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CHAPTER 7
MECHANICAL EQUILIBRIUM
Opener: Alexander Calder, “Little Pierced Disk,” 1947, Sheet metal, wire, paint. Courtesy of Art Resource.
In Chapter 6 we outlined the process for creating a free-body diagram of a system. This diagram describes the key geometric features of the system and the loads acting on it. In this chapter we use the free-body diagram to determine whether or not the system is balanced.
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OBJECTIVES Upon completion of this chapter, you will be able to:
Describe the conditions of equilibrium and their associated component equations
Use the conditions of equilibrium to determine the relationships between applied forces and moments acting on a system
Carry out static analysis using the analysis procedure
Define the unique characteristics of particle, two-force two-force element, three-force three-force element, and frictionless pulley pulley, and be able to identify these elements in analysis
Define statically determinate, determinate, statically statically indeterminate indeterminate, and underconstrained , and be able to ide ntify systems having these characteristics
As an example, consider the system in Figure 7.1 a and its associated free-body diagram (Figure 7.1 b ). If this system is in mechanical equilibrium (where equilibrium is a a and b). state of balance), the loads acting on the system have particular relationships relationships to one another. another. These relationships, called the equilibrium conditions, allow us to determine that the normal force from the floor pushing upward on the person in Figure 7.1 b balances her weight and that the friction force balances the rope tension. This chapter b balances develops the equilibrium conditions and then discusses how to apply them consistently in analyzing a system that is in mechanical equilibrium.
Figure 7.1 (a) Situation: person leans back, holding onto a rope. The end of the rope at D is tied to a door handle. The person weighs 600 N. A is the location of the person's center of gravity. B is where the rope if held. C is is where the person's feet touch the ground. (b) Free-body diagram; moment center (MC) has been placed at C . One reason engineers are interested in determining determining the loads acting on a system in equilibrium is to determine whether any of the loads exceed the capacity of components in the system. For instance, say the manufacturer of the rope in Figure 7.1 specifies that the rope has a maximum capacity capacity of 500 N (meaning that it breaks when a force exceeding 500 N acts on it). An engineer would use the conditions of equilibrium to determine the force in the rope when a person who weighs 600 N leans backward, then compare this force to the 500-N rope capacity. capacity. If the applied force is equal to or greater than the rope capacity, the engineer would warn the person not to attempt to recreate the situation in Figure 7.1, lest there be a serious backward tumble.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
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7.1 CONDITIONS OF MECHANICAL EQUILIBRIUM A system that is in mechanical equilibrium is one that experiences zero linear acceleration and zero angular acceleration about any axis fixed in an inertial reference frame. With no acceleration, the system either does not move at all or, if it does move, has constant translational and angular velocities. For a system in equilibrium there are restrictions, or conditions, on the forces and moments that act on it. The first of these restrictions is that there is no net force acting on the system. This can be described mathematically as
(7.1)
This condition represents the vector sum of all the forces acting on the system and says that this vector sum is zero. It reflects Newton's first and second laws; a system that is not accelerating has no net force acting on it. The second condition for a system to be in equilibrium is that there is no net moment on on the system. This can be described mathematically as
(7.2)
This condition represents the vector sum of the moments acting on the system and says that the vector sum of all moments is zero. Conditions 7.1 and 7.2 are necessary and sufficient conditions for a system to be in mechanical equilibrium. It is generally more convenient to work with conditions in 7.1 and 7.2 in terms of vector components. When we write each force in the summation in condition 7.1 in component form and then group the x, y, and z components, we can rewrite 7.1 as
where F x, F y, and F z represent the force scalar components in the x, y, and z directions of an overall coordinate system, respectively. For this component representation of F of Fnet to be zero, each component must be zero. In other words:
(7.3A)
(7.3B)
(7.3C) These are the force equilibrium equations. equations . Each equation states that if we add up all the force components acting on a system in a particular direction, the sum must be zero if there is equilibrium. Equations (7.3A, 7.3B, 7.3C) comprise the requirements for force balance. balance . Similarly, we can write each moment M moment M in in the summation in condition 7.2 in terms of its vector components and rewrite 7.2 as
where M x, M y, and M z represent the individual moment components about the x, y, and z axes of an overall coordinate system, respectively. For this component representation of M of M net to be zero, each of its components must be zero. In other words:
(7.4A)
(7.4B)
(7.4C)
These are the moment equilibrium equations. equations . Each equation states that if we add up all of the moments about a particular axis, their sum must be zero if there is equilibrium. Equations (7.4A, 7.4B, 7.4C) are the three equilibrium conditions for moment balance. balance . The six equations in 7.3 and 7.4 are true for any system in mechanical equilibrium (i.e., a balanced system). We shall refer to them as the six equilibrium equations. equations . If a system is in equilibrium, the loads acting on it are such that the six equilibrium equations are satisfied—or, in other words, the conditions are met. We can solve the equations for the unknowns they contain, as illustrated in Example 7.1 for the leaning person in Figure 7.1 7.1.. Alternately, we can work directly with equilibrium conditions in 7.1 and 7.2 to ensure that the vector sums of forces and moments acting on the system are zero.
