Modelling of Physical Systems 3. Mechanical Systems

Transcription

1 Karadeniz Technical University Department of Electrical and Electronics Engineering Trabzon, Turkey Chapter Modelling of Physical Systems 3. Bu ders notları sadece bu dersi alan öğrencilerin kullanımına açık olup, üçüncü sahıslara verilmesi, herhangi bir yöntemle çoğaltılıp başka yerlerde kullanılması, yayınlanması Prof. Dr. İsmail H. ALTAŞ ın yazılı iznine tabidir. Aksi durumlarda yasal işlem yapılacaktır. Chapter 3-3-2

2 ÖDEV 1 Değerlendirmeler dersin şifreli web sayfasına atıldı. ÖDEV 2 Sorular dersin şifresiz web sayfasına yüklendi. Ödevler alt başlığından gidilerek indirilebilir. Chapter f f f x M x B x force/velocity force/position Mass f = M dv/dt f = M dx 2 /dt 2 Viscous friction f = B v f = B dx/dt Spring f = k v dt f = k x Chapter 3-3-4

3 J torque/velocity torque/position T, θ T, θ T, θ s B Inertia T = J dω/dt T = J dθ 2 /dt 2 Viscous friction T = B ω T = B dθ/dt Stiffness T = s ωdt T = s θ Chapter i 1 v 1 N 1 N 2 i 2 v 2 f 2, x 2 Transformer v 1 N 1 i 1 N 2 = = v 2 N 2 i 2 N 1 f 1, x 1 L 1 L 2 Lever f 1 L 2 x 1 L 1 = = f 2 L 1 x 2 L 2 T 2, θ 2 N 1 T 1, θ 1 N 2 Gears T 1 N 1 θ 1 N 2 = = T 2 N 2 θ 2 N 1 Chapter 3-3-6

4 Give a mathematical model of the system with f as input and x as output f 0 k x Invoke Newton s second law which states The vector sum of all external forces acting on a rigid body must be equal to the rate of change of momentum, P. m dp =f dt 2 dx m =f 2 dt p=mx& mx=f && Chapter mx=f && f = the vector sum of all external forces acting on the rigid body. f= applied force + force of gravity - resisting spring force f 0 m k x f=f 0+mg-kx mx=f && 0+mg-kx mx=f && 0-kx+mg We can obtain a theoretical system model with f o as input, x as output, x and (dx/dt) as internal variables and mg as the disturbance. mg Chapter 3-3-8

5 mx=f && -kx+mg 0 mg f0 mx&& x&& x& 1 m x k Note: It is important to indicate arrows to denote input and output of black boxes and signs on the summing junction. Note that the system exhibits feedback. Chapter Reminder In SI units, a force of one Newton will impart an acceleration of 1m/sec 2 to a mass of 1 kg. The weight of a mass is mg where g is the acceleration due to gravity ( 9.81 m/sec 2 ). Chapter

6 Comments on example 3.1: A set of differential equations has: inputs, outputs, disturbances and internal variables.. The internal variables are the variables in terms of which the differential equation is expressed. input, outputs or disturbances are not the internal variables In this example, the internal variables are: && x and x& Chapter We can also obtain the block diagram in terms of Laplace transform variables Taking the Laplace transform, the equation (3) becomes (assuming zero initial conditions) mx=f && 0-kx+mg mg 2 s mx(s)=f 0(s)-kX(s)+ s L{ mg } = 1 s mg=constant 1 f(s)= 0 s If the input force f 0 is a constant Chapter

7 mg 1 s mg 2 s mx(s)=f 0(s)-kX(s)+ s 1 2 s 1 1 s s F(s) ms k X(s) Chapter mg mg 2 s mx(s)=f 0(s)-kX(s)+ s 1 s F(s) 0 1 m 1 s 1 s X(s) k Chapter

8 Comments on example 3.1: Block diagrams and the mathematical model of the system are intimately connected in this manner; a block diagram is an elegantly visual way of describing differential equations, seeing, or visualizing a problem makes it easier to understand. Block diagrams show how components of a system are connected together physically without the complication of shafts, gears, dampers, springs etc... The block diagram model involving only integrators, summers and coefficient multipliers is called analog computer representation, since they are in a form suitable for analog computer simulation. Simulation diagrams Chapter General Comments Mechanical systems and electrical systems are analogous. Both have three passive linear components: L <=> M (energy storing elements) 1/C <=> k (energy storage elements) R <=> B (energy dissipating elements) Chapter

9 General Comments Steps required to obtain a mathematical model of a mechanical system. 1. Label internal variables. 2. Consider each mass. 3. For each mass identify the cause or force which causes the motion 4. Choose reference direction of motion. 5. The spring and damping elements oppose the motion. 6. Use Newton s law with acceleration term to the left and all external forces to the right. Chapter General Comments Let x i be the motion of a body m i. Let k i, and b i be the coefficient of the spring and coefficient of damping of the spring and damper attached to the mass m i. x j k i m i b i x i The opposing spring force: ki( xi-xj) The opposing damping force: m j b ( x-x & & ) i i j Chapter

10 General Comments Many mechanical systems are analogous to multiloop and multi-node electrical circuits. They are modelled by a set of simultaneous differential equations and a set of algebraic equations. In mechanical systems, the number of equations of motion (obtained by invoking for example Newton s law of motion for each inertia element) required is equal to the number of linearly independent motions. Chapter General Comments Linearly independent motion implies that a point of motion in a system can still move if all other points of motions are held fixed. A motion is said to be dependent if it can be expressed as a linear combination of the other. Consider a motion θ 1. If it is related to another motion θ 2 as θ 1 = aθ 2 where a is some constant, then θ 1 is said to be dependent on θ 2. A set of motions θ 1, θ 2, θ 3... θ r are linearly dependent if there exists scalars a 1, a 2,..a r such that a1θ 1+a2θ 2+a3θ arθ r=0 Chapter

