Module 5 (CE-409: Introduction to Structural Dynamics and Earthquake Engineering)

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Module 5 (CE-409: Introduction to Structural Dynamics and Earthquake Engineering)

  1. 1. University of Engineering and Technology Peshawar, Pakistan CE-409: Introduction to Structural Dynamics and Earthquake Engineering MODULE 5: UNDAMPED & DAMPED VIBRATIONS IN S.D.O.F SYSTEMS SUBJECTED TO HARMONIC FORCES Prof. Dr. Akhtar Naeem Khan & Prof. Dr. Mohammad Javed drakhtarnaeem@nwfpuet.edu.pk mjaved@nwfpuet.edu.pk 1
  2. 2. Harmonic force A harmonic force is one whose variation which with time is defined by any one of the following equations p(t)  poSin( t) or po Cos(t) Where po is the amplitude or maximum value of force and ω is its frequency also called as exciting frequency or forcing frequency; T=2π/ω is the exciting period or forcing period. The equations used in this module are strictly applicable to po sin (ωt) Time variation of harmonic force CE-409: MODULE 5 (Fall-2013) 2
  3. 3. Harmonic forces A common source of such a sinusoidal force is unbalance in a rotating machines (such as turbines, electric motors and electric generators, as well as fans, or rotating shafts). Unbalance cloth in a rotating drum of a washing machine is also an harmonic force. When the wheels of a car are not balanced, harmonic forces are developed in the rotating wheels. If the rotational speed of the wheels is close to the natural frequency of the car’s suspension system in vertical direction , amplitude of vertical displacement in the car’s suspension system increases and violent shaking occur in car due to match of frequency of the force (due to vertical component of harmonic forces acting at unbalanced mass centre) with natural 3 frequency of car’s suspension system in vertical direction, ωn CE-409: MODULE 5 (Fall-2013) 3
  4. 4. Response of undamped systems subjected to harmonic forces CE-409: MODULE 5 (Fall-2013) 4
  5. 5. Response of undamped systems to harmonic forces The equation of motion for harmonic vibration of undamped system is: m  ku  p oSin( t) u The solution to the equation is made up of two parts. The first part is the solution which correspond to forced vibration and is known as the Particular Solution. The corresponding vibration is known as Steady state vibration, for its present because of the applied force no matter what the initial conditions . The second part is the solution to the free vibration, which does not require any forcing function, this part is known as the Complimentary solution. The corresponding vibration is known as Transient Vibration, which depends on the initial conditions CE-409: MODULE 5 (Fall-2013) 5
  6. 6. Particular solution of undamped harmonic vibrations It can be derived that the particular solution of undamped vibration is as follows: po u p (t)  . k ω 1  1-     n   ωn 2 Sin( t) where   n is termed as frequency ratio For the sack of simplicity, we will use rω in our lectures to represent ω/ωn po  u p (t)  k  1  Sin( t)  2  1 - r  where    n up(t) is the displacements corresponding to the Particular solution (i.e due to forced vibration). CE-409: MODULE 5 (Fall-2013) 6
  7. 7. Complimentary solution of undamped harmonic vibrations Complementary solution of undamped vibration is given as follows: u c (t)  ACos(ωn t)  BSin( ωn t) uc(t) is the displacements corresponding to the Complimentary solution (i.e due to free vibration) and depends on initial conditions. CE-409: MODULE 5 (Fall-2013) 7
  8. 8. Complete solution of undamped harmonic vibrations Complete solution is the sum of complementary solution, uc(t), and particular solution, up(t) po 1 u(t)  ACos(ωn t)  BSin( ωn t)  Sin( t) 2 k 1 - r The constants ‘A’ and ‘B’ are determined by imposing initial conditions i.e.,   u  u(0) and u  u(0)  u(0) p o rω   u(t)  u(0)Cos(ω n t)   Sin(ω n t)  2  ωn  k 1 - rω      Transient po 1 Sin (t) 2 k 1 - rω    CE-409: MODULE 5 (Fall-2013) Steady state 8
