Inertia Forces and Couples

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. It is proportional to an object's mass.

Centre of gravity is the point in or near a body at which the gravitational potential energy of the body is equal to that of a single particle of the same mass located at that point, and through which the resultant of the gravitational forces on the component particles of the body acts.

A force is any influence that causes a free body to undergo a change in speed, a change in direction, or a change in shape. Force can also be described by intuitive concepts such as a push or pull that can cause an object with mass to change its velocity, i.e., to accelerate, or which can cause a flexible object to deform.

Inertia Forces and Couples with particular reference to Reciprocating Engines.

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Introduction

This section on Inertia Forces and Couples should be read in conjunction with those covering Velocity and Acceleration. There you will find details of both velocity and acceleration diagrams and Klein'\b{s construction} and all these are used in the Worked Examples.

Inertia Forces

If the centre of gravity of a body of mass $M$ has a linear acceleration $a$, then the resultant of the external forces acting on the body must be $Ma$. It follows that the external forces would be in equilibrium with a force of $Ma$ in the opposite direction.

This latter force is called the Inertia Force and it is numerically equal to the product of mass and acceleration of the centre of gravity. It acts in the opposite direction to the acceleration.

The system of external forces and inertia forces is treated as if in statical equilibrium. Note that the use of centrifugal force in governor problems is a particular example of this principle.

Inertia Couples

If the angular acceleration of a body is $\alpha$ , then in addition to the Inertia Force at the centre of gravity there is an Inertia Couple $I_G\;\alpha$, where $I_G$ is the moment of Inertia about the centre of gravity. As above, the direction of the inertia couple is opposed to the angular acceleration.

If the body is turning about a fixed axis $O$, then the inertia force and couple can be combined into a couple of magnitude $I_O\;\alpha$

The inertia force and couple may be reduced to a single force of magnitude $Ma$ which acts in a parallel direction at a distance $h$

$$h=\frac{I\;\alpha}{M\;a}$$

(1)

Engine Mechanisms

Inertia of reciprocating parts

It was shown in " Velocity and Acceleration Equation(20)" that the acceleration of the piston is given by:

$a=-r\omega^2\left ( \cos\theta+\frac{\cos2\theta}{n} \right )$

where $\theta$ is measure from the inner dead-centre position and the negative sign indicates that the acceleration is towards the crank. Thus, if $M$ is the mass of the reciprocating parts,

Inertia Force

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

(2)

and the effective force along the line of the crank $P$ is given by,

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

(3)

Where $pA$ is the force of the gas on the piston and is towards the crank.

23287/Inertia-Forces-and-Couples-0001.png

It can be seen from the diagram that $P$ is accompanied by a force in the connecting-rod of $\displaystyle \frac{P}{\cos\phi }$, and the useful turning moment on the crankshaft during the power or outstroke is,

$$\tau =\frac{P}{\cos\phi }\times OM=P\times ON$$

(4)

On the instroke the turning moment in the direction of rotation is $\displaystyle -P\times ON$ with the other senses remaining as before (See example 1)

The Inertia of the Connecting-rod.

The linear acceleration of the centre of gravity, $a$ , and the angular acceleration $\alpha$ can be found graphically by using the method described in "Velocity and Acceleration," or by Klien's construction. The Inertia Force and the Couple can then be calculated and reduced to a single force $Ma$ at a distance $\displaystyle\frac{I\;\alpha }{Ma}$ from the centre of gravity (See paragraph 2).

Assuming that the reaction at the small end is perpendicular to the line of stroke, the reaction at the big end, and hence the turning moment on the crank due to the inertia of the connecting-rod, can be determined (See example 3).

The Equivalent two-mass system

Any body of total mass $M$ can be replaced dynamically by two "point" masses $m_1$ and $m_2$ at distances $a$ and $b$ respectively from the centre of gravity. The choice of the masses and their positions must satisfy the following conditions:

$$m_1+m_2=M$$
(5)

$$m_1\;a=m_2\;b$$
(6)

$$m_1\;a^2+m_2\;b^2=M\;k^2$$
(7)
($k$ is the radius of gyration about $G$)

23287/Inertia-Forces-and-Couples-012.png

i.e., the new system has the same mass, the same position, and the same moment of inertia as the original.

The method of solving these equations is either to :

Fix one of the masses. This allows equations (5) (6) and (7) to be solved and give,

$m_1=\frac{M\;k^2}{a^2+k^2}$

$m_2=\frac{M\;a^2}{a^2+k^2}$

$b=\frac{k^2}{a}$

Fix $a$ and $b$ and calculate the two masses. This allows the calculation of $m_1$ and $m_2$ from equations (5) and(6) only.

