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FREE VIBRATION

Unit 3 <br>Free vibration<br>Dynamics of machines

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FREE VIBRATION

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  1. UNIT 3

  2. Free vibrations

  3. Unit Outcome • At the end of this unit, the students should be able to compute the frequency of free vibrations. SYLLABUS Basic features of vibratory systems – Degrees of freedom – single degree of freedom – Freevibration– Equations of motion – Natural frequency – Types of Damping – Damped vibration– Torsional vibration of shaft – Critical speeds of shafts – Torsional vibration – Two and threerotor torsional systems.

  4. Basic features of vibratory system • Vibrations: When elastic bodies such as a spring, a beam and shaft are displaced from the equilibrium position by the application of external forces, and then released, they execute a vibratory motion.

  5. Basic features of vibratory system • Period of vibration or time period: It is the time interval after which the motion is repeated itself. The period of vibration is usually expressed in seconds. • Cycle: It is the motion completed during one time period. • Frequency: It is the number of cycles described in one second. In S.I. units, the frequency is expressed in hertz (briefly written as Hz) which is equal to one cycle per second.

  6. Basic features of vibratory system • Causes of Vibrations: • Unbalanced forces: Produced within the machine due to wear and tear. • External excitations: Can be periodic or random • Resonance: • When the frequency of the external or applied force is equal to the natural frequency resonance occurs. Vertical Shaking Accident and Cause Investigation of 39-story Office Building

  7. Basic features of vibratory system Components of vibratory system: • Spring/Restoring element: • Its denoted by k or s; • SI unit – N/m • Dashpot/Damping component • Its denoted by c; • SI unit – N/m/s • Mass/Inertia component • Its denoted by m; • SI unit – kg

  8. Degrees of Freedom 1 DOF 2 DOF • The minimum number of independent coordinates required to determine completely the position of all parts of a system at any instant of time defines the degree of freedom of the system.

  9. Types of Vibrations • Free or Natural Vibrations: When no external force acts on the body, after giving it an initial displacement, then the body is said to be under free or natural vibrations. The frequency of the free vibrations is called free or natural frequency. • Forced vibrations When the body vibrates under the influence of external force, the the body is said to be under forced vibrations.The vibrations have the same frequency as the applied force • Damped vibrations: When there is a reduction in amplitude over every cycle of vibration, due to frictional resistance, the motion is said to be damped vibration.

  10. Types of Vibrations Longitudinal Transverse Torsional • Longitudinal Vibrations: Parallel to axis of shaft • Transverse Vibrations: Approx. Perpendicular to axis of shaft • Torsional Vibrations: Moves in circles about axis of shaft

  11. Spring - Stiffness d = Static deflection of spring in meters.

  12. Natural frequency of free vibrations Equilibrium method – Longitud. Vibrations • Restoring force • W – (sd +sx) • - sx • Accelerating force • m(d2x/dt2) • Equating both • m(d2x/dt2) + sx = 0 • d2x/dt2 + (s/m) x = 0 • SHM equation • d2x/dt2 + w2 x = 0 • Angular Velocity: W = mg = sd SF = 0

  13. Natural frequency of free vibrations Equilibrium method • Time period • tp = 2p/w • Natural frequency • A • B • Deflection • s= W/A = Ee = E x (d/l) • d = Wl/ AE

  14. Energy method

  15. Rayleigh’s method In this method, the maximum kinetic energy at the mean position is equal to the maximum potential energy (or strain energy) at the extreme position. Assuming the motion executed by the vibration to be simple harmonic, then

  16. Natural frequency of transverse vibrations Transverse vibrations <= Same as Longitudinal Vibrations

  17. Formula • Natural frequency of longitudinal and transverse vibrations: Where, fn = Natural frequency. (Hz) tp = Time period (s) s = Stiffness (N/m) m = Mass. (kg) g = acceleration due to gravity (m/s2) d = Static deflection. (m)

  18. A cantilever shaft 50 mm diameter and 300 mm long has a disc of mass 100 kg at its free end. The Young's modulus for the shaft material is 200 GN/m 2 . Determine the frequency of longitudinal and transverse vibrations of the shaft

  19. Formula 2. Static deflection in beams, Where, fn = Natural frequency. (Hz) tp = Time period (s) s = Stiffness (N/m) m = Mass. (kg) g = acceleration due to gravity (m/s2) d = Static deflection. (m)

  20. Formula 2. Static deflection in beams, Where, fn = Natural frequency. (Hz) tp = Time period (s) s = Stiffness (N/m) m = Mass. (kg) g = acceleration due to gravity (m/s2) d = Static deflection. (m)

  21. Formula 2. Static deflection in beams,

  22. Formula 2. Static deflection in beams,

  23. A shaft of length 0.75 m, supported freely at the ends, is carrying a body of mass 90 kg at 0.25 m from one end. Find the natural frequency of transverse vibration. AssumeE = 200 GN/m2 and shaft diameter = 50mm. Given l = 0.75 m ; m = 90 kg ; a = AC = 0.25 m ; E = 200 GN/m2 = 200 × 109N/m2 d = 50 mm = 0.05 m

  24. Moment of inertia of shaft • Static deflection • Natural frequency

  25. A flywheel is mounted on a vertical shaft as shown in Fig. 23.8. The both ends of the shaft are fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg. Find the natural frequencies of longitudinal and transverse vibrations. Take E = 200 GN/m2 .

