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Feedback Control Systems (FCS) Lecture-6-7-8 Mathematical Modelling of Mechanical Systems Dr. Imtiaz Hussain email: [email protected] URL :http://imtiazhussainkalwar.weebly.com/ 1 Outline of this Lecture • Part-I: Translational Mechanical System • Part-II: Rotational Mechanical System • Part-III: Mechanical Linkages 2 Basic Types of Mechanical Systems • Translational – Linear Motion • Rotational – Rotational Motion 3 Part-I TRANSLATIONAL MECHANICAL SYSTEMS 4 Basic Elements of Translational Mechanical Systems Translational Spring i) Translational Mass ii) Translational Damper iii) Translational Spring • A translational spring is a mechanical element that can be deformed by an external force such that the deformation is directly proportional to the force applied to it. Translational Spring i) Circuit Symbols Translational Spring Translational Spring • If F is the applied force x2 x1 • Then x1 is the deformation if x2 0 • Or ( x1 x2 ) is the deformation. • The equation of motion is given as F k ( x1 x2 ) • Where k is stiffness of spring expressed in N/m F F Translational Spring • Given two springs with spring constant k1 and k2, obtain the equivalent spring constant keq for the two springs connected in: (1) Parallel (2) Series 8 Translational Spring • The two springs have same displacement therefore: k1 x k 2 x F (1) Parallel ( k1 k 2 ) x F keq x F keq k1 k2 • If n springs are connected in parallel then: keq k1 k 2 k n 9 Translational Spring • The forces on two springs are same, F, however displacements are different therefore: k1 x1 k 2 x2 F F x1 k1 (2) Series F x2 k2 • Since the total displacement is x x1 x2 , and we have F keq x F F F x x1 x2 k eq k1 k 2 10 Translational Spring F F F k eq k1 k 2 • Then we can obtain k eq 1 k1k 2 1 1 k1 k 2 k1 k 2 • If n springs are connected in series then: k eq k1k 2 k n k1 k 2 k n 11 Translational Spring • Exercise: Obtain the equivalent stiffness for the following spring networks. i) k3 ii) k3 12 Translational Mass • Translational Mass is an inertia element. Translational Mass ii) • A mechanical system without mass does not exist. • If a force F is applied to a mass and it is displaced to x meters then the relation b/w force and displacements is given by Newton’s law. F Mx x(t ) F (t ) M Translational Damper • When the viscosity or drag is not negligible in a system, we often model them with the damping force. • All the materials exhibit the property of damping to some extent. • If damping in the system is not enough then extra elements (e.g. Dashpot) are added to increase damping. Translational Damper iii) Common Uses of Dashpots Door Stoppers Bridge Suspension Vehicle Suspension Flyover Suspension Translational Damper F Cx F C( x1 x 2 ) • Where C is damping coefficient (N/ms-1). Translational Damper • Translational Dampers in series and parallel. Ceq C1 C2 C eq C1C 2 C1 C 2 Modelling a simple Translational System • Example-1: Consider a simple horizontal spring-mass system on a frictionless surface, as shown in figure below. or mx kx mx kx 0 18 Example-2 • Consider the following system (friction is negligible) k x M F • Free Body Diagram fk F M fM • Where f k and f M are force applied by the spring and inertial force respectively. 