[Psychology] Mechanical Assemblies Phần 3 doc

96 4 CONSTRAINT IN ASSEMBLY TABLE 4-11. Constraint Analysis of Pin-Hole and Pin-Slot Joint FIGURE 4-32. Illustrating the Equivalence of Three Dif- ferent Positions for the Z Force in WRU in Table 4-11. The arrows for FZ-\ and MX represent the original entries in the first row of WR and are equivalent to Fz2 without MX and Fza also without MX- measured about the global Y axis. Note that there is no cor- responding problem regarding pin axis parallelism about the global X axis because the right pin can rock in the slot about its x axis. 4.F.4.b. Remarks A constraint analysis of a complex assembly may result in numerous reported overconstraints. Several reasons are possible: • The engineer made an error. • The engineer intended that those features be over- constrained. • The overconstraints are there in a mathematical sense but not in a practical sense. Let us consider these cases one at a time. The engineer made an error. In this case, the error may be immediately obvious to the engineer, who can correct it. In a complex assembly, however, this may not be easy, especially in the presence of mathematical overconstraints. The engineer intended that those features be overcon- strained. The engineer does not need to take any action in this case. The overconstraints are mathematical but not practi- cal. This is the most interesting situation. It arises in par- ticular in the case of two-sided constraints. These, in turn, can arise in several ways: 1. The engineer can create a new two-sided feature by intersecting elementary surface contacts. For exam- ple, as we saw in Section 4.F.2.C, a pin-slot feature can be created by intersecting two plane surfaces with a cylinder. The result will contain a two-sided constraint, which will generate an overconstraint report in the analysis. Possibly, the report will be cluttered with overconstraint reports from such fea- tures. We avoided this clutter in the examples above by creating a toolkit feature comprising the allowed motions of a pin-slot. The way we described the allowed motions in the twist matrix suppressed the overconstraint that we know is really inside it. This permitted us to focus on achieving desired con- straints and detecting errors. The engineer may relieve an overconstraint within a feature with two-sided constraint by pro- viding a small amount of clearance in the final de- sign if the resulting location uncertainty, backlash, vibration, or other consequences are tolerable. If the consequences are intolerable, the engineer may pro- vide for a small amount of interference, as long as the resulting compressive stress is tolerable. In the first case, any resulting location uncertainty must be included in the tolerance analysis, while in the other case the resulting stress must be investigated to en- sure that it does not cause damage, cracks, fatigue, and so on. 2. The engineer can combine two elementary features that collectively create two-sided or other means of overconstraint. This occurs, for example, in the pin- hole plus pin-slot. The engineer may relieve such overconstraints by providing a little clearance. All the cautions listed above for single features apply here. »TU = [Tl;T2] TU = 0012-2 0 0016-2 0 0000 1 0 1000 0-6 »WR = recip(TU) WR = 001600 000010 4.F. DESIGN AND ANALYSIS OF ASSEMBLY FEATURES USING SCREW THEORY 97 Alternatively, he can construct a complex new feature containing the geometries of several ele- mentary features, but defined so that interfeature overconstraints among them are suppressed by def- inition. This approach should be avoided for the following reason: It is relatively easy to control the dimensions and variations of a single feature, which usually involves creating surfaces that are near each other relative to the size of the feature. Providing the necessary small clearance to avoid overconstraint without compromising accuracy is also relatively easy. However, controlling interfeature dimensions and variations when these features are far from each other relative to the size of each individual feature is much more difficult and prone to errors that can cause overconstraint. It is better to confront this pos- sibility by using simple individual features rather than defining a new complex feature that suppresses the overconstraint and pretending it will not happen. Defining the overconstraint away will simply keep the engineer from finding constraint errors or non- robust aspects of the design. It is also up to the engineer to decide which toolkit fea- tures best represent