#### Transcript Document

Terms Test 1 - Wednesday, 13/08/2014, 5pm • MCLT101 and 103 • Event Name: ENGR110 Test • Event Type: Terms test • Date(s): Wednesday, 13/08/2014 • Time: 17:10-18:00 • Will alcohol be served? No • Will the VC Attend? No • Other Information: Co345 • FSMs • Physical Modelling • Worth 10% of course mark • 50 Minute duration • Arrive at 5:00 for 5:10 start • No notes allowed • Calculators allowed Physical Modelling: 1. 2. 3. 4. 5. 6. Out there world inside here Modelling Practical example: Passive Dynamic Walkers Base systems and concepts Ideals, assumptions and real life Similarities in systems and responses F1 F2 Design Cycle: https://stillwater.sharepoint.okstate.edu/ENGR1113/default.aspx How to Model Physical Systems • Scale physical model • Mathematical model • Numerical model www.autospeed.com Modelling of Physical Systems • Develop mathematical models, i.e. ordinary differential equations, that describe the relationship between input and output characteristics of a system. • These equations can then be used to forecast the behaviour of the system under specific conditions. • All systems can normally be approximated and modelled by one of several models, e.g.: mechanical, electrical, thermal or fluid. We also find that we can translate a system from one model to another to facilitate the modelling. Lumped Parameter Models • Use standard laws of physics and break a system down into a number of building blocks. • Each of the parameters (property or function) is considered independently. www.brains-minds-media.org http://www.3me.tudelft.nl/en/about-the-faculty/departments/biomechanicalengineering/research/dbl-delft-biorobotics-lab/bipedal-robots/ http://www.cyberphysics.co.uk/ Linear Time Invariant Models • Assume the property of linearity for these models. • A linear system will posses two properties; 1. Superposition 2. Homogeneity. Allows us to use standard mathematical operations to simplify our models www.redlinemotive.com Linear Time Invariant Models • Assume system is time-invariant • Constants stay constant in the time-scales of our model • Proportionality between variables does not change. Our shock absorbers do not wear in our car suspension model! Competition tyres Elements of Systems are Ideal • Each element completely describes a property • Elements are: • Ideal • Linear • Represent only one property The Spring Element Represents the elastic properties (energy storage) in a system. Assume: • No mass • No Fraction • Linear The Spring Element Hooke’s Law: f(t) = K= x1 = f(t) = K(x1 – x2) = Kx(t) force applied to the ends of the spring, spring constant (N.m) or C = 1/K is the spring compliance, Displacement of the one end, x2 = Displacement of the other end, x = x1 – x2 = relative displacement of the two ends. Rotational spring element torque T(t) in terms of the angular displacement : T(t) = K(1 - 2) = K(t) T(t) = Torque (t) = angular displacement The Spring Element Work done for a force, f, to move a spring a distance, dx: W = f . dx The energy then stored in the spring when the ends are displaced a distance x0 from the equilibrium position is: Es xo 0 Kx dx 1 2 Kx 2 0 1 2 Cf 2 0 The Spring Element Real springs show deviation from ideality: • Real springs would have an associated mass, leading to the deviations in the step and sinusoidal response as described before. • Real springs will always contain some friction (energy dissipation) which is exhibited in the non-coincidence of the loading – unloading curves. • Real springs will always exhibit a degree of non-linearity (deviation from f = Kx). See section on linearisation. Energy losses in real springs Several types of practical springs marccardaronella.com Linearisation Many elements show non-linear behaviour. To use in our relatively simple models, we must first linearise this element around an operating point. Get a linear model that should be valid for small excursions from this operating point. Consider a non-linear coil spring that forms part of the suspension system of an automobile. Operating point = Equilibrium position under load The Damper Element Damper element (dashpot) represents the forces opposing motion, i.e. the friction or damping effects. Damping force is proportional to the velocity of the piston. The Damper Element dx 2 dx dx 1 f fv fvv fv dt dt dt http://www.flotronicpumps.co.uk/ where f = force applied to the ends of the damper, dx/dt = v = relative velocity of the plunger fv = coefficient of viscous friction or simply called the damping coefficient www.modified.com The Mass (inertia) Element All real elements will have some mass, so that the mass element in the model will thus represent the physical mass of the system. This element represents the inertia or resistance to acceleration of the system. The mass element will be modelled as concentrated at a point. Car suspension model: • Mechanical system Explain what happens when a car goes over a bump? www.superbike-coach.com Car suspension model: Explain what happens when a car goes over a bump? Simplify to single-inputsingle-output system Form individual component models Determine their relationships (use physical laws!) Combine (and simplify if possible) x This gives us an instantaneous differential equation, but want a time response! finput Car suspension model: Force - Distance Spring x t f f s t Kx t t Damper x t f Mass f c t C t dx t f input t f s t f d t f m t f input t Kx t C dx t dt d x t 2 m dt 2 dt This gives us an instantaneous differential equation, but want a time response! x t f t Source Nise 2004 f m t M d x t 2 dt 2 Integrate: • numerically • theoretically • using tables 9 8 clf; %clear all graphs 7 K = 10 %Spring constant m = 1 %mass (constant) Distance x (m) C = 3 %Damping constant 6 5 4 3 t = [0: 0.01: 20];%set up the time increments 2 1 stept = 1 + 0*t; %graph to show step response plot(t,stept,'m'); 1.5 0 0 2 4 6 8 10 12 Time t (s) 14 16 18 20 xlabel('Time t (s)') ylabel('Distance x (m)') hold on % put each graph on top of each other Distance x (m) 1 0.5 for C = 1.0: 1: 10.0 d = tf(9,[m C K]) 1.5 0 0 2 4 6 8 10 12 Time t (s) 14 16 18 20 2 4 [y,t]=step(d,T);%step response over one second pause(2) end 1 Distance x (m) plot(t,y,'k'); 0.5 0 0 6 8 10 12 Time t (s) 14 16 18 20