Chapter 4 Statics and dynamics of rigid bodies

Transcription

1 Chapter 4 Statics and dynamics of rigid bodies Bachelor Program in AUTOMATION ENGINEERING Prof. Rong-yong Zhao (zhaorongyong@tongji.edu.cn) First Semester,

2 Content of chapter Static equilibrium equations; 4.2 The principle of virtual work; 4.3 d Alembert principle and dynamic equilibrium equations; Energy balance approach; 4.5- Kinematics and dynamics of mechanical systems ; 2

3 4.1 Static equilibrium equations Necessary and sufficient conditions(force and moment balance) of the static equilibrium of a rigid body Vector R is the sum of all the active and reactive (i.e. those exerted by the constraints) forces Vector M o * is the sum of the moments of active and reactive forces with respect of a chosen pole O No possible transmission+ No possible revolution To separate all the active and reactive forces and moments, we have The active force action position The reactive force action position 3

4 Multi-body mechanical system To keep a multi-body system static equilibrium, it s necessary: a) to separate the rigid bodies one from another and apply eqs. (4.2) to every single rigid body composing the mechanical system; b) to consider, besides eqs. (4.2), further equilibrium equations concerning the relative mobility among bodies of the system. 4

5 4.2 The principle of virtual work The virtual work principle: Another description of static equilibrium conditions, or the force and moment balance conditions; Based on energy considerations; The virtual work principle is: Necessary and sufficient condition for the static equilibrium of a mechanical system, with fixed constraints and in absence of friction, is that the virtual work performed by active forces and moments for any virtual displacement is null 5

6 Virtual work principle equation System with fixed constraints and without friction; Composed by rigid bodies; the virtual work equation is: =Wforce + Wcouple=0 represent the virtual displacements of the points of application of the active forces Fj the virtual rotations of the bodies to which active couples Ck are applied Since the constraints are fixed and frictionless, the work done by any reaction force/couple is null by definition, so no constraint forces and couples appear in eq. (4.3). 6

7 Displacement and rotation conditions In order to be virtual, displacements and rotations must satisfy the following conditions: 1 to be infinitesimal(very small), even though of an arbitrary size; 2 to be compatible with the system constraints, i.e. they must not violate the constraint conditions imposed to the system; 3 to be reversible, i.e. to be possible in the negative and in the opposite positive sense, with an assumed sign convention; 4 to be time independent, i.e. time does not vary as virtual displacements are applied. 7

8 Discussion about the virtual work conditions Due to the first and second condition, the displacements δ P j must be tangent to the respective trajectories; Due to the assumed infinitesimal displacements, the system configuration does not vary; As to the multi-body system, the kinematic conditions characterizing the rigid body motions must also be included in the constraints set; virtual displacement of each rigid body will be represented by three independent parameters at most(e.g. two displacement components of one of its points and a rotation); The virtual work principle is particularly useful to study equilibrium problems of one d.of. systems when an external applied force/couple is the only unknown; solving eq. (4.3), in which all the virtual displacements are expressed as a function of one free coordinate, allows us to determine the unknown quantity. Based on static equilibrium equations (4.1) or (4.2), the reaction forces can also be obtained 8

9 4.2.1 Application example: rigid beam on rectilinear guides The rigid beam with both ends constrained to rectilinear guides. Task : To calculate F B by imposing the static equilibrium condition of the beam itself. Analysis :The beam has mass m and is subjected to its own weight and to a horizontal force; If the static equilibrium equations (4.1) are used, the reaction forces exerted in A and B by the constraints have to be included as unknowns. By applying the principle of virtual work the only forces to be considered are the beam weight and the force, since the work done by the reaction forces H A and V B is null. By applying the virtual work principle: where, are the virtual displacement vectors of points B and C 9

10 Continue indicate δx B (positive if directed to the right) as B point displacement ; Indicate δyc (positive if directed upward) as the vertical component of the displacement of point C ; horizontal work is null due to gravity being orthogonal to x axis; 10

11 Continue By substituting in the virtual work expression we have The bigger, Special cases: the bigger F B q = 90 鞍, tan 90 =?, then F 0; mg q = 135 鞍, tan135 = - 1, then FB = ; 2 q = 180 鞍, tan180 = 0, then F =? ; B B 11