the problem at hand, or to design an appropriate feature instead. The overconstraints discussed in the previous subsection will not arise if toolkit features 9, 18, and 19 are used instead of features 2 and 4. One of the thought questions at the end of the chapter asks you to investigate this alternate formulation. Constraint analysis is useful for finding constraint mis- takes. Choosing features like 18 or 19 that optimistically assume away some overconstraint opportunities may re- sult in an optimistic constraint analysis that fails to identify a mistake. A possible design technique is to use the most internally constrained toolkit features like 2 and 4 first, examine the resulting report, and separate the intended constraints from the extraneous ones and the mistakes. Next, eliminate the mistakes. Finally, replace the inter- nally constrained features with similar ones that have a little internal clearance or use some of the less internally constrained features like 18 and 19 and judge whether the intended final constraint arrangement has been achieved. 4.F.5. Graphical Technique for Conducting Twist Matrix Analyses A simple graphical technique can be used to help keep track of which twist matrices should be collected into unions and which should be intersected ([Shukla and Whitney]). The technique is presented here in a series of increasingly complex examples. 4.F.5.a. A Single Feature with a Single Twist Matrix Consider the single feature illustrated in Figure 4-33. To set up the graphical technique, we make a graph that represents parts and the features that join them. A simple graph of this type is shown in Figure 4-34. Then we trace a path or paths in the graph from the moving part to the fixed part, passing through the necessary fea- tures and other parts on the way. In this case, part A is the moving part while part B is the fixed part. The diagram is shown in Figure 4-35. The procedure is: • Identify every path from the moving part to the fixed part. • For each path, construct the twist matrix for the mov- ing part for each feature on the path, using the same reference coordinate frame (such as one attached to the fixed part), and form the union of all these twist matrices. FIGURE 4-33. Single Feature to Illustrate Graphical Technique. FIGURE 4-34. Definition of Terms for Graph Representation of an Assembly. FIGURE 4-35. Diagram for Analyzing the Feature Situation in Figure 4-33. This case is trivial because there is only one path from the moving part (A) to the fixed part (B). 98 4 CONSTRAINT IN ASSEMBLY Form the intersection of all the twist unions using the procedure in 4.E.2.d. A nonempty TR represents underconstraint in the assembly. In this case, there is only one path and on this path there is only one feature, so the procedure is trivial. 4.R5.b. Two Parts Joined by Two Features Next, consider the feature in Figure 4-31. The analysis di- agram is shown in Figure 4-36. In this case, there are two paths from the moving part (A) to the fixed part (B). On each path we find one feature. Fl is chosen arbitrarily to be the pin-hole while F2 is chosen to be the pin-slot. The motion analysis matrix 77? is obtained by intersecting the twist matrices corresponding to Fl and F2, as illustrated in Table 4-10. 4.F.5.C. An Assembly with a Moving Part Finally, consider the 4-bar linkage shown, together with its analysis diagram, in Figure 4-37. The problem is to determine the degrees of freedom of link L3, considered to be the moving part, with respect to LI, considered to be the fixed part. FIGURE 4-36. Diagram for Two Parts Joined by Two Fea- tures. In this case there are two paths. FIGURE 4-37. Four-Bar Linkage and Its Analysis Dia- gram. The problem is to determine the degrees of freedom of link L3 with respect to link L1. We have two paths with two features on each path. The left path connects L3 to L1 via R2, L2, and R1. The right path connects L3 to L1 via R3, l_4, and R4. R1 through R4 are rotary joints, each consisting of a pin-hole feature. Links L1 through L4 are of equal length and all lie nominally in the X-Y plane. The Z axis points out of the paper. The motion analysis goes as follows: For the left path, we find that there are two features, R2, and Rl, between L3 and LI. We need to find how those features generate motion for L3. First, we erase the right path and all its features. Then we form a twist allowing R2 to move L3 while Rl is frozen, and then we form a twist that allows Rl to move L3 while R2 is frozen. Each of these twists is calculated using the same fixed reference