Single degree of freedom system free vibration part -i and ii
- 1. UNIT-I Single Degree of Freedom Systems – Free Vibration Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 1
- 2. Introduction to Vibration When any elastic body such as spring, shaft, or beam is displaced from the equilibrium position by the application of external force and released, it commences cyclic motion. Such cyclic motion of a body or a system , due to elastic deformation under the action of external forces, is known as vibration. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 2
- 3. Phenomenon of Vibration Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 3
- 4. Causes of Vibration Unbalance forces and couples External excitation forces Dry friction between two mating surfaces Wind load may cause vibration in certain systems such as telephone lines, electric lines, etc Earthquakes Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 4
- 5. Disadvantages of Vibration Excessive Stresses Loosening of Assembled parts Failure of machine parts Undesirable noise Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 5
- 6. Advantages of Vibration All musical instrument Vibrating screen, Shakers and conveyors Stress relieving equipment Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 6
- 7. Terminology and Basic Concept Simple Harmonic Motion Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 7 Let x- displacement of point from mean position after time ‘t’ X- maximum displacement of point from mean position • Type equation here.Displacement of point x=Xsin 𝜃=Xsin ω𝑡 ...………………………………………..(1)
- 8. • Velocity of point 𝑥= 𝑑𝑥 𝑑𝑡 𝑥= 𝜔X cos ωt Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 8 • Acceleration of point 𝑥 = 𝑑2 𝑥 𝑑𝑡2 𝑥= -ω2 sin ω𝑡 𝑥= -ω2 x A motion , whose acceleration is proportional to displacement from mean position and is directed towards the mean position, is known as SHM 𝑥 + ω2x = 0…………………fundamental equation of SHM
- 9. Time Period(𝒕 𝒑) is the timerequired to complete one cycle (2𝜋 ). Mathematically, 𝑡 𝑝 = 2 𝜋 𝜔 Frequency (f) The number of cycles per unit time is known as frequency. It is reciprocal of time period. f= 1 𝑡 𝑝 Amplitude (X) It is the maximum displacement of a vibrating body from its mean position. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 9
- 10. Stiffness of spring (K) It is the force required to produce unit displacement in the direction of applied force. K= 𝐹 𝛿 , N/m Where, K= Stiffness of spring , N/m F= force applied on spring, N 𝛿= deflection of spring, m Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 10 Degree of Freedom ( D.O.F) The minimum number of independent co-ordinates required to specify the motion or configuration of a system at any instant is known as DOF Ex.
- 11. Damping Is the resistance to the motion of a vibrating body , which causes a vibrating body to come to rest position. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 11 Damping coefficient (C) Is the damping force or resisting force developed per unit velocity. Mathematically c= 𝐹 𝑣 , N-sec/m Where, F= Force applied on damper v= Velocity of viscous fluid