11 General Comments Usually the number of linearly independent motions is called the number of degrees of freedom. Note the degrees of freedom does not preclude motions which are coupled. For example in a multiloop circuit, each loop current may depend upon other loop currents, but if we open circuit one loop, the other currents will still exist if there is voltage source in that loop. Similarly in a mechanical system with multi-degrees of freedom, one point of motion can be held while other point of motion moves under an applied force. Chapter General Comments The Newton s law of motion is applied to each inertia element as long as the motion associated with the inertia element is independent. In other words, the Newton s law of motion is not applied to an inertia element if its motion is linearly dependent on the motion of other inertia elements. Chapter

12 General Comments We may obtain the equation of motion by using the principle of superposition. Consider each mass and obtain its equation of motion using the Newton s law by assuming all other motions are fixed. If we consider the mass m i then obtain the equation of its motion by assuming that the motion associated with the mass m j is fixed: move m i holding m j still. Similarly for the mass m j obtain the equation of its motion assuming that the motion associated with the mass m i is fixed: move m j holding m i still. Chapter Sign Assignment The assigning of the sign of the forces acting on the mass. I) Assume that positive direction is vertically down Chapter

13 Sign Assignment The assigning of the sign of the forces acting on the mass. ii) Assume that positive direction is vertically up Note that the forces due to the spring and the friction are always kx and b(dx/dt) independent of the assumed positive direction. Chapter Sign Assignment The assigning of the sign of the forces acting on the mass. It is important to distinguish between two types of inputs: one class of input, denoted input, may be manipulated, the other class of input, termed disturbances cannot be manipulated. One has to live with it and suffer the consequence of its action on the system, e.g. gravity force, mg, noise, etc. Chapter

14 Obtain a system theoretic model with f 1 and f 2 as inputs, and x 2 as an output. Note: the hashed regions \\\\\ indicates that they are all at the same fixed reference point similar to the ground connection point in electrical circuits. Chapter Consider the mass m 1 Newton s law of motion (mass)(acceleration) = vector sum of all the forces Chapter

15 Consider the mass m 2 f 2 The above system has two-degrees of freedom as it has two independent motions. Chapter Comment: The block diagram representation is not unique. We will illustrate this considering a mass-spring-damper system with f as input and x as output given by : Chapter

16 1 The following two representations are immediate: Treat (dx/dt) and (d 2 x/dt 2 )as internal variables. Divide the differential equation by m and place the highest derivative (d 2 x/dt 2 ) to the left and all the others to the right. Chapter Treat (dx/dt) and (d 2 x/dt 2 ) as internal variables. Place the highest derivative term, m(d 2 x/dt 2 ), to the left and the rest to the right of the differential equation. Chapter

17 3 The other representation can be obtained from the Laplace transform model of the system. Assuming for simplicity, zero initial conditions, the Laplace transform model becomes: Different representations can be obtained treating terms involving polynomials in s as different entities ( rather than s 2 X(s), sx(s),... as entities) rewriting the equation as : Chapter where the highest order polynomial is on the left-hand side and the rest on the right. Chapter

18 4 Re-writing the equation as : The model is : The block diagram model expressed in terms of integrator blocks can also be expressed in terms of a Laplace transform based model. Chapter Consider the block diagram representation using Chapter

19 Consider the block representation Using We have so far considered the motion of mass in either vertical or horizontal direction. Let us extend the result to motion along a trajectory in a x-y plane Chapter Comment Consider a mass m i which is connected by a spring k i and damper b i to a mass m j. Then the motion of mass m i is opposed by the spring k i and damper b i. This is represented mathematically by assigning negative signs to the resisting spring force k i (x i -x j ) and b x-x & & the damping force: i( i j) ( ) ( ) m && x =... -k x -x -b x& -x& i i 1 i j i i j ( x ) bi( x& i x& j) k x i i j The inertia force mx&& i i representing motion is opposed by spring force k i (x i -x j ) and damping force b ( x-x & & ) i i j m i mx&& i i Chapter

20 günü ders yapamıyorum. Dolayısıyla; 1. Ya yarın 2 gryp birlikte 4 saat, 2. Ya da em yarın, hem de 29 nisanda 2 grup birlikte 2 şer saat birlikte, (26 evet, 8 H) 3. gelecek Cumartesi saat 8-10 ve aresı ayrı ayrı 2 gruba ders yapabilirim. (8 E, 38 H) 4. Hiç telafi yapmayalım. Chapter Rotational systems 15 Nisan 2008 Ders sonu Chapter

21 Rotational systems Obtain the system theoretic model of the following with T as input and θ as output. Using Newton s law of motion, yields Where is mg? Chapter Rotational systems Obtain a block diagram representation of the following with T as input and θ 2 as output. Consider the inertia element J 1, Consider the inertia element J 2 Chapter

22 Rotational systems Chapter Rotational systems Obtain the differential equation and the mechanical system model from the block diagram of the previous example Solution Consider the subsystem 1 Consider the subsystem 2 The mechanical system is given in the previous example. Chapter

23 Rotational systems x 4 x 3 x 2 x 1 x& = x x& = k x b x k x +k x ( ) J2 x& = x x= & kx-kx-bx T ( + ) J1 Chapter Rotational systems System with no inertia elements. Obtain system theoretical model of the above. Solution: This system is composed only of spring and damper. It has no inertia element. We could use the same procedure followed in the previous example by assuming that there is a inertia element J or the inertia is zero. Chapter

24 Rotational systems Consider the inertia element J, Newton s law of motion becomes Express the equation with the highest derivative term on the left hand-side. Chapter Rotational systems Chapter