  9. 9. Complete solution of undamped harmonic vibrations  The transient vibration exist even if u(0)  u(0)  0. In such case the complementary part of solution given on previous slide specializes to:  0 p o rω   u c (t)  0 * Cos(ωn t)   Sin(ωn t) 2 k 1 - rω   ωn po or u c (t)  k  rω  Sin(ωn t)  2  r 1    ω The complete solution is then specialized to the following form po 1 Sin (t)  rωSin(ωn t)  or u(t)  2 k 1 - rω CE-409: MODULE 5 (Fall-2013) 9
  10. 10. Amplitude of ‘Static’ deflection due to harmonic force If the force is applied slowly then   0 and the equation u of motion under harmonic force becomes: m  ku  poSin( t) u ku  p oSin( t) or u st  po k Sin( t) The subscript “st” (standing for static) indicate the elimination of acceleration’s effect The maximum value of static deformation, (ust )o can be interpreted as the deformation corresponding to the amplitude of p of the force po: u st o  po k For brevity we will refer to (ust)o as the static deformation CE-409: MODULE 5 (Fall-2013) 10
  11. 11. Effect of Frequency ratio, rω, on the direction of structural displacements po u p (t)  k  1  Sin( t) can be written as :  2  1 - r   1  p u p (t)  u st o   u st o Sin( t) where o 2 k 1 - r  It can be observed from this equation that up(t) has negative sign when frequency ratio, rω >1 (i.e. ω>ωn), and vice versa. A Graph 1 on next slide is plotted b/w frequency ratio, rω and 1 - r 2 up is positive if this term is positive and vice versa CE-409: MODULE 5 (Fall-2013) 11
  12. 12. Effect of Frequency ratio, rω, on the direction of structural displacements 1 2 1 - r rω  ω ωn CE-409: MODULE 5 (Fall-2013) 12
  13. 13. Effect of Frequency ratio, rω, on the direction of structural displacements Following observation can be made from the plot given on previous slide When rω < 1 ( i.e ω < ωn ), the displacement is positive, indicating that up(t) and p(t) has same directions. The displacement is said to be in phase with the applied force. When rω > 1 ( i.e ω > ωn ), the displacement is negative , indicating that the u(t) and p(t) has apposite direction directions. The displacement is said to be out of phase with the applied force. CE-409: MODULE 5 (Fall-2013) 13
  14. 14. Effect of Frequency ratio, rω, on the direction of structural displacements p oSin( t) u n u p oSin( t) Structure displaces in the direction of force if ω/ωn <1 n Structure displaces opposite to direction of force if ω/ωn >1 CE-409: MODULE 5 (Fall-2013) 14
  15. 15. Deformation response factor, Rd po 1 1 u p (t)  Sin( t)  u st o Sin( t) 2 2 k 1 - r 1 - r u p (t)  u oSin( ωt -  )  u st o R dSin( ωt -  ) Another mathematical form of the above mentioned equation is: uo 1 0o   Where R d  u st o 1 - r 2 and   180 o     n i.e., r  1   n i.e., r  1 Where Rd = Dynamic Magnification factor or Deformation (or displacement) response factor, uo=Amplitude of dynamic displacement, and, φ= Phase angle CE-409: MODULE 5 (Fall-2013) 15
  16. 16. Influence of Frequency ratio, rω, on Deformation response factor, Rd Rd R d  1 at rω  2 R d  1 when rω  Nearly static response . Load may be defined as quasi-static when rω ≤ 0.2 r   2 n CE-409: MODULE 5 (Fall-2013) 16
  17. 17. Influence of Frequency ratio, rω, on Deformation response factor, Rd Following observation can be made from the plot When rω is small ( i.e force is ‘slowly varying’), Rd is only slightly greater than 1 or in the other words amplitude of dynamic deformation, uo, is almost same as amplitude of static deformation, (ust)o. When r  2 (i.e   2n ), Rd  1 and the dynamic deformation amplitude is less than static deformation When r increases beyond 2 , R d become smaller and becomes zero as r   When rω is close to 1.0, Rd is many times larger than 1 CE-409: MODULE 5 (Fall-2013) 17
