Machine Design 3

The effects on the shear and moment diagrams of multiple loads on a beam can also be determined by superposition of the individual loadings. If, for example, the beam of Example 3-3 had two point loads applied to it, each at a different distance a, their combined effect could be found by applying the equations of that example twice, once for each load and position, and then adding (superposing) the two results. Appendix B contains a collection of common beam-loading situations solved for shear and moment functions giving their equations and plots. These solutions can be combined by superposition to accommodate more complicated situations. They can be superposed within your model to obtain and plot the total shear and moment diagrams, their maxima and minima. 3.10 SUMMARY Though the student learning about stress analysis for the first time may not think so, the subject of load analysis can often be more difficult and complicated than that of stress analysis. Ultimately, the accuracy of any stress analysis is limited by the quality of our knowledge about the loads on the system, since the stresses are generally proportional to the loads as will be discussed in Chapter 4. This chapter has presented a review of Newtonian methods of force and moment analysis for both dynamically and statically loaded systems of a few types. It is by no means a complete treatment of the complex subject of load analysis, and the references in the bibliography of this chapter should be consulted for more detail and for cases not covered here. The following factors should be kept in mind when attempting to determine loads on a system: 1 Determine the character of the loading in terms of its load class as defined in Section 3.1 in order to decide on whether a static or dynamic load analysis is in order. 2 Draw complete free-body diagrams (FBD) of the system and of as many subsystems within it as are necessary to define the loads acting on its elements. Include all applied moments and torques as well as forces. The importance of a carefully drawn FBD cannot be overemphasized. Most errors in force analysis occur at this step because the FBD is often incorrectly drawn. 3 Write the relevant equations using Newton's laws to define the unknown forces and moments acting on the system. The solution of these equations for most real problems requires some sort of computer tool such as an equation solver or spreadsheet to get satisfactory results in a reasonable time. This is especially true for dynamic systems, which must be solved for a multiplicity of positions in order to determine the maximum loads. 4 The presence of impact forces can significantly increase the loads on any system. The accurate calculation of forces due to impact is quite difficult. The energy method for impact-force estimation presented in this chapter is crude and should be considered an approximation. Detailed information about the deformations of the bodies at impact is needed for a more accurate result, and this may not be available without testing of the actual system under impact. More sophisticated analysis techniques exist for impact-force analysis but are beyond the scope of this introductory design text. The reader is referred to the bibliography for more information.

∑ x ∑ ∑ = + + ++ = = + + += =+ × ( ) + × ( ) + × ( ) + × ( ) + × ( ) = 21 1 31 21 1 31 21 21 1 1 31 31 0 0 0 ( ) MM R F R F R F RPRF Expanding cross products in the moment equation gives the moment magnitude as M M RF RF RF RF RF RF f R P RP RF RF z h x y y x bx by by bx x y y x Px y Py x dx sheath dy sheath y x ∑ =+ − ( ) + − ( ) + − ( ) + − ( ) + − ( ) = 21 21 21 21 1 1 1 1 31 31 31 31 0 ( ) 5 The total of unknowns at this point (including those listed in step 2 above) is 21: Fb1x, Fb1y, F12x, F12y, F21x, F21y, F32x, F32y, F23x, F23y, F13x, F13y, F31x, F31y, Fcablex , Fcabley , Fsheathx , Fsheathy ,Px, Py, and Mh. We have only nine equations so far, three in equation set (a), one in set (c), two in set (d) and three in set (e). We need twelve more equations to solve this system. We can get seven of them from the Newton's third-law relationships between contacting elements: FF FF FF FF g FF FF F F xx yy xx yy xx y

1 Figure 3-1 shows the handbrake lever assembly, which consists of three subassemblies: the handlebar (1), the lever (2), and the cable (3). The lever is pivoted to the handlebar and the cable is connected to the lever. The cable runs within a plastic-lined sheath (for low friction) down to the brake caliper assembly at the bicycle's wheel rim. The sheath provides a compressive force to balance the tension in the cable (Fsheath = -Fcable). The user's hand applies equal and opposite forces at some points on the lever and handgrip. These forces are transformed to a larger force in the cable by the lever ratio of part 2. Figure 3-1 is a free-body diagram of the entire assembly since it shows all the forces and moments potentially acting on it except for its weight, which is small compared to the applied forces and is thus neglected for this analysis. The "broken away" portion of the handlebar can provide x and y force components and a moment if required for equilibrium. These reaction forces and moments are arbitrarily shown as positive in sign. Their actual signs will "come out in the wash" in the calculations. The known applied forces are shown acting in their actual directions and senses. 2 Figure 3-2 shows the three subassembly elements separated and drawn as free-body diagrams with all relevant forces and moments applied to each element, again neglect

1 Figure 3-9 shows a center-pull brake arm assembly commonly used on bicycles. It consists of six elements or subassemblies, the frame and its pivot pins (1), the two brake arms (2 and 4), the cable spreader assembly (3), the brake pads (5), and the wheel rim (6). This is clearly a three-dimensional device and must be analyzed as such. 2 The cable is the same one that is attached to the brake lever in Figure 3-1. The 267-N (60-lb) hand force is multiplied by the mechanical advantage of the hand lever and transmitted via this cable to the pair of brake arms as was calculated in Case Study

1 Solve for the reaction forces using equations 3.3 (p. 78). Summing moments about the right end and summing forces in the y direction gives M Rl wl a a R wl a l ∑ z == − ( ) − = ( ) − = ( ) − ( ) = 0 2 2 10 10 4 2 10 18 1 2 1 2 2 ( ) F R wl a R b R wl a R ∑ y == − − ( ) + = − ( ) −= − ( ) − = 0 10 10 4 18 42 1 2 2 1 ( ) 2 The shape of the shear diagram can be sketched by graphically integrating the loading diagram shown in Figure 3-24a. As a "device" to visualize this graphical integration process, imagine that you walk backward across the loading diagram of the beam, starting from the left end and taking small steps of length dx. You will record on the shear diagram (Figure 3-24b) the area (force · dx) of the loading diagram that you can see as you take each step. As you take the first step backward from x = 0, the shear diagram rises immediately to the value of R1. As you walk from x = 0 to x = a, no change occurs, since you see no additional forces. As you step beyond x = a, you begin to see strips of area equal to -w · dx, which subtract from the value of R1 on the shear diagram. When you reach x = l, the total area w · (l - a) will have taken the value of the shear diagram to -R2. As you step backward off the beam's loading diagram (Figure 3-24a) and plummet downward, you can now see the reaction force R2 which closes the shear diagram to zero. The largest value of the shear force in this case is then R2 at x = l. 3 If your reflexes are quick enough, you should try to catch the shear diagram (Figure 3-24b) as you fall, climb onto it, and repeat this backward-walking trick across it to create the moment diagram which is the integral of the shear diagram. Note in Figure 3-24c that from x = 0 to x = a this moment function is a straight line with slope = R1. Beyond point a, the shear diagram is triangular, and so integrates to a parabola. The peak moment will occur where the shear diagram crosses zero (i.e., zero slope on the moment diagram). The value of x at V = 0 can be found with a little trigonometry, noting that the slope of the triangle is -w: x a R w c @V . =0 =+ =+ = 1 4 18 10 58 ( ) Positive shear area adds to the moment value and negative area subtracts. So the value of the peak moment can be found by adding the areas of the rectangular and triangular portions of the shear diagram from x = 0 to the point of zero shear at x = 5.8: M Ra R d @ . x . . . =58 1 1 = ( ) + = ( ) + = 1 8 2 18 4 181 8 2 88 2 ( ) The above method gives the magnitudes and locations of the maximum shear and moment on the beam and is useful for a quick determination of those values. However, all that walking and falling can become tiresome, and it would be useful to have a method that can be conveniently computerized to give accurate and complete information on the shear and moment diagrams of any beam-loading case. Such a method will

1A. We will assume no loss of force in the cable guides, thus the full 1 046-N (235-lb) cable force is available at this end. 3 The direction of the normal force between the brake pad and the wheel rim is shown in Figure 3-9 to be at θ = 172° with respect to the positive x axis, and the friction force is directed along the z axis. (See Figures 3-9 and 3-10 for the xyz axis orientations.) 4 Figure 3-10 shows free-body diagrams of the arm, frame, and cable spreader assembly. We are principally interested in the forces acting on the brake arm. However, we first need to analyze the effect of the cable spreader geometry on the force applied to the arm at A. This analysis can be two dimensional if we ignore the small z-offset between the two arms for simplicity. A more accurate analysis would require that the z-directed components of the cable-spreader forces acting on the arms be included. Note that the cable subassembly (3) is a concurrent force system. Writing equations 3.3b in two dimensions for this subassembly, and noting the symmetry about point A, we can write from inspection of the FBD:

