www.getmyuni.com DYNAMICS OF MACHINES
BALANCING OF RECIPROCATING MASSES
SLIDER CRANK MECHANISM:
PRIMARY AND SECONDARY ACCELERATING FORCE:
Acceleration of the reciprocating mass of a slider-crank mechanism is given by,
p = pistonofonAcceleratia
2 cos2θ cosr +θω= −−−−−−−−− 1)( n l nWhere = r
And, the force required to accelerate the mass ‘m’ is
cos2θ 2 +θω= i cosrmF n cos2θ rmcosrm ω+θω= 22 −−−−−−−−− 2)( n
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The first term of the equation (2) , i.e. ω2 cosrm θ is called primary accelerating cos2θ force the second term rm ω2 is called the secondary accelerating force. n
Maximum value of primary accelerating force is rm ω2 rm ω2 And Maximum value of secondary accelerating force is n Generally, ‘n’ value is much greater than one; the secondary force is small compared to primary force and can be safely neglected for slow speed engines.
In Fig (a), the inertia force due to primary accelerating force is shown.
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In Fig (b), the forces acting on the engine frame due to inertia force are shown.
At ‘O’ the force exerted by the crankshaft on the main bearings has two components, h v horizontal F21 and vertical F21 .
h F21 is an horizontal force, which is an unbalanced shaking force.
v v F21 and F41 balance each other but form an unbalanced shaking couple.
The magnitude and direction of these unbalanced force and couple go on changing with angle θ. The shaking force produces linear vibrations of the frame in horizontal direction, whereas the shaking couple produces an oscillating vibration. h The shaking force F21 is the only unbalanced force which may hamper the smooth running of the engine and effort is made to balance the same. However it is not at all possible to balance it completely and only some modifications can be carried out.
BALANCING OF THE SHAKING FORCE:
Shaking force is being balanced by adding a rotating counter mass at radius ‘r’ directly opposite the crank. This provides only a partial balance. This counter mass is in addition to the mass used to balance the rotating unbalance due to the mass at the crank pin. This is shown in figure (c).
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The horizontal component of the centrifugal force due to the balancing mass is ω2 cosrm θ and this is in the line of stroke. This component neutralizes the unbalanced reciprocating force. But the rotating mass also has a component ω2 sinrm θ perpendicular to the line of stroke which remains unbalanced. The unbalanced force is zero at θ = 00 or 1800 and maximum at the middle of the stroke i.e. θ = 900. The magnitude or the maximum value of the unbalanced force remains the same i.e. equal to rm ω2 . Thus instead of sliding to and fro on its mounting, the mechanism tends to jump up and down. 2 To minimize the effect of the unbalance force a compromise is, usually made, is of the 3 1 3 reciprocating mass is balanced or a value between to . 2 4
If ‘c’ is the fraction of the reciprocating mass, then
= 2 θcosωrmcmassthebybalancedforceprimaryThe
and
= − 2 θcosωrmc)1(massthebyunbalancedforceprimaryThe
Vertical component of centrifugal force which remains unbalanced = 2 θsinωrmc
In reciprocating engines, unbalance forces in the direction of the line of stroke are more dangerous than the forces perpendicular to the line of stroke.
Resultant unbalanced force at any instant
[][]2 2 +−= 2 θsinωrmcθcosωrc)m(1 2
1 The resultant unbalanced force is minimum when, c = 2
This method is just equivalent to as if a revolving mass at the crankpin is completely balanced by providing a counter mass at the same radius diametrically opposite to the
crank. Thus if m P is the mass at the crankpin and ‘c’ is the fraction of the reciprocating
mass ‘m’ to be balanced , the mass at the crankpin may be considered as +mmc P which is to be completely balanced.
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Problem 1: A single –cylinder reciprocating engine has a reciprocating mass of 60 kg. The crank rotates at 60 rpm and the stroke is 320 mm. The mass of the revolving parts at 160 mm radius is 40 kg. If two-thirds of the reciprocating parts and the whole of the revolving parts are to be balanced, determine the, (i) balance mass required at a radius of 350 mm and (ii) unbalanced force when the crank has turned 500 from the top-dead centre.
Solution: Given: m = mass of the reciprocating parts = 60 kg
=== mm320stroketheoflengthLrpm,60N 2 ckg,40m === mm350r, 3 cP
(i) Balance mass required at a radius of 350 mm
2πN 2πx60 ω === rad/sπ2 We have, 60 60 L 320 r === mm160 2 2
Mass to be balanced at thecrankpin = M
2 mmcM =+=+= kg804060x P 3
Mr mthereforerMrm ccc == and rc 160x80 mi.e. == 5736. kg c 350
(ii) Unbalanced force when the crank has turned 500 from the top-dead centre.
