FORCED VIBRATION.ppt
- 1. Unit IV- FORCED VIBRATION 1 1.A single cylinder vertical petrol engine of total mass 300 kg is mounted upon a steel chassis frame and causes a vertical static deflection of 2 mm. The reciprocating parts of the engine has a mass of 20 kg and move through a vertical stroke of 150 mm with simple harmonic motion. A dashpot is provided whose damping resistance is directly proportional to the velocity and amounts to 1.5 kN per metre per second. Considering that the steady state of vibration is reached ; determine : 1. the amplitude of forced vibrations, when the driving shaft of the engine rotates at 480 r.p.m., and 2. the speed of the driving shaft at which resonance will occur. Given. m = 300 kg; δ = 2 mm = 2 × 10–3 m ; m1= 20 kg ; l = 150 mm = 0.15 m ; c = 1.5 kN/m/s = 1500 N/m/s ; N = 480 r.p.m. or = 2π×480 / 60 = 50.3 rad/s 1. Amplitude of the forced vibrations
- 2. 2 2. Speed of the driving shaft at which the resonance occurs
- 3. 3 3
- 4. 4 2.A mass of 10 kg is suspended from one end of a helical spring, the other end being fixed. The stiffness of the spring is 10 N/mm. The viscous damping causes the amplitude to decrease to one-tenth of the initial value in four complete oscillations. If a periodic force of 150 cos 50 t N is applied at the mass in the vertical direction, find the amplitude of the forced vibrations. What is its value of resonance ? Solution. Given : m = 10 kg ; s = 10 N/mm = 10 × 103 N/m Since the periodic force, F0 = F cos.t =150cos50t , therefore and angular velocity of the periodic disturbing force, We know that angular speed or natural circular frequency of free vibrations, Amplitude of the forced vibrations
- 6. 6 What is its value of resonance ?
- 7. 7 A body of mass 20 kg is suspended from a spring which deflects 15 mm under this load. Calculate the frequency of free vibrations and verify that a viscous damping force amounting to approximately 1000 N at a speed of 1 m/s is just-sufficient to make the motion aperiodic. If when damped to this extent, the body is subjected to a disturbing force with a maximum value of 125 N making 8 cycles/s, find the amplitude of the ultimate motion. Solution . Given : m = 20 kg ; δ= 15 mm = 0.015 m ; c = 1000 N/m/s ; F = 125 N ; f = 8 cycles/s The critical damping to make the motion aperiodic is such that damped frequency is zero, This means that the viscous damping force is 1023 N at a speed of 1 m/s. Therefore a viscous damping force amounting to approximately 1000 N at a speed of 1 m/s is just sufficient to make the motion aperiodic. Ans
- 8. 8 Amplitude of ultimate motion We know that angular speed of forced vibration, If when damped to this extent, the body is subjected to a disturbing force with a maximum value of 125 N making 8 cycles/s
- 9. 9 A machine part of mass 2 kg vibrates in a viscous medium. Determine the damping coefficient when a harmonic exciting force of 25 N results in a resonant amplitude of 12.5 mm with a period of 0.2 second. If the system is excited by a harmonic force of frequency 4 Hz what will be the percentage increase in the amplitude of vibration when damper is removed as compared with that with damping. Solution . Given : m = 2 kg ; F = 25 N ; Resonant xmax = 12.5 mm = 0.0125 m ; tp = 0.2 s ; f = 4 Hz Damping coefficient “C” Resonant amplitude S={31.42}2X2=1974.43
- 10. 10 C=63.65 N/m/s Percentage increase in amplitude Since the system is excited by a harmonic force of frequency ( f ) = 4 Hz, therefore corresponding circular frequency We know that maximum amplitude of vibration with damping
- 11. 11 and the maximum amplitude of vibration when damper is removed Percentage increase in amplitude
- 12. 12 Vibration Isolation and Transmissibility When an unbalanced machine is installed on the foundation, it produces vibration in the foundation. In order to prevent these vibrations or to minimise the transmission of forces to the foundation, the machines are mounted on springs and dampers or on some vibration isolating material, It may be noted that when a periodic (i.e. simple harmonic) disturbing force F cos t is applied to a machine of mass m supported by a spring of stiffness s, then the force is transmitted by means of the spring and the damper or dashpot to the fixed support or foundation. The ratio of the force transmitted (FT) to the force applied (F) is known as the isolation factor or transmissibility ratio of the spring support. We have discussed above that the force transmitted to the foundation consists of the following two forces :
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- 14. 14 The graph for different values of damping factor c/cc to show the variation of transmissibility ratio ( ) against the ratio /n .