  18. 18. Problem M5.1 A video camera, of mass 2.0 kg, is mounted on the top of a bank building for surveillance. The video camera is fixed at one end of a tubular aluminum rod whose other end is fixed to the building as shown in Fig. The wind-induced force acting on the video camera, is found to be harmonic with p(t) = 25 sin 75t N. Determine the cross-sectional dimensions of the aluminum tube if the maximum amplitude of vibration of the video camera is to be limited to 0.005 m. E Aluminum = 71 GPa Posin ωt CE-409: MODULE 5 (Fall-2013) 18
  19. 19. Response of damped systems under Harmonic forces CE-409: MODULE 5 (Fall-2013) 19
  20. 20. Response of damped systems under Harmonic forces  m  cu  ku  poSin( t) u The equation of motion for harmonic vibration of damped system is: This equation is to be solved subjected to initial conditions   u(t)  u(0) and u(t)  u(0) u p (t)  CSin( ωt)  DCos(ωt) The particular solution of this differential equation is CE-409: MODULE 5 (Fall-2013) 20
  21. 21. Response of damped systems under Harmonic forces Where 2   po 1 - r   C k  1 - r 2 2  2r 2          2r po    D 2 k  1 - r 2  2r 2        CE-409: MODULE 5 (Fall-2013) 21
  22. 22. Response of damped systems under Harmonic forces  ACos(ωD t)  BSin( ωD t) The complementary solution is: u c (t)  e  ζω n t ACos(t) u(t)  e ωD t)  BSin( ωD   The complete solution is:  ζω n t  Transient CSin( ωt)  DCos(ωt)  Steady state CE-409: MODULE 5 (Fall-2013) 22
  23. 23. Steady state response of damped systems under Harmonic forces Transient response (difference b/w total response and steady-state response diminishes after few cycles of forced vibration in damped systems and after this stage i.e., u(t)  u p (t) as u c (t)  0 , u(t)  uc (t)  up (t) , up (t) u(t) ust o t/T CE-409: MODULE 5 (Fall-2013) 23 23
  24. 24. Effect of rω and ζ on the Deformation response factor, Rd u p (t)  u oSin( ωt -  )  u st o R dSin( ωt -  ) The Steady state deformation can be rewritten as:  D C 2  D 2 and   Tan -1     C Substituting the values of C and D (given on slide 20) in above given equation results in: Where u o  uo Rd   u st o 1 - r   2  2r and   Tan  2 1 - r  -1  CE-409: MODULE 5 (Fall-2013)  2r  1 2     2 24
  25. 25. Effect of rω and ζ on the Deformation response factor, Rd , rω CE-409: MODULE 5 (Fall-2013) 25
  26. 26. Effect of rω and ζ on the Deformation response factor, Rd Following observation can be made from the plot Damping reduce Rd for all values of frequency ratio, rω. However rate of reduction highly depend on the magnitude of rω (around 0.5 to 1.5) If the rω is around 0.2 and below, ( i.e force is ‘slowly varying’), Rd is only slightly greater than 1 and thus unaffected by damping. Thus u o  u st o  p o /k provided r  1 If the rω very high, around 2 and above,( i.e force is ‘rapidly varying’), Rd tends to zero. In other words Rd is unaffected by damping. uo can be approximated as: 2 uo  po /m provided r  1 CE-409: MODULE 5 (Fall-2013) 26
  27. 27. Effect of rω and ζ on the Deformation response factor, Rd If the rω≈ 1( i.e frequency of force is close to natural frequency), Rd is very sensitive to damping. If rω =1, Rd as given by: uo Rd   u st o becomes uo Rd   u st o 1 - r   2  2r  1 2 1 -1   2ζ.1 1 2 2 2 2 1  2 This means that for a system with 5% of critical damping the maximum displacement of the dynamic response is 10 times the equivalent static displacement. CE-409: MODULE 5 (Fall-2013) 27
  28. 28. Effect of rω and ζ on the Deformation response factor, Rd The last equation on previous slide can be alternatively written as uo  u st o 2ζ po 1 po 1 . .   2 k 2ζ ω n m  c 2  2mω n      po po mωn . uo   2 c cωn ωn m CE-409: MODULE 5 (Fall-2013) 28
  29. 29. Problem M 5.2 An air-conditioning unit weighing 1200 lb is bolted at the middle of two parallel simply supported steel beams. The clear span of the beams is 8 ft. The second moment of cross-sectional area of each beam is 10 in4. The motor in the unit runs at 300 rpm and produces an unbalanced vertical force of 60 lb at this speed. Neglect the weight of the beams and assume 1% viscous damping in the system; for steel E = 30,000 ksi. Determine the amplitudes of steady-state deflection. CE-409: MODULE 5 (Fall-2013) 29
