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- 1. Mechanical System Modeling K. Craig 1 Mechanical System Modeling Dr. Kevin Craig Professor of Mechanical Engineering Rensselaer Polytechnic Institute
- 2. Mechanical System Modeling K. Craig 2 References for Mechanical Systems • System Dynamics, E. Doebelin, Marcel Dekker, 1998. (This is the finest reference on system dynamics available; many figures in these notes are taken from this reference.) • Modeling, Analysis, and Control of Dynamic Systems, W. Palm, 2nd Edition, Wiley, 1999. • Vector Mechanics for Engineers: Dynamics, 7th Edition, F. Beer, E.R. Johnston, and W. Clausen, McGraw Hill, 2004.
- 3. Mechanical System Modeling K. Craig 3 Mechanical System Elements • Three basic mechanical elements: – Spring (elastic) element – Damper (frictional) element – Mass (inertia) element • Translational and Rotational versions • These are passive (non-energy producing) devices • Driving Inputs – force and motion sources which cause elements to respond
- 4. Mechanical System Modeling K. Craig 4 • Each of the elements has one of two possible energy behaviors: – stores all the energy supplied to it – dissipates all energy into heat by some kind of “frictional” effect • Spring stores energy as potential energy • Mass stores energy as kinetic energy • Damper dissipates energy into heat • Dynamic Response of each element is important – step response – frequency response
- 5. Mechanical System Modeling K. Craig 5 Spring Element • Real-world design situations • Real-world spring is neither pure nor ideal • Real-world spring has inertia and friction • Pure spring has only elasticity - it is a mathematical model, not a real device • Some dynamic operation requires that spring inertia and/or damping not be neglected • Ideal spring: linear • Nonlinear behavior may often be preferable and give significant performance advantages
- 6. Mechanical System Modeling K. Craig 6 • Device can be pure without being ideal (e.g., nonlinear spring with no inertia or damping) • Device can be ideal without being pure (e.g., device which exhibits both linear springiness and linear damping) • Pure and ideal spring element: • Ks = spring stiffness (N/m or N-m/rad) • 1/Ks = Cs = compliance (softness parameter) ( ) ( ) s 1 2 s s 1 2 s f K x x K x T K K = − = = θ − θ = θ s s x C f C T = θ = Ks x f f x Cs
- 7. Mechanical System Modeling K. Craig 7 • Energy stored in a spring • Dynamic Response: Zero-Order Dynamic System Model – Step Response – Frequency Response • Real springs will not behave exactly like the pure/ideal element. One of the best ways to measure this deviation is through frequency response. 2 2 s s s C f K x E 2 2 = =
- 8. Mechanical System Modeling K. Craig 8 Spring Element ( ) ( ) ( ) 0 s 2 2 x s 0 s 0 s 0 Differential Work Done f dx K x dx Total Work Done K x C f K x dx 2 2 = = = = =∫
- 9. Mechanical System Modeling K. Craig 9 Frequency Response Of Spring Elements ( ) ( ) 0 s 0 f f sin t x C f sin t = ω = ω
- 10. Mechanical System Modeling K. Craig 10 Zero-Order Dynamic System Model Step Response Frequency Response
- 11. Mechanical System Modeling K. Craig 11 More Realistic Lumped-Parameter Model for a Spring Ks Ks M B B f, x
- 12. Mechanical System Modeling K. Craig 12 Linearization for a Nonlinear Spring ( ) ( ) ( ) 0 0 0 22 0 0 0 2 x x x x 0 0 x x x xdf d f y f (x ) x x dx dx 2! df y y x x dx = = = − = + − + + ≈ + − ( ) 0 0 0 x x df y y x x dx ˆ ˆy Kx = − ≈ + − =
- 13. Mechanical System Modeling K. Craig 13 • Real Springs – nonlinearity of the force/deflection curve – noncoincidence of the loading and unloading curves (The 2nd Law of Thermodynamics guarantees that the area under the loading f vs. x curve must be greater than that under the unloading f vs. x curve. It is impossible to recover 100% of the energy put into any system.)
