Lecture15-Prof.Ju

Lecture15-Prof.Ju - CEE M237A / MAE M269A Lecture 15:...

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CEE M237A / MAE M269A Lecture 15: Hamilton’s Principle on Dynamic Structural Systems III: Equation of Motion via Hamilton’s Principle Professor J. Woody Ju
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Circular Cylinders –Torsion Members y The primary kinematic variable for torsion of a circular cylinder is the angle of twist Θ( x, t) y The Lagrangian L is: y This functional has the same mathematical characteristic as the axial force member. Thus, its variation will lead to eqn. of the same form, but the primary dependent variable here is Θ . 2 () ( ) ( ) ( ) 2 2 00 0 1 , 11 , 22 , , , LL L xx k k k k xt Jx t d x G J d x x t x tx t T t x x d x = ⎛⎞ =− ⎜⎟ ⎝⎠ ⎧⎫ ++ ⎨⎬ ⎩⎭ ∫∫ ± ρΘ ΘΘ δ L
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Circular Cylinders – Torsion Members (Cont’d) y Carrying out the variation and integrating over time leads to: 3 () 22 11 0 1 + 0 tt L x xk k k dt J GJ t xx Tt x x d x d t = ∂∂ =− + −= ∫∫ ±± Θ δ ρΘ δδ L
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Circular Cylinder – Torsion Member (Cont’d) y Integrating those variations of derivatives by parts gives: y The terms on the two ends of the time interval vanish according to the Hamilton’s principle. As δΘ is arbitrary, the bracketed expression above must vanish, giving the Euler eqn. as: 4 () 2 1 2 2 1 1 0 1 0 0 | 0 tL xx k k t k t L t t J GJ t T t x x dxdt GJ dt J dx x = ⎡⎤ ∂∂ ⎛⎞ −+ + + ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ | = ∫∫ ±± ± Θ ρΘ δ ρ Θ δΘ 1 0 k k k GJ t T t x x J = ++ = δρ
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Circular Cylinder – Torsion Member (Cont’d) y The Natural and Kinematic Boundary Conditions are y Thus, at the ends, either δΘ or must vanish . The kinematic BC = 0 indicates that the angle of twist is prescribed. The natural BC indicates the vanishing torque at the ends. 5 ( ) () ( ) 0 ,0 , , 0 or , at : either 0 or , 0 0, at 0: either or 0, 0 L Lt t GJ GJ L t GJ t xx x xL G J L t x t xG J t x −| = + = ∂∂ == = = Θ δΘ / x /0 GJ x ∂=
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Example 3 –Torsional Shaft System with Rotors 6 y Consider the system of two distinct circular shafts with three rotors. The mechanical, inertial and geometrical properties of the two shafts are The three motors have inertia properties only with no significant torsional stiffness. y Determine the EOM and the boundary and interface conditions for this system. 11 11 1 2 2 2 , , and , , . GJ J L ρ (,,) aa bb cc J JJ ρρ
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Example 3 –Torsional Shaft System with Rotors 7 () 12 1 2 2 22 11 1 1 2 2 2 2 11 1 00 0 1 2 2 2 1 1 1 1 2 0 2 1 2 1 1 0, , , 2 2 LL L L aa bb cc Jd x J d x G J d x x GJ dx J t J L t J x ⎛⎞ =+ ⎜⎟ ⎝⎠ −+ + + ∫∫ ±± ± Θ ρΘ ρ L y The Lagrangian functional for this system is:
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Example 3 –Torsional Shaft System with
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This note was uploaded on 05/18/2011 for the course MAE 269A taught by Professor Ju during the Spring '11 term at UCLA.

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Lecture15-Prof.Ju - CEE M237A / MAE M269A Lecture 15:...

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