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# Chapter 2 free vibration of single degree of freedom

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### Chapter 2 free vibration of single degree of freedom

1. 1. Free vibration of single degree of freedom (SDOF) Chapter 2
2. 2. Introduction •A system is said to undergo free vibration when it oscillates only under an initial disturbance with no external forces acting after the initial disturbance
3. 3. Introduction -SDOF •One coordinate (x) is sufficient to specify the position of the mass at any time •There is no external force applied to the mass •Since there is no element that causes dissipation of energy during the motion of the mass, the amplitude of motion remains constant with time, undampedsystem
4. 4. Introduction -SDOF •If the amplitude of the free vibration diminished gradually over time due to the resistance the resistance offered by the surrounding medium, the system are said to be damped •Examples: oscillations of the pendulum of a grandfather clock, the vertical oscillatory motion felt by a bicyclist after hitting a road bump, and the swing of a child on a swing under an initial push
5. 5. Free Vibration of an UndampedTranslation System •Equation of Motion using Newton’s Second Law ▫Select a suitable coordinate to describe the position of the mass or rigid body ▫Determine the static equilibrium configuration of the system and measure the displacement of the mass or rigid body ▫Draw the free body diagram of the mass or rigid body when a positive displacement and velocity are given ▫Apply Newton’s second law of motion
6. 6. FV of an undamped … • Newton’s second law • Applied to a undamped SDOF system F t mx    ) (    M(t)  J For rigid body undergoing rotational motion
7. 7. FV of an undamped … F t kx mx  ( )    mx kx  0
8. 8. FV of an undamped … • Equation of Motion using other methods ▫ D’Alembert’s Principle ▫ Principle of Virtual Displacements ▫ Principle of Conservation of Energy mx kx  0
9. 9. • Spring-Mass System in Vertical Position mx kx  0 st W  mg  k mx kx  W st    
10. 10. FV of an undamped …Solution! • The solution can be found assuming, substituting st x(t)  Ce ( ) ( ) 0 2 2   st st Ce k Ce dt d m   0 2 C ms  k  n i m k s           2 1 2 1      m k n  characteristic equation eigenvalues 0 2 ms  k 
11. 11. FV of an undamped …Solution! • The general solution, x t C ei nt C e i nt      1 2 ( ) e t i t i t    using,  cos  sin  x t A t A t n n ( ) cos sin 1 2   A and C can be determine from the initial conditions
12. 12. FV of an undamped…Solution! •The initial conditions at t =0, •The solution is 01)(xAtx txtxtxnnn  sincos)(00   02)(xAtxn
13. 13. FV of an undamped…Harmonic •Previous equations are harmonic function of time •The motion is symmetric about the equilibrium position of the mass •The velocity is maximum and the acceleration is zero at the equilibrium position •At the extreme displacement the velocity is zero and the acceleration is a maximum •The quantity is the natural frequency n
14. 14. FV of an undamped …Harmonic sin 2 cos A  A 1 A  A   amplitude x A A A x n                      1/ 2 2 2 0 0 2 2 2 1 2 1   phase angle x x A A n                        0 1 0 1 1 2 tan tan  x t A t A t n n ( ) cos sin 1 2  
15. 15. FV of an undamped …Harmonic 1/ 2 2 2 0 0 0                   n x A x             0 1 0 0 tan x x n    x t  A  t   n ( ) cos Substituting, the solution can be written using the relation, 1 0 0 A  A cos 2 0 0 A  A sin   0 0 x(t)  A sin  t  n
16. 16. • Obtain the free response of a) in the form b) in the form Initial condition are x(0) = 0.05 m and x(0) = - 0.3 m/s Example - Harmonic 2x128x  f (t) x A t A t n n sin cos 1 2   x  A  t  n sin
17. 17. Example -solution ttttxa8cos05.08sin0375.08cos05.08sin83.0)     062.083.005.0)222  Ab806.0062.005.0sin605.0)8(062.03.0cos   rad214.20927.927.)333.1(tan605.0806.0tan11   mtx214.28sin062.0
18. 18. FV of an undamped …Harmonic 1/ 2      m k n  1)Note the following aspect, st st W mg k     1/ 2          st n g   1/ 2 2 1          st n g f   1/ 2 2 1           f g st n n   
