EE/ME/AE324: Dynamical Systems Chapter 2: Modeling Translational Mechanical Systems
Common Variables Used Assumes 1 DoF per mass, i.e., all motion scalar Displacement: x ()[ t [m] Velocity: dx() t vt () [m/s] dt 2 Acceleration: dv() t d x() t 2 at () [m/s] 2 dt dt dmtvt dvt kg m f () t M Ma()[Nor t ] 2 dt dt s Force: ( ( ) ( )) ( ) t Energy (work): wt () wt ( 0) p( ) d J or N m Power: dw () t pt () f() tvt ()[W or J/s] dt t 0
Variable Conventions Position, velocity and force: Supplied power ( pt ( ) 0):
Mass: M [kg] Element Laws: Mass Assumes constant, non relativistic motion with 1 DoF w f Ma 1 2 Kinetic Energy: 2 Potential Energy: k w Mv Mgh g=9.81 [m/s 2 ] @ surface of the earth p h is height above(below) a specified reference point
Element Laws: Viscous Friction f Bv B( v v ) Viscous friction: 2 1
Element Laws: Viscous Friction
Element Laws: Stiffness (Spring) Stiffness: f Kx K( x x ) 2 1 f is a tensile (stretching) force rather than a compressive force when x2 x1 0 (as drawn) d is the natural length of the spring, e.g., when 0 x=0 Potential ti lenergy stored din spring: 1 2 wp K( x) 2
Interconnection Laws D Alembert s Law (restatement of Newton s 2 nd Law): i ( f ) Ma f 0 ext i i i Law of Reaction Forces (Newton s 3 rd Law):
Law of Displacements: Interconnection Laws ( xx ) 0, around any closed path i i
Free body Diagrams Free body diagrams are used as an intermediate step to obtaining system equations of motion (EoM) Assume all elements at equilibrium (EQ) when position and velocity references equal to zero Equilibrium Net force on body = 0, with all inputs constant (zero) Suggested order applying forces to a free body diagram Applied forces, ie i.e., specified inputs Inertial forces, i.e., opposite position reference Spring forces Viscous friction (damping) forces All others, e.g., eg gears, pulleys, levers, etc.
Simple Free body Example The MSD systems below have equivalent dynamics Assume elements at equilibrium (EQ) when x x 0 q ( ) Spring stretched when x 0, as drawn
Equations of Motion To obtain EoM from free body diagram is straight forward left (up) pointing forces = Given the prior free body diagram right (down) pointing forces We havethe followingeom Mx Bx Kx f () t a
Example 2.2 Equations of Motion: M x K x B ( x x ) K ( x x ) 1 1 1 2 1 2 2 1 M x B( x x ) K ( x x ) f ( t) 2 2 2 1 2 2 1 a
Ex. 2.4: Relative Displacements Assume springs in EQ when x z 0 As drawn: Elongation of spring K 2 is x z Inertial forces proportional to absolute (not relative) lti accelerations, e.g., x z
Relative Displacements As drawn x,z>0, which implies K 2 stretched, K 1 compressed
Relative Displacements EoM: M x B x B x K x B z 1 3 1 1 2 M ( x z ) B z K ( x z ) f ( t ) 2 2 2 a
Problem w/absolute References As drawn y>x 2 > x 1 >0, which implies K 1, K 2 stretched x 2 y 1
Problem w/absolute References Free body:
Ex. 2.5: Vertical Motion Assume x is position of the system at EQ Gravitational force on mass is downward F Fg Mg
EoM: Vertical Motion Mx Bx Kx fa () t Mg At EQ with zero applied force, the constant displacement caused by gravity can be calculated Mg Kx Mg x 0 0 K fa ( t) is non-zero x x z When the mass is moving and we can redefine substituting into the EoM yields () 0 Mz Bz K x z f () t Mg 0 a Mz Bz Kz f () t a
