Figure 5.12a,b. The effect of eccentric loading on the relative motion of two rigid members articulating at a ball-and-socket joint (a). The joint can be stabilized and the relative motion coordinated by adding two tension-carrying cords (muscle-tendon complexes) on each side of the joint (b).

Appropriately positioned contact forces are needed to ensure the static balance of an object under the action of gravity. In some cases, the number of unknown contact forces is greater than the number of equations of static equilibrium, and such systems are statically indeterminate. An example of a statically indeterminate system is shown in Fig. 5.13a, where a ballerina is depicted as leaning on her partner. The couple forms contact with the dance floor through all four feet. Because the dance floor supports frictional forces, four vertical and four horizontal contact forces need to be evaluated. Because there are only three independent equations of equilibrium for a planar case, it is not possible to determine uniquely the numerical values of the unknowns using only the conditions of equilibrium. Static human postures such as sitting on a chair with feet grounded and back leaning against the chair constitute examples of statically indeterminate situations.

Next let us consider Fig 5.13b, where a male dancer is carrying the entire weight of the woman dancer. We could evaluate the location of the center of mass of the couple from their measurements and arbitrarily assume that the ground force acts at the center of the sole of the feet. Using the equations of static equilibrium, we can then compute the vertical reaction forces acting at the male dancer's feet uniquely. But what about

Figure 5.13a-e. Snapshots of static equilibrium positions in classical ballet. The couple shown in (a) has contact with the floor at four locations. Their bodies together compose a statically indeterminate structure. The vertical ground forces acting on the couple shown in (b) can be computed using the equations of static equilibrium. The support forces acting on the ballerina in (c) can also be determined with the equations of static equilibrium. The ballerina shown in (d) balances her weight by aligning it with the vertical ground force acting on the toes of her foot. The number of static equations in this case exceeds the number of unknown contact forces. The ballerina is unstable in the sense that a perturbation from her equilibrium position will require finite movement at her ankle and her arms to restore the resting configuration. The dancer (e) is free-falling under the action of gravity. His pose and facial expressions, however, are intended to convince us that he is able to defy gravity by hanging in air.

Figure 5.13a-e. Snapshots of static equilibrium positions in classical ballet. The couple shown in (a) has contact with the floor at four locations. Their bodies together compose a statically indeterminate structure. The vertical ground forces acting on the couple shown in (b) can be computed using the equations of static equilibrium. The support forces acting on the ballerina in (c) can also be determined with the equations of static equilibrium. The ballerina shown in (d) balances her weight by aligning it with the vertical ground force acting on the toes of her foot. The number of static equations in this case exceeds the number of unknown contact forces. The ballerina is unstable in the sense that a perturbation from her equilibrium position will require finite movement at her ankle and her arms to restore the resting configuration. The dancer (e) is free-falling under the action of gravity. His pose and facial expressions, however, are intended to convince us that he is able to defy gravity by hanging in air.

the horizontal reaction forces? Static equilibrium requires that the sum of the horizontal forces must be equal to zero, and so we have two unknown horizontal force components and one equation. The system is statically indeterminate in the horizontal direction. The couple is in a statically stable position because a small alteration in the posture will not lead to larger alterations; all it will do is to change the magnitudes of the reaction forces acting on each foot of the male dancer. If a horizontal force is exerted on this couple, frictional forces at the male dancer's feet could balance the applied force and the couple would remain in static equilibrium.

The ballerina in Fig. 5.13c is supported by vertical and horizontal ground forces and also by her dance partner. If we assume that the force exerted by the partner on the ballerina acts along the direction of the ballerina's arm, then we have a statically determinate situation. There are three unknowns and three equations of equilibrium. Furthermore, the ballerina is in a statically stable pose. The force exerted by her partner could be either tension or compression and thus would prevent her rotating clockwise or counterclockwise.