19. 19. FV of an undamped …Harmonic x t  A  t   n ( ) cos             2 ( ) sin cos  x t  A  t   A  t  n n n n  x t   A  t   A  t    n n n n ( ) cos cos  2 2 2)
20. 20. FV of an undamped …Harmonic 3) If the initial displacement is zero, t x A t x x t n n n n      sin 2 ( ) cos 0 0          If the initial velocity is zero, x t x t n ( ) cos 0 
21. 21. 4)The response of a SDOF system can be represent in the displacement ( ) – velocity ( ), known as the state space or phase plane FV of an undamped …Harmonic x x x t  A  t   n ( ) cos x t   A  t   n n ( ) sin A x A y  1 2 2 2 2   A y A x
22. 22. •The column of the water tank shown is 300 ft. high and is made of reinforced concrete with a tubular cross section of inner diameter 8 ft. and outer diameter 10 ft. The tank weighs 6 x 105lb with water. By neglecting the mass of the column assuming the Young’s modulus of reinforce concrete as 4 x 106psi, determine the following: Example -Harmonic
23. 23. •the natural frequency and the natural time period of transverse vibration of the water tank •the vibration response of the water tank due to an initial transverse displacement of 10 in. •the maximum values of the velocity and acceleration experienced by the water tank Example
24. 24. Example – solution 3 3 l P EI k    a) The stiffness of the beam, l= 3600 in. , E = 4 x 106 psi,  4 4   4 4  4 4 0 120 96 600 10 64 64 I d d in i         lb in l EI k 1545 / 3600 3 3(4 10 )(600 10 ) 3 6 4 3      0.9977 / sec 6 10 1545 386.4 5 rad m k n       6.2977 sec 2   n n   
25. 25. Example –solution b) Using the initial displacement of x0=0 and the initial velocity =0, 00sin)(tAtxninxxxAn1002/120200                 20tan010      nxtttx9977.0cos1029977.0sin10)(     
26. 26. Example –solution c) The velocity and acceleration can be determine by differentiating,      29977.0cos9977.010)(  ttx sec/977.99977.0100maxinAxn      29977.0sin9977.010)(2 ttx 2220maxsec/9540.99977.010inAxn
27. 27. •A simply supported beam of square cross section 5 mm x 5 mm and length 1 m, carrying a mass of 2.3 kg at the middle, is found have a natural frequency of transverse vibration of 30 rad/s. Determine the Young’s modulus of elasticity of the beam. Example -Harmonic
28. 28. Example – solution 3 192 l EI k   3  3  10 4 5 10 5 10 0.5208 10 12 1 I m         m k n   I m l E n 192 2 3   2 3 192 n m l EI k        9 2 10 2 3 207.0132 10 / 192(0.5208 10 ) 2.3 30.0 1.0 E   N m   
29. 29. •An industrial press is mounted on a rubber pad to isolated it from its foundation . If the rubber pad is compressed 5 mm by the self-weight of the press, find the natural frequency of the system Example -Harmonic mst3105 Hzradmgstn0497.7sec/2945.4410581.92/132/1             
30. 30. •An air-conditioning chiller unit weighing 2,000 lb is to be supported by four air springs. Design the air springs such that the natural frequency of vibration of the unit lies between 5 rad/s and 10 rad/s Example -Harmonic
31. 31. Example –solution sradn/5.7 4.3862000 mmkeqn inlbmkneq/1491.2915.74.386200022     inlbkinlbk/78.72/1491.2914
32. 32. Example •An electrical switch gear is supported by a crane through a steel cable of length 4 m and diameter 0.01 m. If the natural time period of the axial vibration of the switch gear is found to be 0.01 s, find the mass of the switch gear
33. 33. Example –solution mN/)10(064.4)10(07.201.04416112      lAEk stiffness cablennnf   211.0 mkn  201.02kgkmn53.1029)20( )10(0644.4262 
34. 34. Example •A bungee jumper weighing 160 lb ties one end of a elastic rope of length 200 ft and stiffness 10 lb/in to bridge and the other to himself and jumps from the bridge. Assuming the bridge to be rigid, determine the vibratory motion of the jumper about his static equilibrium position
35. 35. Example –solution sec/88.361,1)12(200)4.386(22mghjumper theofvelocity 221inghvormv   inlbkinchlbm/10, sec4.3861602    88.362,1)0(,0)0( :position mequilibriu about the00txxtxx )sin()(:jumper theof response00tAtxn
36. 36. Example –solution inkmxxxxnn12.2774.386160101361Awhere, 00220021                     0tanwhere, 010       nnxx   
