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Computational Dynamics edited
- 1. COMPUTATIONAL DYNAMICS Jesan Morales ME 195 Supervisedby Dr. Goyal University of California Merced Dec 22 2013
- 2. Pendulum problem • ForwardEuler Method • Simulink • Linear Statespace • Backward Euler • Newton method Particle problem • Euler methods • Newton method • Non-linear Statespace • Generalized Alpha method Static Rod Model Overview
- 3. Pendulum problem 𝜃 = −𝑔 𝐿 sin(𝜃) 𝜽𝟎 = 𝟏 𝜽 𝟎 = 𝟎 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒇𝒖𝒄𝒕𝒊𝒐𝒏 Ѳ Figure 1. Pendulum.
- 4. Forward Euler Method 𝑦𝑖+1= 𝑦𝑖 + 𝑦𝑖ℎ 𝑦𝑖+1 = 𝑦𝑖 + 𝑦𝑖ℎ Graph 1… ℎ=.2
- 5. Step size • Thestep size h was increased to h=0.002 Smoother and no speed loss Graph 2…
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- 7. Simulink Statespace 𝑥1 𝑥2 = 0 1 −𝑔 𝑙 0 𝑥1 𝑥2 + 0 0 𝑢1 𝑢2 𝑥1= 𝜃 𝑥2 = 𝜃 𝑥1 = 𝑥2 𝑥2 = 𝜃 = −𝑔 𝑙 sin(𝑥1) y= 1 0 𝑥1 𝑥2 +0 𝑢1 𝑢2
- 8. Comparing error withdifferent methods hold on • plot(time,theta,'r'); >>>>>>>>>>>>>>>>>>>>>> euler • plot(timesimulink,pendulumsimulink,'g');>>>> Simulink • plot(time,real,'b');>>>>>>>>>>>>>>>>>>>>>>>> by hand • plot(timesimulink,Statespace,‘dot'); >>>>>>> state space Graph 4. Method Comparison Graph 5. Method comparison (close-up)
- 9. Particle problem • Aparticle is traveling with an acceleration described with this non-linear second order differential equation 𝑦 = ( − 𝑦 3 − 8sin(y) +0.2) 3 The initial conditions of 𝑦(0) = 0 and y(0)=0.2 are given • Find the position of the particle at any given time t Figure 3. www.wpclipart.com
