Cable holds system BUT at t=0 it breaks!! θ=20. Copyright Luis San Andrés (2010) 1

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1 EAMPLE # for MEEN 363 SPRING 6 Objectives: a) To erive EOMS of a DOF system b) To unerstan concept of static equilibrium c) To learn the correct usage of physical units (US system) ) To calculate natural frequencies an natural moe shapes RIGID BODY MODE e) To preict the response of a system using moal coorinates. Case: constant amplitue loa. Explain what a rigi boy means. f) To learn how to combine mathematical statements with explanatory sentences. M K Y Car must pull a heavy block stuck in a hollow an eep mining sha. The front en of car is tie to a big tree with a cable. A flexible cable of stiffness K is connecte to an inextensible cable that in turn, with a pulley system, is connecte to the block. The amping coefficient (C) represents the viscous rag between the block an sha walls. In the figure, Y enote the static equilibrium position (SEP) of the system. θ Cable hols system BUT at t it breaks!! M C SEP means no motion of car an block. Thus, at the SEP spring K is alreay eflecte since it must support 5% of the block weight as easily seen from the cable & pulley constraint. The top cable connecte to a fixe point also hols the system with a force 5% of plus a fraction of the car weight, i.e. sin( ). This knowlege is BASIC, oes not require of elaborate thinking or eriving lengthy equations. Copyright Luis San Anrés ()

2 At time t, the cable BREAKS! an both the car an block start falling. (Notice change of irection of coorinates,y, from prior example). Furthermore, the engine in car is off! Oopps, forgot to set the han brake! a) Ientify the kinematical constraint relating motions Y an. The cable oes NOT slip on the pulley. b) Determine the static eflection (δ s ) of spring element as well as top cable force before it breaks. c) For t>, aer top cable breaks, raw free boy iagrams for the car an block, label all forces an show their constitutive relation in terms of the motion coorinates, if applicable. ) For t>, Derive EOMs for the car an block motion in terms of coorinates &. e) Fin the system (unampe) natural frequencies an moe shapes, i.e. solve for the system eigenvalues an eigenvectors. DISCUSS Significance of rigi boy moes. f) Fin the (unampe) response of the system using moal analysis. Graph responses (moal an physical) versus time. g) Fin the spring- force an graph it vs. time. hat knowlege can be gaine from this force? Copyright Luis San Anrés ()

3 FREE BODY iagrams an kinematic constraints Definitions: F cable force from elastic cable connecte to fixe point (vali for t< only) F s force from elastic cable connecting car to cable on pulley T Tension on cable F D viscous rag force δ s static eflection for spring FREE BODY DIAGRAMS Cable force holing car an block before t. For t>, cable breaks an force holing blocks isappears Fcable M T Assume state of motion to raw FBD Y >, > Y K FsTK (Y-) ½ T T T M FD C /t Copyright Luis San Anrés (6)

4 MEEN 363 EAMPLE DOF- ANALYSIS: block pulls car - RIGID BODY MOTION ORIGIN : K 5 lb : in : 5lb C : 5lb in : lb θ : π 8 (a) kinematic constraint - inextensible cable M : g M : g lb M masses nee be expresse in lb.^/ for consistency in EOM lb M 3.8 The cable length is constant, thus l c l c Y an the kinematic constraint follows as Y (b) Fin the top cable "holing force" an spring- static eflection. For statics, assume NO motion For static equilibrium: force from spring tension T : an T K δ s F s δ s : K Cable attache to fiex point hols fraction of car weight an also balances tension δ s.5 in an F cable : sin θ F cable.84 3 lb (c) Derive EOMS: Cable breaks, for t> Assume a state of motion with Y->, > block pulls car (engine off) Block of mass M: From the FBD iagram, with >, an apply Newton's n law to obtain: M F Damper T t () where F Damper C () is the viscous rag force t T K ( Y ) K δ s F s (3) T is the cable tension Force from spring. (Y-)>, an δs is the static eflection for spring Car of mass M: From the FBD iagram, with Y>, Y, apply Newton's n law to obtain: M sin( θ) T F cable (4) But F cable : lb since it broke! t substitute T an Famper into () to obtain M C K ( Y ) K δ s t t () an since K δ s M C t t K ( ) M C 4K K BLOCK of MASS M: t t (5) an Y