37. 37. Free Vibration of an UndampedTorsionalSystem •If a rigid body oscillate about a specific reference axis, the resulting motion is called torsionalvibration •The displacement of the body is measured in terms of a angular coordinate •The restoring moment may be due to the torsion of an elastic member or to the unbalanced moment of a force or couple
38. 38. lGIMot 324dIo   lGdlGIMkott324  
39. 39. •Equation of Motion ▫The equation of the angular motion of the disc about its axis can be derived by using Newton’s second laws 00tkJ 0Jktn tnkJ02 021Jkftn 
40. 40. •Important aspects of this system ▫If the cross section of the shaft supporting the disc is not circular, an appropriate torsionalspring constant is to be used ▫The polar mass moment of inertia of a disc is given by ▫The general solution gWDDhJ832240  tAtAtnnsincos)(21 01AnA/02 
41. 41. Example •The figure shows a spacecraft with four solar panel. Each panel has the dimension of 5 ft. x 3 ft. x 1 ft. with a density of 0.1 lb/in3 , and is connected to the body by aluminum rods of length 12 in. and diameter 1 in. Determine the natural frequency of vibration each panel about the axis of the connecting rod
42. 42. Example –solution 5820.1386.40.283)112)(12)(3(5panel a of massm      98.170361125820.112m axis- xabout the panel theof inertia momentof mass22220  baJ 4440098175.0132 32 rod of inertia momentofpolarind I  radinlblGIkt/101089.312098175.0108.3480 sec/4841.13210radJktn   
43. 43. Example •Find the equation of motion of the uniform rigid bar OA of length l and mass m shown in the figure. Also find its natural frequency
44. 44. Example –solution llkaakkJt210:motion ofEquation  2220312121 wheremllmmlJ     03122212lkakkmlt  21222213     mllkakktn
45. 45. Rayleigh’s Energy Method •Uses the energy method to find the natural frequencies of a single degree of freedom systems •The principle of conservation of energy, in the context of an undampedvibrating system, can be restated as •Subscript 1 denote the time when the mass is passing through its static equilibrium position (U1=0) 2211UTUT
46. 46. Rayleigh’s Energy Method •Subscript 2 indicate the time corresponding to the maximum displacement of the mass (T2=0) •If the system is undergoing harmonic motion, then T1and U2denote the maximum values 2100UT maxmaxUT
47. 47. Example •Find the natural frequency of the transverse vibration of the water tank considered in the first example by including the mass of the column
48. 48. Example      2 3  3 max 2 3 2 3 6 x l x l y l x EI Px y x     The maximum kinetic energy of the beam, yx dx l m T l 2 0 max 2 1   
49. 49. Example    2 3  3 max 3 2 x l x l y y x      x l x  dx l y l m T l        0 2 3 2 2 3 max max 3 2 2  2 max 7 6 2 max max 140 33 2 1 35 33 2 4 l m y l y l m T            
50. 50. Example 2 max max 2 1 T m y eq   m m eq 140 33  eff eq M  M m meq denotes the equivalent mass of the cantilever at the free end, its maximum kinetic energy The total effective mass acting at the end, M is the mass of the water tank M m k M k eff n 140 33    
51. 51. Free Vibration with Viscous Damping •The viscous damping force F is proportional to the velocity, •c is the damping constant or coefficient of viscous damping •The negative sign indicates that the damping force is opposite to the direction of velocity xcF
52. 52. Free Vibration with Viscous Dampingkxxcxm 0kxxcxm
53. 53. FV with Viscous Damping - Solution mxcx  kx  0 0 2 ms cs  k    st x t  Ce m k m c m c m c c mk s             2 2 1,2 2 2 2 4   s t s t x t C e 1 C e 2 1 2 The general solution  
54. 54. FV with Viscous Damping •Critical Damping Constant and Damping Ratio ▫The critical damping is defined as the damping constant for which the radical becomes zero ▫The damping ratio is defined as the ratio of the damping ratio to the critical damping constant 022     mkmccncmkmmkmc222 ccc/
55. 55. FV with Viscous Damping •Critical Damping Constant and Damping Ratio ▫the solution, nccmcccmc  22 ns122,1 ttnneCeCtx       121122
56. 56. FV with Viscous Damping ▫The nature of the roots and hence the behavior of the solution depends upon the magnitude of damping; ▫Case 1: Underdamped s vibrationundamped toleads,0 nis211 overdamped critical, d,underdampe :cases three,0 mkmcorccorc/2/1 nis221 expressed becan roots theand negative is1 condition, for this2