associated with LI, say one centered on Rl. We then form the union of these two twists to get a representation of the left path. For the right path, the process is similar, except that we erase the left path and consider R3 and R4, and we again use a coordinate reference centered on Rl. Finally, we intersect the left path union and the right path union to find the net motion allowed to L3. The whole process is shown in Table 4-12. We have shown that this simple technique permits anal- ysis of single joints made of several features as well as analysis of several parts connected by several joints. If these joints are made of several features, then the user should analyze each joint separately, finding the net twist allowed by all its constituent features by intersecting their individual twists, and then combine the joints using the method shown here. This method can be used on any assembly or linkage as long as it does not contain cross-coupling. We saw in Fig- ure 4-6 a mechanism that has cross-coupling. The method above will not be able to find the motions of the top hori- zontal link if the bottom horizontal link is fixed. However, the motion of the top link can be found if the left or right vertical link is considered fixed, and the answer can be rewritten to conform to the situation where the bottom link is fixed. 4.F.6. Graphical Technique for Conducting Constraint Analyses 22 Systematic constraint analysis begins the same way that motion analysis does, by drawing the graph and enumerating the paths. However, constraint analysis is considerably more tedious because the intersection method has to be applied to all combinations of features 2 See [Shukla and Whitney 2001b]. 4.F. DESIGN AND ANALYSIS OF ASSEMBLY FEATURES USING SCREW THEORY 99 TABLE 4-12. Motion Analysis of Four-Bar Linkage Motion analysis: The first step is to analyze the left path to find L3's motions as if it is connected to the fixed link only by L2. This is done by considering the motion that each remaining feature could give L3 while considering other remaining motions frozen: Rotate L3 about R2 in Rl coordinates with Rl frozen: »tl = [001100] tl = 001100 Rotate L3 about Rl in Rl coordinates with R2 frozen: »t2 =[001000] t2 = 001000 Form the union of these to get the possible motions of L3 provided by the left path: »tlp = [tl;t2] tip = 001100 001000 Now analyze the right path to find the motions of L3 as if it is connected to the fixed link only by L4. Again, this is done by considering the motion that each remaining feature could give L3 while considering the other remaining motions frozen: Rotate L3 about R3: »t3 =[0011-10] t3 = 0011-10 Rotate L3 about R4: »t4 =[0010-10] t4 = 0010-10 Form the union of these to get the motions of L3 provided by the right path: »trp = [t3;t4] trp = 0011-10 0010-10 Form the intersection of tip and trp to see what the net allowed motion is »wlp = recip(tlp) wlp = 0 1.0000 0000 0 0 1.0000 00 0 000 1.0000 0 0 0000 1.0000 -0.0000 »wrp = recip(trp) wrp= 0 1.0000 000 1.0000 0 0 1.0000 00 0 000 1.0000 0 0 0000 1.0000 0.0000 (continued) 100 4 CONSTRAINT IN ASSEMBLY TABLE 4-12. (Continued) »WU = [wlp;wrp] WU = 0 1.0000 0000 0 0 1.0000 000 000 1.0000 0 0 000 0 1.0000 -0.0000 0 1.0000 000 1.0000 0 0 1.0000 000 000 1.0000 0 0 000 0 1.0000 0.0000 »TR = recip(WU) TR = -0.0000 -0.0000 -0.0000 1.0000 0.0000 0.0000 This says that L3 is permitted to move in the X direction in Rl coordinates. This is the answer we expect based on intuition. The reciprocal of TR, WU, shows what forces and moments can be resisted by the linkage. These are F y , F z , M x , M y , and M z . But we do not know if any of these is overconstrained. This question is resolved in Table 4-13. if we want to identify every feature that contributes to overconstraint. In brief, the process is as follows: • Choose any path and check to see if its twists, formed into a union, overconstrain the parts. • Choose a second path and intersect its twist union with the first path's twist union to find a combined feature that allows only those motions common to those paths. Check to see if this combination over- constrains the parts. • Continue in this way, adding one path's twist union at a time until all have finally been combined and checked for overconstraint. To see how this works, consider a part A joined to an- other part B by four features Fl, F2, F3, and F4, as shown in Figure 4-38. On each path there is one twist, so we have four twists, one for each path: T\, T^, ?3, and T 4 . By using the rela- tionship Wi = recip(Tf), we first find the corresponding