The next case is the figure (Fig. 5.13d) is a ballerina en pointe where she stands on top of the toes on one leg, with weight being taken primarily through the first and second toes. This is just one of the several poses in classical ballet in which the ballerina strikes a delicate balance. In another posture, called an attitude, the body is supported on one leg with the other lifted to the front, side, or back with the knee flexed. In an arabesque, the body is supported on one leg while the other is fully extended behind the dancer. The contact forces acting on a ballerina in such poses are the vertical and horizontal ground forces. There are three equations of equilibrium to be satisfied with two unknown forces. Unless the ballerina can keep her center of mass right on top of the ground force acting on it, the resultant external moment acting on her would not be equal to zero. Keeping in equilibrium in these positions means the accurate positioning of the center of gravity of the body right on the vertical line crossing the point of application of the contact force on the ground. In response to a perturbation, she could realign her center of mass by moving her arms slightly or by bringing her heels down to the floor momentarily. Thus, a small perturbation will lead to movement and even to artistic catastrophe and mean reviews, but she is not helpless in preventing a fall. Even an untrained person can stand on the toes of one foot by rocking on the ankle to keep balance or by slightly bending the knee. The human body is a robust structure that provides various pathways to accomplish a given physical task.

The last picture, in Fig. 5.13e, is that of an airborne ballet dancer. There are no contact forces acting on him. He is free-falling under the action of gravity. His pose and facial expressions, however, create the illusion that he can actually hang in air; this show of apparent defiance of gravity is certainly part of the art of ballet.

The concept of stability of static equilibrium has been the subject of ex tensive investigation in structural engineering. The requirement of stability in structures is concerned with the danger of having unacceptable motions. When a tree is acted upon by a hurricane wind, and is not properly rooted in the ground or balanced by its own weight, it may topple over without disintegrating. The tree is then said to be unstable in rotation. A regular pendulum is in stable equilibrium when the rod connecting the bob to the point of anchor is vertical. Small perturbations of displacements from that configuration will not lead to larger displacements of the pendulum. After a series of oscillations, the pendulum will again reach equilibrium.

In this section we present several examples on the static equilibrium of human-made structures. We discuss a method of computing the internal forces in a structure that arise in response to external loading. Even though internal forces do not contribute to the course of motion of an object, they could lead to material failure, fractures, tearing, or extensive distortion. As such, they are instrumental in the proper functioning of human-made structures. The first example of this section concerns a truss, a structure extensively used in bridges, cranes, roofs, and so on. The function of a truss is to transmit loads applied to it to its points of anchors to the earth.

Example 5.6. Mechanical Analysis of a Truss. Figure 5.14 shows a typical truss that is subject to external forces in the horizontal and vertical direction. Our first aim is to determine the reaction forces at A and E. At point A, the truss is anchored to the ground. The reaction forces acting on the hinge joint A are denoted as A1 and A2, as shown in the figure. The point E may undergo small lateral displacements but is fixed in the vertical direction. Such a joint is called a roller joint. The vertical support force acting on it is denoted as E2. There are three unknown forces in this case and three independent equations of equilibrium. We will use those equations to determine the reaction forces acting on the truss.

The condition of force balance in the horizontal direction leads to

Let us next take the moment of all external forces with respect to point A and set it equal to zero:

-5,000 • 0.8 - 20,000 • 1.8 + E2 • 2.4 = 0 => E2 = 16,667 N (5.16b) The sum of all forces in the vertical direction must also be equal to zero