  30. 30. Dynamic Response Factors Now we introduce deformation (or displacement), velocity, and acceleration response factors that are dimensionless and define the amplitude of these three response quantities. u p (t)  u oSin( ωt -  )  u st o R dSin( ωt -  ) The steady-state displacement from slide 23 is reproduced below By differentiating both sides, a relation can be developed between Velocity response factor, Rv and Rd ,which is given below. R v  rω R d Similarly, another relation can also be developed between Acceleration response factor, Ra and Rd ,which is given below. R a  rω R d  rω R v 2 CE-409: MODULE 5 (Fall-2013) 30
  31. 31. Rd Rv Ra Rd,Rv and RCE-409: MODULE 5 (Fall-2013) harmonic force. a for a damped system excited by 31
  32. 32. Problems 1. A spring-mass-damper system is subjected to a harmonic force. The amplitude is found to be 20 mm at resonance and 10 mm at a frequency 0.75 times the resonant frequency. Find the damping ratio of the system. 2. An air compressor of mass 100 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplitude 100 N is applied to the compressor, the maximum steady-state displacement of 5 mm occurred at a frequency of 300 rpm. Determine the equivalent stiffness and damping constant of the foundation. 3. A 50-kg machine tool is mounted on an elastic foundation. An experiment is run to determine the stiffness and damping properties of the foundation. When the tool is excited with a harmonic force of magnitude 8000 N at a variety of frequencies, the maximum steady– state amplitude obtained is 2.5 mm, occurring at a frequency of 32 Hz. Use this information to determine the stiffness and damping ratio of the foundation. CE-409: MODULE 5 (Fall-2013) 32 32
  33. 33. Problems 15 kips p(t) 4. The steel frame shown in figure supports a rotating machine which exerts a horizontal force at the girder level, p(t)=200 Sin 5.3t lb. Assuming 5% of critical damping, determine: 15 ft (a) The amplitude of the dynamic displacement and (b) The amplitude of equivalent static force. Take E= 29,000 ksi and I= 69.2 in4 HA1M5 Solve problems 1 and 4 CE-409: MODULE 5 (Fall-2013) 33
  34. 34. Vibration Isolation CE-409: MODULE 5 (Fall-2013) 34
  35. 35. Vibration Isolation High vibration levels can cause machinery failure, as well as objectionable noise levels. A common source of objectionable noise in buildings is the vibration of machines that are mounted on floors or walls. A typical problem is a rotating machine (such as a pump, AC compressor, blower, engine, etc) mounted on a roof, or on a floor above the ground floor. The problem is usually most apparent in the immediate vicinity of the vibration source. However, mechanical vibrations can transmit for long distances, and by very circuitous routes through the structure of a building, sometimes resurfacing hundreds of feet from the source. CE-409: MODULE 5 (Fall-2013) 35
  36. 36. Vibration Isolation A related problem is the isolation of vibration-sensitive machines from the normally occurring disturbances in a building (car or bus traffic, slamming doors, foot traffic, elevators…). Examples of sensitive machines include surgical microscopes, electronic equipment, lasers, and computer disk drives. A common example of a vibration source is shown in figure, a large reciprocating air conditioning compressor weighing 20,000 pounds, mounted on a roof. Annoying noise levels at multiples of the compressor rotational frequency, predominantly 60 and 120 Hz, were measured in the rooms directly below the compressor. Also, this type of compressor (reciprocating) is notorious for high vibration levels. Centrifugal or scroll type compressors are much quieter, but more expensive. CE-409: MODULE 5 (Fall-2013) 36