FIGURE 4-38. Path Diagram for Two Parts Joined by Four Features. wrenches: W\, W 2 , W 3 , and W 4 . We then systematically form (the order is arbitrary) T\2,T\23, and ^234, and Wi2, Wi23, and W\234, as follows: Tiz = 0(7,, r 2 ) = recip(U(W } , W 2 )) = net motions allowed by path 1 and path 2 ^123 = n(7i, T 2 , 7s) = ncip(\J(Wi, W 2 , Wj)) = net motions allowed by path 1, path 2, = and path 3 r, 23 4 = n(r,, r 2 , r 3 , r 4 ) = rmp(u(w,, w 2 , w 3 , w 4 )) = net motions allowed by all four paths w }2 = n(Wi, w 2 ) = mcip(u(Ti, r 2 )) = overconstraint provided by path 1 and path 2 W m = n(W 12 , W 3 ) = recip(U(T l2 , r 3 )) = overconstraint provided by path 1, path 2, and path 3 W, 234 = n(W 123 , W 4 ) = recip((J(T m , T 4 )) = overconstraint provided by all four paths Note that W 12 34 / H(Wi, W 2 , W 3 , W 4 ) = mcip(U(Ti, T 2 , T 3 , T 4 )) = the force(s) or moment(s) that can be supported by all four paths at once 4.F. DESIGN AND ANALYSIS OF ASSEMBLY FEATURES USING SCREW THEORY 101 TABLE 4-13. Constraint Analysis of Four-Bar Linkage Constraint analysis: Intersect the wrenches of the two paths: »tlrp = [tlp;trp] tlrp = 0011 00 0010 00 0011-10 0010-10 »WR = recip(tlrp) WR = 0 0 1.0000 000 00 0 1.0000 0 0 00 0 0 1.0000 -0.0000 This says that L3 is overconstrained via force in the Z direction and moments about X and Y. This is due to the presence of two-sided constraints in the pin joints plus the fact that these joints were defined as able to support Z force. In Table 4-12 we found that the linkage could support F y , F,, Mjt , My, and M z . We now know that some of these are overconstrained. This is important because, in case there is no force or moment that is provided by all four features, ^1234 will be empty, but we cannot conclude on this basis that there is no overconstraint. One possible reason is that features Fl, F2, and F3 share the ability to constraint a particular force or moment that F4 cannot constrain. This possible overconstraint is detected by W\ 2 3- For the four-bar linkage, the analysis is shown in Table 4-13. This process can be looked at two ways. First, it can be used as an existence check for overconstraint. As soon as overconstraint is found, the procedure can be stopped. Second, it can be used to identify overconstrained direc- tions and the features that create them. In this case, the procedure must continue until all feature sets have been combined into the growing intersected set. Note that choosing the paths in a different sequence will always result in the same number of degrees of free- dom, if any, being detected as overconstrained, but the WR matrix reporting the overconstraint may appear dif- ferent. The reason for this is that, as features are added to the combination, one such set may properly constrain the parts. Any feature added thereafter will necessarily add overconstraint along the direction(s) it is capable of constraining, and these directions will appear in WR. A different sequence of analysis will eventually arrive at proper constraint with a different subset of the features, and the next one added will be different this time than last time. WR will then report this feature's directions as the overconstrained ones rather than another fea- ture's directions. The engineer can use this information to explore the consequences of establishing joints be- tween parts in different sequences, including deciding which features, if any, to redesign in order to remove the overconstraint. Problem 17 explores this procedure using a simple example. The reader is encouraged to use the proce- dure even if the examples look simple and the answer is easy to predict by intuition. The method will be wel- come indeed when a real industrial strength problem is encountered. We may look ahead at this point to Chapter 7 on as- sembly sequences to see that this procedure will apply to sequences of parts as well as sequences of features within two parts. Choosing which sequence to use will depend on, among other things, which one does a better job of delivering the KC. 4.F.7. Why Are the Motion and Constraint Analyses Different? There appears to be an asymmetry between the motion analysis in Section 4.F.5 and the constraint analysis in Section 4.F.6. Motion analysis requires only intersecting all the twists at once but constraint analysis requires care- ful accumulation of wrench intersections. The reason is as follows. If we intersect n twists and find that the intersection is empty, we know that the parts cannot move. We can intersect any subset of these n twists and may indeed find underconstraint, but we do not care because the full set of twists prevents any motion. On the other hand, if we intersect n wrenches and find an empty intersection, we cannot conclude that there is no overconstraint. We must intersect a series of subsets of these wrenches because one or more of them could cause overconstraint and we want to know if that is the case. Adding more wrenches to the test set will not remove this lurking overconstraint but will just cover it up because the additional wrenches do not share constraining directions with the subset that contains overconstraint. 