Unbalanced force at θ=500 [][]() 2 2 +−= 2 θsinωrmcθcosωrmc1 2
2 2 2 2 0 2 2 0 1 −= () 50cos2πx0.16x60x + () 50sin2πx0.16x60x 3 3 = N209.9
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Problem 2: The following data relate to a single cylinder reciprocating engine: Mass of reciprocating parts = 40 kg Mass of revolving parts = 30 kg at crank radius Speed = 150 rpm, Stroke = 350 mm. If 60 % of the reciprocating parts and all the revolving parts are to be balanced, determine the, (i) balance mass required at a radius of 320 mm and (ii) unbalanced force when the crank has turned 450 from the top-dead centre.
Solution:
Given : m = mass of the reciprocating parts = 40 kg
P ,kg30m ==== mm350stroketheoflengthLrpm,150N
c == mm320r,%60c
(i) Balance mass required at a radius of 350 mm
2πN 2πx150 ω === rad/s15.7 We have, 60 60 L 350 r === mm175 2 2
= Mass to be balanced at thecrank pin M
P =+=+= kg543040x0.60mmcM
Mr mthereforerMrm ccc == and rc 175x54 mi.e. == kg29.53 c 320
(ii) Unbalanced force when the crank has turned 450 from the top-dead centre.
Unbalanced force at θ=450 [][]() 2 2 +−= 2 θsinωrmcθcosωrmc1 2
2 2 []()()2 0 +−= []()2 45sin15.7x0.175x40x0.6045cos15.7x0.175x40x0.601 0 = N880.7
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SECONDARY BALANCING:
cos2θ Secondary acceleration force is equal to rm ω2 −−−−−− 1)( n 1 Its frequency is twice that of the primary force and the magnitude n times the magnitude of the primary force. cos2θ The secondary force is also equal to 2ω)(rm 2 −−−−−− 2)( 4n
Consider, two cranks of an engine, one actual one and the other imaginary with the following specifications.
Actual Imaginary Angular velocity ω 2ω r Length of crank r 4n Mass at the crank pin m m
Thus, when the actual crank has turned through an angle θ = ω t , the imaginary crank would have turned an angle 2θ =2ω t
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(2ωrm )2 Centrifugal force induced in the imaginary crank = 4n
(2ωrm )2 Component of this force along the line of stroke is = 2θcos 4n
r Thus the effect of the secondary force is equivalent to an imaginary crank of length 4n rotating at double the angular velocity, i.e. twice of the engine speed. The imaginary crank coincides with the actual at inner top-dead centre. At other times, it makes an angle with the line of stroke equal to twice that of the engine crank. The secondary couple about a reference plane is given by the multiplication of the secondary force with the distance ‘ l ’ of the plane from the reference plane.
COMPLETE BALANCING OF RECIPROCATING PARTS
Conditions to be fulfilled: 1. Primary forces must balance i.e., primary force polygon is enclosed. 2. Primary couples must balance i.e., primary couple polygon is enclosed. 3. Secondary forces must balance i.e., secondary force polygon is enclosed. 4. Secondary couples must balance i.e., secondary couple polygon is enclosed. Usually, it is not possible to satisfy all the above conditions fully for multi-cylinder engine. Mostly some unbalanced force or couple would exist in the reciprocating engines.
BALANCING OF INLINE ENGINES:
An in-line engine is one wherein all the cylinders are arranged in a single line, one behind the other. Many of the passenger cars such as Maruti 800, Zen, Santro, Honda-city, Honda CR-V, Toyota corolla are the examples having four cinder in-line engines.
In a reciprocating engine, the reciprocating mass is transferred to the crankpin; the axial component of the resulting centrifugal force parallel to the axis of the cylinder is the primary unbalanced force.
Consider a shaft consisting of three equal cranks asymmetrically spaced. The crankpins carry equivalent of three unequal reciprocating masses, then
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Primary force ∑ m rω2 cos θ −−−−−−−−−−−−−= (1)
= ∑ 2 − − − − − − − − − − − − − (2)θcoslωrmcouplePrimary
(2ω)2 ∑ rmforceSecondary −−−−−−−−−−−−−= (3)2θcos n4
(2ω)2 rmcoupleSecondary And = ∑ rmcoupleSecondary 2θcosl n4 ω2 ∑ rm −−−−−−−−−−−−−= (4)2θcosl n
GRAPHICAL SOLUTION:
To solve the above equations graphically, first draw the ∑ m r cos θ polygon ( ω2 is
common to all forces). Then the axial component of the resultant forces r θ)cosF( multiplied by ω2 provides the primary unbalanced force on the system at that moment. This unbalanced force is zero when θ =900 and a maximum when θ =00 .
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If the force polygon encloses, the resultant as well as the axial component will always be zero and the system will be in primary balance. Then,
∑ = ∑ FandF VPhP =00
To find the secondary unbalance force, first find the positions of the imaginary secondary (2ω)2 ω2 cranks. Then transfer the reciprocating masses and multiply the same by or 4n n to get the secondary force. In the same way primary and secondary couple ( m r l ) polygon can be drawn for primary and secondary couples.
Case 1: IN-LINE TWO-CYLINDER ENGINE
Two-cylinder engine, cranks are 1800 apart and have equal reciprocating masses.