- 12. Resonance When the frequency of external excitation force acting on a body is equal to the frequency of a vibrating body, the body starts vibrating with excessively large amplitude. Such state is known as resonance. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 12
- 13. Elements of Vibratory System Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 13
- 14. Equvalent Springs Equivalent Spring Springs in Series Springs in Parallel Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 14
- 15. Spring in Series Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 15 Deflection of Equivalent Spring=Deflection of Spring 1 + Deflection of Spring 2 𝛿 = 𝛿1 + 𝛿2 ……………….(a) Force on Equivalent Spring=Force on Spring 1 + Force on Spring 2 mg = 𝑚1g = 𝑚2g The system of two springs in series is to be replaced by an equivalent spring having stiffness 𝐾𝑒 Deflection of Equivalent Spring= 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑝𝑟𝑖𝑛𝑔 1 + 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑝𝑟𝑖𝑛𝑔 2 𝛿 = 𝛿1 + 𝛿2 ∴ mg 𝐾 𝑒 = 𝑚1g 𝐾1 + 𝑚2g 𝐾2 …….(b) But, mg = 𝑚1g = 𝑚2g … … … . (c) Substituting Equation ( C ) in equation (b), we get 1 𝐾𝑒 = 1 𝐾1 + 1 𝐾2 mg 𝐾𝑒 = mg 𝐾1 + mg 𝐾2 =
- 16. Spring in Parallel Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 16 Deflection of Equivalent Spring=Deflection of Spring 1 = Deflection of Spring 2 𝛿 = 𝛿1 = 𝛿2 ……………….(d) Force on Equivalent Spring=Force on Spring 1 + Force on Spring 2 mg = 𝑚1g + 𝑚2g … … … … … … … … … … … . . (e) The system of two springs in series is to be replaced by an equivalent spring having stiffness 𝐾𝑒 Deflection of Equivalent Spring= 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑝𝑟𝑖𝑛𝑔 1 + 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑝𝑟𝑖𝑛𝑔 2 𝛿 = 𝛿1 = 𝛿2 But, 𝛿 = 𝛿1 = 𝛿2 Substituting Equation ( d ) in equation (f), we get Ke 𝛿 = 𝐾1 𝛿1 + 𝐾2 𝛿2 ……(f) Ke 𝛿 = 𝐾1 𝛿 + 𝐾2 𝛿 𝐾𝑒= 𝐾1 + 𝐾2
- 17. Equvalent Damper Equivalent Damper Dampers in Series Dampers in Parallel Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 17
- 18. Dampers in Series Dampers in Parallel Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 18 1 𝐶𝑒 = 1 𝐶1 + 1 𝐶2 𝐶𝑒= 𝐶1 + 𝐶2
- 19. Introduction to Modeling Physical Modeling Geometric Modeling Mathematical Modeling Combination of Geometric and Mathematical Modeling Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 19
- 20. Practical Example of Mathematical Modeling Motor Bike Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 20
- 21. Practical Example of Mathematical Modeling Bicycle Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 21
- 22. Types of Vibration Types of Vibration According to actuating force Free vibration Forced Vibration According to external resistsance Undamped Vibration Damped Vibration According to motion of system Longitudinal Transverse Torsional According to behavior of vibrating system Linear Vibration Non Linear Vibration According to magnitude of actuating force at a given time Deterministic Random Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 22