  37. 37. Vibration Isolation CE-409: MODULE 5 (Fall-2013) 37
  38. 38. Vibration Isolators A Consider a vibrating machine, bolted to a rigid floor ( see figure a on next slide). The force transmitted to the floor is equal to the force generated in the machine. The transmitted force can be decreased by adding a suspension and damping elements (often called vibration isolaters) Figure b , or by adding what is called an inertia block, a large mass (usually a block of cast concrete), directly attached to the machine (Figure c). Another option is to add an additional level of mass (sometimes called a seismic mass, again a block of cast concrete) and suspension (Figure d). CE-409: MODULE 5 (Fall-2013) 38
  39. 39. Vibration Isolators Input force, p(t) = po sin ωt Transmitted force, fT(t) Vibration isolation systems: a) Machine bolted to a rigid foundation b) Supported on isolation springs, rigid foundation c) machine attached to an inertial block d) Supported on isolation springs, non-rigid foundation (such as a floor); or machine on isolation springs, seismic mass and second level of isolator springs CE-409: MODULE 5 (Fall-2013) 39
  40. 40. Vibration Isolators Typically vibration isolators employ a helical spring to provide stiffness, and an elastomeric layer (such as neoprene) to provide some damping. Other types use a solid elastomeric element for both the stiffness and the damping. Application of elastic sleeper pads for vibration isolation, and adjustment of track CE-409: MODULE 5 (Fall-2013) stiffness 40
  41. 41. Transmission of harmonic forces to base p(t )  po Sint Consider the mass-spring-damper system subjected to a harmonic force. The force transmitted to the base, fT, is:  f T  fS  f D  ku  cu m k c By substituting, solving and rearranging we get: TR   fT o po  1  2rω 2 1 - r  ω 2 2  2rω  fT 2 Where TR is used to represent Transmissibility CE-409: MODULE 5 (Fall-2013) 41
  42. 42. Transmission of harmonic forces to base CE-409: MODULE 5 (Fall-2013) 42
  43. 43. Transmission of harmonic forces to base The magnitude of transmitted force reduces with increase in rω beyond 2 . The force transmitted to base can be decreased by decreasing the value of ωn in such a way so that rω  2 The force transmitted to the base can also be reduced by decreasing damping ratio. Although damping reduces the amplitude of mass for all frequencies, it reduces maximum force transmitted to the foundation only if rω  2 . Below that value, the addition of damping increases the transmitted force CE-409: MODULE 5 (Fall-2013) 43
  44. 44. Transmission of harmonic forces to base If the speed of a machine (forcing frequency) varies, we must compromise in choosing the amount of damping to minimize the transmitted force. The amount of damping should be sufficient to limit the amplitude of displacement and the transmitted force, while passing through the resonance, but not so much to increase unnecessarily the force transmitted at the operating speed (see the effect of damping on force transmission, from the graph, when rω  2 ) Luckily, natural rubber is a very satisfactory material and is often used for the isolation of vibration CE-409: MODULE 5 (Fall-2013) 44
  45. 45. Problem M 5.3 A rotating machine with a 600 kg mass operating at a constant speed produces harmonic force in vertical direction. The harmonic force is expressed as p(t)= 5000 Sin 150t, where p(t) is in N. If the damping ratio of isolators at the foundation of machine is 7.5%, determine the stiffness of isolators so that the Transmissibility at the operating speed does not exceed 0.15. Also determine the amplitude of force transmitted to the foundation CE-409: MODULE 5 (Fall-2013) 45
