Balancing of Inertia Forces
Key Facts
For Rotating Masses In a Single Plane the weight of the balancing mass $W$ is given by:
$Mr\omega^2=\displaystyle\frac{W}{g}b\;\omega^2$ or$W=\displaystyle\frac{Mrg}{b}$ where,
- $M$ = Mass attached
- $r$ = Radius to point of attachment
- $b$ = Radius of rotation
For Reciprocating masses, the Inertia force of a reciprocating mass is: $Mr\omega ^2\left ( \cos\theta+\frac{\cos2\theta}{n} \right )$ where,
- $\displaystyle Mr\omega ^2\cos\theta$ is the Primary Force, and
- $\displaystyle Mr\frac{\omega ^2}{n}\cos2\theta$ the Secondary Force.
The Balancing of Inertia Forces with particular reference 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.
Road transport wheels are routinely balanced by adding small weights around the rim. These improve wheel bearing life, but more importantly prevent vibrations from 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 from "sticking".
Rotating Masses in one Plane
When a mass $M$, attached at a radius $r$ to a shaft, is rotated with an angular velocity $\omega$ , there is an outwards radial force of $M\,r\,\omega^2$. This produces 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 is 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 a direction that is radially inwards. 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 reciprocating 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 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 of the inertia of the reciprocating mass on the engine frame, 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 lines 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 engines in 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 Locomotives
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]
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.
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 $A$.
- 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 $D$.
- 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 2 [imperial]
A shaft 5 ft.long is supported in bearings 6 in. from each end and carries three pulleys, one at each end and one at the mid point. The three pulleys are out of balance to the extent of 6, 9, and 8 lb.in, but are keyed to the shaft so as to give static balance.
Find
- a) The relative angular settings of the three pulleys.
- b) The dynamic load on each bearing when the shaft makes 360 r.p.m.
It is assumed that the central pulley is out of balance by 9 lb.in. and that the angular position of the 6 lb.in is zero.
For static balance a triangle can be drawn whose sides are proportional to 6:8:and 9.
Relative to the 6 lb.in. it is possible to obtain four sets of positions for the 8 and 9 lb.in. The one chosen gives angles of $119^0$ for the 9 lb.in. and $258^0$ for the 8 lb.in. Note that it can easily be verified that the other possible positions will not affect the numerical answer to part b)
To find the out-of-balance couple:
Take the 6 lb.in pulley as the reference plane, lines are drawn proportional to $9\times 2.5=22.5$ and $8\times 5=40$ as shown in diagram (b) The out-of-balance couple is now proportional to the closing line ( shown dotted). This scales at 27.3 in the same units as the other lines (i.e. lb.in.ft.)
But, The centrifugal force = $lb.in \times \displaystyle\frac{\omega^2}{g}\;lb.$
Hence the out-of-balance couple is given by:
$\frac{27.3}{32.2\times 12}\left ( \frac{360\times 2\pi }{60} \right )=101\;lb.ft.$
Since the bearings are 4 ft. apart, the load on each bearing is:
$\frac{101}{4}=25.2\;lb.$
- The relative angular settings: $119^0$ for the $9 lb.in.$ and $258^0$ for the $8 lb.in.$
- The dynamic load is $25.2\;lb.$
Example 3 [imperial]
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 $AC = 1 ft.$, $AD = 3 ft.$, $AB = 4 ft.$
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.
- 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))
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
(c) (d)
$\left ( Mr \right )_{E2}=0.30\;lb.in.$
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 $=\displaystyle\frac{1.25}{2}\left ( \displaystyle\frac{200\times 2\pi }{60} \right )\displaystyle\frac{1}{32.2\times 12}=0.707\;lb.$
- The magnitude of the mass is $M=0.3\;lb.$
- The distance of plane $E$ from plane $A$ is $6.55\;ft.$
- The dynamic loading is $0.707\;lb.$
Example 4 [imperial]
An air compressor has four vertical cylinders 1, 2, 3, 4, in line and the driving cranks, at $90^0$ intervals, reach their uppermost positions in this order. The cranks are 6 in. radius, the connecting rods 20 in. long, and the cylinder centre lines $15\displaystyle\frac{1}{2}$ apart. The reciprocating parts of each for each cylinder weigh 45 lb. and the speed of rotation is 400 r.p.m.
Show that there are no out-of-balance primary or secondary forces and determine the corresponding couples indicating the position of No.1 crank for maximum values. The central plane of the machine may be taken as the reference plane.
The relative positions of the cranks are shown in the diagram. The following table is set out to give quantities proportional to the inertia forces and couples.