- 23. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 23 Examples of Vibration Longitudinal Transverse TorsionalFree Vibration Forced Vibration Damped Vibration
- 24. Examples of Vibration Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 24 Linear Vibration Non- Linear Vibration Deterministic Vibration Random Vibration
- 25. Determination of Natural Frequency Equilibrium Method Energy Method Rayleighs Method Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 25
- 26. Equilibrium Method( D’Alembert Principal) • A body or structure which is not in static equilibrium due to acceleration it possesses can be brought to static equilibrium by introducing the inertia force on it. • The inertia force is equal to the mass times the acceleration direction is opposite to that of acceleration. • The principle is used for developing the equation of motion for vibrating system which is further used to find the natural frequency of the vibrating system. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 26
- 27. Equilibrium Method( D’Alembert Principal) Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 27
- 28. Equilibrium Method( D’Alembert Principal) • The gravitational force must be equal to zero. mg=kδ ------- (1) The force acting on the mass are : 1. inertia force : mẍ(upwards) 2. spring force : K(x+δ) (upwards) 3. gravitational force : mg Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 28
- 29. According to D’Alembert’s principle , 𝑚 𝑥 + K x + δ − 𝑚𝑔 = 0 𝑚 𝑥 + Kx + Kδ − 𝑚𝑔 = 0 ……..(Since 𝑚𝑔 = 𝐾𝛿) ∴ 𝑚 𝑥 + K𝑥 = 0 ∴ 𝑥 + 𝐾 𝑚 x=0 ………………………..(2) We know that the fundamental equation of SHM 𝑥 + ω2 x = 0 ………………………(3) Comparing equation of (2) and (3) ω2 = 𝐾 𝑚 rad/s………………….................(4) Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 29
- 30. The natural frequency f of vibration is , 𝑓 = 𝜔 2𝜋 or 𝑓 = 1 2𝜋 𝐾 𝑚 Hz……………(5) Or from equation (1) 𝑚𝑔 = 𝐾𝛿 ∴ 𝐾 𝑚 = 𝑔 𝛿 Substitute in equation (5) 𝑓 = 1 2𝜋 𝑔 𝛿 Hz and time period 𝑡 𝑝= 1 𝑓 s Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 30
- 31. Energy Method According to law of conservation of energy, energy can neither be created nor be destroyed, but it can be converted from one form to another form. According to law of conservation of energy, Total Energy= Constant K.E+P.E= Constant 𝑑 𝑑𝑡 K. E + P. E = 0………………………………………(a) Considering Spring mass system K.E= 1 2 𝑚 𝑥2 P.E is in the form of strain energy stored in the spring. The strain Energy is given by the area under the force vs deflection diagram. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 31
- 32. PE= Strain Energy=Area under force-deflection diagram PE= 1 2 𝐹𝑥 ∴ PE= 1 2 (𝐾𝑥)𝑥 (Since we know that ,Spring force (F)=Kx ) = 1 2 𝐾𝑥2 Substitute in equation (a) ∴ 𝑑 𝑑𝑡 ( 1 2 𝑚 𝑥2 + 1 2 𝐾𝑥2 )=0 ∴ 1 2 m(2x 𝑥)+ 1 2 K(2x 𝑥)=0 ∴ 𝑚 𝑥 𝑥+(Kx) 𝑥 = 0 ∴ 𝑚 𝑥 + K𝑥 = 0 ∴ 𝑥 + 𝐾 𝑚 x=0 Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 32 x F Strain Energy SpringForce Deflection