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Taking a plane through the centre line as the reference plane,
2 [ ] =++= 0θ)180(cosθcosωrmforcePrimary
2 l l 2 ωrmcouplePrimary θcos −+= + = θcoslωrmθ)180(cos 2 2
Maximum values are mrω2 l at = 180and0θ 00
ωrm 2 ωr2m 2 forceSecondary []θ)2360(cosθ2cos =++= θ2cos n n
2 0000 Maximum values are ωr2m when = 540and360,180,0θ2 n = 270and180,90,0θor 0000
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ωrm 2 l l coupleSecondary 2θcos −+= + = 0)2θ360(cos n 2 2
ANALYTICAL METHOD OF FINDING PRIMARY FORCES AND COUPLES
• First the positions of the cranks have to be taken in terms of θ 0 • The maximum values of these forces and couples vary instant to instant and are equal to the values as given by the equivalent rotating masses at the crank pin.
If a particular position of the crank shaft is considered, the above expressions may not give the maximum values. For example, the maximum value of primary couple is mrω2 l and this value is obtained at crank positions 00 and 1800. However, if the crank positions are assumed at 900 and 2700, the values obtained will be zero.
• If any particular position of the crank shaft is considered, then both X and Y components of the force and couple can be taken to find the maximum values.
For example, if the crank positions considered as 1200 and 3000, the primary couple can be obtained as
l l 0002 ωrmcomponentX 120cos −+=− ()+ 120180cos 2 2 1 −= 2 lωrm 2
l l 0002 ωrmcomponentY 120sin −+=− ()+120180sin 2 2 3 = 2 lωrm 2
2 2 1 2 3 2 Therefore, couplePrimary −= lωrm + lωrm 2 2 = 2 lωrm
Case 2: IN-LINE FOUR-CYLINDER FOUR-STROKE ENGINE
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This engine has tow outer as well as inner cranks (throws) in line. The inner throws are at 0 180 to the outer throws. Thus the angular positions for the cranks are θ 0 for the first, +θ180 00 for the second, +θ180 00 for the third and θ 0 for the fourth.
DYNAMICS OF MACHINES 13 VIJAYAVITHAL BONGALE
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FINDING PRIMARY FORCES, PRIMARY COUPLES, SECONDARY FORCES AND SECONDARY COUPLES:
Choose a plane passing through the middle bearing about which the arrangement is symmetrical as the reference plane.
002 [ +++++= θcosθ)180(cosθ)180(cosθcosωrmforcePrimary ] = 0
3l l θcos 0 ++ θ)180(cos 2 2 = ωrmcouplePrimary 2 l 0 3l −+ θ)180(cos −++ θcos 2 2 = 0
ωrm 2 0 ++ θ)2360(cosθ2cos forceSecondary = n 0 +++ θ2cosθ)2360(cos ωr4m 2 = θ2cos n
ωrm 2 valueMaximum = n 0000 = or054and360,018,02θat = 027and180,09,0θ 0000
3l l 2θcos 0 ++ )2θ360(cos ωrm 2 2 2 coupleSecondary = = 0 n l 3l −+ 0 )2θ360(cos −++ 2θcos 2 2
Thus the engine is not balanced in secondary forces.
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Problem 1:
A four-cylinder oil engine is in complete primary balance. The arrangement of the reciprocating masses in different planes is as shown in figure. The stroke of each piston is 2 r mm. Determine the reciprocating mass of the cylinder 2 and the relative crank position.
Solution:
Given:
4321 ==== kg480mkg,590m,?mkg,380m L r2 lengthcrank === r 2 2
Cent. Distance Couple/ ω2 Mass (m) Radius (r) Force/ω2 from Ref Plane ( m r l ) kg m (m r ) plane ‘2’ kg m2 kg m m 1 380 r 380 r -1.3 -494 r 2(RP) m2 r m2 r 0 0 3 590 r 590 r 2.8 1652 r 4 480 r 480 r 4.1 1968 r
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Analytical Method:
0 Choose plane 2 as the reference plane and θ3 =0 . . Step 1: Resolve the couples into their horizontal and vertical components and take their sums.
Sum of the horizontal components gives
− θ + 0 + cosrcosrcosr θ = 0196801652494 1 4 41 −−−−−−−−−θ+=θ+ 119681652494 )(coscos.,e.i
Sum of the vertical components gives
− θ + 0 + sinrsinrsinr θ = 0196801652494 1 4 41 −−−−−−−−−θ=θ 21968494 )(sinsin.,e.i
Squaring and adding (1) and (2), we get
2 2 2 ( ) ( 4 ) ++= ( θsin1968θcos19681652494 4 ) .,e.i
22 2 2 ()()()()44 +++= θsin1968θcos1968θcos1968x1652x21652494 4 get,wesolvingOn
00 44 =−= 192.1or167.9θand0.978θcos
0 Choosing one value, say 4 =167.9θ
Dividing (2) by (1), we get
1968 sin(167.9 0 ) +412.53 θtan = = −= 1.515 1 + 0 )(167.9cos19681652 − 272.28
0 1 = 123.4θi.e.,
Step 2:
Resolve the forces into their horizontal and vertical components and take their sums.