- 14. Mechanical System Modeling K. Craig 14 • Several Types of Practical Springs: – coil spring – hydraulic (oil) spring – cantilever beam spring – pneumatic (air) spring – clamped-end beam spring – ring spring – rubber spring (shock mount) – tension rod spring – torsion bar spring
- 15. Mechanical System Modeling K. Craig 15 • Spring-like Effects in Unfamiliar Forms – aerodynamic spring – gravity spring (pendulum) – gravity spring (liquid column) – buoyancy spring – magnetic spring – electrostatic spring – centrifugal spring
- 16. Mechanical System Modeling K. Craig 16 Damper Element • A pure damper dissipates all the energy supplied to it, i.e., converts the mechanical energy to thermal energy. • Various physical mechanisms, usually associated with some form of friction, can provide this dissipative action, e.g., – Coulomb (dry friction) damping – Material (solid) damping – Viscous damping
- 17. Mechanical System Modeling K. Craig 17 • Pure / ideal damper element provides viscous friction. • All mechanical elements are defined in terms of their force/motion relation. (Electrical elements are defined in terms of their voltage/current relations.) • Pure / Ideal Damper – Damper force or torque is directly proportional to the relative velocity of its two ends. 1 2dx dx dx f B B dt dt dt ⎛ ⎞ = − =⎜ ⎟ ⎝ ⎠ 1 2d d d T B B dt dt dt θ θ θ⎛ ⎞ = − =⎜ ⎟ ⎝ ⎠
- 18. Mechanical System Modeling K. Craig 18 – Forces or torques on the two ends of the damper are exactly equal and opposite at all times (just like a spring); pure springs and dampers have no mass or inertia. This is NOT true for real springs and dampers. – Units for B to preserve physical meaning: • N/(m/sec) • (N-m)/(rad/sec) – Transfer Function ( ) 2 2 2 2 dx d x Dx D x dt dt x x (x)dt x dt dt D D ⎡ ⎤ ⎣ ⎦∫ ∫ ∫ Differential Operator Notation
- 19. Mechanical System Modeling K. Craig 19 • Operational Transfer Functions • We assume the initial conditions are zero. – Damper element dissipates into heat all mechanical energy supplied to it. • Force applied to damper causes a velocity in same direction. f BDx T BD = = θ ( ) ( ) ( ) ( ) f T D BD D BD x x 1 1 D D f BD T BD θ θ ( )( ) 2 dx dx Power force velocity f B dt dt ⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
- 20. Mechanical System Modeling K. Craig 20 • Power input to the device is positive since the force and velocity have the same sign. • It is impossible for the applied force and resulting velocity to have opposite signs. • Thus, a damper can never supply power to another device; Power is always positive. • A spring absorbs power and stores energy as a force is applied to it, but if the force is gradually relaxed back to zero, the external force and the velocity now have opposite signs, showing that the spring is delivering power. • Total Energy Dissipated ( ) ( ) 2 dx dx P dt B dt B dx f dx dt dt ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫ ∫ ∫ ∫
- 21. Mechanical System Modeling K. Craig 21 Damper Element Step Input Force causes instantly (a pure damper has no inertia) a Step of dx/dt and a Ramp of x
- 22. Mechanical System Modeling K. Craig 22 Frequency Response of Damper Elements ( ) ( ) ( ) 0 t 0 0 0 0 f f sin t dx B dt 1 x x f sin t dt B f 1 cos t B = ω = − = ω ⎡ ⎤= − ω⎣ ⎦ω ∫ 0 x f 0 f A 1B A f B ω= = ω
- 23. Mechanical System Modeling K. Craig 23 • Sinusoidal Transfer Function – M is the amplitude ratio of output over input – φ is the phase shift of the output sine wave with respect to the input sine wave (positive if the output leads the input, negative if the output lags the input) ( ) x 1 D f BD = D i⇒ ω ( ) x 1 i M f i B ω = = ∠φ ω ( ) x 1 1 i M 90 f i B B ° ω = = ∠φ = ∠ − ω ω
- 24. Mechanical System Modeling K. Craig 24 • Real Dampers – A damper element is used to model a device designed into a system (e.g., automotive shock absorbers) or for unavoidable parasitic effects (e.g., air drag). – To be an energy-dissipating effect, a device must exert a force opposite to the velocity; power is always negative when the force and velocity have opposite directions. – Let’s consider examples of real intentional dampers.