- 33. We know that the fundamental equation of SHM 𝑥 + ω2 x = 0 Comparing above equation with equation of SHM ω2 = 𝐾 𝑚 rad/s The natural frequency f of vibration is , 𝑓 = 𝜔 2𝜋 or 𝑓 = 1 2𝜋 𝐾 𝑚 Hz Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 33
- 34. Rayleigh’s Method This is the extension of energy method , which is developed by Lord Rayleigh According to principal of conservation of energy TE= Constant (𝑇𝐸) 𝑎𝑡 𝑚𝑒𝑎𝑛 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛= 𝑇𝐸 𝑎𝑡 𝑒𝑥𝑡𝑟𝑒𝑚𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 ∴ (𝐾𝐸) 𝑚𝑎𝑥= 𝑃𝐸 𝑚𝑎𝑥 ………………………………………………..(a) Let body having mass m is moving with SHM , therefore the displacement of the body is given by 𝒙 = 𝑿 𝒔𝒊𝒏 𝝎 𝒏 𝒕 x = Displacement of a body from the mean position after time t sec. X= Maximum displacement of a body from mean positon 𝜔 𝑛= Circular natural frequency 𝒙= 𝒅𝒙 𝒅𝒕 =𝝎 𝒏X𝐜𝐨𝐬 𝝎 𝒏 𝒕 𝒙 𝒎𝒂𝒙= 𝝎 𝒏 𝐗 (at t=0) Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 34
- 35. At mean position , the maximum KE is (𝑲𝑬) 𝒎𝒂𝒙= 𝟏 𝟐 𝒎( 𝒙 𝒎𝒂𝒙 𝟐 ) or (𝑲𝑬) 𝒎𝒂𝒙= 𝟏 𝟐 𝒎𝝎 𝒏 𝟐 𝑿 𝟐 And PE is , PE = 𝟏 𝟐 𝑲𝒙 𝟐 At extreme position ( at x=X) , the max PE is 𝑷𝑬 𝒎𝒂𝒙= 𝟏 𝟐 𝑲𝑿 𝟐 Substituting in equation (a) 𝟏 𝟐 𝒎𝝎 𝒏 𝟐 𝑿 𝟐= 𝟏 𝟐 𝑲𝑿 𝟐 𝝎 𝟐 = 𝑲 𝒎 rad/s The natural frequency f of vibration is , 𝒇 = 𝝎 𝟐𝝅 or 𝒇 = 𝟏 𝟐𝝅 𝑲 𝒎 Hz Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 35
- 36. Undamped Free Transverse Vibration Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 36 Consider a cantilever beam of negligible mass carrying a concentrated mass “m” at the free end, as shown in fig,
- 37. • The gravitational force must be equal to zero. mg=kδ ------- (1) The force acting on the mass are : 1. inertia force : mẍ(upwards) 2. spring force : K(x+δ) (upwards) 3. gravitational force : mg Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 37
- 38. According to D’Alembert’s principle , 𝑚 𝑥 + K x + δ − 𝑚𝑔 = 0 𝑚 𝑥 + Kx + Kδ − 𝑚𝑔 = 0 ……..(Since 𝑚𝑔 = 𝐾𝛿) ∴ 𝑚 𝑥 + K𝑥 = 0 ∴ 𝑥 + 𝐾 𝑚 x=0 ………………………..(2) We know that the fundamental equation of SHM 𝑥 + ω2 x = 0 ………………………(3) Comparing equation of (2) and (3) ω2 = 𝐾 𝑚 rad/s………………….................(4) Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 38
- 39. The natural frequency f of vibration is , 𝑓 = 𝜔 2𝜋 or 𝑓 = 1 2𝜋 𝐾 𝑚 Hz……………(5) Or from equation (1) 𝑚𝑔 = 𝐾𝛿 ∴ 𝐾 𝑚 = 𝑔 𝛿 Substitute in equation (5) 𝑓 = 1 2𝜋 𝑔 𝛿 Hz and time period 𝑡 𝑝= 1 𝑓 s Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 39
- 40. Torsional Stiffness Torsional Stiffness of Shaft (𝐾𝑡) is defined as the torque required to produce unit angular deflection in the direction of applied torque. Mathematically, 𝐾𝑡= 𝑇 𝜃 𝐾𝑡= 𝐺𝐽 𝐿 G=Modulus of rigidity N/𝑚𝑚2 J= Polar Moment of Inertia ( 𝜋 32 𝑑4 ) = 𝑚𝑚4 l= Length of Shaft D = Diameter of the shaft Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 40
- 41. Parameters for linear & torsional vibration Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 41