In applying to a connecting-rod it is normal to fix $m_1$ and place it at the small end where it can be added to the reciprocating parts. $m_2$ will lie near to the big end and may be added to the rotating parts for a first approximation.

Example 1 [imperial]

Problem

A horizontal steam engine running at 240 r.p.m. has a bore of 15 in. and a stroke of 30 in. The connecting rod is 52.5 in. long and the reciprocating parts weigh 120 lb. When the crank is at $60^0$ past its inner dead-centre, the steam pressure on the covered side of the piston is 90 p lb/sq.in. while that on the crank side is 10 lb/sq/in.

Neglecting the area of the piston rod, determine:

a) The force in the piston rod.
b) The turning moment on the crankshaft.

Workings

From the given dimensions $r=15\;in.$, $N=\displaystyle\frac{52.5}{15}=3.5$, and $\theta=60^0$

a) Using equation (3)

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

$=(90-10)\frac{\pi }{4}\times 15^2-\frac{120\times 15}{32.2\times 12}\left ( \frac{240\times 2\,\pi }{60^2} \right )^2\left ( 0.5-\frac{0.5}{3.5} \right )$

$=14,100-1050=13,050\;lb.$

b) This part of the question has a graphical solution.

23287/Inertia-Forces-and-Couples-0010.png

From the drawing $ON = 14.8 in.$

$\tau =P\times ON =13050\times \frac{14.8}{12}=16,100\;ft.lb.$

Solution

a) The force in the piston rod is $13,050\;lb.$
b) The turning moment on the crankshaft is $16,100\;ft.lb.$

Example 2 [imperial]

Problem

A vertical internal combustion engine has a cylinder bore of 7 in. and a stroke of 8 in. The speed is 500 r.p.m., the connecting rod is 16 in. long and the weight of the parts moving with the piston is 45 lb. On the working stroke the gas pressure is 176 lb./sq.in. when the piston has moved downwards a distance corresponding to a rotation of $30^0$ of the crank.

Determine graphically the velocity and acceleration of the piston for this position. Find also the turning moment exerted on the crankshaft taking into account the weight and inertia of the piston.

Workings

For notes on velocity and acceleration diagrams see " Theory of Machines - Velocity and acceleration. Paragraphs 2 and 4"

In diagram (a) let $I$ be the instantaneous centre of the connecting-rod $CP$. The velocity of the crank pin is then given by:

23287/Inertia-Forces-and-Couples-0002.png

$v_c=4\times 500\times \frac{2\,\pi }{60}=209.2\;in./sec.$

The angular velocity of $CP$ is:

$\omega _{cP}=\frac{v_c}{IC}=\frac{209.2}{18.2}=11.5\;rad./sec$

And the velocity of the piston =$\omega _{cp}\times IP=11.5\times 11.08=2120\;in./sec.$

Now from (b) the acceleration diagram:

23287/Inertia-Forces-and-Couples-0003.png

The acceleration of $C$ is given by:

$oc=4\left ( 500\times \frac{2\,\pi }{60} \right )^2=10,950\;in./sec^2$

And the centripetal acceleration of $P$ to $C$ is:

$cp'=CP\times \omega _{cp}^2=16\times 11.5^2=2120\;in./sec^2$

Note. $pp$' is the tangential component (perpendicular to $CP$) and $op$ is along the line of the stroke.

Piston Acceleration, $op =10950\;in./sec^2$ downwards.

Piston Inertia Force =$\left ( \displaystyle\frac{45}{32.2\times 12} \right )\times 10,950$ upwards.

The effective downwards force is the sum of the gas and weight forces less the inertia force.

i.e. The effective Downwards Force =$176\times \pi \times \displaystyle\frac{7^2}{4}+45-1275=5530\; lb.$

Using the principles of work:

Turning Moment $X$ Angular Velocity = Force $X$ Linear velocity

Hence, Turning Moment =$\displaystyle\frac{5530\times 127.5}{500\times \displaystyle\frac{2\,\pi }{60}}=13,500\;lb.in.$

Alternatively, the turning moment = the force in $CP\;X\;OM$

$=\frac{\text{vertical force at P}}{\cos OPC}\times OM$

$=(\text {Force at P})\times ON$ ($ON$ is perpendicular to $OP$)

$=5530\times 2.46=13600\;lb.in.$

Solution

The turning moment is $13600\;lb.in.$

Example 4 [imperial]

Problem

A connecting- rod is 4 ft.long and 3 in. in diameter which is assumed to be uniform throughout its length. The crank is 1 ft. long and the engine speed is 240 r.p.m.

