Final Exam Solution Dynamics 2 191157140 31-01-2013 8:45 12:15 Problem 1 Bateau Bateau is a trapeze act by Cirque du Soleil in which artists perform aerial maneuvers on a boat shaped structure. The boat is modeled as a rigid uniform bar with mass and length. The boat is connected to the ceiling of the theater by means of two suspension bars with length that are considered massless. The gravitational constant is. The act consists of two phases: A and B, shown in the figure below. Phase A: The artists with combined mass are on top of the boat, which is in its equilibrium position. They exert a force on the boat to start up its motion. The generalized coordinate describing the motion of the boat is the angle between the suspension bars and the vertical. Phase B: The artists are suspended underneath the boat. This is modelled as if the artists form a pendulum of length. The generalized coordinate describing the motion of the artists is the angle between the artists and the vertical. Bateau Phase A Phase B a) Consider phase A. Derive the equation of motion of the system for arbitrary large angles using the force and moment balances method. In order to do so, make an accurate free-body diagram of the boat. Show that the linearized equation of motion is equal to (i). (i) First, the constraint equations are derived that are used to express the position, velocity and acceleration of the center of mass of the boat in terms of (t). Position: Velocity:
Acceleration: The free-body diagram of the boat in a positive deflected state must contain the gravity, the forces in the suspension bars and along the direction of the suspension bars and the external force. The force balances are: Substitution of the constraint equations: [ ] [ ] Multiply the first equation by and the second by : [ ] [ ] Adding yields the equation of motion for arbitrary large angles: Linearizing the equation of motion using and, yields (i): (i) Constraint equations: Free-body diagram: Force balances: Correct non-linear equation: Correct linearization: 2p ( for the velocities, for the accelerations) 2p (-1 for every missing or incorrect force) 2p ( for each correct balance) 3p 10p b) Assume that all artists are distributed evenly over the boat. Derive the expressions for the forces and in the suspension bars for arbitrary large motions of the boat. Prove that both forces are equal. Comment on the physical meaning of all terms in your result. 2
Moment balance about the center of gravity of the boat: From this it follows that both forces are equal: Now multiply the first equation of a) by and the second by : [ ] [ ] Subtract the first from the second: So: [ ] It can be seen that both suspension bars carry half of the total weight, the fictitious centripetal force and the component of the excitation force in the direction of the suspension bars. Forces equal by using : Solving from and : 2p Interpretation: 4p c) Calculate the steady state amplitude of the boat when the force applied by the artists is harmonic:. Sketch the magnitude plot of the frequency response function in which you indicate important characteristics. Write the equation of motion in standard form: Trial function and its second time derivative can be cosines because there is no damping: 3
Substitution and rewriting: Solving for : The magnitude plot of is equal to at and tends to zero for. The magnitude is equal to infinity for. EoM in standard form: Trial function equal to RHS: Solving amplitude : Correct plot: 2p (-1 for every incorrect plot characteristic) 5p d) What is the expression of the frequency, with which the artists should apply the force, such that the amplitude of the boat increases the fastest? Derive the response of the boat for this specific case, using initial conditions. Prove that the response can be written as (ii). Sketch the response and indicate its important characteristics. (ii) The amplitude of the motion increases the fastest when the system is excited in its natural frequency, because in that case resonance will occur. The homogenous solution of the differential equation is simply: In resonance, the trial solution for the particular part must be multiplied by : Substitution in the equation of motion make that all terms multiplied by cancel: 4
From this it can easily be seen that and that for holds: So the total solution becomes: Apply the initial conditions: Response thus becomes equal to (ii): The plot of this response must clearly contain the linear increase in time. Fastest increase in resonance: Homogenous solution: Trial function with : Solving: 2p Initial conditions: Plot linear amplitude increase: 7p e) The system may be subjected to two types of damping: viscous damping or Coulomb damping (dry friction). Sketch the free vibration response of the system in both cases. Indicate and explain three important differences. In case of viscous damping, the decay of the amplitude in time is exponential. The frequency of the oscillation is in the case of underdamping and has no oscillation for critical of overdamping. The system will reach its equilibrium position for. In the case of Coulomb damping the decay of the amplitude in time is linear. The frequency of the oscillation is for all cases. The system will never reach its equilibrium position, but will come to rest as soon as the static friction force is larger than the restoring force due to gravity. Moreover, the response is only piecewise harmonic and not continuously harmonic due to the non-linear nature of the EoM. Three correct differences: 3p ( for every difference that is also sketched) 3p 5
f) Consider case B. Derive the equations of motion of the system for arbitrary large angles and, using Lagrange s equation (iii). You may need the goniometric formula (iv). ( ) (iii) (iv) First, the constraint equations are derived that are used to express the position and velocity of the center of mass of the artists in terms of and (t). Position: Velocity: Kinetic energy: ( ) ( ) Substitution of the constraint equations: ( [ ]) Rewrite: Potential energy: No damping and no external forces so. 6
