Scale physical model Mathematical model Numerical model How to Model 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. 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. Modelling of Physical Systems

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. engineering/research/dbl-delft-biorobotics-lab/bipedal-robots/

Lumped Parameter Models Millennium wobbly bridge

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

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 Mechanical Systems Model using a lumped element model Linear, ordinary differential equations describe system Three elements to describe properties of a system Each element completely describe a property No spatial variation in properties Elements are ideal; linear and represent only one property n7064/fig_tab/438043a_F1.html

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

Hooke’s Law: f(t) = K(x 1 – x 2 ) = Kx(t) f(t) = force applied to the ends of the spring, K = spring constant (N.m) or C = 1/K is the spring compliance, x 1 = Displacement of the one end, x 2 = Displacement of the other end, x = x 1 – x 2 = 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 x 0 from the equilibrium position is: The Spring Element

Real springs show deviation from ideal: 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. The Spring Element

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

Example: Consider f(x) = 2x + 5x 3 f(x) = f(2 (x o )+5(x o ) 3 ) + f’( x o 2 ).(x-1) f(x) = f(2 (1)+5(1) 3 ) + ( )(x-1) Linearise: Intuitively with tangent Or Taylor series: A finite number of terms will give an approximation of the function – the first two terms will give a linear approximation Example: Consider f(x) = 2x + 5x 3 Operating point x = 1: f(x) ~ x Model the spring force as f(x) = x around the point x = 1.0 and the linearised spring forced constant would be given by df/dx = 17 N/m. Linearisation

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.

wheref = force applied to the ends of the damper, dx/dt = v = relative velocity of the plunger f v = coefficient of viscous friction or simply called the damping coefficient The Damper Element

For a rotational damper: where T = applied torque D = Rotational damping coefficient  = relative angular displacement  = angular velocity. A pure damper would be dissipated all energy into heat: Energy dissipated = Dv 2 The Damper Element

Real dampers Desired: shock absorber in a suspension system Parasitic: Air friction. Classic intentional dampers: piston/cylinder filled with a viscous fluid (oil) The damping coefficient will be determined by the geometry of the system and the viscosity of the liquid. Rotational damperAir filled damper Eddy current damping: Any movement of the copper cup in the magnetic field would induce eddy currents to flow in the cup that would induce a force that opposes the original motion of the cup. Small effect, but useful in instrumentation

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.

The three solutions stiffening the structure, dampening structural vibration, limiting/modifying bridge traffic Estimates place the cost of the stabilization work in the neighborhood of £5 million. (Source: Millennium Bridge London

Car suspension model: Mechanical system Explain what happens when a car goes over a bump?

Car suspension model: Explain what happens when a car goes over a bump? x f input Simplify to single-input- single-output system Form individual component models Determine their relationships (use physical laws!) Combine (and simplify if possible) This gives us an instantaneous differential equation, but want a time response!

Spring Damper Mass Source Nise 2004 Force - Distance Car suspension model: This gives us an instantaneous differential equation, but want a time response! Integrate: numerically theoretically using tables

clf; %clear all graphs K = 10 %Spring constant C = 3 %Damping constant m = 1 %mass (constant) t = [0: 0.01: 20];%set up the time increments stept = 1 + 0*t; %graph to show step response plot(t,stept,'m'); xlabel('Time t (s)') ylabel('Distance x (m)') hold on % put each graph on top of each other for C = 1.0: 1: 10.0 d = tf(9,[m C K]) [y,t]=step(d,T);%step response over one second plot(t,y,'k'); pause(2) end