Solution Manual for Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th Edition

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DESIGN OF MACHINERY - 5th EdSOLUTION MANUAL 2-1-1PROBLEM 2-1Statement:Find three (or other number as assigned) of the following common devices. Sketch carefulkinematic diagrams and find their total degrees of freedom.a.An automobile hood hinge mechanismb.An automobile hatchback lift mechanismc.An electric can openerd.A folding ironing boarde.A folding card tablef.A folding beach chairg.A baby swingh.A folding baby walkeri.A fancy corkscrew as shown in Figure P2-9j.A windshield wiper mechanismk.A dump-truck dump mechanisml.A trash truck dumpster mechanismm.A pickup tailgate mechanismn.An automobile jacko.A collapsible auto radio antennaSolution:See Mathcad file P0201.Equation 2.1c is used to calculate the mobility (DOF) of each of the models below.a.An automobile hood hinge mechanism.The hood (3) is linked to the body (1) through two rocker links (2 and 4).3BODY124HOODNumber of linksL4Number of full jointsJ14Number of half jointsJ20M3L1()2J1J2M1b.An automobile hatchback lift mechanism.The hatch (2) is pivoted on the body (1) and is linked to the body by the lift arm, which can be modeled as twolinks (3 and 4) connected through a translating slider joint.2BODY1143HATCHNumber of linksL4Number of full jointsJ14Number of half jointsJ20M3L1()2J1J2M1c.An electric can opener has 2DOF.d.A folding ironing board.The board (1) itself has one pivot (full) joint and one pin-in-slot sliding (half) joint. The two legs (2 and 3) hava common pivot. One leg connects to the pivot joint on the board and the other to the slider joint.

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DESIGN OF MACHINERY - 5th EdSOLUTION MANUAL 2-1-2Number of linksL3231Number of full jointsJ12Number of half jointsJ21M3L1()2J1J2M1e.A folding card table has 7DOF: One for each leg, 2 for location inxyspace, and one for angular orientation.f.A folding beach chair.The seat (3) and the arms (6) are ternary links. The seat is linked to the front leg(2), the back (5) and a couplinglink (4). The arms are linked to the front leg (2), the rear leg (1), and the back (5). Links 1, 2, 4, and 5 are binarlinks. The analysis below is appropriate when the chair is not fully opened. When fully opened, one or morelinks are prevented from moving by a stop. Subtract 1 DOF when forced against the stop.125643Number of linksL6Number of full jointsJ17Number of half jointsJ20M3L1()2J1J2M1g.A baby swing has 4DOF: One for the angular orientation of the swing with respect to the frame, and 3 for thelocation and orientation of the frame with respect to a 2-D frame.h.A folding baby walker has 4DOF: One for the degree to which it is unfolded, and 3 for the location andorientation of the walker with respect to a 2-D frame.i.A fancy corkscrew has 2 DOF: The screw can be rotated and the arms rotate to translate the screw.j.A windshield wiper mechanism has 1DOF: The position of the wiper blades is defined by a single input.k.A dump-truck dump mechanism has 1DOF: The angle of the dump body is determined by the length of thehydraulic cylinder that links it to the body of the truck.l.A trash truck dumpster mechanism has 2DOF: These are generally a rotation and a translation.m.A pickup tailgate mechanism has 1DOF:n.An automobile jack has 4DOF: One is the height of the jack and the other 3 are the position and orientation othe jack with respect to a 2-D frame.o.A collapsible auto radio antenna has as manyDOFas there are sections, less one.

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-2-1PROBLEM 2-2Statement:How many DOF do you have in your wrist and hand combined?Solution:See Mathcad file P0202.1.Holding the palm of the hand level and facing toward the floor, the hand can be rotated about an axisthrough the wrist that is parallel to the floor (and perpendicular to the forearm axis) and one perpendicularto the floor (2DOF). The wrist can rotate about the forearm axis (1DOF).2.Each finger (and thumb) can rotate up and down and side-to-side about the first joint. Additionally, eachfinger can rotate about each of the two remaining joints for a total of 4DOFfor each finger (and thumb).3.Adding allDOF, the total isWrist1Hand2Thumb4Fingers 4x416TOTAL23

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-3-1PROBLEM 2-3Statement:How manyDOFdo the following joints have?a.Your kneeb.Your anklec.Your shoulderd.Your hipe.Your knuckleSolution:See Mathcad file P0203.a.Your knee.1DOF: A rotation about an axis parallel to the ground.b.Your ankle.3DOF: Three rotations about mutually perpendicular axes.c.Your shoulder.3DOF: Three rotations about mutually perpendicular axes.d.Your hip.3DOF: Three rotations about mutually perpendicular axes.eYour knuckle.2DOF: Two rotations about mutually perpendicular axes.

