Balancing of Inertia Forces

Inertia is the resistance of any physical object to a change in its state of motion or rest, or the tendency of an object to resist any change in its motion.

Linear velocity is the velocity along a straight line, and its magnitude is commonly measured in such units as meters per second (m/s), feet per second (ft/s), and miles per hour (mi/h).

The angular velocity is a vector quantity (more precisely, a pseudovector) which specifies the angular speed of an object and the axis about which the object is rotating.

The Balancing of Inetia Forces with particular referance to Engines; Air Compresssors and Steam Locomotives.

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Introduction

The high speeds of rotation at which modern machines and engines are required to operate, has made it increasingly important that all revolving and reciprocating parts are as completely balanced as possible.

Not only are the bearing loads and stresses in components increased by out of balance dynamic forces but there is also the possibility of causing significant harmful vibrations. On the railways these were called Hammer Blows and they were reduced by casting Balancing weights into the driving wheels.

These can be clearly seen in the photograph at the bottom of the wheels

On road transport wheels are routinely balanced by adding small weighs around the rim. These improve wheel bearing life but more importantly prevent vibrations being set up in the steering system which are at best unpleasant and at worst very dangerous.

It would be wrong to assume that all vibrations caused by out-of balance forces are undesirable since they are used to agitate liquids or to prevent granular solids for "sticking".

Rotating Masses in one Plane

When a mass $M$ , which is attached at a radius $r$ to a shaft, is rotated with an angular velocity $\omega$ , there is an outwards radial force of $M\,r\,\omega^2$ . which will produce a bending moment in the shaft. To counteract the effects of this inertia force ,a balance weight may be introduced into the plane of rotation of the original mass such that the inertia forces of the two masses are equal.

If $W$ is the weight of the balancing mass and $b$ its radius of rotation. Then:

$Mr\omega^2=\displaystyle\frac{W}{g}b\;\omega^2$ or $W=\displaystyle\frac{Mrg}{b}$

Note: It is normal to make $b$ as large as possible since this reduces the value of $W$ .

If a system of several rotating masses, attached to the same shaft, the effect is equivalent to a set of concurrent forces. Since $\omega$ is the same for all the masses attached to one shaft, the centrifugal forces are proportional to $M r$ (or $W r$ ) and the problem can be solved graphically by means of a polygon of forces or analytically by resolving in two directions.

If the masses are going to be balanced, the polygon must close. Note that the polygon can be drawn for any particular angular position of the shaft carrying the masses and will rotate as a whole with the shaft.

Rotating Masses in Parallel Planes

Again each mass exerts an outwards centrifugal force proportional to $M r$ . The forces are equivalent to a set of exactly similar forces in one chosen plane ( The reference Plane) together with a set of couples proportional to $M r x$ where $x$ is the distance from the plane of $M$ to the reference plane. It is necessary to reckon positive to one side of the reference plane and negative to the other.

The forces, $M r$ are balanced as before by drawing a force polygon. This tests for Static Balance. Similarly the couples can be represented by vectors drawn in the radial direction and can be balanced by drawing a Couple Polygon $M r x$ . Negative couples are drawn in the radially inwards direction. If the polygon closes, then the system is in dynamic balance. If it does not close,then the gap represents the out-of balance moment about the reference plane.

Note. It is only a pure couple if there is no out-of-balance force. Static balance can be obtained without dynamic balance but the reverse is not true. (See Example 3)

Reciprocating Masses

The Inertia Force of a rotating mass $M$ is:

$Mr\omega ^2\left ( \cos\theta+\frac{\cos2\theta}{n} \right )$

This force is provided by the pull of the connecting-rod and with reference to the following diagram:

The connecting-rod is in tension and the force $Q$ applied by the rod to the crank pin $C$ is equivalent to an equal and parallel force through $O$ together with a couple $Q$ . $x$ .

The Couple $Q$ . $x$ . tends to retard the rotation of the crankshaft and its effect is taken into account when finding the net turning moment on the crankshaft.

The force at $O$ is transmitted from the crankshaft through the main bearings and into the engine frame.

Both the force at $O$ and that at $P$ may be resolved parallel and perpendicular to the line of stroke. The horizontal components are equal and opposite.

The one acting through $P$ accelerates the reciprocating parts

The other through $O$ is an unbalanced force applied to the frame and this causes the frame to slide backwards and forwards on its mountings as the crank rotates

The two vertical components are equal and opposite and constitute a couple applied to the frame which attempts to rotate the frame in a clockwise sense.