5,000N

20,000N H

5,000N

SSSA\vsss

H | |||||||||||||

HG / | |||||||||||||

fhc/ |
Figure 5.14a-d. A truss loaded with a vertical and horizontal force (a). Also shown in the figure are the free-body diagrams of joints E (b) and D (c) and the free-body diagram of the section of the truss consisting of members that connect at joints D, E, and H (d). All horizontal elements of the truss have equal lengths. Thus, we were able to compute the reaction forces uniquely. The choice of the particular equations of static equilbrium was somewhat arbitrary. For example, instead of using Eqn. 5.16c, the condition of force balance in the vertical direction, we could have determined the resultant external moment with respect to point E and set it equal to zero: -5,000 • 0.8 + 20,000 • 0.6 - A2 • 2.4 = 0 => A2 = 3,333 N Thus, one could determine the resultant moment with respect to two distinct points and set them equal to zero. Together with the condition of force balance in the horizontal direction, this set of equations is sufficient to ensure planar static equilibrium. How much force does each element of the truss bear in transmitting the external loads to the ground at points A and E? This we can also evaluate using the equations of static equilibrium. in a truss, elements are pinned to each other. Thus, if we neglect their own weight in comparison with the external forces acting on the truss, a truss element is essentially a bar loaded only at its ends. These forces must be equal in magnitude, opposite in direction. Further, they must have the same line of action. This means that truss elements carry only compression or tension. We can compute the values of these axial loads either by considering the static equilibrium of each joint of the truss (method of joints) or by using the method of sections, where one isolates a portion of the truss and considers its static equilibrium. We will illustrate these methods here. Let us consider, for example the static equilibrium of joint E. Joint E is virtually a point (negligibly small in size in comparison with the truss structure) rather than an object with moment of inertia, and thus the static equilibrium corresponds to the condition of force balance. A free-body diagram of this joint is presented in Fig. 5.14b. We have shown the directions of unknown forces as if they were in tension. In tension, the truss member pulls the joint in the direction of the member. If the member is in compression, it pushes against the joint. Members that are in tension and those in compression can be identified with the symbols [T], and [C], respectively. If it turns out that our initial assignment of tension to both members is not correct for a particular member, the value of the member force will appear in negative. Static equilibrium of joint E yields the following equations: Let us next consider the equilibrium of joint D: Note that if there are three truss elements coming together in a joint of a truss and two of the three share the same line of action, then the force on the third must be equal to zero. Let us next compute the internal force in the bar member CH using the method of sections. We introduce an imaginary cut on the truss and consider the portion shown in Fig. 5.14d. Let us consider the equation of equilibrium in the vertical direction: -20,000 N + 16,667 N - Fch • (0.8/1.0) = 0 => Fch = -4,166 N Fch = 4,166 N [C] Thus, it is possible to evaluate the internal forces (forces carried by the truss elements) by considering the conditions of static equilibrium of truss joints and or of various sections of the truss. This is of course true if each joint (section) is statically determinate. For example, if we had connected points B and G or point D and G with a truss element, parts of the truss would be overbuilt or rigidified and would be statically indeterminate. The question that comes to mind is how do we determine the internal forces in a statically indeterminate system? This question has a rather complex answer. The following example, although admittedly simple, captures the essence of a statically indeterminate system. Example 5.7. Axial Loading of a Rod That Is Fixed at Both Ends. Consider a rod whose ends are embedded into large concrete structures (Fig. 5.15). The length of the rod is L. A horizontal force P is exerted on the rod at a distance of L/3 from the fixed support B. The force P is much greater than the weight of the rod. Determine the reaction forces at the fixed ends B and D. Ignore the effect of the rod's weight on the support forces. Solution: The condition of force balance in the horizontal direction requires that This is the only equation of static equilibrium that relates these two forces. We have one equation and two unknowns. Clearly we need another equation to determine the two reaction forces uniquely. This additional equation concerns with the deformation of the beam. Because the beam is embedded in large and heavy structures at both ends, the length of the beam should not change even after the imposition of load P. That means part of the beam must be in compression and shorten a small amount and the other part be in tension and extend in length. Furthermore, the extent of compression must be equal to the extent of stretching. It has been experimentally observed that, in ordinary construction materials, the length change in response to an axial load is proportional to the length of the structural element and the