Sum of the horizontal components gives
0 + θ + + 00 ).(cosrcosrcosrm).(cosr =0916748005904123380 22 22 −−−−−−−−−−−−−=θ 3588 )(.cosmor 16 www.getmyuni.com
Sum of the vertical components gives
0 + θ + + 00 ).(sinrsinrsinrm).(sinr =0916748005904123380 22 22 −−−−−−−−−−−−−−=θ 49417 )(.sinmor
Squaring and adding (3) and (4), we get
2 = 1427 Anskg.m
− 417.9 θtan = −= 4.72 Dividing (4) by (3), we get 2 88.5 0 2 = 282θo Ansr
Graphical Method:
Step 1: Draw the couple diagram taking a suitable scale as shown.
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mandm,m This diagram provides the relative direction of the masses 431 .
Step 2: Now, draw the force polygon taking a suitable scale as shown.
This gives the direction and magnitude of mass m2.
The results are:
θ = 0 θ = 0 ,, θ =282123168 0 4 1 2 22 == 427427 Anskgmorrrm
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Problem 2:
Each crank of a four- cylinder vertical engine is 225 mm. The reciprocating masses of the first, second and fourth cranks are 100 kg, 120 kg and 100 kg and the planes of rotation are 600 mm, 300 mm and 300 mm from the plane of rotation of the third crank. Determine the mass of the reciprocating parts of the third cylinder and the relative angular positions of the cranks if the engine is in complete primary balance.
Solution: Given: = mm225r
421 === kg100mandkg120mkg,100m
Cent. Distance Couple/ ω2 Mass (m) Radius (r) Force/ω2 from Ref Plane ( m r l ) kg m (m r ) plane ‘2’ kg m2 kg m m 1 100 0.225 22.5 -0.600 -13.5 2 120 0.225 27.0 -0.300 -8.1 3(RP) m3 0.225 0.225 m3 0 0 4 100 0.225 22.5 0.300 6.75
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Analytical Method:
0 Choose plane 3 as the reference plane and θ1 =0 . Step 1: Resolve the couples into their horizontal and vertical components and take their sums. Sum of the horizontal components gives
0 cos.cos.cos. 42 =θ+θ−− 0756180513
42 +θ−=θ− .cos.cos..,e.i 51375618
42 −−−−−−−−−−θ=θ 151375618 )(.cos.cos..,e.i
Sum of the vertical components gives
− 0 − θ + sin.sin.sin. θ = 0756180513 42 42 −−−−−−−−−θ=θ 275618 )(sin.sin..,e.i
Squaring and adding (1) and (2), we get
2 2 2 4 +−= 4 )θsin(6.7513.5)θcos(6.75(8.1) 2 2 44 ++−= θsin45.563182.25θcos182.25θcos45.56365.61 4 2 2 4 44 +−+= 182.25θcos182.25)θsinθ45.563(cos
4 += 182.25θcos182.25-45.563
4 =+= 162.20365.61-182.2545.563θcos182.25i.e.,
162.203 Therefore, θcos == 0 Ans27.13θand 182.25 44
Dividing (2) by (1), we get
0 )(27.13sin6.75 3.078 θtan == −= 1.515 2 0 13.5-)(27.13cos6.75 − 7.493
000 2 =+= 157.6718022.33-θi.e.,
Step 2: Resolve the forces into their horizontal and vertical components and take their sums.
Sum of the horizontal components gives
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00 0 33 =+++ 0)27.13(cos22.5θcosm0.225)(157.67cos27)0(cos22.5
33 =++− 020.02θcosm0.22524.97522.5i.e.,
33 −−−−−−−−−−−−−−= (3)17.545θcosm0.225i.e.,
And sum of the vertical components gives
00 0 33 =+++ 0)27.13(sin22.5θsinm0.225)(157.67sin27)0(sin22.5
33 =++ 010.26θsinm0.22510.258i.e.,
33 −−−−−−−−−−−−−−= (4)-20.518θsinm0.225i.e.,
Squaring and adding (3) and (4), we get
2 222 (0.225) m3 =(−17.545) +(− 20.518)
22 −17.545 − 20.518 mi.e., = + 3 0.225 0.225 ≈= Anskg120kg119.98
− 20.518 θtan = Dividing (4) by (3), we get 3 17.545- 0 3 = Ans229.5θor
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Problem 3:
The cranks of a four cylinder marine oil engine are at angular intervals of 900. The engine speed is 70 rpm and the reciprocating mass per cylinder is 800 kg. The inner cranks are 1 m apart and are symmetrically arranged between outer cranks which are 2.6 m apart. Each crank is 400 mm long. Determine the firing order of the cylinders for the best balance of reciprocating masses and also the magnitude of the unbalanced primary couple for that arrangement.
Analytical Solution:
Given: Nπ2 ωm,0.4r,rpm70Nkg,800m ===== s/rad7.33 60 22 == 17195(7.33)x0.4x800ωrm
Note:
There are four cranks. They can be used in six different arrangements as shown. It can be observed that in all the cases, primary forces are always balanced. Primary couples in each case will be as under.