As the triangles $Oba$ and $POM$ are similar:

Since $\displaystyle\frac{ba}{Ob}=\displaystyle\frac{S}{F}=\displaystyle\frac{OM}{OP}$ . Therefore $S\times OP=F\times OM$

Also, $\displaystyle\frac{ba}{Oa}=\displaystyle\frac{S}{Q}=\displaystyle\frac{OM}{PM}$

And $\displaystyle\frac{x}{OM}=\cos\phi =\displaystyle\frac{OP}{PM}$ . So that $\displaystyle\frac{OM}{PM}=\displaystyle\frac{x}{OP}$

Therefore, $\displaystyle\frac{S}{Q}=\displaystyle\frac{x}{OP}$ and $S\times OP=Q\times x$

$S\times OP = F\times OM = Q\times x$

The full effect on the engine frame of the inertia of the reciprocating mass is equivalent to a force $F$ along the line of stroke at $O$ and to the clockwise couple of magnitude $S$ , $OP$ .

The Inertia Force can be separated into two parts:

$\displaystyle Mr\omega ^2\cos\theta$ is called The Primary Force.

$\displaystyle Mr\frac{\omega ^2}{n}\cos2\theta$ which is called The Secondary Force

It is clear that the primary force is equivalent to the component along the line of stroke of the centrifugal force due to an equal mass $M$ rotating with the crank and at crank radius. Consequently, in the case of a single-cylinder engine, the primary reciprocating force could be balanced by a rotating mass on the other side of the crank pin. However this would introduce an unbalanced component of the centrifugal force of magnitude $Mr\omega^2\sin\theta$ perpendicular to the line of stroke. A compromise solution (partial balance) is usually applied, the inertia force being reduced to a minimum when 50% of the reciprocating mass is balanced.

The secondary force is similarly equivalent to the component of the centrifugal force of mass $M$ at radius of $\displaystyle\frac{r}{4n}$ rotating at $2\omega$ being coincident with the crank at inner dead-centre.

Multi-cylinder In-line Engines

The usual arrangement for multi-cylinder engines is to have the cylinder centre line all in the same plane and on the same side of the crankshaft centre line. This constitutes an "In-line" engine.

Notable exceptions to this rule are Vee enginesin which there are in effect two banks of in-line cylinders and Flat engines in which half the cylinders are arranged on opposite sides of the crankshaft.

Assuming that $\omega$ and $n$ are the same for all cranks, they can be omitted from all considerations of engine balance but they must be included when actual values are required.

Primary Balance

For couples and forces to be in balance:

$\displaystyle\sum Mr\cos\theta=0$ And $\displaystyle\sum Mrx\cos\theta=0$ (For all values of $\theta$ .)

The equations can be solved analytically or by polygons drawn in the relative crank directions. This is similar to those used for rotating balance.

Any gap remaining in the force or couple polygons represents ( to a certain scale) the maximum out-of-balance value. This occurs twice per revolution of the crank-shaft when its direction lies along the line of stroke.

Secondary Balance

For complete balance:

$\displaystyle\sum Mr\cos2\theta=0$ And $\displaystyle\sum Mrx\cos2\theta=0$

To solve graphically it is only necessary to draw vectors in the directions $2\theta$ (i.e. relative to any one crank taken as zero) and repeat it for primary balance.

The Partial Balance of Two-cylinder Locomatives

It is normal for the cranks to be at right angles and as a result the secondary forces are small and in opposite directions. As a result they are usually neglected and only the primary forces and couples are considered.

It is usual to balance about two-thirds of the reciprocating parts with masses fixed to the wheels.

The unbalanced vertical components of the reciprocating masses give rise to a variation of rail pressure known as Hammer Blow and a Rocking Couple about a fore and aft horizontal axis.

The unbalanced reciprocating masses cause a variation in draw-bar pull and a swaying couple about a vertical axis ( See examples 12 and 13)

Radial Engines - Direct and Reverse Cranks

The primary force for a reciprocating mass $M$ is equivalent to the resultant of the centrifugal forces of two masses $\displaystyle\frac{M}{2}$ rotating at a crank radius $r$ and at a speed $\omega$ , one in the forward direction of motion and the other in the reverse direction. Note that the "direct" and "reverse" cranks are equally inclined to the dead centre position.

Similarly the secondary force can be represented by direct and reverse cranks inclined at $2\theta$ to the inner dead centre and each carrying a mass $\displaystyle\frac{M}{2}$ at a radius of $\displaystyle\frac{r}{4n}$ rotating at a speed of $2\omega$ .

This method is particularly useful for examining the balance of radial engines with a number of connecting rods attached to the same crank. It is usually assumed that the crank and connecting rod lengths are the same for each cylinder, though from a practical consideration of design this is not generally true. (see Example 11)

Example 1 [imperial]

Problem

A motor armature is in running balance when weights of 0.130 oz. and 0.075 oz. ( There are 16 oz. in 1 lb.) are added temporarily in the positions shown in the planes $A$ and $D$ in the diagram.