axial force exerted on it (Fig. 5.15b): Figure 5.15a-c. A beam under the influence of external load P (a). The segment BP is under tension (b) and PD under compression. The horizontal reaction forces can be computed by using the requirement that the length of the beam remains constant despite the load applied on it (c). in which Al denotes the change in length, and l and A denote the original length and cross-sectional area of the axial member, respectively. The symbol F denotes the axial force exerted on the structural element; F is positive if it is tensile and F is negative if it is compressive. The parameter E is called the Young's modulus and has the units of force per unit area. It is a material property. A material is said to be stiff or compliant, depending on the value of E. When E is large, AL is small and therefore the material is stiff. Equation 5.19a is not a law of nature, but it is an experimental observation that appears to be valid for most materials undergoing small stretches or compressions. Typically, when a material is loaded gradually, the change in length is proportional to the load initially and the material will instantaneously go back to its unloaded configuration when the load is released. This is the elastic phase of loading when most materials act like a linear spring. Further increases in loading could complicate the relationship between the force exerted and the resulting change in length. Equation 5.19 is called the basic law of elasticity or Hooke's law after the British scientist Robert Hooke (1635-1703). Hooke was an exceptional scientist and designer, with major contributions to the wave theory of light, theory of elasticity, thermodynamics, crystal structures, and gravity and the motion of planets. He was also involved in a bitter controversy with Newton, complaining that he was not given sufficient credit for his contributions to the formulation of nature's laws of motion. He wrote Eqn. 5.19b in a slightly different form: The term on the left is called the force intensity, force per unit area, or axial stress. It is usually denoted by the symbol a. The term AL/L is the dimensionless change in length or axial strain e. Thus, a = E e (5.19c) Ligaments and tendons of the human body obey Hooke's law only at very low levels of strain. At higher levels of stress, the following nonlinear equation may be a better fit to describe the correlation between axial stress and strain: This equation shows that for a material to behave elastically (in a springlike fashion) it is not necessary to have a linear relationship between stress and strain. All that is required is that for a given stress there is a unique strain. When stress goes to zero, strain also goes to zero. Tendons and ligaments are nearly elastic in the physiological loading range because the stress-strain curve obtained during the loading phase is approximately the same as the corresponding curve during the relaxation phase. Stress in a bone is typically three dimensional and direction dependent; that is, E would depend on the direction in which the force is applied. The skele tal muscle, when relaxed, can elastically be deformed. When activated, on the other hand, although it still stores elastic energy it will contract while in tension. The muscle property of being able to shorten under tension is a feature no nonliving material can replicate. Returning to the problem at hand, we can now write the zero-elongation equation for the beam loaded with a point force P: — (L/3) • Hb/(AE) - (2L/3) • Hd/(AE) = 0 (5.20) Solving Eqns. 5.18 and 5.20 for HB and HD, we find: Because HB turned out to be negative, it means that our initial supposition about its direction was wrong. The correct directions of the these support forces are shown in Fig. 5.15c. Note that the portion of the beam from B to P is in tension and the rest is in compression. Example 5.8. A Cantilever Beam Carrying Its Own Weight. Let us consider a cantilever beam (Fig. 5.16a) that is embedded at one end in a body much larger in mass than the mass of the beam itself. The weight of the beam is equal to W and the span of the beam is L. Determine the support forces at point B. Solution: The free-body diagram of the beam under consideration is shown in Fig. 5.16b. The equations of static equilibrium require that the following conditions be met: Thus, the surroundings at B exert not only a vertical force of W to the beam but also a counterclockwise external moment equal to W (L/2). When the material of the beam is homogeneous, this moment is the result of the linear distribution of axial forces in the cross section (Fig. 5.16d). The top layers of the beam are under tension because they stretch as the beam deforms under its own weight. The bottom layers, on the other hand, contract and therefore are under compression. The axial force intensity (stress) a at a vertical distance y from the central line of the beam is then given by the relation: in which ao is the maximum axial stress that occurs at y = (h/2). This equation holds for small deformations of the beam where planar cross sections normal to the axis of the beam remain planar after deformation. Similarly, one can show that the moment created by a is given by the following relation: dM = a (b • dy) y = (ao) [y 2/(h/2)] (b • dy) (5.22)
Was this article helpful? ## Getting Started With DumbbellsThe use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers. |

## Post a comment