5.2. Transmission of Forces in Pairs

In design of the machine one must know the forces which act on the individual links of the machine so that these links will not fail under the applied loads. In order to do this we must know the manner in which the forces are transmitted through the individual links. These forces are transmitted through the various pairs which connect the links, and the links themselves. For the present we will disregard the effects of friction in the various types of pairing and consider it in a later section.

In general the force transmitted from one link to another is always normal to the surfaces in contact, if friction is not considered. We will consider each type of pair individually and see how the forces are transmitted through these pairs. The turning pair or pin joint in machines we recognize as a bearing composed of a circular pin or shaft fitting in a hole in the link. Ball and roller bearings are also turning pairs. Since the forces transmitted through these bearings are always normal to the surfaces in contact, the resultant always passes through the center of the bearing. This can be seen from the examples shown in Fig. 5.1.

T

Fig. 5.1. Transmission of forces in a turning pair

Fig. 5.2. Transmission of forces in a sliding pair

he lower sliding pair is represented by a piston, or any link sliding relative to another with surface contact. The resultant force transmitted through the sliding pair is always normal to the surface of contact, but its line of action cannot be determined from a consideration of the joint alone. The other pairs on the sliding link usually determine its line of action. In the case of a sliding link with a single pin joint as the only other pair, as shown in Fig. 5.2, the resultant can be considered to pass through the pin joint.

This will be discussed in more detail when we discuss the force analysis of a slider crank mechanism.

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Fig. 5.3. Transmission of forces in higher pairs

n higher pairs (rolling pairs and higher sliding pairs), the resultant of the force transmitted through the pair must be normal to the surfaces at the point or line of contact and will pass through the center of the arc defining the surface at the point of contact. The individual elements in a ball or roller bearing represent rolling contact (Fig. 5.3) as well as a wheel rolling along a flat surface as shown schematically in Fig. 5.3 (a). Higher sliding pairs are represented by two gear teeth in mesh or a cam with flat faced follower as shown in Fig. 5.3 (b). In actual machine members point or line contact never exists so long as a force is transmitted through the pair. The members in contact deform under the load so that an area of contact exists, and we really have surface contact for either rolling or higher sliding pairs. The contact area is extremely small, however, and the forces can be assumed to act at a point.

5.3. Transmission of Forces in Pairs with Friction Included

Friction forces are encountered in machines at all points where there is a relative sliding of two members or where there is rolling contact between two members. In a large majority of machines the friction forces are so small in relation to the static or dynamic forces that they can be ignored in a force analysis. However in some cases the friction forces are of such magnitude that they must be considered in the force analysis. Friction forces are always associated with relative motion between links, or impending relative motion, and the question might arise as to why they are included in static analysis. There are two primary reasons for including a discussion of friction forces in the study of static forces:(1) although friction forces are present only when there is relative motion, or impending motion, between links of a mechanism, the magnitudes of the friction forces are usually considered as independent of the velocity or acceleration and they can therefore be considered as static forces, and (2) in cases where the dynamic forces arc the significant forces, the friction forces arc usually negligible in relation to the dynamic forces. When the dynamic forces are relatively large, the relative velocities are generally of such a magnitude that proper lubrication must be provided, thereby reducing the friction forces to a small value.

In the discussion of friction forces we will assume that the coefficient of friction is known. Consider a block sliding along a rough surface as shown in Fig. 5.4 (a)

Fig. 5.4. Reaction of a rough surface

The normal force between the two bodies isand the force, in the direction of sliding, is just sufficient to balance the friction force and keep the block sliding with a constant velocity. It is well known that the relation between the forcesandN is expressed by the equation

,

where is the coefficient of friction, a dimensionless quantity. The coefficient of friction is dependent on the materials in contact, the lubrication at the contact surfaces, the temperature, the humidity and other factors, and must be determined experimentally for any given case. Also it should be pointed out that the static coefficient of friction is higher than the sliding coefficient of friction. Its value is usually small where it is encountered in machines, ranging from a maximum of 0.3 for rough, dry, metal-to-metal sliding, to 0.002 for well lubricated bearings.

The friction force acts at the surfaces in contact and always in a direction tending to resist the motion. A free-body diagram of the block is shown in Fig. 5.4 (b). The friction force is a component of the force of link 2 on 1, and for equilibrium, the friction force must be equal and opposite to the external force F. The normal reaction must be equal and opposite to N. The resultant of the friction force and the normal component is shown in Fig. 5.4 (c) and we see that it makes an anglewith the normal component. This angle is dependent only on the coefficient of friction as can be determined from the relation

.

The angle is called the friction angle and we will make use of it in our analysis. It should be born in mind that friction angle is independent of the magnitudes of the forces so that the direction of the forceis fixed once the coefficient of friction is known, and it must pass through the intersection of the lines of action ofand.

Fig. 5.5. Forces in a rough turning pair

In the above discussion the effect of the friction force on lower sliding pairs was shown, and the friction force can be handled in the same manner for higher sliding pairs and for rolling pairs, in each case making use of the friction angle.

Consider Fig. 5.5 (a) which shows a pin, link 1, in a bearing which is part of link 2. The clearance between the pin and bearing is greatly exaggerated so that the force vectors can be shown. Assuming that motion or impending motion of link 2 is clockwise relative to the pin, the friction forces will be as shown in the Fig. 5.5 (a), tending to oppose the motion. A large view of the pin is shown in Fig. 5.5 (b) with the summation of the normal forces and the friction forces as single force vectors, and the resultant of the two, The resultantmust cause the same moment about the center of the pin that the friction force causes. If the moment arm of the resultant isr, the following condition must exist

where R is the radius of the pin, or.

Then

. (5.1)

From this we see that the resultant must be tangent to a circle about the center of the pin of radius . This circle is called the friction circle and we note that the friction circle radius is independent of the force, being a function of the bearing radius and the friction angle only. In an analysis where we are considering friction, we will always draw in the friction circle at each pin joint and the forces at the pin joint must then be tangent to the friction circle instead of passing through the center of the pin. Since the friction angle is small and the friction coefficient is never exact, we use an approximation of the friction circle radius in our analysis. For small angles the sine of the angle is very nearly equal to the tangent of the angle. Making this substitution in Eq. (5.1) we have

. (5.2)

This gives us a quick means of determining the friction circle radius sufficient accuracy once we have selected the coefficient of friction.