Draw the inertia - load diagram for the connecting-rod when the crank is at $60^0$ from the inner dead-centre position. Determine the value of the maximum bending moment and state its position.

Workings

Klien's construction (See Theory of Machines Velocity and Acceleration)is used to find $CM$, the acceleration image of the connecting-rod.

23287/Inertia-Forces-and-Couples-0005.png

For any point $Q$ on the rod, a line parallel to $OP$ is drawn to cut $CM$ in $q\;\,\q\,O\,.\,\omega^2$ is the acceleration of $Q$. The component acceleration perpendicular to the rod is $r\,O\,.\omega^2$ which is found by projecting $qO$ onto $OL$, which in itself perpendicular to $CP$.

It can be seen from the construction that the value of $rO$ varies uniformly from $LO$ (0.95 ft. to scale) at $C$ to $mO$(0.08) at $P$. The rate of inertia load will therefore vary from:

$\frac{\pi }{4}\left ( \frac{1}{4} \right )^2\times \frac{480}{32.2}\times 0.95\left ( \frac{240\times 2\,\pi }{60} \right )^2=438\;lb,/ft.\;\;\;\;\;\text{at C}$

$\frac{\pi }{4}\left ( \frac{1}{4} \right )^2\times \frac{480}{32.2}\times 0.08\left ( \frac{240\times 2\,\pi }{60} \right )^2=37\;lb,/ft.\;\;\;\;\;\text{at P}$

This loading is shown diagrammatically as a "trapezium" and may be broken down into a uniform rate of 37 lb./ft. throughout, together with a uniformly varying rate having a value of $O$ at $P$ and 401 lb./ft.at $C$.

Treating $CP$ as a "Beam" by taking moments about $C$ the reaction at $P$ is given by:

$\frac{37\times 4\times 2+\displaystyle\frac{1}{2}\times 401\times 4\times\displaystyle\frac{4}{3}}{4}=341\;lb.$

If x ft. is the distance of the point of maximum bending moment from $P$ then:

$\displaystyle\frac{1}{2}\times \displaystyle\frac{401x}{4}\times x+37x=341$ (i.e. the shearing force is zero)

from which $x = 2.65 ft.$ and the maximum bending moment is :

$341\times 2.65-37\times\frac{2.65^2}{2}-\frac{1}{2}\left ( 401\times \frac{2.65}{4} \right )2.65\left ( \frac{2.65}{3} \right )$

$=900-130-310=460\;lb.ft.$

Solution

The value of the maximum bending moment is $460\;lb.ft.$

Example 5 [imperial]

Problem

The connecting-rod for an internal combustion engine has a length between centres of 9 in. and a total weight of $3\displaystyle\frac{1}{2}\;in.$. Its centre of gravity is $6\displaystyle\frac{1}{2}\;in.$ from the small end and its radius of gyration about the centre of gravity (for oscillations in the plane of swing of the connecting-rod) is $3\displaystyle\frac{3}{4}\;in.$ . The weight of the piston and gudgeon pin is $4\displaystyle\frac{1}{2}\;lb.$; the stroke is $5\displaystyle\frac{1}{2}\;in.$ and the cylinder bore is 4 in.

Determine the magnitude and direction of the resultant force acting on the crank pin when the crank is at $30^0$ after the inner dead-centre and speed is 1,600 r.p.m.and if the effective gas pressure on the piston is 250 lb./sq.in.

Workings

Klien's construction produces the acceleration diagram OCLM shown in the diagram (a). (A description of Klein's construction will be found in " Theory of Machines - Velocity and Acceleration")

23287/Inertia-Forces-and-Couples-0006.png

Piston acceleration =$MO\times \omega ^2=2.85\left ( 1600\times\displaystyle\frac{2\,\pi }{60} \right )^2=80,000\;in./sec.^2$

The effective Force on the piston is given by (Compare this with equation (1))

$P=250\times 4\,\pi -\left ( \frac{4.4}{32.2} \right )\left ( \frac{80,000}{12} \right )=2210\;lb.$

For the connecting-rod the acceleration of $G$ is given by:

$gO\times \omega ^2=2.7\left ( \frac{1600\times 2\,\pi }{60} \right )^2=75,000\;in./sec.^2$

And the angular acceleration is given by:

$LM\times \frac{\omega ^2}{CP}=1.29\times \frac{\omega ^2}{9}=4010\;rad./sec.^2$

The Inertia Force for the connecting-rod, $R$, is given by:

$R=\left ( \frac{3.25}{32.2} \right )\left ( \frac{75,500}{12} \right )=635\;lb.$

In the reverse direction to the acceleration and acting at a distance :

$h=\frac{i\,\alpha}{R}=\frac{3.25}{32.2}\times \frac{3.75^2}{12}\times \frac{4010}{635}=0.75\;in.\;\;\;\text{from G}$

Let the components of the reaction at the crank pin be $F_t$ and $F_r$. Then by moments about $I$:

$F_t=\frac{P\times 6.5-R\times 2}{10.22}=1285\;lb.$

The Force diagram can now be drawn

23287/Inertia-Forces-and-Couples-0007.png

From the above diagram

$F_r=1120\;lb.$

And the Force on the crank pin $Q$ is shown to be:

$Q=1720\;lb.$

Example 6 [imperial]

Problem

A two cylinder vertical stem engine has cranks at right angles. The crank radius is 4 in. and the length of the connecting rod is 16 in. The reciprocating parts for each cylinder (i.e. Piston; piston rod; and crosshead ) weigh 50 lb. Each connecting-rod weighs 28 lb., the centre of gravity being 6 in. from the centre of the big end and the radius of gyration about an axis through the centre of gravity parallel to the crank shaft being $5\displaystyle\frac{1}{2}\;in.$. The rotating parts of the engine weigh 150 lb. with a radius of gyration of 7 in.

Calculate the total kinetic energy of the moving parts at the instant when one piston is at top dead centre, the speed then being 180 r.p.m.

Determine the error which would result from making the common assumption that one-third of the mass of the connecting-rod may be treated as being concentrated at the cross head pin and two thirds at the crank-pin.

Workings

a) Exact Analysis

For the Crank on top dead centre. The velocity of the crank pin is given by: $4\times 180\times \displaystyle\frac{2\,\pi }{60}=75.3\;in./sec.$

At the same moment the velocity of the piston is zero and thus the instantaneous centre of the connecting rod is at the piston. Therefore the angular velocity of the connecting-rod is: $\displaystyle\frac{75.3}{16}=4.71 \;rad./sec.$

The velocity of the centre of gravity of the connecting-rod is:

$\left ( 16-6 \right )\times 4.71=47.1\;ft./sec.$

NOTE The piston has NO velocity and the centre of gravity of the connecting rod has both linear and angular velocity. Hence the Kinetic Energy of the piston and connecting-rod is:

$\text{K.E.}=\frac{1}{22\times 32.2\times 12}\left ( 50\times 0+28\times 47.1^2+5.5\times 4.71^2 \right )=105\;in.lb.$

Now considering the other cylinder. The Crank is at $90^0$. The velocity of the crank pin for this cylinder is equal to the velocity of the piston. i.e. The Instantaneous Centre of the connecting-rod is at infinity and the angular velocity of its centre of gravity is zero.

Hence the Kinetic Energy is given by:

$\text{K.E.}=\frac{1}{22\times 32.2\times 12}\left ( 50\times 75.3^2+28\times 75.3^2 \right )=572\;in.lb.$

The Kinetic Energy of the moving parts is:

$\text{K.E.}=\frac{150\times 7^2}{2\times 32.2\times 12}\left ( \frac{180\times 2\,\pi }{60} \right )^2=3380\;in.lb.$

Total Kinetic Energy =$1105+572+3380=4057\;in.lb.$

b) Approximate Analysis

For the piston with crank at top dead centre, the Kinetic Energy becomes:

$\frac{1}{2\times 32.2\times 12}\left ( \frac{2}{3} \times 28\times 75.3^2 \right )=137\;in.lb.$

For the piston with crank at $90^0$

$\frac{1}{2\times 32.2\times 12}\left [ \left ( 50+\frac{28}{3} \right )75.3^2+\frac{2\times 28}{3}\times75.3^2 \right ]=572\;in.lb.$

The Kinetic Energy of the rotating parts remains the same.

The Total Kinetic Energy is now: $137+572+3380=4090\;in.lb.$

c) The error

The Error = $\displaystyle\frac{4090-4057}{4057}\times 100 = 0.82\%$

Solution

The error is $0.82\%$

Inertia Forces and Couples

Introduction

Inertia Forces

Inertia Couples

Engine Mechanisms

Inertia of reciprocating parts

The Inertia of the Connecting-rod.

The Equivalent two-mass system

Example 1 [imperial]

Example 2 [imperial]

Example 3 [imperial]

Example 4 [imperial]

Example 5 [imperial]

Example 6 [imperial]