12 4.2.2 Application example: slider crank mechanism the value of the torque M m has to be calculated in order to have the static equilibrium conditions when a force F r is applied to the slider, 12

13 Continue 13

14 Continue 14

15 Continue 15

16 4.3 d Alembert principle and dynamic equilibrium equations Focus on the equations expressing the dynamic behavior of mechanical systems; To reflect the relationship between the motion of a system and the forces applied to it; two dynamic behavior approaches: 1) d Alèmbert equations (or dynamic equilibrium equations), the static equilibrium equations; 2) Energy approach, the virtual work principle, base on the work done by inertia forces, or power balance equation inertial characteristics of the system play a fundamental role in the relationship between forces and resulting accelerations every notion illustrated in Chapter 3 related to mass geometry will be used. 16

17 Introduction of D'Alembert's massive principle D'Alembert's principle, also known as the Lagrange d'alembert principle, is a statement of the fundamental classical laws of motion. It is named after its discoverer, the French physicist and mathematician Jean le Rond d'alembert. The principle states that the sum of the differences between the forces acting on a system of mass particles and the time derivatives of the momenta of the system itself along any virtual displacement consistent with the constraints of the system, is zero. Jean d'alembert French ( ) Paris 17

18 Introduction of D'Alembert's principle Original expression of D'Alembert's principle 18

19 4.3.1 d Alembert principle: dynamic equilibrium of a particle Giving a particle of mass m; Newton s law states that the particle acceleration vector depends on the resultant of all the acting forces; forces :active and reactive, i.e. those due to constraints dynamic equilibrium By defining as force of inertia the vector 19

20 Discussion dynamic equilibrium has the same form of a static equilibrium equation; We add inertia force(a virtual force) the applied forces; To transfer a dynamic problem into a static problem; (4.9) and (4.10), constitutes the d Alembert principle for a massive particle. this principles is much more meaningful and useful from the engineering point of view when dealing with rigid bodies and rigid body systems 20

21 4.3.2 d Alembert principle: dynamic equilibrium of a rigid body Consider a very simple rigid body composed by two particles of masses m 1 and m 2 ; m1 and m2 keeping a constant distance to one another; the modulus P2 P1 of the vector joining the two particles must be constant By applying d Alembert principle 21

22 Continue Based on eq. (3.1) The coordinates,gx,gy of the centre of gravity G: By deriving twice with respect to time 22

23 Continue the sum of the each particle inertia forces equals a force of inertia due to the acceleration of the center of gravity G multiplied by the total mass m of the body. motion of the gravity center principle a fictitious force (the force of inertia) 23

24 Continue to undergo to a translation and rotation motion, with an acceleration of its center of gravity of Cartesian components (dotted lines) along x,y axes; an angular speed and acceleration ( (dotted curve) 24

25 Continue With reference to eq. (2.31) in Chapter 2, the accelerations of points P 1 and P 2 have the general form (2.31) 25

26 Continue Eq. (4.10b) defining the position of the center of gravity implies also (see Fig. 4.3a) The reasoning progress is rearranging eq. (4.10b) by squaring, summing the two sides of the last expressions and taking the square roots 26

27 Continue The force of inertia in vector form Writing the equilibrium at rotation around G in scalar form we have (positive clockwise): a moment of inertia Note : this moment of inertia is opposed to the angular acceleration the moment of inertia Recalling the equations (4.1), (4.2) of static equilibrium, by means of D Alèmbert principle dynamic equilibrium equations become: 27

28 Continue By separating the above said contributions Dynamic equilibrium conditions for a single rigid body are fully expressed by eqs. (4.15). 28

29 Continue If a mechanical system is composed of more than one rigid body, eqs. (4.15) represents only a necessary condition, not sufficient; Necessary conditions: 1) to separate the rigid bodies one from another and apply eqs. (4.15) to every single rigid body composing the mechanical system; 2) to consider, besides eqs. (4.15), further dynamic equilibrium equations concerning the relative mobility among bodies of the system. 29

30 Example :a continuous rigid body a beam with a uniformly distributed mass m per unit length;, rotating around a pinned end, as shown in Fig D Alèmbert principle can be applied to every single infinitesimal particle; Then a continuous distribution of forces of inertia, each given by the motion of the body must satisfy the dynamic equilibrium conditions under the action exerted by the applied forces (both active and reactive) together with the of inertia. 30