- 25. Mechanical System Modeling K. Craig 25 Viscous (Piston/Cylinder) Damper A relative velocity between the cylinder and piston forces the viscous oil through the clearance space h, shearing the fluid and creating a damping force. 2 2 2 2 2 1 2 13 2 6 L h R R B R R h hh 2 R 2 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥πμ −⎛ ⎞ = − − −⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦ −⎢ ⎥ ⎣ ⎦ μ = fluid viscosity
- 26. Mechanical System Modeling K. Craig 26 Simple Shear Damper And Viscosity Definition fluid viscosity shearing stress F / A velocity gradient V / t μ = 2A F V t F 2A B V t μ = μ = =
- 27. Mechanical System Modeling K. Craig 27 Examples of Rotary Dampers 3 D L B 4t π μ = 4 0D B 16t π μ =
- 28. Mechanical System Modeling K. Craig 28 Commercial Air Damper laminar flow linear damping turbulent flow nonlinear damping (Data taken with valve shut) Air Damper • much lower viscosity • less temperature dependent • no leakage or sealing problem
- 29. Mechanical System Modeling K. Craig 29 Eddy-Current Damper • Motion of the conducting cup in the magnetic field generates a voltage in the cup. • A current is generated in the cup’s circular path. • A current-carrying conductor in a magnetic field experiences a force proportional to the current. • The result is a force proportional to and opposing the velocity. • The dissipated energy shows up as I2R heating of the cup.
- 30. Mechanical System Modeling K. Craig 30 Temperature Sensitivity Of Damping Methods
- 31. Mechanical System Modeling K. Craig 31 Other Examples of Damper Forms
- 32. Mechanical System Modeling K. Craig 32 • The damper element can also be used to represent unavoidable parasitic energy dissipation effects in mechanical systems. – Frictional effects in moving parts of machines – Fluid drag on vehicles (cars, ships, aircraft, etc.) – Windage losses of rotors in machines – Hysteresis losses associated with cyclic stresses in materials – Structural damping due to riveted joints, welds, etc. – Air damping of vibrating structural shapes
- 33. Mechanical System Modeling K. Craig 33 Hydraulic Motor Friction and its Components
- 34. Mechanical System Modeling K. Craig 34 Coulomb Friction: Modeling and Simulation • In most control systems, Coulomb friction is a nuisance. • Coulomb friction is difficult to model and troublesome to deal with in control system design. • It is a nonlinear phenomenon in which a force is produced that tends to oppose the motion of bodies in contact in a mechanical system. • Undesirable effects: “hangoff” and limit cycling
- 35. Mechanical System Modeling K. Craig 35 • Hangoff (or dc limit cycle) prevents the steady- state error from becoming zero with a step command input. • Limit Cycling is behavior in which the steady-state error oscillates or hunts about zero. • What Should the Control Engineer Do? – Minimize friction as much as possible in the design – Appraise the effect of friction in a proposed control system design by simulation – If simulation predicts that the effect of friction is unacceptable, you must do something about it!
- 36. Mechanical System Modeling K. Craig 36 – Remedies can include simply modifying the design parameters (gains), using integral control action, or using more complex measures such as estimating the friction and canceling its effect. – Modeling and simulation of friction should contribute significantly to improving the performance of motion control systems.
- 37. Mechanical System Modeling K. Craig 37 Modeling Coulomb Friction V Ff Fslip Fstick "Stiction" Coulomb Friction Model
- 38. Mechanical System Modeling K. Craig 38 Case Study to Evaluate Friction Model m k Ff V0 V m = 0.1 kg k = 100 N/m Fstick = 0.25 N Fslip = 0.20 N (assumed independent of velocity) V0 = step of 0.002 m/sec at t = 0 sec
- 39. Mechanical System Modeling K. Craig 39 Friction Model in Simulink
- 40. Mechanical System Modeling K. Craig 40 Simulink Block Diagram
- 41. Mechanical System Modeling K. Craig 41 Example with Friction Model
- 42. Mechanical System Modeling K. Craig 42 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.005 0.01 0.015 0.02 0.025 time (sec) 2*position,velocity,0.1*FrictionForce Position, Velocity, Friction Force vs. Time
- 43. Mechanical System Modeling K. Craig 43 Inertia Element • A designer rarely inserts a component for the purpose of adding inertia; the mass or inertia element often represents an undesirable effect which is unavoidable since all materials have mass. • There are some applications in which mass itself serves a useful function, e.g., accelerometers and flywheels.