57. 57. FV with Viscous Damping   i nt i nt x t C e C e                       2 2 1 2 1 1 xt e C t C t n n nt      ' 2 2 ' 2 1  cos 1  sin 1  Case 1: Underdamped n nx x C x and C    2 ' 0 0 0 2 ' 1 1     For the initial condition, 0 0 x(t  0)  x and x(t  0)  x
58. 58.                t x x x t e x t n n n n nt         2 2 2 0 0 0 sin 1 1 cos 1  the solution, d n    2  1 The frequency of damped vibration is
59. 59. FV with Viscous Damping    ' 2 2 ' 2 0 1 X  X  C  C  '  2 ' 1 1 tan C /C    The constants are,  '  1 ' 2 1 0  tan C /C               x t Xe t n nt 2 sin 1     0 2 0 cos 1         x t X e t n nt the solution can be expressed as ,
60. 60. FV with Viscous Damping ▫ Case 2: Critical damped n c m c s  s     2 1 2 or c c or c m k m c  1  / 2  / In this case the two roots and are equal, 1 2 s s the solution,   x t C C t e nt    1 2 ( ) For the initial condition, 0 0 x(t  0)  x and x(t  0)  x 1 0 C  x 2 0 0 C x x n   
61. 61. tnnetxxxtx000)( zero odiminish t eventually lmotion wil the,0 as 0Since . isequation by the drepresentemotion the teaperiodictn
62. 62. FV with Viscous Damping ▫ Case 3: Overdamped or c c or c m k m c  1  / 2  / In this case the two roots and are real and distint, 1 2 s s the solution, nt nt x t C e C e                       1 2 1 1 2 2 ( )  1 0 2 1      n s     1 0 2 2      n s   
63. 63. FV with Viscous Damping ▫ Case 3: Overdamped For the initial condition, 0 0 x(t  0)  x and x(t  0)  x   2 1 1 2 0 2 0 1           n n x x C    2 1 1 2 0 2 0 2            n n x x C  the roots are negative , the motion diminishes exponentally with time the motion represented by the equation is aperiodic. Since
64. 64. FV with Viscous Damping
65. 65. FV with Viscous Damping •The nature of the roots with varying values of damping can be shown in a complex plane. The semicircle represent the locus of the roots for different values of damping ratio in the range of 0 to 1
66. 66. FV with Viscous Damping •A critically damped system will have the smallest damping required for aperiodicmotion: hence the mass returns to the position of rest in the shortest possible time without overshooting •The figure represent the phase plane or state space of a damped system
67. 67. FV with Viscous Damping •Logarithmic Decrement ▫Represent the rate at which the amplitude of a free damped vibration decreases. It is defined as the natural logarithm of the ratio of any successive amplitude        nnneteXteXxxntnt      02001021coscos21mcxxdnndn221212ln2221             damping smallfor 2222      or
68. 68. FV with Viscous Damping •Energy Dissipated in Viscous Damping ▫The rate of change of energy with time is given ▫The negative sign denotes that energy dissipate with time. Assuming a simple harmonic motion 22 velocity force     dtdxccvFvdtdW tdcXdtdtdxcWdddtn        22022/20cos2XcWd
69. 69. FV with Viscous Damping •Energy Dissipated in Viscous Damping ▫The fraction of the total energy of the vibrating system that is dissipated in each cycle of motion is called the specific damping capacity ▫Another quantity used to compare damping capacity of engineering materials is called loss coefficient and is define as the ratio of energy dissipated per radian and the total energy constant4222222212                  mcXmXcWWddd  WWWW   22/ tcoefficien loss    
70. 70. FV with Viscous Damping •TorsionalSystems with Viscous Damping  tcT d,underdampe for the case;linear in the as00ttkcJ is,motion ofequation thend21 0Jktn 0022JkcJcccttnttct  
71. 71. Example-FV viscous damping •The human leg has a measured natural frequency of around 20 Hz when in its rigid (knee-locked) position in the longitudinal direction ( i.e., along the length of the bone) with damping ratio of ξ= 0.224. Calculate the response of the tip of the leg bone to an initial velocity of v0= 0.6m/s and zero initial displacement ( this would correspond to the vibration induced while landing on your feet, with your knee locked from a height of 18 mm. What is the maximum acceleration experience by the leg assuming no damping?
