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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 Ed SOLUTION MANUAL 2-1-1 PROBLEM 2-1 Statement: Find three (or other number as assigned) of the following common devices. Sketch careful kinematic diagrams and find their total degrees of freedom. a. An automobile hood hinge mechanism b. An automobile hatchback lift mechanism c. An electric can opener d. A folding ironing board e. A folding card table f. A folding beach chair g. A baby swing h. A folding baby walker i. A fancy corkscrew as shown in Figure P2-9 j. A windshield wiper mechanism k. A dump-truck dump mechanism l. A trash truck dumpster mechanism m. A pickup tailgate mechanism n. An automobile jack o. A collapsible auto radio antenna Solution: 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). 3 BODY 1 2 4 HOOD Number of links L 4  Number of full joints J 1 4  Number of half joints J 2 0  M 3 L 1  ( )  2 J 1   J 2   M 1  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 two links (3 and 4) connected through a translating slider joint. 2 BODY 1 1 4 3 HATCH Number of links L 4  Number of full joints J 1 4  Number of half joints J 2 0  M 3 L 1  ( )  2 J 1   J 2   M 1  c. An electric can opener has 2 DOF . 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) hav a 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 Ed SOLUTION MANUAL 2-1-2 Number of links L 3  2 3 1 Number of full joints J 1 2  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 1  e. A folding card table has 7 DOF : One for each leg, 2 for location in xy space, 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 coupling link (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 binar links. The analysis below is appropriate when the chair is not fully opened. When fully opened, one or more links are prevented from moving by a stop. Subtract 1 DOF when forced against the stop. 1 2 5 6 4 3 Number of links L 6  Number of full joints J 1 7  Number of half joints J 2 0  M 3 L 1  ( )  2 J 1   J 2   M 1  g. A baby swing has 4 DOF : One for the angular orientation of the swing with respect to the frame, and 3 for the location and orientation of the frame with respect to a 2-D frame. h. A folding baby walker has 4 DOF : One for the degree to which it is unfolded, and 3 for the location and orientation 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 1 DOF : The position of the wiper blades is defined by a single input. k. A dump-truck dump mechanism has 1 DOF : The angle of the dump body is determined by the length of the hydraulic cylinder that links it to the body of the truck. l. A trash truck dumpster mechanism has 2 DOF : These are generally a rotation and a translation. m. A pickup tailgate mechanism has 1 DOF : n. An automobile jack has 4 DOF : One is the height of the jack and the other 3 are the position and orientation o the jack with respect to a 2-D frame. o. A collapsible auto radio antenna has as many DOF as there are sections, less one.

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-2-1 PROBLEM 2-2 Statement: 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 axis through the wrist that is parallel to the floor (and perpendicular to the forearm axis) and one perpendicular to the floor (2 DOF ). The wrist can rotate about the forearm axis (1 DOF ). 2. Each finger (and thumb) can rotate up and down and side-to-side about the first joint. Additionally, each finger can rotate about each of the two remaining joints for a total of 4 DOF for each finger (and thumb). 3. Adding all DOF , the total is Wrist 1 Hand 2 Thumb 4 Fingers 4x4 16 TOTAL 23

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

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-4-1 PROBLEM 2-4 Statement: How many DOF do the following have in their normal environment? a. A submerged submarine b. An earth-orbit satellite c. A surface ship d. A motorcycle (road bike) e. A two-button mouse f. 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 6 DOF : 3 linear coordinates and 3 angles. b. An earth-orbit satellite. If the satellite was just a particle it would have 3 DOF . But, since it probably needs to be oriented with respect to the earth, sun, etc., it has 6 DOF . c. A surface ship. There is no difference between a submerged submarine and a surface ship, both have 6 DOF . One might argue 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 and latitude). For a given position, a ship can also have pitch, yaw, and roll angles. Thus, for all practical purposes, 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 some heading angle (turning a corner) and roll angle (if turning). Thus, there are 4 DOF . e. A two-button mouse. A two-button mouse has 4 DOF . It can move in the x and y directions and each button has 1 DOF . f. A computer joy stick. The joy stick has 2 DOF (x and y) and orientation, for a total of 3 DOF .

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-5-1 PROBLEM 2-5 Statement: 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 2-6-1 PROBLEM 2-6 Statement: Describe the motion of the following items as pure rotation, pure translation, or complex planar motion. a. A windmill b. A bicycle (in the vertical plane, not turning) c. A conventional "double-hung" window d. The keys on a computer keyboard e. The hand of a clock f. A hockey puck on the ice g. A "casement" window Solution: 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 2-7-1 PROBLEM 2-7 Statement: 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. 5 1 6 2 1 4 3 a. Number of links L 6  Number of full joints J 1 7  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 0  (a) 2 1 3 1 b. Number of links L 3  Number of full joints J 1 2  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 1  (b) 3 2 1 1 4 c. Number of links L 4  Number of full joints J 1 4  Number of half joints J 2 0  M 3 L 1  ( )  2 J 1   J 2   M 1  (c) 1 2 5 3 4 6 1 7 d. Number of links L 7  Number of full joints J 1 7  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 3  (d)