- 44. Mechanical System Modeling K. Craig 44 Useful Applications of Inertia Flywheels are used as energy-storage devices or as a means of smoothing out speed fluctuations in engines or other machines. Accelerometer
- 45. Mechanical System Modeling K. Craig 45 – Newton’s Law defines the behavior of mass elements and refers basically to an idealized “point mass”: – The concept of rigid body is introduced to deal with practical situations. For pure translatory motion, every point in a rigid body has identical motion. – Real physical bodies never display ideal rigid behavior when being accelerated. – The pure / ideal inertia element is a model, not a real object. ( )( )forces mass acceleration=∑
- 46. Mechanical System Modeling K. Craig 46 Rigid and Flexible Bodies: Definitions and Behavior
- 47. Mechanical System Modeling K. Craig 47 – Newton’s Law in rotational form for bodies undergoing pure rotational motion about a single fixed axis: – The concept of moment of inertia J also considers the rotating body to be perfectly rigid. – Note that to completely describe the inertial properties of any rigid body requires the specification of: • Its total mass • Location of the center of mass • 3 moments of inertia and 3 products of inertia ( )( )torques moment of inertia angular acceleration=∑
- 48. Mechanical System Modeling K. Craig 48 Rotational Inertia J (kg-m2) ( )( ) ( )( ) ( ) tangential force mass acceleration 2 rL dr r = ⎡ ⎤= π ρ α⎣ ⎦ ( ) R 2 2 3 2 0 R MR total torque 2 L r dr R L J 2 2 = πρ α = π ρ = α = α∫
- 49. Mechanical System Modeling K. Craig 49 Moments of Inertia For Some Common Shapes
- 50. Mechanical System Modeling K. Craig 50 – How do we determine J for complex shapes with possibly different materials involved? • In the design stage, where the actual part exists only on paper, estimate as well as possible! • Once a part has been constructed, use experimental methods for measuring inertial properties. How?
- 51. Mechanical System Modeling K. Craig 51 Experimental Measurement Of Moment of Inertia ( ) 2 2 2 s 2 2 s 2 0 n 0 s n n n d torques J J dt d K J dt Kd 0 dt J cos t ( 0) K rad/sec J f cycles/sec 2 θ = α = θ − θ = θ + θ = θ = θ ω θ = ω ω π ∑ s 22 n K J 4 f = π
- 52. Mechanical System Modeling K. Craig 52 – Actually the oscillation will gradually die out due to the bearing friction not being zero. – If bearing friction were pure Coulomb friction, it can be shown that the decay envelope of the oscillations is a straight line and that friction has no effect on the frequency. – If the friction is purely viscous, then the decay envelope is an exponential curve, and the frequency of oscillation does depend on the friction but the dependence is usually negligible for the low values of friction in typical apparatus.
- 53. Mechanical System Modeling K. Craig 53 Inertia Element Real inertias may be impure (have some springiness and friction) but are very close to ideal. ( ) ( )2 2 x 1 1 D D f MD T JD θ = = Inertia Element stores energy as kinetic energy: 2 2 Mv J or 2 2 ω
- 54. Mechanical System Modeling K. Craig 54 – A step input force applied to a mass initially at rest causes an instantaneous jump in acceleration, a ramp change in velocity, and a parabolic change in position. – The frequency response of the inertia element is obtained from the sinusoidal transfer function: • At high frequency, the inertia element becomes very difficult to move. • The phase angle shows that the displacement is in a direction opposite to the applied force. ( ) ( ) 2 2 x 1 1 i 180 f MM i ° ω = = ∠ − ωω
- 55. Mechanical System Modeling K. Craig 55 Useful Frequency Range for Rigid Model of a Real Flexible Body A real flexible body approaches the behavior of a rigid body if the forcing frequency is small compared to the body’s natural frequency.