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- 11. Damping • 𝑚 𝑥+ 𝑐 𝑥 + 𝑘𝑥 = 0 • 𝜁 = 𝑐 2 𝑚𝑘 • Critical damping (ζ = 1) • Over-damping (ζ > 1) • Under-damping (0 ≤ ζ < 1) 0 = 3 𝑦 + 𝑦 3 + 8 sin 𝑦 + .2 𝜁= 1/3 2 3∗8 =.034021 • Under-damped
- 12. Under-Damped Graph 7. UnderdampedOscillations. http://commons.wikimedia.org
- 13. Forward Euler Method •𝑦𝑖+1 = 𝑦𝑖 + 𝑦𝑖ℎ • 𝑦𝑖+1 = 𝑦𝑖 + ( ,2−8sin(𝑦)− 𝑦 𝑖 𝑖 3 )ℎ • 𝑦𝑖+1 = 𝑦𝑖 + 𝑦𝑖ℎ • 𝑦𝑖+1 = 𝑦𝑖 + ( 𝑦𝑖 + ( ,2−8sin(𝑦)− 𝑦 𝑖 𝑖 3 )ℎ)ℎ Image 4. Forward Euler Method
- 14. Forward Euler Method Results: h=2 Graph 8. Step 2
- 15. Forward Euler Method h=1 Graph9. Step 1
- 16. h=.02 Results (cont.) h=0.2 ForwardEuler Method Graph 10.Step 0.2
- 17. Results (Cont.) h=0.02 Forward Euler Method Graph 11. Step 0.02
- 18. Results (cont.) h=0 .002 Forward Euler Method Graph 12. Step 0.002
- 19. Forward Euler Method Results(cont.) h= 0 .002 Graph 13. Step 0.002 Zoomed-out
- 20. Backwards Euler method •𝑦𝑖+1 = 𝑦𝑖 + 𝑦𝑖+1ℎ • 𝑦𝑖+1 = 𝑦𝑖 + ,2−8sin(𝑦 𝑖+1)− 𝑦 𝑖+1 3 3 ℎ • 𝑦𝑖+1 = ( 𝑦 𝑖+ .2−8 sin 𝑦 𝑖+1 3 ℎ) (1+ ℎ 9 )
- 21. Backwards Euler Method(cont.) • 𝑦𝑖+1 = 𝑦𝑖 + 𝑦𝑖+1ℎ • 𝑦𝑖+1 = 𝑦𝑖 + 𝑦 𝑖+ .2−8 sin 𝑦 𝑖+1 3 ℎ 1+ ℎ 9 ℎ • 0 = −𝑦𝑖+1 + 𝑦𝑖 + 𝑦 𝑖+ .2−8 sin 𝑦 𝑖+1 3 ℎ 1+ ℎ 9 ℎ • 0 = −𝑥 + 𝑦𝑖 + 𝑦 𝑖+ .2−8 sin 𝑥 3 ℎ 1+ ℎ 9 ℎ
- 22. Newton Method f(x)= −𝑥 + 𝑦𝑖 + 𝑦 𝑖+ .2−8 sin 𝑥 3 ℎ 1+ ℎ 9 ℎ f’(x)=−1 − 8 cos 𝑥 ℎ2 3+ ℎ 3 • Guess a value of 𝑥 𝑘 = 𝑥0 • 𝑥 𝑘+1 = 𝑥 𝑘 − 𝑓(𝑥 𝑘) 𝑓′(𝑥 𝑘) Iterate with a tolerance of 10−7
- 23. Symbolic vs. Discretize •Symbolic functions • Takes about 5 minutes • Anonymous functions • About 20 seconds • Discretized • Takes a few seconds Graph 12.
- 24. Non-Linear Statespace 𝑥1 =𝑦 , 𝑥2 = 𝑦′ y’’ =( -y’/3 - 8sin(y) +.2)/3 𝑥1 𝑥2 = 𝑥2 −𝑥2 3 − 8 sin 𝑥1 + .2 3