31 Example :a continuous rigid body The acceleration of a point at distance ξ from the pinned end presents a tangent component, orthogonal to the beam, and a normal component, directed towards the axis of rotation. the trajectories of all points of the rotating beam are circles with center in the fixed point, two components are given: t and n are unit vectors indicating the tangent and normal directions. 31

32 Example :a continuous rigid body the force of inertia acting on an element of length dξ are: calculate the force resulting from all the distributed elementary forces of inertia: 32

33 Continue To understand the meaning of the results above, observe that : ml is the total mass M of the beam; gravity G center is found at a distance L/2; therefore,the tangent and normal acceleration components of G are: recalling the motion of the gravity center principle Vector sum 33

34 Conclusion as already seen for the two particles rigid body, the resulting force of inertia of a continuous rigid body is equal to the product of the overall mass of the body multiplied by the acceleration of its center of gravity and changed in sign. 34

35 Consider the moment Consider the moment of the elementary forces of tangent inertia, let us choose as a pole the center of gravity G; the moment of the normal components is null,, since they all pass through G ; The moment of the tangential forces: 35

36 Continue recalling the mass moment of inertia of a homogeneous bar (3.15) around its gravity center the moment of the inertia forces can be written as 36

37 Conclusion As a general conclusion : the overall system of inertia acting on a rigid body can always be reduced to a resulting force, applied to the center of gravity, equal to the product of the body mass times the acceleration vector of the center of gravity and changed in sign (as already expressed by eq. (4.17)) and a moment (called moment of inertia) equal to the product of the mass moment of inertia around the center of gravity times the angular acceleration vector of the body and changed in sign (as stated in eq.(4.18)). 37

38 Dynamic equilibrium of the rotating beam Recalling the symbols used in this example, the dynamic equilibrium of the rotating beam reads 38

39 Question : Application example: thin rigid body rotating around a fixed axis a thin rigid body of mass m and mass moment of inertia J G around its gravity center is rotating around the pin axis in O at a given angular speed ω. The external forces applied to the body are given by a force F applied in P point; The gravity force mg (body weigh) applied in the center of gravity G and a moment M; The unknown quantity to be determine by the dynamic equilibrium approach (D Alèmbert principle) is the angular acceleration ω; Fig. 4.5 a 39

40 Application example: thin rigid body rotating around a fixed axis Solution: first calculate Inertia forces (force and moment) ; three unit vectors : t :normal to the line OG n :along OG, (see Fig. 4.5 b) k :normal to the directory plane By applying d Alembert principle we have: Fig. 4.5 b 40

41 Solution continue By removing the constraining pin, two components R n and R t of the reaction force applied in O point; Assuming as a sign convention positive the forces along t and n, and positive the moments if clockwise the dynamic equilibrium of the moments around O: Fig. 4.5 b the requested angular acceleration can be directly obtained 41

42 Further discussion By projecting the first equation of the system (4.19) along t and n directions represent the dynamic equilibrium of the body at translation in two distinguished directions allow for the reaction forces Rn and Rt to be determined 42

43 4.3.3 Mechanical system composed of rigid bodies a mechanical system composed of n rigid bodies it can written n systems of equations (4.19); i.e. the 2n vector equations of the form: j=1,2,,n 43

44 Mechanical system composed of rigid bodies From vector form (4.20), 3n scalar equations can be obtained: 2n from the first of (4.20) for the dynamic equilibrium at translation; n related to moment equations; the resulting force R j acting on the jth body includes: 1 the external forces applied to the jth body; 2 the reaction forces, corresponding to the constraints connecting the jth to the ground; 3 the reaction forces corresponding to the constraints connecting the jth body to other bodies; 44

45 Mechanical system composed of rigid bodies These equations allow for 3n unknowns to be determined, generally include: 1) A certain number of kinematic parameters (linear acceleration components, angular accelerations of bodies) equal in number to the number of degrees of freedom (d.o.f.); this is the case called direct dynamic problem which can be stated briefly as given the force,determine the acceleration ; the case of inverse dynamic problem vice versa can be indicated as given the acceleration determine the force necessary to produce it ; 2) A number of unknown active forces or moments corresponding to the d.o.f. of the system: this is a kinetic-static problem. 45