EXAMPLE 7.1 LEANING PERSON If the rope that the person in Figure 7.1 a is using to lean back with will break if its tension exceeds 500 N, should a is the person be concerned about it breaking?
a,, then compare this Goal We are to find the tension in t he rope for the configuration shown in Figure 7.1 a tension to 500 N. If it exceeds 500 N, the rope exceeds the design capacity and may break.
Given We are given the dimensions associated with the situation, along with the weight of the person and the location of her center of gravity (point A).
Assume We assume that the rope is taut and horizontal, and that there is sufficient friction between the person's feet and the floor for there to be no sliding. We also assume that the person is in equilibrium.
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Draw The free-body diagram of the system (taken as the person) is shown in Figure 7.1 b.
Formulate Equations We set up the force equilibrium equations based on the information in the free-body diagram. Force (a) equilibrium:
Next we set up the moment equilibrium equations. We start by establishing a moment center (MC). The (b) moment center could be placed anywhere, but placing it at a location where lines of action of forces intersect at C make the algebra easier. Moment equilibrium with moment center at C is:
Solve The equations in (a) and (b) for the unknowns: From 7.3B, F Cy = 600 N From 7.3C, F Cz = 0 N From 7.4C, F Bx = 420 N From 7.3A, F Cx = 420 N The tension in the rope is 420 N when a person weighing 600 N leans backward in the configuration shown. And the further back she leans, the greater the tension. This is less than the 500-N breaking force of the rope. Answer 420 N (tension in rope) < 500 N (tension that will break the rope). Therefore the person should not be concerned that the rope will break under the given conditions.
Check We add the forces found above to the free-body diagram of the person in mechanical equilibrium, as shown in Figure 7.2. Notice that F Bx and FCx form a couple that is countered by the couple created by W and FCy. This is a useful visual check of the calculation.
Figure 7.2 Free-body diagram of person, including calculated forces
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
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7.2 APPLICATION OF THE CONDITIONS—PLANAR SYSTEMS The Basic Equations Any system that is in equilibrium must satisfy all six equilibrium equations. Three of these equations, however, are automatically satisfied for a planar system. In Chapter 6 a planar system was defined as one where all the forces and moments involved are contained in the same plane. For example, consider the system consisting of triangular frame ABC in Figure 7.3 a, with its associated free-body diagram in Figure 7.3 b. Notice that there are no forces in the z direction in this s ystem—therefore, 7.3C is automatically satisfied for this planar system. Furthermore, because all the forces are contained in the xy plane, they create no moments about the x or y axis—therefore, 7.4A and 7.4B are also automatically satisfied.
Figure 7.3 (a) Frame ABC ; (b) free-body diagram of frame ABC We must still confirm that the loads acting on the s ystem satisfy the three remaining equilibrium equations:
(7.5A)
(7.5B)
(7.5C) We refer to these three equations as the planar equilibrium equations .
The Analysis Procedure Revisited Now we consider how these planar equilibrium equations fit into the analysis procedure. Recall that a free-body diagram, created as part of the Draw step of our analysis procedure, shows the loads acting on a system as well as key dimensions and establishes a Cartesian coordinate system. The free-body diagram serves as the starting point for using the equilibrium equations to determine the magnitudes and/or directions of loads acting on a balanced system. The Draw step is followed by the Formulate Equations , Solve, and Check steps, as we discuss next. Consider the situation and free-body diagram in Figure 7.4 a. The weight W of the block is known and we wish to determine the loads acting on the block at A and C .
Figure 7.4 (a) The block is connected to its surroundings via a pin connection at A and a slot on pin at C ; (b) the associated fre diagram including calculated forces; ( c) checking the solution with graphical force addition
Formulate Equations. Reading force information from the free-body diagram, write the x component force equilibrium equation 7.5A. Make sure to include all x component forces. Repeat for the y component force equilibrium equation 7.5B. These two equations may contain unknown forces, represented by variable labels:
Notice, for example, that in the s econd term on the right-hand side of 7.5A we have written − F C sin 60°. This term is the component of FC in the x direction. The negative sign indicates that this component is in the negative x direction (based on how FC is drawn in Figure 7.4 a). This point is important when it comes to interpreting answers.
Now write the z component moment equation 7.5C by first deciding which point to use as the moment center. By choosing a point through which force lines of action pass, you reduce the number of simultaneous equations you deal with in the Solve step (because a force does not create a moment at a point that its line of action passes through). Be sure to include all moments about the z axis created by forces and by moments. The equation you write may contain unknowns (moments, forces, and/or dimensions), represented by variable labels. For the situation in Figure 7.4 a we choose the moment center to be at A. Therefore 7.5C becomes
Solve. You now have three equations of equilibrium that can be solved for (at most) three unknowns. Furthermore, these three equations are independent of one another. 1 Most of the systems presented in this book are such that the equilibrium equations can reasonably be solved by hand for the unknowns. (However, if you have the opportunity to learn to use a numerical software package such as MatLab, we urge you to do so; solving more complex and realistic problems by setting them up in matrix form and using a computer-based solver makes solving a lot easier.) From the three equations 7.5A, 7.5B, and 7.5C written based on Figure 7.4 a, we find that Answer
In presenting answers be sure to make them stand out, so that someone reviewing your work can readily identify them. These values can also be presented as part of the free-body diagram in Figure 7.4 b.