Taking 1 as the reference plane,
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2 2 222 p1 ( 3 ) ( 42 ) ( ) ( −+−=−+−ω= ...lllrmC 62808117195 ) = 43761 mN
pp == 43761 4216 erchangedintarelandlcesin,onlymNCC
2 2 222 p2 ( 4 ) ( 32 ) ( ) ( −+−=−+−ω= ...lllrmC 81806217195 ) = 47905 mN
pp == 47905 3225 erchangedintarelandlcesin,onlymNCC
2 2 222 p3 ( 2 ) ( 34 ) ( ) ( −+−=−+−ω= ...lllrmC 81628017195 ) = 19448 mN
pp == 19448 3434 erchangedintarelandlcesin,onlymNCC
Thus the best arrangement is of 3rd and 4th. The firing orders are 1423 and 1324 respectively. Unbalanced couple = 19448 N m.
Graphical solution:
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Case 3: SIX – CYLINDER, FOUR –STROKE ENGINE
Crank positions for different cylinders for the firing order 142635 for clockwise rotation of the crankshaft are, for
0 0 And First θ1 =0 Second θ2 =240 ===== mmmmmm Third θ =1200 Fourth =θ 1200 654321 3 4 ===== rrrrrr 0 0 654321 Fifth θ5 =240 Sixth θ6 =0
Since all the force and couple polygons close, it is inherently balanced engine for primary and secondary forces and couples.
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Problem 1:
Each crank and the connecting rod of a six-cylinder four-stroke in-line engine are 60 mm and 240 mm respectively. The pitch distances between the cylinder centre lines are 80 mm, 80 mm, 100 mm, 80 mm and 80 mm respectively. The reciprocating mass of each cylinder is 1.4 kg. The engine speed is 1000 rpm. Determine the out-of-balance primary and secondary forces and couples on the engine if the firing order be 142635. Take a plane midway between the cylinders 3 and 4 as the reference plane.
Solution:
Given : === ,mm240lengthrodconnectingl,mm60r == ,kg1.4cylindereachofmassingreciprocatm = rpm1000N Nπ2 1000xπ2 ωhave,We === /srad72104. 60 60
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Cent. Distance Couple/ ω2 Radius (r) Force/ω2 from Ref Plane Mass (m) kg ( m r l ) m (m r ) plane ‘2’ kg m2 kg m m 1 1.4 0.06 0.084 0.21 0.01764 2 1.4 0.06 0.084 0.13 0.01092 3 1.4 0.06 0.084 0.05 0.0042 4 1.4 0.06 0.084 -0.05 -0.0042 5 1.4 0.06 0.084 -0.13 -0.01092 6 1.4 0.06 0.084 -0.21 -0.01764
Graphical Method:
Step 1: Draw the primary force and primary couple polygons taking some convenient scales. Note: For drawing these polygons take primary cranks position as the reference
NO UNBALANCED PRIMARY FORCE
NO UNBALANCED PRIMARY COUPLE
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Step 2: Draw the secondary force and secondary couple polygons taking some convenient scales. Note: For drawing these polygons take secondary cranks position as the reference
NO UNBALANCED SECONDARY FORCE
NO UNBALANCED SECONDARY COUPLE Problem 2:
The firing order of a six –cylinder vertical four-stroke in-line engine is 142635. The piston stroke is 80 mm and length of each connecting rod is 180 mm. The pitch distances between the cylinder centre lines are 80 mm, 80 mm, 120 mm, 80 mm and 80 mm respectively. The reciprocating mass per cylinder is 1.2 kg and the engine speed is 2400 rpm. Determine the out-of-balance primary and secondary forces and couples on the engine taking a plane midway between the cylinders 3 and 4 as the reference plane.
Solution: Given : L 80 r ===== ,mm180lengthrodconnectingl,mm40 2 2 == ,kg1.2cylindereachofmassingreciprocatm = rpm2400N Nπ2 2400xπ2 ωhave,We === /srad251.33 60 60
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Cent. Distance Couple/ ω2 Radius (r) Force/ω2 from Ref Plane Mass (m) kg ( m r l ) m (m r ) plane ‘2’ kg m2 kg m m 1 1.2 0.04 0.048 0.22 0.01056 2 1.2 0.04 0.048 0.14 0.00672 3 1.2 0.04 0.048 0.06 0.00288 4 1.2 0.04 0.048 -0.06 -0.00288 5 1.2 0.04 0.048 -0.14 -0.00672 6 1.2 0.04 0.048 -0.22 -0.01056
Graphical Method:
Step 1: Draw the primary force and primary couple polygons taking some convenient scales. Note: For drawing these polygons take primary cranks position as the reference
Note: No primary unbalanced force or couple
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Step 2: Draw the secondary force and secondary couple polygons taking some convenient scales. Note: For drawing these polygons take secondary cranks position as the reference
Note: No secondary unbalanced force or couple
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Problem 3:
The stroke of each piston of a six-cylinder two-stroke inline engine is 320 mm and the connecting rod is 800 mm long. The cylinder centre lines are spread at 500 mm. The cranks are at 600 apart and the firing order is 145236. The reciprocating mass per cylinder is 100 kg and the rotating parts are 50 kg per crank. Determine the out of balance forces and couples about the mid plane if the engine rotates at 200 rpm.