- 56. Mechanical System Modeling K. Craig 56 – Analysis: ( ) ( ) ( ) i o o 2 o o i 2 o i n2 2 n i i 2 2 2 o o 2 n n n 2AE x x ALx L L x x x 2E D 2E 1 x x L x x1 1 1 D i Dx x i1 1 1 − = ρ ρ + = ⎛ ⎞ + = ω⎜ ⎟ ω ρ⎝ ⎠ = ω = = ⎛ ⎞ ⎛ ⎞ω ω+ + −⎜ ⎟ ⎜ ⎟ω ω ω⎝ ⎠ ⎝ ⎠
- 57. Mechanical System Modeling K. Craig 57 – ωmax is the highest frequency for which the real body behaves almost like an ideal rigid body. • Frequency response is unmatched as a technique for defining the useful range of application for all kinds of dynamic systems. ( )o 2 i max n max n x 1 i 1.05 x 1 0.308 E 0.218 L ω = = ⎛ ⎞ω −⎜ ⎟ω⎝ ⎠ ω = ω = ρ 96200 cycles/min for a 6-inch steel rod
- 58. Mechanical System Modeling K. Craig 58 Motion Transformers • Mechanical systems often include mechanisms such as levers, gears, linkages, cams, chains, and belts. • They all serve a common basic function, the transformation of the motion of an input member into the kinematically-related motion of an output member. • The actual system may be simplified in many cases to a fictitious but dynamically equivalent one.
- 59. Mechanical System Modeling K. Craig 59 • This is accomplished by “referring” all the elements (masses, springs, dampers) and driving inputs to a single location, which could be the input, the output, or some selected interior point of the system. • A single equation can then be written for this equivalent system, rather than having to write several equations for the actual system. • This process is not necessary, but often speeds the work and reduces errors.
- 60. Mechanical System Modeling K. Craig 60 Motion Transformers Gear Train Relations: θ θ m m m m N N N T T N N N ′ = ≡ ′ = ≡ 2 1 1 2 1 Tm N1 N2 θm ′Tm ′θm
- 61. Mechanical System Modeling K. Craig 61 Translational Equivalent for A Complex System x1, x2, θ are kinematically related Refer all elements and inputs to the x1 location and define a fictitious equivalent system whose motion will be the same as x1 but will include all the effects in the original system.
- 62. Mechanical System Modeling K. Craig 62 – Define a single equivalent spring element which will have the same effect as the three actual springs. – Mentally apply a static force f1 at location x1 and write a torque balance equation: ( ) 1 s2 1 1 s1 1 1 1 s2 2 1 1 1 se 1 2 2 se s1 s2 s2 1 1 x KL f L K x L x K L L L f K x L 1 K K K K L L ⎛ ⎞ = + +⎜ ⎟ ⎝ ⎠ = ⎡ ⎤⎛ ⎞ + +⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠⎣ ⎦
- 63. Mechanical System Modeling K. Craig 63 – The equivalent spring constant Kse refers to a fictitious spring which, if installed at location x1, would have exactly the same effect as all the springs together in the actual system. – To find the equivalent damper, mentally remove the inertias and springs and again apply a force f1 at x1: ( ) ( )1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 1 1 1 e 1 2 2 e 1 2 2 1 1 f L x B L x B L B L x x B L x B B L L f B x L 1 B B B B L L = + + θ = + + = ⎡ ⎤⎛ ⎞ + +⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠⎣ ⎦
- 64. Mechanical System Modeling K. Craig 64 – Finally, consider only the inertias present. – While the definitions of equivalent spring and damping constants are approximate due to the assumption of small motions, the equivalent mass has an additional assumption which may be less accurate; we have treated the masses as point masses, i.e., J = ML2. ( ) ( ) ( )2 21 1 1 1 1 1 1 2 2 1 1 1 1 e 1 2 2 e 1 2 2 1 1 x x x f L M L M L J L L L f M x L 1 M M M J L L ≈ + + ≈ ⎡ ⎤⎛ ⎞ + +⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠⎣ ⎦
- 65. Mechanical System Modeling K. Craig 65 – To refer the driving inputs to the x1 location we note that a torque T is equivalent to a force T/L1 at the x1 location, and a force f2 is equivalent to a force (L2/L1)f2. – If we set up the differential equation of motion for this system and solve for its unknown x1, we are guaranteed that this solution will be identical to that for x1 in the actual system. – Once we have x1, we can get x2 and/or θ immediately since they are related to x1 by simple proportions.