3.0 INTRODUCTION This chapter provides a review of the fundamentals of static and dynamic force analysis, impact forces, and beam loading. The reader is assumed to have had first courses in statics and dynamics. Thus, this chapter presents only a brief, general overview of those topics but also provides more powerful solution techniques, such as the use of singularity functions for beam calculations. The Newtonian solution method of force analysis is reviewed and a number of case-study examples are presented to reinforce understanding of this subject. The case studies also set the stage for analysis of these same systems for stress, deflection, and failure modes in later chapters. Table 3-0 shows the variables used in this chapter and references the equations, sections, or case studies in which they are used. At the end of the chapter, a summary section is provided which groups all the significant equations from this chapter for easy reference and identifies the chapter section in which their discussion can be found. 3.1 LOADING CLASSES The type of loading on a system can be divided into several classes based on the character of the applied loads and the presence or absence of system motion. Once the general configuration of a mechanical system is defined and its kinematic motions calculated, the next task is to determine the magnitudes and directions of all the forces and couples present on the various elements. These loads may be constant or may be varying over time. The elements in the system may be stationary or moving. The most general class is that of a moving system with time-varying loads. The other combinations are subsets of the general class.

3.3 LOAD ANALYSIS This section presents a brief review of Newton's laws and Euler's equations as applied to dynamically loaded and statically loaded systems in both 3-D and 2-D. The method of solution presented here may be somewhat different than that used in your previous statics and dynamics courses. The approach taken here in setting up the equations for force and moment analysis is designed to facilitate computer programming of the solution. This approach assumes all unknown forces and moments on the system to be positive in sign, regardless of what one's intuition or an inspection of the free-body diagram might indicate as to their probable directions. However, all known force components are given their proper signs to define their directions. The simultaneous solution of the set of equations that results will cause all the unknown components to have the proper signs when the solution is complete. This is ultimately a simpler approach than the one often taught in statics and dynamics courses which requires that the student assume directions for all unknown forces and moments (a practice that does help the student develop some intuition, however). Even with that traditional approach, an incorrect assumption of direction results in a sign reversal on that component in the solution. Assuming all unknown forces and moments to be positive allows the resulting computer program to be simpler than would otherwise be the case. The simultaneous equation solution method used is extremely simple in concept, though it requires the aid of a computer to solve. Software is provided with the text to solve the simultaneous equations. See program MATRIX on the CD-ROM. Real dynamic systems are three dimensional and thus must be analyzed as such. However, many 3-D systems can be analyzed by simpler 2-D methods. Accordingly, we will investigate both approaches. Three-Dimensional Analysis Since three of the four cases potentially require dynamic load analysis, and because a static force analysis is really just a variation on the dynamic analysis, it makes sense to start with the dynamic case. Dynamic load analysis can be done by any of several methods, but the one that gives the most information about internal forces is the Newtonian approach based on Newton's laws. NEWTON'S FIRST LAW A body at rest tends to remain at rest and a body in motion at constant velocity will tend to maintain that velocity unless acted upon by an external force. NEWTON'S SECOND LAW The time rate of change of momentum of a body is equal to the magnitude of the applied force and acts in the direction of the force. Newton's second law can be written for a rigid body in two forms, one for linear forces and one for moments or torques: ∑ ∑ Fa M = = m G G a H˙ (. ) 3 1 where F = force, m = mass, a = acceleration, MG = moment about the center of gravity, and H˙ G = the time rate of change of the moment of momentum, or the angular mo

Class 2 describes a stationary system with time-varying loads. An example is a bridge which, though essentially stationary, is subjected to changing loads as vehicles drive over it and wind impinges on its structure. Class 3 defines a moving system with constant loads. Even though the applied external loads may be constant, any significant accelerations of the moving members can create time-varying reaction forces. An example might be a powered rotary lawn mower. Except for the case of mowing the occasional rock, the blades experience a nearly constant external load from mowing the grass. However, the accelerations of the spinning blades can create high loads at their fastenings. A dynamic load analysis is necessary for Classes 2 and 3. Note however that, if the motions of a Class 3 system are so slow as to generate negligible accelerations on its members, it could qualify as a Class 1 system and then would be called quasi-static. An automobile scissors jack (see Figure 3-5, p. 88) can be considered to be a Class 1 system since the external load (when used) is essentially constant, and the motions of the links are slow with negligible accelerations. The only complexity introduced by the motions of the elements in this example is that of determining in which position the internal loads on the jack's elements will be maximal, since they vary as the jack is raised, despite the essentially constant external load. Class 4 describes the general case of a rapidly moving system subjected to timevarying loads. Note that even if the applied external loads are essentially constant in a given case, the dynamic loads developed on the elements from their accelerations will still vary with time. Most machinery, especially if powered by a motor or engine, will be in Class 4. An example of such a system is the engine in your car. The internal parts (crankshaft, connecting rods, pistons, etc.) are subjected to time-varying loads from the gasoline explosions, and also experience time-varying inertial loads from their own accelerations. A dynamic load analysis is necessary for Class 4. 3.2 FREE-BODY DIAGRAMS In order to correctly identify all potential forces and moments on a system, it is necessary to draw accurate free-body diagrams (FBDs) of each member of the system. These FBDs should show a general shape of the part and display all the forces and moments that are acting on it. There may be external forces and moments applied to the part from outside the system, and there will be interconnection forces and/or moments where each part joins or contacts adjacent parts in the assembly or system. In addition to the known and unknown forces and couples shown on the FBD, the dimensions and angles of the elements in the system are defined with respect to local coordinate systems located at the centers of gravity (CG) of each element.* For a dynamic load analysis, the kinematic accelerations, both angular and linear (at the CG), need to be known or calculated for each element prior to doing the load analysis.

E mv mgh W h i = = ηη η 2 2 = (3.13a) If the deflection at impact is small compared to the drop distance h, this equation is sufficient. But, if the deflection is significant compared to h, the energy of impact needs to include an amount due to the deflection through which the weight falls beyond h. The total potential energy given up by the mass on impact is then E Wh W W h = η δ ηδ += + i i ( ) (3.13b) Equate this potential energy to the elastic energy stored in the struck member and substitute equation 3-9b and the expression W=kδst F k W h F kW h W W h F W h h F W F W F W h i i i i st i i st i st st i i i st 2 2 2 2 2 2 2 2 2 2 2 0 2 = = η δ η δ δ η δ η δ δ δ η δ η δ ( ) + ( ) + = + ( ) ⎛ ⎝ ⎞ ⎠ = += + ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ − ⎛ ⎝ ⎞ ⎠ − = (3.14a) giving a quadratic equation in Fi/W whose solution is: F W i i h st st = =+ + δ δ η δ 1 1 2 (3.14b) The right side expression is the impact ratio for the falling weight case. Equation 3-14b can be used for any impact case involving a falling weight. For example, if the weight is dropped on a beam, the static deflection of the beam at the point of impact is used. If the distance h to which the mass is raised is set to zero, equation 3.14b becomes equal to 2. This says that if the mass is held in contact with the "struck" member (with the weight of the mass separately supported) and then allowed to suddenly impart its weight to that member, the dynamic force will be twice the weight. This is the case of "force impact" described earlier, in which there is no actual collision between the objects. A more accurate analysis, using wave methods, predicts that the dynamic force will be more than doubled even in this noncollision case of sudden application of load.[6] Many designers use 3 or 4 as a more conservative estimate of this dynamic factor for the case of sudden load application. This is only a crude estimate, however, and, if possible, experimental measurements or a wave-method analysis should be made to determine more suitable dynamic factors for any particular design. Burr derives the correction factors η for several impact cases in reference 6. Roark and Young provide factors for additional cases in reference 7. For the case of a mass axially impacting a rod as shown in Figure 3-18, the correction factor is[6] η = + 1 1 3 m m