23287/Balancing-of-Inertia-Forces-0003.png

If the actual balancing is to be carried out by the permanent addition of masses in the planes $B$ and $C$ each at 4 in radius, find their respective magnitudes and angular positions to the radius shown in plane $A$ .

Workings

The problem is to determine the masses in the planes $B$ and $C$ which will provide the same resultant force and couple , when rotating, as the given masses in the planes $A$ and $D$ .

It is possible to eliminate one of the "unknowns" (say $B$ ) by taking $B$ as the reference plane. The following table can now be constructed. The figures in bracket are added as and when they are calculated.

The $Mrx$ polygon can now be drawn. Note that $A$ is negative.

The angular position and magnitude of $Mrx$ for $C$ can now be measured since $C$ is the resultant of $A$ and $D$

The $Mr$ value of $C$ is now $Mrx$ for $C$ divided by the appropriate value of $x$ . With this information the Force polygon $b$ can be constructed in which $B$ and $C$ give the same resultant as $A$ and $D$ .

Solution

The $Mr$ value and angular position of $B$ can now be added to the table.

The required magnitudes of $B$ and $C$ are calculated by dividing $Mr$ by the corresponding value for $r$ .

Example 3 [imperial]

Problem

A shaft turning at a uniform speed carries two uniform discs $A$ and $B$ of mass 10 lb. and 8 lb. respectively. The mass centres of the discs are each 0.1 in. from the axis of rotation. The radii to the mass centres are at right angles. The shaft is carried in bearings $C$ and $D$ between $A$ and $B$ such that $AC = 1 ft.$ , $AD = 3 ft.$ , $AB = 4 ft.$

23287/Balancing-of-Inertia-Forces-0006.png

It is required to make the dynamic loading on the bearings equal and a minimum for any shaft speed by adding a mass at a radius of 1 in. in a plane $E$ .

Determine:

a) The magnitude of the mass in plane $E$ and its angular position relative to the radial through the mass centre in plane $A$ .

b) The distance of plane $E$ from plane $A$ .

c) The dynamic loading on each bearing when the mass in plane $E$ has been attached to the shaft which is turning at 200 r.p.m.

Workings

a) The above diagram shows the positions of the planes $A$ and $B$ and the bearings at $C$ and $D$ . The angular positions of $A$ and $B$ are also shown.

There are two possible solutions to investigate.

1. The bearing forces are equal and opposite. The masses $A$ , $B$ , and $E_1$ are in static balance, as shown in diagram (a) and a pure couple remains.

Choosing $A$ as the reference plane

$\left ( Mrx \right )_B=8\times 0.1\times 4=3.2\;lb.in.ft.$

The minimum out-of-balance couple is then obtained by drawing a perpendicular to the $E_1$ vector whose direction is known. (see diagram (b))

23287/Balancing-of-Inertia-Forces-0007.png

From the force triangle (diagram (a))

$\theta_1=\tan^{-1}0.8=38^0\;40'$

Therefore , The out-of-balance couple $=3.2\cos38^0\;40'=2.5\;lb.in.ft$

Dividing by the distance between the bearings gives a force on each bearing of:

$\left ( \frac{2.5}{2} \right )\left ( \frac{\omega ^2}{g} \right ) \;lb.$

2. The bearing forces are equal and in the same direction, so there is no moment about the central plane. Taking this as the reference plane, the couple triangle (c) is drawn ( Note that $A$ is negative) giving the direction of the mass $E_2$ . The minimum out-of-balance force can now be found by dropping a perpendicular on to the $E_2$ vector as shown in diagram (d).

By measurement the out-of-balance force = 1.25 lb.in.

This is a force of $\left ( \displaystyle\frac{1.25}{2} \right )\left ( \displaystyle\frac{\omega ^2}{g} \right )$ on each bearing

This is only half the value given in case (1.) and hence is the required solution.

fig 11.7 modified to change letters

23287/Balancing-of-Inertia-Forces-0008.png

$\left ( Mr \right )_{E2}=0.30\;lb.in.$

and since $r$ is given as 1 in. $M=0.3 lb.$

$\theta_2=\cot^{-1}0.8=51^0\;20'$ , or $141^0\;20'$ to $A$

b) From diagram (c)

$(Mrx)_{E2}=12.565$ in the opposite direction to the force

$\therefore \;\;\;\;\;x=-\frac{2.565}{0.3}=8.55\;ft.$

which is to the left of the central plane or 6.55 ft to the left of plane $A$

c) The dynamic load $=\displaystyle\frac{1.25}{2}\left ( \displaystyle\frac{200\times 2\pi }{60} \right )\displaystyle\frac{1}{32.2\times 12}=0.707\;lb.$