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4.3 Free-Body Diagrams A free-body diagram is a drawing of a part of a complete system, isolated in order to determine the forces acting on that rigid body. The following force convention is de®ned: F ij represents the force exerted by link i on link j. Figure 4.4 shows various free-body diagrams that can be considered in the analysis of a crank slider mechanism (Fig. 4.4a). In Fig. 4.4b, the free body consists of the three moving links isolated from the frame 0. The forces acting on the system include a driving torque M, an external driven force F, and the forces transmitted from the frame at kinematic pair A, F 01 , and at kinematic pair C, F 03 . Figure 4.4c is a free-body diagram of the two links 1 and 2. Figure 4.4d is a free-body diagram of a single link. Figure 4.4 Used with permission from Ref. 15. 4. Kinetostatics 227 Mechanisms The force analysis can be accomplished by examining individual links or subsystems of links. In this way the reaction forces between links as well as the required input force or moment for a given output load are computed. 4.4 Reaction Forces Figure 4.5a is a schematic diagram of a crank slider mechanism comprising of a crank 1, a connecting rod 2, and a slider 3. The center of mass of link 1 is C 1 , the center of mass of link 2 is C 2 , and the center of mass of slider 3 is C. The mass of the crank is m 1 , the mass of the connecting road is m 2 , and the mass of the slider is m 3 . The moment of inertia of link i is I Ci , i 1Y 2Y 3. The gravitational force is G i Àm i g , i 1Y 2Y 3, where g 9X81 mas 2 is the acceleration of gravity. For a given value of the crank angle f and a known driven force F ext , the kinematic pair reactions and the drive moment M on the crank can be computed using free-body diagrams of the individual links. Figures 4.5b, 4.5c, and 4.5d show free-body diagrams of the crank 1, the connecting rod 2, and the slider 3. For each moving link the dynamic equilibrium equations are applied. j Figure 4.5 Used with permission from Ref. 15. 228 Theory of Mechanisms Mechanisms For the slider 3 the vector sum of the all the forces (external forces F ext , gravitational force G 3 , inertia forces F in 3 , reaction forces F 23 , F 03 ) is zero (Fig. 4.5d): F 3 F 23 F in 3 G 3 F ext F 03 0X Projecting this force onto the x and y axes gives F 3 ÁF 23x Àm 3 x C F ext 0 4X19 F 3 ÁF 23y À m 3 g F 03y 0X 4X20 For the connecting rod 2 (Fig. 4.5c), two vertical equations can be written: F 2 F 32 F in 2 G 2 F 12 0 M 2 B r C À r B ÂF 32 r C 2 À r B ÂF in 2 G 2 M in 2 0Y or F 2 ÁF 32x Àm 2 x C 2 F 12x 0 4X21 F 2 ÁF 32y Àm 2 y C 2 Àm 2 g F 12y 0 4X22 k x C À x B y C À y B 0 F 32x F 32y 0 k x C 2 À x B y C 2 À y B 0 Àm 2 x C 2 Àm 2 y C 2 À m 2 g 0 À I C 2 a 2 k 0X 4X23 For the crank 1 (Fig. 4.5b), there are two vectorial equations, F 1 F 21 F in 1 G 1 F 01 0 M 1 A r B Â F 21 r C 1 ÂF in 1 G 1 M in 1 M 0 or F 1 ÁF 21x Àm 1 x C 1 F 01x 0 4X24 F 1 ÁF 21y Àm 1 y C 1 Àm 1 g F 01y 0 4X25 k x B y B 0 F 21x F 21y 0 k x C 1 y C 1 0 Àm 1 x C 1 Àm 1 y C 1 À m 1 g 0 À I C 1 a 1 k M k 0Y 4X26 where M jMj is the magnitude of the input torque on the crank. The eight scalar unknowns F 03y , F 23x ÀF 32x , F 23y ÀF 32y , F 12x ÀF 21x , F 12y ÀF 21y , F 01x , F 01y , and M are computed from the set of eight equations (4.19), (4.20), (4.21), (4.22), (4.23), (4.24), (4.25), and (4.26). 