Equation 3.4 simplifies this model even further by assuming the damping d to be zero. If damping is included, the expressions for the fundamental, damped natural frequency ωd with units of radians/sec, or fd, with units of Hz, become ω π ω d d d k m d m f = − ⎛ ⎝ ⎞ ⎠ = 2 1 2 2 (3.7) This damped frequency ωd will be slightly lower than the undamped frequency ωn. EFFECTIVE VALUES Determining the effective mass for a lumped model is straightforward and requires only summing all the values of the connected, moving masses in appropriate mass units. Determining the values of the effective spring constant and the effective damping coefficient is more complicated and will not be addressed here. See reference 2 for an explanation. RESONANCE A condition called resonance can be experienced if the operating or forcing frequency applied to the system is the same as any one of its natural frequencies. That is, if the input angular velocity applied to a rotating system is the same as, or close to, ωn, the vibratory response will be very large. This can create large forces and cause failure. Thus it is necessary to avoid operation at or near the natural frequencies if possible. Dynamic Forces If we write equation 3.1 for the simple, one-DOF model of the dynamic system in Figure 3-15 and substitute equations 3.5 and 3.6, we get F ma my F F F my F my dy ky y cam spring damper cam ∑ = −− = = ++ = ˙˙ ˙˙ ˙˙ ˙ (3.8) If the kinematic parameters of displacement, velocity, and acceleration are known for the system, this equation can be solved directly for the force on the cam as a function of time. If the cam force is known and the kinematic parameters are desired, then the well-known solution to this linear, constant-coefficient differential equation can be applied. See reference 3 for a detailed derivation of that solution. Though the coordinate system used for a dynamic analysis can be arbitrarily chosen, it is important to note that both the kinematic parameters (displacement, velocity, and acceleration) and the forces in equation 3.8 must be defined in the same coordinate system. As an example of the effect of vibration on the dynamic forces of a system, we now revisit the fourbar linkage of Case Study 5A and see the results of actual measurements of dynamic forces under operating conditions.

Equations (a) through (e) comprise a set of 16 simultaneous equations that can be solved either by matrix reduction or by iterative root-finding methods. Putting them in standard form for a matrix solution gives: FFF FFF RF RF RF RF RF RF FF P FF P RF RF RF R xxx yyy xy yx xy yx xy yx x xx y yy xy yx xy y 12 32 42 12 32 42 12 12 12 12 32 32 32 32 42 42 42 42 23 43 23 43 23 23 23 23 43 43 43 0 0 0 ++= ++= −+−+ − = + =− + =− −+− F RP RP FFF FFF RF RF RF RF RF RF f F F F F F F F x Px y Py x xxx yyy xy yx xy yx xy yx x x y y x x 43 14 24 34 14 24 34 14 14 14 14 24 24 24 24 34 34 34 34 32 23 32 23 34 43 34 0 0 0 0 0 0 =− + ++= ++= −+ −+− = + = + = + = ( ) y y x x y y y x F F F F F F F + = + = + = − = 43 42 24 42 24 24 24 0 0 0 tanθ 0 11 Substituting the given data from Table 3-4 part 1 gives: FFF FFF FFFFF F F F F F FFF xxx yyy yx yxy x x x y y yxy 12 32 42 12 32 42 12 12 32 32 42 42 23 43 23 43 23 23 43 0 0 3 12 1 80 2 08 1 20 2 71 0 99 0 0 0 1 000 0 78 0 78 0 78 0 ++= ++= −+ + − + − = + = + = −+ + + .... .. . . . . .78 500 0 0 3 12 1 80 2 58 1 04 2 08 1 20 0 0 0 0 43 14 24 34 14 24 34 14 14 24 24 34 34 32 23 32 23 34 43 34 43 F FFF FFF FFFF FF g F F F F F F F F x xxx yyy yx yx yx x x y y x x y y = − ++= ++= +−−−−= + = + = + = + = . . . . . . () 0 0 0 10 0 42 24 42 24 24 24 F F F F F F x x y y y x + = + = + = . 12 Put these equations into matrix form.

Euler's equations 3.1d show that if the rotational motions (ω, α) and applied moments or couples exist about only one axis (say the z axis), then that set of three equations reduces to one equation, ∑ MI a z zz = α (. ) 3 2 because the ω and α terms about the x and y axes are now zero. Equation 3.1b is reduced to ∑ ∑ F ma F ma b xx yy = = (. ) 3 2 Equations 3.2 can be written for all the connected bodies in a two-dimensional system and the entire set solved simultaneously for forces and moments. The number of second-law equations will now be up to three times the number of elements in the system plus the necessary reaction equations at connecting points, again resulting in large systems of equations for even simple systems. Note that even though all motion is about one (z) axis in a 2-D system, there may still be loading components in the z direction due to external forces or couples. Static Load Analysis The difference between a dynamic loading situation and a static one is the presence or absence of accelerations. If the accelerations in equations 3.1 and 3.2 are all zero, then for the three-dimensional case these equations reduce to FFF a MMM xyz xyz ∑∑∑ ∑∑∑ === === 000 3 3 000 (. ) and for the two-dimensional case, FFM b ∑∑∑ xy z == = 0 0 0 33 (. ) Thus, we can see that the static loading situation is just a special case of the dynamic loading one, in which the accelerations happen to be zero. A solution approach based on the dynamic case will then also satisfy the static one with appropriate substitutions of zero values for the absent accelerations. 3.4 TWO-DIMENSIONAL, STATIC LOADING CASE STUDIES This section presents a series of three case studies of increasing complexity, all limited to two-dimensional static loading situations. A bicycle handbrake lever, a crimping tool, and a scissors jack are the systems analyzed. These case studies provide examples of the simplest form of force analysis, having no significant accelerations and having forces acting in only two dimensions.

Figure 3-3 shows the tool in the closed position in the process of crimping a metal connector onto a wire. The user's hand provides the input forces between links 1 and 2, shown as the reaction pair Fh. The user can grip the handle anywhere along its length, but we are assuming a nominal moment arm of Rh for the application of the resultant of the user's grip force (see Figure 3-4). The high mechanical advantage of the tool transforms the grip force to a large force at the crimp. Figure 3-3 is a free-body diagram of the entire assembly, neglecting the weight of the tool which is small compared to the crimp force. There are four elements, or links, in the assembly, all pinned together. Link 1 can be considered to be the "ground" link, with the other links moving with respect to it as the jaw is closed. The desired magnitude of the crimp force Fc is defined and its direction will be normal to the surfaces at the crimp. The third law relates the action-reaction pair acting on links 1 and 4: F F a F F cx c x cy c y 1 4 1 4 = − = − ( ) 2 Figure 3-4 shows the elements of the crimping-tool assembly separated and drawn as free-body diagrams with all forces applied to each element, again neglecting their weights as being insignificant compared to the applied forces. The centers of gravity of the respective elements are used as the origins of the local, nonrotating coordinate systems in which the points of application of all forces on the elements are located.* 3 We will consider link 1 to be the ground plane and analyze the remaining moving links. Note that all unknown forces and moments are initially assumed positive. Link 2 has three forces acting on it: Fh is the unknown force from the hand, and F12 and F32 are

Figure 3-5 shows a schematic of a simple scissors jack used to raise a car. It consists of six links which are pivoted and/or geared together and a seventh link in the form of a lead screw which is turned to raise the jack. While this is clearly a three-dimensional device, it can be analyzed as a two-dimensional one if we assume that the applied load (from the car) and the jack are exactly vertical (in the z direction). If so, all forces will be in the xy plane. This assumption is valid if the car is jacked from a level surface. If not, then there will be some forces in the yz and xz planes as well. The jack designer needs to consider the more general case, but for our simple example we will initially assume two-dimensional loading. For the overall assembly as shown in Figure 3-5, we can solve for the reaction force Fg, given force P, by summing forces: Fg = -P. 2 Figure 3-6 shows a set of free-body diagrams for the entire jack. Each element or subassembly of interest has been separated from the others and the forces and moments shown acting on it (except for its weight, which is small compared to the applied forces and is thus neglected for this analysis). The forces and moments can be either internal reactions at interconnections with other elements or external loads from the "out