46 Mechanical system composed of rigid bodies the system composed of several bodies, the three scalar equations of dynamic equilibrium will contain: 1) the external forces applied to all the bodies composing the considered sub-system; 2) the reaction forces, corresponding to the constraints connecting the sub-system bodies to the ground; 3) the reaction forces corresponding to the constraints connecting the considered sub-system to the remaining part of the system; Note: forces exerted between any two bodies belonging to the considered sub-system (indicated as internal forces) will not appear. 46

47 Discussion the overall number of independent equations that can be written for a system in plane motion is always 3n(the maximum number of equations); n being the number of bodies composing the mechanical system; In fact, not necessary to write all of them, depending on the specific problem; 47

48 4.4 - Energy balance approach Based on virtual work, an energy approach can be used also in Dynamics; Based on D Alèmbert principle, by introducing an opportune system of forces and moments of inertia to treat a dynamic problem as a static one; Therefore, to apply the principle of virtual work by including also the work done by forces and moments of inertia; to write as many equations as the number of d.o.f. of the system; Particularly, for a one d.o.f. system, one equation will be obtained; 48

49 Virtual work method consider again the case of the rigid body represented in Fig. 4.5, already analyzed by means of the dynamic equilibrium approach; Given the pin constraint in O, the virtual displacement of the body can only be a rotation around O, described by a virtual rotation δϑ (positive if anti-clockwise). virtual work: where 49

50 Virtual work method By solving the dot products and grouping terms : Since eq (4.22) has to be valid for any rotation δϑ, the following pure equation same as that found previously writing the moment dynamic equilibrium around O. in the case of fixed and frictionless constraints, we substitute to the virtual displacements the actual displacements, 50

51 Virtual work method actually executed in an infinitesimal time dt, which depend on the actual velocities of the points of the body and from its angular speed: the overall sum of the power of all the forces and moments acting on the body (including those of inertia) must be zero. 51

52 Power balance The equation here obtained represents the so called power balance, and, for a mechanical system, can be summarized by the symbolic equation : states that the sum of the powers of every force and moment acting on the system (including those of forces and moments of inertia) must be zero. In dynamic problems the power balance equation is of more spontaneous use since it is more closely related to the motion of the bodies composing the system. 52

53 Kinetic energy of a rigid body Given the gravity center and mass moment of inertia, it is possible to calculate the kinetic energy for a rigid body translation and rotation; Assuming the gravity center G as a reference, the velocity vector of a point P of a rigid body is: kinetic energy: Keeping account of eq (4.25) 53

54 Continue By executing the products indicated in eq. (4.26) and after reordering them following the definition of center of gravity, =0 Then, the second and third term=0 54

55 König theorem The final expression of the kinetic energy (4.27 )becomes: translation motion rotation around of gravity center Eq. (4.28) represents the König theorem, the simplest form to express the kinetic energy of a rigid body in translation and rotation motion. It has to be noted that this two-terms expression is valid only if referred to the center G. Reason: Choosing any other point will cause the second and third term of (4.27) to be not zero. 55

56 Kinetic energy theorem deriving the kinetic energy (4.28), then comparison to the expression obtained by the power balance leads to the equivalence eq. (4.24) becomes 56

57 Kinetic energy theorem the sum of the powers of all the active forces equals the first derivative of the kinetic energy with respect to time; this is known also as the kinetic energy theorem; The validity of this theorem is not limited to a single body but holds true for a system with any number of bodies; an important physical meaning: during the motion of the system, when the sum of the powers of the active forces is positive the kinetic energy has a positive first derivative; vice versa, with a negative powers sum, the kinetic energy decreases, exhibiting a negative first derivative. 57

58 4.5- Kinematics and dynamics of mechanical systems To study the dynamic behaviour of mechanical systems composed of several rigid bodies the dynamic equilibrium equations (4.20) can be used. For one degree of freedom systems without constrains, internal forces, and frictionless, with n bodies, generalized expression of eq. (4.29): 58