Check. Just solving equations and getting numbers is not enough. The results should be checked using technical knowledge, engineering judgment, and common sense. Do the results match your expectations? Verify that the results make sense by looking at the order of magnitude, units, and directions of loads. A negative answer for a load value means that the direction in which the load acts is opposite the direction depicted in the free-body diagram. Also, check your algebra by substituting numerical answers into the
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equilibrium equations to verify that the equations are satisfied. Still another option for checking your results is to select an alternate moment center and write the moment equilibrium equation, substituting in your numerical answers. Finally, make sure the answers are clearly communicated in the write-up. Compare values with design requirements (if available) and draw conclusions about the adequacy of the design. For the situation in Figure 7.4 a, we check that we have correctly calculated the values of F , F , and F C by selecting a moment center Ax Ay at C , writing out 7.5C, and substituting in the calculated values:
Indeed, the calculated values give us zero moment at a moment center at C . An alternate check is also presented in Figure 7.4 c that involves application of force equilibrium condition 7.1 using graphical force addition.
EXAMPLE 7.2 CANTILEVER BEAM WITH TWO FORCES AND A MOMENT A 500-kg uniform beam subjected to the loads shown in Figure 7.5 is in mechanical equilibrium. Find the loads acting on the beam at support O for the given configuration.
Figure 7.5
Goal We are to find the loads acting on the beam at O for a beam that is in mechanical equilibrium.
Given We are given information about the geometry and mass of the beam and its mass distribution, as well as the size of the loads acting at A, B, and C .
Assume We will assume that the support at O is fixed and that we can treat the system (the beam) as planar. Also, the beam is on earth, oriented so that gravity acts in the − y direction.
Draw Based on the information given in the problem and our assumptions, we draw a free-body diagram ( Figure 7.6). The loads at a fixed support are represented by two forces and a moment; see Table 6.1. Also, the weight of the beam is found by converting its mass to its weight on earth [(500 kg)(9.81 m/s 2) = 4,905 N]. Since the beam is uniform, this weight acts at the center of the beam.
Formulate Equations and Solve We set up the equations for planar equilibrium to find the unknown loads at O: Based on 7.5A:
Based on 7.5B:
Based on 7.5C, with the moment center at O:
Answer
Check The directions and magnitudes of the loads at the fixed connection are as expected. F Ox balances F Cx. All other forces are acting in the y direction. As an additional check, 7.5C could be written with the results above for an alternate moment center.
Figure 7.6
EXAMPLE 7.3 A SIMPLE STRUCTURE
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The structure shown in Figure 7.7 is pinned to its surroundings at A and rests against a roller at B. Force P acts at C . The weight of the structure can be ignored because it is much smaller than P. Find the loads acting at supports A and B as a function of the geometry and loading.
Figure 7.7
Goal We are to find the reactions (loads) at A and B as a function of geometry (dimensions a and b) and force ( P).
Given We are given information about the geometry of the structure, and are told that A is a pin connection and B is a roller. We are also told that the weight of the structure is negligible compared to P.
Assume We will assume that we can treat the system as planar.
Draw Based on the information given in the problem and our assumption, we draw a free-body diagram and arbitrarily draw the unknown loads in the directions of the positive axes ( Figure 7.8). (At the pin connection the loads are two forces, and at the roller a force pushes on the system; see Table 6.1.)
Formulate Equations and Solve We set up the equations for planar equilibrium to find the unknown loads at A and B. Based on 7.5A:
(1)
The minus sign in equation 1 means F will be in the opposite direction of F . Ax Bx Based on 7.5B:
(2)
Based on 7.5C, with the moment center at A:
(3)
Substituting 3 into 1, we find that
We show three possible ways the answers could be presented. Answer (Approach 1: Scalar components)
(the minus sign means that it acts in the direction opposite to that defined in Figure 7.8)
(Approach 2: Vector notation) The loads acting on the structure are:
(Approach 3: Narrative description)
The loads acting on the structure: @ A; a force with a component of
in the x direction and a component of P in
the y direction. @ B; a force with a component of
in the x direction.
Check Notice that F Ay balances P, and F Ax balances F Bx. Also, F Ay and P form a clockwise couple that balances the counterclockwise couple created by F Ax and F . We can also check that dimension b in the numerators of F and F Bx Bx Ax makes sense; as b increases, the magnitude of the couple created by F and P increases. If dimension a is fixed, the Ay magnitudes of F Ax and F must increase to balance this couple. A similar argument can be made for dimension a being in Bx the denominator.
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