Figure 3-7 shows the upper half of the jack assembly. Because of the mirror symmetry between the upper and lower portions, the lower half can be removed to simplify the analysis. The forces calculated for this half will be duplicated in the other. If we wished, we could solve for the reaction forces at A and B using equations 3.3b from this free-body diagram of the half-jack assembly. 4 Figure 3-8a shows the free-body diagrams for the upper half of the jack assembly which are essentially the same as those of Figure 3-6. We now have four elements but can consider the subassembly labeled 1 to be the "ground," leaving three elements on which to apply equations 3.3. Note that all unknown forces and moments are initially assumed positive in the equations. 5 Link 2 has three forces acting on it: F42 is the unknown force at the gear tooth contact with link 4; F12 and F32 are the unknown reaction forces from links 1 and 3 respectively. Force F12 is provided by part 1 on part 2 at the pivot pin, and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and directions of these pin forces and the magnitude of F42 are unknown. The direction of F42 is along the common normal shown in Figure 3-8b. Write equations 3.3b for this element to sum forces in the x and y directions and sum moments about the CG (with the cross products expanded):* FF F F FF F F a M RF RF RF RF RF RF xxxx yyyy z xy yx xy yx xy yx ∑ ∑ ∑ =++= =++= =−+−+− = 12 32 42 12 32 42 12 12 12 12 32 32 32 32 42 42 42 42 0 0 0 ( ) 6 Link 3 has three forces acting on it: the applied load P, F23, and F43. Only P is known. Writing equations 3.3b for this element gives FF F P FF F P b M R F R F R F R F RP RP x x xx y y yy z x y y x x y y x Px y Py x ∑ ∑ ∑ = + += = + += = − + − +− = 23 43 23 43 23 23 23 23 43 43 43 43 0 0 0

Figures 3-11 and 3-12 show the fourbar linkage demonstrator model. It consists of three moving elements (links 2, 3, and 4) plus the frame or ground link (1). The motor drives link 2 through a gearbox. The two fixed pivots are instrumented with piezoelectric force transducers to measure the dynamic forces acting in x and y directions on the ground plane. A pair of accelerometers is mounted to a point on the floating coupler (link 3) to measure its accelerations. 2 Figure 3-12 shows a schematic of the linkage. The links are designed with lightening holes to reduce their masses and mass moments of inertia. The input to link 2 can be an angular acceleration, a constant angular velocity, or an applied torque. Link 2 rotates fully about its fixed pivot at O2. Even though link 2 may have a zero angular acceleration α2, if run at constant angular velocity ω2, there will still be time-varying angular accelerations on links 3 and 4 since they oscillate back and forth. In any case, the CGs of the links will experience time-varying linear accelerations as the linkage moves. These angular and linear accelerations will generate inertia forces and torques

Link 4 has three forces acting on it: F24 is the unknown force from link 2; F14 and F34 are the unknown reaction forces from links 1 and 3 respectively. FF F F FF F F c M RF RF RF RF RF RF xxxx yyyy z xy yx xy yx xy yx ∑ ∑ ∑ =++= =++= =−+ − +− = 14 24 34 14 24 34 14 14 14 14 24 24 24 24 34 34 34 34 0 0 0 ( ) 8 The nine equations in sets a through c have 16 unknowns in them, F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y, F14x, F14y, F34x, F34y, F24x, F24y, F42x, and F42y . We can write the third-law relationships between action-reaction pairs at each of the joints to obtain six of the seven additional equations needed: FF FF FF FF d FF FF xx yy xx yy xx yy 32 23 32 23 34 43 34 43 42 24 42 24 =− =− =− =− =− =− ( ) 9 The last equation needed comes from the relationship between the x and y components of the force F24 (or F42) at the tooth/tooth contact point. Such a contact (or half) joint can transmit force (excepting friction force) only along the common normal[4] which is perpendicular to the joint's common tangent as shown in Figure 3-8b. The common normal is also called the axis of transmission. The tangent of the angle of this common normal relates the two components of the force at the joint:

For link 4: F F F mA F F F mA c M RF RF RF RF I xxx G yyy G z xy yx xy yx G x y ∑ ∑ ∑ =+= =+= = − ( ) + − ( ) = 14 34 4 4 14 34 4 4 14 14 14 14 34 34 34 34 4 4 ( ) α 8 There are 13 unknowns in these nine equations: F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y, F14x, F14y, F34x, F34y, and T2. Four third-law equations can be written to equate the action-reaction pairs at the joints. F F F F FF d F F x x y y x x y y 32 23 32 23 34 43 34 43 = − = − = − = − ( ) 9 The set of thirteen equations in a through d can be solved simultaneously to determine the forces and driving torque either by matrix reduction or by iterative root-finding methods. This case study was solved by both techniques and files for both are on the CD. Note that the masses and mass moments of inertia of the links are constant with time and position, but the accelerations are time-varying. Thus, a complete analysis requires that equations a-d be solved for all positions or time steps of interest. The models use lists or arrays to store the calculated values from equations a-d for 13 values of the input angle θ2 of the driving link ( 0 to 360° by 30° increments). The model also calculates the kinematic accelerations of the links and their CGs which are needed for the force calculations. The largest and smallest forces present on each link during the cycle can then be determined for use in later stress and deflection analyses. The given data and results of this force analysis for one crank position (θ2 = 30°) are shown in Table 3-6, parts 1 and 2. Plots of the forces at the fixed pivots for one complete revolution of the crank are shown in Figure 3-14. 3.7 VIBRATION LOADING In systems that are dynamically loaded, there will usually be vibration loads superimposed on the theoretical loads predicted by the dynamic equations. These vibration loads can be due to a variety of causes. If the elements in the system were infinitely stiff, then vibrations would be eliminated. But all real elements, of any material, have elasticity and thus act as springs when subjected to forces. The resulting deflections can cause additional forces to be generated from the inertial forces associated with the vibratory movements of elements or, if clearances allow contact of mating parts, to generate impact (shock) loads (see below) during their vibrations. A complete discussion of vibration phenomena is beyond the scope of this text and will not be attempted here. References are provided in the bibliography at the end of this chapter for further study. The topic is introduced here mainly to alert the machine designer to the need to consider vibration as a source of loading. Often the only way to get an accurate measure of the effects of vibration on a system is to do testing of prototypes or production systems under service conditions. The discussion of safety fac

Table 3-1 shows the four possible classes. Class 1 is a stationary system with constant loads. One example of a Class 1 system is the base frame for an arbor press used in a machine shop. The base is required to support the dead weight of the arbor press which is essentially constant over time, and the base frame does not move. The parts brought to the arbor press (to have something pressed into them) temporarily add their weight to the load on the base, but this is usually a small percentage of the dead weight. A static load analysis is all that is necessary for a Class 1 system.

The 13 equations b-e can be solved simultaneously by matrix reduction or by iteration with a root-finding algorithm. For a matrix solution, the unknown terms are placed on the left and known terms on the right of the equal signs. F F FFF RF R F R F R F R F F F F F RF RF RF RF FF F x x y y h h h xy yx xy yx x x y y xy yx xy yx x x cx 12 32 12 32 12 12 12 12 32 32 32 32 23 43 23 43 23 23 23 23 43 43 43 43 14 34 4 0 0 0 0 0 0 + = + += +−+− = + = + = −+−= + =− ( f FF F RF RF RF RF RF RF F F F F F F F F y y cy x y y x x y y x cxcy cycx x x x x y y y y ) 14 34 4 14 14 14 14 34 34 34 34 4 4 4 4 32 23 34 43 32 23 34 43 0 0 0 0 + =− − + − =− + + = + = + = + = 8 Substitute the given data from Table 3-3 part 1

The loading considered so far has either been static or, if time-varying, has been assumed to be gradually and smoothly applied, with all mating parts continually in contact. Many machines have elements that are subjected to sudden loads or impacts. One example is the crank-slider mechanism which forms the heart of an automobile engine. The piston head is subjected to an explosive rise in pressure every two crank revolutions when the cylinder fires, and the clearance between the circumference of the piston and the cylinder wall can allow an impact of these surfaces as the load is reversed each cycle. A more extreme example is a jackhammer, whose purpose is to impact pavement and break it up. The loads that result from impact can be much greater than those that would result from the same elements contacting gradually. Imagine trying to drive a nail by gently placing the hammer head on the nail rather than by striking it. What distinguishes impact loading from static loading is the time duration of the application of the load. If the load is applied slowly, it is considered static; if applied rapidly, then it is impact. One criterion used to distinguish the two is to compare the time of load application tl (defined as the time it takes the load to rise from zero to its peak value) to the period of the natural frequency Tn of the system. If tl is less than half Tn, it is considered to be impact. If tl is greater than three times Tn, it is considered static. Between those limits is a gray area in which either condition can exist. Two general cases of impact loading are considered to exist, though we will see that one is just a limiting case of the other. Burr[6] calls these two cases striking impact and