- 66. Mechanical System Modeling K. Craig 66 – Rules for calculating the equivalent elements without deriving them from scratch: • When referring a translational element (spring, damper, mass) from location A to location B, where A’s motion is N times B’s, multiply the element’s value by N2. This is also true for rotational elements coupled by motion transformers such as gears, belts, and chains. • When referring a rotational element to a translational location, multiply the rotational element by 1/R2, where the relation between translation x and rotation θ (in radians) is x = R θ. For the reverse procedure (referring a translational element to a rotational location) multiply the translational element by R2.
- 67. Mechanical System Modeling K. Craig 67 • When referring a force at A to get an equivalent force at B, multiply by N (holds for torques). Multiply a torque at θ by 1/R to refer it to x as a force. A force at x is multiplied by R to refer it as a torque to θ. – These rules apply to any mechanism, no matter what its form, as long as the motions at the two locations are linearly related.
- 68. Mechanical System Modeling K. Craig 68 Mechanical Impedance • When trying to predict the behavior of an assemblage of subsystems from their calculated or measured individual behavior, impedance methods have advantages. • Mechanical impedance is defined as the transfer function (either operational or sinusoidal) in which force is the numerator and velocity the denominator. The inverse of impedance is called mobility.
- 69. Mechanical System Modeling K. Craig 69 Mechanical Impedance for the Basic Elements ( ) ( ) ( ) ( ) ( ) ( ) s S B M Kf Z D D v D f Z D D B v f Z D D MD v = = =
- 70. Mechanical System Modeling K. Craig 70 • Measurement of impedances of subsystems can be used to analytically predict the behavior of the complete system formed when the subsystems are connected. We can thus discover and correct potential design problems before the subsystems are actually connected. • Impedance methods also provide “shortcut” analysis techniques. – When two elements carry the same force they are said to be connected in parallel and their combined impedance is the product of the individual impedances over their sum.
- 71. Mechanical System Modeling K. Craig 71 – For impedances which have the same velocity, we say they are connected in series and their combined impedance is the sum of the individual ones. – Consider the following systems: Parallel Connection Series Connection f, v x1 , v1 B K K f, v B
- 72. Mechanical System Modeling K. Craig 72 – Parallel Connection – Series Connection ( ) K Bf KBDD Kv BD KB D = = ++ ( ) f K BD K D B v D D + = + =
- 73. Mechanical System Modeling K. Craig 73 Force and Motion Sources • The ultimate driving agency of any mechanical system is always a force not a motion; force causes acceleration, acceleration does not cause force. • Motion does not occur without a force occurring first. • At the input of a system, what is known, force or motion? If motion is known, then this motion was caused by some (perhaps unknown) force and postulating a problem with a motion input is acceptable.
- 74. Mechanical System Modeling K. Craig 74 • There are only two classes of forces: – Forces associated with physical contact between two bodies – Action-at-a-distance forces, i.e., gravitational, magnetic, and electrostatic forces. • There are no other kinds of forces! (Inertia force is a fictitious force.) • The choice of an input form to be applied to a system requires careful consideration, just as the choice of a suitable model to represent a component or system. • Here are some examples of force and motion sources.
- 75. Mechanical System Modeling K. Craig 75 Force and Motion Inputs acting on a Multistory Building
- 76. Mechanical System Modeling K. Craig 76 A Mechanical Vibration Shaker: Rotating Unbalance as a Force Input
- 77. Mechanical System Modeling K. Craig 77 Electrodynamic Vibration Shaker as a Force Source
- 78. Mechanical System Modeling K. Craig 78 Force Source Constructed from a Motion Source and a Soft Spring
- 79. Mechanical System Modeling K. Craig 79 • Energy Considerations – A system can be caused to respond only by the source supplying some energy to it; an interchange of energy must occur between source and system. – If we postulate a force source, there will be an associated motion occurring at the force input point. – The instantaneous power being transmitted through this energy port is the product of instantaneous force and velocity. – If the force applied by the source and the velocity caused by it are in the same direction, power is supplied by the source to the system. If force and velocity are opposed, the system is returning power to the source.
