1.1 Unbalancing

Unbalance is most common source of vibration in machine with rotating parts. It is a very important factor to be considered in modern machine design, especially where high speed and reliability are significant considerations. Balancing of rotors prevents excessive loading of bearings and avoids fatigue failure, thus increasing the useful life of machinery.

1.2 Types Of Unbalance

 1.2.1 Static Unbalance –

Static Unbalance is defined as the eccentricity of the center of gravity of a rotor, caused by a point mass at a certain radius from the center of rotation.

Static Unbalance

 1.2.2 Couple (Moment) Unbalance –

In the case of a cylinder, shown in Fig.- 13.2, it is possible to have two equal masses placed symmetrically about the centre of gravity, but positioned at 180° from each other. The rotor is in static balance, i.e. there is no eccentricity of the centre of gravity, but when the rotor turns , the two masses cause a shift in the inertia axis, so that it is no longer aligned with the rotation axis, leading to strong vibrations in the bearings. The unbalance can only be corrected by taking vibration measurements with the rotor turning and adding correction masses in two planes. The difference between static balance and couple balance is illustrated in Fig. It can be seen that when the rotor is stationary, the end masses balance each other. However, when it rotates, strong unbalance is experienced.

 Couple Unbalance

 

 1.2.3 Dynamic Unbalance -

Dynamic Unbalance is a combination of static and couple unbalance and is the most common type of unbalance found in rotors. To correct dynamic unbalance, it is necessary to make vibration measurements while the machine is running and to add balancing masses in two planes.

                                                                           Dynamic Unbalance

1.3 Balancing

Balancing is the technique of eliminating unwanted inertia forces and moments in rotating and reciprocating masses and is achieved by changing the location of the mass centers. In rotor or reciprocating machines with the moving masses, if the parts are not properly balanced, the dynamic forces are set up. These forces not only increase the loads on the bearings and stresses in the various components but also it may cause unpleasant noise, vibration and cause wear and tear of the system.

When a mass moves along a circular path, it experiences a centripetal acceleration which generates a force and acts inwards. An equal and opposite force called centrifugal force acts radially outwards and it is a disturbing force on the axis of rotation. The magnitude of this remains constant but the direction changes with the rotation of mass.

The centrifugal disturbing force is given by      $F_c=m\omega^2r - - - - - - -Eqn 13.1$

 

Where ,
     m = mass of the rotating mass
     ω = angular velocity of the component, rad/sec
     r = distance of the centre of gravity of mass from the axis of rotation, m

 1.3.1 Types of Balancing

(i) Static Balancing: A rotating mass is said to be statically balanced if the rotating mass can rest without turning at any angular position in its bearings. The condition is attained when the sum of the centrifugal forces on the rotating mass due to unbalanced masses is zero at any radial direction. Static balancing is valid for a rotor with only one disc or balancing is required in single plane only.

(ii) Dynamic Balancing: A rotating mass is said to be dynamically balanced when it does not vibrate in its running state. In dynamic balancing, resultant moments or couple which involved in the acceleration of different moving parts is equal to zero. To make a rotating mass dynamically balanced, it must first be statically balanced.

 1.3.2 Balancing Of Single Revolving Mass By A Single Mass In The Same Plane :

Let,
     m = disturbing mass, kg      B = balancing mass

     $m\omega^2r=B\omega^2b => mr=Bb- - - - - -eqn13.2$ 

Balancing Of Single Revolving Mass By A Single Mass In The Same Plane (Fig.- 13.4)

Dynamic reactions at the bearings will be zero. Thus the shaft will be free from any dynamic bending stress.

 1.3.3 Balancing Of A Single Mass By Two Masses Rotating In Different Planes:

By this type of arrangement for balancing gives rise to a couple which tends to rock the shaft in its bearings. So two balancing masses are placed in two different planes parallel to the plane of rotation of the disturbing mass.