Primary cranks position
Relative positions of Cranks in degrees Firing θ θ θ θ θ θ order 1 2 3 4 5 6 142635 0 240 120 120 240 0 145236 0 180 240 60 120 300
Secondary cranks position
Relative positions of Cranks in degrees Firing θ θ θ θ θ θ order 1 2 3 4 5 6 142635 0 120 240 240 120 0 145236 0 0 120 120 240 240
Calculation of primary forces and couples:
Total mass at the crank pin = 100 kg + 50 kg = 150 kg
Cent. Distance Couple/ ω2 Mass (m) Radius (r) Force/ω2 from Ref Plane ( m r l ) kg m (m r ) plane kg m2 kg m m 1 150 0.16 24 1.25 30 2 150 0.16 24 0.75 18 3 150 0.16 24 0.25 6 4 150 0.16 24 -0.25 -6 5 150 0.16 24 -0.75 -18 6 150 0.16 24 -1.25 -30
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Calculation of secondary forces and couples:
Since rotating mass does not affect the secondary forces as they are only due to second harmonics of the piston acceleration, the total mass at the crank is taken as 100 kg.
Cent. Distance Couple/ ω2 Mass (m) Radius (r) Force/ω2 from Ref Plane ( m r l ) kg m (m r ) plane kg m2 kg m m 1 100 0.16 16 1.25 20 2 100 0.16 16 0.75 12 3 100 0.16 16 0.25 4 4 100 0.16 16 -0.25 -4 5 100 0.16 16 -0.75 -12 6 100 0.16 16 -1.25 -20
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BALANCING OF V – ENGINE
Two Cylinder V-engine:
A common crank OA is operated by two connecting rods. The centre lines of the two – cylinders are inclined at an angle α to the X-axis. Let θ be the angle moved by the crank from the X-axis.
Determination of Primary force: 2 Primary force of 1 along line of stroke OB1 = ω θ−α − − − − − − 1)()(cosrm
Primary force of 1 along X - axis = ω2 θ−α α − − − 2)(cos)(cosrm
2 Primary force of 2 along line of stroke OB2 = ω θ+α − − − − − 3)()(cosrm
Primary force of 2 along X-axis = ω2 θ+α α − − − 4)(cos)(cosrm
Total primary force along X - axis 2 []++−= )αθ(cos)αθ(cosαcosωrm 2 []−++= αsinθsinαcosθcosαsinθsinαcosθcosαcosωrm = 2 αcosθcos2xαcosωrm 22 −−−−−−−−−−−−= (5)θcosαcosωrm2
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Similarly,
Total primary force along Z - axis 2 []+−= α)sinαθ(cos-αsin)αθ(cosωrm 2 []−−+= αsinθsinαcosθcos(α)sinθsinαcosθ(cosαsinωrm = 2 αsinθsin2xαsinωrm 22 −−−−−−−−−−−−= (6)θsinαsinωrm2
Resultant Primary force ()()22 2 += 22 θsinαsinωrm2θcosαcosωrm2 2 ()()22 2 2 2 −−−−−+= (7)θsinαsinθcosαcosωrm2
and this resultant primary force will be at angle β with the X – axis, given by,
2 θsinαsin βtan −−−−−−= (8) 2 θcosαcos
If α = 902 0 , the resultant force will be equal to
()()022 2 + 02 θsin45sinθcos45cosωrm2 2 2 −−−−−= (9)ωrm
02 θsin45sin and βtan −−−−−−== (10)θtan 02 θcos45cos
i.e., β = θ or it acts along the crank and therefore, can be completely balanced by a mass at a suitable radius diametrically opposite to the crank, such that,
m rrr =mr -----(11)
For a given value of α, the resultant force is maximum (Primary force), when
( 2 )2 + ( 2 )2 maximumisθsinαsinθcosαcos or ()+ 2424 maximumisθsinαsinθcosαcos
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Or
d ()2424 =+ 0θsinαsinθcosαcos dθ 44 =+ 0θcosθsin2xαsinθsinθcos2xαcos-i.e.,
i.e., - cos α x sin2θ+ sin44 α x sin2θ =0
0αcos-αsinθ2sin i.e., []44 =0αcos-αsinθ2sin
As α is not zero, therefore for a given value of α , the resultant primary force is maximum when θ=0 0 .