The loading function q(x) is typically known and the shear V and moment M distributions can be found by integrating equation 3.16a: dV qdx V V b V V x x B A A B A B ∫ ∫ = =− (. ) 3 16 Equation 3.16b shows that the difference in the shear forces between any two points, A and B, is equal to the area under the graph of the loading function, equation 3.16a. Integrating the relationship between shear and moment gives dM Vdx M M c M M x x B A A B A B ∫ ∫ = =− (. ) 3 16 showing that the difference in the moment between any two points, A and B, is equal to the area under the graph of the shear function, equation 3.16b. SIGN CONVENTION The usual (and arbitrary) sign convention used for beams is to consider a moment positive if it causes the beam to deflect concave downward (as if to collect water). This puts the top surface in compression and the bottom surface in tension. The shear force is considered positive if it causes a clockwise rotation of the section on which it acts. These conventions are shown in Figure 3-23 and result in positive moments being created by negative applied loads. All the applied loads shown in Figure 3-22 are negative. For example, in Figure 3-22a, the distributed load magnitude from a to l is q = -w. EQUATION SOLUTION The solution of equations 3.3 (p. 78) and 3.16 for any beam problem may be carried out by one of several approaches. Sequential and graphical solutions are described in many textbooks on statics and mechanics of materials. One classical approach to these problems is to find the reactions on the beam using equations 3.3 and then draw the shear and moment diagrams using a graphical integration approach combined with calculations for the significant values of the functions. This approach has value from a pedagogical standpoint as it is easily followed, but it is cumbersome to implement. The approach most amenable to computer solution uses a class of mathematical functions called singularity functions to represent the loads on the beam. We present the classical approach as a pedagogical reference and also introduce the use of singularity functions which offer some computational advantages. While this approach may be new to some students, when compared to the methods usually learned in other courses, it has significant advantages in computerizing the solution. Singularity Functions Because the loads on beams typically consist of collections of discrete entities, such as point loads or segments of distributed loads that can be discontinuous over the beam length, it is difficult to represent these discrete functions with equations that are valid over the entire continuum of beam length. A special class of functions called singularity functions was invented to deal with these mathematical situations. Singularity functions are often denoted by a binomial in angled brackets as shown in equations 3.17. The first quantity in the brackets is the variable of interest, in our case x, the distance along the beam length. The second quantity a is a user-defined parameter that denotes where in x the singularity function either acts or begins to act. For example, for a point

This equation set can be easily solved to yield F F F F F b F F y y cable x y x x 23 43 23 23 43 23 2 1 046 2 523 56 523 1 483 353 353 = =− =− =− = ( )° = − = − =− = N N () N tan . Newton's third law relates these forces to their reactions on the brake arm at point A: F F FF c F x x y y z 32 23 32 23 32 353 523 0 =− = =− = = N N () 5 We can now write equations 3.3a for the arm (link 2). For the forces: FF F F F F FF F F F F d FF F F F F xxx x x x yyy y y y zzz z z z ∑ ∑ ∑ = + + = + =− = + + = + =− =++= += 12 32 52 12 52 12 32 52 12 52 12 32 52 12 52 0 353 0 523 0 0 ; ; ; ( ) For the moments: M M RF RF RF RF RF RF M M RF RF RF R F RF RF e x x yz zy yz zy yz zy y y zx xz zx xz zx xz ∑ ∑ =+ − ( ) + − ( ) + − ( ) = =+ − ( ) + − ( ) + − ( ) = 12 12 12 12 12 32 32 32 32 52 52 52 52 12 12 12 12 12 32 32 32 32 52 52 52 52 0 0 () M RF RF RF RF RF RF z xy yx xy yx xy yx ∑ = − ( ) + − ( ) + − ( ) = 12 12 12 12 32 32 32 32 52 52 52 52 0 Note that all unknown forces and moments are initially assumed positive in the equations, regardless of their apparent directions on the FBDs. The moments M12x and M12y are due to the fact that there is a moment joint between the arm (2) and the pivot pin (1) about the x and y axes. We assume negligible friction about the z axis, thus allowing M12z to be zero. 6 The joint between the brake pad (5) and the wheel rim (6) transmits a force normal to the plane of contact. The friction force magnitude, Ff, in the contact plane is related to the normal force by the Coulomb friction equation, FN f f = μ ( ) where μ is the coefficient of friction and N is the normal force. The velocity of the point on the rim below the center of the brake pad is in the z direction. The force components F52x and F52y are due entirely to the normal force being transmitted through the pad to the arm and are therefore related by Newton's third law.

This example of deviations from theoretical forces in a very simple dynamic system is presented to point out that our best calculations of forces (and thus the resulting stresses) on a system may be in error due to factors not included in a simplified force analysis. It is common for the theoretical predictions of forces in a dynamic system to understate reality, which is, of course, a nonconservative result. Wherever feasible, the testing of physical prototypes will give the most accurate and realistic results. The effects of vibration on a system can cause significant loadings and are difficult to predict without test data of the sort shown in Figure 3-16, where the actual loads are seen to be double their predicted values; this will obviously double the stresses. A traditional, and somewhat crude, approach used in machine design has been to apply overload factors to the theoretical calculated loads based on experience with the same or similar equipment. As an example, see Table 12-17 (p. 715) in the chapter on spurgear design. This table lists industry-recommended overload factors for gears subjected to various types of shock loading. These sorts of factors should be used only if one cannot develop more accurate test data of the type shown in Figure 3-16 (p. 106).

This section presents a case study involving three-dimensional static loading on a bicycle brake caliper assembly. The same techniques used for two-dimensional load analysis also work for the three-dimensional case. The third dimension requires more equations, which are available from the summation of forces in the z direction and the summation of moments about the x and y axes as defined in equations 3.1 and 3.3 for the dynamic and static cases, respectively. As an example, we will now analyze the bicycle brake arm that is actuated by the handbrake lever that was analyzed in Case Study 1A.

Two more equations come from the assumption (shown in Figure 3-1) that the two forces provided by the hand on the brake lever and handgrip are equal and opposite:* F F h F F bx b x by b y 1 2 1 2 = − = − ( ) The remaining three equations come from the given geometry and the assumptions made about the system. The direction of the forces Fcable and Fsheath are known to be in the same direction as that end of the cable. In the figure it is seen to be horizontal, so we can set FF i cable sheath y y = = 0 0 ; () Because of our no-friction assumption, the force F31 can be assumed to be normal to the surface of contact between the cable and the hole in part 1. This surface is horizontal in this example, so F31 is vertical and F j 31x = 0 ( ) 6 This completes the set of 21 equations (equation sets a, c, d, e, g, h, i, and j), and they can be solved for the 21 unknowns simultaneously "as is," that is, all 21 equations could be put into matrix form and solved with a matrix-reduction computer program. However, the problem can be simplified by manually substituting equations c, g, h, i, and j into the others to reduce them to a set of eight equations in eight unknowns. The known or given data are as shown in Table 3-2, part 1. 7 As a first step, for link 2, substitute equations b and c in equation a to get: FFF FFF k RF RF R F RF RF RF x bx x y by x x y y x bxby bybx x x y x 12 2 32 12 2 32 12 12 12 12 2 2 2 2 32 32 32 32 0 0 0 ++= ++ = ( ) − + − ( ) + − ( ) = tan tan θ θ ( ) 8 Next, take equations d for link 3 and substitute equation c and also -F32x for F23x, and -F32y for F23y from equation g to eliminate those variables. F FF l F FF cable x x cable y x x y +− = +− = 13 32 13 32 0 0 ( ) tanθ 9 For link 1, substitute equation f in e and replace F21x with -F12x, F21y with -F12y, F31x with -F13x , F31y with -F13y, and Fsheathx with -Fcablex from equation g, − + − +− = − + − += +− + ( ) + − ( ) +− + ( ) + − ( ) + F F F PF FFF P m M RF RF RF RF R F R F RP RP R x bx x x cable y by x y h x y y x bx by by bx x y y x Px y Py x dy 12 1 13 x 12 1 32 21 12 21 12 1 1 1 1 31 13 31 13 0 tanθ 0 ( ) Fcablex = 0 10 Finally, substitute equations h, i, and j into equations k, l, and m to yield the following set of eight simultaneous equations in the eight remaining unknowns: F12x, F12y, F32x , F13y, Fcablex , Px, Py, and Mh. Put them in the standard form which has all un- * But not necessarily colinear. known terms on the left and all known terms to the right of the equal signsq

