5.5. Illustrative Problems

Problem 5.1. Consider the slider crank mechanism shown in Fig. 5.7 (a) with an external force , acting on the slider3.The mechanism is kept in equilibrium by an external couple applied to the crank 1. Determine the forces acting on all the pin joints, the normal force between the slider and the fixed link 4 and the external torque applied to the crank to hold the mechanism in equilibrium.

Solution. Isolate each link and make a free-body diagram of the links with the forces acting thereon as shown in Fig. 5.7 (b). For each link all the forces acting on the link are indicated, regardless of what is known about these forces. If the forces are completely known, they are shown in solid lines along the proper line of action with the arrowhead, indicating sense, at the point of application of the force, such as force F acting on link 3. If only the direction of the force is known, the force is represented by a straight dashed line in the direction of the force, but not in contact with the link, with arrowheads at each end of the dashed line indicating the sense is not known such as the force of link on link 3. Note that when the line of action or point of application of the force is not known, the force is not drawn in contact with the link.

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Fig. 5.7. Static force analysis for slider-crank mechanism

f only the point of application of a force on a link is known, it is represented by a curved dashed line with the arrowhead at the point of application, as the force of link2 on link 1 which must be applied at the pin joint. An unknown couple acting on a link will be indicated by a dashed arc with arrowheads at each end, as the moment acting on link 1.

External forces will be labeled with capital letters and a subscript to indicate the link on which they act. Forces imposed by one link on another will be labeled with a capital F and subscripts with a solidus between them, as , indicating the force exerted by link2 on link 3. Couples will be labeled with a capital .

Now returning to our free-body diagrams we find that link 3 has three forces acting on it; the external force F which is completely known, the force applied by link 4, known in direction only, and the force exerted by link2, which is completely unknown except for its point of application. Link2 has two forces acting on it, and.The only thing known about these forces is the point of application at the pin joints. Link1 is acted upon by the two forces, and, and the external moment.

Each of these links must be in equilibrium under the influence of the forces applied to it. We also know from Newton's third law that the forces exerted by two links on each other have the same magnitude and direction but opposite sense. The condition of equilibrium and Newton's law give us sufficient information to determine the forces acting on all the links. In some cases the forces on a link can be determined by considering only the single link in equilibrium, but in many cases we must deal with two or more links simultaneously in order to determine the forces, making use thereby of Newton's third law.

Consider the free-body diagram of link 3. There are four unknown quantities which must be determined in order to completely define the forces acting on the link. These are: the magnitude of , its direction, the magnitude of, and the line of action of. The equations of equilibrium can be used to determine only three unknown quantities for the plane force system. Therefore we cannot determine the forces acting on link3 by considering only equilibrium of this link.

Link 2 also has four unknowns, the magnitude and direction of and the magnitude and direction of, and therefore cannot be analyzed by equilibrium considerations of this link alone. Link1 has five unknowns, the magnitude and direction of the magnitude and direction of and the moment of. So we find that none of the links can be analyzed by considering equilibrium alone, and our analysis must be made by considering two of the links simultaneously.

If we take links 1 and 2 together we know from Newton's third law that must be equal and opposite to. This eliminates two of the unknowns from the combination leaving a total of seven unknowns for the two and the equilibrium equations for the two give us only six equations, not sufficient for the determination of the seven unknowns. However if we consider links2 and 3 simultaneously, since is equal and opposite to we have a total of only six unknowns for the combination and the six equilibrium equations are sufficient to determine these six unknowns.

Knowing that we can analyze links 2 and 3 by considering them simultaneously, we will start our analysis with these two. Looking at link 2 we see that it is a two force member and for it to be in equilibrium under the action of two forces, they must be collinear. Then, since the point of application of the two forces is at the pin joints, their direction must be as shown in the free-body diagram shown in Fig. 5.7 (c). It should be noted that in the absence of friction a torque cannot be applied to a link through a pin joint; therefore, link 2 has no couple acting on it. The magnitude of the forces acting on link 2 cannot be determined from equilibrium conditions since the only condition necessary for equilibrium is that the two forces are equal and opposite and forces of any magnitude will satisfy these conditions.