4.5 Contour Method An analytical method to compute reaction forces that can be applied for both planar and spatial mechanisms will be presented. The method is based on the decoupling of a closed kinematic chain and writing the dynamic i j i j i j i j i j i j i j 4. Kinetostatics 229 Mechanisms equilibrium equations. The kinematic links are loaded with external forces and inertia forces and moments. A general monocontour closed kinematic chain is considered in Fig. 4.6. The reaction force between the links i À1 and i (kinematic pair A i ) will be determined. When these two links i À 1 and i are separated (Fig. 4.6b), the reaction forces F iÀ1Yi and F iYiÀ1 are introduced and F iÀ1Yi F iYiÀ1 0X 4X27 Table 4.1 shows the reaction forces for several kinematic pairs. The following notations have been used: M D is the moment with respect to the axis D, and F D is the projection of the force vector F onto the axis D. It is helpful to ``mentally disconnect'' the two links (i À 1) and i, which create the kinematic pair A i , from the rest of the mechanism. The kinematic pair at A i will be replaced by the reaction forces F iÀ1Yi , and F iYiÀ1 . The closed kinematic chain has been transformed into two open kinematic chains, and two paths I and II can be associated. The two paths start from A i . For the path I (counterclockwise), starting at A i and following I the ®rst kinematic pair encountered is A iÀ1 . For the link i À 1 left behind, dynamic equilibrium equations can be written according to the type of kinematic pair Figure 4.6 230 Theory of Mechanisms Mechanisms at A iÀ1 . Following the same path I, the next kinematic pair encountered is A iÀ2 . For the subsystem (i À 1 and i À 2), equilibrium conditions correspond- ing to the type of the kinematic pair at A iÀ2 can be speci®ed, and so on. A similar analysis can be performed for the path II of the open kinematic chain. The number of equilibrium equations written is equal to the number of unknown scalars introduced by the kinematic pair A i (reaction forces at this kinematic pair). For a kinematic pair, the number of equilibrium conditions is equal to the number of relative mobilities of the kinematic pair. Table 4.1 Reaction Forces for Several Kinematic Pairs Type of joint Joint force or moment Unknowns Equilibrium condition F x F y F F c DD jF x jF x jF y jF y M D 0 F c DD jFjF x F D 0 F x F y F F c DD jF x jF x jF y jF y x F D 0 M D 0 F c DD Fkn jFjF x F D 0 M D 0 F x F y F z F jF x jF x jF y jF y jF z jF z M D 1 0 M D 2 0 M D 3 0 4. Kinetostatics 231 Mechanisms The ®ve-link ( j 1Y 2Y 3Y 4Y 5) mechanism shown in Fig. 4.7a has the center of mass locations designated by C j x C j Y y C j Y 0. The following analysis will consider the relationships of the inertia forces F in j , the inertia moments M in j , the gravitational force G j , the driven force, F ext , to the joint reactions F ij , and the drive torque M on the crank 1 [15]. To simplify the notation, the total vector force at C j is written as F j F in j G j and the inertia torque of link j is written as M j M in j . The diagram representing the mechanism is depicted in Fig. 4.7b and has two contours 0-1-2-3-0 and 0-3-4-5-0. Remark The kinematic pair at C represents a rami®cation point for the mechanism and the diagram, and the dynamic force analysis will start with this kinematic pair. The force computation starts with the contour 0-3-4-5-0 because the driven load F ext on link 5 is given. 4.5.1 (I) CONTOUR 0-3-4-5-0 Reaction F 34 The rotation kinematic pair at C (or C R Y where the subscript R means rotation), between 3 and 4, is replaced with the unknown reaction (Fig. 4.8) F 34 ÀF 43 F 34x F 34y Xi j Figure 4.7 Used with permission from Ref. 15. 