Condition 1. Net force acting on the shaft must be equal to zero

 

therefor, $m\omega^2r=B_1b_1\omega^2+B_mb_m\omega^2 \quad or \quad mr=B_1b_1+B_mb_m - - - - - - - eqn13.3$

 

Balancing Of A Single Mass By Two Masses Rotating In Different Planes (Fig.- 13.5)

Condition 2. The algebraic sum of the moments about any point in the plane must be zero. Taking moment about L and M axis

$m\omega^2rl=B_m\omega^2b_md-------eqn13.4$  , and 

$m\omega^2rx=B_1\omega^2b_1d - - - - - - - - - - -eqn13.5$

Two possibilities:

     (a) The distributing mass may be in between the planes of the balancing masses
     (b) The distributing mass be on the left or right.

 1.3.4 Balancing Of Several Masses Rotating In The Same Plane

Balancing Of Several Masses Rotating In The Same Plane (Fig.- 13.6)

Consider any number of masses of magnitude m1, m2,m3 and m4 at distances of r1, r2, r3 and r4 from the axis of the rotating shaft. Let α , β and γ be the angles between the masses. Let these masses rotate about an axis through o and perpendicular to the plane of the paper with constant angular velocity, ω rad/\sec. The magnitude and position of the balancing mass may be found out analytically or graphically.

 1.3.5 Analytical Method:

$F_h=m_1\omega^2r_1Cos0+m_2\omega^2r_2Cos\alpha+m_3\omega^2r_3Cos(\alpha+\beta)+m_4\omega^2r_4Cos(\alpha+\beta+Y)$

$F_v=m_1\omega^2r_1Sin0+m_2\omega^2r_2Sin\alpha+m_3\omega^2r_3Sin(\alpha+\beta)+m_4\omega^2r_4Sin(\alpha+\beta+Y)$

Resultant $Bb= \sqrt (F_h^2+F_v^2)$ and $Tan\theta={F_v\over F_h} - - - - - - -eqn 13.6$

Balancing force should be in opposite direction

 1.3.6 Graphical Method:

Graphical Method (Fig.- 13.7)

The effect of transferring the rotating mass in the refrence plane will be to produce a centrifugal force $F=m\omega^2r$ and ac couple $C=Fl=m\omega^2rl$.L in the refrence plane.

Couple is represented by a vector perpendicular to the plane in which.

1.4 Balancing Of Several Masses In Different Planes

Balancing of several masses in different planes (Fig.- 13.8)

Plane	Mass (m)	Radius (r)	cent for a - ω2 (mr)	Dist from pl (l)	Couple - ω2 (mrl)
A	$m_a$	$r_a$	$m_ar_a$	$-l_a$	$-m_ar_al_a$
L	$B_l$	$b_l$	$B_lb_l$	0	0
B	$m_b$	$r_b$	$m_br_b$	$l_b$	$m_br_bl_b$
C	$m_c$	$r_c$	$m_cr_c$	$l_c$	$m_cr_cl_c$
M	$B_m$	$b_m$	$B_mb_m$	d	$B_mb_md$
D	$m_d$	$r_d$	$m_dr_d$	$l_d$	$m_dr_dl_d$

 

$B_l$ and $B_m$ are obtained from force polygon and couple polygon

1.5 Balancing Of Reciprocating Masses

   

 1.5.1 Analytical Determination Of Acceleration Of The Piston Of A Reciprocating Engine:

Analytical Determination Of Acceleration Of The Piston Of A Reciprocating Engine (Fig.- 13.9)

Let;$n={L \over r}$

Now by geometry we see that $x= PP_r=P_rO-PO=L+r-(Lcos\phi+r cos\theta) - - - - - - - -eqn13.7$,

Our interest is to present the equation 13.7 in terms of one variable i.e either $\theta \quad or\quad \phi$. Let us present it in terms of $\theta$ only.

the common side $CR=Lsin\phi=rsin\theta \quad or \quad sin\phi={sin\theta\over n}$, thus 