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-7-2 7 2 1 5 4 3 6 1 2 1 8 8 9 10 1 4 2 1 3 1 1 5 5 6 1 (e) (f) e. Number of links L 10  f. Number of links L 6  Number of full joints J 1 13  Number of full joints J 1 6  Number of half joints J 2 0  Number of half joints J 2 2  M 3 L 1  ( )  2 J 1   J 2   M 3 L 1  ( )  2 J 1   J 2   M 1  M 1  2 1 4 3 1 1 5 4 2 1 7 6 1 1 8 7 g. Number of links L 8  Number of full joints J 1 9  Number of half joints J 2 2  M 3 L 1  ( )  2 J 1   J 2   M 1  2 1 3 4 1 (g) h. Number of links L 4  Number of full joints J 1 4  Number of half joints J 2 0  M 3 L 1  ( )  2 J 1   J 2   M 1  (h)

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-8-1 PROBLEM 2-8 Statement: 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 the text to classify the linkages. 5 1 6 2 1 4 3 a. Number of links L 6  Number of full joints J 1 7  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 0  Structure (a) 2 1 3 1 b. Number of links L 3  Number of full joints J 1 2  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 1  Mechanism (b) 3 2 1 1 4 c. Number of links L 4  Number of full joints J 1 4  Number of half joints J 2 0  M 3 L 1  ( )  2 J 1   J 2   M 1  Mechanism (c) 1 2 5 3 4 6 1 7 d. Number of links L 7  Number of full joints J 1 7  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 3  Mechanism (d)

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-9-1 PROBLEM 2-9 Statement: Use linkage transformation on the linkage of Figure P2-1a to make it a 1- DOF mechanism. Solution: See Figure P2-1a and Mathcad file P0209. 1. The mechanism in Figure P2-1a has mobility: 5 1 6 2 1 4 3 Number of links L 6  Number of full joints J 1 7  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 0  2. Use rule 2, which states: "Any full joint can be replaced by a half joint, but this will increase the DOF by one." One way to do this is to replace one of the pin joints with a pin-in-slot joint such as that shown in Figure 2-3c. Choosing the joint between links 2 and 4, we now have mobility: 5 1 6 2 1 4 3 Number of links L 6  Number of full joints J 1 6  Number of half joints J 2 2  M 3 L 1  ( )  2 J 1   J 2   M 1 

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-10-1 PROBLEM 2-10 Statement: Use linkage transformation on the linkage of Figure P2-1d to make it a 2- DOF mechanism. Solution: See Figure P2-1d and Mathcad file P0210. 1 2 5 3 4 6 1 7 1. The mechanism in Figure P2-1d has mobility: Number of links L 7  Number of full joints J 1 7  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 3  2. Use rule 3, which states: "Removal of a link will reduce the DOF by one." One way to do this is to remove link 7 such that link 6 pivots on the fixed pin attached to the ground link (1). We now have mobility: 1 2 1 5 3 4 6 Number of links L 6  Number of full joints J 1 6  Number of half joints J 2 1  M 3 L 1  ( )  2 J 1   J 2   M 2 

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-11-1 PROBLEM 2-11 Statement: Use number synthesis to find all the possible link combinations for 2- DOF , up to 9 links, to hexagonal order, using only revolute joints. Solution: See Mathcad file P0211. 1. Use equations 2.4a and 2.6 with DOF = 2 and iterate the solution for valid combinations. Note that the number of links must be odd to have an even DOF (see Eq. 2.4). The smallest possible 2- DOF mechanism is then 5 links since three will give a structure (the delta triplet, see Figure 2-7). L B T  Q  P  H   L 3  M  T 2 Q   3 P   4 H    M 2  L 5  T 2 Q   3 P   4 H    2. For L 5  0 T 2 Q   3 P   4 H    0 T = Q = P = H = B 5  3. For L 7  2 T 2 Q   3 P   4 H    H 0  P 0  Case 1: Q 0  T 2 2 Q   3 P   4 H    T 2  B L T  Q  P  H   B 5  Case 2: Q 1  T 2 2 Q   3 P   4 H    T 0  B L T  Q  P  H   B 6  4. For L 9  4 T 2 Q   3 P   4 H    Case 1: H 1  T 0  Q 0  P 0  B L T  Q  P  H   B 8  Case 2a: H 0  4 T 2 Q   3 P    9 B T  Q  P   Case 2b: P 1  1 T 2 Q    Q 0  T 1  B L T  Q  P  H   B 7  Case 2c: P 0  4 T 2 Q    9 B T  Q   Case 2c1: Q 2  T 4 2 Q    T 0  B 9 T  Q   B 7  Case 2c2: Q 1  T 4 2 Q    T 2  B 9 T  Q   B 6  Case 2c3: Q 0  T 4 2 Q    T 4  B 9 T  Q   B 5 

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-12-1 PROBLEM 2-12 Statement: Find all of the valid isomers of the eightbar 1- DOF link combinations in Table 2-2 (p. 38) having a. 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 five possible link sets, four of which are listed above. Therefore, we expect that the 16 valid isomers are distributed among the five link sets and that there will be fewer than 16 isomers among the four link sets listed above. 2. One method that is helpful in finding isomers is to represent the linkage in terms of molecules as defined in Franke's Condensed Notations for Structural Synthesis. A summary of the rules for obtaining Franke's molecules 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 must be 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 in connecting 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 that can be drawn that connect the circles such that there are exactly three lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 4. In this case, there are three valid isomers as depicted by Franke's molecules and kinematic chains below. 6 8 2 1 3 4 5 7 1 3 3 1 0 0 1 3 3 1

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    DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-12-2 6 1 2 3 4 5 8 7 2 3 3 1 0 1 0 3 3 0 6 1 2 3 4 5 8 7 3 3 3 3 0 0 0 2 2 0 The 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 unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines 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 valid isomers as depicted by Franke's molecules and kinematic chains below. 6 8 1 5 7 4 2 3 3 3 4 2 0 1 2 0

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