- 25. Euler Statespace • 𝑦𝑖+1= 𝑦𝑖 + .2−8 sin 𝑦 𝑖+1 − 𝑦 𝑖+1 3 3 ℎ • 𝑥1 𝑖+1 𝑥2 𝑖+1 = 𝑥1 𝑖 + ℎ𝑥2 𝑖+1 𝑥2 𝑖 + .2−8 sin 𝑥1 𝑖+1 − 𝑥2 𝑖+1 3 ℎ 3 • g(x)= 𝑔1(𝑥) 𝑔2(𝑥) = −𝑥1 𝑖+1 + 𝑥1 𝑖 + ℎ𝑥2 𝑖+1 −𝑥2 𝑖+1 + 𝑥2 𝑖 + .2−8 sin 𝑥1 𝑖+1 − 𝑥2 𝑖+1 3 ℎ 3
- 26. Euler Statespace • g’(x)= 𝑑𝑔1 𝑑𝑥1𝑖+1 𝑑𝑔1 𝑑𝑥2 𝑖+1 𝑑𝑔2 𝑑𝑥1 𝑖+1 𝑑𝑔2 𝑑𝑥2 𝑖+1 • g’(x)= −1 ℎ −8ℎ𝑐𝑜𝑠(𝑥1) 3 −(1 + ℎ 9 ) • 𝑥 𝑘+1 = 𝑥 𝑘 − 𝑔(𝑥) 𝑔′(𝑥) • 𝑥 𝑘+1 = 𝑥 𝑘 − 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑔′ 𝑥 ∗ 𝑔(𝑥)
- 27. General Alpha Method •𝑦 = 𝑓 𝑦, 𝑡 , 𝑦 𝑡0 = 𝑦0 • 𝑦𝑖+1 = 𝑦𝑖 + ∆𝑡 1 − 𝛾 𝑦𝑖 + 𝛾 𝑦𝑖+1 • 𝛼 𝑦𝑖 + 1 − 𝛼 𝑦𝑖+1 = 𝛽𝑓𝑖 + (1 − 𝛽)𝑓𝑖+1 • 𝛼 − 𝛽 + 𝛾 = 1 2
- 28. Euler Statespace Image 5.Euler Satespace h=.002
- 29. General Alpha Method(cont.) • 𝑦𝑖+1 = 𝑦𝑖 + ∆𝑡 1 − 𝛾 𝑦𝑖 + 𝛾 𝑦𝑖 • 𝑥1 = 𝑦 • 𝑥2 = 𝑦′ • 𝑥3 = 𝑦′′= −𝑥2 3 −8 sin 𝑥1 +.2 3 • 𝑔1 𝑥 = 𝐺1(𝑥) 𝐺2(𝑥) = −𝑥1 𝑖+1 + 𝑥1 𝑖 + ∆𝑡 1 − 𝛾 𝑥2 𝑖 + 𝛾𝑥2 𝑖 𝑥2 𝑖 + ∆𝑡 1 − 𝛾 𝑥3 𝑖 + 𝛾𝑥3 𝑖 • 𝑔2 𝑥 = 𝐺3(𝑥) 𝐺4(𝑥) = −𝛼𝑥2 𝑖 − 1 − 𝛼 𝑥2 𝑖+1 + 𝛽𝑓1 𝑖 + (1 − 𝛽)𝑓1 𝑖+1 −𝛼𝑥3 𝑖 − 1 − 𝛼 𝑥3 𝑖+1 + 𝛽𝑓2 𝑖 + (1 − 𝛽)𝑓2 𝑖+1
- 30. Non-Linear Statespace 𝑥1 =𝑦 , 𝑥2 = 𝑦′ y’’ =( -y’/3 - 8sin(y) +.2)/3 f(x)= 𝑥1 𝑥2 = 𝑥2 −𝑥2 3 −8 sin 𝑥1 +.2 3
- 31. General Alpha Method(Cont.) g(x)= 𝑔1(𝑥) 𝑔2(𝑥) = −𝑥1 𝑖+1 + 𝑥1 𝑖 + ∆𝑡 1 − 𝛾 𝑥2 𝑖 + 𝛾𝑥2 𝑖+1 −𝑥2 𝑖+1 + 𝑥2 𝑖 + ∆𝑡 1 − 𝛾 𝑥3 𝑖 + 𝛾𝑥3 𝑖+1 −𝛼𝑥2 𝑖 − 1 − 𝛼 𝑥2 𝑖+1 + 𝛽𝑓1 𝑖 + (1 − 𝛽)𝑓1 𝑖+1 −𝛼𝑥3 𝑖 − 1 − 𝛼 𝑥3 𝑖+1 + 𝛽𝑓2 𝑖 + (1 − 𝛽)𝑓2 𝑖+1 𝑔′ 𝑥 = 𝑑𝑔1 𝑑𝑥1 𝑖+1 𝑑𝑔1 𝑑𝑥2 𝑖+1 𝑑𝑔2 𝑑𝑥1 𝑖+1 𝑑𝑔2 𝑑𝑥2 𝑖+1 =
- 32. General Alpha Method(Cont.) • 𝑔′ x = −1 0 0 −1 𝛾ℎ 0 0 𝛾ℎ 0 (1 − 𝛾) −8cos(𝑥)(1−𝛾) 3 −(1−𝛾) 9 (𝛼 − 1) 0 0 (𝛼 − 1)
- 33. Newton with GeneralAlpha Method • 𝑥 𝑘+1 = 𝑥 𝑘 − 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑔′ 𝑥 ∗ 𝑔 𝑥 • h=.001 Image 6.Newton with General Alpha Method
- 34. Different 𝛼, 𝛽𝑎𝑛𝑑 𝛾 ( 𝛼,𝛽,𝛾) Magenta=(0.5,0.5,0.5) Red=(0.3,0.1,0.3) Graph 14. Different 𝛼, 𝛽, 𝛾.