  46. 46. Base Excitations (Transmission of harmonic displacements from base) If the ground motion is defined as u g  u gosin( t) , it can be shown that the amplitude u o t of the total displacement u t (t) of the mass can be calculated from the same formula that is used for transmission of force from a system to its foundation. i.e., t u o  TR  u go t 1  2rω  1 - r  ω 2 2 CE-409: MODULE 5 (Fall-2013)  2rω  2 2 46
  47. 47. Base Excitations (Transmission of harmonic displacements from base) ut m k ug CE-409: MODULE 5 (Fall-2013) c Base 47
  48. 48. Base Excitations (Transmission of harmonic displacements from base) ug ut k m Base  c CE-409: MODULE 5 (Fall-2013) 48
  49. 49. Problem M 5.4 An automobile is modeled as SDOF system vibrating in Vertical Direction. It is driven along a road where the elevation varies sinusoidally. The distance from peak to trough is 0.2m and the distance along the road b/w the peaks is 35 m. If the natural frequency of automobile is 2 Hz and the damping ratio of the shock absorbers is 0.15, determine the maximum displacement by which the automobile jump while moving at a speed of 60km/hr. Do you think that the shock absorbers have appropriate damping ratio.? If the speed of the automobile is varied, find the most unfavorable speed for passengers. CE-409: MODULE 5 (Fall-2013) 49
  50. 50. Base Excitations (Transmission of harmonic accelerations from base)  ut m k c ut TR   o go u  ug Base  ug  ut k m Base rω  ω c CE-409: MODULE 5 (Fall-2013) ωn 50
  51. 51. Base Excitations (Transmission of harmonic accelerations from base) If accelerations acting at the base of a system varies sinusoidally, it can be proved that:  o u TR    go u t 1  2rω  1 - r  ω 2 2 CE-409: MODULE 5 (Fall-2013)  2rω  2 2 51
  52. 52. Problems 15 kips 15 ft 1. The steel frame shown in figure is subjected to a sinusoidal ground motion ug(t)=0.2 Sin 5.3t inches. Determine the amplitudes of displacement , the equivalent static force and acceleration at the top end. Take E= 29,000 ksi and I= 69.2 in4 and ζ = 0.05 CE-409: MODULE 5 (Fall-2013) 52
  53. 53. Problems 2: A delicate instrument is to be spring mounted to the floor of a test laboratory where it has been determined that the floor vibrates freely with harmonic motion of amplitude 0.1in at 10 cycles per second, If the instrument weighs 100lb, determine the stiffness of isolation springs required to reduce the vertical motion amplitude of the instrument to 0.01 in. neglect damping. 3. When the person stands in the centre of the floor system shown, he causes a Deflection of 0.2 in. of floor under his feet. He walks (or runs quickly) in the same area , how many steps per second would cause the floor to vibrate with the greatest vertical amplitude CE-409: MODULE 5 (Fall-2013) 53 53
  54. 54. Problems 4. What is the required column stiffness of single one-story structure to limit its acceleration amplitude to 2.1 m/s2 during an earthquake whose acceleration amplitude is 150 mm/s2 at a frequency of 50 rad/s? The mass of structure is 1800 kg. Assume a damping ratio of 0.05. 5. A 10-kg laser flow-measuring device is used on a table in a laboratory. Because of operation of other equipment, the table is subject to vibration. Accelerometer measurements show that the dominant component of the table vibrations is at 300 Hz and has an amplitude of 4.3 m/s2. For effective operation, the laser can be subject to an acceleration amplitude of 0.7 m/s2. (a) Design an undamped isolator to reduce the transmitted acceleration, to an acceptable amplitude. (b) Design the isolator such that it has a damping ratio of 0.04. CE-409: MODULE 5 (Fall-2013) 54 54
  55. 55. Problems 6. A 150-kg engine operates at speeds between 1000 and 2000 rpm. It is desired to achieve at least 85 percent isolation at all speeds. The only readily available isolator has a stiffness of 5x 105 N/m. How much mass must be added to the engine to achieve the desired isolation? HA2M5 Solve problems 3,4 and 6 CE-409: MODULE 5 (Fall-2013) 55

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