## Forces in the Valve Train of an Internal Combustion Engine

1. Introduction
To allow the maximum amount of mixture (or air for a diesel) to be drawn into the cylinders it is desirable that the inlet valve(s) open as quickly as possible, with as high a lift as possible and remain open for as long as possible followed by rapid closing. To achieve these aims the components in the valve train are subjected to high accelerations (and hence forces) as the inertia of the components has to be overcome.
During the valve opening phase the cam provides the necessary force and during the closing phase the valve spring(s) provides the forces (except in an engine with desmodromic valve operation - where a second cam provides the closing forces. Currently only Ducati motorcycles use this system).
To simplify the analysis it will be assumed that the cam profile is symmetrical, ie, where the closing phase is the 'mirror image' of the opening phase. In practice this may not be the case.

2. Practical Details
Although high accelerations are needed to give rapid opening and closing, too rapid a change in acceleration - the 'jerk' or 'jerk rate' - will give rough operation due to the sudden changes in forces. For this reason cam profiles are designed not to give very rapid changes in accelerations.
It may also be noted that as higher forces can more easily be provided by the cam than by the valve springs, it is common to us higher accelerations when starting the opening of the valves and when slowing their closing at the end of the closing phase. These aspects are controlled by the cam, whereas the slowing of the valve at the end of the opening phase and the acceleration of the valve at the start of the closing phase are controlled by the valve springs.

3. Analysis - Acceleration of Cam Follower
The analysis of anything other than a simple configuration can be quite complex. The analysis will depend upon the type of follower and the detailed geometry. Because of these difficulties with the analysis it was common for accelerations to be determined graphically.

In this section we will make some simplifying assumptions that a knife edge follower is being used. This will not be very accurate, but will give some idea of values.

i) The most simple assumption for analysis is to assume that the opening and closing is simple harmonic motion (SHM). Assume that the engine speed is 4000 revs/minute, this gives a cam shaft rotation speed of 2000 revs/minute (in a 4 stroke engine the cam turns at half the crankshaft speed) so the time taken for 1 revolution of the cam shaft is 0.03 seconds
Also assume that the cam is opening and closing the valve for 120o of its rotation.
Hence the complete valve cycle is completed in 1/3 camshaft revolution, or 0.01 seconds.
The equation describing SHM is: displacement = amplitude x cos(omega x t)

where t = time and
omega is the 'angular velocity' of the system in rad/s, and is equal to (2 x 3.14159)/(the time for 1 cycle)

So the omega value here is 2 x 3.14159/0.01 = 628.3 rad/sec.

Differentiating the expression for displacement gives:

velocity = - omega x amplitude x sin(omega t), then differentiating again:

acceleration = - (omega)2 x amplitude x cos(omega t)

The maximum acceleration occurs when the term cos(omega t) has the maximum value of 1, this occurs at the extremes of the motion.

If the cam has a lift of 20 mm, the amplitude of the motion is 10 mm, and the maximum acceleration is given by: accelerationmax = (628.3)2 x 0.01 = 3948 m/s2

ii) An alternative assumption is to assume that that the peak of the cam is finished with specific constant radius with a centre at a distance 'e' from the centre of the cam. The term 'e' may be called the eccentricity. The peak radius is often blended to the the base circle by a large radius curve.
In this case the acceleration of a 'mushroom' - flat foot - follower in the vicinity of the maximum cam lift is given by: (cam shaft angular velocity)2 x eccentricity. If the cam base circle is blended to the nose radius by a large radius, the maximum acceleration occurs when the follower is on this larger radius and is equal to the large blend radius x ( cam shaft angular velocity)2

4. Analysis - Valve Train Forces.
If the cam shaft is a direct acting overhead unit then all the components in the (short) valve train will undergo the same acceleration. The exception to this is the spring, for which an allowance may be made by using 1/3 of its mass as an effective mass.

Maximum force will then be:

F = max. acceleration x (masses of tappet + retainer + collets + valve + 1/3 of spring)

At least this much force must be generated by the valve spring at its maximum compression. Use the spring design applet with measurements and / or data from the workshop manual to check this.

The analysis becomes more complex when the engine does not have an overhead cam shaft, or when there is an overhead cam shaft that operates the valves by rockers. When this is the case the intertia of the rocker needs to be considered and the 'lever effect' of the rocker needs to be considered - as the two side arms of the rocker are frequently unequal.

As an example consider the layout shown below:

Let us assume the following values and dimensions:

• 1) tappet 200 gm
• 2) push rod 250 gm
• 3) valve, collet, retainer and 1/3 spring, total 300gm
• 4) rocker arm 200 gm
• 5) L1 20 mm
• 6) L2 40 mm

The easiest way to carry out the calculations is to determine the equivalent mass of the valve train at the valve side, then use:

Fvalve side = mtotal equiv. on valve side x"valve side.

Components in group (3) above, are the items on the valve side with a mass of 300 gm.

The rocker now needs to be represented as an equivalent mass at the valve side.
First the moment of inertia needs to be estimated. Depending upon the design of the rocker it may be possible to perform a 'compound pendulum' experiment to determine this. If this is not possible then an estimate of the radius of gyration of the rocker about the rocker shaft (kp) can be made. The radius of gyration is the radius at which the mass would have to be concentrated to give the same moment of inertia. For a uniform slender rod, length, L, pivoted about its centre, this is L/(12)0.5
For this rocker it will be assumed that the radius of gyration about the rocker shaft is 10mm, hence the moment of inertia about the rocker shaft is mass x (kp)2

This gives Irocker = 0.2 x (0.01)2 kg m2 = 0.000020 kg m2

To determine the equivalent mass at the distance of the valve axis from the rocker shaft pivot, L2, start with the free body diagram below:

Let the linear acceleration (upwards) of the point on the rocker where F is acting be x"

Then writing x"/L2 = alpha into the equation above gives (NB I is upper case i, NOT 1):

F L2 = I x"/L2 and re-arranging gives:

F = I x"/(L2)2

So the equivalent mass of the rocker on the valve side is: I/(L2)2 kg

In the analysis below F on the valve side will be referred to as Fvalve side. From equilibrium of the rocker, take moments about the rocker shaft, so Fvalve side L2 = Fcam side L1.

Let the mass of the components on the cam side (1 + 2 from list above) be mcam side and let the force required to give this mass a (downward) acceleration x"cam side be Fcam side so:

Fcam side = mcam side x"cam side

Both sides of the rocker arm have the same angular acceleration, alpha:

alpha = x"cam side/L1 = x"valve side/L2

The force on the valve side to accelerate the cam side at x"cam side is given by

Fvalve side L2/L1 = mcam side x"cam side

and the force on the valve side to give the mass on the cam side an acceleration of x"valve side is given by:

Fvalve side L2/L1 = mcam side x"valve side L1/L2 or

Fvalve side = mcam side x"valve side (L1/L2)2

So the equivalent mass of the cam side at the valve side = mcam side (L1/L2)2

For the example above:

Equivalent mass at valve side = mass of valve + retainer + collet + 1/3 of spring + Irocker /(L2)2 + (mass tappet + mass push rod)(L1/L2)2

Putting in values:

meq valve side = 0.3 + 0.000020/(0.04)2 + 0.450(20/40)2

meq valve side = 0.425 kg

If the maximum acceleration required by the valve when being driven by the spring is 3000 m/s2, then the spring force needed is 3000 x 0.425 = 1275 N.

Dr David J Grieve, 23rd February 2004.