72. 72. Example-FV viscous damping •Highway crash barrier are design to absorb a vehicle’s kinetic energy without bringing the vehicle to such an abrupt stop that the occupants are injured. Knowledge of the barrier’s materials provide the spring constant k and the damping coefficient c; the mass m is the vehicle mass. For this application , t=0 denotes the time at which the moving vehicle contacts the barrier at x=0; thus v(0) is the speed of the vehicle at the time of contact and x(0) =0. The applied force is zero. Most of the barrier’s resistance is due to the term cv, and it stops resisting after the vehicle comes to rest; so the barrier does reverse the vehicle motion. Cont……
73. 73. Example-FV viscous damping •A particular barrier’s construction gives k=18000 N/m and c = 20000 N s/m. A vehicle 1800 kg vehicle strikes the barrier at 22 m/s. Determine how long it takes for the vehicle to come to rest, how far the vehicle compress the barrier, and the maximum deceleration of the vehicle
74. 74. Example-FV viscous damping •An underdampedshock absorber is to be design for a motorcycle of mass 200 kg. When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as shown. Find the necessary stiffness and damping constants of the shock absorber if the damped period is to be 2 s and the amplitude x1is to be reduced to one-fourth in one half cycle (i.e., x1.5 = x1/4 ). Also find the minimum initial velocity that leads to a maximum displacement of 250 mm
75. 75. Example-Solution •Approach: We use the equation for the logarithm decrement in term of the damping ratio, equation for the damped period of vibration, time corresponding to maximum displacement for an underdampedsystem, and envelop passing through the maximum points of an underdampedsystem 221127726.2)16ln(ln      xx16/4/,4/15.1215.1xxxxx 4037.0
76. 76. Example -Solution 21222        nddsradn/4338.34037.01222      msNmcnc/54.13734338.320022 msNccc/4981.55454.13734037.0 mNmkn/2652.23584338.320022
77. 77. Example - Solution 2 1 sin t  1 d   0.3678 sec sin 0.9149 1 1     t x Xe nt      2 1 The displacement of the mass attain its maximum value at time t, sin sin 1 0.4037 0.9149 2 1 1 t  t    d   The envelop passing through the maximum points,
78. 78. Example - Solution x Xe nt      2 1 The envelop passing through the maximum points,  2 0.40374.4330.3678 0.25 1 0.4037    Xe X  0.4550m The velocity of the mass can be obtained by differentiating, xt Xe t d nt   sin  
79. 79. Example -Solution The velocity of the mass, ttXetxnddntncossin 2010ndXXxtx  sm/4294.14037.01)4338.3)(4550.0(2  
80. 80. Example •The maximum permissible recoil distance of a gun is specified as 0.5 m. If the initial recoil velocity is to be 8 m/s and 10 m/s, find the mass of the gun and the spring stiffness of the recoil mechanism. Assume that a critically damped dashpot is used in the recoil mechanism and the mass of the gun has to be at least 500 kg
81. 81. Example - Solution        1 0 0 0 x t x x x t e E t n n            2 0 2 0 0 x t e x x t x t E n n nt           Let tm = time at which x=xmax and v=0 occur. Here x0 = 0 and v0 =0 initial recoil velocity. By setting v(t)=0,    3 1 0 0 0 0 0 E x x x x x t n n n n m            
82. 82. Example -Solution ntmexetxxmn   100max   7178.25.0max0nnexx smx/10 when,0 sec/3575.77178.2*5.0/10radn kg, 500 isgun of masswhen mNmkn/403.066,275003575.722
83. 83. FV with Coulomb Damping •In vibrating structure, whenever the components slide relative to each other, dry friction damping appears internally •Coulomb’s law of dry friction states that, when two bodies are in contact, the force required to produce sliding is proportional to the normal force acting in the plane of contact mgWNF