19 Example-2 fk F M fM F fk fM • Then the differential equation of the system is: F Mx kx • Taking the Laplace Transform of both sides and ignoring initial conditions we get F ( s ) Ms2 X ( s ) kX ( s ) 20 Example-2 F ( s ) Ms2 X ( s ) kX ( s ) • The transfer function of the system is X (s) 1 F(s) Ms2 k • if M 1000kg k 2000Nm 1 X (s) 0.001 2 F(s) s 2 21 Example-2 X (s) 0.001 2 F(s) s 2 • The pole-zero map of the system is Pole-Zero Map 40 30 Imaginary Axis 20 10 0 -10 -20 -30 -40 -1 -0.5 0 Real Axis 0.5 1 22 Example-3 • Consider the following system k F x M C • Free Body Diagram fk F M fC fM F f k f M fC 23 Example-3 Differential equation of the system is: F Mx Cx kx Taking the Laplace Transform of both sides and ignoring Initial conditions we get F ( s ) Ms2 X ( s ) CsX ( s ) kX ( s ) X ( s) 1 F ( s ) Ms2 Cs k 24 Example-3 X ( s) 1 F ( s ) Ms2 Cs k • if Pole-Zero Map 2 1.5 M 1000kg k 2000Nm 1 C 1000N / ms 1 Imaginary Axis 1 0.5 0 -0.5 -1 X ( s) 0.001 2 F ( s ) s s 1000 -1.5 -2 -1 -0.5 0 0.5 1 Real Axis 25 Example-4 • Consider the following system • Free Body Diagram (same as example-3) fk F M fB fM F fk fM fB X ( s) 1 F ( s ) Ms2 Bs k 26 Example-5 • Consider the following system x2 x1 k B F M • Mechanical Network x1 F ↑ k x2 M B 27 Example-5 • Mechanical Network x1 F At node ↑ k x2 M B x1 F k ( x1 x2 ) At node x2 2 0 k ( x2 x1 ) Mx2 Bx 28 Example-6 • Find the transfer function X2(s)/F(s) of the following system. k M1 B M2 Example-7 x1 x2 B3 k f (t ) B4 M1 M2 B1 B3 x1 f (t ) ↑ k M1 B2 B1 B2 x2 M2 B4 30 Example-8 • Find the transfer function of the mechanical translational system given in Figure-1. Free Body Diagram fk fB Figure-1 M f (t ) f (t ) f k f M f B fM X ( s) 1 F ( s ) Ms2 Bs k 31 Example-9 • Restaurant plate dispenser 32 Example-10 • Find the transfer function X2(s)/F(s) of the following system. Free Body Diagram f k1 f k f B 2 M2 k2 F (t ) f M 2 f k1 fB M1 f M1 F(t ) f k1 f k2 f M 2 f B 0 f k1 f M1 f B 33 Example-11 x1 u (t ) x2 k1 B1 x3 B4 B3 M1 k2 B2 M2 k3 B5 34 Example-12: Automobile Suspension 35 Automobile Suspension 36 Automobile Suspension mxo b( xo xi ) k ( xo xi ) 0 (eq .1) mxo bxo kxo bxi kxi eq. 2 Taking Laplace Transform of the equation (2) 2 ms X o ( s ) bsXo ( s ) kX o ( s ) bsXi ( s ) kX i ( s ) X o (s) bs k X i ( s ) ms2 bs k 37 Example-13: Train Suspension Car Body Bogie-2 Bogie-1 Secondary Suspension Wheelsets Primary Bogie Frame Suspension 38 Example: Train Suspension 39 Part-I ROTATIONAL MECHANICAL SYSTEMS 40 Basic Elements of Rotational Mechanical Systems Rotational Spring 2 1 T k (1 2 ) Basic Elements of Rotational Mechanical Systems Rotational Damper C 2 1 T C(1 2 ) T Basic Elements of Rotational Mechanical Systems Moment of Inertia J T J T Example-1 1 k1 B1 2 3 T J1 1 T ↑ k1 2 J1 k2 J2 B1 3 J2 k2 Example-2 1 k1 B2 2 T 3 J2 J1 B3 B1 1 T ↑ k1 2 J1 B4 B2 B1 B3 3 J2 B4 Example-3 1 k1 J1 B2 2 T J2 k2 Example-4 Part-III MECHANICAL LINKAGES 48 Gear • Gear is a toothed machine part, such as a wheel or cylinder, that meshes with another toothed part to transmit motion or to change speed or direction. 