$Cos\phi = \sqrt(1-Sin^2\phi)={1\over n}\sqrt(n^2-Sin^2\theta) - - - - -eqn13.8$

Applying (Eqn.-13.8) in (Eqn.-13.7)

$x=r(1-Cos\theta)+L(1-{1\over n}\sqrt(n^2-Sin^2\theta))$

or, $x=r(1-Cos\theta)+r(n-\sqrt(n^2-Sin^2\theta)) \quad or\quad x=r(1-Cos\theta+n-\sqrt(n^2-Sin^2\theta))$

Thus any instantaneous position of the piston can be found out by$x=r(1-Cos\theta+n-\sqrt(n^2-Sin^2\theta))$

the instantaneous velocity of piston can be found by differentiating the above equation with time.

$V={dx\over dt} ={dx\over d\theta}.{d\theta\over dt}$, but  ${dx \over d\theta} =\omega, \quad angular \quad velocity.$ 

Thus velocity of piston at any point will be $V=-\omega r(Sin\theta+{Sin2\theta \over 2\sqrt {n^2-Sin^2\theta}})$

Since $Sin^2\theta<<n^2,$ we neglect $Sin^2\theta$, so the expresion for velocity change to

$V=-\omega r(Sin\theta+{Sin2\theta \over 2n})$

Similarly for the knowing the instantaneous acceleration of piston we differentiate the velocity once again w.r.t time.

${dV\over dt}={d\theta\over dt}.{dV\over d\theta}=\omega.{dV\over d\theta}$

therefore instantaneous acceleration of the piton is  ,$f=-\omega^2r(cos\theta+{cos2\theta\over n})$

 1.5.2 The Effect Of The Inertia Force Of The Reciprocating Mass On The Engine Frame :

We now know that acceleration of the piston is given  by

$f=-\omega^2r(cos\theta+{cos2\theta\over n})$

where $f$ is the acceleration of the piston , θ is the inclination of the crank with IDC., ω is the annular speed of the crank , r is the radius of the crank., n is the ratio of length of connecting rod L to the crank radius r i.e. $n={L\over r}$

 

The Effect Of The Inertia Force Of The Reciprocating Mass On The Engine Frame (Fig.- 13.12)

Force to accelerate the reciprocating mass, $F=mf=-m\omega^2r(cos\theta+{cos2\theta\over n})$

where m = mass of the reciprocating parts.

The force applied to the frame is equal and opposite to the force required to accelerate the reciprocating mass. It is equal to the reversed effective force or the inertia force.

$F=m\omega^2rcos\theta+m\omega^2r{cos2\theta\over n}$, the first part of the equation is the primary unbalanced force and the second part is the secondary unbalanced force. Thus $F=F_p+F_s$

Fp→ primary unbalanced force = $m\omega^2rcos\theta$

FS→ secondary unbalanced force = $m\omega^2r{cos2\theta\over n}$

Secondary unbalanced force is 1/n times the primary unbalanced force and may be neglected. The primary unbalanced force is maximum when θ = 0 or 180. The primary force is maximum force in one revolution of the crank.The secondary unbalanced force is maximum when θ = 0 , 90, 180 and 270, 360. Hence it is maximum four times in one revolution of the crank. Thus secondary force has twice the frequency of the primary force.The unbalanced force due to reciprocating masses varies in magnitude but constant in direction. (i.e. it acts along OP).The unbalanced force due to revolving masses is constant in magnitude but varies in direction.

1.6 Partial Primary Balance

$F=m\omega^2rcos\theta$

The primary unbalance force acts from O to P which may be considered as the component of the centrifugal force produced by a rotating mass 'm' placed at the crank pin . This is balanced by having a mass B at a radius b placed diametrically opposite to the crank pin C. It has two component.