- 42. Undamped free torsional vibration Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 42 Consider a disc having mass moment of inertia ‘I’ suspended on a shaft with negligible mass, as shown in fig, For angular displacement of disc ‘θ’ in clockwise direction, the torques acting on the disc are : Inertia torque Restoring troque
- 43. Therefor according to D’Alembert’s priciple, ∑ (Inertia Torque + External Torque)= 0 ∴ 𝐼 𝜃 + 𝐾𝑡 𝜃 = 0 𝜃+ 𝐾𝑡 𝐼 𝜃 = 0……………………(1) We know that the fundamental equation of SHM 𝜃 + ω2 θ = 0……………….(2) Comparing eqn (1) and (2) 𝜔 𝑛= 𝐾𝑡 𝐼 𝑓 = 𝜔 𝑛 2𝜋 = 1 2𝜋 𝐾𝑡 𝐼 = 1 2𝜋 𝐺𝐽 𝐿 Hz. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 43
- 44. Problem 1) Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 44 Find W such that the system has natural frequency 10 cps. Solution: If 𝑘 𝑒1 is the effective spring stiffness of the top three springs in series, then 1 𝑘 𝑒1 = 1 𝑘1 + 1 𝑘2 + 1 𝑘3 = 1 2.0 + 1 1.5 + 1 3.0 = 1.5 Or 𝑘 𝑒1=0.667 kg/cm. If 𝑘 𝑒2 is the effective spring stiffness of the lower two springs in parallel, then 𝑘 𝑒2 = 𝑘4 + 𝑘5 = 0.5 + 0.5 = 1.0 Or 𝑘 𝑒2 =1.0 kg/cm Now 𝑘 𝑒1 and 𝑘 𝑒2 are two springs in parallel, therefore effective stiffness 𝑘 𝑒= 𝑘 𝑒1+𝑘 𝑒2 = 0.667 + 1.0 = 1.667 𝑘𝑔/𝑐𝑚 𝑓𝑛 = 𝜔 2𝜋 = 1 2𝜋 𝑘 𝑒 𝑔 𝑊 = 10 𝑐𝑝𝑠 𝑔𝑖𝑣𝑒𝑛 Therefore 𝑊 = 𝑘 𝑒 𝑔 4𝜋2×102 = 1.667×980 4𝜋2×102 = 0.414 𝑘𝑔. 𝑊 = 0.414 𝑘𝑔 … … … … 𝐴𝑛𝑠
- 45. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 45 Problem no. 2) Find natural frequency for the mathematical model as shown in figure K=𝟐 × 𝟏𝟎 𝟓 N/m , M=20 kg. Solution :K=2 × 105 N/m M=20 kg. We know that when two springs are in parallel 𝑘 𝑒 = 𝑘1 + 𝑘2 and when in series 1 𝑘 𝑒1 = 1 𝑘1 + 1 𝑘2 From Fig (g) 𝑓 = 1 2𝜋 𝐾 𝑚 Hz = 1 2𝜋 13 21 𝐾 𝑚 = 1 2𝜋 13 × 2 × 102 21 × 20 = 𝟏𝟐. 𝟓𝟐 𝑯𝒛. Ans
- 46. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 46 Problem no. 3 ) A mass of 1 kg is suspended by a spring passing over the pulley , as shown in fig. The system is supported horizontally by a spring of stiffness 1 KN/m. Determine the natural frequency of vibration of a system . Using following data Mass of pulley M= 10 kg Radius of pulley R= 50 mm Distance of spring from centre of pulley r= 35 mm Solution: The mass and spring K are not attached on one cord or string. Therefore , consider 𝑥1 be the displacement of mass in downward direction & 𝑥2 be the deflection of spring. The pulley will rotate through an angle 𝜃 . Angular displacement of pulley = 𝜃 Linear displacement of mass = 𝑥1= 𝑅𝜃 Linear velocity of mass = 𝑥1= 𝑅 𝜃 Linear acceleration of mass = 𝑥1= 𝑟𝜃 Deflection of spring = 𝑥2= 𝑟𝜃 Energy Method K.E of the mass = 𝟏 𝟐 𝒎 𝒙 𝟏 𝟐 = 𝟏 𝟐 𝒎𝑹 𝟐 𝜽 𝟐 K.E of the pulley = 𝟏 𝟐 𝑰 𝟎 𝝎 𝟐 = 𝟏 𝟐 𝑰 𝟎 𝜽 𝟐 = 𝟏 𝟒 𝑴𝑹 𝟐 𝜽 𝟐 PE of the spring = 𝟏 𝟐 𝑲𝒙 𝟐 𝟐 = 𝟏 𝟐 𝑲𝒓 𝟐 𝜽 𝟐
- 47. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 47 Total K.E = 𝟏 𝟐 𝒎𝑹 𝟐 𝜽 𝟐 + 𝟏 𝟒 𝑴𝑹 𝟐 𝜽 𝟐 Total P.E = 𝟏 𝟐 𝑲𝒓 𝟐 𝜽 𝟐 According to Energy Method 𝑑 𝑑𝑡 K. E + P. E = 0 ∴ 𝑑 𝑑𝑡 𝟏 𝟐 𝒎𝑹 𝟐 𝜽 𝟐 + 𝟏 𝟒 𝑴𝑹 𝟐 𝜽 𝟐 + 𝟏 𝟐 𝑲𝒓 𝟐 𝜽 𝟐 = 𝟎 ∴ 1 2 2𝑚𝑅2 𝜃 𝜃 + 1 4 2𝑀𝑅2 𝜃 𝜃 + 1 2 2𝐾𝑟2 𝜃 𝜃 = 0 ∴ 𝑚𝑅2 𝜃 + 1 2 𝑀𝑅2 𝜃 + 𝐾𝑟2 𝜃 = 0 ∴ 𝑚𝑅2 + 𝑀𝑅2 2 𝜃 + 𝐾𝑟2 𝜃 = 0 ∴ 𝜃 + 𝐾𝑟2 𝑀𝑅2 2 + 𝑚𝑅2 𝜃 = 0 Comparing this with SHM Equation 𝜔 𝑛 = 𝐾𝑟2 𝑀 2 + 𝑚 𝑅2