Turning now to link 3, we know by Newton's third law that the direction of must be the same as the direction of, and a free-body diagram can be drawn for3 as shown in Fig. 5.7 (c). Now the forces on link 3 contain only three unknown quantities and they can be determined from the equations of equilibrium. A force polygon can be drawn to determine the magnitudes of andas shown in Fig. 5.7 (d) by remembering that for a body in equilibrium the force polygon must be closed.

The magnitude and direction of one force and the direction of the other two is enough information to draw the closed polygon. The line of action of must pass through the intersection of the other two forces,and, since link3 is a three force member.

The magnitude and sense of the forces acting on link 2 are now readily determined since is equal and opposite toandis equal and opposite to, and we see that link2 is in compression.

Turning now to link 1, is equal and opposite to. Since only three unknowns remain, they can be determined from the equilibrium equations. Since the external momentis a couple, the resultant ofandmust be a couple which balances the external couple. The two forces must balance, sois equal and opposite toand the external couple necessary to hold the mechanism in equilibrium will be equal and opposite to the couple which is the resultant of these two forces. The magnitude ofthen is equal to the forcemultiplied by the distancex as shown in the free-body diagram; its direction will be clockwise.

In general we must start the analysis with the link on which the known external forces are acting, or this link in combination with an adjoining link.

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Fig. 5.8. A six-bar mechanism

roblem 5.2. The six-bar mechanism shown in Fig. 5.8 (a) with the external known force acting on links3, force acting on link4 and an external couple acting on link5. Determine all the forces acting on the pin joints and the external couple which must be applied to link 1 to hold the mechanism in equilibrium.

Solution.The free-body diagrams of all the links are shown in Fig. 5.8 (b).

We know that our analysis must start with links 3 or 5. Considering link 3, we find that it contains only three unknowns. But in this case we know the line of action, or the point of application, of force only because the line of action of the external forcepasses through the pin jointS which is also the point of application of . Therefore, since link3 is a three-force member, all forces must intersect at a common point. Then we cannot analyze link 3 from the equations of equilibrium alone, even though there are three unknowns, because the moment equation has been used in determining the line of action of . If we take links3 and 2 together, we find that there are seven unknowns and they cannot be analyzed simultaneously. If we consider links 4 and 5 together we find that the two contain only six unknowns so that we can analyze these links. The analysis of these links is shown in Fig. 5.8 (c). Force is resolved into the two componentsandand moments are taken with respect to pointQ to find the magnitude of . Transferring this known component to link4 the forces acting on this link can then be determined with a force polygon.

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Fig. 5.8. Force polygons

Fig. 5.8 (e) Link 1

ransferring the known forceto link2 as find that the two links2 and 3 in combination contain only five unknowns and we have five equilibrium equations from which to determine them. (The summation of moments for link 3 was used to determine the line of action of ) The analysis is shown in Fig. 5.8 (d) where the free-body diagrams are redrawn for the two links. Forceis resolved into two components, one in the direction ofand one normal to it. Forcemust be equal and opposite tosince they are the only forces in the horizontal direction. Then by transferringto link2 we can determine the forces on link 2 with force polygon. The force can now be determined by transferringto link3, and we see that it must be equal and opposite to .

The analysis of link 1 follows the procedure used in the preceding example. It is shown in Fig. 5.8 (e). Thus .

Problem 5.3. Make force analysis of the slider crank mechanism considered in problem 5.1, Fig. 5.9 (a), taking into account friction. The coefficient of friction and the radii of the pins are known.