232 Theory of Mechanisms Mechanisms If the path I is followed (Fig. 4.8a), for the rotation kinematic pair at E (E R )a moment equation can be written as M 4 E r C À r E ÂF 32 r C 4 À r D ÂF 4 M 4 0Y or k x C À x E y C À y E 0 F 34x F 34y 0 k x C 4 À x E y C 4 À y E 0 F 4x F 4y 0 M 4 k 0X 4X28 Continuing on path I, the next kinematic pair is the translational kinematic pair at D (D T ). The projection of all the forces that act on 4 and 5 onto the sliding direction D (x axis) should be zero: F 45 D F 45 ÁF 34 F 4 F 5 F ext Á F 34x F 4x F 5x F ext 0X 4X29 After the system of Eqs. (4.28) and (4.29) are solved, the two unknowns F 34x and F 34y are obtained. Reaction F 45 The rotation kinematic pair at E (E R ), between 4 and 5, is replaced with the unknown reaction (Fig. 4.9) F 45 ÀF 54 F 45x F 45y X i j i j & & i i i j Figure 4.8 Used with permission from Ref. 15. 4. Kinetostatics 233 Mechanisms If the path I is traced (Fig. 4.9a), for the pin kinematic pair at C (C R )a moment equation can be written, M 4 C r E À r C ÂF 54 r C 4 À r C ÂF 4 M 4 0Y or k x E À x C y E À y C 0 ÀF 45x ÀF 45y 0 k x C 4 À x C y C 4 À y C 0 F 4x F 4y 0 M 4 k 0X 4X30 For the path II the slider kinematic pair at E (E T ) is encountered. The projection of all forces that act on 5 onto the sliding direction D (x axis) should be zero: F 5 D F 5 ÁF 45 F 5 F ext Á F 45x F 5x F ext 0X 4X31 The unknown force components F 45x and F 45y are calculated from Eqs. (4.30) and (4.31). i j i j i i Figure 4.9 Used with permission from Ref. 15. 234 Theory of Mechanisms Mechanisms Reaction F 05 The slider kinematic pair at E (E T ), between 0 and 5, is replaced with the unknown reaction (Fig. 4.10) F 05 F 05y X The reaction kinematic pair introduced by the translational kinematic pair is perpendicular to the sliding direction, F 05 c D. The application point P of the force F 05 is unknown. If the path I is followed, as in Fig. 4.10a, for the pin kinematic pair at E (E R ) a moment equation can be written for link 5, M 5 E r P À r E ÂF 05 0Y or xF 05y 0 A x 0X 4X32 The application point is at E (P E). Continuing on path I, the next kinematic pair is the pin kinematic pair C (C R ): M 45 C r E À r C ÂF 05 F 5 F ext r C 4 À r C ÂF 4 M 4 0Y j & Figure 4.10 Used with permission from Ref. 15. 4. Kinetostatics 235 Mechanisms or k x E À x C y E À y C 0 F 5x F ext F 05y 0 k x C 4 À x C y C 4 À y C 0 F 4x F 4y 0 M 4 k 0X 4X33 The kinematic pair reaction force F 05y can be computed from Eq. (4.33). 4.5.2 (II) CONTOUR 0-1-2-3-0 For this contour the kinematic pair force F 43 ÀF 34 at the rami®cation point C is considered as a known external force. Reaction F 03 The pin kinematic pair D R , between 0 and 3, is replaced with unknown reaction force (Fig. 4.11) F 03 F 03x F 03y X If the path I is followed (Fig. 4.11a), a moment equation can be written for the pin kinematic pair C R for the link 3, M 3 C r D À r C ÂF 03 r C 3 À r C ÂF 3 M 3 0Y i j i j i j Figure 4.11 Used with permission from Ref. 15. 236 Theory of Mechanisms Mechanisms [...]... Vassaia scola, Minsc, Russia, 1970 Mechanisms 5 Machine Components DAN B MARGHITU, CRISTIAN I DIACONESCU, AND NICOLAE CRACIUNOIU Department of Mechanical Engineering, Auburn University, Auburn, Alabama 36849 Inside 1 Screws 1.1 1.2 2 Gears 2.1 2.2 2.3 2.4 2 .5 2.6 2.7 2.8 253 Introduction 253 Geometry and Nomenclature 253 Interference and Contact Ratio 258 Ordinary Gear Trains 261 Epicyclic Gear Trains... mmX The theoretical center distance is c dp dg a2 rp rg 47X5 70 117X5 mmX The base circle radii of pinion and gear are rbp rp cos f 47X5 cos 20 44X6 35 mm rbg rg cos f 70 cos 20 65X778 mmX The addendum circle radii of pinion and gear are rap rp a m Np 2a2 52 X5 mm rag rg a m Ng 2a2 75 mmX The maximum possible addendum circle radii of pinion and gear,... Static Loading 303 Standard Dimensions 304 Bearing Selection 308 5 Lubrication and Sliding Bearings 5. 