Lagrange s terms for : ( ) ( ) First equation of motion: Lagrange s terms for : ( ) ( ) Second equation of motion: Constraint equations: 2p ( for the position, for the velocity) Kinetic energy is : Rewrite kinetic energy: Potential energy: No damping and virtual work: Correct EoMs: 4p (2p for each correct EoM) 10p g) Show that the linearized equations of motion are equal to (v). Assume that and. Derive the expressions for the natural frequencies of the system. Prove that the corresponding natural mode shapes are (vi). [ ] { } [ ] { } { } (v) 7
{ } { } { } { } (vi) Linearizing means, and neglecting higher order terms. By doing so, (v) is obtained. With the simplifications given, the system reduces to: [ ] { Rewrite: [ ] { Trial function: } [ ] { } { } } [ ] { } { } { } { } Eigenvalue problem: [ ] { } { } Characteristic equation is obtained by equating the determinant to zero: Roots are obtained using the quadratic formula: ( ) Since both roots are negative. This means that the natural frequencies are: First natural mode is obtained by back substitution of ( ) : [ ] { } { } 8
Using the first equation it follows that: ( ) ( ) So the first natural mode is: { } { } Second natural mode is obtained by back substitution of ( ) : [ ] { } { } Using the first equation it follows that: ( ) ( ) So the second natural mode is: { } { } Linearizing: Correct eigenvalue problem: Characteristic equation: Natural frequencies: Natural modes: 2p ( for each correct natural frequency) 2p ( for each correct natural mode) 7p h) Apply a coordinate transformation towards the natural coordinates and. Normalize the natural modes with respect to the mass matrix and derive the uncoupled equations of motion. Explain the benefit of the uncoupled equations of motion over the coupled equations of motion. Modal matrix: [ ] [ ] Coordinate transformation: [ ] [ ][ ]{ } [ ] [ ][ ]{ } { } 9
For the mass matrix: [ ] [ ][ ] [ [ ] [ ] [ ] ] [( ) ( ) ( ) ( ) ] [ ( ) ( ) ] Normalized if: For the stiffness matrix: [ ] [ ][ ] [ ] [ ] [ ] [ ] [ ] [ ] Substitution of the values of and : [ ] [ ][ ] [ ] [ ] The uncoupled equations of motion become: [ ]{ } [ ]{ } { } The advantage of uncoupling the equations of motion is that the response of the system to either initial conditions or harmonic excitations can be solved for each natural mode independently. This means that the response of the 2DOF system can be written as a combination of the solutions of two 1DOF problems for which the solution is obtained with relative ease. Modal matrix: Mass matrix: 2p Normalizing, and : Stiffness matrix: 2p Benefit of uncoupling: 7p 10
i) In the case that the artists make continuous loops at constant angular velocity, their motion can be seen as prescribed in time:. Apply this on the non-linear equation of motion of the boat as derived in f). Prove that when and, the linearized equation of motion is equal to (vii). (vii) Recall the non-linear equation of motion as determined in f). Apply : Rewrite : [ ] Simplified case: Rewrite: Linearizing for small angles this can be rewritten to (vii): (vii) Prescribing : Rewriting : Simplifying: Linearzing: 4p 11
j) Because of gravity, it is very hard to make loops at a constant angular velocity. This means that in practice the boat is not excited harmonically in this way, but periodic instead. Explain extensively how this influences the procedure required to obtain the solution as a result of this periodic excitation. Transform the excitation to a series of harmonic functions using a Fourier series. Then solve for all individual components. The response is the superposition of all individual responses. Fourier series: Procedure: 2p 3p Overview of points problem 1: a b c d e f g h i j tot 10 4 5 7 3 10 7 7 4 3 60 12
Problem 2 Railway track Some structures consisting of beams are supported by an elastic foundation. A good example is the railway track, which consists of steel rails that are connected to the fixed world by means of wooden or concrete crossbars (also known as ties or sleepers) approximately every meters. The crossbars and rails are supported by their foundation which is typically made up from crushed stone. The figure below shows a schematic side view of the railway track in between two crossbars. The rail is modeled as an uniform beam of length, linear density and bending stiffness which is simply supported (hinged) at the locations of both crossbars. The supporting effect of the foundation is modeled as a distributed series of translational springs, which have stiffness per unit beam length. The transverse deformation of the beam is denoted by. As a result of passing trains, the beam may be subjected to an external loading. The effect of gravity is neglected. a) Derive the partial differential equation of motion for transverse vibrations in beams on an elastic foundation (i) using Hooke s law for bending (ii). In order to do so, make an accurate free-body diagram of an internal element of the beam. (i) (ii) The free-body diagram must contain an internal element in positive deflected state, moment and, shear forces and, foundation force and external loading. Force balance in transverse direction and moment balance about the left end of the internal element: [ ] [ ] [ ] 13
Neglecting higher order terms in the moment balance yields: Substitution in the force balance and rewriting yields: Apply Hooke s law and rearrange the terms yields the equation of motion (i): Free-body diagram: Balances: Solving: Total 2p (- for every incorrect force or moment) 2p ( for each balance) 5p b) Derive the general solution of the equation of motion (i) for free standing vibrations in the beam. Use the technique of separation of variables. Introduce the wave number as (iii). What is the benefit of using separation of variables to solve the equation of motion? (iii) Free vibrations: Standing waves gives us the opportunity to separate variables: Substitution in the equation of motion: Divide by and, rearrange and introduce the separation constant : 14