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-4-1PROBLEM 2-4Statement:How many DOF do the following have in their normal environment?a. A submerged submarineb. An earth-orbit satellitec. A surface shipd. A motorcycle (road bike)e. A two-button mousef. A computer joy stick.Solution:See Mathcad file P0204.a.A submerged submarine.Using a coordinate frame attached to earth, or an inertial coordinate frame, a submarine has 6DOF: 3 linearcoordinates and 3 angles.b.An earth-orbit satellite.If the satellite was just a particle it would have 3DOF. But, since it probably needs to be oriented withrespect to the earth, sun, etc., it has 6DOF.c.A surface ship.There is no difference between a submerged submarine and a surface ship, both have 6DOF. One mightargue that, for an earth-centered frame, the depth of the ship with respect to mean sea level is constant,however that is not strictly true. A ship's position is generally given by two coordinates (longitude andlatitude). For a given position, a ship can also have pitch, yaw, and roll angles. Thus, for all practicalpurposes,a surface ship has 5 DOF.d.A motorcycle.At an intersection, the motorcycle's position is given by two coordinates. In addition, it will have someheading angle (turning a corner) and roll angle (if turning). Thus, there are 4DOF.e.A two-button mouse.A two-button mouse has4 DOF. It can move in the x and y directions and each button has1 DOF.f.A computer joy stick.The joy stick has 2DOF(x and y) and orientation, for a total of 3DOF.

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-5-1PROBLEM 2-5Statement:Are the joints in Problem 2-3 force closed or form closed?Solution:See Mathcad file P0205.They are force closed by ligaments that hold them together. None are geometrically closed.

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-6-1PROBLEM 2-6Statement:Describe the motion of the following items as pure rotation, pure translation, or complex planarmotion.a.A windmillb.A bicycle (in the vertical plane, not turning)c.A conventional "double-hung" windowd.The keys on a computer keyboarde.The hand of a clockf.A hockey puck on the iceg.A "casement" windowSolution:See Mathcad file P0206.a.A windmill.Pure rotation.b.A bicycle (in the vertical plane, not turning).Pure translation for the frame, complex planar motion for the wheels.c.A conventional "double-hung" window.Pure translation.d.The keys on a computer keyboard.Pure translation.e.The hand of a clock.Pure rotation.f.A hockey puck on the ice.Complex planar motion.g.A "casement" window.Pure rotation.

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-7-1PROBLEM 2-7Statement:Calculate the mobility of the linkages assigned from Figure P2-1 part 1 and part 2.Solution:See Figure P2-1 and Mathcad file P0207.1.Use equation 2.1c (Kutzbach's modification) to calculate the mobility.5162143a.Number of linksL6Number of full jointsJ17Number of half jointsJ21M3L1()2J1J2M0(a)2131b.Number of linksL3Number of full jointsJ12Number of half jointsJ21M3L1()2J1J2M1(b)32114c.Number of linksL4Number of full jointsJ14Number of half jointsJ20M3L1()2J1J2M1(c)12534617d.Number of linksL7Number of full jointsJ17Number of half jointsJ21M3L1()2J1J2M3(d)

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-7-272154361218891014213115561(e)(f)e.Number of linksL10f.Number of linksL6Number of full jointsJ113Number of full jointsJ16Number of half jointsJ20Number of half jointsJ22M3L1()2J1J2M3L1()2J1J2M1M12143115421761187g.Number of linksL8Number of full jointsJ19Number of half jointsJ22M3L1()2J1J2M121341(g)h.Number of linksL4Number of full jointsJ14Number of half jointsJ20M3L1()2J1J2M1(h)

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-8-1PROBLEM 2-8Statement:Identify the items in Figure P2-1 as mechanisms, structures, or preloaded structures.Solution:See Figure P2-1 and Mathcad file P0208.1.Use equation 2.1c (Kutzbach's modification) to calculate the mobility and the definitions in Section 2.5 of thetext to classify the linkages.5162143a.Number of linksL6Number of full jointsJ17Number of half jointsJ21M3L1()2J1J2M0Structure(a)2131b.Number of linksL3Number of full jointsJ12Number of half jointsJ21M3L1()2J1J2M1Mechanism(b)32114c.Number of linksL4Number of full jointsJ14Number of half jointsJ20M3L1()2J1J2M1Mechanism(c)12534617d.Number of linksL7Number of full jointsJ17Number of half jointsJ21M3L1()2J1J2M3Mechanism(d)

