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# Modelling Workshop General 1: Solution

## by TU Delft OpenCourseWare on Jan 27, 2014

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The Modelling Workshop General 1 Solution of the Modelling Course of Industrial Design of the TU Delft

The Modelling Workshop General 1 Solution of the Modelling Course of Industrial Design of the TU Delft

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## Modelling Workshop General 1: SolutionDocument Transcript

• Workshop G-W-1: The Seesaw(Wipwap) Analysis Modeling Always use a restart at begining to avoid errors; > restart; Here we inlcude the plot library for later plot function. We use ":" to ask Maple NOT to display the response (make it looks clean) > with plots : Variable Symbo Value l Moment of inertia MI Point mass:
• m\$ L 2 2 kg m3 Density ρ (rho) 700 Friction torque constant f 8 N·m Spring constant k 3000 N·m/Radian Distance from the pivoting point to one end of the board L 2m Width of board W 0.2 m Thickness of board T 0.1 m Mass of a child: each mChild 20 kg Mass of seating board mBoard V\$ρ Volume V Gravitational acceleration g Input force F t 2\$ L\$W\$T 9.81 m 2 s 100 N (from 0 s to 0.2 s) The input force is applied just for an instant (during 0.2 s) at the beginning. We can express this with the sum of two heaviside functions, as shown below: > F d t/100\$ Heaviside t K100\$Heaviside t K0.2 ; F := t/100 Heaviside t K100 Heaviside t K0.2 Have a look at the foce, which is constructed from the heaviside function (from 0 to 10 seconds) > plot F t , t = 0 ..10 (3.1)
• 100 80 60 40 20 0 0 2 4 6 8 10 t The model Start from Sum(F) = 0 We have: 1. Inertia force (2 children, 2 sides of board) 2. Gravity (2 children, 2 board) 3. Spring 4. Friction 5. Input Note that the gravity contribution of each children cancel out (they have the same mass, they are at the same distance from the pivoting point, but they exert angular momentum in opposite directions),
• thus they don't appear in the equation. The same applies to each side of the seating board. MI is the moment of inertia of one side of the wipwap, we multiply it by 2 for the system. > equ dK 2\$MI\$diff theta t , t\$2 K k\$theta t CF t \$LK signum diff theta t , t \$f = 0 d2 θ t 2 dt d Ksignum θ t dt equ := K MI 2 Kk θ t C 100 Heaviside t K100 Heaviside t K0.2 L (3.2) f=0 For the force, you can also apply a more precise definition as F(t)*L*cos(theta(t)) as: > equPrecise dK 2\$MI\$diff theta t , t\$2 K k\$theta t CF t \$L\$cos theta t K signum diff theta t , t \$f = 0 d2 equPrecise := K MI 2 θ t Kk θ t C 100 Heaviside t K100 Heaviside t dt2 d K0.2 L cos θ t Ksignum θ t f=0 dt (3.3) Mass moment of inertia considering a point mass, which is mass*(rototion radius)^2 L ; 2 RChild d L; MI d mBoardside\$ RBoard 2 CmChild\$ RChild 2; 1 RBoard := L 2 RChild := L 1 mBoardside L2 CmChild L2 MI := 4 > RBoard d Solution (Trial & error) Values of parameters > L d 2 : W d 0.20 : T d 0.1 : rho d 700 : g d 9.8 : mChild d 20 : f d 8; mBoardside d rho\$L\$W\$T; (3.4)
• f := 8 mBoardside := 28.000 (4.1) k := 3000 (4.2) Trial and error approach > k d 3000; Mass moment of inertia considering a point mass > MI; 108.0000000 (4.3) Initial conditions: the displacement is 0, the speed is zero as well > ics d theta 0 = 0, D theta 0 = 0; ics := θ 0 = 0, D θ 0 = 0 Always have a look at the equation before solve it to avoid input errors > equ; 2 d θ t dt2 d K8 signum θ t dt K 216.0000000 K3000 θ t C200 Heaviside t K200 Heaviside t K0.2 (4.4) (4.5) =0 Solve it numerically, take care the listprocedure output, which will make your numerical output simular to a function > sol d dsolve equ, ics , theta t , type = numeric, output = listprocedure ; d θ t = proc t (4.6) sol := t = proc t ... end proc, θ t = proc t ... end proc, dt ... end proc Save the resulting angle in a variable > angle d rhs sol 2 ; angle := proc t ... end proc (4.7) Save the resulting angular speed in a variable > angularSpeed d rhs sol 3 ; angularSpeed := proc t ... end proc (4.8) Plot the angle and the angular speed (the angle is given in radians), here we specify the output graph is constructed by 100 sample points > plot angle t , angularSpeed t , t = 0 ..10, numpoints = 100, color = red, blue ;
• 0.15 0.10 0.05 0 2 4 6 t K 0.05 K 0.10 K 0.15 Plot the angle in degrees in red color angle t \$180 > plot , t = 0 ..10, color = red ; 3.1415926 8 10
