Read textbook CHAPTER 1.4, Apps B&D

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1 Lecture 2 Read textbook CHAPTER 1.4, Apps B&D Today: Derive EOMs & Linearization undamental equation of motion for mass-springdamper system (1DO). Linear and nonlinear system. Examples of derivation of EOMs Appendix A Equivalence of principles of conservation of mechanical energy and conservation of linear momentum. Appendix B: Linearization ork problems: Chapter 1: 5, 8, 13, 14, 15, 20, 43, 44, 56 1

2 Kinetics of 1-DO Lecture 2 ME617 mechanical systems The fundamental elements in a mechanical system and the process to set a coordinate system and derive an equation of motion. LINEARIZATION included. Luis San Andres Texas A&M University M K C 2014 Luis San Andres 2

3 A system with an elastic element (linear spring)

4 Linear elastic element (spring-like) ree length l s static deflection Applied force on spring K=d/d s deflection Notes: a) Spring element is regarded as massless b) Applied force is STATIC c) Spring reacts with a force proportional to deflection, s = -K s and stores potential energy d) K = stiffness coefficient [N/m or lbf/in] is constant Balance of static forces s =-K s s 4

5 Statics of system with elastic & mass elements

6 Linear spring + added weight ree length l s static deflection Reaction force from spring K=d/d static deflection s Notes: a) Block has weight = Mg Balance of static forces b) Block is regarded as a point mass =s = K s s 6

7 Dynamics of system with elastic & mass elements Derive the equation of motion (EOM) for the system

8 Linear spring + weight (mass x g ) + force (t) (t) Reaction force from spring ree length l + (t) s SEP s+x(t) + K=d/d dynamic deflection X+s Notes: a) Coordinate X describing motion has origin at Static Equilibrium Position (SEP) b) or free body diagram, assume state of motion, for example X(t)>0 c) Then, state Newton s equation of motion d) Assume no lateral (side motions) ree Body diagram acceleration M Ax = + - spring + M spring 8

9 Derive the equation of motion ree length l + s SEP X(t) s+x(t) M Ax = + - spring spring = K(X+s ) M Ax = + K(X+s) BD: X>0 + M spring M Ax = ( -Ks) + KX Cancel terms from force balance at SEP to get M Ax = + KX M Ax + K X= (t) Note: Motions from SEP A X 2 d X dt 2 X 9

10 Another choice of coordinate system + Coordinate y has origin at free length of spring element. ree length l BD: y>0 + y(t) M Ay = + - spring spring = Ky M Ay = + K y eight (static force) remains in equation M spring M Ay + K y= + (t) A y 2 d y 2 dt y 10

11 EOM in two coordinate systems + M Ay + K y = + (t) (1) ree length l y(t) s X(t) SEP A y 2 d y y 2 dt M Ax + K X= (t) (2) y = X+ s y X spring = Ky = K(X+ s ) Coordinate y has origin at free length of spring element. X has origin at SEP. A X 2 d X 2 dt X (?) Are Eqs. (1) and (2) the same? (?) Do Eqs. (1) and (2) represent the same system? (?) Does the weight disappear? (?) Can I just cancel the weight? 11

12 K-M system with dissipative element (a viscous dashpot) Derive the equation of motion (EOM) for the system

13 Recall: a linear elastic element ree length l s s s static deflection s Applied force on spring K=d/d s deflection Notes: a) Spring element is regarded as massless b) K = stiffness coefficient is constant c) Spring reacts with a force proportional to deflection and stores potential energy 13

14 (Linear) viscous damper element V velocity Applied force on damper D=d/dV velocity V Notes: a) Dashpot element (damper) is regarded as massless b) D = damping coefficient [N/(m/s) or lbf/(in/s) ] is constant c) A damper needs velocity to work; otherwise it can not dissipate mechanical energy d) Under static conditions (no motion), a damper does NOT react with a force 14

15 Spring + Damper + Added weight ree length l static deflection s SEP Reaction force from spring K=d/d s static deflection Notes: a) Block has weight = Mg and is regarded as a point mass. b) eight applied very slowly static condition. c) Damper does NOT react to a static action, i.e. D =0 SEP: static equilibrium position Balance of static forces =s = K s s 15

16 Spring + Damper + eight + force (t) (t) Reaction force from spring L-s + s X(t) SEP spring K=d/d dynamic deflection X+s Notes: a) Coordinate X describes motion from Static Equilibrium Position (SEP) b) or free body diagram, assume a state of motion X(t)>0 c) State Newton s equation of motion d) Assumed no lateral (side motions) ree Body diagram M spring acceleration + damper M Ax = + - damper -spring 16

17 Equation of motion: spring-damper-mass + M Ax = + (t) - damper -spring L-s K D X(t) SEP A V X X 2 d X dt 2 X dx X dt damper = D VX spring = K(X+ s ) M Ax = + K(X+s) D VX BD: X>0 + M M Ax = ( -Ks) + KX - DVX Cancel terms from force balance at SEP to get M Ax = + KX D VX spring damper Notes: Motions from SEP M Ax + DVX + K X = (t) EOM: M X DX K X t () 17

18 Simple nonlinear mechanical system Derive the equation of motion (EOM) for the system and linearize EOM about SEP

19 Recall a linear spring element ree length l s s s static deflection s Applied force on spring K=d/d s deflection Notes: a) Spring element is regarded as massless b) K = stiffness coefficient is constant c) Spring reacts with a force proportional to deflection and stores potential energy 19

20 A nonlinear spring element ree length l s static deflection Applied force on spring s K K 1 3 static deflection 3 Notes: a) Spring element is massless b) orce vs. deflection curve is NON linear c) K1 and K3 are material parameters [N/m, N/m 3 ]

21 Linear spring + added weight ree length l s static deflection Reaction force from spring s static deflection Notes: a) Block has weight = Mg b) Block is regarded as a point mass c) Static deflection s found from solving nonlinear equation: K K Balance of static forces 3 spring 1 s 3 s spring

22 Nonlinear spring + weight + force (t) ree length l (t) + SEP s+x(t) +s Reaction force from spring deflection X+s Notes: a) Coordinate X describing motion has origin at Static Equilibrium Position (SEP) b) or free body diagram, assume state of motion, for example X(t)>0 c) Then, state Newton s equation of motion d) Assume no lateral (side motions) ree Body diagram acceleration M Ax = + - spring + M spring

23 Nonlinear Equation of motion ree length l + SEP s+x(t) A MX () t spring 2 d X X X 2 dt 3 K X K X spring 1 s 3 s 3 M X K X K X () t 1 s 3 s BD: X>0 + M spring Expand RHS: M X () t K1 s X K3 s 3 s X 3 s X X M X K K K K X K X X () t 1 s 3 s 1 3 s 3 s Cancel terms from force balance at SEP to get M X K K X K X X s 3 s ( t) Notes: Motions are from SEP

24 Linearize equation of motion SEP ree length l + EOM is nonlinear: M X K K X K X X s+x(t) s 3 s ( t) Assume motions X(t) << s which means (t) << BD: X>0 + M spring to obtain the linear EOM: M X K X where the linearized stiffness is a function of the static condition e d K and set X 2 ~0, X 3 ~0 K () t K K 3K e s dspring Ke K13K3 s d d 2 s

25 Linear equation of motion SEP ree length l BD: X>0 + M spring + s+x(t) Stiffness (linear) Ke e () t 2 K e K 13K3 s M X K X s d K K 3 dspring Ke K13K3 s d deflection s d X s deflection

26 Read & rework Examples of derivation of EOMS for physical systems available on class URL site(s) 2014 Luis San Andres