- 80. Mechanical System Modeling K. Craig 80 – The concept of mechanical impedance is of some help here. – The transfer function relating force and velocity at the input port of a system is called the driving-point impedance Zdp. – We can write an expression for power: dp dp f Z (D) (D) v f Z (i ) (i ) v = ω = ω 2 dp dp f f P fv f Z Z = = =
- 81. Mechanical System Modeling K. Craig 81 – If we apply a force source to a system with a high value of driving-point impedance, not much power will be taken from the source, since the force produces only a small velocity. The extreme case of this would the application of a force to a perfectly rigid wall (driving- point impedance is infinite, since no motion is produced no matter how large a force is applied). In this case the source would not supply any energy. – The higher the driving-point impedance, the more a real force source behaves like an ideal force source. – The lower the driving-point impedance, the more a real motion source behaves like an ideal motion source.
- 82. Mechanical System Modeling K. Craig 82 – Real sources may be described accurately as combinations of ideal sources and an output impedance characteristic of the physical device. – A complete description of the situation thus requires knowledge of two impedances: • The output impedance of the real source • The driving-point impedance of the driven system
- 83. Mechanical System Modeling K. Craig 83 Mechanical System Examples Problem Statement Develop the equivalent rotational model of the rack-and-pinion gear system shown. The applied torque T is the input variable, and the angular displacement θ is the output variable. Neglect any twist in the shaft. Bearings are frictionless. The pinion gear mass moment of inertia about its CG (geometric center) is Ip. ( )2 2 2 m s p rI I I m R cR kR T+ + + θ + θ + θ = Rack-and-Pinion Gear System
- 84. Mechanical System Modeling K. Craig 84 Problem Statement A load inertia I5 is driven through a double-gear pair by a motor with inertia I4, as shown. The shaft inertias are negligible. The gear inertias are I1, I2, and I3. The speed ratios are ω1/ω2 = 2 and ω2/ω3 = 5. The motor torque is T1 and the viscous damping coefficient c = 4 lb-ft-sec/rad. Neglect elasticity in the system, and use the following inertia values (sec2-ft-lb/rad): I1 = 0.1, I2 = 0.2, I3 = 0.4, I4 = 0.3, I5 = 0.7. Derive the mathematical model for the motor shaft speed ω1 with T1 as the input. ( ) ( ) 2 2 2 2 4 1 5 3 2 1 1 1 1 1 1 1 I I I I I c T 5 2 5 2 ⎧ ⎫⎡ ⎤⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + + + + ω + ω =⎨ ⎬⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭ Multi-Gear System
- 85. Mechanical System Modeling K. Craig 85 Physical System Physical Model Problem Statement A dynamic vibration absorber consists of a mass and an elastic element that is attached to another mass in order to reduce its vibration. The figure is a representation of a vibration absorber attached to the cantilever support. For a cantilever beam with a force at its end, k = Ewh3/4L3 where L = beam length, w = beam width, and h = beam thickness. (a) Obtain the equation of motion for the system. The force f is a specified force acting on the mass m, and is due to the rotating unbalance of the motor. The displacements x and x2 are measured from the static equilibrium positions when f = 0. (b) Obtain the transfer functions x/f and x2/f. ( )[ ] ( )[ ] 2 2 2 4 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 m D kx F mm D m k k mk D kk x k F mm D m k k mk D kk + = + + + + = + + + + Dynamic Vibration Absorber