102 4 CONSTRAINT IN ASSEMBLY 4.G. ADVANCED CONSTRAINT ANALYSIS TECHNIQUE On the CD-ROM packaged with this book is an appendix to this chapter, written by J. Michael Gray, explaining a more general method of determining the state of mobility and constraint of an assembly. It is not as simple to apply as the one explained here, but it suffers none of the limita- tions. It is based on work in [Davies 1981, 1983a, 1983b, and 1983c]. 4.H. COMMENT The reader may have detected that we have dealt exten- sively with assembly concepts in the last two chapters without talking much about parts in the usual sense. Most books on engineering design deal with parts such as shafts, gears, and bearings. The detailed shape of these parts is important to such studies. We have said virtually noth- ing about the shape of parts, and deliberately so, for the following reasons. First, the concepts we need, such as location and orien- tation in space, and degree of constraint, can be described with mathematical precision and very few symbols (and correspondingly few bytes of memory) using a few num- bers in a 4 x 4 matrix or a twist matrix. To capture the equivalent information, such as the location and orienta- tion of a shaft axis, or the fact that one part can rotate with respect to another, using purely geometric data, would en- tail thousands or millions of bytes and could possibly be less precise. 23 A major point of the last two chapters is that the main information we need to define a kinematic assem- bly is not geometric. It amounts to coordinate frames and twist matrices. We can add the geometry later. Second, we are dealing with only the geometric rela- tionships between parts, not any forces, loads, or deforma- tions that they might experience. We addressed force and deformation only when we showed why pure kinematic constraints consisting of sharp point contacts are not used in practice. A complete engineering design of an assem- bly must include forces and deformations. Such factors will provide the engineer with most of the information to decide what shape the parts must have. The size and other details of the assembly features will also be influenced by such factors. Nevertheless, the scheme by which the parts will be located in space prior to experiencing loads must be designed with care using the methods described in this book. Whether the loads are considered first and used to influence the locating scheme, or the locating scheme is decided first and then the parts are sized to suit the loads, depends on the engineer's style of working, the materials used, and the degree to which the structure is stressed as a percentage of the yield stresses of its materials. 4.1. CHAPTER SUMMARY This chapter is one of the most important in this book. It presents a way to design competent assemblies us- ing the principles of kinematic constraint. We distinguish between kinematically constrained assemblies, deliber- ately overconstrained or underconstrained assemblies, and 23 A student once asked the author, "Dr. Whitney, what do you do about the facets?" "What facets?" I asked. "I built a pin-and-hole model in my CAD system, and I sometimes find that my motion algorithm says that the pin cannot turn in the hole because a vertex on the pin interferes with a facet on the hole." Real round pins and holes do not have facets and vertices, of course. Only approximate geometric models of them in CAD systems do. Faceted models are used for approximate interference analysis and to create screen dis- plays. They are appropriate for modeling assembly drawings but not for modeling assemblies. assemblies that contain constraint errors. Kinematic as- semblies are capable of achieving rapid, accurate, and re- peatable assembly at reasonable cost. Both Whitehead and Kamm make this point in their books. The car seat exam- ple shows this vividly. The method of Screw Theory permits us to define as- sembly features as geometric entities capable of establish- ing