- 35. General Alpha Method Graph15. Generalized Alpha Method ( 𝛼,𝛽,𝛾) Magenta=(0.5,0.5,0.5) Red=(0.3,0.1,0.3)
- 36. Origin error Graph 17.Origin Error ( 𝛼,𝛽,𝛾) Magenta=(0.5,0.5,0.5) Red=(0.3,0.1,0.3)
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- 41. Step h= 0.001 (𝛼,𝛽,𝛾) Magenta=(0.5,0.5,0.5) Red=(0.3,0.1,0.3)
- 42. General Alpha method Thesecond-order accuracy for the generalized-α method requires 𝛼 − 𝛽 + 𝛾 = 1 2 • Unconditionally stable 𝛼, 𝛽 ≤ 1 2 ≤ 𝛾
- 43. General Alpha method •𝑦 = 𝑓 𝑦, 𝑡 , 𝑦 𝑡0 = 𝑦0 • 𝑦𝑖+1 = 𝑦𝑖 + ∆𝑡 1 − 𝛾 𝑦𝑖 + 𝛾 𝑦𝑖+1 • 𝛼 𝑦𝑖 + 1 − 𝛼 𝑦𝑖+1 = 𝛽𝑓𝑖 + (1 − 𝛽)𝑓𝑖+1 • 𝛼 − 𝛽 + 𝛾 = 1 2
- 44. Forward Euler andGeneralized Alpha Method • If 𝛾 =0 • 𝑦𝑖+1 = 𝑦𝑖 + ∆𝑡 1 − 𝛾 𝑦𝑖 + 𝛾 𝑦𝑖+1 • 𝑦𝑖+1 = 𝑦𝑖 + ∆𝑡 𝑦𝑖 • and 𝛼 = 𝛽
- 45. Forward Euler andGeneralized Alpha Method • If 𝛾 = 0 and 𝛼 = 𝛽 • Then 𝛼 − 𝛽 + 𝛾 ≠ 1 2 • Therefore it is not second order accurate • Since 𝛼, 𝛽 ≤ 1 2 ≤ 𝛾 • Is not true then it is not unconditionally stable
- 46. Backward Euler andGeneralized Alpha Method • If 𝛾 = 1 • 𝑦𝑖+1 = 𝑦𝑖 + ∆𝑡 1 − 𝛾 𝑦𝑖 + 𝛾 𝑦𝑖+1 • 𝑦𝑖+1 = 𝑦𝑖 + ∆𝑡 𝑦𝑖+1 • and if 𝛼 = 𝛽 • 𝑦 = 𝑓
- 47. Backward Euler andGeneralized Alpha Method • If 𝛾 = 1 and 𝛼 = 𝛽 • Then 𝛼 − 𝛽 + 𝛾 ≠ 1 2 • Therefore it is not second order accurate • If 𝛼 ≤ 𝛽 ≤ 1 2 • Then 𝛼, 𝛽 ≤ 1 2 ≤ 𝛾 is satisfied and • Backward Euler is unconditionally stable
- 48. Static Rod Model •The following equation describe the rod model • Non-linear differential equations govern the formation of the beam and lead to loop deformation Image 6.
- 49. • These equationrepresent the following system • Where s is along the rod • Unshearable and inextensible Image 7.