59 Discussion about (4.29a) Case 1: direct dynamic problem, the active forces are known and eq. (4.29a) the acceleration is to be calculated; Being a one d.o.f. problem, the accelerations of the rigid bodies composing the system are all connected to the one of a reference body. A reference body velocity has to be known, if any active force is velocity dependent. Case 2: inverse dynamic problem, the acceleration is known and an active force/moment has to be determined by eq. (4.29a). The first solution stage of this kind of problems is the kinematic analysis of the system, in order to find the relationships existing between the kinematic quantities of interest. 59

60 Example 1: Four-bar linkage mechanism an inverse dynamic problem is proposed in Fig. 4.6a; a known force F B is applied at point B and the moment M m necessary to keep the motion with certain imposed angular velocity In complex system 60

61 Add geometry quantities: Position analysis 61

62 Position analysis 62

63 Position analysis The two values of the roots of (4.35) correspond to the two possible positions B, B', to be chosen by the user. Once the value of α 3 is determined, eq. (4.32) give us the value of α 2 for a given value of the crank rotation α 1 the position of the mechanism is found By updating the value of α 1, the previous expressions can be used in a recursive form Then to determine a complete rotation of the crank AO, as well as the trajectory of any point belonging to the connecting rod AB. 63

64 Velocity analysis deriving the closure equation (4.31) with respect to time, then By separating the real and the imaginary part of eq. (4.31) 64

65 Velocity analysis- graphical scale of representation Equation (4.36) can also be solved graphically (see Fig. 4.6c) : to represent the known vectors on the base of a chosen graphical scale of representation. Velocity of point A and the trajectory of point B are known; 65

66 Discussion Once the vector V A is represented, we can draw the known direction of the two unknown vectors V B,V BA so closing the velocity triangle over V A ; The two unknown vectors are such as to respect the vector equation (4.38); The previous velocity expressions can be used in a recursive form, thus allowing us to determine the velocities (linear and angular) of the system components for any given value of the position and velocity of the crank AO 1. 66

67 Acceleration analysis By deriving the velocity equation (4.36) with respect to time: By separating the real and the imaginary part i.e. the derivatives of (4.37) with respect to time. 67

68 Graphical acceleration analysis Equation (4.39) can also be solved graphically graphical scale of representation for the known acceleration components. With reference to the rigid body AB the following relation where 68

69 Graphical acceleration analysis First we draw the known vectors draw the known directions of the two unknown vectors closing the accelerations polygon the previous expressions for the acceleration components can be used in a recursive form, thus to determine their values (linear and angular) for any given value of the position and velocity of the crank AO1. 69

70 Dynamic analysis Considering the moment M m as the only unknown; the dynamic analysis can now be performed by means of eq. (4.29a) all the kinematic quantities inertia characteristics of the bodies the general form masses mass moments of inertia 70

71 Note The angular accelerations are already known; To determine: therefore therefore 71

72 Example 2: Slider-slotted arm mechanism consider a kinematic and dynamic problem To determine the moment M, necessary to make the component AB to rotate at a given constant angular speed a force F is applied at point P of the slotted arm. Based on eq. (4.29a) Fig. 4.7 a 72

73 Position analysis Given the geometry of the system through the quantities h, r, α; first determine the position of the body OB as a function of α (independent variable); closing an appropriate loop of vectors (Fig. 4.7 b) separating the real and imaginary part Fig. 4.7 b 73

74 Solution the solution of (4.44): Represent the position of point B of the slider 74

75 Velocity analysis deriving the closure equation (4.43) The angular velocity of the body AB is supposed to be assigned. By separating the real and the imaginary part of eq. (4.46) 75

76 Graphical analysis Equation (4.47) can also be solved graphically (see Fig. 4.7a) ; the known vectors on the base of a chosen graphical scale of representation. referred to the rigid body AB (we know the velocity of point B) 76

77 Graphical analysis 77

78 Acceleration analysis deriving the velocity equation (4.46) with respect to time : separating the real and the imaginary part in (4.49) 78

79 Graphical solution graphical solution of equation (4.50), referred to an instantaneous situation; reference to the rigid body AB 79

80 Dynamic analysis considering the moment M as the only unknown, the dynamic analysis can now be performed by means of eq. (4.42), all the kinematic quantities previously found are to be introduced. By solving the dot products, the power balance becomes 80