constraint relations between the parts they join. Screw Theory also permits us to build up a joint between parts using arbitrary combinations of simpler features and then to examine the state of constraint that is established by that joint. These concepts and tools permit us to use features and the connective assembly models defined in Chapter 3 to build kinematically constrained assemblies of rigid parts 4.J. PROBLEMS AND THOUGHT QUESTIONS 103 that lie at particular desired places and orientations in space so that they will achieve key characteristics as de- fined in Chapter 2. These concepts are mathematically pre- cise and consistent, and they capture the most fundamental properties of assemblies. They can be used as the basis for computer models of assemblies, for motion and constraint analyses, and for other analyses, such as variation, that are discussed in later chapters. 4.J. PROBLEMS AND THOUGHT QUESTIONS Write down the twist matrices for the plate resting on each of the other two hemispheres, then combine the three twist matri- ces according to the twist matrix intersection algorithm in Section 4.E.2.d.2. You should get (give or take some minus signs that are not significant) ' "-plane — This says, row by row, that the plate can slide in the X direc- tion, it can slide in the Y direction, and it can rotate about Z. This is consistent with the properties of a plane. FIGURE 4-39. Figure for Problem 1. FIGURE 4-40. Figure for Problem 2. 3. Use toolkit features 9, 18, and 19 to analyze the situation shown in Figure 4-41. Follow the methodology in Section 4.F.4.a. You should be able to show that the upper plate cannot move and it is not overconstrained. FIGURE 4-41. Figure for Problem 3. 2. Figure 4-40 shows an arrangement in which part 1 has three hemispherical features under part 2 and two such features at its right. Prove that this configuration leaves part 2 with exactly one unconstrained degree of freedom relative to part 1. Confirm that the result makes sense in terms of the coordinates shown in Figure 4-40. 4. Consider the part pair in Figure 4-42, consisting of plate 1 with two pins, mating to plate 2 with one hole and one slot. Analyze the state of motion and the state of constraint using the methods in Section 4.F.4. Compare your answers with those in Table 4-10 and Table 4-11 and explain every similarity or difference between the matrices row by row. 1. Prove that three points define a plane using three hemispher- ical features touching a plate, as shown in Figure 4-39. Hint: The twist matrix for the plate resting on hemisphere number 1, referred to the lower left corner of the plate, is 1000 0 -1 01000 1 0011-1 0 00010 0 00001 0 THS, = "000100" 000010 _0 0 1 0 0 0. 104 4 CONSTRAINT IN ASSEMBLY FIGURE 4-44. Figure for Problem 9. FIGURE 4-45. Figure for Problem 10. features will be needed between A and C and B and C, respec- tively. These he has sketched in as irregular polygons for the time being just as reminders. But he has not yet chosen their final shape. Assume that he wants A and B to firmly locate C, and therefore that A and B must be joined first. Is the hole he chose for the A-B mate sufficient, and what are his alternatives for the remaining features? What should he take into consideration when making these choices? 11. Consider the three-bar linkage shown in Figure 4-46. Link A is fixed, while links B and C have pin joints with link A and with each other. Use the twist matrix intersection algorithm to prove that this linkage is rigid and cannot move. 12. Consider the five-bar linkage shown below. Show that the diagram in Figure 4-47 is correct and use it to set up the necessary twist matrices for determining the net motion of L3. 13. Consider the five-bar linkage shown in Figure 4-48. Assume link L2 to be fixed and find the state of motion and constraint of link LI. Repeat this analysis for link L5, again assuming L2 is fixed. Note that the path method as outlined in this chapter cannot FIGURE 4-46. Figure for Problem 11. FIGURE 4-43. Figure for Problem 6. 7. Consider the car seat example in Section 4.C.5.b. Reproduce the joint used in the original design by using four instances of toolkit feature 2. Form matrices 77? and WR for this case and explain each resulting matrix row by row. 8. Returning to the car seat example, reproduce the revised de- sign by using the appropriate toolkit features. Explain the resulting matrices row by row and compare them to the results in Problem 4. Discuss any overconstraints that remain. 