  26. Transverse vibrations

  27. Effect of multiple loads/Energy (or Rayleigh’s) method Multiple loads on SSB • To find the natural frequency of a beam on which multiple loads are acting, we find the deflection caused by each load separately and find the total effect.

  28. Deflection in point loads

  29. Effect of self weight/Dunkerley’s method • The weight of a beam is taken as an UDL acting on top of existing loads.

  30. Deflection under self weight/UDL

  31. A shaft 50 mm diameter and 3 metres long is simply supported at the ends and carries three loads of 1000 N, 1500 N and 750 N at 1 m, 2 m and 2.5 m from the left support.The Young's modulus for shaft material is 200 GN/m2. Find the frequency of transverse vibration. Given d = 50 mm = 0.05 m ; l = 3 m, W1 = 1000 N ; W2 = 1500 N ; W3 = 750 N; E = 200 GN/m2 = 200 × 109 N/m2

  32. Moment of inertia of shaft • Static deflections due to point loads,. • 1000 N load • 1500 N load

  33. Static deflections due to point loads • 750 N load • Frequency of transverse vibrations

  34. 2. Calculate the natural frequency of a shaft 20 mm diameter and 0.6 m long carrying a mass of 1 kg a its mid-point. The density of the shaft material is 40 Mg/m3 and Young’s modulus is 200 GN/m2. Assume the shaft to be freely supported. Given d = 20 mm = 0.02 m ; l = 0.6 m ; m1 = 1 kg ; ρ = 40 Mg/m3 = 40 × 106 g/m3 = 40 × 103 kg/m3 E = 200 GN/m2 = 200 × 109 N/m2

  35. Moment of inertia of shaft • Mass per unit length • Static deflection due to point load

  36. Static deflection due to self weight-UDL • Natural frequency of shaft

  37. Assignment Calculate the natural frequency of a shaft 30 mm diameter and 0.6 m long carrying a mass of 1 kg a itsmid-point and 2kg at 0.4m from the left end. The density of the shaft material is 30 Mg/m3 and Young’s modulus is 200 GN/m2. Assume the shaft to be freely supported.

  38. Critical speed of shaft

  39. Critical/whirling speed of shaft • The centre of gravity of the pulley or gear mounted on a shaft is at certain distance from the axis of rotation. • Due to this, the shaft is subjected to centrifugal force which willbend the shaft which will further increase the distance of centre of gravity from the axis of rotation. • The bending of shaft not only depends upon the distance between C.G of the shaft and gears, but alsodepends upon the speed at which the shaft rotates. • The speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite, is known as critical or whirling speed.

  40. 1. A shaft 1.5 m long, supported in flexible bearings at the ends carries two wheels each of 50 kg mass. One wheel is situated at the centre of the shaft and the other at a distance of 375 mm from the centre towards left. The shaft is hollow of external diameter 75 mmand internal diameter 40 mm. The density of the shaft materialis7700 kg/m3 and itsmodulusofelasticity is 200 GN/m2. Find the lowest whirling speed of the shaft, taking into account the mass of the shaft. Given: l = 1.5 m ; m1 = m2 = 50 kg ; d1 = 75 mm = 0.075 m ; d2 = 40 mm = 0.04 m ; ρ = 7700 kg/m3 E = 200 GN/m2 = 200 × 109 N/m2

  41. Moment of inertia of shaft • Mass per unit length

  42. Deflection of beam at C • Deflection of beam at D

  43. Static deflection due to weight of shaft • Natural frequency of shaft • Critical speed of shaft • Nc = 32.4 x 60 = 1944 rpm

  44. Given • d = 5 mm = 0.005 m ; • l = 200 mm = 0.2 m ; • m = 50 kg ; • E = 200 GN/m2 = 200 × 109 N/m2 • Moment of inertia of shaft 2. . A vertical shaft of 5 mm diameter is 200 mm long and is supported in long bearings at its ends. A disc of mass 50 kg is attached to the centre of the shaft. Neglecting any increase in stiffness due to the attachment of the disc to the shaft, find the critical speed of rotation

  45. Static deflection of shaft (fixed shaft-Long bearings) • d = Wa3b3/3EIl3 • = 50 x 9.81 x (0.1)3 x(0.1)3 / 3 x200 × 109x 30.7 x 10-12 • = 3.33 x 10-3 m • Frequency of the shaft • fn • Critical speed of shaft • Nc = 8.64 x 60 = 518.4 rpm ~ 520 rpm

  46. A vertical steel shaft 15 mm diameter is held in long bearings 1 metre apart and carries at its middle a disc of mass 15 kg. The eccentricity of the centre of gravity of the disc from the centre of the rotor is 0.30 mm. The modulus of elasticity for the shaft material is 200 GN/m2 and the permissible stress is 70 MN/m2. Determine : 1. The critical speed of the shaft and 2. The range of speed over which it is unsafe to run the shaft. Neglect the mass of the shaft

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