Two more equations come from the assumption (shown in Figure 3-1) that the two forces provided by the hand on the brake lever and handgrip are equal and opposite:* F F h F F bx b x by b y 1 2 1 2 = − = − ( ) The remaining three equations come from the given geometry and the assumptions made about the system. The direction of the forces Fcable and Fsheath are known to be in the same direction as that end of the cable. In the figure it is seen to be horizontal, so we can set FF i cable sheath y y = = 0 0 ; () Because of our no-friction assumption, the force F31 can be assumed to be normal to the surface of contact between the cable and the hole in part 1. This surface is horizontal in this example, so F31 is vertical and F j 31x = 0 ( ) 6 This completes the set of 21 equations (equation sets a, c, d, e, g, h, i, and j), and they can be solved for the 21 unknowns simultaneously "as is," that is, all 21 equations could be put into matrix form and solved with a matrix-reduction computer program. However, the problem can be simplified by manually substituting equations c, g, h, i, and j into the others to reduce them to a set of eight equations in eight unknowns. The known or given data are as shown in Table 3-2, part 1. 7 As a first step, for link 2, substitute equations b and c in equation a to get: FFF FFF k RF RF R F RF RF RF x bx x y by x x y y x bxby bybx x x y x 12 2 32 12 2 32 12 12 12 12 2 2 2 2 32 32 32 32 0 0 0 ++= ++ = ( ) − + − ( ) + − ( ) = tan tan θ θ ( ) 8 Next, take equations d for link 3 and substitute equation c and also -F32x for F23x, and -F32y for F23y from equation g to eliminate those variables. F FF l F FF cable x x cable y x x y +− = +− = 13 32 13 32 0 0 ( ) tanθ 9 For link 1, substitute equation f in e and replace F21x with -F12x, F21y with -F12y, F31x with -F13x , F31y with -F13y, and Fsheathx with -Fcablex from equation g, − + − +− = − + − += +− + ( ) + − ( ) +− + ( ) + − ( ) + F F F PF FFF P m M RF RF RF RF R F R F RP RP R x bx x x cable y by x y h x y y x bx by by bx x y y x Px y Py x dy 12 1 13 x 12 1 32 21 12 21 12 1 1 1 1 31 13 31 13 0 tanθ 0 ( ) Fcablex = 0 10 Finally, substitute equations h, i, and j into equations k, l, and m to yield the following set of eight simultaneous equations in the eight remaining unknowns: F12x, F12y, F32x , F13y, Fcablex , Px, Py, and Mh. Put them in the standard form which has all un- * But not necessarily colinear. known terms on the left and all known terms to the right of the equal signs

Vibration loading can also severely increase the actual loading above the theoretically calculated levels as shown in Case Study 5B and Figure 3-16 (p. 106). Experimental measurements made under real loading conditions are the best way to develop information in these cases. The case studies in this chapter are designed to set up realistic problems for stress and failure analysis in the following chapters. Though their complexity may be a bit daunting to the student at first encounter, much benefit can be gained from time spent studying them. This effort will be rewarded with a better understanding of the stressanalysis and failure-theory topics in succeeding chapters. Important Equations Used in This Chapter See the referenced sections for information on the proper use of these equations. Newton's Second Law (Section 3.3): F ma F ma F ma b ∑∑∑ xx yy zz === (. ) 3 1 Euler's Equations (Section 3.3): MI II M I II d MI II x xx y z yz y yy z x zx z zz x y xy ∑ ∑ ∑ = −− ( ) = −− ( ) = −− ( ) α ωω α ωω α ωω (. ) 3 1 Static Loading (Section 3.3): FFF a MMM xyz xyz ∑∑∑ ∑∑∑ === === 000 3 3 000 (. ) Undamped Natural Frequency (Section 3.7): ω π ω n n n k m f = (3.4) = 1 2 Damped Natural Frequency (Section 3.7): ω π ω d d d k m d m f = − ⎛ ⎝ ⎞ ⎠ = 2 1 2 2 (3.7) Spring Constant (Section 3.7): k F = a δ

Viscous Damping (Section 3.7): d F y = ˙ (3.6) Dynamic Force Ratio (Section 3.8): F W i i h st st = =+ + δ δ η δ 1 1 2 (3.14b) Beam-Loading, Shear, and Moment Functions (Section 3.9): q x dV dx d M dx ( ) = = a 2 2 (. ) 3 16 Integrals of Singularity Functions (Section 3.9): λ λ − = − −∞ ∫ a d x a a x 2 3 3 (. ) 3 18 λ λ − = − −∞ ∫ a d x a b x 1 2 2 (. ) 3 18 λ λ − =− −∞ ∫ ad x a c x 0 1 (. ) 3 18 λ λ − =− − −∞ ∫ a d xa d x 1 0 (. ) 3 18 λ λ − =− − −∞ − ∫ a d xa e x 2 1

also allow us to obtain the beam's deflection curve with little additional work. The simple method shown above is not as useful for determining deflection curves, as will be seen in the next chapter. We will now repeat this example using singularity functions to determine the loading, shear, and moment diagrams.

as defined by Newton's second law. Thus, even with no external forces or torques applied to the links, the inertial forces will create reaction forces at the pins. It is these forces that we wish to calculate. 3 Figure 3-13 shows the free-body diagrams of the individual links. The local, nonrotating, coordinate system for each link is set up at its CG. The kinematic equations of motion must be solved to determine the linear accelerations of the CG of each link and the link's angular acceleration for every position of interest during the cycle. (See reference 1 for an explanation of this procedure.) These accelerations, AGn and αn, are shown acting on each of the n links. The forces at each pin connection are shown as xy pairs, numbered as before, and are initially assumed to be positive. 4 Equations 3.1 can be written for each moving link in the system. The masses and the mass moments of inertia of each link about its CG must be calculated for use in these equations. In this case study, a solid modeling CAD system was used to design the links' geometries and to calculate their mass properties. 5 For link 2: F F F mA F F F mA a M T RF RF RF RF I xxx G yyy G z xy yx xy yx G x y ∑ ∑ ∑ =+= =+= =+ − ( ) + − ( ) = 12 32 2 2 12 32 2 2 2 12 12 12 12 32 32 32 32 2 2 ( ) α 6 For link 3: F F F mA F F F mA a M T RF RF RF RF I xxx G yyy G z xy yx xy yx G x y ∑ ∑ ∑ =+= =+= =+ − ( ) + − ( ) = 12 32 2 2 12 32 2 2 2 12 12 12 12 32 32 32 32 2 2

force impact. Striking impact refers to an actual collision of two bodies, such as in hammering or the taking up of clearance between mating parts. Force impact refers to a suddenly applied load with no velocity of collision, as in a weight suddenly being taken up by a support. This condition is common in friction clutches and brakes (see Chapter 17). These cases can occur independently or in combination. Severe collisions between moving objects can result in permanent deformation of the colliding bodies as in an automobile accident. In such cases the permanent deformation is desirable in order to absorb the large amount of energy of the collision and protect the occupants from more severe harm. We are concerned here only with impacts that do not cause permanent deformation; that is, the stresses will remain in the elastic region. This is necessary to allow continued use of the component after impact. If the mass of the striking object m is large compared to that of the struck object mb and if the striking object can be considered rigid, then the kinetic energy possessed by the striking object can be equated to the energy stored elastically in the struck object at its maximum deflection. This energy approach gives an approximate value for the impact loading. It is not exact because it assumes that the stresses throughout the impacted member reach peak values at the same time. However, waves of stress are set up in the struck body, travel through it at the speed of sound, and reflect from the boundaries. Calculating the effects of these longitudinal waves on the stresses in elastic media gives exact results and is necessary when the ratio of mass of the striking object to that of the struck object is small. The wave method will not be discussed here. The reader is directed to reference 6 for further information. Energy Method The kinetic energy of the striking body will be converted to stored potential energy in the struck body, assuming that no energy is lost to heat. If we assume that all particles of the combined bodies come to rest at the same instant, then just before rebound, the force, stress, and deflection in the struck body will be maximal. The elastic energy stored in the struck body will be equal to the area under the force-deflection curve defined by its particular spring constant. A generalized force-deflection curve for a linear spring element is shown in Figure 3-17. The elastic energy stored is the area under the curve between zero and any combination of force and deflection. Because of the linear relationship, this is the area of a triangle, A = 1/2bh. Thus, the energy stored at the point of peak impact deflection, δi, is EF a = i i 1 2 δ (. ) 3 9 Substituting equation 3.5 gives E F k b i = 2 2 (. ) 3 9 HORIZONTAL IMPACT Figure 3-18a shows a mass about to impact the end of a horizontal rod. This device is sometimes called a slide hammer and is used to remove dents from automobile sheet metal among other uses. At the point of impact, the portion of the kinetic energy of the moving mass that is imparted to the struck mass is

hese examples of beam systems represent only a small fraction of all the possible combinations of beam loadings and constraints that one will encounter in practice. Rather than having to write and integrate the loading functions for every new beam situation from scratch, the particular problem can often be solved by using superposition, which simply means adding the individual results together. For small deflections, it is safe to assume linearity in these problems, and linearity is a requirement for superposition to be valid. For example, the load due to the weight of a beam (ignored in the above examples) can be accounted for by superposing a uniform load over the beam's entire length on whatever other applied loads may be present.