- 86. Mechanical System Modeling K. Craig 86 Rigid Body Dynamics: Kinematics Reference Frames R - Ground xyz R1 - Body x1y1z1 ( )1 1 1 1 1 1 R R RR P R A R R AP R AP R R RP R P a a r r a 2 v ⎡ ⎤ ⎡ ⎤= + ω × ω × + α ×⎣ ⎦⎣ ⎦ ⎡ ⎤+ + ω ×⎣ ⎦ y z O P xR x1 y1 z1 R1 A ( )1 1R RR P R A R AP P v v r v= + ω × + Note: For any vector q 1 1 RR RRdq dq q dt dt = + ω ×
- 87. Mechanical System Modeling K. Craig 87 R R1 R2 O θ = 30º r = 0.06 m Rigid-Body Kinematics Example Given: Find: Reference Frames: R → ground: xyz R1 → shaft: x1y1z1 R2 → disk: x2y2z2 φ x1 y1 x2 y2 O z1 y z y1 O α 1 1 2 RR R R 1 ˆ5i constant ˆ4k constant ω = = ω = = R P a 1 1 1 ˆ ˆi i1 0 0 ˆ ˆj 0 cos sin j ˆ ˆ0 sin cosk k ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥= α α⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− α α⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
- 88. Mechanical System Modeling K. Craig 88 ( )2 2 2 2 2 2 R R RR P R O R R OP R OP R R RP R P a a r r a 2 v ⎡ ⎤ ⎡ ⎤= + ω × ω × + α ×⎣ ⎦⎣ ⎦ ⎡ ⎤+ + ω ×⎣ ⎦ 2 2 R O R P R P a 0 a 0 v 0 = = = Point O at end of rotating shaft fixed in R Point P fixed in R2 (disk) ( ) ( ) ( ) 2 1 1 2 2 2 1 R R R RR R 1 RR R R RR 1 R RR1 1 1 1 1 ˆ ˆ5i 4k d d ˆ ˆ5i 4k dt dt dk ˆ0 4 4 k dt ˆ ˆ ˆ4 5i k 20j ω = ω + ω = + ω ⎡ ⎤α = = +⎣ ⎦ = + = ω × = × = − ( )ˆ ˆ20 jcos ksin= − α + α ( ) ( )OP 1 1 ˆ ˆr rcos i rsin j= θ + θ
- 89. Mechanical System Modeling K. Craig 89 After Substitution and Simplification: ( ) ( ) ( )R P 1 1 1 ˆ ˆ ˆa 16rcos i 41rsin j 40rcos k= − θ + − θ + θ Alternate Solution: ( )1 1 1 1 1 1 R R RR P R O R R OP R OP R R RP R P a a r r a 2 v ⎡ ⎤ ⎡ ⎤= + ω × ω × + α ×⎣ ⎦⎣ ⎦ ⎡ ⎤+ + ω ×⎣ ⎦ 1 1 1 R O RR RR R RR a 0 ˆ5i constant d 0 dt = ω = = ω α = = ( ) ( )OP 1 1 ˆ ˆr rcos i rsin j= θ + θ
- 90. Mechanical System Modeling K. Craig 90 ( )1 1 1 2 1 2 1 2R R R R R R R RP O OP OP a a r r⎡ ⎤ ⎡ ⎤= + ω × ω × + α ×⎣ ⎦⎣ ⎦ (P is fixed in R2) ( ) 1 1 2 1 1 2 1 1 2 1 1 1 2 1 R O R R 1 R R R R R R 1 R R R RP O OP R O a 0 ˆ4k d d ˆ4k 0 dt dt v v r v 0 = ω = ω ⎡ ⎤α = = =⎣ ⎦ = + ω × = ( ) ( )OP 1 1 ˆ ˆr rcos i rsin j= θ + θ After Substitution and Simplification: ( ) ( ) ( )R P 1 1 1 ˆ ˆ ˆa 16rcos i 41rsin j 40rcos k= − θ + − θ + θ (same result)
- 91. Mechanical System Modeling K. Craig 91 Rigid Body Dynamics: Kinetics Linear Momentum Angular Momentum about point C Equations of Motion Point C: mass center of a rigid body of mass m. Reference Frames R - Ground xyz R1 - Body x1y1z1 R C L m v= y1 y z O ’ xR x1 z1 R1 A C y1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 RR x x x x y x z x RR y y x y y y z y RR z z x z y z z z H I I I H I I I H I I I ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ω ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ω⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ω⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 1 1x 1 y 1 z 1 ˆ ˆ ˆH H i H j H k= + + R R C R d v F m dt dH M dt ∑ = ∑ =
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