Modelling Workshop Mechanics 2: Solution
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Modelling Workshop Mechanics 2: Solution

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The Modelling Workshop Mechanics 2 Solution of the Modelling Course of Industrial Design of the TU Delft

The Modelling Workshop Mechanics 2 Solution of the Modelling Course of Industrial Design of the TU Delft

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Modelling Workshop Mechanics 2: Solution Modelling Workshop Mechanics 2: Solution Document Transcript

  • P-W-M-2: The Uno single wheel bike Analysis 1st case Analysis the system See Extended Study Guide Modeling The sum of all foce is zero > restart; Input displacement > Road d t/0.05$Heaviside t Road := t/0.05 Heaviside t The sum of all foce is zero (3.2.1)
  • > equ d K m$diff x t , t$2 Kc$ diff x t , t Kdiff Road t , t K x t K k$ Road t 2 equ := K m d x t dt2 d x t K0.05 Dirac t dt Kc Kk x t K0.05 Heaviside t =0 (3.2.2) =0 Solve it we suppose c = 5 in the begining > m d 80 : k d 15000 : c d 20 c := 20 (3.3.1) Initial conditions > ics d x 0 = 0, D x 0 = 0; ics := x 0 = 0, D x 0 = 0 (3.3.2) Solve it > sol d dsolve equ, ics , x t ; 1 K 1 8 sol := x t = e 18460 1 K t 1 8 K e sin 18460 t 13 8 71 sin 13 8 71 t 71 t Heaviside t 71 C (3.3.3) 1 Heaviside t 20 1 1 K8 t 13 K e Heaviside t cos 20 8 > assign sol ; Speed > v d t/diff x t , t ; 71 t v := t/ d x t dt (3.3.4) a := t/ d v t dt (3.3.5) Acceleteation > a d t/diff v t , t ; Plot it > with plots : > p1 d plot x t , t = 0 ..20, numpoints = 100, color = red, style = line : > p2 d plot v t , t = 0 ..20, numpoints = 100, color = green, style = line : > p3 d plot a t , t = 0 ..20, numpoints = 100, color = blue, style = line : > display p1, p2, p3 ;
  • 8 6 4 2 0 5 K 2 10 t 15 20 K 4 K 6 K 8 > Speed > v d t/diff x t , t ; v := t/ d x t dt (3.3.6) a := t/ d v t dt (3.3.7) v := t/ d x t dt (3.3.8) Acceleteation > a d t/diff v t , t ; Acceleteation > v d t/diff x t , t ; Evaluate it
  • How can we make the vibration stop quicker? Advanced - Numerical solutions Design parameter c=20 (find a better C by yourself) > restart; > msd d K m$diff x t , t$2 Kc$ diff x t , t Kdiff Road t , t K x t K k$ Road t =0 d2 d d msd := K m x t Kc x t K Road t Kk x t KRoad t = 0 (3.5.1) 2 dt dt dt > m d 80 : k d 15000 : c d 20 : g d 9.8 g := 9.8 (3.5.2) > Road d t/0.05$Heaviside t : > > ics d x 0 = 0, D x 0 = 0; ics := x 0 = 0, D x 0 = 0 > sol d dsolve msd, ics , x t , numeric, output = listprocedure ; d sol := t = proc t ... end proc, x t = proc t ... end proc, x t = proc t dt ... end proc (3.5.3) (3.5.4) > s d rhs sol 2 ; s := proc t ... end proc (3.5.5) v := proc t ... end proc 0.662489511968130218 (3.5.6) a := k/fdiff v t , t = k (3.5.7) > > v d rhs sol 3 ; v 0.1 > a d k/fdiff v t , t = k ; > > for i from 0 to 2000 by 1 do p1 i p4 i d a i$0.01 : end do: > with plots : > pl 1 d plot seq p1 i , p2 i > pl 2 d plot seq p1 i , p3 i > pl 3 d plot seq p1 i , p4 i > display pl 1 , pl 2 , pl 3 ; d i$0.01 : p2 i d s i$0.01 : p3 i d v i$0.01 : , i = 0 ..2000 , i = 0 ..2000 , i = 0 ..2000 , color = red, style = line : , color = blue, style = line : , color = green, style = line :
  • 8 6 4 2 0 5 10 15 20 K 2 K 4 K 6 K 8 Green line - Congraluations, you are qualified for workshop PW-M-2 Here you get the full grade of this workshop, but don't stop... 2nd case Analysis the system See Extended Study Guide Modeling The sum of all foce is zero > restart; Input displacement > y d t/0.05$Heaviside t C0.003$ cos 50$ t C0.001$cos 100$t ;
  • y := t/0.05 Heaviside t C0.003 cos 50 t C0.001 cos 100 t (5.2.1) The sum of all foce is zero > msd1 d K m1$diff x1 t , t$2 Kks$ x1 t Kx2 t Kcs$ diff x1 t , t K diff x2 t , t = 0; d2 d d msd1 := K m1 x1 t Kks x1 t Kx2 t Kcs x1 t K x2 t = 0 (5.2.2) 2 dt dt dt > msd2 dK m2$diff x2 t , t$2 Cks$ x1 t Kx2 t Ccs$ diff x1 t , t K diff x2 t , t Kct$ diff x2 t , t Kdiff y t , t Kkt$ x2 t Ky t = 0; d2 d d msd2 := K m2 x2 t Cks x1 t Kx2 t Ccs x1 t K x2 t (5.2.3) 2 dt dt dt d Kct x2 t K0.05 Dirac t C0.150 sin 50 t C0.100 sin 100 t Kkt x2 t dt K0.05 Heaviside t K0.003 cos 50 t K0.001 cos 100 t = 0 > Solve it Parametervalues > m1 d 77 : ks d 10000 : cs d 200 : > m2 d 3 : kt d 15000 : ct d 20 : Initial conditions > ics d x1 0 = 0, D x1 0 = 0, x2 0 = 0, D x2 0 = 0; ics := x1 0 = 0, D x1 0 = 0, x2 0 = 0, D x2 0 = 0 (5.3.1) Solve it > dsol d dsolve msd1, msd2, ics , numeric, method = rkf45, maxfun = 100000, x1 t , x2 t , output = listprocedure ; d dsol := t = proc t ... end proc, x1 t = proc t ... end proc, x1 t = proc t (5.3.2) dt ... end proc, x2 t = proc t ... end proc, d x2 t = proc t ... end proc dt Plot it > x1 d rhs dsol 2 ; x1 := proc t ... end proc (5.3.3) x2 := proc t ... end proc (5.3.4) > x2 d rhs dsol 4 ; > plot y t , x1 t , x2 t , t = 0 ..10, color = red, blue, green ;
  • 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0 2 4 6 t > Evaluate it How can we make the vibration stop quicker? 8 10