Determination of Secondary force:
Secondary force of 1 along line of stroke OB1 is equal to
rm ω2 −−−−−−α−θ 12 )()(cos n
rm ω2 Secondary force of 1 along X - axis = −−−αα−θ 22 )(cos)(cos n
Secondary force of 2 along line of stroke OB2 =
rm ω2 −−−−−α+θ 32 )()(cos n
rm ω2 Primary force of 2 along X-axis = −−−αα+θ 42 )(cos)(cos n Therefore,
Total secondary force along X -axis ωrm 2 []++−= )αθ(2cos)αθ(2cosαcos n ωrm 2 [ −++= α2sinθ2sinα2cosθ2(cos)α2sinθ2sinα2cosθ2(cosαcos ] n ωrm2 2 −−−−−−−−−−−−= (5)α2cosθ2cosαcos n
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Similarly,
Total secondary force along Z - axis
ωrm2 2 −−−−−−−−−= (6)α2sinθ2sinαsin n
Resultant Secondary force 2 ωrm2 22 ()() −−−−−+= (7)α2sinθ2sinαsinα2cosθ2cosαcos n
sinα sin2θsin2α And βtan ' −−−−−−= (8) α2cosθ2cosαcos
If 2α 90 or α == 4500 ,
rm 2 22 θω 2 rmsin ω2 Secondary force = = −−−−θ 922 )(sin n 2 n
And tanβ and β 900'' −−−−−−=∞= (10) i.e., the force acts along Z- axis and is a harmonic force and special methods are needed to balance it.
Problem 1:
The cylinders of a twin V-engine are set at 600 angle with both pistons connected to a single crank through their respective connecting rods. Each connecting rod is 600 mm long and the crank radius is 120 mm. The total rotating mass is equivalent to 2 kg at the crank radius and the reciprocating mass is 1.2 kg per piston. A balance mass is also fitted opposite to the crank equivalent to 2.2 kg at a radius of 150 mm. Determine the maximum and minimum values of the primary and secondary forces due to inertia of the reciprocating and the rotating masses if the engine speed is 800 mm.
Solution:
Given : == kg1.2pistoneachofmassingreciprocatm kg2 mass rotatingequivalentM mass == kg2
C C === mm150rkg,2.2massbalancingm == mm600lengthrodconnectingl == mm120radiuscrankr = rpm800N Nπ2 800xπ2 l 600 ωhave,We nand/srad83.78 ======5 60 60 r 120
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Primary Force:
22 Total primary force along X - axis =2 mr ω cos α cos θ − − − − − − − − − − − −(1)
Centrifugal force due to rotating mass along X − axis 2 −−−−−−−−−−−−= (2)θcosωrM
Centrifugal force due to balancing mass along X − axis 2 CC −−−−−−−−−−−−−= (3)θcosωrm
Therefore total unbalance force along X –axis = (1) + (2) + (3)
That is Total Unbalance force along X axis
222 2 −+= CC θcosωrmθcosωrMθcosαcosωrm2 22 []+= rm-rMαcosrm2θcosω CC ()2 []02 −+= 2.2x0.152x0.1230xcos2x1.2x0.12θcos83.78 ()2 [] =+= (4)------Nθcos884.410.33-0.240.216θcos83.78
Total primary force along Z - axis =2 mr ω s 22 α sinin θ − − − − − − − − − − − −(5)
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Centrifugal force due to rotating mass along Z − axis 2 in −−−−−−−−−−−−= (6)θsωrM
Centrifugal force due to balancing mass along Z − axis 2 CC in −−−−−−−−−−−−−= (7)θsωrm
Therefore total unbalance force along Z –axis = (5) + (6) + (7)
That is
Total Unbalance force along Z - axis
222 2 −+= CC θsinωrmθsinωrMθsinαsinωrm2 22 []+= rm-rMαsinrm2θsinω CC ()2 []02 −+= 2.2x0.152x0.1230xsin2x1.2x0.12θsin83.78 ()2 [] =+= (8)------Nθsin126.34-0.33-0.240.072θsin83.78
Resultant Primary force
22 ()()+= θsin126.34-θcos884.41 += 22 θsin15961.8θcos782181.05 2 −−−−−+= (9)15961.8θcos766219.25
This is maximum, when θ =00 and minimum, when θ =900
0θwhen i.e., force, Primary Maximum Primary force, i.e., = 0θwhen 0
−−−−−=+= (10)N884.4115961.8766219.25
9 0θwhen i.e., force, Primary MinimumAnd Primary force, i.e., = 90θwhen 0
02 −−−−−=+= (11)N126.3415961.890cos766219.25
Secondary force:
The rotating masses do not affect the secondary forces as they are only due to second harmonics of the piston acceleration.
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Resultant Secondary force
2 ωrm2 22 ()()+= α2sinθ2sinαsinα2cosθ2cosαcos n 2 78)x0.12x(83.x1.22 2 ()60cosθ2cos30cos 22 = 5 + ()60sinθ2sin30sin 00 2 []22 −−−−−+= (12)2θθsin(0.18752θθcos(0.1875404.3
This is maximum, when θ =00 and minimum, when θ =1800
0θwhen i.e., force, secondary Maximum secondary force, i.e., = 0θwhen 0
[]2020 −−−−−=−+= (13)N175.07)0sin(0.1875)0cos(0.1875404.3
18 0θwhen i.e., force, secondary MinimumAnd secondary force, i.e., =180θwhen 0
[]2020 −−−−−=+= (14)N175.07)018sin(0.1875)180cos(0.1875404.3
BALANCING OF W, V-8 AND V-12 – ENGINES
BALANCING OF W ENGINE
In this engine three connecting rods are operated by a common crank.