Partial Primary Balance (Fig.- 13.11)

$B\omega^2bcos\theta$ along the line of stroke  and $B\omega^2bsin\theta$ perpendicular to the line of stroke.The primary force is balanced if

$m\omega^2rcos\theta=B\omega^2bcos\theta$ or $mr=Bb$

But in doing so another unbalanced vertical force of magnitude $B\omega^2rsin\theta$ has been introduced . The maximum value of this force is equal to $B\omega^2b$ when $\theta$ = 90 and 270. Thus balancing the force along the line of stroke, another unbalanced vertical force of the same magnitude has been introduced. As a compromise let a fraction 'e' of the reciprocating masses is balanced . i.e. $Bb=emr$.

Thus the new unbalanced force along the line of stroke, =$m\omega^2rcos\theta-B\omega^2bcos\theta$ 

Unbalanced force = $m\omega^2rcos\theta-em\omega^2rcos\theta =(1-e)m\omega^2rcos\theta$ 

The unbalanced force along the perpendicular to the line of stroke, =$B\omega^2bsin\theta =em\omega^2rsin\theta$

∴ Resultant unbalanced force at any instant

=$m\omega^2r\sqrt((1-e)^2cos^2\theta+e^2sin^2\theta)$

If the balancing mass is requested to balance the revolving masses as well as reciprocating masses then

$B.b=m_1r+emr =r(m_1+em)$  

$m_1$= magnitude of the revolving masses  

$m$= magnitude of the reciprocating masses  

1.7 Partial Balancing Of Locomotives

The locomotives, usually it have two cylinder with cranks placed at right angles to each oher in order to have uniformity in turning moment diagram. The two Cylinder locomotives may be, 
(a) inside cylinder locomotives 
(b) outside cylinder locomotives

In inside cylinder locomotives, the two cylinder are placed inside the driving wheels, where as the outside cylinder locomotives , the two cylinder are placed outside the driving wheels.

Partial Balancing Of Locomotives (Fig.- 13.12)

The locomotives may be single or uncoupled locomotive if the efforts is transmitted to one pair of the wheels.

 1.7.1 Coupled locomotive :

The driving wheels are connected to the loading and troiling wheels by an outside coupling rod.

Since the cranks are placed at right angle to each other. which facilitates starting of the locomotive in any position. Because one crank will always be away from IDC. Position or the position of no torque on the crank. Further, the secondary forces for one set of reciprocating parts are equal and opposite to those for the other.

 1.7.2 Effect Of Partial Balancing Of Locomotive

Due to this partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke which produces.
(a) Variation of tractive forces
(b) Hammer Blow
(c) Swaying couple and an unbalanced primary force perpendicular to the line of stroke which produce.

(a) Variation Of Tractive Forces

The variation of tractive force is caused by unbalanced part of the primary disturbing force due to reciprocating masses . the resultant unbalanced force due to the two cylinders. along the line of stroke known as tractive force.

Let,

m = mass of the reciprocating parts/cylinder

e = fraction of the reciprocating parts to be balanced

θ = inclination of the crank for 1st cylinder

(90+ θ) = inclination of the crank for 2nd cylinder

∴ For cylinder 1 . the unbalanced force along the line of stroke = $(1-e)m\omega^2rcos\theta$

∴ For cylinder 2 . the unbalanced force along the line of stroke=$(1-e)m\omega^2rcos(90+\theta)$, IF 'a' is the wheel to wheel distance then the swayin couple,C =$(1-e)m\omega^2r(cos\theta+sin\theta).{a\over 2}$

Maximum value of swaying 

${dC\over d\theta}=0$ or ${d[(1-e)m\omega^2r(cos\theta+sin\theta)]\over d\theta}=0 => (1-e)m\omega^2r {d(cos\theta+sin\theta)\over d\theta}=0$, which implies that $sin\theta+cos\theta =0$, this is possibble at $\theta =45 \quad 225 \quad degrees$

thus, $C_{max}=(1-e)m\omega^2r{a\over 2}[Cos45+Sin45]$, and $C_{max}=(1-e)m\omega^2r{a\over 2}[Cos225+Sin225]$

therefore $C_{max,min}=\pm(1-e)m\omega^2r{a\over \sqrt2}$

(b) Hammer Blow

When vertical unbalanced force is considered in the plane of the wheel, it gives rise to the lifting of the wheels from the rails. The maximum value occurs when the center if gravity of the balance. Mass is either directly above the wheel or directly blow the wheel. The maximum value of the unbalanced force in hammer blow.