1 5. 2 5. 3 5. 4 Viscosity 318 Petroff's Equation 323 Hydrodynamic Lubrication Theory Design Charts 328 318 326 References 336 243 244 Machine Components 1 Screws T hreaded fasteners such as screws, nuts, and bolts are important components of mechanical structures and machines Screws may be used as removable... module 5, with 19 and 28 teeth, operate at a pressure angle of 20 Determine whether there will be interference when standard full-depth teeth are used Find the contact ratio Solution Machine Components A standard full-depth tooth has on addendum of a m 5 mm The gears will mesh at their pitch circles, and the pitch circle radii of pinion and gear are rp mNp a2 5 19a2 47X5 mm rg mNg a2 5 28a2... of module m are as follows: 0.2 to 1.0 by increments of 0.1 1.0 to 4.0 by increments of 0. 25 4.0 to 5. 0 by increments of 0 .5 Addendum, minimum dedendum, and clearance for standard full-depth involute teeth (pressure angle is 20 ) with English units in common use are Addendum a 1aPd Minimum dedendum b 1X 157 aPd X Machine Components For stub involute teeth with a pressure angle equal to 20 , the... q 2 ra max p rbp c 2 sin2 f 60X061 mm b rap 52 X5 mm q 2 ra max g rbg c 2 sin2 f 77X083 mm b rag 75 mmX Clearly, the use of standard teeth would not cause interference 261 2 Gears The contact ratio is q q 2 2 2 2 rap À rbp rag À rbg À c sin f CR 1X590Y pm cos f which should be a suitable value CR b 1X2 m... 12 teeth per inch Machine Components Figure 2.4 258 Machine Components of pitch diameter) With SI units ``pitch'' means circular pitch (a ``gear of pitch 3.14 mm'' refers to a gear having a circular pitch p of 3.14 mm) Standard diametral pitches Pd (English units) in common use are as follows: 1 to 2 by increments of 0. 25 2 to 4 by increments of 0 .5 4 to 10 by increments of 1 10 to 20 by increments... Vol 1 Academic Press, New York, 1 959 11 T R Kane, Analytical Elements of Mechanics, Vol 2 Academic Press, New York, 1961 12 T R Kane and D A Levinson, Dynamics McGraw-Hill, New York, 19 85 13 J T Kimbrell, Kinematics Analysis and Synthesis McGraw-Hill, New York, 1991 14 N I Manolescu, F Kovacs, and A Oranescu, The Theory of Mechanisms and Machines EDP, Bucharest, 1972 15 D B Marghitu and M J Crocker,... Interference will occur, preventing rotation of the mating gears, if either of the addendum circles extends beyond tangent points A and B (Fig 2 .5) , which are called interference points In Fig 2 .5 both addendum circles extend beyond the interference points Figure 2 .5 The maximum possible addendum circle radius of a pinion or gear without interference is q 2 2X7 ra max rb... Force Analysis 270 Strength of Gear Teeth 2 75 3 Springs 3.1 3.2 3.3 3.4 3 .5 3.6 3.7 3.8 244 Screw Thread 244 Power Screws 247 283 Introduction 283 Materials for Springs 283 Helical Extension Springs 284 Helical Compression Springs 284 Torsion Springs 290 Torsion Bar Springs 292 Multileaf Springs 293 Belleville Springs 296 4 Rolling Bearings 4.1 4.2 4.3 4.4 4 .5 4.6 297 Generalities 297 Classi®cation 298 . act on 5 onto the sliding direction D (x axis) should be zero: F 5 D F 5 ÁF 45 F 5 F ext Á F 45x F 5x F ext 0X 4X31 The unknown force components F 45x and F 45y are. F 34y are obtained. Reaction F 45 The rotation kinematic pair at E (E R ), between 4 and 5, is replaced with the unknown reaction (Fig. 4.9) F 45 ÀF 54 F 45x F 45y X i j i j & & i i i. r C ÂF 05 F 5 F ext r C 4 À r C ÂF 4 M 4 0Y j & Figure 4.10 Used with permission from Ref. 15. 4. Kinetostatics 2 35 Mechanisms or k x E À x C y E À y C 0 F 5x F ext F 05y 0 k x C