Total primary force along X -axis
()22 αcos2θcosωrm −−−−−−−−−−−−+= (1)1
Total primary force along Z-axis will be same as in the V − twin engine, − zero)isaxisZalong3offorceprimarythe(since 22 −−−−−−−−−−−−= (2)θsinαsinωrm2
Resultant Primary force []()()22 2 2 2 −−−−−++= (3)θαsin2sin1α2cosθcosωrm
and this resultant primary force will be at angle β with the X – axis, given by,
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2 θsinα2sin βtan = −−−−−− (4) ()2 +1α2cosθcos
If α = 600 , Resultant Primary force 3 and 2 −−−−−= (5)ωrm 2
tanβ = = tanθ − − − − − −(6)
i.e., β = θ or it acts along the crank and therefore, can be completely balanced by a mass at a suitable radius diametrically opposite to the crank, such that,
m rrr =mr ---- -(7)
Total secondary force along X -axis
ωr2m 2 2θcos += −−−−−−−−−−−− (8)1α2cosαcos n
Total secondary force along Z –direction will be same as in the V-twin engine.
Resultant secondary force
2 ωrm 22 []()() −−−−−++= (9)2θsin2αsinα2sin12αcosα2cos2θcos n
2sinα sin2θsin2α βtan ' = −−−−−− (10) ()+1α2cosα2cosθ2cos
If α = 600 ,
Secondary force along X -axis
ωrm 2 −−−−−−−−−−−−= (11)2θcos 2n
Secondary force along Z-axis
ωr3m 2 −−−−−−−−−−−−= (12)2θsin 2n
It is not possible to balance these forces simultaneously 40 www.getmyuni.com
V-8 ENGINE
It consists of two banks of four cylinders each. The two banks are inclined to each other in the shape of V. The analysis will depend on the arrangement of cylinders in each bank.
V-12 ENGINE
It consists of two banks of six cylinders each. The two banks are inclined to each other in the shape of V. The analysis will depend on the arrangement of cylinders in each bank.
If the cranks of the six cylinders on one bank are arranged like the completely balanced six cylinder, four stroke engine then, there is no unbalanced force or couple and thus the engine is completely balanced.
BALANCING OF RADIAL ENGINES:
It is a multicylinder engine in which all the connecting rods are connected to a common crank.
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Direct and reverse crank method of analysis:
In this all the forces exists in the same plane and hence no couple exist.
In a reciprocating engine the primary force is given by, ω2 cosrm θ which acts along the line of stroke.
In direct and reverse crank method of analysis, a force identical to this force is generated by two masses as follows.
1. A mass m/2, placed at the crank pin A and rotating at an angular velocity ω in the counter clockwise direction. 2. A mass m/2, placed at the crank pin of an imaginary crank OA’ at the same angular position as the real crank but in the opposite direction of the line of stroke. It is assumed to rotate at an angular velocity ω in the clockwise direction (opposite). 3. While rotating, the two masses coincide only on the cylinder centre line.
The components of the centrifugal forces due to rotating masses along the line of stroke are,
m AatmasstoDue 2 cosr θω= 2
m AatmasstoDue ' 2 cosr θω= 2
Thus, total force along the line of stroke = ω2 cosrm θ which is equal to the primary force. At any instant, the components of the centrifugal forces of these masses normal to the line of stroke will be equal and opposite. The crank rotating in the direction of engine rotation is known as the direct crank and the imaginary crank rotating in the opposite direction is known as the reverse crank.
Now,
Secondary accelerating force is cos2θ cos2θ rm ω 2ω= )(rm 22 n 4n
r m 2 cos)( 22 θω= 4n
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This force can also be generated by two masses in a similar way as follows.
r 1. A mass m/2, placed at the end of direct secondary crank of length at an angle 2θ 4n and rotating at an angular velocity 2ω in the counter clockwise direction.
r 2. A mass m/2, placed at the end of reverse secondary crank of length at an angle -2θ 4n and rotating at an angular velocity 2ω in the clockwise direction.
The components of the centrifugal forces due to rotating masses along the line of stroke are,
m r mrω2 CatmasstoDue 2 2θcosω)(2 == 2θcos 2 4n 2n
m r mrω2 C'atmasstoDue 2 2θcosω)(2 == 2θcos 2 4n 2n
Thus, total force along the line of stroke =
m r mrω2 2x 2 2θcosω)(2 = 2θcos which is equal to the secondary force. 2 4n n
References:
1. Theory of Machines by S.S.Rattan, Third Edition, Tata McGraw Hill Education Private Limited. 2. Kinematics and Dynamics of Machinery by R. L. Norton, First Edition in SI units, Tata McGraw Hill Education Private Limited.
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