• 2 1 0 2 4 6 8 10 t K 1 K 2 Save the plot > p1 d plot angle t \$180 , t = 0 ..10, color = red ; 3.1415926 p1 := PLOT ... (4.9) Green line - Congraluations, you are qualified for workshop GW-1 Solve the more precise model Now we try to solve the more precise model Have a look at the model > equPrecise (6 1)
• K 216.0000000 K0.2 d2 θ t dt2 cos θ t K3000 θ t C2 100 Heaviside t K100 Heaviside t K8 signum d θ t dt (6.1) =0 Solve it > solPrecise d dsolve equPrecise, ics , theta t , type = numeric, output = listprocedure ; d solPrecise := t = proc t ... end proc, θ t = proc t ... end proc, θ t = proc t dt ... end proc Save the resulting angle in a function named anglePrecise > anglePrecise d rhs solPrecise 2 ; anglePrecise := proc t ... end proc (6.2) (6.3) Save the plot > pPrecise d plot anglePrecise t \$180 , t = 0 ..10, color = blue ; 3.1415926 pPrecise := PLOT ... Compare the graph with previous model > display p1, pPrecise ; (6.4)
• 2 1 0 2 4 6 8 10 t K 1 K 2 It is found that due to the very short time (0.2) second, the abgular displacement is very small, thus cos(theta(t)) is very close to 1, the two solutions donot have any differences (overlap eachother) Treat the board as a box (mass moment of inertia) Compute the moment of intertia for a box > MI d 2 mBoardside\$T2 mBoardside\$ L C CmChild\$ RChild 2; 3 12 MI := 117.3566667 (7.1) Solve it numerically > solBox d dsolve equ, ics , type = numeric, output = listprocedure ; d θ t = proc t solBox := t = proc t ... end proc, θ t = proc t ... end proc, dt ... end proc (7.2)
• Save the resulting angle in a variable > angleBox d rhs solBox 2 ; angleBox := proc t ... end proc (7.3) Save the resulting angular speed in a variable > angularSpeedBox d rhs solBox 3 ; angularSpeedBox := proc t ... end proc (7.4) Plot the angle and the angular speed (the angle is given in radians) > plot angleBox t , angularSpeedBox t , t = 0 ..10, numpoints = 100, color = red, blue ; 0.10 0.05 0 2 4 6 t K 0.05 K 0.10 Plot the angle in degrees angleBox t \$180 > plot , t = 0 ..10, color = green ; 3.1415926 8 10
• 2 1 0 2 4 6 8 10 t K 1 K 2 Compare both results angleBox t \$180 > p2 d plot , t = 0 ..10, color = green ; 3.1415926 p2 := PLOT ... > display p1, p2 ; (7.5)
• 2 1 0 2 4 6 8 10 t K 1 K 2 It is evident that both results are qualitatively the same. Therefore, it would be convenient to use the procedure for the point mass in this particular case, because it is simpler. However, one should take into account that the solution may eventually diverge. Try to change the mass of the child to 0 to simulate one push to the the wip-wap, and see what happens What would happen if only one child sits on one side? Compute the moment of intertia for a box, note here we use MI to descibe 3 moment of inertia, board*2 + 1 child 2 mBoardside\$L2 mBoardside\$T > MI d 2\$ C CmChild\$ RChild 2; 3 12 (8.1) MI := 154.7133333 K k\$theta t > equOneChild dK MI\$diff theta t , t\$2 KmChild\$g\$ L\$cos theta t CF t \$L Ksignum diff theta t , t \$f = 0 d2 K 154.7133333 θ t K392.0 cos θ t K3000 θ t (8.2) equOneChild := dt2
• C200 Heaviside t K200 Heaviside t K0.2 K8 signum d θ t dt =0 The inital angle of the board > equ1 d mChild\$g\$L\$cos angle = k\$angle equ1 := 392.0 cos angle = 3000 angle > resa d solve equ1, angle resa := 0.1295713366 Initial conditions > icsOneChild d theta 0 =K resa, D theta 0 = 0; icsOneChild := θ 0 = K 0.1295713366, D θ Solve it numerically > solOneChild d dsolve = listprocedure ; (8.3) (8.4) 0 =0 (8.5) equOneChild, icsOneChild , type = numeric, output solOneChild := t = proc t ... end proc, θ t = proc t ... end proc, d θ t = proc t dt (8.6) ... end proc Save the resulting angle in a variable > angleOneChild d rhs solOneChild 2 ; angleOneChild := proc t ... end proc (8.7) Save the resulting angular speed in a variable > angularSpeedOneChild d rhs solOneChild 3 ; angularSpeedOneChild := proc t ... end proc (8.8) Plot the angle and the angular speed (the angle is given in radians) > plot angleOneChild t , angularSpeedOneChild t , t = 0 ..10, numpoints = 100, color = red, blue ;
• 0.2 0.1 0 2 4 6 8 t K 0.1 K 0.2 Plot the angle in degrees angleOneChild t \$180 > plot , t = 0 ..10, color = green ; 3.1415926 10
• t 0 2 4 K 5 K 6 K 7 K 8 K 9 K 10 > This is the story of a wipwap, Enjoy. 6 8 10