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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'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.

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

(a) Inertia of reciprocating parts

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

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

(2)

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:-

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

(3)

and the effective force along the line of the crank P is given by:-

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

(4)

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:-

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

(5)

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)

(b) 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 Coouple can then be calculated and reduced to a single force Ma at a distance $\frac{I\.\alpha }{Ma}$ 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:-

$m_1+m_2=M$
(6)

$m_1\;a=m_2\;b$
(7)

$m_1\;a^2+m_2\;b^2=M\;k^2$
(8)
(k is the radius of gyration about G)

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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 (6) (7) and (8) to be solved and give:-

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

(9)

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

(10)

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

(11)

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Fix a and b and calculate the two masses. This allows the calculation of $m_1$ and $m_2$ from equations (6) and(7) 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.

Examples

Example 1

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. (U.L.)

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Example 2

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.

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Example 3

The single cylinder engine shown in the diagram, has a crank BC of 5 in. and a connecting-rod AB of 20 in. long. The piston weights 60 lb. The connecting -rod weighs 40 lb. with a centre of gravity G 6 in. from B and a radius of gyration about G of 8 in. The acceleration of A and that of G and the angular acceleration of the rod are shown for the position shown in the diagram.

Find the turning moment which must be applied at the crankshaft to overcome the inertia of the piston. Find also the single force required for the acceleration of the rod and considering this as the resultant of the reactions at its two ends, find the corresponding turning moment at the crankshaft.

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Example 4

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. (U.L.)

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Example 5

The connecting-rod for an internal combustion engine has a length between centres of 9 in. and a total weight of $3\tfrac{1}{2}\;in.$ . Its centre of gravity is $6\tfrac{1}{2}\;in.$ 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 $3\tfrac{3}{4}\;in.$ . The weight of the piston and gudgeon pin is $4\tfrac{1}{2}\;lb.$ ; the stroke is $5\tfrac{1}{2}\;in.$ 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. (U.L.)

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Example 6

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 $5\frac{1}{2}\;in.$ . 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. (U.L.)

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Example 7

The crank OA is shown in the diagram. It is 4 in.long and rotates clockwise at 200 r.p.m.. A rod AB passes through a slot in a pin C which can be swiveled about an axis parallel to the crankshaft. OC=6 in. AB is of uniform cross-section 12 in. long and weighs 14 lb.

When COA = $120^0$ find the torque required on the crankshaft to overcome the inertia of the rod AB. Do not include gravitational forces and neglect friction (U.L.)

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Example 8

The crank AB of the linkwork ABCD, shown in the diagram, rotates at a speed of 120 r.p.m. clockwise. The mass of BC is 10 lb. and its centre of gravity is 6 in. from B The radius of gyration of BC about its centre of gravity is 4.5 in. For the position when AB is 60^0 from the horizontal direction, as indicated, determine the force acting at the hinges B anad C due to the inertia of BC (U.L.)

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