For the time part: Solution: For the spatial part: Introduce the wave number as given: Solution: The benefit of separating the variables is that the temporal behavior and spatial behavior can be studied separately. Hence, in this we it is known exactly what the shape of the beam is and how this shape is behaving in time. Separation of variables: Correct time part: Correct spatial part: 2p Benefit separation of variables: Total 6p c) Determine the boundary conditions at and. Use them to determine the expressions for the natural frequencies and the corresponding natural mode shapes. How many natural frequencies does the system have? Hint: Look carefully at your boundary conditions. You can reduce the matrix problem easily to a matrix problem. At both ends, the displacement and bending moment are zero. For the following equations: this yields 15
At this yields the following equations: In matrix vector form: [ ] { } { } Note that in order to satisfy both the first and the second equation reduces the system to:, which [ ] { } { } Characteristic equation is obtained by equating the determinant zero: For non-zero values of, this is case for: So: Using the definition of the wave number (iii): Solving for the natural frequencies: Back substitution of in the matrix vector equation yields: [ ] { } { } 16
From this it follows that, which leaves for the natural mode shapes: Boundary conditions: 2p Eliminate time part: Creating matrix: reducing matrix: Characteristic determinant: Natural frequencies: Natural modes: 8p d) With the natural frequencies and natural mode shapes known, explain extensively the procedure with which you would determine the response of the system to an initial excitation. In this procedure, what determines the expression of the mode shape coefficients? The response of the system can be written as a linear combination of the motion of all natural modes over time: The initial excitation of all modes is determined by multiplying the above expression by and integrating over the length of the beam. This integral can be simplified using the orthogonality conditions. Such that equations in terms of natural coordinates are obtained: The solution of this is simply: The total solution is obtained by back substitution. The coefficients are determined such that the orthogonality conditions hold and that the integral encountered in the above derivation is equal to 1. is combination of modes: Initial excitation of all modes: Natural coordinates: Orthogonality conditions: Determination of : 5p 17
e) In the case that the system would have been so complicated that its natural frequencies cannot be obtained analytically, an estimation of the lowest natural frequency can be obtained using Rayleigh s quotient. Proof that in this case the expression for Rayleigh s quotient is equal to (iv). ( ) (iv) Look at the spatial part of the equation of motion and separate the parts with and without : Multiply by and integrate over the length of the beam: The first integral must be integrated by parts twice: [ ] [ ] [ ] ( ) In view of the boundary conditions, the terms between square brackets are zero and the equation reduces to: ( ) Solving for yields Rayleigh s quotient (iv): ( ) ( ) Multiply by, integrate: 2p Integration by parts: 3p BCs for [ ] terms Final expression: 7p 18
f) Two possible trial functions that can be used in Rayleigh s quotient are (v). Determine for both trial functions whether it is a comparison function or an admissible function. Based on this, explain which trial function you think will result in the best approximation of the lowest natural frequency. Suppose that you would calculate Rayleigh s quotient using both trial functions. How would you know which of the two estimation is best, without knowing the exact solution? (v) For the first trial function, consider the geometric boundary conditions: The first and second derivatives are: Because the second derivative is not zero at and, the trial function does not fulfil the natural boundary conditions. Hence, the first trial function is an admissible function. For the second trial function, consider the geometric boundary conditions: The first and second derivatives are: Because the second derivative is zero at and, the trial function does fulfil the natural boundary conditions. Hence, the second trial function is a comparison function. Since a comparison function is a better approximation of the real behavior of the beam than an admissible function, using this trial function results in a better approximation of the lowest natural frequency. After solving Rayleigh s quotient of both trial function, the lowest value is the best approximation. 19
Check geometric BCs: Check natural BCs: Conclusion comp. / admis.: Comparisson is better: Lowest value is best: 2p ( for each trial function) 6p g) Suppose the beam is given an initial deflection according to the first trial function of (v). Using the procedure referred to in d), it is possible to calculate the response of the system on this initial excitation (you do not need to do this!). Which of all natural mode shapes will be in visible in this response? Why? Since the trial function is a polynomial and the natural modes are sines, an infinite number of modes are excited. Because the function is symmetric, only the oddnumbered modes are excited. Hence, only the odd-numbered natural modes will be visible in the response. Polynomial vs sines: Infinte modes: Symmetry / Odd-numbered: 3p Overview of points problem 2: a b c d e f g tot 5 6 8 5 7 6 3 40 20