- 48. ∴ 𝜔 𝑛 = 𝑟 𝑅 𝐾 𝑀 2 + 𝑚 𝑓𝑛 = 1 2𝜋 × 𝑟 𝑅 𝐾 𝑀 2 + 𝑚 , Hz ∴ 𝑓𝑛 = 0.035 2𝜋 × 0.05 × 1000 10 2 + 1 𝑓𝑛 = 1.4382 𝐻𝑧 𝑡 𝑝 = 0.69 𝑠 Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 48
- 49. Problem no. 4 Determine the natural frequency of oscillation of the simple pendulum shown in fig. Solution: Let, 𝐼0 = 𝑚𝑎𝑠𝑠 𝑀. 𝐼. 𝑜𝑓 𝑠𝑖𝑚𝑝𝑙𝑒 𝑝𝑒𝑛𝑑𝑢𝑙𝑢𝑚 𝑎𝑏𝑜𝑢𝑡 𝑂 𝐼0 = 𝑚𝐿2 , kg-𝑚2 Equilibrium Method Angular motion about O: 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑡𝑜𝑟𝑞𝑢𝑒 + 𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑜𝑟𝑞𝑢𝑒 = 0 ∴ 𝐼0 𝜃 + 𝑚𝑔 sin 𝜃 × 𝐿 = 0 ∴ 𝐼0 𝜃 + 𝑚𝑔𝐿𝜃 = 0 (𝑠𝑖𝑛𝑐𝑒 𝜃 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙, sin 𝜃 ≅ 𝜃) ∴ 𝜃 + 𝑚𝑔𝐿 𝐼0 𝜃 = 0 ……………………..(a) Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 49
- 50. Natural circular frequency comparing equation (a) with the equation of SHM 𝜔 𝑛 2 = 𝑚𝑔𝐿 𝐼0 ∴ 𝜔 𝑛 = 𝑚𝑔𝐿 𝐼0 𝜔 𝑛 = 𝑚𝑔𝐿 𝑚𝐿2 Or 𝜔 𝑛= 𝑔 𝐿 , rad/s Natural Frequency : 𝒇 𝒏 = 𝟏 𝟐𝝅 × 𝒈 𝑳 ……………………………….Ans. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 50
- 51. Damped Free Vibration Damping Vibration: In a vibratory system , if an external resistance is provided so as to reduce the amplitude of vibrations, the vibration is known as damped vibration Damping : The external resistance which is provided to reduce the amplitude of vibrations is known as damping. Damper : The damper is a unit which absorbs the energy of vibratory system, thereby reducing the amplitude of vibration. Important Parameter : 1. Frequency of damped vibration 2. Rate of decay of amplitude Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 51
- 52. Types of damping Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 52 Types of damping Viscous Damping Fluid Dashpot Damping Eddy Current Damping Coulomb or Dry Friction Damping Material or Solid or Structural or Hysteresis Damping
- 53. Viscous Damping When the system is allowed to vibrate in a viscous medium , the damping is called as viscous damping. Viscosity is the property of a fluid by virtue of which it offers resistance to the motion of one layer over the adjacent one. Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 53 t Moving plate u= 𝒙 Fixed Plate It can be explained from fig. where two plates are seperateed by fluid film of thickness (t) . The upper plate is allowed to move parallel to the fixed plate with a velocity ( 𝒙).The next force (F) required for maintaining the velocity ( 𝒙) is 𝐹 = 𝜇𝐴 𝑡 𝑥
- 54. The force can also be written as 𝐹 = 𝐶 𝑥 𝐶 = 𝜇𝐴 𝑡 Where C= Viscous Damping Coefficient Fluid Dashpot Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 54
- 55. Eddy Current Damping Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 55
- 56. Coulomb Damping or Dry Friction Damping Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 56 𝐹𝑟 = 𝜇𝑅 𝑛 𝜇 = 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑅 𝑛 = 𝑁𝑜𝑟𝑚𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛
- 57. Material or Solid or Structural or Hysteresis Damping Prof. S. S. Patil (ME DESIGN) TSSM’s PVPIT Bavdhan Pune 57