MEEN 363. EAMPLE of ANALYSIS (1 DOF) Luis San Anrés Objectives: a) To erive EOM for a 1-DOF (one egree of freeom) system b) To unerstan concept of static equilibrium c) To learn the correct usage of physical units (US system) ) To calculate natural frequency an amping ratio e) To preict the time response (steay state an full transient) of a system f) To learn how to combine mathematical statements with explanatory sentences. Problem statement A pulley an cable system are assemble to pull a heavy block stuck in a hollow mining shaft. A motorcar imposes the known motion Z(t) require to pull the block. The stiffness K represents a flexible connection to the rive motorcar. The amping coefficient (C) represents the viscous rag between the block an shaft walls. Z K Y In the figure, YZ enote the static equilibrium position (SEP) of the system. a) Draw free boy iagrams for the block, label all forces an show their constitutive relation in terms of the motion coorinates, if applicable. b) Ientify the kinematical constraint relating motions Y an. The cable oes NOT slip on the pulley. c) Fin the static eflection ( ) of the spring element. ) Derive a single EOM for the block motion in terms of coorinate. θ M C For items (e) - (h) use K 1 5 lb/in, Mg5 lb, an C15 lb.s/in e) Fin the system natural frequency [Hz] an viscous amping ratio (ζ). f) The motorcar moves with Z(t)v t, where v 1 ft/. What is the terminal or final velocity of the block (M) in (ft/)? g) Fin the complete solution to the problem. That is fin (t) STATIC EQUILIBRIUM POSITION (SEP) SEP means no motion of block or external agent holing spring - cable. Thus, at the SEP spring K is alreay eflecte since it must support 5% of the block weight (W) as easily seen from the cable & pulley constraint. This knowlege is BASIC, oes not require of elaborate thinking or eriving lengthy equations. Important: Z(t) is KNOWN for all times, i.e. a function of time impose on the system by an external agent (motorcar)
FREE BODY iagrams an kinematic constraints Definitions: T Tension from cable connecting block to spring. Cable is NOT extensible F s force in spring connecting external DRIVE agent to inextensible cable F D viscous rag force static eflection for spring To raw FBDs, assume a state of motion >, (Z-Y)>. These mathematical statements mean: block moves UP, an spring is transmitting a force also upwars an to the left, see iagrams where F rive F spring T K + K (Z-Y) Note that (K W/ ) is the static force in the spring. This the static force necessary to hol the system statically, i.e. without motion. Hence.5 W/K
W : 5 lb K 1 5 lb in : C 15 lb (a) Assume a state of motion with Z-Y>, > ie. motorcar pulls block : in Z M : W g Y Drive force From the FBD iagram, assume >, an apply Newton's n law to obtain: M t W F Damper + T (1) where F Damper C t T K ( Z Y) + K F Drive () is the viscous rag force (3) T is the cable tension. (Z-Y)>, an δs is the spring static eflection (b) kinematic constraint - inextensible cable K W T M T FD C /t T The cable length is constant, thus an the kinematic constraint follows as (c) Static eflection of spring l c l c + Y Y (4) By efinition of SEP (Static equilibrium position), i.e. when ZY an at rest (without motion): W + K (5) (K δs ) is the static force neee to HOLD the block w/o motion Static eflection of the spring is: : W K.5 in (c) Derive single EOM for block motion Substituting (), (3) an (4) into EOM (1) reners Note: EOM cannot contain internal forces (Tension for example). The tension is DETERMINED by the motion. M t W C + K ( Z ) + K δ t s (6)
an thus the final EOM is: + C + 4 K K Zt () Ft () t t M (7) Note K Zt ()"appears" as a (ynamic) external force riving the block into motion. (e) Calculate natural frequency an viscous amping ratio: ω n : 4K W g.5 ω n 175.747 1 f n : ω n π f n 7.971 Hz ω : ω n 1 ζ 1 The amping ratio is rather large - motion will be oscillatory but quickly ampe! T : f The ampe natural frequency an perio of motion are: f ζ C 1 : T n :.5 f W ζ.33 n 4 K g ( ).5 ω 165.931 ra 6.49 Hz T.38 f : ω π
(f) pulling motocar moves with constant spee v, FIND terminal spee of block Let Hence: v : 1 ft zt () vt : (8) + C + 4 K K zt () Ft () t t M (7) What is steay-state motion? Since z(t) is linear in time, the particular solution to eqn (7) is: i.e block ALSO moves with constant SPEED p a + b t t p (8) b t p To fin the en or terminal velocity, take the time erivative of (7) 3 M 3 P t + C P + 4 K t t P K v Using the knowlege from erivatives of p 4 K t P K v Terminal velocity of block: t P v v.5 ft A more elaborate way follows from fining the whole particular solution: Substitute (8) into EOM (7) to fin Cb + 4 Ka ( + bt ) K v t From: equating like-powers of t Cb 4 K + K a b K p () t : a+ bt terminal velocity at which block moves is b.5 ft v Hence: v b : b 6 in C b a : K a.45 in
(g) fin the full transient response - ynamic motion of block Particular solution: P + C P + K P A + B t t t p a + b t M 1 a A C B b K K Complete solution: t () H + P B K..15 p () t.1 zt ().5.5.1.15 t t () e ζ ( ( ) ( )) ω n t C 1 cos ω t + C sin ω t + ( a + b t) (11) t e ζ ω n t ( D 1 cos( ω t) + D sin( ω t) ) + b (1) Where D 1 ζ ω n C 1 + C ω D ζ ω n C (13) C ω satisfy initial conitions at t: from (11) an (1) at time t an from (13) o : ft o C 1 + a V o D 1 + b D 1 + ζω n C 1 C : ω V o : ft D C 1 : o a D 1 : V o b : ζ ω n C motion starts from rest C ω C 1.45 in C. in D 1 6 in D 4.578 in
Let's graph the response for time values up to 4 x ampe perio (my choice) t (): e ζ ω n t ( C 1 cos( ω t) + C sin( ω t) ) + ( a + b t) ( ( ) ( )) Vt () e ζ ω n t : D 1 cos ω t + D sin ω t + b T max : 4 T velocity (ft/).8.5.1.15 time (s) V(t) V final (ft) t () a+ b t.8.6.4..5.1.15 t time (s) motion - relative to steay state.4 t () ( a+ bt )...43.65.87.11.13.15 t
Spring (cable) force (ynamic+static) F s () t : K ( zt () t ()) W 5 1 3 lb + K K.5 1 3 lb Spring force (lbf). 4 1 1.5. 4 1 1. 4 1 5 ( ) 1.15 1 4 F s T max lb.5.1.15 time (s) at steay state, spring (cable) force approaches ( a).115 in since: is the final eflection of spring F SS () t : K vt ( a + b t) + v 1 ft b ( ) F SS : K a 1 ft F SS 1.15 1 4 lb as the graph shows!