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-9-1PROBLEM 2-9Statement:Use linkage transformation on the linkage of Figure P2-1a to make it a 1-DOFmechanism.Solution:See Figure P2-1a and Mathcad file P0209.1.The mechanism in Figure P2-1a has mobility:5162143Number of linksL6Number of full jointsJ17Number of half jointsJ21M3L1()2J1J2M02.Use rule 2, which states: "Any full joint can be replaced by a half joint, but this will increase theDOFbyone." One way to do this is to replace one of the pin joints with a pin-in-slot joint such as that shown inFigure 2-3c. Choosing the joint between links 2 and 4, we now have mobility:5162143Number of linksL6Number of full jointsJ16Number of half jointsJ22M3L1()2J1J2M1

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-10-1PROBLEM 2-10Statement:Use linkage transformation on the linkage of Figure P2-1d to make it a 2-DOFmechanism.Solution:See Figure P2-1d and Mathcad file P0210.125346171.The mechanism in Figure P2-1d has mobility:Number of linksL7Number of full jointsJ17Number of half jointsJ21M3L1()2J1J2M32.Use rule 3, which states: "Removal of a link will reduce theDOFby one." One way to do this is to removelink 7 such that link 6 pivots on the fixed pin attached to the ground link (1). We now have mobility:1215346Number of linksL6Number of full jointsJ16Number of half jointsJ21M3L1()2J1J2M2

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-11-1PROBLEM 2-11Statement:Use number synthesis to find all the possible link combinations for 2-DOF, up to 9 links, tohexagonal order, using only revolute joints.Solution:See Mathcad file P0211.1.Use equations 2.4a and 2.6 withDOF= 2 and iterate the solution for valid combinations. Note that thenumber of links must be odd to have an evenDOF(see Eq. 2.4). The smallest possible 2-DOFmechanism is then 5 links since three will give a structure (the delta triplet, see Figure 2-7).LBTQPHL3MT2Q3P4HM2L5T2Q3P4H2.ForL50T2Q3P4H0T=Q=P=H=B53.ForL72T2Q3P4HH0P0Case 1:Q0T22Q3P4HT2BLTQPHB5Case 2:Q1T22Q3P4HT0BLTQPHB64.ForL94T2Q3P4HCase 1:H1T0Q0P0BLTQPHB8Case 2a:H04T2Q3P9BTQPCase 2b:P11T2QQ0T1BLTQPHB7Case 2c:P04T2Q9BTQCase 2c1:Q2T42QT0B9TQB7Case 2c2:Q1T42QT2B9TQB6Case 2c3:Q0T42QT4B9TQB5

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-12-1PROBLEM 2-12Statement:Find all of the valid isomers of the eightbar 1-DOFlink combinations in Table 2-2 (p. 38) havinga. Four binary and four ternary links.b. Five binaries, two ternaries, and one quaternary link.c. Six binaries and two quaternary links.d. Six binaries, one ternary, and one pentagonal link.Solution:See Mathcad file P0212.1.Table 2-3 lists 16 possible isomers for an eightbar chain. However, Table 2-2 shows that there are fivepossible link sets, four of which are listed above. Therefore, we expect that the 16 valid isomers aredistributed among the five link sets and that there will be fewer than 16 isomers among the four link sets listedabove.2.One method that is helpful in finding isomers is to represent the linkage in terms of molecules as defined inFranke's Condensed Notations for Structural Synthesis. A summary of the rules for obtaining Franke'smolecules follows:(1) The links of order greater than 2 are represented by circles.(2) A number is placed within each circle (the "valence" number) to describe the type (ternary, quaternary,etc.) of link.(3) The circles are connected using straight lines. The number of straight lines emanating from a circle mustbe equal to its valence number.(4) Numbers (0, 1, 2, etc.) are placed on the straight lines to correspond to the number of binary links used inconnecting the higher order links.(5) There is one-to-one correspondence between the molecule and the kinematic chain that it represents.a.Four binary and four ternary links.Draw 4 circles with valence numbers of 3 in each. Then find all unique combinations of straight lines thatcan be drawn that connect the circles such that there are exactly three lines emanating from each circle andthe total of the numbers written on the lines is exactly equal to 4. In this case, there are three valid isomersas depicted by Franke's molecules and kinematic chains below.682134571331001331

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DESIGN OF MACHINERY - 5th Ed.SOLUTION MANUAL 2-12-2612345872331010330612345873333000220The mechanism shown in Figure P2-5b is the same eightbar isomer as that depicted schematically above.b.Five binaries, two ternaries, and one quaternary link.Draw 2 circles with valence numbers of 3 in each and one with a valence number of 4. Then find all uniquecombinations of straight lines that can be drawn that connect the circles such that there are exactly threelines emanating from each circle with valence of three and four lines from the circle with valence of four;and the total of the numbers written on the lines is exactly equal to 5. In this case, there are five validisomers as depicted by Franke's molecules and kinematic chains below.6815742333420120

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