Since the inertia force at each crank is the same, the primary forces form a square:-
And the secondary forces al fall in the same line:-
The primary couples are shown on the following diagram.
- 1 and 2 have been taken as negative since the reference plane is to be taken as the central plane.
- The out-of-balance couple is proportional to the dotted line which has a length by symmetry of $(1+3)\;\sqrt{2}=11,830\;lb.in^2$ units from the table.
- The actual couple is obtained by multiplying by $\displaystyle \frac{\omega ^2}{g}$
i.e. $\displaystyle\frac{11,830}{32.2\times 12}\left ( \displaystyle\frac{400\times 2\pi }{60} \right )^2=53,700\;lb.in.$
- This will occur when crank 1 is at $45^0$ and $225^0$ i.e the dotted line to be along the line of stroke.
Similarly the following diagram shows the secondary couples:
- The out-of-balance couple is proportional to $1+4-(2+3=8360)\;lb.in.^2\;units$
- The actual couple is given by $\displaystyle\frac{8360\;\omega ^2}{gn}=\displaystyle\frac{8360}{32.2\times12}\left ( \displaystyle\frac{400\times2\pi }{60} \right )^2\times \displaystyle\frac{6}{20}=11,400\;lb.in.$
- This will occur when crank 1 is at $0^0$ $\;90^0$,$\;180^0$,and $\;270^0$
- This will occur when crank 1 is at $0^0$ $\;90^0$,$\;180^0$,and $\;270^0$.
Example 5 [imperial]
A four-crank engine has the two outer cranks set at $120^0$ to each other and their reciprocating masses are each 800 lb. The distances between the planes of rotation of adjacent cranks are 18, 30, and 24 in.
If the engine is in complete primary balance, find the reciprocating mass and the relative angular position for each of the inner cranks.
If the length of each crank is 12 in., the length of the connecting-rods 48 in. and the speed of rotation 240 r.p.m., what is the maximum secondary unbalanced force?
Taking No.2 Crank as the reference plane, the following table can be constructed.
- For complete balance the $Wr$ and $Wrx$ polygons , drawn in the direction of $\theta$ , must close.
- The couple polygon $Wrx$ is drawn first ( diag. a) from which $W_3$ is found to be 1730;lb. and $\theta=314^0$
- The force polygon can now be constructed. Note that $12\,W_3=20,700$ and for ease of construction this can be written as $2.16\times 9600$.
- From the above polygon $12\;W_3=2.12\times 9600$ and $\theta_2=161^0$ i.e. $W_3=1700\;lb.$
- To construct the secondary force polygon vectors proportional to $Wr$ are drawn in the directions$2\theta$.
- The closing dotted line scales at $4.85\times 9600\;lb.in.\,units$
$\;\;\;\;\;\;\;=\frac{4.85\times 9600}{32.2\times 12}\left ( \frac{2400\times2\pi }{60} \right )^2\times \frac{12}{48}=19,000\;lb.$
- The couple polygon: $W_3=1730;lb.$ and $\theta=314^0$
- The force polygon: $W_3=1700\;lb.$ and $\theta_2=161^0$
- The maximum secondary unbalanced force is $19,000\;lb.$
Example 6 [imperial]
The diagram shows the arrangement of the cranks in a four-crank symmetrical engine in which the weights of the reciprocating parts at cranks 1 and 4 are each equal to $W_1$ and at cranks 2 and 3 are each equal to $W_2$.
Show that the arrangement is balanced for primary forces and couples and for secondary forces provided that:
$\frac{W_1}{W_2}=\frac{\cos\theta_2}{\cos\theta_1}\;:\;\;\;\frac{a_1}{a_2}=\frac{\tan\theta_2}{\tan\theta_1}\;;\;\;\;\;\cos\theta_1\times\cos\theta_2=\frac{1}{2}$
Find also the value of the out-of-balance secondary couple when the system rotates at $\omega\;rad./sec.$
The primary forces are show in diagram (a)
- The lines for cranks 1 and 4 are proportional to $W_1$
- Similarly the lines for 2 and 3 are proportional to $W_2$
- For closure along the horizontal datum
$W_1\cos\theta_1=W_2\cos\theta_2$
Or
Diagram (b) shows the primary couples.
- 1 and 4 are proportional to $W_1a_1.$
- 2 and 3 are proportional to $W_2a_2.$
- Note that the directions of 1 and 2 are reversed as the central plane has been taken as the reference.
- For closure along the vertical datum
Combining equations (1) and (2)
The secondary angles are shown on the following diagram and assume:
- Anticlockwise rotation.
- $\theta$ is measured from zero.