5 substitute an T into (4) M sin( θ) K ( Y ) Sub Y t M t sin( θ) K ( ) CAR of MASS M M K K sin( θ) (6) t Eqns. (5) an (6) are the esire equations of motion for the car an block. Please recall that motions are from the static equilibrium position In matrix form, the EOMs are: M M t K K K 4K C t sin( θ) (7) Notes * for t>, weights of block an (fraction of ) car PULL the system ownwars! * mass & stiffness matrices are symmetric.the amping matrix is NOT * Stiffness matrix is singular, i.e. its eterminant equals zero, A RIGID BODY MODE EPECTED () Fin natural frequencies an natural moe shapes of UNDAMPED system. Disregaing amping, an letting the force RHS, eq. (7) becomes M M t Let a cos ωt K K K 4K (8) a cos ωt (9) Δ( K) 4K 4K Substitution of (9) into (8) reners the homogeneous system of eqns K M ω K K 4K M ω a a () eqn () has a non-trivial solution if the eterminant of the system of equations equals zero, i.e.: K M ω Δω 4K M ω 4K Let λ ω an expaning the proucts in the eqn. above λ M M λ K M 4K M K 4K 4K Let: aλ bλ c with: a : M M b : K M 4K M c : K 4K 4K Note c!!!

6 a lb 4 b lb c lb NOTE That the coefficients a,b,c have consistent physical units. That is, since the physical unit for λ is (/^), the physical units of the eterminant are (lbf/)^ The roots (eigenvalues) of the characteristic equation are b b 4a c b b 4a c λ : λ : a a.5 an the natural frequencies are: ω ( λ ).5 for ω K M ω a a : K.5 φ : a.5 λ φ is the n eigenvector (natural moe) (3).4 DOF () an DOF () move 8 eg OUT of phase, with > : ω : λ ω () 63.6 Now, fin the eigenvectors The two equations in () are linearly epenent. Thus, one cannot solve for a an a. Set a : arbitrarily; an from the first equation for ω a : K M ω K a φ : a ra φ is the first eigenvector (natural moe) ().5 DOF () an DOF () move in phase, with Y,.5 Y (recall kinematic constraint) RIGID BODY MOTION (e) Response in MODAL coorinates: Buil the physical matrices: M M : K : M Use the MODAL transformation x Aq K K K 4K Disregar DAMPING Make moal matrix using eigenvectors A : augment φ, φ check orthogonality property of natural moes A (4).5.4 M M : A T MA K M : A T KA M lb M K M lb non-iagonal elements are very small non zero b/c of rounoff with computer

7 efine moal masses an stiffnesses: M m : M M M m : M M,, K m : K M K m : K M,, check K m M m.5 ra K m M m.5 ω 63.6 ra OK 63.6 ra efine initial conitions: isplacements an velocities in moal coorinates At time ts, the system is at its static equilibrium position, hence the initial conitions are null isplacements an null velocities. Of course, the same applies to moal space, i.e. null initial isplacements an velocities for generality, efine: o : V o : Calculate inverse of A matrix an in moal coorinates A inv : A A.5.4 q o : A inv o q o_ot : A inv V o q o q o_ot (5) as expecte - Do not perform this step unless Initial conitions are ifferent from null Define moal force Q: A T F Fin moal responses F sin( θ) : Physical force vector Q lb F Both natural moes will be excite (6) first moe: Rigi Boy Moe M m q t.84 3 Q since K m lb lb (7a) on moe: vibratory moe M m q t K m q Q (7b) an since the initial conitions are null - using cheat sheet: q () t Q Q t : q M m () t : ( cos( ω t) ) 8) an K m π GRAPH moal responses T large : 3 ω arbitrary scale for plot

8 moal isplacements () q () t.5.75 q () t q q t time () T large.7 Moal response shows quaratic increase in time - RIGID BODY MODE Moal response is too small to be seen (will iscuss later) The response in physical coorinates, an, equals (from transformation xaq) t (): q () t q () t t :.5q () t.4q () t (9) A.5.4 isplacements () t ().5.75 t () T ( large ).4 T ( large ).5 Note that for t>>, Y One question remains: here is the on moe? i.e oscillatory behavior with perio equal to Let's graph the relative motion -Y.3 - car -block t time () rel () t : t () t () Motion shows car () an block () falling by the pull of gravity, i.e. their weights T : ( π) ω T.4 isplacements () rel () t.. which clearly shows the oscillatory response of the on moe t time () - relative

9 So why is this important? Recall that the connecting spring force is, eqn. (3) F s () t : K ( t () t ()) t> Force (lb) F s () t lb thus, although the ynamic eflections in spring maybe small, the spring force is rather large!!! t time () Spring force Spring force varies from F s ( ).5 3 T lb to F s lb