        Hammer Blow

Hammer Blow =$B\omega^2b=em\omega^2r$

The effect of hammer blow is to cause the variation in pressure between the wheel and the rail and cause oscillation of the locomotive in a vertical plane about horizontal axis.

Let $P$ = static wheel load. This will act in downward direction.

Net pressure between the wheel and the rail = $P \pm B\omega^2b$

If ($P-B\omega^2b$)is negative then the wheel will be lifted from the rail. In order to prevent this lifting of the wheel , the limiting speed is given by 

$P=B\omega^2b$ => $\omega_{limiting}=\sqrt({P \over (Bb)} \quad rad/sec$

 

 1.7.3 Balancing Of Multi Cylinder In Line Engine

The multi-cylinder engines with the cylinder center lines in the same plane and on the same side of the center line of the crank shaft, are known as In-line engine

                                                       Balancing of Multi Cylinder Inline Engines

For primary balance the following two conditions must be satisfied in order to given the primary balance or the reciprocating points

1.The algebraic sum of the primary forces must be zero i.e $\Sigma m_i\omega^2r_icos\theta =0$

 2.The algebraic sum of the couples about any point in the plane of the primary forces must be zero i.e $\Sigma m_i\omega^2r_ix_icos\theta=0$

Here line PQ is parallel to the line of stroke ab+bc = cd+da

The algebraic sum of these four forces is zero, the engine is balanced for primary forces. When the crank shaft are in the position shown. But suppose the crank shaft turns clockwise through an angle $\alpha$ in the new position the algebraic sum is not zero. Therefore the primary forces can only balance for this new position, if the force polygon is a closed one. In a similar way it would be shown that the primary couples can only balanced if the couple polygon for the corresponding centrifugal force is closed.

For the primary force balance of the reciprocating parts of a multi-cylinder in line engine , it is convenient to imagine the reciprocating masses to be transferred & their respective crankpin and to treat the problem as one of revolving masses.

For Secondary Balance

$F_s=m\omega^2r{cos2\theta \over n}=m(2\omega)^2r{cos2\theta \over 4n}$

Secondary Balance (Fig.- 13.15)

Second force Fs is equivalent to the component parallel to the line of stroke , of the centrifugal crank force produced by an equal mass placed at an imaginary crank length r/4n and revolving at twice the speed of the actual crank. Force secondary balance

$\Sigma m(2\omega)^2{r\over 4n}cos2\theta = 0$ (secondary force polygon must close) , and 

$\Sigma m(2\omega)^2{r\over 4n}x cos2\theta = 0$ (secondary couple polygon must close)  

 1.7.4 Direct And Reverse Crank Method

This method is used in balancing radial or V-engines in which the connecting rods are connected to a common crank, The plane of rotation of the various cranks is same, therefore there is no unbalanced primary or secondary couple.

Direct And Reverse Crank Method (Fig.- 13.16)

The direct crank revolves in a clockwise direction . The reversed crank will revolve in the anticlockwise direction.

Primary Force

$F_p=m\omega^2rcos\theta$

It is assumed that a mass m/2 is fixed at the direct crankpin C and m/2 at the reverse crankpin C'.

centrifugal force acting on the direct reverse crank=$m\omega^2r \over2$

∴ Total component of the centrifugal force along the stroke =${2 m\omega^2rcos\theta \over2}=m\omega^2rcos\theta=F_p$

Hence the primary effect the mass 'm' of the reciprocating parts at P may be replaced by two masses and C and C' each of magnitude m/2.

Secondary Force

$F_s=m (2\omega)^2{r \over 4n}cos2\theta$

Secondary Force (Fig.- 13.17)

1.8 V Engine And Redial Engine