- For the secondary forces to balance:
$W_1\cos\2\theta_1=-W_2\cos\theta_2$
Substituting from equation (2) and expanding:
$\cos\theta_2(2\cos^2\theta_1-1)=-\cos\theta_1(2\cos^2\theta_2-1)$
$\therefore \;\;\;\;\cos\theta_1\times \cos\theta_2(2\cos\theta_1+2\cos\theta_2)=\cos\theta_1+\cos\theta_2$
Thus, $\cos\theta_1\times\cos\theta_2=\displaystyle\frac{1}{2}$
The unbalanced secondary couple is shown in the diagram.
$=\left ( \frac{4r\omega^2}{gn} \right )\left [ W_1a_1\sin\theta_1\cos\theta_1+W_1\left ( \frac{\cos\theta_1}{\cos\theta_2 }\right )a_1\left ( \frac{\tan\theta_1}{\tan\theta_2} \right )\sin\theta_2\cos\theta_2 \right ]$
$=\left ( \frac{4r\omega_2}{gn} \right )W_1a_1\sin\theta_1(\cos\theta_1+\cos\theta_2)$
combing with equation (4)
$=\left ( \frac{4r\omega^2}{gn} \right )W_1a_1\sin\theta_1\left ( \cos\theta_1+\frac{1}{2\cos\theta_1} \right )$
- The value of the out-of-balance secondary couple is $=\left ( \displaystyle\frac{4r\omega^2}{gn} \right )W_1a_1\sin\theta_1\left ( \cos\theta_1+\displaystyle\frac{1}{2\cos\theta_1} \right )$
Example 7 [imperial]
The five cylinders of a vertical in-line engine are similar in detail and symmetrical about the central one. The crank settings are as shown in the diagram.
By sketching suitable vector diagrams verify that the engine is in balance for primary and secondary forces.
Show in a similar manner that the engine is in balance for primary couples if the distances of the outer and inner cylinders from the central cylinder are in the ratio $k=1.618$ and that the out-of-balance secondary couple then has the following magnitude :
$4.252\left ( \frac{W}{g} \right )\omega^2\left ( \frac{r_2}{l} \right )a$
Where $W$ is the weight of each reciprocating mass; $l$ is the connecting-rod length; $a$ is the distance between the central and inner cylinders and $\omega$ the speed of the engine.
- The relative angular position of the cranks is shown in diagram (a) The inside numbers refer to the primary forces and the outside numbers to the secondary forces (The angels relative to crank 3 have been doubled)
- Since the inertia force for each cylinder is proportional to $\cos\theta$ (primary) or$\cos2\theta$ (secondary), each force polygon closes as shown in (b)
The primary couples are represented by the following diagram.
- The central plane if the reference plane
- Distances to the left are negative (i.e. the directions on the diagram for cranks 1 and 2 are reversed)
- The polygon will close if $ln = mn$ i.e. $ka\cos45^0=a\cos18^0$
$\therefore \;\;\;\;k=1.618$
The out-of-balance secondary couple is shown as $pq$ on the diagram.
$pq=\left ( \frac{W}{g} \right )r\omega^2\left ( \frac{r}{l} \right )^2\left ( a\cos45^0+ka\cos18^0 \right )$
$=4.252\left ( \frac{W}{g} \right )\omega^2\left ( \frac{r^2}{l} \right )a$
Example 8 [imperial]
A waterworks pumping engine has two cranks at right angles, radius 2 ft. in planes 10 ft. apart. Each crank has two connecting rods attached to it, one driven from a vertical high pressure cylinder in tandem with a pumping cylinder, the total reciprocating mass being 10,000 lb. and the other one from a horizontal low pressure cylinder in tandem with another pumping cylinder. The total reciprocating mass of this is 12,000 lb. The engine runs at 30 r.p.m.
Find in terms of the angle made by the leading crank with the vertical, expressions for the vertical and horizontal components of the primary inertia forces and of the couples referred to the central plane. Hence calculate the maximum values of the resultant primary force and couple.
The arrangement of the crank is shown in the diagram.
The leading crank is shown at $\theta$ to the vertical and the following crank at $theta$ to the horizontal. Each crank will initially be considered separately:
The radius of both cranks is 2 ft. and they revolve at $\pi$ radians /second.