84. 84. FV with Coulomb Damping •Equation of Motion ▫Case 1 When x is positive and dx/dtis positive or when x is negative and dx/dtis positive (half cycle) Nkxxm
85. 85. FV with Coulomb Damping •Equation of Motion ▫Case 1 This is a second-order nonhomogeneousDFQ. The solution is  kNtAtAtxnn  sincos21kNxA 30102AkNxxt   4/2/ position, extreme At the02nn 0)0(/2)0(0txandkNxtx
86. 86. FV with Coulomb Damping
87. 87. FV with Coulomb Damping • Equation of Motion ▫ Case 2  When x is positive and dx/dt is negative or when x is negative and dx/dt is negative (half cycle) mx kx N
88. 88. FV with Coulomb Damping ▫Case 2 The solution is  kNtAtAtxnn  sincos43kNxA  0304A      kNxx   2/ tposition, extreme At the01n0)0()0(0txandxtx
89. 89. FV with Coulomb Damping             kNkNxr   2cycles ofnumber 0kNxn/at stopsmotion 
90. 90. FV with Coulomb Damping ▫The previous solutions can be expressed as a single equation ▫Where sgn(y) is called the signumfunction and it is define as 0)sgn(kxxmgxm 101111    yforyforyfor
91. 91. FV with Coulomb Damping •The equation of motion is nonlinear with Coulomb damping, while it is linear with viscous damping •The natural frequency of the system is unaltered with the addition of Coulomb damping, while it is reduced with addition of viscous damping •The motion is periodic with Coulomb damping, while it can be nonperiodicin a viscoulydamped (overdamped) system
92. 92. FV with Coulomb Damping •The system comes to rest some time with Coulomb damping, whereas the motion theoretically continues forever with viscous and hysteresis damping •The amplitude reduces linearly with Coulomb damping, whereas it reduces exponentially with viscous damping
93. 93. FV with Coulomb Damping •In each successive cycle, the amplitude of motion is reduced by the amount 4μN/k, so the amplitude at the end of any two consecutive cycles are related: The slope of the enveloping straight lines is kNXXmm/41                kNkNnn   224
94. 94. FV with Coulomb Damping
95. 95. Example •A metal block, placed on a rough surface, is attached to a spring and is given an initial displacement of 10 cm from its equilibrium position. It is found that the natural time period of motion is 1 s and the amplitude reduces by 5 cm in each cycle. Find (a) the kinetic coefficient between the metal block and the surface and (b) the number of cycles of motion executed by the block before it stops.
96. 96. FV with Coulomb Damping •TorsionalSystems with Coulomb Damping ▫If a constant frictional torque acts on a torsionalsystem, the equation governing the angular oscillations can be derived similar to that of the linear, torquedampingconstant thedenoted TTkJt 00Jktn trkTr2 amplitude, the0 TkJt 0             kTkTr2ceased,motion 0
97. 97. FV with Hysteretic Damping •The damping caused by the friction between the internal planes that slip or slide as the material deforms is called hysteretic damping •This causes a loop to be formed in the stress- strain of force-displacement curve •The energy loss in one loading and unloading cycle is equal to the area enclosed by the loop •The energy loss per cycle is independent of the frequency but proportional to the square of the amplitude
98. 98. FV with Hysteretic Damping
99. 99. FV with Hysteretic Damping Damping coefficient c and h is the hysteretic damping constant  h c  The energy dissipated, 2 W hX The force -displacement relation, F  k ihx complex stiffness, k ih  k1i 
100. 100. FV with Hysteretic Damping The hysteresis logarithm decrement,                 ln ln 1 j 1 j X X          1 2 2 j 1 j X X
101. 101. FV with Hysteretic Damping ratio damping viscousequivalent The  kheq  2is,constant damping equivalent The      hkmkmkcceqceq 22isfrequency Themk  kheq22   