Ex. 2.6: Vertical Motion Assume springs relaxed when x x 0 1 2 Determine the static EQ positions due to gravity Note: x is relative to M 2 1 total displacement of M is x +x As drawn: x, x 0 2 1 springs K 1, K 2 are stretched t spring K 3 is compressed 2 1 2 friction B can be thought of as a damper acting parallel to spring K 2 damper acting parallel to spring K 2 with forces in the same direction
Ex. 2.6: Vertical Motion
Ex. 2.6: Vertical Motion EoM: M x K x f () t Bx ( K K ) x M g 1 1 1 1 a 2 2 3 2 1 M ( x x ) Bx ( K K ) x M g 2 1 2 2 2 3 2 2 To find displacement due to gravity set applied force and all derivatives to zero to obtain: Kx ( K K) x Mg 1 10 2 3 20 1 ( K K ) x M g 2 3 20 2 x x 20 10 Mg 2 K K 2 3 ( M1 M2 ) g K 1
Ex. 2.7: Ideal Pulley An ideal pulley is a mass and friction less fi i element that changes direction of motion, e.g., horizontal to vertical, without t cable slippage or stretch t h( (assume in tension)
Ex. 2.7: Ideal Pulley The pulley system is the same as the one bl below, except that is includes the gravitational force M 2 g
Ex. 2.8: Parallel Combinations For springs with a common natural llength, the variablex ibl can be used to describe their displacement; otherwise use d(t) as shown in the free body bl below:
Ex. 2.8: Parallel Combinations The resulting EoM: Mx Bx ( K K ) x Mx Bx K x f ( t ) where K K K eq 1 2 1 2 eq a
Ex. 2.8: Series Combinations
Ex. 2.8: Series Combinations EoM at A: K ( x x ) K x x EoM at Mass: 1 1 2 2 2 2 Kx 1 1 ( K K ) 1 2 Kx Mx 1 1 1 Bx 1 K1( x 1 x2) Mx 1 Bx 1 K1( x 1 ) ( K K ) 1 2 Mx Bx KK x Mx Bx K x f t KK 1 2 where Keq K K 1 2 1 1 1 1 1 eq 1 a () K1 K2 1 2
Ex. 2.11: Parallel Series Combinations B eq B B ( BB) ( B ) 1 2 3 B B 2 3 eq 1 B B 1 2 3 2 3 B B B B B BB B B BB B B 1 2 1 3 1 2 1 3 2 3 B B B B B 2 3 1 2 3
P2.21 Assume system at EQ when all 0, and x x 1 2 0; then, as drawn: K compressed by amount 1 1 K stretched by amount x x x 2 1 2 K stretched by amount 3 2 B stretched by amount 1 1 2 B stretched by amount x 2 2 x x x x i
P2.21 M x 2 2 3 2 K x B2x 2 M 2 B ( x x ) 1 1 2 f () a t M1x1 Mg 2 K ( x x ) 2 1 2 M 1 f () b t Kx 1 1 M1g
P2.21 M 2x2 K3x2 B2x2 B1( x1 x2 ) M 1x1 K 2 ( x1 x2 ) M 2 M 1 B1( x 1 x 2) () a M g ( ) f () b t Kx 1 1 M g 1 EOM: f t 2 K2 x1 x2 M x B x K x B ( x x ) K ( x x ) f ( t ) M g 2 2 2 2 3 2 1 1 2 2 1 2 a 2 Mx B( x x ) K( xx) Kx f( t) Mg 1 1 1 1 2 2 1 2 1 1 b 1
P2.24 Assume system at EQ when all 0, and x x 1 2 0; then, as drawn: K stretched by amount 1 1 2 K stretched by amount 2 2 B stretched by amount 1 1 B stretched by amount x 2 2 x x x x x i
P2.24 2 2 M x Bx 2 2 K 2x2 M 2 K ( x x ) 1 1 2 M 2g M1x1 fa () t M 1 K ( x x ) 1 1 2 B x 1 1 EOM: M x B x K x K ( x x ) M g 2 2 2 2 2 2 1 1 2 2 M x B x K ( x x ) f ( t ) 1 1 1 1 1 1 2 a
P2.14 Displacements x, x are relative to x 1 2 3 x i Assume system at EQ when all K compressed by amount 1 1 2 x x 0; therefore, as drawn: K,B, and B stretched by amount x, x and x, respectively 2 2 1 2 2 1 An absolute ref erence to M 1 is y1 x3 x1 y10 An absolute reference to M is y x x y 2 2 3 2 20
P2.14 My M x x M 1 Bx 1 1 K ( x x ) 1 1 2 K2x2 M 2 Bx 1 1 M 2 2 1( 3 1) 2y2 M ( x x ) 2 3 2 Bx K2x 1 1 2 Bx 2 2 M x M 3 f () a t 3 3
My 1 1 M ( x x ) 1 3 1 M 1 Bx 1 1 P2.14 K ( x x ) 1 1 2 M 2y2 M ( x x ) 2 3 2 K x M 2 2 2 Bx 2 2 M 3x M 3 3 Bx K2x 1 1 2 Bx 2 2 f () a t EoM: M ( x x ) Bx K ( x x ) 0 1 3 1 1 1 1 1 2 M ( x x ) K ( x x ) K x B x 2 3 2 1 1 2 2 2 2 2 3 3 2 2 2 2 1 1 a () Mx Bx Kx Bx f t
Questions?