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.

It can be seen from the diagram that $P$ is accompanied by a foirce 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 by 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 there 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 $G$ )

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 (#4) (#5) and (#6) 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 (#4) and(#5) 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 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.