his section presents a case study involving two-dimensional dynamic loading on a fourbar linkage designed as a dynamic-load demonstration device. A photograph of this machine is shown in Figure 3-11. This machine can be analyzed in two dimensions since all elements move in parallel planes. The presence of significant accelerations on the moving elements in a system requires that a dynamic analysis be done with equations 3.1. The approach is identical to that used in the preceding static load analyses except for the need to include the mA and Iα terms in the equations.

ide world." The centers of gravity of the respective elements are used as the origins of the local, nonrotating coordinate systems in which the points of application of all forces on the elements are located. In this design, stability is achieved by the mating of two pairs of crude (noninvolute) gear segments acting between links 2 and 4 and between links 5 and 7. These interactions are modeled as forces acting along a common normal shared by the two teeth. This common normal is perpendicular to the common tangent at the contact point. There are 3 second-law equations available for each of the seven elements, allowing 21 unknowns. An additional 10 third-law equations will be needed for a total of 31. This is a cumbersome system to solve for such a simple device, but we can use its symmetry to advantage in order to simplify the problem.

which is defined as 0 when x ≤ a, and equal to (x - a) when x > a. A uniformly distributed load over a portion of a beam can be represented mathematically by a unit step function, xa c − 0 (. ) 3 17 which is defined as 0 when x < a, unity when x > a, and is undefined at x = a. A concentrated force can be represented by the unit impulse function, xa d − −1 (. ) 3 17 which is defined as 0 when x < a, ∞ when x = a, and 0 when x > a. Its integral evaluates to unity at a. A concentrated moment can be represented by the unit doublet function, xa e − −2 (. ) 3 17 which is defined as 0 when x < a, indeterminate when x = a, and 0 when x > a. It generates a unit couple moment at a. This process can be extended to obtain polynomial singularity functions of any order n to fit distributed loads of any shape. Four of the five singularity functions described here are shown in Figure 3-22, as applied to various beam types. A computer program is needed to evaluate these functions. Table 3-7 shows five of the singularity functions implemented in a "BASIC-like" pseudocode. The For loops run the variable x from zero to the beam length l. The If test determines whether x has reached the value of a which is the location of the start of the singularity function. Depending on this test, the value of y(x) is set either to zero or to the specified magnitude of the singularity function. This type of code can easily be implemented in any computer language (e.g., C+, Fortran, BASIC) or in an equation solver (e.g., Mathcad, MATLAB, TK Solver, EES). The integrals of these singularity functions have special definitions that, in some cases, defy common sense but nevertheless provide the desired mathematical results. For example, the unit impulse function (Eq. 3.17d) is defined in the limit as having zero width and infinite magnitude, yet its area (integral) is defined as equal to one (Eq. 3.18d). (See reference 8 for a more complete discussion of singularity functions.) The integrals of the singularity functions in equations 3.17 are defined as λ λ − = − −∞ ∫ a d x a a x 2 3 3 (. ) 3 18 λ λ − = − −∞ ∫ a d x a b x 1 2 2 (. ) 3 18 λ λ − = − −∞ ∫ ad x a c x 0 1 (. ) 3 18 λ λ − = − − −∞ ∫ a d xa d x 1 0

ing the weights of the parts. The lever (part 2) has three forces on it, Fb2, F32, and F12. The two-character subscript notation used here should be read as, force of element 1 on 2 (F12) or force at B on 2 (Fb2), etc. This defines the source of the force (first subscript) and the element on which it acts (second subscript). This notation will be used consistently throughout this text for both forces and position vectors such as Rb2, R32, and R12 in Figure 3-2 which serve to locate the above three forces in a local, nonrotating coordinate system whose origin is at the center of gravity (CG) of the element or subassembly being analyzed.* On this brake lever, Fb2 is an applied force whose magnitude and direction are known. F32 is the force in the cable. Its direction is known but not its magnitude. Force F12 is provided by part 1 on part 2 at the pivot pin. Its magnitude and direction are both unknown. We can write equations 3.3b for this element to sum forces in the x and y directions and sum moments about the CG. Note that all unknown forces and moments are initially assumed positive in the equations. Their true signs will come out in the calculation.† However, all known or given forces must carry their proper signs. FF F F FF F F a x x bx x y y by y z bb ∑ ∑ ∑ =++= =++= = × ( ) + × ( ) + × ( ) = 12 2 32 12 2 32 12 12 2 2 32 32 0 0 0 ( ) MRF RF RF The cross products in the moment equation represent the "turning forces" or moments created by the application of these forces at points remote from the CG of the element. Recall that these cross products can be expanded to M RF RF R F RF b RF RF z x y y x bx by bybx xy yx ∑ = − ( ) + − ( ) + − ( ) = 12 12 12 12 2 2 2 2 32 32 32 32 0 ( ) We have three equations and four unknowns (F12x, F12y, F32x, F32y) at this point, so we need another equation. It is available from the fact that the direction of F32 is known. (The cable can pull only along its axis.) We can express one component of the cable force F32 in terms of its other component and the known angle, θ of the cable. FF c 32 32 y x = tan ( ) θ We could now solve the four unknowns for this element, but will wait to do so until the equations for the other two links are defined. 3 Part 3 in Figure 3-2 is the cable which passes through a hole in part 1. This hole is lined with a low-friction material which allows us to assume no friction at the joint between parts 1 and 3. We will further assume that the three forces F13, F23, and Fcable form a concurrent system of forces acting through the CG and thus create no moment. With this assumption only a summation of forces is necessary for this element. FF F F d FF F F x cable x x y cable y y x y ∑ ∑ = ++= = ++= 13 23 13 23 0 0 ( ) 4 The assembly of elements labeled part 1 in Figure 3-2 can have both forces and moments on it (i.e., it is not a concurrent system), so the three equations 3.3b are needed.

mentum about the CG. The left sides of these equations respectively sum all the forces and moments that act on the body, whether from known applied forces or from interconnections with adjacent bodies in the system. For a three-dimensional system of connected rigid bodies, this vector equation for the linear forces can be written as three scalar equations involving orthogonal components taken along a local x, y, z axis system with its origin at the CG of the body: F ma F ma F ma b ∑∑∑ xx yy zz === (. ) 3 1 If the x, y, z axes are chosen coincident with the principal axes of inertia of the body,* the angular momentum of the body is defined as H i jk G xx yy zz =++ III c ωωω ˆ ˆ ˆ (. ) 3 1 where Ix, Iy, and Iz are the principal centroidal mass moments of inertia (second moments of mass) about the principal axes. This vector equation can be substituted into equation 3.1a to yield the three scalar equations known as Euler's equations: MI II M I II d MI II x xx y z yz y yy z x zx z zz x y xy ∑ ∑ ∑ = −− ( ) = −− ( ) = −− ( ) α ωω α ωω α ωω (. ) 3 1 where Mx, My, Mz are moments about those axes and αx, αy, αz are the angular accelerations about the axes. This assumes that the inertia terms remain constant with time, i.e., the mass distribution about the axes is constant. NEWTON'S THIRD LAW states that when two particles interact, a pair of equal and opposite reaction forces will exist at their contact point. This force pair will have the same magnitude and act along the same direction line, but have opposite sense. We will need to apply this relationship as well as applying the second law in order to solve for the forces on assemblies of elements that act upon one another. The six equations in equations 3.1b and 3.1d can be written for each rigid body in a 3-D system. In addition, as many (third-law) reaction force equations as are necessary will be written and the resulting set of equations solved simultaneously for the forces and moments. The number of second-law equations will be up to six times the number of individual parts in a three-dimensional system (plus the reaction equations), meaning that even simple systems result in large sets of simultaneous equations. A computer is needed to solve these equations, though high-end pocket calculators will solve large sets of simultaneous equations also. The reaction (third-law) equations are often substituted into the second-law equations to reduce the total number of equations to be solved simultaneously. Two-Dimensional Analysis All real machines exist in three dimensions but many three-dimensional systems can be analyzed two dimensionally if their motions exist only in one plane or in parallel planes.