- 1. LEADING CRANK
The Primary force is $Mr\omega^2\cos\theta$ and hence :
- The Vertical Force = $\displaystyle\frac{10,000}{32.2}\times 2\times \pi ^2\cos\theta=6130\cos\theta =a\cos\theta$
Say, where $\;a=6130$
- The vertical Couple = $5a\cos\theta$ About the central plane
- The Horizontal Force =$\displaystyle\frac{12,000}{32.2}\times 2\pi ^2\cos(90^0+\theta)=-b\sin\theta$
(where $b=7360$)
- The Horizontal Couple =$-5b\sin\theta$
- 2. FOLLOWING CRANK
- The vertical Force =$a\cos(270^0+\theta)=a\sin\theta$
- The vertical Couple =$-5a\sin\theta$
- The Horizontal Force =$b\cos\theta$
- The Horizontal Couple =$-5b\cos\theta$
The two sets of forces and couples can now be combined as follows. To simplify the calculations the substitution $\phi =\theta+45^0$ has been used.
- 3. COMBINED
- Vertical Force =$a(\cos\theta+\sin\theta)=a\sqrt{2}\times \sin\phi$
- Vertical Couple =$5a(\cos\theta-\sin\theta)=a\sqrt{2}\times \cos\phi$
- Horizontal Force =$b(\cos\theta-\sin\theta)=b\sqrt{2}\times \cos\phi$
- Horizontal Couple =$-5b(\cos\theta+\sin\theta)=-5\sqrt{2}\times b \sin\phi$
- 4. THE RESULTANTS
- The Force $=\sqrt{2a^2\sin^2\phi+2b^2\cos^2\phi}=\sqrt{2a^2+2(b^2-a^2)\cos^2\phi}$
- The maximum force will occur when $\cos\phi=1$ and since $b>a$
Maximum Force =$10,400\;lb$
- The Couple $=\sqrt{50a^2\cos^2\phi+50b^2\sin^2\phi}=\sqrt{50a^2+50(b^2-a^2)\sin^2\phi}$
- The Maximum Couple will occur when $\sin\phi=1$ and thus
Maximum Couple =$b\sqrt{50}=52,000\;ft.lb.$
- The Maximum Force =$10,400\;lb$
- The Maximum Couple = $52,000\;ft.lb.$
Example 10 [imperial]
An internal combustion engine is arranged with three cylinders opening into a common combustion chamber as shown in the diagram. The centre lines of the cylinders are spaced at $120^0$ to one another. The reciprocating mass per cylinder is 4.6 lb., the piston stroke is 3.5 in. and the connecting-rods are 6 in. long.
Show that when $\alpha= \beta=\gamma$, the engine is in complete balance for both primary and secondary forces.
Calculate the maximum out-of-balance primary and secondary forces when crank 3 is advanced $15^0$ relative to cranks 1 and 2 so that $y=\alpha+15^0=\beta+15^0$ for all positions of the cranks. The engine speed is 1200 r.p.m.
- The primary forces in cylinders 1, 2, and 3 are proportional to $\cos\alpha,\;\;\cos\beta,\;\;\;and\;\;\;\cos\gamma$ respectively and act along their lines of stroke.
If $\aphha=\beta=\gamma$, the primary forces are all equal and since the lines of stroke are concurrent and equally spaced, they are in equilibrium.
- The secondary forces, being proportional to $\cos2\alpha,\;\;\cos2\beta,\;\;\cos2\gamma$ are equal and as they act along the line of stroke, in balance.
When crank 3 is advanced by $15^0$:
Each Primary Force for cylinders 1 and 2 is equal to: $\frac{4.6}{32.2}\times \frac{3.5}{24}\left ( \frac{1200\times 2\pi }{60} \right )^2\;\cos\alpha=328\cos\alpha$
These two forces act along their respective lines of stroke. Together they form a resultant in the direction of the line of stroke of cylinder 3
Hence the out-of-balance-primary force is given by:
$328[\cos\alpha-\cos(\alpha+15^0)$ (When $\gamma=\alpha+15^0$) $=328\times 2\sin\;\frac{1}{2}(2\alpha+15^0)\sin7 \displaystyle\frac{1}{2}^0$
The maximum Primary Force occurs when $\alpha=82.5^0$ and is given by: $656\;\sin7\displaystyle\frac{1}{2}^0=85.6\;lb.$
- $n=\displaystyle\frac{6}{1.75}=3.43$ and the secondary forces, which act along their lines of stroke, for cylinders 1 and 2 are:
$\frac{328}{3.43}\cos2\alpha=95.7\cos\alpha$
- The resultant secondary force which acts in the direction of the line of stroke of 3 is:
$95.7[\cos2\alpha-\cos2(\alpha+15^0)]=95.7\times 2(2\alpha+15^0)\sin15^0$
- The Maximum Secondary force therefore =$191.4\sin15^0=49.5\;lb.$
- The out-of-balance-primary force is $328\times 2\sin\;\displaystyle\frac{1}{2}(2\alpha+15^0)\sin7 \displaystyle\frac{1}{2}^0$
- The Maximum Secondary force =$49.5\;lb.$