- 50. Vectors • Internal forcealong the cross section fixed reference • Moment vector applied to the cross section • Curvature third component is twist
- 51. Constitutive Relationship • Theseequations show the relationship between the moment and the curvature which will be helpful in solving for the linear and non-linear equations:
- 52. Pure Torsion Image 8.Pure Torsion
- 53. Pure Moment Image 9.Pure Moment
- 54. Pure Shear Force Image10. Pure Shear Force
- 55. All Applied Equally Image11. All applied equally
- 56. Static Rod Model •Here are the step taken to derive the equations. 𝜅 = 0 𝑑𝜃 𝑑𝑠 0 𝑓 = 𝑓1 0 𝑓3 𝑡 = 0 0 1 𝑞 = 0 𝑞2 0 • Linearized equations about 𝑎2: 𝑑 𝑑𝑠 𝑓1 𝑓2 𝑓3 + 𝜅1 𝜅2 𝜅3 X 𝑓1 𝑓2 𝑓3 = −𝐹1 −𝐹2 −𝐹3 >>>>>>> 𝑑 𝑑𝑠 𝑓1 𝑓2 𝑓3 + 𝜅2 𝑓3 − 𝜅3 𝑓2 𝜅3 𝑓1 − 𝜅1 𝑓3 𝜅1 𝑓2 − 𝜅2 𝑓1 = −𝐹1 −𝐹2 −𝐹3 𝑑 𝑑𝑠 𝑞1 𝑞2 𝑞3 + 𝜅1 𝜅2 𝜅3 X 𝑞1 𝑞2 𝑞3 = 𝑓1 𝑓2 𝑓3 𝑋 0 0 1 >>>>>>> 𝑑 𝑑𝑠 𝑞1 𝑞2 𝑞3 + 𝜅2 𝑞3 − 𝜅3 𝑞2 𝜅3 𝑞1 − 𝜅1 𝑞3 𝜅1 𝑞2 − 𝜅2 𝑞1 = 𝑓2 𝑓1 0
- 57. Linear Rod Model 𝑑 𝑑𝑠 𝑓1 0 𝑓3 + 𝜅2𝑓3 0 −𝜅2 𝑓1 = −𝐹1 0 −𝐹3 𝑑 𝑑𝑠 𝑞1 𝑞2 𝑞3 + 0 0 0 = 0 𝑓1 0 0 𝑞2 0 = 𝐸𝐼1 𝜅1 𝐸𝐼2 𝜅2 𝐺𝐽𝜅3
- 58. Static Rod Model •𝜅2 = 𝑞2 𝐸𝐼2 • 𝑑𝑞2 𝑑𝑠 = 𝑓1 • 𝑑𝑓1 𝑑𝑠 + 𝑞2 𝑓3 𝐸𝐼2 = −𝐹1 • 𝑑𝑓3 𝑑𝑠 − 𝑞2 𝑓1 𝐸𝐼2 = −𝐹3
- 59. Static Rod Model •𝑥 = 𝑓1 𝑓3 𝑞2 , 𝑑𝑥 𝑑𝑠 = 𝑑 𝑑𝑠 𝑓1 𝑓3 𝑞2 , 𝑓 𝑥 = −𝑓1 −𝑞2 𝑓3 𝐸𝐼2 𝑞2 𝑓1 𝐸𝐼2 , 𝑢 = 0 −𝐹1 −𝐹3 • 𝑑𝑥 𝑑𝑠 = 𝑓 𝑥 + 𝑢 • 𝑑 𝑑𝑠 𝑓1 𝑓3 𝑞2 − −𝑞2 𝑓3 𝐸𝐼2 𝑞2 𝑓1 𝐸𝑓2 −𝑓1 = −𝐹1 −𝐹3 0
- 60. Static Rod Model • 𝑑𝑓 𝑑𝑥 = 0−1 0 −𝑓3 𝐸𝐼2 0 −𝑞2 𝐸𝐼2 𝑓1 𝐸𝐼2 𝑞2 𝐸𝐼2 0 = 𝑑𝑓(𝑥 𝑒) 𝑑𝑥 = 0 −1 0 0 0 0 0 0 0 • 𝑥 𝑒 = 0 0 0 , 𝑓 𝑥 𝑒 = 0 0 0 • 𝑑𝑥 𝑑𝑠 = 𝑑𝑓(𝑥 𝑒) 𝑑𝑥 𝑥 + 𝑢
- 61. Linear Rod Model • 𝑑𝑥 𝑑𝑠 = 𝑑 𝑑𝑠 𝑞2 𝑓3 𝑓1 • 𝑑𝑥 𝑑𝑠 = 0−1 0 0 0 0 0 0 0 𝑞2 𝑓1 𝑓3 + 0 −𝐹1 −𝐹3 • 𝑑 𝑑𝑠 𝑓1 𝑓3 𝑞2 = −𝐹3 −𝐹1 −𝑓1
- 62. Results • 𝑑𝑞2 𝑑𝑠 = −𝑓1 𝑄= 𝑑(−𝑚) 𝑑𝑥 • 𝑑𝑓1 𝑑𝑠 = −𝐹1 • 𝑑𝑓1 𝑑𝑠 = −𝐹3
- 63. Thank you foryour time Any Questions?