49 Fundamental Properties • The two gears turn in opposite directions: one clockwise and the other counterclockwise. • Two gears revolve at different speeds when number of teeth on each gear are different. Gearing Up and Down • Gearing up is able to convert torque to velocity. • The more velocity gained, the more torque sacrifice. • The ratio is exactly the same: if you get three times your original angular velocity, you reduce the resulting torque to one third. • This conversion is symmetric: we can also convert velocity to torque at the same ratio. • The price of the conversion is power loss due to friction. Why Gearing is necessary? • A typical DC motor operates at speeds that are far too high to be useful, and at torques that are far too low. • Gear reduction is the standard method by which a motor is made useful. 52 Gear Trains 53 Gear Ratio • You can calculate the gear ratio by using the number of teeth of the driver divided by the number of teeth of the follower. • We gear up when we increase velocity and decrease torque. Ratio: 3:1 Driver Follower • We gear down when we increase torque and reduce velocity. Ratio: 1:3 Gear Ratio = # teeth input gear / # teeth output gear = torque in / torque out = speed out / speed in Example of Gear Trains • A most commonly used example of gear trains is the gears of an automobile. 55 Mathematical Modelling of Gear Trains • Gears increase or reduce angular velocity (while simultaneously decreasing or increasing torque, such that energy is conserved). Energy of Driving Gear = Energy of Following Gear N11 N1 1 N2 2 N 2 2 Number of Teeth of Driving Gear Angular Movement of Driving Gear Number of Teeth of Following Gear Angular Movement of Following Gear 56 Mathematical Modelling of Gear Trains • In the system below, a torque, τa, is applied to gear 1 (with number of teeth N1, moment of inertia J1 and a rotational friction B1). • It, in turn, is connected to gear 2 (with number of teeth N2, moment of inertia J2 and a rotational friction B2). • The angle θ1 is defined positive clockwise, θ2 is defined positive clockwise. The torque acts in the direction of θ1. • Assume that TL is the load torque applied by the load connected to Gear-2. N1 N2 B1 B2 57 Mathematical Modelling of Gear Trains • For Gear-1 a J11 B11 T1 Eq (1) • For Gear-2 T2 J 22 B22 TL Eq (2) N1 N2 B1 • Since B2 N11 N 2 2 • therefore N1 2 1 N2 Eq (3) 58 Mathematical Modelling of Gear Trains • Gear Ratio is calculated as T2 N2 N1 T1 T2 T1 N1 N2 N2 N1 • Put this value in eq (1) B1 N a J 11 B11 1 T2 N2 B2 • Put T2 from eq (2) N1 a J 11 B11 ( J 22 B22 TL ) N2 • Substitute θ2 from eq (3) N1 N1 N1 N1 a J 11 B11 (J2 1 B2 2 TL ) N2 N2 N2 N 2 59 Mathematical Modelling of Gear Trains N1 N1 N1 N1 a J 11 B11 (J2 1 B2 2 TL ) N2 N2 N2 N2 • After simplification 2 2 N1 N1 N J 21 B11 B21 1 TL a J11 N2 N2 N2 2 2 N1 N N1 1 J 2 1 B1 B2 1 a J1 TL N2 N2 N2 2 J eq N J1 1 J 2 N2 Beq N B1 1 N2 N1 a J eq1 Beq1 TL N2 2 B2 60 Mathematical Modelling of Gear Trains • For three gears connected together J eq Beq N1 J1 N2 N1 B1 N2 2 2 2 2 N1 J 2 N2 N1 B2 N2 N3 N4 N3 N4 2 J 3 2 B3 61 Home Work • Drive Jeq and Beq and relation between applied torque τa and load torque TL for three gears connected together. 1 2 3 N1 N2 J1 τa J2 B2 B1 N3 J3 TL B3 62 To download this lecture visit http://imtiazhussainkalwar.weebly.com/ END OF LECTURES-6-7-8 63