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- 16. 16 1.The mass of an electric motor is 120 kg and it runs at 1500 r.p.m. The armature mass is 35 kg and its C.G. lies 0.5 mm from the axis of rotation. The motor is mounted on five springs of negligible damping so that the force transmitted is one-eleventh of the impressed force. Assume that the mass of the motor is equally distributed among the five springs. Determine : 1. stiffness of each spring; 2. dynamic force transmitted to the base at the operating speed; and 3. natural frequency of the system. 1. Stiffness of Each Spring
- 17. 17 Since these are five springs, therefore stiffness of each spring 2. Dynamic force transmitted to the base at the operating speed (i.e. 1500 r.p.m. or 157.1 rad/s) 3. Natural frequency of the system
- 18. 18 A machine has a mass of 100 kg and unbalanced reciprocating parts of mass 2 kg which move through a vertical stroke of 80 mm with simple harmonic motion. The machine is mounted on four springs, symmetrically arranged with respect to centre of mass, in such a way that the machine has one degree of freedom and can undergo vertical displacements only. Neglecting damping, calculate the combined stiffness of the spring in order that the force transmitted to the foundation is 1 / 25 th of the applied force, when the speed of rotation of machine crank shaft is 1000 r.p.m. When the machine is actually supported on the springs, it is found that the damping reduces the amplitude of successive free vibrations by 25%. Find : 1. the force transmitted to foundation at 1000 r.p.m., 2. the force transmitted to the foundation at resonance, and 3. the amplitude of the forced vibration of the machine at resonance. 1. Combined stiffness of the spring
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- 20. 20 1. the force transmitted to foundation at 1000 r.p.m Since the damping reduces the amplitude of successive free vibrations by 25%, therefore final amplitude of vibration,
- 21. 21 2.the force transmitted to the foundation at resonance,
- 22. 22 3. the amplitude of the forced vibration of the machine at resonance.
- 23. 23 3.A single-cylinder engine of total mass 200 kg is to be mounted on an elastic support which permits vibratory movement in vertical direction only. The mass of the piston is 3.5 kg and has a vertical reciprocating motion which may be assumed simple harmonic with a stroke of 150 mm. It is desired that the maximum vibratory force transmitted through the elastic support to the foundation shall be 600 N when the engine speed is 800 r.p.m. and less than this at all higher speeds. 1. Find the necessary stiffness of the elastic support, and the amplitude of vibration at 800 r.p.m., and 2. If the engine speed is reduced below 800 r.p.m. at what speed will the transmitted force again becomes 600 N? Solution. Given : m1 = 200 kg ; m2 = 3.5 kg ; l = 150 mm = 0.15 mm or r = l/2 = 0.075 m ; FT = 600 N ; N = 800 r.p.m. or= 2π×800 / 60 = 83.8 rad/s FO = Centrifugal force on the piston
- 24. 24 1. Stiffness of elastic support and amplitude of vibration Since the max. vibratory force transmitted to the foundation is equal to the force on the elastic support (neglecting damping), therefore Max. vibratory force transmitted to the foundation,
- 25. 25 2. Speed at the which the transmitted force again becomes 600 N The transmitted force will rise as the speed of the engine falls and passes through resonance. There will be a speed below resonance at which the transmitted force will again equal to 600 N. Let this speed be 1 rad/s (or N1 r.p.m.).
- 26. 26 Since the engine speed is reduced below N1 = 800 r.p.m., therefore in this case, max, amplitude of vibration,