sin 0mr mr mrl mrl = = = =

    Taking reference as the crank B for the phase measurement (i.e. B = 0) and for the moment the

    cylinder A as reference and are as follows

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    cos cos 0; sin sin 0

    cos cos cos 0; sin sin sin 0

    B B B C C C C d d d C C C C d d d d

    A A A B B C C C d d d A A A C C C d d d

    m r l m r l m r m r l m r l

    m r m r m r m r n m r m r m r

    + + = + =

    + + + = + + =

    Hence,

    1080 390.7 8 cos 0; 2215.82 8 sin 0

    7.88 108 13 8 cos 0; 1.39 73.86 8 sin 0

    d d d d d d

    A d d A d d

    m l m l

    m m m m

    + = + =

    + + = + + =

    or 8 cos 689.3 (1)

    8 sin 2215.82

    d d d

    d d d

    m l

    m l

    =

    = (2)

    7.88 8 cos 95 (3)

    1.39 8 sin 73.86 (4)

    A d d

    A d d

    m m

    m m

    + =

    + =

    Equations (1) and (2) will give tan 3.2146d = 252.72 so in third quadrant.d = So that

    90=ddlm (5)

    Equations (3) and (4) gives

    7.88 2.3763 95; 1.39 7.639 73.86

    9.67 kg; 7.91 kg; 36.66 cm; 6.66 cm

    A d A d

    A d d cd

    m m m m

    m m l l

    = =

    = = = =

    Using Tabular method calculation are much easier. Take moments about plane A and plane D with

    phase reference as crank B. Choosing plane D as first reference as advantage that mD and D will not

    appear in two moment equation and mA and lCD can be calculated from this two equation directly. So

    choice of reference plane should be such that maximum number of unknown can be eliminated (at

    least two). Second reference plane can be chosen as A to eliminate mA from equation.

    Example 2: By direct and reverse cranks method, consider the W-engine in which there are three

    rows of cylinders. An engine of this type has four cylinders in each row and the crankshaft is of the

    normal flat type with one connecting rod from each row coupled directly to each crankpin. The

    middle row of cylinders is vertical and the other two rows are inclined at 600 to the vertical. The

    weight of the reciprocating parts is 26.69 N per cylinder; the cranks are 7.6 cm long. The connecting

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    rods 27.9 cm long and the rpm 2000 cw. Find the maximum and minimum values of the secondary

    disturbing force on the engine (Figure 15).

    Figure-15 W-engine Solution:

    26.69 N; 7.6 cm; 3.67; 2.72 kg; 27.9 cm; 2000 rpm 209.43 rad/secrec recl

    w r n m l cr

    = = = = = = = =

    2

    1 3

    600 600

    Figure-16 W-engine

    Secondary cracks

    2

    (m/2) 2 (m/2)

    600 600

    (a) Direct (b) reverse

    2 2 (m/2) (m/2) Resultant Unbalance

    mmm

    =

    + 5.02

    22

    424

    )2(2

    22 r

    m

    n

    r

    m=

    +

    n

    m

    n

    rm

    22

    4)2( =

    m/2

    2 2

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    Total 4 rows: 2

    4 9882.2 Nm r

    n= Total 4 rows:

    2

    2 4941.1 Nm r

    n=

    (i). At present position Resultant is minimum:

    2

    2

    Minimum unbalance 24941.1 N (2 )2 4

    m r

    n

    = = +

    ;

    0 0 0

    2 2 2 20, 90 ; 180 , 270 = = = =

    (ii). For 0 0 0 0 02 1 345 (or 60 45 or 300 45 ) = = + = + the resultant is maximum.

    2

    m

    m/2

    2

    maximum unbalance 23

    (2 ) 14823.3 N2 4

    m r

    n= =

    0 0 0 0

    4 4 4 445 , 135 ; 225 , 315 = = = =

    Exercise Problems:

    1. A four-cylinder marine oil engine has the cranks arranged at angular intervals of 900. The inner

    crank are 1.2 m apart and are placed symmetrically between the outer cranks, which are 3.0m apart.

    Each crank is 45.7 cm long, the engine runs at 90 rpm, and the weight of the reciprocating parts for

    each cylinder is 8006 N. In which order should the cranks be arranged for the best balance of the

    reciprocating masses, and what will then be the magnitude of the unbalanced primary couple?

    2. The reciprocating masses for three cylinders of a four-crank engine weight 3047, 5078 and 8125

    kgs and the centerlines of the se cylinders are 3.66 m, 2.59 m. and 1.07 m. respectively from that of

    the fourth cylinder. Find the fourth reciprocating mass and the angles between the cranks so that they

    may be mutually balanced for primary forces and couples. If the cranks are each 0.61 m long, the

    connecting distributing rods 2.75 m long and the rpm 60, find the maximum value of the secondary

    distributing force and crank positions at which it occurs.

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