Solution. Determine the position of the forces relative to the friction circles. The rule used to determine the position is that the friction force always tends to oppose the relative motion between the two links; therefore the forces must be tangent to the friction circles at a point such that the moment of the force about the center of the pin will oppose the relative angular motion. Fig.5.9 (b) shows the forces in the free-body diagrams corresponding to the case in which the motion, or impending motion, of link 3 is towards the left. The approximate direction and sense of the forces are obtained from the previous analysis (the friction forces will cause a slight change in direction) and the relative angular motion of the links can be obtained by inspection of the mechanism, assuming motion of 3 to the left as stated.

If we look at point R on link 2 we see that since the link is in compression and is clockwise, the forcemust be tangent to the friction circle as shown in order to cause an opposing counterclockwise torque. Also, the forcemust be as s

Fig. 5.9. Force analysis with friction forces

hown since is clockwise. Since the link2 is a two-force member, the two forces must be collinear as shown, thereby fixing the direction of the forces. The forces on link 3 are shown in their correct positions and directions where is at an angle, the friction angle, relative to the normal, so as to oppose the motion of the slider. Forcemust also pass through the intersection of the other two forces. Now going to link1 we see that the two forces acting on it are closer together because of the friction forces than they were when friction was not considered. This decrease in the distance x is caused by the forces being tangent to the friction circles and also by the change in direction of the forces. This means that the external moment applied to the link in order to hold the mechanism in equilibrium is smaller than in the problem 5.1, where friction was neglected. This would be expected since the friction forces tend to oppose the motion and this motion was assumed in the direction of the external force

If we assume that link 3 is moving towards the right, the free-body diagram will be as shown in Fig. 5.9 (c). Note that in this case the external moment acting on link 3 is greater than in the previous case because the moment must overcome not only the force but also the friction forces, which now have the effect of an increased external force.

Problem 5.4. Determine the magnitude of the force , acting vertically at the midpoint of the crank, necessary to hold the mechanism, shown in Fig. 5.10 (a), in equilibrium with a known force,acting horizontally on the slider. The crank and connecting rod are of the same length,, and the crank angle is 45°.

Solution. Let the slider move a small amount, , to the right doing negative work under the action of the force. The forcewill move downward and towards the right. Pointwill move h

Fig. 5.10. Application of the principle of virtual work

orizontally to the right a distance and point S, the point of application of force , will move horizontally to the right half the distance ofP, or . Since the displacement was small, pointS will also move downward a distance and this downward displacement is the work absorbing component of the displacement of, and the work is positive. Then the work done on the mechanism by the external forces must be zero, or

whence

.

Now if we draw a velocity polygon for the mechanism, as in Fig. 5.10 (b) we see by observation that the vertical component of the velocity of point S is one-fourth the velocity of point R in the direction of and the magnitude of forcecould have been determined from the velocity image by using Eq. (5.5), giving us the relation

.

The scale of the velocity image is unimportant since only velocity ratios are needed. If determination of is the only information desired, this method is much quicker than making a complete force analysis of the mechanism.

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Fig. 5.11. A rock crusher mechanism

roblem 5.5. Fig. 5.11 (a) shows a rock crusher mechanism with a stone between the jaws which are links 5 and 6. The stone is in contact with link 5 at point W. Determine the relation between the force normal to the jaw at W and the torque applied to the crank, link 1.

Solution. If a force were applied at point P normal to the crank the torque on the crank would be the force times the length of the crank. The ratio of this force to the normal force on the jaw at W will be inversely proportional to the velocity of P and the component of the velocity of W normal to the jaw. From the velocity image of the mechanism, shown in Fig. 5.11 (b), we can get these velocities and thereby determine the desired relationship. Note that there is a component of velocity of W parallel to the jaws and this component tends to move the stone upward. Another advantage of the virtual work method, making use of a velocity image, is the ease with which we can rearrange the mechanism to give us the desired ratio if the force relationship is not satisfactory. By observation of the velocity image and the mechanism we can readily see what various changes in the dimensions of the links will do to the force ratio.