9. Analyze the state of motion and constraint for the two situa- tions shown in Figure 4-44. In each case, the part to be analyzed contains two slots, through each of which there is a pin. Explain the resulting matrices row by row. 10. An engineer is considering how to join the three parts A, B, and C shown in Figure 4-45. He has decided that he needs a peg and hole to join A and B, and he knows that some kind of FIGURE 4-42. Figure for Problem 4. 5. Form a joint between two plates using two pin-hole joints (toolkit feature 2). Analyze the state of motion and state of con- straint for this joint by forming TR and WR. Explain all the re- sulting matrices row by row. 6. Figure 4-43 represents, in two dimensions, a common way of supporting the deck of a bridge. Use motion and constraint analysis to show that this arrange- ment has one degree of freedom. Now assume that the bridge deck expands due to rising temperature. Show that it is able to do this without encountering overconstraint. 4.J. PROBLEMS AND THOUGHT QUESTIONS 105 FIGURE 4-47. Figure for Problem 12. FIGURE 4-48. Figure for Problem 13. analyze the state of constraint of link L5 if link L3 is considered fixed. What conclusions can you draw concerning the state of constraint of link L5? Does it matter which link is assumed fixed? 14. Consider the aircraft structure example in Section 4.C.5.C and explain why there should not be locating holes for the longi- tudinal location of the ribs in both the upper and lower spars. 15. Consider the copier example in Section 4.C.5.a. Assume that the two side panels form a rigid unit. Model the joints between the curved panel and the two side panels using toolkit features. Form matrices TR and WR for the curved panel joined to the side panel unit and explain the resulting matrices row by row. 16. Suggest a redesign for the joints in problem 15. Form ma- trices TR and WR and prove that your design is an improvement. 17. Consider the example given in Figure 4-49, in which two plates are joined by four toolkit hemisphere-slot features. First, decide intuitively whether the plates are underconstrained, fully constrained, or overconstrained. Then find the wrench intersection WR considering all four features at once. Explain the answer, row by row. Then find the intersection of features 1 and 2 and deter- mine how or if they constrain the parts. Then intersect feature 3 with the previously created intersection of features 1 and 2 and determine how or if they constrain the parts. Last, intersect feature 4 with the previously calculated intersection involving features 1, 2, and 3. This last step should reveal the true state of constraint of these parts. Repeat this process using the features in the sequence 2, 3, 4, 1. Explain any differences you observe. 18. Figure 4-50 shows two plates joined by a pin-hole feature. Write the twist matrix for this feature. Now consider the two situations in Figure 4-51. In each case, a plate is joined to another via a slot in one plate and a tiny pin on the other. Assuming that the pin always stays in contact with the same side of the slot as shown in the figure, prove that a combination of these two situations has the same twist matrix as the pin-hole feature. Does it matter how big the slots are? FIGURE 4-50. First Figure for Problem 18. FIGURE 4-51. Second Figure for Problem 18. FIGURE 4-49. Figure for Problem 17. [...]... TOLERANCING TABLE 5 -3 Percent of Normal Random Variables That Fall Within and Outside a Range Around the Mean fi, Measured in Units of One Standard Deviation a #0fff % within ±# of 2 "3 where u>\ = (R(ox)T U>2 = (RCOV)T o >3 = (Ra>,)T V] = r x a>\ V2 = r X ft>2 D3 = r x 0 )3 11—Pin in oversize hole Tu... R TABLE 5-4 Values for Coefficients A2, D3, and D4 for X-bar and R Chart Limits n A2 0 3 04 5 10 0.577 0 .30 8 0 0.2 23 2.114 1.777 Note: Values of these factors for other values of n may be found in [Swift] It should be noted that a sample size of 5 or 10 is not large enough to capture process variations that evolve slowly For example, a cutting tool may last 30 0 parts, and, as it wears, the parts drift . each remaining feature could give L3 while considering the other remaining motions frozen: Rotate L3 about R3: »t3 =[0011-10] t3 = 0011-10 Rotate L3 about R4: »t4 =[0010-10] t4 = 0010-10 Form . path 1 and path 2 W m = n(W 12 , W 3 ) = recip(U(T l2 , r 3 )) = overconstraint provided by path 1, path 2, and path 3 W, 234 = n(W 1 23 , W 4 ) = recip((J(T m , T 4 )) = . a>. C0 V) 0)2 l>2 3 " ;3 TQ = where u> = (R(o x ) T U>2 = (RCO V ) T o> ;3 = (Ra>,) T V] = r x a> V2 = r X ft>2 D 3 = r x 0 )3 co 0 0 i>i 0 v 2 where