oad, the quantity a represents the particular value of x at which the load acts (see Figure 3-22b). The definition of this singularity function, called the unit impulse or Dirac delta function, is given in equation 3.17d. Note that all singularity functions involve a conditional constraint. The unit impulse evaluates to ∞ if x = a and is 0 at any other value of x. The unit step function or Heaviside step function (Eq. 3.17c) evaluates to 0 for all values of x less than a and to 1 for all other x. Since these functions are defined to evaluate to unity, multiplying them by a coefficient creates any magnitude desired. Their application is shown in the following three examples and is explained in the most detail in Example 3-2B. If a loading function both starts and stops within the range of x desired, it needs two singularity functions to describe it. The first defines the value of a1 at which the function begins to act and has a positive or negative coefficient as appropriate to its direction. The second defines the value a2 at which the function ceases to act and has a coefficient of the same magnitude but opposite sign as the first. These two functions will cancel beyond a2, making the load zero. Such a case is shown in Example 4-6 in the next chapter. Quadratically distributed loads can be represented by a unit parabolic function, xa a − 2 (. ) 3 17 which is defined as 0 when x ≤ a, and equal to (x - a)2 when x > a. Linearly distributed loads can be represented by a unit ramp function,

the reaction forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the pivot pin and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and directions of both these pin forces are unknown. We can write equations 3.3b for this element to sum forces in the x and y directions and sum moments about the CG (with cross products expanded). FF F FF F F b M FR R F R F R F R F xxx yyy h z h h xy yx xy yx ∑ ∑ ∑ =+= = + += =+ − ( ) + − ( ) = 12 32 12 32 12 12 12 12 32 32 32 32 0 0 0 ( ) 4 Link 3 has two forces on it, F23 and F43. Write equations 3.3b for this element: FF F FF F c M RF RF RF RF xxx yyy z xy yx xy yx ∑ ∑ ∑ =+= =+= = − ( ) + − ( ) = 23 43 23 43 23 23 23 23 43 43 43 43 0 0 0 ( ) 5 Link 4 has three forces acting on it: Fc4 is the known (desired) force at the crimp, and F14 and F34 are the reaction forces from links 1 and 3, respectively. The magnitudes and directions of both these pin forces are unknown. Write equations 3.3b for this element: FF F F FF F F d M RF RF RF RF RF RF x x x cx y y y cy z xy yx xy yx cxcy cycx ∑ ∑ ∑ =++= =++= = − ( ) + − ( ) + − ( ) = 14 34 4 14 34 4 14 14 14 14 34 34 34 34 44 44 0 0 0 ( ) 6 The 9 equations in sets b through d have 13 unknowns: F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y, F14x, F14y, F34x, F34y, and Fh. We can write the third-law relationships between action-reaction pairs at each of the joints to obtain the four additional equations needed: FF FF FF FF e 32 23 34 43 32 23 34 43 xx xx yy yy =− =− =− =− ; ;

tors in Section 1.7 mentioned that many industries (automotive, aircraft, etc.) engage in extensive test programs to develop realistic loading models of their equipment. This topic will be discussed further in Section 6.4 when fatigue loading is introduced. Modern finite element (FEA) and boundary element (BEA) analysis techniques also allow vibration effects on a system or structure to be modeled and calculated. It is still difficult to obtain a computer model of a complex system that is as accurate as a real, instrumented prototype. This is especially true when clearances (gaps) between moving parts allow impacts to occur in the joints when loads reverse. Impacts create nonlinearities which are very difficult to model mathematically. Natural Frequency When designing machinery, it is desirable to determine the natural frequencies of the assembly or subassemblies in order to predict and avoid resonance problems in operation. Any real system can have an infinite number of natural frequencies at which it will readily vibrate. The number of natural frequencies that are necessary or desirable to calculate will vary with the situation. The most complete approach to the task is to use Finite Element Analysis (FEA) to break the assembly into a large number of discrete elements. See Chapter 8 for more information on FEA. The stresses, deflections, and number of natural frequencies that can be calculated by this technique are mainly limited by time and the computer resources available. If not using FEA, we would like to determine, at a minimum, the system's lowest, or fundamental, natural frequency, since this frequency will usually create the largest magnitude of vibrations. The undamped fundamental natural frequency ωn, with units of rad/sec, or fn, with units of Hz, can be computed from the expressions ω π ω n n n

where m is its mass and vi its velocity at impact. We need to modify the kinetic energy term by a correction factor η to account for the energy dissipation associated with the particular type of elastic member being struck. If the dissipation is negligible, η will be 1. Assuming that all the kinetic energy transferred from the moving mass is converted to elastic energy stored in the struck member allows us to equate equations 3.9 and 3.10: F k mv F v mk i i i i 2 2 2 2 = η η (3.11) = If the mass were allowed to statically load the struck member, the resulting static deflection would be δst = W / k where W = mg. Substituting these into equation 3.11 gives a ratio of dynamic force to static force or dynamic deflection to static deflection: F W v g i i st i st = = δ δ η δ (3.12) The term on the right side of this equation is called the impact factor which provides a ratio of impact to static force or deflection. Thus if the static deflection can be calculated for the application of a force equal to the weight of the mass, an estimate of the dynamic force and dynamic deflection can be obtained. Note that equations 3.11 and 3.12 are valid for any case of horizontal impact, whether the object is loaded axially as shown here, in bending, or in torsion. Methods for calculating the deflections of various cases are addressed in the next chapter. The spring rate k for any object can be found by rearranging its deflection equation according to equation 3.5. VERTICAL IMPACT For the case of a mass falling through a distance h onto a rod as shown in Figure 3-18b, equation 3.11 also applies with the impact velocity vi 2 = 2gh. The potential energy for a drop through distance h is:

where m is the mass of the striking object, and mb is the mass of the struck object. As the ratio of the striking mass to the struck mass increases, the correction factor η asymptotically approaches one. Figure 3-19 shows η as a function of mass ratio for three cases: an axial rod (equation 3.15) plotted in color, a simply supported beam struck at midspan (black-solid), and a cantilever beam struck at the free end (black-dotted).[6] The correction factor η is always less than one (and > 0.9 for mass ratios > 5) , so assuming it to be one is conservative. However, be aware that this energy method gives approximate and nonconservative results in general, and it needs to be used with larger-than-usual safety factors applied.

where ωn is the fundamental natural frequency, m is the moving mass of the system in true mass units (e.g., kg, g, blob, or slug, not lbm), and k is the effective spring constant of the system. (The period of the natural frequency is its reciprocal in seconds, Tn = 1/fn.) Equation 3.4 is based on a single-degree-of-freedom, lumped model of the system. Figure 3-15 shows such a model of a simple cam-follower system consisting of a cam, a sliding follower, and a return spring. The simplest lumped model consists of a mass connected to ground through a single spring and a single damper. All the moving mass in the system (follower, spring) is contained in m and all the "spring" including the physical spring and the springiness of all other parts is lumped in the effective spring constant k. SPRING CONSTANT A spring constant k is an assumed linear relationship between the force, F, applied to an element and its resulting deflection δ (see Figure 3-17): k F = a δ (. ) 3 5 If an expression for the deflection of an element can be found or derived, it will provide this spring-constant relationship. This topic is revisited in the next chapter. In the example of Figure 3-15, the spring deflection δ is equal to the displacement y of the mass. k F y = (. ) 3 5b DAMPING All the damping, or frictional, losses are lumped in the damping coefficient d. For this simple model, damping is assumed to be inversely proportional to the velocity ydot of the mass. d F y

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